4
Partial Differential
Equations
12. Orthogonal Functions and Fourier Series
13. Boundary-Value Problems in Rectangular Coordinates
14. Boundary-Value Problems in Other Coordinate Systems
15. Integral Transform Method
16. Numerical Solutions of Partial Differential Equations
© mariakraynova/Shutterstock
PART
© science photo/Shutterstock
CHAPTER
12
Our goal in Part 4 of this text is
solve certain kinds of linear
partial differential equations in
an applied context. Although we
do not solve any PDEs in this
chapter, the material covered sets
the stage for the procedures
discussed later.
In calculus you saw that a
sufficiently differentiable
function f could often be
expanded in a Taylor series, which
essentially is an infinite series
consisting of powers of x. The
principal concept examined in
this chapter also involves
expanding a function in an
infinite series. In the early 1800s,
the French mathematician Joseph
Fourier (1768–1830) advanced
the idea of expanding a function f
in a series of trigonometric
functions. It turns out that
Fourier series are just special
cases of a more general type of
series representation of a
function using an infinite set of
orthogonal functions. The
notion of orthogonal functions
leads us back to eigenvalues and
the corresponding set of
eigenfunctions. Since eigenvalues
and eigenfunctions are the
linchpins of the procedures in the
two chapters that follow, you are
encouraged to review Example 2
in Section 3.9.
Orthogonal Functions
and Fourier Series
CHAPTER CONTENTS
12.1
12.2
12.3
12.4
12.5
12.6
Orthogonal Functions
Fourier Series
Fourier Cosine and Sine Series
Complex Fourier Series
Sturm–Liouville Problem
Bessel and Legendre Series
12.6.1 Fourier–Bessel Series
12.6.2 Fourier–Legendre Series
Chapter 12 in Review
12.1
Orthogonal Functions
INTRODUCTION In certain areas of advanced mathematics, a function is considered to be a
generalization of a vector. In this section we shall see how the two vector concepts of inner, or
dot, product and orthogonality of vectors can be extended to functions. The remainder of the
chapter is a practical application of this discussion.
Inner Product Recall, if u u1i u2 j u3k and v v1i v2 j v3k are two vectors
in R3 or 3-space, then the inner product or dot product of u and v is a real number, called a scalar,
defined as the sum of the products of their corresponding components:
(u, v) u1v1 u2v2 u3v3 a ukvk.
3
In Chapter 7, the inner
product was denoted by u v.
k1
The inner product (u, v) possesses the following properties:
(i)
(ii)
(iii)
(iv)
(u, v) (v, u)
(ku, v) k(u, v), k a scalar
(u, u) 0 if u 0 and (u, u) 0 if u Z 0
(u v, w) (u, w) (v, w).
We expect any generalization of the inner product to possess these same properties.
Suppose that f1 and f2 are piecewise-continuous functions defined on an interval [a, b].* Since
a definite integral on the interval of the product f1(x) f2(x) possesses properties (i)(iv) of the inner
product of vectors, whenever the integral exists we are prompted to make the following definition.
Definition 12.1.1
Inner Product of Functions
The inner product of two functions f1 and f2 on an interval [a, b] is the number
#
( f1, f2) b
a
f1(x) f2(x ) dx.
Orthogonal Functions Motivated by the fact that two vectors u and v are orthogonal
whenever their inner product is zero, we define orthogonal functions in a similar manner.
Definition 12.1.2
Orthogonal Functions
Two functions f1 and f2 are said to be orthogonal on an interval [a, b] if
( f1, f2) EXAMPLE 1
#
b
a
f1(x) f2(x) dx 0.
(1)
Orthogonal Functions
The functions f1(x) x2 and f2(x) x3 are orthogonal on the interval [1, 1]. This fact follows
from (1):
( f1, f2) #
1
1
x 2 x 3 dx #
1
1
x 5 dx 1 6 1
x d 0.
6
1
Unlike vector analysis, where the word orthogonal is a synonym for perpendicular, in this
present context the term orthogonal and condition (1) have no geometric significance.
Orthogonal Sets We are primarily interested in infinite sets of orthogonal functions.
*The interval could also be (q, q), [0, q), and so on.
672
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CHAPTER 12 Orthogonal Functions and Fourier Series
Definition 12.1.3
Orthogonal Set
A set of real-valued functions {f0(x), f1(x), f2(x), …} is said to be orthogonal on an interval
[a, b] if
b
# f (x)f (x) dx 0,
(fm, fn) a
m
m Z n.
n
(2)
Orthonormal Sets The norm, or length i ui , of a vector u can be expressed in terms of
the inner product. The expression (u, u) i ui 2 is called the square norm, and so the norm is
i ui "(u, u). Similarly, the square norm of a function fn is i fn(x)i 2 (fn, fn), and so the
norm, or its generalized length, is i fn(x)i "(fn, fn). In other words, the square norm and
norm of a function fn in an orthogonal set {fn(x)} are, respectively,
ifn(x)i2 #
b
a
b
f2n(x) dx
ifn(x)i and
Å
# f (x) dx.
a
2
n
(3)
If {fn(x)} is an orthogonal set of functions on the interval [a, b] with the property that i fn(x) i 1
for n 0, 1, 2, …, then {fn(x)} is said to be an orthonormal set on the interval.
EXAMPLE 2
Orthogonal Set of Functions
Show that the set {1, cos x, cos 2x, …} is orthogonal on the interval [p, p].
SOLUTION If we make the identification f0(x) 1 and fn(x) cos nx, we must then show
p
p
that ep
f0(x)fn(x) dx 0, n Z 0, and ep
fm(x)fn(x) dx 0, m Z n. We have, in the first
case, for n Z 0,
(f0, fn) #
p
f0(x)fn(x) dx p
#
p
cos nx dx
p
p
1
1
sin nxd
fsin np 2 sin (np)g 0,
n
n
p
and in the second, for m Z n,
f(m, fn) #
1
2
EXAMPLE 3
p
fm(x)fn(x) dx p
#
#
p
cos mx cos nx dx
p
p
p
f cos (m n)x cos (m 2 n)xg dx d trigonometric
identity
sin (m 2 n)x p
1 sin (m n)x
d 0.
c
m2n
2
mn
p
Norms
Find the norms of each function in the orthogonal set given in Example 2.
SOLUTION
For f0(x) 1 we have from (3)
i f0(x)i 2 #
p
p
dx 2p
so that i f0(x)i !2p. For fn(x) cos nx, n 0, it follows that
ifn(x)i2 #
p
p
cos2 nx dx Thus for n 0, i fn(x)i !p.
1
2
#
p
p
f1 cos 2nxg dx p.
12.1 Orthogonal Functions
|
673
An orthogonal set can be
made into an orthonormal set.
Any orthogonal set of nonzero functions {fn(x)}, n 0, 1, 2, …, can be normalized—that is,
made into an orthonormal set—by dividing each function by its norm. It follows from Examples 2
and 3 that the set
e
1
cos x cos 2x
,
, pf
"2p "p "p
,
is orthonormal on the interval [p, p].
Vector Analogy We shall make one more analogy between vectors and functions. Suppose
v1, v2, and v3 are three mutually orthogonal nonzero vectors in 3-space. Such an orthogonal set
can be used as a basis for 3-space; that is, any three-dimensional vector can be written as a linear
combination
u c1v1 c2v2 c3v3,
(4)
where the ci , i 1, 2, 3, are scalars called the components of the vector. Each component ci can
be expressed in terms of u and the corresponding vector vi . To see this we take the inner product
of (4) with v1:
(u, v1) c1(v1, v1) c2(v2, v1) c3(v3, v1) c1 i v1 i 2 c2 0 c3 0.
c1 Hence
(u, v1)
.
iv1 i2
In like manner we find that the components c2 and c3 are given by
c2 (u, v2)
iv2 i2
and
c3 (u, v3)
.
iv3 i2
Hence (4) can be expressed as
3 (u, v )
(u, v3)
(u, v1)
(u, v2)
n
v
v
v
vn.
1
2
3
a
2
iv1 i2
iv2 i2
iv3 i2
n 1 ivn i
u
(5)
Orthogonal Series Expansion Suppose {fn(x)} is an infinite orthogonal set of functions on an interval [a, b]. We ask: If y f (x) is a function defined on the interval [a, b], is it
possible to determine a set of coefficients cn , n 0, 1, 2, …, for which
f (x) c0f0(x) c1f1(x) p cnfn(x) p ?
(6)
As in the foregoing discussion on finding components of a vector, we can find the coefficients cn
by utilizing the inner product. Multiplying (6) by fm(x) and integrating over the interval [a, b] gives
#
b
a
f (x)fm(x) dx c0
#
b
f0(x)fm(x) dx c1
a
#
b
f1(x)fm(x) dx p cn
a
b
# f (x)f (x) dx p
n
m
a
c0(f0, fm) c1(f1, fm) p cn(fn , fm) p .
By orthogonality, each term on the right-hand side of the last equation is zero except when m n.
In this case we have
#
b
a
b
f (x)fn(x) dx cn
# f (x) dx.
a
2
n
It follows that the required coefficients cn are given by
b
cn 674
|
CHAPTER 12 Orthogonal Functions and Fourier Series
ea f (x)fn(x) dx
b
ea f2n(x) dx
, n 0, 1, 2, p .
f (x) a cnfn(x) ,
q
In other words,
(7)
n0
b
ea f (x)fn(x) dx
.
ifn(x)i2
(8)
q
( f, fn)
f (x).
f (x) a
2 n
n 0 ifn(x)i
(9)
cn where
With inner product notation, (7) becomes
Thus (9) is seen to be the function analogue of the vector result given in (5).
Definition 12.1.4
Orthogonal Set/Weight Function
A set of real-valued functions {f0(x), f1(x), f2(x), …} is said to be orthogonal with respect
to a weight function w(x) on an interval [a, b] if
b
# w(x)f (x)f (x) dx 0,
m
a
n
m Z n.
The usual assumption is that w(x) 0 on the interval of orthogonality [a, b]. The set
{1, cos x, cos 2x, …} in Example 2 is orthogonal with respect to the weight function w(x) 1
on the interval [p, p].
If {fn(x)} is orthogonal with respect to a weight function w(x) on the interval [a, b], then
multiplying (6) by w(x)fn(x) and integrating yields
b
cn ea f (x) w (x) fn(x) dx
,
ifn(x)i2
(10)
b
where
ifn(x)i2 # w(x) f (x) dx.
a
2
n
(11)
The series (7) with coefficients cn given by either (8) or (10) is said to be an orthogonal series
expansion of f or a generalized Fourier series.
Complete Sets The procedure outlined for determining the coefficients cn was formal;
that is, basic questions on whether an orthogonal series expansion such as (7) is actually possible
were ignored. Also, to expand f in a series of orthogonal functions, it is certainly necessary that
f not be orthogonal to each fn of the orthogonal set {fn(x)}. (If f were orthogonal to every fn,
then cn 0, n 0, 1, 2, ….) To avoid the latter problem we shall assume, for the remainder of
the discussion, that an orthogonal set is complete. This means that the only continuous function
orthogonal to each member of the set is the zero function.
REMARKS
Suppose that { f0(x), f1(x), f2(x), …} is an infinite set of real-valued functions that are continuous
on an interval [a, b]. If this set is linearly independent on [a, b] (see page 357 for the definition
of an infinite linearly independent set), then it can always be made into an orthogonal set and,
as described earlier in this section, can be made into an orthonormal set. See Problem 27 in
Exercises 12.1.
12.1 Orthogonal Functions
|
675
Exercises
12.1
Answers to selected odd-numbered problems begin on page ANS-29.
A real-valued function is said to be periodic with period
T 2 0 if f (x T) f (x) for all x in the domain of f. If T is
the smallest positive value for which f (x T) f (x) holds,
then T is called the fundamental period of f. In Problems
21–26, determine the fundamental period T of the given
function.
4
21. f (x) cos 2px
22. f (x) sin x, L . 0
L
23. f (x) sin x sin 2x
24. f (x) sin 2x cos 4x
In Problems 1–6, show that the given functions are orthogonal
on the indicated interval.
1. f1(x) x, f2(x) x 2; [2, 2]
2. f1(x) x3, f2(x) x 2 1;
x
3. f1(x) e , f2(x) xe
x
[1, 1]
x
e ;
2
4. f1(x) cos x, f2(x) sin x;
5. f1(x) x, f2(x) cos 2x;
6. f1(x) ex, f2(x) sin x;
[0, 2]
[0, p]
[p/2, p/2]
[p/4, 5p/4]
25. f (x) sin 3x cos 2x
In Problems 7–12, show that the given set of functions is orthogonal on the indicated interval. Find the norm of each function in
the set.
7. {sin x, sin 3x, sin 5x, …}; [0, p/2]
8. {cos x, cos 3x, cos 5x, …}; [0, p/2]
9. {sin nx}, n 1, 2, 3, …; [0, p]
10. e sin
np
x f , n 1, 2, 3, …; [0, p]
p
11. e 1, cos
np
mp
x, sin
x f , n 1, 2, 3, …,
12. e 1, cos
p
p
m 1, 2, 3, …; [p, p]
In Problems 13 and 14, verify by direct integration that the
functions are orthogonal with respect to the indicated weight
function on the given interval.
(q, q)
[0, q)
2
w(x) ex,
15. Let {f n(x)} be an orthogonal set of functions on [a, b]
b
such that f0(x) 1. Show that ea fn(x) dx 0 for n 1, 2, ….
16. Let {fn(x)} be an orthogonal set of functions on [a, b] such
b
that f0(x) 1 and f1(x) x. Show that ea (ax b)fn(x) dx 0
for n 2, 3, … and any constants a and b.
17. Let {fn(x)} be an orthogonal set of functions on [a, b]. Show
that i fm(x) fn(x)i 2 i fm(x)i 2 i fn(x)i 2, m Z n.
18. From Problem 1 we know that f1(x) x and f2(x) x 2 are
orthogonal on [2, 2]. Find constants c1 and c2 such that
f3(x) x c1x 2 c2x 3 is orthogonal to both f1 and f2 on the
same interval.
19. The set of functions {sin nx}, n 1, 2, 3, …, is orthogonal
on the interval [p, p]. Show that the set is not complete.
20. Suppose f1, f2, and f3 are functions continuous on the interval
[a, b]. Show that ( f1 f2, f3) ( f1, f3) ( f2, f3).
|
27. The Gram–Schmidt process for constructing an orthogonal
set that was discussed in Section 7.7 carries over to a linearly
independent set { f0(x), f1(x), f2(x), …} of real-valued functions
continuous on an interval [a, b]. With the inner product
b
( fn, fn) ea fn(x)fn(x) dx, define the functions in the set
B9 5f0(x), f1(x), f2(x), p 6 to be
f1(x) f1(x) 2
( f1, f0)
f (x)
(f0, f0) 0
f2(x) f2(x) 2
( f2, f0)
( f2, f1)
f (x) 2
f (x)
(f0, f0) 0
(f1, f1) 1
o
w(x) e x ,
14. L0(x) 1, L1(x) x 1, L2(x) 12 x 2 2x 1;
676
Discussion Problems
f0(x) f0(x)
np
x f , n 1, 2, 3, …; [0, p]
p
13. H0(x) 1, H1(x) 2x, H2(x) 4x 2 2;
26. f (x) sin2 px
CHAPTER 12 Orthogonal Functions and Fourier Series
o
and so on.
(a) Write out f3(x) in the set.
(b) By construction, the set B9 5f0(x), f1(x), f2(x), p 6
is orthogonal on [a, b]. Demonstrate that f0(x), f1(x),
and f2(x) are mutually orthogonal.
28. (a) Consider the set of functions {1, x, x 2, x3, …} defined on
the interval [1, 1]. Apply the Gram–Schmidt process
given in Problem 27 to this set and find f0(x), f1(x), f2(x),
and f3(x) of the orthogonal set B.
(b) Discuss: Do you recognize the orthogonal set?
29. Verify that the inner product ( f1, f2) in Definition 12.1.1
satisfies properties (i)(iv) given on page 672.
30. In R3, give an example of a set of orthogonal vectors that
is not complete. Give a set of orthogonal vectors that is
complete.
31. The function
q
np
np
f (x) A0 a aAn cos
x Bn sin
xb,
p
p
n51
where the coefficients An and Bn depend only on n, is periodic.
Find the period T of f.
12.2 Fourier Series
INTRODUCTION We have just seen in the preceding section that if {f0(x), f1(x), f2(x), …}
is a set of real-valued functions that is orthogonal on an interval [a, b] and if f is a function defined
on the same interval, then we can formally expand f in an orthogonal series c0f0(x) c1f1(x) c2f2(x) … . In this section we shall expand functions in terms of a special orthogonal set of
trigonometric functions.
Trigonometric Series In Problem 12 in Exercises 12.1, you were asked to show that
the set of trigonometric functions
e 1, cos
2p
3p
p
2p
3p
p
x, cos
x, cos
x, p , sin x, sin
x, sin
x, p f
p
p
p
p
p
p
(1)
is orthogonal on the interval [p, p]. This set will be of special importance later on in the solution
of certain kinds of boundary-value problems involving linear partial differential equations. In
those applications we will need to expand a function f defined on [p, p] in an orthogonal series
consisting of the trigonometric functions in (1); that is,
f (x) This is why 12 a0 is used
instead of a0.
q
a0
np
np
x bn sin
xb.
a aan cos
p
p
2
n1
(2)
The coefficients a0, a1, a2, …, b1, b2, …, can be determined in exactly the same manner as in the
general discussion of orthogonal series expansions on pages 674 and 675. Before proceeding, note
that we have chosen to write the coefficient of 1 in the set (1) as 12 a0 rather than a0; this is for
convenience only because the formula of an will then reduce to a0 for n 0.
Now integrating both sides of (2) from p to p gives
#
p
p
f (x) dx a0
2
#
p
dx a aan
q
p
n1
#
p
cos
p
np
x dx bn
p
#
p
sin
p
np
x dxb.
p
(3)
Since cos(npx/p) and sin(npx/p), n 1, are orthogonal to 1 on the interval, the right side of
(3) reduces to a single term:
#
p
p
f (x) dx a0
2
#
p
p
dx Solving for a0 yields
a0 1
p
#
a0 p
x d pa0.
2 p
p
f (x) dx.
(4)
p
Now we multiply (2) by cos(mpx/p) and integrate:
#
p
p
f (x) cos
a0
mp
x dx p
2
#
p
cos
p
a aan
q
n1
mp
x dx
p
#
p
p
cos
mp
np
x cos
x dx bn
p
p
#
p
p
cos
np
mp
x sin
x dxb. (5)
p
p
By orthogonality we have
#
p
p
cos
mp
x dx 0,
p
m . 0,
#
p
p
cos
np
mp
x sin
x dx 0
p
p
12.2 Fourier Series
|
677
#
and
p
cos
p
np
0, m 2 n
mp
x cos
x dx e
p
p
p, m n.
#
Thus (5) reduces to
p
np
x dx an p,
p
f (x) cos
p
an and so
1
p
#
p
f (x) cos
p
np
x dx.
p
(6)
Finally, if we multiply (2) by sin(mpx/p), integrate, and make use of the results
#
p
#
mp
x dx 0, m . 0,
sin
p
p
#
and
p
sin
p
sin
p
np
mp
x cos
x dx 0
p
p
np
0, m 2 n
mp
x sin
x dx e
p
p
p, m n,
1
bn p
we find that
p
#
p
f (x) sin
p
np
x dx.
p
(7)
The trigonometric series (2) with coefficients a0, an , and bn defined by (4), (6), and (7),
respectively, is said to be the Fourier series of the function f. The coefficients obtained from (4),
(6), and (7) are referred to as Fourier coefficients of f.
In finding the coefficients a0, an, and bn , we assumed that f was integrable on the interval
and that (2), as well as the series obtained by multiplying (2) by cos (mpx/p), converged in such
a manner as to permit term-by-term integration. Until (2) is shown to be convergent for a given
function f, the equality sign is not to be taken in a strict or literal sense. Some texts use the
symbol in place of . In view of the fact that most functions in applications are of a type that
guarantees convergence of the series, we shall use the equality symbol. We summarize the results:
Fourier Series
Definition 12.2.1
The Fourier series of a function f defined on the interval (p, p) is given by
f (x) where
a0 1
p
an 1
p
bn 1
p
Expand
in a Fourier series.
π
–π
π
x
FIGURE 12.2.1 Function f in Example 1
678
|
#
p
#
p
#
p
(8)
f (x) dx
(9)
p
f (x) cos
np
x dx
p
(10)
f (x) sin
np
x dx.
p
(11)
p
p
Expansion in a Fourier Series
EXAMPLE 1
y
q
a0
np
np
x bn sin
xb,
a aan cos
p
p
2
n51
f (x) e
p , x , 0
000 # x , p
0,
p 2 x,
(12)
SOLUTION The graph of f is given in FIGURE 12.2.1. With p p we have from (9) and (10)
that
a0 1
p
#
p
f (x) dx p
CHAPTER 12 Orthogonal Functions and Fourier Series
1
c
p
#
0
p
0 dx p
x2
# (p 2 x) dxd p cpx 2 2 d
0
1
p
0
p
2
an 1
p
#
p
f (x) cos nx dx p
1
c
p
1
sin nx p
1
2 c(p 2 x)
p
n 0
n
#
0
p
0 dx p
#
# (p 2 x) cos nx dxd
0
p
sin nx dxd
0
1 cos nx p
d
np
n
0
cos np 1
n2p
1 2 (1)n
.
n2p
d cos np (1)n
In like manner we find from (11) that
bn Note that
0, n even
1 (1) e
2, n odd.
f (x) Therefore
n
1
p
p
# (p 2 x) sin nx dx n.
1
0
q
1 2 (1)n
1
p
cos nx sin nx f .
ae
2
n
4
np
n1
(13)
Note that an defined by (10) reduces to a0 given by (9) when we set n 0. But as Example 1
shows, this may not be the case after the integral for an is evaluated.
Convergence of a Fourier Series The following theorem gives sufficient conditions
for convergence of a Fourier series at a point.
Conditions for Convergence
Theorem 12.2.1
Let f and f be piecewise continuous on the interval [p, p]; that is, let f and f be continuous
except at a finite number of points in the interval and have only finite discontinuities at these
points. Then for all x in the interval (p, p) the Fourier series of f converges to f (x) at a point
of continuity. At a point of discontinuity, the Fourier series converges to the average
f (x1) f (x)
,
2
where f (x) and f (x) denote the limit of f at x from the right and from the left, respectively.*
For a proof of this theorem you are referred to the classic text by Churchill and Brown. †
Convergence of a Point of Discontinuity
EXAMPLE 2
The function (12) in Example 1 satisfies the conditions of Theorem 12.2.1. Thus for every x
in the interval (p, p), except at x 0, the series (13) will converge to f (x). At x 0 the
function is discontinuous, and so the series (13) will converge to
f (01) f (0)
p0
p
.
2
2
2
* In other words, for x a point in the interval and h 0,
f (x) lim f (x h),
hS0
†
f (x) lim f (x h).
hS0
Ruel V. Churchill and James Ward Brown, Fourier Series and Boundary Value Problems (New York:
McGraw-Hill, 2000).
12.2 Fourier Series
|
679
We may assume that
the given function f is
periodic.
Periodic Extension Observe that each of the functions in the basic set (1) has a different
fundamental period,* namely, 2p/n, n 1, but since a positive integer multiple of a period is also
a period we see that all of the functions have in common the period 2p (verify). Hence the righthand side of (2) is 2p-periodic; indeed, 2p is the fundamental period of the sum. We conclude that
a Fourier series not only represents the function on the interval (p, p) but also gives the periodic
extension of f outside this interval. We can now apply Theorem 12.2.1 to the periodic extension
of f, or we may assume from the outset that the given function is periodic with period T 2p; that
is, f (x T ) f (x). When f is piecewise continuous and the right- and left-hand derivatives exist
at x p and x p, respectively, then the series (8) converges to the average [ f (p) f (p)]/2
at these endpoints and to this value extended periodically to 3p, 5p, 7p, and so on. The
Fourier series in (13) converges to the periodic extension of (12) on the entire x-axis. At 0, 2p,
4p, …, and at p, 3p, 5p, …, the series converges to the values
f (01) f (0)
p
2
2
f (p) f (p1)
0,
2
and
respectively. The solid dots in FIGURE 12.2.2 represent the value p/2.
y
π
–4π –3π –2π
π
–π
2π
3π
x
4π
FIGURE 12.2.2 Periodic extension of the function f
shown in Figure 12.2.1
Sequence of Partial Sums It is interesting to see how the sequence of partial sums {SN(x)}
of a Fourier series approximates a function. For example, the first three partial sums of (13) are
S1 (x) p
p
2
p
2
1
, S2(x) cos x sin x, S3(x) cos x sin x sin 2x.
p
p
4
4
4
2
In FIGURE 12.2.3 we have used a CAS to graph the partial sums S5(x), S8(x), and S15(x) of (13) on
the interval (p, p). Figure 12.2.3(d) shows the periodic extension using S15(x) on (4p, 4p).
y
y
3
3
2
2
1
1
x
0
–3
–2
–1
0
1
2
3
–3
1
y
2
2
1
1
x
–1
0
1
2
3
(c) S 15 (x) on (– , )
FIGURE 12.2.3 Partial sums of a Fourier series
* See Problems 21–26 in Exercises 12.1.
|
0
y
0
680
–1
(b) S 8 (x) on (– , )
3
–2
–2
(a) S 5 (x) on (– , )
3
–3
x
0
CHAPTER 12 Orthogonal Functions and Fourier Series
2
3
x
0
–10
–5
5
0
(d) S 15 (x) on (–4 , 4 )
10
Exercises
12.2
Answers to selected odd-numbered problems begin on page ANS-29.
In Problems 1–16, find the Fourier series of the function f on the
given interval. Give the number to which the Fourier series
converges at a point of discontinuity of f.
0, p , x , 0
1,
0#x,p
1, p , x , 0
e
2,
0#x,p
1, 1 , x , 0
e
x,
0#x,1
0, 1 , x , 0
e
x,
0#x,1
0, p , x , 0
e 2
x,
0#x,p
p2,
p , x , 0
e 2
2
p 2 x , 0 # x , p
x p, p , x , p
3 2 2x, p , x , p
0,
p , x , 0
e
sin x, 0 # x , p
0,
p>2 , x , 0
e
cos x, >20 # x , p>2
0,
2 , x , 1
2, 1 # x , 0
μ
1,
0#x,1
0,
1#x,2
0,
2 , x , 0
• x, 0 # x , 1
1, 1 # x , 2
1,
5 , x , 0
e
1 x, 0 # x , 5
2 x, 2 , x , 0
e
2,
0 # x , 2
e x, p , x , p
0,
p , x , 0
e x
e 2 1, 0 # x , p
1. f (x) e
2. f (x) 3. f (x) 4. f (x) 5. f (x) 6. f (x) 7. f (x) 8. f (x) 9. f (x) 10. f (x) 11. f (x) 12. f (x) 13. f (x) 14. f (x) 15. f (x) 16. f (x) In Problems 17 and 18, sketch the periodic extension of the
indicated function.
17. The function f in Problem 9
18. The function f in Problem 14
19. Use the result of Problem 5 to show
p2
1
1
1
1 2 2 2p
6
2
3
4
and
p2
1
1
1
1 2 2 2 2 2 p.
12
2
3
4
20. Use Problem 19 to find a series that gives the numerical value
of p2 /8.
21. Use the result of Problem 7 to show
p
1
1
1
1 2 2 p.
4
3
5
7
22. Use the result of Problem 9 to show
p
1
1
1
1
1
2
2
p.
4
2
13
35
57
79
23. The root–mean–square value of a function f (x) defined over
an interval (a, b) is given by
b
RMS( f ) ea f 2(x) dx
.
É b2a
If the Fourier series expansion of f is given by (8), show that
the RMS value of f over the interval (p, p) is given by
RMS( f ) 1 2
1 q
a 0 a (a 2n b 2n ),
Å4
2 n51
where a0, an , and bn are the Fourier coefficients in (9), (10),
and (11), respectively.
12.3 Fourier Cosine and Sine Series
INTRODUCTION The effort expended in the evaluation of coefficients a0, an , and bn in
expanding a function f in a Fourier series is reduced significantly when f is either an even or an
odd function. A function f is said to be
even if f (x) f (x)
and
odd if f (x) f (x).
On a symmetric interval such as (p, p), the graph of an even function possesses symmetry with
respect to the y-axis, whereas the graph of an odd function possesses symmetry with respect to
the origin.
Even and Odd Functions It is likely the origin of the words even and odd derives from
the fact that the graphs of polynomial functions that consist of all even powers of x are symmetric
12.3 Fourier Cosine and Sine Series
|
681
y
with respect to the y-axis, whereas graphs of polynomials that consist of all odd powers of x are
symmetric with respect to the origin. For example,
y = x2
f(– x)
–x
x
f (x) x 2 is even since f (x) (x)2 x 2 f (x)
x
T
f(–x)
See FIGURES 12.3.1 and 12.3.2. The trigonometric cosine and sine functions are even and odd functions, respectively, since cos (x) cos x and sin (x) sin x. The exponential functions
f (x) e x and f (x) ex are neither even nor odd.
y = x3
f(x )
–x
x
x
Properties The following theorem lists some properties of even and odd functions.
Theorem 12.3.1
FIGURE 12.3.2 Odd function
odd integer
f (x) x3 is odd since f (x) (x)3 x3 f (x).
FIGURE 12.3.1 Even function
y
even integer
T
f(x )
(a)
(b)
(c)
(d)
(e)
(f )
(g)
Properties of Even/Odd Functions
The product of two even functions is even.
The product of two odd functions is even.
The product of an even function and an odd function is odd.
The sum (difference) of two even functions is even.
The sum (difference) of two odd functions is odd.
a
a
If f is even, then ea f (x) dx 2 e0 f (x) dx.
a
If f is odd, then ea f (x) dx 0.
PROOF OF (b): Let us suppose that f and g are odd functions. Then we have f (x) f (x)
and g(x) g(x). If we define the product of f and g as F(x) f (x)g(x), then
F(x) f (x)g(x) (f (x))(g(x)) f (x)g(x) F(x).
This shows that the product F of two odd functions is an even function. The proofs of the remaining properties are left as exercises. See Problem 56 in Exercises 12.3.
Cosine and Sine Series If f is an even function on the interval (p, p), then in view of
the foregoing properties, the coefficients (9), (10), and (11) of Section 12.2 become
a0 1
p
an 1
p
#
p
#
p
p
f (x) dx f (x) cos
p
2
p
p
# f (x) dx
0
2
np
x dx p
p
p
# f (x) cos
0
np
x dx
p
even
bn 1
p
#
p
p
f (x) sin
np
x dx 0.
p
odd
Similarly, when f is odd on the interval (p, p),
an 0,
n 0, 1, 2, …,
bn We summarize the results in the following definition.
682
|
CHAPTER 12 Orthogonal Functions and Fourier Series
2
p
#
p
0
f (x) sin
np
x dx.
p
Definition 12.3.1
Fourier Cosine and Sine Series
(i) The Fourier series of an even function on the interval (p, p) is the cosine series
f (x) where
q
a0
np
x,
a an cos
p
2
n51
a0 2
p
an 2
p
(1)
p
# f (x) dx
(2)
0
p
# f (x) cos
0
np
x dx.
p
(3)
(ii) The Fourier series of an odd function on the interval (p, p) is the sine series
q
np
f (x) a bn sin
x,
p
n51
bn where
p
# f (x) sin
0
np
x dx.
p
(5)
Expansion in a Sine Series
EXAMPLE 1
Expand f (x) x, 2
y
2
p
(4)
x
2, in a Fourier series.
SOLUTION Inspection of FIGURE 12.3.3 shows that the given function is odd on the interval
(2, 2), and so we expand f in a sine series. With the identification 2p 4, we have p 2.
Thus (5), after integration by parts, is
x
bn y = x, –2 < x < 2
FIGURE 12.3.3 Odd function f in
Example 1
#
2
x sin
0
f (x) Therefore
4(1)n 1
np
.
x dx np
2
np
4 q (1)n 1
sin
x.
p na
n
2
1
(6)
The function in Example 1 satisfies the conditions of Theorem 12.2.1. Hence the series (6)
converges to the function on (2, 2) and the periodic extension (of period 4) given in FIGURE 12.3.4.
y
x
–10
–8
–6
–4
–2
2
4
6
8
10
FIGURE 12.3.4 Periodic extension of the function f shown in Figure 12.3.3
EXAMPLE 2
y
1
–π
π
x
1, , p , x , 0
shown in FIGURE 12.3.5 is odd on the interval
1,
0#x,p
(p, p). With p p we have from (5)
The function f (x) e
bn –1
FIGURE 12.3.5 Odd function f in
Example 2
Expansion in a Sine Series
and so
2
p
#
p
0
(1) sin nx dx f (x) 2 1 2 (1)n
,
p
n
2 q 1 2 (1)n
sin nx.
a
p n51
n
12.3 Fourier Cosine and Sine Series
(7)
|
683
Gibbs Phenomenon With the aid of a CAS we have plotted in FIGURE 12.3.6 the graphs
S1(x), S2(x), S3(x), S15(x) of the partial sums of nonzero terms of (7). As seen in Figure 12.3.6(d)
the graph of S15(x) has pronounced spikes near the discontinuities at x 0, x p, x p, and
so on. This “overshooting” by the partial sums SN from the function values near a point of discontinuity does not smooth out but remains fairly constant, even when the value N is taken to be
large. This behavior of a Fourier series near a point at which f is discontinuous is known as the
Gibbs phenomenon.
y
y
1
1
0.5
0.5
x
0
–0.5
–0.5
–1
–1
–3
–2
–1
0
1
(a) S1(x)
2
x
0
–3
3
–2
–1
0
1
(b) S2(x)
2
3
y
y
1
1
0.5
0.5
x
0
–0.5
–0.5
–1
–1
–3
–2
–1
0
1
(c) S3(x)
2
x
0
–3
3
–2
–1
0
1
(d) S15(x)
2
3
FIGURE 12.3.6 Partial sums of sine series (7) on the interval (–p, p)
The periodic extension of f in Example 2 onto the entire x-axis is a meander function (see
page 247).
Half-Range Expansions Throughout the preceding discussion it was understood that
a function f was defined on an interval with the origin as midpoint, that is, (p, p). However, in
many instances we are interested in representing a function that is defined on an interval (0, L)
by a trigonometric series. This can be done in many different ways by supplying an arbitrary
definition of the function on the interval (L, 0). For brevity we consider the three most important
cases. If y f (x) is defined on the interval (0, L), then:
(i)
reflect the graph of the function about the y-axis onto (L, 0); the function is now even on
the interval (L, L) (see FIGURE 12.3.7); or
(ii) reflect the graph of the function through the origin onto (L, 0); the function is now odd
on the interval (L, L) (see FIGURE 12.3.8); or
(iii) define f on (L, 0) by f (x) f (x L) (see FIGURE 12.3.9).
y
y
x
–L
L
FIGURE 12.3.7 Even reflection
684
|
CHAPTER 12 Orthogonal Functions and Fourier Series
y
–L
L
x
FIGURE 12.3.8 Odd reflection
–L
L
x
FIGURE 12.3.9 Identity reflection
Note that the coefficients of the series (1) and (4) utilize only the definition of the function
on (0, p), that is, for half of the interval (p, p). Hence in practice there is no actual need to make
the reflections described in (i) and (ii). If f is defined on (0, L), we simply identify the half-period
as the length of the interval p L . The coefficient formulas (2), (3), and (5) and the corresponding series yield either an even or an odd periodic extension of period 2L of the original function.
The cosine and sine series obtained in this manner are known as half-range expansions. Last,
in case (iii) we are defining the function values on the interval (L, 0) to be the same as the
values on (0, L). As in the previous two cases, there is no real need to do this. It can be shown
that the set of functions in (1) of Section 12.2 is orthogonal on [a, a 2p] for any real number
a. Choosing a p, we obtain the limits of integration in (9), (10), and (11) of that section. But
for a 0 the limits of integration are from x 0 to x 2p. Thus if f is defined over the interval
(0, L), we identify 2p L or p L/2. The resulting Fourier series will give the periodic extension of f with period L. In this manner the values to which the series converges will be the same
on (L, 0) as on (0, L).
Expansion in Three Series
EXAMPLE 3
2
Expand f (x) x , 0
series.
y
y = x2, 0 < x < L
L, (a) in a cosine series, (b) in a sine series, (c) in a Fourier
SOLUTION The graph of the function is given in FIGURE 12.3.10.
(a) We have
2
a0 L
x
L
FIGURE 12.3.10 Function f in Example 3
x
#
L
2
x dx L2,
3
2
an L
2
0
L
#x
2
0
cos
4L2(1)n
np
x dx ,
L
n2p2
where integration by parts was used twice in the evaluation of an . Thus
f (x) L2
4L2 q (1)n
np
2a
cos
x.
3
L
p n 1 n2
(8)
(b) In this case we must again integrate by parts twice:
bn Hence
2
L
L
#x
0
2
sin
2L2(1)n 1
4L2
np
3 3 f(1)n 2 1g.
x dx np
L
np
2
2L2 q (1)n 1
np
3 2 f(1)n 2 1g f sin
e
x.
p na
n
L
np
1
f (x) (9)
(c) With p L/2, 1/p 2/L, and np/p 2np/L, we have
a0 2
L
#
L
0
x 2 dx bn and
Therefore
f (x) 2 2
L,
3
2
L
#
an L
0
x 2 sin
2
L
#
L
0
x 2 cos
2np
L2
x dx 2 2
L
np
L2
2np
x dx .
np
L
2np
2np
L2
L2 q
1
1
a e 2 cos
x 2 sin
xf.
p
n
3
L
L
n1 n p
(10)
The series (8), (9), and (10) converge to the 2L-periodic even extension of f, the 2L-periodic
odd extension of f, and the L-periodic extension of f, respectively. The graphs of these periodic
extensions are shown in FIGURE 12.3.11.
12.3 Fourier Cosine and Sine Series
|
685
y
– 4L
–3L
–2L
–L
L
2L
3L
4L
2L
3L
4L
2L
3L
4L
x
(a) Cosine series
y
– 4L
–3L
–2L
L
–L
x
(b) Sine series
y
– 4L
–3L
–2L
–L
L
x
(c) Fourier series
FIGURE 12.3.11 Different periodic extensions of the function f in Example 3
Periodic Driving Force Fourier series are sometimes useful in determining a particular
solution of a differential equation describing a physical system in which the input or driving force
f (t) is periodic. In the next example we find a particular solution of the differential equation
m
d 2x
kx f (t)
dt 2
(11)
by first representing f by a half-range sine expansion and then assuming a particular solution of
the form
q
np
t.
xp(t) a Bn sin
p
n1
Particular Solution of a DE
EXAMPLE 4
f(t)
π
1
2
3
4
5
t
An undamped spring/mass system, in which the mass m 161 slug and the spring constant
k 4 lb/ft, is driven by the 2-periodic external force f (t) shown in FIGURE 12.3.12. Although
the force f (t) acts on the system for t 0, note that if we extend the graph of the function in
a 2-periodic manner to the negative t-axis, we obtain an odd function. In practical terms this
means that we need only find the half-range sine expansion of f (t) pt, 0 t 1. With
p 1 it follows from (5) and integration by parts that
#
1
bn 2 pt sin npt dt –π
0
FIGURE 12.3.12 Periodic forcing function
f in Example 4
(12)
2(1)n 1
.
n
From (11) the differential equation of motion is seen to be
q 2(1)n 1
1 d 2x
sin npt.
4x
a
n
16 dt 2
n1
(13)
To find a particular solution xp(t) of (13), we substitute the series (12) into the differential
equation and equate coefficients of sin npt. This yields
a
686
|
2(1)n 1
32(1)n 1
1 2 2
.
n p 4b Bn or Bn n
16
n(64 2 n2p2)
CHAPTER 12 Orthogonal Functions and Fourier Series
Thus
q
32(1)n 1
sin npt.
xp(t) a
2 2
n 1 n(64 2 n p )
(14)
Observe in the solution (14) that there is no integer n 1 for which the denominator 64 n2p2
of Bn is zero. In general, if there is a value of n, say, N, for which Np/p v, where v !k>m,
then the system described by (11) is in a state of pure resonance. In other words, we have pure
resonance if the Fourier series expansion of the driving force f (t) contains a term sin(Np/L)t
(or cos(Np/L)t) that has the same frequency as the free vibrations.
Of course, if the 2p-periodic extension of the driving force f onto the negative t-axis yields an
even function, then we expand f in a cosine series.
12.3
Exercises
Answers to selected odd-numbered problems begin on page ANS-29.
In Problems 1–10, determine whether the given function is even,
odd, or neither.
1. f (x) sin 3x
2. f (x) x cos x
2
3. f (x) x x
4. f (x) x3 4x
|x|
5. f (x) e
6. f (x) ex ex
x 2,
1 , x , 0
7. f (x) e
2
x ,
0#x,1
x 5,
2 , x , 0
8. f (x) e
x 5,
0#x,2
9. f (x) x3, 0 x 2
10. f (x) |x5|
In Problems 11–24, expand the given function in an appropriate
cosine or sine series.
11.
12.
13.
14.
15.
16.
17.
18.
19.
p,
1 , x , 0
f (x) e
p,
0#x,1
1, 2 , x , 1
f (x) • 0, 1 , x , 1
1, 1 , x , 2
f (x) |x|, p x p
f (x) x, p x p
f (x) x 2, 1 x 1
f (x) x|x|, 1 x 1
f (x) p2 x 2, p x p
f (x) x 3, p x p
x 2 1, p , x , 0
f (x) e
x 1,
0#x,p
x 1,
20. f (x) e
x 2 1,
21.
22.
23.
24.
In Problems 25–34, find the half-range cosine and sine expansions
of the given function.
27.
29.
30.
31.
32.
33.
34.
In Problems 35–38, expand the given function in a Fourier series.
35. f (x) x 2, 0 x 2p
36. f (x) x, 0 x p
37. f (x) x 1, 0 x 1
38. f (x) 2 x, 0 x 2
In Problems 39–42, suppose the function y f (x), 0 x L,
given in the figure is expanded in a cosine series, in a sine series,
and in a Fourier series. Sketch the periodic extension to which
each series converges.
39.
y
y = f (x)
y
40.
y = f(x)
x
x
L
L
1 , x , 0
0#x,1
1,
2 , x , 1
x, 1 # x , 0
f (x) μ
x,
0 # x , 1
1,
1 # x , 2
p, 2p , x , p
f (x) • x,
p # x , p
p,
2p # x , 2p
f (x) |sin x|, p x p
f (x) cos x, p/2 x p/2
1, 0 , x , 12
0, 0 , x , 12
26.
f
(x)
e
0, 12 # x , 1
1, 12 # x , 1
f (x) cos x, 0 x p/2 28. f (x) sin x, 0 x p
x,
>p0 , x , p>2
f (x) e
p 2 x, p>2 # x , p
0,
0,x,p
f (x) e
x 2 p, p # x , 2p
x, 0 , x , 1
f (x) e
1, 1 # x , 2
1,
0,x,1
f (x) e
2 2 x, 1 # x , 2
f (x) x 2 x, 0 x 1
f (x) x(2 x), 0 x 2
25. f (x) e
FIGURE 12.3.13 Graph for
Problem 39
41.
FIGURE 12.3.14 Graph for
Problem 40
y
42.
y = f (x)
y
y = f(x)
x
x
L
FIGURE 12.3.15 Graph for
Problem 41
L
FIGURE 12.3.16 Graph for
Problem 42
12.3 Fourier Cosine and Sine Series
|
687
In Problems 43 and 44, proceed as in Example 4 to find a
particular solution xp(t) of equation (11) when m ⫽ 1, k ⫽ 10,
and the driving force f (t) is as given. Assume that when f (t) is
extended to the negative t-axis in a periodic manner, the
resulting function is odd.
L/3
2L/3
L
FIGURE 12.3.17 Graph for Problem 50
In Problems 45 and 46, proceed as in Example 4 to find a particular
solution xp(t) of equation (11) when m ⫽ 14 , k ⫽ 12, and the driving
force f (t) is as given. Assume that when f (t) is extended to the negative
t-axis in a periodic manner, the resulting function is even.
45. f (t) ⫽ 2pt ⫺ t 2, 0 ⬍ t ⬍ 2p; f (t ⫹ 2p) ⫽ f (t)
t,
0 , t , 12
46. f (x) ⫽ e
; f (t ⫹ 1) ⫽ f (t)
1 2 t, 12 , t , 1
47. (a) Solve the differential equation in Problem 43,
x⬙ ⫹ 10x ⫽ f (t), subject to the initial conditions x(0) ⫽ 0,
x⬘(0) ⫽ 0.
(b) Use a CAS to plot the graph of the solution x(t) in part (a).
48. (a) Solve the differential equation in Problem 45,
1
4 x⬙ ⫹ 12x ⫽ f (t), subject to the initial conditions x(0) ⫽ 1,
x⬘(0) ⫽ 0.
(b) Use a CAS to plot the graph of the solution x(t) in part (a).
49. Suppose a uniform beam of length L is simply supported at
x ⫽ 0 and at x ⫽ L. If the load per unit length is given by
w(x) ⫽ w0 x/L, 0 ⬍ x ⬍ L, then the differential equation for
the deflection y(x) is
w0 x
d 4y
⫽
,
L
dx 4
where E, I, and w0 are constants. See (4) in Section 3.9.
(a) Expand w(x) in a half-range sine series.
(b) Use the method of Example 4 to find a particular solution
y(x) of the differential equation.
50. Proceed as in Problem 49 to find a particular solution y(x)
when the load per unit length is as given in FIGURE 12.3.17.
12.4
w0
x
⫺
5,
0 ,t,p
; f (t ⫹ 2p) ⫽ f (t)
⫺5, p , t , 2p
44. f (t) ⫽ 1 ⫺ t, 0 ⬍ t ⬍ 2; f (t ⫹ 2) ⫽ f (t)
43. f (t) ⫽ e
EI
w(x)
Computer Lab Assignments
In Problems 51 and 52, use a CAS to graph the partial sums
{SN (x)} of the given trigonometric series. Experiment with different values of N and graphs on different intervals of the x-axis.
Use your graphs to conjecture a closed-form expression for a
function f defined for 0 ⬍ x ⬍ L that is represented by the series.
51. f (x) ⫽ ⫺
52. f (x) ⫽ ⫺
1 2 2(⫺1)n
p q (⫺1)n 2 1
cos
nx
⫹
sin nxd
⫹a c
n
4 n⫽1
n2p
1
4 q 1
np
np
⫹ 2 a 2 a1 2 cos
b cos
x
4
2
2
p n⫽1 n
Discussion Problems
53. Is your answer in Problem 51 or in Problem 52 unique?
Give a function f defined on a symmetric interval about the
origin (⫺a, a) that has the same trigonometric series as in
Problem 51; as in Problem 52.
54. Discuss why the Fourier cosine series expansion of f (x) ⫽ ex,
0 ⬍ x ⬍ p converges to e⫺x on the interval (⫺p, 0).
55. Suppose f (x) ⫽ e x, 0 ⬍ x ⬍ p is expanded in a cosine series,
and then f (x) ⫽ e x, 0 ⬍ x ⬍ p is expanded in a sine series. If
the two series are added and then divided by 2 (that is, the
average of the two series) we get a series with cosines and
sines that also represents f (x) ⫽ e x on the interval (0, p). Is
this a full Fourier series of f ? [Hint: What does the averaging
of the cosine and sine series represent on the interval (⫺p, 0)?]
56. Prove properties (a), (c), (d), (e), (f ), and (g) in Theo rem
12.3.1.
Complex Fourier Series
INTRODUCTION As we have seen in the preceding two sections, a real function f can be
represented by a series of sines and cosines. The functions cos nx, n ⫽ 0, 1, 2, … and sin nx,
n ⫽ 1, 2, … are real-valued functions of a real variable x. The three different real forms of Fourier
series given in Definitions 12.2.1 and 12.3.1 will be exceedingly important in Chapters 13 and
14 when we set about to solve linear partial differential equations. However, in certain applications, for example, the analysis of periodic signals in electrical engineering, it is actually more
convenient to represent a function f in an infinite series of complex-valued functions of a real
variable x such as the exponential functions einx, n ⫽ 0, 1, 2, …, and where i is the imaginary unit
defined by i2 ⫽ ⫺1. Recall for x a real number, Euler’s formula
eix ⫽ cos x ⫹ i sin x
gives
e⫺ix ⫽ cos x ⫺ i sin x.
(1)
In this section we are going to use the results in (1) to recast the Fourier series in Definition 12.2.1
into a complex form or exponential form. We will see that we can represent a real function by
688
|
CHAPTER 12 Orthogonal Functions and Fourier Series
a complex series: a series in which the coefficients are complex numbers. To that end, recall that
a complex number is a number z a ib, where a and b are real numbers, and i2 1. The
number z a ib is called the conjugate of z.
Complex Fourier Series If we first add the two expressions in (1) and solve for cos x
and then subtract the two expressions and solve for sin x, we arrive at
cos x e ix e ix
2
sin x and
e ix 2 e ix
.
2i
(2)
Using (2) to replace cos(npx/p) and sin(npx/p) in (8) of Section 12.2, the Fourier series of a
function f can be written
q
a0
e inpx>p e inpx>p
e inpx>p 2 e inpx>p
a can
bn
d
2
2
2i
n1
q
a0
1
1
a c (an 2 ibn)e inpx>p (an ibn)e inpx>p d
2
2
n1 2
c0 a cne inpx>p a cne inpx>p,
q
q
n1
n1
(3)
where c0 12 a0, cn 12 (an ibn), and cn 12 (an ibn). The symbols a0, an , and bn are the coefficients (9), (10), and (11) respectively, in Definition 12.2.1. When the function f is real, cn and
cn are complex conjugates and can also be written in terms of complex exponential functions:
c0 cn 1 1
2 p
f (x) dx,
(4)
p
1
1 1
(a 2 ibn) a
2 n
2 p
1
2p
1
2p
cn #
p
#
p
#
p
p
f (x) c cos
#
p
f (x) cos
p
1
np
x dx 2 i
p
p
#
p
f (x) sin
p
np
x dxb
p
np
np
x 2 i sin
xd dx
p
p
f (x) e inpx>p dx,
(5)
p
1
1 1
(an ibn) a
2
2 p
p
1
2p
#
1
2p
p
p
#
f (x) c cos
#
p
f (x) cos
p
1
np
x dx i
p
p
#
p
f (x) sin
p
np
x dxb
p
np
np
x i sin
xd dx
p
p
f (x) e inpx>p dx.
(6)
p
Since the subscripts of the coefficients and exponents range over the entire set of integers…3,
2, 1, 0, 1, 2, 3, …, we can write the results in (3), (4), (5), and (6) in a more compact manner
by summing over both the negative and nonnegative integers. In other words, we can use one
summation and one integral that defines all three coefficients c0, cn , and cn.
Definition 12.4.1
Complex Fourier Series
The complex Fourier series of functions f defined on an interval (p, p) is given by
f (x) a cne inpx>p,
q
(7)
n q
where
cn 1
2p
#
p
p
f (x) e inpx>p dx,
n 0,
1,
2, p .
12.4 Complex Fourier Series
(8)
|
689
If f satisfies the hypotheses of Theorem 12.2.1, a complex Fourier series converges to f (x) at
a point of continuity and to the average
f (x1) f (x)
2
at a point of discontinuity.
EXAMPLE 1
Complex Fourier Series
Expand f (x) ex, p
p, in a complex Fourier series.
x
SOLUTION With p p, (8) gives
cn 1
2p
#
p
e xe inx dx p
1
2p
#
p
e (in 1)x dx
p
1
ce (in 1)p 2 e (in 1)p d .
2p(in 1)
We can simplify the coefficients cn somewhat using Euler’s formula:
e(in1)p ep(cos np i sin np) (1)nep
e(in1)p ep(cos np i sin np) (1)nep,
and
since cos np (1)n and sin np 0. Hence
cn (1)n
(e p 2 e p)
sinh p 1 2 in
.
(1)n
p n2 1
2(in 1)p
(9)
The complex Fourier series is then
f (x) sinh p q
1 2 in inx
e .
(1)n 2
p n a
n
1
q
(10)
The series (10) converges to the 2p-periodic extension of f.
You may get the impression that we have just made life more complicated by introducing a complex version of a Fourier series. The reality of the situation is that in areas of engineering, the form
(7) given in Definition 12.4.1 is sometimes more useful than that given in (8) of Definition 12.2.1.
Fundamental Frequency The Fourier series in Definitions 12.2.1 and 12.4.1 define a
periodic function and the fundamental period of that function (that is, the periodic extension
of f ) is T 2p. Since p T/2, (8) of Section 12.2 and (7) become, respectively,
q
a0
a (an cos nvx bn sin nvx)
2
n1
and
invx
,
a cne
q
(11)
n q
where number v 2p/T is called the fundamental angular frequency. In Example 1 the periodic extension of the function has period T 2p; the fundamental angular frequency is
v 2p/2p 1.
Frequency Spectrum In the study of time-periodic signals, electrical engineers find
it informative to examine various spectra of a wave form. If f is periodic and has fundamental
period T, the plot of the points (nv, |cn|), where v is the fundamental angular frequency and the
cn are the coefficients defined in (8), is called the frequency spectrum of f.
EXAMPLE 2
Frequency Spectrum
In Example 1, v 1 so that nv takes on the values 0, 1, 2, …. Using | a ib | "a2 b2 ,
we see from (9) that
Zcn Z 690
|
CHAPTER 12 Orthogonal Functions and Fourier Series
sinh p
1
.
2
p
"n 1
The following table shows some values of n and corresponding values of cn.
n
3
2
1
0
1
2
|cn|
3.5
3
2.5
2
1.5
1
0.5
–3ω –2ω – ω
Zcn Z
1.162
1.644
2.599
3.676
2.599
3
1.644
1.162
The graph in FIGURE 12.4.1, lines with arrowheads terminating at the points, is a portion of the
frequency spectrum of f.
Frequency Spectrum
EXAMPLE 3
ω
0
2ω
3ω
frequency
Find the frequency spectrum of the periodic square wave or periodic pulse shown in FIGURE 12.4.2.
The wave is the periodic extension of the function f :
FIGURE 12.4.1 Frequency spectrum of f in
Example 2
y
12 , x , 14
14 , x , 14
14 , x , 12.
0,
f (x) • 1,
0,
SOLUTION Here T 1 2p so p 12 . Since f is 0 on the intervals (12 , 14 ) and (14 , 12 ), (8) becomes
–1
x
1
cn FIGURE 12.4.2 Periodic pulse in
Example 3
|cn|
0.5
0.4
cn That is,
0.3
#
1>2
1>2
f (x) e 2inpx dx #
1>4
1>4
1 e 2inpx dx
e 2inpx 1>4
d
2inp 1>4
1 e inp>2 2 e inp>2
.
np
2i
1
np
sin
.
np
2
d by (2)
0.2
Since the last result is not valid at n 0, we compute that term separately:
0.1
c0 –5ω –4ω –3ω –2ω – ω 0 ω 2ω 3ω 4ω 5ω
frequency
FIGURE 12.4.3 Frequency spectrum of f in
Example 3
#
1>4
1
dx .
2
1>4
The following table shows some of the values of |cn|, and FIGURE 12.4.3 shows the
n
–5
–4
–3
–2
–1
0
1
2
3
4
5
Zcn Z
1
5p
0
1
3p
0
1
p
1
2
1
p
0
1
3p
0
1
5p
frequency spectrum of f. Since the fundamental frequency is v 2p/T 2p, the units nv on the
horizontal scale are 2p, 4p, 6p, …. The curved dashed lines were added in Figure 12.4.3
to emphasize the presence of the zero values of |cn| when n is an even nonzero integer.
12.4
Exercises
Answers to selected odd-numbered problems begin on page ANS-30.
In Problems 1–6, find the complex Fourier series of f on the
given interval.
1. f (x) e
1,
1,
0,
2. f(x) e
1,
0,
3. f (x) • 1,
0,
4. f (x) e
0,
x,
5. f (x) x, 0
periodic extension of the function f in Problem 3.
0,x,1
1,x,2
12 , x , 0
0 , x , 14
1
1
4 , x , 2
p , x , 0
0,x,p
2p
6. f (x) e|x|, 1
periodic extension of the function f in Problem 1.
8. Find the frequency spectrum of the periodic wave that is the
2 , x , 0
0,x,2
x
7. Find the frequency spectrum of the periodic wave that is the
x
1
In Problems 9 and 10, sketch the given periodic wave. Find the
frequency spectrum of f.
9. f (x) 4 sin x, 0 x p; f (x p) f (x) [Hint: Use (2).]
cos x,
0 , x , p>2
10. f (x) e
; f (x p) f (x)
0,
p>2 , x , p
11. (a) Show that an cn cn and bn i(cn cn).
(b) Use the results in part (a) and the complex Fourier series
in Example 1 to obtain the Fourier series expansion of f.
12. The function f in Problem 1 is odd. Use the complex Fourier
series to obtain the Fourier sine series expansion of f.
12.4 Complex Fourier Series
|
691
12.5
Sturm–Liouville Problem
INTRODUCTION For convenience we present here a brief review of some of the ordinary
differential equations that will be of importance in the sections and chapters that follow.
Linear equations
General solutions
y c1e ax
y9 ay 0,
y0 a2y 0,
a.0
y0 2 a2y 0,
a.0
y c1 cos ax c2 sin ax
e
Cauchy–Euler equation
y c1e ax c2e ax, or
y c1 cosh ax c2 sinh ax
General solutions, x + 0
x 2y0 xy9 2 a2y 0,
e
a$0
Parametric Bessel equation (n ⴝ 0)
y c1x a c2x a,
y c1 c2ln x,
aZ0
a0
General solution, x + 0
2
xy0 y9 a xy 0
y c1J0(ax) c2Y0(ax)
Legendre’s equation
(n ⴝ 0, 1, 2, …)
Particular solutions
are polynomials
(1 2 x 2)y0 2 2xy9 n(n 1)y 0
y P0(x) 1,
y P1(x) x,
y P2(x) 12(3x 2 2 1), p
Regarding the two forms of the general solution of y a2y 0, we will, in the future, employ
the following informal rule:
This rule will be useful in
Chapters 13 and 14.
Use the exponential form y c1eax c2eax when the domain of x is an infinite or
semi-infinite interval; use the hyperbolic form y c1 cosh ax c2 sinh ax when the domain
of x is a finite interval.
Eigenvalues and Eigenfunctions Orthogonal functions arise in the solution of differential equations. More to the point, an orthogonal set of functions can be generated by solving
a two-point boundary-value problem involving a linear second-order differential equation containing a parameter l. In Example 2 of Section 3.9 we saw that the boundary-value problem
y ly 0,
y(0) 0,
y(L) 0,
(1)
possessed nontrivial solutions only when the parameter l took on the values ln n2p2/L2,
n 1, 2, 3, … called eigenvalues. The corresponding nontrivial solutions y c2 sin(npx/L) or
simply y sin(npx/L) are called the eigenfunctions of the problem. For example, for (1) we have
BVP:
not an eigenvalue
T
y 5y 0,
Solution is trivial:
y(0) 0,
y(L) 0
y 0.
is an eigenvalue (n 2)
T
BVP:
y 4p2
y 0,
L2
Solution is nontrivial:
692
|
CHAPTER 12 Orthogonal Functions and Fourier Series
y(0) 0,
y(L) 0
y sin(2px/L).
For our purposes in this chapter it is important to recognize the set of functions generated by this
BVP; that is, {sin(npx/L)}, n 1, 2, 3, …, is the orthogonal set of functions on the interval [0, L]
used as the basis for the Fourier sine series.
EXAMPLE 1
Eigenvalues and Eigenfunctions
It is left as an exercise to show, by considering the three possible cases for the parameter l
(zero, negative, or positive; that is, l 0, l a2 0, a 0, and l a2 0, a 0), that
the eigenvalues and eigenfunctions for the boundary-value problem
y ly 0,
y(0) 0,
y(L) 0
(2)
are, respectively, ln a2n n2p2 /L2, n 0, 1, 2, …, and y c1 cos (npx/L), c1 Z 0. In contrast
to (1), l0 0 is an eigenvalue for this BVP and y 1 is the corresponding eigenfunction. The
latter comes from solving y 0 subject to the same boundary conditions y(0) 0, y(L) 0.
Note also that y 1 can be incorporated into the family y cos (npx/L) by permitting n 0.
The set {cos (npx/L)}, n 0, 1, 2, 3, …, is orthogonal on the interval [0, L]. See Problem 3
in Exercises 12.5.
Regular Sturm–Liouville Problem The problems in (1) and (2) are special cases of an
important general two-point boundary-value problem. Let p, q, r, and r be real-valued functions
continuous on an interval [a, b], and let r (x) 0 and p(x) 0 for every x in the interval. Then
Solve:
d
[r(x)y] (q(x) lp(x))y 0
dx
(3)
Subject to: A1y(a) B1y(a) 0
(4)
A2 y(b) B2 y(b) 0
(5)
is said to be a regular Sturm–Liouville problem. The coefficients in the boundary conditions (4)
and (5) are assumed to be real and independent of l. In addition, A1 and B1 are not both zero, and A2
and B2 are not both zero. The boundary-value problems in (1) and (2) are regular Sturm–Liouville
problems. From (1) we can identify r(x) 1, q(x) 0, and p(x) 1 in the differential equation (3);
in boundary condition (4) we identify a 0, A1 1, B1 0, and in (5), b L, A2 1, B2 0. From
(2) the identifications would be a 0, A1 0, B1 1 in (4), and b L, A2 0, B2 1 in (5).
The differential equation (3) is linear and homogeneous. The boundary conditions in (4) and (5),
both a linear combination of y and y equal to zero at a point, are also called homogeneous. A
boundary condition such as A2 y(b) B2 y(b) C2, where C2 is a nonzero constant, is nonhomogeneous. Naturally, a boundary-value problem that consists of a homogeneous linear differential equation and homogeneous boundary conditions is said to be homogeneous; otherwise
it is nonhomogeneous. The boundary conditions (4) and (5) are said to be separated because
each condition involves only a single boundary point. Boundary conditions are referred to as
mixed if each condition involves both boundary points x a and x b. For example, the periodic
boundary conditions y(a) y(b), y(a) y(b) are mixed boundary conditions.
Because a regular Sturm–Liouville problem is a homogeneous BVP, it always possesses the trivial
solution y 0. However, this solution is of no interest to us. As in Example 1, in solving such a problem
we seek numbers l (eigenvalues) and nontrivial solutions y that depend on l (eigenfunctions).
Properties Theorem 12.5.1 is a list of some of the more important of the many properties
of the regular Sturm–Liouville problem. We shall prove only the last property.
Theorem 12.5.1
Properties of the Regular Sturm–Liouville Problem
(a) There exist an infinite number of real eigenvalues that can be arranged in increasing
order l1 l2 l3 p ln p such that ln S q as n S q.
(b) For each eigenvalue there is only one eigenfunction (except for nonzero constant
multiples).
(c) Eigenfunctions corresponding to different eigenvalues are linearly independent.
(d) The set of eigenfunctions corresponding to the set of eigenvalues is orthogonal with
respect to the weight function p(x) on the interval [a, b].
12.5 Sturm–Liouville Problem
|
693
PROOF OF (d): Let ym and yn be eigenfunctions corresponding to eigenvalues lm and ln ,
respectively. Then
d
[r (x)ym ] (q(x) lm p(x))ym 0
dx
(6)
d
[r (x)yn ] (q(x) ln p(x))yn 0.
dx
(7)
Multiplying (6) by yn and (7) by ym and subtracting the two equations gives
(lm 2 ln) p(x)ymyn ym
d
d
fr (x) yn9g 2 yn
fr (x)ym9g.
dx
dx
Integrating this last result by parts from x a to x b then yields
b
(lm ln)
# p(x)y y dx r (b) [y (b) y9(b) y (b)y 9(b)]
a
m n
m
n
n
m
r (a)[ ym(a)yn9(a) yn(a)ym9(a)]. (8)
Now the eigenfunctions ym and yn must both satisfy the boundary conditions (4) and (5). In
particular, from (4) we have
A1ym(a) B1 ym9(a) 0
A1yn(a) B1 yn9(a) 0.
For this system to be satisfied by A1 and B1, not both zero, the determinant of the coefficients
must be zero:
ym(a)yn9(a) yn(a)ym9(a) 0.
A similar argument applied to (5) also gives
ym(b)yn9(b) yn(b)ym9(b) 0.
Using these last two results in (8) shows that both members of the right-hand side are zero. Hence
we have established the orthogonality relation
#
b
a
p(x)ym(x)yn(x) dx 0, lm Z ln.
(9)
It can also be proved that the orthogonal set of eigenfunctions {y1(x), y2(x), y3(x), …} of a
regular Sturm–Liouville problem is complete on [a, b]. See page 675.
EXAMPLE 2
A Regular Sturm–Liouville Problem
Solve the boundary-value problem
y
y = tan x
y ly 0,
y(0) 0,
y(1) y(1) 0.
(10)
2
x1
x2
x3
x4
y = –x
x
SOLUTION You should verify that for l 0 and for l a
0, where a 0, the BVP
in (10) possesses only the trivial solution y 0. For l a2 0, a 0, the general solution
of the differential equation y a2y 0 is y c1cos ax c2sin ax. Now the condition
y(0) 0 implies c1 0 in this solution and so we are left with y c2 sin ax. The second
boundary condition y(1) y(1) 0 is satisfied if
c2 sin a c2a cos a 0.
Choosing c2 Z 0, we see that the last equation is equivalent to
tan a a.
FIGURE 12.5.1 Positive roots of
tan x x in Example 2
694
|
(11)
If we let x a in (11), then FIGURE 12.5.1 shows the plausibility that there exists an infinite
number of roots of the equation tan x x, namely, the x-coordinates of the points where
CHAPTER 12 Orthogonal Functions and Fourier Series
the graph of y x intersects the branches of the graph of y tan x. The eigenvalues of
problem (10) are then ln a2n, where an, n 1, 2, 3, …, are the consecutive positive roots
a1, a2, a3, … of (11). With the aid of a CAS it is easily shown that, to four rounded decimal
places, a1 2.0288, a2 4.9132, a3 7.9787, and a4 11.0855, and the corresponding
solutions are y1 sin 2.0288x, y2 sin 4.9132x, y3 sin 7.9787x, and y4 sin 11.0855x.
In general, the eigenfunctions of the problem are {sin an x}, n 1, 2, 3, ….
With identifications r (x) 1, q(x) 0, p(x) 1, A1 1, B1 0, A2 1, and B2 1 we
see that (10) is a regular Sturm–Liouville problem. Thus {sin an x}, n 1, 2, 3, … is an orthogonal set with respect to the weight function p(x) 1 on the interval [0, 1].
In some circumstances we can prove the orthogonality of the solutions of (3) without the
necessity of specifying a boundary condition at x a and at x b.
Singular Sturm–Liouville Problem There are several other important conditions
under which we seek nontrivial solutions of the differential equation (3):
• r (a) 0 and a boundary condition of the type given in (5) is specified at x b;
(12)
• r (b) 0 and a boundary condition of the type given in (4) is specified at x a;
(13)
• r (a) r (b) 0 and no boundary condition is specified at either x a or at x b; (14)
• r (a) r (b) and boundary conditions y(a) y(b), y(a) y(b).
(15)
The differential equation (3) along with one of conditions (12) or (13) is said to be a singular
boundary-value problem. Equation (3) with the conditions specified in (15) is said to be a
periodic boundary-value problem because the boundary conditions are periodic. Observe that
if, say, r (a) 0, then x a may be a singular point of the differential equation, and consequently
a solution of (3) may become unbounded as x S a. However, we see from (8) that if r (a) 0,
then no boundary condition is required at x a to prove orthogonality of the eigenfunctions
provided these solutions are bounded at that point. This latter requirement guarantees the existence
of the integrals involved. By assuming the solutions of (3) are bounded on the closed interval [a, b]
we can see from inspection of (8) that
• If r (a) 0, then the orthogonality relation (9) holds with no boundary
condition at x a;
(16)
• If r (b) 0, then the orthogonality relation (9) holds with no boundary
condition at x b*;
(17)
• If r (a) r (b) 0, then the orthogonality relation (9) holds with no boundary
conditions specified at either x a or x b;
(18)
• If r (a) r (b), then the orthogonality relation (9) holds with the periodic
boundary conditions y(a) y(b), y(a) y(b).
(19)
d
Self-Adjoint Form If we carry out the differentiation fr (x)y9g , the differential
dx
equation in (3) is the same as
r (x)y0 r9(x)y9 (q(x) lp(x))y 0.
(20)
For example, Legendre’s differential equation (1 2 x 2 )y0 2 2xy9 n(n 1)y 0 is exactly
of the form given in (20) with r (x) 1 2 x 2 and r9(x) 2x. In other words, another way of
writing Legendre’s DE is
d
f(1 2 x 2)y9g n(n 1)y 0.
(21)
dx
But if you compare other second-order DEs (say, Bessel’s equation, Cauchy–Euler equations,
and DEs with constant coefficients) you might believe, given the coefficient of y is the derivative
of the coefficient of y , that few other second-order DEs have the form given in (3). On the contrary, if the coefficients are continuous and a(x) Z 0 for all x in some interval, then any secondorder differential equation
a(x)y b(x)y (c(x) ld(x))y 0
(22)
*Conditions (16) and (17) are equivalent to choosing A1 0, B1 0 in (4), and A2 0, B2 0 in (5),
respectively.
12.5 Sturm–Liouville Problem
|
695
can be recast into the so-called self-adjoint form (3). To see this, we proceed as in Section 2.3
d
where we rewrote a linear first-order equation a1(x)y9 ⫹ a0(x)y ⫽ 0 in the form fµyg ⫽ 0 by
dx
dividing the equation by a1(x) and then multiplying by the integrating factor µ ⫽ e eP(x) dx where,
assuming no common factors, P(x) ⫽ a0(x)/a1(x). So first, we divide (22) by a(x). The first two
b(x)
terms are then Y9 ⫹
Y ⫹ p where, for emphasis, we have written Y ⫽ y⬘. Second, we mula(x)
tiply this equation by the integrating factor e e(b(x)>a(x)) dx , where a(x) and b(x) are assumed to have
no common factors
e e(b(x)>a(x)) dx Y9 ⫹
b(x) e(b(x)>a(x)) dx
d e(b(x)>a(x)) dx
d e(b(x)>a(x)) dx
e
Y⫹p⫽
fe
Yg ⫹ p ⫽
fe
y9g ⫹ p .
a(x)
dx
dx
derivative of a product
In summary, by dividing (22) by a(x) and then multiplying by e e(b(x)>a(x)) dx we get
e e(b>a) dx y0 ⫹
d(x) e(b>a) dx
b(x) e(b>a) dx
c(x) e(b>a) dx
e
y9 ⫹ a
e
⫹l
e
b y ⫽ 0.
a(x)
a(x)
a(x)
(23)
Equation (23) is the desired form given in (20) and is the same as (3):
d(x) e(b>a) dx
c(x) e(b>a) dx
d e(b>a) dx
ce
y9 d ⫹ a
e
⫹l
e
by ⫽ 0.
dx
a(x)
a(x)
r (x)
q(x)
p(x)
For example, to express 3y⬙ ⫹ 6y⬘ ⫹ ly ⫽ 0 in self-adjoint form, we write y⬙ ⫹ 2y⬘ ⫹ l 13 y ⫽ 0
and then multiply by e e2 dx ⫽ e 2x . The resulting equation is
r(x)
T
r⬘(x)
T
p(x)
T
1
e 2xy0 ⫹ 2e 2xy9 ⫹ l e 2xy ⫽ 0
3
Note.
d 2x
1
fe y9g ⫹ l e 2x y ⫽ 0.
dx
3
or
It is certainly not necessary to put a second-order differential equation (22) into the self-adjoint
form (3) in order to solve the DE. For our purposes we use the form given in (3) to determine the
weight function p(x) needed in the orthogonality relation (9). The next two examples illustrate
orthogonality relations for Bessel functions and for Legendre polynomials.
EXAMPLE 3
Parametric Bessel Equation
In Section 5.3 we saw that the general solution of the parametric Bessel differential equation
x 2y⬙ ⫹ xy⬘ ⫹ (a2 x 2 ⫺ n2)y ⫽ 0, n ⫽ 0, 1, 2, … is y ⫽ c1Jn(ax) ⫹ c2Yn(ax). After dividing the
parametric Bessel equation by the lead coefficient x 2 and multiplying the resulting equation
by the integrating factor e e(1>x) dx ⫽ e ln x ⫽ x, x . 0, we obtain the self-adjoint form
n2
by⫽0
x
xy0 ⫹ y9 ⫹ aa2x 2
or
d
n2
fxy9g ⫹ aa2x 2 b y ⫽ 0,
x
dx
where we identify r (x) ⫽ x, q(x) ⫽ ⫺n2 /x, p(x) ⫽ x, and l ⫽ a2. Now r(0) ⫽ 0, and of the
two solutions Jn(ax) and Yn(a x) only Jn(a x) is bounded at x ⫽ 0. Thus in view of (16) above,
the set {Jn(ai x)}, i ⫽ 1, 2, 3, …, is orthogonal with respect to the weight function p(x) ⫽ x on
an interval [0, b]. The orthogonality relation is
#
b
0
x Jn(ai x)Jn(aj x) dx ⫽ 0,
li Z lj,
(24)
provided the ai, and hence the eigenvalues li ⫽ a2i , i ⫽ 1, 2, 3, …, are defined by means of
a boundary condition at x ⫽ b of the type given in (5):
A2Jn(lb) ⫹ B2aJ⬘n(ab) ⫽ 0.
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CHAPTER 12 Orthogonal Functions and Fourier Series
(25)
The extra factor of a in (25) comes from the Chain Rule:
d
d
Jn(ax) Jn9(ax)
ax aJn9(ax).
dx
dx
For any choice of A2 and B2, not both zero, it is known that (25) has an infinite number of roots
xi ai b. The eigenvalues are then li a2i (xi /b)2 . More will be said about eigenvalues in the
next chapter.
EXAMPLE 4
Legendre’s Equation
From the result given in (21) we can identify q(x) 0, p(x) 1, and l n(n 1). Recall
from Section 5.3 when n 0, 1, 2, … Legendre’s DE possesses polynomial solutions Pn(x).
Now we can put the observation that r (1) r (1) 0 together with the fact that the
Legendre polynomials Pn(x) are the only solutions of (21) that are bounded on the closed
interval [1, 1], to conclude from (18) that the set {Pn(x)}, n 0, 1, 2, …, is orthogonal with
respect to the weight function p(x) 1 on [1, 1]. The orthogonality relation is
#
1
1
Pm(x)Pn(x) dx 0,
m Z n.
REMARKS
(i) A Sturm–Liouville problem is also considered to be singular when the interval under
consideration is infinite. See Problems 9 and 10 in Exercises 12.5.
(ii) Even when the conditions on the coefficients p, q, r, and r are as assumed in the regular
Sturm–Liouville problem, if the boundary conditions are periodic, then property (b) of
Theorem 12.5.1 does not hold. You are asked to show in Problem 4 of Exercises 12.5 that
corresponding to each eigenvalue of the BVP
y0 ly 0,
y(L) y(L),
y9(L) y9(L)
there exist two linearly independent eigenfunctions.
12.5
Exercises
Answers to selected odd-numbered problems begin on page ANS-30.
In Problems 1 and 2, find the eigenfunctions and the equation
that defines the eigenvalues for the given boundary-value
problem. Use a CAS to approximate the first four eigenvalues
l1, l2, l3, and l4. Give the eigenfunctions corresponding to
these approximations.
1. y ly 0, y(0) 0, y(1) y(1) 0
2. y ly 0, y(0) y(0) 0, y(1) 0
3. Consider y ly 0 subject to y(0) 0, y(L) 0. Show
that the eigenfunctions are
p
2p
x, p f .
e 1, cos x, cos
L
L
This set, which is orthogonal on [0, L], is the basis for the
Fourier cosine series.
4. Consider y ly 0 subject to the periodic boundary conditions y(L) y(L), y(L) y(L). Show that the eigenfunctions are
e 1, cos
p
2p
p
2p
3p
x, cos
x, p , sin x, sin
x, sin
x, p f .
L
L
L
L
L
This set, which is orthogonal on [L, L], is the basis for the
Fourier series.
5. Find the square norm of each eigenfunction in Problem 1.
6. Show that for the eigenfunctions in Example 2,
i sin anx i2 1
f1 cos 2 ang.
2
7. (a) Find the eigenvalues and eigenfunctions of the boundary-
value problem
x 2y xy ly 0, y(1) 0, y(5) 0.
(b) Put the differential equation in self-adjoint form.
(c) Give an orthogonality relation.
8. (a) Find the eigenvalues and eigenfunctions of the boundaryvalue problem
y y ly 0, y(0) 0, y(2) 0.
(b) Put the differential equation in self-adjoint form.
(c) Give an orthogonality relation.
9. Laguerre’s differential equation
xy0 (1 2 x)y9 ny 0,
n 0, 1, 2, p ,
has polynomial solutions Ln(x). Put the equation in self-adjoint
form and give an orthogonality relation.
12.5 Sturm–Liouville Problem
|
697
10. Hermite’s differential equation
y0 2 2xy9 2ny 0,
Discussion Problem
n 0, 1, 2, p ,
has polynomial solutions Hn(x). Put the equation in self-adjoint
form and give an orthogonality relation.
11. Consider the regular Sturm–Liouville problem:
d
l
f(1 x 2)y9g y 0,
dx
1 x2
y(0) 0,
y(1) 0.
(a) Find the eigenvalues and eigenfunctions of the boundaryvalue problem. [Hint: Let x tan u and then use the
Chain Rule.]
(b) Give an orthogonality relation.
12. (a) Find the eigenfunctions and the equation that defines the
eigenvalues for the boundary-value problem
x 2y0 xy9 (lx 2 2 1)y 0,
y is bounded at x 0,
y(3) 0.
(b) Use Table 5.3.1 of Section 5.3 to find the approximate
values of the first four eigenvalues l1, l2, l3, and l4.
12.6
13. Consider the special case of the regular Sturm–Liouville
problem on the interval [a, b]:
d
fr (x)y9g lp(x)y 0,
dx
y9(a) 0,
y9(b) 0.
Is l 0 an eigenvalue of the problem? Defend your answer.
Computer Lab Assignments
14. (a) Give an orthogonality relation for the Sturm–Liouville
problem in Problem 1.
(b) Use a CAS as an aid in verifying the orthogonality relation for the eigenfunctions y1 and y2 that correspond to
the first two eigenvalues l1 and l2, respectively.
15. (a) Give an orthogonality relation for the Sturm–Liouville
problem in Problem 2.
(b) Use a CAS as an aid in verifying the orthogonality relation for the eigenfunctions y1 and y2 that correspond to
the first two eigenvalues l1 and l2, respectively.
Bessel and Legendre Series
INTRODUCTION Fourier series, Fourier cosine series, and Fourier sine series are three ways
of expanding a function in terms of an orthogonal set of functions. But such expansions are by
no means limited to orthogonal sets of trigonometric functions. We saw in Section 12.1 that a
function f defined on an interval (a, b) could be expanded, at least in a formal manner, in terms
of any set of functions {fn(x)} that is orthogonal with respect to a weight function on [a, b].
Many of these orthogonal series expansions or generalized Fourier series derive from Sturm–
Liouville problems that, in turn, arise from attempts to solve linear partial differential equations
serving as models for physical systems. Fourier series and orthogonal series expansions (the
latter includes the two series considered in this section) will appear in the subsequent consideration of these applications in Chapters 13 and 14.
12.6.1 Fourier–Bessel Series
We saw in Example 3 of Section 12.5 that for a fixed value of n the set of Bessel functions
{Jn(ai x)}, i 1, 2, 3, … , is orthogonal with respect to the weight function p(x) x on an interval [0, b] when the ai are defined by means of a boundary condition of the form
A2Jn(ab) B2aJn9(ab) 0.
(1)
The eigenvalues of the corresponding Sturm–Liouville problem are li a2i . From (7) and (8)
of Section 12.1 the orthogonal series expansion or generalized Fourier series of a function f
defined on the interval (0, b) in terms of this orthogonal set is
f (x) a ci Jn(ai x),
q
(2)
i 1
b
where
ci e0 x Jn(ai x) f (x) dx
.
iJn(ai x)i2
(3)
The square norm of the function Jn(ai x) is defined by (11) of Section 12.1:
b
i Jn(ai x)i2 # x J (a x) dx.
0
2
n
i
The series (2) with coefficients (3) is called a Fourier–Bessel series.
698
|
CHAPTER 12 Orthogonal Functions and Fourier Series
(4)
Differential Recurrence Relations The differential recurrence relations that were
given in (22) and (23) of Section 5.3 are often useful in the evaluation of the coefficients (3). For
convenience we reproduce those relations here:
d
fx nJn(x)g x nJn 2 1(x)
dx
(5)
d n
fx Jn(x)g x nJn 1(x).
dx
(6)
Square Norm The value of the square norm (4) depends on how the eigenvalues
li a2i are defined. If y Jn(ax), then we know from Example 3 of Section 12.5 that
d
n2
fxy9g aa2x 2 b y 0.
x
dx
After we multiply by 2xy, this equation can be written as
d
d
fxy9g 2 (a2x 2 2 n2 )
fyg 2 0.
dx
dx
Integrating the last result by parts on [0, b] then gives
#
b
b
2a2 xy 2 dx ¢ fxy9g 2 (a2x 2 2 n2)y 2 ≤ d .
0
0
Since y Jn(ax), the lower limit is zero for n 0 because Jn(0) 0. For n 0, the quantity
[xy]2 a2 x 2y2 is zero at x 0. Thus
#
b
2a2 x J n2(ax) dx a2b 2 fJn9(ab)g 2 (a2b 2 2 n2)fJn(ab)g 2,
(7)
0
where we have used the Chain Rule to write y aJn(ax).
We now consider three cases of the boundary condition (1).
Case I: If we choose A2 1 and B2 0, then (1) is
Jn(ab) 0.
(8)
There are an infinite number of positive roots xi aib of (8) (see Figure 5.3.1) that
define the ai as ai xi /b. The eigenvalues are positive and are then li a2i x 2i /b2 .
No new eigenvalues result from the negative roots of (8) since Jn(x) (1)nJn(x).
(See page 284.) The number 0 is not an eigenvalue for any n since Jn(0) 0 for
n 1, 2, 3, … and J0(0) 1. In other words, if l 0, we get the trivial function
(which is never an eigenfunction) for n 1, 2, 3, … , and for n 0, l 0 (or
equivalently, a 0) does not satisfy the equation in (8). When (6) is written in the
form x Jn(x) n Jn(x) x Jn1(x), it follows from (7) and (8) that the square norm
of Jn(ai x) is
b2 2
J n 1(ai b).
2
Case II: If we choose A2 h 0, B2 b, then (1) is
i Jn(ai x)i2 (9)
h Jn(a b) a bJn(a b) 0.
(10)
Equation (10) has an infinite number of positive roots xi aib for each positive integer n 1, 2, 3, …. As before, the eigenvalues are obtained from li a2i x 2i /b2 .
l 0 is not an eigenvalue for n 1, 2, 3, …. Substituting aibJn(aib) hJn(aib)
into (7), we find that the square norm of Jn(ai x) is now
iJn(ai x)i2 a2i b 2 2 n2 h 2 2
J n(ai b).
2a2i
(11)
Case III: If h 0 and n 0 in (10), the ai are defined from the roots of
J0(a b) 0.
12.6 Bessel and Legendre Series
(12)
|
699
Even though (12) is just a special case of (10), it is the only situation for which l 0
is an eigenvalue. To see this, observe that for n 0, the result in (6) implies that
J0(ab) 0 is equivalent to J1(ab) 0. Since x1 aib 0 is a root of the last equation, a1 0, and because J0(0) 1 is nontrivial, we conclude from l1 a21 x 21>b 2
that l1 0 is an eigenvalue. But obviously we cannot use (11) when a1 0, h 0,
and n 0. However, from the square norm (4) we have
i1i2 #
b
0
x dx b2
.
2
(13)
For ai 0 we can use (11) with h 0 and n 0:
iJ0(ai x)i2 b2 2
J 0 (ai b).
2
(14)
The following definition summarizes three forms of the series (2) corresponding to the square
norms in the three cases.
Definition 12.6.1
Fourier–Bessel Series
The Fourier–Bessel series of a function f defined on the interval (0, b) is given by
f (x) a ci Jn(ai x)
q
(i)
(15)
i 1
ci b
2
b 2J 2n 1(ai b)
# x J (a x) f (x) dx,
n
0
i
(16)
where the ai are defined by Jn(ab) 0.
f (x) a ci Jn(ai x)
q
(ii)
(17)
i 1
ci 2a2i
(a2i b 2 2 n2 h 2) J n2(ai b)
b
# x J (a x) f (x) dx,
n
0
i
(18)
where the ai are defined by hJn(ab) abJn(ab) 0.
f (x) c1 a ci J0(ai x)
q
(iii)
(19)
i 2
2
c1 2
b
b
# x f (x) dx,
2
ci 2 2
b J 0(ai b)
0
b
# x J (a x) f (x) dx,
0
0
i
(20)
where the ai are defined by J0(ab) 0.
Convergence of a Fourier–Bessel Series Sufficient conditions for the convergence
of a Fourier–Bessel series are not particularly restrictive.
Theorem 12.6.1
Conditions for Convergence
Let f and f be piecewise continuous on the interval [0, b]. Then for all x in the interval (0, b),
the Fourier–Bessel series of f converges to f (x) at a point where f is continuous and to the
average
f (x1) f (x)
2
at a point where f is discontinous.
700
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CHAPTER 12 Orthogonal Functions and Fourier Series
EXAMPLE 1
Expansion in a Fourier–Bessel Series
Expand f (x) x, 0 x 3, in a Fourier–Bessel series, using Bessel functions of order one
that satisfy the boundary condition J1(3a) 0.
SOLUTION We use (15) where the coefficients ci are given by (16) with b 3:
ci 2
2 2
3 J 2 (3ai)
#
3
0
x 2J1(ai x) dx.
To evaluate this integral we let t ai x, dx dt/ai, x 2 t 2 /a2i , and use (5) in the form
d 2
[t J2(t)] t 2 J1(t):
dt
3ai
2
d 2
2
ci 3 2
ft J2(t)g dt .
ai J2(3ai)
9ai J 2 (3ai) 0 dt
#
Therefore the desired expansion is
q
1
f (x) 2 a
J1(ai x).
a
J
i 1 i 2(3ai)
You are asked to find the first four values of the ai for the foregoing Bessel series in Problem 1
in Exercises 12.6.
EXAMPLE 2
Expansion in a Fourier–Bessel Series
If the ai in Example 1 are defined by J1(3a) aJ91 (3a) 0, then the only thing that changes
in the expansion is the value of the square norm. Multiplying the boundary condition by 3
gives 3 J1(3a) 3aJ91 (3a) 0, which now matches (10) when h 3, b 3, and n 1. Thus
(18) and (17) yield, in turn,
ci y
and
2.5
18ai J2(3ai)
(9a2i 8) J 21 (3ai)
q
ai J2(3ai)
f (x) 18 a
J1(ai x).
2
2
i 1 (9ai 8) J 1(3ai)
2
1.5
1
0.5
0
3
0
y
0.5
1
1.5
2
2.5
(a) S5(x), 0 < x < 3
3
x
2
1
x
0
Use of Computers Since Bessel functions are “built-in functions” in a CAS, it is a straightforward task to find the approximate values of the ai and the coefficients ci in a Fourier–Bessel series.
For example, in (9) we can think of xi aib as a positive root of the equation h Jn(x) x Jn(x) 0.
Thus in Example 2 we have used a CAS to find the first five positive roots xi of 3J1(x) x J91 (x) 0
and from these roots we obtain the first five values of ai: a1 x1 /3 0.98320, a2 x2/3 1.94704,
a3 x3/3 2.95758, a4 x4 /3 3.98538, and a5 x5 /3 5.02078. Knowing the roots xi 3ai
and the ai , we again use a CAS to calculate the numerical values of J2(3ai), J 21 (3ai), and finally
the coefficients ci. In this manner we find that the fifth partial sum S5(x) for the Fourier–Bessel
series representation of f (x) x, 0 x 3 in Example 2 is
S5(x) 4.01844 J1(0.98320x) 1.86937 J1(1.94704x)
1.07106 J1(2.95758x) 0.70306 J1(3.98538x) 0.50343 J1(5.02078x).
–1
0
10
20
30
40
(b) S10(x), 0 < x < 50
FIGURE 12.6.1 Partial sums of a
Fourier–Bessel series
50
The graph of S5(x) on the interval (0, 3) is shown in FIGURE 12.6.1(a). In Figure 12.6.1(b) we have
graphed S10(x) on the interval (0, 50). Notice that outside the interval of definition (0, 3) the series
does not converge to a periodic extension of f because Bessel functions are not periodic functions.
See Problems 11 and 12 in Exercises 12.6.
12.6.2 Fourier–Legendre Series
From Example 4 of Section 12.5 we know that the set of Legendre polynomials {Pn(x)},
n 0, 1, 2, …, is orthogonal with respect to the weight function p(x) 1 on the interval [1, 1].
12.6 Bessel and Legendre Series
|
701
Furthermore, it can be proved that the square norm of a polynomial Pn(x) depends on n in the
following manner:
2
iPn(x)i #
1
1
P 2n (x) dx 2
.
2n 1
The orthogonal series expansion of a function in terms of the Legendre polynomials is summarized in the next definition.
Fourier–Legendre Series
Definition 12.6.2
The Fourier–Legendre series of a function f defined on the interval (1, 1) is given by
f (x) a cnPn(x),
q
(21)
n0
2n 1
2
cn where
#
1
1
f (x)Pn(x) dx.
(22)
Convergence of a Fourier–Legendre Series Sufficient conditions for convergence
of a Fourier–Legendre series are given in the next theorem.
Conditions for Convergence
Theorem 12.6.2
Let f and f be piecewise continuous on the interval [1, 1]. Then for all x in the interval
(1, 1), the Fourier–Legendre series of f converges to f (x) at a point where f is continuous and
to the average
f (x1) f (x)
2
at a point where f is discontinuous.
Expansion in a Fourier–Legendre Series
EXAMPLE 3
Write out the first four nonzero terms in the Fourier–Legendre expansion of
f (x) e
1 , x , 0
0 # x , 1.
0,
1,
SOLUTION The first several Legendre polynomials are listed on page 289. From these and
(22) we find
1
c0 2
c1 3
2
c2 5
2
c3 7
2
9
c4 2
c5 Hence
702
|
#
1
#
1
#
1
#
1
#
1
1
f (x)P0(x) dx 2
1
1
1
1
f (x)P1(x) dx 3
2
f (x)P2(x) dx 5
2
f (x)P3(x) dx 7
2
9
f (x)P4(x) dx 2
1
11
2
#
1
1
f (x)P5(x) dx f (x) 1
# 1 1 dx 2
1
0
1
# 1 x dx 4
3
0
1
# 1 2 (3x
1
2
2 1) dx 0
3
2 3x) dx 0
1
# 1 2 (5x
1
0
1
# 1 8 (35x
1
0
11
2
1
4
2 30x 2 3) dx 0
# 1 8 (63x
0
1
7
16
5
2 70x 3 15x) dx 1
3
7
11
P0(x) P1(x) 2
P3(x) P (x) p .
2
4
16
32 5
CHAPTER 12 Orthogonal Functions and Fourier Series
11
.
32
y
1
0.8
0.6
0.4
0.2
x
0
–1
–0.5
0
0.5
1
FIGURE 12.6.2 Partial sum S5(x) of
Fourier–Legendre series in Example 3
Like the Bessel functions, Legendre polynomials are built-in functions in computer algebra
systems such as Mathematica and Maple, and so each of the coefficients just listed can be found
using the integration application of such a program. Indeed, using a CAS, we further find that
65
. The fifth partial sum of the Fourier–Legendre series representation of the
c6 0 and c7 256
function f defined in Example 3 is then
1
3
7
11
65
S5(x) P0(x) P1(x) 2
P3(x) P5(x) 2
P (x).
2
4
16
32
256 7
The graph of S5(x) on the interval (1, 1) is given in FIGURE 12.6.2.
Alternative Form of Series In applications, the Fourier–Legendre series appears in
an alternative form. If we let x cos u, then x 1 implies u 0, whereas x 1 implies u p.
Since dx sin u du, (21) and (22) become, respectively,
F(u) a cnPn(cos u)
q
(23)
n0
2n 1
2
where f (cos u) has been replaced by F(u).
cn 12.6
Exercises
p
# F(u)P (cos u) sin u du,
(24)
n
0
Answers to selected odd-numbered problems begin on page ANS-30.
12.6.1 Fourier–Bessel Series
In Problems 1 and 2, use Table 5.3.1 in Section 5.3.
1. Find the first four ai 0 defined by J1(3a) 0.
2. Find the first four ai 0 defined by J0(2a) 0.
In Problems 3–6, expand f (x) 1, 0 x 2, in a Fourier–
Bessel series using Bessel functions of order zero that satisfy the
given boundary condition.
3. J0(2a) 0
4. J0(2a) 0
5. J0(2a) 2a J0(2a) 0
6. J0(2a) a J0(2a) 0
In Problems 7–10, expand the given function in a Fourier–
Bessel series using Bessel functions of the same order as in the
indicated boundary condition.
7. f (x) 5x, 0 x 4
8. f (x) x 2, 0 x 1
3J1(4a) 4a J1(4a) 0
J2(a) 0
9. f (x) x 2, 0 x 3
10. f (x) 1 x 2, 0 x 1
J0(3a) 0
J0(a) 0
[Hint: t 3 t 2 t.]
Computer Lab Assignments
11. (a) Use a CAS to graph y 3J1(x) x J1(x) on an interval so
that the first five positive x-intercepts of the graph are shown.
(b) Use the root-finding capability of your CAS to approximate the first five roots xi of the equation
3J1(x) x J1(x) 0.
(c) Use the data obtained in part (b) to find the first five
positive values of ai that satisfy
3J1(4a) 4a J1(4a) 0.
See Problem 7.
(d) If instructed, find the first 10 positive values of ai.
12. (a) Use the values of ai in part (c) of Problem 11 and a CAS
to approximate the values of the first five coefficients ci
of the Fourier–Bessel series obtained in Problem 7.
(b) Use a CAS to graph the partial sums SN (x), N 1, 2, 3, 4, 5,
of the Fourier–Bessel series in Problem 7.
(c) If instructed, graph the partial sum S10(x) for 0 x 4
and for 0 x 50.
Discussion Problems
13. If the partial sums in Problem 12 are plotted on a symmetric
interval such as (30, 30), would the graphs possess any symmetry? Explain.
14. (a) Sketch, by hand, a graph of what you think the Fourier–
Bessel series in Problem 3 converges to on the interval
(2, 2).
(b) Sketch, by hand, a graph of what you think the Fourier–
Bessel series would converge to on the interval
(4, 4) if the values ai in Problem 7 were defined by
3J2(4a) 4aJ2(4a) 0.
12.6.2 Fourier–Legendre Series
In Problems 15 and 16, write out the first five nonzero terms in
the Fourier–Legendre expansion of the given function. If
instructed, use a CAS as an aid in evaluating the coefficients.
Use a CAS to graph the partial sum S5(x).
15. f (x) e
1 , x , 0
0,x,1
0,
x,
16. f (x) e x, 1
x
1
17. The first three Legendre polynomials are P0(x) 1, P1(x) x,
and P2(x) 12 (3x 2 1). If x cos u, then P0(cos u) 1 and
P1(cos u) cos u. Show that P2(cos u) 14 (3 cos 2u 1).
12.6 Bessel and Legendre Series
|
703
18. Use the results of Problem 17 to find a Fourier–Legendre
expansion (23) of F(u) ⫽ 1 ⫺ cos 2u.
19. A Legendre polynomial Pn(x) is an even or odd function, depending on whether n is even or odd. Show that if f is an even
function on the interval (⫺1, 1), then (21) and (22) become,
respectively,
f (x) ⫽ a c2nP2n(x)
q
(25)
n⫽0
#
1
c2n ⫽ (4n ⫹ 1) f (x)P2n(x) dx.
(26)
0
20. Show that if f is an odd function on the interval (⫺1, 1), then
(21) and (22) become, respectively,
f (x) ⫽ a c2n ⫹ 1P2n ⫹ 1(x)
(27)
n⫽0
1
c2n ⫹ 1 ⫽ (4n ⫹ 3) f (x)P2n ⫹ 1(x) dx.
(28)
0
Chapter in Review
12
tion that is defined on the interval (⫺1, 1) necessarily a finite
series?
24. Use your conclusion from Problem 23 to find the finite
Fourier–Legendre series of f (x) ⫽ x 2. The series of f (x) ⫽ x3.
Do not use (21) and (22).
Answers to selected odd-numbered problems begin on page ANS-30.
In Problems 1–10, fill in the blank or answer true/false without
referring back to the text.
1. The functions f (x) ⫽ x 2 ⫺ 1 and g(x) ⫽ x 5 are orthogonal on
the interval [⫺p, p].
2. The product of an odd function f with an odd function g is an
function.
3. To expand f (x) ⫽ |x| ⫹ 1, ⫺p ⬍ x ⬍ p, in an appropriate
trigonometric series we would use a
Discussion Problems
23. Why is a Fourier–Legendre expansion of a polynomial func-
q
#
The series (25) and (27) can also be used when f is defined on only
the interval (0, 1). Both series represent f on (0, 1); but on the interval (⫺1, 0), (25) represents an even extension, whereas (27)
represents an odd extension. In Problems 21 and 22, write out the
first four nonzero terms in the indicated expansion of the given
function. What function does the series represent on the interval
(⫺1, 1)? Use a CAS to graph the partial sum S4(x).
21. f (x) ⫽ x, 0 ⬍ x ⬍ 1; (25)
22. f (x) ⫽ 1, 0 ⬍ x ⬍ 1; (27)
series.
10. The set {Pn (x)}, n ⫽ 0, 1, 2, … of Legendre polynomials is
orthogonal with respect to the weight function p(x) ⫽ 1 on the
1
interval [⫺1, 1]. Hence, for n ⬎ 0, e⫺1Pn (x) dx ⫽
.
11. Without doing any work, explain why the cosine series
of f (x) ⫽ cos2 x, 0 ⬍ x ⬍ p, is the finite series
1 1
f (x) ⫽ ⫹ cos 2x.
2 2
12. (a) Show that the set
4. y ⫽ 0 is never an eigenfunction of a Sturm–Liouville
e sin
problem.
5. l ⫽ 0 is never an eigenvalue of a Sturm–Liouville problem.
6. If the function
f (x) ⫽ e
x ⫹ 1,
⫺x,
⫺1 , x , 0
0,x,1
is expanded in a Fourier series, the series will converge to
at x ⫽ ⫺1, to
at x ⫽ 0, and to
at x ⫽ 1.
7. Suppose the function f (x) ⫽ x 2 ⫹ 1, 0 ⬍ x ⬍ 3, is expanded
in a Fourier series, a cosine series, and a sine series. At x ⫽ 0,
the Fourier series will converge to
, the cosine series
will converge to
, and the sine series will converge to
.
8. The corresponding eigenfunction for the boundary-value
problem
y⬙ ⫹ ly ⫽ 0,
for l ⫽ 25 is
y⬘(0) ⫽ 0,
y(p/2) ⫽ 0
.
9. The set {P2n (x)}, n ⫽ 0, 1, 2, … of Legendre polynomials of
even degree is orthogonal with respect to the weight function
p(x) ⫽ 1 on the interval [0, 1].
704
|
CHAPTER 12 Orthogonal Functions and Fourier Series
13.
14.
15.
16.
17.
p
3p
5p
x, sin
x, sin
x, p f
2L
2L
2L
is orthogonal on the interval [0, L].
(b) Find the norm of each function in part (a). Construct an
orthonormal set.
Expand f (x) ⫽ |x| ⫺ x, ⫺1 ⬍ x ⬍ 1, in a Fourier series.
Expand f (x) ⫽ 2x 2 ⫺ 1, ⫺1 ⬍ x ⬍ 1, in a Fourier series.
Expand f (x) ⫽ e⫺x, 0 ⬍ x ⬍ 1, in a cosine series. In a sine series.
In Problems 13, 14, and 15, sketch the periodic extension of
f to which each series converges.
Find the eigenvalues and eigenfunctions of the boundary-value
problem
x 2y0 ⫹ xy9 ⫹ 9ly ⫽ 0, y9(1) ⫽ 0, y(e) ⫽ 0.
18. Give an orthogonality relation for the eigenfunctions in
Problem 17.
19. Chebyshev’s differential equation
(1 ⫺ x 2)y⬙ ⫺ xy⬘ ⫹ n2y ⫽ 0
has a polynomial solution y ⫽ Tn(x) for n ⫽ 0, 1, 2, …. Specify
the weight function p(x) and the interval over which the set
of Chebyshev polynomials {Tn(x)} is orthogonal. Give an
orthogonality relation.
20. Expand the periodic function shown in FIGURE 12.R.1 in an
appropriate Fourier series.
(a) Verify the identity f (x) ⫽ fe(x) ⫹ fo(x), where
y
fe(x) ⫽
2
–4
–2
0
2
4
6
x
FIGURE 12.R.1 Graph for Problem 20
1, 0 , x , 2
in a Fourier–Bessel series,
0, 2 , x , 4
using Bessel functions of order zero that satisfy the boundary
condition J0(4a) ⫽ 0.
21. Expand f (x) ⫽ e
22. Expand f (x) ⫽ x4, ⫺1 ⬍ x ⬍ 1, in a Fourier–Legendre series.
23. Suppose the function y ⫽ f (x) is defined on the interval (⫺⬁, ⬁).
f(x) ⫹ f(⫺x)
f(x) 2 f(⫺x)
and fo(x) ⫽
.
2
2
(b) Show that fe is an even function and fo an odd function.
24. The function f (x) ⫽ ex is neither even nor odd. Use Problem
23 to write f as the sum of an even function and an odd function. Identify fe and fo.
25. Suppose f is an integrable 2p-periodic function. Prove that for
any real number a,
#
2p
0
f(x) dx ⫽
#
a ⫹ 2p
f(x) dx.
a
CHAPTER 12 in Review
|
705
© Corbis Premium RF/Alamy Images
CHAPTER
13
In this and the next two chapters,
the emphasis will be on two
procedures that are frequently
used in solving problems
involving temperatures,
oscillatory displacements, and
potentials. These problems, called
boundary-value problems (BVPs)
are described by relatively simple
linear second-order partial
differential equations (PDEs). The
thrust of both procedures is to
find particular solutions of a PDE
by reducing it to one or more
ordinary differential equations
(ODEs).
Boundary-Value
Problems in
Rectangular Coordinates
CHAPTER CONTENTS
13.1
13.2
13.3
13.4
13.5
13.6
13.7
13.8
Separable Partial Differential Equations
Classical PDEs and Boundary-Value Problems
Heat Equation
Wave Equation
Laplace’s Equation
Nonhomogeneous Boundary-Value Problems
Orthogonal Series Expansions
Fourier Series in Two Variables
Chapter 13 in Review
13.1
Separable Partial Differential Equations
INTRODUCTION Partial differential equations (PDEs), like ordinary differential equations
(ODEs), are classified as linear or nonlinear. Analogous to a linear ODE (see (6) of Section 1.1),
the dependent variable and its partial derivatives appear only to the first power in a linear PDE. In
this and the chapters that follow, we are concerned only with linear partial differential equations.
Linear Partial Differential Equation If we let u denote the dependent variable and
x and y the independent variables, then the general form of a linear second-order partial differential equation is given by
A
0 2u
0 2u
0 2u
0u
0u
B
C 2D
E
Fu G,
2
0x0y
0x
0y
0x
0y
(1)
where the coefficients A, B, C, … , G are constants or functions of x and y. When G(x, y) 0,
equation (1) is said to be homogeneous; otherwise, it is nonhomogeneous.
Linear Second-Order PDEs
EXAMPLE 1
The equations
0 2u
0 2u
0
0x 2
0y 2
and
0 2u
0u
2
xy
2
0y
0x
are examples of linear second-order PDEs. The first equation is homogeneous and the second
is nonhomogeneous.
We are interested only in
particular solutions of PDEs.
Solution of a PDE A solution of a linear partial differential equation (1) is a function
u(x, y) of two independent variables that possesses all partial derivatives occurring in the equation
and that satisfies the equation in some region of the xy-plane.
It is not our intention to examine procedures for finding general solutions of linear partial
differential equations. Not only is it often difficult to obtain a general solution of a linear secondorder PDE, but a general solution is usually not all that useful in applications. Thus our focus
throughout will be on finding particular solutions of some of the important linear PDEs, that is,
equations that appear in many applications.
Separation of Variables Although there are several methods that can be tried to find
particular solutions of a linear PDE, the one we are interested in at the moment is called the
method of separation of variables. In this method if we are seeking a particular solution of, say,
a linear second-order PDE in which the independent variables are x and y, then we seek to find
a particular solution in the form of product of a function x and a function of y:
u(x, y) X(x)Y( y).
With this assumption, it is sometimes possible to reduce a linear PDE in two variables to two
ODEs. To this end we observe that
0u
X9Y,
0x
0u
XY9,
0y
0 2u
X0Y,
0x 2
0 2u
XY0,
0y 2
where the primes denote ordinary differentiation.
EXAMPLE 2
Using Separation of Variables
Find product solutions of
SOLUTION
0u
0 2u
4 .
2
0y
0x
Substituting u(x, y) X(x)Y( y) into the partial differential equation yields
X Y 4XY .
After dividing both sides by 4XY, we have separated the variables:
X0
Y9
.
4X
Y
708
|
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
Since the left-hand side of the last equation is independent of y and is equal to the right-hand
side, which is independent of x, we conclude that both sides of the equation are independent
of x and y. In other words, each side of the equation must be a constant. As a practical matter
it is convenient to write this real separation constant as l. From the two equalities,
X0
Y9
l
4X
Y
we obtain the two linear ordinary differential equations
X 4lX 0
See Example 2, Section 3.9
and Example 1, Section 12.5.
and
Y lY 0.
(2)
2
For the three cases for l: zero, negative, or positive; that is, l 0, l a 0, and
l a2 0, where a 0, the ODEs in (2) are, in turn,
X 0
2
X 4a X 0
2
X 4a X 0
Case I (l 0):
and
Y 0,
(3)
and
2
Y a Y 0,
(4)
and
2
(5)
Y a Y 0.
The DEs in (3) can be solved by integration. The solutions are
X c1 c2 x and Y c3. Thus a particular product solution of the
given PDE is
u XY (c1 c2x)c3 A1 B1x,
(6)
where we have replaced c1c3 and c2c3 by A1 and B1, respectively.
Case II (l a 2): The general solutions of the DEs in (4) are
X c4 cosh 2ax c5 sinh 2ax
and
2
Y c6 e a y ,
respectively. Thus, another particular product solution of the PDE is
2
u XY (c4 cosh 2ax c5 sinh 2ax) c6e a y
or
2
2
u A2e a y cosh 2ax B2e a y sinh 2ax,
(7)
where A2 c4c6 and B2 c5c6.
Case III (l a 2):
Finally, the general solutions of the DEs in (5) are
2
X c7 cos 2ax c8 sin 2ax and Y c9e a y,
respectively. These results give yet another particular solution
2
2
u A3e a y cos 2ax B3e a y sin 2ax,
(8)
where A3 c7c9 and B3 c8c9.
It is left as an exercise to verify that (6), (7), and (8) satisfy the given partial differential equation uxx 4uy. See Problem 29 in Exercises 13.1.
Separation of variables is not a general method for finding particular solutions; some linear
partial differential equations are simply not separable. You should verify that the assumption
u XY does not lead to a solution for 0 2u/0x2 0u/0y x.
Superposition Principle The following theorem is analogous to Theorem 3.1.2 and
is known as the superposition principle.
Theorem 13.1.1
Superposition Principle
If u1, u2, … , uk are solutions of a homogeneous linear partial differential equation, then the
linear combination
u c1u1 c2u2 p ckuk ,
where the ci , i 1, 2, … , k are constants, is also a solution.
13.1 Separable Partial Differential Equations
|
709
Throughout the remainder of the chapter we shall assume that whenever we have an infinite
set u1, u2, u3, … of solutions of a homogeneous linear equation, we can construct yet another
solution u by forming the infinite series
u a ckuk ,
q
k51
where the ck , k 1, 2, … , are constants.
Classification of Equations A linear second-order partial differential equation in two
independent variables with constant coefficients can be classified as one of three types. This
classification depends only on the coefficients of the second-order derivatives. Of course, we
assume that at least one of the coefficients A, B, and C is not zero.
Definition 13.1.1
Classification of Equations
The linear second-order partial differential equation
A
0 2u
0 2u
0 2u
0u
0u
B
C
D
E
Fu G,
2
2
0x0y
0x
0y
0x
0y
where A, B, C, D, E, F, and G are real constants, is said to be
hyperbolic if B2 4AC 0,
parabolic if
B2 4AC 0,
elliptic if
B2 4AC 0.
Classifying Linear Second-Order PDEs
EXAMPLE 3
Classify the following equations:
(a) 3
0u
0 2u
0y
0x 2
SOLUTION
(b)
0 2u
0 2u
2
2
0x
0y
(c)
0 2u
0 2u
2 0.
2
0x
0y
(a) By rewriting the given equation as
3
0 2u
0u
2
0
0y
0x 2
we can make the identifications A 3, B 0, and C 0. Since B2 4AC 0, the equation is parabolic.
(b) By rewriting the equation as
0 2u
0 2u
2
0,
0x 2
0y 2
we see that A 1, B 0, C 1, and B2 4AC 4(1)(1) 0. The equation is
hyperbolic.
(c) With A 1, B 0, C 1, and B2 4AC 4(1)(1) 0, the equation is elliptic.
REMARKS
(i) Separation of variables is not a general method for finding particular solutions of linear
partial differential equations. Some equations are simply not separable. You are encouraged
to verify that the assumption u(x, y) X(x)Y(y) does not lead to a solution of the linear secondorder PDE 0 2u> 0x 2 0u> 0x y.
(ii) A detailed explanation of why the classifications given in Definition 13.1.1 are important
is beyond the scope of this text. But you should at least be aware that these classifications do
have a practical importance. Beginning in Section 13.3 we are going to solve some PDEs
subject to both boundary and initial conditions. The kinds of side conditions appropriate for
a given equation depend on whether the equation is hyperbolic, parabolic, or elliptic. Also,
we shall see in Chapter 16 that numerical solution methods for linear second-order PDEs
differ in conformity with the classification of the equation.
710
|
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
13.1
Exercises
Answers to selected odd-numbered problems begin on page ANS-31.
In Problems 1–16, use separation of variables to find, if
possible, product solutions for the given partial differential
equation.
1.
3.
5.
7.
9.
11.
12.
13.
14.
16.
0u
0u
0u
0u
2.
3
0
0x
0y
0x
0y
ux uy u
4. ux uy u
0u
0u
0u
0u
x
y
6. y
x
0
0x
0y
0x
0y
0 2u
0 2u
0 2u
0 2u
0
8.
y
u0
0x0y
0x0y
0x 2
0y 2
0 2u
0u
0 2u
0u
k 2 2 u , k . 0 10. k 2 , k . 0
0t
0t
0x
0x
2
2
0u
0u
a2 2 2
0x
0t
2
2
0
u
0
u
0u
a 2 2 2 2k , k . 0
0t
0x
0t
0 2u
0 2u
0u
2 2k , k . 0
0t
0x 2
0y
2
2
0
u
0
u
x2 2 2 0
15. uxx uyy u
0x
0y
2
a uxx 2 g utt, g a constant
0 2u
0 2u
0 2u
0u
0u
2
2
26
0
2
0x0y
0x
0y
0x
0y
0 2u
0 2u
24.
2u
2
0x
0y
2
0
u
0 2u
0 2u
0u
25. a 2 2 2
26. k 2 , k.0
0t
0x
0t
0x
23.
In Problems 27 and 28, show that the given partial differential
equation possesses the indicated product solution.
27. k a
0 2u
1 0u
0u
b ;
2
r 0r
0t
0r
2
u e ka t(c1J0(ar) c2Y0(ar))
0 2u
1 0u
1 0 2u
28.
2 2 0;
2
r 0r
0r
r 0u
u (c1 cos au c2 sin au)(c3r a c4r a)
29. Verify that each of the products u X(x)Y( y) in (6), (7), and
(8) satisfies the second-order PDE in Example 2.
30. Definition 13.1.1 generalizes to linear PDEs with coefficients
that are functions of x and y. Determine the regions in the
xy-plane for which the equation
(xy 1)
In Problems 17–26, classify the given partial differential
equation as hyperbolic, parabolic, or elliptic.
17.
18.
19.
20.
21.
0 2u
0 2u
0 2u
20
2
0x0y
0x
0y
0 2u
0 2u
0 2u
3 25
20
0x0y
0x
0y
0 2u
0 2u
0 2u
6
9 20
0x0y
0x 2
0y
0 2u
0 2u
0 2u
2
23 20
0x0y
0x 2
0y
2
2
0u
0u
0 2u
0 2u
0u
9
22.
2 22
0
2
0x0y
0x0y
0x
0x
0y
13.2
0 2u
0 2u
0 2u
(x 2y)
2 x y 2u 0
2
0x0y
0x
0y
is hyperbolic, parabolic, or elliptic.
Discussion Problems
In Problems 31 and 32, discuss whether product solutions
u X(x)Y( y) can be found for the given partial differential
equation. [Hint: Use the superposition principle.]
31.
0 2u
2u0
0x 2
32.
0 2u
0u
0
0x0y
0x
Classical PDEs and Boundary-Value Problems
INTRODUCTION For the remainder of this and the next chapter we shall be concerned with
finding product solutions of the second-order partial differential equations
k
0 2u
0u
, k.0
2
0t
0x
(1)
0 2u
0 2u
0x 2
0t 2
(2)
0 2u
0 2u
20
2
0x
0y
(3)
a2
13.2 Classical PDEs and Boundary-Value Problems
|
711
or slight variations of these equations. These classical equations of mathematical physics are
known, respectively, as the one-dimensional heat equation, the one-dimensional wave equation,
and Laplace’s equation in two dimensions. “One-dimensional” refers to the fact that x denotes
a spatial dimension whereas t represents time; “two dimensional” in (3) means that x and y are
both spatial dimensions. Laplace’s equation is abbreviated 2u 0, where
=2u 0 2u
0 2u
0x 2
0y 2
is called the two-dimensional Laplacian of the function u. In three dimensions the Laplacian
of u is
=2u 0 2u
0 2u
0 2u
2 2.
2
0x
0y
0z
By comparing equations (1)(3) with the linear second-order PDE given in Definition 13.1.1,
with t playing the part of y, we see that the heat equation (1) is parabolic, the wave equation (2)
is hyperbolic, and Laplace’s equation (3) is elliptic. This classification is important in
Chapter 16.
cross section of area A
0
x
x + Δx
L
x
FIGURE 13.2.1 One-dimensional flow
of heat
Heat Equation Equation (1) occurs in the theory of heat flow—that is, heat transferred
by conduction in a rod or thin wire. The function u(x, t) is temperature. Problems in mechanical
vibrations often lead to the wave equation (2). For purposes of discussion, a solution u(x, t) of
(2) will represent the displacement of an idealized string. Finally, a solution u(x, y) of Laplace’s
equation (3) can be interpreted as the steady-state (that is, time-independent) temperature distribution throughout a thin, two-dimensional plate.
Even though we have to make many simplifying assumptions, it is worthwhile to see how
equations such as (1) and (2) arise.
Suppose a thin circular rod of length L has a cross-sectional area A and coincides with the
x-axis on the interval [0, L]. See FIGURE 13.2.1. Let us suppose:
• The flow of heat within the rod takes place only in the x-direction.
• The lateral, or curved, surface of the rod is insulated; that is, no heat escapes from this
surface.
• No heat is being generated within the rod by either chemical or electrical means.
• The rod is homogeneous; that is, its mass per unit volume r is a constant.
• The specific heat g and thermal conductivity K of the material of the rod are constants.
To derive the partial differential equation satisfied by the temperature u(x, t), we need two
empirical laws of heat conduction:
(i) The quantity of heat Q in an element of mass m is
Q g m u,
(4)
where u is the temperature of the element.
(ii) The rate of heat flow Qt through the cross section indicated in Figure 13.2.1 is proportional to the area A of the cross section and the partial derivative with respect to x of the
temperature:
Qt K Aux.
(5)
Since heat flows in the direction of decreasing temperature, the minus sign in (5) is used to ensure
that Qt is positive for ux 0 (heat flow to the right) and negative for ux 0 (heat flow to the left).
If the circular slice of the rod shown in Figure 13.2.1 between x and x x is very thin, then
u(x, t) can be taken as the approximate temperature at each point in the interval. Now the mass
of the slice is m r(A x), and so it follows from (4) that the quantity of heat in it is
Q grA
x u.
(6)
Furthermore, when heat flows in the positive x-direction, we see from (5) that heat builds up in
the slice at the net rate
K Aux(x, t) [K Aux(x 712
|
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
x, t)] K A[ux(x x, t) ux(x, t)].
(7)
By differentiating (6) with respect to t we see that this net rate is also given by
Qt grA x ut .
(8)
K ux(x Dx, t) 2 ux(x, t)
ut .
gr
Dx
(9)
Equating (7) and (8) gives
Taking the limit of (9) as x S 0 finally yields (1) in the form*
K
u ut .
gr xx
It is customary to let k K/gr and call this positive constant the thermal diffusivity.
u
Δs
0
u(x, t)
L x
x x + Δx
(a) Segment of string
T2
u
θ2
Δs
θ1
T1
0
x
x + Δx
x
(b) Enlargement of segment
Wave Equation Consider a string of length L, such as a guitar string, stretched taut between two points on the x-axis—say, x 0 and x L. When the string starts to vibrate, assume
that the motion takes place in the xy-plane in such a manner that each point on the string moves
in a direction perpendicular to the x-axis (transverse vibrations). As shown in FIGURE 13.2.2(a), let
u(x, t) denote the vertical displacement of any point on the string measured from the x-axis for
t 0. We further assume:
• The string is perfectly flexible.
• The string is homogeneous; that is, its mass per unit length r is a constant.
• The displacements u are small compared to the length of the string.
• The slope of the curve is small at all points.
• The tension T acts tangent to the string, and its magnitude T is the same at all points.
• The tension is large compared with the force of gravity.
• No other external forces act on the string.
Now in Figure 13.2.2(b) the tensions T1 and T2 are tangent to the ends of the curve on the
interval [x, x x]. For small values of u1 and u2 the net vertical force acting on the corresponding element s of the string is then
FIGURE 13.2.2 Taut string anchored at
two points on the x-axis
T sin u2 T sin u1 ⬇ T tan u2 T tan u1
T [ux(x x, t) ux(x, t)],†
where T |T1| |T2|. Now r s ⬇ r x is the mass of the string on [x, x x], and so Newton’s
second law gives
temperature as a
function of position
on the hot plate
T [ux(x thermometer
or
y
x, t) ux(x, t)] r
x utt
r
ux (x Dx, t) 2 ux(x, t)
utt.
Dx
T
If the limit is taken as x S 0, the last equation becomes uxx (r/T )utt. This of course is (2) with
a2 T/r.
(x, y)
W
x
H
O
FIGURE 13.2.3 Steady-state temperatures
in a rectangular plate
Laplace’s Equation Although we shall not present its derivation, Laplace’s equation
in two and three dimensions occurs in time-independent problems involving potentials such as
electrostatic, gravitational, and velocity in fluid mechanics. Moreover, a solution of Laplace’s
equation can also be interpreted as a steady-state temperature distribution. As illustrated in
FIGURE 13.2.3, a solution u(x, y) of (3) could represent the temperature that varies from point to
point—but not with time—of a rectangular plate.
We often wish to find solutions of equations (1), (2), and (3) that satisfy certain side
conditions.
ux(x Dx, t) 2 ux(x, t)
.
Dx
†
tan u2 ux(x x, t) and tan u1 ux(x, t) are equivalent expressions for slope.
*Recall from calculus that uxx lim
DxS0
13.2 Classical PDEs and Boundary-Value Problems
|
713
Initial Conditions Since solutions of (1) and (2) depend on time t, we can prescribe
what happens at t 0; that is, we can give initial conditions (IC). If f (x) denotes the initial
temperature distribution throughout the rod in Figure 13.2.1, then a solution u(x, t) of (1) must
satisfy the single initial condition u(x, 0) f (x), 0 x L. On the other hand, for a vibrating
string, we can specify its initial displacement (or shape) f (x) as well as its initial velocity g(x).
In mathematical terms we seek a function u(x, t) satisfying (2) and the two initial conditions:
u
0
u=0
at x = 0
0u
2
g(x),
0t t 0
u(x, 0) f (x),
h
u=0
at x = L
FIGURE 13.2.4 Plucked string
L x
0 , x , L.
(10)
For example, the string could be plucked, as shown in FIGURE 13.2.4, and released from rest
(g(x) 0).
Boundary Conditions The string in Figure 13.2.4 is secured to the x-axis at x 0 and
x L for all time. We interpret this by the two boundary conditions (BC):
u(0, t) 0,
u(L, t) 0,
t 0.
Note that in this context the function f in (10) is continuous, and consequently f (0) 0 and
f (L) 0. In general, there are three types of boundary conditions associated with equations (1),
(2), and (3). On a boundary we can specify the values of one of the following:
(i) u,
(ii)
0u
,
0n
or
(iii)
0u
hu, h a constant.
0n
Here 0u/0n denotes the normal derivative of u (the directional derivative of u in the direction
perpendicular to the boundary). A boundary condition of the first type (i) is called a Dirichlet
condition, a boundary condition of the second type (ii) is called a Neumann condition, and a
boundary condition of the third type (iii) is known as a Robin condition. For example, for t 0
a typical condition at the right-hand end of the rod in Figure 13.2.1 can be
(i)
u(L, t) u0 , u0 a constant,
(ii)
0u
2
0, or
0x x L
(iii)
0u
2
h(u(L, t) um), h 0 and um constants.
0x x L
Condition (i) simply states that the boundary x L is held by some means at a constant temperature u0 for all time t 0. Condition (ii) indicates that the boundary x L is insulated. From
the empirical law of heat transfer, the flux of heat across a boundary (that is, the amount of heat
per unit area per unit time conducted across the boundary) is proportional to the value of the
normal derivative 0u/0n of the temperature u. Thus when the boundary x L is thermally insulated, no heat flows into or out of the rod and so
0u
2
0.
0x x L
We can interpret (iii) to mean that heat is lost from the right-hand end of the rod by being in
contact with a medium, such as air or water, that is held at a constant temperature. From Newton’s
law of cooling, the outward flux of heat from the rod is proportional to the difference between
the temperature u(L, t) at the boundary and the temperature um of the surrounding medium. We
note that if heat is lost from the left-hand end of the rod, the boundary condition is
0u
2
h(u (0, t) 2 um).
0x x 0
The change in algebraic sign is consistent with the assumption that the rod is at a higher temperature than the medium surrounding the ends so that u(0, t) um and u(L, t) um. At x 0
and x L, the slopes ux(0, t) and ux(L, t) must be positive and negative, respectively.
714
|
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
Of course, at the ends of the rod we can specify different conditions at the same time. For
example, we could have
0u
2
0
0x x 0
and
u(L, t) u0, t 0.
We note that the boundary condition in (i) is homogeneous if u0 0; if u0 0, the boundary
condition is nonhomogeneous. The boundary condition (ii) is homogeneous; (iii) is homogeneous if um 0 and nonhomogeneous if um 0.
Boundary-Value Problems Problems such as
Solve:
a2
0 2u
0 2u
, 0 x L, t 0
0x 2
0t 2
Subject to: (BC) u(0, t) 0, u(L, t) 0, t 0
(IC) u(x, 0) f (x),
(11)
0u
2
g(x), 0 x L
0t t 0
and
Solve:
0 2u
0 2u
0, 0 x a, 0 y b
0x 2
0y 2
0u
0u
2
0,
2
0, 0 , y , b
0x
0x x a
x0
c
Subject to: (BC)
u(x, 0) 0, u(x, b) f (x), 0 , x , a
(12)
are called boundary-value problems. The problem in (11) is classified as a homogeneous BVP
since the partial differential equation and the boundary conditions are homogeneous.
Variations The partial differential equations (1), (2), and (3) must be modified to take
into consideration internal or external influences acting on the physical system. More general
forms of the one-dimensional heat and wave equations are, respectively,
k
a2
and
0 2u
0u
F(x, t, u, ux) 2
0t
0x
(13)
0 2u
0 2u
F(x,
t,
u,
u
)
.
t
0x 2
0t 2
(14)
For example, if there is heat transfer from the lateral surface of a rod into a surrounding medium
that is held at a constant temperature um, then the heat equation (13) is
k
0 2u
0u
2 h(u 2 um) ,
0t
0x 2
where h is a constant. In (14) the function F could represent the various forces acting on the
string. For example, when external, damping, and elastic restoring forces are taken into account,
(14) assumes the form
external force
T
a2
damping
T
restoring force
T
0 2u
0u
0 2u
f
(x,
t)
2
c
2
ku
.
0t
0x 2
0t 2
(15)
F (x, t, u, ut )
13.2 Classical PDEs and Boundary-Value Problems
|
715
REMARKS
The analysis of a wide variety of diverse phenomena yields the mathematical models (1), (2), or
(3) or their generalizations involving a greater number of spatial variables. For example, (1) is
sometimes called the diffusion equation since the diffusion of dissolved substances in solution is
analogous to the flow of heat in a solid. The function c(x, t) satisfying the partial differential equation in this case represents the concentration of the dissolved substance. Similarly, equation (2) and
its generalization (15) arise in the analysis of the flow of electricity in a long cable or transmission
line. In this setting (2) is known as the telegraph equation. It can be shown that under certain assumptions the current i(x, t) and the voltage v(x, t) in the line satisfy two partial differential equations
identical to (2) (or (15)). The wave equation (2) also appears in fluid mechanics, acoustics, and
elasticity. Laplace’s equation (3) is encountered in determining the static displacement of membranes.
Exercises
13.2
Answers to selected odd-numbered problems begin on page ANS-31.
7. The ends are secured to the x-axis. The string is released from
In Problems 1– 6, a rod of length L coincides with the interval
[0, L] on the x-axis. Set up the boundary-value problem for the
temperature u(x, t).
rest from the initial displacement x(L x).
8. The ends are secured to the x-axis. Initially the string is
undisplaced but has the initial velocity sin(px/L).
1. The left end is held at temperature zero, and the right end is
9. The left end is secured to the x-axis, but the right end moves in
insulated. The initial temperature is f (x) throughout.
a transverse manner according to sin pt. The string is released
from rest from the initial displacement f (x). For t 0 the
transverse vibrations are damped with a force proportional to
the instantaneous velocity.
10. The ends are secured to the x-axis, and the string is initially at
rest on that axis. An external vertical force proportional to the
horizontal distance from the left end acts on the string for t 0.
2. The left end is held at temperature u0, and the right end is held
at temperature u1. The initial temperature is zero throughout.
3. The left end is held at temperature 100 , and there is heat trans-
fer from the right end into the surrounding medium at temperature zero. The initial temperature is f(x) throughout.
4. There is heat transfer from the left end into a surrounding
medium at temperature 20 , and the right end is insulated. The
initial temperature is f (x) throughout.
5. The left end is at temperature sin(pt/L), the right end is held
at zero, and there is heat transfer from the lateral surface of
the rod into the surrounding medium held at temperature zero.
The initial temperature is f (x) throughout.
6. The ends are insulated, and there is heat transfer from the
lateral surface of the rod into the surrounding medium held
at temperature 50 . The initial temperature is 100 throughout.
In Problems 11 and 12, set up the boundary-value problem for
the steady-state temperature u(x, y).
11. A thin rectangular plate coincides with the region in the xy-plane
defined by 0 x 4, 0 y 2. The left end and the bottom of
the plate are insulated. The top of the plate is held at temperature
zero, and the right end of the plate is held at temperature f ( y).
12. A semi-infinite plate coincides with the region defined by
0 x p, y 0. The left end is held at temperature ey, and
the right end is held at temperature 100 for 0 y 1 and
temperature zero for y 1. The bottom of the plate is held at
temperature f (x).
In Problems 7–10, a string of length L coincides with the
interval [0, L] on the x-axis. Set up the boundary-value problem
for the displacement u(x, t).
13.3
u=0
u=0
0
L
x
FIGURE 13.3.1 Find the temperature u in a
finite rod
Heat Equation
INTRODUCTION Consider a thin rod of length L with an initial temperature f (x) throughout
and whose ends are held at temperature zero for all time t 0. If the rod shown in FIGURE 13.3.1
satisfies the assumptions given on page 712, then the temperature u(x, t) in the rod is determined
from the boundary-value problem
0 2u
0u
, 0 , x , L, t . 0
2
0t
0x
(1)
u(0, t) 0, u(L, t) 0, t . 0
(2)
u(x, 0) f (x), 0 , x , L.
(3)
k
In the discussion that follows next we show how to solve this BVP using the method of separation
of variables introduced in Section 13.1.
716
|
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
Solution of the BVP Using the product u(x, t) ⫽ X(x)T(t), and ⫺l as the separation
constant, leads to
X0
T9
⫽
⫽ ⫺l
X
kT
(4)
X ⬙ ⫹ lX ⫽ 0
(5)
T⬘ ⫹ klT ⫽ 0.
(6)
and
Now the boundary conditions in (2) become u(0, t) ⫽ X(0)T(t) ⫽ 0 and u(L, t) ⫽ X(L)T(t) ⫽ 0.
Since the last equalities must hold for all time t, we must have X(0) ⫽ 0 and X(L) ⫽ 0. These
homogeneous boundary conditions together with the homogeneous ODE (5) constitute a regular
Sturm–Liouville problem:
X ⬙ ⫹ lX ⫽ 0,
X(0) ⫽ 0,
X(L) ⫽ 0.
(7)
The solution of this BVP was discussed in detail in Example 2 of Section 3.9 and on page 692
of Section 12.5. In that example, we considered three possible cases for the parameter l: zero,
negative, and positive. The corresponding general solutions of the DEs are
X(x) ⫽ c1 ⫹ c2 x,
l⫽0
X(x) ⫽ c1 cosh ax ⫹ c2 sinh ax,
l ⫽ ⫺a2 , 0
(8)
(9)
2
X(x) ⫽ c1 cos ax ⫹ c2 sin ax,
l ⫽ a . 0.
(10)
Recall, when the boundary conditions X(0) ⫽ 0 and X(L) ⫽ 0 are applied to (8) and (9) these
solutions yield only X(x) ⫽ 0 and so we are left with the unusable result u ⫽ 0. Applying the first
boundary condition X(0) ⫽ 0 to the solution in (10) gives c1 ⫽ 0. Therefore X(x) ⫽ c2 sin ax.
The second boundary condition X(L) ⫽ 0 now implies
X(L) ⫽ c2 sin a L ⫽ 0.
(11)
If c2 ⫽ 0, then X ⫽ 0 so that u ⫽ 0. But (11) can be satisfied for c2 ⫽ 0 when sin aL ⫽ 0. This
last equation implies that aL ⫽ np or a ⫽ np/L, where n ⫽ 1, 2, 3, … . Hence (7) possesses
nontrivial solutions when ln ⫽ a2n ⫽ n2p2 /L2, n ⫽ 1, 2, 3, … . The values ln and the corresponding solutions
X(x) ⫽ c2 sin
np
x,
L
n ⫽ 1, 2, 3, p
(12)
are the eigenvalues and eigenfunctions, respectively, of the problem in (7).
2 2 2
The general solution of (6) is T(t) ⫽ c3e ⫺k (n p >L ) t , and so
2
2
2
un ⫽ X(x)T(t) ⫽ An e ⫺k (n p >L ) t sin
np
x,
L
(13)
where we have replaced the constant c2c3 by An. The products un(x, t) given in (13) satisfy the
partial differential equation (1) as well as the boundary conditions (2) for each value of the positive integer n. However, in order for the functions in (13) to satisfy the initial condition (3), we
would have to choose the coefficient An in such a manner that
un(x, 0) ⫽ f (x) ⫽ An sin
np
x.
L
(14)
In general, we would not expect condition (14) to be satisfied for an arbitrary, but reasonable,
choice of f. Therefore we are forced to admit that un(x, t) is not a solution of the problem given
in (1)⫺(3). Now by the superposition principle the function
q
q
np
2 2 2
u(x, t) ⫽ a un ⫽ a An e ⫺k (n p >L ) t sin
x
L
n⫽1
n⫽1
13.3 Heat Equation
(15)
|
717
must also, although formally, satisfy equation (1) and the conditions in (2). If we substitute t 0
into (15), then
q
np
u(x, 0) f (x) a An sin
x.
L
n1
This last expression is recognized as the half-range expansion of f in a sine series. If we make
the identification An bn, n 1, 2, 3, … , it follows from (5) of Section 12.3 that
2
An L
u
100
80
t = 0.05
t = 0.35
60
t = 0.6
40
t=1
20
t = 1.5
0
0.5
1
1.5
t=0
2.5
3
x
and that the series (17) is
80
x = π /4
60
x = π /6
40
x = π /12
u(x, t) 20
0
1
2
3
4
x=0
t
5
6
(b) u(x, t) graphed as a
function of t for
various fixed positions
FIGURE 13.3.2 Graphs obtained using
partial sums of (18)
Exercises
13.3
1,
0 , x , L>2
0, L>2 , x , L
2. u(0, t) 0, u(L, t) 0
u(x, 0) x(L x)
3. Find the temperature u(x, t) in a rod of length L if the initial
temperature is f (x) throughout and if the ends x 0 and x L
are insulated.
4. Solve Problem 3 if L 2 and
u(x, 0) e
# f (x) sin
0
np
np
2 2 2
x dxbe k (n p >L ) t sin
x.
L
L
(17)
200 1 2 (1)n
d,
c
p
n
200 q 1 2 (1)n n2t
d e sin nx.
c
p na
n
1
(18)
takes on the form
0 2u
0u
k 2 2 hu , 0 , x , L, t . 0,
0t
0x
h a constant. Find the temperature u(x, t) if the initial
temperature is f (x) throughout and the ends x 0 and x L
are insulated. See FIGURE 13.3.3.
insulated
length L into a surrounding medium at temperature zero. If
the linear law of heat transfer applies, then the heat equation
0°
0
0°
insulated
L
x
heat transfer from
lateral surface of
the rod
0,x,1
1 , x , 2.
5. Suppose heat is lost from the lateral surface of a thin rod of
|
L
Answers to selected odd-numbered problems begin on page ANS-31.
1. u(0, t) 0, u(L, t) 0
x,
f (x) e
0,
(16)
Use of Computers The solution u in (18) is a function of two variables and as such its
graph is a surface in 3-space. We could use the 3D-plot application of a computer algebra system
to approximate this surface by graphing partial sums Sn(x, t) over a rectangular region defined
by 0 x p, 0 t T. Alternatively, with the aid of the 2D-plot application of a CAS we plot
the solution u(x, t) on the x-interval [0, p] for increasing values of time t. See FIGURE 13.3.2(a).
In Figure 13.3.2(b) the solution u(x, t) is graphed on the t-interval [0, 6] for increasing values
of x (x 0 is the left end and x p/2 is the midpoint of the rod of length L p). Both sets of
graphs verify that which is apparent in (18)—namely, u(x, t) S 0 as t S q.
In Problems 1 and 2, solve the heat equation (1) subject to the
given conditions. Assume a rod of length L.
718
2 q
a
L na
1
An u
x = π /2
0
np
x dx.
L
In the special case when the initial temperature is u(x, 0) 100, L p, and k 1, you should
verify that the coefficients (16) are given by
(a) u(x, t) graphed as a
function of x for
various fixed times
100
# f (x) sin
We conclude that a solution of the boundary-value problem described in (1), (2), and (3) is given
by the infinite series
u(x, t) 2
L
FIGURE 13.3.3 Rod in Problem 5
6. Solve Problem 5 if the ends x 0 and x L are held at tem-
perature zero.
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
7. A thin wire coinciding with the x-axis on the interval [L, L]
is bent into the shape of a circle so that the ends x L
and x L are joined. Under certain conditions the
temperature u(x, t) in the wire satisfies the boundary-value
problem
Computer Lab Assignments
9. (a) Solve the heat equation (1) subject to
u(0, t) 0, u(100, t) 0, t 0
u(x, 0) e
2
k
0u
0u
, L x L, t 0,
2
0t
0x
0.8x,
0.8(100 2 x),
50 # x # 50
50 , x # 100.
(b) Use the 3D-plot application of your CAS to graph the
partial sum S5(x, t) consisting of the first five nonzero
terms of the solution in part (a) for 0 x 100,
0 t 200. Assume that k 1.6352. Experiment with
various three-dimensional viewing perspectives of the
surface (called the ViewPoint option in Mathematica).
u(L, t) u(L, t), t 0
0u
0u
2
2 , t0
0x x L
0x x L
Discussion Problems
u(x, 0) f (x), L x L.
Find the temperature u(x, t).
8. Find the temperature u(x, t) for the boundary-value
problem (1) – (3) when L 1 and f (x) 100 sin 6px. [Hint:
Look closely at (13) and (14).]
10. In Figure 13.3.2(b) we have the graphs of u(x, t) on the interval
[0, 6] for x 0, x p/12, x p/6, x p/4, and x p/2.
Describe or sketch the graphs of u(x, t) on the same time interval but for the fixed values x 3p/4, x 5p/6, x 11p/12,
and x p.
13.4 Wave Equation
INTRODUCTION We are now in a position to solve the boundary-value problem (11) dis-
cussed in Section 13.2. The vertical displacement u(x, t) of a string of length L that is freely
vibrating in the vertical plane shown in Figure 13.2.2(a) is determined from
a2
0 2u
0 2u
, 0 , x , L, t . 0
0x 2
0t 2
(1)
u(0, t) 0,
u(L, t) 0, t . 0
(2)
u(x, 0) f (x),
0u
2
g(x), 0 , x , L.
0t t 0
(3)
Solution of the BVP With the usual assumption that u(x, t) X(x)T(t), separating
variables in (1) gives
T0
X0
2 l
X
aT
so that
X0 lX 0
(4)
T0 a 2lT 0.
(5)
As in Section 13.3, the boundary conditions (2) translate into X(0) 0 and X(L) 0. The ODE
in (4) along with these boundary-conditions is the regular Sturm–Liouville problem
X0 lX 0, X(0) 0, X(L) 0.
(6)
Of the usual three possibilities for the parameter l: l 0, l a2 0, and l a2 0, only
the last choice leads to nontrivial solutions. Corresponding to l a2, a 0, the general solution
of (4) is
X(x) c1 cos ax c2 sin ax.
13.4 Wave Equation
|
719
X(0) 0 and X(L) 0 indicate that c1 0 and c2 sin aL 0. The last equation again implies
that aL np or a np/L. The eigenvalues and corresponding eigenfunctions of (6) are
np
ln n2p2 /L2 and X(x) c2 sin
x, n 1, 2, 3, … . The general solution of the second-order
L
equation (5) is then
T(t) c3 cos
npa
npa
t c4 sin
t.
L
L
By rewriting c2c3 as An and c2c4 as Bn, solutions that satisfy both the wave equation (1) and
boundary conditions (2) are
un aAn cos
and
npa
npa
np
t Bn sin
tb sin
x
L
L
L
q
npa
npa
np
u(x, t) a aAn cos
t Bn sin
tb sin
x.
L
L
L
n1
(7)
(8)
Setting t 0 in (8) and using the initial condition u(x, 0) f (x) gives
q
np
u(x, 0) f (x) a An sin
x.
L
n1
Since the last series is a half-range expansion for f in a sine series, we can write An bn:
An 2
L
L
# f (x) sin
0
np
x dx.
L
(9)
To determine Bn we differentiate (8) with respect to t and then set t 0:
q
0u
npa
npa
npa
npa
np
a aAn
sin
t Bn
cos
tb sin
x
0t
L
L
L
L
L
n1
q
0u
npa
np
2
g(x) a aBn
b sin
x.
0t t 0
L
L
n1
In order for this last series to be the half-range sine expansion of the initial velocity g on the
interval, the total coefficient Bnnpa/L must be given by the form bn in (5) of Section 12.3—
that is,
Bn
npa
2
L
L
L
# g(x) sin
0
np
x dx
L
from which we obtain
2
Bn npa
L
# g(x) sin
0
np
x dx.
L
(10)
The solution of the boundary-value problem (1)(3) consists of the series (8) with coefficients
An and Bn defined by (9) and (10), respectively.
We note that when the string is released from rest, then g(x) 0 for every x in the interval [0, L]
and consequently Bn 0.
Plucked String A special case of the boundary-value problem in (1)(3) when g(x) 0
is a model of a plucked string. We can see the motion of the string by plotting the solution or
displacement u(x, t) for increasing values of time t and using the animation feature of a CAS.
Some frames of a movie generated in this manner are given in FIGURE 13.4.1. You are asked to
emulate the results given in the figure by plotting a sequence of partial sums of (8). See Problems
7, 8, and 27 in Exercises 13.4.
720
|
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
u
u
1
1
x
0
x
0
–1
–1
u
1
2
(a) t = 0 initial shape
1
3
u
2
(b) t = 0.2
3
1
1
x
0
x
0
–1
–1
1
u
2
(c) t = 0.7
3
1
u
2
(d) t = 1.0
3
1
1
x
0
x
0
–1
–1
1
2
(e) t = 1.6
1
3
2
(f) t = 1.9
3
FIGURE 13.4.1 Frames of plucked-string movie
Standing Waves Recall from the derivation of the wave equation in Section 13.2 that
the constant a appearing in the solution of the boundary-value problem in (1) – (3) is given by
"T>r, where r is mass per unit length and T is the magnitude of the tension in the string. When
T is large enough, the vibrating string produces a musical sound. This sound is the result of
standing waves. The solution (8) is a superposition of product solutions called standing waves
or normal modes:
u(x, t) u1(x, t) u2(x, t) u3(x, t) p .
In view of (6) and (7) of Section 3.8, the product solutions (7) can be written as
un(x, t) Cn sin a
npa
np
t fn b sin
x,
L
L
(11)
where Cn "A2n B 2n and fn is defined by sin fn An/Cn and cos fn Bn /Cn. For n 1, 2, 3, …
the standing waves are essentially the graphs of sin(npx/L), with a time-varying amplitude given by
L x
0
(a) First standing wave
node
0
L x
L
2
(b) Second standing wave
nodes
0
L
2L
3
3
(c) Third standing wave
L x
FIGURE 13.4.2 First three standing waves
Cn sin a
npa
t fn b.
L
Alternatively, we see from (11) that at a fixed value of x each product function un(x, t) represents
simple harmonic motion with amplitude Cn u sin(npx/L) u and frequency fn na/2L. In other
words, each point on a standing wave vibrates with a different amplitude but with the same
frequency. When n 1,
u1(x, t) C1 sin a
pa
p
t f1 b sin x
L
L
is called the first standing wave, the first normal mode, or the fundamental mode of vibration. The first three standing waves, or normal modes, are shown in FIGURE 13.4.2. The dashed
graphs represent the standing waves at various values of time. The points in the interval (0, L),
for which sin(np/L)x 0, correspond to points on a standing wave where there is no motion.
These points are called nodes. For example, in Figures 13.4.2(b) and (c) we see that the second
standing wave has one node at L/2 and the third standing wave has two nodes at L/3 and 2L/3.
In general, the nth normal mode of vibration has n 1 nodes.
13.4 Wave Equation
|
721
The frequency
f1 a
1 T
2L
2LÅ r
of the first normal mode is called the fundamental frequency or first harmonic and is directly
related to the pitch produced by a stringed instrument. It is apparent that the greater the tension
on the string, the higher the pitch of the sound. The frequencies fn of the other normal modes,
which are integer multiples of the fundamental frequency, are called overtones. The second
harmonic is the first overtone, and so on.
Superposition Principle The superposition principle, Theorem 13.1.1, is the key in
making the method of separation of variables an effective means of solving certain kinds of
boundary-value problems involving linear partial differential equations. Sometimes a problem
can also be solved by using a superposition of solutions of two easier problems. If we can solve
each of the problems,
Problem 1
Problem 2
0 2u1
0 2u1
, 0 , x , L, t . 0
0x 2
0t 2
u1(0, t) 0,
u1(L, t) 0, t . 0
0u1
u1(x, 0) f(x),
`
0, 0 , x , L
0t t 0
0 2u2
0 2u2
, 0 , x , L, t . 0
0x 2
0t 2
u2(0, t) 0,
u2(L, t) 0, t . 0
0u2
u2(x, 0) 0,
`
g(x), 0 , x , L
0t t 0
a2
a2
(12)
then a solution of (1)–(3) is given by u(x, t) u1(x, t) u2(x, t). To see this we know that
u(x, t) u 1(x, t) u 2(x, t) is a solution of the homogeneous equation in (1) because of
Theorem 13.1.1. Moreover, u(x, t) satisfies the boundary condition (2) and the initial conditions (3) because, in turn,
BC e
and
u(0, t) u1(0, t) u2(0, t) 0 0 0
u(L, t) u1(L, t) u2(L, t) 0 0 0,
u(x, 0) u1(x, 0) u2(x, 0) f(x) 0 f(x)
0u1
0u2
IC • 0u
`
`
`
0 g(x) g(x).
0t t 0
0t t 0
0t t 0
You are encouraged to try this method to obtain (8), (9), and (10). See Problems 5 and 14 in
Exercises 13.4.
Exercises
13.4
Answers to selected odd-numbered problems begin on page ANS-31.
In Problems 1–6, solve the wave equation (1) subject to the
given conditions.
1. u(0, t) 0,
u(x, 0) u(L, t) 0, t . 0
1
x(L 2 x),
4
2. u(0, t) 0,
0u
`
0, 0 , x , L
0t t 0
u(L, t) 0, t . 0
0u
`
x(L 2 x), 0 , x , L
0t t 0
3. u(0, t) 0, u(p, t) 0, t . 0
u(x, 0) 0,
u(x, 0) 0,
722
|
0u
`
sin x, 0 , x , p
0t t 0
4. u(0, t) 0,
u(p, t) 0, t . 0
0u
1
`
0, 0 , x , p
u(x, 0) x(p2 2 x 2),
6
0t t 0
5. u(0, t) 0, u(1, t) 0, t . 0
0u
`
x(1 2 x), 0 , x , 1
u(x, 0) x(1 2 x),
0t t 0
6. u(0, t) 0, u(p, t) 0, t . 0
0u
`
0, 0 , x , p
u(x, 0) 0.01 sin 3px,
0t t 0
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
In Problems 7–10, a string is tied to the x-axis at x 0 and at
x L and its initial displacement u(x, 0) f(x), 0 x L, is
shown in the figure. Find u(x, t) if the string is released from rest.
7. f (x)
12. A model for the motion of a vibrating string whose ends are
allowed to slide on frictionless sleeves attached to the vertical
axes x 0 and x L is given by the wave equation (1) and
the conditions
h
0u
`
0,
0x x 0
L
L/2
u(x, 0) f(x),
x
0u
`
0, t . 0
0x x L
0u
`
g(x), 0 , x , L.
0t t 0
See FIGURE 13.4.8. The boundary conditions indicate that the
motion is such that the slope of the curve is zero at its ends
for t 0. Find the displacement u(x, t).
FIGURE 13.4.3 Initial displacement for Problem 7
8. f (x)
h
u
L
L/3
x
0
FIGURE 13.4.4 Initial displacement for Problem 8
L
x
FIGURE 13.4.8 String whose ends are attached to
frictionless sleeves in Problem 12
9. f (x)
h
13. In Problem 10, determine the value of u(L/2, t) for t 0.
14. Rederive the results given in (8), (9), and (10), but this time
use the superposition principle discussed on page 722.
2L/3
L/3
L
x
FIGURE 13.4.5 Initial displacement for Problem 9
10. f (x)
15. A string is stretched and secured on the x-axis at x 0 and
x p for t 0. If the transverse vibrations take place in a
medium that imparts a resistance proportional to the instantaneous velocity, then the wave equation takes on the form
0 2u
0 2u
0u
5 2 1 2b ,
2
0t
0x
0t
h
2L/3
L
L/3
x
0 , b , 1,
t . 0.
Find the displacement u(x, t) if the string starts from rest from
the initial displacement f (x).
16. Show that a solution of the boundary-value problem
0 2u
0 2u
u, 0 , x , p, t . 0
0x 2
0t 2
u(0, t) 0, u(p, t) 0, t . 0
–h
u(x, 0) e
FIGURE 13.4.6 Initial displacement for Problem 10
11. The longitudinal displacement of a vibrating elastic bar
shown in FIGURE 13.4.7 satisfies the wave equation (1) and
the conditions
0u
`
0, t . 0
0x x L
0u
`
0, 0 , x , L.
u(x, 0) x,
0t t 0
The boundary conditions at x 0 and x L are called
free-end conditions. Find the displacement u(x, t).
0u
`
0,
0x x 0
u(x, t)
x
0
L
FIGURE 13.4.7 Elastic bar in Problem 11
is
u(x, t) x,
p 2 x,
p>0 , x , p>2
p>2 # x , p
0u
`
0, 0 , x , p
0t t 0
4 q (1)k 1
sin (2k 2 1) x cos "(2k 2 1)2 1t.
2
p ka
(2k
2
1)
1
17. Consider the boundary-value problem given in (1) – (3) of this
section. If g(x) 0 on 0 x L, show that the solution of
the problem can be written as
u(x, t) 1
[ f (x at) f (x at)].
2
[Hint: Use the identity
2 sin u1 cos u2 sin(u1 u2) sin(u1 u2).]
13.4 Wave Equation
|
723
18. The vertical displacement u(x, t) of an infinitely long string
is determined from the initial-value problem
a2
0 2u
0 2u
, q , x , q, t . 0
0x 2
0t 2
u(x, 0) f (x),
0u
2
g(x).
0t t 0
(13)
This problem can be solved without separating variables.
(a) Show that the wave equation can be put into the form
02u/0 h 0 j 0 by means of the substitutions j x at
and h x at.
(b) Integrate the partial differential equation in part (a), first
with respect to h and then with respect to j, to show
that u(x, t) F(x at) G(x at), where F and G are
arbitrary twice differentiable functions, is a solution of
the wave equation. Use this solution and the given initial
conditions to show that
F(x) and
1
1
f (x) 2
2a
#
#
u(0, t) 0,
u(L, t) 0, t . 0
0 2u
2
0,
0x 2 x 0
0 2u
2
0, t . 0
0x 2 x L
u(x, 0) f (x),
0u
2
g(x), 0 , x , L.
0t t 0
Solve for u(x, t). [Hint: For convenience use l a4 when
separating variables.]
u
x
0
g(s) ds c
Computer Lab Assignments
x
1
1
g(s) ds 2 c,
G(x) f (x) 2
2
2a x0
24. If the ends of the beam in Problem 23 are embedded at x 0
and x L, the boundary conditions become, for t 0,
where x0 is arbitrary and c is a constant of integration.
(c) Use the results in part (b) to show that
1
1
u(x, t) [ f (x at) f (x at)] 2
2a
#
x 1 at
g(s) ds. (14)
1
[ f (x at) f (x at)], q x q.
2
The last solution can be interpreted as a superposition
of two traveling waves, one moving to the right (that
is, 12 f (x at)) and one moving to the left ( 12 f (x at)).
Both waves travel with speed a and have the same basic
shape as the initial displacement f (x). The form of u(x, t)
given in (14) is called d’Alembert’s solution.
In Problems 19–21, use d’Alembert’s solution (14) to solve the
initial-value problem in Problem 18 subject to the given initial
conditions.
f (x) sin x, g(x) 1
f (x) sin x, g(x) cos x
f (x) 0, g(x) sin 2x
Suppose f (x) 1/(1 x2), g (x) 0, and a 1 for the initialvalue problem given in Problem 18. Graph d’Alembert’s
solution in this case at the time t 0, t 1, and t 3.
23. The transverse displacement u(x, t) of a vibrating beam of
length L is determined from a fourth-order partial differential
equation
0 4u
0 2u
a 2 4 2 0, 0 , x , L, t . 0.
0x
0t
19.
20.
21.
22.
724
|
u(0, t) 0,
u(L, t) 0
0u
2
5 0,
0x x 5 0
0u
2
5 0.
0x x 5 L
x 2 at
Note that when the initial velocity g(x) 0 we obtain
u(x, t) L
FIGURE 13.4.9 Simply supported beam in Problem 23
x
x0
If the beam is simply supported, as shown in FIGURE 13.4.9,
the boundary and initial conditions are
(a) Show that the eigenvalues of the problem are l x 2n /L2
where xn, n 1, 2, 3, … , are the positive roots of the
equation cosh x cos x 1.
(b) Show graphically that the equation in part (a) has an
infinite number of roots.
(c) Use a CAS to find approximations to the first four
eigenvalues. Use four decimal places.
25. A model for an infinitely long string that is initially held at the
three points (1, 0), (1, 0), and (0, 1) and then simultaneously
released at all three points at time t 0 is given by (13) with
f (x) e
1 2 ZxZ,
0,
ZxZ # 1
and g(x) 0.
ZxZ . 1
(a) Plot the initial position of the string on the interval [6, 6].
(b) Use a CAS to plot d’Alembert’s solution (14) on [6, 6]
for t 0.2k, k 0, 1, 2, … , 25. Assume that a 1.
(c) Use the animation feature of your computer algebra system to make a movie of the solution. Describe the motion
of the string over time.
26. An infinitely long string coinciding with the x-axis is
struck at the origin with a hammer whose head is 0.2 inch
in diameter. A model for the motion of the string is given
by (13) with
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
f (x) 0 and g(x) e
1,
0,
ZxZ # 0.1
ZxZ . 0.1.
(a) Use a CAS to plot d’Alembert’s solution (14) on [6, 6]
for t 0.2k, k 0, 1, 2, … , 25. Assume that a 1.
(b) Use the animation feature of your computer algebra
system to make a movie of the solution. Describe the
motion of the string over time.
27. The model of the vibrating string in Problem 7 is called a
plucked string.
(a) Use a CAS to plot the partial sum S6(x, t); that is, the first
six nonzero terms of your solution u(x, t), for t 0.1k,
k 0, 1, 2, … , 20. Assume that a 1, h 1, and L p.
(b) Use the animation feature of your computer algebra
system to make a movie of the solution to Problem 7.
28. Consider the vibrating string in Problem 10. Use a CAS to plot
the partial sum S6(x, t); that is, the first six nonzero terms of your
solution u(x, t) for t 0.25k, k 0, 2, 3, 4, 6, 8, 10, 14. Assume
that a 1, h 1, and L ␲. Then superimpose the eight graphs
on the same coordinate system.
13.5 Laplace’s Equation
y
u = f (x)
insulated
INTRODUCTION Suppose we wish to find the steady-state temperature u(x, y) in a rectangular
(a, b)
insulated
u=0
x
FIGURE 13.5.1 Find the temperature u in
a rectangular plate
plate whose vertical edges x 0 and x a are insulated, and whose upper and lower edges y b
and y 0 are maintained at temperatures f (x) and 0, respectively. See FIGURE 13.5.1. When no
heat escapes from the lateral faces of the plate, we solve the following boundary-value problem:
0 2u
0 2u
0, 0 , x , a, 0 , y , b
0x 2
0y 2
(1)
0u
2
0,
0x x 0
0u
2
0, 0 , y , b
0x x a
(2)
u(x, 0) 0,
u(x, b) f (x), 0 x a.
(3)
Solution of the BVP With u(x, y) X(x)Y( y), separation of variables in (1) leads to
X0
Y0
l
X
Y
X0 lX 0
(4)
Y0 2 lY 0.
(5)
The three homogeneous boundary conditions in (2) and (3) translate into X (0) 0, X (a) 0,
and Y(0) 0. The Sturm–Liouville problem associated with the equation in (4) is then
X0 lX 0, X9(0) 0, X9(a) 0.
(6)
Examination of the cases corresponding to l 0, l a2 0, and l a2 0, where a 0,
has already been carried out in Example 1 in Section 12.5. For convenience a shortened version
of that analysis follows.
For l 0, (6) becomes
X0 0, X9(0) 0, X9(a) 0.
The solution of the ODE is X c1 c2x. The boundary condition X(0) 0 then implies c2 0,
and so X c1. Note that for any c1, this constant solution satisfies the second boundary condition
X (a) 0. By imposing c1 0, X c1 is a nontrivial solution of the BVP (6). For l a2 0,
(6) possesses no nontrivial solution. For l a2 0, (6) becomes
X0 a2X 0, X9(0) 0, X9(a) 0.
Applying the boundary condition X (0) 0 the solution X c1 cos ax c2 sin ax implies c2 0
and so X c1cos ax. The second boundary condition X(a) 0 applied to this last expression
then gives c1a sin aa 0. Because a 0, the last equation is satisfied when aa np or
a np/a, n 1, 2, … . The eigenvalues of (6) are then l0 and ln a2n n2p2 /a2, n 1, 2, ….
13.5 Laplace’s Equation
|
725
By corresponding l0 0 with n 0, the eigenfunctions of (6) are
X 5 c1, n 5 0,
and
X 5 c1 cos
np
x, n 5 1, 2, p .
a
We must now solve equation (5) subject to the single homogeneous boundary condition
Y(0) 0. First, for l0 0 the DE in (5) is simply Y 0, and thus its solution is Y c3 c4y. But
n2p2
Y(0) 0 implies c3 0 so Y c4y. Second, for ln n2p2 /a2, the DE in (5) is Y 2 Y 0.
a
Because 0 y b is a finite interval, we write the general solution in terms of hyperbolic
functions:
Why hyperbolic functions?
See pages 122 and 692.
Y( y) c3 cosh(npy/a) c4 sinh(npy/a).
From this solution we see Y(0) 0 again implies c3 0 so Y c4 sinh(npy/a).
Thus product solutions un X(x)Y( y) that satisfy the Laplace’s equation (1) and the three
homogeneous boundary conditions in (2) and (3) are
A0 y,
n 5 0,
and
An sinh
np
np
y cos
x, n 5 1, 2, p ,
a
a
where we have rewritten c1c4 as A0 for n 0 and as An for n 1, 2, ….
The superposition principle yields another solution
q
np
np
u(x, y) A0 y a An sinh
y cos
x.
a
a
n1
(7)
Finally, by substituting y b in (7) we see
q
np
np
bb cos
x,
u(x, b) f (x) A0b a aAn sinh
a
a
n1
is a half-range expansion of f in a Fourier cosine series. If we make the identifications A0b a0/2
and An sinh(npb/a) an, n 1, 2, … , it follows from (2) and (3) of Section 12.3 that
2A0b A0 and
An sinh
2
a
0
a
# f (x) dx
np
b
a
(8)
0
a
# f (x) cos
0
a
2
a sinh
# f (x) dx
1
ab
np
2
b
a
a
An a
# f (x) cos
0
np
x dx
a
np
x dx.
a
(9)
The solution of the boundary-value problem (1) – (3) consists of the series in (7), with coefficients A0 and An defined in (8) and (9), respectively.
Dirichlet Problem A boundary-value problem in which we seek a solution to an elliptic
partial differential equation such as Laplace’s equation 2u 0 within a region R (in the plane
or 3-space) such that u takes on prescribed values on the entire boundary of the region is called
a Dirichlet problem. In Problem 1 in Exercises 13.5 you are asked to show that the solution of
the Dirichlet problem for a rectangular region
0 2u
0 2u
1 2 5 0, 0 , x , a, 0 , y , b
2
0x
0y
726
|
u(0, y) 0,
u(a, y) 0
u(x, 0) 0,
u(x, b) f (x)
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
100
80
u(x, y) 60
40
20
01
0.8 0.6
0.8 1
0.4
0.4 0.6
y 0.2 0 0.2 x
(a) Surface
1
y
80
0.8
60
0.6
40
0.4
20
0.2
10
0
0
0.2
0.4
0.6
0.8
1
x
(b) Isotherms
FIGURE 13.5.2 Surface is graph of partial
sums when f (x) 100 and a b 1
in (10)
is
q
np
np
u(x, y) a An sinh
y sin
x where An a
a
n1
2
a sinh
a
f (x) sin
npb #
0
np
x dx.
a
(10)
a
In the special case when f (x) 100, a 1, b 1, the coefficients A n are given by
1 2 (1)n
An 200
. With the help of a CAS the plot of the surface defined by u(x, y) over
np sinh np
the region R: 0 x 1, 0 y 1 is given in FIGURE 13.5.2(a). You can see in the figure that
boundary conditions are satisfied; especially note that along y 1, u 100 for 0 x 1.
The isotherms, or curves, in the rectangular region along which the temperature u(x, y) is
constant can be obtained using the contour plotting capabilities of a CAS and are illustrated
in Figure 13.5.2(b). The isotherms can also be visualized as the curves of intersection (projected into the xy-plane) of horizontal planes u 80, u 60, and so on, with the surface in
Figure 13.5.2(a). Notice that throughout the region the maximum temperature is u 100 and
occurs on the portion of the boundary corresponding to y 1. This is no coincidence. There
is a maximum principle that states a solution u of Laplace’s equation within a bounded
region R with boundary B (such as a rectangle, circle, sphere, and so on) takes on its maximum
and minimum values on B. In addition, it can be proved that u can have no relative extrema
(maxima or minima) in the interior of R. This last statement is clearly borne out by the surface
shown in Figure 13.5.2(a).
Superposition Principle A Dirichlet problem for a rectangle can be readily solved by
separation of variables when homogeneous boundary conditions are specified on two parallel
boundaries. However, the method of separation of variables is not applicable to a Dirichlet
problem when the boundary conditions on all four sides of the rectangle are nonhomogeneous.
To get around this difficulty we break the boundary-value problem
0 2u
0 2u
1
5 0,
0x 2
0y2
0 , x , a, 0 , y , b
u(0, y) F( y),
u(a, y) G( y), 0 y b
u(x, 0) f (x),
u(x, b) g(x), 0 x a
(11)
into two problems, each of which has homogeneous boundary conditions on parallel boundaries,
as shown.
Problem 1
Problem 2
0 2u1
0 2u1
1
5 0, 0 , x , a, 0 , y , b
0x2
0y2
0 2u2
0 2u2
1
5 0, 0 , x , a, 0 , y , b
0x2
0y2
u1(0, y) 0,
u1(a, y) 0, 0 , y , b
u2(0, y) F(y),
u2(a, y) G(y), 0 , y , b
u1(x, 0) f (x),
u1(x, b) g(x), 0 , x , a
u2(x, 0) 0,
u2(x, b) 0, 0 , x , a
Suppose u 1 and u 2 are the solutions of Problems 1 and 2, respectively. If we define
u(x, y) u1(x, y) u2(x, y), it is seen that u satisfies all boundary conditions in the original
problem (11). For example,
u(0, y) u1(0, y) u2(0, y) 0 F( y) F( y)
u(x, b) u1(x, b) u2(x, b) g(x) 0 g(x)
and so on. Furthermore, u is a solution of Laplace’s equation by Theorem 13.1.1. In other
words, by solving Problems 1 and 2 and adding their solutions we have solved the original
13.5 Laplace’s Equation
|
727
problem. This additive property of solutions is known as the superposition principle. See
FIGURE 13.5.3.
y
F(y)
g (x)
(a, b)
∇2u = 0
G(y)
y
=
g (x)
0
0
∇2u1 = 0
x
f (x)
y
(a, b)
+
F(y)
x
f (x)
0
(a, b)
∇2u2 = 0
G(y)
0
x
FIGURE 13.5.3 Solution u Solution u1 of Problem 1 Solution u2 of Problem 2
We leave as exercises (see Problems 13 and 14 in Exercises 13.5) to show that a solution of
Problem 1 is
q
np
np
np
y Bn sinh
y f sin
x,
u1(x, y) a e An cosh
a
a
a
n1
2
a
An where
Bn 2
1
a
a
np
sinh
b
a
a
# f (x) sin
np
x dx
a
0
a
# g(x) sin
0
np
np
x dx 2 An cosh
bb,
a
a
and that a solution of Problem 2 is
q
np
np
np
x Bn sinh
x f sin
y,
u2(x, y) a e An cosh
b
b
b
n1
Bn Exercises
13.5
2
1
a
np
b
sinh
a
b
0u
0u
2
2
5 0,
50
0x x 5 0
0x x 5 a
u(x, 0) x, u(x, b) 0
728
|
# G( y) sin
0
np
np
y dy 2 An cosh
ab.
b
b
6. u(0, y) g( y),
0u
2
50
0x x 5 1
0u
0u
2
5 0,
2
50
0y y 5 0
0y y 5 p
0u
2
5 0, u(x, b) f (x)
0y y 5 0
4.
b
0
np
y dy
b
0u
0u
2
5 0,
2
50
0y y 5 0
0y y 5 1
2. u(0, y) 0, u(a, y) 0
u(x, 0) f (x), u(x, b) 0
# F( y) sin
5. u(0, y) 0, u(1, y) 1 y
1. u(0, y) 0, u(a, y) 0
3. u(0, y) 0, u(a, y) 0
b
Answers to selected odd-numbered problems begin on page ANS-32.
In Problems 1–10, solve Laplace’s equation (1) for a rectangular
plate subject to the given boundary conditions.
u(x, 0) 0, u(x, b) f (x)
2
b
An where
7.
0u
2
u(0, y), u(p, y) 1
0x x 0
u(x, 0) 0, u(x, p) 0
8. u(0, y) 0, u(1, y) 0
0u
2
u(x, 0), u(x, 1) f (x)
0y y 0
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
9. u(0, y) ⫽ 0, u(1, y) ⫽ 0
19. (a) Use the contour-plot application of your CAS to graph the
u(x, 0) ⫽ 100, u(x, 1) ⫽ 200
0u
10. u(0, y) ⫽ 10y,
2
5 21
0x x 5 1
u(x, 0) ⫽ 0, u(x, 1) ⫽ 0
In Problems 11 and 12, solve Laplace’s equation (1) for the
semi-infinite plate extending in the positive y-direction. In each
case assume that u(x, y) is bounded at y S q.
11.
y
12.
y
isotherms u ⫽ 170, 140, 110, 80, 60, 30 for the solution
of Problem 9. Use the partial sum S5(x, y) consisting of
the first five nonzero terms of the solution.
(b) Use the 3D-plot application of your CAS to graph the
partial sum S5(x, y).
20. Use the contour-plot application of your CAS to graph the
isotherms u ⫽ 2, 1, 0.5, 0.2, 0.1, 0.05, 0, ⫺0.05 for the solution of Problem 10. Use the partial sum S5(x, y) consisting of
the first five nonzero terms of the solution.
Discussion Problems
u=0
u=0
0
π
insulated
x
0
u = f(x)
FIGURE 13.5.4 Semi-infinite
Plate in Problem 11
insulated
π
x
u = f (x)
FIGURE 13.5.5 Semi-infinite
Plate in Problem 12
In Problems 13 and 14, solve Laplace’s equation (1) for a
rectangular plate subject to the given boundary conditions.
13. u(0, y) ⫽ 0, u(a, y) ⫽ 0
u(x, 0) ⫽ f (x), u(x, b) ⫽ g(x)
14. u(0, y) ⫽ F( y), u(a, y) ⫽ G( y)
u(x, 0) ⫽ 0, u(x, b) ⫽ 0
In Problems 15 and 16, use the superposition principle to solve
Laplace’s equation (1) for a square plate subject to the given
boundary conditions.
15. u(0, y) ⫽ 1, u(p, y) ⫽ 1
u(x, 0) ⫽ 0, u(x, p) ⫽ 1
16. u(0, y) ⫽ 0, u(2, y) ⫽ y(2 ⫺ y)
x,
0,x,1
u(x, 0) ⫽ 0, u(x, 2) ⫽ e
2 2 x, 1 # x , 2
17. In Problem 16, what is the maximum value of the temperature
u for 0 ⱕ x ⱕ 2, 0 ⱕ y ⱕ 2?
Computer Lab Assignments
18. (a) In Problem 1 suppose a ⫽ b ⫽ p and f (x) ⫽ 100x(p ⫺ x).
Without using the solution u(x, y) sketch, by hand, what
the surface would look like over the rectangular region
defined by 0 ⱕ x ⱕ p, 0 ⱕ y ⱕ p.
(b) What is the maximum value of the temperature u for
0 ⱕ x ⱕ p, 0 ⱕ y ⱕ p?
(c) Use the information in part (a) to compute the coefficients
for your answer in Problem 1. Then use the 3D-plot
application of your CAS to graph the partial sum S5(x, y)
consisting of the first five nonzero terms of the solution
in part (a) for 0 ⱕ x ⱕ p, 0 ⱕ y ⱕ p. Use different
perspectives and then compare with part (a).
21. Solve the Neumann problem for a rectangle:
0 2u
0 2u
1 2 5 0, 0 , x , a, 0 , y , b
2
0x
0y
0u
0u
2
5 0,
2
5 0, 0 , x , a
0y y 5 0
0y y 5 b
0u
0u
2
⫽ g(y), 0 , y , b.
2
5 0,
0x x 5 0
0x x ⫽ a
(a) Explain why a necessary condition for a solution u to
exist is that g satisfy
b
# g(y) dy ⫽ 0.
0
This is sometimes called a compatibility condition. Do
some extra reading and explain the compatibility condition on physical grounds.
(b) If u is a solution of the BVP, explain why u ⫹ c, where
c is an arbitrary constant, is also a solution.
22. Consider the boundary-value problem
0 2u
0 2u
1 2 5 0, 0 , x , 1, 0 , y , p
2
0x
0y
u(0, y) ⫽ u0 cos y, u(1, y) ⫽ u0(1 ⫹ cos 2y)
0u
0u
2
5 0,
2
5 0.
0y y 5 0
0y y 5 p
Discuss how the following answer was obtained
u(x, y) ⫽ u0 x ⫹
u0
u0
sinh(1 2 x) cos y ⫹
sinh 2x cos 2y.
sinh 1
sinh 2
Carry out your ideas.
13.5 Laplace’s Equation
|
729
13.6 Nonhomogeneous Boundary-Value Problems
INTRODUCTION A boundary-value problem is said to be nonhomogeneous if either the
partial differential equation or the boundary conditions are nonhomogeneous. The method of
separation of variables employed in the preceding three sections may not be applicable to a
nonhomogeneous boundary-value problem directly. In the first of the two techniques examined
in this section we employ a change of dependent variable u v ␺ that transforms a nonhomogeneous boundary-value problem into two BVPs: one involving an ODE and the other involving a PDE. The latter problem is homogeneous and solvable by separation of variables. The
second technique may also start with a change of a dependent variable, but is basically a frontal
attack on the BVP using orthogonal series expansions.
The two solution methods that follow are distinguished by different types of nonhomogeneous
boundary-value problems.
Time Independent PDE and BCs We first consider a BVP involving a time-independent
nonhomogeneous equation and time-independent boundary conditions. An example of such a
problem is
k
0u
0 2u
F(x) , 0 , x , L, t . 0
2
0t
0x
u(0, t) u0, u(L, t) u1, t . 0
(1)
u(x, 0) f(x), 0 , x , L,
where k 0 is a constant. We can interpret (1) as a model for the temperature distribution
u(x, t) within a rod of length L where heat is being generated internally throughout the rod by
either electrical or chemical means at a rate F(x) and the boundaries x 0 and x L are held at
constant temperatures u0 and u1, respectively. When heat is generated at a constant rate r within
the rod, the heat equation in (1) takes on the form
k
0u
0 2u
r .
0t
0x 2
(2)
Now equation (2) is readily shown not to be separable. On the other hand, suppose we wish to
solve the usual homogeneous heat equation kuxx ut when the boundaries x 0 and x L are
held at, say, nonzero temperatures. Even though the substitution u(x, t) X(x)T(t) separates the
PDE, we quickly find ourselves at an impasse in determining eigenvalues and eigenfunctions because no conclusion about the values of X(0) and X(L) can be drawn from u(x, 0) X(0)T(t) u0
and u(x, L) X(L)T(t) u1.
By changing the dependent variable u to a new dependent variable v by the substitution
u(x, t) v(x, t) ␺(x), (1) can be reduced to two problems:
Problem A: 5kc0 F(x) 0, c(0) u0, c(L) u1
0 2v
0v
,
0t
0x 2
Problem B: μ
v(0, t) 0, v(L, t) 0
v(x, 0) f (x) 2 c(x)
k
Observe in Problem A that the simple ODE k␺⬙ F(x) 0 can be solved by integration. Moreover,
Problem B is a homogeneous BVP that can be solved straight away by the method of separation
of variables. A solution of the given nonhomogeneous problem is then a superposition of solutions:
Solution u Solution c of Problem A Solution v of Problem B.
There is nothing in the above discussion that should be memorized; you should work through
the substitution u(x, t) v(x, t) ␺(x) each time as outlined in the next example.
730
|
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
Time-Independent PDE and BCs
EXAMPLE 1
Solve
k
0 2u
0u
⫹ r ⫽ , 0 , x , 1, t . 0
0t
0x 2
u(0, t) ⫽ 0, u(1, t) ⫽ u1, t . 0
(3)
u(x, 0) ⫽ f(x), 0 , x , 1,
where r and u1 are nonzero constants.
SOLUTION Both the partial differential equation and the boundary condition at x ⫽ 1 are
nonhomogeneous. If we let u(x, t) ⫽ v(x, t) ⫹ c(x), then
0 2v
0 2u
⫽
⫹ c0(x)
0x 2
0x 2
and
0u
0v
⫽ .
0t
0t
After substituting these results, the PDE in (3) then becomes
k
0 2v
0v
⫹ kc0(x) ⫹ r ⫽ .
2
0t
0x
(4)
Equation (4) reduces to a homogeneous equation if we demand that ␺ satisfy
kc0(x) ⫹ r ⫽ 0
r
c0(x) ⫽ ⫺ .
k
or
Integrating the last equation twice reveals that
c(x) ⫽ ⫺
r 2
x ⫹ c1x ⫹ c2.
2k
(5)
u(0, t) ⫽ v(0, t) ⫹ c(0) ⫽ 0
Furthermore,
u(1, t) ⫽ v(1, t) ⫹ c(1) ⫽ u1.
We have v(0, t) ⫽ 0 and v(1, t) ⫽ 0 provided that c also satisfies
c (0) ⫽ 0
and
c (1) ⫽ u1.
By applying the latter two conditions to (5) we obtain, in turn, c2 ⫽ 0 and c1 ⫽ r/2k ⫹ u1.
Consequently
c(x) ⫽ ⫺
r 2
r
x ⫹ a ⫹ u1 bx.
2k
2k
Finally, the initial condition in (3) implies v(x, 0) ⫽ u(x, 0) ⫺ ␺(x) ⫽ f(x) ⫺ ␺(x).
Thus to determine v(x, t) we solve the new boundary-value problem
k
0 2v
0v
5 , 0 , x , 1, t . 0
2
0t
0x
v(0, t) ⫽ 0,
v(1, t) ⫽ 0, t ⬎ 0
v(x, 0) ⫽ f (x) ⫹
r 2
r
x 2 a ⫹ u1 bx, 0 , x , 1
2k
2k
by separation of variables. In the usual manner we find
v(x, t) ⫽ a An e ⫺kn p tsin npx,
q
2
2
(6)
n51
1
where
An ⫽ 2
# cf(x) ⫹ 2kx
0
r
2
2 a
r
⫹ u1 bxd sin npx dx.
2k
13.6 Nonhomogeneous Boundary-Value Problems
(7)
|
731
A solution of the original problem is u(x, t) v(x, t) ␺(x), or
u(x, t) q
r 2
r
2 2
x a u1 bx a An e kn p tsin npx,
2k
2k
n1
(8)
where the coefficients An are defined in (7).
Inspection of (6) shows that v(x, t) S 0 as t S , and so v(x, t) is called a transient solution.
But observe in (8) that u(x, t) S ␺(x) as t S . In the context of solving forms of the heat equation, ␺(x) is called a steady-state solution.
Time-Dependent PDE and BCs We turn now to a method for solving some kinds of
BVPs that involve a time-dependent nonhomogeneous equation and time-dependent boundary
conditions. A problem similar to (1),
k
0 2u
0u
F(x, t) , 0 , x , L, t . 0
2
0t
0x
u(0, t) u0(t), u(L, t) u1(t), t . 0
(9)
u(x, 0) f(x), 0 , x , L,
describes the temperatures of a rod of length L but in this case the heat-source term F and the temperatures at the two ends of the rod can vary with time t. Intuitively one might expect that the line
of attack for this problem would be a natural extension of the procedure that worked in Example 1,
namely, seek a solution of the form u(x, t) v(x, t) c(x, t). While this form of the solution is
correct in some instances, it is usually not possible to find a function of two variables ␺(x, t) that
reduces a problem for v(x, t) to a homogeneous one. To understand why this is so, let’s see what
happens when u(x, t) v(x, t) c(x, t) is substituted into the PDE in (9). Because
0 2c
0 2u
0 2v
2 2
2
0x
0x
0x
and
0c
0u
0v
,
0t
0t
0t
(10)
the BVP (9) becomes
k
0 2c
0c
0 2v
0v
k 2 F(x, t) 2
0t
0t
0x
0x
v(0, t) c(0, t) u0(t), v(L, t) c(L, t) u1(t)
(11)
v(x, 0) f(x) 2 c(x, 0).
The boundary conditions on v in (11) will be homogeneous if we demand that
c(0, t) u0(t), c(L, t) u1(t).
(12)
Were we, at this point, to follow the same steps in the method used in Example 1, we would try
to force the problem in (11) to be homogeneous by requiring kcxx F(x, t) ct and then imposing the conditions in (12) on the solution ␺. But in view of the fact that the defining equation
for ␺ is itself a nonhomogeneous PDE, this is an unrealistic expectation. So we try an entirely
different tack by simply constructing a function ␺ that satisfies both conditions given in (12).
One such function is given by
c(x, t) u0(t) x
fu (t) 2 u0(t)g.
L 1
(13)
Reinspection of (11) shows that we have gained some additional simplification with this choice
of ␺ because cxx 0. We now start over. This time if we substitute
u(x, t) v(x, t) u0(t) 732
|
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
x
fu (t) 2 u0(t)g
L 1
(14)
the boundary-value problem (11) then becomes
k
0v
0 2v
⫹ G(x, t) ⫽ , 0 , x , L, t . 0
2
0t
0x
(15)
v(0, t) ⫽ 0, v(L, t) ⫽ 0, t . 0
v(x, 0) ⫽ f (x) 2 c(x, 0), 0 , x , L,
where G(x, t) ⫽ F(x, t) 2 ct. While the problem (15) is still nonhomogeneous (the boundary
conditions are homogeneous but the partial differential equation is nonhomogeneous) it is a
problem that we can solve.
The Basic Strategy The solution method for (15) is a bit involved, so before illustrating
with a specific example, we first outline the basic strategy:
Make the assumption that time-dependent coefficients vn(t) and Gn(t) can be found such
that both v(x, t) and G(x, t) in (15) can be expanded in the series
q
q
np
np
v(x, t) ⫽ a vn(t) sin x and G(x, t) ⫽ a Gn(t) sin x,
L
L
n⫽1
n⫽1
(16)
where sin(npx>L), n ⫽ 1, 2, 3, p, are the eigenfunctions of X0 ⫹ lX ⫽ 0, X(0) ⫽ 0,
X(L) ⫽ 0 corresponding to the eigenvalues ln ⫽ a2n ⫽ n2p2>L2. This Sturm-Liouville
problem would have been obtained had separation of variables been applied to the associated
q
np
homogeneous BVP of (15). In (16), observe that the assumed series v(x, t) ⫽ a vn(t)sin x
L
n⫽1
already satisfies the boundary conditions in (15). Now substitute this series for v(x, t) into the
nonhomogeneous PDE in (15), collect terms, and equate the resulting series with the actual
series expansion found for G(x, t).
In the next example we illustrate this method by solving a special case of (9).
EXAMPLE 2
Solve
Time-Dependent Boundary Condition
0 2u
0u
⫽ , 0 , x , 1, t . 0
2
0t
0x
u(0, t) ⫽ cos t, u(1, t) ⫽ 0, t . 0
u(x, 0) ⫽ 0, 0 , x , 1.
SOLUTION We match this problem with (9) by identifying k ⫽ 1, L ⫽ 1, F(x, t) ⫽ 0,
u 0(t) ⫽ cos t, u1 (t) ⫽ 0, and f(x) ⫽ 0. We begin with the construction of ␺. From (13)
we get
c(x, t) ⫽ cos t ⫹ xf0 2 cos tg ⫽ (1 2 x)cos t,
and then, as indicated in (14), we use
u(x, t) ⫽ v(x, t) ⫹ (1 2 x)cos t
(17)
and substitute the quantities
0 2v
0 2u
⫽ 2,
2
0x
0x
0u
0v
⫽
⫹ (1 2 x)(⫺sin t),
0t
0t
u(0, t) ⫽ v(0, t) ⫹ cos t, u(1, t) ⫽ v(1, t) and u(x, 0) ⫽ v(x, 0) ⫹ 1 2 x
13.6 Nonhomogeneous Boundary-Value Problems
|
733
into the given problem to obtain the BVP for v(x, t):
0 2v
0v
(1 2 x) sin t , 0 , x , 1, t . 0
0t
0x 2
v(0, t) 0, v(1, t) 0, t . 0
(18)
v(x, 0) x 2 1, 0 , x , 1.
The eigenvalues and eigenfunctions of the Sturm-Liouville problem
X0 lX 0, X(0) 0, X(1) 0
are found to be ln a2n n2p2 and sin n␲x, n 1, 2, 3, . . . . With G(x, t) (1 x)sin t we
assume from (18) that for fixed t, v and G can be written as Fourier sine series:
v(x, t) a vn(t) sin npx
(19)
(1 2 x)sin t a Gn(t) sin npx.
(20)
q
n1
q
and
n1
By treating t as a parameter, the coefficients Gn in (20) can be computed:
Gn(t) 2
1
#
1
0
#
1
(1 2 x)sin t sin npx dx 2sin t (1 2 x)sin npx dx 0
2
sin t.
np
q
2
(1 2 x)sin t a
sin t sin npx.
np
Hence,
(21)
n1
We can determine the coefficients vn(t) by substituting (20) and (21) back into the PDE in
(18). To that end, the partial derivatives of v are
q
0 2v
2 2
a vn(t)(n p )sin npx and
0x 2
n1
q
0v
a v9n(t)sin npx.
0t
n1
(22)
Writing the PDE in (18) as vt 2 vxx (1 2 x)sin t and using (21) and (22) we get
q
2sin t
2 2
fv9
(t)
n
p
v
(t)gsin
npx
n
a n
a np sin npx.
n1
n1
q
We then equate the coefficients of sin n␲x on each side of the equality to get the linear firstorder ODE
v9n(t) n2p2vn(t) 2sin t
.
np
2
2
Proceeding as in Section 2.3, we multiply the last equation by the integrating factor e n p t and
rewrite it as
d n2p2t
2 n2p2t
fe vn(t)g e sin t.
np
dt
Integrating both sides, we find that the general solution of this equation is
vn(t) 2
n2p2sin t 2 cos t
2 2
Cne n p t,
np(n4p4 1)
where Cn denotes the arbitrary constant. Therefore the assumed form of v(x, t) in (19) can be written
q
n2p2 sin t 2 cos t
2 2
Cne n p t d sin npx.
v(x, t) a c2
4 4
np(n p 1)
n1
734
|
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
(23)
The coefficients Cn can be found by applying the initial condition v(x, 0) to (23).
From the Fourier sine series,
q
2
Cn d sin npx
x21 ac
4 4
n 1 np(n p 1)
(24)
we see that the quantity in the brackets represents the Fourier sine coefficients bn for x 1.
That is,
1
2
Cn 2 (x 2 1) sin npx dx
4 4
np(n p 1)
0
#
Cn Therefore,
2
2
Cn .
4
np
np(n p 1)
or
4
2
2
2
.
np
np(n4p4 1)
By substituting the last result into (23) we obtain a solution of (18):
2 q n2p2sin t 2 cos t e n p t
e n p t
v(x, t) a c
2
d sin npx.
4
4
pn1
n
n(n p 1)
2
2
2
2
At long last, then, it follows from (17) that the desired solution u(x, t) is
en p t
2 q n2p2 sin t 2 cos t en p t
2
d sin npx.
c
p na
n
n(n4p4 1)
1
2
u(x, t) (1 2 x) cos t 2
2
2
If the boundary-value problem has homogeneous boundary conditions and a time-dependent
term F(x, t) in the PDE, then there is no actual need to change the dependent variable through
the substitution u(x, t) v(x, t) c(x, t). For example, if both u0 and u1 are 0 in a problem such
as (9), then it follows from (13) that c(x, t) 0. The method of solution then begins by assuming
an appropriate orthogonal series expansions for u(x, t) and F(x, t) as in (16), where the symbols
v and G in (16) are naturally replaced by u and F, respectively.
EXAMPLE 3
Time-Dependent PDE and Homogeneous BCs
0 2u
0u
(1 2 x)sin t , 0 , x , 1, t . 0
2
0t
0x
Solve
u(0, t) 0, u(1, t) 0, t . 0
(25)
u(x, 0) 0, 0 , x , 1.
SOLUTION Except for the initial condition, the BVP (25) is basically (18). As pointed out
in the paragraph preceding this example, because the boundary conditions are both homogeneous we have c(x, t) 0. Thus all steps in Example 2 used in the solution of (18) are the
same except the initial condition u(x, 0) 0 indicates that the analogue of (24) is then
q
2
Cn d sin npx.
0 ac
4 4
np(n
p 1)
n1
We conclude from this identity that the coefficient of sin n␲x must be 0 and so
Cn 2
.
np(n p4 1)
4
Hence a solution of (25) is
2 q n2p2 sin t 2 cos t
en p t
c
d sin npx.
p na
n(n4p4 1)
n(n4p4 1)
1
2
u(x, t) 2
13.6 Nonhomogeneous Boundary-Value Problems
|
735
In Problems 13–16 in Exercises 13.6 you are asked to construct c(x, t) as illustrated in
Example 2. In Problems 17–20 of Exercises 13.6 the given boundary conditions are homogenous
and so you can start as we did in Example 3 with the assumption that c(x, t) 0.
REMARKS
Don’t put any special emphasis on the fact that we used the heat equation throughout the
foregoing discussion. The method outlined in Example 1 can be applied to the wave equation
and Laplace’s equation as well. See Problems 1–12 in Exercises 13.6. The method outlined
in Example 2 is predicated on time dependence in the problem and so is not applicable to
BVPs involving Laplace’s equation.
Exercises
13.6
Answers to selected odd-numbered problems begin on page ANS-32.
In Problems 1–12, proceed as in Example 1 of this section to
solve the given boundary-value problem.
In Problems 1 and 2, solve the heat equation kuxx ut , 0 x 1,
t 0 subject to the given conditions.
1. u(0, t) 100,
The PDE is a form of the heat equation when heat is lost by
radiation from the lateral surface of a thin rod into a medium
at temperature zero.
7. Find a steady-state solution c(x) of the boundary-value
problem
u(1, t) 100
k
u(x, 0) 0
2. u(0, t) u0, u(1, t) 0
u(x, 0) f (x)
u(0, t) u0,
In Problems 3 and 4, solve the heat equation (2) subject to the
given conditions.
3. u(0, t) u0,
u(1, t) u0
u(x, 0) 0
4. u(0, t) u0, u(1, t) u1
u(x, 0) f (x)
5. Solve the boundary-value problem
0u
0u
1 Ae2bx 5 , b 0, 0 x 1, t 0
0t
0x 2
u(0, t) 0,
u(1, t) 0, t 0
u(1, t) 0, t 0
8. Find a steady-state solution c (x) if the rod in Problem 7
is semi-infinite extending in the positive x-direction,
radiates from its lateral surface into a medium at temperature zero, and
u(0, t) u0 ,
9. When a vibrating string is subjected to an external vertical
force that varies with the horizontal distance from the left end,
the wave equation takes on the form
a2
0 2u
0 2u
1 Ax 5 2 ,
2
0x
0t
where A is constant. Solve this partial differential equation
subject to
u(0, t) 0, u(1, t) 0, t . 0
2
k
0u
0u
2 hu 5 ,
2
0t
0x
u(0, t) 0,
0 x p, t 0
u(p, t) u0 , t 0
u(x, 0) 0, 0 x p.
736
|
lim u(x, t) 0, t 0
xS q
u(x, 0) f (x), x 0.
u(x, 0) f (x), 0 x 1,
where A is a constant. The PDE is a form of the heat equation
when heat is generated within a thin rod due to radioactive
decay of the material.
6. Solve the boundary-value problem
0 x 1, t 0
u(x, 0) f (x), 0 x 1.
2
k
0 2u
0u
2 h(u 2 u0) ,
0t
0x 2
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
u(x, 0) 0,
0u
2
0, 0 , x , 1.
0t t 0
10. A string initially at rest on the x-axis is secured on the x-axis
at x ⫽ 0 and x ⫽ 1. If the string is allowed to fall under its
own weight for t ⬎ 0, the displacement u(x, t) satisfies
15.
u(0, t) ⫽ 0, u(1, t) ⫽ sin t, t . 0
0 2u
0 2u
a2 2 2 g 5 2 , 0 , x , 1, t . 0,
0x
0t
u(x, 0) ⫽ 0,
where g is the acceleration of gravity. Solve for u(x, t).
11. Find the steady-state temperature u(x, y) in the semi-infinite
plate shown in FIGURE 13.6.1. Assume that the temperature is
16.
bounded as x S q. [Hint: Use u(x, y) ⫽ v(x, y) ⫹ c ( y).]
u = u0
In Problems 17–20, proceed as in Example 3 to solve the given
boundary-value problem.
x
u = u1
FIGURE 13.6.1 Semi-infinite plate in Problem 11
17.
0 2u
0 2u
1 2 5 2h,
2
0x
0y
u(x, 0) ⫽ 0, 0 , x , p.
18.
where h ⬎ 0 is a constant, occurs in many problems involving
electric potential and is known as Poisson’s equation. Solve
the above equation subject to the conditions
u(x, 0) ⫽ 0, 0 ⬍ x ⬍ p.
0 2u
0u
⫹ xe 2 3t ⫽ , 0 , x , p, t . 0
2
0t
0x
0u
0u
`
⫽ 0,
`
⫽ 0, t . 0
0x x ⫽ 0
0x x ⫽ p
u(x, 0) ⫽ 0, 0 , x , p
u(0, y) ⫽ 0, u(p, y) ⫽ 1, y ⬎ 0
19.
In Problems 13–16, proceed as in Example 2 to solve the given
boundary-value problem.
0 2u
0u
2 1 ⫹ x 2 xcos t ⫽ , 0 , x , 1, t . 0
2
0t
0x
u(0, t) ⫽ 0, u(1, t) ⫽ 0, t . 0
u(x, 0) ⫽ x(1 2 x), 0 , x , 1
0 2u
0u
⫽ , 0 , x , 1, t . 0
0t
0x 2
20.
u(0, t) ⫽ sin t, u(1, t) ⫽ 0, t . 0
u(x, 0) ⫽ 0, 0 , x , 1
0 2u
0u
14.
⫹ 2t ⫹ 3tx ⫽ , 0 , x , 1, t . 0
2
0t
0x
u(0, t) ⫽ t 2, u(1, t) ⫽ 1, t . 0
0 2u
0u
⫹ xe 2 3t ⫽ , 0 , x , p, t . 0
2
0t
0x
u(0, t) ⫽ 0, u(p, t) ⫽ 0, t . 0
12. The partial differential equation
13.
0 2u
0u
⫽ , 0 , x , 1, t . 0
2
0t
0x
u(x, 0) ⫽ 0, 0 , x , 1
u=0
0
0u
`
⫽ 0, 0 , x , 1
0t t⫽ 0
u(0, t) ⫽ 1 2 e ⫺t, u(1, t) ⫽ 1 2 e ⫺t, t . 0
y
1
0 2u
0 2u
⫽ 2 , 0 , x , 1, t . 0
2
0x
0t
0 2u
0 2u
⫹
sin
x
cos
t
⫽
, 0 , x , p, t . 0
0x 2
0t 2
u(0, t) ⫽ 0, u(p, t) ⫽ 0, t . 0
u(x, 0) ⫽ 0,
0u
`
⫽ 0, 0 , x , p
0t t⫽ 0
u(x, 0) ⫽ x 2, 0 , x , 1
13.7 Orthogonal Series Expansions
INTRODUCTION For certain types of boundary conditions, the method of separation of vari-
ables and the superposition principle lead to an expansion of a function in an infinite series that
is not a Fourier series. To solve the problems in this section we shall utilize the concept of orthogonal series expansions or generalized Fourier series developed in Section 12.1.
13.7 Orthogonal Series Expansions
|
737
Using Orthogonal Series Expansions
EXAMPLE 1
The temperature in a rod of unit length in which there is heat transfer from its right boundary
into a surrounding medium kept at a constant temperature zero is determined from
k
0 2u
0u
5 , 0 , x , 1, t . 0
0t
0x 2
0u
2
hu(1, t), h . 0, t . 0
0x x 1
u(0, t) 0,
u(x, 0) 1, 0 x 1.
Solve for u(x, t).
SOLUTION Proceeding exactly as we did in Section 13.3, with u(x, t) X(x)T(t) and l as
the separation constant, we find the separated ODEs and boundary conditions to be, respectively,
X(0) 0
X lX 0
(1)
T klT 0
(2)
X(1) hX(1).
and
(3)
Equation (1) along with the homogeneous boundary conditions (3) comprise the regular
Sturm–Liouville problem:
X0 lX 0, X(0) 0, X9(1) hX(1) 0.
(4)
Except for the presence of the symbol h, the BVP in (4) is essentially the problem solved in
Example 2 of Section 12.5. As in that example, (4) possesses nontrivial solutions only in the
case l a2 0, a 0. The general solution of the DE in (4) is X(x) c1 cos ax c2 sin ax.
The first boundary condition in (4) immediately gives c1 0. Applying the second boundary
condition in (4) to X(x) c2 sin ax yields
a cos a 1 h sin a 5 0
a
tan a 5 2 .
h
or
(5)
Because the graphs of y tan x and y x/h, h 0, have an infinite number of points of
intersection for x 0 (Figure 12.5.1 illustrates the case h 1), the last equation in (5) has an
infinite number of roots. Of course, these roots depend on the value of h. If the consecutive
positive roots are denoted an, n 1, 2, 3, … , then the eigenvalues of the problem are ln a2n ,
and the corresponding eigenfunctions are X(x) c2 sin an x, n 1, 2, 3, … . The solution of
2
the first-order DE (2) is T(t) c3 e2kan t and so
2
un X(x)T(t) Ane kan t sin anx
and
u(x, t) a Ane kan t sin an x.
q
2
n1
Now at t 0, u(x, 0) 1, 0 x 1, so that
1 5 a An sin an x.
q
(6)
n51
The series in (6) is not a Fourier sine series; rather, it is an expansion of u(x, 0) 1 in terms
of the orthogonal functions arising from the Sturm–Liouville problem (4). It follows that the
set of eigenfunctions {sinan x}, n 1, 2, 3, … , where the a’s are defined by tan a a/h is
orthogonal with respect to the weight function p(x) 1 on the interval [0, 1]. With f (x) 1 and
fn(x) sin an x, it follows from (8) of Section 12.1, that the coefficients An in (6) are
1
An e0 sin an x dx
1
e0 sin2 an x dx
.
(7)
To evaluate the square norm of each of the eigenfunctions we use a trigonometric identity:
#
1
0
738
|
1
sin an x dx 2
2
1
# (1 2 cos 2a x) dx 2 a1 2 2a
0
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
1
n
1
n
sin 2an b.
(8)
With the aid of the double angle formula sin 2an ⫽ 2 sin an cos an and the first equation in (5)
in the form an cos an ⫽ ⫺h sin an, we can simplify (8) to
1
# sin a x dx ⫽ 2h (h ⫹ cos a ).
2
0
1
Also
1
# sin a x dx ⫽ ⫺a
0
2
n
1
n
n
Consequently (7) becomes
An ⫽
1
cos an x d ⫽
0
n
1
(1 2 cos an).
an
2h(1 2 cos an)
.
an(h ⫹ cos2an)
Finally, a solution of the boundary-value problem is
q
1 2 cos an
2
u(x, t) ⫽ 2h a
e ⫺kant sin an x.
2
n ⫽ 1 an (h ⫹ cos an)
EXAMPLE 2
Using Orthogonal Series Expansions
The twist angle u(x, t) of a torsionally vibrating shaft of unit length is determined from
a2
θ
0
twisted shaft
1
FIGURE 13.7.1 The twist angle u in
Example 2
0 2u
0 2u
5 2 , 0 , x , 1, t . 0
2
0x
0t
u(0, t) ⫽ 0,
0u
2
⫽ 0, t . 0
0x x ⫽ 1
u(x, 0) ⫽ x,
0u
2
⫽ 0, 0 , x , 1.
0t t⫽ 0
See FIGURE 13.7.1. The boundary condition at x ⫽ 1 is called a free-end condition. Solve
for u(x, t).
SOLUTION Proceeding as in Section 13.4 with u(x, t) ⫽ X(x)T(t) and using ⫺l once again
as the separation constant, the separated equations and boundary conditions are
X0 ⫹ lX ⫽ 0
(9)
T0 ⫹ a 2lT ⫽ 0.
(10)
X(0) ⫽ 0 and X9(1) ⫽ 0.
(11)
Equation (9) together with the homogeneous boundary conditions in (11),
X0 ⫹ lX ⫽ 0, X(0) ⫽ 0, X9(1) ⫽ 0,
(12)
is a regular Sturm–Liouville problem. You are encouraged to verify that for l ⫽ 0 and for
l ⫽ ⫺a2, a ⬎ 0, the only solution of (12) is X ⫽ 0. For l ⫽ a2 ⬎ 0, a ⬎ 0, the boundary
conditions X(0) ⫽ 0 and X⬘(1) ⫽ 0 applied to the general solution X(x) ⫽ c1 cos ax ⫹ c2 sin ax
give, in turn, c1 ⫽ 0 and c2 cos a ⫽ 0. Since cos a is zero only when a is an odd integer multiple
of p/2 we write an ⫽ (2n ⫺ 1)p/2. The eigenvalues of (12) are ln ⫽ a2n ⫽ (2n ⫺ 1)2p2 /4, and
2n 2 1
the corresponding eigenfunctions are X(x) ⫽ c2 sin an x ⫽ c2 sin a
b px, n ⫽ 1, 2, 3, … .
2
Since the rod is released from rest, the initial condition ut(x, 0) ⫽ 0 translates into
X(x) T⬘(0) ⫽ 0 or T⬘(0) ⫽ 0. When applied to the general solution T(t) ⫽ c3 cos aant ⫹
c4 sin aant of the second-order DE (10), T⬘(0) ⫽ 0 implies c4 ⫽ 0 leaving T(t) ⫽ c3 cos aant ⫽
2n 2 1
b pt. Therefore,
c3 cos a a
2
un ⫽ X(x)T(t) ⫽ An cos a a
2n 2 1
2n 2 1
b pt sina
b px.
2
2
13.7 Orthogonal Series Expansions
|
739
In order to satisfy the remaining initial condition we form the superposition of the un,
u(x, t) ⫽ a An cos aa
q
n⫽1
2n 2 1
2n 2 1
b pt sina
b px.
2
2
(13)
When t ⫽ 0, we must have, for 0 ⬍ x ⬍ 1,
u(x, 0) ⫽ x ⫽ a An sina
q
n⫽1
2n 2 1
b px.
2
(14)
2n 2 1
b px f , n ⫽ 1, 2, 3, … , is orthogonal
2
with respect to the weight function p(x) ⫽ 1 on the interval [0, 1]. Even though the trigonometric series in (14) looks more like a Fourier series than (6), it is not a Fourier sine series
because the argument of the sine function is not an integer multiple of px/L (here L ⫽ 1). The
series is again an orthogonal series expansion or generalized Fourier series. Hence from (8)
of Section 12.1 the coefficients An in (14) are given by
As in Example 1, the set of eigenfunctions e sina
1
An ⫽
# x sina
0
#
1
0
2n 2 1
b px dx
2
2n 2 1
sin a
b px dx
2
2
.
Carrying out the two integrations, we arrive at
An ⫽
8(⫺1)n ⫹ 1
.
(2n 2 1)2p 2
The twist angle is then
u(x, t) ⫽
Exercises
13.7
2n 2 1
2n 2 1
8 q (⫺1)n ⫹ 1
cos aa
b pt sina
b px.
2
2
2
p2 na
(2n
2
1)
⫽1
Answers to selected odd-numbered problems begin on page ANS-33.
1. In Example 1 find the temperature u(x, t) when the left end of
5. Find the temperature u(x, t) in a rod of length L if the initial
the rod is insulated.
2. Solve the boundary-value problem
0 2u
0u
k 2 ⫽ , 0 , x , 1, t . 0
0t
0x
temperature is f (x) throughout and if the end x ⫽ 0 is maintained at temperature zero and the end x ⫽ L is insulated.
6. Solve the boundary-value problem
0 2u
0 2u
a 2 2 ⫽ 2 , 0 , x , L, t . 0
0x
0t
u(0, t) ⫽ 0,
0u
2
⫽ ⫺h(u(1, t) 2 u0), h . 0, t . 0
0x x ⫽ 1
u(0, t) ⫽ 0, E
u(x, 0) ⫽ f (x), 0 ⬍ x ⬍ 1.
u(x, 0) ⫽ 0,
3. Find the steady-state temperature for a rectangular plate for
which the boundary conditions are
0u
u(0, y) ⫽ 0,
2
⫽ ⫺hu(a, y), h . 0, 0 , y , b,
0x x ⫽ a
u(x, 0) ⫽ 0,
u(x, b) ⫽ f (x), 0 ⬍ x ⬍ a.
4. Solve the boundary-value problem
0 2u
0 2u
⫹ 2 ⫽ 0, x . 0, 0 , y , 1
2
0x
0y
xSq
0u
0u
2
⫽ 0,
2
⫽ ⫺hu(x, 1), h . 0, x . 0.
0y y ⫽ 0
0y y ⫽ 1
|
0u
2
⫽ g(x), 0 , x , L.
0t t⫽ 0
The solution u(x, t) represents the longitudinal displacement
of a vibrating elastic bar that is anchored at its left end and is
subjected to a constant force F0 at its right end. See Figure 13.4.7
on page 723. E is called the modulus of elasticity.
7. Solve the boundary-value problem
0 2u
0 2u
⫹ 2 ⫽ 0, 0 , x , 1, 0 , y , 1
2
0x
0y
u(0, y) ⫽ u0, lim u(x, y) ⫽ 0, 0 , y , 1
740
0u
2
⫽ F0, t . 0
0x x ⫽ L
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
0u
2
⫽ 0, u(1, y) ⫽ u0 , 0 , y , 1
0x x ⫽ 0
u(x, 0) ⫽ 0,
0u
2
⫽ 0, 0 , x , 1.
0y y ⫽ 1
8. Solve the boundary-value problem
0 2u
0 4u
1 2 5 0, 0 , x , 1, t . 0
4
0x
0t
0u
0 2u
2 u ⫽ , 0 , x , 1, t . 0
0t
0x 2
0u
2
⫽ 0, u(1, t) ⫽ 0, t . 0
0x x ⫽ 0
u(x, 0) ⫽ 1 2 x 2, 0 , x , 1.
9. The initial temperature in a rod of unit length is f (x) throughout.
There is heat transfer from both ends, x ⫽ 0 and x ⫽ 1, into a surrounding medium kept at a constant temperature zero. Show that
u(x, t) ⫽ a Ane ⫺kant(an cos an x ⫹ h sin an x),
q
2
n⫽1
where
An ⫽
2
2
(an ⫹ 2h ⫹ h 2)
1
# f (x) (a cos a x ⫹ h sin a x) dx.
n
0
n
n
The eigenvalues are ln ⫽ a2n , n ⫽ 1, 2, 3, … , where the an are
the consecutive positive roots of tan a ⫽ 2ah/(a2 ⫺ h2).
10. Use the method discussed in Example 3 of Section 13.6 to
solve the nonhomogeneous boundary-value problem
k
u(0, t) ⫽ 0,
0u
2
5 0, t . 0
0x x 5 0
0 2u
2
5 0,
0x2 x 5 1
0 3u
2
5 0, t . 0
0x3 x 5 1
u(x, 0) ⫽ f (x),
0u
2
⫽ g(x), 0 , x , 1.
0t t⫽ 0
This boundary-value problem could serve as a model for the
displacements of a vibrating airplane wing.
(a) Show that the eigenvalues of the problem are determined
from the equation cos a cosh a ⫽ ⫺1.
(b) Use a CAS to find approximations to the first two positive
eigenvalues of the problem. [Hint: See Problem 23 in
Exercises 13.4.]
u
x
0 2u
0u
1 xe22t 5 , 0 , x , 1, t . 0
2
0t
0x
0u
2
⫽ ⫺u(1, t), t . 0
0x x ⫽ 1
u(0, t) ⫽ 0,
1
FIGURE 13.7.2 Cantilever beam in Problem 11
u(x, 0) ⫽ 0, 0 , x , 1.
12. (a) Find an equation that defines the eigenvalues when the
Computer Lab Assignments
11. A vibrating cantilever beam is embedded at its left end (x ⫽ 0)
and free at its right end (x ⫽ 1). See FIGURE 13.7.2. The trans-
verse displacement u(x, t) of the beam is determined from
y
c
(b, c)
ends of the beam in Problem 11 are embedded at x ⫽ 0
and x ⫽ 1.
(b) Use a CAS to find approximations to the first two
eigenvalues of the problem.
13.8 Fourier Series in Two Variables
INTRODUCTION In the preceding sections we solved one-dimensional forms of the heat
b
and wave equations. In this section we extend the method of separation of variables to certain
problems involving the two-dimensional heat and wave equations.
x
(a)
u
c
y
b
Heat and Wave Equations in Two Dimensions Suppose the rectangular region
in FIGURE 13.8.1(a) is a thin plate in which the temperature u is a function of time t and position
(x, y). Then, under suitable conditions, u(x, y, t) can be shown to satisfy the two-dimensional
heat equation
x
k¢
(b)
FIGURE 13.8.1 (a) Rectangular plate
(b) Rectangular membrane
0 2u
0 2u
0u
⫹
≤ ⫽ .
2
2
0t
0x
0y
(1)
On the other hand, suppose Figure 13.8.1(b) represents a rectangular frame over which a thin
flexible membrane has been stretched (a rectangular drum). If the membrane is set in motion,
then its displacement u, measured from the xy-plane (transverse vibrations), is also a function of
13.8 Fourier Series in Two Variables
|
741
time t and position (x, y). When the displacements are small, free, and undamped, u(x, y, t) satisfies
the two-dimensional wave equation
a2 ¢
0 2u
0 2u
0 2u
≤
.
0x 2
0y 2
0t 2
(2)
As the next example will show, solutions of boundary-value problems involving (1) and (2)
lead to the concept of a Fourier series in two variables. Because the analysis of problems involving (1) and (2) are quite similar, we illustrate a solution only of the heat equation (1).
EXAMPLE 1
Temperatures in a Plate
Find the temperature u(x, y, t) in the plate shown in Figure 13.8.1(a) if the initial temperature
is f (x, y) throughout and if the boundaries are held at temperature zero for time t 0.
SOLUTION We must solve
k¢
subject to
0 2u
0 2u
0u
≤ , 0 , x , b, 0 , y , c, t . 0
2
2
0t
0x
0y
u(0, y, t) 0,
u(b, y, t) 0,
0 y c, t 0
u(x, 0, t) 0,
u(x, c, t) 0,
0 x b, t 0
u(x, y, 0) f (x, y),
0 x b, 0 y c.
To separate variables for the PDE in three independent variables x, y, and t we try to find a
product solution u(x, y, t) X(x)Y( y)T(t). Substituting, we get
k(X0YT XY0T ) XYT9
X0
Y0
T9
.
X
Y
kT
or
(3)
Since the left side of the last equation in (3) depends only on x and the right side depends only
on y and t, we must have both sides equal to a constant l:
X0
Y0
T9
l
X
Y
kT
and so
X lX 0
(4)
Y0
T9
l.
Y
kT
(5)
By the same reasoning, if we introduce another separation constant µ in (5), then
Y0
T9
µ and
l µ
Y
kT
Y µY 0 and
T k(l µ)T 0.
(6)
Now the homogeneous boundary conditions
u(0, y, t) 0,
u(x, 0, t) 0,
u(b, y, t) 0
r
u(x, c, t) 0
b
imply
X(0) 0,
Y(0) 0,
X(b) 0
Y(c) 0.
Thus we have two Sturm–Liouville problems, one in the variable x,
X0 lX 0, X(0) 0, X(b) 0
(7)
and the other in the variable y,
Y0 µY 0, Y(0) 0, Y(c) 0.
2
(8)
2
2
The usual consideration of cases (l 0, l a 0, l a 0, µ 0, l b 0,
and so on) leads to two independent sets of eigenvalues defined by sin lb 0 and
sin µc 0.
742
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CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
These equations in turn imply
lm 5
m2p2
,
b2
and mn 5
n2p2
.
c2
(9)
The corresponding eigenfunctions are
X(x) c2 sin
mp
x, m 1, 2, 3, p
b
and Y(y) c4 sin
np
y, n 1, 2, 3, p .
c
(10)
After substituting the values in (9) into the first-order DE in (6), its general solution is
2
2
T(t) c5e kf(mp>b) (np>c) gt . A product solution of the two-dimensional heat equation that
satisfies the four homogeneous boundary conditions is then
2
2
umn(x, y, t) Amn e kf(mp>b) (np>c) gt sin
mp
np
y,
x sin
c
b
where Amn is an arbitrary constant. Because we have two sets of eigenvalues, we are prompted
to try the superposition principle in the form of a double sum
q
q
mp
np
2
2
y.
u(x, y, t) a a Amn e kf(mp>b) (np>c) gt sin
x sin
c
b
m 1 n 1
(11)
At t 0 we want the temperature f (x, y) to be represented by
q
q
mp
np
u(x, y, 0) f (x, y) a a Amn sin
y.
x sin
c
b
m 1 n 1
(12)
Finding the coefficients Amn in (12) really does not pose a problem; we simply multiply the
double sum (12) by the product sin (mpx/b) sin (npy/c) and integrate over the rectangle defined
by 0 x b, 0 y c. It follows that
Amn 4
bc
c
b
# # f (x, y) sin
0 0
mp
np
y dx dy.
x sin
c
b
(13)
Thus, the solution of the boundary-value problem consists of (11) with the Amn defined
by (13).
The series (11) with coefficients (13) is called a sine series in two variables, or a double sine
series. The cosine series in two variables of a function f (x, y) is a little more complicated.
If the function f is defined over a rectangular region defined by 0 x b, 0 y c, then the
double cosine series is given by
q
q
mp
np
y
f (x, y) A00 a Am0 cos
x a A0n cos
c
b
m 1
n1
q
q
mp
np
1 a a Amn cos
y,
x cos
c
b
m51 n51
where
A00 1
bc
Am0 2
bc
A0n 2
bc
Amn 4
bc
c
b
# # f (x, y) dx dy
0 0
c
b
# # f (x, y) cos
0 0
c
b
# # f (x, y) cos
0 0
c
b
# # f (x, y) cos
0 0
mp
x dx dy
b
np
y dx dy
c
mp
np
y dx dy.
x cos
c
b
See Problem 2 in Exercises 13.8 for a boundary-value problem leading to a double cosine series.
13.8 Fourier Series in Two Variables
|
743
Exercises
13.8
Answers to selected odd-numbered problems begin on page ANS-33.
In Problems 1 and 2, solve the heat equation (1) subject to the
given conditions.
In Problems 5–7, solve Laplace’s equation
0 2u
0 2u
0 2u
1
1
50
0x 2
0y2
0z2
1. u(0, y, t) 0, u(p, y, t) 0
u(x, 0, t) 0, u(x, p, t) 0
u(x, y, 0) u0
0u
2. 0u
2
5 0,
2
50
0x x 5 0
0x x 5 1
0u
0u
2
5 0,
2
50
0y y 5 0
0y y 5 1
for the steady-state temperature u(x, y, z) in the rectangular
parallelepiped shown in FIGURE 13.8.2.
z
u(x, y, 0) xy
(a, b, c)
In Problems 3 and 4, solve the wave equation (2) subject to the
given conditions.
3. u(0, y, t) 0, u(p, y, t) 0
y
x
u(x, 0, t) 0, u(x, p, t) 0
u(x, y, 0) xy(x p)( y p)
FIGURE 13.8.2 Rectangular parallelepiped in Problems 5–7
0u
2
50
0t t5 0
5. The top (z c) of the parallelepiped is kept at temperature
f (x, y) and the remaining sides are kept at temperature zero.
6. The bottom (z 0) of the parallelepiped is kept at temperature
4. u(0, y, t) 0, u(b, y, t) 0
f (x, y) and the remaining sides are kept at temperature zero.
u(x, 0, t) 0, u(x, c, t) 0
u(x, y, 0) f (x, y)
7. The parallelepiped is a unit cube (a b c 1) with the top
(z 1) and bottom (z 0) kept at constant temperatures
u0 and u0, respectively, and the remaining sides kept at
temperature zero.
0u
2
g(x, y)
0t t 0
Chapter in Review
13
Answers to selected odd-numbered problems begin on page ANS-33.
In Problems 1 and 2, use separation of variables to find product
solutions u X(x)Y( y) of the given partial differential equation.
0 2u
0 2u
0 2u
0u
0u
1.
2.
5u
1
12
12
50
2
2
0x0y
0x
0y
0x
0y
3. Find a steady-state solution c(x) of the boundary-value problem
0 2u
0u
k 2 5 , 0 , x , p, t . 0
0t
0x
0u
u(p, t) 2 u1, t . 0
u(0, t) u0, 2
0x x p
u(x, 0) 0, 0 x p.
4. Give a physical interpretation for the boundary conditions in
6. The partial differential equation
0 2u
0 2u
2
1
x
5
0x 2
0t 2
is a form of the wave equation when an external vertical force
proportional to the square of the horizontal distance from the
left end is applied to the string. The string is secured at x 0
one unit above the x-axis and on the x-axis at x 1 for t 0.
Find the displacement u(x, t) if the string starts from rest from
the initial displacement f (x).
7. Find the steady-state temperature u(x, y) in the square plate
shown in FIGURE 13.R.2.
Problem 3.
5. At t 0 a string of unit length is stretched on the positive
x-axis. The ends of the string x 0 and x 1 are secured on
the x-axis for t 0. Find the displacement u(x, t) if the initial
velocity g(x) is as given in FIGURE 13.R.1.
g(x)
h
u=0
( π, π )
u = 50
u=0
x
FIGURE 13.R.2 Square plate in Problem 7
x
744
y
u=0
1
4
|
(14)
1
2
3
4
1
FIGURE 13.R.1 Initial velocity
in Problem 5
CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates
8. Find the steady-state temperature u(x, y) in the semi-infinite
plate shown in FIGURE 13.R.3.
y
insulated
π
u = 50
x
0
insulated
FIGURE 13.R.3 Semi-infinite plate in Problem 8
9. Solve Problem 8 if the boundaries y ⫽ 0 and y ⫽ p are held
at temperature zero for all time.
10. Find the temperature u(x, t) in the infinite plate of width 2L
shown in FIGURE 13.R.4 if the initial temperature is u0 through-
out. [Hint: u(x, 0) ⫽ u0, ⫺L ⬍ x ⬍ L is an even function of x.]
y
u=0
u=0
–L
L
x
14. The concentration c(x, t) of a substance that both diffuses in
a medium and is convected by the currents in the medium
satisfies the partial differential equation
0 2c
0c
0c
k 22h
5 , 0 , x , 1, t . 0,
0x
0t
0x
where k and h are constants. Solve the PDE subject to
c(0, t) ⫽ 0, c(1, t) ⫽ 0, t ⬎ 0
c(x, 0) ⫽ c0, 0 ⬍ x ⬍ 1,
where c0 is a constant.
15. Solve the boundary-value problem
0u
0 2u
⫽ , 0 , x , 1, t . 0
2
0t
0x
0u
u(0, t) ⫽ u0,
`
⫽ ⫺u(1, t) ⫹ u1, t . 0
0x x ⫽ 1
u(x, 0) ⫽ u0, 0 , x , 1,
where u0 and u1 are constants.
16. Solve Laplace’s equation for a rectangular plate subject to the
boundary-value conditions
u(0, y) ⫽ 0, u(p, y) ⫽ 0, 0 , y , p
u(x, 0) ⫽ x 2 2 px, u(x, p) ⫽ x 2 2 px, 0 , x , p.
17. Use the substitution u(x, y) ⫽ v(x, y) ⫹ c(x) and the result
FIGURE 13.R.4 Infinite plate in Problem 10
11. (a) Solve the boundary-value problem
0 2u
0u
5 , 0 , x , p, t . 0
2
0t
0x
u(0, t) ⫽ 0, u(p, t) ⫽ 0, t ⬎ 0
u(x, 0) ⫽ sin x, 0 ⬍ x ⬍ p.
(b) What is the solution of the BVP in part (a) if the initial
temperature is
u(x, 0) ⫽ 100 sin 3x ⫺ 30 sin 5x?
12. Solve the boundary-value problem
0 2u
0u
1 sin x 5 , 0 , x , p, t . 0
2
0t
0x
u(0, t) ⫽ 400, u(p, t) ⫽ 200, t ⬎ 0
u(x, 0) ⫽ 400 ⫹ sin x, 0 ⬍ x ⬍ p.
13. Find a series solution of the problem
0 2u
0u
0 2u
0u
⫹
2
⫽
⫹2
⫹ u, 0 , x , p, t . 0
2
2
0x
0t
0x
0t
u(0, t) ⫽ 0,
u(p, t) ⫽ 0, t ⬎ 0
0u
2
⫽ 0, 0 , x , p.
0t t⫽ 0
Do not attempt to evaluate the coefficients in the series.
of Problem 16 to solve the boundary-value problem
0 2u
0 2u
⫹
⫽ ⫺2, 0 , x , p, 0 , y , p
0x 2
0y 2
u(0, y) ⫽ 0, u(p, y) ⫽ 0, 0 , y , p
u(x, 0) ⫽ 0, u(x, p) ⫽ 0, 0 , x , p.
18. Solve the boundary-value problem
0 2u
0u
⫹ e x ⫽ , 0 , x , p, t . 0
2
0t
0x
0u
u(0, t) ⫽ 0,
`
⫽ 0, t . 0
0x x ⫽ p
u(x, 0) ⫽ f(x), 0 , x , p.
19. A rectangular plate is described by the region in the xy-plane
defined by 0 # x # a, 0 # y # b. In the analysis of the deflection w(x, y) of the plate under a sinusoidal load, the following
linear fourth-order partial differential equation is encountered:
q0
py
0 4w
0 4w
0 4w
px
⫹
2
⫹
⫽
sin
sin ,
4
2 2
4
a
D
b
0x
0x 0y
0y
where q0 and D are constants. Find a constant C so that the product
px py
w(x, y) ⫽ C sin sin is a particular solution of the PDE.
a
b
20. If the four edges of the rectangular plate in Problem 19 are
simply supported, then show that the particular solution satisfies the boundary conditions
w(0, y) ⫽ 0, w(a, y) ⫽ 0, 0 , y , b
w(x, 0) ⫽ 0, w(x, b) ⫽ 0, 0 , x , a
0 2w
0 2w
`
⫽
0,
`
⫽ 0, 0 , y , b
0x 2 x ⫽ 0
0x 2 x ⫽ a
0 2w
`
⫽ 0,
0y 2 y ⫽ 0
0 2w
`
⫽ 0, 0 , x , a.
0y 2 y ⫽ b
CHAPTER 13 in Review
|
745
© Palette7/ShutterStock, Inc.
CHAPTER
14
In the previous chapter we
utilized Fourier series to solve
boundary-value problems that
were described in the Cartesian or
rectangular coordinate system. In
this chapter we will finally put to
practical use the theory of
Fourier–Bessel series (Section
14.2) and Fourier–Legendre series
(Section 14.3) in the solution of
boundary-value problems
described, respectively in
cylindrical coordinates and in
spherical coordinates.
Boundary-Value
Problems in Other
Coordinate Systems
CHAPTER CONTENTS
14.1
14.2
14.3
Polar Coordinates
Cylindrical Coordinates
Spherical Coordinates
Chapter 14 in Review
14.1
Polar Coordinates
INTRODUCTION All the boundary-value problems that have been considered so far have
been expressed in terms of rectangular coordinates. If, however, we wish to find temperatures in
a circular disk, a circular cylinder, or in a sphere, we would naturally try to describe the problems
in polar coordinates, cylindrical coordinates, or spherical coordinates, respectively.
Because we consider only steady-state temperature problems in polar coordinates in this section, the first thing that must be done is to convert the familiar Laplace’s equation in rectangular
coordinates to polar coordinates.
Laplacian in Polar Coordinates The relationships between polar coordinates in the
plane and rectangular coordinates are given by
y
x r cos u,
y r sin u
and
r 2 x2 y2 ,
tan u .
x
(x, y) or
(r, θ )
y
r
y
θ
x
x
FIGURE 14.1.1 Polar coordinates of a
point (x, y) are (r, u)
See FIGURE 14.1.1. The first pair of equations transform polar coordinates (r, u) into rectangular
coordinates (x, y); the second pair of equations enable us to transform rectangular coordinates
into polar coordinates. These equations also make it possible to convert the two-dimensional
Laplacian of the function u, 2u 2u/x 2 2u/y 2, into polar coordinates. You are encouraged to work through the details of the Chain Rule and show that
0u
0u 0r
0u 0 u
0u
sin u 0u
cos u
2
r 0u
0x
0r 0x
0 u 0x
0r
0u
0u 0r
0u 0 u
0u
cos u 0u
sin u
r 0u
0y
0r 0y
0 u 0y
0r
2
0 2u
2 sin u cos u 0 2u
sin 2 u 0 2u
sin 2 u 0u
2 sin u cos u 0u
2 0 u
5
cos
u
2
1
1
1
2
2
2
2
r
r
0r0
u
0r
0u
0x
0r
r
0u
r2
(1)
2
2 sin u cos u 0 2u
cos 2 u 0 2u
cos2 u 0u
2 sin u cos u 0u
0 2u
2 0 u
5
sin
u
1
1
1
2
.
2
2
2
2
r
r
0r0
u
0r
0u
0y
0r
r
0u
r2
(2)
Adding (1) and (2) and simplifying yields the Laplacian of u in polar coordinates:
=2u u = f(θ )
y
0 2u
1 0u
1 0 2u
.
r 0r
0r 2
r 2 0 u2
In this section we shall concentrate only on boundary-value problems involving Laplace’s
equation in polar coordinates:
c
0 2u
1 0u
1 0 2u
0.
r 0r
0r 2
r 2 0 u2
x
FIGURE 14.1.2 Dirichlet problem for
a circular plate
(3)
Our first example is the Dirichlet problem for a circular disk. We wish to solve Laplace’s
equation (3) for the steady-state temperature u (r, u) in a circular disk or plate of radius c when
the temperature of the circumference is u (c, u) f (u), 0 u 2p. See FIGURE 14.1.2. It is assumed that the two faces of the plate are insulated. This seemingly simple problem is unlike any
we have encountered in the previous chapter.
EXAMPLE 1
Steady Temperatures in a Circular Plate
Solve Laplace’s equation (3) subject to u (c, u) f (u),
0 u 2p.
SOLUTION Before attempting separation of variables we note that the single boundary condition is nonhomogeneous. In other words, there are no explicit conditions in the statement
of the problem that enable us to determine either the coefficients in the solutions of the separated ODEs or the required eigenvalues. However, there are some implicit conditions.
First, our physical intuition leads us to expect that the temperature u(r, u) should be continuous and therefore bounded inside the circle r c. In addition, the temperature u(r, u)
748
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CHAPTER 14 Boundary-Value Problems in Other Coordinate Systems
should be single-valued; this means that the value of u should be the same at a specified point
in the plate regardless of the polar description of that point. Since (r, u 2p) is an equivalent
description of the point (r, u), we must have u(r, u) u(r, u 2p). That is, u(r, u) must be
periodic in u with period 2p. If we seek a product solution u R(r)(u), then (u) needs to
be 2p-periodic.
With all this in mind, we choose to write the separation constant in the separation of
variables as l:
Q0
r 2R0 rR9
l.
R
Q
The separated equations are then
r 2R0 rR9 2 lR 0
(4)
Q0 l Q 0
(5)
We are seeking a solution of the problem
Q0 l Q 0,
Q(u) Q( u 2p).
(6)
Although (6) is not a regular Sturm–Liouville problem, nonetheless the problem generates
eigenvalues and eigenfunctions. The latter form an orthogonal set on the interval [0, 2p].
Of the three possible general solutions of (5),
Q(u) c1 c2 u,
l50
(7)
Q(u) c1 cosh a u c2 sinh a u,
l 5 2a2 , 0
(8)
Q(u) c1 cos a u c2 sin a u,
l 5 a2 . 0
(9)
we can dismiss (8) as inherently non-periodic unless c1 c2 0. Similarly, solution (7) is
non-periodic unless we define c2 0. The remaining constant solution (u) c1, c1 0, can be
assigned any period and so l 0 is an eigenvalue. Finally, solution (9) will be 2p-periodic
if we take a n, where n 1, 2, ….* The eigenvalues of (6) are then l0 0 and ln n2,
n 1, 2, …. If we correspond l0 0 with n 0, the eigenfunctions of (6) are
Q(u) c1, n 0,
Q(u) c1 cos n u c2 sin n u, n 1, 2, p .
and
When ln n2, n 0, 1, 2, … the solutions of the Cauchy–Euler DE (4) are
R(r) c3 c4 ln r, n 0,
(10)
R(r) c3r n c4r n, n 1, 2, p .
(11)
n
n
Now observe in (11) that r 1/r . In either of the solutions (10) or (11), we must define
c4 0 in order to guarantee that the solution u is bounded at the center of the plate (which is
r 0). Thus product solutions un R(r)(u) for Laplace’s equation in polar coordinates are
u0 A0, n 0,
and
un r n(An cos n u Bn sin n u), n 1, 2, p ,
where we have replaced c3c1 by A0 for n 0 and by An for n 1, 2, …; the product c3c2 has
been replaced by Bn. The superposition principle then gives
u(r, u) A0 a r n(An cos n u Bn sin n u).
q
(12)
n1
By applying the boundary condition at r c to the result in (12) we recognize
f ( u) A0 a c n(An cos n u Bn sin n u)
q
n1
* For example, note that cos n(u 2p) cos(nu 2np) cos nu.
14.1 Polar Coordinates
|
749
as an expansion of f in a full Fourier series. Consequently we can make the identifications
A0 5
a0
,
2
cnAn 5 an,
1
A0 2p
That is,
An #
1
n
cp
1
Bn n
cp
and cnBn 5 bn.
2p
f ( u) d u
(13)
0
#
2p
#
2p
f ( u) cos n u d u
(14)
f ( u) sin n u d u.
(15)
0
0
The solution of the problem consists of the series given in (12), where the coefficients A0, An,
and Bn are defined in (13), (14), and (15), respectively.
Observe in Example 1 that corresponding to each positive eigenvalue, ln n2, n 1, 2, …,
there are two different eigenfunctions—namely, cos n u and sin n u. In this situation the eigenvalues
are sometimes called double eigenvalues.
y
u = u0
EXAMPLE 2
Steady Temperatures in a Semicircular Plate
Find the steady-state temperature u(r, u) in the semicircular plate shown in FIGURE 14.1.3.
SOLUTION The boundary-value problem is
c
θ =π
u = 0 at
θ =π
θ
0 2u
1 0u
1 0 2u
0, 0 , u , p, 0 , r , c
r 0r
0r 2
r 2 0 u2
x
u = 0 at
θ =0
u(c, u) u0, 0 u p
FIGURE 14.1.3 Semicircular plate in
Example 2
u(r, 0) 0,
u(r, p) 0, 0 r c.
Defining u R(r)(u) and separating variables gives
r 2R0 rR9
Q0
l
R
Q
and
r 2 R rR lR 0
(16)
l 0.
(17)
The homogeneous conditions stipulated at the boundaries u 0 and u p translate into
(0) 0 and (p) 0. These conditions together with equation (17) constitute a regular
Sturm–Liouville problem:
Q0 l Q 0,
Q(0) 0, Q(p) 0.
(18)
This familiar problem* possesses eigenvalues ln n2 and eigenfunctions (u) c2 sin n u,
n 1, 2, …. Also, by replacing l by n2 the solution of (16) is R(r) c3r n c4rn. The reasoning used in Example 1, namely, that we expect a solution u of the problem to be bounded
at r 0, prompts us to define c4 0. Therefore un R(r)(u) Anr n sin n u and
u (r, u) a Anr n sin n u.
q
n1
The remaining boundary condition at r c gives the Fourier sine series
u0 5 a Ancn sin n u.
q
n51
Consequently
Ancn 5
2
p
p
# u sin n u d u,
0
0
*The problem in (18) is Example 2 of Section 3.9 with L p.
750
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CHAPTER 14 Boundary-Value Problems in Other Coordinate Systems
An and so
2u0 1 2 (1)n
.
n
pc n
Hence the solution of the problem is given by
u (r, u) 14.1
Exercises
Answers to selected odd-numbered problems begin on page ANS-33.
In Problems 1–4, find the steady-state temperature u(r, u) in a
circular plate of radius 1 if the temperature on the circumference
is as given.
u0,
1. u(1, u) e
0,
2u0 q 1 2 (1)n r n
a b sin n u.
p na
n
c
1
9. Find the steady-state temperature u(r, u) in the portion of a
circular plate shown in FIGURE 14.1.5.
y
0,u,p
p , u , 2p
u,
0,u,p
p 2 u, p , u , 2p
3. u(1, u ) 2pu u 2, 0 u 2p
4. u(1, u ) u, 0 u 2p
5. If the boundaries u 0 and u p of a semicircular plate of
radius 2 are insulated, we then have
u=0
at q = b
u = f (q )
at r = c
2. u(1, u) e
0u
`
0, 0 , r , 2.
0u u p
0u
`
0,
0u u 0
Find the steady-state temperature u(r, u) if
u0 ,
u(2, u) e
0,
0 , u , p>2
p>2 , u , p,
x
u=0
at q = 0
FIGURE 14.1.5 Portion of a circular plate in Problem 9
10. Find the steady-state temperature u(r, u) in the infinite wedgeshaped plate shown in FIGURE 14.1.6. [Hint: Assume that the
temperature is bounded as r S 0 and as r S q.]
y
y=x
u = 30
where u0 is a constant.
6. Find the steady-state temperature u(r, u) in a semicircular plate
of radius 1 if the boundary-conditions are
u(1, u) u0, 0 , u , p
u(r, 0) 0, u(r, p) u0, 0 , r , 1,
where u0 is a constant.
7. Find the steady-state temperature u(r, u) in the quarter-circular
plate shown in FIGURE 14.1.4.
y
x
u=0
FIGURE 14.1.6 Wedge-shaped plate in Problem 10
11. Find the steady-state temperature u(r, u) in the plate in the
form of an annulus bounded between two concentric circles
of radius a and b, a , b, shown in FIGURE 14.1.7. [Hint: Proceed
as in Example 1.]
u = f (θ )
y
u=0
u = f (θ )
c
a
x
b
x
u=0
FIGURE 14.1.4 Quarter-circular plate in Problem 7
8. Find the steady-state temperature u(r, u) in the quarter-circular
plate shown in Figure 14.1.4 if the boundaries u 0 and
u p>2 are insulated, and
u(c, u) e
1,
0,
0 , u , p/4
p/4 , u , p.
u=0
FIGURE 14.1.7 Annular plate in Problem 11
12. If the boundary-conditions for the annular plate in Figure
14.1.7 are
u(a, u) u0, u(b, u) u1, 0 , u , 2p,
14.1 Polar Coordinates
|
751
where u0 and u1 are constants, show that the steady-state
temperature is given by
u(r, u) u0 ln(r>b) 2 u1 ln(r>a)
ln(a>b)
.
y=x
y
u=0
u=0
a
[Hint: Try a solution of the form u(r, u) v(r, u) c(r).]
13. Find the steady-state temperature u(r, u) in the annular plate
shown in Figure 14.1.7 if the boundary conditions are
0u
`
0r
ra
0, u(b, u) f (u), 0 , u , 2p.
14. Find the steady-state temperature u(r, u) in the annular plate
shown in Figure 14.1.7 if a 1, b 2, and
u(1, u) 75sin u, u(2, u) 60 cos u, 0 , u , 2p.
u = 100
u=0
b
x
FIGURE 14.1.10 One-eighth annular plate in Problem 18
19. Solve the exterior Dirichlet problem for a circular disk of
radius c shown in FIGURE 14.1.11. In other words, find the
steady-state temperature u(r, u) in an infinite plate that coincides with the entire xy-plane in which a circular hole of radius c
has been cut out around the origin and the temperature on
the circumference of the hole is f(u). [Hint: Assume that the
temperature u is bounded as r S q.]
y
15. Find the steady-state temperature u(r, u) in the semiannular
plate shown in FIGURE 14.1.8 if the boundary conditions
u = f (θ )
are
c
u(a, u) u(p 2 u), u(b, u) 0, 0 , u , p
u(r, 0) 0, u(r, p) 0, a , r , b.
x
y
FIGURE 14.1.11 Infinite plate in Problem 19
a
20. Consider the steady-state temperature u(r, u) in the
b
semiannular plate shown in Figure 14.1.8 with a 1, b 2,
and boundary conditions
x
FIGURE 14.1.8 Semiannular plate in Problem 15
16. Find the steady-state temperature u(r, u) in the semiannular
plate shown in Figure 14.1.8 if a 1, b 2, and
u(1, u) 0, u(2, u) u0, 0 , u , p
u(r, 0) 0, u(r, p) 0, 1 , r , 2,
where u0 is a constant.
17. Find the steady-state temperature u(r, u) in the quarter-annular
plate shown in FIGURE 14.1.9.
u(1, u) 0, u(2, u) 0, 0 , u , p
u(r, 0) 0, u(r, p) r, 1 , r , 2.
Show that in this case the choice of l a2 in (4) and (5)
leads to eigenvalues and eigenfunctions. Find the steady-state
temperature u(r, u).
Computer Lab Assignments
21. (a) Find the series solution for u(r, u) in Example 1 when
u(1, u) e
y
insulated
u = f(q )
at r = 2
u=0
at r = 1
x
insulated
FIGURE 14.1.9 Quarter-annular plate in Problem 17
18. The plate in the first quadrant shown in FIGURE 14.1.10 is
one-eighth of the annular plate in Figure 14.1.7. Find the
steady-state temperature u(r, u).
752
|
100,
0,
0,u,p
p , u , 2p.
See Problem 1.
(b) Use a CAS or a graphing utility to plot the partial sum S5(r, u)
consisting of the first five nonzero terms of the solution in
part (a) for r 0.9, r 0.7, r 0.5, r 0.3, r 0.1.
Superimpose the graphs on the same coordinate axes.
(c) Approximate the temperatures u(0.9, 1.3), u(0.7, 2),
u(0.5, 3.5), u(0.3, 4), u(0.1, 5.5). Then approximate
u(0.9, 2p 2 1.3), u(0.7, 2p 2 2), u(0.5, 2p 2 3.5),
u(0.3, 2p 2 4), u(0.1, 2p 2 5.5).
(d) What is the temperature at the center of the circular plate?
Why is it appropriate to call this value the average
temperature in the plate? [Hint: Look at the graphs in
part (b) and look at the numbers in part (c).]
CHAPTER 14 Boundary-Value Problems in Other Coordinate Systems
23. Consider the annular plate shown in Figure 14.1.7. Discuss
Discussion Problems
how the steady-state temperature u(r, u) can be found when
the boundary conditions are
22. Solve the Neumann problem for a circular plate:
1 0u
1 0 2u
0 2u
2 2 0, 0 , u , 2p, 0 , r , c
2
r 0r
0r
r 0u
u(a, u) f(u), u(b, u) g(u), 0 # u # 2p.
24. Verify that u(r, u) 34 r sin u 2 14r 3 sin 3u is a solution of the
0u
`
f(u), 0 , u , 2p.
0r r c
boundary-value problem
1 0u
1 0 2u
0 2u
0, 0 , u , 2p, 0 , r , 1
r 0r
0r 2
r 2 0u2
Give the compatibility condition. [Hint: See Problem 21 of
Exercises 13.5]
14.2
u(1, u) sin3u, 0 , u , 2p.
Cylindrical Coordinates
INTRODUCTION In this section we are going to consider boundary-value problems involving forms of the heat and wave equation in polar coordinates and a form of Laplace’s equation
in cylindrical coordinates. There is a commonality throughout the examples and most of the
exercises—the boundary-value problem possesses radial symmetry.
Radial Symmetry The two-dimensional heat and wave equations
ka
2
0 2u
0 2u
0u
0 2u
0 2u
2 0 u
1
b
5
and
a
a
1
b
5
0t
0x 2
0y2
0x 2
0y2
0t 2
expressed in polar coordinates are, in turn,
ka
2
0 2u
1 0u
1 0 2u
0u
1 0u
1 0 2u
0 2u
2 0 u
1
1
b
5
and
a
a
1
1
b
5
,
r 0r
r 0r
0t
0r 2
r 2 0u2
0r 2
r 2 0u2
0t 2
(1)
where u u(r, u, t). To solve a boundary-value problem involving either of these equations by
separation of variables we must define u R(r)(u)T(t). As in Section 13.8, this assumption
leads to multiple infinite series. See Problem 18 in Exercises 14.2. In the discussion that follows
we shall consider the simpler, but still important, problems that possess radial symmetry—that
is, problems in which the unknown function u is independent of the angular coordinate u. In this
case the heat and wave equations in (1) take, in turn, the forms
ka
2
0 2u
1 0u
0u
1 0u
0 2u
2 0 u
b
and
a
a
b
,
r 0r
r 0r
0t
0r 2
0r 2
0t 2
(2)
where u u(r, t). Vibrations described by the second equation in (2) are said to be radial vibrations.
The first example deals with the free undamped radial vibrations of a thin circular membrane.
We assume that the displacements are small and that the motion is such that each point on the
membrane moves in a direction perpendicular to the xy-plane (transverse vibrations)—that is,
the u-axis is perpendicular to the xy-plane. A physical model to keep in mind while studying this
example is a vibrating drumhead.
u
EXAMPLE 1
u = f (r) at t = 0
y
x
u = 0 at r = c
FIGURE 14.2.1 Initial displacement of
circular membrane in Example 1
Radial Vibrations of a Circular Membrane
Find the displacement u(r, t) of a circular membrane of radius c clamped along its circumference if its initial displacement is f (r) and its initial velocity is g(r). See FIGURE 14.2.1.
SOLUTION The boundary-value problem to be solved is
a2 a
0 2u
1 0u
0 2u
b 2 , 0 , r , c, t . 0
2
r 0r
0r
0t
u(c, t) 0,
t
u(r, 0) f (r),
0
0u
2
g(r), 0 , r , c.
0t t 0
14.2 Cylindrical Coordinates
|
753
Substituting u R(r)T(t) into the partial differential equation and separating variables gives
1
R9
r
T0
2 l.
R
aT
R0 (3)
Note in (3) we have returned to our usual separation constant l. The two equations obtained
from (3) are
and
rR0 R9 lrR 0
(4)
T0 a 2lT 0.
(5)
Because of the vibrational nature of the problem, equation (5) suggests that we use only
l a 2 0, a 0. Now (4) is not a Cauchy–Euler equation but is the parametric Bessel
differential equation of order n 0; that is, rR R a2rR 0. From (13) of Section 5.3
the general solution of the last equation is
R(r) c1J0(ar) c2Y0(ar).
(6)
The general solution of the familiar equation (5) is
T(t) c3 cos aat c4 sin aat.
Now recall, the Bessel function of the second kind of order zero has the property that
Y0(ar) S q as r S 0, and so the implicit assumption that the displacement u(r, t) should
be bounded at r 0 forces us to define c2 0 in (6). Thus R(r) c1J0(ar).
Since the boundary condition u(c, t) 0 is equivalent to R(c) 0, we must have c1J0(ac) 0.
We rule out c1 0 (this would lead to a trivial solution of the PDE), so consequently
J0(ac) 0.
(7)
If xn anc are the positive roots of (7), then an xn /c and so the eigenvalues of the problem
are ln a2n x2n /c2 and the eigenfunctions are c1J0(anr). Product solutions that satisfy the
partial differential equation and the boundary condition are
un R(r)T (t) (An cos aant Bn sin aant)J0(anr),
(8)
where we have done the usual relabeling of the constants. The superposition principle then gives
u(r, t) a (An cos aant Bn sin aant) J0(anr).
q
(9)
n1
The given initial conditions determine the coefficients An and Bn.
Setting t 0 in (9) and using u(r, 0) f (r) gives
f (r) a An J0(anr).
q
(10)
n1
This last result is recognized as the Fourier–Bessel expansion of the function f on the interval
(0, c). Hence by a direct comparison of (7) and (10) with (8) and (15) of Section 12.6 we can
identify the coefficients An with those given in (16) of Section 12.6:
An #
c
2
rJ0(anr ) f (r) dr.
2 2
c J 1 (anc) 0
(11)
Next, we differentiate (9) with respect to t, set t 0, and use ut (r, 0) g(r):
g(r) a aan Bn J0(anr).
q
n1
This is now a Fourier–Bessel expansion of the function g. By identifying the total coefficient
aanBn with (16) of Section 12.6 we can write
Bn #
c
2
rJ0(anr)g(r) dr.
2 2
aanc J 1 (anc) 0
(12)
Finally, the solution of the given boundary-value problem is the series (9) with coefficients
An and Bn defined in (11) and (12), respectively.
754
|
CHAPTER 14 Boundary-Value Problems in Other Coordinate Systems
Standing Waves Analogous to (11) of Section 13.4, the product solutions (8) are called
standing waves. For n 1, 2, 3, …, the standing waves are basically the graph of J0(anr) with
the time varying amplitude
An cos aant Bn sin aant.
The standing waves at different values of time are represented by the dashed graphs in FIGURE 14.2.2.
The zeros of each standing wave in the interval (0, c) are the roots of J0(anr) 0 and correspond
to the set of points on a standing wave where there is no motion. This set of points is called a nodal
line. If (as in Example 1) the positive roots of J0(anc) 0 are denoted by xn, then xn anc implies
an xn /c and consequently the zeros of the standing waves are determined from
n=1
(a)
J0(anr) J0 a
xn
rb 0.
c
Now from Table 5.3.1, the first three positive zeros of J0 are (approximately) x1 2.4, x2 5.5,
and x3 8.7. Thus for n 1, the first positive root of
J0 a
x1
rb 5 0
c
is
2.4
r 5 2.4 or r 5 c.
c
Since we are seeking zeros of the standing waves in the open interval (0, c), the last result means
that the first standing wave has no nodal line. For n 2, the first two positive roots of
J0 a
n=2
(b)
x2
rb 5 0
c
are determined from
5.5
5.5
r 5 5.5.
r 5 2.4 and
c
c
Thus the second standing wave has one nodal line defined by r x1c/x2 2.4c/5.5. Note that
r ⬇ 0.44c c. For n 3, a similar analysis shows that there are two nodal lines defined by
r x1c/x3 2.4c/8.7 and r x2c/x3 5.5c/8.7. In general, the nth standing wave has n 1 nodal
lines r x1c/xn, r x2c/xn, . . . , r xn1c/xn. Since r constant is an equation of a circle in polar
coordinates, we see in Figure 14.2.2 that the nodal lines of a standing wave are concentric circles.
Use of Computers It is possible to see the effect of a single drumbeat for the model
solved in Example 1 by means of the animation capabilities of a computer algebra system. In
Problem 20 in Exercises 14.2 you are asked to find the solution given in (9) when
c 1,
n=3
(c)
f (r) 0,
and
g(r) e
v0 ,
0,
0#r,b
b # r , 1.
Some frames of a “movie” of the vibrating drumhead are given in FIGURE 14.2.3.
FIGURE 14.2.2 Standing waves
(x, y, z) or
(r, θ , z)
z
FIGURE 14.2.3 Frames of a CAS “movie”
Laplacian in Cylindrical Coordinates From FIGURE 14.2.4 we can see that the relationship between the cylindrical coordinates of a point in space and its rectangular coordinates is given by
z
θ
r
y
x
FIGURE 14.2.4 Cylindrical coordinates
of a point (x, y, z) are (r, u, z)
x r cos u,
y r sin u,
z z.
It follows immediately from the derivation of the Laplacian in polar coordinates (see Section 14.1)
that the Laplacian of a function u in cylindrical coordinates is
=2 u 0 2u
1 0u
1 0 2u
0 2u
2 2 2.
2
r 0r
0r
r 0u
0z
14.2 Cylindrical Coordinates
|
755
z
u = u0 at z = 4
EXAMPLE 2
Steady Temperatures in a Circular Cylinder
Find the steady-state temperature in the circular cylinder shown in FIGURE 14.2.5.
u=0
at r = 2
SOLUTION The boundary conditions suggest that the temperature u has radial symmetry.
Accordingly, u (r, z) is determined from
1 0u
0 2u
0 2u
0, 0 , r , 2, 0 , z , 4
r 0r
0r 2
0z 2
y
x
u(2, z) 0, 0 z 4
u = 0 at z = 0
FIGURE 14.2.5 Finite cylinder in
Example 2
u(r, 0) 0,
u(r, 4) u0,
0 r 2.
Using u R(r)Z(z) and separating variables gives
R0 1
R9
r
R
and
For the choice l a2
0, a
Z0
l
Z
(13)
rR R lrR 0
(14)
Z lZ 0.
(15)
0, the general solution of (14) is
R(r) c1J0(ar) c2Y0(ar),
and since a solution of (15) is defined on the finite interval [0, 2], we write its general
solution as
Z(z) c3 cosh az c4 sinh az.
As in Example 1, the assumption that the temperature u is bounded at r 0 demands that
c2 0. The condition u(2, z) 0 implies R(2) 0. This equation,
J0(2a) 0,
(16)
defines the positive eigenvalues ln a2n of the problem. Last, Z(0) 0 implies c3 0. Hence
we have R c1J0(anr), Z c4 sinh anz,
un R(r) Z( z) An sinh anz J0(anr)
and
u(r, z) a An sinh anz J0(anr).
q
n51
The remaining boundary condition at z 4 then yields the Fourier–Bessel series
u0 a An sinh 4an J0(anr),
q
n51
so that in view of (16) the coefficients are defined by (16) of Section 12.6,
An sinh 4an 2u0
22 J 21(2an)
2
# rJ (a r) dr.
0
0
n
d
To evaluate the last integral we first use the substitution t anr, followed by ftJ1(t)g dt
tJ0( t). From
An sinh 4an 2an
u0
d
2 2
2an J 1(2an) 0 dt
#
we obtain
An 756
|
ft J1(t)g dt u0
.
an sinh 4an J1(2an)
CHAPTER 14 Boundary-Value Problems in Other Coordinate Systems
u0
,
an J1(2an)
Finally, the temperature in the cylinder is
u(r, z) u0 a
q
n1
sinh anz
J (a r).
an sinh 4an J1(2an) 0 n
Do not conclude from two examples that every boundary-value problem in cylindrical
coordinates gives rise to a Fourier–Bessel series.
EXAMPLE 3
Steady Temperatures in a Circular Cylinder
Find the steady-state temperatures u(r, z) in the circular cylinder defined by 0 # r # 1, 0 # z # 1
if the boundary conditions are
u(1, z) 1 2 z, 0 , z , 1
u(r, 0) 0, u(r, 1) 0, 0 , r , 1.
SOLUTION Because of the nonhomogeneous condition specified at r 1 we do not expect
the eigenvalues of the problem to be defined in terms of zeros of a Bessel function of the first
kind. As we did in Section 14.1 it is convenient in this problem to use l as the separation
constant. Thus from (13) of Example 2 we see that separation of variables now gives the two
ordinary differential equations
rR0 R9 2 lrR 0
Z0 lZ 0.
and
You should verify that the two cases l 5 0 and l 5 2a2 , 0 lead only to the trivial solution
u 0. In the case l 5 a2 . 0 the DEs are
rR0 R9 2 a2rR 0
Review pages 282–283
of Section 5.3. See also
Figures 5.3.3 and 5.3.4.
Z0 a2Z 0.
and
The first equation is the parametric form of Bessel’s modified DE of order n 0. The
solution of this equation is R(r) c1I0(ar) c2K0(ar). We immediately define c2 5 0
because the modified Bessel function of the second kind K0(ar) is unbounded at r 0.
Therefore, R(r) c1I0(ar).
Now the eigenvalues and eigenfunctions of the Sturm–Liouville problem
Z0 a2Z 0,
Z (0) 0,
Z (1) 0
are ln 5 n2p2, n 5 1, 2, 3, . . . and Z(z) c3 sin npz. Thus product solutions that satisfy the
PDE and the homogeneous boundary conditions are
un R(r)Z(z) An I0(npr) sin npz.
Next we form
u(r, z) a An I0 (npr) sin npz.
q
n1
The remaining condition at r 1 yields the Fourier sine series
u(1, z) 1 2 z a An I0(np) sin npz.
q
n1
From (5) of Section 12.3 we can write
#
1
An I0(np) 2 (1 2 z) sin npz dz 0
and
An The steady-state temperature is then
u(r, z) 2
np
d integration by parts
2
.
npI0(np)
2 q I0(npr)
sin npz.
p na
1 nI0(np)
14.2 Cylindrical Coordinates
|
757
REMARKS
Because Bessel functions appear so frequently in the solutions of boundary-value problems
expressed in cylindrical coordinates, they are also referred to as cylinder functions.
Exercises
14.2
Answers to selected odd-numbered problems begin on page ANS-34.
1. Find the displacement u(r, t) in Example 1 if f (r) 0 and the
circular membrane is given an initial unit velocity in the
upward direction.
2. A circular membrane of radius 1 is clamped along its circumference. Find the displacement u(r, t) if the membrane starts
from rest from the initial displacement f (r) 1 r 2, 0 r 1.
[Hint: See Problem 10 in Exercises 12.6.]
3. Find the steady-state temperature u(r, z) in the cylinder in
Example 2 if the boundary conditions are u(2, z) 0, 0 z 4,
u(r, 0) u0, u(r, 4) 0, 0 r 2.
4. If the lateral side of the cylinder in Example 2 is insulated,
then
0u
2
5 0, 0 , z , 4.
0r r 5 2
10. Solve Problem 9 if the edge r c of the plate is insulated.
11. When there is heat transfer from the lateral side of an infinite
circular cylinder of radius 1 (see FIGURE 14.2.6) into a surround-
ing medium at temperature zero, the temperature inside the
cylinder is determined from
ka
0u
2
hu(1, t), h . 0, t . 0
0r r 1
u(r, 0) f (r),
Solve for u(r, t).
In Problems 5–8, find the steady-state temperature u(r, z) in a
finite cylinder defined by 0 # r # 1, 0 # z # 1 if the boundary
conditions are as given.
5. u(1, z) z, 0 , z , 1
6. u(1, z) z, 0 , z , 1
0u
2 0, 0 , r , 1
0z z 0
0u
2 0, 0 , r , 1
0z z 0
0u
2 0, 0 , r , 1
0z z 1
u(r, 1) 0,
0,r,1
7. u(1, z) u0 , 0 , z , 1
8. u(1, z) 0,
0,z,1
0u
2 0,
0z z 0
0,r,1
0,r,1
0u
2 0, 0 , r , 1
0z z 1
u(r, 1) u0 , 0 , r , 1
9. The temperature in a circular plate of radius c is determined
from the boundary-value problem
ka
0 2u
1 0u
0u
1
b 5
, 0 , r , c, t . 0
2
r
0r
0t
0r
u(c, t) 0,
t
u(r, 0) f (r),
y
1
x
FIGURE 14.2.6 Infinite cylinder in Problem 11
12. Find the steady-state temperature u(r, z) in a semi-infinite
cylinder of radius 1 (z 0) if there is heat transfer from its
lateral side into a surrounding medium at temperature zero
and if the temperature of the base z 0 is held at a constant
temperature u0.
In Problems 13 and 14, use the substitution u(r, t) v(r, t) c(r)
to solve the given boundary-value problem. [Hint: Review
Section 13.6.]
13. A circular plate is a composite of two different materials
in the form of concentric circles. See FIGURE 14.2.7. The
temperature u(r, t) in the plate is determined from the
boundary-value problem
0
0 r c.
Solve for u(r, t).
758
|
0 r 1.
z
(a) Find the steady-state temperature u(r, z) when u(r, 4) f (r),
0 r 2.
(b) Show that the steady-state temperature in part (a) reduces
to u(r, z) u0 z/4 when f (r) u0. [Hint: Use (12) of
Section 12.6.]
u(r, 0) 0,
0 2u
1 0u
0u
1
b 5 , 0 , r , 1, t . 0
2
r
0r
0t
0r
CHAPTER 14 Boundary-Value Problems in Other Coordinate Systems
0 2u
1 0u
0u
1
5 ,
2
r
0r
0t
0r
u(2, t) 100, t
200,
u(r, 0) e
100,
0 , r , 2, t . 0
0
0,r,1
1 , r , 2.
(a) Use the substitution u(r, t) v(r, t) Bt in the preceding
problem to show that v(r, t) satisfies
y
u = 100
2
1
0v
1 0v
0 2v
B, 0 , r , 1, t . 0
r 0r
0t
0r 2
x
0v
`
1, t . 0
0r r 1
v(r, 0) 0, 0 , r , 1.
FIGURE 14.2.7 Circular plate in Problem 13
14.
0 2u 1 0u
0u
b , 0 , r , 1, t . 0 , b a constant
2
r
0r
0t
0r
u(1, t) 0, t 0
u(r, 0) 0, 0 r 1.
15. The horizontal displacement u(x, t) of a heavy chain of length L
oscillating in a vertical plane satisfies the partial differential
equation
0u
0 2u
0
ax b 5 2 ,
g
0x
0x
0t
t
u(x, 0) f (x),
1 0u
1
0 2u
0 2u
2 2 u 2 0, 0 , r , 1, 0 , z , 1
2
r 0r
0r
r
0z
u(1, z) 0, 0 , z , 1
u(r, 0) 0, u(r, 1) r 2 r 3, 0 , r , 1.
0 , x , L, t . 0.
See FIGURE 14.2.8.
(a) Using l as a separation constant, show that the ordinary
differential equation in the spatial variable x is
xX X lX 0. Solve this equation by means of the
substitution x t2 /4.
(b) Use the result of part (a) to solve the given partial
differential equation subject to
u(L, t) 0,
Here B is a constant to be determined.
(b) Now use the substitution v(r, t) w(r, t) c(r) to solve
the boundary-value problem in part (a). [Hint: You may
need to review Section 3.5.]
(c) What is the solution u(r, t) of the first problem?
17. Solve the boundary-value problem
0
0u
2
0, 0 , x , L.
0t t 0
[Hint: Assume the oscillations at the free end x 0 are
finite.]
x
[Hint: See equation (12) in Section 5.3.]
18. In this problem we consider the general case—that is, with u
dependence—of the vibrating circular membrane of
radius c:
a2 a
0 2u
1 0u
1 0 2u
0 2u
1
1
b
5
, 0 , r , c, t . 0
r 0r
0r 2
r 2 0u2
0t 2
u(c, u, t) 0, 0 , u , 2p, t . 0
u(r, u, 0) f (r, u), 0 , r , c, 0 , u , 2p
0u
2
g(r, u), 0 , r , c, 0 , u , 2p.
0t t 0
(a) Assume that u R(r)(u)T(t) and the separation constants are l and n. Show that the separated differential equations are
T0 a 2lT 0,
Q0 nQ 0
r 2R0 rR9 (lr 2 2 n)R 0.
L
u
0
FIGURE 14.2.8 Oscillating chain in Problem 15
(b) Let l a2 and n b2 and solve the separated equations
in part (a).
(c) Determine the eigenvalues and eigenfunctions of the
problem.
(d) Use the superposition principle to determine a multiple
series solution. Do not attempt to evaluate the
coefficients.
16. Consider the boundary-value problem
1 0u
0u
0 2u
, 0 , r , 1, t . 0
2
r
0r
0t
0r
0u
`
1, t . 0
0r r 1
u(r, 0) 0, 0 , r , 1.
Computer Lab Assignments
19. (a) Consider Example 1 with a 1, c 10, g(r) 0, and
f (r) 1 r/10, 0 r 10. Use a CAS as an aid in
finding the numerical values of the first three eigenvalues
l1, l2, l3 of the boundary-value problem and the first
14.2 Cylindrical Coordinates
|
759
three coefficients A1, A2, A3 of the solution u(r, t) given
in (9). Write the third partial sum S3(r, t) of the series
solution.
(b) Use a CAS to plot the graph of S3(r, t) for t 0, 4, 10, 12, 20.
20. Solve Problem 9 with boundary conditions u(c, t) 200,
u(r, 0) 0. With these imposed conditions, one would expect
intuitively that at any interior point of the plate, u(r, t) S 200
as t S q. Assume that c 10 and that the plate is cast iron
so that k 0.1 (approximately). Use a CAS as an aid in finding the numerical values of the first five eigenvalues
l1, l2, l3, l4, l5 of the boundary-value problem and the five
coefficients A1, A2, A3, A4, A5 in the solution u(r, t). Let the
corresponding approximate solution be denoted by S5(r, t).
Plot S5(5, t) and S5(0, t) on a sufficiently large time interval
[0, T ]. Use the plots of S5(5, t) and S5(0, t) to estimate the times
(in seconds) for which u(5, t) ⬇ 100 and u(0, t) ⬇ 100. Repeat
for u(5, t) ⬇ 200 and u(0, t) ⬇ 200.
21. Consider an idealized drum consisting of a thin membrane
stretched over a circular frame of radius 1. When such a drum
is struck at its center, one hears a sound that is frequently
described as a dull thud rather than a melodic tone. We can
model a single drumbeat using the boundary-value problem
solved in Example 1.
(a) Find the solution u(r, t) given in (9) when c 1, f (r) 0,
and
14.3
g(r) e
v0,
0,
0#r,b
b # r , 1.
(b) Show that the frequency of the standing wave un(r, t) is
fn aln /2p, where ln is the nth positive zero of J0(x).
Unlike the solution of the one-dimensional wave equation
in Section 13.4, the frequencies are not integer multiples
of the fundamental frequency f1. Show that f2 ⬇ 2.295f1
and f3 ⬇ 3.598f1. We say that the drumbeat produces
anharmonic overtones. As a result the displacement
function u(r, t) is not periodic, and so our ideal drum
cannot produce a sustained tone.
(c) Let a 1, b 14 , and v0 1 in your solution in part (a).
Use a CAS to graph the fifth partial sum S5(r, t) at the
times t 0, 0.1, 0.2, 0.3, …, 5.9, 6.0 on the interval
[1, 1]. Use the animation capabilities of your CAS to
produce a movie of these vibrations.
(d) For a greater challenge, use the 3D plotting capabilities
of your CAS to make a movie of the motion of the circular drumhead that is shown in cross section in part (c).
[Hint: There are several ways of proceeding. For a fixed
time, either graph u as a function of x and y using
r "x 2 y 2 or use the equivalent of Mathematica’s
RevolutionPlot3D.]
Spherical Coordinates
INTRODUCTION In this section we continue our examination of boundary-value problems
z
in different coordinate systems. This time we are going to consider problems involving the heat,
wave, and Laplace’s equation in spherical coordinates.
(x, y, z) or
(r, θ , φ )
θ
r
y
φ
Laplacian in Spherical Coordinates As shown in FIGURE 14.3.1, a point in 3-space
is described in terms of rectangular coordinates and in spherical coordinates. The rectangular
coordinates x, y, and z of the point are related to its spherical coordinates r, u, and f through the
equations
x
x r sin u cos f,
FIGURE 14.3.1 Spherical coordinates of a
point (x, y, z) are (r, u, f)
y r sin u sin f,
z r cos u.
By using the equations in (1) it can be shown that the Laplacian u in the spherical coordinate
system is
0 2u
0 2u
2 0u
1
1 0 2u
cot u 0u
=2u 2 2
.
(2)
2
2
2
2
r
0r
0r
r sin u 0f
r 0u
r 2 0u
As you might imagine, problems involving (1) can be quite formidable. Consequently we shall
consider only a few of the simpler problems that are independent of the azimuthal angle f.
Our first example is the Dirichlet problem for a sphere.
z
c
EXAMPLE 1
y
Steady Temperatures in a Sphere
Find the steady-state temperature u(r, u) in the sphere shown in FIGURE 14.3.2.
SOLUTION The temperature is determined from
u = f (θ )
at r = c
x
FIGURE 14.3.2 Dirichlet problem for a
sphere in Example 1
760
|
(1)
2
0 2u
2 0u
1 0 2u
cot u 0u
2
0, 0 , r , c, 0 , u , p
2
2
2
r
0r
0r
r 0u
r 0u
u(c, u) f (u), 0 u p.
CHAPTER 14 Boundary-Value Problems in Other Coordinate Systems
If u R(r)(u), the partial differential equation separates as
Q0 cot u Q9
r 2R0 2rR9
l,
R
Q
r2 R 2rR lR 0
(3)
sin u cos u l sin u 0.
(4)
and so
After we substitute x cos u, 0 u p, (4) becomes
(1 2 x 2)
d 2Q
dQ
2 2x
lQ 0,
2
dx
dx
1 # x # 1.
(5)
The latter equation is a form of Legendre’s equation (see Problems 52 and 53 in Exercises 5.3).
Now the only solutions of (5) that are continuous and have continuous derivatives on the
closed interval [1, 1] are the Legendre polynomials Pn(x) corresponding to l n(n 1),
n 0, 1, 2, …. Thus we take the solutions of (4) to be
(u) Pn(cos u).
Furthermore, when l n(n 1), the general solution of the Cauchy–Euler equation (3) is
R(r) c1r n c2r(n1).
Since we again expect u(r, u ) to be bounded at r 0, we define c2 0. Hence un Anr nPn(cos u ),
and
u(r, u) a An r nPn(cos u).
q
n0
f (u) a An c nPn(cos u).
q
At r c,
n0
Therefore Ancn are the coefficients of the Fourier–Legendre series (23) of Section 12.6:
An 2n 1
2c n
p
# f (u)P (cos u) sin u du.
0
n
It follows that the solution is
u(r, u) a a
q
n0
14.3
Exercises
#
p
0
r n
f (u)Pn(cos u) sin u dub a b Pn(cos u).
c
Answers to selected odd-numbered problems begin on page ANS-34.
1. Solve the problem in Example 1 if
50,
f (u) e
0,
2n 1
2
0 , u , p>2
p>2 , u , p.
Write out the first four nonzero terms of the series solution.
[Hint: See Example 3, Section 12.6.]
2. The solution u(r, u) in Example 1 could also be interpreted as
the potential inside the sphere due to a charge distribution f (u)
on its surface. Find the potential outside the sphere.
3. Find the solution of the problem in Example 1 if f (u) cos u,
0 u p. [Hint: P1(cos u) cos u. Use orthogonality.]
4. Find the solution of the problem in Example 1 if f (u) 1 cos 2u, 0 u p. [Hint: See Problem 18, Exercises 12.6.]
14.3 Spherical Coordinates
|
761
5. Find the steady-state temperature u(r, u) within a hollow
0 2u
2 0u
0u
1
5 , 0 , r , 1, t . 0
r 0r
0t
0r 2
sphere a r b if its inner surface r a is kept at
temperature f (u) and its outer surface r b is kept at temperature zero. The sphere in the first octant is shown in
FIGURE 14.3.3.
0u
2
h(u(1, t) 2 u1), 0 , h , 1
0r r 1
u(r, 0) u0,
u = f(θ ) z
at r = a
0 r 1.
Solve for u(r, t). [Hint: Proceed as in Problem 9.]
1
y
u=0
at r = b
x
u1
FIGURE 14.3.3 Hollow sphere in Problem 5
6. The steady-state temperature in a hemisphere of radius c is
determined from
0 2u
2 0u
1 0 2u
cot u 0u
1
1
1 2
5 0, 0 , r , c, 0 , u , p>2
2
2
2
r 0r
0r
r 0u
r 0u
u (r, p>2) 0, 0 , r , c
FIGURE 14.3.4 Container in Problem 10
11. Solve the boundary-value problem involving spherical
vibrations:
u(c, u) f (u), 0 , u , p>2.
a2 a
Solve for u(r, u ). [Hint: Pn(0) 0 only if n is odd. Also see
Problem 20, Exercises 12.6.]
7. Solve Problem 6 when the base of the hemisphere is insulated;
that is,
0u
2
5 0,
0u u 5 p>2
0 , r , c.
8. Solve Problem 6 for r c.
9. The time-dependent temperature within a sphere of radius 1
is determined from
0 2u
2 0u
0u
1
5 , 0 , r , 1, t . 0
r 0r
0t
0r 2
u(1, t) 100,
t
0
u(r, 0) 0, 0 r 1.
Solve for u(r, t). [Hint: Verify that the left side of the partial
1 02
(ru). Let
differential equation can be written as
r 0r 2
ru(r, t) v(r, t) c(r). Use only functions that are
bounded as r S 0.]
10. A uniform solid sphere of radius 1 at an initial constant tem-
perature u0 throughout is dropped into a large container of
fluid that is kept at a constant temperature u1 (u1 u0) for all
time. See FIGURE 14.3.4. Since there is heat transfer across the
boundary r 1, the temperature u(r, t) in the sphere is determined from the boundary-value problem
762
|
0 2u
2 0u
0 2u
1
b
5
, 0 , r , c, t . 0
r 0r
0r 2
0t 2
u(c, t) 0,
t
u(r, 0) f (r),
0
0u
2
g(r), 0 , r , c.
0t t 0
[Hint: Write the left side of the partial differential equation as
1 02
(ru). Let v(r, t) ru(r, t).]
a2
r 0r 2
12. A conducting sphere of radius c is grounded and placed in a
uniform electric field that has intensity E in the z-direction.
The potential u(r, u) outside the sphere is determined from
the boundary-value problem
0 2u
2 0u
1 0 2u
cot u 0u
1
1 2 21 2
5 0, r . c, 0 , u , p
2
r 0r
0r
r 0u
r 0u
u(c, u) 0, 0 u p
lim u(r, u) Ez Er cos u.
rSq
Show that
u(r, u) Er cos u E
c3
cos u.
r2
[Hint: Explain why e0p cos u Pn(cos u) sin u du 0 for all
nonnegative integers except n 1. See (24) of Section 12.6.]
CHAPTER 14 Boundary-Value Problems in Other Coordinate Systems
In Problems 13 and 14, you are asked to find a product solution u(r, u, f) R(r)Q(u)(f) of Helmholtz’s partial differential equation =2u k 2u 0 where the Laplacian =2u is
defined in (2).
13. (a) Proceed as in Example 1 but using u(r, u, f) R(r)Q(u)(f) and the separation constant n(n 1) to
show that the radial dependence of the solution u is defined by the equation
d 2R
dR
fk 2r 2 2 n(n 1)gR 0.
2r
dr
dr 2
(b) Now use the second separation constant m2 to show that
the remaining separated equations are
r2
d 2
m2 0
df2
(c) Use the substitution x cos u to show that the second
differential equation in part (b) becomes
(1 2 x 2)
d 2Q
dQ
m2
2 2x
cn(n 1) 2
d Q 0.
2
dx
dx
1 2 x2
14. (a) Assume that m and n are nonnegative integers. Then find
a product solution u(r, u, f) R(r)Q(u)(f) of
Helmholtz’s PDE using the general solution of the ODE
in part (a), the general solution of the first ODE in part
(b), and a particular solution of the second ODE in part
(b) of Problem 13. [Hint: See Problems 41, 42(c), and
54 in Exercises 5.3.]
(b) What product solution in part (a) would be bounded at
the origin?
cos u dQ
m2
d 2Q
d Q 0.
cn(n
1)
2
sin u du
du2
sin2 u
Chapter in Review
14
Answers to selected odd-numbered problems begin on page ANS-34.
In Problems 1 and 2, find the steady-state temperature u(r, u) in
a circular plate of radius c if the temperature on the circumference is as given.
u0 ,
0 , u , p2
u0, p , u , 2p
1, 2p>0 , u , p>2
2. u(c, u) • 0,
2p>2 , u , 3p>2
1, 3p>2 , u , 2p
In Problems 3 and 4, find the steady-state temperature u(r, u) in a
semicircular plate of radius 1 if boundary conditions are as given.
1. u(c, u) e
3. u(1, u) u0(pu 2 u2),
0,u,p
u(r, 0) 0, u(r, p) 0, 0 , r , 1
4. u(1, u) sin u, 0 , u , p
u(r, 0) 0, u(r, p) 0, 0 , r , 1
5. Find the steady-state temperature u(r, u ) in a semicircular
plate of radius c if the boundaries u 0 and u p are insulated
and u(c, u ) f (u ), 0 u p.
6. Find the steady-state temperature u(r, u) in a semicircular plate
of radius c if the boundary u 0 is held at temperature zero,
the boundary u p is insulated, and u(c, u) f (u), 0 u p.
In Problems 7 and 8, find the steady-state temperature u(r, u) in
the plate shown in the figure.
7.
y=x
y
8. y
u = u1
at θ = β
u = f(θ )
at r = b
u=0
at r = a
x
u = u0
at θ = 0
FIGURE 14.R.2 Plate in Problem 8
9. If the boundary conditions for an annular plate defined by
1 r 2 are
0u
2
0, 0 , u , 2p,
0r r 2
u(1, u) sin2u,
show that the steady-state temperature is
u(r, u) 1
1 2
8 2
a
r r b cos 2u.
2
34
17
[Hint: See Figure 14.1.4. Also, use the identity sin2 u 1
2 (1 cos 2u).]
10. Find the steady-state temperature u(r, u) in the infinite plate
shown in FIGURE 14.R.3.
u=0
y
insulated
at r = 1
u = u0
at r = 12
u = f (θ )
x
u=0
at θ = 0
FIGURE 14.R.1 Plate in Problem 7
1
u=0
x
u=0
FIGURE 14.R.3 Infinite plate in Problem 10
CHAPTER 14 in Review
|
763
11. Suppose heat is lost from the flat surfaces of a very thin circular
plate of radius 1 into a surrounding medium at temperature
zero. If the linear law of heat transfer applies, the heat equation
assumes the form
0 2u
1 0u
0u
1
2 hu 5 ,
r 0r
0t
0r 2
h . 0, 0 , r , 1, t . 0.
16. Solve the boundary-value problem
0 2u
2 0u
0 2u
1
5 2 , 0 , r , 1, t . 0
2
r 0r
0r
0t
0u
2
5 0,
0r r 5 1
t.0
u(r, 0) f (r),
See FIGURE 14.R.4. Find the temperature u(r, t) if the edge r 1
is kept at temperature zero and if initially the temperature of
the plate is unity throughout.
0u
2
g(r), 0 , r , 1.
0t t 0
[Hint: Proceed as in Problems 9 and 10 in Exercises 14.3, but
let v(r, t) ru(r, t). See Section 13.7.]
17. The function u(x) Y0(aa)J0(ax) J0(a a)Y0(a x), a
0 is
a solution of the parametric Bessel equation
0°
x2
u=0
d 2u
du
1x
1 a2x2u 5 0
2
dx
dx
on the interval [a, b]. If the eigenvalues ln a2n are defined
by the positive roots of the equation
Y0(aa)J0(ab) J0(aa)Y0(ab) 0,
1
0°
show that the functions
FIGURE 14.R.4 Circular plate in Problem 11
12. Suppose xk is a positive zero of J0 . Show that a solution of the
boundary-value problem
um(x) Y0(ama)J0(amx) J0(ama)Y0(amx)
un(x) Y0(ana)J0(anx) J0(ana)Y0(anx)
are orthogonal with respect to the weight function p(x) x
on the interval [a, b]; that is,
b
0 2u
1 0u
0 2u
a2 a 2 1
b 5 2 , 0 , r , 1, t . 0
r 0r
0r
0t
u(1, t) 0,
t
0
u(r, 0) u0 J0(xkr),
0u
2
0, 0 , r , 1
0t t 0
is u(r, t) u0 J0(xkr) cos axkt.
# x u (x)u (x) dx 0,
a
0 2u
1 0u
0 2u
1
1
5 0, 0 , r , 1, 0 , z , 1
r 0r
0r 2
0z 2
n
m
n.
[Hint: Follow the procedure on pages 693 and 694.]
18. Use the results of Problem 17 to solve the following boundary-
value problem for the temperature u(r, t) in an annular plate:
0 2u
1 0u
0u
1
5 , a , r , b, t . 0
r 0r
0t
0r 2
u(a, t) 0, u(b, t) 0, t 0
13. Find the steady-state temperature u(r, z) in the cylinder in
Figure 14.2.5 if the lateral side is kept at temperature zero,
the top z 4 is kept at temperature 50, and the base z 0
is insulated.
14. Solve the boundary-value problem
m
u(r, 0) f (r),
a r b.
19. Discuss how to solve
0 2u
1 0u
0 2u
1
1 2 5 0, 0 , r , c, 0 , z , L
2
r 0r
0r
0z
with the boundary conditions given in FIGURE 14.R.5.
u = f (r)
at z = L
0u
2
5 0, 0 , z , 1
0r r 5 1
u(r, 0) f (r), u(r, 1) g(r),
0 r 1.
u = h(z)
at r = c
∇2u = 0
15. Find the steady-state temperature u(r, u) in a sphere of unit
radius if the surface is kept at
u(1, u) e
100,
100,
0 , u , p>2
p>2 , u , p.
[Hint: See Problem 22 in Exercises 12.6.]
u = g(r)
at z = 0
FIGURE 14.R.5 Cylinder in Problem 19
20. Carry out your ideas and find u (r, z) in Problem 19.
[Hint: Review (11) of Section 13.5.]
764
|
CHAPTER 14 Boundary-Value Problems in Other Coordinate Systems
In Problems 21–24, solve the given boundary-value problem.
21.
0 2u
1 0u
0 2u
0, 0 , r , 1, 0 , z , 1
r 0r
0r 2
0z 2
u(1, z) 100, 0 , z , 1
0u
2
0, 0 , r , 1
0z z 0
u(r, 1) 200, 0 , r , 1.
0 2u
1 0u
0 2u
22.
0, 0 , r , 3, 0 , z , 1
r 0r
0r 2
0z 2
23.
0 2u
1 0u
0 2u
2 0, 0 , r , 1, z . 0
2
r 0r
0r
0z
u(1, z) 0, z . 0
u(r, 0) 100, 0 , r , 1
24.
0 2u
1 0u
0 2u
0, 0 , r , 1, z . 0
r 0r
0r 2
0z 2
u(1, z) 0, z . 0
u(r, 0) u0(1 2 r 2), 0 , r , 1
u(3, z) u0, 0 , z , 1
u(r, 0) 0, 0 , r , 3
u(r, 1) 0, 0 , r , 3.
CHAPTER 14 in Review
|
765
© enigmatico/ShutterStock, Inc.
CHAPTER
15
The method of separation of
variables that we employed in
Chapters 13 and 14 is a powerful,
but not universally applicable
method for solving boundaryvalue problems. If the partial
differential equation in question
is nonhomogeneous, or if the
boundary conditions are time
dependent, or if the domain of
the spatial variable is infinite:
(q, q), or semi-infinite:
(a, q), we may be able to use an
integral transform to solve the
problem. In Section 15.2 we will
solve problems that involve the
heat and wave equations by
means of the familiar Laplace
transform. In Section 15.4 three
new integral transforms, Fourier
transforms, will be introduced
and used.
Integral Transform
Method
CHAPTER CONTENTS
15.1
15.2
15.3
15.4
15.5
Error Function
Applications of the Laplace Transform
Fourier Integral
Fourier Transforms
Fast Fourier Transform
Chapter 15 in Review
15.1
Error Function
INTRODUCTION There are many functions in mathematics that are defined by means of an
integral. For example, in many traditional calculus texts the natural logarithm is defined in the
following manner: ln x ex1 dt>t, x 0. In earlier chapters we have already seen, albeit briefly,
the error function erf (x), the complementary error function erfc(x), the sine integral function Si(x),
the Fresnel sine integral S(x), and the gamma function (a); all of these functions are defined
in terms of an integral. Before applying the Laplace transform to boundary-value problems, we
need to know a little more about the error function and the complementary error function. In this
section we examine the graphs and a few of the more obvious properties of erf (x) and erfc(x).
Properties and Graphs Recall from (14) of Section 2.3 that the definitions of the error
function erf (x) and complementary error function erfc(x) are, respectively,
#
"p
2
erf (x) See Appendix II.
x
2
e u du
and
0
erfc(x) With the aid of polar coordinates, it can be demonstrated that
#
q
0
2
e u du "p
2
#
"p
2
or
q
0
#
"p
2
q
2
e u du.
(1)
x
2
e u du 1.
(2)
We have already seen in (15) of Section 2.3 that when the second integral in (2) is written as e0q e0x exq
we obtain an identity that relates the error function and the complementary error function:
erf (x) erfc(x) 1.
(3)
For x 0 it is seen in FIGURE 15.1.1 that erf(x) can be interpreted as the area of the blue region
2
under the graph of f(t) (2> !p)e t on the interval [0, x] and erfc(x) is the area of the red region
on [x, q). The graph of the function f is often referred to as a bell curve.
Because of the importance of erf(x) and erfc(x) in the solution of partial differential equations
and in the theory of probability and statistics, these functions are built into computer algebra systems.
So with the aid of Mathematica we get the graphs of erf(x) (in blue) and erfc(x) (in red) given in
FIGURE 15.1.2. The y-intercepts of the two graphs give the values
• erf(0) 0, erfc(0) 1.
y
erfc(x)
2
y
1
1
f(t) =
2 –t 2
e
p
x
–4
−2
x
−1
t
1
erf(x)
2
–2
2
4
–1
FIGURE 15.1.2 Graphs of erf(x) and erfc(x)
FIGURE 15.1.1 Bell curve
Other numerical values of erf(x) and erfc(x) can be obtained directly from a CAS. Further inspection of the two graphs shows that:
• the domains of erf(x) and erfc(x) are (q, q),
• erf(x) and erfc(x) are continuous functions,
• lim erf(x) 1, lim erf(x) S 1,
xSq
• lim erfc(x) S 0,
xSq
xSq
lim erfc(x) S 2.
xSq
It should also be apparent that the graph of the error function is symmetric with respect to the
origin and so erf(x) is an odd function:
erf (x) erf(x).
You are asked to prove (4) in Problem 14 of Exercises 15.1.
768
|
CHAPTER 15 Integral Transform Method
(4)
Table 15.1.1 contains Laplace transforms of some functions involving the error and complementary error functions. These results will be useful in the exercises in the next section.
TABLE 15.1.1
f (t), a 0
1.
2.
1
+{ f (t)} F(s)
e a"s
2
"pt
e a >4t
a
3. erfc a
e a >4t
a
2"t
4. 2
"s
2
2"pt 3
f (t), a 0
e
5. e ab e b t erfc ab"t a"s
a
2"t
b
a
2
6. e ab e b t erfc ab"t s
e a"s
a
t a2>4t
e
2 a erfc a
b
Åp
2"t
2
e a"s
b
+{ f (t)} F(s)
2"t
s "s
e a"s
b erfc a
a
2"t
"s ("s b)
be a"s
b
s( "s b)
REMARKS
The proofs of the results in Table 15.1.1 will not be given because they are long and somewhat
complicated. For example, the proofs of entries 2 and 3 of the table require several changes of
variables and the use of the convolution theorem. For those who are curious, see Introduction to
the Laplace Transform, by Holl, Maple, and Vinograde, Appleton-Century-Crofts, 1959, pages
142–143. A flavor of these kinds of proofs can be gotten by working Problem 1 in Exercises 15.1.
15.1
Exercises
1. (a) Show that erf ("t ) Answers to selected odd-numbered problems begin on page ANS-35.
#
"p
1
t
e t
7. Let C, G, R, and x be constants. Use Table 15.1.1 to show that
dt.
0 "t
(b) Use part (a), the convolution theorem, and the result of
Problem 43 in Exercises 4.1 to show that
1
+5erf ("t )6 s "s 1
2. Use the result of Problem 1 to show that
+ 1 e
8. Let a be a constant. Show that
.
+ 1 e
1
1
c1 2
d.
s
"s 1
3. Use the result of Problem 1 to show that
+5erfc ("t )6 +5e t erf ("t )6 1
"s (s 2 1)
"s ("s 1)
5. Use the result of Problem 4 to show that
+e
1
2 e t erfc ("t ) f "pt
6. Find the inverse transform
+1 e
1
"s 1
f a cerf a
q
.
1
f.
s sinh "s
2n 1 a
2"t
b 2 erf a
2n 1 2 a
2"t
bd.
[Hint: Use the exponential definition of the hyperbolic sine.
Expand 1/(1 e 2"s ) in a geometric series.]
9. Use the Laplace transform and Table 15.1.1 to solve the
integral equation
.
1
sinh a"s
n0
4. Use the result of Problem 2 to show that
+5e t erfc ("t )6 C
x RC
(1 2 e x"RCs RG ) f e Gt>Cerf a
b.
Cs G
2Å t
y(t) 1 2
#
t
0
y(t)
"t 2 t
dt.
10. Use the third and fifth entries in Table 15.1.1 to derive the
sixth entry.
.
1 "s 1
[Hint: Rationalize a denominator followed by a rationalization
of a numerator.]
"p
[erf (b) erf (a)].
2
2
a
Show that ea e u du "p erf (a).
erf(x)
2
Show that lim
.
x
xS0
!p
Prove that erf(x) is an odd function.
Show that erfc(x) 1 erf(x).
b
2
11. Show that ea e u du 12.
13.
14.
15.
15.1 Error Function
|
769
15.2
Applications of the Laplace Transform
INTRODUCTION In Chapter 4 we defined the Laplace transform of a function f (t), t 0, to be
+5 f (t)6 #
q
e st f (t) dt,
0
whenever the improper integral converges. This integral transforms a function f (t) into another
function F of the transform parameter s, that is, +{ f (t)} F(s). The main application of the
Laplace transform in Chapter 4 was to the solution of certain types of initial-value problems
involving linear ordinary differential equations with constant coefficients. Recall, the Laplace
transform of such an equation reduces the ODE to an algebraic equation. In this section we are
going to apply the Laplace transform to linear partial differential equations. We will see that this
transform reduces a PDE to an ODE.
Transform of Partial Derivatives The boundary-value problems considered in this
section will involve either the one-dimensional wave and heat equations or slight variations of these
equations. These PDEs involve an unknown function of two independent variables u(x, t), where
the variable t represents time t 0. We define the Laplace transform of u(x, t) with respect to t by
+5u(x, t)6 #
q
0
e stu(x, t) dt U(x, s),
where x is treated as a parameter. Throughout this section we shall assume that all the operational
properties of Sections 4.2, 4.3, and 4.4 apply to functions of two variables. For example, by
Theorem 4.2.2, the transform of the partial derivative 0u/0t is
+e
+e
that is,
+e
Similarly,
0u
f s+5u(x, t)6 2 u(x, 0);
0t
0u
f sU(x, s) 2 u(x, 0).
0t
(1)
0 2u
f s 2U(x, s) 2 su(x, 0) 2 ut (x, 0).
0t 2
(2)
Since we are transforming with respect to t, we further suppose that it is legitimate to interchange
integration and differentiation in the transform of 02u/0x 2:
+e
0 2u
f 0x 2
#
q
0 2u
dt 0x 2
e st
0
d2
dx 2
#
q
0
02
fe stu(x, t)g dt
0x 2
e stu(x, t) dt 0
+e
that is,
#
q
d2
+5u(x, t)6;
dx 2
0 2u
d 2U
f
.
0x 2
dx 2
(3)
In view of (1) and (2) we see that the Laplace transform is suited to problems with initial
conditions—namely, those problems associated with the heat equation or the wave equation. We
will see in Section 15.4 that boundary-value problems involving Laplace’s equation in which one
(or both) of the spatial variables is defined on an unbounded interval can often be solved using
different integral transforms.
EXAMPLE 1
Laplace Transform of a PDE
Find the Laplace transform of the wave equation a 2
0 2u
0 2u
, t . 0.
0x 2
0t 2
SOLUTION From (2) and (3),
+ e a2
770
|
CHAPTER 15 Integral Transform Method
0 2u
0 2u
f +e 2 f
2
0x
0t
a2
becomes
d2
+5u(x, t)6 s 2+5u(x, t)6 2 su(x, 0) 2 ut (x, 0)
dx 2
a2
or
d 2U
2 s 2U su(x, 0) 2 ut (x, 0).
dx 2
(4)
The Laplace transform with respect to t of either the wave equation or the heat equation eliminates that variable, and for the one-dimensional equations the transformed equations are then
ordinary differential equations in the spatial variable x. In solving a transformed equation, we
treat s as a parameter.
Using the Laplace Transform to Solve a BVP
EXAMPLE 2
2
Solve
0u
0 2u
, 0 , x , 1, t . 0
0x 2
0t 2
subject to
u(0, t) 0,
u(1, t) 0, t 0
u(x, 0) 0,
0u
2 sin px, 0 x 1.
0t t 0
SOLUTION The partial differential equation is recognized as the wave equation with a 1.
From (4) and the given initial conditions, the transformed equation is
d 2U
2 s 2U sin px,
dx 2
(5)
where U(x, s) +{u(x, t)}. Since the boundary conditions are functions of t, we must also
find their Laplace transforms:
+{u(0, t)} U(0, s) 0
and
+{u(1, t)} U(1, s) 0.
(6)
The results in (6) are boundary conditions for the ordinary differential equation (5). Since (5)
is defined over a finite interval, its complementary function is
Uc(x, s) c1 cosh sx c2 sinh sx.
The method of undetermined coefficients yields a particular solution
Up(x, s) 1
sin px.
s p2
2
U(x, s) c1 cosh sx c2 sinh sx Hence
1
sin px.
s p2
2
But the conditions U(0, s) 0 and U(1, s) 0 yield, in turn, c1 0 and c2 0. We conclude that
U(x, s) 1
sin px
s 2 p2
u(x, t) +1 e
Therefore
EXAMPLE 3
1
1
p
sin px f sin px +1 e 2
f.
p
s 2 p2
s p2
u(x, t) 1
sin px sin pt.
p
Using the Laplace Transform to Solve a BVP
A very long string is initially at rest on the nonnegative x-axis. The string is secured at x 0,
and its distant right end slides down a frictionless vertical support. The string is set in motion
by letting it fall under its own weight. Find the displacement u(x, t).
SOLUTION Since the force of gravity is taken into consideration, it can be shown that the
wave equation has the form
a2
0 2u
0 2u
2 g 2 , x . 0, t . 0,
2
0x
0t
15.2 Applications of the Laplace Transform
|
771
where g is the acceleration due to gravity. The boundary and initial conditions are,
respectively,
0u
0, t . 0
u(0, t) 0, lim
xSq 0x
u(x, 0) 0,
0u
2
0, x . 0.
0t t 0
The second boundary condition limxS q 0u/0x 0 indicates that the string is horizontal at a
great distance from the left end. Now from (2) and (3),
+ e a2
becomes
a2
0 2u
0 2u
f 2 +5g6 + e 2 f
2
0x
0t
g
d 2U
2 s 2U 2 su(x, 0) 2 ut (x, 0)
s
dx 2
or, in view of the initial conditions,
g
d 2U
s2
2 2U 2 .
2
dx
a
a s
The transforms of the boundary conditions are
+{u(0, t)} U(0, s) 0
and
+ e lim
xSq
0u
dU
f lim
0.
0x
xSq dx
With the aid of undetermined coefficients, the general solution of the transformed equation
is found to be
g
U(x, s) c1e(x/a)s c2e(x/a)s 3 .
s
The boundary condition limxS q dU/dx 0 implies c2 0, and U(0, s) 0 gives c1 g/s 3.
Therefore
g
g
U(x, s) 3 e (x>a)s 2 3 .
s
s
Now by the second translation theorem we have
u
u(x, t) +1 e
vertical
support
“at ∞”
at
x
(at, – 1 gt2)
2
FIGURE 15.2.1 A long string falling under
its own weight in Example 3
or
g (x>a)s
g
1
x 2
x
1
e
2
f
g
at
2
b 8 at 2 b 2 gt 2
3
3
a
a
2
2
s
s
1
gt 2,
2
u(x, t) μ
g
2 (2axt 2 x 2),
2a
0#t,
x
a
x
t$ .
a
To interpret the solution, let us suppose t 0 is fixed. For 0 x at, the string is the shape
of a parabola passing through the points (0, 0) and (at, 12 gt 2). For x at, the string is
described by the horizontal line u 12 gt 2. See FIGURE 15.2.1.
Observe that the problem in the next example could be solved by the procedure in Section 13.6.
The Laplace transform provides an alternative solution.
EXAMPLE 4
A Solution in Terms of erf (x)
Solve the heat equation
0 2u
0u
,
2
0t
0x
subject to
0 , x , 1, t . 0
u(0, t) 0, u(1, t) u0, t 0
u(x, 0) 0, 0 x 1.
772
|
CHAPTER 15 Integral Transform Method
SOLUTION From (1) and (3) and the given initial condition,
+e
0 2u
0u
f +e f
2
0t
0x
d 2U
2 sU 0.
dx 2
becomes
(7)
The transforms of the boundary conditions are
U(0, s) 0
and
U(1, s) u0
.
s
(8)
Since we are concerned with a finite interval on the x-axis, we choose to write the general
solution of (7) as
U(x, s) c1 cosh (!sx) c2 sinh (!sx).
Applying the two boundary conditions in (8) yields, respectively, c1 0 and c2 u0 (s sinh !s).
Thus
U(x, s) u0
sinh (!sx)
.
s sinh !s
Now the inverse transform of the latter function cannot be found in most tables. However,
by writing
sinh (!sx)
e !sx 2 e!sx
e (x 2 1)!s 2 e(x 1)!s
s sinh !s
s(e !s 2 e!s)
s(1 2 e2!s)
and using the geometric series
1
1 2 e2!s
a e2n!s
q
n0
we find
q
sinh (!sx)
e(2n 1 x)!s
e(2n 1 2 x)!s
2
d.
ac
s
s
s sinh !s
n0
If we assume that the inverse Laplace transform can be done term by term, it follows from
entry 3 of Table 15.1.1 that
Also see Problem 8 in
Exercises 15.1
u(x, t) u0 +1 e
sinh (!sx)
f
s sinh !s
u0 a c+1 e
q
n0
u0 a cerfc a
q
n0
e(2n 1 2 x)"s
e(2n 1 x)"s
f 2 +1 e
fd
s
s
2n 1 2 x
2n 1 x
b 2 erfc a
bd.
2!t
2!t
(9)
The solution (9) can be rewritten in terms of the error function using erfc(x) 1 erf (x):
u(x, t) u0 a cerf a
q
n0
2n 1 x
2n 1 2 x
b 2 erf a
bd.
2!t
2!t
(10)
FIGURE 15.2.2(a), obtained with the aid of the 3D plot function in a CAS, shows the surface over
the rectangular region 0 x 1, 0 t 6 defined by the partial sum S10(x, t) of the solution
(10). It is apparent from the surface and the accompanying two-dimensional graphs that at a fixed
value of x (the curve of intersection of a plane slicing the surface perpendicular to the x-axis) on
the interval [0, 1], the temperature u(x, t) increases rapidly to a constant value as time increases.
See Figure 15.2.2(b) and 15.2.2(c). For a fixed time (the curve of intersection of a plane slicing
15.2 Applications of the Laplace Transform
|
773
the surface perpendicular to the t-axis), the temperature u(x, t) naturally increases from 0 to 100.
See Figure 15.2.2(d) and 15.2.2(e).
u(x, t)
100
75
50
25
0
0
6
4
0.2
t
0.4
x
2
0.6
0.8
1 0
(a) u0 = 100
u(0.2, t)
100
u(0.7, t)
100
80
80
60
60
40
40
20
20
0
1
2
3
4
(b) x = 0.2
5
6
t
0
u(x, 0.1)
120
100
80
60
40
20
0
1
2
3
4
(c) x = 0.7
5
6
0.8
1
t
u(x, 4)
120
100
80
60
40
20
0.2
0.4 0.6
(d) t = 0.1
0.8
1
x
0
0.2
0.4 0.6
(e) t = 4
x
FIGURE 15.2.2 Graph of solution given in (10). In (b) and (c), x is held constant.
In (d) and (e), t is held constant.
Exercises
15.2
Answers to selected odd-numbered problems begin on page ANS-35.
In the following problems use tables as necessary.
1. A string is secured to the x-axis at (0, 0) and (L, 0). Find the
displacement u(x, t) if the string starts from rest in the initial
position A sin(px/L).
2. Solve the boundary-value problem
0 2u
0 2u
2,
2
0x
0t
0 , x , 1, t . 0
u(0, t) 0,
u(1, t) 0
u(x, 0) 0,
0u
2
2 sin px 4 sin 3px.
0t t 0
3. The displacement of a semi-infinite elastic string is determined
from
a2
0 2u
0 2u
,
0x 2
0t 2
u(0, t) f (t),
u(x, 0) 0,
x . 0, t . 0
lim u(x, t) 0, t . 0
xSq
0u
2
0, x . 0.
0t t 0
Solve for u(x, t).
4. Solve the boundary-value problem in Problem 3 when
f (t) e
sin pt,
0,
0#t#1
0 # t . 1.
Sketch the displacement u(x, t) for t 1.
774
|
CHAPTER 15 Integral Transform Method
5. In Example 3, find the displacement u(x, t) when the left end
10. Solve the boundary-value problem
of the string at x 0 is given an oscillatory motion described
by f (t) A sin vt.
6. The displacement u(x, t) of a string that is driven by an external
force is determined from
2
0 2u
0 2u
2,
2
0x
0t
u(0, t) 1,
2
0u
0u
sin px sin vt 2 ,
0x 2
0t
u(0, t) 0,
0 , x , 1, t . 0
u(x, 0) ex,
u(1, t) 0, t 0
0u
2
0, 0 , x , 1.
0t t 0
u(x, 0) 0,
7. A uniform bar is clamped at x 0 and is initially at rest. If a
constant force F0 is applied to the free end at x L, the
longitudinal displacement u(x, t) of a cross section of the bar
is determined from
2
2
0u
0u
a
2,
2
0x
0t
2
u(0, t) 0,
0u
2
0, 0 , x , L.
0t t 0
u(x, 0) 0,
Solve for u(x, t). [Hint: Expand 1/(1 e2sL/a ) in a geometric
series.]
8. A uniform semi-infinite elastic beam moving along the x-axis
with a constant velocity v0 is brought to a stop by hitting a
wall at time t 0. See FIGURE 15.2.3. The longitudinal
displacement u(x, t) is determined from
a2
12. u(0, t) u0,
0 2u
0 2u
2,
2
0x
0t
u(0, t) 0,
x . 0, t . 0
lim
xSq
0u
0, t . 0
0x
0u
2
v0, x . 0.
0t t 0
u(x, 0) 0,
Solve for u(x, t).
0u
0, x . 0.
2
0t t 0
lim u(x, t) u1,
u(x, 0) u1
xSq
lim
xSq
u(x, t)
u1,
x
13.
0u
2
u(0, t),
0x x 0
14.
0u
2
u(0, t) 2 50,
0x x 0
0 , x , L, t . 0
0u
2
F0, E a constant, t . 0
E
0x x L
lim u(x, t) 0, t . 0
xSq
In Problems 11–18, use the Laplace transform to solve the heat
equation uxx ut, x 0, t 0 subject to the given conditions.
11. u(0, t) u0,
Solve for u(x, t).
x . 0, t . 0
15. u(0, t) f (t),
u(x, 0) u1x
lim u(x, t) u0,
xSq
u(x, 0) u0
lim u(x, t) 0,
xSq
lim u(x, t) 0,
u(x, 0) 0
u(x, 0) 0
xSq
[Hint: Use the convolution theorem.]
16.
0u
2
f (t),
0x x 0
lim u(x, t) 0, u(x, 0) 0
xSq
17. u(0, t) 60 40 8(t 2 2),
lim u(x, t) 60,
xSq
u(x, 0) 60
18. u(0, t) e
20,
0,
0,t,1
,
t$1
lim u(x, t) 100,
xSq
u(x, 0) 100
19. Solve the boundary-value problem
0u
0 2u
, q , x , 1, t . 0
2
0t
0x
0u
2
100 2 u(1, t),
0x x 1
u(x, 0) 0,
lim u(x, t) 0, t . 0
xSq
q x 1.
20. Show that a solution of the boundary-value problem
wall
beam
x
x=0
FIGURE 15.2.3 Moving elastic beam in Problem 8
9. Solve the boundary-value problem
0 2u
0 2u
2,
2
0x
0t
u(0, t) 0,
0 2u
0u
r , x . 0, t . 0
0t
0x 2
0u
u(0, t) 0, lim
0, t . 0
xSq 0x
k
v0
x . 0, t . 0
lim u(x, t) 0, t . 0
xSq
u(x, 0) xex,
0u
2
0, x . 0.
0t t 0
u(x, 0) 0, x 0,
where r is a constant, is given by
#
t
u(x, t) rt 2 r erfc a
x
b dt.
2"kt
21. A rod of length L is held at a constant temperature u0 at its
ends x 0 and x L. If the rod’s initial temperature is
u0 u0 sin(xp/L), solve the heat equation uxx ut, 0 x L,
t 0 for the temperature u(x, t).
0
15.2 Applications of the Laplace Transform
|
775
22. If there is a heat transfer from the lateral surface of a thin wire
of length L into a medium at constant temperature um, then the
heat equation takes on the form
temperatures outside the sphere are described by the boundaryvalue problem
2 0u
0u
0 2u
,
r 0r
0t
0r 2
2
0u
0u
k 2 2 h(u 2 um) , 0 , x , L, t . 0,
0t
0x
where h is a constant. Find the temperature u(x, t) if the initial
temperature is a constant u0 throughout and the ends x 0
and x L are insulated.
23. A rod of unit length is insulated at x 0 and is kept at
temperature zero at x 1. If the initial temperature of the rod
is a constant u0, solve kuxx ut, 0 x 1, t 0 for the
temperature u(x, t). [Hint: Expand 1/(1 e2"s>k ) in a
geometric series.]
24. An infinite porous slab of unit width is immersed in a solution
of constant concentration c0. A dissolved substance in the
solution diffuses into the slab. The concentration c(x, t) in the
slab is determined from
u(1, t) 100, t 0
lim u(r, t) 0
rSq
u(r, 0) 0, r 1.
Use the Laplace transform to find u(r, t). [Hint: After
transforming the PDE, let v(r, s) r U(r, s), where
+{u(r, t)} U(r, s).]
28. Show that a solution of the boundary-value problem
0 2u
0u
2 hu , x . 0, t . 0, h constant
2
0t
0x
u(0, t) u0,
2
D
0c
0c
, 0 , x , 1, t . 0
2
0t
0x
c (0, t) c0 ,
c (1, t) c0 , t . 0
c(x, 0) 0, 0 x 1,
where D is a constant. Solve for c(x, t).
25. A very long telephone transmission line is initially at a constant
potential u0. If the line is grounded at x 0 and insulated at
the distant right end, then the potential u(x, t) at a point x along
the line at time t is determined from
0 2u
0u
2 RC
2 RGu 0, x . 0, t . 0
2
0t
0x
0u
u(0, t) 0, lim
0, t 0
xSq 0x
u(x, 0) u0 , x 0,
where R, C, and G are constants known as resistance, capacitance, and conductance, respectively. Solve for u(x, t).
[Hint: See Problem 7 in Exercises 15.1.]
26. Starting at t 0, a concentrated load of magnitude F0 moves
with a constant velocity v0 along a semi-infinite string. In this
case the wave equation becomes
0 2u
0 2u
x
a 2 2 2 F0 d at 2 b ,
v
0x
0t
0
where d(t x/v0) is the Dirac delta function. Solve this PDE
subject to
u(0, t) 0,
u(x, 0) 0,
lim u(x, t) 0, t . 0
xSq
0u
2
0, x . 0
0t t 0
(a) when v0 a, and
(b) when v0 a.
27. In Problem 9 of Exercises 14.3 you were asked to find the
time-dependent temperatures u(r, t) within a unit sphere. The
776
|
CHAPTER 15 Integral Transform Method
r 1, t 0
u(x, 0) 0,
lim u(x, t) 0, t . 0
xSq
x0
t
2
e ht 2 x >4t
is
u(x, t) dt.
t3>2
2"p 0
29. The temperature in a semi-infinite solid is modeled by the
boundary-value problem
u0 x
#
0 2u
0u
, x . 0, t . 0
0t
0x 2
u(0, t) u0, lim u(x, t) 0, t . 0
k
xSq
u(x, 0) 0, x . 0
Solve for u(x, t). Use the solution to determine analytically
the value of lim tSq u(x, t), x . 0.
30. In Problem 29, if there is a constant flux of heat into the solid
at its left-hand boundary, then the boundary condition is
0u
u0, u0 . 0, t . 0. Solve for u(x, t). Use the soluP
0x x 0
tion to determine analytically the value of lim tSq u(x, t), x . 0.
Computer Lab Assignments
31. Use a CAS to obtain the graph of u(x, t) in Problem 29 over
the rectangular region defined by 0 # x # 10, 0 # t # 15.
Assume u0 100 and k 1. Indicate the two boundary conditions and initial condition on your graph. Use 2D and 3D
plots of u(x, t) to verify the value of lim tSq u(x, t).
32. Use a CAS to obtain the graph of u(x, t) in Problem 30 over
the rectangular region defined by 0 # x # 10, 0 # t # 15.
Assume u0 100 and k 1. Use 2D and 3D plots of u(x, t)
to verify the value of lim tSq u(x, t).
33. Humans gather most of their information on the outside world
through sight and sound. But many creatures use chemical
signals as their primary means of communication; for
example, honeybees, when alarmed, emit a substance and fan
their wings feverishly to relay the warning signal to the bees
that attend to the queen. These molecular messages between
members of the same species are called pheromones. The
(a) Solve the boundary-value problem if it is further known
that c(x, 0) ⫽ 0, x ⬎ 0, and limxS q c(x, t) ⫽ 0, t ⬎ 0.
(b) Use a CAS to plot the graph of the solution in part (a) for
x ⬎ 0 at the fixed times t ⫽ 0.1, t ⫽ 0.5, t ⫽ 1, t ⫽ 2,
t ⫽ 5.
(c) For a fixed time t, show that e0qc(x, t) dx ⫽ Ak. Thus Ak
represents the total amount of chemical discharged.
signals may be carried by moving air or water or by a
diffusion process in which the random movement of gas
molecules transports the chemical away from its source.
FIGURE 15.2.4 shows an ant emitting an alarm chemical into
the still air of a tunnel. If c(x, t) denotes the concentration
of the chemical x centimeters from the source at time t, then
c(x, t) satisfies
k
0 2c
0c
⫽ , x . 0, t . 0,
2
0t
0x
x
0
and k is a positive constant. The emission of pheromones as
a discrete pulse gives rise to a boundary condition of the
form
FIGURE 15.2.4 Ants in Problem 33
0u
2
⫽ ⫺Ad(t),
0x x ⫽ 0
where d(t) is the Dirac delta function.
15.3
Fourier Integral
INTRODUCTION In preceding chapters, Fourier series were used to represent a function f
defined on a finite interval (⫺p, p) or (0, L). When f and f ⬘ are piecewise continuous on such an
interval, a Fourier series represents the function on the interval and converges to the periodic
extension of f outside the interval. In this way we are justified in saying that Fourier series are
associated only with periodic functions. We shall now derive, in a nonrigorous fashion, a means
of representing certain kinds of nonperiodic functions that are defined on either an infinite interval
(⫺q, q) or a semi-infinite interval (0, q).
From Fourier Series to Fourier Integral Suppose a function f is defined on (⫺p, p).
If we use the integral definitions of the coefficients (9), (10), and (11) of Section 12.2 in (8) of that
section, then the Fourier series of f on the interval is
1
2p
f (x) ⫽
#
p
⫺p
f (t) dt ⫹
1 q
ca
p na
⫽1
#
p
f (t) cos
⫺p
np
np
t dtb cos
x⫹ a
p
p
#
p
np
np
t dtb sin
xd .
p
p
(1)
f (t) sin ant dtb sin an xd Da.
(2)
f (t) sin
⫺p
If we let an ⫽ np/p, ⌬a ⫽ an⫹1 ⫺ an ⫽ p/p, then (1) becomes
f (x) ⫽
1
a
2p
#
p
⫺p
f (t) dtb Da ⫹
1 q
ca
p na
⫽1
#
p
⫺p
f (t) cos ant dtb cos an x ⫹ a
#
p
⫺p
We now expand the interval (⫺p, p) by letting p S q. Since p S q implies that ⌬a S 0, the
q
limit of (2) has the form lim⌬aS 0 g n ⫽ 1F(an)⌬a, which is suggestive of the definition of the
q
integral e0q F(a) da. Thus if e⫺q
f (t) dt exists, the limit of the first term in (2) is zero and the limit
of the sum becomes
f (x) ⫽
1
p
q
# #
0
ca
q
⫺q
f (t) cos at dtb cos ax ⫹ a
#
q
f (t) sin at dtb sin axd da.
(3)
⫺q
The result given in (3) is called the Fourier integral of f on the interval (⫺q, q). As the following summary shows, the basic structure of the Fourier integral is reminiscent of that of a
Fourier series.
15.3 Fourier Integral
|
777
Definition 15.3.1
Fourier Integral
The Fourier integral of a function f defined on the interval (q, q) is given by
f (x) #
1
p
q
fA(a) cos ax B(a) sin axg da,
0
A(a) where
B(a) #
q
#
q
(4)
f (x) cos ax dx
(5)
f (x) sin ax dx.
(6)
q
q
Convergence of a Fourier Integral Sufficient conditions under which a Fourier integral
converges to f (x) are similar to, but slightly more restrictive than, the conditions for a Fourier series.
Theorem 15.3.1
Conditions for Convergence
Let f and f be piecewise continuous on every finite interval, and let f be absolutely integrable
on (q, q).* Then the Fourier integral of f on the interval converges to f (x) at a point of
continuity. At a point of discontinuity, the Fourier integral will converge to the average
f (x1) f (x)
,
2
where f (x) and f (x) denote the limit of f at x from the right and from the left, respectively.
Fourier Integral Representation
EXAMPLE 1
Find the Fourier integral representation of the piecewise-continuous function
0,
f (x) • 1,
0,
x,0
0,x,2
x . 2.
SOLUTION The function, whose graph is shown in FIGURE 15.3.1, satisfies the hypotheses of
Theorem 15.3.1. Hence from (5) and (6) we have at once
y
1
A(a) 2
x
FIGURE 15.3.1 Function f in Example 1
#
q
#
0
f (x) cos ax dx
q
q
f (x) cos ax dx 2
B(a) # cos ax dx 0
#
0
f (x) cos ax dx #
q
f (x) cos ax dx
2
sin 2a
a
2
q
q
#
2
f (x) sin ax dx # sin ax dx 0
1 2 cos 2a
.
a
Substituting these coefficients into (4) then gives
f (x) 1
p
#
q
ca
0
sin 2a
1 2 cos 2a
b cos ax a
b sin axd da.
a
a
When we use trigonometric identities, the last integral simplifies to
f (x) *This means that the integral
#
|
CHAPTER 15 Integral Transform Method
#
q
0
sin a cos a(x 2 1)
da.
a
q
q
778
2
p
Z f (x)Z dx converges.
(7)
The Fourier integral can be used to evaluate integrals. For example, at x 1 it follows from
Theorem 15.3.1 that (7) converges to f (1); that is,
#
q
0
sin a
p
da .
a
2
The latter result is worthy of special note since it cannot be obtained in the “usual” manner; the
integrand (sin x)/x does not possess an antiderivative that is an elementary function.
Cosine and Sine Integrals When f is an even function on the interval (q, q), then
the product f (x) cos ax is also an even function, whereas f (x) sin ax is an odd function. As a
consequence of property (g) of Theorem 12.3.1, B(a) 0, and so (4) becomes
f (x) q
# #
2
p
a
0
q
f (t) cos at dtb cos ax da.
0
Here we have also used property (f ) of Theorem 12.3.1 to write
#
q
q
f (t) cos at dt 2
#
q
f (t) cos at dt.
0
Similarly, when f is an odd function on (q, q) the products f (x) cos ax and f (x) sin ax are odd
and even functions, respectively. Therefore A(a) 0 and
f (x) 2
p
q
# #
0
a
q
f (t) sin at dtb sin ax da.
0
We summarize in the following definition.
Definition 15.3.2
Fourier Cosine and Sine Integrals
(i) The Fourier integral of an even function on the interval (q, q) is the cosine integral
f (x) A(a) where
q
#
2
p
A(a) cos ax da,
(8)
0
q
#
f (x) cos ax dx.
(9)
0
(ii) The Fourier integral of an odd function on the interval (q, q) is the sine integral
2
f (x) p
B(a) where
#
#
q
B(a) sin ax da,
(10)
0
q
f (x) sin ax dx.
(11)
0
Cosine Integral Representation
EXAMPLE 2
Find the Fourier integral representation of the function
f (x) e
ZxZ , a
ZxZ . a.
SOLUTION It is apparent from FIGURE 15.3.2 that f is an even function. Hence we represent f
by the Fourier cosine integral (8). From (9) we obtain
y
1
A(a) –a
1,
0,
a
x
#
q
0
f (x) cos ax dx #
a
f (x) cos ax dx 0
a
FIGURE 15.3.2 Function f in Example 2
and so
f (x) # cos ax dx 0
2
p
#
q
0
#
q
f (x) cos ax dx
a
sin aa
,
a
sin aa cos ax
da.
a
15.3 Fourier Integral
(12)
|
779
The integrals (8) and (10) can be used when f is neither odd nor even and defined only on the
half-line (0, q). In this case (8) represents f on the interval (0, q) and its even (but not periodic)
extension to (⫺q, 0), whereas (10) represents f on (0, q) and its odd extension to the interval
(⫺q, 0). The next example illustrates this concept.
y
1
Cosine and Sine Integral Representations
EXAMPLE 3
Represent f (x) ⫽ e⫺x, x ⬎ 0 (a) by a cosine integral; (b) by a sine integral.
x
SOLUTION The graph of the function is given in FIGURE 15.3.3.
FIGURE 15.3.3 Function f in Example 3
(a) Using integration by parts, we find
y
A(a) ⫽
x
#
q
0
e⫺x cos ax dx ⫽
Therefore from (8) the cosine integral of f is
f (x) ⫽
(a) Cosine integral
1
.
1 ⫹ a2
2
p
#
q
0
cos ax
da.
1 ⫹ a2
(13)
(b) Similarly, we have
y
B(a) ⫽
x
0
e⫺x sin ax dx ⫽
a
.
1 ⫹ a2
From (10) the sine integral of f is then
f (x) ⫽
(b) Sine integral
FIGURE 15.3.4 In Example 3, (a) is the
even extension of f ; (b) is the odd
extension of f
#
q
2
p
#
q
0
a sin ax
da.
1 ⫹ a2
(14)
FIGURE 15.3.4 shows the graphs of the functions and their extensions represented by the two
integrals.
Complex Form The Fourier integral (4) also possesses an equivalent complex form, or
exponential form, that is analogous to the complex form of a Fourier series (see Section 12.4).
If (5) and (6) are substituted into (4), then
1
p
##
q
f (x) ⫽
q
1
p
q
q
⫽
1
2p
# #
q
⫽
1
2p
q
# #
q
⫽
1
2p
q
# #
q
⫽
1
2p
q
⫽
⫺q
0
##
f (t)f cos at cos ax ⫹ sin at sin axg dt da
f (t) cos a(t 2 x) dt da
⫺q
0
q
f (t) cos a(t 2 x) dt da
(15)
f (t)f cos a(t 2 x) ⫹ i sin a(t 2 x)g dt da
(16)
⫺q ⫺q
⫺q ⫺q
f (t) e ia(t2 x) dt da
⫺q ⫺q
# #
⫺q
q
a
⫺q
q
q
f (t) e iat dtb e⫺iax da.
(17)
We note that (15) follows from the fact that the integrand is an even function of a. In (16) we
have simply added zero to the integrand,
i
# #
⫺q ⫺q
780
|
CHAPTER 15 Integral Transform Method
f (t) sin a(t 2 x) dt da ⫽ 0,
because the integrand is an odd function of a. The integral in (17) can be expressed as
f (x) where
0.5
x
0
–0.5
–1
–3
–2
0
–1
1
2
3
Fb(x) y
Gb(x) x
–0.5
–1
0
–1
1
2
3
(b) G20(x)
FIGURE 15.3.5 Graphs of partial integrals
15.3
Exercises
(19)
2
p
#
b
0
cos ax
da,
1 a2
1
1
0
0
,
,
,
,
x
x
x
x
,
,
,
.
1
0
1
1
#
b
0
a sin ax
da.
1 a2
0,
5,
7. f (x) μ
5,
0,
1
1
0
0
,
,
,
,
x
x
x
x
ZxZ , 1
1 , ZxZ , 2
0,
ZxZ . 2
ZxZ, ZxZ , p
0, ZxZ . p
11. f (x) e ZxZ sin x
6. f (x) e
1
0
1
1
0,
9. f (x) e
0,
0#x,0
4. f (x) • sin x, 0 # x # p
0,
0#x.p
,
,
,
.
8. f (x) • p,
0, 0 , x , 0
3. f (x) • x, 0 , x , 3
0, 0 , x . 3
0,
x,0
x
e , x.0
2
p
Answers to selected odd-numbered problems begin on page ANS-35.
0, p , x , p
2. f (x) • 4, p , x , 2p
0, p , x . 2p
5. f (x) e
f (x) e iax dx.
Because the Fourier integrals (13) and (14) converge, the graphs of the partial integrals Fb(x)
and Gb(x) for a specified value of b 0 will be an approximation to the graph of f and its
even and odd extensions shown in Figure 15.3.4(a) and 15.3.4(b), respectively. The graphs of
Fb(x) and Gb(x) for b 20 given in FIGURE 15.3.5 were obtained using Mathematica and its
NIntegrate application. See Problem 21 in Exercises 15.3.
In Problems 1–6, find the Fourier integral representation of the
given function.
0,
1,
1. f (x) μ
2,
0,
q
and x is treated as a parameter. Similarly, the Fourier sine integral representation of f (x) ex,
x 0 in (14) can be written as f (x) limb S qGb(x), where
0
–2
(18)
q
q
0.5
–3
C(a) eiax da,
Use of Computers The convergence of a Fourier integral can be examined in a manner
that is similar to graphing partial sums of a Fourier series. To illustrate, let’s use the results in
parts (a) and (b) of Example 3. By definition of an improper integral, the Fourier cosine integral
representation of f (x) ex, x 0 in (13) can be written as f (x) limb S q Fb(x), where
(a) F20(x)
1.5
1
#
q
This latter form of the Fourier integral will be put to use in the next section when we return
to the solution of boundary-value problems.
y
1.5
1
C(a) #
1
2p
10. f (x) e
x, ZxZ , p
0, ZxZ . p
12. f (x) xe ZxZ
In Problems 13–16, find the cosine and sine integral representations of the given function.
13. f (x) ekx, k 0, x 0 14. f (x) ex e3x, x 0
15. f (x) xe2x, x 0
e x, ZxZ , 1
0, ZxZ . 1
In Problems 7–12, represent the given function by an
appropriate cosine or sine integral.
16. f (x) ex cos x, x 0
In Problems 17 and 18, solve the given integral equation for the
function f.
17.
#
q
0
f (x) cos ax dx ea
15.3 Fourier Integral
|
781
18.
#
q
0
f (x) sin ax dx e
1,
0,
19. (a) Use (7) to show that
(a) Use a trigonometric identity to show that an alternative
form of the Fourier integral representation (12) of the
function f in Example 2 (with a 1) is
0,a,1
0,a.1
#
q
0
p
sin 2x
dx .
x
2
[Hint: a is a dummy variable of integration.]
(b) Show in general that, for k 0,
#
q
0
1
Fb(x) p
20. Use the complex form (19) to find the Fourier integral
21. While the integral (12) can be graphed in the same manner
discussed on page 781 to obtain Figure 15.3.5, it can also be
expressed in terms of a special function that is built into
a CAS.
0
sin a(x 1) 2 sin a(x 2 1)
da.
a
#
b
0
bSq
sin a(x 1) 2 sin a(x 2 1)
da.
a
Show that the last integral can be written as
representation of f (x) e ZxZ . Show that the result is the same
as that obtained from (8) and (9).
Computer Lab Assignment
#
q
(b) As a consequence of part (a), f (x) lim Fb(x), where
p
sin kx
dx .
x
2
15.4
1
p
f (x) Fb(x) 1
fSi(b (x 1)) 2 Si(b (x 2 1))g,
p
where Si(x) is the sine integral function. See Problem 43
in Exercises 2.3.
(c) Use a CAS and the sine integral form obtained in part (b)
to graph Fb(x) on the interval [3, 3] for b 4, 6, and 15.
Then graph Fb(x) for larger values of b 0.
Fourier Transforms
INTRODUCTION Up to now we have studied and used only one integral transform: the
Laplace transform. But in Section 15.3 we saw that the Fourier integral had three alternative
forms: the cosine integral, the sine integral, and the complex or exponential form. In the present
section we shall take these three forms of the Fourier integral and develop them into three new
integral transforms naturally called Fourier transforms. In addition, we shall expand on the
concept of a transform pair, that is, an integral transform and its inverse. We shall also see that
the inverse of an integral transform is itself another integral transform.
Transform Pairs The Laplace transform F(s) of a function f (t) is defined by an integral,
but up to now we have been using the symbolic representation f (t) + 1{F(s)} to denote the
inverse Laplace transform of F(s). Actually, the inverse Laplace transform is also an integral
transform. If
q
#
+5 f (t)6 0
estf (t) dt F(s),
(1)
then the inverse Laplace transform is
+1 5F(s)6 #
1
2pi
g iq
g 2 iq
e stF(s) ds f (t).
(2)
The last integral is called a contour integral; its evaluation requires the use of complex variables
and is beyond the scope of this discussion. The point here is this: Integral transforms appear in
transform pairs. If f (x) is transformed into F(a) by an integral transform
b
F(a) # f (x) K(a, x) dx,
(3)
a
then the function f can be recovered by another integral transform
b
f (x) # F(a) H(a, x) da,
(4)
c
called the inverse transform. The functions K and H in the integrands of (3) and (4) are called
the kernels of their respective transforms. We identify K(s, t) est as the kernel of the Laplace
transform and H(s, t) est /2pi as the kernel of the inverse Laplace transform.
782
|
CHAPTER 15 Integral Transform Method
Fourier Transform Pairs The Fourier integral is the source of three new integral transforms. From (8) and (9), (10) and (11), and (18) and (19) of the preceding section, we are prompted
to define the following Fourier transform pairs.
Definition 15.4.1
Fourier Transform Pairs
#
^5 f (x)6 (i) Fourier transform:
q
q
^1 5F(a)6 Inverse Fourier transform:
#
^s 5 f (x)6 (ii) Fourier sine transform:
Inverse Fourier
sine transform:
^c 5 f (x)6 (iii) Fourier cosine transform:
Inverse Fourier
cosine transform:
^1
c 5F(a)6
1
2p
#
#
q
q
(5)
F(a) e iax da f (x)
(6)
q
f (x) sin ax dx F(a)
0
^1
s 5F(a)6 f (x) e iax dx F(a)
2
p
#
(7)
q
0
F(a) sin ax da f (x)
(8)
q
f (x) cos ax dx F(a)
0
2
p
#
(9)
q
0
F(a) cos ax da f (x)
(10)
Existence The conditions under which (5), (7), and (9) exist are more stringent than those
for the Laplace transform. For example, you should verify that ^{1}, ^ s{1}, and ^ c{1} do not
exist. Sufficient conditions for existence are that f be absolutely integrable on the appropriate
interval and that f and f be piecewise continuous on every finite interval.
Operational Properties Since our immediate goal is to apply these new transforms to
boundary-value problems, we need to examine the transforms of derivatives.
Fourier Transform Suppose that f is continuous and absolutely integrable on the interval (q, q) and f is piecewise continuous on every finite interval. If f (x) S 0 as x S q, then
integration by parts gives
^5 f 9(x)6 that is,
#
q
f 9(x) e iax dx
q
q
#
q
f (x) e iax d
q
ia
f (x) e iax dx;
#
2 ia
f (x) e iax dx
q
q
q
^{ f (x)} iaF(a).
(11)
Similarly, under the added assumptions that f is continuous on (q, q), f (x) is piecewise
continuous on every finite interval, and f (x) S 0 as x S q, we have
^{ f (x)} (ia)2 ^{ f (x)} a2F(a).
(12)
In general, under conditions analogous to those leading to (12), we have
^{ f (n)(x)} (ia)n ^{ f (x)} (ia)nF(a),
where n 0, 1, 2, … .
It is important to be aware that the sine and cosine transforms are not suitable for transforming
the first derivative (or, for that matter, any derivative of odd order). It is readily shown that
^ s{ f (x)} a^ c{ f (x)}
and
^ c{ f (x)} a^ s{ f (x)} f (0).
15.4 Fourier Transforms
|
783
The difficulty is apparent; the transform of f (x) is not expressed in terms of the original integral
transform.
Fourier Sine Transform Suppose that f and f are continuous, f is absolutely integrable
on the interval [0, q), and f is piecewise continuous on every finite interval. If f S 0 and f S 0
as x S q, then
^s 5 f 0(x)6 #
q
f 0(x) sin ax dx
0
f 9(x) sin axd
q
2a
0
a cf (x) cos ax `
#
q
f 9(x) cos ax dx
0
q
0
a
#
q
f (x) sin ax dxd
0
af (0) 2 a2^s 5 f (x)6;
^ s{ f (x)} a2F(a) af (0).
that is,
(13)
Fourier Cosine Transform Under the same assumptions that lead to (9), we find the
Fourier cosine transform of f (x) to be
^ c{ f (x)} a2F(a) f (0).
(14)
The nature of the transform properties (12), (13), and (14) indicate, in contrast to the Laplace
transform, that Fourier transforms are suitable for problems in which the spatial variable x (or y)
is defined on an infinite or semi-infinite interval. But a natural question then arises:
How do we know which transform to use on a given boundary-value problem?
These assumptions are
sometimes used during the
actual solution process. See
Problems 13, 14, and 26 in
Exercises 15.4.
Clearly, to use the Fourier transform (5), the domain of the variable to eliminate must be
(q, q). To utilize a sine or cosine transform, the domain of at least one of the spatial variables
in the problem must be [0, q). However, the determining factor in choosing between the sine
transform (7) and the cosine transform (9) is the type of boundary condition specified at x 0
(or y 0), that is, whether u or its first partial derivative is given at this boundary.
In solving boundary-value problems using integral transforms most solutions are formal.
In the language of mathematics, this means assumptions about the solution u and its partial
derivatives go unstated. But one assumption should be kept in the back of your mind. In the
examples that follow, it will be assumed without further mention that u and 0u> 0x (or 0u> 0y)
approach 0 as x S q (or y S q). These are not major restrictions since these conditions
hold in most applications.
EXAMPLE 1
Using the Fourier Transform
Solve the heat equation k
0 2u
0u
, q x q, t 0, subject to
2
0t
0x
u(x, 0) f (x)
where
f (x) e
u0,
0,
ZxZ , 1,
ZxZ . 1.
SOLUTION The problem can be interpreted as finding the temperature u(x, t) in an infinite
rod. Because the domain of x is the infinite interval (q, q) we use the Fourier transform
(5) and define the transform of u(x, t) to be
^5u(x, t)6 #
q
q
u(x, t) e iax dx U(a, t).
If we write
^e
784
|
CHAPTER 15 Integral Transform Method
0 2u
f a2U(a, t)
0x 2
and
^e
0u
d
dU
f ^5u(x, t)6 ,
0t
dt
dt
then the Fourier transform of the partial differential equation,
f
0u
0 2u
f ^e f ,
2
0t
0x
^ek
u0
becomes the ordinary differential equation
ka2U(a, t) x
–1
1
dU
dt
or
dU
ka2U(a, t) 0.
dt
2
FIGURE 15.4.1 Initial temperature f
in Example 1
Solving the last equation by the method of Section 2.3 gives U(a, t) ce ka t. The initial
temperature u(x, 0) f(x) in the rod is shown in FIGURE 15.4.1 and its Fourier transform is
^5u(x, 0)6 U(a, 0) #
q
f (x) e iax dx q
#
1
1
u0 e iax dx u0
e ia 2 eia
.
ia
By Euler’s formula
e ia cos a i sin a
e ia cos a 2 i sin a.
e ia 2 e ia
Subtracting these two results and solving for sin a gives sin a . Hence we can
2i
sin a
rewrite the transform of the initial condition as U(a, 0) 2u0
. Applying this condition
a
sin a
2
to the solution U(a, t) ce ka t gives U(a, 0) c 2u0
and so
a
U(a, t) 2u0
sin a ka2t
.
e
a
It then follows from the inverse Fourier transform (6) that
u0
u(x, t) p
#
q
q
sin a ka2t iax
e
da.
e
a
This integral can be simplified somewhat by using Euler’s formula again as eiax cos ax sin ax and noting that
#
q
q
sin a ka2t
sin ax da 0
e
a
because the integrand is an odd function of a. Hence we finally have the solution
u(x, t) u0
p
#
q
q
sin a cos ax ka2t
da.
e
a
(15)
It is left to the reader to show that the solution (15) in Example 1 can be expressed in terms
of the error function. See Problem 23 in Exercises 15.4.
EXAMPLE 2
Two Useful Fourier Transforms
It is a straightforward exercise in integration by parts to show that Fourier sine and cosine
transforms of f(x) ebx, x 0, b 0, are, in turn,
^s 5e bx6 ^c 5e bx6 #
q
0
#
q
0
e bx sin ax dx a
,
b a2
(16)
e bx cos ax dx b
.
b a2
(17)
2
2
Another way to quickly obtain (and remember) these two results is to identify the two integrals
with the more familiar Laplace transform in (2) of Section 4.1. With the symbols x, b, and
a playing the part of t, s, and k, respectively, it follows that (16) and (17) are identical to (d)
and (e) in Theorem 4.1.1.
15.4 Fourier Transforms
|
785
EXAMPLE 3
Using the Cosine Transform
The steady-state temperature in a semi-infinite plate is determined from
0 2u
0 2u
0, 0 , x , p, y . 0
0x 2
0y 2
u(0, y) 0, u(p, y) ey, y . 0
0u
2
0, 0 , x , p.
0y y 0
Solve for u(x, y).
SOLUTION The domain of the variable y and the prescribed condition at y 0 indicate that
the Fourier cosine transform is suitable for the problem. We define
#
^ c{u(x, y)} In view of (14),
becomes
^c e
q
0
u (x, y) cos ay dy U(x, a).
0 2u
0 2u
f
^
e
f ^ c{0}
c
0x 2
0y 2
d 2U
2 a2U(x, a) 2 uy(x, 0) 0
dx 2
d 2U
2 a 2U 0.
dx 2
or
Since the domain of x is a finite interval, we choose to write the solution of the ordinary
differential equation as
U(x, a) c1 cosh ax c2 sinh ax.
(18)
Now ^ c{u(0, y)} ^ c{0} and ^ c{u(p, y)} ^ c{ey} are in turn equivalent to
U(0, a) 0
and
U(p, a) 1
.
1 a2
(19)
Note that the value U(p, a) in (19) is (17) of Example 2 with b 1. When we apply the two
conditions in (19) to the solution (18) we obtain c1 0 and c2 1[(1 a 2) sinh a]. Therefore,
U(x, a) sinh ax
,
(1 a2 ) sinh ap
and so from (10) we arrive at
u(x, y) 2
p
#
q
0
sinh ax
cos ay da.
(1 a2 ) sinh ap
(20)
Had u(x, 0) been given in Example 3 rather than uy(x, 0), then the sine transform would have
been appropriate.
Exercises
15.4
Answers to selected odd-numbered problems begin on page ANS-36.
In Problems 1–18 and 24–26, use an appropriate Fourier
transform to solve the given boundary-value problem. Make
assumptions about boundedness where necessary.
0 2u
0u
, q , x , q, t . 0
1. k 2 0t
0x
u(x, 0) e|x|, q x q
786
|
CHAPTER 15 Integral Transform Method
2. k
0 2u
0u
, q , x , q, t . 0
2
0t
0x
0,
100,
u(x, 0) μ
100,
0,
1
1
0
0
,
,
,
,
x
x
x
x
,
,
,
.
1
0
1
1
3. Find the temperature u(x, t) in a semi-infinite rod if u(0, t) u0,
t 0 and u(x, 0) 0, x 0.
#
16.
q
sin ax
p
da , x 0, to show that the
a
2
0
solution in Problem 3 can be written as
4. Use the result
u(x, t) u0 2
2u0
p
#
q
0
sin ax ka2t
da.
e
a
5. Find the temperature u(x, t) in a semi-infinite rod if u(0, t) 0,
t 0, and
u(x, 0) e
1,
0,
0 , x , 1.
x . 1.
0 2u
0 2u
2 0, 0 , x , p, y . 0
2
0x
0y
0u
2
0, y . 0
u(0, y) f (y),
0x x p
0u
2
0, 0 , x , p
0y y 0
In Problems 17 and 18, find the steady-state temperature u(x, y)
in the plate given in the figure. [Hint: One way of proceeding is
to express Problems 17 and 18 as two and three boundary-value
problems, respectively. Use the superposition principle (see
Section 13.5).]
y
17.
6. Solve Problem 3 if the condition at the left boundary is
u = e –y
0u
2
A, t . 0.
0x x 0
7. Solve Problem 5 if the end x 0 is insulated.
8. Find the temperature u(x, t) in a semi-infinite rod if u(0, t) 1,
t 0, and u(x, 0) ex, x 0.
9. (a) a 2
0 2u
0u
, q , x , q,
0t
0x 2
u(x, 0) f (x),
(b) If g(x) 0, show that the solution of part (a) can be written
as u(x, t) 12 [ f (x at) f (x at)].
10. Find the displacement u(x, t) of a semi-infinite string if
u(0, t) 0,
y
18.
u=0
u = e –y
1
u = 100
0
u = f (x)
FIGURE 15.4.3 Semi-infinite plate in Problem 18
0u
2
0, x . 0.
0t t 0
2
2 2
19. Use the result ^ 5ex >4p 6 2"ppep a to solve the
boundary-value problem
k
11. Solve the problem in Example 3 if the boundary conditions
at x 0 and x p are reversed:
u(0, y) ey,
x
π
2
t0
u(x, 0) xex,
FIGURE 15.4.2 Infinite plate in Problem 17
t.0
0u
2
g(x), q , x , q
0t t 0
x
u = e –x
0 2u
0u
, q , x , q, t . 0
0t
0x 2
2
u(x, 0) ex , q , x , q.
u(p, y) 0, y 0.
20. If ^{ f (x)} F(a) and ^{g(x)} G(a), then the convolution
theorem for the Fourier transform is given by
12. Solve the problem in Example 3 if the boundary condition at
y 0 is u(x, 0) 1, 0 x p.
13. Find the steady-state temperature u(x, y) in a plate defined by
x 0, y 0 if the boundary x 0 is insulated and, at y 0,
#
q
q
f (t)g(x 2 t) dt ^1 5F(a)G(a)6.
2
50,
u(x, 0) e
0,
0 , x , 1.
x . 1.
14. Solve Problem 13 if the boundary condition at x 0 is
k
u(0, y) 0, y 0.
0 2u
0 2u
15.
2 0, x . 0, 0 , y , 2
2
0x
0y
u(0, y) 0, 0 y 2
u(x, 0) f (x), u(x, 2) 0, x 0
2
Use this result and the transform ^ 5ex >4p 6 given in
Problem 19 to show that a solution of the boundary-value
problem
0 2u
0u
, q , x , q, t . 0
2
0t
0x
u(x, 0) f (x), q , x , q
is
u(x, t) #
2"kpt
1
q
q
2
f (t) e(x 2 t) >4kt dt.
15.4 Fourier Transforms
|
787
2
2
21. Use the transform ^ 5ex >4p 6 given in Problem 19 to find the
steady-state temperature u(x, y) in the infinite strip shown in
FIGURE 15.4.4.
y
1
u = e –x2
25. Find the steady-state temperature u(r, z) in the semi-infinite
cylinder in Problem 24 if the base of the cylinder is insulated and
u0, 0 , z , 1
0,
z . 1.
26. Find the steady-state temperature u(x, y) in the infinite plate
defined by q , x , q, y . 0 if the boundary condition
at y 0 is
u(1, z) e
u(x, 0) e
x
FIGURE 15.4.4 Infinite plate in Problem 21
22. The solution of Problem 14 can be integrated. Use entries 46
and 47 of the table in Appendix III to show that
1
1
100
x
x21
x21
2 arctan
d.
u(x, y) carctan 2 arctan
p
y
y
y
2
2
23. Use Problem 20, the change of variables v (x t)2 !kt,
and Problem 11 in Exercises 15.1 to show that the solution
of Example 1 can be expressed as
u0
x1
x21
cerf a
b 2 erf a
bd.
2
2"kt
2"kt
Computer Lab Assignment
27. Assume u0 100 and k 1 in the solution of Problem 23.
Use a CAS to graph u(x, t) over the rectangular region
4 x 4, 0 t 6. Use a 2D plot to superimpose the
graphs of u(x, t) for t 0.05, 0.125, 0.5, 1, 2, 4, 6, and 15 for
4
x
4. Use the graphs to conjecture the values of
limtS q u(x, t) and limxS q u(x, t). Then prove these results
analytically using the properties of erf (x).
Discussion Problem
28. (a) Suppose
#
24. Find the steady-state temperature u(r, z) in a semi-infinite
cylinder described by the boundary-value problem
2
0u
1 0u
2
r 0r
0r
u(1, z) 0,
u(r, 0) u0,
y
f0
y = f(x)
f1
f2
F(a) e
z0
fn
2p
T
FIGURE 15.5.1 Sampling of a continuous
function
788
|
1 2 a,
0,
0#a#1
0 # a . 1.
Find f (x).
(b) Use part (a) to show that
#
q
0
sin2 x
p
dx .
2
x2
Fast Fourier Transform
INTRODUCTION Consider a function f that is defined and continuous on the interval [0, 2p].
f (nT)
...
f (x) cos ax dx F(a),
where
[Hint: Use the integral in Problem 4 and the parametric form
of the modified Bessel equation on page 283.]
15.5
q
0
2
0u
0, 0 r 1,
0z 2
z0
0 r 1.
ZxZ , 1
ZxZ . 1.
[Hint: Consider the two cases a . 0 and a , 0 when you
solve the resulting ordinary differential equation.]
insulated
u(x, t) u0 ,
0,
x
If x0, x1, x2, … , xn, … are equally spaced points in the interval, then the corresponding function
values f0, f1, f2, … , fn, … shown in FIGURE 15.5.1 are said to represent a discrete sampling of the
function f. The notion of discrete samplings of a function is important in the analysis of continuous signals.
In this section, the complex or exponential form of a Fourier series plays an important role in
the discussion. A review of Section 12.4 is recommended.
Discrete Fourier Transform Consider a function f defined on the interval [0, 2p]. From
(11) of Section 12.4 we saw that f can be written in a complex Fourier series,
CHAPTER 15 Integral Transform Method
q
1
f (x) a cne invx where cn 2p
n q
#
2p
0
f (x)einvx dx,
(1)
where the v 2p/2p p/p is the fundamental angular frequency and 2p is the fundamental
period. In the discrete case, however, the input is f0, f1, f2, … , which are the values of the function f
at equally spaced points x nT, n 0, 1, 2, … . The number T is called the sampling rate or
the length of the sampling interval.* If f is continuous at T, then the sample of f at T is defined
to be the product f (x)d(x T ), where d(x T ) is the Dirac delta function (see Section 4.5). We
can then represent this discrete version of f, or discrete signal, as the sum of unit impulses acting
on the function at x nT:
a f (x) d (x 2 nT ).
q
(2)
n q
If we apply the Fourier transform to the discrete signal (2), we have
#
iax
a f (x) d (x 2 nT )e dx.
q
q
(3)
q n q
By the sifting property of the Dirac delta function (see the Remarks at the end of Section 4.5),
(3) is the same as
F(a) a f (nT )e ianT.
q
(4)
n q
The expression F(a) in (4) is called the discrete Fourier transform (DFT) of the function f.
We often write the coefficients f (nT ) in (4) as f (n) or fn. It is also worth noting that since eiax is
periodic in a and eiaT ei(aT2p) ei(a2p/T)T, we only need to consider the function for a in
[0, 2p/T ]. Let N 2p/T. This places x in the interval [0, 2p]. So, because we sample over one
period, the sum in (4) is actually finite.
Now consider the function values f (x) at N equally spaced points, x nT, n 0, 1, 2, … ,
N 1, in the interval [0, 2p]; that is, f0, f1, f2, … , fN1. The (finite) discrete Fourier series
q
f (x) g n q cne inx using these N terms gives us
f0 c0 c1ei10
c2ei20 p
cN1ei(N1)0
f1 c0 c1ei2p/N
c2ei4p/N p
cN1ei2(N1)p/N
f2 c0 c1ei4p/N
c2ei8p/N p
cN1ei4(N1)p/N
(
(
fN1 c0 c1ei2(N1)p/N c2ei4(N1)p/N p cN1ei2(N1)2p/N.
2p
2p
i sin
and use the usual laws of exponents, this system of
If we let vn ei2p/n cos
n
n
equations is the same as
f0 c0 c1
c2 p
cN 2 1
f1 c0 c1vN
c2v2N p
cN 2 1vNN 2 1
f2 c0 c1v2N
c2v4N p
2 1)
cN 2 1v2(N
N
(
(5)
(
2 1)
2 1)2
p cN 2 1v(N
.
fN 2 1 c0 c1vNN 2 1 c2v2(N
N
N
*Note that the symbol T used here does not have the same meaning as in Section 12.4.
15.5 Fast Fourier Transform
|
789
If we use matrix notation (see Sections 8.1 and 8.2), then (5) is
f0
1
f1
1
• f2 μ • 1
(
(
fN 2 1
1
1
vN
v2N
1
v2N
v4N
p
vNN 2 1
2 1)
v2(N
N
p
1
c0
vNN 2 1
c1
2 1)
v2(N
μ • c2 μ .
N
(
(
(N 2 1)2
cN 2 1
vN
(6)
Let the N N matrix in (6) be denoted by the symbol FN. Given the inputs f0, f1, f2, … , fN1, is
there an easy way to find the Fourier coefficients c0, c1, c2, … , cN1? If F N is the matrix consisting of the complex conjugates of the entries of FN and if I denotes the N N identity matrix,
then we have
FN FN FN FN N I
and so
F 1
N 1
F .
N N
It follows from (6) and the last equation that
c0
f0
c1
f1
1
• c2 μ FN • f2 μ .
N
(
(
cN 2 1
fN 2 1
Discrete Transform Pair Recall from Section 15.4 that in the Fourier transform pair
we use a function f (x) as input and compute the coefficients that give the amplitude for each
frequency k (ck in the case of periodic functions of period 2p) or we compute the coefficients
that give the amplitude for each frequency a (F(a) in the case of nonperiodic functions).
Also, given these frequencies and coefficients, we could reconstruct the original function f (x).
In the discrete case, we use a sample of N values of the function f (x) as input and compute the
coefficients that give the amplitude for each sampled frequency. Given these frequencies and
coefficients, it is possible to reconstruct the n sampled values of f (x). The transform pair, the
discrete Fourier transform pair, is given by
c
where
EXAMPLE 1
1
F f
N N
c0
c1
c • c2 μ
(
cN 2 1
and
f FN c
and
f0
f1
f • f2 μ .
(
fN 2 1
(7)
Discrete Fourier Transform
Let N 4 so that the input is f0, f1, f2, f3 at the four points x 0, p/2, p, 3p/2. Since
v4 eip/2 cos (p>2) i sin (p>2) i, the matrix F4 is
1
1
F4 ±
1
1
1
i
1
i
1
1
1
1
1
i
≤.
1
i
Hence from (7), the Fourier coefficients are given by c 14 F4 f :
c0
1
c1
1 1
± ≤ ±
c2
4 1
c3
1
790
|
CHAPTER 15 Integral Transform Method
1
i
1
i
1
1
1
1
f0
1
i
f
≤ ± 1≤.
1
f2
i
f3
|F(α )|
3
If we let f0, f1, f2, f3 be 0, 2, 4, 6, respectively, we find from the preceding matrix product that
c0
3
c1
1 i
c ± ≤ ±
≤.
c2
1
c3
1 2 i
2.5
2
1.5
Note that we obtain the same result using (4); that is, F(a) g n 0 f (nT )eianT, with T p/2,
a 0, 1, 2, 3. The graphs of |cn|, n 0, 1, 2, 3, or equivalently |F(a)| for a 0, 1, 2, 3, are
given in FIGURE 15.5.2.
3
1
0.5
1
1.5
2
2.5
FIGURE 15.5.2 Graph of |F(a)| in
Example 1
3
α
Finding the coefficients involves multiplying by matrices Fn and F n. Because of the nature of
these matrices, these multiplications can be done in a very computationally efficient manner,
using the Fast Fourier Transform (FFT), which is discussed later in this section.
Heat Equation and Discrete Fourier Series If the function f in the initial-value
problem
k
0 2u
0u
, q , x , q, t . 0
0t
0x 2
u(x, 0) f (x),
(8)
q , x , q
is periodic with period 2p, the solution can be written in terms of a Fourier series for f (x). We
can also approximate this solution with a finite sum
u(x, t) a ck(t) e ikx.
n21
k0
If we examine both sides of the one-dimensional heat equation in (8), we see that
n 2 1 dc
0u
j ijx
a
e
0t
dt
j 0
and
k
n21
0 2u
k a cj (t)(i j)2e ijx,
2
0x
j 0
d 2e ijx
(i j)2e ijx .
dx 2
Equating these last two expressions, we have the first-order DE
since
dcj
dt
2
k j 2cj (t) with solution cj (t) cj (0) e k j t.
The final task is to find the values c j(0). However, recall that u(x, t) g k 0 ck (t)eikx and
u(x, 0) f (x), so cj(0) are the coefficients of the discrete Fourier series of f (x). Compare this
with Section 13.3.
n21
Heat Equation and Discrete Fourier Transform The initial-value problem (8)
can be interpreted as the mathematical model for the temperature u(x, t) in an infinitely long bar.
In Section 15.4 we saw that we can solve (8) using the Fourier transform and that the solution
u(x, t) depends on the Fourier transform F(a) of f (x) (see pages 784–785). We can approximate
F(a) by taking a different look at the discrete Fourier transform.
First we approximate values of the transform by discretizing the integral ^{ f (x)} F(a) q
f (x) e iax dx. Consider an interval [a, b]. Let f (x) be given at n equally spaced points
eq
xj a b2a
j,
n
j 0, 1, 2, … ,
n 1.
15.5 Fast Fourier Transform
|
791
Now approximate:
F(a) <
b 2 a n21
iaxj
a f (xj) e
n j
0
b 2 a n21
b2a
jb e iaxj
f aa a
n j 0
n
b2a
b 2 a n21
b2a
jb e iaa e ia n j
f aa a
n j 0
n
b2a
b 2 a iaa n 2 1
b2a
e a f aa jb e ia n j.
n
n
j 0
If we now choose a convenient value for a, say
Fa
2pM
with M an integer, we have
b2a
2pjM
2pM
b2a
b 2 a i 2pMa n 2 1
e b 2 a a f aa jbe i n
b<
n
n
b2a
j 0
b 2 a i 2pMa n 2 1
b2a
e b 2 a a f aa jb v jM
n ,
n
n
j 0
where, recall that vn ei2p/n. This is a numerical approximation to the Fourier transform of f (x)
2pM
evaluated at points
with M an integer.
b2a
Example 1, Section 15.4—Revisited
EXAMPLE 2
Recall from Example 1 in Section 15.4 (with u0 1) that the Fourier transform of a rectangular
pulse defined by
f (x) e
1,
0,
F(a) is
ZxZ , 1
ZxZ . 1
2 sin a
.
a
The frequency spectrum is the graph of |F(a)| versus a given in FIGURE 15.5.3(a). Using n 16
equally spaced points between a 2 and b 2, and M running from 6 to 6, we get the
discrete Fourier transform of f (x), superimposed over the graph of |F(a)|, in Figure 15.5.3(b).
|F(α )|
2
|F(α )|
2
1.75
1.75
1.5
1.5
1.25
1.25
1
1
0.75
0.75
0.5
0.5
0.25
–10
–5
(a)
0.25
5
10
α
–10
–5
(b)
5
10
α
FIGURE 15.5.3 In Example 2 (a) is the graph of |F(a)|; (b) is the discrete Fourier
transform of f
Aliasing A problem known as aliasing may appear whenever one is sampling data at
equally spaced intervals. If you have ever seen a motion picture where rotating wheels appear to
be rotating slowly (or even backwards!), you have experienced aliasing. The wheels may rotate
792
|
CHAPTER 15 Integral Transform Method
(a) y = sin 20π x; x range: [0, 1]; y range: [–1, 1]
(b) y = sin 100π x; x range: [0, 1]; y range: [–1, 1]
FIGURE 15.5.4 TI-92
(a) y = sin 20π x; x range: [0, 1]; y range: [–1, 1]
(b) y = sin 100π x; x range: [0, 1]; y range: [–1, 1]
FIGURE 15.5.5 TI-83
at a high rate, but because the frames in a motion picture are “sampled” at equally spaced intervals, we see a low rate of rotation.
Graphing calculators also suffer from aliasing due to the way they sample points to create
graphs. For example, plot the trigonometric function y sin 20px with frequency 10 on a Texas
Instruments TI-92 and you get the nice graph in FIGURE 15.5.4(a). At higher frequencies, say
y sin 100px with frequency 50, you get the correct amount of cycles, but the amplitudes of
the graph in Figure 15.5.4(b) are clearly not 1.
On a calculator such as the Texas Instruments TI-83, the graphs in FIGURE 15.5.5 show aliasing
much more clearly.
The problem lies in the fact that e2npi cos 2np i sin 2np 1 for all integer values of n.
The discrete Fourier series cannot distinguish einx from 1 as these functions are equal at sampled
points x 2kp>n. The higher frequency is seen as the lower one. Consider the functions cos (pn>2)
and cos (7pn>2). If we sample at the points n 0, 1, 2, … , these two functions appear the same,
the lower frequency is assumed, and the amplitudes (Fourier coefficients) associated with the
higher frequencies are added in with the amplitude of the low frequency. If these Fourier coefficients at large frequencies are small, however, we do not have a big problem. In the Sampling
Theorem below, we will see what can take care of this problem.
Signal Processing Beyond solving PDEs as we have done earlier, the ideas of this section are useful in signal processing. Consider the functions we have been dealing with as signals
from a source. We would like to reconstruct a signal transmitted by sampling it at discrete points.
The problem of calculating an infinite number of Fourier coefficients and summing an infinite
series to reconstruct a signal (function) is not practical. A finite sum could be a decent approximation, but certain signals can be reconstructed by a finite number of samples.
Theorem 15.5.1
Sampling Theorem
If a signal f (x) is band-limited; that is, if the range of frequencies of the signal lie in a band
A k A, then the signal can be reconstructed by sampling two times for every cycle of
the highest frequency present; in fact,
q
np sin (Ax 2 np)
f (x) a f a b
.
A
Ax 2 np
n q
15.5 Fast Fourier Transform
|
793
To justify Theorem 15.5.1, consider the Fourier transform F(a) of f (x) as a periodic extension
so that F(a) is defined for all values of a, not just those in A k A. Using the Fourier transform, we have
F(a) 1
f (x) 2p
#
q
q
#
f (x)e iax dx
(9)
q
iax
F(a)e
q
1
da 2p
#
A
F(a)eiax da.
(10)
A
Treating F(a) as a periodic extension, the Fourier series for F(a) is
F(a) a cne inpa>A,
q
(11)
n q
1
cn 2A
where
#
A
F(a)einpa>A da.
(12)
A
Using (10), note that
p np
p 1
fa b A
A
A 2p
which by (12) is equal to cn. Substituting cn #
A
F(a)einpa>A da,
A
p np
f a b into (11) yields
A
A
q
p np inpa>A
F(a) a
f a be
.
A
n q A
Substituting this expression for F(a) back into (10), we have
1
f (x) 2p
#
A
A
q
p np inpa>A iax
a a
f a be
be
da
A
n q A
np
1 q
fa b
a
2A n q
A
1 q
np
fa b
a
2A n q
A
1 q
np
fa b
a
2A n q
A
1 q
np
fa b
a
2A n q
A
#
A
#
A
e inpa>A eiax da
A
np
ia
e A
x
np
1
ae iAa A
np
ia
2 xb
A
q
np sin (Ax 2 np)
a fa b
.
A
Ax 2 np
n q
|
CHAPTER 15 Integral Transform Method
2 xb
np
eiAa A
1
2i sin (np 2 Ax)
np
ia
2 xb
A
q
np sin (np 2 Ax)
a fa b
A
np 2 Ax
n q
794
da
A
2 xb
b
Note that we used, in succession, an interchange of summation and integration (not always
e iu 2 eiu
, and
allowed, but is acceptable here), integration of an exponential function, sin u 2i
the fact that sin(u) sin u.
So, from samples at intervals of p兾A, all values of f can be reconstructed. Note that if we
allow eiAx (in other words, we allow k A), then the Sampling Theorem will fail. If, for
example, f (x) sin Ax, then all samples will be 0 and f cannot be reconstructed, as aliasing
appears again.
Band-Limited Signals A signal that contains many frequencies can be filtered so that
only frequencies in an interval survive, and it becomes a band-limited signal. Consider the signal
f (x). Multiply the Fourier transform F(a) of f by a function G(a) that is 1 on the interval containing the frequencies a you wish to keep, and 0 elsewhere. This multiplication of two Fourier
transforms in the frequency domain is a convolution of f (x) and g(x) in the time domain. Recall
that Problem 20 in Exercises 15.4 states that
^ 1{F(a)G(a)} #
q
f (t) g (x 2 t) dt.
q
The integral on the right-hand side is called the convolution of f and g and is written f *g. The
last statement can be written more compactly as
^{ f *g} F(a)G(a).
The analogous idea for Laplace transforms is in Section 4.4. The function g(x) its Fourier transform the pulse function
G(a) e
1,
0,
sin Ax
has as
px
A , a , A
elsewhere.
This implies that the function ( f * g)(x) is band-limited, with frequencies between A
and A.
Computing with the Fast Fourier Transform Return to the discrete Fourier transform of f (x), where we have f sampled at n equally spaced points a distance of T apart, namely,
0, T, 2T, 3T, … , (n 1)T. (We used T p/n at the beginning of this section.) Substituting this,
the discrete Fourier transform
Fa
becomes
2pM
b 2 a i 2pMa n 2 1
b2a
e b 2 a a f aa jb vnjM
b n
n
b2a
j 0
Fa
n21
2pk
b T a f ( jT )vkjn, k 0, 1, 2, p , n 2 1.
nT
j 0
For simplicity of notation, write this instead as
ck a fj vkjn, k 0, 1, 2, p , n 2 1.
n21
j 0
15.5 Fast Fourier Transform
|
795
This should remind you of (6), where we had
f0
1
f1
1
• f2 μ • 1
(
(
fn 2 1
1
1
vn
v2n
(
n21
vn
p
p
p
1
v2n
v4n
(
1
c0
c1
vnn 2 1
2(n 2 1)
μ • c2 μ ,
vn
(
(
(n 2 1)2
cn 2 1
vn
p
2 1)
v2(n
n
or f Fnc. The key to the FFT is properties of vn and matrix factorization. If n 2N, we can
write Fn in the following way (which we will not prove):
F2N a
I 2N 2 1
I 2N 2 1
D2N 2 1 F2N 2 1
ba
D2N 2 1
0
0
F2N 2 1
b P,
(13)
where Ik is the k k identity matrix and P is the permutation matrix that rearranges the matrix c
so that the even subscripts are ordered on the top and the odd ones are ordered on the bottom.
The matrix D is a diagonal matrix defined by
1
v 2N
(v2N)2
D 2N 2 1 •
μ.
f
(v2N)2
N21
21
Note that each of the F2N 2 1 matrices can, in turn, be factored. In the end, the matrix Fn with n2
nonzero entries is factored into the product of n simpler matrices at a great savings to the number
of computations needed on the computer.
EXAMPLE 3
The FF T
2
Let n 2 4 and let F4 be the matrix in Example 1:
1
1
F4 ±
1
1
1
i
1
i
1
1
1
1
1
i
≤.
1
i
From (13), the desired factorization of F4 is
1
0
F4 ±
1
0
0
1
0
1
1
0
1
0
A
0
1
i
1
≤ ±
0
0
i
0
1
1
0
0
0
0
1
1
B
0
1
0
0
≤ ±
1
0
1
0
0
0
1
0
0
1
0
0
0
0
≤.
0
1
(14)
P
We have inserted dashed lines in the matrices marked A and B so that you can identify the
submatrices I2, D2, D2, and F2 by comparing (14) directly with (13). You are also encour3
5
aged to multiply out the right side of (14) and verify that you get F4. Now if c ± ≤ , then
8
20
796
|
CHAPTER 15 Integral Transform Method
1
0
F4c ±
1
0
0
1
0
1
1 0
1
0
i
1 1 0
≤ ±
0 1
0
0
0 1
1
0 i
0
0 1
0
1 0
0
0 0
≤ ±
1
0 1
1
0 0
0
1
0
0
1
0
±
1
0
1
1 0
0
1
0
1 1 0
1
0
i
≤ ±
0
0 1
0 1
0
0
0 1
1
0 i
0
3
0
8
≤ ± ≤
1
5
1
20
1
0
±
1
0
11
36
0
1
0
5
5 2 15i
1
0
i
≤ ±
≤ f.
≤ ±
25
14
0 1
0
15
5 15i
1
0 i
0
3
0
5
≤ ± ≤
0
8
1
20
Without going into details, a computation of F n requires n 2 computations, while
using the matrix factorization (the FFT) means the number of computations is reduced
to one proportional to n ln n. Try a few larger values of n and you will see substantial
savings.
15.5
Exercises
Answers to selected odd-numbered problems begin on page ANS-36.
1
1. Show that F 1
4 4 F4 .
2. Prove the sifting property of the Dirac delta function:
#
q
q
f (x) d(x 2 a) dx f (a).
[Hint: Consider the function
1
,
2e
de(x 2 a) •
0,
3.
4.
5.
6.
Zx 2 aZ , e
elsewhere.
Use the mean value theorem for integrals and then let P S 0.]
Find the Fourier transform of the Dirac delta function d(x).
Show that the Dirac delta function is the identity under the
convolution operation; that is, show f *d d * f f. [Hint:
Use Fourier transforms and Problem 3.]
Show that the derivative of the Dirac delta function d (x a)
has the property that it sifts out the derivative of a function f
at a. [Hint: Use integration by parts.]
Use a CAS to show that the Fourier transform of the function
sin Ax
is the pulse function
g(x) px
G(a) e
1,
0,
7. Write the matrix F8 and then write it in factored form (13).
Verify that the product of the factors is F8. If instructed, use
a CAS to verify the result.
8. Let vn ei2p/n cos (2p>n) i sin (2p>n). Since ei2pk 1,
the numbers vkn , k 0, 1, 2, … , n 1, all have the property
that (vkn )n 1. Because of this, vkn , k 0, 1, 2, … , n 1, are
called the nth roots of unity and are solutions of the equation
zn 1 0. Find the eighth roots of unity and plot them in the
xy-plane where a complex number is written z x iy. What
do you notice?
Computer Lab Assignments
2
9. Use a CAS to verify that the function f *g, where f (x) e5x
sin 2x
, is band-limited. If your CAS can handle
px
it, plot the graphs of ^{ f *g} and F(a)G(a) to verify the
result.
10. If your CAS has a discrete Fourier transform command, choose
any six points and compare the result obtained using this
command with that obtained from c 16 F 6f.
and g(x) A , a , A
elsewhere.
15.5 Fast Fourier Transform
|
797
Chapter in Review
15
Answers to selected odd-numbered problems begin on page ANS-36.
In Problems 1–20, solve the given boundary-value problem by
an appropriate integral transform. Make assumptions about
boundedness where necessary.
1.
9.
0 2u
0 2u
0, x . 0, 0 , y , p
0x 2
0y 2
u(x, 0) e
0u
2
0, 0 , y , p
0x x 0
u(x, 0) 0,
0u
2
ex, x . 0
0y y p
10.
3.
4.
0u
0u
2 2 e ZxZ, q , x , q, t . 0
0t
0x
11.
0 2u
0u
r , 0 , x , 1, t . 0
2
0t
0x
0 2u
0 2u
0, x . 0, 0 , y , p
0x 2
0y 2
u(0, y) A, 0 y p
0u
0u
2
0,
2
Bex, x . 0
0y y 0
0y y p
12.
u(x, 0) u0 , q x q
5.
0 2u
0u
, x . 1, t . 0
2
0t
0x
0 2u
0u
, 0 , x , 1, t . 0
2
0t
0x
u(0, t) u0, u(1, t) u0, t 0
u(x, 0) 0, 0 x 1
[Hint: Use the identity
sinh(x y) sinh x cosh y cosh x sinh y,
and then use Problem 8 in Exercises 15.1.]
u(0, t) t, lim u(x, t) 0
xSq
u(x, 0) 0, x 0 [Hint: Use Theorem 4.4.2.]
6.
0 2u
0u
7. k 2 , q , x , q, t . 0
0t
0x
0,
u(x, 0) • u0,
0,
8.
13. k
0 2u
0 2u
, 0 , x , 1, t . 0
0x 2
0t 2
u(0, t) 0, u(1, t) 0, t 0
0u
2
sin px, 0 , x , 1
u(x, 0) sin px,
0t t 0
x,0
0,x,p
x.p
0 2u
0 2u
2 0, 0 , x , p, y . 0
2
0x
0y
0, 0 , y , 1
u(0, y) 0, u(p, y) • 1, 1 , y , 2
0,
y.2
0u
2
0, 0 , x , p
0y y 0
798
|
CHAPTER 15 Integral Transform Method
0,x,1
x.1
u(x, 0) 0, 0 x 1
0 2u
0u
2 hu , h . 0, x . 0, t . 0
0t
0x 2
0u
0, t 0
u(0, t) 0, lim
xSq 0x
u(x, 0) u0, x 0
2
100,
0,
0u
2
0, u(1, t) 0, t . 0
0x x 0
0 2u
0u
2.
, 0 , x , 1, t . 0
2
0t
0x
u(0, t) 0, u(1, t) 0, t 0
u(x, 0) 50 sin 2px, 0 x 1
0 2u
0 2u
2 0, x . 0, y . 0
2
0x
0y
50, 0 , y , 1
u(0, y) e
0,
y.1
0 2u
0u
, q , x , q, t . 0
2
0t
0x
u(x, 0) e
14.
0,
ex,
x,0
x.0
0 2u
0u
, x . 0, t . 0
2
0t
0x
0u
2
50, lim u(x, t) 100, t . 0
0x x 0
xSq
u(x, 0) 100, x 0
15. Show that a solution of the BVP
0 2u
0 2u
2 0, q , x , q, 0 , y , 1
2
0x
0y
0u
2
0, u(x, 1) f (x), q , x , q
0y y 0
is
u(x, y) 1
p
q
# #
0
q
q
f (t)
cosh ay cos a(t 2 x)
dt da.
cosh a
16. 0.5
0 2u
0u
⫽ , x . 0, t . 0
0t
0x 2
0u
2
⫽ 0,
0x x ⫽ 0
t.0
u(x, 0) ⫽ e⫺2x ,
x.0
2
17.
0u
0u
⫽ , 0 , x , p,
2
0t
0x
u(0, t) ⫽ 1, u(p, t) ⫽ 1,
u(x, 0) ⫽ 1 ⫹ sin 2x,
18.
t.0
t.0
0,x,p
0 2u
0u
⫽ , x . 0, t . 0
0t
0x 2
u(0, t) ⫽ 100f8(t 2 5) 2 8(t 2 10)g, t . 0
lim u(x, t) ⫽ 50, t . 0
xSq
u(x, 0) ⫽ 50,
19.
x.0
0 2u
0u
2 A 2 Bu ⫽ 0, A, B constants, x . 0, t . 0
0t
0x 2
0u
u(0, t) ⫽ u1e ⫺10t, lim
⫽ 0, t . 0
xSq 0x
u(x, 0) ⫽ u0, x . 0
21. Solve the boundary-value problem
20.
0u
0 2u
⫽ , x . 0, t . 0
2
0t
0x
0u
`
⫽ ⫺1, lim u(x, t) ⫽ 0, t . 0
0x x ⫽ 0
xSq
u(x, 0) ⫽ 0, x . 0
using the Laplace transform. Give two different forms of the
solution u(x, t).
22. (a) Solve the BVP in Problem 21 using a Fourier transform.
(b) Use a CAS to carry out an integration to show that the
answer in part (a) is equivalent to one of the answers in
Problem 21.
0 2u
0 2u
⫹ 2 ⫽ hu, h constant, x . 0, 0 , y , p
2
0x
0y
u(0, y) ⫽ 0, 0 , y , p
u(x, 0) ⫽ 0, u(x, p) ⫽ f(x), x . 0
CHAPTER 15 in Review
|
799
© Design Pics Inc./Alamy Images
CHAPTER
16
In Section 6.5, we saw that one
way of approximating a solution
of a second-order boundary-value
problem was to work with a finite
difference equation replacement
of the linear ordinary differential
equation. The difference equation
was constructed by replacing the
ordinary derivatives d 2y/dx2 and
dy/dx by difference quotients.
We will see in this chapter that
the same idea carries over to
boundary-value problems involving
linear partial differential equations.
Numerical Solutions
of Partial
Differential Equations
CHAPTER CONTENTS
16.1
16.2
16.3
Laplace’s Equation
Heat Equation
Wave Equation
Chapter 16 in Review
16.1
Laplace’s Equation
INTRODUCTION Recall from Section 13.1 that linear second-order PDEs in two independent
variables are classified as elliptic, parabolic, and hyperbolic. Roughly, elliptic PDEs involve
partial derivatives with respect to spatial variables only and as a consequence solutions of such
equations are determined by boundary conditions alone. Parabolic and hyperbolic equations involve
partial derivatives with respect to both spatial and time variables, and so solutions of such equations generally are determined from boundary and initial conditions. A solution of an elliptic PDE
(such as Laplace’s equation) can describe a physical system whose state is in equilibrium (steady
state), a solution of a parabolic PDE (such as the heat equation) can describe a diffusional state,
whereas a hyperbolic PDE (such as the wave equation) can describe a vibrational state.
In this section we begin our discussion with approximation methods appropriate for elliptic
equations. Our focus will be on the simplest but probably the most important PDE of the elliptic
type: Laplace’s equation.
y
Difference Equation Replacement Suppose that we are seeking a solution u(x, y)
C
of Laplace’s equation
R
02 u
02 u
2 0
2
0x
0y
∇2u = 0
in a planar region R that is bounded by some curve C. See FIGURE 16.1.1. Analogous to (6) of
Section 6.5, using the central differences
u(x h, y) 2u(x, y) u(x h, y)
x
FIGURE 16.1.1 Planar region R with
boundary C
C
6h
R
5h
4h
2h
x
3h 4h
(a)
5h
6h
Pi, j + 1
Pi – 1, j
Pi j
Pi + 1, j
(b)
FIGURE 16.1.2 Region R overlaid with
rectangular grid
|
u(x h, y) ui1, j,
u(x, y h) ui, j1
u(x h, y) ui1, j,
u(x, y h) ui, j1,
then (3) becomes
Pi, j – 1
802
(3)
If we adopt the notation u(x, y) uij and
h
h
(2)
u(x h, y) u(x, y h) u(x h, y) u(x, y h) 4u(x, y) 0.
P20
h
0 2u
1
< 2 fu(x, y h) 2 2u(x, y) u(x, y 2 h)g.
2
0y
h
Hence we can replace Laplace’s equation by the difference equation
P11 P21 P31
h
(1)
0 2u
0 2u
1
< 2 [u(x h, y) u(x, y h) u(x h, y) u(x, y h) 4u(x, y)].
2
2
0x
0y
h
P12 P22
2h
0 2u
1
< 2 fu(x h, y) 2 2u(x, y) u(x 2 h, y)g
2
0x
h
Now by adding (1) and (2) we obtain a five-point approximation to the Laplacian:
P13
3h
u(x, y h) 2u(x, y) u(x, y h),
approximations for the second partial derivatives uxx and uyy can be obtained using the difference
quotients
y
7h
and
ui1, j ui, j1 ui1, j ui, j1 4uij 0.
(4)
To understand (4) a little better, suppose a rectangular grid consisting of horizontal lines spaced
h units apart and vertical lines spaced h units apart is placed over the region R. The number h is
called the mesh size. See FIGURE 16.1.2(a). The points Pij P(ih, jh), i and j integers, of intersection of the horizontal and vertical lines, are called mesh points or lattice points. A mesh point is
an interior point if its four nearest neighboring mesh points are points of R. Points in R or on C
that are not interior points are called boundary points. For example, in Figure 16.1.2(a) we have
P20 P(2h, 0),
P11 P(h, h),
CHAPTER 16 Numerical Solutions of Partial Differential Equations
P21 P(2h, h),
P22 P(2h, 2h),
and so on. Of the points listed, P21 and P22 are interior points, whereas P20 and P11 are boundary
points. In Figure 16.1.2(a) interior points are the dots shown in red and the boundary points are
shown in black. Now from (4) we see that
uij 14 [ui1, j ui, j1 ui1, j ui, j1],
(5)
and so, as shown in Figure 16.1.2(b), the value of uij at an interior mesh point of R is the average
of the values of u at four neighboring mesh points. The neighboring points Pi1, j, Pi, j1, Pi1, j,
and Pi, j1 correspond, respectively, to the four points on a compass: E, N, W, and S.
Dirichlet Problem Recall that in the Dirichlet problem for Laplace’s equation 2u 0,
the values of u(x, y) are prescribed on the boundary C of a region R. The basic idea is to find an
approximate solution to Laplace’s equation at interior mesh points by replacing the partial differential equation at these points by the difference equation (4). Hence the approximate values
of u at the mesh points—namely, the uij —are related to each other and, possibly, to known values
of u if a mesh point lies on the boundary C. In this manner we obtain a system of linear algebraic
equations that we solve for the unknown uij. The following example illustrates the method for a
square region.
EXAMPLE 1
A Boundary-Value Problem Revisited
In Problem 16 of Exercises 13.5 you were asked to solve the boundary-value problem
y
0
0
2
3
P12 P22
P11 P21
0
0 2u
0 2u
0,
0x 2
0y 2
2
3
0
8
9
8
9
x
FIGURE 16.1.3 Square region R for
Example 1
0 , x , 2, 0 , y , 2
u(0, y) 0,
u(2, y) y(2 y), 0 y 2
u(x, 0) 0,
u(x, 2) e
x,
0,x,1
2 2 x, 1 # x , 2
utilizing the superposition principle. To apply the present numerical method, let us start with a
mesh size of h 23 . As we see in FIGURE 16.1.3, that choice yields four interior points and eight
boundary points. The numbers listed next to the boundary points are the exact values of u obtained
from the specified condition along that boundary. For example, at P31 P(3h, h) P(2, 23 ) we
have x 2 and y 23 , and so the condition u(2, y) gives u(2, 23 ) 23 (2 23 ) 89 . Similarly, at
P13 P( 23 , 2), the condition u(x, 2) gives u( 23 , 2) 23 . We now apply (4) at each interior point.
For example, at P11 we have i 1 and j 1, so (4) becomes
u21 u12 u01 u10 4u11 0.
Since u01 u(0, 23 ) 0 and u10 u( 23 , 0) 0, the foregoing equation becomes 4u11 u21 u12 0.
Repeating this, in turn, at P21, P12, and P22, we get three additional equations:
4u11 u21 u12
u11 4u21 0
u22 89
(6)
4u12 u22 23
u11
u21 u12 4u22 149 .
Using a computer algebra system to solve the system, we find the approximate temperatures
at the four interior points to be
u11 367 0.1944,
u21 125 0.4167,
u12 13
36 0.3611,
u22 127 0.5833.
16.1 Laplace’s Equation
|
803
As in the discussion of ordinary differential equations, we expect that a smaller value of h
will improve the accuracy of the approximation. However, using a smaller mesh size means, of
course, that there are more interior mesh points, and correspondingly there is a much larger
system of equations to be solved. For a square region whose length of side is L, a mesh size of
h L /n will yield a total of (n 1) 2 interior mesh points. In Example 1, for n 8, the mesh
size is a reasonable h 28 14 , but the number of interior points is (8 1) 2 49. Thus we have
49 equations in 49 unknowns. In the next example we use a mesh size of h 12 .
Example 1 with More Mesh Points
EXAMPLE 2
y
1
2
1
P13 P23 P33
0
P12 P22 P32
0
P11 P21 P31
0
0
0
As we see in FIGURE 16.1.4, with n 4, a mesh size h 24 12 for the square in Example 1 gives
32 9 interior mesh points. Applying (4) at these points and using the indicated boundary
conditions, we get nine equations in nine unknowns. So that you can verify the results, we
give the system in an unsimplified form:
1
2
0
3
4
1
u21 u12 0 0 4u11 0
3
4
u31 u22 u11 0 4u21 0
x
3
4
FIGURE 16.1.4 Region R in Example 1
with additional mesh points
u32 u21 0 4u31 0
u22 u13 u11 0 4u12 0
u32 u23 u12 u21 4u22 0
(7)
1 u33 u22 u31 4u32 0
u23 1
2
0 u12 4u13 0
u33 1 u13 u22 4u23 0
3
4
1
2
u23 u32 4u33 0.
In this case, a CAS yields
u11 647 0.1094,
51
u21 224
0.2277,
u31 177
448 0.3951
47
u12 224
0.2098,
u22 13
32 0.4063,
u32 135
224 0.6027
u13 145
448 0.3237,
u23 131
224 0.5848,
u33 39
64 0.6094.
After we simplify (7), it is interesting to note that the 9 9 matrix of coefficients is
4
1
0
1
© 0
0
0
0
0
1
4
1
0
1
0
0
0
0
0
1
4
0
0
1
0
0
0
1
0
0
4
1
0
1
0
0
0
1
0
1
4
1
0
1
0
0
0
1
0
1
4
0
0
1
0
0
0
1
0
0
4
1
0
0
0
0
0
1
0
1
4
1
0
0
0
0
0π .
1
0
1
4
(8)
This is an example of a sparse matrix in that a large percentage of the entries are zeros. The
matrix (8) is also an example of a banded matrix. These kinds of matrices are characterized by
the properties that the entries on the main diagonal and on diagonals (or bands) parallel to the
main diagonal are all nonzero. The bands shown in red in (8) are separated by diagonals consisting of all zeros or not.
804
|
CHAPTER 16 Numerical Solutions of Partial Differential Equations
Gauss–Seidel Iteration Problems requiring approximations to solutions of partial
differential equations invariably lead to large systems of linear algebraic equations. It is not
uncommon to have to solve systems involving hundreds of equations. Although a direct method
of solution such as Gaussian elimination leaves unchanged the zero entries outside the bands
in a matrix such as (8), it does fill in the positions between the bands with nonzeros. Since
storing very large matrices uses up a large portion of computer memory, it is usual practice
to solve a large system in an indirect manner. One popular indirect method is called Gauss–
Seidel iteration.
We shall illustrate this method for the system in (6). For the sake of simplicity we replace the
double-subscripted variables u11, u21, u12, and u22 by x1, x2, x3, and x4, respectively.
Gauss–Seidel Iteration
EXAMPLE 3
Step 1: Solve each equation for the variables on the main diagonal of the system. That is,
in (6), solve the first equation for x1, the second equation for x2, and so on:
x1 0.25x2 0.25x3
x2 0.25x1 0.25x4 0.2222
(9)
x3 0.25x1 0.25x4 0.1667
x4 0.25x2 0.25x3 0.3889.
These equations can be obtained directly by using (5) rather than (4) at the interior points.
Step 2: Iterations. We start by making an initial guess for the values of x1, x2, x3, and x4. If this
were simply a system of linear equations and we knew nothing about the solution, we could
start with x1 0, x2 0, x3 0, x4 0. But since the solution of (9) represents approximations to a solution of a boundary-value problem, it would seem reasonable to use as the initial
guess for the values of x1 u11, x2 u21, x3 u12, and x4 u22 the average of all the boundary conditions. In this case the average of the numbers at the eight boundary points shown in
Figure 16.1.2 is approximately 0.4. Thus our initial guess is x1 0.4, x2 0.4, x3 0.4, and
x4 0.4. Iterations of the Gauss–Seidel method use the x values as soon as they are computed.
Note that the first equation in (9) depends only on x2 and x3; thus substituting x2 0.4 and
x3 0.4 gives x1 0.2. Since the second and third equations depend on x1 and x4, we use the
newly calculated values x1 0.2 and x4 0.4 to obtain x2 0.3722 and x3 0.3167. The
fourth equation depends on x2 and x3, so we use the new values x2 0.3722 and x3 0.3167
to get x4 0.5611. In summary, the first iteration has given the values
x1 0.2,
x2 0.3722,
x3 0.3167,
x4 0.5611.
Note how close these numbers are already to the actual values given at the end of Example 1.
The second iteration starts with substituting x2 0.3722 and x3 0.3167 into the first
equation. This gives x1 0.1722. From x1 0.1722 and the last computed value of x4 (namely,
x4 0.5611), the second and third equations give, in turn, x2 0.4055 and x3 0.3500.
Using these two values, we find from the fourth equation that x4 0.5678. At the end of the
second iteration we have
x1 0.1722,
x2 0.4055,
x3 0.3500,
x4 0.5678.
The third through seventh iterations are summarized in Table 16.1.1.
TABLE 16.1.1 Iteration
x1
x2
x3
x4
3rd
4th
5th
6th
7th
0.1889
0.4139
0.3584
0.5820
0.1931
0.4160
0.3605
0.5830
0.1941
0.4165
0.3610
0.5833
0.1944
0.4166
0.3611
0.5833
0.1944
0.4166
0.3611
0.5833
16.1 Laplace’s Equation
|
805
Note.
To apply Gauss–Seidel iteration to a general system of n linear equations in n unknowns, the
variable xi must actually appear in the ith equation of the system. Moreover, after each equation
is solved for xi, i 1, 2, …, n, the resulting system has the form X AX B, where all the
entries on the main diagonal of A are zero.
REMARKS
x=1
y
y = 12
0
0
0
P11 P21 P31
0
0
100 100 100
FIGURE 16.1.5 Rectangular region R
Exercises
16.1
x
(i) In the examples given in this section, the values of uij were determined using known values
of u at boundary points. But what do we do if the region is such that boundary points do not
coincide with the actual boundary C of the region R ? In this case the required values can be
obtained by interpolation.
(ii) It is sometimes possible to cut down the number of equations to solve by using symmetry.
Consider the rectangular region defined by 0 x 2, 0 y 1, shown in FIGURE 16.1.5. The
boundary conditions are u 0 along the boundaries x 0, x 2, y 1, and u 100 along
y 0. The region is symmetric about the lines x 1 and y 12 , and the interior points P11 and
P31 are equidistant from the neighboring boundary points at which the specified values of u are
the same. Consequently we assume that u11 u31, and so the system of three equations in three
unknowns reduces to two equations in two unknowns. See Problem 2 in Exercises 16.1.
(iii) In the context of approximating a solution to Laplace’s equation, the iteration technique
illustrated in Example 3 is often referred to as Liebman’s method.
(iv) It may not be noticeable on a computer, but convergence of Gauss–Seidel iteration (or
Liebman’s method) may not be particularly fast. Also, in a more general setting, Gauss–Seidel
iteration may not converge at all. For conditions that are sufficient to guarantee convergence
of Gauss–Seidel iteration you are encouraged to consult texts on numerical analysis.
Answers to selected odd-numbered problems begin on page ANS-36.
In Problems 1–8, use a computer as a computational aid.
In Problems 1–4, use (4) to approximate the solution of
Laplace’s equation at the interior points of the given region.
Use symmetry when possible.
1. u(0, y) 0,
u(3, y) y(2 y), 0 y 2
u(x, 0) 0, u(x, 2) x(3 x), 0 x 3
mesh size: h 1
2. u(0, y) 0, u(2, y) 0, 0 y 1
u(x, 0) 100, u(x, 1) 0, 0 x 2
mesh size: h 12
3. u(0, y) 0, u(1, y) 0, 0 y 1
u(x, 0) 0, u(x, 1) sin px, 0 x 1
mesh size: h 13
4. u(0, y) 108y2 (1 y), u(1, y) 0, 0 y 1
u(x, 0) 0, u(x, 1) 0, 0 x 1
mesh size: h 13
In Problems 5 and 6, use (5) and Gauss–Seidel iteration to
approximate the solution of Laplace’s equation at the interior
points of a unit square. Use the mesh size h 14 . In Problem 5,
the boundary conditions are given; in Problem 6, the values of u
at boundary points are given in FIGURE 16.1.6.
5. u(0, y) 0,
u(x, 0) 0,
806
|
u(1, y) 100y, 0 y 1
u(x, 1) 100x, 0 x 1
6.
y
10
20
40
P13 P23 P33
20
P12 P22 P32
40
P11 P21 P31
20
10
20
30
70
60
50
x
FIGURE 16.1.6 Region in Problem 6
7. (a) In Problem 12 of Exercises 13.6, you solved a potential
problem using a special form of Poisson’s equation
0 2u
0 2u
2 f (x, y). Show that the difference equation
2
0x
0y
replacement for Poisson’s equation is
ui1, j ui, j1 ui1, j ui, j1 4uij h2f (x, y).
(b) Use the result in part (a) to approximate the solution of the
0 2u
0 2u
Poisson equation 2 2 2 at the interior points
0x
0y
of the region in FIGURE 16.1.7. The mesh size is h 12 , u 1
at every point along ABCD, and u 0 at every point along
DEFGA. Use symmetry and, if necessary, Gauss–Seidel
iteration.
CHAPTER 16 Numerical Solutions of Partial Differential Equations
y
F
G
A
interior points of the region in FIGURE 16.1.8. The mesh is h 18 ,
and u 0 at every point on the boundary of the region. If
necessary, use Gauss–Seidel iteration.
y
B
C
D
E
x
FIGURE 16.1.7 Region in Problem 7
x
8. Use the result in part (a) of Problem 7 to approximate the
solution of the Poisson equation
FIGURE 16.1.8 Region in Problem 8
0 2u
0 2u
2 64 at the
2
0x
0y
16.2 Heat Equation
INTRODUCTION The basic idea in the following discussion is the same as in Section 16.1;
we approximate a solution of a PDE—this time a parabolic PDE—by replacing the equation
with a finite difference equation. But unlike the preceding section, we shall consider two finitedifference approximation methods for parabolic partial differential equations: one called an
explicit method and the other an implicit method. For the sake of definiteness, we consider only
the one-dimensional heat equation.
Difference Equation Replacement To approximate the solution u(x, t) of the one-
dimensional heat equation
c
02 u
0u
2
0t
0x
(1)
we again replace the derivatives by difference quotients. Using the central difference approximation (2) of Section 16.1,
0 2u
1
< 2 [u(x h, t) 2u(x, t) u(x h, t)]
2
0x
h
and the forward difference approximation (3) of Section 6.5,
0u
1
< [u(x, t h) u(x, t)]
0t
h
equation (1) becomes
c
1
[u(x h, t) 2u(x, t) u(x h, t)] [u(x, t k) u(x, t)].
2
k
h
(2)
If we let l ck/h2 and
u(x, t) uij, u(x h, t) ui1, j, u(x h, t) ui1, j, u(x, t k) ui, j1,
then, after simplifying, (2) is
ui, j1 lui1, j (1 2l)uij lui1, j .
(3)
In the case of the heat equation (1), typical boundary conditions are u(0, t) u1, u(a, t) u2,
t 0, and an initial condition is u(x, 0) f (x), 0 x a. The function f can be interpreted as
the initial temperature distribution in a homogeneous rod extending from x 0 to x a ; u1 and
u2 can be interpreted as constant temperatures at the endpoints of the rod. Although we shall not
16.2 Heat Equation
|
807
prove it, the boundary-value problem consisting of (1) and these two boundary conditions
and one initial condition has a unique solution when f is continuous on the closed interval [0, a].
This latter condition will be assumed, and so we replace the initial condition by u(x, 0) f (x),
0 x a. Moreover, instead of working with the semi-infinite region in the xt-plane defined by
the inequalities 0 x a, t 0, we use a rectangular region defined by 0 x a, 0 t T,
where T is some specified value of time. Over this region we place a rectangular grid consisting
of vertical lines h units apart and horizontal lines k units apart. See FIGURE 16.2.1. If we choose
two positive integers n and m and define
...
t
T
3k
2k
k
0
h
2h
3h
...
a
x
FIGURE 16.2.1 Rectangular region in
xt-plane
(j + 1)st time
line
jth time line
k
uij
ui + 1, j
h
FIGURE 16.2.2 u at t j 1 is determined
from three values of u at t j
a
T
and k ,
n
m
then the vertical and horizontal grid lines are defined by
xi ih,
ui, j + 1
ui – 1, j
h
i 0, 1, 2, …, n
and
tj jk,
j 0, 1, 2, …, m.
As illustrated in FIGURE 16.2.2, the plan here is to use formula (3) to estimate the values of
the solution u(x, t) at the points on the ( j 1)st time line using only values from the jth time
line. For example, the values on the first time line ( j 1) depend on the initial condition
ui, 0 u(xi, 0) f (xi) given on the zeroth time line ( j 0). This kind of numerical procedure is
called an explicit finite difference method.
EXAMPLE 1
Using the Finite Difference Method
Consider the boundary-value problem
0 2u
0u
, 0 , x , 1, 0 , t , 0.5
2
0t
0x
u(0, t) 0,
u(1, t) 0, 0 t 0.5
u(x, 0) sin px,
0 x 1.
First we identify c 1, a 1, and T 0.5. If we choose, say, n 5 and m 50, then
h 15 0.2, k 0.5
50 0.01, l 0.25,
1
xi i ,
5
i 0, 1, 2, 3, 4, 5, and
tj j
1
,
100
j 0, 1, 2, …, 50.
Thus (3) becomes
ui, j1 0.25(ui1, j 2uij ui1, j).
By setting j 0 in this formula, we get a formula for the approximations to the temperature
u on the first time line:
ui, 1 0.25(ui1, 0 2ui, 0 ui1, 0 ).
If we then let i 1, …, 4 in the last equation, we obtain, in turn,
u11 0.25(u20 2u10 u00)
u21 0.25(u30 2u20 u10)
u31 0.25(u40 2u30 u20)
u41 0.25(u50 2u40 u30).
The first equation in this list is interpreted as
u11 0.25(u(x2, 0) 2u(x1, 0) u(0, 0))
0.25(u(0.4, 0) 2u(0.2, 0) u(0, 0)).
From the initial condition u(x, 0) sin px, the last line becomes
u11 0.25(0.951056516 2(0.587785252) 0) 0.531656755.
This number represents an approximation to the temperature u(0.2, 0.01).
808
|
CHAPTER 16 Numerical Solutions of Partial Differential Equations
Since it would require a rather large table of over 200 entries to summarize all the approximations over the rectangular grid determined by h and k, we give only selected values in Table 16.2.1.
TABLE 16.2.1
TABLE 16.2.2
Exact
Approximation
u(0.4, 0.05) 0.5806
u(0.6, 0.06) 0.5261
u(0.2, 0.10) 0.2191
u(0.8, 0.14) 0.1476
u25 0.5758
u36 0.5208
u1, 10 0.2154
u4, 14 0.1442
Explicit Difference Equation Approximation with
h 0.2, k 0.01, l 0.25
Time
x 0.20
x 0.40
x 0.60
x 0.80
0.00
0.10
0.20
0.30
0.40
0.50
0.5878
0.2154
0.0790
0.0289
0.0106
0.0039
0.9511
0.3486
0.1278
0.0468
0.0172
0.0063
0.9511
0.3486
0.1278
0.0468
0.0172
0.0063
0.5878
0.2154
0.0790
0.0289
0.0106
0.0039
You should verify, using the methods of Chapter 13, that an exact solution of the boundary2
value problem in Example 1 is given by u(x, t) ep t sin px. Using this solution, we compare
in Table 16.2.2 a sample of exact values with their corresponding approximations.
Stability These approximations are comparable to the exact values and accurate enough
for some purposes. But there is a problem with the foregoing method. Recall that a numerical
method is unstable if round-off errors or any other errors grow too rapidly as the computations proceed. The numerical procedure in Example 1 can exhibit this kind of behavior. It can
be proved that the procedure is stable if l is less than or equal to 0.5, but unstable otherwise.
To obtain l 0.25 0.5 in Example 1, we had to choose the value k 0.01; the necessity
of using very small step sizes in the time direction is the principal fault of this method. You
are urged to work Problem 12 in Exercises 16.2 and witness the predictable instability when
l 1.
Crank–Nicholson Method There are implicit finite difference methods for solving
parabolic partial differential equations. These methods require that we solve a system of equations to determine the approximate values of u on the ( j 1)st time line. However, implicit
methods do not suffer from instability problems.
The algorithm introduced by J. Crank and P. Nicholson in 1947 is used mostly for solving the
0u
0 2u
heat equation. The algorithm consists of replacing the second partial derivative in c 2 0t
0x
by an average of two central difference quotients, one evaluated at t and the other at t k:
u(x h, t k) 2 2u(x, t k) u(x 2 h, t k)
c u(x h, t) 2 2u(x, t) u(x 2 h, t)
1
c
d fu(x, t k) 2 u(x, t)g. (4)
2
2
2
k
h
h
If we again define l ck/h2, then after rearranging terms we can write (4) as
ui1, j1 aui, j1 ui1, j1 ui1, j buij ui1, j ,
(5)
where a 2(1 1/l) and b 2(1 1/l), j 0, 1, …, m 1, and i 1, 2, …, n 1.
For each choice of j, the difference equation (5) for i 1, 2, …, n 1 gives n 1 equations
in n 1 unknowns ui, j1. Because of the prescribed boundary conditions, the values of ui, j1
are known for i 0 and for i n. For example, in the case n 4, the system of equations for
determining the approximate values of u on the ( j 1)st time line is
u0, j1 au1, j1 u2, j1 u2, j bu1, j u0, j
u1, j1 au2, j1 u3, j1 u3, j bu2, j u1, j
u2, j1 au3, j1 u4, j1 u4, j bu3, j u2, j
or
au1, j1 u2, j1
b1
u1, j1 au2, j1 u3, j1 b2
(6)
u2, j1 au3, j1 b3,
16.2 Heat Equation
|
809
b1 u2, j bu1, j u0, j u0, j1
where
b2 u3, j bu2, j u1, j
b3 u4, j bu3, j u2, j u4, j1.
In general, if we use the difference equation (5) to determine values of u on the ( j 1)st time line,
we need to solve a linear system AX B, where the coefficient matrix A is a tridiagonal matrix,
a
1
0
Aß 0
(
0
0
1
a
1
0
0
1
a
1
0
0
1
a
0
0
0
1
0
0
0
0
0
0
0
0
p
f
p
a
1
0
0
0
0∑
(
1
a
and the entries of the column matrix B are
b1 u2, j bu1, j u0, j u0, j1
b2 u3, j bu2, j u1, j
b3 u4, j bu3, j u2, j
o
bn1 un, j bun1, j un2, j un, j1.
EXAMPLE 2
Using the Crank–Nicholson Method
Use the Crank–Nicholson method to approximate the solution of the boundary-value
problem
0.25
0 2u
0u
,
2
0t
0x
u(0, t) 0,
0 , x , 2,
0 , t , 0.3
u(2, t) 0, 0 t 0.3
u(x, 0) sin px, 0 x 2,
using n 8 and m 30.
1
0.01, and
SOLUTION From the identifications a 2, T 0.3, h 14 0.25, k 100
c 0.25, we get l 0.04. With the aid of a computer we get the results in Table 16.2.3. As
in Example 1, the entries in this table represent only a selected number from the 210 approximations over the rectangular grid determined by h and k.
TABLE 16.2.3 Crank–Nicholson Method with h 0.25, k 0.01, l 0.25
TABLE 16.2.4
Exact
Approx.
u(0.75, 0.05) 0.6250
u(0.50, 0.20) 0.6105
u(0.25, 0.10) 0.5525
u35 0.6289
u2, 20 0.6259
u1, 10 0.5594
810
|
Time
x 0.25
x 0.50
x 0.75
x 1.00
x 1.25
x 1.50
x 1.75
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.7071
0.6289
0.5594
0.4975
0.4425
0.3936
0.3501
1.0000
0.8894
0.7911
0.7036
0.6258
0.5567
0.4951
0.7071
0.6289
0.5594
0.4975
0.4425
0.3936
0.3501
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.7071
0.6289
0.5594
0.4975
0.4425
0.3936
0.3501
1.0000
0.8894
0.7911
0.7036
0.6258
0.5567
0.4951
0.7071
0.6289
0.5594
0.4975
0.4425
0.3936
0.3501
Like Example 1, the boundary-value problem in Example 2 also possesses an exact solution
2
given by u(x, t) ep t>4 sin px. The sample comparisons listed in Table 16.2.4 show that the absolute
errors are of the order 102 or 103. Smaller errors can be obtained by decreasing either h or k.
CHAPTER 16 Numerical Solutions of Partial Differential Equations
16.2
Exercises
Answers to selected odd-numbered problems begin on page ANS-37.
In Problems 1–12, use a computer as a computational aid.
1. Use the difference equation (3) to approximate the solution
of the boundary-value problem
0 2u
0u
,
2
0t
0x
0 , x , 2, 0 , t , 1
u(0, t) 0,
u(2, t) 0, 0 t 1
u(x, 0) e
1,
0,
0#x#1
1 , x # 2.
5.
6.
Exercises 13.3 with L 2, one can sum the first 20 terms
to estimate the values for u(0.25, 0.1), u(1, 0.5), and
u(1.5, 0.8) for the solution u(x, t) of Problem 1 above. A
student wrote a computer program to do this and obtained
the results u(0.25, 0.1) 0.3794, u(1, 0.5) 0.1854, and
u(1.5, 0.8) 0.0623. Assume these results are accurate for
all digits given. Compare these values with the approximations obtained in Problem 1 above. Find the absolute errors
in each case.
Solve Problem 1 by the Crank–Nicholson method with n 8
and m 40. Use the values for u(0.25, 0.1), u(1, 0.5), and
u(1.5, 0.8) given in Problem 2 to compute the absolute
errors.
Repeat Problem 1 using n 8 and m 20. Use the values
for u(0.25, 0.1), u(1, 0.5), and u(1.5, 0.8) given in Problem 2
to compute the absolute errors. Why are the approximations
so inaccurate in this case?
Solve Problem 1 by the Crank–Nicholson method with n 8
and m 20. Use the values for u(0.25, 0.1), u(1, 0.5), and
u(1.5, 0.8) given in Problem 2 to compute the absolute errors. Compare the absolute errors with those obtained in
Problem 4.
It was shown in Section 13.2 that if a rod of length L is made
of a material with thermal conductivity K, specific heat g, and
density r, the temperature u(x, t) satisfies the partial differential equation
K 0 2u
0u
, 0 , x , L.
gr 0x 2
0t
Consider the boundary-value problem consisting of the foregoing equation and conditions
u(0, t) 0,
u(x, 0) f (x),
0.8x,
0.8(100 2 x),
0 # x # 50
50 , x # 100.
7. Solve Problem 6 by the Crank–Nicholson method with n 10
and m 10.
Use n 8 and m 40.
4.
f (x) e
8. Repeat Problem 6 if the endpoint temperatures are u(0, t) 0,
2. Using the Fourier series solution obtained in Problem 1 of
3.
Use the difference equation (3) in this section with n 10
and m 10 to approximate the solution of the boundaryvalue problem when
(a) L 20, K 0.15, r 8.0, g 0.11, f (x) 30
(b) L 50, K 0.15, r 8.0, g 0.11, f (x) 30
(c) L 20, K 1.10, r 2.7, g 0.22, f (x) 0.5x (20 x)
(d) L 100, K 1.04, r 10.6, g 0.06,
u(L, t) 0, 0 t 10
0 x L.
u(L, t) 20, 0 t 10.
9. Solve Problem 8 by the Crank–Nicholson method.
10. Consider the boundary-value problem in Example 2. Assume
that n 4.
(a) Find the new value of l.
(b) Use the Crank–Nicholson difference equation (5) to find
the system of equations for u11, u21, and u31, that is, the
approximate values of u on the first time line. [Hint: Set
j 0 in (5), and let i take on the values 1, 2, 3.]
(c) Solve the system of three equations without the aid of a
computer program. Compare your results with the corresponding entries in Table 16.2.3.
11. Consider a rod whose length is L 20 for which K 1.05,
r 10.6, and g 0.056. Suppose
u(0, t) 20, u(20, t) 30
u(x, 0) 50.
(a) Use the method outlined in Section 13.6 to find the steadystate solution c(x).
(b) Use the Crank–Nicholson method to approximate the
temperatures u(x, t) for 0 t Tmax. Select Tmax large
enough to allow the temperatures to approach the steadystate values. Compare the approximations for t Tmax
with the values of c(x) found in part (a).
12. Use the difference equation (3) to approximate the solution
of the boundary-value problem
0u
0 2u
,
2
0t
0x
0 , x , 1, 0 , t , 1
u(0, t) 0,
u(1, t) 0, 0 t 1
u(x, 0) sin px,
0 x 1.
Use n 5 and m 25.
16.2 Heat Equation
|
811
16.3 Wave Equation
INTRODUCTION In this section we approximate a solution of the one-dimensional wave
equation using the finite difference method used in the preceding two sections. The one-dimensional
wave equation is the prototype hyperbolic partial differential equation.
Difference Equation Replacement Suppose u(x, t) represents a solution of the
one-dimensional wave equation
c2
0 2u
0 2u
.
2 0x
0t 2
(1)
Using two central differences,
0 2u
1
< 2 fu(x h, t) 2 2u(x, t) u(x 2 h, t)g
2
0x
h
0 2u
1
< 2 fu(x, t k) 2 2u(x, t) u(x, t 2 k)g
2
0t
k
we replace equation (1) by
c2
1
[u(x h, t) 2u(x, t) u(x h, t)] 2 [u(x, t k) 2u(x, t) u(x, t k)].
h2
k
(2)
We solve (2) for u(x, t k), which is ui, j1. If l ck/h, then (2) yields
ui, j1 l2ui1, j 2(1 l2)uij l2ui1, j ui, j1
ui – 1, j
jth time line
k
( j – 1)st time
line
for i 1, 2, …, n 1 and j 1, 2, …, m 1.
In the case when the wave equation (1) is a model for the vertical displacements u(x, t) of a vibrating string, typical boundary conditions are u(0, t) 0, u(a, t) 0, t 0, and initial conditions
are u(x, 0) f (x), u/ t |t0 g(x), 0 x a. The functions f and g can be interpreted as the initial
position and the initial velocity of the string. The numerical method based on equation (3), like the first
method considered in Section 16.2, is an explicit finite difference method. As before, we use the
difference equation (3) to approximate the solution u(x, t) of (1), using the boundary and initial
conditions, over a rectangular region in the xt-plane defined by the inequalities 0 x a, 0 t T,
where T is some specified value of time. If n and m are positive integers and
ui, j + 1
( j + 1)st time
line
uij
(3)
h
ui + 1, j
ui, j – 1
and
k
T
,
m
the vertical and horizontal grid lines on this region are defined by
xi ih,
h
FIGURE 16.3.1 u at t j 1 is determined
from three values of u at t j and one
value at t j 1
a
n
i 0, 1, 2, …, n
and
tj jk,
j 0, 1, 2, …, m.
As shown in FIGURE 16.3.1, (3) enables us to obtain the approximation ui, j1 on the ( j 1)st time
line from the values indicated on the jth and ( j 1)st time lines. Moreover, we use
u0, j u(0, jk) 0,
and
un, j u(a, jk) 0
ui, 0 u(xi, 0) f (xi).
d boundary conditions
d initial condition
There is one minor problem in getting started. You can see from (3) that for j 1 we need to
know the values of ui, 1 (that is, the estimates of u on the first time line) in order to find ui, 2. But
from Figure 16.3.1, with j 0, we see that the values of ui, 1 on the first time line depend on the
values of ui, 0 on the zeroth time line and on the values of ui, 1. To compute these latter values
we make use of the initial-velocity condition ut(x, 0) g(x). At t 0, it follows from (5) of
Section 6.5 that
g(xi) ut (xi, 0) <
812
|
CHAPTER 16 Numerical Solutions of Partial Differential Equations
u(xi, k) 2 u(xi, k)
.
2k
(4)
In order to make sense of the term u(xi, k) ui, 1 in (4), we have to imagine u(x, t) extended
backward in time. It follows from (4) that
u(xi, k) < u(xi, k) 2 2kg(xi).
This last result suggests that we define
ui, 1 ui, 1 2kg(xi)
(5)
in the iteration of (3). By substituting (5) into (3) when j 0, we get the special case
ui, 1 EXAMPLE 1
l2
(u
ui1, 0) (1 l2) ui, 0 kg(xi).
2 i1, 0
(6)
Using the Finite Difference Method
Approximate the solution of the boundary-value problem
4
0 2u
0 2u
2,
2
0x
0t
u(0, t) 0,
0 , x , 1, 0 , t , 1
u(1, t) 0, 0 t 1
u(x, 0) sin px,
0u
2
0, 0 # x # 1,
0t t 0
using (3) with n 5 and m 20.
SOLUTION We make the identifications c 2, a 1, and T 1. With n 5 and m 20
we get h 15 0.2, k 201 0.05, and l 0.5. Thus, with g(x) 0, equations (6) and (3)
become, respectively,
ui, 1 0.125(ui1, 0 ui1, 0) 0.75ui, 0
(7)
ui, j1 0.25ui1, j 1.5uij 0.25ui1, j ui, j1.
(8)
For i 1, 2, 3, 4, equation (7) yields the following values for the ui, 1 on the first time line:
u11 0.125(u20 u00) 0.75u10 0.55972100
u21 0.125(u30 u10) 0.75u20 0.90564761
(9)
u31 0.125(u40 u20) 0.75u30 0.90564761
u41 0.125(u50 u30) 0.75u40 0.55972100.
Note that the results given in (9) were obtained from the initial condition u(x, 0) sin px.
For example, u20 sin(0.2p), and so on. Now j 1 in (8) gives
ui, 2 0.25ui1, 1 1.5ui, 1 0.25ui1, 1 ui, 0,
and so for i 1, 2, 3, 4 we get
u12 0.25u21 1.5u11 0.25u01 u10
u22 0.25u31 1.5u21 0.25u11 u20
u32 0.25u41 1.5u31 0.25u21 u30
u42 0.25u51 1.5u41 0.25u31 u40.
Using the boundary conditions, the initial conditions, and the data obtained in (9), we get
from these equations the approximations for u on the second time line. These last results and
an abbreviation of the remaining calculations are summarized in Table 16.3.1 on page 814.
16.3 Wave Equation
|
813
TABLE 16.3.1 Explicit Difference Equation Approximation
with h 0.2, k 0.05, l 0.5
Time
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
TABLE 16.3.2
Exact
Approx.
u(0.4, 0.25) 0
u(0.6, 0.3) 0.2939
u(0.2, 0.5) 0.5878
u(0.8, 0.7) 0.1816
u25 0.0185
u36 0.2727
u1, 10 0.5873
u4, 14 0.2119
TABLE 16.3.3
Exact
Approx.
u(0.25, 0.3125) 0.2706 u25 0.2706
u(0.375, 0.375) 0.6533 u36 0.6533
u(0.125, 0.625) 0.2706 u1,10 0.2706
Exercises
16.3
u(a, t) 0, 0 t T
0u
2
0,
0t t 0
0 # x # a,
when
(a) c 1, a 1, T 1, f (x) x(1 x); n 4 and m 10
2
(b) c 1, a 2, T 1, f (x) e16(x 2 1) ; n 5 and m 10
(c) c "2, a 1, T 1,
0,
.50 # x # 0.5
f (x) e
0.5, 0.5 , x # 1;
n 10 and m 25.
2. Consider the boundary-value problem
0 2u
0 2u
2 , 0 , x , 1, 0 , t , 0.5
2
0x
0t
u(0, t) 0, u(1, t) 0, 0 t 0.5
|
x 0.80
0.5878
0.4782
0.1903
0.1685
0.4645
0.5873
0.4912
0.2119
0.1464
0.4501
0.5860
0.9511
0.7738
0.3080
0.2727
0.7516
0.9503
0.7947
0.3428
0.2369
0.7283
0.9482
0.9511
0.7738
0.3080
0.2727
0.7516
0.9503
0.7947
0.3428
0.2369
0.7283
0.9482
0.5878
0.4782
0.1903
0.1685
0.4645
0.5873
0.4912
0.2119
0.1464
0.4501
0.5860
Answers to selected odd-numbered problems begin on page ANS-38.
of the boundary-value problem
0 2u
0 2u
c 2 2 2 , 0 , x , a, 0 , t , T
0x
0t
814
x 0.60
Stability We note in conclusion that this explicit finite difference method for the wave
equation is stable when l 1 and unstable when l 1.
1. Use the difference equation (3) to approximate the solution
u(x, 0) f (x),
x 0.40
It is readily verified that the exact solution of the BVP in Example 1 is u(x, t) sin px cos 2pt.
With this function we can compare the exact results with the approximations. For example,
some selected comparisons are given in Table 16.3.2. As you can see in the table, the approximations are in the same “ball park” as the exact values, but the accuracy is not particularly impressive.
We can, however, obtain more accurate results. The accuracy of this algorithm varies with the
choice of l. Of course, l is determined by the choice of the integers n and m, which in turn determine the values of the step sizes h and k. It can be proved that the best accuracy is always
obtained from this method when the ratio l kc/h is equal to one—in other words, when the
step in the time direction is k h /c. For example, the choice n 8 and m 16 yields h 18 ,
k 161 , and l 1. The sample values listed in Table 16.3.3 clearly show the improved
accuracy.
In Problems 1, 3, 5, and 6, use a computer as a computational aid.
u(0, t) 0,
x 0.20
u(x, 0) sin px,
0u
2
0, 0 # x # 1.
0t t 0
(a) Use the methods of Chapter 13 to verify that the solution
of the problem is u(x, t) sin px cos pt.
(b) Use the method of this section to approximate the solution
of the problem without the aid of a computer program.
Use n 4 and m 5.
(c) Compute the absolute error at each interior grid point.
3. Approximate the solution of the boundary-value problem in
Problem 2 using a computer program with
(a) n 5, m 10
(b) n 5, m 20.
4. Given the boundary-value problem
0 2u
0 2u
2,
2
0x
0t
0 x 1,
u(0, t) 0,
u(1, t) 0, 0 t 1
u(x, 0) x(1 2 x),
0t1
0u
2
0,
0t t 0
0x1
use h k 15 in equation (6) to compute the values of ui, 1 by
hand.
CHAPTER 16 Numerical Solutions of Partial Differential Equations
5. It was shown in Section 13.2 that the equation of a vibrating
string is
0 2u
T 0 2u
,
r 0x 2
0t 2
Use the difference equation (3) in this section to approximate
the solution of the boundary-value problem when h 10,
k 5 "r>T and where r 0.0225 g/cm, T 1.4 107 dynes.
Use m 50.
6. Repeat Problem 5 using
where T is the constant magnitude of the tension in the string
and r is its mass per unit length. Suppose a string of length
60 centimeters is secured to the x-axis at its ends and is released
from rest from the initial displacement
f (x) •
16
0.01x,
30 # x # 30
x 2 30
0.30 2
,
100
30 , x # 60.
Chapter in Review
f (x) •
0.02x,
10 # x # 15
x 2 15
0.30 2
,
150
15 , x # 60
and h 10, k 2.5 "r>T . Use m 50.
Answers to selected odd-numbered problems begin on page ANS-39.
u(0, t) 0, u(1, t) 0, t
1. Consider the boundary-value problem
0 2u
0 2u
2 0, 0 , x , 2, 0 , y , 1
2
0x
0y
0
u(x, 0) x, 0 x 1.
u(0, y) 0, u(2, y) 50, 0 y 1
u(x, 0) 0, u(x, 1) 0, 0 x 2.
Approximate the solution of the differential equation at the
interior points of the region with mesh size h 12 . Use Gaussian
elimination or Gauss–Seidel iteration.
2. Solve Problem 1 using mesh size h 14 . Use Gauss–Seidel
iteration.
3. Consider the boundary-value problem
0 2u
0u
, 0 , x , 1, 0 , t , 0.05
2
0t
0x
(a) Note that the initial temperature u(x, 0) x indicates that
the temperature at the right boundary x 1 should be
u(1, 0) 1, whereas the boundary conditions imply that
u(1, 0) 0. Write a computer problem for the explicit
finite difference method so that the boundary conditions
prevail for all times considered, including t 0. Use the
program to complete Table 16.R.1.
(b) Modify your computer program so that the initial condition prevails at the boundaries at t 0. Use this program
to complete Table 16.R.2.
(c) Are Tables 16.R.1 and 16.R.2 related in any way? Use a
larger time interval if necessary.
TABLE 16.R.1
Time
x 0.00
x 0.20
x 0.40
x 0.60
x 0.80
x 1.00
0.00
0.01
0.02
0.03
0.04
0.05
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.2000
0.4000
0.6000
0.8000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
TABLE 16.R.2
Time
x 0.00
x 0.20
x 0.40
x 0.60
x 0.80
x 1.00
0.00
0.01
0.02
0.03
0.04
0.05
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.2000
0.4000
0.6000
0.8000
1.0000
0.0000
0.0000
0.0000
0.0000
0.0000
CHAPTER 16 in Review
|
815
5
Complex Analysis
17. Functions of a Complex Variable
18. Integration in the Complex Plane
19. Series and Residues
20. Conformal Mappings
© BABAROGA/Shutterstock
PART
© Peteri/ShutterStock, Inc.
CHAPTER
17
In elementary algebra courses
you learned about the existence
and some of the properties of
complex numbers. But in courses
such as calculus, it is likely that
you did not even see a complex
number. Introductory calculus is
basically the study of functions
of a real variable. In advanced
courses, you may have seen
complex numbers occasionally
(see Sections 3.3, 8.8, and 10.2).
However, in the next four
chapters we are going to
introduce you to complex
analysis; that is, the study of
functions of a complex variable.
Although there are many
similarities between complex
analysis and real analysis, there
are many interesting differences
and some surprises.
Functions of a
Complex Variable
CHAPTER CONTENTS
17.1
17.2
17.3
17.4
17.5
17.6
17.7
17.8
Complex Numbers
Powers and Roots
Sets in the Complex Plane
Functions of a Complex Variable
Cauchy–Riemann Equations
Exponential and Logarithmic Functions
Trigonometric and Hyperbolic Functions
Inverse Trigonometric and Hyperbolic Functions
Chapter 17 in Review
17.1
Complex Numbers
INTRODUCTION You have undoubtedly encountered complex numbers in your earlier
courses in mathematics. When you first learned to solve a quadratic equation ax2 bx c 0
by the quadratic formula, you saw that the roots of the equation are not real, that is, complex,
whenever the discriminant b2 4ac is negative. So, for example, simple equations such as
x2 5 0 and x2 x 1 0 have no real solutions. For example, the roots of the last equation
"3
1
"3
1
and . If it is assumed that "3 "3"1, then the roots
are 2
2
2
2
1
"3
1
"3
are written "1 and 2
"1.
2
2
2
2
A Definition Two hundred years ago, around the time that complex numbers were gaining some respectability in the mathematical community, the symbol i was originally used as a
disguise for the embarrassing symbol "1. We now simply say that i is the imaginary unit
and define it by the property i2 1. Using the imaginary unit, we build a general complex
number out of two real numbers.
Definition 17.1.1
Complex Number
A complex number is any number of the form z a ib where a and b are real numbers
and i is the imaginary unit.
Note: The imaginary part of
z 4 9i is 9, not 9i.
Terminology The number i in Definition 17.1.1 is called the imaginary unit. The real
number x in z x iy is called the real part of z; the real number y is called the imaginary
part of z. The real and imaginary parts of a complex number z are abbreviated Re(z) and Im(z),
respectively. For example, if z 4 9i, then Re(z) 4 and Im(z) 9. A real constant multiple
of the imaginary unit is called a pure imaginary number. For example, z 6i is a pure imaginary
number. Two complex numbers are equal if their real and imaginary parts are equal. Since this
simple concept is sometimes useful, we formalize the last statement in the next definition.
Definition 17.1.2
Equality
Complex numbers z1 x1 iy1 and z2 x2 iy2 are equal, z1 z2, if
Re(z1) Re(z2)
and
Im(z1) Im(z2).
A complex number x iy 0 if x 0 and y 0.
Arithmetic Operations Complex numbers can be added, subtracted, multiplied, and
divided. If z1 x1 iy1 and z2 x2 iy2, these operations are defined as follows.
Addition:
z1 z2 (x1 iy1) (x2 iy2) (x1 x2) i( y1 y2)
Subtraction:
z1 z2 (x1 iy1) (x2 iy2) (x1 x2) i( y1 y2)
Multiplication:
z1z2 (x1 iy1)(x2 iy2)
x1x2 y1 y2 i( y1x2 x1 y2)
Division:
x1 iy1
z1
z2
x2 iy2
820
|
CHAPTER 17 Functions of a Complex Variable
x1x2 y1y2
y1x2 2 x1y2
i
x 22 y 22
x 22 y 22
The familiar commutative, associative, and distributive laws hold for complex numbers.
Commutative laws:
e
Associative laws:
e
Distributive law:
z1 z2 z2 z1
z1z2 z2z1
z1 (z2 z3) (z1 z2) z3
z1(z2z3) (z1z2)z3
z1(z2 z3) z1z2 z1z3
In view of these laws, there is no need to memorize the definitions of addition, subtraction,
and multiplication. To add (subtract) two complex numbers, we simply add (subtract) the corresponding real and imaginary parts. To multiply two complex numbers, we use the distributive
law and the fact that i 2 1.
EXAMPLE 1
Addition and Multiplication
If z1 2 4i and z2 3 8i, find (a) z1 z2 and (b) z1z2.
SOLUTION
(a) By adding the real and imaginary parts of the two numbers, we get
(2 4i) (3 8i) (2 3) (4 8)i 1 12i.
(b) Using the distributive law, we have
(2 4i)(3 8i) (2 4i)(3) (2 4i)(8i)
6 12i 16i 32i 2
(6 32) (16 12)i 38 4i.
There is also no need to memorize the definition of division, but before discussing that we
need to introduce another concept.
Conjugate If z is a complex number, then the number obtained by changing the sign of
its imaginary part is called the complex conjugate or, simply, the conjugate of z. If z x iy,
then its conjugate is
z x 2 iy.
For example, if z 6 3i, then z 6 3i; if z 5 i, then z 5 i. If z is a real number,
say z 7, then z 7. From the definition of addition it can be readily shown that the conjugate
of a sum of two complex numbers is the sum of the conjugates:
z1 z2 z1 z2 .
Moreover, we have the additional three properties
z1 2 z2 z1 2 z2 ,
z1z2 z1z2 ,
z1
z1
a b .
z2
z2
The definitions of addition and multiplication show that the sum and product of a complex
number z and its conjugate z are also real numbers:
z z (x iy) (x iy) 2x
(1)
zz (x iy)(x iy) x2 i2y2 x2 y2.
(2)
The difference between a complex number z and its conjugate z is a pure imaginary number:
z z (x iy) (x iy) 2iy.
(3)
Since x Re(z) and y Im(z), (1) and (3) yield two useful formulas:
Re(z) zz
2
and
Im(z) z2z
.
2i
17.1 Complex Numbers
|
821
However, (2) is the important relationship that enables us to approach division in a more practical
manner: To divide z1 by z2, we multiply both numerator and denominator of z1/z2 by the conjugate
of z2. This procedure is illustrated in the next example.
Division
EXAMPLE 2
If z1 2 3i and z2 4 6i, find (a)
z1
1
and (b) .
z2
z1
SOLUTION In both parts of this example we shall multiply both numerator and denominator
by the conjugate of the denominator and then use (2).
2 2 3i
2 2 3i 4 2 6i
8 2 12i 2 12i 18i 2
4 6i
4 6i 4 2 6i
16 36
(a)
10 2 24i
5
6
2
i.
52
26
13
1
1 2 3i
2 3i
2
3
i.
2 2 3i
2 2 3i 2 3i
49
13
13
(b)
y
z = x + iy
x
FIGURE 17.1.1 z as a position vector
Geometric Interpretation A complex number z x iy is uniquely determined by an
ordered pair of real numbers (x, y). The first and second entries of the ordered pairs correspond,
in turn, with the real and imaginary parts of the complex number. For example, the ordered pair
(2, 3) corresponds to the complex number z 2 3i. Conversely, z 2 3i determines the
ordered pair (2, 3). In this manner we are able to associate a complex number z x iy with
a point (x, y) in a coordinate plane. But, as we saw in Section 7.1, an ordered pair of real numbers
can be interpreted as the components of a vector. Thus, a complex number z x iy can also
be viewed as a vector whose initial point is the origin and whose terminal point is (x, y). The
coordinate plane illustrated in FIGURE 17.1.1 is called the complex plane or simply the z-plane. The
horizontal or x-axis is called the real axis and the vertical or y-axis is called the imaginary axis.
The length of a vector z, or the distance from the origin to the point (x, y), is clearly "x 2 y 2 .
This real number is given a special name.
Modulus or Absolute Value
Definition 17.1.3
The modulus or absolute value of z x iy, denoted by ZzZ, is the real number
ZzZ "x 2 y 2 "z z.
(4)
Modulus of a Complex Number
EXAMPLE 3
If z 2 3i, then Zz Z "22 (3)2 "13.
y
As FIGURE 17.1.2 shows, the sum of the vectors z1 and z2 is the vector z1 z2. For the triangle
given in the figure, we know that the length of the side of the triangle corresponding to the vector
z1 z2 cannot be longer than the sum of the remaining two sides. In symbols this is
z1 + z2
z1
z1
Zz1 z2 Z Zz1 Z Zz2 Z.
z2
x
FIGURE 17.1.2 Sum of vectors
(5)
The result in (5) is known as the triangle inequality and extends to any finite sum:
Zz1 z2 z3 p zn Z Zz1 Z Zz2 Z Zz3 Z p Zzn Z.
(6)
Using (5) on z1 z2 (z2), we obtain another important inequality:
Zz1 z2 Z Zz1 Z Zz2 Z.
822
|
CHAPTER 17 Functions of a Complex Variable
(7)
REMARKS
Many of the properties of the real system hold in the complex number system, but there are
some remarkable differences as well. For example, we cannot compare two complex numbers
z1 x1 iy1, y1 0, and z2 x2 iy2, y2 0, by means of inequalities. In other words, statements such as z1 z2 and z2 z1 have no meaning except in the case when the two numbers
z1 and z2 are real. We can, however, compare the absolute values of two complex numbers.
Thus, if z1 3 4i and z2 5 i, then |z1| 5 and |z2| !26, and consequently |z1| |z2|.
This last inequality means that the point (3, 4) is closer to the origin than is the point (5, 1).
Exercises
17.1
Answers to selected odd-numbered problems begin on page ANS-39.
In Problems 1–26, write the given number in the form a ib.
3
1. 2i 3i 5i
2. 3i 5 i 4 7i 3 10i 2 9
3. i 8
4. i 11
5. (5 9i) (2 4i)
6. 3(4 i) 3(5 2i)
7. i (5 7i)
8. i (4 i) 4i(1 2i)
9. (2 3i)(4 i)
10. ( 12 14 i)( 23 53 i)
11. (2 3i)2
12. (1 i)3
2
13.
i
2 2 4i
15.
3 5i
i
14.
1i
10 2 5i
16.
6 2i
17.
(3 2 i) (2 3i)
1i
18.
24. (2 3i) a
25. a
22i 2
b
1 2i
i
1
b a
b
32i
2 3i
33. 2z i(2 9i)
35. z 2 i
34. z 2z 7 6i 0
36. z 2 4z
39. 10 8i, 11 6i
40. 12 14 i, 23 16 i
41. Prove that |z1 z2| is the distance between the points z1 and
(1 i) (1 2 2i)
(2 i) (4 2 3i)
z2 in the complex plane.
42. Show for all complex numbers z on the circle x 2 y 2 4 that
Zz 6 8iZ 12.
Discussion Problems
43. For n a nonnegative integer, in can be one of four values: i, 1,
1
(1 i) (1 2 2i) (1 3i)
17.2
In Problems 33–38, use Definition 17.1.2 to find a complex
number z satisfying the given equation.
22i
z
38.
3 4i
1 3i
1z
In Problems 39 and 40, determine which complex number is
closer to the origin.
20.
26.
28. Re(z 2)
30. Im( z 2 z 2)
32. Zz 5z Z
37. z 2z (4 5i) 2i 3
(2 i)2
22. (1 i )2 (1 i)3
1
23. (3 6i) (4 i )(3 5i ) 22i
(5 2 4i) 2 (3 7i)
(4 2i) (2 2 3i)
21. i (1 i)(2 i)(2 6i)
19.
In Problems 27–32, let z x iy. Find the indicated expression.
27. Re(1/z)
29. Im(2z 4z 4i)
31. Zz 1 3iZ
2
i, and 1. In each of the following four cases express the integer
exponent n in terms of the symbol k, where k 0, 1, 2, . . . .
(a) in i (b) in 1 (c) in i (d) in 1
44. (a) Without doing any significant work such as multiplying
out or using the binomial theorem, think of an easy way
of evaluating (1 i)8.
(b) Use your method in part (a) to evaluate (1 i)64.
Powers and Roots
INTRODUCTION Recall from calculus that a point (x, y) in rectangular coordinates can also
be expressed in terms of polar coordinates (r, u). We shall see in this section that the ability to
express a complex number z in terms of r and u greatly facilitates finding powers and roots of z.
Polar Form Rectangular coordinates (x, y) and polar coordinates (r, u) are related by the
equations x r cos u and y r sin u (see Section 14.1). Thus a nonzero complex number
z x iy can be written as z (r cos u) i(r sin u) or
z r (cos u i sin u).
17.2 Powers and Roots
(1)
|
823
y
z = x + iy
r
r sin θ
θ
x
r cos θ
FIGURE 17.2.1 Polar coordinates
We say that (1) is the polar form of the complex number z. We see from FIGURE 17.2.1 that the
polar coordinate r can be interpreted as the distance from the origin to the point (x, y). In other
words, we adopt the convention that r is never negative so that we can take r to be the modulus
of z; that is, r ZzZ. The angle u of inclination of the vector z measured in radians from the positive
real axis is positive when measured counterclockwise and negative when measured clockwise.
The angle u is called an argument of z and is written u arg z. From Figure 17.2.1 we see that
an argument of a complex number must satisfy the equation tan u y/x. The solutions of this
equation are not unique, since if u0 is an argument of z, then necessarily the angles u0 2p,
u0 4p, . . ., are also arguments. The argument of a complex number in the interval p u p
is called the principal argument of z and is denoted by Arg z. For example, Arg(i) p/2.
A Complex Number in Polar Form
EXAMPLE 1
Express 1 "3i in polar form.
y
5π /3
– π /3
1 – √3i
x
SOLUTION With x 1 and y "3, we obtain r ZzZ #(1)2 ("3)2 2. Now since
the point (1, "3) lies in the fourth quadrant, we can take the solution of tan u "3/1 "3
to be u arg z 5p/3. It follows from (1) that a polar form of the number is
z 2 acos
As we see in FIGURE 17.2.2, the argument of 1 "3i that lies in the interval (p, p], the principal argument of z, is Arg z p/3. Thus, an alternative polar form of the complex number is
p
p
z 2 ccos a b i sin a b d .
3
3
FIGURE 17.2.2 Two arguments of
z 1 "3i in Example 1
5p
5p
i sin
b.
3
3
Multiplication and Division The polar form of a complex number is especially convenient to use when multiplying or dividing two complex numbers. Suppose
z1 r1(cos u1 i sin u1)
z2 r2(cos u2 i sin u2),
and
where u1 and u2 are any arguments of z1 and z2, respectively. Then
z1z2 r1r2[(cos u1 cos u2 sin u1 sin u2) i(sin u1 cos u2 cos u1 sin u2)]
(2)
and for z2 0,
z1
r1
[(cos u1 cos u2 sin u1 sin u2) i(sin u1 cos u2 cos u1 sin u2)].
z2
r2
(3)
From the addition formulas from trigonometry, (2) and (3) can be rewritten, in turn, as
and
z1z2 r1r2 [cos(u1 u2) i sin(u1 u2)]
(4)
z1
r1
[cos(u1 u2) i sin(u1 u2)].
z2
r2
(5)
Inspection of (4) and (5) shows that
Zz1z2 Z Zz1 Z Zz2 Z ,
arg(z1z2) arg z1 arg z2,
and
EXAMPLE 2
2
Zz1 Z
z1
2 ,
z2
Zz2 Z
z1
arga b arg z1 arg z2.
z2
(6)
(7)
Argument of a Product and of a Quotient
We have seen that Arg z1 p/2 for z1 i. In Example 1 we saw that Arg z2 p/3 for
z2 1 "3i. Thus, for
z1z2 i(1 2 "3i ) "3 i
824
|
CHAPTER 17 Functions of a Complex Variable
and
z1
"3
i
1
i
z2
4
4
1 2 "3i
it follows from (7) that
arg(z1z2) p
p
p
2
2
3
6
and
z1
p
p
5p
arga b 2 a b .
z2
2
3
6
In Example 2 we used the principal arguments of z1 and z2 and obtained arg(z1z2) Arg(z1z2) and
arg(z1/z2) Arg(z1/z2). It should be observed, however, that this was a coincidence. Although (7)
is true for any arguments of z1 and z2, it is not true, in general, that Arg(z1z2) Arg z1 Arg z2
and Arg(z1/z2) Arg z1 Arg z2. See Problem 39 in Exercises 17.2.
Integer Powers of z We can find integer powers of the complex number z from the
results in (4) and (5). For example, if z r (cos u i sin u), then with z1 z and z2 z, (4) gives
z 2 r 2 [cos (u u) i sin (u u)] r 2 (cos 2u i sin 2u).
Since z 3 z2z, it follows that
z3 r3 (cos 3u i sin 3u).
Moreover, since arg(1) 0, it follows from (5) that
1
z2 r2 [cos(2u) i sin(2u)].
z2
Continuing in this manner, we obtain a formula for the nth power of z for any integer n:
z n r n (cos nu i sin nu).
EXAMPLE 3
(8)
Power of a Complex Number
3
Compute z for z 1 "3i.
SOLUTION In Example 1 we saw that
p
p
z 2 c cos a b i sin a b d .
3
3
Hence from (8) with r 2, u p/3, and n 3, we get
p
p
(1 2 "3i)3 23 c cos a3a b b i sin a3a b b d
3
3
8[cos(p) i sin(p)] 8.
DeMoivre’s Formula When z cos u i sin u, we have ZzZ r 1 and so (8) yields
(cos u i sin u)n cos nu i sin nu.
(9)
This last result is known as DeMoivre’s formula and is useful in deriving certain trigonometric
identities. See Problems 37 and 38 in Exercises 17.2.
Roots A number w is said to be an nth root of a nonzero complex number z if w n z . If
we let w r(cos f i sin f) and z r (cos u i sin u) be the polar forms of w and z , then in
view of (8) , w n z becomes
rn (cos nf i sin nf) r (cos u i sin u).
From this we conclude that rn r or r r1/n and
cos nf i sin nf cos u i sin u.
By equating the real and imaginary parts, we get from this equation
cos nf cos u
and
sin nf sin u.
17.2 Powers and Roots
|
825
These equalities imply that nf u 2kp, where k is an integer. Thus,
f
u 2kp
.
n
As k takes on the successive integer values k 0, 1, 2, . . ., n 1, we obtain n distinct roots with
the same modulus but different arguments. But for k n we obtain the same roots because the
sine and cosine are 2p-periodic. To see this, suppose k n m, where m 0, 1, 2, . . . . Then
f
u 2(n m)p
u 2mp
2p
n
n
sin f sin a
and so
u 2mp
b,
n
cos f cos a
u 2mp
b.
n
We summarize this result. The n nth roots of a nonzero complex number z r (cos u i sin u)
are given by
wk r 1>n c cos a
where k 0, 1, 2, . . ., n 1.
EXAMPLE 4
u 2kp
u 2kp
b i sin a
bd,
n
n
(10)
Roots of a Complex Number
Find the three cube roots of z i.
SOLUTION With r 1, u arg z p/2, the polar form of the given number is
z cos(p/2) i sin(p/2). From (10) with n 3 we obtain
wk (1)1>3 ccos a
p>2 2kp
3
Hence, the three roots are
p>2 2kp
3
b d , k 0, 1, 2.
k 0, w0 cos
p
p
"3
1
i sin i
6
6
2
2
k 1, w1 cos
5p
5p
"3
1
i sin
i
6
6
2
2
k 2, w2 cos
3p
3p
i sin
i.
2
2
The root w of a complex number z obtained by using the principal argument of z with
k 0 is sometimes called the principal nth root of z. In Example 4, since Arg(i) p/2,
y
w1
b i sin a
w0
x
w2
FIGURE 17.2.3 Three cube roots of i
w0 "3/2 (1/2)i is the principal third root of i.
Since the roots given by (8) have the same modulus, the n roots of a nonzero complex
number z lie on a circle of radius r1/n centered at the origin in the complex plane. Moreover,
since the difference between the arguments of any two successive roots is 2p/n, the nth roots
of z are equally spaced on this circle. FIGURE 17.2.3 shows the three cube roots of i equally spaced
on a unit circle; the angle between roots (vectors) wk and wk 1 is 2p/3.
As the next example will show, the roots of a complex number do not have to be “nice”
numbers as in Example 3.
EXAMPLE 5
Roots of a Complex Number
Find the four fourth roots of z 1 i.
SOLUTION In this case, r "2 and u arg z p/4. From (10) with n 4, we obtain
wk ( "2)1>4 ccos a
826
|
CHAPTER 17 Functions of a Complex Variable
p>4 2kp
4
b i sin a
p>4 2kp
4
b d , k 0, 1, 2, 3.
The roots, rounded to four decimal places, are
k 0, w0 ( "2 )1>4 ccos
k 1, w1 ( "2)1>4 ccos
k 2, w2 ( "2)1>4 ccos
k 3, w3 ( "2)1>4 ccos
17.2
Exercises
1. 2
2. 10
3. 3i
4. 6i
5. 1 i
6. 5 5i
7. "3 i
8. 2 2 "3i
10.
12
"3 i
In Problems 11–14, write the number given in polar form in the
form a ib.
11. z 5 acos
7p
7p
i sin
b
6
6
12. z 8 "2 acos
13. z 6 acos
11p
11p
i sin
b
4
4
p
p
i sin b
8
8
14. z 10 acos
p
p
i sin b
5
5
In Problems 15 and 16, find z1z2 and z1/z2. Write the number in
the form a ib.
15. z1 2 acos
p
p
3p
3p
i sin b, z2 4 acos
i sin
b
8
8
8
8
16. z1 "2 acos
z2 "3 acos
p
p
i sin b,
4
4
p
p
i sin b
12
12
In Problems 17–20, write each complex number in polar form.
Then use either (4) or (5) to obtain a polar form of the given
number. Write the polar form in the form a ib.
17. (3 3i)(5 5 "3i)
9p
9p
i sin
d 0.2127 1.0696i
16
16
17p
17p
i sin
d 1.0696 2 0.2127i
16
16
25p
25p
i sin
d 0.2127 2 1.0696i.
16
16
Answers to selected odd-numbered problems begin on page ANS-39.
In Problems 1–10, write the given complex number in polar form.
3
9.
1 i
p
p
i sin d 1.0696 0.2127i
16
16
18. (4 4i)(1 i)
19.
i
2 2 2i
20.
"2 "6i
1 "3i
In Problems 21–26, use (8) to compute the indicated power.
21. (1 "3i)9
23. ( 12 12 i)10
25. acos
p
p 12
i sin b
8
8
26. c "3 acos
22. (2 2i)5
24. ("2 "6i)4
2p
2p 6
i sin
bd
9
9
In Problems 27–32, use (10) to compute all roots. Sketch these
roots on an appropriate circle centered at the origin.
27. (8)1/3
28. (1)1/8
29. (i)1/2
30. (1 i)1/3
31. (1 "3i)1/2
32. (1 "3i)1/4
33. z4 1 0
34. z8 2z4 1 0
In Problems 33 and 34, find all solutions of the given equation.
In Problems 35 and 36, express the given complex number first
in polar form and then in the form a ib.
35. acos
p
p
p 12
p 5
i sin b c2 acos i sin b d
9
9
6
6
3p
3p 3
i sin
bd
8
8
36.
p
p 10
c2 acos
i sin b d
16
16
c8 acos
37. Use the result (cos u i sin u)2 cos 2u i sin 2u to find
trigonometric identities for cos 2u and sin 2u.
38. Use the result (cos u i sin u)3 cos 3u i sin 3u to find
trigonometric identities for cos 3u and sin 3u.
17.2 Powers and Roots
|
827
39. (a) If z1 1 and z2 5i, verify that
40. For the complex numbers given in Problem 39, verify in both
parts (a) and (b) that
Arg(z1z2) Arg(z1) Arg(z2).
arg(z1z2) arg(z1) arg(z2)
(b) If z1 1 and z2 5i, verify that
Arg(z1/z2) Arg(z1) Arg(z2).
17.3
z1
arg a b arg(z1) arg(z2).
z2
and
Sets in the Complex Plane
INTRODUCTION In the preceding sections we examined some rudiments of the algebra and
geometry of complex numbers. But we have barely scratched the surface of the subject known
as complex analysis; the main thrust of our study lies ahead. Our goal in the sections and chapters
that follow is to examine functions of a single complex variable z x iy and the calculus of
these functions.
Before introducing the notion of a function of a complex variable, we need to state some
essential definitions and terminology about sets in the complex plane.
Terminology Before discussing the concept of functions of a complex variable, we need
to introduce some essential terminology about sets in the complex plane.
z0
ρ
Suppose z0 x0 iy0. Since Zz z0 Z "(x 2 x0)2 ( y 2 y0)2 is the distance between the
points z x iy and z0 x0 iy0, the points z x iy that satisfy the equation
Zz z0 Z r,
|z – z0| = ρ
FIGURE 17.3.1 Circle of radius r
r
0, lie on a circle of radius r centered at the point z0. See FIGURE 17.3.1.
EXAMPLE 1
Circles
(a) Zz Z 1 is the equation of a unit circle centered at the origin.
(b) Zz 1 2i Z 5 is the equation of a circle of radius 5 centered at 1 2i.
z0
FIGURE 17.3.2 Open set
The points z satisfying the inequality Zz z0 Z r, r 0, lie within, but not on, a circle of
radius r centered at the point z0. This set is called a neighborhood of z0 or an open disk. A
point z0 is said to be an interior point of a set S of the complex plane if there exists some neighborhood of z0 that lies entirely within S. If every point z of a set S is an interior point, then S is
said to be an open set. See FIGURE 17.3.2. For example, the inequality Re(z) 1 defines a right
half-plane, which is an open set. All complex numbers z x iy for which x 1 are in this
set. If we choose, for example, z0 1.1 2i, then a neighborhood of z0 lying entirely in the set
is defined by Zz (1.1 2i)Z 0.05. See FIGURE 17.3.3. On the other hand, the set S of points in
the complex plane defined by Re(z) 1 is not open, since every neighborhood of a point on the
line x 1 must contain points in S and points not in S. See FIGURE 17.3.4.
|z – (1.1 + 2i)| < 0.05
y
y
in S
z = 1.1 + 2i
not in S
x
x
x=1
FIGURE 17.3.3 Open set magnified view
of a point near x 1
828
|
CHAPTER 17 Functions of a Complex Variable
x=1
FIGURE 17.3.4 Set S is not open
EXAMPLE 2
Open Sets
FIGURE 17.3.5 illustrates some additional open sets.
y
y
x
x
Im(z) < 0
lower half-plane
(a)
–1 < Re(z) < 1
infinite strip
(b)
y
y
x
x
|z| > 1
exterior of unit circle
(c)
1 < |z| < 2
circular ring
(d)
FIGURE 17.3.5 Four examples of open sets
The set of numbers satisfying the inequality
r1
z2
z1
FIGURE 17.3.6 Connected set
Zz z0 Z
r2,
such as illustrated in Figure 17.3.5(d), is called an open annulus.
If every neighborhood of a point z0 contains at least one point that is in a set S and at least
one point that is not in S, then z0 is said to be a boundary point of S. The boundary of a set S
is the set of all boundary points of S. For the set of points defined by Re(z) 1, the points on
the line x 1 are boundary points. The points on the circle Z z i Z 2 are boundary points for
the disk Z z i Z 2.
If any pair of points z1 and z2 in an open set S can be connected by a polygonal line that lies
entirely in the set, then the open set S is said to be connected. See FIGURE 17.3.6. An open connected
set is called a domain. All the open sets in Figure 17.3.5 are connected and so are domains. The
set of numbers satisfying Re(z) 4 is an open set but is not connected, since it is not possible
to join points on either side of the vertical line x 4 by a polygonal line without leaving the set
(bear in mind that the points on x 4 are not in the set).
A region is a domain in the complex plane with all, some, or none of its boundary points.
Since an open connected set does not contain any boundary points, it is automatically a region.
A region containing all its boundary points is said to be closed. The disk defined by Zz iZ 2
is an example of a closed region and is referred to as a closed disk. A region may be neither open
nor closed; the annular region defined by 1 Zz 5Z 3 contains only some of its boundary
points and so is neither open nor closed.
REMARKS
Often in mathematics the same word is used in entirely different contexts. Do not confuse the
concept of “domain” defined in this section with the concept of the “domain of a function.”
17.3 Sets in the Complex Plane
|
829
Exercises
17.3
Answers to selected odd-numbered problems begin on page ANS-40.
In Problems 1–8, sketch the graph of the given equation.
1. Re(z) 5
2. Im(z) 2
17.
3. Im(z 3i) 6
19.
4. Im(z i) Re(z 4 3i)
21.
5. Zz 3i Z 2
6. Z2z 1 Z 4
23.
7. Zz 4 3i Z 5
8. Zz 2 2i Z 2
24.
In Problems 9–22, sketch the set of points in the complex plane
satisfying the given inequality. Determine whether the set is a
domain.
1
11. Im(z) 3
13. 2 Re(z 1)
10. ZRe(z)Z
9. Re(z)
26.
5
14. 1 Im(z)
4
25.
2
12. Im(z i)
17.4
1
0
16. Im(1/z)
2
0 arg (z) 2p/3
18. Zarg (z) Z
p/4
Zz iZ 1
20. Zz iZ
0
2 Zz i Z 3
22. 1 Zz 1 iZ
2
Describe the set of points in the complex plane that satisfies
Zz 1 Z Zz iZ.
Describe the set of points in the complex plane that satisfies
ZRe(z)Z ZzZ.
Describe the set of points in the complex plane that satisfies
z2 z 2 2.
Describe the set of points in the complex plane that satisfies
Zz iZ Zz iZ 1.
15. Re(z 2)
4
Functions of a Complex Variable
INTRODUCTION One of the most important concepts in mathematics is that of a function. You
may recall from previous courses that a function is a certain kind of correspondence between two
sets; more specifically: A function f from a set A to a set B is a rule of correspondence that assigns
to each element in A one and only one element in B. If b is the element in the set B assigned to the
element a in the set A by f, we say that b is the image of a and write b f (a). The set A is called
the domain of the function f (but is not necessarily a domain in the sense defined in Section 17.3).
The set of all images in B is called the range of the function. For example, suppose the set A is a set
of real numbers defined by 3 x q and the function is given by f (x) !x 2 3; then f (3) 0,
f (4) 1, f (8) !5, and so on. In other words, the range of f is the set given by 0 y q. Since
A is a set of real numbers, we say f is a function of a real variable x.
Functions of a Complex Variable When the domain A in the foregoing definition
of a function is a set of complex numbers z, we naturally say that f is a function of a complex
variable z or a complex function for short. The image w of a complex number z will be some
complex number u iv; that is,
w f (z) u(x, y) iv(x, y),
(1)
where u and v are the real and imaginary parts of w and are real-valued functions. Inherent in the
mathematical statement (1) is the fact that we cannot draw a graph of a complex function w f (z)
since a graph would require four axes in a four-dimensional coordinate system.
Some examples of functions of a complex variable are
f (z) z2 4z,
y
w = f (z)
z
domain of f
v
f (z) range of f
w
x
u
z
,
z 1
2
f (z) z Re(z),
z any complex number
z 2 i and z 2 i
z any complex number.
Each of these functions can be expressed in form (1). For example,
f (z) z2 4z (x iy)2 4(x iy) (x2 y2 4x) i(2xy 4y).
(a) z-plane
(b) w-plane
FIGURE 17.4.1 Mapping from z-plane to
w-plane
830
|
Thus, u(x, y) x2 y2 4x, and v(x, y) 2xy 4y.
Although we cannot draw a graph, a complex function w f (z) can be interpreted as a mapping
or transformation from the z-plane to the w-plane. See FIGURE 17.4.1.
CHAPTER 17 Functions of a Complex Variable
v
y
EXAMPLE 1
u=1–
x
Find the image of the line Re(z) ⫽ 1 under the mapping f (z) ⫽ z2.
u
SOLUTION For the function f (z) ⫽ z2 we have u(x, y) ⫽ x 2 ⫺ y 2 and v(x, y) ⫽ 2xy. Now,
Re(z) ⫽ x and so by substituting x ⫽ 1 into the functions u and v, we obtain u ⫽ 1 ⫺ y2 and
v ⫽ 2y. These are parametric equations of a curve in the w-plane. Substituting y ⫽ v/2 into
the first equation eliminates the parameter y to give u ⫽ 1 ⫺ v 2 /4. In other words, the image
of the line in FIGURE 17.4.2(a) is the parabola shown in Figure 17.4.2(b).
x=1
(a) z-plane
Image of a Vertical Line
v2/4
(b) w-plane
FIGURE 17.4.2 Image of x ⫽ 1 is a
parabola
We shall pursue the idea of f (z) as a mapping in greater detail in Chapter 20.
It should be noted that a complex function is completely determined by the real-valued
functions u and v. This means a complex function w ⫽ f (z) can be defined by arbitrarily specifying
u(x, y) and v(x, y), even though u ⫹ iv may not be obtainable through the familiar operations on the
symbol z alone. For example, if u(x, y) ⫽ xy2 and v(x, y) ⫽ x2 ⫺ 4y3, then f (z) ⫽ xy2 ⫹ i(x2 ⫺ 4y3)
is a function of a complex variable. To compute, say, f (3 ⫹ 2i), we substitute x ⫽ 3 and y ⫽ 2
into u and v to obtain f (3 ⫹ 2i) ⫽ 12 ⫺ 23i.
y
i
x
–i
FIGURE 17.4.3 f1(z) ⫽ z (normalized)
Complex Functions as Flows We also may interpret a complex function w ⫽ f (z)
as a two-dimensional fluid flow by considering the complex number f (z) as a vector based
at the point z. The vector f (z) specifies the speed and direction of the flow at a given point z.
FIGURES 17.4.3 and 17.4.4 show the flows corresponding to the complex functions f1(z) ⫽ z and
f2(z) ⫽ z 2, respectively.
If x(t) ⫹ iy(t) is a parametric representation for the path of a particle in the flow, the tangent
vector T ⫽ x⬘(t) ⫹ iy⬘(t) must coincide with f (x(t) ⫹ iy(t)). When f (z) ⫽ u(x, y) ⫹ iv(x, y), it
follows that the path of the particle must satisfy the system of differential equations
dx
⫽ u(x, y)
dt
y
dy
⫽ v(x, y).
dt
We call the family of solutions of this system the streamlines of the flow associated
with f (z).
x
EXAMPLE 2
Streamlines
Find the streamlines of the flows associated with the complex functions (a) f1(z) ⫽ z and
(b) f2(z) ⫽ z2.
SOLUTION (a) The streamlines corresponding to f1(z) ⫽ x ⫺ iy satisfy the system
FIGURE 17.4.4 f2(z) ⫽ z 2 (normalized)
dx
⫽x
dt
dy
⫽ ⫺y
dt
and so x(t) ⫽ c1et and y(t) ⫽ c2e⫺t. By multiplying these two parametric equations, we see
that the point x(t) ⫹ iy(t) lies on the hyperbola xy ⫽ c1c2.
(b) To find the streamlines corresponding to f2(z) ⫽ (x2 ⫺ y2) ⫹ i 2xy, note that dx/dt ⫽ x2 ⫺ y2,
dy/dt ⫽ 2xy, and so
dy
2xy
⫽ 2
.
dx
x 2 y2
This homogeneous differential equation has the solution x2 ⫹ y2 ⫽ c2 y, which represents a
family of circles that have centers on the y-axis and pass through the origin.
Limits and Continuity The definition of a limit of a complex function f (z) as z S z0
has the same outward appearance as the limit in real variables.
17.4 Functions of a Complex Variable
|
831
Limit of a Function
Definition 17.4.1
Suppose the function f is defined in some neighborhood of z0, except possibly at z0 itself. Then
f is said to possess a limit at z0, written
lim f (z) L
zSz0
if, for each e
y
v
z0
f(z)
z
δ
ε
L
D
x
(a) δ -neighborhood
u
(b) ε -neighborhood
FIGURE 17.4.5 Geometric meaning of a
complex limit
0, there exists a d
0 such that Z f (z) LZ
e whenever 0
Zz z0 Z
d.
In words, lim zSz0 f (z) L means that the points f (z) can be made arbitrarily close to the
point L if we choose the point z sufficiently close to, but not equal to, the point z0. As shown in
FIGURE 17.4.5, for each e-neighborhood of L (defined by Z f (z) LZ e) there is a d-neighborhood
of z0 (defined by Zz z0 Z d) so that the images of all points z z0 in this neighborhood lie in
the e-neighborhood of L.
The fundamental difference between this definition and the limit concept in real variables
lies in the understanding of z S z0. For a function f of a single real variable x, lim xSx0 f (x) L
means f (x) approaches L as x approaches x0 either from the right of x0 or from the left of
x0 on the real number line. But since z and z0 are points in the complex plane, when we say
that lim zSz0 f (z) exists, we mean that f (z) approaches L as the point z approaches z0 from
any direction.
The following theorem summarizes some properties of limits:
Limit of Sum, Product, Quotient
Theorem 17.4.1
Suppose lim zSz0 f (z) L1 and lim zSz0 g(z) L2. Then
(i) lim [ f (z) g(z)] L1 L2
zSz0
(ii) lim f (z)g(z) L1L2
zSz0
f (z)
L1
,
zSz0 g(z)
L2
(iii) lim
L2 0.
Continuity at a Point
Definition 17.4.2
A function f is continuous at a point z0 if
lim f (z) f (z0).
zSz0
As a consequence of Theorem 17.4.1, it follows that if two functions f and g are continuous
at a point z0, then their sum and product are continuous at z0. The quotient of the two functions
is continuous at z0 provided g(z0) 0.
A function f defined by
f (z) anzn an1zn1 p a2z2 a1z a0,
an 0,
(2)
where n is a nonnegative integer and the coefficients ai, i 0, 1, . . ., n, are complex constants,
is called a polynomial of degree n. Although we shall not prove it, the limit result lim z z0
zSz0
indicates that the simple polynomial function f (z) z is continuous everywhere—that is, on
the entire z-plane. With this result in mind and with repeated applications of Theorem 17.4.1 (i)
and (ii), it follows that a polynomial function (2) is continuous everywhere. A rational
function
f (z) g(z)
,
h(z)
where g and h are polynomial functions, is continuous except at those points at which h(z) is zero.
832
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CHAPTER 17 Functions of a Complex Variable
Derivative The derivative of a complex function is defined in terms of a limit. The symbol z
used in the following definition is the complex number x i y.
Definition 17.4.3
Derivative
Suppose the complex function f is defined in a neighborhood of a point z0. The derivative of
f at z0 is
f 9(z0) lim
DzS0
f (z0 Dz) 2 f (z0)
Dz
(3)
provided this limit exists.
If the limit in (3) exists, the function f is said to be differentiable at z0. The derivative of a
function w f (z) is also written dw/dz.
As in real variables, differentiability implies continuity:
If f is differentiable at z0, then f is continuous at z0.
Moreover, the rules of differentiation are the same as in the calculus of real variables. If f and g
are differentiable at a point z, and c is a complex constant, then
Constant Rules:
d
d
c 0,
cf (z) c f 9(z)
dz
dz
(4)
Sum Rule:
d
f f (z) g(z)g f 9(z) g9(z)
dz
(5)
Product Rule:
d
f f (z)g(z)g f (z)g9(z) g(z) f 9(z)
dz
(6)
Quotient Rule:
Chain Rule:
g(z) f 9(z) 2 f (z)g9(z)
d f (z)
c
d dz g(z)
fg(z)g 2
(7)
d
f (g(z)) f 9(g(z))g9(z).
dz
(8)
The usual Power Rule for differentiation of powers of z is also valid:
d n
z nz n 2 1 , n an integer.
dz
EXAMPLE 3
Using the Rules of Differentiation
Differentiate (a) f (z) 3z4 5z3 2z and (b) f (z) SOLUTION
(9)
z2
.
4z 1
(a) Using the Power Rule (9) along with the Sum Rule (5), we obtain
f (z) 3 4z3 5 3z2 2 12z3 15z2 2.
(b) From the Quotient Rule (7),
f 9(z) (4z 1) 2z 2 z 2 4
4z 2 2z
.
(4z 1)2
(4z 1)2
f (z0 Dz) 2 f (z0)
DzS0
Dz
must approach the same complex number from any direction. Thus in the study of complex
variables, to require the differentiability of a function is a greater demand than in real variables.
If a complex function is made up, such as f (z) x 4iy, there is a good chance that it is not
differentiable.
In order for a complex function f to be differentiable at a point z0, lim
17.4 Functions of a Complex Variable
|
833
EXAMPLE 4
A Function That Is Nowhere Differentiable
Show that the function f (z) x 4iy is nowhere differentiable.
SOLUTION With z x i y, we have
f (z z) f (z) (x x) 4i( y y) x 4iy x 4i y
and so
lim
DzS0
f (z Dz) 2 f (z)
Dx 4iDy
lim
.
Dz
DzS0 Dx iDy
(10)
Now, if we let z S 0 along a line parallel to the x-axis, then y 0 and the value of (10) is 1.
On the other hand, if we let z S 0 along a line parallel to the y-axis, then x 0 and the value
of (10) is seen to be 4. Therefore, f (z) x 4iy is not differentiable at any point z.
Analytic Functions While the requirement of differentiability is a stringent demand,
there is a class of functions that is of great importance whose members satisfy even more severe
requirements. These functions are called analytic functions.
Definition 17.4.4
Analyticity at a Point
A complex function w f (z) is said to be analytic at a point z0 if f is differentiable at z0 and
at every point in some neighborhood of z0.
A function f is analytic in a domain D if it is analytic at every point in D.
The student should read Definition 17.4.4 carefully. Analyticity at a point is a neighborhood
property. Analyticity at a point is, therefore, not the same as differentiability at a point. It is left
as an exercise to show that the function f (z) Zz Z 2 is differentiable at z 0 but is differentiable
nowhere else. Hence, f (z) ZzZ 2 is nowhere analytic. In contrast, the simple polynomial f (z) z 2
is differentiable at every point z in the complex plane. Hence, f (z) z 2 is analytic everywhere.
A function that is analytic at every point z is said to be an entire function. Polynomial functions
are differentiable at every point z and so are entire functions.
REMARKS
Recall from algebra that a number c is a zero of a polynomial function if and only if x c
is a factor of f (x). The same result holds in complex analysis. For example, since
f (z) z4 5z2 4 (z2 1)(z2 4), the zeros of f are i, i, 2i, and 2i. Hence,
f (z) (z i)(z i)(z 2i)(z 2i). Moreover, the quadratic formula is also valid. For
example, using this formula, we can write
f (z) z2 2z 2 (z (1 i))(z (1 i))
(z 1 i)(z 1 i).
See Problems 21 and 22 in Exercises 17.4.
Exercises
17.4
Answers to selected odd-numbered problems begin on page ANS-40.
In Problems 1–6, find the image of the given line under the mapping f (z) z2.
1. y 2
3. x 0
5. y x
2. x 3
4. y 0
6. y x
In Problems 7–14, express the given function in the form
f (z) u iv.
834
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CHAPTER 17 Functions of a Complex Variable
7. f (z) 6z 5 9i
8. f (z) 7z 9i z 3 2i
9. f (z) z 2 3z 4i
10. f (z) 3z 2 2z
11. f (z) z3 4z
12. f (z) z4
13. f (z) z 1/z
14. f (z) z
z1
In Problems 15–18, evaluate the given function at the indicated
points.
2
3
2
15. f (z) 2x y i(xy 2x 1)
(a) 2i
(b) 2 i
(c)
16. f (z) (x 1 1/x) i(4x2 2y2 4)
(a) 1 i
(b) 2 i
(c)
17. f (z) 4z i z Re(z)
(a) 4 6i
(b) 5 12i
(c)
18. f (z) ex cos y iex sin y
(a) pi/4
(b) 1 pi
(c)
33. f (z) 1 4i
z
z
z 2 2 2z 2
zS1 i
z 2 2 2i
zS1
xy21
z21
In Problems 25 and 26, use (3) to obtain the indicated derivative
of the given function.
25. f (z) z2, f (z) 2z
26. f (z) 1/z, f (z) 1/z2
In Problems 27–34, use (4)–(8) to find the derivative f (z) for
the given function.
27. f (z) 4z3 (3 i)z2 5z 4
z
z 2 3i
36. f (z) 2i
z 2 2z 5iz
2
z3 z
z 2 4 3i
38. f (z) 2
2
z 4
z 2 6z 25
39. Show that the function f (z) z is nowhere differentiable.
40. The function f (z) |z|2 is continuous throughout the entire
complex plane. Show, however, that f is differentiable only at
the point z 0. [Hint: Use (3) and consider two cases: z 0
and z 0. In the second case let z approach zero along a
line parallel to the x-axis and then let z approach zero along
a line parallel to the y-axis.]
37. f (z) lim
24. lim
5z 2 2 z
z3 1
35. f (z) In Problems 23 and 24, show that the given limit does not exist.
zS0
34. f (z) 3 pi/3
lim
23. lim
3z 2 4 8i
2z i
In Problems 35–38, give the points at which the given function
will not be analytic.
zSi
22.
32. f (z) (2z 1/z)6
2 7i
19. lim (4z3 5z2 4z 1 5i)
5z 2 2 2z 2
zS1 2 i
z1
4
z 21
21. lim
zSi z 2 i
30. f (z) (z5 3iz3)(z4 iz3 2z2 6iz)
31. f (z) (z2 4i)3
5 3i
In Problems 19–22, the given limit exists. Find its value.
20.
29. f (z) (2z 1)(z2 4z 8i)
In Problems 41–44, find the streamlines of the flow associated
with the given complex function.
41. f (z) 2z
43. f (z) 1/ z
42. f (z) iz
44. f (z) x2 iy2
In Problems 45 and 46, use a graphics calculator or computer
to obtain the image of the given parabola under the mapping
f (z) z2.
45. y 12 x2
46. y (x 1)2
28. f (z) 5z4 iz3 (8 i)z2 6i
17.5
Cauchy–Riemann Equations
INTRODUCTION In the preceding section we saw that a function f of a complex variable z is
analytic at a point z when f is differentiable at z and differentiable at every point in some neighborhood of z. This requirement is more stringent than simply differentiability at a point because
a complex function can be differentiable at a point z but yet be differentiable nowhere else. A
function f is analytic in a domain D if f is differentiable at all points in D. We shall now develop
a test for analyticity of a complex function f (z) u(x, y) iv(x, y).
A Necessary Condition for Analyticity In the next theorem we see that if a function
f (z) u(x, y) iv(x, y) is differentiable at a point z, then the functions u and v must satisfy a pair
of equations that relate their first-order partial derivatives. This result is a necessary condition
for analyticity.
Theorem 17.5.1
Cauchy–Riemann Equations
Suppose f (z) u(x, y) iv(x, y) is differentiable at a point z x iy. Then at z the first-order
partial derivatives of u and v exist and satisfy the Cauchy–Riemann equations
0u
0v
0x
0y
and
0u
0v
.
0y
0x
17.5 Cauchy–Riemann Equations
(1)
|
835
PROOF: Since f (z) exists, we know that
f 9(z) lim
DzS0
f (z Dz) 2 f (z)
.
Dz
(2)
By writing f (z) u(x, y) iv(x, y) and z x i y, we get from (2)
f 9(z) lim
DzS0
u(x Dx, y Dy) iv(x Dx, y Dy) 2 u(x, y) 2 iv(x, y)
.
Dx iDy
(3)
Since this limit exists, z can approach zero from any convenient direction. In particular, if
z S 0 horizontally, then z x and so (3) becomes
f 9(z) lim
DxS0
u(x Dx, y) 2 u(x, y)
v(x Dx, y) 2 v(x, y)
i lim
.
Dx
DxS0
Dx
(4)
Since f (z) exists, the two limits in (4) exist. But by definition the limits in (4) are the first partial
derivatives of u and v with respect to x. Thus, we have shown that
f 9(z) 0u
0v
i .
0x
0x
(5)
Now if we let z S 0 vertically, then z i y and (3) becomes
f 9(z) lim
DyS0
u(x, y Dy) 2 u(x, y)
v(x, y Dy) 2 v(x, y)
i lim
,
iDy
DyS0
iDy
(6)
which is the same as
f 9(z) i
0u
0v
.
0y
0y
(7)
Equating the real and imaginary parts of (5) and (7) yields the pair of equations in (1).
If a complex function f (z) u(x, y) iv(x, y) is analytic throughout a domain D, then the real
functions u and v must satisfy the Cauchy–Riemann equations (1) at every point in D.
EXAMPLE 1
Using the Cauchy–Riemann Equations
The polynomial f (z) z2 z is analytic for all z and f (z) x2 y2 x i(2xy y). Thus,
u(x, y) x2 y2 x and v(x, y) 2xy y. For any point (x, y), we see that the Cauchy–
Riemann equations are satisfied:
0u
0v
2x 1 0x
0y
EXAMPLE 2
and
0u
0v
2y .
0y
0x
Using the Cauchy–Riemann Equations
Show that the function f (z) (2x2 y) i( y2 x) is not analytic at any point.
SOLUTION We identify u(x, y) 2x2 y and v(x, y) y2 x. Now from
0u
4x
0x
and
0v
2y
0y
0u
1x
0y
and
0v
1
0x
we see that 0u/0y 0v/0x but that the equality 0u/0x 0v/0y is satisfied only on the line
y 2x. However, for any point z on the line, there is no neighborhood or open disk about z in
which f is differentiable. We conclude that f is nowhere analytic.
836
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CHAPTER 17 Functions of a Complex Variable
Important.
By themselves, the Cauchy–Riemann equations are not sufficient to ensure analyticity.
However, when we add the condition of continuity to u and v and the four partial derivatives,
the Cauchy–Riemann equations can be shown to imply analyticity. The proof is long and complicated and so we state only the result.
Theorem 17.5.2
Criterion for Analyticity
Suppose the real-valued functions u(x, y) and v(x, y) are continuous and have continuous
first-order partial derivatives in a domain D. If u and v satisfy the Cauchy–Riemann equations
at all points of D, then the complex function f (z) u(x, y) iv(x, y) is analytic in D.
EXAMPLE 3
Using Theorem 17.5.2
For the function f (z) y
x
2i 2
we have
2
x y
x y2
2
y2 2 x 2
0u
0v
2
2 2
0x
0y
(x y )
and
2xy
0u
0v
2
.
2 2
0y
0x
(x y )
In other words, the Cauchy–Riemann equations are satisfied except at the point where
x2 y2 0; that is, at z 0. We conclude from Theorem 17.5.2 that f is analytic in any
domain not containing the point z 0.
The results in (5) and (7) were obtained under the basic assumption that f was differentiable
at the point z. In other words, (5) and (7) give us a formula for computing f (z):
f 9(z) 0u
0v
0v
0u
i
2i .
0x
0x
0y
0y
(8)
For example, we know that f (z) z2 is differentiable for all z. With u(x, y) x2 y2, 0u/0x 2x,
v(x, y) 2xy, and 0v/0x 2y, we see that
f (z) 2x i2y 2(x iy) 2z.
Recall that analyticity implies differentiability but not vice versa. Theorem 17.5.2 has an analogue
that gives sufficient conditions for differentiability:
If the real-valued functions u(x, y) and v(x, y) are continuous and have continuous firstorder partial derivatives in a neighborhood of z, and if u and v satisfy the Cauchy–Riemann
equations at the point z, then the complex function f (z) u(x, y) iv(x, y) is differentiable
at z and f (z) is given by (8).
The function f (z) x2 y2 i is nowhere analytic. With the identifications u(x, y) x2 and
v(x, y) y2, we see from
0u
0v
0u
0v
2x,
2y
and
0,
0
0x
0y
0y
0x
that the Cauchy–Riemann equations are satisfied only when y x. But since the functions u,
0u/0x, 0u/0y, v, 0v/0x, and 0v/0y are continuous at every point, it follows that f is differentiable on
the line y x and on that line (8) gives the derivative f (z) 2x 2y.
Harmonic Functions We saw in Chapter 13 that Laplace’s equation 02u/0x2 02u/0y2 0
occurs in certain problems involving steady-state temperatures. This partial differential equation
also plays an important role in many areas of applied mathematics. Indeed, as we now see, the
real and imaginary parts of an analytic function cannot be chosen arbitrarily, since both u and v
must satisfy Laplace’s equation. It is this link between analytic functions and Laplace’s equation
that makes complex variables so essential in the serious study of applied mathematics.
Definition 17.5.1
Harmonic Functions
A real-valued function f(x, y) that has continuous second-order partial derivatives in a domain D
and satisfies Laplace’s equation is said to be harmonic in D.
17.5 Cauchy–Riemann Equations
|
837
A Source of Harmonic Functions
Theorem 17.5.3
Suppose f (z) u(x, y) iv(x, y) is analytic in a domain D. Then the functions u(x, y) and
v(x, y) are harmonic functions.
We will see in Chapter 18 that
an analytic function possesses
derivatives of all orders.
PROOF: In this proof we shall assume that u and v have continuous second-order partial
derivatives. Since f is analytic, the Cauchy–Riemann equations are satisfied. Differentiating both
sides of 0u/0x 0v/0y with respect to x and differentiating both sides of 0u/0y 0v/0x with respect
to y then give
0 2u
0 2v
2
0x 0y
0x
and
0 2u
0 2v
.
2
0y 0x
0y
With the assumption of continuity, the mixed partials are equal. Hence, adding these two equations gives
0 2u
0 2u
2 0.
2
0x
0y
This shows that u(x, y) is harmonic.
Now differentiating both sides of 0u/0x 0v/0y with respect to y and differentiating both sides
of 0u/0y 0v/0x with respect to x and subtracting yield
0 2v
0 2v
2 0.
2
0x
0y
Harmonic Conjugate Functions If f (z) u(x, y) iv(x, y) is analytic in a domain D,
then u and v are harmonic in D. Now suppose u(x, y) is a given function that is harmonic in D.
It is then sometimes possible to find another function v(x, y) that is harmonic in D so that
u(x, y) iv(x, y) is an analytic function in D. The function v is called a harmonic conjugate
function of u.
Harmonic Function/Harmonic Conjugate Function
EXAMPLE 4
(a) Verify that the function u(x, y) x3 3xy2 5y is harmonic in the entire complex plane.
(b) Find the harmonic conjugate function of u.
SOLUTION
(a) From the partial derivatives
0u
3x 2 2 3y 2,
0x
0 2u
6x,
0x 2
0u
6xy 2 5,
0y
0 2u
6x
0y 2
we see that u satisfies Laplace’s equation:
0 2u
0 2u
6x 2 6x 0.
0x 2
0y 2
(b) Since the harmonic conjugate function v must satisfy the Cauchy–Riemann equations,
we must have
0v
0u
3x 2 2 3y 2
0y
0x
and
0v
0u
6xy 5.
0x
0y
(9)
Partial integration of the first equation in (9) with respect to y gives v(x, y) 3x2y y3 h(x).
From this we get
0v
6xy h9(x).
0x
Substituting this result into the second equation in (9) gives h (x) 5, and so h(x) 5x C.
Therefore, the harmonic conjugate function of u is v(x, y) 3x2y y3 5x C. The analytic
function is f (z) x3 3xy2 5y i(3x2y y3 5x C ).
838
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CHAPTER 17 Functions of a Complex Variable
REMARKS
Suppose u and v are the harmonic functions that comprise the real and imaginary parts of an
analytic function f (z). The level curves u(x, y) c1 and v(x, y) c2 defined by these functions
form two orthogonal families of curves. See Problem 32 in Exercises 17.5. For example, the
level curves generated by the simple analytic function f (z) z x iy are x c1 and y c2.
The family of vertical lines defined by x c1 is clearly orthogonal to the family of horizontal
lines defined by y c2. In electrostatics, if u(x, y) c1 defines the equipotential curves, then
the other, and orthogonal, family v (x, y) c2 defines the lines of force.
Exercises
17.5
Answers to selected odd-numbered problems begin on page ANS-40.
In Problems 1 and 2, the given function is analytic for all z.
Show that the Cauchy–Riemann equations are satisfied at every
point.
1. f (z) z
3
2
2. f (z) 3z 5z 6i
In Problems 3–8, show that the given function is not analytic at
any point.
3. f (z) Re(z)
4. f (z) y ix
5. f (z) 4z 6z 3
7. f (z) x 2 y 2
6. f (z) z 2
8. f (x) In Problems 9–14, use Theorem 17.5.2 to show that the given
function is analytic in an appropriate domain.
9. f (z) ex cos y iex sin y
10. f (z) x sin x cosh y i(y cos x sinh y)
2
2
f (z) x 2 y 2 2xyi; x-axis
f (z) 3x 2y 2 6x 2y 2i; coordinate axes
f (z) x 3 3xy 2 x i( y 3 3x 2y y); coordinate axes
f (z) x 2 x y i( y 2 5y x); y x 2
Use (8) to find the derivative of the function in Problem 9.
Use (8) to find the derivative of the function in Problem 11.
23. u(x, y) x
24. u(x, y) 2x 2xy
2
25. u(x, y) x y
2
26. u(x, y) 4xy3 4x3y x
27. u(x, y) loge(x2 y2)
28. u(x, y) ex (x cos y y sin y)
2
11. f (z) e x 2 y cos 2xy ie x 2 y sin 2xy
12. f (z) 4x2 5x 4y2 9 i(8xy 5y 1)
29. Sketch the level curves u(x, y) c1 and v(x, y) c2 of the
y
x21
13. f (z) 2i
2
2
(x 2 1) y
(x 2 1)2 y 2
3
2
x xy x
x 2y y 3 2 y
14. f (x) i
x 2 y2
x 2 y2
In Problems 15 and 16, find real constants a, b, c, and d so that
the given function is analytic.
15. f (z) 3x y 5 i(ax by 3)
16. f (z) x2 axy by2 i(cx2 dxy y2)
17.6
17.
18.
19.
20.
21.
22.
In Problems 23–28, verify that the given function u is harmonic.
Find v, the harmonic conjugate function of u. Form the corresponding analytic function f (z) u iv.
y
x
i 2
x 2 y2
x y2
2
In Problems 17–20, show that the given function is not analytic
at any point but is differentiable along the indicated curve(s).
analytic function f (z) z2.
30. Consider the function f (z) 1/z. Describe the level curves.
31. Consider the function f (z) z 1/z. Describe the level curve
v(x, y) 0.
32. Suppose u and v are the harmonic functions forming the real
and imaginary parts of an analytic function. Show that the
level curves u(x, y) c1 and v(x, y) c2 are orthogonal. [Hint:
Consider the gradient of u and the gradient of v. Ignore the
case where a gradient vector is the zero vector.]
Exponential and Logarithmic Functions
INTRODUCTION In this and the next section, we shall examine the exponential, logarithmic,
trigonometric, and hyperbolic functions of a complex variable z. Although the definitions of
these complex functions are motivated by their real variable analogues, the properties of these
complex functions will yield some surprises.
Exponential Function Recall that in real variables the exponential function f (x) ex
has the properties
f (x) f (x)
and
f (x1 x2) f (x1)f (x2).
17.6 Exponential and Logarithmic Functions
(1)
|
839
We certainly want the definition of the complex function f (z) ez, where z x iy, to reduce
ex for y 0 and to possess the same properties as in (1).
We have already used an exponential function with a pure imaginary exponent. Euler’s
formula,
eiy cos y i sin y,
y a real number,
(2)
played an important role in Section 3.3. We can formally establish the result in (2) by using the
Maclaurin series for ex and replacing x by iy and rearranging terms:
q (iy)k
(iy)2
(iy)3
(iy)4
e iy a
1 iy p
2!
3!
4!
k 0 k!
Maclaurin series for
cos y and sin y.
a1 2
y2
y4
y6
y3
y5
y7
2
p b i ay 2
2
pb
2!
4!
6!
3!
5!
7!
cos y i sin y.
For z x iy, it is natural to expect that
e xiy e xe iy
e xiy ex (cos y i sin y).
and so by (2),
Inspired by this formal result, we make the following definition.
Exponential Function
Definition 17.6.1
ez e xiy ex (cos y i sin y).
(3)
The exponential function ez is also denoted by the symbol exp z. Note that (3) reduces to ex when
y 0.
Complex Value of the Exponential Function
EXAMPLE 1
Evaluate e1.74.2i.
SOLUTION With the identifications x 1.7 and y 4.2 and the aid of a calculator, we have,
to four rounded decimal places,
e1.7 cos 4.2 2.6837
e1.7 sin 4.2 4.7710.
and
It follows from (3) that e1.74.2i 2.6837 4.7710i.
The real and imaginary parts of e z, u(x, y) e x cos y and v(x, y) e x sin y, are continuous
and have continuous first partial derivatives at every point z of the complex plane. Moreover, the
Cauchy–Riemann equations are satisfied at all points of the complex plane:
0u
0v
e x cos y 0x
0y
and
0u
0v
e x sin y .
0y
0x
It follows from Theorem 17.5.2 that f (z) e z is analytic for all z; in other words, f is an entire
function.
Properties We shall now demonstrate that e z possesses the two desired properties given
in (1). First, the derivative of f is given by (5) of Section 17.5:
f (z) e x cos y i(e x sin y) e x (cos y i sin y) f (z).
As desired, we have established that
d z
e e z.
dz
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|
CHAPTER 17 Functions of a Complex Variable
Second, if z1 ⫽ x1 ⫹ iy1 and z2 ⫽ x2 ⫹ iy2, then by multiplication of complex numbers and the
addition formulas of trigonometry, we obtain
y
z + 4π i
3π i
f (z1) f (z2) ⫽ e x1 (cos y1 ⫹ i sin y1) e x 2 (cos y2 ⫹ i sin y2)
z + 2π i
⫽ e x1 ⫹ x2 f( cos y1 cos y2 2 sin y1 sin y2) ⫹ i ( sin y1 cos y2 ⫹ cos y1 sin y2)g
πi
z
x
z – 2πi
⫽ e x1 ⫹ x2 f( cos( y1 ⫹ y2) ⫹ i sin( y1 ⫹ y2)g ⫽ f (z1 ⫹ z2).
e z1e z2 ⫽ e z1 ⫹ z2.
In other words,
–π i
(4)
It is left as an exercise to prove that
e z1
⫽ e z1 2 z2.
e z2
–3π i
FIGURE 17.6.1 Values of f (z) ⫽ e z at the
four points are the same
Periodicity Unlike the real function ex, the complex function f (z) ⫽ ez is periodic with
y
πi
x
–πi
the complex period 2pi. Since e2pi ⫽ cos 2p ⫹ i sin 2p ⫽ 1 and, in view of (4), ez⫹2pi ⫽ eze2pi ⫽ ez
for all z, it follows that f (z ⫹ 2pi) ⫽ f (z). Because of this complex periodicity, all possible
functional values of f (z) ⫽ ez are assumed in any infinite horizontal strip of width 2p. Thus,
if we divide the complex plane into horizontal strips defined by (2n ⫺ 1)p ⬍ y ⱕ (2n ⫹ 1)p,
n ⫽ 0, ⫾1, ⫾2, . . ., then, as shown in FIGURE 17.6.1, for any point z in the strip ⫺p ⬍ y ⱕ p, the
values f (z), f (z ⫹ 2pi), f (z ⫺ 2pi), f (z ⫹ 4pi), and so on, are the same. The strip ⫺p ⬍ y ⱕ p
is called the fundamental region for the exponential function f (z) ⫽ ez. The corresponding flow
over the fundamental region is shown in FIGURE 17.6.2.
Polar Form of a Complex Number In Section 17.2, we saw that the complex number z
could be written in polar form as z ⫽ r (cos u ⫹ i sin u). Since eiu ⫽ cos u ⫹ i sin u, we can now
write the polar form of a complex number as
z ⫽ reiu.
FIGURE 17.6.2 Flow over the fundamental
region
For example, in polar form z ⫽ 1 ⫹ i is z ⫽ "2e pi>4 .
Circuits In applying mathematics, mathematicians and engineers often approach the same
problem in completely different ways. Consider, for example, the solution of Example 10 in
Section 3.8. In this example we used strictly real analysis to find the steady-state current ip(t) in
an LRC-series circuit described by the differential equation
L
d 2q
dq
1
⫹R
⫹ q ⫽ E0 sin gt.
dt
C
dt 2
Electrical engineers often solve circuit problems such as this using complex analysis. To
illustrate, let us first denote the imaginary unit !⫺1 by the symbol j to avoid confusion with
the current i. Since current i is related to charge q by i ⫽ dq/dt, the differential equation is the
same as
L
1
di
⫹ Ri ⫹ q ⫽ E0 sin gt.
dt
C
Moreover, the impressed voltage E0 sin gt can be replaced by Im(E0e jgt ), where Im means the
“imaginary part of.” Because of this last form, the method of undetermined coefficients suggests
that we assume a solution in the form of a constant multiple of complex exponential—that is,
ip(t) ⫽ Im(Ae jgt). We substitute this expression into the last differential equation, use the fact that
q is an antiderivative of i, and equate coefficients of e jgt:
a jLg ⫹ R ⫹
1
b A ⫽ E0 gives A ⫽
jCg
E0
1
R ⫹ j aLg 2
b
Cg
.
17.6 Exponential and Logarithmic Functions
|
841
The quantity Z R j(Lg 1/Cg) is called the complex impedance of the circuit. Note that the
modulus of the complex impedance, ZZZ "R 2 (Lg 2 1>Cg)2 , was denoted in Example 10
of Section 3.8 by the letter Z and called the impedance.
Now, in polar form the complex impedance is
Lg 2
Z ZZ Ze ju
tan u where
1
Cg
R
.
Hence, A E0 /Z E0 /(ZZZe ju), and so the steady-state current can be written as
ip(t) Im
E0 ju jgt
e e .
ZZZ
The reader is encouraged to verify that this last expression is the same as (35) in Section 3.8.
Logarithmic Function The logarithm of a complex number z x iy, z 0, is defined
as the inverse of the exponential function—that is,
w ln z
z ew.
if
(5)
In (5) we note that ln z is not defined for z 0, since there is no value of w for which ew 0. To
find the real and imaginary parts of ln z , we write w u iv and use (3) and (5):
x iy euiv eu (cos v i sin v) eu cos v ieu sin v.
The last equality implies x eu cos v and y eu sin v. We can solve these two equations for u
and v. First, by squaring and adding the equations, we find
e2u x2 y2 r 2 ZzZ 2
and so
u loge ZzZ,
where loge ZzZ denotes the real natural logarithm of the modulus of z. Second, to solve for v, we
divide the two equations to obtain
y
tan v .
x
This last equation means that v is an argument of z; that is, v u arg z. But since there is
no unique argument of a given complex number z x iy, if u is an argument of z, then so is
u 2np, n 0, 1, 2, . . . .
Definition 17.6.2
Logarithm of a Complex Number
For z 0, and u arg z,
ln z loge|z| i(u 2np),
n 0,
1,
2, . . . .
(6)
As is clearly indicated in (6), there are infinitely many values of the logarithm of a
complex number z. This should not be any great surprise since the exponential function
is periodic.
In real calculus, logarithms of negative numbers are not defined. As the next example will
show, this is not the case in complex calculus.
Complex Values of the Logarithmic Function
EXAMPLE 2
Find the values of (a) ln(2), (b) ln i, and (c) ln(1 i).
SOLUTION
(a) With u arg(2) p and loge|2| 0.6932, we have from (6)
ln(2) 0.6932 i(p 2np).
842
|
CHAPTER 17 Functions of a Complex Variable
(b) With u arg(i) p/2 and loge|i| loge 1 0, we have from (6)
ln i ia
p
2npb.
2
In other words, ln i pi/2, 3pi/2, 5pi/2, 7pi/2, and so on.
(c) With u arg(1 i) 5p/4 and loge|1 i| loge "2 0.3466, we have from (6)
ln(1 i) 0.3466 i a
EXAMPLE 3
5p
2npb .
4
Solving an Exponential Equation
Find all values of z such that ez !3 i.
SOLUTION From (5), with the symbol w replaced by z, we have z ln( !3 i). Now
| !3 i | 2 and tan u 1/ !3 imply that arg( !3 i) p/6, and so (6) gives
z log e2 ia
p
2npb
6
or z 0.6931 ia
p
2npb.
6
Principal Value It is interesting to note that as a consequence of (6), the logarithm
of a positive real number has many values. For example, in real calculus, loge5 has only one
value: loge 5 1.6094, whereas in complex calculus, ln 5 1.6094 2npi. The value of ln 5
corresponding to n 0 is the same as the real logarithm loge 5 and is called the principal value
of ln 5. Recall that in Section 17.2 we stipulated that the principal argument of a complex number,
written Arg z, lies in the interval (p, p]. In general, we define the principal value of ln z as
that complex logarithm corresponding to n 0 and u Arg z. To emphasize the principal value
of the logarithm, we shall adopt the notation Ln z. In other words,
Ln z loge|z| i Arg z.
(7)
Since Arg z is unique, there is only one value of Ln z for each z 0.
EXAMPLE 4
Principal Values
The principal values of the logarithms in Example 2 are as follows:
(a) Since Arg(2) p, we need only set n 0 in the result given in part (a) of Example 2:
Ln(2) 0.6932 pi.
(b) Similarly, since Arg(i) p/2, we set n 0 in the result in part (b) of Example 2 to
obtain
Ln i p
i.
2
(c) In part (c) of Example 2, arg(1 i) 5p/4 is not the principal argument of z 1 i.
The argument of z that lies in the interval (p, p] is Arg(1 i) 3p/4. Hence, it follows
from (7) that
Ln(1 i) 0.3466 3p
i.
4
Up to this point we have avoided the use of the word function for the obvious reason that
ln z defined in (6) is not a function in the strictest interpretation of that word. Nonetheless, it is
customary to write f (z) ln z and to refer to f (z) ln z by the seemingly contradictory phrase
multiple-valued function. Although we shall not pursue the details, (6) can be interpreted as an
infinite collection of logarithmic functions (standard meaning of the word). Each function in the
collection is called a branch of ln z. The function f (z) Ln z is then called the principal branch
of ln z, or the principal logarithmic function. To minimize the confusion, we shall hereafter
simply use the words logarithmic function when referring to either f (z) ln z or f (z) Ln z .
17.6 Exponential and Logarithmic Functions
|
843
Some familiar properties of the logarithmic function hold in the complex case:
ln(z1z2) ⫽ ln z1 ⫹ ln z2
z1
ln a b ⫽ ln z1 ⫺ ln z2.
z2
and
(8)
Equations (8) and (9) are to be interpreted in the sense that if values are assigned to two of the
terms, then a correct value is assigned to the third term.
Properties of Logarithms
EXAMPLE 5
Suppose z1 ⫽ 1 and z2 ⫽ ⫺1. Then if we take ln z1 ⫽ 2pi and ln z2 ⫽ pi, we get
ln(z1z2) ⫽ ln(⫺1) ⫽ ln z1 ⫹ ln z2 ⫽ 2pi ⫹ pi ⫽ 3pi
z1
ln a b ⫽ ln(⫺1) ⫽ ln z1 ⫺ ln z2 ⫽ 2pi ⫺ pi ⫽ pi.
z2
Just as (7) of Section 17.2 was not valid when arg z was replaced with Arg z , so too (8) is
not true, in general, when ln z is replaced by Ln z. See Problems 45 and 46 in Exercises 17.6.
y
Analyticity The logarithmic function f (z) ⫽ Ln z is not continuous at z ⫽ 0 since f (0) is
not defined. Moreover, f (z) ⫽ Ln z is discontinuous at all points of the negative real axis. This
is because the imaginary part of the function, v ⫽ Arg z, is discontinuous only at these points.
To see this, suppose x0 is a point on the negative real axis. As z S x0 from the upper half-plane,
Arg z S p, whereas if z S x0 from the lower half-plane, then Arg z S ⫺p. This means that
f (z) ⫽ Ln z is not analytic on the nonpositive real axis. However, f (z) ⫽ Ln z is analytic throughout
the domain D consisting of all the points in the complex plane except those on the nonpositive
real axis. It is convenient to think of D as the complex plane from which the nonpositive real
axis has been cut out. Since f (z) ⫽ Ln z is the principal branch of ln z, the nonpositive real axis
is referred to as a branch cut for the function. See FIGURE 17.6.3. It is left as exercises to show
that the Cauchy–Riemann equations are satisfied throughout this cut plane and that the derivative of Ln z is given by
branch
cut
x
FIGURE 17.6.3 Branch cut for Ln z
y
d
1
Ln z ⫽
z
dz
for all z in D.
i
FIGURE 17.6.4 shows w ⫽ Ln z as a flow. Note that the vector field is not continuous along the
x
–i
(9)
branch cut.
Complex Powers Inspired by the identity xa ⫽ ea ln x in real variables, we can define com-
plex powers of a complex number. If a is a complex number and z ⫽ x ⫹ iy, then za is defined by
za ⫽ ea ln z,
z ⫽ 0.
(10)
a
FIGURE 17.6.4 w ⫽ Ln z as a flow
In general, z is multiple-valued since ln z is multiple-valued. However, in the special case when
a ⫽ n, n ⫽ 0, ⫾1, ⫾2, . . ., (10) is single-valued since there is only one value for z2, z3, z⫺1, and
so on. To see that this is so, suppose a ⫽ 2 and z ⫽ reiu, where u is any argument of z. Then
e2 ln z ⫽ e 2 (loger ⫹ iu) ⫽ e 2 loger ⫹ 2iu ⫽ e 2 loger e2iu ⫽ r 2 eiueiu ⫽ (reiu )(reiu ) ⫽ z2.
If we use Ln z in place of ln z, then (10) gives the principal value of z a.
Complex Power
EXAMPLE 6
2i
Find the value of i .
SOLUTION With z ⫽ i, arg z ⫽ p/2, and a ⫽ 2i , it follows from (10) that
i2i ⫽ e2i[loge1⫹i(p/2⫹2np)] ⫽ e⫺(1⫹4n)p
where n ⫽ 0, ⫾1, ⫾2, . . . . Inspection of the equation shows that i2i is real for every value of n.
Since p/2 is the principal argument of z ⫽ i, we obtain the principal value of i2i for n ⫽ 0. To
four rounded decimal places, this principal value is i 2i ⫽ e⫺p ⫽ 0.0432.
844
|
CHAPTER 17 Functions of a Complex Variable
Exercises
17.6
Answers to selected odd-numbered problems begin on page ANS-41.
In Problems 1–10, express e z in the form a ib.
p
p
1. z i
2. z i
6
3
p
p
3. z 1 i
4. z 2 2
i
4
2
3p
5. z p pi
6. z p i
2
7. z 1.5 2i
8. z 0.3 0.5i
9. z 5i
10. z 0.23 i
In Problems 35–38, find all values of z satisfying the given equation.
38. e2z ez 1 0
40. 3i/p
(1 i)
42. (1 "3i)3i
41. (1 i)
e 2 3pi
12. 3 pi>2
e
In Problems 43 and 44, find the principal value of the given
quantity. Express answers in the form a ib.
43. (1)(2i/p)
44. (1 i)2i
45. If z1 i and z2 1 i, verify that
iz
14. f (z) e
z2
16. f (z) e1/z
15. f (z) e
37. ez1 ie2
39. (i)4i
Ln(z1z2) Ln z1 Ln z2.
In Problems 13–16, use Definition 17.6.1 to express the given
function in the form f (z) u iv.
13. f (z) e
36. e1/z 1
In Problems 39–42, find all values of the given quantity.
In Problems 11 and 12, express the given number in the form a ib.
11. e15pi/4 e1pi/3
35. ez 4i
46. Find two complex numbers z1 and z2 such that
2z
Ln(z1/z2) Ln z1 Ln z2.
47. Determine whether the given statement is true.
In Problems 17–20, verify the given result.
e z1
17. |ez| ex
18. z2 e z1 2 z2
e
19. ezpi ezpi
20. (ez)n enz, n an integer
21. Show that f (z) e z is nowhere analytic.
2
22. (a) Use the result in2 Problem 15 to show that f (z) e z is an
entire function.
2
(b) Verify that u(x, y) Re(e z ) is a harmonic function.
In Problems 23–28, express ln z in the form a ib.
23. z 5
24. z ei
25. z 2 2i
26. z 1 i
27. z "2 "6i
28. z "3 i
In Problems 29–34, express Ln z in the form a ib.
29. z 6 6i
30. z e3
31. z 12 5i
32. z 3 4i
5
33. z (1 "3i)
34. z (1 i)4
17.7
(a) Ln(1 i)2 2 Ln(1 i)
(b) Ln i 3 3 Ln i
(c) ln i 3 3 ln i
48. The laws of exponents hold for complex numbers a and b:
zazb zab,
za
zab,
zb
(za)n zna,
n an integer.
However, the last law is not valid if n is a complex number.
Verify that (i i)2 i 2i, but (i 2)i i 2i.
49. For complex numbers z satisfying Re(z)
0, show that (7)
can be written as
Ln z y
1
loge(x2 y2) i tan1 .
x
2
50. The function given in Problem 49 is analytic.
(a) Verify that u(x, y) loge(x2 y2) is a harmonic function.
(b) Verify that v(x, y) tan1( y/x) is a harmonic function.
Trigonometric and Hyperbolic Functions
INTRODUCTION In this section we define the complex trigonometric and hyperbolic func-
tions. Analogous to the complex functions ez and Ln z defined in the previous section, these
functions will agree with their real counterparts for real values of z. In addition, we will show
that the complex trigonometric and hyperbolic functions have the same derivatives and satisfy
many of the same identities as the real trigonometric and hyperbolic functions.
Trigonometric Functions If x is a real variable, then Euler’s formula gives
eix cos x i sin x
and
eix cos x i sin x.
By subtracting and then adding these equations, we see that the real functions sin x and cos x can
be expressed as a combination of exponential functions:
sin x e ix 2 eix
,
2i
cos x e ix eix
.
2
17.7 Trigonometric and Hyperbolic Functions
(1)
|
845
Using (1) as a model, we now define the sine and cosine of a complex variable:
Definition 17.7.1
Trigonometric Sine and Cosine
For any complex number z x iy,
sin z e iz 2 eiz
2i
and
cos z e iz eiz
.
2
(2)
As in trigonometry, we define four additional trigonometric functions in terms of sin z
and cos z:
sin z
1
1
1
, cot z , csc z , sec z .
cos z
cos
z
tan z
sin z
tan z (3)
When y 0, each function in (2) and (3) reduces to its real counterpart.
Analyticity Since the exponential functions eiz and eiz are entire functions, it follows
that sin z and cos z are entire functions. Now, as we shall see shortly, sin z 0 only for the real
numbers z np, n an integer, and cos z 0 only for the real numbers z (2n 1)p/2, n an
integer. Thus, tan z and sec z are analytic except at the points z (2n 1)p/2, and cot z and
csc z are analytic except at the points z np.
Derivatives Since (d/dz)ez ez, it follows from the Chain Rule that (d/dz)eiz ieiz and
(d /dz)eiz ieiz. Hence,
d
d e iz 2 eiz
e iz eiz
sin z cos z.
dz
dz
2i
2
In fact, it is readily shown that the forms of the derivatives of the complex trigonometric functions are the same as the real functions. We summarize the results:
d
sin z cos z
dz
d
cos z sin z
dz
d
tan z sec 2z
dz
d
cot z csc 2z
dz
d
sec z sec z tan z
dz
d
csc z csc z cot z.
dz
(4)
Identities The familiar trigonometric identities are also the same in the complex case:
sin(z) sin z
cos(z) cos z
cos2z sin2z 1
sin(z1
z2) sin z1 cos z2
cos(z1
z2) cos z1 cos z2 sin z1 sin z2
sin 2z 2 sin z cos z
cos z1 sin z2
cos 2z cos2z sin2z.
Zeros To find the zeros of sin z and cos z we need to express both functions in the form
u iv. Before proceeding, recall from calculus that if y is real, then the hyperbolic sine and
hyperbolic cosine are defined in terms of the real exponential functions ey and ey:
sinh y e y 2 ey
2
and
cosh y e y ey
.
2
(5)
Now from Definition 17.7.1 and Euler’s formula we find, after simplifying,
sin z e i(x iy) 2 ei(x iy)
e y ey
e y 2 ey
sin x a
b i cos x a
b.
2i
2
2
Thus from (5) we have
sin z sin x cosh y i cos x sinh y.
846
|
CHAPTER 17 Functions of a Complex Variable
(6)
It is left as an exercise to show that
cos z cos x cosh y i sin x sinh y.
(7)
From (6), (7), and cosh2y 1 sinh2y, we find
|sin z|2 sin2 x sinh2y
2
2
(8)
2
|cos z| cos x sinh y.
and
(9)
Now a complex number z is zero if and only if |z|2 0. Thus, if sin z 0, then from (8) we must
have sin2 x sinh2y 0. This implies that sin x 0 and sinh y 0, and so x np and y 0.
Thus the only zeros of sin z are the real numbers z np 0i np, n 0, 1, 2, . . . . Similarly,
it follows from (9) that cos z 0 only when z (2n 1)p/2, n 0, 1, 2, . . . .
EXAMPLE 1
Complex Value of the Sine Function
From (6) we have, with the aid of a calculator,
sin(2 i) sin 2 cosh 1 i cos 2 sinh 1 1.4031 0.4891i.
In ordinary trigonometry we are accustomed to the fact that |sin x| 1 and |cos x| 1. Inspection
of (8) and (9) shows that these inequalities do not hold for the complex sine and cosine, since
sinh y can range from q to q. In other words, it is perfectly feasible to have solutions for
equations such as cos z 10.
EXAMPLE 2
Solving a Trigonometric Equation
Solve the equation cos z 10.
SOLUTION From (2), cos z 10 is equivalent to (eiz eiz)/2 10. Multiplying the last
equation by eiz then gives the quadratic equation in eiz:
e2iz 20e iz 1 0.
From the quadratic formula we find eiz 10 3 !11. Thus, for n 0, 1, 2, … , we
have iz loge(10
3 !11) 2npi. Dividing by i and utilizing loge(10 3 !11) loge(10 3 "11), we can express the solutions of the given equation as z 2np
i loge(10 3 "11).
Hyperbolic Functions We define the complex hyperbolic sine and cosine in a manner
analogous to the real definitions given in (5).
Definition 17.7.2
Hyperbolic Sine and Cosine
For any complex number z x iy,
sinh z e z 2 ez
2
and
cosh z e z ez
.
2
(10)
The hyperbolic tangent, cotangent, secant, and cosecant functions are defined in terms of
sinh z and cosh z:
tanh z sinh z
1
1
1
, coth z , sech z , csch z .
cosh z
tanh z
cosh z
sinh z
(11)
The hyperbolic sine and cosine are entire functions, and the functions defined in (11) are analytic except at points where the denominators are zero. It is also easy to see from (10) that
d
sinh z cosh z
dz
and
d
cosh z sinh z .
dz
(12)
It is interesting to observe that, in contrast to real calculus, the trigonometric and hyperbolic functions are related in complex calculus. If we replace z by iz everywhere in (10) and compare the results
with (2), we see that sinh(iz) i sin z and cosh(iz) cos z. These equations enable us to express
17.7 Trigonometric and Hyperbolic Functions
|
847
sin z and cos z in terms of sinh(iz) and cosh(iz), respectively. Similarly, by replacing z by iz in (2) we
can express, in turn, sinh z and cosh z in terms of sin(iz) and cos(iz). We summarize the results:
sin z i sinh(iz),
sinh z i sin(iz),
cos z cosh(iz)
(13)
cosh z cos(iz).
(14)
Zeros The relationships given in (14) enable us to derive identities for the hyperbolic functions utilizing results for the trigonometric functions. For example, to express sinh z in the form
u iv we write sinh z i sin(iz) in the form sinh z i sin(y ix) and use (6):
sinh z i [sin(y) cosh x i cos(y) sinh x].
Since sin(y) sin y and cos(y) cos y, the foregoing expression simplifies to
Similarly,
sinh z sinh x cos y i cosh x sin y.
(15)
cosh z cosh x cos y i sinh x sin y.
(16)
It also follows directly from (14) that the zeros of sinh z and cosh z are pure imaginary and are,
respectively,
pi
z npi and z (2n 1) ,
n 0, 1, 2, . . . .
2
Periodicity Since sin x and cos x are 2p-periodic, we can easily demonstrate that sin z
and cos z are also periodic with the same real period 2p. For example, from (6), note that
sin(z 2p) sin(x 2p iy)
sin(x 2p) cosh y i cos(x 2p) sinh y
sin x cosh y i cos x sinh y;
that is, sin(z 2p) sin z. In exactly the same manner, it follows from (7) that cos(z 2p) cos z.
In addition, the hyperbolic functions sinh z and cosh z have the imaginary period 2pi. This last
result follows from either Definition 17.7.2 and the fact that ez is periodic with period 2pi, or
from (15) and (16) and replacing z by z 2pi.
Exercises
17.7
Answers to selected odd-numbered problems begin on page ANS-41.
In Problems 1–12, express the given quantity in the form a ib.
1. cos(3i)
2. sin(2i)
p
3. sina ib
4
4. cos(2 4i)
p
3ib
2
8. csc(1 i)
3p
10. sinha
ib
2
5. tan(i)
6. cota
7. sec(p i)
9. cosh(pi)
11. sinha1 p
ib
3
12. cosh(2 3i)
In Problems 13 and 14, verify the given result.
17. sinh z i
19. cos z sin z
18. sinh z 1
20. cos z i sin z
In Problems 21 and 22, use the definition of equality of complex
numbers to find all values of z satisfying the given equation.
21. cos z cosh 2
22. sin z i sinh 2
23. Prove that cos z cos x cosh y i sin x sinh y.
24. Prove that sinh z sinh x cos y i cosh x sin y.
25. Prove that cosh z cosh x cos y i sinh x sin y.
26. Prove that |sinh z|2 sin2y sinh2 x.
27. Prove that |cosh z|2 cos2y sinh2 x.
28. Prove that cos2z sin2z 1.
p
5
i ln 2b 2
4
p
3
14. cosa i ln 2b i
2
4
29. Prove that cosh2z sinh2z 1.
In Problems 15–20, find all values of z satisfying the given
equation.
31. Prove that tanh z is periodic with period pi.
13. sina
15. sin z 2
848
|
30. Show that tan z u iv, where
u
16. cos z 3i
CHAPTER 17 Functions of a Complex Variable
sinh 2y
sin 2x
and v .
cos 2x cosh 2y
cos 2x cosh 2y
32. Prove that (a) sin z sin z and (b) cos z cos z.
17.8
Inverse Trigonometric and Hyperbolic Functions
INTRODUCTION As functions of a complex variable z, we have seen that both the trigonometric and hyperbolic functions are periodic. Consequently, these functions do not possess inverses
that are functions in the strictest interpretation of that word. The inverses of these analytic functions
are multiple-valued functions. As we did in Section 17.6, in the examination of the logarithmic
function, we shall drop the adjective multiple-valued throughout the discussion that follows.
Inverse Sine The inverse sine function, written as sin1z or arcsin z, is defined by
w sin1z
z sin w.
if
(1)
The inverse sine can be expressed in terms of the logarithmic function. To see this we use (1)
and the definition of the sine function:
e iw 2 eiw
z
2i
or
e2iw 2izeiw 1 0.
From the last equation and the quadratic formula, we then obtain
eiw iz (1 z2)1/2.
(2)
2
Note in (2) we did not use the customary symbolism "1 2 z , since we know from Section 17.2
that (1 z2)1/2 is two-valued. Solving (2) for w then gives
sin1z i ln[iz (1 z2)1/2].
(3)
Proceeding in a similar manner, we find the inverses of the cosine and tangent to be
cos1z i ln[z i(1 z2)1/2]
(4)
i iz
ln
.
2 i2z
(5)
tan1z EXAMPLE 1
Values of an Inverse Sine
Find all values of sin1 "5.
SOLUTION From (3) we have
sin 1 "5 i ln f "5i (1 2 ( "5 )2 )1>2g.
With (1 ( "5)2)1/2 (4)1/2 sin1 "5 i ln f("5
i clog e("5
2i, the preceding expression becomes
2)ig
2) a
p
2npbid , n 0,
2
1,
2, . . . .
The foregoing result can be simplified a little by noting that loge( "5 2) loge(1/( "5 2)) loge( "5 2). Thus for n 0, 1, 2, . . .,
sin1 "5 p
2np
2
i log e("5 2).
(6)
To obtain particular values of, say, sin1z, we must choose a specific root of 1 z2 and a
specific branch of the logarithm. For example, if we choose (1 ( "5)2)1/2 (4)1/2 2i and
the principal branch of the logarithm, then (6) gives the single value
sin1 "5 p
2 i log e("5 2).
2
17.8 Inverse Trigonometric and Hyperbolic Functions
|
849
Derivatives The derivatives of the three inverse trigonometric functions considered above
can be found by implicit differentiation. To find the derivative of the inverse sine function
w sin1z, we begin by differentiating z sin w:
d
d
z
sin w
dz
dz
gives
dw
1
.
cos w
dz
Using the trigonometric identity cos2w sin2w 1 (see Problem 28 in Exercises 17.7) in the
form cos w (1 sin2w)1/2 (1 z2)1/2, we obtain
d
1
sin1z .
dz
(1 2 z 2 )1>2
(7)
d
1
cos 1z dz
(1 2 z 2 )1>2
(8)
d
1
tan 1z .
dz
1 z2
(9)
Similarly, we find that
It should be noted that the square roots used in (7) and (8) must be consistent with the square
roots used in (3) and (4).
Evaluating a Derivative
EXAMPLE 2
Find the derivative of w sin1z at z "5.
SOLUTION In Example 1, if we use (1 ( "5)2)1/2 (4)1/2 2i, then that same root must
be used in (7). The value of the derivative consistent with this choice is given by
dw
1
1
1
1
2
i.
1>2
2 1>2
dz z "5
2i
2
(4)
(1 2 ("5) )
Inverse Hyperbolic Functions The inverse hyperbolic functions can also be expressed
in terms of the logarithm. We summarize these results for the inverse hyperbolic sine, cosine,
and tangent along with their derivatives:
sinh1z ln fz (z 2 1)1>2g
(10)
cosh1z ln fz (z 2 2 1)1>2g
(11)
1 1z
ln
2 12z
(12)
d
1
sinh1z 2
dz
(z 1)1>2
(13)
d
1
cosh1z 2
dz
(z 2 1)1>2
(14)
d
1
tanh1z .
dz
1 2 z2
(15)
tanh1z EXAMPLE 3
Values of an Inverse Hyperbolic Cosine
Find all values of cosh1(1).
SOLUTION From (11) with z 1, we get
cosh1(1) ln(1) loge1 (p 2np)i.
Since loge1 0 we have for n 0,
1,
2, . . .,
cosh1(1) (2n 1)pi.
850
|
CHAPTER 17 Functions of a Complex Variable
Exercises
17.8
Answers to selected odd-numbered problems begin on page ANS-41.
In Problems 1–14, find all values of the given quantity.
1. sin1(i)
3. sin1 0
5. cos1 2
2. sin1 "2
4. sin1 13
5
6. cos1 2i
Chapter in Review
17
7. cos1 12
9. tan1 1
11. sinh1 43
8. cos1 53
10. tan1 3i
12. cosh1 i
13. tanh1(1 2i)
14. tanh1( "3i)
Answers to selected odd-numbered problems begin on page ANS-41.
Answer Problems 1–16 without referring back to the text. Fill in
the blank or answer true/false.
26. Let z and w be complex numbers such that |z| 1 and |w| 1.
Prove that
1. Re(1 i )10 _____ and Im(1 i )10 _____.
2. If z is a point in the third quadrant, then iz is in the _____
quadrant.
z
z
i127 5i 9 2i1 _____
4i
If z , then |z| _____.
3 2 4i
Describe the region defined by 1 |z 2| 3. _____
Arg(z z) 0 _____
5
, then Arg z _____.
If z "3 i
If ez 2i, then z _____.
If |ez| 1, then z is a pure imaginary number. _____
The principal value of (1 i)(2i) is _____.
If f (z) x 2 3xy 5y 3 i(4x 2y 4x 7y), then
f (1 2i) _____.
If the Cauchy–Riemann equations are satisfied at a point, then
the function is necessarily analytic there. _____
f (z) e z is periodic with period _____.
Ln(ie3) _____
f (z) sin(x iy) is nowhere analytic. _____
3. If z 3 4i, then Re a b _____.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
In Problems 17–20, write the given number in the form a ib.
17. i (2 3i)2 (4 2i)
19.
(1 2 i)10
(1 i)3
18.
32i
2 2 2i
2 3i
1 5i
21. Im(z ) 2
22. Im(z 5i) 3
1
23.
1
24. Im(z) Re(z)
ZzZ
25. Look up the definitions of conic sections in a calculus text.
Now describe the set of points in the complex plane that satisfy the equation |z 2i| |z 2i| 5.
z2w
2 1.
1 2 zw
In Problems 27 and 28, find all solutions of the given equation.
1
22i
1i
29. If f (z) z24 3z20 4z12 5z6, find f a
b.
"2
30. Write f (z) Im(z 3z) z Re(z 2 ) 5z in the form
f (z) u(x, y) iv(x, y).
27. z4 1 i
28. z 3>2 In Problems 31 and 32, find the image of the line x 1 in the
w-plane under the given mapping.
31. f (z) x2 y i ( y2 x)
32. f (z) 1
z
In Problems 33–36, find all complex numbers for which the
given statement is true.
33. z z1
35. z z
1
z
2
36. z ( z )2
34. z 37. Show that the function f (z) (2xy 5x) i(x2 5y y2)
is analytic for all z. Find f (z).
38. Determine whether the function
f (z) x3 xy2 4x i(4y y3 x2y)
20. 4epi/3 epi/4
In Problems 21–24, sketch the set of points in the complex plane
satisfying the given inequality.
2
2
is differentiable. Is it analytic?
In Problems 39 and 40, verify the given equality.
39. Ln[(1 i)(1 i)] Ln(1 i) Ln(1 i)
40. Lna
1i
b Ln(1 i) Ln(1 i)
12i
CHAPTER 17 in Review
|
851
© hofhauser/Shutterstock, Inc.
CHAPTER
18
To define an integral of a
complex function f, we start with
f defined along some curve C or
contour in the complex plane. We
will see in this chapter that the
definition of a complex integral,
its properties, and method of
evaluation are quite similar to
those of a real line integral in the
plane (Section 9.8).
Integration in the
Complex Plane
CHAPTER CONTENTS
18.1
18.2
18.3
18.4
Contour Integrals
Cauchy–Goursat Theorem
Independence of the Path
Cauchy’s Integral Formulas
Chapter 18 in Review
18.1
Contour Integrals
INTRODUCTION In Section 9.8 we saw that the definition of the definite integral eab f (x) dx
starts with a real function y f (x) that is defined on an interval [a, b] on the x-axis. Because a
planar curve is the two-dimensional analogue of an interval, we then generalized the definition
of the definite integral to integrals of real functions of two variables defined on a curve C in the
Cartesian plane. We shall see in this section that a complex integral is defined in a manner that
is quite similar to that of a line integral in the Cartesian plane. In case you have not studied
Sections 9.8 and 9.9, a review of those sections is recommended.
z*n
zn –1
zn
C
z0
z*1
z1
z*2
z2
FIGURE 18.1.1 Sample points are the
red dots
A Definition Integration in the complex plane is defined in a manner similar to that of a
line integral in the plane. In other words, we shall be dealing with an integral of a complex function f (z) that is defined along a curve C in the complex plane. These curves are defined in terms
of parametric equations x x(t), y y(t), a t b, where t is a real parameter. By using x(t) and
y(t) as real and imaginary parts, we can also describe a curve C in the complex plane by means of
a complex-valued function of a real variable t: z(t) x(t) iy(t), a t b. For example, x cos t,
y sin t, 0 t 2p, describes a unit circle centered at the origin. This circle can also be described
by z(t) cos t i sin t, or even more compactly by z(t) eit, 0 t 2p. The same definitions
of smooth curve, piecewise-smooth curve, closed curve, and simple closed curve given in Section
9.8 carry over to this discussion. As before, we shall assume that the positive direction on C corresponds to increasing values of t. In complex variables, a piecewise-smooth curve C is also called
a contour or path. An integral of f (z) on C is denoted by C f (z) dz or C f (z) dz if the contour C
is closed; it is referred to as a contour integral or simply as a complex integral.
1. Let f (z) u(x, y) iv(x, y) be defined at all points on a smooth curve C defined by x x(t),
y y(t), a t b.
2. Divide C into n subarcs according to the partition a t0 t1 . . . tn b of [a, b].
The corresponding points on the curve C are z0 x0 iy0 x(t0) iy(t0), z1 x1 iy1 x(t1) iy(t1), . . ., zn xn iyn x(tn) iy(tn). Let zk zk zk1, k 1, 2, . . ., n.
3. Let P be the norm of the partition, that is, the maximum value of |zk|.
4. Choose a sample point z*k 5 x*k 1 iy*k on each subarc. See FIGURE 18.1.1.
5. Form the sum a f (z *k ) zk.
n
k1
Definition 18.1.1
Contour Integral
Let f be defined at points of a smooth curve C defined by x x(t), y y(t), a t b. The
contour integral of f along C is
# f (z) dz C
lim a f (z *k ) Dzk.
iPiS0
n
(1)
k1
The limit in (1) exists if f is continuous at all points on C and C is either smooth or piecewise
smooth. Consequently we shall, hereafter, assume these conditions as a matter of course.
Method of Evaluation We shall turn now to the question of evaluating a contour
integral. To facilitate the discussion, let us suppress the subscripts and write (1) in the
abbreviated form
# f (z) dz lim (u iv)(x i y)
C
lim [(ux vy) i (vx uy)].
This means
# f (z) dz # u dx v dy i # v dx u dy.
C
854
|
CHAPTER 18 Integration in the Complex Plane
C
C
(2)
In other words, a contour integral C f (z) dz is a combination of two real-line integrals C u dx v dy
and C v dx u dy. Now, since x x(t) and y y(t), a t b, the right side of (2) is the
same as
#
b
a
b
# [v(x(t), y(t))x (t) u(x(t), y(t))y (t)] dt.
[u(x(t), y(t))x (t) v(x(t), y(t))y (t)] dt i
a
But if we use z(t) x(t) iy(t) to describe C, the last result is the same as ba f (z(t))z (t) dt
when separated into two integrals. Thus we arrive at a practical means of evaluating a contour
integral:
Theorem 18.1.1
Evaluation of a Contour Integral
If f is continuous on a smooth curve C given by z(t) x(t) iy(t), a t b, then
#
C
f (z) dz #
b
f (z(t)) z9(t) dt.
(3)
a
If f is expressed in terms of the symbol z, then to evaluate f (z(t)) we simply replace the symbol
z by z(t). If f is not expressed in terms of z, then to evaluate f (z(t)), we replace x and y wherever
they appear by x(t) and y(t), respectively.
Evaluating a Contour Integral
EXAMPLE 1
Evaluate
# z dz, where C is given by x 3t, y t , 1 t 4.
2
C
SOLUTION We write z(t) 3t it 2 so that z (t) 3 2it and f (z(t)) 3t 1 it 2 3t it 2.
Thus,
#
C
z dz #
4
#
4
(3t it 2)(3 2it) dt
21
(2t 3 9t) dt i
21
BC
䉲
3t 2 dt 195 65i.
21
Evaluating a Contour Integral
EXAMPLE 2
Evaluate
#
4
1
dz, where C is the circle x cos t, y sin t, 0 t 2p.
z
SOLUTION In this case z(t) cos t i sin t eit, z (t) ieit, and f (z) 1/z eit. Hence,
BC
䉲
1
dz z
#
2p
0
(e it )ie itdt i
#
2p
0
dt 2pi.
For some curves, the real variable x itself can be used as the parameter. For example, to evaluate C (8x2 iy) dz on y 5x, 0 x 2, we write C (8x2 iy) dz 20 (8x2 5ix)(1 5i) dx
and integrate in the usual manner.
Properties The following properties of contour integrals are analogous to the properties
of line integrals.
18.1 Contour Integrals
|
855
Properties of Contour Integrals
Theorem 18.1.2
Suppose f and g are continuous in a domain D and C is a smooth curve lying entirely in D. Then
# kf (z) dz k # f (z) dz, k a constant
(ii) # [ f (z) g(z)] dz # f (z) dz # g(z) dz
(iii) # f (z) dz # f (z) dz # f (z) dz, where C is the union of the smooth curves C and C
(iv) # f (z) dz # f (z) dz, where C denotes the curve having the opposite orientation of C.
(i)
C
C
C
C
C
C
C1
1
C2
2
C
2C
The four parts of Theorem 18.1.2 also hold when C is a piecewise-smooth curve in D.
y
Evaluating a Contour Integral
EXAMPLE 3
2
2
1 + 2i
Evaluate C (x iy ) dz, where C is the contour shown in FIGURE 18.1.2.
C2
SOLUTION In view of Theorem 18.1.2(iii) we write
1+i
# (x iy ) dz #
2
C1
2
C
x
C1
(x2 iy2) dz #
C2
(x2 iy2) dz.
Since the curve C1 is defined by y x, it makes sense to use x as a parameter. Therefore,
z(x) x ix, z (x) 1 i, f (z(x)) x 2 ix 2, and
#
FIGURE 18.1.2 Piecewise-smooth contour
in Example 3
C1
1
(x2 iy2) dz # (x ix )(1 i) dx
2
2
0
#
1
(1 i)2 x 2dx 0
(1 i)2
2
i.
3
3
The curve C2 is defined by x 1, 1 y 2. Using y as a parameter, we have z( y) 1 iy,
z ( y) i, and f (z( y)) 1 iy2. Thus,
#
C2
(x2 iy2) dz #
2
1
(1 iy2) i dy #
2
1
y2 dy i
#
2
1
7
dy i.
3
Finally, we have eC (x 2 iy 2) dz 23– i (73– i) 73– 53– i.
There are times in the application of complex integration that it is useful to find an upper
bound for the absolute value of a contour integral. In the next theorem we shall use the fact that
b
the length of a plane curve is s ea "fx9(t)g 2 fy9(t)g 2 dt. But if z (t) x (t) iy (t), then
|z (t)| "fx9(t)g 2 fy9(t)g 2 and consequently s ab |z (t)| dt.
Theorem 18.1.3
A Bounding Theorem
If f is continuous on a smooth curve C and if | f (z)| M for all z on C, then Z eC f (z) dz Z # ML ,
where L is the length of C.
PROOF: From the triangle inequality (6) of Section 17.1 we can write
2 a f (z *k )Dzk 2 # a Z f (z *k )Z ZDzk Z # M a ZDzk Z.
856
|
CHAPTER 18 Integration in the Complex Plane
n
n
n
k1
k1
k1
(4)
Now, |zk| can be interpreted as the length of the chord joining the points zk and zk1. Since the
sum of the lengths of the chords cannot be greater than the length of C, (4) becomes
|nk1 f (z*k ) zk| ML. Hence, as ||P|| S 0, the last inequality yields |C f (z) dz| ML.
Theorem 18.1.3 is used often in the theory of complex integration and is sometimes referred
to as the ML-inequality.
A Bound for a Contour Integral
EXAMPLE 4
Find an upper bound for the absolute value of
ez
dz, where C is the circle |z| 4.
BC z 1
䉲
SOLUTION First, the length s of the circle of radius 4 is 8p. Next, from the inequality (7) of
Section 17.1, it follows that |z 1| |z| 1 4 1 3, and so
2
Zez Z
Zez Z
ez
2#
5
.
z11
ZzZ 2 1
3
(5)
In addition, |ez| |ex (cos y i sin y)| ex. For points on the circle |z| 4, the maximum that
x can be is 4, and so (5) becomes
2
ez
e4
2# .
z11
3
Hence from Theorem 18.1.3 we have
2
ez
8p e 4
dz 2 #
.
3
BC z 1
䉲
Circulation and Net Flux Let T and N denote the unit tangent vector and the unit
normal vector to a positively oriented simple closed contour C. When we interpret the complex
function f (z) u(x, y) iv(x, y) as a vector, the line integrals
BC
䉲
BC
and
䉲
f T ds f N ds BC
䉲
u dx v dy
(6)
BC
u dy 2 v dx
(7)
䉲
have special interpretations. The line integral in (6) is called the circulation around C and measures the tendency of the flow to rotate the curve C. See Section 9.8 for the derivation. The net
flux across C is the difference between the rate at which fluid enters and the rate at which fluid
leaves the region bounded by C. The net flux across C is given by the line integral in (7), and a
nonzero value for C f N ds indicates the presence of sources or sinks for the fluid inside the
curve C. Note that
䉲
and so
a
BC
䉲
f T dsb i a
BC
䉲
f N dsb BC
䉲
(u 2 iv)(dx i dy) circulation Re a
net flux Ima
BC
䉲
f (z) dz
BC
䉲
f (z) dzb
(8)
BC
䉲
f (z) dzb.
(9)
Thus, both of these key quantities may be found by computing a single complex integral.
18.1 Contour Integrals
|
857
y
Net Flux
EXAMPLE 5
Given the flow f (z) (1 i)z, compute the circulation around, and the net flux across, the
circle C: |z| 1.
SOLUTION
BC
x
C
䉲
Since f (z) (1 i)z and z(t) eit, 0 t 2p, we have
f (z) dz #
2p
0
(1 2 i) eitie it dt (1 i)
#
2p
0
dt 2p(1 i) 2p 2pi.
Using (8) and (9), the circulation around C is 2p and the net flux across C is 2p. See
FIGURE 18.1.3.
FIGURE 18.1.3 Flow f (z) (1 i)z
Exercises
18.1
Answers to selected odd-numbered problems begin on page ANS-41.
In Problems 1–16, evaluate the given integral along the
indicated contour.
19.
1. C (z 3) dz, where C is x 2t, y 4t 1, 1 t 3
BC
䉲
z 2 dz
20.
BC
䉲
z 2 dz
y
2. C (2z z) dz, where C is x t, y t 2 2, 0 t 2
1+i
3. C z 2 dz, where C is z(t) 3t 2it, 2 t 2
4. C (3z 2 2z) dz, where C is z(t) t it 2, 0 t 1
1 1 z dz, where C is the right half of the circle |z| 1
z
from z i to z i
5. C
x
1
6. C |z|2 dz, where C is x t 2, y 1/t, 1 t 2
7. C Re(z) dz, where C is the circle |z| 1
䉲
8.
9.
10.
11.
12.
13.
14.
15.
16.
1
5
2
8b dz, where C is the circle | z i | 1,
C a
3
z
i
(z i)
0 t 2p
C (x2 iy3) dz, where C is the straight line from z 1 to z i
C (x3 iy3) dz, where C is the lower half of the circle |z| 1
from z 1 to z 1
C ez dz, where C is the polygonal path consisting of the line
segments from z 0 to z 2 and from z 2 to z 1 pi
C sin z dz, where C is the polygonal path consisting of the line
segments from z 0 to z 1 and from z 1 to z 1 i
C Im(z i) dz, where C is the polygonal path consisting of
the circular arc along |z| 1 from z 1 to z i and the line
segment from z i to z 1
C dz, where C is the left half of the ellipse x 2 /36 y 2 /4 1
from z 2i to z 2i
C zez dz, where C is the square with vertices z 0, z 1,
z 1 i, and z i
2,
x,0
C f (z) dz, where f (z) e
and C is the parabola
6x, x . 0
y x 2 from z 1 i to z 1 i
䉲
In Problems 21–24, evaluate C (z2 z 2) dz from i to 1 along
the indicated contours.
21.
y
22.
y
i
i
x
x
1
1
FIGURE 18.1.6 Contour in
Problem 22
FIGURE 18.1.5 Contour in
Problem 21
23.
1+i
24.
y
y
䉲
In Problems 17–20, evaluate the given integral along the
contour C given in FIGURE 18.1.4.
17.
FIGURE 18.1.4 Contour in Problems 17–20
BC
䉲
x dz
858
|
18.
BC
䉲
(2z 2 1) dz
CHAPTER 18 Integration in the Complex Plane
i
i
y = 1 – x2
x2 + y2 = 1
x
1
FIGURE 18.1.7 Contour in
Problem 23
x
1
FIGURE 18.1.8 Contour in
Problem 24
In Problems 25–28, find an upper bound for the absolute value
of the given integral along the indicated contour.
30. Use Definition 18.1.1 to show for any smooth curve C between
z0 and zn that C z dz 12 (z n2 z02). [Hint: The integral exists,
so choose z*k zk and z*k zk1.]
31. Use the results of Problems 29 and 30 to evaluate C (6z 4) dz
where C is
(a) the straight line from 1 i to 2 3i, and
(b) the closed contour x4 y4 4.
ez
dz, where C is the circle |z| 5
BC z 2 1
1
dz, where C is the right half of the circle |z| 6
26.
2
z
2
2i
C
from z 6i to z 6i
25.
䉲
䉲
#
In Problems 32–35, compute the circulation and net flux for the
given flow and the indicated closed contour.
27. C (z2 4) dz, where C is the line segment from z 0 to
z1i
1
28.
dz, where C is one quarter of the circle |z| 4 from
3
z
C
z 4i to z 4
32. f (z) 1>z, where C is the circle |z| 2
33. f (z) 2z, where C is the circle |z| 1
#
34. f (z) 1>(z 2 1), where C is the circle |z 1| 2
35. f (z) z, where C is the square with vertices z 0, z 1,
29. (a) Use Definition 18.1.1 to show for any smooth curve C
z 1 i, z i
between z0 and zn that C dz zn z0.
(b) Use the result in part (a) to verify the answer to Problem 14.
18.2 Cauchy–Goursat Theorem
INTRODUCTION In this section we shall concentrate on contour integrals where the contour
C is a simple closed curve with a positive (counterclockwise) orientation. Specifically, we shall
see that when f is analytic in a special kind of domain D, the value of the contour integral C f (z) dz
is the same for any simple closed curve C that lies entirely within D. This theorem, called the
Cauchy–Goursat theorem, is one of the fundamental results in complex analysis. Preliminary
to discussing the Cauchy–Goursat theorem and some of its ramifications, we first need to distinguish two kinds of domains in the complex plane: simply connected and multiply connected.
䉲
D
(a) Simply connected domain
D
(b) Multiply connected domain
FIGURE 18.2.1 Two kinds of domains
Simply and Multiply Connected Domains In the discussion that follows, we shall
concentrate on contour integrals where the contour C is a simple closed curve with a positive
(counterclockwise) orientation. Before doing this, we need to distinguish two kinds of domains.
A domain D is said to be simply connected if every simple closed contour C lying entirely in D
can be shrunk to a point without leaving D. In other words, in a simply connected domain, every
simple closed contour C lying entirely within it encloses only points of the domain D. Expressed
yet another way, a simply connected domain has no “holes” in it. The entire complex plane is an
example of a simply connected domain. A domain that is not simply connected is called a multiply
connected domain; that is, a multiply connected domain has “holes” in it. See FIGURE 18.2.1. As
in Section 9.9, we call a domain with one “hole” doubly connected, a domain with two “holes”
triply connected, and so on.
Cauchy’s Theorem In 1825, the French mathematician Louis-Augustin Cauchy proved
one of the most important theorems in complex analysis. Cauchy’s theorem says:
Suppose that a function f is analytic in a simply connected domain D and that f is
continuous in D. Then for every simple closed contour C in D, C f (z) dz 0.
䉲
The proof of this theorem is an immediate consequence of Green’s theorem and the Cauchy–Riemann
equations. Since f is continuous throughout D, the real and imaginary parts of f (z) u iv and
their first partial derivatives are continuous throughout D. By (2) of Section 18.1 we write C f (z) dz
in terms of real-line integrals and use Green’s theorem on each line integral:
䉲
BC
䉲
f (z) dz BC
䉲
u(x, y) dx v(x, y) dy i
BC
䉲
v(x, y) dx u(x, y) dy
0v
0u
0u
0v
6 a 2 b dA i 6 a
2 b dA.
0x
0y
0x
0y
D
(1)
D
18.2 Cauchy–Goursat Theorem
|
859
Now since f is analytic, the Cauchy–Riemann equations, u/ x v/ y and u/ y v/ x,
imply that the integrands in (1) are identically zero. Hence, we have C f (z) dz 0.
In 1883, the French mathematician Edouard Goursat (1858–1936) proved Cauchy’s theorem
without the assumption of continuity of f . The resulting modified version of Cauchy’s theorem
is known as the Cauchy–Goursat theorem.
䉲
Cauchy–Goursat Theorem
Theorem 18.2.1
Suppose a function f is analytic in a simply connected domain D. Then for every simple closed
contour C in D, C f (z) dz 0.
䉲
Since the interior of a simple closed contour is a simply connected domain, the Cauchy–Goursat
theorem can be stated in the slightly more practical manner:
y
If f is analytic at all points within and on a simple closed contour C, then
C f (z) dz 0.
䉲
C
Applying the Cauchy–Goursat Theorem
EXAMPLE 1
Evaluate
x
BC
䉲
ez dz, where C is the curve shown in FIGURE 18.2.2.
SOLUTION The function f (z) ez is entire and C is a simple closed contour. It follows from
the form of the Cauchy–Goursat theorem given in (2) that C ez dz 0.
FIGURE 18.2.2 Contour in Example 1
䉲
y
EXAMPLE 2
i
Evaluate
x
1
–1
–i
Applying the Cauchy–Goursat Theorem
( y 2 5)2
dz
2
,
where
C
is
the
ellipse
(x
2)
1.
4
BC z 2
䉲
SOLUTION The rational function f (z) 1/z2 is analytic everywhere except at z 0. But z 0
is not a point interior to or on the contour C. Thus, from (2) we have C dz/z2 0.
䉲
EXAMPLE 3
FIGURE 18.2.3 Flow f (z) cos z
(2)
Applying the Cauchy–Goursat Theorem
Given the flow f (z) cos z, compute the circulation around and net flux across C, where C
is the square with vertices z 1, z i, z 1, and z i.
SOLUTION We must compute C f (z) dz C cos z dz and then take the real and imaginary
parts of the integral to find the circulation and net flux, respectively. The function cos z is
analytic everywhere, and so C f (z) dz 0 from (2). The circulation and net flux are therefore
both 0. FIGURE 18.2.3 shows the flow f (z) cos z and the contour C.
䉲
䉲
䉲
C1
D
Cauchy–Goursat Theorem for Multiply Connected Domains If f is analytic in
a multiply connected domain D, then we cannot conclude that C f (z) dz 0 for every simple
closed contour C in D. To begin, suppose D is a doubly connected domain and C and C1
are simple closed contours such that C1 surrounds the “hole” in the domain and is interior to C.
See FIGURE 18.2.4(a). Suppose, also, that f is analytic on each contour and at each point interior
to C but exterior to C1. When we introduce the cut AB shown in Figure 18.2.4(b), the region
bounded by the curves is simply connected. Now the integral from A to B has the opposite value
of the integral from B to A, and so from (2) we have C f (z) dz C1 f (z) dz 0 or
䉲
C
(a)
A
䉲
B
BC
䉲
D
C
(b)
FIGURE 18.2.4 Doubly connected
domain D
860
|
f (z) dz BC
䉲
f (z) dz.
䉱
(3)
1
The last result is sometimes called the principle of deformation of contours, since we can
think of the contour C1 as a continuous deformation of the contour C. Under this deformation
of contours, the value of the integral does not change. Thus, on a practical level, (3) allows us
to evaluate an integral over a complicated simple closed contour by replacing that contour with
one that is more convenient.
CHAPTER 18 Integration in the Complex Plane
y
–2 + 4i
Evaluate
2 + 3i
C
Applying Deformation of Contours
EXAMPLE 4
4i
SOLUTION In view of (3), we choose the more convenient circular contour C1 in the figure. By
taking the radius of the circle to be r ⫽ 1, we are guaranteed that C1 lies within C. In other words,
C1 is the circle |z ⫺ i| ⫽ 1, which can be parameterized by x ⫽ cos t, y ⫽ 1 ⫹ sin t, 0 ⱕ t ⱕ 2p,
or equivalently by z ⫽ i ⫹ eit, 0 ⱕ t ⱕ 2p. From z ⫺ i ⫽ eit and dz ⫽ ieit dt we obtain
C1
i
dz
dz
⫽
⫽
BC z 2 i BC1 z 2 i
x
–2
dz
, where C is the outer contour shown in FIGURE 18.2.5.
z
BC 2 i
䉲
䉲
2 – 2i
–2i
FIGURE 18.2.5 We use the simpler
contour C1 in Example 4
䉲
#
2p
0
ie it
dt ⫽ i
e it
#
2p
dt ⫽ 2pi.
0
The result in Example 4 can be generalized. Using the principle of deformation of contours (3)
and proceeding as in the example, we can show that if z0 is any constant complex number interior
to any simple closed contour C, then
dz
2pi,
n ⫽ e
0,
BC (z 2 z0)
n⫽1
n an integer 2 1.
䉲
(4)
The fact that the integral in (4) is zero when n is an integer ⫽ 1 follows only partially from
the Cauchy–Goursat theorem. When n is zero or a negative integer, 1/(z ⫺ z0)n is a polynomial
(for example, n ⫽ ⫺3, 1/(z ⫺ z0)⫺3 ⫽ (z ⫺ z0)3) and therefore entire. Theorem 18.2.1 then implies
养C dz/(z ⫺ z0)n ⫽ 0. It is left as an exercise to show that the integral is still zero when n is a positive integer different from one. See Problem 22 in Exercises 18.2.
䉲
Applying Formula (4)
EXAMPLE 5
Evaluate
5z ⫹ 7
dz, where C is the circle |z ⫺ 2| ⫽ 2.
BC z ⫹ 2z 2 3
䉲
2
SOLUTION Since the denominator factors as z2 ⫹ 2z ⫺ 3 ⫽ (z ⫺ 1)(z ⫹ 3), the integrand
fails to be analytic at z ⫽ 1 and z ⫽ ⫺3. Of these two points, only z ⫽ 1 lies within the contour
C, which is a circle centered at z ⫽ 2 of radius r ⫽ 2. Now by partial fractions,
3
5z 1 7
2
5
1
z21
z13
z 1 2z 2 3
2
and so
BC z 2 ⫹ 2z 2 3
5z ⫹ 7
䉲
dz ⫽ 3
BC z 2 1
dz
䉲
⫹2
BC z ⫹ 3
䉲
dz
.
(5)
In view of the result given in (4), the first integral in (5) has the value 2pi. By the Cauchy–
Goursat theorem, the value of the second integral is zero. Hence, (5) becomes
BC z 2 ⫹ 2z 2 3
D
5z ⫹ 7
䉲
C1
C2
C
If C, C1, and C2 are the simple closed contours shown in FIGURE 18.2.6 and if f is analytic on each
of the three contours as well as at each point interior to C but exterior to both C1 and C2, then by introducing cuts, we get from Theorem 18.2.1 that 养C f (z) dz ⫹ 养C1 f (z) dz ⫹ 养C2 f (z) dz ⫽ 0. Hence,
BC
䉲
FIGURE 18.2.6 Triply connected domain D
dz ⫽ 3(2pi) ⫹ 2(0) ⫽ 6pi.
f (z) dz ⫽
BC
䉲
䉲
BC
f (z) dz ⫹
䉲
1
䉱
䉱
f (z) dz.
2
The next theorem summarizes the general result for a multiply connected domain with n “holes”:
Theorem 18.2.2
Cauchy–Goursat Theorem for Multiply Connected Domains
Suppose C, C1, . . ., Cn are simple closed curves with a positive orientation such that C1, C2, . . ., Cn
are interior to C but the regions interior to each Ck, k ⫽ 1, 2, . . ., n, have no points in common. If f is analytic on each contour and at each point interior to C but exterior to all the
Ck, k ⫽ 1, 2, . . ., n, then
f (z) dz ⫽ a
B
n
䉲
C
k⫽1
BC
䉲
f (z) dz.
(6)
k
18.2 Cauchy–Goursat Theorem
|
861
Applying Theorem 18.2.2
EXAMPLE 6
Evaluate
y
C
i
dz
, where C is the circle |z| 3.
2
z
BC 1
䉲
SOLUTION In this case the denominator of the integrand factors as z2 1 (z i)(z i).
Consequently, the integrand 1/(z2 1) is not analytic at z i and z i. Both of these points
lie within the contour C. Using partial fraction decomposition once more, we have
C1
1>2i
1>2i
1
5
2
z2i
z1i
z 11
2
x
–i
1
dz
1
1
c
2
d dz.
BC z 1 2i BC z 2 i z i
C2
and
FIGURE 18.2.7 Contour in Example 6
䉲
䉲
2
We now surround the points z i and z i by circular contours C1 and C2, respectively,
that lie entirely within C. Specifically, the choice |z i| 12 for C1 and |z i| 12 for C2 will
suffice. See FIGURE 18.2.7. From Theorem 18.2.2 we can then write
1
dz
1
1
1
1
1
c
2
d dz c
2
d dz
2i B
zi
BC z 1 2i BC1 z 2 i z i
C2 z 2 i
䉲
䉲
2
䉲
1
dz
1
dz
1
dz
1
dz
2
2
.
2i B
z
2
i
2i
z
i
2i
z
2
i
2i
z
B
B
B
C1
C1
C2
C2 i
䉲
䉲
䉲
䉲
(7)
Because 1/(z i) is analytic on C1 and at each point in its interior and because 1/(z i) is
analytic on C2 and at each point in its interior, it follows from (4) that the second and third
integrals in (7) are zero. Moreover, it follows from (4), with n 1, that
BC z 2 i
dz
䉲
2pi
and
1
Thus (7) becomes
C
FIGURE 18.2.8 Contour C is closed but
not simple
Exercises
dz
p 2 p 0.
BC z 1
䉲
2
䉲
Answers to selected odd-numbered problems begin on page ANS-41.
䉲
|
2pi.
2
Throughout the foregoing discussion we assumed that C was a simple closed contour; in other
words, C did not intersect itself. Although we shall not give the proof, it can be shown that the
Cauchy–Goursat theorem is valid for any closed contour C in a simply connected domain D.
As shown in FIGURE 18.2.8, the contour C is closed but not simple. Nevertheless, if f is analytic
in D, then C f (z) dz 0.
In Problems 1–8, prove that C f (z) dz 0, where f is the given
function and C is the unit circle |z| 1.
1
1. f (z) z3 1 3i
2. f (z) z2 z24
z
z23
3. f (z) 4. f (z) 2
2z 1 3
z 1 2z 1 2
862
dz
䉲
REMARKS
D
18.2
BC z i
CHAPTER 18 Integration in the Complex Plane
5. f (z) sin z
(z 2 25) (z 2 9)
2
7. f (z) tan z
ez
2z 1 11z 1 15
z2 2 9
8. f (z) cosh z
6. f (z) 2
1
dz, where C is the contour shown in
BC z
FIGURE 18.2.9.
9. Evaluate
䉲
14.
15.
y
16.
C
17.
x
2
18.
19.
FIGURE 18.2.9 Contour in Problem 9
5
dz, where C is the contour shown in
BC z 1 i
FIGURE 18.2.10.
10. Evaluate
20.
䉲
21.
y
x 4 + y 4 = 16
10
dz; Zz iZ 1
(z
i)4
BC
2z 1
dz; (a) |z| 12 , (b) |z| 2, (c) |z 3i| 1
BC z 2 z
2z
dz; (a) |z| 1, (b) |z 2i| 1, (c) |z| 4
BC z 2 3
3z 2
dz; (a) |z 5| 2, (b) |z| 9
2
BC z 2 8z 12
3
1
a
2
b dz; (a) |z| 5, (b) |z 2i| 12
BC z 2 z 2 2i
z21
dz; Zz 2 iZ 12
BC z(z 2 i)(z 2 3i)
1
dz; ZzZ 1
3
BC z 2iz 2
8z 2 3
dz, where C is the closed contour shown
Evaluate
BC z 2 2 z
in FIGURE 18.2.11. [Hint: Express C as the union of two closed
curves C1 and C2.]
䉲
䉲
䉲
䉲
䉲
䉲
䉲
䉲
y
C
C
x
1
x
FIGURE 18.2.11 Contour in Problem 21
22. Suppose z0 is any constant complex number interior to any
FIGURE 18.2.10 Contour in Problem 10
simple closed contour C. Show that
dz
2pi,
n e
0,
BC (z 2 z0)
In Problems 11–20, use any of the results in this section to
evaluate the given integral along the indicated closed contour(s).
1
az b dz; ZzZ 2
z
BC
1
12.
az 2 b dz; ZzZ 2
z
BC
z
13.
dz; ZzZ 3
2
BC z 2 p2
11.
䉲
n1
n a positive integer 2 1.
In Problems 23 and 24, evaluate the given integral by any means.
䉲
ez
2 3zb dz, C is the unit circle |z| 1
BC z 3
24. C (z 3 z 2 Re(z)) dz, C is the triangle with vertices
z 0, z 1 2i, z 1
23.
䉲
䉲
䉲
䉲
18.3
a
Independence of the Path
INTRODUCTION In real calculus when a function f possesses an elementary antiderivative,
that is, a function F for which F (x) f (x), a definite integral can be evaluated by the Fundamental
Theorem of Calculus:
b
# f (x) dx F(b) F(a).
(1)
a
b
Note that ea f (x) dx depends only on the numbers a and b at the initial and terminal points of the
interval of integration. In contrast, the value of a real-line integral C P dx Q dy generally
depends on the curve C. However, we saw in Section 9.9 that there exist line integrals whose
value depends only on the initial point A and terminal point B of the curve C, and not on C
itself. In this case we say that the line integral is independent of the path. These integrals can
18.3 Independence of the Path
|
863
be evaluated by the Fundamental Theorem of Line Integrals (Theorem 9.9.1). It seems natural
then to ask:
Is there a complex version of the Fundamental Theorem of Calculus?
Can a contour integral eC f (z) dz be independent of the path?
In this section we will see that the answer to both of these questions is “yes.”
A Definition As the next definition shows, the definition of path independence for a
contour integral C f (z) dz is essentially the same as for a real-line integral C P dx Q dy.
Definition 18.3.1
Independence of the Path
Let z0 and z1 be points in a domain D. A contour integral C f (z) dz is said to be independent of the
path if its value is the same for all contours C in D with an initial point z0 and a terminal point z1.
At the end of the preceding section we noted that the Cauchy–Goursat theorem also holds for
closed contours, not just simple closed contours, in a simply connected domain D. Now suppose,
as shown in FIGURE 18.3.1, that C and C1 are two contours in a simply connected domain D, both
with initial point z0 and terminal point z1. Note that C and C1 form a closed contour. Thus, if
f is analytic in D, it follows from the Cauchy–Goursat theorem that
z1
C1
C
# f (z) dz #
D
C
z0
f (z) dz 0.
(2)
f (z) dz.
(3)
2C1
But (2) is equivalent to
# f (z) dz #
FIGURE 18.3.1 If f is analytic in D,
integrals on C and C1 are equal
C
C1
The result in (3) is also an example of the principle of deformation of contours introduced in (3)
of Section 18.2. We summarize the last result as a theorem.
Analyticity Implies Path Independence
Theorem 18.3.1
If f is an analytic function in a simply connected domain D, then C f (z) dz is independent
of the path C.
y
Choosing a Different Path
EXAMPLE 1
Evaluate C 2z dz, where C is the contour with initial point z 1 and terminal point
z 1 i shown in FIGURE 18.3.2.
–1 + i
C1
SOLUTION Since the function f (z) 2z is entire, we can replace the path C by any convenient
contour C1 joining z 1 and z 1 i. In particular, by choosing C1 to be the straight
line segment x 1, 0 y 1, shown in red in Figure 18.3.2, we have z 1 iy,
dz i dy. Therefore,
x
–1
C
# 2z dz 5 #
FIGURE 18.3.2 Contour in Example 1
C
2z dz 5 22
C1
#
1
0
y dy 2 2i
#
1
dy 5 21 2 2i.
0
A contour integral C f (z) dz that is independent of the path C is usually written
f (z) dz, where z0 and z1 are the initial and terminal points of C. Hence in Example 1 we can
1 i
write e1 2z dz.
There is an easier way to evaluate the contour integral in Example 1, but before proceeding
we need another definition.
z
ez01
Definition 18.3.2
Antiderivative
Suppose f is continuous in a domain D. If there exists a function F such that F (z) f (z) for
each z in D, then F is called an antiderivative of f.
864
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CHAPTER 18 Integration in the Complex Plane
For example, the function F(z) ⫽ ⫺cos z is an antiderivative of f (z) ⫽ sin z, since F⬘(z) ⫽ sin z.
As in real calculus, the most general antiderivative, or indefinite integral, of a function f (z) is
written 兰 f (z) dz ⫽ F(z) ⫹ C, where F⬘(z) ⫽ f (z) and C is some complex constant.
Since an antiderivative F of a function f has a derivative at each point in a domain D, it is
necessarily analytic and hence continuous in D (recall that differentiability implies continuity).
We are now in a position to prove the complex analogue of (1).
Fundamental Theorem for Contour Integrals
Theorem 18.3.2
Suppose f is continuous in a domain D and F is an antiderivative of f in D. Then for any contour C
in D with initial point z0 and terminal point z1,
# f (z) dz ⫽ F(z ) ⫺ F(z ).
1
C
(4)
0
PROOF: We will prove (4) in the case when C is a smooth curve defined by z ⫽ z(t), a ⱕ t ⱕ b.
Using (3) of Section 18.1 and the fact that F⬘(z) ⫽ f (z) for each z in D, we have
#
C
f (z) dz ⫽
⫽
#
b
#
b
a
a
f (z(t)) z9(t) dt ⫽
d
F(z(t)) dt
dt
⫽ F(z(t)) d
#
b
F9(z(t)) z9(t) dt
a
d Chain Rule
b
a
⫽ F(z(b)) 2 F(z(a)) ⫽ F(z1) 2 F(z0).
Using an Antiderivative
EXAMPLE 2
In Example 1 we saw that the integral 兰C 2z dz, where C is shown in Figure 18.3.2, is independent
of the path. Now since f (z) ⫽ 2z is an entire function, it is continuous. Moreover, F(z) ⫽ z2
is an antiderivative of f, since F⬘(z) ⫽ 2z. Hence by (4) we have
#
⫺1 ⫹ i
⫺1
2z dz ⫽ z 2 d
⫺1 ⫹ i
⫺1
⫽ (⫺1 ⫹ i)2 2 (⫺1)2 ⫽ ⫺1 2 2i.
Using an Antiderivative
EXAMPLE 3
Evaluate 兰C cos z dz, where C is any contour with initial point z ⫽ 0 and terminal point z ⫽ 2 ⫹ i.
SOLUTION F(z) ⫽ sin z is an antiderivative of f (z) ⫽ cos z, since F⬘(z) ⫽ cos z. Therefore
from (4) we have
#
C
cos z dz ⫽
#
2⫹i
0
cos z dz ⫽ sin z d
2⫹i
0
⫽ sin (2 ⫹ i) 2 sin 0 ⫽ sin (2 ⫹ i).
If we desired a complex number of the form a ⫹ ib for an answer, we can use sin(2 ⫹ i) ⫽
1.4031 ⫺ 0.4891i (see Example 1 in Section 17.7). Hence,
# cos z dz ⫽ 1.4031 ⫺ 0.4891i.
C
We can draw several immediate conclusions from Theorem 18.3.2. First, observe that if the
contour C is closed, then z0 ⫽ z1 and consequently
BC
䉲
f (z) dz ⫽ 0.
18.3 Independence of the Path
(5)
|
865
Next, since the value of 兰C f (z) dz depends on only the points z0 and z1, this value is the same for
any contour C in D connecting these points. In other words:
If a continuous function f has an antiderivative F in D, then eC f (z) dz is
independent of the path.
(6)
In addition we have the following sufficient condition for the existence of an antiderivative:
If f is continuous and eC f (z) dz is independent of the path in a domain D,
then f has an antiderivative everywhere in D.
D
z
s
z + Δz
z0
FIGURE 18.3.3 Contour used in proof of (7)
(7)
The last statement is important and deserves a proof. Assume that f is continuous, 兰C f (z) dz
z
is independent of the path in a domain D, and F is a function defined by F(z) ez0 f (s) ds where s
denotes a complex variable, z0 is a fixed point in D, and z represents any point in D. We wish to
show that F(z) f (z); that is, F is an antiderivative of f in D. Now,
F(z Dz) 2 F(z) #
z Dz
#
f (s) ds 2
z0
z
z0
f (s) ds #
z Dz
f (s) ds.
(8)
z
Because D is a domain we can choose z so that z z is in D. Moreover, z and z z can be
joined by a straight segment lying in D, as shown in FIGURE 18.3.3. This is the contour we use in
the last integral in (8). With z fixed, we can write*
f (z) Dz f (z)
#
z Dz
z
ds #
z Dz
f (z) ds
f (z) and
z
1
Dz
#
z Dz
f (z) ds.
(9)
z
From (8) and (9) it follows that
F(z Dz) 2 F(z)
1
2 f (z) Dz
Dz
#
z Dz
f f (s) 2 f (z)g ds.
z
Now f is continuous at the point z. This means that for any e 0 there exists a d 0 so that
| f (s) f (z)| e whenever |s z| d. Consequently, if we choose z so that | z| d, we have
F(z Dz) 2 F(z)
1
2
2 f (z) 2 2
Dz
Dz
#
1
2
22
Dz
z Dz
f f (s) 2 f (z)g ds 2
z
#
z Dz
f f (s) 2 f (z)g ds 2 #
z
1
eZDzZ e.
ZDzZ
Hence, we have shown that
lim
DzS0
F(z Dz) 2 F(z)
f (z)
Dz
or
F9(z) f (z).
If f is an analytic function in a simply connected domain D, it is necessarily continuous
throughout D. This fact, when put together with the results in Theorem 18.3.1 and (7), leads to
a theorem that states that an analytic function possesses an analytic antiderivative.
Theorem 18.3.3
Existence of an Antiderivative
If f is analytic in a simply connected domain D, then f has an antiderivative in D; that is, there
exists a function F such that F(z) f (z) for all z in D.
In (9) of Section 17.6 we saw that 1/z is the derivative of Ln z. This means that under some circumstances Ln z is an antiderivative of 1/z. Care must be exercised in using this result. For example, suppose
D is the entire complex plane without the origin. The function 1/z is analytic in this multiply
*See Problem 29 in Exercises 18.1.
866
|
CHAPTER 18 Integration in the Complex Plane
connected domain. If C is any simple closed contour containing the origin, it does not follow from
(5) that C dz /z 0. In fact, from (4) of Section 18.2 with the identification z0 0, we see that
䉲
1
dz 2pi.
BC z
䉲
In this case, Ln z is not an antiderivative of 1/z in D, since Ln z is not analytic in D. Recall that Ln z
fails to be analytic on the nonpositive real axis (the branch cut off the principal branch of the logarithm).
Using the Logarithmic Function
EXAMPLE 4
y
Evaluate
2i
C
#
C
1
dz, where C is the contour shown in FIGURE 18.3.4.
z
SOLUTION Suppose that D is the simply connected domain defined by x Re(z) 0,
y Im(z) 0. In this case, Ln z is an antiderivative of 1/z, since both these functions are
analytic in D. Hence by (4),
x
3
#
FIGURE 18.3.4 Contour in Example 4
2i
3
2i
1
dz Ln z d Ln 2i 2 Ln 3.
z
3
From (7) of Section 17.6, we have
Ln 2i loge2 #
and so
2i
3
p
i
2
and
Ln 3 loge3
1
2
p
dz 5 log e 1 i 5 20.4055 1 1.5708i.
z
3
2
REMARKS
Suppose f and g are analytic in a simply connected domain D that contains the contour C. If z0 and
z1 are the initial and terminal points of C, then the integration by parts formula is valid in D:
#
z1
z0
z1
f (z) g9(z) dz f (z) g(z)d
2
z0
#
z1
f 9(z) g(z) dz.
z0
This can be proved in a straightforward manner using Theorem 18.3.2 on the function
(d/dz)( fg). See Problems 21–24 in Exercises 18.3.
Exercises
18.3
Answers to selected odd-numbered problems begin on page ANS-41.
In Problems 1 and 2, evaluate the given integral, where C is the
contour given in the figure, by (a) finding an alternative path of
integration and (b) using Theorem 18.3.2.
1.
# (4z 1) dz
2.
C
3.
# e dz
z
y
4.
3 + 3i
3
i(t 4 4t 3 2), 1 t 1
# 6z dz, where C is z(t) 2 cos pt i sin
2
3
C
2
p
t, 0 t 2
4
In Problems 5–24, use Theorem 18.3.2 to evaluate the given
integral. Write each answer in the form a ib.
x
3+i
|z| = 1
FIGURE 18.3.5 Contour in
Problem 1
# 2z dz, where C is z(t) 2t
C
C
y
i
–i
In Problems 3 and 4, evaluate the given integral along the
indicated contour C.
0
x
FIGURE 18.3.6 Contour in
Problem 2
5.
#
31i
#
11i
z2 dz
6.
12i
#
2i
(3z 2 2 4z 5i ) dz
2i
0
7.
#
1
z3 dz
8.
(z 3 2 z) dz
3i
18.3 Independence of the Path
|
867
9.
#
12i
#
i
#
p 1 2i
⫺i>2
11.
(2z ⫹ 1)2 dz
epz dz
10.
p
#
1 1 2i
#
pi
1
12.
z
sin dz
2
14.
2
zez dz
20.
cos z dz
1 ⫹ (p>2)i
16. #
sinh 3z dz
# cosh z dz
1
# z dz, C is the arc of the circle z ⫽ 4e , ⫺p/2 ⱕ t ⱕ p/2
1
# z dz, C is the straight line segment between z ⫽ 1 ⫹ i and
i
pi
17.
19.
1 2 2i
2pi
15.
(iz ⫹ 1)3 dz
12i
i>2
13.
#
i
21.
#
4i
#
1 ⫹ "3i
#
i
#
11i
1
dz, C is any contour not passing through the origin
2
24i z
1
1
⫹ 2 b dz, C is any contour in the right half-plane
z
z
12i
Re(z) ⬎ 0
a
i
ez cos z dz
22.
23.
z
ze dz
24.
#
pi
z2ez dz
0
i
it
# z sin z dz
0
p
C
18.
C
z ⫽ 4 ⫹ 4i
18.4
Cauchy’s Integral Formulas
INTRODUCTION In the last two sections we saw the importance of the Cauchy–Goursat
theorem in the evaluation of contour integrals. In this section we are going to examine several
more consequences of the Cauchy–Goursat theorem. Unquestionably, the most significant of
these is the following result:
The value of an analytic function f at any point z0 in a simply connected domain can
be represented by a contour integral.
After establishing this proposition we shall use it to further show that
An analytic function f in a simply connected domain possesses derivatives of all orders.
The ramifications of these two results alone will keep us busy not only for the remainder of this
section but in the next chapter as well.
First Formula We begin with the Cauchy integral formula. The idea in the next theorem
is this: If f is analytic in a simply connected domain and z0 is any point D, then the quotient
f (z)/(z ⫺ z0) is not analytic in D. As a consequence, the integral of f (z)/(z ⫺ z0) around a simple
closed contour C that contains z0 is not necessarily zero but has, as we shall now see, the value
2pi f (z0). This remarkable result indicates that the values of an analytic function f at points inside
a simple closed contour C are determined by the values of f on the contour C.
Theorem 18.4.1
Cauchy’s Integral Formula
Let f be analytic in a simply connected domain D, and let C be a simple closed contour lying
entirely within D. If z0 is any point within C, then
f (z0) ⫽
f (z)
1
dz.
z
2 z0
2pi B
C
䉲
(1)
PROOF: Let D be a simply connected domain, C a simple closed contour in D, and z0 an interior
point of C. In addition, let C1 be a circle centered at z0 with radius small enough that it is interior to C.
By the principle of deformation of contours, we can write
f (z)
f (z)
dz ⫽
dz.
z
2
z
z
BC
BC1 2 z0
0
䉲
868
|
CHAPTER 18 Integration in the Complex Plane
䉲
(2)
We wish to show that the value of the integral on the right is 2pi f (z0). To this end we add and
subtract the constant f (z0) in the numerator:
f (z0) 2 f (z0) ⫹ f (z)
f (z)
dz ⫽
dz
z 2 z0
BC1 z 2 z0
BC1
䉲
䉲
⫽ f (z0)
f (z) 2 f (z0)
dz
⫹
dz.
BC1 z 2 z0 BC1 z 2 z0
䉲
(3)
䉲
Now from (4) of Section 18.2 we know that
dz
⫽ 2pi.
BC1 z 2 z0
䉲
Thus, (3) becomes
f (z) 2 f (z0)
f(z)
dz ⫽ 2pi f (z0) ⫹
dz.
z
2
z
BC1
BC1 z 2 z0
0
䉲
(4)
䉲
Since f is continuous at z0 for any arbitrarily small e ⬎ 0, there exists a d ⬎ 0 such that
| f (z) ⫺ f (z0)| ⬍ e whenever |z ⫺ z0| ⬍ d. In particular, if we choose the circle C1 to be
|z ⫺ z0| ⫽ d/2 ⬍ d, then by the ML-inequality (Theorem 18.1.3) the absolute value of the integral on the right side of (4) satisfies
2
f (z) 2 f (z0)
e
d
dz 2 #
2p a b ⫽ 2pe.
z
2
z
d>2
2
BC1
0
䉲
In other words, the absolute value of the integral can be made arbitrarily small by taking the
radius of the circle C1 to be sufficiently small. This can happen only if the integral is zero. The
Cauchy integral formula (1) follows from (4) by dividing both sides by 2pi.
The Cauchy integral formula (1) can be used to evaluate contour integrals. Since we often
work problems without a simply connected domain explicitly defined, a more practical restatement of Theorem 18.4.1 is
If f is analytic at all points within and on a simple closed contour C, and z 0 is
f (z)
1
any point interior to C, then f (z0) ⫽
dz.
2pi B
C z 2 z0
(5)
䉲
Using Cauchy’s Integral Formula
EXAMPLE 1
z 2 2 4z ⫹ 4
Evaluate
dz, where C is the circle |z| ⫽ 2.
BC z ⫹ i
䉲
SOLUTION First, we identify f (z) ⫽ z2 ⫺ 4z ⫹ 4 and z0 ⫽ ⫺i as a point within the circle C.
Next, we observe that f is analytic at all points within and on the contour C. Thus by the Cauchy
integral formula we obtain
y
z 2 2 4z ⫹ 4
dz ⫽ 2pi f (⫺i) ⫽ 2pi(3 ⫹ 4i) ⫽ 2p(⫺4 ⫹ 3i).
BC z ⫹ i
C
䉲
3i
x
EXAMPLE 2
Evaluate
–3i
FIGURE 18.4.1 Contour in Example 2
BC z 2 ⫹ 9
䉲
z
Using Cauchy’s Integral Formula
dz, where C is the circle |z ⫺ 2i| ⫽ 4.
SOLUTION By factoring the denominator as z2 ⫹ 9 ⫽ (z ⫺ 3i)(z ⫹ 3i), we see that 3i is the
only point within the closed contour at which the integrand fails to be analytic. See FIGURE 18.4.1.
18.4 Cauchy’s Integral Formulas
|
869
Now by writing
z
z
z 1 3i
5
,
z 2 3i
z2 1 9
y
we can identify f (z) ⫽ z/(z ⫹ 3i). This function is analytic at all points within and on the
contour C. From the Cauchy integral formula we then have
z1
z
z
z ⫹ 3i
3i
dz ⫽
dz ⫽ 2pi f (3i) ⫽ 2pi ⫽ pi.
2
6i
BC z ⫹ 9
BC z 2 3i
䉲
䉲
x
Flux and Cauchy’s Integral Formula
EXAMPLE 3
(a) Source: k > 0
The complex function f (z) ⫽ k/(z 2 z1 ), where k ⫽ a ⫹ ib and z1 are complex numbers, gives
rise to a flow in the domain z ⫽ z1. If C is a simple closed contour containing z ⫽ z1 in its
interior, then from the Cauchy integral formula we have
y
BC
z1
䉲
x
(b) Sink: k < 0
FIGURE 18.4.2 Vector fields in Example 3
f (z) dz ⫽
a 2 ib
dz ⫽ 2pi(a 2 ib).
BC z 2 z1
䉲
Thus the circulation around C is 2pb and the net flux across C is 2pa. If z1 were in the exterior
of C, both the circulation and net flux would be zero by Cauchy’s theorem.
Note that when k is real, the circulation around C is zero but the net flux across C
is 2pk. The complex number z 1 is called a source for the flow when k ⬎ 0 and a sink
when k ⬍ 0. Vector fields corresponding to these two cases are shown in FIGURE 18.4.2(a)
and 18.4.2(b).
Second Formula We can now use Theorem 18.4.1 to prove that an analytic function
possesses derivatives of all orders; that is, if f is analytic at a point z0, then f ⬘, f ⬙, f ⵮, and so on,
are also analytic at z0. Moreover, the values of the derivatives f (n)(z0), n ⫽ 1, 2, 3, … , are given
by a formula similar to (1).
Cauchy’s Integral Formula for Derivatives
Theorem 18.4.2
Let f be analytic in a simply connected domain D, and let C be a simple closed contour lying
entirely within D. If z0 is any point interior to C, then
f (n)(z0) ⫽
f (z)
n!
dz.
n⫹1
2pi B
C (z 2 z0)
(6)
䉲
PARTIAL PROOF: We will prove (6) only for the case n ⫽ 1. The remainder of the proof can
be completed using the principle of mathematical induction.
We begin with the definition of the derivative and (1):
f 9(z0) ⫽ lim
DzS0
⫽ lim
DzS0
⫽ lim
DzS0
870
|
CHAPTER 18 Integration in the Complex Plane
f (z0 ⫹ Dz) 2 f (z0)
Dz
f (z)
f (z)
1
c
dz 2
dz d
z
2pi Dz B
BC 2 z0
C z 2 (z0 ⫹ Dz)
䉲
䉲
f (z)
1
dz.
2pi B
C (z 2 z0 2 Dz) (z 2 z0)
䉲
Before proceeding, let us set up some preliminaries. Since f is continuous on C, it is bounded;
that is, there exists a real number M such that | f (z)| M for all points z on C. In addition,
let L be the length of C and let d denote the shortest distance between points on C and the point
z0. Thus for all points z on C, we have
Zz 2 z0 Z $ d
or
1
1
# 2.
Zz 2 z0 Z2
d
Furthermore, if we choose |z| d/2, then
Zz 2 z0 2 DzZ $ iz 2 z0Z 2 ZDzi $ d 2 Z DzZ $
d
1
2
and so
# .
2
Zz 2 z0 2 DzZ
d
Now,
2
2MLZDzZ
f (z)
f (z)
Dz f (z)
dz 2 #
dz 2
dz 2 2
.
2
2
d3
BC (z 2 z0)
BC (z 2 z0 2 Dz) (z 2 z0)
BC (z 2 z0) (z 2 z0 2 Dz)
䉲
䉲
䉲
Because the last expression approaches zero as z S 0, we have shown that
f 9(z0) lim
f (z0 Dz) 2 f (z0)
f (z)
1
dz.
Dz
2pi B
(z
2
z0)2
C
䉲
DzS0
If f (z) u(x, y) iv(x, y) is analytic at a point, then its derivatives of all orders exist at that
point and are continuous. Consequently, from
f 9 (z) 0u
0v
0v
0u
i
2i
0x
0x
0y
0y
f 0 (z) 0 2u
0 2v
0 2v
0 2u
i 2
2i
2
0y 0x
0y 0x
0x
0x
we can conclude that the real functions u and v have continuous partial derivatives of all orders
at a point of analyticity.
Like (1), (6) can sometimes be used to evaluate integrals.
Using Cauchy’s Integral Formula for Derivatives
EXAMPLE 4
Evaluate
BC z 4 4z 3
䉲
z1
dz, where C is the circle |z| 1.
SOLUTION Inspection of the integrand shows that it is not analytic at z 0 and z 4, but
only z 0 lies within the closed contour. By writing the integrand as
z1
z4
z1
,
z 4 4z 3
z3
we can identify z0 0, n 2, and f (z) (z 1)/(z 4). By the Quotient Rule, f (z) 6/(z 4)3 and so by (6) we have
BC z 4 4z 3
䉲
z1
dz 2pi
3p
f 0 (0) i.
2!
32
18.4 Cauchy’s Integral Formulas
|
871
y
Evaluate
i
x
0
Using Cauchy’s Integral Formula for Derivatives
EXAMPLE 5
C2
z3 3
dz, where C is the contour shown in FIGURE 18.4.3.
BC z(z 2 i)2
䉲
SOLUTION Although C is not a simple closed contour, we can think of it as the union of two
simple closed contours C1 and C2 as indicated in Figure 18.4.3. By writing
z3 3
z3 3
z3 3
dz dz dz
2
2
BC z(z 2 i)
BC1 z(z 2 i)
BC2 z(z 2 i)2
䉱
䉲
C1
FIGURE 18.4.3 Contour in Example 5
䉲
BC
䉲
1
z3 3
z3 3
2
z
(z 2 i)
dz dz I1 I2,
z
BC2 (z 2 i)2
䉲
we are in a position to use both (1) and (6).
To evaluate I1, we identify z0 0 and f (z) (z3 3)/(z i)2. By (1) it follows that
I1 BC
䉲
1
z3 3
(z 2 i)2
dz 2pi f (0) 6pi.
z
To evaluate I2 we identify z0 i, n 1, f (z) (z3 3)/z, and f (z) (2z3 3)/z2. From (6)
we obtain
I2 z3 3
z
BC (z 2 i)2
䉲
dz 2
2pi
f 9(i) 2pi(3 2i) 2p(2 3i).
1!
Finally we get
z3 3
dz I1 I2 6pi 2p(2 3i) 4p(1 3i).
BC z(z 2 i)2
䉲
Liouville’s Theorem If we take the contour C to be the circle |z z0| r, it follows
from (6) and the ML-inequality that
Z f (n)(z0)Z f (z)
n!
n!
1
n!M
2
dz 2 #
M n 1 2pr n ,
n1
2p B
2p r
r
C (z 2 z0)
䉲
(7)
where M is a real number such that | f (z)| M for all points z on C. The result in (7), called
Cauchy’s inequality, is used to prove the next result.
Theorem 18.4.3
Liouville’s Theorem
The only bounded entire functions are constants.
PROOF: Suppose f is an entire function and is bounded; that is, | f (z)| M for all z. Then for
any point z0, (7) gives | f (z0)| M/r. By making r arbitrarily large, we can make | f (z0)| as
small as we wish. This means f (z0) 0 for all points z0 in the complex plane. Hence f must
be a constant.
Fundamental Theorem of Algebra Liouville’s theorem enables us to prove, in turn,
a result that is learned in elementary algebra:
If P(z) is a nonconstant polynomial, then the equation P(z) 0 has at least one root.
872
|
CHAPTER 18 Integration in the Complex Plane
This result is known as the Fundamental Theorem of Algebra. To prove it, let us suppose that
P(z) 0 for all z. This implies that the reciprocal of P, f (z) 1/P(z), is an entire function. Now
since | f (z)| S 0 as |z| S q, the function f must be bounded for all finite z. It follows from
Liouville’s theorem that f is a constant and therefore P is a constant. But this is a contradiction
to our underlying assumption that P was not a constant polynomial. We conclude that there must
exist at least one number z for which P(z) 0.
Exercises
18.4
Answers to selected odd-numbered problems begin on page ANS-41.
In Problems 1–24, use Theorems 18.4.1 and 18.4.2, when
appropriate, to evaluate the given integral along the indicated
closed contour(s).
1.
2.
4
dz; ZzZ 5
BC z 2 3i
17.
18.
䉲
z2
dz; ZzZ 5
BC (z 2 3i)2
䉲
19.
ez
3.
dz; ZzZ 4
BC z 2 pi
䉲
20.
1 2e z
dz; ZzZ 1
4.
BC z
䉲
5.
6.
z 2 2 3z 4i
dz; ZzZ 3
BC z 2i
䉲
8.
9.
10.
22.
cos z
dz; ZzZ 1.1
BC 3z 2 p
䉲
z
dz;
BC z 4
2
7.
21.
䉲
2
(a) Zz 2 iZ 2,
z 2 3z 2i
dz;
BC z 2 3z 2 4
䉲
(a) ZzZ 2,
23.
(b) Zz 2iZ 1
(b) Zz 5Z BC
䉲
a
a
BC z 2 p2
䉲
sin z
(a) ZzZ 1,
(a) ZzZ 1,
(b) Zz 2 1 2 iZ 1
(b) Zz 2 2Z 1
e 2iz
z4
2
b dz; ZzZ 6
4
z
(z 2 i)3
cosh z
sin 2z
2
b dz; ZzZ 3
3
(z 2 p)
(2z 2 p)3
1
dz; Zz 2 2Z 5
BC z (z 2 1)2
䉲
3
1
dz; Zz 2 iZ BC z (z 1)
䉲
2
2
3
2
3z 1
dz; C is given in FIGURE 18.4.4
BC z (z 2 2)2
䉲
y
C
z2 4
dz; Zz 2 3iZ 1.3
BC z 2 2 5iz 2 4
0
2
x
dz; Zz 2 2iZ 2
ez
dz; Zz 2 iZ 1
BC (z 2 i)3
FIGURE 18.4.4 Contour in Problem 23
䉲
z
12.
dz; ZzZ 2
BC (z i)4
14.
BC
䉲
3
䉲
24.
䉲
13.
1
dz;
BC z (z 2 4)
䉲
3
2
2
11.
z2
dz;
2
z
(z
2 1 2 i)
BC
䉲
BC
䉲
e iz
dz; C is given in FIGURE 18.4.5
BC (z 2 1)2
䉲
cos 2z
dz; ZzZ 1
z5
y
e z sin z
dz; Zz 2 1Z 3
BC z 3
i
䉲
2z 5
dz; (a) ZzZ 12, (b) Zz 1Z 2,
15.
BC z 2 2 2z
(c) Zz 2 3Z 2, (d) Zz 2iZ 1
x
䉲
16.
z
dz; (a) ZzZ 12,
BC (z 2 1)(z 2 2)
(c) Zz 2 1Z 12, (d) ZzZ 4
䉲
(b) Zz 1Z 1,
–i
C
FIGURE 18.4.5 Contour in Problem 24
18.4 Cauchy’s Integral Formulas
|
873
Chapter in Review
18
Answers to selected odd-numbered problems begin on page ANS-41.
Answer Problems 1–12 without referring back to the text. Fill in
the blank or answer true/false.
14.
# (x iy) dz; C is the contour shown in Figure 18.R.1
C
1. The sector defined by p/6 arg z p/6 is a simply
connected domain. _____
2. If
BC
䉲
y
C
f (z) dz 0 for every simple closed contour C, then f is
analytic within and on C. _____
x
#
z22
dz is the same for any path C in the
z
C
right half-plane Re(z) 0 between z 1 i and
3. The value of
FIGURE 18.R.1 Contour in Problems 13 and 14
z 10 8i. _____
g(z)
g(z)
4. If g is entire, then
dz dz, where C is the
BC z 2 i
BC1 z 2 i
circle |z| 3 and C1 is the ellipse x2 y2 /9 1. _____
䉲
15.
䉲
f (z) dz _____.
j2 6j 2 2
dj, where C is |z| 3, then
BC j 2 z
f (1 i) _____.
6. If f (z) 16.
f (z)
dz _____.
BC (z pi)3
17.
18.
20.
f 9 (z)
f (z)
dz dz _____ .
z
2
z
(z
2
z0)2
BC
BC
0
0,
if n _____
2pi, if n _____
BC
where n is an integer and C is |z| 1.
z ndz e
䉲
C
|
CHAPTER 18 Integration in the Complex Plane
4
i(1 t 3)2, 1 t 1
# (4z 3z 2z 1) dz; C is the line segment from 0 to 2i
3
2
23.
e 2z
dz; C is the circle |z 1| 3
BC z 4
24.
cos z
dz; C is the circle |z| 12
BC z 2 z 2
25.
1
dz; C is the ellipse x2 /4 y2 1
BC 2z 7z 3
26.
BC
f (z) dz 2 # _____ .
(x iy) dz; C is the contour shown in FIGURE 18.R.1
874
# sin z dz; C is z(t) t
BC z 2 2 1
In Problems 13–28, evaluate the given integral using the
techniques considered in this chapter.
#
(4z 2 6) dz
22.
䉲
BC
12i
BC
z0 is a point within C, then
12. If |f (z)| 2 on |z| 3, then 2
#
epz dz; C is the ellipse x2 /100 y2 /64 1
21.
10. If f is analytic within and on the simple closed contour C and
13.
䉲
C
䉲
䉲
BC
C
1
9.
dz 0 for every simple closed contour C
BC (z 2 z0)(z 2 z1)
that encloses the points z0 and z1. _____
䉲
dz; C is the line segment from z i to z 1 i
pz
3i
19.
䉲
2
#e
䉲
8. If f is entire and | f (z)| 10 for all z, then f (z) _____.
11.
2
C
7. If f (z) z3 ez and C is the contour z 8eit, 0 t 2p,
then
# |z | dz; C is z(t) t it , 0 t 2
C
䉲
5. If f is a polynomial and C is a simple closed curve, then
BC
3
–4
䉲
䉲
(z2 z1 z z2) dz; C is the circle |z| 1
3z 4
dz; C is the circle |z| 2
䉲
䉲
䉲
䉲
3
2
z csc z dz; C is the rectangle with vertices 1 i, 1 i,
2 i, 2 i
27.
z
dz; C is the contour shown in FIGURE 18.R.2
z
⫹
i
BC
䉲
y
C
e ipz
dz; C is (a) |z| ⫽ 1, (b) |z ⫺ 3| ⫽ 2,
BC 2z 2 5z ⫹ 2
(c) |z ⫹ 3| ⫽ 2
䉲
2
29. Let f (z) ⫽ zng(z), where n is a positive integer, g(z) is entire,
x
–2
28.
3
and g(z) ⫽ 0 for all z. Let C be a circle with center at the
f 9(z)
origin. Evaluate
dz.
BC f (z)
䉲
30. Let C be the straight line segment from i to 2 ⫹ i. Show that
FIGURE 18.R.2 Contour in Problem 27
#
2 Ln (z ⫹ 1) dz 2 # log e10 ⫹
C
p
.
2
CHAPTER 18 in Review
|
875
Dennis K. Johnson/Getty Images
CHAPTER
19
Cauchy’s integral formula for
derivatives indicates that if a
function f is analytic at a point
z0, then it possesses derivatives
of all orders at that point. As a
consequence of this result, we
will see that f can always be
expanded in a power series at
that point. On the other hand, if
f fails to be analytic at a point z0,
then we may still be able to
expand it in a different kind of
series known as a Laurent
series. The notion of Laurent
series leads to the concept of a
residue, and this, in turn, leads
to yet another way of evaluating
complex integrals.
Series and Residues
CHAPTER CONTENTS
19.1
19.2
19.3
19.4
19.5
19.6
Sequences and Series
Taylor Series
Laurent Series
Zeros and Poles
Residues and Residue Theorem
Evaluation of Real Integrals
Chapter 19 in Review
19.1
Sequences and Series
INTRODUCTION Much of the theory of complex sequences and series is analogous to that
encountered in real calculus. In this section we explore the definitions of convergence and divergence for complex sequences and complex infinite series. In addition, we give some tests for
convergence of infinite series. You are urged to pay special attention to what is said about
geometric series since this type of series will be important in the later sections of this chapter.
Sequences A sequence {zn} is a function whose domain is the set of positive integers; in
other words, to each integer n 1, 2, 3, . . ., we assign a complex number zn. For example, the
sequence {1 in} is
y
ε
L
1 i,
c
n 1,
x
FIGURE 19.1.1 If {zn} converges to L, all
but a finite number of terms are in any
e-neighborhood of L
0,
c
n 2,
1 i,
c
n 3,
2,
1 i, . . .
c
c
n 4, n 5, . . . .
If limn S q zn L we say the sequence {zn} is convergent. In other words, {zn} converges to the
number L if, for each positive number e, an N can be found such that |zn L| e whenever
n N. As shown in FIGURE 19.1.1, when a sequence {zn} converges to L, all but a finite number
of the terms of the sequence are within every e-neighborhood of L. The sequence {1 i n} illustrated in (1) is divergent, since the general term zn 1 i n does not approach a fixed complex
number as n S q. Indeed, the first four terms of this sequence repeat endlessly as n increases.
y
1
3
–1
5
A Convergent Sequence
EXAMPLE 1
i
4
x
–1
The sequence e
i
n1
n
f converges, since
lim
nSq
– i
2
FIGURE 19.1.2 The terms of the sequence
spiral toward 0 in Example 1
i n1
0.
n
i 1 i
1
1, , , , , . . .,
2 3 4 5
As we see from
and FIGURE 19.1.2, the terms of the sequence spiral toward the point z 0.
The following theorem should make intuitive sense.
Theorem 19.1.1
Criterion for Convergence
A sequence {zn} converges to a complex number L if and only if Re(zn) converges to Re(L)
and Im(zn) converges to Im(L).
EXAMPLE 2
The sequence e
Illustrating Theorem 19.1.1
ni
f converges to i. Note that Re(i) 0 and Im(i) 1. Then from
n 1 2i
zn n2
ni
2n
i 2
,
2
n 2i
n 4
n 4
we see that Re(zn) 2n/(n2 4) S 0 and Im(zn) n2 /(n2 4) S 1 as n S q.
Series An infinite series of complex numbers
a z k z1 z2 z3 q
k1
878
|
(1)
CHAPTER 19 Series and Residues
… zn …
is convergent if the sequence of partial sums {Sn}, where
Sn z1 z2 z3 … zn,
converges. If Sn S L as n S q, we say that the sum of the series is L.
Geometric Series For the geometric series
a az
q
k21
k1
a az az2 … azn1 …
(2)
the nth term of the sequence of partial sums is
Sn a az az2 … azn1.
(3)
By multiplying Sn by z and subtracting this result from Sn, we obtain Sn zSn a azn. Solving
for Sn gives
Sn a(1 2 z n)
.
12z
(4)
Since zn S 0 as n S q, whenever |z| 1 we conclude from (4) that (2) converges to
a
12z
when |z| 1; the series diverges when |z| 1. The special geometric series
1
1 z z 2 z3 …
12z
(5)
1
1 z z 2 z3 …
1z
(6)
valid for |z| 1, will be of particular usefulness in the next two sections. In addition, we shall use
1 2 zn
1 z z2 z3 … zn1
12z
(7)
1
zn
1 z z2 z3 … zn1 12z
12z
(8)
in the alternative form
in the proofs of the two principal theorems of this chapter.
Convergent Geometric Series
EXAMPLE 3
The series
(1 2i)k
(1 2i)2
(1 2i)3
1 2i
p
a
5
52
53
5k
k1
q
is a geometric series with a (1 2i)/5 and z (1 2i)/5. Since |z| "5/5 1, the series
converges and we can write
(1 2i)k
a
5k
k1
q
Theorem 19.1.2
If
q
gk 1
1 2i
5
i
.
1 2i
2
12
5
Necessary Condition for Convergence
zk converges, then limnS q zn 0.
19.1 Sequences and Series
|
879
An equivalent form of Theorem 19.1.2 is the familiar nth term test for divergence of an
infinite series.
The nth Term Test for Divergence
Theorem 19.1.3
If limnS q zn 0, then the series g k 5 1 zk diverges.
q
For example, the series g k 5 1 (k 5i)/k diverges since zn (n 5i)/n S 1 as n S q. The
geometric series (2) diverges when |z| 1, since, in this case, limnS q |zn| does not exist.
q
Definition 19.1.1
Absolute Convergence
An infinite series g k 5 1 zk is said be absolutely convergent if g k 5 1 Zzk Z converges.
q
EXAMPLE 4
q
gk5 1
q
Absolute Convergence
The series
(i /k2) is absolutely convergent since |i k /k2 | 1/k2 and the real series
q
q
2
g k 5 1 (1/k ) converges. Recall from calculus that a real series of the form g k 5 1 (1/k p) is called
a p-series, p a real number, and converges for p 1 and diverges for p 1.
k
As in real calculus,
Absolute convergence implies convergence.
Thus in Example 4, because the series
q
ik
1
i
p
a k2 5 i 2 22 2 32 1
k51
converges absolutely, it is also convergent.
The following two tests are the complex versions of the ratio and root tests that are encountered
in calculus:
Theorem 19.1.4
Suppose
q
gk5 1
Ratio Test
zk is a series of nonzero complex terms such that
lim 2
nSq
zn 1
2 L.
zn
(9)
(i) If L 1, then the series converges absolutely.
(ii) If L 1 or L q, then the series diverges.
(iii) If L 1, the test is inconclusive.
Theorem 19.1.5
Suppose
q
gk5 1
Root Test
zk is a series of complex terms such that
n
lim "Zzn Z L.
nSq
(i) If L 1, then the series converges absolutely.
(ii) If L 1 or L q, then the series diverges.
(iii) If L 1, the test is inconclusive.
We are interested primarily in applying these tests to power series.
880
|
CHAPTER 19 Series and Residues
(10)
Power Series The notion of a power series is important in the study of analytic functions.
An infinite series of the form
k
2
…
a ak (z ⫺ z0) ⫽ a0 ⫹ a1(z ⫺ z0) ⫹ a2(z ⫺ z0) ⫹ ,
q
(11)
k⫽0
y
where the coefficients ak are complex constants, is called a power series in z ⫺ z0. The power
series (11) is said to be centered at z0, and the complex point z0 is referred to as the center of
the series. In (11), it is also convenient to define (z ⫺ z0)0 ⫽ 1 even when z ⫽ z0.
|z–z0| = R
Circle of Convergence Every complex power series has radius of convergence R, where
R is a real number. Analogous to the concept of an interval of convergence in real calculus, when
0 ⬍ R ⬍ q, a complex power series (11) has a circle of convergence defined by |z ⫺ z0| ⫽ R.
The power series converges absolutely for all z satisfying |z ⫺ z0| ⬍ R and diverges for |z ⫺ z0| ⬎ R.
See FIGURE 19.1.3. The radius R of convergence can be
convergence
z0
R
divergence
FIGURE 19.1.3 A power series
converges at all points within the
circle of convergence
x
(i) zero (in which case (11) converges at only z ⫽ z0),
(ii) a finite number (in which case (11) converges at all interior points of the circle |z ⫺ z0| ⫽ R), or
(iii) q (in which case (11) converges for all z).
A power series may converge at some, all, or none of the points on the actual circle of convergence.
Consider the power series g k ⫽ 1 (zk⫹1/k). By the ratio test (9),
EXAMPLE 5
Circle of Convergence
q
z n⫹2
n⫹1
n
lim 4 n ⫹ 1 4 ⫽ lim
ZzZ ⫽ ZzZ.
nSq
nSq n ⫹ 1
z
n
Thus the series converges absolutely for |z| ⬍ 1. The circle of convergence is |z| ⫽ 1 and the
radius of convergence is R ⫽ 1. Note that on the circle of convergence, the series does not
converge absolutely, since the series of absolute values is the well-known divergent harmonic
q
series g k ⫽ 1 (1/k). Bear in mind this does not say, however, that the series diverges on the
q
circle of convergence. In fact, at z ⫽ ⫺1, g k ⫽ 1 ((⫺1)k⫹1/k) is the convergent alternating
harmonic series. Indeed, it can be shown that the series converges at all points on the circle
|z| ⫽ 1 except at z ⫽ 1.
It should be clear from Theorem 19.1.4 and Example 5 that for a power series g k ⫽ 0 ak(z ⫺ z0)k,
the limit (9) depends on only the coefficients ak. Thus, if
an ⫹ 1
(i) lim 2
2 ⫽ L ⫽ 0, the radius of convergence is R ⫽ 1/L;
an
nSq
q
(ii)
lim 2
nSq
an ⫹ 1
2 ⫽ 0, the radius of convergence is q;
an
an ⫹ 1
2 ⫽ q, the radius of convergence is R ⫽ 0.
an
nSq
n
Similar remarks can be made for the root test (10) by utilizing lim nSq "Zan Z.
(iii) lim 2
EXAMPLE 6
Radius of Convergence
Consider the power series a
q
k⫽1
(⫺1)k ⫹ 1(z 2 1 2 i)k
. Identifying an ⫽ (⫺1)n⫹1/n!, we have
k!
(⫺1)n ⫹ 2
(n ⫹ 1)!
1
4 ⫽ lim
⫽ 0.
lim 4
n⫹1
nSq
nSq n ⫹ 1
(⫺1)
n!
19.1 Sequences and Series
|
881
Thus the radius of convergence is q; the power series with center 1 i converges absolutely
for all z.
EXAMPLE 7
Radius of Convergence
q
6k 1 k
6n 1 n
Consider the power series a a
b (z 2 2i)k . With an a
b , the root test in
2n 5
k 1 2k 5
the form
6n 1
n
lim "Zan Z lim
3
nSq
nSq 2n 5
shows that the radius of convergence of the series is R 13 . The circle of convergence is
|z 2i| 13 ; the series converges absolutely for |z 2i| 13 .
Exercises
19.1
Answers to selected odd-numbered problems begin on page ANS-41.
In Problems 1–4, write out the first five terms of the given sequence.
1. {5in}
2. {2 (i)n}
npi
3. {1 e }
4. {(1 i)n} [Hint: Write in polar form.]
In Problems 5–10, determine whether the given sequence
converges or diverges.
3ni 2
ni 2n
5. e
f
6. e
f
n ni
3ni 5n
(ni 2)2
n(1 i n)
f
7. e
8.
e
f
n1
n2i
n in
9. e
f
10. {e1/n 2(tan1n)i}
"n
In Problems 11 and 12, show that the given sequence {zn}
converges to a complex number L by computing limnS q Re(zn)
and limnS q Im(zn).
11. e
4n 3ni
f
2n i
12. e a
1i n
b f
4
In Problems 13 and 14, use the sequence of partial sums to show
that the given series is convergent.
q
1
1
13. a c
2
d
k
2i
k
1
2i
k1
q
i
14. a
k 2 k(k 1)
In Problems 15–20, determine whether the given geometric
series is convergent or divergent. If convergent, find its sum.
q
q
1 k21
15. a (1 2 i)k
16. a 4i a b
3
k0
k1
17. a a b
k1 2
q
i
k
q
19. a 3 a
b
1 2i
k0
q
18. a i k
k0 2
2
k
1
20. a
k21
k 2 (1 i)
q
ik
In Problems 21–28, find the circle and radius of convergence of
the given power series.
q
1
21. a
(z 2 2i)k
k1
(1
2
2i)
k0
k
q
1
i
22. a a
b zk
1i
k1 k
q (1)k
q
1
k
23. a
(z
2
1
2
i)
24.
(z 3i)k
a
k
2
k
k 1 k2
k 1 k (3 4i)
q
q
zk
25. a (1 3i)k(z 2 i)k
26. a k
k0
k1 k
q (z 2 4 2 3i)k
27. a
52k
k0
q
1 2i k
28. a (1)k a
b (z 2i)k
2
k0
q (z 2 i)k
29. Show that the power series a
is not absolutely conk2k
k1
vergent on its circle of convergence. Determine at least one
point on the circle of convergence at which the power series
converges.
q
zk
30. (a) Show that the power series a 2 converges at every
k1 k
point on its circle of convergence.
(b) Show that the power series a kz k diverges at every point
k1
on its circle of convergence.
q
19.2 Taylor Series
INTRODUCTION The correspondence between a complex number z within the circle of
q
convergence and the number to which the series g k 1ak(z 2 z0)k converges is single-valued.
In this sense, a power series defines or represents a function f ; for a specified z within the
circle of convergence, the number L to which the power series converges is defined to be the
882
|
CHAPTER 19 Series and Residues
value of f at z; that is, f (z) L. In this section we present some important facts about the
nature of this function f.
In the preceding section we saw that every power series has a radius of convergence R.
q
Throughout the discussion in this section, we will assume that a power series g k 1ak(z 2 z0)k
has either a positive or an infinite radius R of convergence. The next three theorems will give
some important facts about the nature of a power series within its circle of convergence |z z0| R,
R 0.
Continuity
Theorem 19.2.1
A power series g k 0 ak(z z0)k represents a continuous function f within its circle of
convergence |z z0| R, R 0.
q
Theorem 19.2.2
Term-by-Term Integration
q
g k 0 ak(z
z0)k can be integrated term by term within its circle of converA power series
gence |z z0| R, R 0, for every contour C lying entirely within the circle of convergence.
Term-by-Term Differentiation
Theorem 19.2.3
A power series g k 0 ak(z z0)k can be differentiated term by term within its circle of
convergence |z z0| R, R 0.
q
Taylor Series Suppose a power series represents a function f for |z z0| R, R 0; that is,
f (z) a ak(z z0)k a0 a1(z z0) a2(z z0)2 a3(z z0)3 ….
q
(1)
k0
It follows from Theorem 19.2.3 that the derivatives of f are
f (z) a kak (z z0)k1 a1 2a2(z z0) 3a3(z z0)2 …
(2)
f (z) a k(k 1)ak(z z0)k2 2 1a2 3 2a3(z z0) …
(3)
f (z) a k(k 1)(k 2)ak(z z0)k3 3 2 1a3 …
(4)
q
k1
q
k2
q
k3
and so on. Each of the differentiated series has the same radius of convergence as the original
series. Moreover, since the original power series represents a differentiable function f within its
circle of convergence, we conclude that when R 0:
A power series represents an analytic function within its circle of convergence.
There is a relationship between the coefficients ak and the derivatives of f. Evaluating
(1), (2), (3), and (4) at z z0 gives
f (z0) a0,
f (z0) 1!a1,
f (z0) 2!a2,
and
f (z0) 3!a3,
respectively. In general, f (n)(z0) n!an or
an f (n)(z0)
, n $ 0.
n!
(5)
When n 0, we interpret the zeroth derivative as f (z0) and 0! 1. Substituting (5) into (1) yields
f (z) a
q
k0
f (k)(z0)
(z 2 z0)k.
k!
(6)
19.2 Taylor Series
|
883
This series is called the Taylor series for f centered at z0. A Taylor series with center z0 0,
f (z) a
q
k0
f (k)(0) k
z ,
k!
(7)
is referred to as a Maclaurin series.
We have just seen that a power series with a nonzero radius of convergence represents an
analytic function. On the other hand, if we are given a function f that is analytic in some domain D,
can we represent it by a power series of the form (6) and (7)? Since a power series converges in
a circular domain, and a domain D is generally not circular, the question becomes: Can we expand
f in one or more power series that are valid in circular domains that are all contained in D? The
question will be answered in the affirmative by the next theorem.
Taylor’s Theorem
Theorem 19.2.4
Let f be analytic within a domain D and let z0 be a point in D. Then f has the series
representation
f (z) a
q
k0
f (k)(z0)
(z 2 z0)k
k!
(8)
valid for the largest circle C with center at z0 and radius R that lies entirely within D.
PROOF: Let z be a fixed point within the circle C and let s denote the variable of integration. The
circle C is then described by |s z0| R. See FIGURE 19.2.1. To begin, we use the Cauchy integral
formula to obtain the value of f at z:
s
z0
z
R
C
f (z) 1
2pi
1
2pi
D
FIGURE 19.2.1 Circular contour C used
in proof of Theorem 19.2.4
f (s)
1
ds s
2
z
2pi
BC
䉲
f (s)
s
BC 2 z0 c
䉲
f (s)
ds
(s
2
z
)
BC
0 2 (z 2 z0)
䉲
(9)
1
ds.
z 2 z0 s
12
s 2 z0
By replacing z by (z z0)/(s z0) in (8) of Section 19.1, we have
z 2 z0
z 2 z0 2
z 2 z0 n 2 1
(z 2 z0)n
1
1
a
b p a
b
,
z 2 z0
s 2 z0
s 2 z0
s 2 z0
(s 2 z)(s 2 z0)n 2 1
12
s 2 z0
and so (9) becomes
f (z) 1
2pi
z 2 z0
f (s)
ds 2pi
BC s 2 z0
䉲
(z 2 z0)n 2 1
2pi
(z 2 z0)2
f (s)
ds 2
2pi
BC (s 2 z0)
䉲
(z 2 z0)n
f (s)
n ds 2pi
BC (s 2 z0)
䉲
BC (s 2 z0)3
䉲
f (s)
ds p
f (s)
ds.
BC (s 2 z)(s 2 z0)n
(10)
䉲
Utilizing Cauchy’s integral formula for derivatives, we can write (10) as
f (z) f (z0) f 9(z0)
f 0(z0)
(z 2 z0) (z 2 z0)2 p
1!
2!
f (n 2 1)(z0)
(z 2 z0)n 2 1 Rn(z),
(n 2 1)!
where
884
|
CHAPTER 19 Series and Residues
Rn(z) (z 2 z0)n
f (s)
n ds.
2pi B
C (s 2 z)(s 2 z0)
䉲
(11)
Equation (11) is called Taylor’s formula with remainder Rn. We now wish to show that Rn(z) S 0
as n S q. Since f is analytic in D, | f (z)| has a maximum value M on the contour C. In addition,
since z is inside C, we have |z z0| R, and, consequently,
|s z| |s z0 (z z0)| |s z0| |z z0| R d,
where d |z z0| is the distance from z and z0. The ML-inequality then gives
ZRn(z)Z 2
(z 2 z0)n
f (s)
dn
M
MR
d n
ds
2
#
2pR
a
b .
n
2pi B
2p (R 2 d )R n
R2d R
C (s 2 z)(s 2 z0)
䉲
Because d R, (d/R)n S 0 as n S q, we conclude that |Rn(z)| S 0 as n S q. It follows that the
infinite series
f 9(z0)
f 0(z0)
(z 2 z0) (z 2 z0)2 p
1!
2!
converges to f (z). In other words, the result in (8) is valid for any point z interior to C.
f (z0) We can find the radius of convergence of a Taylor series in exactly the same manner illustrated
in Examples 5–7 of Section 19.1. However, we can simplify matters even further by noting that the
radius of convergence is the distance from the center z0 of the series to the nearest isolated singularity
of f. We shall elaborate more on this concept in the next section, but an isolated singularity is a point
at which f fails to be analytic but is, nonetheless, analytic at all other points throughout some neighborhood of the point. For example, z 5i is an isolated singularity of f (z) 1/(z 5i). If the function
f is entire, then the radius of convergence of a Taylor series centered at any point z0 is necessarily
infinite. Using (8) and the last fact, we can say that the Maclaurin series representations
ez 1 q
z
z2
zk
p a
1!
2!
k 0 k!
(12)
q
z3
z5
z 2k 1
2 p a (1)k
3!
5!
(2k 1)!
k0
q
z4
z 2k
z2
2 p a (1)k
cos z 1 2
2!
4!
(2k)!
k0
sin z z 2
are valid for all z.
If two power series with center z0:
k
a ak(z 2 z0)
q
k0
and
(13)
(14)
k
a bk(z 2 z0)
q
k0
represent the same function and have the same nonzero radius of convergence, then ak bk ,
k 0, 1, 2, . . . . Stated in another way, the power series expansion of a function with center z0 is
unique. On a practical level, this means that a power series expansion of an analytic function f
centered at z0, irrespective of the method used to obtain it, is the Taylor series expansion of the
function. For example, we can obtain (14) by simply differentiating (13) term by term. The
2
Maclaurin series for ez can be obtained by replacing the symbol z in (12) by z2.
EXAMPLE 1
Maclaurin Series
1
.
(1 2 z)2
SOLUTION We could, of course, begin by computing the coefficients using (8). However,
we know from (5) of Section 19.1 that for |z| 1,
1
1 z z2 z3 ….
(15)
12z
Differentiating both sides of the last result with respect to z then yields
q
1
1 2z 3z 2 p a kz k 2 1.
2
(1 2 z)
k1
Find the Maclaurin expansion of f (z) Since we are using Theorem 19.2.3, the radius of convergence of this last series is the same
as the original series, R 1.
19.2 Taylor Series
|
885
EXAMPLE 2
Taylor Series
1
in a Taylor series with center z0 2i.
12z
SOLUTION We shall solve this problem in two ways. We begin by using (8). From the first
several derivatives,
1
2 1
3 2
, f 0(z) , f -(z) ,
f 9(z) 2
3
(1 2 z)
(1 2 z)
(1 2 z)4
Expand f (z) we conclude that f (n)(z) n!/(1 z)n1 and so f (n)(2i) n!/(1 2i)n1. Thus from (8) we
obtain the Taylor series
q
1
1
a
(z 2 2i)k.
(16)
k1
12z
(1
2
2i)
k0
Since the distance from the center z0 2i to the nearest singularity z 1 is !5, we conclude
that the circle of convergence for the power series in (16) is |z 2i| !5. This can be verified by the ratio test of the preceding section.
ALTERNATIVE SOLUTION In this solution we again use the geometric series (15). By adding
and subtracting 2i in the denominator of 1/(1 z), we can write
1
1
1
1
1
.
12z
1 2 z 2i 2 2i
1 2 2i 2 (z 2 2i)
1 2 2i
z 2 2i
12
1 2 2i
1
We now write
as a power series by using (15) with the symbol z replaced
z 2 2i
12
1 2 2i
by (z 2i)/(1 2i):
1
z 2 2i
z 2 2i 2
z 2 2i 3
1
c1 a
b a
b pd
12z
1 2 2i
1 2 2i
1 2 2i
1 2 2i
1
1
1
1
(z 2 2i) (z 2 2i)2 (z 2 2i)3 p .
1 2 2i
(1 2 2i)2
(1 2 2i)3
(1 2 2i)4
The reader should verify that this last series is exactly the same as that in (16).
y
|z – 2i| = 5
In (15) and (16) we represented the same function 1/(1 z) by two different power series.
The first series
1
1 z z 2 z3 …
12z
has center zero and radius of convergence one. The second series
z*
x
|z| = 1
FIGURE 19.2.2 Series (15) and (16) both
converge within the shaded region
Exercises
19.2
1
1
1
1
1
(z 2 2i) (z 2 2i)2 (z 2 2i)3 p
12z
1 2 2i
(1 2 2i)2
(1 2 2i)3
(1 2 2i)4
has center 2i and radius of convergence !5. The two different circles of convergence are illustrated in FIGURE 19.2.2. The interior of the intersection of the two circles (shaded) is the region
where both series converge; in other words, at a specified point z* in this region, both series
converge to the same value f (z*) 1/(1 z*). Outside the shaded region, at least one of the two
series must diverge.
Answers to selected odd-numbered problems begin on page ANS-41.
In Problems 1–12, expand the given function in a Maclaurin
series. Give the radius of convergence of each series.
z
1z
1
3. f (z) (1 2z)2
1. f (z) 886
|
1
4 2 2z
z
4. f (z) (1 2 z)3
2. f (z) CHAPTER 19 Series and Residues
5. f (z) e2z
7. f (z) sinh z
6. f (z) zez
8. f (z) cosh z
2
z
10. f (z) sin 3z
2
11. f (z) sin z2
12. f (z) cos2z [Hint: Use a trigonometric identity.]
9. f (z) cos
In Problems 13–22, expand the given function in a Taylor series
centered at the indicated point. Give the radius of convergence
of each series.
13. f (z) 1/z, z0 1
14. f (z) 1/z, z0 1 i
1
1
15. f (z) , z0 2i
16. f (z) , z i
32z
1z 0
z21
1z
17. f (z) , z 1
18. f (z) , z i
32z 0
12z 0
19. f (z) cos z, z0 p/4
20. f (z) sin z, z0 p/2
21. f (z) ez, z0 3i
22. f (z) (z 1)e2z, z0 1
In Problems 23 and 24, use (7) to find the first three nonzero
terms of the Maclaurin series of the given function.
23. f (z) tan z
24. f (z) e1/(1z)
In Problems 25 and 26, use partial fractions as an aid in
obtaining the Maclaurin series for the given function. Give
the radius of convergence of the series.
i
z27
25. f (z) 26. f (z) 2
(z 2 i)(z 2 2i)
z 2 2z 2 3
In Problems 27 and 28, without actually expanding, determine
the radius of convergence of the Taylor series of the given
function centered at the indicated point.
4 5z
27. f (z) , z0 2 5i
1 z2
28. f (z) cot z, z0 pi
In Problems 29 and 30, expand the given function in the Taylor
series centered at the indicated points. Give the radius of convergence of each series. Sketch the region within which both series
converge.
1
29. f (z) , z 1, z0 i
2z 0
1
30. f (z) , z0 1 i, z0 3
z
31. (a) Suppose the principal branch of the logarithm f (z) Ln z loge|z| i Arg z is expanded in a Taylor series with center
z0 1 i. Explain why R 1 is the radius of the largest
circle centered at z0 1 i within which f is analytic.
(b) Show that within the circle |z (1 i)| 1 the Taylor
series for f is
q
1
3p
1 1i k
Ln z log e2 i2 a a
b (z 1 2 i)k.
2
4
2
k1 k
(c) Show that the radius of convergence for the power series
in part (b) is R !2. Explain why this does not contradict the result in part (a).
32. (a) Consider the function f (z) Ln(1 z). What is the radius
of the largest circle centered at the origin within which f
is analytic?
(b) Expand f in a Maclaurin series. What is the radius of
convergence of this series?
(c) Use the result in part (b) to find a Maclaurin series for
Ln(1 z).
1z
b.
(d) Find a Maclaurin series for Ln a
12z
In Problems 33 and 34, approximate the value of the given
expression using the indicated number of terms of a Maclaurin
series.
1i
33. e(1i)/10, three terms
34. sin a
b , two terms
10
35. In Section 15.1 we defined the error function as
erf (z) z
#e
"p
2
t 2
dt.
0
Find a Maclaurin series for erf (z).
36. Use the Maclaurin series for eiz to prove Euler’s formula for
complex z:
eiz cos z i sin z.
19.3 Laurent Series
INTRODUCTION If a complex function f fails to be analytic at a point z z0, then this point
is said to be a singularity or a singular point of the function. For example, the complex numbers
z 2i and z 2i are singularities of the function f (z) z /(z2 4) because f is discontinuous
at each of these points. Recall from Section 17.6 that the principal value of the logarithm, Ln z,
is analytic at all points except those points on the branch cut consisting of the nonpositive x-axis;
that is, the branch point z 0 as well as all negative real numbers are singular points of Ln z. In
this section we will be concerned with a new kind of “power series” expansion of f about an
isolated singularity z0. This new series will involve negative as well as nonnegative integer
powers of z z0.
Isolated Singularities Suppose that z z0 is a singularity of a complex function f.
The point z z0 is said to be an isolated singularity of the function f if there exists some deleted
neighborhood, or punctured open disk, 0 | z z0 | R of z0 throughout which f is analytic. For
example, we have just seen that z 2i and z 2i are singularities of f (z) z /(z2 4). Both
2i and 2i are isolated singularities since f is analytic at every point in the neighborhood defined
by |z 2i| 1 except at z 2i and at every point in the neighborhood defined by | z (2i)| 1
except at z 2i. In other words, if f is analytic in the deleted neighborhood, 0 |z 2i| 1
19.3 Laurent Series
|
887
and 0 | z 2i | 1. On the other hand, the branch point z 0 is not an isolated singularity of
Ln z since every neighborhood of z 0 must contain points on the negative x-axis. We say that
a singular point z z0 of a function f is nonisolated if every neighborhood of z0 contains at least
one singularity of f other than z0. For example, the branch point z 0 is a nonisolated singularity of Ln z since every neighborhood of z 0 contains points on the negative real axis.
A New Kind of Series If z z0 is a singularity of a function f, then certainly f cannot be
expanded in a power series with z0 as its center. However, about an isolated singularity z z0 it
is possible to represent f by a new kind of series involving both negative and nonnegative integer
powers of z z0; that is,
f (z) … a2
a1
a0 a1(z z0) a2(z z0)2 ….
2
z
2 z0
(z 2 z0)
Using summation notation, the last expression can be written as the sum of two series
f (z) a ak(z 2 z0)k a ak(z 2 z0)k.
q
q
k1
k0
(1)
The two series on the right-hand side in (1) are given special names. The part with negative
powers of z z0; that is,
q
q
ak
k
a ak(z 2 z0) a (z 2 z )k
k1
k1
0
is called the principal part of the series (1) and will converge for |1/(z z0)| r* or equivalently
for |z z0 | 1/r* r. The part consisting of the nonnegative powers of z z0,
k
a ak(z 2 z0)
q
k0
is called the analytic part of the series (1) and will converge for |z z0| R. Hence, the sum of
these parts converges when z both |z z0| r and |z z0| R; that is, when z is a point in an
annular domain defined by r |z z0| R.
By summing over negative and nonnegative integers, (1) can be written compactly as
f (z) a ak(z 2 z0)k.
q
k q
The next example illustrates a series of the form (1) in which the principal part of the series
consists of a finite number of nonzero terms, but the analytic part consists of an infinite number
of nonzero terms.
EXAMPLE 1
A New Kind of Series
The function f (z) (sin z)/z3 is not analytic at z 0 and hence cannot be expanded in a
Maclaurin series. However, sin z is an entire function, and from (13) of Section 19.2 we know
that its Maclaurin series,
sin z z 2
z3
z5
z7
2
p,
3!
5!
7!
converges for all z. Dividing this power series by z3 gives the following series with negative
and nonnegative integer powers of z:
f (z) sin z
1
1
z2
z4
2
2
p.
3!
5!
7!
z3
z2
(2)
This series converges for all z except z 0; that is, for 0 |z|.
A series representation of a function f that has the form given in (1), and (2) is such an example,
is called a Laurent series or a Laurent expansion of f.
888
|
CHAPTER 19 Series and Residues
Laurent’s Theorem
Theorem 19.3.1
C
Let f be analytic within the annular domain D defined by r |z z0| R. Then f has the
series representation
f (z) a ak(z 2 z0)k
q
R
z0
(3)
k q
r
valid for r |z z0| R. The coefficients ak are given by
D
ak FIGURE 19.3.1 Contour in Theorem 19.3.1
f (s)
1
ds, k 0, 1, 2, . . .,
2pi B
(s
2
z0)k 1
C
(4)
䉲
where C is a simple closed curve that lies entirely within D and has z0 in its interior (see
FIGURE 19.3.1).
PROOF: Let C1 and C2 be concentric circles with center z0 and radii r1 and R2, where
C2
r r1 R2 R. Let z be a fixed point in D that also satisfies r1 |z z0| R2. See FIGURE 19.3.2.
By introducing a cross cut between C2 and C1, we find from Cauchy’s integral formula that
C1
z0
f (z) f (s)
f (s)
1
1
ds 2
ds.
s
2
z
s
2pi B
2pi
BC1 2 z
C2
䉲
(5)
䉲
Proceeding as in the proof of Theorem 19.2.4, we can write
q
f (s)
1
ds a ak(z 2 z0)k,
2pi B
C2 s 2 z
k0
z
FIGURE 19.3.2 C1 and C2 are concentric
circles
(6)
䉲
ak where
f (s)
1
ds, k 0, 1, 2, . . . .
k1
2pi B
C2 (s 2 z0)
(7)
䉲
Now using (5) and (8) of Section 19.1, we have
1
2pi
BC s 2 z
f (s)
䉲
1
2pi
BC (z 2 z0) 2 (s 2 z0)
1
2pi
f (s)
z
BC1 2 z0 •
s 2 z0
s 2 z0 2
f (s)
1
e1 a
b p
z 2 z0
z 2 z0
2pi B
C1 z 2 z0
ds 1
f (s)
䉲
ds
1
䉲
1
ds
s 2 z0 ¶
12
z 2 z0
䉲
(8)
s 2 z0 n 2 1
(s 2 z0)n
a
b
r ds
z 2 z0
(z 2 s)(z 2 z0)n 2 1
n
ak
a
Rn(z),
(z
2
z0)k
k1
where
and
ak f (s)
1
ds, k 1, 2, 3, . . .,
k 1
2pi B
C1 (s 2 z0)
Rn(z) (9)
䉲
f (s)(s 2 z0)n
1
ds.
z2s
2pi(z 2 z0)n B
C1
䉲
Now let d denote the distance from z to z0; that is, |z z0| d, and let M denote the maximum
value of | f (z)| on the contour C1. Since |s z0| r1,
|z s| |z z0 (s z0)| |z z0| |s z0| d r1.
19.3 Laurent Series
|
889
The ML-inequality then gives
ZRn(z)Z 2
1
2pi(z 2 z0)n
f (s) (s 2 z0)n
Mr n1
Mr1
r1 n
1
ds 2 #
2pr
a
b .
1
2pd n d 2 r1
d 2 r1 d
BC1 z 2 s
䉲
Because r1 d, (r1/d)n S 0 as n S q and so |Rn(z)| S 0 as n S q. Thus we have shown that
1
2pi
q
ak
f (s)
ds a
,
(z
2
z0)k
BC1 s 2 z
k1
䉲
(10)
where the coefficients ak are given in (9). Combining (6) and (10), we see that (5) yields
q
q
ak
f (z) a ak(z 2 z0)k a
.
(11)
k
k0
k 1 (z 2 z0)
Finally, by summing over nonnegative and negative integers, we can write (11) as
q
f (z) g k q ak (z z0)k. However, (7) and (9) can be written as a single integral:
ak 1
2pi
f (s)
ds, k 0, 1, 2, . . .,
BC (s 2 z0)k 1
䉲
where, in view of (3) of Section 18.2, we have replaced the contours C1 and C2 by any simple
closed contour C in D with z0 in its interior.
In the case when ak 0 for k 1, 2, 3, . . ., the Laurent series (3) is a Taylor series. Because
of this, a Laurent expansion is a generalization of a Taylor series.
The annular domain in Theorem 19.3.1 defined by r |z z0| R need not have the “ring”
shape illustrated in Figure 19.3.2. Some other possible annular domains are (i) r 0, R finite;
(ii) r 0, R S q; and (iii) r 0, R S q. In the first case, the series converges in the annular
domain defined by 0 |z z0| R. This is the interior of the circle |z z0| R except the
point z0. In the second case, the annular domain is defined by r |z z0|; in other words, the
domain consists of all points exterior to the circle |z z0| r. In the third case, the domain is
defined by 0 |z z0|. This represents the entire complex plane except the point z0. The series
we obtained in (2) is valid on this last type of domain.
In actual practice, the formula in (4) for the coefficients of a Laurent series is seldom used. As
a consequence, finding the Laurent series of a function in a specified annular domain is generally
not an easy task. We often use the geometric series (5) and (6) of Section 19.1 or, as we did in
Example 1, other known series. But regardless of how a Laurent expansion of a function f is obtained
in a specified annular domain, it is the Laurent series; that is, the series we obtain is unique.
EXAMPLE 2
y
y
Laurent Expansions
1
in a Laurent series valid for (a) 0 |z| 1, (b) 1 |z|, (c) 0 |z 1| 1,
z(z 2 1)
and (d) 1 |z 1|.
Expand f (z) 0
1
x
x
0
(a)
1
(b)
y
y
SOLUTION The four specified annular domains are shown in FIGURE 19.3.3. In parts (a) and (b),
we want to represent f in a series involving only negative and nonnegative integer powers of z,
whereas in parts (c) and (d) we want to represent f in a series involving negative and nonnegative
integer powers of z 1.
(a) By writing
f (z) 1 1
,
z 12z
we can use (5) of Section 19.1:
0
x
1
(c)
0
1
(d)
FIGURE 19.3.3 Annular domains for
Example 2
890
|
x
1
[1 z z2 z3 …].
z
The series in the brackets converges for |z| 1, but after this expression is multiplied by 1/z,
the resulting series
1
f (z) 1 z z2 …
z
converges for 0 |z| 1.
CHAPTER 19 Series and Residues
f (z) (b) To obtain a series that converges for 1 |z| we start by constructing a series that converges
for |1/z| 1. To this end we write the given function f as
f (z) 1
z2
1
12
1
z
and again use (5) of Section 19.1 with z replaced by 1/z:
f (z) 1
1
1
1
c1 2 3 p d .
z
z2
z
z
The series in the brackets converges for |1/z| 1 or equivalently for 1 |z|. Thus the required
Laurent series is
f (z) 1
1
1
1
3 4 5 p.
z2
z
z
z
(c) This is basically the same problem as part (a) except that we want all powers of z 1. To
that end we add and subtract 1 in the denominator and use (6) of Section 19.1 with z replaced
by z 1:
1
f (z) (1 2 1 z) (z 2 1)
1
1
z 2 1 1 (z 2 1)
1
f1 2 (z 2 1) (z 2 1)2 2 (z 2 1)3 p g
z21
1
2 1 (z 2 1) 2 (z 2 1)2 p .
z21
The series in brackets converges for |z 1| 1, and so the last series converges for
0 |z 1| 1.
(d) Proceeding as in part (b), we write
f (z) 1
1
1
z 2 1 1 (z 2 1)
(z 2 1)2
1
1
z21
1
1
1
1
c1 2
2
pd
2
2
z21
(z 2 1)
(z 2 1)
(z 2 1)3
1
1
1
1
1
2
2
p.
(z 2 1)2
(z 2 1)3
(z 2 1)4
(z 2 1)5
Because the series within the brackets converges for |1/(z 1)| 1, the final series converges
for 1 |z 1|.
EXAMPLE 3
Laurent Expansions
1
in a Laurent series valid for (a) 0 |z 1| 2 and
(z 2 1)2(z 2 3)
(b) 0 |z 3| 2.
Expand f (z) SOLUTION (a) As in parts (c) and (d) of Example 2, we want only powers of z 1, and so we
need to express z 3 in terms of z 1. This can be done by writing
f (z) 1
1
1
2
2
(z 2 1) (z 2 3)
(z 2 1) 2 (z 2 1)
1
2(z 2 1)2
1
z21
12
2
19.3 Laurent Series
|
891
and then using (5) of Section 19.1 with z replaced by (z 1)/2:
f (z) (z 2 1)2
(z 2 1)3
1
z21
c1
pd
2
2(z 2 1)2
22
23
1
1
1
1
2
2 2
(z 2 1) 2 p .
2
4(z 2 1)
8
16
2(z 2 1)
(12)
(b) To obtain powers of z 3 we write z 1 2 (z 3) and
f (z) 1
1
1
z 2 3 2
f2 (z 2 3)g 2 c1 d .
2
z23
4(z 2 3)
2
(z 2 1) (z 2 3)
At this point we can expand c1 theorem:
f (z) z 2 3 2
d in a power series using the general binomial
2
(2) z 2 3
(2)(3) z 2 3 2
(2)(3) (4) z 2 3 3
1
c1 a
b a
b a
b pd.
4(z 2 3)
1!
2
2!
2
3!
2
The binomial series is valid for |(z 3)/2| 1 or |z 3| 2. Multiplying this series by
1/4(z 3) gives a Laurent series that is valid for 0 |z 3| 2:
f (z) 1
1
3
1
2 (z 2 3) 2 (z 2 3)2 p .
4(z 2 3)
4
16
8
A Laurent Expansion
EXAMPLE 4
Expand f (z) 8z 1
in a Laurent series valid for 0 |z| 1.
z(1 2 z)
SOLUTION By (5) of Section 19.1 we can write
f (z) 8z 1
8z 1 1
1
a8 b (1 z z 2 z 3 p ).
z
z
z(1 2 z)
12z
We then multiply the series by 8 1/z and collect like terms:
1
f (z) 9 9z 9z2 ….
z
The geometric series converges for |z| 1. After multiplying by 8 1/z, the resulting Laurent
series is valid for 0 |z| 1.
In the preceding examples, the point at the center of the annular domain of validity for each
Laurent series was an isolated singularity of the function f. A reexamination of Theorem 19.3.1
shows that this need not be the case.
y
EXAMPLE 5
Expand f (z) 0
1
2
x
|
1
in a Laurent series valid for 1 |z 2| 2.
z(z 2 1)
SOLUTION The specified annular domain is shown in FIGURE 19.3.4. The center of this domain,
z 2, is a point of analyticity of the function f. Our goal now is to find two series involving
integer powers of z 2: one converging for 1 |z 2| and the other converging for |z 2| 2.
To accomplish this, we start with the decomposition of f into partial fractions:
FIGURE 19.3.4 Annular domain for
Example 5
892
A Laurent Expansion
CHAPTER 19 Series and Residues
1
1
f (z) f1(z) f2(z).
z
z21
(13)
1
1
f1(z) z
2z22
Now,
1
2
1
z22
1
2
(z 2 2)2
(z 2 2)3
1
z22
c1 2
2
pd
2
2
22
23
(z 2 2)2
(z 2 2)3
1
z22
2
2 p.
2
22
23
24
This series converges for |(z 2)/2| 1 or |z 2| 2. Furthermore,
f2(z) 1
1
1
z21
1z22
z22
1
1
z22
1
1
1
1
c1 2
2
pd
z22
z22
(z 2 2)2
(z 2 2)3
1
1
1
1
1
2
2
p
2
3
z22
(z 2 2)
(z 2 2)
(z 2 2)4
converges for |1/(z 2)| 1 or 1 |z 2|. Substituting these two results in (13) then gives
f (z) p 2
(z 2 2)2
(z 2 2)3
1
1
1
1
1
z22
2
2
2
2 p.
z22
2
(z 2 2)4
(z 2 2)3
(z 2 2)2
22
23
24
This representation is valid for 1 |z 2| 2.
A Laurent Expansion
EXAMPLE 6
Expand f (z) e
3/z
in a Laurent series valid for 0 |z|.
SOLUTION From (12) of Section 19.2 we know that for all finite z,
ez 1 z z2
z3
p.
2!
3!
(14)
By replacing z in (14) by 3/z, z 0, we obtain the Laurent series
e 3>z 1 32
3
33
p.
2
z
2!z
3!z 3
This series is valid for 0 |z|.
REMARKS
In conclusion, we point out a result that will be of particular importance to us in Sections 19.5
and 19.6. Replacing the complex variable s with the usual symbol z, we see that when k 1,
(4) for the Laurent series coefficients yields
a1 1
f (z) dz
2pi B
C
or, more importantly, the integral can be written as
BC
f (z) dz 2pi a1.
(15)
19.3 Laurent Series
|
893
Exercises
19.3
Answers to selected odd-numbered problems begin on page ANS-42.
15. 0 |z 1| 1
In Problems 1–6, expand the given function in a Laurent series
valid for the indicated annular domain.
1.
2.
3.
4.
5.
6.
16. 0 |z 2| 1
z
in a Laurent
(z 1)(z 2 2)
series valid for the indicated annular domain.
17. 0 |z 1| 3
18. |z 1| 3
19. 1 |z| 2
20. 0 |z 2| 3
1
In Problems 21 and 22, expand f (z) in a Laurent
z(1 2 z)2
series valid for the indicated annular domain.
21. 0 |z| 1
22. |z| 1
In Problems 17–20, expand f (z) cos z
, 0 |z|
f (z) z
z 2 sin z
f (z) , 0 |z|
z5
2
f (z) e21>z , 0 |z|
1 2 ez
f (z) , 0 |z|
z2
z
e
f (z) , 0 |z 1|
z21
1
f (z) z cos , 0 |z|
z
1
in a
(z 2 2)(z 2 1)3
Laurent series valid for the indicated annular domain.
23. 0 |z 2| 1
24. 0 |z 1| 1
In Problems 23 and 24, expand f (z) 1
in a Laurent series
z(z 2 3)
valid for the indicated annular domain.
7. 0 |z| 3
8. |z| 3
9. 0 |z 3| 3
10. |z 3| 3
11. 1 |z 4| 4
12. 1 |z 1| 4
1
in a Laurent
In Problems 13–16, expand f (z) (z 2 1)(z 2 2)
series valid for the indicated annular domain.
13. 1 |z| 2
14. |z| 2
In Problems 7–12, expand f (z) 7z 2 3
in a Laurent
z(z 2 1)
series valid for the indicated annular domain.
25. 0 |z| 1
26. 0 |z 1| 1
In Problems 25 and 26, expand f (z) z2 2 2z 1 2
in a Laurent
z22
series valid for the indicated annular domain.
27. 1 |z 1|
28. 0 |z 2|
In Problems 27 and 28, expand f (z) 19.4 Zeros and Poles
INTRODUCTION Suppose that z z0 is an isolated singularity of a function f and that
f (z) a ak(z 2 z0)k a ak(z 2 z0)k a ak(z 2 z0)k
q
q
q
k q
k1
k0
(1)
is the Laurent series representation of f valid for the punctured open disk 0 |z z0| R. We saw
in the preceding section that a Laurent series (1) consists of two parts. That part of the series in
(1) with negative powers of z z0, namely,
q
q
ak
k
a ak(z 2 z0) a (z 2 z )k
k1
k1
0
(2)
is the principal part of the series. In the discussion that follows we will assign different names
to the isolated singularity z z0 according to the number of terms in the principal part.
Classification of Isolated Singular Points An isolated singular point z z0 of a
complex function f is given a classification depending on whether the principal part (2) of its
Laurent expansion (1) contains zero, a finite number, or an infinite number of terms.
If the principal part is zero; that is, all the coefficients ak in (2) are zero, then z z0 is
called a removable singularity.
(ii) If the principal part contains a finite number of nonzero terms, then z z0 is called a pole.
If, in this case, the last nonzero coefficient in (2) is an, n 1, then we say that z z0 is a
pole of order n. If z z0 is a pole of order 1, then the principal part (2) contains exactly
one term with coefficient a1. A pole of order 1 is commonly called a simple pole.
(i)
894
|
CHAPTER 19 Series and Residues
(iii) If the principal part (2) contains infinitely many nonzero terms, then z z0 is called an
essential singularity.
The following table summarizes the form of the Laurent series for a function f when z z0
is one of the above types of isolated singularities. Of course, R in the table could be q.
z z0
Removable singularity
Laurent Series
a0 a1(z z0) a2(z z0)2 …
Pole of order n
a(n 2 1)
an
a1
p
a0 a1(z 2 z0) p
n21
z
2 z0
(z 2 z0)n
(z 2 z0)
Simple pole
a1
a0 a1(z 2 z0) a2(z 2 z0)2 p
z 2 z0
Essential singularity
p
EXAMPLE 1
a2
a1
a0 a1(z 2 z0) a2(z 2 z0)2 p
2
z
2 z0
(z 2 z0)
Removable Discontinuity
Proceeding as we did in (2) of Section 19.3, we see from
z2
sin z
z4
12
2p
z
3!
5!
(3)
that z 0 is a removable singularity of the function f (z) (sin z)/z.
If a function f has a removable singularity at the point z z0, then we can always supply an
appropriate definition for the value of f (z0) so that f becomes analytic at the point. For instance,
since the right side of (3) is 1 at z 0, it makes sense to define f (0) 1. With this definition, the
function f (z) (sin z)/z in Example 1 is now analytic at z 0.
EXAMPLE 2
Poles and Essential Singularity
(a) From
principal part
T
z
sin z
1
z3
2
2 p,
z
3!
5!
z2
0 |z|, we see that a1 0, and so z 0 is a simple pole of the function f (z) (sin z)/z2. The
function f (z) (sin z)/z3 represented by the series in (2) of Section 19.3 has a pole of order 2
at z 0.
(b) In Example 3 of Section 19.3 we showed that the Laurent expansion of f (z) 1/(z 1)2(z 3)
valid for 0 |z 1| 2 was
principal part
f (z) 1
1
1
z21
2
2 2
2 p.
4(z 2 1)
8
16
2(z 2 1)2
Since a2 0, we conclude that z 1 is a pole of order 2.
(c) From Example 6 of Section 19.3 we see from the Laurent series that the principal part
of the function f (z) e3/z contains an infinite number of terms. Thus z 0 is an essential
singularity.
In part (b) of Example 2 in Section 19.3, we showed that the Laurent series representation of
f (z) 1/z(z 1) valid for 1 |z| is
f (z) 1
1
1
1
3 4 5 p.
z2
z
z
z
19.4 Zeros and Poles
|
895
The point z 0 is an isolated singularity of f and the Laurent series contains an infinite number
of terms involving negative integer powers of z. Does this mean that z 0 is an essential singularity of f ? The answer is “no.” For this particular function, a reexamination of (1) shows that
the Laurent series we are interested in is the one with the annular domain 0 |z| 1. From
part (a) of that same example we saw that
1
f (z) 1 z z2 …
z
was valid for 0 |z| 1. Thus we see that z 0 is a simple pole.
Zeros Recall that z0 is a zero of a function f if f (z0) 0. An analytic function f has a zero
of order n at z z0 if
f (z0) 0, f (z0) 0, f (z0) 0, …, f (n1)(z0) 0, but f (n)(z0) 0.
(4)
For example, for f (z) (z 5)3 we see that f (5) 0, f (5) 0, f (5) 0, but f (5) 6. Thus
z 5 is a zero of order 3. If an analytic function f has a zero of order n at z z0, it follows
from (4) that the Taylor series expansion of f centered at z0 must have the form
f (z) an(z z0)n an1(z z0)n1 an2(z z0)n2 …
(z z0)n [an an1(z z0) an2(z z0)2 …],
(5)
where an 0.
EXAMPLE 3
Order of a Zero
The analytic function f (z) z sin z2 has a zero at z 0. By replacing z by z2 in (13) of
Section 19.2, we obtain
sin z 2 z 2 2
and so
z6
z 10
2p
3!
5!
f (z) z sin z 2 z 3 c1 2
z4
z8
2 pd.
3!
5!
Comparing the last result with (5) we see that z 0 is a zero of order 3.
A zero z0 of a nontrivial analytic function f is isolated in the sense that there exists some
neighborhood of z0 for which f (z) 0 at every point z in that neighborhood except at z z0. As
a consequence, if z0 is a zero of a nontrivial analytic function f , then the function 1/f (z) has an
isolated singularity at the point z z0. The following result enables us, in some circumstances,
to determine the poles of a function by inspection.
Theorem 19.4.1
Pole of Order n
If the functions f and g are analytic at z z0 and f has a zero of order n at z z0 and g(z0) 0,
then the function F (z) g(z)/f (z) has a pole of order n at z z0.
EXAMPLE 4
Order of Poles
(a) Inspection of the rational function
F(z) 2z 5
(z 2 1)(z 5)(z 2 2)4
shows that the denominator has zeros of order 1 at z 1 and z 5, and a zero of order 4 at
z 2. Since the numerator is not zero at these points, it follows from Theorem 19.4.1 that F
has simple poles at z 1 and z 5, and a pole of order 4 at z 2.
(b) In Example 3 we saw that z 0 is a zero of order 3 of f (z) z sin z2. From Theorem 19.4.1,
we conclude that the function F (z) 1/(z sin z2) has a pole of order 3 at z 0.
From the preceding discussion, it should be intuitively clear that if a function has a pole at
z z0, then | f (z)| S q as z S z0 from any direction.
896
|
CHAPTER 19 Series and Residues
19.4
Exercises
Answers to selected odd-numbered problems begin on page ANS-42.
In Problems 1 and 2, show that z 0 is a removable singularity
of the given function. Supply a definition of f (0) so that f is
analytic at z 0.
1. f (z) e2z 2 1
z
2. f (z) sin 4z 2 4z
z2
In Problems 3–8, determine the zeros and their orders for the
given function.
3. f (z) (z 2 i)2
4. f (z) z4 16
9
5. f (z) z4 z2
6. f (z) z z
7. f (z) e2z ez
8. f (z) sin2z
In Problems 9–12, the indicated number is a zero of the given
function. Use a Maclaurin or Taylor series to determine the
order of the zero.
9. f (z) z(1 cos z2); z 0
10. f (z) z sin z; z 0
11. f (z) 1 ez1; z 1
12. f (z) 1 pi z ez; z pi
In Problems 13–24, determine the order of the poles for the
given function.
3z 2 1
6
13. f (z) 2
14. f (z) 5 2
z 1 2z 1 5
z
1 4i
z21
15. f (z) 16. f (z) (z 2)(z i)4
(z 1)(z 3 1)
cot pz
17. f (z) tan z
18. f (z) z2
z
1 2 cosh z
e
19. f (z) 20. f (z) 2
z4
z
1
ez 2 1
21. f (z) 22.
f
(z)
1 2 ez
z4
sin z
cos z 2 cos 2z
23. f (z) 2
24. f (z) z 2z
z6
25. Determine whether z 0 is an isolated or nonisolated
singularity of f (z) tan (1/z).
26. Show that z 0 is an essential singularity of f (z) z3 sin (1/z).
19.5 Residues and Residue Theorem
INTRODUCTION We saw in the last section that if the complex function f has an isolated
singularity at the point z0, then f has a Laurent series representation
q
a2
a1
f (z) a ak(z 2 z0)k p a0 a1(z 2 z0) p ,
2
z
2 z0
(z
2
z
)
k q
0
which converges for all z near z0. More precisely, the representation is valid in some deleted
neighborhood of z0, or punctured open disk, 0 |z z0| R. In this section our entire focus will
be on the coefficient a1 and its importance in the evaluation of contour integrals.
Residue The coefficient a1 of 1/(z z0) in the Laurent series given above is called the
residue of the function f at the isolated singularity z0. We shall use the notation
a1 Res ( f (z), z0)
to denote the residue of f at z0. Recall, if the principal part of the Laurent series valid for
0 |z z0| R contains a finite number of terms with an the last nonzero coefficient, then z0
is a pole of order n; if the principal part of the series contains an infinite number of terms with
nonzero coefficients, then z0 is an essential singularity.
EXAMPLE 1
Residues
(a) In Example 2 of Section 19.4 we saw that z 1 is a pole of order 2 of the function
f (z) 1/(z 1)2 (z 3). From the Laurent series given in that example we see that the coefficient of 1/(z 1) is a1 Res ( f (z), 1) 14 .
(b) Example 6 of Section 19.3 showed that z 0 is an essential singularity of f (z) e3/z.
From the Laurent series given in that example we see that the coefficient of 1/z is
a1 Res ( f (z), 0) 3.
19.5 Residues and Residue Theorem
|
897
Later on in this section we will see why the coefficient a1 is so important. In the meantime
we are going to examine ways of obtaining this complex number when z0 is a pole of a function
f without the necessity of expanding f in a Laurent series at z0. We begin with the residue at a
simple pole.
Residue at a Simple Pole
Theorem 19.5.1
If f has a simple pole at z z0, then
Res ( f (z), z0) lim (z z0) f (z).
zSz0
(1)
PROOF: Since z z0 is a simple pole, the Laurent expansion of f about that point has the
form
f (z) a1
a0 a1(z z0) a2(z z0)2 ….
z 2 z0
By multiplying both sides by z z 0 and then taking the limit as z S z 0, we obtain
lim (z z0) f (z) lim [a1 a0(z z0) a1(z z0)2 …] a1 Res ( f (z), z0).
zSz0
zSz0
Residue at a Pole of Order n
Theorem 19.5.2
If f has a pole of order n at z z0, then
Res ( f (z), z0) 1
d n21
lim n 2 1 (z 2 z0)n f (z).
(n 2 1)! zSz0 dz
(2)
PROOF: Since f is assumed to have a pole of order n, its Laurent expansion for 0 |z z0| R
must have the form
f (z) an
p a2 a1 a0 a1(z 2 z0) p .
n z 2 z0
(z 2 z0)
(z 2 z0)2
We multiply the last expression by (z z0)n :
(z z0)nf (z) an . . . a2(z z0)n2 a1(z z0)n1 a0(z z0)n a1(z z0)n1 . . .
and then differentiate n 1 times:
d n21
(z z0)nf (z) (n 1)!a1 n!a0(z z0) . . . .
dz n 2 1
(3)
Since all the terms on the right side after the first involve positive integer powers of z z0, the
limit of (3) as z S z0 is
d n21
(z 2 z0)nf (z) (n 2 1)!a1.
n21
zSz0 dz
lim
Solving the last equation for a1 gives (2).
Note that (2) reduces to (1) when n 1.
EXAMPLE 2
Residue at a Pole
1
has a simple pole at z 3 and a pole of order 2 at z 1.
(z 2 1)2 (z 2 3)
Use Theorems 19.5.1 and 19.5.2 to find the residue at each pole.
The function f (z) 898
|
CHAPTER 19 Series and Residues
SOLUTION
Since z 3 is a simple pole, we use (1):
Res ( f (z), 3) lim (z 2 3) f (z) lim
zS3
zS3
1
1
.
4
(z 2 1)2
Now at the pole of order 2 it follows from (2) that
Res ( f (z), 1) 1
d
lim (z 2 1)2 f (z)
1! zS1 dz
lim
d
1
dz z 2 3
lim
1
1
.
2
4
(z 2 3)
zS1
zS1
When f is not a rational function, calculating residues by means of (1) can sometimes be
tedious. It is possible to devise alternative residue formulas. In particular, suppose a function f
can be written as a quotient f (z) g(z)/h(z), where g and h are analytic at z z0. If g(z0) 0 and
if the function h has a zero of order 1 at z0, then f has a simple pole at z z0 and
An alternative method for
computing a residue at a
simple pole.
Res ( f (z), z0) g(z0)
.
h9(z0)
(4)
To see this last result, we use (1) and the facts that h(z0) 0 and that limzSz0(h(z) h(z0))/(z z0)
is a definition of the derivative h (z0):
Res ( f (z), z0) lim (z 2 z0)
zSz0
g(z0)
g(z)
g(z)
lim
.
h(z)
zSz0 h(z) 2 h(z0)
h9(z0)
z 2 z0
Analogous formulas for residues at poles of order greater than 1 are complicated and will not
be given.
EXAMPLE 3
Using (4) to Compute a Residue
The polynomial z4 1 can be factored as (z z1)(z z2)(z z3)(z z4), where z1, z2, z3,
and z4 are the four distinct roots of the equation z4 1 0. It follows from Theorem 19.4.1
that the function
f (z) 1
z4 1
has four simple poles. Now from (10) of Section 17.2 we have z1 epi/4, z2 e3pi/4, z3 e5pi/4,
z4 e7pi/4. To compute the residues, we use (4) and Euler’s formula:
Res ( f (z), z1) 1
1
1
1
e3pi>4 2
i
3
4
4z 1
4"2
4"2
Res ( f (z), z2) 1
1
1
1
e9pi>4 2
i
3
4
4z 2
4"2
4"2
Res ( f (z), z3) 1
1
1
1
e15pi>4 i
3
4
4z 3
4"2
4"2
Res ( f (z), z4) 1
1
1
1
e21pi>4 i.
3
4
4z 4
4"2
4"2
Residue Theorem We come now to the reason for the importance of the residue concept.
The next theorem states that under some circumstances, we can evaluate complex integrals 养C f (z) dz
by summing the residues at the isolated singularities of f within the closed contour C.
䉲
19.5 Residues and Residue Theorem
|
899
Let D be a simply connected domain and C a simple closed contour lying entirely within D.
If a function f is analytic on and within C, except at a finite number of singular points
z1, z2, . . ., zn within C, then
Cn
C
Cauchy’s Residue Theorem
Theorem 19.5.3
zn
f (z) dz 2pi a Res ( f (z), zk).
B
n
䉲
C2
z1
D
z2
C1
(5)
k1
C
PROOF: Suppose C1, C2, . . ., Cn are circles centered at z1, z2, . . ., zn, respectively. Suppose further
that each circle Ck has a radius rk small enough so that C1, C2, . . ., Cn are mutually disjoint and are
interior to the simple closed curve C. See FIGURE 19.5.1. Recall that (15) of Section 19.3 implies
养Ck f (z) dz 2pi Res( f (z), zk), and so Theorem 18.2.2 gives
䉲
f (z) dz a
B
n
FIGURE 19.5.1 n singular points within
contour C
䉲
k1
C
f (z) dz 2pi a Res ( f (z), zk).
B
n
䉲
k1
Ck
Evaluation by the Residue Theorem
EXAMPLE 4
1
dz , where
BC (z 2 1)2 (z 2 3)
(a) the contour C is the rectangle defined by x 0, x 4, y 1, y 1, and
(b) the contour C is the circle |z| 2.
Evaluate
䉲
SOLUTION
(a) Since both poles z 1 and z 3 lie within the square, we have from (5) that
1
dz 2pi fRes ( f (z), 1) Res ( f (z), 3)g.
BC (z 2 1)2 (z 2 3)
䉲
We found these residues in Examples 2 and 3, and so
1
1
1
dz 2pi c d 0.
2
4
4
BC (z 2 1) (z 2 3)
䉲
(b) Since only the pole z 1 lies within the circle |z| 2, we have from (5) that
1
1
p
dz 2pi Res ( f (z), 1) 2pi a b i.
2
4
2
BC (z 2 1) (z 2 3)
䉲
Evaluation by the Residue Theorem
EXAMPLE 5
Evaluate
BC z 2 4
䉲
2z 6
dz , where the contour C is the circle |z i| 2.
SOLUTION By writing z2 4 (z 2i)(z 2i), we see that the integrand has simple poles
at 2i and 2i. Now since only 2i lies within the contour C, it follows from (5) that
BC z 2 4
䉲
But
2z 6
Res ( f (z), 2i) lim (z 2 2i)
zS2i
Hence,
900
|
CHAPTER 19 Series and Residues
dz 2pi Res ( f (z), 2i).
BC z 2 4
䉲
2z 6
2z 6
(z 2 2i)(z 2i)
6 4i
3 2i
.
4i
2i
dz 2pi a
3 2i
b p(3 2i).
2i
Evaluation by the Residue Theorem
EXAMPLE 6
e
dz , where the contour C is the circle |z| 2.
BC z 5z 3
z
Evaluate
䉲
4
SOLUTION Since z4 5z3 z3 (z 5) we see that the integrand has a pole of order 3 at
z 0 and a simple pole at z 5. Since only z 0 lies within the given contour, we have
from (5) and (2)
ez
dz 2pi Res ( f (z), 0)
BC z 4 5z 3
1
d2
ez
2pi lim 2 z 3 3
2! zS0 dz
z (z 5)
䉲
pi lim
zS0
Evaluation by the Residue Theorem
EXAMPLE 7
Evaluate
BC
䉲
(z 2 8z 17)e z
17p
i.
3
125
(z 5)
tan z dz, where the contour C is the circle |z| 2.
SOLUTION The integrand tan z sin z/cos z has simple poles at the points where cos z 0.
We saw in Section 17.7 that the only zeros for cos z are the real numbers z (2n 1)p/2,
n 0, 1, 2, . . . . Since only p/2 and p/2 are within the circle |z| 2, we have
p
p
tan z dz 2pi cRes af (z), b Res af (z), b d .
2
2
BC
䉲
Now from (4) with g(z) sin z, h(z) cos z, and h (z) sin z, we see that
sin (p>2)
sin (p>2)
p
p
Res af (z), b 1 and Res af (z), b 1.
2
sin (p>2)
2
sin (p>2)
BC
Therefore,
EXAMPLE 8
Evaluate
BC
䉲
䉲
tan z dz 2pif1 2 1g 4pi.
Evaluation by the Residue Theorem
e 3>z dz, where the contour C is the circle |z| 1.
SOLUTION As we have seen, z 0 is an essential singularity of the integrand f (z) e3/z and
so neither formula (1) nor (2) is applicable to find the residue of f at that point. Nevertheless,
we saw in part (b) of Example 1 that the Laurent series of f at z 0 gives Res( f (z), 0) 3.
Hence from (5) we have
BC
䉲
e 3>z dz 2pi Res ( f (z), 0) 6pi.
REMARKS
In the application of the limit formulas (1) and (2) for computing residues, the indeterminate
form 0/0 may result. Although we are not going to prove it, it should be pointed out that
L’Hôpital’s rule is valid in complex analysis. If f (z) g(z)/h(z), where g and h are analytic
at z z0, g(z0) 0, h(z0) 0, and h (z0) 0, then
lim
zSz0
g9(z0)
g(z)
.
h(z)
h9(z0)
19.5 Residues and Residue Theorem
|
901
Exercises
19.5
Answers to selected odd-numbered problems begin on page ANS-42.
In Problems 1–6, use a Laurent series to find the indicated residue.
2
1. f (z) ; Res ( f (z), 1)
(z 2 1)(z 4)
1
2. f (z) 3
; Res ( f (z), 0)
z (1 2 z)3
4z 2 6
3. f (z) ; Res ( f (z), 0)
z(2 2 z)
2
4. f (z) (z 3)2 sin
; Res ( f (z), 3)
z3
2
5. f (z) e 2>z ; Res ( f (z), 0)
e z
6. f (z) ; Res ( f (z), 2)
(z 2 2)2
19.
20.
21.
In Problems 17–20, use Cauchy’s residue theorem, where
appropriate, to evaluate the given integral along the indicated
contours.
29.
z1
dz
18.
2
z
(z
2 2i)
BC
(b) |z| 32
(c) |z| 3
(b) |z 2i| 3
(c) |z| 5
dz
23.
24.
25.
26.
27.
28.
1
dz, C: |z 3i| 3
2
z
4z
13
BC
1
dz, C: |z 2| 32
BC z 3(z 2 1)4
z
dz, C: |z| 2
4
BC z 2 1
z
dz, C is the ellipse 16x2 y2 4
BC (z 1)(z 2 1)
ze z
dz, C: |z| 2
2
BC z 2 1
ez
dz, C: |z| 3
3
BC z 2z 2
tan z
dz, C: |z 1| 2
BC z
cot pz
dz, C: |z| 12
BC z 2
䉲
䉲
䉲
䉲
䉲
䉲
䉲
䉲
BC
䉲
cot pz dz, C is the rectangle defined by x 12 , x p, y 1,
y1
2z 2 1
dz, C is the rectangle defined by x 2, x 1,
30.
BC z 2(z 3 1)
y 12 , y 1
e iz sin z
31.
dz, C: |z 3| 1
BC (z 2 p)4
cos z
dz, C: |z 1| 1
32.
BC (z 2 1)2(z 2 9)
䉲
䉲
(a) |z| 1
1
䉲
(c) |z 3| 1
䉲
䉲
(a) |z| 12
BC z sin z
(a) |z 2i| 1
(b) |z i| 2
In Problems 21–32, use Cauchy’s residue theorem to evaluate
the given integral along the indicated contour.
22.
1
dz
BC (z 2 1)(z 2)2
2
z 3e 1>z dz
(a) |z| 5
In Problems 7–16, use (1), (2), or (4) to find the residue at each
pole of the given function.
z
4z 8
7. f (z) 2
8. f (z) 2z 2 1
z 16
1
1
9. f (z) 4
10. f (z) 2
3
2
z z 2 2z
(z 2 2z 2)2
2
5z 2 4z 3
11. f (z) (z 1)(z 2)(z 3)
2z 2 1
12. f (z) (z 2 1)4(z 3)
cos z
ez
13. f (z) 2
14. f (z) z
3
e 21
z (z 2 p)
1
15. f (z) sec z
16. f (z) z sin z
17.
BC
䉲
䉲
(b) |z 2i| 1
(c) |z 2i| 4
19.6 Evaluation of Real Integrals
INTRODUCTION In this section we shall see how residue theory can be used to evaluate real
integrals of the forms
#
2p
F(cos u, sin u) du,
(1)
0
#
q
q
902
|
CHAPTER 19 Series and Residues
f (x) dx,
(2)
#
q
f (x) cos ax dx
and
q
#
q
f (x) sin ax dx,
(3)
q
where F in (1) and f in (2) and (3) are rational functions. For the rational function f (x) p(x)/q(x)
in (2) and (3), we will assume that the polynomials p and q have no common factors.
Integrals of the Form e02p F(cos u, sin u ) du The basic idea here is to convert
an integral of form (1) into a complex integral where the contour C is the unit circle centered at
the origin. This contour can be parameterized by z cos u i sin u eiu, 0 u 2p. Using
e iu e iu
e iu 2 e iu
, sin u ,
2
2i
dz ie iu du, cos u we replace, in turn, du, cos u, and sin u by
du dz
1
1
, cos u (z z 1), sin u (z 2 z 1).
iz
2
2i
(4)
The integral in (1) then becomes
1
1
dz
F a (z z 1), (z 2 z 1)b ,
2i
iz
BC 2
䉲
where C is |z| 1.
A Real Trigonometric Integral
EXAMPLE 1
Evaluate
#
2p
0
1
du.
(2 cos u)2
SOLUTION Using the substitutions in (4) and simplifying yield the contour integral
4
i
BC (z 2 4z 1)2
z
䉲
dz.
With the aid of the quadratic formula we can write
z
z
f (z) 2
,
(z 4z 1)2
(z 2 z0)2 (z 2 z1)2
where z0 2 "3 and z1 2 "3. Since only z1 is inside the unit circle C, we have
BC (z 2 4z 1)2
䉲
z
dz 2pi Res ( f (z), z1).
Now z1 is a pole of order 2 and so from (2) of Section 19.5,
Res ( f (z), z1) lim
zSz1
d
d
z
(z 2 z1)2 f (z) lim
2
dz
zSz1 dz (z 2 z0)
z z0
1
.
3
zSz1 (z 2 z0)
6"3
lim
Hence,
4
i
4
4
1
z
dz 2pi Res ( f (z), z1) 2pi
2
i
i
BC (z 4z 1)
6"3
䉲
2
#
and finally
2p
0
1
4p
du .
2
(2 cos u)
3"3
q
Integrals of the Form eq
f(x) dx When f is continuous on (q, q), recall from
q
calculus that the improper integral eq
f (x) dx is defined in terms of two distinct limits:
#
q
q
f (x) dx lim
#
0
rSq r
R
f (x) dx lim
# f (x) dx.
(5)
RSq 0
19.6 Evaluation of Real Integrals
|
903
If both limits in (5) exist, the integral is said to be convergent; if one or both of the limits fail to
q
exist, the integral is divergent. In the event that we know (a priori) that an integral e⫺q
f (x) dx
converges, we can evaluate it by means of a single limiting process:
#
q
⫺q
f (x) dx ⫽ lim
#
R
f (x) dx.
(6)
RSq ⫺R
It is important to note that the symmetric limit in (6) may exist even though the improper integral is
q
divergent. For example, the integral e⫺q
x dx is divergent since limRS⬁ e0R x dx ⫽ limRS⬁ 12 R2 ⫽ q.
However, using (6), we obtain
lim
#
R
x dx ⫽ lim c
RSq ⫺R
RSq
(⫺R)2
R2
2
d ⫽ 0.
2
2
(7)
The limit in (6) is called the Cauchy principal value of the integral and is written
P.V.
y
CR
zn
z2
z1
z3
z4
0
–R
R
x
#
q
⫺q
f (x) dx ⫽ lim
#
R
RSq ⫺R
f (x) dx.
q
In (7) we have shown that P.V. e⫺q
x dx ⫽ 0. To summarize, when an integral of form (2) converges, its Cauchy principal value is the same as the value of the integral. If the integral diverges,
it may still possess a Cauchy principal value.
q
To evaluate an integral e⫺q
f (x) dx, where f (x) ⫽ P(x)/Q(x) is continuous on (⫺q, q), by
residue theory we replace x by the complex variable z and integrate the complex function f over
a closed contour C that consists of the interval [⫺R, R] on the real axis and a semicircle CR of
radius large enough to enclose all the poles of f (z) ⫽ P(z)/Q(z) in the upper half-plane Re(z) ⬎ 0.
See FIGURE 19.6.1. By Theorem 19.5.3 we have
BC
FIGURE 19.6.1 Closed contour C consists
of a semicircle CR and the interval [⫺R, R]
䉲
f (z) dz ⫽
#
CR
#
f (z) dz ⫹
R
f (x) dx ⫽ 2pi a Res ( f (z), zk),
n
⫺R
k⫽1
where zk, k ⫽ 1, 2, . . ., n, denotes poles in the upper half-plane. If we can show that the integral
eCR f (z) dz S 0 as R S q, then we have
P.V.
#
q
⫺q
EXAMPLE 2
#
f (x) dx ⫽ lim
R
RSq ⫺R
f (x) dx ⫽ 2pi a Res ( f (z), zk).
n
(8)
k⫽1
Cauchy P.V. of an Improper Integral
Evaluate the Cauchy principal value of
#
q
⫺q
SOLUTION
y
CR
Let f (z) ⫽ 1/(z2 ⫹ 1)(z2 ⫹ 9). Since
(z2 ⫹ 1)(z2 ⫹ 9) ⫽ (z ⫺ i)(z ⫹ i)(z ⫺ 3i)(z ⫹ 3i),
3i
we let C be the closed contour consisting of the interval [⫺R, R] on the x-axis and the semicircle CR of radius R ⬎ 3. As seen from FIGURE 19.6.2,
i
–R
1
dx.
(x ⫹ 1) (x 2 ⫹ 9)
2
R
1
dz ⫽
2
BC (z ⫹ 1)(z 2 ⫹ 9)
x
䉲
FIGURE 19.6.2 Closed contour C for
Example 2
#
R
1
dx ⫹
2
⫺R (x ⫹ 1)(x ⫹ 9)
2
#
1
dz
2
CR (z ⫹ 1)(z ⫹ 9)
2
⫽ I1 ⫹ I2
and
I1 ⫹ I2 ⫽ 2pi[Res ( f (z), i) ⫹ Res ( f (z), 3i)].
At the simple poles z ⫽ i and z ⫽ 3i we find, respectively,
Res ( f (z), i) ⫽
904
|
CHAPTER 19 Series and Residues
1
16i
and
Res ( f (z), 3i) ⫽ ⫺
1
,
48i
I1 I2 2pi c
so that
1
1
p
2
d .
16i
48i
12
(9)
We now want to let R S q in (9). Before doing this, we note that on CR,
|(z2 1)(z2 9)| |z2 1| |z2 9| ||z|2 1| ||z|2 9| (R2 1)(R2 9),
and so from the ML-inequality of Section 18.1 we can write
ZI2 Z 2
#
pR
1
dz 2 # 2
.
2
(R 2 1)(R 2 2 9)
CR (z 1)(z 9)
2
This last result shows that |I2| S 0 as R S q, and so we conclude that limRS q I2 0. It follows
from (9) that limRS q I1 p/12; in other words,
lim
RSq
#
R
p
1
dx or P.V.
2
2
12
R (x 1)(x 9)
#
q
q
p
1
dx .
2
12
(x 1)(x 9)
2
It is often tedious to have to show that the contour integral along CR approaches zero as R S q.
Sufficient conditions under which this is always true are given in the next theorem.
Behavior of Integral as R S q
Theorem 19.6.1
Suppose f (z) P(z)/Q(z), where the degree of P(z) is n and the degree of Q(z) is m n 2.
If CR is a semicircular contour z Reiu, 0 u p, then eCR f (z) dz S 0 as R S q.
In other words, the integral along CR approaches zero as R S q when the denominator of f is
of a power at least 2 more than its numerator. The proof of this fact follows in the same manner
as in Example 2. Notice in that example that the conditions stipulated in Theorem 19.6.1 are
satisfied, since the degree of P(z) 1 is 0 and the degree of Q(z) (z2 1)(z2 9) is 4.
Cauchy P.V. of an Improper Integral
EXAMPLE 3
Evaluate the Cauchy principal value of
#
q
q
1
dx.
x 1
4
SOLUTION By inspection of the integrand, we see that the conditions given in Theorem
19.6.1 are satisfied. Moreover, we know from Example 3 of Section 19.5 that f has simple
poles in the upper half-plane at z1 epi/4 and z2 e3pi/4. We also saw in that example that
the residues at these poles are
Res ( f (z), z1) 1
4"2
2
1
4"2
i
and
Res ( f (z), z2) 1
4"2
2
1
4"2
i.
Thus, by (8),
P.V.
#
q
q
1
p
dx 2pi fRes ( f (z), z1) Res ( f (z), z2)g .
x 1
"2
4
q
q
Integrals of the Forms eq
f(x) cos ax dx or eq
f(x) sin ax dx
We encountered integrals of this type in Section 15.4 in the study of Fourier transforms.
q
q
Accordingly, eq
f (x) cos ax dx and eq
f (x) sin ax dx, a 0, are referred to as Fourier integrals.
q
Fourier integrals appear as the real and imaginary parts in the improper integral eq
f (x)eiax dx.
iax
Using Euler’s formula e cos ax i sin ax, we get
#
q
q
f (x)e iax dx #
q
q
f (x) cos ax dx i
#
q
f (x) sin ax dx
(10)
q
whenever both integrals on the right side converge. When f (x) P(x)/Q(x) is continuous on
(q, q) we can evaluate both Fourier integrals at the same time by considering the integral
兰C f (z)eiaz dz, where a 0 and the contour C again consists of the interval [R, R] on the real
axis and a semicircular contour CR with radius large enough to enclose the poles of f (z) in the
upper half-plane.
19.6 Evaluation of Real Integrals
|
905
Before proceeding we give, without proof, sufficient conditions under which the contour
integral along CR approaches zero as R S q:
Behavior of Integral as R S q
Theorem 19.6.2
Suppose f (z) P(z)/Q(z), where the degree of P(z) is n and the degree of Q(z) is m n 1. If CR
is a semicircular contour z Reiu, 0 u p, and a 0, then eCR(P(z)/Q(z))eiaz dz S 0 as R S q.
Using Symmetry
EXAMPLE 4
#
q
x sin x
dx.
2
x
9
0
SOLUTION First note that the limits of integration are not from q to q as required by the
method. This can be rectified by observing that since the integrand is an even function of x,
we can write
Evaluate the Cauchy principal value of
#
q
0
1
x sin x
dx 2
x2 9
#
q
x sin x
dx.
2
x
9
q
(11)
With a 1, we now form the contour integral
BC z 2 9
z
䉲
e iz dz,
where C is the same contour shown in Figure 19.6.2. By Theorem 19.5.3,
#
z
e iz dz 2
CR z 9
#
R
x
e ix dx 2pi Res ( f (z) e iz, 3i),
R x 9
2
where f (z) z/(z2 9). From (4) of Section 19.5,
Res ( f (z) e iz, 3i) ze iz
e 3
2
.
2z z 3i
2
Hence, in view of Theorem 19.6.2 we conclude eCR f (z)eiz dz S 0 as R S q and so
But by (10),
#
q
x
e3
p
e ix dx 2pi a
b 3 i.
2
x
9
e
q
#
P.V.
q
2
x
e ix dx 2
x
9
q
#
q
x cos x
dx i
2
x
9
q
#
q
p
x sin x
dx 3 i.
2
x
9
e
q
Equating real and imaginary parts in the last line gives the bonus result
P.V.
#
q
x cos x
dx 0 along with P.V.
2
q x 9
#
q
p
x sin x
dx 3 .
2
e
q x 9
Finally, in view of (11) we obtain the value of the prescribed integral:
#
y
q
0
CR
–Cr
–R
c
FIGURE 19.6.3 Indented contour
906
|
R
x
1
x sin x
dx P.V.
2
x2 9
#
q
p
x sin x
dx 3 .
2
x
9
2e
q
Indented Contours The improper integrals of form (2) and (3) that we have considered
up to this point were continuous on the interval (q, q). In other words, the complex function
f (z) P(z)/Q(z) did not have poles on the real axis. In the event f has poles on the real axis,
q
we must modify the procedure used in Examples 2– 4. For example, to evaluate eq
f (x) dx by
residues when f (z) has a pole at z c, where c is a real number, we use an indented contour
as illustrated in FIGURE 19.6.3. The symbol Cr denotes a semicircular contour centered at z c
oriented in the positive direction. The next theorem is important to this discussion.
CHAPTER 19 Series and Residues
Behavior of Integral as r S 0
Theorem 19.6.3
Suppose f has a simple pole z c on the real axis. If Cr is the contour defined by z c reiu,
0 u p, then
lim
#
rS0 C
r
f (z) dz pi Res ( f (z), c).
PROOF: Since f has a simple pole at z c, its Laurent series is
a1
g(z),
z2c
where a1 Res( f (z), c) and g is analytic at c. Using the Laurent series and the parameterization of Cr , we have
f (z) #
#
f (z) dz a1
Cr
p
0
ire iu
du ir
re iu
p
# g(c re ) e
iu
0
iu
du I1 I2.
(12)
First, we see that
I1 a1
#
p
0
ire iu
du a1
re iu
#
p
0
i du pia1 pi Res ( f (z), c).
Next, g is analytic at c and so it is continuous at this point and bounded in a neighborhood of the
point; that is, there exists an M 0 for which |g(c reiu )| M. Hence,
ZI2 Z 2 ir
#
p
0
p
g(c re iu) e iu du 2 # r
# M du prM.
0
It follows from this last inequality that limrS0|I2| 0 and consequently limrS 0 I2 0. By taking
the limit of (12) as r S 0, we have proved the theorem.
Using an Indented Contour
EXAMPLE 5
Evaluate the Cauchy principal value of
#
q
q
SOLUTION Since the integral is of form (3), we consider the contour integral 养C eiz dz /z(z2 2z 2).
The function f (z) 1/z(z2 2z 2) has simple poles at z 0 and at z 1 i in the upper half-plane.
The contour C shown in FIGURE 19.6.4 is indented at the origin. Adopting an obvious notation, we have
y
䉲
CR
–r
BC
1+i
–Cr
–R
sin x
dx.
x(x 2 2 2x 2)
r
䉲
R
x
# #
CR
r
R
#
Cr
#
r
R
2pi Res ( f (z) e iz, 1 i),
(13)
where eCr eCr . Taking the limits of (13) as R S q and as r S 0, we find from
Theorems 19.6.2 and 19.6.3 that
FIGURE 19.6.4 Indented contour C for
Example 5
P.V.
#
q
q
e ix
dx 2 pi Res ( f (z) e iz, 0) 2pi Res ( f (z) e iz, 1 i).
x(x 2 2 2x 2)
Now,
Res ( f (z) e iz, 0) 1
2
and
Res ( f (z) e iz, 1 i) e 1 i
(1 i).
4
Therefore,
P.V.
#
q
q
1i
e ix
1
e 1 i
dx
pi
a
a
b
2pi
(1 i)b.
2
4
x(x 2 2 2x 2)
e1(cos 1 i sin 1), simplifying, and then equating real and imaginary parts,
Using e
we get from the last equality
P.V.
#
q
q
and
P.V.
#
q
q
p
cos x
dx e 1(sin 1 cos 1)
2
x(x 2 2x 2)
2
p
sin x
dx f1 e 1 (sin 1 2 cos 1)g.
2
x(x 2 2x 2)
2
19.6 Evaluation of Real Integrals
|
907
Exercises
19.6
Answers to selected odd-numbered problems begin on page ANS-43.
In Problems 1–10, evaluate the given trigonometric integral.
1.
3.
5.
6.
8.
10.
#
#
#
#
#
#
2p
#
#
2p
1
1
du
2.
du
1
0.5
sin
u
10
2
6 cos u
0
0
2p
2p
cos u
1
du
du
4.
3
sin
u
1
3
cos 2 u
0
0
p
1
du [Hint: Let t 2p u.]
2
2
cos u
0
p
2p
1
sin2 u
du
7.
du
2
5 4 cos u
0 1 sin u
0
2p
2p
cos 2 u
cos 2u
du
9.
du
3
2
sin
u
5
2
4 cos u
0
0
2p
1
du
cos
u
2
sin u 3
0
#
#
In Problems 11–30, evaluate the Cauchy principal value of the
given improper integral.
q
q
1
1
dx
dx
11.
12.
2
2
q x 2 2x 2
q x 2 6x 25
q
q
1
x2
13.
dx
14.
dx
2
2
2
2
q (x 4)
q (x 1)
q
q
1
x
15.
dx
16.
dx
2
3
2
(x
1)
(x
4)3
q
q
q
q
2x 2 2 1
dx
dx
17.
18.
4
2
2
2
2
q x 5x 4
q (x 1) (x 9)
q 2
q
x 1
1
dx
dx
19.
20.
4
6
0 x 1
0 x 1
q
q
cos x
cos 2x
dx
dx
21.
22.
2
2
x
1
x
1
q
q
q
q
x sin x
cos x
dx
23.
24.
dx
2
2
2
q x 1
0 (x 4)
q
q
cos 3x
sin x
dx
25.
dx
26.
2
2
2
0 (x 1)
q x 4x 5
q
q
cos 2x
x sin x
dx
dx
27.
28.
4
4
x
1
x
1
0
0
q
cos x
dx
29.
2
(x
1)(x 2 9)
q
q
x sin x
dx
30.
2
2
0 (x 1)(x 4)
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
Chapter in Review
19
1. A function f is analytic at a point z0 if f can be expanded in a
convergent power series centered at z0. _____
2. A power series represents a continuous function at every point
within and on its circle of convergence. _____
|
CHAPTER 19 Series and Residues
#
#
#
p
0
du
ap
, a.1
(a cos u)2
("a 2 2 1)3
and use this formula to verify the answer in Example 1.
34. Establish the general result
#
2p
0
sin2u
2p
du 2 (a 2 "a 2 2 b 2 ), a . b . 0
a b cos u
b
and use this formula to verify the answer to Problem 7.
35. Use the contour shown in FIGURE 19.6.5 to show that
P.V.
#
q
q
e ax
p
dx , 0 , a , 1.
1 ex
sin ap
y
2π i
πi
C
r
–r
x
FIGURE 19.6.5 Contour in Problem 35
36. The steady-state temperature u(x, y) in a semi-infinite plate is
determined from
0 2u
0 2u
2 0, 0 , x , p, y . 0
2
0x
0y
2y
, y.0
u(0, y) 0, u(p, y) 4
y 4
u(x, 0) 0, 0 , x , p.
Use a Fourier transform and the residue method to show that
u(x, y) #
q
0
e a sin a sinh ax
sin ay da.
sinh ap
Answers to selected odd-numbered problems begin on page ANS-43.
Answer Problems 1–12 without referring back to the text. Fill in
the blank or answer true/false.
908
In Problems 31 and 32, use an indented contour and residues to
establish the given result.
q
sin x
dx p
31. P.V.
x
q
q
sin x
dx p(1 2 e 1)
32. P.V.
2
q x(x 1)
33. Establish the general result
3. For f (z) 1/(z 3), the Laurent series valid for |z| 3 is
z1 3z2 9z3 … . Since there are an infinite number
of negative powers of z z 0, z 0 is an essential
singularity. _____
4. The only possible singularities of a rational function are poles.
_____
5. The function f (z) e1/(z1) has an essential singularity at
6.
7.
8.
9.
10.
z 1. _____
The function f (z) z/(ez 1) has a removable singularity at
z 0. _____
The function f (z) z(ez 1) possesses a zero of order 2 at
z 0. _____
The function f (z) (z 5)/(z3 sin2z) has a pole of order _____
at z 0.
If f (z) cot p z, then Res( f (z), 0) _____.
The Laurent series of f valid for 0 |z 1| is given by
(z 2 1)1
(z 2 1)
(z 2 1)3
(z 1)3 2
p.
3!
5!
7!
From this series we see that f has a pole of order _____ at
z 1 and Res( f (z), 1) _____.
q
(z 2 i)k
11. The circle of convergence of the power series a
k1
k 1 (2 i)
is _____.
12. The power series a k 1 converges at z 2i. _____
k1 2
13. Find a Maclaurin expansion of f (z) ez cos z. [Hint: Use the
q
zk
identity cos z (eiz eiz)/2.]
14. Show that the function f (z) 1/sin(p/z) has an infinite number
of singular points. Are any of these isolated singular points?
In Problems 15–18, use known results as an aid in expanding
the given function in a Laurent series valid for the indicated
annular region.
1 2 e iz
15. f (z) , 0 |z| 16. f (z) ez/(z2), 0 |z 2|
z4
1
17. f (z) (z i)2 sin
, 0 |z i|
z2i
1 2 cos z 2
18. f (z) , 0 |z|
z5
2
in an appropriate series valid for
19. Expand f (z) 2
z 2 4z 3
(a) |z| 1
(b) 1 |z| 3
(c) |z| 3
(d) 0 |z 1| 2.
1
20. Expand f (z) in an appropriate series valid for
(z 2 5)2
(a) |z| 5
(b) |z| 5
(c) 0 |z 5|.
In Problems 21–30, use Cauchy’s residue theorem to evaluate
the given integral along the indicated contour.
2z 5
21.
dz, C: |z 2| 52
BC z(z 2)(z 2 1)4
z2
dz, C is the ellipse x2 /4 y2 1
22.
BC (z 2 1)3(z 2 4)
1
23.
dz, C: |z 12 | 13
2
sin
z
21
BC
z1
24.
dz, C is the rectangle defined by x 1, x 1,
BC sinh z
y 4, y 1
䉲
䉲
䉲
䉲
25.
26.
27.
28.
29.
30.
e 2z
dz, C: |z| 4
BC z 2z 3 2z 2
1
dz, C is the square defined by x 2,
BC z 4 2 2z 2 4
x 2, y 0, y 1
1
dz, C: |z| 1 [Hint: Use the Maclaurin series
z
BC z(e 2 1)
for z(ez 1).]
z
dz, C: |z 1| 3
BC (z 1)(z 2 1)10
sin z
cze 3>z 2
d dz, C: |z| 6
z (z 2 p)3
BC
䉲
4
䉲
䉲
䉲
䉲
BC
csc pz dz, C is the rectangle defined by x 12 , x 52 ,
䉲
y 1, y 1
In Problems 31 and 32, evaluate the Cauchy principal value of
the given improper integral.
q
x2
31.
dx
2
2
2
q (x 2x 2)(x 1)
32.
#
#
q
q
a cos x x sin x
dx, a 0 [Hint: Consider eiz /(z ai).]
x 2 a2
In Problems 33 and 34, evaluate the given trigonometric
integral.
2p
2p
cos 2u
cos 3u
du
34.
du
2
sin
u
5
2
4 cos u
0
0
35. Use an indented contour to show that
33.
#
#
P.V.
#
q
0
36.
1 2 cos x
p
dx .
2
x2
2 2
Show that e0qea x
2
2
cos bx dx eb >4a !p>2a by considering
2 2
the complex integral 养Cea z e ibz dz along the contour C shown
q a2x 2
in FIGURE 19.R.1. Use the known result eq
e
dx !p>a.
䉲
y
b i
2a2
C
r
–r
x
FIGURE 19.R.1 Contour in Problem 36
can be shown to be f (z) g k q Jk (u)zk, where Jk(u) is the
Bessel function of the first kind of order k. Use (4) of
Section 19.3 and the contour C: |z| 1 to show that the
coefficients Jk(u) are given by
37. The Laurent expansion of f (z) e(u/2)(z1/z) valid for 0 |z|
q
Jk(u) 1
2p
#
2p
cos (kt 2 u sin t) dt.
0
CHAPTER 19 in Review
|
909
© Andy Sacks/Getty Images
© Takeshi Takahara/Photo Researchers/Getty Images
CHAPTER
20
In this chapter we will study the
mapping properties of the
elementary functions introduced
in Chapter 17 and develop two
new classes of special mappings
called the linear fractional
transformations and the
Schwarz–Christoffel
transformations.
In earlier chapters we used
Fourier series and integral
transforms to solve boundaryvalue problems involving Laplace’s
equation. Conformal mapping
methods discussed in this chapter
can be used to transfer known
solutions to Laplace’s equation
from one region to another. In
addition, fluid flows around
obstacles and through channels
can be determined using
conformal mappings.
Conformal Mappings
CHAPTER CONTENTS
20.1
20.2
20.3
20.4
20.5
20.6
Complex Functions as Mappings
Conformal Mappings
Linear Fractional Transformations
Schwarz–Christoffel Transformations
Poisson Integral Formulas
Applications
Chapter 20 in Review
20.1
Complex Functions as Mappings
INTRODUCTION In Chapter 17 we emphasized the algebraic definitions and properties of
complex functions. In order to give a geometric interpretation of a complex function w f (z),
we place a z-plane and a w-plane side by side and imagine that a point z x iy in the domain of
the definition of f has mapped (or transformed) to the point w f (z) in the second plane. Thus the
complex function w f (z) u(x, y) iv(x, y) may be considered as the planar transformation
u u(x, y),
v v(x, y)
and w f (z) is called the image of z under f.
FIGURE 20.1.1 indicates the images of a finite number of complex numbers in the region R. More
useful information is obtained by finding the image of the region R together with the images of
a family of curves lying inside R. Common choices for the curves are families of lines, families
of circles, and the system of level curves for the real and imaginary parts of f .
y
v
f
R
z2
w1
z1
w2
z3
x
u
w3
(a) z-plane
(b) w-plane
FIGURE 20.1.1 w1, w2, w3 are images of z1, z2, z3
Images of Curves Note that if z(t) x(t) iy(t), a t b, describes a curve C in the
region, then w f (z(t)), a t b, is a parametric representation of the corresponding curve C
in the w-plane. In addition, a point z on the level curve u(x, y) a will be mapped to a point w
that lies on the vertical line u a, and a point z on the level curve v(x, y) b will be mapped to
a point w that lies on the horizontal line v b.
y
πi
x
(a)
v
Arg w = π
Arg w = 0
u
(b)
FIGURE 20.1.2 Images of vertical and
horizontal lines in Example 1
The Mapping f (z) e z
EXAMPLE 1
The horizontal strip 0 y p lies in the fundamental region of the exponential function
f (z) ez. A vertical line segment x a in this region can be described by z(t) a it, 0 t p,
and so w f (z(t)) eaeit. Thus the image is a semicircle with center at w 0 and with radius
r ea. Similarly, a horizontal line y b can be parametrized by z(t) t ib, q t q,
and so w f (z(t)) eteib. Since Arg w b and ZwZ et, the image is a ray emanating from
the origin, and since 0 Arg w p, the image of the entire horizontal strip is the upper halfplane v 0. Note that the horizontal lines y 0 and y p are mapped onto the positive and
negative u-axis, respectively. See FIGURE 20.1.2 for the mapping by f (z) ez.
From w exeiy, we can conclude that ZwZ e x and y Arg w. Hence, z x iy loge ZwZ i Arg w Ln w. The inverse function f 1(w) Ln w therefore maps the upper half-plane
v 0 back to the horizontal strip 0 y p.
The Mapping f (z) 1/z
EXAMPLE 2
The complex function f (z) 1/z has domain z 0 and real and imaginary parts u(x, y) x/(x2 y2) and v(x, y) y/(x2 y2), respectively. When a 0, a level curve u(x, y) a
can be written as
x2 2
912
|
CHAPTER 20 Conformal Mappings
1
x y2 0
a
or
ax 2
1 2
1 2
b y2 a b .
2a
2a
y
a= – 1
2
The level curve is therefore a circle with its center on the x-axis and passing through the origin.
A point z on this circle other than zero is mapped to a point w on the line u a. Likewise, the
level curve v(x, y) b, b 0, can be written as
b=– 1
2
a= 1
2
x 2 ay x
and a point z on this circle is mapped to a point w on the line v b. FIGURE 20.1.3 shows the
mapping by f (z) 1/z. Figure 20.1.3(a) shows the two collections of circular level curves,
and Figure 20.1.3(b) shows their corresponding images in the w-plane.
Since w 1/z, we have z 1/w. Thus f 1(w) 1/w, and so f f 1. We can therefore
conclude that f maps the horizontal line y b to the circle u2 (v 12 b)2 ( 12 b)2, and f maps
the vertical line x a to the circle (u 2 12a)2 v 2 (12a)2 .
b= 1
2
(a)
v
2
–2
1 2
1 2
b a b ,
2b
2b
2
u
–2
(b)
FIGURE 20.1.3 Images of circles in
Example 2
y
Translation and Rotation The elementary linear function f (z) z z0 may be interpreted as a translation in the z-plane. To see this, we let z x iy and z0 h ik. Since
w f (z) (x h) i( y k), the point (x, y) has been translated h units in the horizontal direction and k units in the vertical direction to the new position at (x h, y k). In particular, the
origin O has been mapped to z0 h ik.
The elementary function g(z) e iu0 z may be interpreted as a rotation through u0 degrees, for
if z reiu, then w g(z) re i(u u0) . Note that if the complex mapping h(z) e iu0 z z0 is applied
to a region R that is centered at the origin, the image region R can be obtained by first rotating
R through u0 degrees and then translating the center to the new position z0. See FIGURE 20.1.4 for
the mapping by h(z) e iu0 z z0.
Rotation and Translation
EXAMPLE 3
Find a complex function that maps the horizontal strip 1 y 1 onto the vertical strip
2 x 4.
R
x
SOLUTION If the horizontal strip 1 y 1 is rotated through 90 , the vertical strip 1 x 1
results, and the vertical strip 2 x 4 can be obtained by shifting this vertical strip 3 units to the
right. See FIGURE 20.1.5. Since eip/2 i, we obtain h(z) iz 3 as the desired complex mapping.
(a)
R'
v
y
4
4
2
2
v
z0
–4
–2
2
4
x
–4
u
2
u
4
–2
–2
θ0
–2
–4
–4
(a)
(b)
FIGURE 20.1.5 Image of horizontal strip in Example 3
(b)
FIGURE 20.1.4 Translation and rotation
Magnification A magnification is a complex function of the form f (z) az, where a
is a fixed positive real number. Note that ZwZ ZazZ aZzZ, and so f changes the length (but not
the direction) of the complex number z by a fixed factor a. If g(z) az b and a r0e iu0 , then the
vector z is rotated through u0 degrees, magnified by a factor of r0, and then translated using b.
EXAMPLE 4
Contraction and Translation
Find a complex function that maps the disk ZzZ 1 onto the disk Zw (1 i)Z 12 .
SOLUTION We must first contract the radius of the disk by a factor of 12 and then translate
its center to the point 1 i. Therefore, w f (z) 12 z (1 i) maps Z zZ 1 to the disk
Zw (1 i)Z 12 .
20.1 Complex Functions as Mappings
|
913
Power Functions A complex function of the form f (z) za, where a is a fixed positive
y
θ0
R
x
real number, is called a real power function. FIGURE 20.1.6 shows the effect of the complex function f (z) za on the angular wedge 0 Arg z u0. If z reiu, then w f (z) raeiau. Hence,
0 Arg w au0 and the opening of the wedge is changed by a factor of a. It is not difficult to
show that a circular arc with center at the origin is mapped to a similar circular arc, and rays
emanating from the origin are mapped to similar rays.
EXAMPLE 5
Find a complex function that maps the upper half-plane y 0 onto the wedge 0 Arg w p/4.
(a)
v
The Power Function f (z) z 1/4
SOLUTION The upper half-plane y 0 can also be described by the inequality 0 Arg z p.
We must therefore find a complex mapping that reduces the angle u0 p by a factor of a 14 .
Hence, f (z) z1/4.
R′
αθ 0
u
Successive Mappings To find a complex mapping between two regions R and R, it is
often convenient to first map R onto a third region R and then find a complex mapping from R onto R.
More precisely, if z f (z) maps R onto R , and w g(z) maps R onto R, then the composite
function w g( f (z)) maps R onto R. See FIGURE 20.1.7 for a diagram of successive mappings.
R
R′
(b)
FIGURE 20.1.6 R is the image of the
angular wedge R
w-plane
z-plane
f
g
R ′′
ζ -plane
FIGURE 20.1.7 R is image of R under successive mappings
EXAMPLE 6
Successive Mappings
Find a complex function that maps the horizontal strip 0 y p onto the wedge
0 Arg w p/4.
SOLUTION We saw in Example 1 that the complex function f (z) ez mapped the horizontal strip 0 y p onto the upper half-plane 0 Arg z p. From Example 5, the upper
half-plane 0 Arg z p is mapped onto the wedge 0 Arg w p/4 by g(z) z1/4. It
therefore follows that the composite function w g( f (z)) g(ez) ez/4 maps the horizontal
strip 0 y p onto the wedge 0 Arg w p/4.
EXAMPLE 7
Successive Mappings
Find a complex function that maps the wedge p/4 Arg z 3p/4 onto the upper half-plane
v 0.
SOLUTION We first rotate the wedge p/4 Arg z 3p/4 so that it is in the standard position
shown in Figure 20.1.6. If z f (z) eip/4z, then the image of this wedge is the wedge R
defined by 0 Arg z p/2. The real power function w g(z) z2 expands the opening of
R by a factor of two to give the upper half-plane 0 Arg w p as its image. Therefore,
w g( f (z)) (eip/4z)2 iz2 is the desired mapping.
In Sections 20.2–20.4, we will expand our knowledge of complex mappings and show how
they can be used to solve Laplace’s equation in the plane.
914
|
CHAPTER 20 Conformal Mappings
20.1
Exercises
Answers to selected odd-numbered problems begin on page ANS-43.
In Problems 1–10, a curve in the z-plane and a complex mapping
w f (z) are given. In each case, find the image curve in the
w-plane.
1. y x under w 1/z
2. y 1 under w 1/z
3. Hyperbola xy 1 under w z2
4. Hyperbola x2 y2 4 under w z2
5. Semicircle ZzZ 1, y 0, under w Ln z
6. Ray u p/4 under w Ln z
7. Ray u u0 under w z1/2
8. Circular arc r 2, 0 u p/2, under w z1/2
9. Curve ex cos y 1 under w ez
10. Circle ZzZ 1 under w z 1/z
In Problems 11–20, a region R in the z-plane and a complex
mapping w f (z) are given. In each case, find the image
region R in the w-plane.
11. First quadrant under w 1/z
12. Strip 0 y 1 under w 1/z
13. Strip p/4 y p/2 under w ez
14. Rectangle 0 x 1, 0 y p, under w ez
15. Circle ZzZ 1 under w z 4i
16. Circle ZzZ 1 under w 2z 1
17. Strip 0 y 1 under w iz
18. First quadrant under w (1 i)z
19. Wedge 0 Arg z p/4 under w z3
20. Wedge 0 Arg z p/4 under w z1/2
y
v
y= π
v= π
R′
R
x
u
FIGURE 20.1.9 Regions R and R for Problem 30
The mapping in Problem 10 is a special case of the
mapping w z k2 /z, where k is a positive constant, called
the Joukowski transformation.
(a) Show that the Joukowski transformation maps any circle
x2 y2 R2 into the ellipse
31. Project
u2
v2
R2, k
2 2
k
k2 2
a1 2 b
a1 2 2 b
R
R
R.
(b) What is the image of the circle when R k?
(c) The importance of the transformation w z k2 /z
does not lie in its effect on circles ZzZ R centered at the
origin but on off-centered circles with center on the
real axis. Show that the Joukowski transformation can
be written
w 2 2k
z2k 2
a
b .
w 2k
zk
With k 1, this particular transformation maps a circle
passing through z 1 and containing the point z 1
into a closed curve with a sharply pointed trailing edge.
This kind of curve, which resembles a cross section of an
airplane wing, is known as a Joukowski airfoil.
Write a report on the use of the Joukowski transformation in the study of the flow of air around an airfoil. There
is a lot of information on this topic on the Internet.
In Problems 21–30, find a complex mapping from the given
region R in the z-plane to the image region R in the w-plane.
21. Strip 1 y 4 to the strip 0 u 3
22. Strip 1 y 4 to the strip 0 v 3
23. Disk Zz 1Z 1 to the disk ZwZ 2
24. Strip 1 x 1 to the strip 1 v 1
25. Wedge p/4 Arg z p/2 to the upper half-plane v 0
26. Strip 0 y 4 to the upper half-plane v 0
27. Strip 0 y p to the wedge 0 Arg w 3p/2
28. Wedge 0 Arg z 3p/2 to the half-plane u 2
29.
y
30.
v
i
R
v=1
R′
x
u
FIGURE 20.1.8 Regions R and R for Problem 29
20.1 Complex Functions as Mappings
|
915
20.2 Conformal Mappings
INTRODUCTION
In Section 20.1 we saw that a nonconstant linear mapping f (z) az b,
a and b complex numbers, acts by rotating, magnifying, and translating points in the complex plane.
As a result, it is easily shown that the angle between any two intersecting curves in the z-plane is
equal to the angle between the images of the arcs in the w-plane under a linear mapping. Other complex mappings that have this angle-preserving property are the subject of our study in this section.
z2′
y
C2
θ
z1′
C1
x
(a) z-plane
C2′
v
w2′
Angle-Preserving Mappings A complex mapping w f (z) defined on a domain D
is called conformal at z ⴝ z0 in D when f preserves the angles between any two curves in D that
intersect at z0. More precisely, if C1 and C2 intersect in D at z0, and C19 and C29 are the corresponding
images in the w-plane, we require that the angle u between C1 and C2 equal the angle f between
C19 and C29. See FIGURE 20.2.1.
These angles can be computed in terms of tangent vectors to the curves. If z19 and z29 denote
tangent vectors to curves C1 and C2, respectively, then, applying the law of cosines to the triangle
determined by z19 and z29, we have
Zz19 z29Z 2 Zz19Z 2 Zz29Z 2 2Zz19Z Zz29Z cos u
φ
u cos 1 a
or
C1′
w1′
Zz19Z2 Zz29Z2 2 Zz 19 2 z29Z2
2Zz19ZZz29Z
b.
(1)
Likewise, if w19 and w29 denote tangent vectors to curves C19 and C29, respectively, then
u
(b) w-plane
FIGURE 20.2.1 Conformal mapping if
uf
f cos 1 a
Zw19Z2 Zw29Z2 2 Zw19 2 w29Z2
2Zw19ZZw29Z
b.
(2)
The next theorem gives a simple condition that guarantees that u f.
Theorem 20.2.1*
Conformal Mapping
If f (z) is analytic in the domain D and f (z0)
0, then f is conformal at z z0.
PROOF: If a curve C in D is parameterized by z z(t), then w f (z(t)) describes the image
curve in the w-plane. Applying the Chain Rule to w f (z(t)) gives w f (z(t))z(t). If curves
C1 and C2 intersect in D at z0, then w19 f (z0)z19 and w29 f (z0)z29. Since f (z0) 0, we can
use (2) to obtain
f cos1 a
Z f 9(z0)z19Z2 Z f 9(z0)z29Z2 2 Z f 9(z0)z19 2 f 9(z0)z29 Z2
2Z f 9(z0)z19ZZ f 9(z0)z29Z
b.
We can apply the laws of absolute value to factor out Z f (z0)Z 2 in the numerator and denominator
and obtain
f cos1 a
Zz19Z2 Zz29Z2 2 Zz19 2 z29Z2
Therefore, from (1), f u.
EXAMPLE 1
2Zz19ZZz29Z
b.
Conformal Mappings
(a) The analytic function f (z) ez is conformal at all points in the z-plane, since f (z) ez
is never zero.
(b) The analytic function g(z) z2 is conformal at all points except z 0 since g(z) 2z 0
for z 0. From Figure 20.1.6 we see that g(z) doubles the angle formed by the two rays at
the origin.
* It is also possible to prove that f preserves the sense of direction between the tangent vectors.
916
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CHAPTER 20 Conformal Mappings
If f (z0) 0 but f (z0) 0, it is possible to show that f doubles the angle between any two
curves in D that intersect at z z0. The next two examples will introduce two important complex
mappings that are conformal at all but a finite number of points in their domains.
EXAMPLE 2
f (z) sin z as a Conformal Mapping
The vertical strip p/2 x p/2 is called the fundamental region of the trigonometric
function w sin z. A vertical line x a in the interior of this region can be described by
z(t) a it, q t q. From (6) in Section 17.7, we have
sin z sin x cosh y i cos x sinh y
and so
y
B
From the identity cosh2t sinh2t 1, it follows that
E
A
π
–
2
D
π
2
C
F
v2
u2
2
1.
2
sin a
cos 2a
x
(a)
v
B′
C′
u
F′
D′
The image of the vertical line x a is therefore a hyperbola with sin a as u-intercepts, and
since p/2 a p/2, the hyperbola crosses the u-axis between u 1 and u 1. Note
that if a p/2, then w cosh t, and so the line x p/2 is mapped onto the interval
(q, 1] on the negative u-axis. Likewise, the line x p/2 is mapped onto the interval
[1, q) on the positive u-axis.
A similar argument establishes that the horizontal line segment described by z(t) t ib,
p/2 t p/2, is mapped onto either the upper portion or the lower portion of the ellipse
u2
v2
1
2
cosh b
sinh2b
E′
A′
(b)
FIGURE 20.2.2 Image of vertical strip
in Example 2
according to whether b 0 or b 0. These results are summarized in FIGURE 20.2.2(b), which
shows the mapping by f (z) sin z. Note that we have carefully used capital letters to indicate where portions of the boundary are mapped. Thus, for example, boundary segment AB
is transformed to AB.
Since f (z) cos z, f is conformal at all points in the region except z p/2. The hyperbolas and ellipses are therefore orthogonal since they are images of the orthogonal families
of horizontal segments and vertical lines. Note that the 180 angle at z p/2 formed by
segments AB and AC is doubled to form a single line segment at w 1.
EXAMPLE 3
B
A
C
E
x
(a)
v
D′
A′
f (z) z 1/z as a Conformal Mapping
The complex mapping f (z) z 1/z is conformal at all values of z except z 1 and z 0.
In particular, the function is conformal for all values of z in the upper half-plane that satisfy
ZzZ 1. If z reiu, then w reiu (1/r)eiu, and so
y
D
u iv sin(a it) sin a cosh t i cos a sinh t.
B′
C′
E′
u
(b)
FIGURE 20.2.3 Images of rays and circles
in Example 3
1
1
u ar b cos u, v ar 2 b sin u.
r
r
(3)
Note that if r 1, then v 0 and u 2 cos u. Therefore, the semicircle z eit, 0 t p,
is mapped to the segment [2, 2] on the u-axis. It follows from (3) that if r 1, then the
semicircle z reit, 0 t p, is mapped onto the upper half of the ellipse u2 /a2 v2 /b2 1,
where a r 1/r and b r 1/r. See FIGURE 20.2.3 for the mapping by f (z) z 1/z.
For a fixed value of u, the ray z teiu, for t 1, is mapped to the portion of the hyperbola
2
u /cos2 u v2 /sin2 u 4 in the upper half-plane v 0. This follows from (3), since
v2
1 2
u2
1 2
2
at
b
2
at
2
b 4.
t
t
cos 2u
sin 2u
Since f is conformal for ZzZ 1 and a ray u u0 intersects a circle ZzZ r at a right angle, the
hyperbolas and ellipses in the w-plane are orthogonal.
20.2 Conformal Mappings
|
917
Conformal Mappings Using Tables Conformal mappings are given in Appendix IV.
The mappings have been categorized as elementary mappings (E-1 to E-9), mappings to halfplanes (H-1 to H-6), mappings to circular regions (C-1 to C-5), and miscellaneous mappings
(M-1 to M-10). Some of these complex mappings will be derived in Sections 20.3 and 20.4.
The entries indicate not only the images of the region R but also the images of various portions
of the boundary of R. This will be especially useful when we attempt to solve boundary-value
problems using conformal maps. You should use the appendix much like you use integral tables
to find antiderivatives. In some cases a single entry can be used to find a conformal mapping
between two given regions R and R. In other cases, successive transformations may be required
to map R to R.
EXAMPLE 4
Using a Table of Conformal Mappings
Use the conformal mappings in Appendix IV to find a conformal mapping between the strip
0 y 2 and the upper half-plane v 0. What is the image of the negative x-axis?
SOLUTION An appropriate mapping may be obtained directly from entry H-2. Letting a 2
then f (z) epz/2 and noting the positions of E, D, E, and D in the figure, we can map the
negative x-axis onto the interval (0, 1) on the u-axis.
EXAMPLE 5
Using a Table of Conformal Mappings
Use the conformal mappings in Appendix IV to find a conformal mapping between the strip
0 y 2 and the disk ZwZ 1. What is the image of the negative x-axis?
SOLUTION Appendix IV does not have an entry that maps the strip 0 y 2 directly onto the
disk. In Example 4, the strip was mapped by f (z) epz/2 onto the upper half-plane and, from
i2z
entry C-4, the complex mapping w maps the upper half-plane to the disk ZwZ 1.
i
z
i 2 e pz>2
maps the strip 0 y 2 onto the disk ZwZ 1.
Therefore, w g( f (z)) i e pz>2
The negative x-axis is first mapped to the interval (0, 1) in the z-plane, and from the
position of points C and C in C-4, the interval (0, 1) is mapped to the circular arc w eiu,
0 u p/2, in the w-plane.
Harmonic Functions and the Dirichlet Problem A bounded harmonic function
u u(x, y) that takes on prescribed values on the entire boundary of a region R is called a solution to a Dirichlet problem on R. In Chapters 13–15 we introduced a number of techniques for
solving Laplace’s equation in the plane, and we interpreted the solution to a Dirichlet problem
as the steady-state temperature distribution in the interior of R that results from the fixed temperatures on the boundary.
There are at least two disadvantages to the Fourier series and integral transform methods
presented in Chapters 13–15. The methods work only for simple regions in the plane and the
solutions typically take the form of either infinite series or improper integrals. As such, they are
difficult to evaluate. In Section 17.5 we saw that the real and imaginary parts of an analytic function are both harmonic. Since we have a large stockpile of analytic functions, we can find closedform solutions to many Dirichlet problems and use these solutions to sketch the isotherms and
lines of flow of the temperature distribution.
We will next show how conformal mappings can be used to solve a Dirichlet problem in a
region R once the solution to the corresponding Dirichlet problem in the image region R is
known. The method depends on the following theorem.
Theorem 20.2.2
Transformation Theorem for Harmonic Functions
Let f be an analytic function that maps a domain D onto a domain D. If U is harmonic in D,
then the real-valued function u(x, y) U( f (z)) is harmonic in D.
PROOF: We will give a proof for the special case in which D is simply connected. If U has a
harmonic conjugate V in D, then H U iV is analytic in D, and so the composite function
918
|
CHAPTER 20 Conformal Mappings
H( f (z)) U( f (z)) iV( f (z)) is analytic in D. By Theorem 17.5.3, it follows that the real part
U( f (z)) is harmonic in D, and the proof is complete.
To establish that U has a harmonic conjugate, let g(z) U/x i U/y. The first Cauchy–
Riemann equation (/x)(U/x) (/y)(U/y) is equivalent to Laplace’s equation
2U/x2 2U/y2 0, which is satisfied because U is harmonic in D. The second Cauchy–
Riemann equation (/y)(U/x) (/x)(U/y) is equivalent to the equality of the secondorder mixed partial derivatives. Therefore, g(z) is analytic in the simply connected domain D and so, by
Theorem 18.3.3, has an antiderivative G(z). If G(z) U1 iV1, then g(z) G(z) U1/x i U1/y.
Since g(z) U/x i U/y, it follows that U and U1 have equal first partial derivatives. Therefore,
H U iV1 is analytic in D, and so U has a harmonic conjugate in D.
Theorem 20.2.2 can be used to solve a Dirichlet problem in a region R by transforming the
problem to a region R in which the solution U either is apparent or has been found by prior
methods (including the Fourier series and integral transform methods of Chapters 13–15). The
key steps are summarized next.
Solving Dirichlet Problems Using Conformal Mapping
1. Find a conformal mapping w f (z) that transforms the original region R onto the image
region R. The region R may be a region for which many explicit solutions to Dirichlet
problems are known.
2. Transfer the boundary conditions from the boundary of R to the boundary of R. The value
of u at a boundary point j of R is assigned as the value of U at the corresponding boundary
point f (j). See FIGURE 20.2.4 for an illustration of transferring boundary conditions.
B
u=1
U=1
f
B′
f(ξ )
ξ
A
U( f(ξ )) = u(ξ )
A′
C
R
u=0
R′
C′
U=0
u=2
E′
E
U=2
D
u = –1
D′
U = –1
FIGURE 20.2.4 R is image of R under a conformal mapping f
3. Solve the corresponding Dirichlet problem in R. The solution U may be apparent from
B
the simplicity of the problem in R or may be found using Fourier or integral transform
methods. (Additional methods will be presented in Sections 20.3 and 20.5.)
4. The solution to the original Dirichlet problem is u(x, y) U( f (z)).
E
y
R
u=1
u=0
A
–π u=1 O u=0
2
(a)
R'
B′
A′
D
π
2
EXAMPLE 6
x
v
O′
D′
E′
u
U=1 U=1 U=0 U=0
(b)
FIGURE 20.2.5 Image of semi-infinite
vertical strip in Example 6
Solving a Dirichlet Problem
The function U(u, v) (1/p) Arg w is harmonic in the upper half-plane v 0 since it is the
imaginary part of the analytic function g(w) (1/p) Ln w. Use this function to solve the
Dirichlet problem in FIGURE 20.2.5(a).
SOLUTION The analytic function f (z) sin z maps the original region to the upper half-plane
v 0 and maps the boundary segments to the segments shown in Figure 20.2.5(b). The harmonic function U(u, v) (1/p) Arg w satisfies the transferred boundary conditions U(u, 0) 0
for u 0 and U(u, 0) 1 for u 0. Therefore, u(x, y) U(sin z) (1/p) Arg(sin z) is the
solution to the original problem. If tan1(v/u) is chosen to lie between 0 and p, the solution
can also be written as
u (x, y) cos x sinh y
1
b.
tan1 a
p
sin x cosh y
20.2 Conformal Mappings
|
919
Solving a Dirichlet Problem
EXAMPLE 7
From C–1 in Appendix IV of conformal mappings, the analytic function f (z) (z a)/(az 1),
where a (7 2 "6)/5, maps the region outside the two open disks ZzZ 1 and Zz 52 Z 12
onto the annular region r0 ZwZ 1, where r0 5 2 !6. FIGURE 20.2.6(a) shows the original
Dirichlet problem, and Figure 20.2.6(b) shows the transferred boundary conditions.
U=0
v
y
u=0
A
u=1
B
B′
x
1
A′
2
3
u
U=1
R'
R
(a)
(b)
FIGURE 20.2.6 Image of Dirichlet problem in Example 7
In Problem 12 in Exercises 14.1, we discovered that U(r, u) (loger)/(loger0) is the solution
to the new Dirichlet problem. From Theorem 20.2.2 we can conclude that the solution to the
original boundary-value problem is
u (x, y) z 2 (7 2!6)>5
1
log e 2
2.
log e (5 2 2!6)
(7 2!6)z>5 2 1
R'
z
Arg(z – a)
U=π
a
U=0
A favorite image region R for a simply connected region R is the upper half-plane y 0. For
any real number a, the complex function Ln(z a) loge Zz aZ i Arg(z a) is analytic in R.
Therefore, Arg(z a) is harmonic in R and is a solution to the Dirichlet problem shown in
FIGURE 20.2.7.
It follows that the solution in R to the Dirichlet problem with
FIGURE 20.2.7 Image of a Dirichlet
problem
U(x, 0) e
c0,
0,
a,x,b
otherwise
is the harmonic function U(x, y) (c0 /p)(Arg(z b) Arg(z a)). A large number of Dirichlet
problems in the upper half-plane y 0 can be solved by adding together harmonic functions of
this form.
Exercises
20.2
Answers to selected odd-numbered problems begin on page ANS-43.
In Problems 1–6, determine where the given complex mapping
is conformal.
1. f (z) z3 3z 1
2. f (z) cos z
3. f (z) z ez 1
4. f (z) z Ln z 1
5. f (z) (z2 1)1/2
6. f (z) pi 12 [Ln(z 1) Ln(z 1)]
In Problems 7–10, use the results in Examples 2 and 3.
7. Use the identity cos z sin(p/2 z) to find the image of the
strip 0 x p under the complex mapping w cos z. What
is the image of a horizontal line in the strip?
920
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CHAPTER 20 Conformal Mappings
8. Use the identity sinh z i sin(iz) to find the image of the
strip p/2 y p/2, x 0, under the complex mapping
w sinh z. What is the image of a vertical line segment in
the strip?
9. Find the image of the region defined by p/2 x p/2,
y 0, under the complex mapping w (sin z)1/4. What is the
image of the line segment [p/2, p/2] on the x-axis?
10. Find the image of the region ZzZ 1 in the upper half-plane
under the complex mapping w z 1/z. What is the image
of the line segment [1, 1] on the x-axis?
In Problems 11–18, use the conformal mappings in Appendix IV
to find a conformal mapping from the given region R in the
z-plane onto the target region R in the w-plane, and find the
image of the given boundary curve.
v
R
B
y
A
11.
y
16.
v
R
R′
A
1
R′
x
u
B
2
u
x
FIGURE 20.2.13 Regions R and R for Problem 16
v
y
17.
FIGURE 20.2.8 Regions R and R for Problem 11
R′
v
y
12.
A
B
R
R
πi
R′
x
A
u
1
18.
y
B
u
1
x
FIGURE 20.2.14 Regions R and R for Problem 17
FIGURE 20.2.9 Regions R and R for Problem 12
13.
–1
v
y
v
v=u
B
A
1
y=π
B
R′
R
u
x
A
πi
R′
R
u
x
FIGURE 20.2.10 Regions R and R for Problem 13
FIGURE 20.2.15 Regions R and R for Problem 18
14.
v
y
In Problems 19–22, use an appropriate conformal mapping and
the harmonic function U (1/p) Arg w to solve the given
Dirichlet problem.
B
R′
i
R
A
1
u
x
19.
y
u=1
R
FIGURE 20.2.11 Regions R and R for Problem 14
y
15.
v
FIGURE 20.2.16 Dirichlet problem in Problem 19
20.
y
u=1
R′
C
B
u=1
R
i
A
x
u=0
R
x
FIGURE 20.2.12 Regions R and R for Problem 15
u
u=0
u=0
1
x
FIGURE 20.2.17 Dirichlet problem in Problem 20
20.2 Conformal Mappings
|
921
y
21.
y
25.
u=0
u=0
u = 10
1
u=1
x
R
u=1
u = 10
FIGURE 20.2.18 Dirichlet problem in Problem 21
22.
y
1 u=0
u=1
u=0
u=0
u=4
In Problems 23–26, use an appropriate conformal mapping and
the harmonic function U (c0 /p)[Arg(w 1) Arg(w 1)]
to solve the given Dirichlet problem.
23.
R
x
FIGURE 20.2.19 Dirichlet problem in Problem 22
x
y
R
u=0
u=0
FIGURE 20.2.22 Dirichlet problem in Problem 25
26.
u=1
u=0
i
R
2
x
FIGURE 20.2.23 Dirichlet problem in Problem 26
27. A real-valued function f(x, y) is called biharmonic in a
domain D when the fourth-order differential equation
0 4f
0 4f
0 4f
2
0
0x 4
0x 2 0y 2
0y 4
y
u=0
at all points in D. Examples of biharmonic functions are the
Airy stress function in the mechanics of solids and velocity
potentials in the analysis of viscous fluid flow.
(a) Show that if f is biharmonic in D, then u 2 f/x2 2 f/y2 is harmonic in D.
(b) If g(z) is analytic in D and f(x, y) Re(z g(z)), show that
f is biharmonic in D.
R
i
u=1
u=1
1
u=0
x
FIGURE 20.2.20 Dirichlet problem in Problem 23
24.
y
u=0
R
i
u=5
1
u=0
x
FIGURE 20.2.21 Dirichlet problem in Problem 24
20.3 Linear Fractional Transformations
INTRODUCTION
In many applications that involve boundary-value problems associated
with Laplace’s equation, it is necessary to find a conformal mapping that maps a disk onto the
half-plane v 0. Such a mapping would have to map the circular boundary of the disk to the
boundary line of the half-plane. An important class of elementary conformal mappings that map
circles to lines (and vice versa) are the fractional transformations. In this section we will define
and study this special class of mappings.
922
|
CHAPTER 20 Conformal Mappings
Linear Fractional Transformation If a, b, c, and d are complex constants with
ad bc
0, then the complex function defined by
T (z) az b
cz d
is called a linear fractional transformation. Since
T 9(z) ad 2 bc
,
(cz d)2
T is conformal at z provided ad bc 0 and z d/c. (If 0, then T(z) 0 and T(z)
would be a constant function.) Linear fractional transformations are circle preserving in a sense
that we will make precise in this section, and, as we saw in Example 7 of Section 20.2, they can
be useful in solving Dirichlet problems in regions bounded by circles.
Note that when c 0, T(z) has a simple pole at z0 d/c and so
lim ZT (z)Z q.
zSz0
We will write T(z0) q as shorthand for this limit. In addition, if c
a b>z
lim T (z) lim
ZzZSq
ZzZSq
c d>z
0, then
a
,
c
and we write T(q) a/c.
EXAMPLE 1
A Linear Fractional Transformation
If T(z) (2z 1)/(z i), compute T(0), T(q), and T(i).
SOLUTION Note that T(0) 1/(i) i and T(q) lim|z|S q T(z) 2. Since z i is a simple
pole for T(z), we have limzSi ZT(z)Z q and we write T(i) q.
Circle-Preserving Property If c 0, the linear fractional transformation reduces to a
linear function T(z) Az B. We saw in Section 20.1 that such a complex mapping can be
considered as the composite of a rotation, magnification, and translation. As such, a linear function will map a circle in the z-plane to a circle in the w-plane. When c 0, we can divide cz d
into az b to obtain
w
1
az b
bc 2 ad
a
.
c
c
cz d
cz d
(1)
If we let A (bc ad)/c and B a/c, T(z) can be written as the composite of transformations:
z1 cz d,
z2 1
,
z1
w Az2 B.
(2)
A general linear fractional transformation can therefore be written as the composite of two linear
functions and the inversion w 1/z. Note that if Zz z1 Z r and w 1/z, then
2
Zw 2 w1 Z
1
1
2
r
2
w
w1
ZwZZw1 Z
or
Zw 2 w1 Z (r Zw1 Z)Zw 2 0Z.
(3)
It is not hard to show that the set of all points w that satisfy
Zw w1 Z lZw w2 Z
(4)
is a line when l 1 and is a circle when l 0 and l 1. It follows from (3) that the image of
the circle Zz z1 Z r under the inversion w 1/z is a circle except when r 1/Zw1 Z Zz1 Z. In the
20.3 Linear Fractional Transformations
|
923
latter case, the original circle passes through the origin and the image is a line. See Figure 20.1.3.
From (2), we can deduce the following theorem.
Theorem 20.3.1
Circle-Preserving Property
A linear fractional transformation maps a circle in the z-plane to either a line or a circle in
the w-plane. The image is a line if and only if the original circle passes through a pole of
the linear fractional transformation.
PROOF: We have shown that a linear function maps a circle to a circle, whereas an inver-
sion maps a circle to a circle or a line. It follows from (2) that a circle in the z-plane will be
mapped to either a circle or a line in the w-plane. If the original circle passes through a pole z0,
then T(z0) ⫽ q, and so the image is unbounded. Therefore, the image of such a circle must be a line.
If the original circle does not pass through z0, then the image is bounded and must be a circle.
EXAMPLE 2
Images of Circles
Find the images of the circles ZzZ ⫽ 1 and ZzZ ⫽ 2 under T(z) ⫽ (z ⫹ 2)/(z ⫺ 1). What are the
images of the interiors of these circles?
SOLUTION The circle Z zZ ⫽ 1 passes through the pole z0 ⫽ 1 of the linear fractional
transformation and so the image is a line. Since T(⫺1) ⫽ ⫺ 12 and T(i) ⫽ ⫺ 12 ⫺ 32 i, we can
conclude that the image is the line u ⫽ ⫺ 12 . The image of the interior Z zZ ⬍ 1 is either the
half-plane u ⬍ ⫺ 12 or the half-plane u ⬎ ⫺ 12 . Using z ⫽ 0 as a test point, T(0) ⫽ ⫺2, and so
the image is the half-plane u ⬍ ⫺ 12 .
The circle ZzZ ⫽ 2 does not pass through the pole and so the image is a circle. For ZzZ ⫽ 2,
ZzZ ⫽ 2
and
T (z) ⫽
z⫹2
z⫹2
⫽
⫽ T ( z ).
z21
z21
Therefore, T(z) is a point on the image circle and so the image circle is symmetric with respect
to the u-axis. Since T(⫺2) ⫽ 0 and T(2) ⫽ 4, the center of the circle is w ⫽ 2 and the image
is the circle Z w ⫺ 2Z ⫽ 2. See FIGURE 20.3.1. The image of the interior Z zZ ⬍ 2 is either the
interior or the exterior of the image circle Zw ⫺ 2Z ⫽ 2. Since T(0) ⫽ ⫺2, we can conclude
that the image is Zw ⫺ 2Z ⬎ 2.
v
T(z)
u
–2
2
T(z)
FIGURE 20.3.1 Images of test points in Example 2
Constructing Special Mappings In order to use linear fractional transformations to
solve Dirichlet problems, we must construct special functions that map a given circular region
R to a target region R⬘ in which the corresponding Dirichlet problem is solvable. Since a circular
boundary is determined by three of its points, we must find a linear fractional transformation
w ⫽ T(z) that maps three given points z1, z2, and z3 on the boundary of R to three points w1, w2,
and w3 on the boundary of R⬘. In addition, the interior of R⬘ must be the image of the interior of R.
See FIGURE 20.3.2.
924
|
CHAPTER 20 Conformal Mappings
R
R′
z1
z2
w = T(z)
z3
w1
w2
w3
FIGURE 20.3.2 R is image of R under T
Matrix Methods Matrix methods can be used to simplify many of the computations. We
can associate the matrix
A a
a
c
b
b
d
with T(z) (az b)/(cz d).* If T1(z) (a1z b1)/(c1z d1) and T2(z) (a2z b2) /(c2z d2),
then the composite function T(z) T2(T1(z)) is given by T(z) (az b)/(cz d ), where
a
a
c
b
a
b a 2
d
c2
b2 a1
ba
d2 c1
b1
b.
d1
(5)
If w T(z) (az b)/(cz d ), we can solve for z to obtain z (dw b) /(cw a). Therefore
the inverse of the linear fractional transformation T is T 1(w) (dw b) /(cw a) and we
associate the adjoint of the matrix A,
adj A a
d
c
b
b,
a
(6)
with T 1. The matrix adj A is the adjoint matrix of A (see Section 8.6), the matrix for T.
EXAMPLE 3
Using Matrices to Find an Inverse Transform
2z 2 1
z2i
If T(z) and S(z) , find S1(T(z)).
z2
iz 2 1
SOLUTION From (5) and (6), we have S1(T(z)) (az b)/(cz d ), where
a
a
c
b
1
b adj a
d
i
a
Therefore,
1
i
i 2
ba
1 1
i 2
ba
1 1
1
b
2
1
2 i2
b a
2
1 2 2i
S1(T(z)) 1 2i
b.
2 i2
(2 i) z 1 2i
.
(1 2 2i) z 2 i
Triples to Triples The linear fractional transformation
T (z) z 2 z 1 z2 2 z 3
z 2 z3 z2 2 z1
has a zero at z z1, a pole at z z3, and T(z2) 1. Therefore, T(z) maps three distinct complex
z 2 z 1 z2 2 z 3
numbers z1, z2, and z3 to 0, 1, and q, respectively. The term
is called the cross-ratio
z 2 z3 z2 2 z1
of the complex numbers z, z1, z2, and z3.
* The matrix A is not unique since the numerator and denominator in T(z) can be multiplied by a nonzero
constant.
20.3 Linear Fractional Transformations
|
925
w 2 w1 w2 2 w3
sends w1, w2, and w3 to 0, 1, and q,
w 2 w3 w2 2 w1
1
and so S maps 0, 1, and q to w1, w2, and w3. It follows that the linear fractional transformation
w S1(T(z)) maps the triple z1, z2, and z3 to w1, w2, and w3. From w S1(T(z)), we have
S(w) T(z) and we can conclude that
Likewise, the complex mapping S(w) w 2 w1 w2 2 w3
z 2 z 1 z2 2 z 3
.
z 2 z3 z2 2 z1
w 2 w3 w2 2 w1
(7)
In constructing a linear fractional transformation that maps the triple z1, z2, and z3 to w1, w2,
and w3, we can use matrix methods to compute w S1(T(z)). Alternatively, we can substitute
into (7) and solve the resulting equation for w.
Constructing a Linear Fractional Transformation
EXAMPLE 4
Construct a linear fractional transformation that maps the points 1, i, and 1 on the circle
ZzZ 1 to the points 1, 0, and 1 on the real axis.
SOLUTION Substituting into (7), we have
w1 021
z21i1
w 2 1 0 2 (1)
z1i21
or
w1
z21
i
.
w21
z1
Solving for w, we obtain w i(z i)/(z i). Alternatively, we could use the matrix method
to compute w S1(T(z)).
When zk q plays the role of one of the points in a triple, the definition of the cross-ratio is
changed by replacing each factor that contains zk by 1. For example, if z2 q, both z2 z3 and
z2 z1 are replaced by 1, giving (z z1)/(z z3) as the cross-ratio.
Constructing a Linear Fractional Transformation
EXAMPLE 5
Construct a linear fractional transformation that maps the points q, 0, and 1 on the real axis
to the points 1, i, and 1 on the circle ZwZ 1.
y
A
u=0
SOLUTION Since z1 q, the terms z z1 and z2 z1 in the cross product are replaced by 1.
It follows that
u=0
w21i1
1 021
w1i21
z21 1
C
B
1
D
x
u=0
a
(a)
v
U = 1 C′
and so w U = 1 D′ U = 1
EXAMPLE 6
B′
A′
U=0
U=0
U=0
(b)
u
|
a
c
b
i
b adj a
d
1
i 0
ba
1 1
iz 2 1 i
z212i
.
iz 1 i
z21i
1
i
b a
1
i
1 i
b
1i
Solving a Dirichlet Problem
Solve the Dirichlet problem in FIGURE 20.3.3(a) using conformal mapping by constructing a
linear fractional transformation that maps the given region into the upper half-plane.
SOLUTION The boundary circles ZzZ 1 and Zz 12 Z 12 each pass through z 1. We can
therefore map each boundary circle to a line by selecting a linear fractional transformation
that has z 1 as a pole. If we further require that T(i) 0 and T(1) 1, then
FIGURE 20.3.3 Image of Dirichlet
problem in Example 6
926
w21
1
T (z).
w1
z21
If we use the matrix method to find w S1(T(z)), then
u=1
u=0
S(w) i
or
CHAPTER 20 Conformal Mappings
T (z) z 2 i 1 2 1
z2i
(1 2 i)
.
z 2 1 1 2 i
z21
Since T(0) 1 i and T( 12 12 i) 1 i, T maps the interior of the circle ZzZ 1 onto the
upper half-plane and maps the circle Zz 12 Z 12 onto the line v 1. Figure 20.3.3(b) shows
the transferred boundary conditions.
The harmonic function U(u, v) v is the solution to the simplified Dirichlet problem in the
w-plane, and so, by Theorem 20.2.2, u(x, y) U(T(z)) is the solution to the original Dirichlet
problem in the z-plane.
1 2 x 2 2 y2
z2i
Since the imaginary part of T(z) (1 i)
is
, the solution is given by
z 2 1 (x 2 1)2 y 2
y
u(x, y) 0.2
0.4
1
x
The level curves u(x, y) c can be written as
0.6
0.8
FIGURE 20.3.4 Circles are level curves
in Example 6
20.3
Exercises
1 2 x 2 2 y2
.
(x 2 1)2 y 2
ax 2
2
2
c
1
b y2 a
b
1c
1c
and are therefore circles that pass through z 1. See FIGURE 20.3.4. These level curves may
be interpreted as the isotherms of the steady-state temperature distribution induced by the
boundary temperatures.
Answers to selected odd-numbered problems begin on page ANS-43.
In Problems 1–4, a linear fractional transformation is given.
(a) Compute T(0), T(1), and T(q).
(b) Find the images of the circles ZzZ 1 and Zz 1Z 1.
(c) Find the image of the disk ZzZ 1.
i
1
1. T(z) 2. T(z) z
z21
z1
z2i
3. T(z) 4. T(z) z
z21
In Problems 5–8, use the matrix method to compute S1(w) and
S1(T(z)) for each pair of linear fractional transformations.
z
iz 1
5. T(z) and
S(z) iz 2 1
z21
2z 1
iz
and
S(z) 6. T(z) z 2 2i
z1
2z 2 3
z22
7. T(z) and S(z) z23
z21
(2 2 i)z
z21i
8. T(z) and
S(z) iz 2 2
z212i
In Problems 9–16, construct a linear fractional transformation
that maps the given triple z1, z2, and z3 to the triple w1, w2,
and w3.
9. 1, 0, 2 to 0, 1, q
10. i, 0, i to 0, 1, q
11. 0, 1, q to 0, i, 2
12. 0, 1, q to 1 i, 0, 1 i
13. 1, 0, 1 to i, q, 0
14. 1, 0, 1 to q, i, 1
15. 1, i, i to 1, 0, 3
16. 1, i, i to i, i, 1
17. Use the results in Example 2 and the harmonic function
U (loge r)/(loge r0) to solve the Dirichlet problem in FIGURE 20.3.5.
Explain why the level curves must be circles.
y
u=0
u=1
R
2
– 0.5
x
2
FIGURE 20.3.5 Dirichlet problem in Problem 17
18. Use the linear fractional transformation that maps 1, 1, 0 to
0, 1, q to solve the Dirichlet problem in FIGURE 20.3.6. Explain
why, with one exception, all level curves must be circles.
Which level curve is a line?
y
u=0
u=1
–1
1
x
R
FIGURE 20.3.6 Dirichlet problem in Problem 18
19. Derive the conformal mapping H-1 in the conformal mappings
in Appendix IV.
20.3 Linear Fractional Transformations
|
927
20. Derive the conformal mapping H-5 in the conformal mappings
in Appendix IV by first mapping 1, i, 1 to q, i, 0.
21. Show that the composite of two linear fractional transformations is a linear fractional transformation and verify (5).
22. If w1
w2 and l 0, show that the set of all points w that
satisfy Zw w1 Z lZw w2 Z is a line when l 1 and is a
circle when l 1. [Hint: Write as Zw w1 Z 2 l2 Zw w2 Z 2
and expand.]
20.4 Schwarz–Christoffel Transformations
z4
α4
z5
z3
α3
α2
α1
z2
z1
(a) Bounded region
z4
α4
α3
f (z) A(z x1 )(a1 /p)1(z x2 )(a2 /p)1,
(1)
where x1 x2. In determining the images of line segments on the x-axis, we will use the fact that
a curve w w(t) in the w-plane is a line segment when the argument of its tangent vector w(t)
is constant. From (1), an argument of f (t) is given by
z2
z1
Special Cases To motivate the general Schwarz–Christoffel formula, we first examine
the effect of the mapping f (z) (z x1)a/p, 0 a 2p, on the upper half-plane y 0 shown
in FIGURE 20.4.2(a). This mapping is the composite of the translation z x1 and the real power
function w a/p. Since w a/p changes the angle in a wedge by a factor of a/p, the interior
angle in the image region is (a/p)p a. See Figure 20.4.2(b).
Note that f (z) A(z x1)(a/p)1 for A a/p. Next assume that f (z) is a function that is analytic in the upper half-plane and that has the derivative
z3
α2
α1
INTRODUCTION If D is a simply connected domain with at least one boundary point, then
the famous Riemann mapping theorem asserts the existence of an analytic function g that
conformally maps the unit open disk ZzZ 1 onto D. The Riemann mapping theorem is a pure
existence theorem that does not specify a formula for the conformal mapping. Since the upper
half-plane y 0 can be conformally mapped onto this disk using a linear fractional transformation, it follows that there exists a conformal mapping f between the upper half-plane and D. In
particular, there are analytic functions that map the upper half-plane onto polygonal regions of
the types shown in FIGURE 20.4.1. Unlike the Riemann mapping theorem, the Schwarz–Christoffel
formula specifies a form for the derivative f (z) of a conformal mapping from the upper half-plane
to a bounded or unbounded polygonal region.
(b) Unbounded region
arg f (t) Arg A a
FIGURE 20.4.1 Polygonal regions
a1
a2
2 1b Arg (t x1) a 2 1b Arg (t x2).
p
p
(2)
Since Arg(t x) p for t x, we can find the variation of arg f (t) along the x-axis. The results
are shown in the following table.
π
A
B
x1
(a)
α
0
Change in Argument
(q, x1)
(x1, x2)
(x2, q)
Arg A (a1 p) (a2 p)
Arg A (a2 p)
Arg A
0
p a1
p a2
Since arg f (t) is constant on the intervals in the table, the images are line segments, and FIGURE 20.4.3
shows the image of the upper half-plane. Note that the interior angles of the polygonal image region
are a1 and a2. This discussion generalizes to produce the Schwarz–Christoffel formula.
Schwarz–Christoffel Formula
Let f (z) be a function that is analytic in the upper half-plane y
(a1 /p)1
f (z) A(z x1 )
B
(b)
FIGURE 20.4.2 Image of upper half-plane
|
arg f (t)
Theorem 20.4.1
A′
928
Interval
(a2 /p)1
(z x2 )
0 and that has the derivative
… (z xn )(an /p)1,
(3)
where x1 x2 . . . xn and each ai satisfies 0 ai 2p. Then f (z) maps the upper
half-plane y 0 to a polygonal region with interior angles a1, a2, . . . , an.
CHAPTER 20 Conformal Mappings
w = f(t), t > x2
π – α2
w = f (t), x1< t < x2
α2
π – α1
α1
In applying this formula to a particular polygonal target region, the reader should carefully
note the following comments:
One can select the location of three of the points xk on the x-axis. A judicious choice can
simplify the computation of f (z). The selection of the remaining points depends on the
shape of the target polygon.
(ii) A general formula for f (z) is
(i)
#
f (z) A¢ (z 2 x1)(a1>p) 2 1(z 2 x2)(a2>p) 2 1 p (z 2 xn)(an>p) 2 1 dz ≤ B,
w = f(t), t < x1
FIGURE 20.4.3 Image of upper half-plane
and therefore f (z) may be considered as the composite of the conformal mapping
#
g(z) (z 2 x1)(a1>p) 2 1(z 2 x2)(a2>p) 2 1 p (z 2 xn)(an>p) 2 1 dz
y
A
x
1 B
–1
and the linear function w Az B. The linear function w Az B allows us to magnify,
rotate, and translate the image polygon produced by g(z). (See Section 20.1.)
(iii) If the polygonal region is bounded, only n 1 of the n interior angles should be included
in the Schwarz–Christoffel formula. As an illustration, the interior angles a1, a2, a3, and
a4 are sufficient to determine the Schwarz–Christoffel formula for the pentagon shown in
Figure 20.4.1(a).
(a)
v
EXAMPLE 1
i
Use the Schwarz–Christoffel formula to construct a conformal mapping from the upper halfplane to the strip ZvZ 1, u 0.
B′
π
2
–i
u
π
2
A′
Constructing a Conformal Mapping
SOLUTION We may select x1 1 and x2 1 on the x-axis, and we will construct a conformal mapping f with f (1) i and f (1) i. See FIGURE 20.4.4. Since a1 a2 p/2, the
Schwarz–Christoffel formula (3) gives
f 9(z) A(z 1)1>2(z 2 1)1>2 A
(b)
FIGURE 20.4.4 Image of upper half-plane
in Example 1
1
A
1
.
i (1 2 z 2 )1>2
(z 2 2 1)1>2
Therefore, f (z) Ai sin1z B. Since f (1) i and f (1) i, we obtain, respectively,
i Ai
y
p
B
2
and
i Ai
p
B
2
and conclude that B 0 and A 2/p. Thus, f (z) (2/p)i sin1z.
EXAMPLE 2
A
1 B
–1
x
Constructing a Conformal Mapping
Use the Schwarz–Christoffel formula to construct a conformal mapping from the upper halfplane to the region shown in FIGURE 20.4.5(b).
SOLUTION We again select x1 1 and x2 1, and we will require that f (1) ai
and f (1) 0. Since a1 3p/2 and a2 p/2, the Schwarz–Christoffel formula (3) gives
(a)
v
f (z) A(z 1)1/2 (z 1)1/2.
A′
If we write f (z) as A(z/(z2 1)1/2 1/(z2 1)1/2), it follows that
ai
B′
u
(b)
FIGURE 20.4.5 Image of upper half-plane
in Example 2
f (z) A[(z2 1)1/2 cosh1z] B.
Note that cosh1(1) pi and cosh1 1 0, and so ai f (1) A(pi) B and 0 f (1) B.
Therefore, A a/p and f (z) (a/p)[(z2 1)1/2 cosh1z].
The next example will show that it may not always be possible to find f (z) in terms of
elementary functions.
20.4 Schwarz–Christoffel Transformations
|
929
Constructing a Conformal Mapping
EXAMPLE 3
Use the Schwarz–Christoffel formula to construct a conformal mapping from the upper halfplane to the interior of the equilateral triangle shown in FIGURE 20.4.6(b).
y
v
A′
A
0
x
1 B
B′
1
0
u
(b)
(a)
FIGURE 20.4.6 Image of upper half-plane in Example 3
SOLUTION Since the polygonal region is bounded, only two of the three 60 interior angles
should be included in the Schwarz–Christoffel formula. If x 1 0 and x 2 1, we
obtain f (z) Az2/3 (z 1)2/3. It is not possible to evaluate f (z) in terms of elementary functions; however, we can use Theorem 18.3.3 to construct the antiderivative
f (z) A
#
z
1
ds B.
s 2>3(s 2 1)2>3
0
If we require that f (0) 0 and f (1) 1, it follows that B 0 and
1A
#
1
0
s
2>3
1
ds.
(s 2 1)2>3
It can be shown that this last integral is G(13), where denotes the gamma function. Therefore,
the required conformal mapping is
f (z) 1
G(13)
#
z
0
1
ds.
s 2>3(s 2 1)2>3
The Schwarz–Christoffel formula can sometimes be used to suggest a possible conformal
mapping from the upper half-plane onto a nonpolygonal region R. A key first step is to approximate R by polygonal regions. This will be illustrated in the final example.
y
EXAMPLE 4
Constructing a Conformal Mapping
Use the Schwarz–Christoffel formula to construct a conformal mapping from the upper halfplane to the upper half-plane with the horizontal line v p, u 0, deleted.
A
0
–1
B
x
(a)
SOLUTION The nonpolygonal target region can be approximated by a polygonal region by
adjoining a line segment from w pi to a point u0 on the negative u-axis. See FIGURE 20.4.7(b).
If we require that f (1) pi and f (0) u0, the Schwarz–Christoffel transformation satisfies
v
A′
v=π
f 9(z) A(z 1)(a1>p) 2 1z (a2>p) 2 1.
α1
α2
B′
u0
u
(b)
FIGURE 20.4.7 Image of upper half-plane
in Example 4
930
|
Note that as u0 approaches q, the interior angles a1 and a2 approach 2p and 0, respectively.
This suggests we examine conformal mappings that satisfy w A(z 1)1z1 A(1 1/z)
or w A(z Ln z) B.
We will first determine the image of the upper half-plane under g(z) z Ln z and then
translate the image region if needed. For t real,
g(t) t loge ZtZ i Arg t.
If t 0, Arg t p and u(t) t loge ZtZ varies from q to 1. It follows that w g(t) moves
along the line v p from q to 1. When t 0, Arg t 0 and u(t) varies from q to q.
Therefore, g maps the positive x-axis onto the u-axis. We can conclude that g(z) z Ln z maps
CHAPTER 20 Conformal Mappings
the upper half-plane onto the upper half-plane with the horizontal line v p, u 1, deleted.
Therefore, w z Ln z 1 maps the upper half-plane onto the original target region.
Many of the conformal mappings in Appendix IV can be derived using the Schwarz–Christoffel
formula, and we will show in Section 20.6 that these mappings are especially useful in analyzing
two-dimensional fluid flows.
Exercises
20.4
Answers to selected odd-numbered problems begin on page ANS-44.
In Problems 1–4, use (2) to describe the image of the upper
half-plane y 0 under the conformal mapping w f (z) that
satisfies the given conditions. Do not attempt to find f (z).
1. f (z) (z 1)1/2, f (1) 0
2. f (z) (z 1)1/3, f (1) 0
3. f (z) (z 1)1/2 (z 1)1/2,
f (1) 0
4. f (z) (z 1)1/2 (z 1)3/4, f (1) 0
In Problems 5–8, find f (z) for the given polygonal region using
x1 1, x2 0, x3 1, x4 2, and so on. Do not attempt to
find f (z).
5. f (1) 0, f (0) 1
8. f (1) i,
f (0) 0
v
i
π
4
u
FIGURE 20.4.11 Polygonal region for Problem 8
9. Use the Schwarz–Christoffel formula to construct a conformal
v
mapping from the upper half-plane y 0 to the region in
FIGURE 20.4.12. Require that f (1) pi and f (1) 0.
i
v
πi
u
1
FIGURE 20.4.8 Polygonal region for Problem 5
6. f (1) 1,
u
f (0) 0
FIGURE 20.4.12 Image of upper half-plane in Problem 9
v
10. Use the Schwarz–Christoffel formula to construct a conformal
mapping from the upper half-plane y 0 to the region in
FIGURE 20.4.13. Require that f (1) ai and f (1) ai.
v
π
3
u
–1
ai
FIGURE 20.4.9 Polygonal region for Problem 6
7. f (1) 1,
f (0) 1
u
– ai
v
FIGURE 20.4.13 Image of upper half-plane in Problem 10
2π
3
–1
11. Use the Schwarz–Christoffel formula to construct a conformal
2π
3
1
u
FIGURE 20.4.10 Polygonal region for Problem 7
mapping from the upper half-plane y 0 to the horizontal
strip 0 v p by first approximating the strip by the polygonal region shown in FIGURE 20.4.14 . Require that
f (1) pi, f (0) w2 w1 , and f (1) 0, and let w1 S q
in the horizontal direction.
20.4 Schwarz–Christoffel Transformations
|
931
13. Verify M-4 in Appendix IV by first approximating the region
R⬘ by the polygonal region shown in FIGURE 20.4.16. Require
v
πi
w2
that f (⫺1) ⫽ ⫺u1, f (0) ⫽ ai, and f (1) ⫽ u1 and let u1 S 0
along the u-axis.
w1
v
u
FIGURE 20.4.14 Image of upper half-plane in Problem 11
R′
12. Use the Schwarz–Christoffel formula to construct a conformal
mapping from the upper half-plane y ⱖ 0 to the wedge
0 ⱕ Arg w ⱕ p/4 by first approximating the wedge by the
region shown in FIGURE 20.4.15. Require that f (0) ⫽ 0 and
f (1) ⫽ 1 and let u S 0.
v
ai
–u1
u1
u
FIGURE 20.4.16 Image of upper half-plane in Problem 13
14. Show that if a curve in the w-plane is parameterized by w ⫽ w(t),
a ⱕ t ⱕ b, and arg w⬘(t) is constant, then the curve is a line
segment. [Hint: If w(t) ⫽ u(t) ⫹ iv(t), then tan(arg w⬘(t)) ⫽ dv/du.]
θ
u
1
FIGURE 20.4.15 Image of upper half-plane in Problem 12
20.5 Poisson Integral Formulas
INTRODUCTION The success of the conformal mapping method depends on the recognition
of the solution to the new Dirichlet problem in the image region R⬘. It would therefore be helpful
if a general solution could be found for Dirichlet problems in either the upper half-plane y ⱖ 0
or the unit disk ZzZ ⱕ 1. The Poisson integral formula for the upper half-plane provides such a
solution by expressing the value of a harmonic function u(x, y) at a point in the interior of the
upper half-plane in terms of its values on the boundary y ⫽ 0.
z
θ (z)
Arg(z – b)
Arg(z – a)
u = ui
u=0 a
b u=0
x
Formulas for the Upper Half-Plane To develop the formula, we first assume that the
boundary function is given by u(x, 0) ⫽ f (x), where f (x) is the step function indicated in FIGURE 20.5.1.
The solution of the corresponding Dirichlet problem in the upper half-plane is
u(x, y) ⫽
FIGURE 20.5.1 Boundary conditions
on y ⫽ 0
θn
a = x0 x1
x2
u = 0 u = u1 u = u2
|
The superposition principle can be used to solve the more general Dirichlet problem in
FIGURE 20.5.2. If u(x, 0) ⫽ ui for xi⫺1 ⱕ x ⱕ xi and u(x, 0) ⫽ 0 outside the interval [a, b], then
•• •
x
from (1),
xn–1
xn = b
u = un
u=0
FIGURE 20.5.2 General boundary
conditions on y ⫽ 0
932
(1)
Since Arg(z ⫺ b) is an exterior angle in the triangle formed by z, a, and b, Arg(z ⫺ b) ⫽ u(z) ⫹
Arg(z ⫺ a), where 0 ⬍ u(z) ⬍ p, and we can write
ui
ui
z2b
b.
(2)
u(x, y) ⫽ u(z) ⫽ Arg a
p
p
z2a
z
θ1 θ2
ui
[Arg (z ⫺ b) ⫺ Arg (z ⫺ a)].
p
n u
1 n
i
u(x, y) ⫽ a
fArg (z 2 xi) 2 Arg (z 2 xi2 1)g ⫽
a uiui(z).
p i⫽
i⫽ 1 p
1
(3)
Note that Arg (z ⫺ t) ⫽ tan⫺1( y/(x ⫺ t)), where tan⫺1 is selected between 0 and p, and therefore
d/dt Arg (z ⫺ t) ⫽ y/((x ⫺ t)2 ⫹ y2). From (3),
CHAPTER 20 Conformal Mappings
1 n
u(x, y) a
p i
1
d
1 n
ui Arg (z 2 t) dt a
p i
dt
xi 2 1
1
#
xi
xi
ui y
dt.
(x
2
t)2 y 2
xi 2 1
#
Since u(x, 0) 0 outside of the interval [a, b], we have
y
u(x, y) p
#
q
q
u(t, 0)
dt.
(x 2 t)2 y 2
(4)
A bounded piecewise-continuous function can be approximated by step functions, and therefore
our discussion suggests that (4) is the solution to the Dirichlet problem in the upper half-plane.
This is the content of Theorem 20.5.1.
Theorem 20.5.1
Poisson Integral Formula for the Upper Half-Plane
Let u(x, 0) be a piecewise-continuous function on every finite interval and bounded on
q x q. Then the function defined by
u(x, y) y
p
#
q
q
u(t, 0)
dt
(x 2 t)2 y 2
is the solution of the corresponding Dirichlet problem on the upper half-plane y 0.
There are a few functions for which it is possible to evaluate the integral in (4), but in general,
numerical methods are required to evaluate the integral.
EXAMPLE 1
Solving a Dirichlet Problem
Find the solution of the Dirichlet problem in the upper half-plane that satisfies the boundary
condition u(x, 0) x when ZxZ 1, and u(x, 0) 0 otherwise.
SOLUTION By the Poisson integral formula,
u(x, y) y
p
#
1
1
t
dt.
(x 2 t)2 y 2
Using the substitution s x t, we can show that
u(x, y) 1 y
x 2 t t 1
bd
,
c log e ((x 2 t)2 y 2) 2 x tan1 a
p 2
y
t 1
which can be simplified to
u(x, y) y
In most of the examples and exercises u(x, 0) is a step function, and we will use the integrated
solution (3) rather than (4). If the first interval is (q, x1), then the term Arg(z x1) Arg(z a)
in the sum should be replaced by Arg(z x1). Likewise, if the last interval is (xn1, q), then
Arg(z b) Arg(z xn1) should be replaced by p Arg(z xn1).
u=0
u=1
u=1
–2
2
x
(a)
v
U=1
U=0
–2
y
(x 2 1)2 y 2
x
x1
x21
b 2 tan1 a
b d.
log e c
d ctan1 a
p
y
y
2p
(x 1)2 y 2
EXAMPLE 2
Solving a Dirichlet Problem
The conformal mapping f (z) z 1/z maps the region in the upper half-plane and outside
the circle ZzZ 1 onto the upper half-plane v 0. Use this mapping and the Poisson integral
formula to solve the Dirichlet problem shown in FIGURE 20.5.3(a).
U=1
2
(b)
FIGURE 20.5.3 Image of Dirichlet
problem in Example 2
u
SOLUTION Using the results of Example 3 in Section 20.2, we can transfer the boundary
conditions to the w-plane. See Figure 20.5.3(b). Since U(u, 0) is a step function, we will use the
integrated solution (3) rather than the Poisson integral. The solution to the new Dirichlet problem is
U(u, v) 1
1
w2
1
b,
Arg (w 2) fp 2 Arg (w 2 2)g 1 Arg a
p
p
p
w22
20.5 Poisson Integral Formulas
|
933
and therefore
z 1>z 2
1
1
b,
u(x, y) U az b 1 Arg a
p
z
z 1>z 2 2
z1 2
1
b .
Arg a
p
z21
which can be simplified to u(x, y) 1 Formula for the Unit Disk A Poisson integral formula can also be developed to solve
the general Dirichlet problem for the unit disk.
Theorem 20.5.2
Poisson Integral Formula for the Unit Disk
iu
Let u(e ) be bounded and piecewise continuous for p u p. Then the solution to the
corresponding Dirichlet problem on the open unit disk ZzZ 1 is given by
u(x, y) 1
2p
u = u(x, y)
frame
u(e it )
p
1 2 ZzZ2
Ze it 2 zZ2
dt.
(5)
Geometric Interpretation FIGURE 20.5.4 shows a thin membrane (such as a soap film)
that has been stretched across a frame defined by u u(eiu). The displacement u in the direction
perpendicular to the z-plane satisfies the two-dimensional wave equation
a2 a
|z| < 1
FIGURE 20.5.4 Thin membrane on
a frame
#
p
0 2u
0 2u
0 2u
b
,
0x 2
0y 2
0t 2
and so at equilibrium, the displacement function u u(x, y) is harmonic. Formula (5) provides
an explicit solution for the displacement u and has the advantage that the integral is over the finite
interval [p, p]. When the integral cannot be evaluated, standard numerical integration procedures can be used to estimate u(x, y) at a fixed point z x iy with ZzZ 1.
EXAMPLE 3
Displacement of a Membrane
A frame for a membrane is defined by u(eiu) ZuZ for p u p. Estimate the equilibrium
displacement of the membrane at (0.5, 0), (0, 0), and (0.5, 0).
#
1
SOLUTION From (5), we get u(x, y) 2p
u(0, 0) 1
2p
p
p
#
ZtZ
1 2 ZzZ2
Ze it 2 zZ2
p
ZtZ dt p
dt. When (x, y) (0, 0), we get
p
.
2
For the other two values of (x, y), the integral is not elementary and must be estimated using
a numerical integration procedure. Using Simpson’s rule, we obtain (to four decimal places)
u(0.5, 0) 2.2269 and u(0.5, 0) 0.9147.
Fourier Series Form The Poisson integral formula for the unit disk is actually a compact
way of writing the Fourier series solution to Laplace’s equation that we developed in Chapter
14. To see this, first note that un(r, u) r n cos nu and vn(r, u) r n sin nu are each harmonic,
since these functions are the real and imaginary parts of zn. If a0, an, and bn are chosen to be the
Fourier coefficients of u(eiu) for p u p, then, by the superposition principle,
u(r, u) 934
|
CHAPTER 20 Conformal Mappings
q
a0
a (anr n cos nu bnr n sin nu)
2
n1
(6)
frame u(eiθ ) = sin 4θ
is harmonic and u(1, u) (a0 /2) g n 1 (an cos nu bn sin nu) u(eiu). Since the solution of
the Dirichlet problem is also given by (5), we have
q
1
u(r, u) 2p
#
level curves
FIGURE 20.5.5 Level curves in
Example 4
Exercises
20.5
u=0
1
u = –1 u = 1
u=0
Solving a Dirichlet Problem
SOLUTION Rather than working with the Poisson integral (5), we will use the Fourier series
solution (6), which reduces to u(r, u) r4 sin 4u. Therefore, u 0 if and only if sin 4u 0.
This implies u 0 on the lines x 0, y 0, and y x.
If we switch to rectangular coordinates, u(x, y) 4xy(x2 y2). The surface u(x, y) 4xy(x2 y2),
the frame u(eiu) sin 4u, and the system of level curves were sketched using graphics software
and are shown in FIGURE 20.5.5.
Answers to selected odd-numbered problems begin on page ANS-44.
y
–1
q
a0
1 2 r2
dt a (anr n cos nu bnr n sin nu).
iu 2
2
Ze 2 re Z
n1
it
Find the solution of the Dirichlet problem in the unit disk satisfying the boundary condition
u(eiu) sin 4u. Sketch the level curve u 0.
In Problems 1–4, use the integrated solution (3) to the Poisson
integral formula to solve the given Dirichlet problem in the
upper half-plane.
1.
u(e it )
p
EXAMPLE 4
z-plane
p
5. Find the solution of the Dirichlet problem in the upper half-
plane that satisfies the boundary condition u(x, 0) x2 when
0 x 1, and u(x, 0) 0 otherwise.
6. Find the solution of the Dirichlet problem in the upper halfplane that satisfies the boundary condition u(x, 0) cos x.
[Hint: Let s t x and use the Section 19.6 formulas
#
x
q
for a
FIGURE 20.5.6 Dirichlet problem in Problem 1
1
–2
u=0
u=5
cos s
pea
ds ,
2
2
a
s a
u=1 u=0
7.
x
y
8.
i
R
1
–1
u = 0 u = –1 u = 1 u = 0 u = 5
9.
u=0
–1
R
x
y
u = 1 u = –1 u = 1 u = 1 u = 0
u=0
10.
u=1
u=1
y
u=0
R
u=1
x
FIGURE 20.5.9 Dirichlet problem in Problem 4
x
FIGURE 20.5.11 Dirichlet
problem in Problem 8
x
1
u=1
u=0 3
R
y
–2
u=0
FIGURE 20.5.10 Dirichlet
problem in Problem 7
x
FIGURE 20.5.8 Dirichlet problem in Problem 3
1
y
u=1
u=0
u=1
4.
sin s
ds 0
s 2 a2
0.]
y
–2
q
u=5
FIGURE 20.5.7 Dirichlet problem in Problem 2
3.
#
q
In Problems 7–10, solve the given Dirichlet problem by finding
a conformal mapping from the given region R onto the upper
half-plane v 0.
y
2.
q
u=0
u=1 1
FIGURE 20.5.13 Dirichlet
problem in Problem 10
FIGURE 20.5.12 Dirichlet
problem in Problem 9
20.5 Poisson Integral Formulas
|
935
x
11. A frame for a membrane is defined by u(eiu) u2 /p2 for
p u p. Use the Poisson integral formula for the unit
disk to estimate the equilibrium displacement of the membrane
at (0.5, 0), (0, 0), and (0.5, 0).
12. A frame for a membrane is defined by u(eiu) e|u| for
p u p. Use the Poisson integral formula for the unit
disk to estimate the equilibrium displacement of the membrane
at (0.5, 0), (0, 0), and (0.5, 0).
13. Use the Poisson integral formula for the unit disk to show that
u(0, 0) is the average value of the function u u(eiu ) on the
boundary ZzZ 1.
In Problems 14 and 15, solve the given Dirichlet problem for the
unit disk using the Fourier series form of the Poisson integral
formula, and sketch the system of level curves.
14. u(eiu ) cos 2u
15. u(eiu ) sin u cos u
20.6 Applications
INTRODUCTION In Sections 20.2, 20.3, and 20.5 we demonstrated how Laplace’s partial
differential equation can be solved with conformal mapping methods, and we interpreted a solution u u(x, y) of the Dirichlet problem as either the steady-state temperature at the point (x, y)
or the equilibrium displacement of a membrane at the point (x, y). Laplace’s equation is a
fundamental partial differential equation that arises in a variety of contexts. In this section we
will establish a general relationship between vector fields and analytic functions and use
our conformal mapping techniques to solve problems involving electrostatic force fields and
two-dimensional fluid flows.
Vector Fields A vector field F(x, y) P(x, y)i Q(x, y)j in a domain D can also be
expressed in the complex form
F(x, y) P(x, y) iQ(x, y)
and thought of as a complex function. Recall from Chapter 9 that div F P/x Q/y and
curl F (Q/x P/y)k. If we require that both div F 0 and curl F 0, then
0Q
0P
0x
0y
and
0Q
0P
.
0y
0x
(1)
This set of equations is reminiscent of the Cauchy–Riemann criterion for analyticity presented
in Theorem 17.5.2 and suggests that we examine the complex function g(z) P(x, y) iQ(x, y).
Theorem 20.6.1
Vector Fields and Analyticity
(i) Suppose that F(x, y) P(x, y) iQ(x, y) is a vector field in a domain D and P(x, y) and
Q(x, y) are continuous and have continuous first partial derivatives in D. If div F 0 and
curl F 0, then the complex function
g(z) P(x, y) iQ(x, y)
is analytic in D.
(ii) Conversely, if g(z) is analytic in D, then F(x, y) g(z) defines a vector field in D for
which div F 0 and curl F 0.
PROOF: If u(x, y) and v(x, y) denote the real and imaginary parts of g(z), then u P and v Q.
Therefore the equations in (1) are equivalent to the equations
0(v)
0u
0x
0y
936
|
CHAPTER 20 Conformal Mappings
and
0(v)
0u
;
0y
0x
that is,
0v
0u
0x
0y
0v
0u
.
0y
0x
and
(2)
The equations in (2) are the Cauchy–Riemann equations for analyticity.
EXAMPLE 1
Vector Field Gives an Analytic Function
The vector field defined by F(x, y) (kq/Zz z0 Z 2)(z z0) may be interpreted as the electric
field produced by a wire that is perpendicular to the z-plane at z z0 and carries a charge of
q coulombs per unit length. The corresponding complex function is
g(z) Since g(z) is analytic for z
EXAMPLE 2
kq
kq
(z 2 z0) .
z 2 z0
Zz 2 z0 Z2
z0, div F 0 and curl F 0.
Analytic Function Gives a Vector Field
The complex function g(z) Az, A 0, is analytic in the first quadrant and therefore gives
rise to the vector field V(x, y) g(z) Ax iAy, which satisfies div V 0 and curl V 0.
We will show toward the end of this section that V(x, y) may be interpreted as the velocity of
a fluid that moves around the corner produced by the boundary of the first quadrant.
The physical interpretation of the conditions div F 0 and curl F 0 depends on the setting.
If F(x, y) represents the force in an electric field that acts on a unit test charge placed at (x, y),
then, by Theorem 9.9.2, curl F 0 if and only if the field is conservative. The work done in
transporting a test charge between two points in D must be independent of the path.
If C is a simple closed contour that lies in D, Gauss’s law asserts that the line integral 养C (F n) ds
is proportional to the total charge enclosed by the curve C. If D is simply connected and all the
electric charge is distributed on the boundary of D, then 养C (F n) ds 0 for any simple closed
contour in D. By the divergence theorem in the form (1) of Section 9.16,
BC
(F n) ds 6 div F dA,
(3)
R
where R is the region enclosed by C, and we can conclude that div F 0 in D. Conversely, if
div F 0 in D, the double integral is zero and therefore the domain D contains no charge.
Potential Functions Suppose that F(x, y) is a vector field in a simply connected domain
D with both div F 0 and curl F 0. By Theorem 18.3.3, the analytic function g(z) P(x, y) iQ(x, y)
has an antiderivative
G(z) f(x, y) ic(x, y)
(4)
in D, which is called a complex potential for the vector field F. Note that
g(z) G9(z) and so
0f
P
0x
0c
0f
0f
0f
i
2i
0x
0x
0x
0y
and
0f
Q.
0y
(5)
Therefore, F f and, as in Section 9.9, the harmonic function f is called a (real) potential
function for F.* When the potential f is specified on the boundary of a region R, we can use
conformal mapping techniques to solve the resulting Dirichlet problem. The equipotential lines
f(x, y) c can be sketched and the vector field F can be determined using (5).
*If F is an electric field, the electric potential function is defined to be f and F .
20.6 Applications
|
937
y
EXAMPLE 3
Complex Potential
The potential f in the half-plane x 0 satisfies the boundary conditions f(0, y) 0 and
f(x, 0) 1 for x 1. See FIGURE 20.6.1(a). Determine a complex potential, the equipotential
lines, and the force field F.
φ =0
1
x
φ =1
φ =0
SOLUTION We saw in Example 2 of Section 20.2 that the analytic function z sin w maps
the strip 0 u p/2 in the w-plane to the region R in question. Therefore, f (z) sin1z maps
R onto the strip, and Figure 20.6.1(b) shows the transferred boundary conditions. The simplified Dirichlet problem has the solution U(u, v) (2/p)u, and so f(x, y) U(sin1z) Re ((2/p) sin1z) is the potential function on D, and G(z) (2/p) u sin1z is a complex potential
for the force field F.
Note that the equipotential lines f c are the images of the equipotential lines U c in
the w-plane under the inverse mapping z sin w. In Example 2 of Section 20.2 we showed
that the vertical line u a is mapped onto a branch of the hyperbola
0.75
0.25
0.5
(a)
v
U=1
U=0
π
2
U=1
U=0
0.25 0.5 0.75
y2
x2
2
1.
sin 2a
cos 2a
u
Since the equipotential line U c, 0 c 1, is the vertical line u p/2c, it follows that the
equipotential line f c is the right branch of the hyperbola
y2
x2
2
1.
sin2 (pc>2)
cos 2 (pc>2)
(b)
FIGURE 20.6.1 Images of boundary
conditions in Example 3
Since F G9(z) and d/dz sin1z 1/(1 z2)1/2, the force field is given by
F
2
2
1
1
.
p (1 2 z 2 )1>2
p (1 2 z 2 )1>2
Steady-State Fluid Flow The vector V(x, y) P(x, y) iQ(x, y) may also be interpreted
as the velocity vector of a two-dimensional steady-state fluid flow at a point (x, y) in a domain D.
The velocity at all points in the domain is therefore independent of time, and all movement takes
place in planes that are parallel to a z-plane.
The physical interpretation of the conditions div V 0 and curl V 0 was discussed in
Section 9.7. Recall that if curl V 0 in D, the flow is called irrotational. If a small circular
paddle wheel is placed in the fluid, the net angular velocity on the boundary of the wheel is zero,
and so the wheel will not rotate. If div V 0 in D, the flow is called incompressible. In a simply
connected domain D, an incompressible flow has the special property that the amount of fluid
in the interior of any simple closed contour C is independent of time. The rate at which fluid
enters the interior of C matches the rate at which it leaves, and consequently there can be no fluid
sources or sinks at points in D.
If div V 0 and curl V 0, V has a complex velocity potential
G(z) f(x, y) ic(x, y)
that satisfies G9(z) V. In this setting, special importance is placed on the level curves c(x, y) c.
If z(t) x(t) iy(t) is the path of a particle (such as a small cork) that has been placed in the
fluid, then
dx
P(x, y)
dt
dy
Q(x, y).
dt
(6)
Hence, dy/dx Q(x, y)/P(x, y) or Q(x, y) dx P(x, y) dy 0. This differential equation is
exact, since div V 0 implies (Q)/y P/x. By the Cauchy–Riemann equations,
c/x f/y Q and c/y f/x P, and therefore all solutions of (6) satisfy
c(x, y) c. The function c(x, y) is therefore called a stream function and the level curves
c(x, y) c are streamlines for the flow.
938
|
CHAPTER 20 Conformal Mappings
y
EXAMPLE 4
V
V
x
(a)
The uniform flow in the upper half-plane is defined by V(x, y) A(1, 0), where A is a fixed
positive constant. Note that ZVZ A, and so a particle in the fluid moves at a constant speed.
A complex potential for the vector field is G(z) Az Ax iAy, and so the streamlines are
the horizontal lines Ay c. See FIGURE 20.6.2(a). Note that the boundary y 0 of the region is
itself a streamline.
EXAMPLE 5
y
Uniform Flow
Flow Around a Corner
The analytic function G(z) z2 gives rise to the vector field V(x, y) G9(z) (2x, 2y) in
the first quadrant. Since z2 x2 y2 i(2xy), the stream function is c(x, y) 2xy and the
streamlines are the hyperbolas 2xy c. This flow, called flow around a corner, is depicted in
Figure 20.6.2(b). As in Example 4, the boundary lines x 0 and y 0 in the first quadrant
are themselves streamlines.
Constructing Special Flows The process of constructing an irrotational and incompressible flow that remains inside a given region R is called streamlining. Since the streamlines are described by c(x, y) c, two distinct streamlines do not intersect. Therefore, if the boundary is itself a
streamline, a particle that starts inside R cannot leave R. This is the content of the following theorem:
V
V
x
(b)
FIGURE 20.6.2 (a) Uniform flow in
Example 4; (b) Flow around a corner
in Example 5
Theorem 20.6.2
Suppose that G(z) f(x, y) ic(x, y) is analytic in a region R and c(x, y) is constant on the
boundary of R. Then V(x, y) G9(z) defines an irrotational and incompressible fluid flow in R.
Moreover, if a particle is placed inside R, its path z z(t) remains in R.
EXAMPLE 6
Flow Around a Cylinder
The analytic function G(z) z 1/z maps the region R in the upper half-plane and outside
the circle ZzZ 1 onto the upper half-plane v 0. The boundary of R is mapped onto the uaxis, and so v c(x, y) y y/(x2 y2) is zero on the boundary of R. FIGURE 20.6.3 shows
the streamlines of the resulting flow. The velocity field is given by G9(z) 1 1/ z 2 , and so
y
V
Streamlining
–1
1
1 2iu
e .
r2
It follows that V 艐 (1, 0) for large values of r, and so the flow is approximately uniform at large
distances from the circle ZzZ 1. The resulting flow in the region R is called flow around a cylinder.
The mirror image of the flow can be adjoined to give a flow around a complete cylinder.
G9(re iu ) 1 2
x
FIGURE 20.6.3 Flow around a cylinder
in Example 6
If R is a polygonal region, we can use the Schwarz–Christoffel formula to find a conformal
mapping z f (w) from the upper half-plane R onto R. The inverse function G(z) f 1(z) maps
the boundary of R onto the u-axis. Therefore, if G(z) f(x, y) ic(x, y), then c(x, y) 0 on
the boundary of R. Note that the streamlines c(x, y) c in the z-plane are the images of the horizontal lines v c in the w-plane under z f (w).
EXAMPLE 7
The analytic function f (w) w Ln w 1 maps the upper half-plane v 0 to the upper
half-plane y 0 with the horizontal line y p, x 0, deleted. See Example 4 in Section 20.4.
If G(z) f 1(z) f(x, y) ic (x, y), then G(z) maps R onto the upper half-plane and maps
the boundary of R onto the u-axis. Therefore, c (x, y) 0 on the boundary of R.
It is not possible to find an explicit formula for the stream function c (x, y). The streamlines,
however, are the images of the horizontal lines v c under z f (w). If we write w t ic,
c 0, then the streamlines can be represented in the parametric form
y
y =π
z f (t ic) t ic Ln(t ic) 1;
x
FIGURE 20.6.4 Flow in Example 7
Streamlines Defined Parametrically
1
loge (t 2 c2), y c Arg(t ic).
2
Graphing software was used to generate the streamlines in FIGURE 20.6.4.
that is,
xt1
20.6 Applications
|
939
A stream function c (x, y) is harmonic but, unlike a solution to a Dirichlet problem, we do not
require c (x, y) to be bounded (see Examples 4–6) or to assume a fixed set of constants on the
boundary. Therefore, there may be many different stream functions for a given region that satisfy
Theorem 20.6.2. This will be illustrated in the final example.
EXAMPLE 8
Streamlines Defined Parametrically
The analytic function f (w) w ew 1 maps the horizontal strip 0 v p onto the
region R shown in Figure 20.6.4. Therefore, G(z) f 1(z) f(x, y) ic(x, y) maps R back
to the strip and, from M-1 in the conformal mappings in Appendix IV, maps the boundary
line y 0 onto the u-axis and maps the boundary line y p, x 0, onto the horizontal line
v p. Therefore, c(x, y) is constant on the boundary of R.
The streamlines are the images of the horizontal lines v c, 0 c p, under z f (w).
As in Example 7, a parametric representation of the streamlines is
y
y =π
z f (t ic) t ic etic 1
x
Exercises
5. The potential f on the wedge 0 Arg z p/4 satisfies the
boundary conditions f(x, 0) 0 and f(x, x) 1 for x 0.
Determine a complex potential, the equipotential lines, and
the corresponding force field F.
6. Use the conformal mapping f (z) 1/z to determine a complex
potential, the equipotential lines, and the corresponding force
field F for the potential f that satisfies the boundary conditions
shown in FIGURE 20.6.6.
y
i
φ =0
|
boundary conditions f(x, 0) 0, 1 x 1, and
f(eiu ) 1, 0 u p. Show that
1
z21 2
f(x, y) Arg a
b
p
z1
and use the mapping properties of linear fractional transformations to explain why the equipotential lines are arcs of circles.
8. Use the conformal mapping C-1 in Appendix IV to find the
potential f in the region outside the two circles ZzZ 1 and
Zz 3Z 1 if the potential is kept at zero on ZzZ 1 and one
on Zz 3Z 1. Use the mapping properties of linear fractional
transformations to explain why the equipotential lines are,
with one exception, circles.
In Problems 9–14, a complex velocity potential G(z) is defined
on a region R.
(a) Find the stream function and verify that the boundary of R is
a streamline.
(b) Find the corresponding velocity vector field V(x, y).
(c) Use a graphing utility to sketch the streamlines of the flow.
9. G(z) z4
y=x
R
1
x
x
FIGURE 20.6.6 Boundary conditions in Problem 6
940
7. The potential f on the semicircle ZzZ 1, y 0, satisfies the
y
φ =1
φ =0
y c et sin c.
Answers to selected odd-numbered problems begin on page ANS-44.
In Problems 1–4, verify that div F 0 and curl F 0 for the
given vector field F(x, y) by examining the corresponding
complex function g(z) P(x, y) iQ(x, y). Find a complex
potential for the vector field and sketch the equipotential lines.
1. F(x, y) (cos u0) i (sin u0) j
2. F(x, y) y i x j
y
x
3. F(x, y) 2
i 2
j
x y2
x y2
x 2 2 y2
2xy
4. F(x, y) 2
i 2
j
2 2
(x y )
(x y 2)2
φ =1
x t 1 et cos c,
The streamlines are shown in FIGURE 20.6.5. Unlike the flow in Example 7, the fluid appears
to emerge from the strip 0 y p, x 0.
FIGURE 20.6.5 Flow in Example 8
20.6
or
CHAPTER 20 Conformal Mappings
FIGURE 20.6.7 Region R for Problem 9
10. G(z) z2/3
14. G(z) ez
y
πi
y
R
R
x
x
FIGURE 20.6.12 Region R for Problem 14
FIGURE 20.6.8 Region R for Problem 10
11. G(z) sin z
y
In Problems 15–18, a conformal mapping z f (w) from the upper
half-plane v 0 to a region R in the z-plane is given and the flow
in R with complex potential G(z) f 1(z) is constructed.
(a) Verify that the boundary of R is a streamline for the flow.
(b) Find a parametric representation for the streamlines of the
flow.
(c) Use a graphing utility to sketch the streamlines of the flow.
15.
16.
17.
18.
19.
R
π
2
–π
2
x
20.
FIGURE 20.6.9 Region R for Problem 11
12. G(z) i sin1z
y
21.
R
–1
x
1
22.
FIGURE 20.6.10 Region R for Problem 12
13. G(z) z2 1/z2
y
23.
M-9 in Appendix IV
M-4 in Appendix IV; use a 1
M-2 in Appendix IV; use a 1
M-5 in Appendix IV
A stagnation point in a flow is a point at which V 0. Find
all stagnation points for the flows in Examples 5 and 6.
For any two real numbers k and x1, the function G(z) k Ln(z x1) is analytic in the upper half-plane and therefore
is a complex potential for a flow. The real number x1 is called
a sink when k 0 and a source for the flow when k 0.
(a) Show that the streamlines are rays emanating from x1.
(b) Show that V (k/Z z x1 Z 2)(z x1) and conclude that
the flow is directed toward x1 precisely when k 0.
If f (z) is a conformal mapping from a domain D onto the upper half-plane, a flow with a source at a point 0 on the boundary of D is defined by the complex potential
G(z) k Ln( f (z) f (0)), where k 0. Determine the streamlines for a flow in the first quadrant with a source at 0 1
and k 1.
(a) Construct a flow on the horizontal strip 0 y p with a
sink at the boundary point 0 0. [Hint: See Problem 21.]
(b) Use a graphing utility to sketch the streamlines of the
flow.
The complex potential G(z) k Ln(z 1) k Ln(z 1) with
k 0 gives rise to a flow on the upper half-plane with a single
source at z 1 and a single sink at z 1. Show that the
streamlines are the family of circles x 2 ( y c)2 1 c2.
See FIGURE 20.6.13.
R
i
1
x
x
–1
FIGURE 20.6.11 Region R for Problem 13
1
FIGURE 20.6.13 Streamlines in Problem 23
20.6 Applications
|
941
(c) Conclude that the logarithmic spirals in part (b) spiral
inward if and only if a 0, and the curves are traversed
clockwise if and only if b 0. See FIGURE 20.6.14.
24. The flow with velocity vector V (a ib)> z is called a
vortex at z 0, and the geometric nature of the streamlines
depends on the choice of a and b.
(a) Show that if z x(t) iy(t) is the path of a particle, then
y
ax 2 by
dx
2
dt
x y2
dy
bx ay
2
.
dt
x y2
x
(b) Change to polar coordinates to establish that dr/dt a/r
and du/dt b/r 2, and conclude that r ceau/b for b 0.
[Hint: See (2) of Section 11.1.]
Chapter in Review
20
FIGURE 20.6.14 Logarithmic spiral in Problem 24
Answers to selected odd-numbered problems begin on page ANS-45.
Answer Problems 1–10 without referring back to the text. Fill in
the blank or answer true/false.
1. Under the complex mapping f (z) z2, the curve xy 2 is
mapped onto the line _____.
v
12. y
A
i
R
2. The complex mapping f (z) iz is a rotation through _____
B
degrees.
R′
u
x
3. The image of the upper half-plane y 0 under the complex
mapping f (z) z2/3 is _____.
FIGURE 20.R.1 Regions R and R for Problem 12
4. The analytic function f (z) cosh z is conformal except at
z _____.
5. If w f (z) is an analytic function that maps a domain D onto
v
y
13.
A
the upper half-plane v 0, then the function u Arg( f (z))
is harmonic in D. _____
R′
R
6. Is the image of the circle Zz 1Z 1 under the complex map-
ping T(z) (z 1)/(z 2) a circle or a line? _____
7. The linear fractional transformation T(z) maps the triple z1, z2, and z3 to _____.
1
B
z 2 z 1 z2 2 z 3
z 2 z3 z2 2 z1
1
8. If f (z) z1/2 (z 1)1/2 (z 1)1/2, then f (z) maps the upper
half-plane y
with div F 0 and curl F 0, then the complex function
g(z) P(x, y) iQ(x, y) is analytic in D. _____
10. If G(z) f(x, y) ic(x, y) is analytic in a region R and
V(x, y) iG9(z), then the streamlines of the corresponding
flow are described by f(x, y) c. _____
x
FIGURE 20.R.2 Regions R and R for Problem 13
0 onto the interior of a rectangle. _____
9. If F(x, y) P(x, y) i Q(x, y) j is a vector field in a domain D
In Problems 14 and 15, use an appropriate conformal mapping
to solve the given Dirichlet problem.
14.
y
u=0
11. Find the image of the first quadrant under the complex map-
ping w Ln z loge Z zZ i Arg z. What are images of the
rays u u0 that lie in the first quadrant?
In Problems 12 and 13, use the conformal mappings in
Appendix IV to find a conformal mapping from the given
region R in the z-plane onto the target region R in the w-plane,
and find the image of the given boundary curve.
942
|
CHAPTER 20 Conformal Mappings
u
u=1
eiπ /4
R
u=1
1
u=0
x
FIGURE 20.R.3 Dirichlet problem in Problem 14
v
y
15.
u=1
2i
u1 + π i
u=1
R'
–u1 + π i
2
R
i
u=0
πi
2
u=0
u1
x
FIGURE 20.R.5 Image of upper half-plane in Problem 17
FIGURE 20.R.4 Dirichlet problem in Problem 15
16. Derive conformal mapping C-4 in Appendix IV by construct-
ing the linear fractional transformation that maps 1, 1, q
to i, i, 1.
17. (a) Approximate the region R in M-9 in Appendix IV by the
polygonal region shown in FIGURE 20.R.5. Require that
f (1) u1, f (0) pi/2, and f (1) u1 pi.
(b) Show that when u1 S q,
f 9(z) Az(z 1)1(z 2 1)1 1
1
1
Ac
d.
2 z1
z21
(c) If we require that Im( f (t)) 0 for t 1, Im( f (t)) p
for t 1, and f (0) pi/2, conclude that
f (z) pi u
18. (a) Find the solution u(x, y) of the Dirichlet problem in the
upper half-plane y 0 that satisfies the boundary condition u(x, 0) sin x. [Hint: See Problem 6 in Exercises
20.5.]
(b) Find the solution u(x, y) of the Dirichlet problem in the
unit disk Z zZ 1 that satisfies the boundary condition
u(eiu ) sin u.
19. Explain why the streamlines in Figure 20.6.5 may also be interpreted as the equipotential lines of the potential f that satisfies f(x, 0) 0 for q x q and f(x, p) 1 for x 0.
20. Verify that the boundary of the region R defined by y2 4(1 x)
is a streamline for the fluid flow with complex potential
G(z) i(z1/2 1). Sketch the streamlines of the flow.
1
[Ln(z 1) Ln(z 1)].
2
CHAPTER 20 in Review
|
943
Appendices
I. Derivative and Integral Formulas
II. Gamma Function
III. Table of Laplace Transforms
© Songchai W/Shutterstock
IV. Conformal Mappings
Appendix I
Derivative and Integral
Formulas
Differentiation Rules
1. Constant:
3. Sum:
d
c⫽0
dx
2. Constant Multiple:
d
f f(x) ⫾ g(x)g ⫽ f 9(x) ⫾ g9(x)
dx
5. Quotient:
7. Power:
4. Product:
d
cf(x) ⫽ c f 9(x)
dx
d
f (x)g(x) ⫽ f(x)g9(x) ⫹ g(x) f 9(x)
dx
g(x)f 9(x) 2 f(x)g9(x)
d f(x)
d
⫽
6. Chain:
f (g(x)) ⫽ f 9(g(x))g9(x)
dx g(x)
dx
fg(x)g 2
d n
x ⫽ nx n 2 1
dx
8. Power:
d
fg(x)g n ⫽ nfg(x)g n 2 1g9(x)
dx
Derivatives of Functions
Trigonometric:
d
sin x ⫽ cos x
dx
d
12.
cot x ⫽ ⫺csc 2 x
dx
9.
d
cos x ⫽ ⫺sin x
dx
d
13.
sec x ⫽ sec x tan x
dx
10.
d
tan x ⫽ sec 2 x
dx
d
14.
csc x ⫽ ⫺csc x cot x
dx
11.
Inverse trigonometric:
d
1
d
1
sin 2 1x ⫽
16.
cos 2 1x ⫽ ⫺
2
dx
dx
"1 2 x
"1 2 x 2
d
1
d
1
18.
cot 2 1x ⫽ ⫺
19.
sec 2 1x ⫽
dx
dx
1 ⫹ x2
ZxZ"x 2 2 1
15.
17.
d
1
tan 2 1x ⫽
dx
1 ⫹ x2
20.
d
1
csc 2 1x ⫽ ⫺
dx
ZxZ"x 2 2 1
Hyperbolic:
d
sinh x ⫽ cosh x
dx
d
24.
coth x ⫽ ⫺csch2 x
dx
21.
d
d
cosh x ⫽ sinh x
23.
tanh x ⫽ sech2x
dx
dx
d
d
25.
sech x ⫽ ⫺sech x tanh x 26.
csch x ⫽ ⫺csch x coth x
dx
dx
22.
Inverse hyperbolic:
d
1
d
1
d
1
sinh 2 1x ⫽
28.
cosh 2 1 x ⫽
29.
tanh 2 1 x ⫽
, ZxZ , 1
2
2
dx
dx
dx
1
2
x2
"x ⫹ 1
"x 2 1
d
1
d
1
d
1
30.
coth21x ⫽
, ZxZ .1 31.
sech 2 1 x ⫽ ⫺
32.
csch 2 1 x ⫽ ⫺
2
dx
dx
dx
12x
x"1 2 x 2
ZxZ"x 2 ⫹ 1
27.
APP-2
Exponential:
d x
e ⫽ ex
dx
Logarithmic:
33.
d
1
lnZxZ ⫽
x
dx
Of an integral:
35.
37.
#
34.
d x
b ⫽ b x(ln b)
dx
36.
d
1
log b x ⫽
dx
x(ln b)
38.
d
dx
x
d
g(t) dt ⫽ g(x)
dx a
#
b
b
g(x, t) dt ⫽
a
# 0x g(x, t) dt
0
a
Integration Formulas
1.
4.
7.
10.
13.
16.
u n⫹1
#u du ⫽ n ⫹ 1 ⫹ C, n 2 ⫺1
n
2.
#b du ⫽ ln b b ⫹ C
#sec u du ⫽ tan u ⫹ C
#csc u cot u du ⫽ ⫺csc u ⫹ C
#sec u du ⫽ lnZsec u ⫹ tan uZ ⫹ C
#u cos u du ⫽ cos u ⫹ u sin u ⫹ C
1
u
u
5.
2
19.
#sin au sin bu du ⫽
21.
#e
23.
8.
11.
14.
17.
# u du ⫽ lnZuZ ⫹ C
1
#sin u du ⫽ ⫺cos u ⫹ C
#csc u du ⫽ ⫺cot u ⫹ C
#tan u du ⫽ ⫺ln Zcos uZ ⫹ C
#csc u du ⫽ lnZcsc u 2 cot uZ ⫹ C
#sin u du ⫽ u 2 sin 2u ⫹ C
2
2
sin(a 2 b)u
sin(a ⫹ b)u
2
⫹C
2(a 2 b)
2(a ⫹ b)
1
2
#e du ⫽ e
u
3.
1
4
u
⫹C
#cos u du ⫽ sin u ⫹ C
#sec u tan u du ⫽ sec u ⫹ C
#cot u du ⫽ ln Zsin uZ ⫹ C
#u sin u du ⫽ sin u 2 u cos u ⫹ C
#cos u du ⫽ u ⫹ sin 2u ⫹ C
6.
9.
12.
15.
2
18.
1
2
1
4
sin(a 2 b)u
sin(a ⫹ b)u
⫹
⫹C
2(a 2 b)
2(a ⫹ b)
20.
#cos au cos bu du ⫽
22.
#e
#sinh u du ⫽ cosh u ⫹ C
24.
#cosh u du ⫽ sinh u ⫹ C
25.
#sech u du ⫽ tanh u ⫹ C
26.
#csch u du ⫽ ⫺coth u ⫹ C
27.
#tanh u du ⫽ ln(cosh u) ⫹ C
28.
#coth u du ⫽ lnZsinh uZ ⫹ C
29.
#ln u du ⫽ u ln u 2 u ⫹ C
30.
#u ln u du ⫽
1 2
2 u ln
# "a
du ⫽ ln P u ⫹ "a 2 ⫹ u 2 P ⫹ C
31.
33.
35.
37.
au
sin bu du ⫽
e au
(a sin bu 2 b cos bu) ⫹ C
a ⫹ b2
2
2
# "a
#
1
2
u
du ⫽ sin 2 1 ⫹ C
a
2 u2
"a 2 2 u 2du ⫽
#a
2
32.
2
u
a
u
"a 2 2 u 2 ⫹
sin 2 1 ⫹ C
a
2
2
1
1 a⫹u
du ⫽ ln P
P ⫹C
a a2u
2 u2
# "u
1
2
2 a2
du ⫽ ln P u ⫹ "u 2 2 a 2 P ⫹ C
34.
au
cos bu du ⫽
e au
(a cos bu ⫹ b sin bu) ⫹ C
a ⫹ b2
2
2
#
1
2
⫹u
2
"a 2 ⫹ u 2du ⫽
36.
#a
38.
#
2
u 2 14u 2 ⫹ C
u
a2
"a 2 ⫹ u 2 ⫹
ln P u ⫹ "a 2 ⫹ u 2 P ⫹ C
2
2
1
1
u
du ⫽ tan 2 1 ⫹ C
a
a
⫹ u2
"u 2 2 a 2du ⫽
u
a2
"u 2 2 a 2 2
ln P u ⫹ "u 2 2 a 2 P ⫹ C
2
2
APPENDIX I Derivative and Integral Formulas
APP-3
Appendix II Gamma Function
The Gamma Function Euler’s integral definition of the gamma function* is
G(x) ⫽
#
q
t x 2 1e⫺t dt.
(1)
0
Convergence of the integral requires that x ⫺ 1 ⬎ ⫺1, or x ⬎ 0. The recurrence relation
G(x ⫹ 1) ⫽ x G(x)
(2)
that we saw in Section 5.3 can be obtained from (1) by employing integration by parts. Now
when x ⫽ 1,
G(1) ⫽
and thus (2) gives
#
q
0
e⫺tdt ⫽ 1,
G(2) ⫽ 1G(1) ⫽ 1
G(3) ⫽ 2G(2) ⫽ 2 ⭈ 1
G(4) ⫽ 3G(3) ⫽ 3 ⭈ 2 ⭈ 1,
and so on. In this manner it is seen that when n is a positive integer,
G(n ⫹ 1) ⫽ n!.
For this reason the gamma function is often called the generalized factorial function.
Although the integral form (1) does not converge for x ⬍ 0, it can be shown by means of
alternative definitions that the gamma function is defined for all real and complex numbers except
x ⫽ ⫺n, n ⫽ 0, 1, 2, … . As a consequence, (2) is actually valid for x ⫽ ⫺n. Considered as
a function of a real variable x, the graph of ⌫(x) is as given in FIGURE A.1. Observe that the
nonpositive integers correspond to the vertical asymptotes of the graph.
In Problems 31 and 32 in Exercises 5.3, we utilized the fact that ⌫(12 ) ⫽ !p. This result can
be derived from (1) by setting x ⫽ 12 :
G(12) ⫽
#
q
t ⫺1>2e⫺t dt.
(3)
0
By letting t ⫽ u2, we can write (3) as
G(12) ⫽ 2
#
q
2
e⫺u du.
0
*This function was first defined by Leonhard Euler in his text Institutiones Calculi Integralis published in 1768.
APP-4
#
But
q
0
fG(12)g 2 ⫽ a2
and so
#
q
0
2
#
2
e⫺u du ⫽
e⫺u dub a2
#
q
0
q
Γ(x)
2
e⫺v dv
0
2
e⫺v dvb ⫽ 4
q
##
0
q
2
2
e⫺(u ⫹ v ) du dv.
0
Switching to polar coordinates u ⫽ r cos u, v ⫽ r sin u enables us to evaluate the double integral:
q
4
##
0
q
0
p>2
2
2
e⫺(u ⫹ v ) du dv ⫽ 4
fG(12)g 2 ⫽ p
Hence,
# #
0
or
q
0
2
e⫺r r dr du ⫽ p.
⌫( 21– ) ⫽ !p.
(4)
In view of (2) and (4) we can find additional values of the gamma function. For example, when
x ⫽ ⫺ 12, it follows from (2) that ⌫( 12 ) ⫽ ⫺ 12 ⌫(⫺ 12). Therefore, ⌫(⫺ 12 ) ⫽ ⫺2⌫( 12 ) ⫽ ⫺2!p.
II
Exercises
x
FIGURE A.1 Graph of gamma function
Answers to selected odd-numbered problems begin on page ANS-46.
1. Evaluate the following.
(a) ⌫(5)
(c) ⌫(⫺ 32 )
(b) ⌫(7)
(d) ⌫(⫺ 52 )
⫽ 0.92 to evaluate
#
3. Use (1) and the fact that ⌫( 53 ) ⫽ 0.89 to evaluate
#
2. Use (1) and the fact that
1
4. Evaluate
⌫( 65 )
q
0
q
5
x 5e⫺x dx. [Hint: Let t ⫽ x5.]
3
x 4e⫺x dx.
0
3
# x aln x b dx. [Hint: Let t ⫽ ⫺ln x.]
3
1
0
1
5. Use the fact that ⌫(x) ⬎
#t
x 2 1 ⫺t
e
0
dt to show that ⌫(x) is unbounded as x S 0⫹.
6. Use (1) to derive (2) for x ⬎ 0.
7. A definition of the gamma function due to Carl Friedrich Gauss that is valid for all real
numbers, except x ⫽ 0, x ⫽ ⫺1, x ⫽ ⫺2, p, is given by
n! nx
.
n Sq x (x ⫹ 1)(x ⫹ 2) p (x ⫹ n)
G(x) ⫽ lim
Use this definition to show that ⌫(x ⫹ 1) ⫽ x ⌫(x).
APPENDIX II Gamma Function
APP-5
Appendix III Table of Laplace Transforms
f (t)
1. 1
2. t
3. t n
4. t21/2
5. t1/2
APP-6
+{ f (t)} ⴝ F(s)
1
s
1
s2
n!
s
n11
, n positive integer
p
Ä s
!p
2s3/2
6. t a
G(a ⫹ 1)
, a . ⫺1
s a⫹ 1
7. sin kt
k
s 1 k2
8. cos kt
s
s 1 k2
9. sin2kt
2k 2
s(s ⫹ 4k 2)
10. cos2kt
s 2 ⫹ 2k 2
s(s 2 ⫹ 4k 2)
11. eat
1
s2a
12. sinh kt
k
s 2 k2
13. cosh kt
s
s 2 k2
14. sinh 2kt
2k 2
s(s 2 4k 2)
2
2
2
2
2
2
15. cosh 2kt
s 2 2 2k 2
s(s 2 2 4k 2)
16. eatt
1
(s 2 a)2
17. eatt n
n!
, n a positive integer
(s 2 a)n ⫹ 1
18. eat sin kt
k
(s 2 a)2 ⫹ k 2
19. eat cos kt
s2a
(s 2 a)2 ⫹ k 2
20. eat sinh kt
k
(s 2 a)2 2 k 2
21. eat cosh kt
s2a
(s 2 a)2 2 k 2
22. t sin kt
2ks
(s ⫹ k 2)2
23. t cos kt
s2 2 k2
(s 2 ⫹ k 2)2
24. sin kt 1 kt cos kt
2ks 2
(s ⫹ k 2)2
25. sin kt 2 kt cos kt
2k 3
(s 2 ⫹ k 2)2
26. t sinh kt
2ks
(s 2 2 k 2)2
27. t cosh kt
s2 ⫹ k2
(s 2 2 k 2)2
2
2
28.
eat 2 ebt
a2b
1
(s 2 a)(s 2 b)
29.
aeat 2 bebt
a2b
s
(s 2 a)(s 2 b)
30. 1 2 cos kt
k2
s(s 2 ⫹ k 2)
31. kt 2 sin kt
k3
s (s ⫹ k 2)
32. a sin bt 2 b sin at
ab(a 2 2 b 2)
(s 2 ⫹ a 2)(s 2 ⫹ b 2)
33. cos at 2 cos bt
s(b 2 2 a 2)
(s ⫹ a 2)(s 2 ⫹ b 2)
34. sin kt sinh kt
2k2s
s 1 4k4
35. sin kt cosh kt
k(s 2 ⫹ 2k 2)
s 4 ⫹ 4k 4
36. cos kt sinh kt
k(s 2 2 2k 2)
s 4 ⫹ 4k 4
37. sin kt cosh kt ⫹ cos kt sinh kt
2ks 2
s 4 ⫹ 4k 4
2
2
2
4
APPENDIX III Table of Laplace Transforms
APP-7
38. sin kt cosh kt 2 cos kt sinh kt
4k 3
s ⫹ 4k 4
39. cos kt cosh kt
s3
s 4 ⫹ 4k 4
40. sinh kt 2 sin kt
2k 3
s4 2 k4
41. cosh kt 2 cos kt
2k 2s
s4 2 k4
1
42. J0(kt)
bt
4
2
"s 1 k2
s2a
ln
s2b
at
43.
e 2e
t
44.
2(1 2 cos at)
t
ln
s2 1 a2
s2
45.
2(1 2 cosh at)
t
ln
s2 2 a2
s2
46.
sin at
t
47.
sin at cos bt
t
a
arctan a b
s
1
1
a1b
a2b
1 arctan
arctan
s
s
2
2
e2a!s
!s
1 2a2/4t
e
!pt
a
2
49.
e2a /4t
3
2"pt
48.
50. erfc a
e2a!s
e2a!s
s
a
b
2!t
t ⫺a2>4t
a
e
2 a erfc a
b
Äp
2!t
51. 2
2
52. eabeb t erfc ab!t 1
2
a
b
2!t
53. 2eabeb t erfc ab!t 1
54. e atf (t)
a
a
b 1 erfc a
b
2!t
2!t
e2a!s
s !s
e ⫺a!s
!s(!s ⫹ b)
be ⫺!s
s( !s ⫹ b)
F(s 2 a)
55. 8(t 2 a)
e2as
s
56. f (t 2 a)8(t 2 a)
e⫺asF(s)
57. g(t)8 (t 2 a)
e⫺as+ 5g(t ⫹ a)6
58. f (n)(t)
59. t nf (t)
t
60.
# f (t)g(t 2 t) dt
s nF(s) 2 s n 2 1f (0) 2 p 2 f (n 2 1)(0)
dn
(⫺1)n n F(s)
ds
F(s)G(s)
0
APP-8
61. d(t)
1
62. d(t 2 a)
e2as
APPENDIX III Table of Laplace Transforms
Appendix IV Conformal Mappings
Elementary Mappings
E-1
y
v
z0
w = z + z0
x
E-2
u
v
y
w = eiθ z
θ
x
E-3
u
v
y
w = α z, α > 0
u
x
E-4
y
v
C
B
E-5
θ0
w = z α, α > 0
C′
αθ0
x
A
B′
v
y
B
πi
u
A′
A
w = ez
z = Ln w
C
D
x
A′
B′
C′
D′
u
APP-9
E-6
y
v
B
E
A
– π2
D
C
w = sin z
z = sin–1 w
x
π
2
A′ D′
B′
C′
E′
u
F′
–1 1
F
E-7
v
y
C′
B
w = 1z
x
A
u
A′
C
B′
E-8
v
πi
y
C
E-9
E
a
D′
F′
C′
w = loge|z| + i Arg z
a>1
x
F
D
E′
b
ln a
ln b
v
y
D
πi C
B
A
w = cosh z
x
–1
D′
1
C′ B′
A′
u
Mappings to Half-Planes
H-1
v
y
B
A
1
C
H-2
–1
1
D′A′
B′
y
v
E
H-3
w = eπ z /a
F
A′
–1
1
B′ C′ D′E′
F′
u
v
(
B
APP-10
–1
x
y
A
u
A
width = a
D
w = i 11–+ zz
D
ai B
C
x
1
C
w = a2 z + 1z
x
APPENDIX IV Conformal Mappings
(
A′
–a
B′
a
C′
u
u
H-4
y
v
A
D
width = a
w = cos
C′
C
a x
B
H-5
(πaz(
D′
B′
A′
1
–1
u
v
y
B
w=
x
A
C
1
D
H-6
(11 –+ zz(
2
A′
1
–1
B′ C′ D′
u
v
y
C
B
π /z
1 D
A
E
–π /z
w= eπ /z + e–π /z
e –e
x
1
–1
A′
B′ C′ D′
u
E′
Mappings to Circular Regions
C-1
y
v
A′
A
B
1 b
c
2
2
a= bc + 1 +b√+(bc –1)(c
C-2
x
z–a
w = az – 1
B′
r0
–1)
r0 =
y
v
B′
B
b
1
c
x
(1– b2)(1– c2)
D
u
(1– b2)(1– c2)
v
πi
E′
E
C
r0
1
r0 = 1 – bc + √c – b
y
B
A′
z–a
w = az – 1
a = 1 + bc + √c + b
A
u
bc – 1 – √ (b2 –1)(c2 –1)
c–b
A
C-3
1
w=e z
x
1
B′ A′ C′ D′
u
APPENDIX IV Conformal Mappings
APP-11
C-4
y
v
C′
A
–1
1
B
C
C-5
w=
D
1
D′
A′
i–z
i+z
u
B′
x
y
v
A′
2
w=i zz 2 +2iz+1
–2iz+1
A
B
C
1
D
1 u
D′
B′
x
C′
Miscellaneous Mappings
v
y
M-1
A
B
y =π
C
x
w=z+e z +1
y =– π
D
E
B′ π i
A′
C′
F
u
F′
D′
E′
y
M-2
v
A′
A
–1
1
B
C
D
w = πa
B′
x
–1
u
v
–a
B′
a
u
C′ D′
1
x
C
D
2
1/2
–1
w = 2a
π [(z –1) + sin (1/z)]
B
y
M-4
D′
[(z2–1)1/2 + cosh–1 z]
A′
A
ai
C′
y
M-3
–π i
v
C′ ai
–1
A
APP-12
B
C
1
D
x
A′
E
w = a(z2–1)1/2
APPENDIX IV Conformal Mappings
B′ D′
E′
u
M-5
y
v
A′
C′
–1
B C D
A
x
E
B′ π i
D′
u
E′
( (
–1
w = 2ζ + Ln ζζ +1
1/2
ζ = (z + 1)
M-6
v
y
πi
A′
–π
E′
F′
C′
D′
B′
–1
1
D
E
C
B
A
x
F
(
(
u
( (
w = i Ln 1+ iζζ + Ln 1+ ζζ
1–
1– i
– 1 1/2
ζ = zz +
1
(
(
y
M-7
v
A′
–1
1
B C D E
A
M-8
y
y =π
B
B′
πi
C′
x
D′
F
w = z + Ln z +1
u
E′ F′
v
E′
A
C
D
E
F
x
G
y = –π
F′ G′ H′
z + 1 C′ B′ A′ 1
e
w= z
e –1
D′
H
v
y
M-9
u
πi
G′
π i/2
–1
A
B
D′
1
x
A′
C D E F G
1
w= π i– 2 [Ln(z +1) + Ln(z–1)]
D′
D
E
E′
C′
B′
u
v
yA
M-10
B
F′
a
F
C
0 < a< 1
1
v = a/(1 – a)
E′ F′
x
–i
w = (1– i) zz –1
A′
B′
C′
u
APPENDIX IV Conformal Mappings
APP-13
Answers to Selected
Odd-Numbered Problems
Exercises 1.1, Page 11
1. linear, second order
3. linear, fourth order
5. nonlinear, second order
7. linear, third order
9. linear in x but nonlinear in y
15. domain of function is [2, q); largest interval of definition for solution is (2, q).
17. domain of function is the set of real numbers except x 2 and x 2; largest
intervals of definition for solution are
(q, 2) (2, 2) or (2, q).
et 2 1
19. X t
defined on (q, ln 2) or on
e 22
( ln 2, q)
31. m 2
33. m 2, m 3
35. m 0, m 1
37. y 2
39. no constant solutions
Exercises 1.2, Page 17
y 1/(1 4ex)
y 1/(x2 1); (1, q)
y 1/(x2 1); (q, q)
x cos t 8 sin t
"3
9. x 4 cos t 14 sin t
1.
3.
5.
7.
y 32 e x 2 12 e x
y 5ex 1
y 0, y x3
half-planes defined by either y 0 or
y0
19. half-planes defined by either x 0 or
x0
21. the regions defined by y 2, y 2, or
2 y 2
11.
13.
15.
17.
23.
25.
27.
29.
31.
39.
41.
43.
any region not containing (0, 0)
yes
no
(a) y cx
(b) any rectangular region not touching
the y-axis
(c) No, the function is not differentiable
at x 0.
(b) y 1/(1 x) on (q, 1);
y 1/(x 1) on (1, q)
y sin 3x
y0
no solution
Exercises 1.3, Page 25
dP
dP
1.
kP r ;
kP 2 r
dt
dt
dP
3.
k1P 2 k2P 2
dt
dx
kx (1000 2 x)
7.
dt
dA
1
9.
A 0; A(0) 50
dt
100
dA
7
11.
A6
dt
600 2 t
dh
cp
"h
13.
dt
450
di
15. L Ri E(t)
dt
dv
17. m
mg 2 kv 2
dt
d 2x
kx
dt 2
dv
dm
v
kv mg R
21. m
dt
dt
19. m
ANS-1
23.
27.
gR 2
d 2r
2 0
2
dt
r
dx
kx r, k . 0
dt
25.
dA
k (M 2 A), k . 0
dt
2
29.
dy
x "x y
y
dx
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 2
Chapter 1 in Review, Page 30
dy
1.
ky
3. y0 k 2 y 0
dx
5. y0 2 2y9 y 0
7. (a), (d)
9. (b)
11. (b)
13. y c1 and y c2ex, c1 and c2 constants
15. y x2 y2
17. (a) The domain is the set of all real numbers.
(b) either (q, 0) or (0, q)
19. For x0 1, the interval is (q, 0), and for x0 2,
the interval is (0, q).
x 2, x , 0
21. (c) y e 2
23. (q, q)
x ,
x$0
25. (0, q)
35. y e 3x 2 e x 4x
41. y 1 1
10
tan ( 101 x)
45. (a) y "x 2 x 2 1 (c) (q, 12 2 12 "5)
53. y(x) (4h/L2)x2 a
Exercises 2.3, Page 57
3. y 14 e3x cex, (q, q)
1. y ce5x, (q, q)
3
5. y 13 ce x , (q, q)
9. y cx x cos x, (0, q)
7. y x1 ln x cx1, (0, q)
11. y 17 x3 15 x cx4, (0, q)
37. y 32e 3x 3 2 92e x 2 1 4x
y 12 x2 ex cx2 ex, (0, q)
x 2y6 cy4, (0, q)
y sin x c cos x, (p/2, p/2)
(x 1)e x y x2 c, (1, q)
(sec u tan u)r u cos u c, (p/2, p/2)
y e3x cx1e3x, (0, q)
y x1ex (2 e)x1, (0, q)
E
E
27. i ai0 2 b eRt>L , (q, q)
R
R
29. (x 1)y x ln x x 21, (0, q)
41. y0 3, y1 0
31. y (2!x e 2 2 2)e2!x , (0, q)
2
43. x
2 e2x),
2x
,
2 (e 2 1)e
1
2 (1
Exercises 2.1, Page 40
21. 0 is asymptotically stable (attractor); 3 is unstable
(repeller).
23. 2 is semi-stable.
25. 2 is unstable (repeller); 0 is semi-stable; 2 is
asymptotically stable (attractor).
27. 1 is asymptotically stable (attractor); 0 is unstable
(repeller).
39. 0 , P0 , h>k
41. "mg>k
Exercises 2.2, Page 48
1. y 15
4
cos 5x c
11.
13.
15.
19.
1 3x
3e
2y
3. y 5. y cx
9.
13.
15.
17.
19.
21.
23.
25.
2
d x
dx
a b 32x 160
2
dt
dt
7. 3e
c
2e3x c
1 3
3x
ln x 19 x3 12 y2 2y ln | y| c
4 cos y 2x sin 2x c
(ex 1)2 2(ey 1)1 c
ce t
S cekr
17. P 1 ce t
5 x
5 y
(y 3) e c(x 4) e
21. y sin( 12 x2 c)
25. y 23. x tan(4t 34 p)
e (1 1>x)
x
27. y 12 x 12 "3"1 2 x 2
33. y e 1
35. y e
37. y e
33. y ln(2 2 e x), (q, ln 2)
35. (a) y 2, y 2, y 2
"5
2 )
3 2 e 4x 2 1
3 e 4x 2 1
Answers to Selected Odd-Numbered Problems
2
32ex ,
2
(12e 32)ex ,
0#x,1
x$1
2x 2 1 4e2x,
4x 2 ln x (1 4e2)x 2,
2
21
0#x#1
x.1
2
12 "pe x (erf (x) 2 erf (1))
x
41. y e
1 2 ex
e
ex
# e dt
et
0
43. y 10x 2 fSi(x) 2 Si(1)g
53. E(t) E0e(t2 4)>RC
Exercises 2.4, Page 64
1. x2 x 32 y2 7 y c
5. x2y2 3x 4y c
9.
13.
17.
19.
21.
23.
25.
27.
31.
35.
31. y "x 2 x 2 1, (q,12
0#x#3
x.3
6
1
2
39. y e x
2
exe t dt
29. y(x) e 4
ANS-2
2
39. y 1
3. 52 x2 4xy 2y4 c
7. not exact
xy3 y2 cos x 12 x2 c 11. not exact
xy 2xex 2ex 2x3 c 15. x3y3 tan1 3x c
ln | cos x | cos x sin y c
t 4y 5t 3 ty y3 c
1 3
4
2
2
3 x x y xy y 3
2
2
4ty t 5t 3y y 8
y2 sin x x3y x2 y ln y y 0
k 10
29. x2 y2 cos x c
x2y2 x3 c
33. 3x2y 3 y4 c
2
10 3x
3x
2ye 3 e x c 37. e y (x 2 4) 20
39. (c) y1(x) x 2 2 "x 4 2 x 3 4,
y2(x) x 2 "x 4 2 x 3 4
x
9
2 2
Å3
x
45. (a) v(x) 8
(b) 12.7 ft>s
Exercises 2.5, Page 68
1. y x ln |x| cx
3. (x y) ln |x y| y c(x y)
5. x y ln |x| cy
7. ln(x2 y2) 2 tan1 (y/x) c
9. 4x y(ln | y| c)2
11. y3 3x3 ln |x| 8x3
y/x
13. ln |x| e 1
15. y3 1 cx3
17. y3 x 21. y3 1
3x
3 ce
95 x1 495 x6
19. et/y ct
23. y x 1 tan(x c)
27. 4(y 2x 3) (x c)2
29. cot (x y) csc (x y) x "2 1
2
( 14 x cx3)1
x
a
37. P b c1e at
35. (b) y Exercises 2.6, Page 73
1. y2 2.9800, y4 3.1151
3. y10 2.5937, y20 2.6533; y ex
5. y5 0.4198, y10 0.4124
7. y5 0.5639, y10 0.5565
9. y5 1.2194, y10 1.2696
13. Euler: y10 3.8191, y20 5.9363
RK4: y10 42.9931, y20 84.0132
Exercises 2.7, Page 79
1. 7.9 years; 10 years
3. 760; approximately 11 persons/yr
5. 11 h
7. 136.5 h
9. I(15) 0.00098I0 or approximately 0.1% of I0
11. 15,963 years
13. T(1) 36.76F; approximately 3.06 min
15. approximately 82.1 s; approximately 145.7 s
17. 390F
19. approximately 1.6 h
t/50
21. A(t) 200 170e
23. A(t) 1000 1000et/100
25. A(t) 1000 2 10t 2
1
10 (100
2 t)2; 100 min
31. q(t) 3
5
35 e500t; i S 35 as t S q
1
100
33. i (t) e
43. (a) As t S q, x(t) S r/k.
(b) x(t) r/k (r/k)ekt; (ln 2)/k
45. (a) tb 50 s
(b) 70 m/s
dv
1
(c) 1250 m
(e)
v 9.8
dt
50
49. (a) v(0) 5 m3, v(t) 0.8 4.2e0.2t, approximately
0.117%
(b) in approximately 11.757 min or at approximately
9:12 A.M.
(c) approximately 829.114 m3/min
Exercises 2.8, Page 88
1. (a) N 2000
2000e t
; N(10) 1834
(b) N(t) 1999 e t
3. 1,000,000; 52.9 mo
5. (b) P(t) 4(P0 2 1) 2 (P0 2 4)e3t
(P0 2 1) 2 (P0 2 4)e3t
(c) For 0 P0 1, time of extinction is
t 13 ln
4(P0 2 1)
.
P0 2 4
2P0 2 5
5
"3
"3
tan c
t tan 1 a
bd;
2
2
2
"3
time of extinction is
7. P(t) t
2
ctan 1
5
tan 1 a
2P0 2 5
bd
"3
"3
"3
bt
9. P(t) e a>bece , where c (a/b) ln P0
11. 29.3 g; X S 60 as t S q; 0 g of A and 30 g of B
4Ah 2
13. (a) h(t) a"H 2
tb ; I is [0, "HAw>4Ah ]
Aw
(b) 576 "10 s or 30.36 min
15. (a) approximately 858.65 s or 14.31 min
27. 64.38 lb
29. i(t) 41. (a) P(t) P0 e (k1 2 k2)t
1 50t
;
100 e
i(t) t>10
60 2 60e
,
60 (e 2 2 1)e t>10,
1 50t
2e
0 # t # 20
t . 20
mg
mg kt>m
35. (a) v (t) av0 2
be
k
k
mg
(b) v(t) S
as t S q
k
(c)
mg kt>m
mg
mg
m
m
be
b s0
t 2 av0 2
av0 2
s(t) k
k
k
k
k
(b) 243 s or 4.05 min
17. (a) v (t) (b)
mg
kg
tanh a
t c1 b
Å k
Å m
where c1 tanh1 a
mg
Å k
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 2
25. 2y 2x sin 2(x y) c
3
rg k
rgr0
r0
a t r0 b 2
4k r
4k ° k
¢
t r0
r
(b) 33 13 min
39. (a) v (t) k
vb
Å mg 0
kg
m
ln cosh a
t c1 b c2
k
Å m
where c2 (m/k) ln (cosh c1)
(c) s (t) Answers to Selected Odd-Numbered Problems
ANS-3
dv
mg 2 kv 2 2 rV, where r is the weight
dt
density of water
19. (a) m
(b) v (t) Å
mg 2 rV
"kmg 2 krV
tanh a
t c1b
m
k
mg 2 rV
Å
k
21. (a) W 0 and W 2
(b) W(x) 2 sech2 (x c1)
(c) W(x) 2 sech2 x
(c)
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 3
23. x(t) "2t, t 25. x(t) e
0 # t # 14
1
1, t 4 , t # 2
1
2
1. x (t) x0e l1t
x0l1
y (t) (e l1t 2 e l2t )
l2 2 l1
5.
7.
9.
l2
l1
e l1t e l2t b
l2 2 l1
l2 2 l1
5, 20, 147 days. The time when y(t) and z(t) are the
same makes sense because most of A and half of B are
gone, so half of C should have been formed.
(a) P(t) P0e(lA lC)t
(b) approximately 1.25 109 years
lA
(c) A(t) P f1 2 e (lA lC)tg
lA lC 0
lC
C(t) P f1 2 e (lA lC)tg
lA lC 0
(d) 10.5% of P0, 89.5% of P0
dx1
6 252 x1 501 x2
dt
dx2
252 x1 252 x2
dt
dx1
x2
x1
(a)
3
22
dt
100 2 t
100 t
dx2
x1
x2
2
23
dt
100 t
100 2 t
z (t) x0 a1 2
(b) x1(t) x2(t) 150; x2(30) 艐 47.4 lb
di2
15. L1
(R1 R2)i2 R1i3 E(t)
dt
di3
L2
R1i2 (R1 R3)i3 E(t)
dt
17. i(0) i0, s(0) n i0, r (0) 0; because the population
is assumed to be constant
Chapter 2 in Review, Page 99
1. A/k, a repeller for k 0, an attractor for k 0
dy
3.
(y 1)2 (y 3)2
dx
ANS-4
13. Q ct1 15. y 1
4
1 4
25 t (1
2
4
5 ln t)
c(x 4)
#
Answers to Selected Odd-Numbered Problems
x
x
# te
2 sin t
dt
0
3
1
2
2 t 2e t dt
x2
x 1
xex 5ex, 0 # x , 1
21. y e x
6e ,
x$1
23. y csc x, (p, 2p)
19. y Exercises 2.9, Page 97
3.
stable for n even and asymptotically stable for n odd
9. 2x sin 2x 2 ln(y2 1) c
11. (6x 1)y3 3x3 c
17. y e2 sin x 12e2 sin x
1
2
!2t,
!2 2 !1 2 2t,
5. semi-stable for n even and unstable for n odd; semi-
25. (b) y 14 (x 2"y0 2 x0)2, [x0 2 2"y0, q)
29. P(45) 8.99 billion
31. (b) approximately 3257 BC
10 "100 2 y 2
b 2 "100 2 y 2
y
BT1 T2 BT1 T2
35. (a)
,
1B
1B
BT1 T2
T1 2 T2 k (1 B)t
e
(b) T (t) 1B
1B
1>Ck2
k1
37. q E0C (q0 2 E0C)a
b
k1 k2t
33. x 10 ln a
39. h(t) ( "2 2 0.00000163t)2
41. no
ac1e ak1t
, y(t) c2(1 c1e ak1t )k2>k1
1 c1e ak1t
0.02948t
P(t) 450e 2.4204e
; 166
2
2
x y c2
x y 1 c2ey
(a) k 0.083 seems to work well;
k 0.1063 and k 0.0823
43. x (t) 45.
47.
49.
51.
Exercises 3.1, Page 116
1. y 12 ex 12 ex
9. (q, 2)
sinh x
e
(e x 2 ex) (b) y sinh 1
e 21
(a) y ex cos x ex sin x
(b) no solution
(c) y ex cos x ep/2 ex sin x
(d) y c2ex sin x, where c2 is arbitrary
dependent
17. dependent
dependent
21. independent
The functions satisfy the DE and are linearly independent on the interval since W(e3x, e4x) 7ex 0;
y c1e3x c2e4x.
The functions satisfy the DE and are linearly independent on the interval since W(ex cos 2x, ex sin 2x) 2e2x 0; y c1ex cos 2x c2ex sin 2x.
11. (a) y 13.
15.
19.
23.
25.
3. y 3x 4x ln x
2
27. The functions satisfy the DE and are linearly inde3
4
6
pendent on the interval since W(x , x ) x
0;
y c1x3 c2x4.
29. The functions satisfy the DE and are linearly independent on the interval since W(x, x2, x2 ln x) 9x6 0; y c1x c2x2 c3x2 ln x.
35. (b) yp x2 3x 3e2x ; yp 2x2 6x 13 e2x
3. y2 sin 4x
7. y2 xe2x/3
11. y2 1
15. y2 x2 x 2
19. y2 e2x, yp 52 e3x
17. y2 e2x, yp 12
21. y2 x
#
x t
e
x0
t
dt, x0 . 0
13. y e x>3(c1 cos 13 !2x c2 sin 13 !2x)
15. y c1 c2ex c3e5x
17. y c1ex c2e3x c3xe3x
19. u c1et et (c2 cos t c3 sin t)
21. y c1ex c2xex c3x2 ex
23. y c1 c2x e x>2(c3 cos 12 !3x c4 sin 12 !3x)
25. y c1 cos 12 !3x c2 sin 12 !3x c3x cos 12 !3x
c4x sin 12 !3x
27. u c1er c2rer c3er c4rer c5e5r
29. y 2 cos 4x 12 sin 4x 31. y 13 e(t 1) 13 e5(t 1)
5
35. y 36
365 e6x 16 xe6x
39. y 0
33. y 0
37. y e5x xe5x
41. y 12 (1 2
5
!3x
!3 ) e
y cosh !3x 5
!3
49. y0 2 7y9 6y 0
53. y0 64y 0
12 (1 sinh !3x
5
!3x
;
!3 ) e
51. y0 2 3y9 0
55. y0 2 2y9 2y 0
57. y- 2 7y0 0
c2xe
2x
x2 2x 3
27. y !2 sin 2x 2
1
2
x/5
29. y 200 200e
3x2 30x
2x
31. y 10e
cos x 9e2x sin x 7e4x
F0
F0
sin vt 2
t cos vt
2
2v
2v
35. y 11 11ex 9xex 2x 12x2 ex 12 e5x
37. y 6 cos x 6(cot 1) sin x x2 1
33. x 4 sin !3x
2x
sin !3 !3 cos !3
cos 2x 56 sin 2x 13 sin x,
5
3 cos 2x 6 sin 2x,
0 # x # p>2
x . p>2
41. y e 2
Exercises 3.5, Page 140
1. y c1 cos x c2 sin x x sin x cos x ln | cos x |
3. y c1 cos x c2 sin x 12 x cos x
5. y c1 cos x c2 sin x 12 16 cos 2x
7. y c1ex c2ex 12 x sinh x
9. y c1e 3x c3e 3x 2 14xe 3x (1 3x)
11. y c1ex c2e2x (ex e2x) ln (1 ex)
13. y c1e2x c2ex e2x sin ex
15. y c1et c2tet 12 t 2 et ln t 34 t 2 et
17. y c1ex sin x c2ex cos x 13 xex sin x
13 ex cos x ln | cos x |
19. y 14 ex/2 34 ex/2 18 x2 ex/2 14 xex/2
1 2x
2x
21. y 49 e4x 25
19 ex
36 e 4 e
23. y c1 cos x c2 sin x
2 cos x
#
x
x0
#
2
#
x
x0
2
x0
25. y c1e 2x c2e x
2 13e 2x
x
e t sin t dt sin x e t cos t dt
#
x
e 2t ln t dt 13e x e t ln t dt, x0 . 0
x0
1/2
ln | cos x | sin x ln | sec x tan x |
3
5
x 4x 72
2
7. y c1 cos !3x c2 sin !3x
(4x 2 4x 2 43)e 3x
9. y c1 c2ex 3x
11. y c1ex/2 c2xex/2 12 12 x2 ex/2
13. y c1 cos 2x c2 sin 2x 34 x cos 2x
15. y c1 cos x c2 sin x 12 x2 cos x 12 x sin x
2x
23. y c1ex c2xex 25. y c1 cos x c2 sin x c3x cos x c4x sin x
27. y c1x
cos x c2x
sin x x1/2
29. y c1 c2 cos x c3 sin x
3. y c1e5x c2xe5x 65 x 5. y c1e
21. y c1 c2x c3e
1/2
Exercises 3.4, Page 135
1. y c1ex c2e2x 3
2x
1
2
6x
39. y Exercises 3.3, Page 125
1. y c1 c2ex/4
3. y c1e3x c2e2x
5. y c1e4x c2xe4x
7. y c1e2x/3 c2ex/4
9. y c1 cos 3x c2 sin 3x
11. y e2x (c1 cos x c2 sin x)
9
cos x 12
25 sin 2x 25 cos
6
1 2
1
4 x 37 cos x 37 sin x
c3x2 ex x 3 23 x3ex
19. y c1ex c2xex ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 3
Exercises 3.2, Page 119
1. y2 xe2x
5. y2 sinh x
9. y2 x4 ln | x |
13. y2 x cos (ln x)
17. y c1ex cos 2x c2ex sin 2x 14 xex sin 2x
Exercises 3.6, Page 146
1. y c1x1 c2x2
3. y c1 c2 ln x
5. y c1 cos(2 ln x) c2 sin(2 ln x)
7. y c1x (2 2 !6) c2x (2 !6)
9. y c1 cos( 15 ln x) c2 sin( 15 ln x)
11. y c1x2 c2x2 ln x
13. y x 1>2 [c1 cos ( 16 !3 ln x) c2 sin ( 16 !3 ln x)]
Answers to Selected Odd-Numbered Problems
ANS-5
15. y c1x 3 c2 cos (!2 ln x) c3 sin ( !2 ln x)
3
2
11. (a) x(t) 23 cos 10t sin 10t
sin(10t 0.927)
17. y c1 c2x c3x c4x
19. y c1 c2x5 15 x5 ln x
(b) 56 ft; p5
(c) 15 cycles
(d) 0.721 s
21. y c1x c2x ln x x(ln x)2
23. y c1x1 c2x ln x
25. y 2 2x2
27. y cos(ln x) 2 sin(ln x)
37. x 2y0 xy9 y 0
(2n 1)p
0.0927, n 0, 1, 2, …
20
(f ) x(3) 0.597 ft
(g) x(3) 5.814 ft /s
(h) x (3) 59.702 ft /s2
39. y c1(x 3)2 c2(x 3)7
(i)
29. y 3
4
ln x (e)
1 2
4x
31. y c1(x 2 2 x 6), c1 any real constant
33. x 2y0 2 xy9 2 8y 0
35. x 2y0 7xy9 9y 0
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 3
10
1
2
43. y c1x
c2x
45. y c1x
2
47. y x [c1 cos(3 ln x) c2 sin(3 ln x)] 49. y 2(x)
1/2
5(x)
2
51. T(r) 5T0(r r 1)
1/2
(b) w(r) 4
13
301 x2
3
10 x
ln(x), x 0
53. (a) w(r) c1 c2 ln r c3r 2 q
(a 2 2 r 2)2
64D
q 4
r
64D
Exercises 3.7, Page 150
3. y ln | cos(c1 x) | c2
1
1
5. y 2 ln Zc1x 1Z 2
x c2
c
c1
1
7. (x c2)2 y 2 c 21
9.
11. (b) y tan ( 14p 2 12x)
1 3
3y
17. y 1 x
13. keff 160 lb>ft; x(t) 18 sin 16t
15. keff 30 lb>ft; x(t) c1 y x c2
(c) (p>2, 3p>2)
12 x2 12 x3 16 x4 101 x5 p
12 x2 23 x3 14 x4 607 x5 p
19. y "1 2 x 2
s; 12 s, x( 12 ) e2; that is, the weight is approximately
0.14 ft below the equilibrium position.
1
4
27. (a) x(t) 43 e2t 13 e8t
(b) x(t) 23 e2t 53 e8t
29. (a) x(t) e2t (cos 4t (b) x(t) (b)
(c)
7. (a)
x (p>4) 12 ; x (9p>32) "2
4
4 ft /s; downward
(2n 1)p
t
, n 0, 1, 2, …
16
the 20-kg mass
the 20-kg mass; the 50-kg mass
t np, n 0, 1, 2, …; at the equilibrium position; the 50-kg mass is moving upward whereas
the 20-kg mass is moving upward when n is even
and downward when n is odd.
9. (a) x(t) 12 cos 2t 34 sin 2t
(b) x(t) (c) x(t) !13
4 sin(2t 0.588)
!13
4 cos(2t 2 0.983)
Answers to Selected Odd-Numbered Problems
!5
2
e
(c) t 1.294 s
31. (a) b 5
2
2t
35. x(t) sin 4t)
(b) b 52
10
3
1
4
4t
e
1
2
sin(4t 4.249)
33. x(t) et>2 (43 cos
Exercises 3.8, Page 163
!2p
1.
3. x(t) 14 cos 4 !6t
8
5. (a) x (p>12) 14 ; x (p>8) 12 ; x (p>6) 14 ;
"3
6 sin
4"3t
17. Compared to a single-spring system with spring
constant k, the parallel-spring system is more stiff.
21. (a) above
(b) heading upward
23. (a) below
(b) heading upward
25.
1
13. y "1 2 c 21x 2 c2
c1
15. y 1 x
(j) 0.1451 8
c2x
8 13 ft /s
np
np
; 0.3545 , n 0, 1, 2, …
5
5
np
(k) 0.3545 , n 0, 1, 2, …
5
41. y c1(x 2) c2(x 2) ln (x 2)
(b)
(c)
ANS-6
1
2
5
6
!47
2
t2
(c) 0 b 52
64
3 !47
sin
(cos 3t sin 3t)
!47
2
t)
te4t 14 cos 4t
37. x(t) 12 cos 4t 94 sin 4t 12 e2t cos 4t 2e2t sin 4t
39. (a) m
d 2x
dx
k(x h) b
or
2
dt
dt
dx
d 2x
2l v2x v2h(t),
dt
dt 2
where 2l b/m and v2 k/m
(b) x(t) e2t ( 56
13 cos 2t 32
13 sin t
41. x(t) cos 2t 1
8
72
13
sin 2t) cos t
sin 2t 34 t sin 2t 54 t cos 2t
F0
t sin vt
2v
49. 4.568 C; 0.0509 s
51. q(t) 10 10e3t (cos 3t sin 3t)
i(t) 60e3t sin 3t; 10.432 C
43. (b)
56
13
53. qp 100
13
57. q(t) 100
150
sin t 150
13 cos t, ip 13 cos t 13 sin t
3
2
12 e10t (cos
3
2
10t sin 10t) ; C
E0C
t
b cos
61. q(t) aq0 2
1 2 g2LC
"LC
"LCi0 sin
t
E0C
cos gt
1 2 g2LC
(b) vn n2p2 EI
, n 1, 2, 3, p
L2 Å r
Exercises 3.10, Page 186
x
1. yp(x) 14 ex0 sinh 4(x 2 t) f (t) dt
x
3. yp(x) ex0(x 2 t)e (x 2 t)f (t) dt
x
Exercises 3.9, Page 173
w0
1. (a) y(x) (6L2 x2 4Lx3 x4)
24EI
w0
3. (a) y(x) (3L2 x2 5Lx3 2x4)
48EI
w0
5. (a) y(x) (7L4x 2 10L2x 3 3x 5)
360EI
(c) x < 0.51933, ymax < 0.234799
w0 EI
P
7. y(x) 2 cosh
x
Ä EI
P
w0 EI
w0 L !EI
P
a 2 sinh
L2
b
Ä EI
P
P!P
w0 2
w0 EI
x 2P
P2
9. (a) EIy-(x) e
12w0L,
w0(x 2 L),
P
x
Ä EI
P
cosh
L
Ä EI
sinh
0 # x , L>2
L>2 # x # L
(b)
w0
(4Lx 3 9L2x 2), 0 # x , L>2
48EI
y(x) e w0 (16x 4 2 64Lx 3 96L2x 2 2 8L3x L4),
384EI
L>2 # x # L
11.
13.
15.
17.
19.
21.
23.
41w0L4
(c) y(L) 384EI
ln n2, n 1, 2, 3, …; yn sin nx
(2n 2 1)2p2
ln , n 1, 2, 3, …;
4L2
(2n 2 1)px
y cos
2L
2
ln n , n 0, 1, 2, …; yn cos nx
n2p2
npx
ln , n 1, 2, 3, …; yn ex sin
25
5
ln n2, n 1, 2, 3, …; yn sin(n ln x)
ln n4p4, n 1, 2, 3, p; yn sin npx
x L/4, x L/2, x 3L/4
27. vn np"T
L "r
, n 1, 2, 3, …; yn sin
npx
L
5. yp(x) 13 ex0 sin 3(x 2 t) f (t) dt
x
7. y c1e 4x c2e 4x 14 ex0 sinh 4(x 2 t) te 2tdt
x
9. y c1e x c2xe x ex0(x 2 t)e (x 2 t)e tdt
x
11. y c1cos 3x c2 sin 3x 13 ex0 sin 3(x 2 t)(t sin t) dt
13. yp(x) 14 xe 2x 2
1 2x
16 e
1 2x
16 e
15. yp(x) 12 x 2e 5x
p
sin x 2 x sin x 2 cos x ln Zsin xZ
2
9 2x
2 16 e 14 xe 2x
17. yp(x) cos x 19. y 25 2x
16 e
5x
6xe 5x 12 x 2e 5x
23. y x sin x cos x ln Zsin xZ
25. y (cos 1 2)ex (1 sin 1 cos 1)e2x
e2x sin ex
27. y 4x 2x2 x ln x
21. y e
46 3
45 x
2 201 x 2 361 2 16 ln x
31. y(x) 5e x 3e x yp(x),
1 2 cosh x, x , 0
where yp(x) e
1 cosh x, x $ 0
33. y cos x 2 sin x yp(x),
29. y 0,
x,0
where yp(x) • 10 2 10 cos x, 0 # x # 3p
20 cos x,
x . 3p
x
1
35. yp(x) (x 2 1) e0 t f (t) dt xex (t 2 1) f (t) dt
37. yp(x) 12 x 2 2 12 x
sin (x 2 1)
sin x
2
1
sin 1
sin 1
41. yp(x) ex cos x ex sin x ex
39. yp(x) 43. yp(x) 12 (ln x)2 12 ln x
Exercises 3.11, Page 193
d 2x
7.
x0
dt 2
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 3
"LC
E0C
1
t
bsin
i(t) i0 cos
2
aq0 2
2
1 2 g LC
"LC "LC
"LC
E0Cg
sin gt
2
1 2 g2LC
t
u1b 2 u0 a
u0 2 u1 ab
b
r
b2a
b2a
4 4
np
npx
31. (a) ln , n 1, 2, 3, p; yn(x) sina
b
4
L
L
29. u(r) a
(c) 4!10 < 12.65 ft/s
17. (a) u1(t) u0 cos v1t, v1 "g>l
(b) T p>2v1 (p>2)"l>g
(c) u2(t) 12 u0 sin v2t, v2 2"g>l; the amplitude
and period of the shorter pendulum are half that
of the longer pendulum
15. (a) x(t) 5
Answers to Selected Odd-Numbered Problems
ANS-7
19. (a) xy r "1 (y9)2. When t 0, x a, y 0,
dy/dx 0.
(b) When r 1,
1
ar
1
x 1r
x 12r
a
2
.
a b
a b d y(x) c
2 1r a
12r a
1 2 r2
When r 1,
1 a
1 1
y(x) c (x 2 2 a 2) ln d .
a
x
2 2a
(c) The paths intersect when r 1.
19. y e3x/2 (c2 cos
21.
23.
25.
27.
29.
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 4
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
Answers to Selected Odd-Numbered Problems
31.
33.
35.
37.
41.
43.
45.
49.
53.
55.
!11
2 x
x 222
625
y c1 c2e2x c3e3x 15 sin x 15 cos x 43 x
y ex (c1 cos x c2 sin x) ex cos x ln | sec x tan x |
y c1x1/3 c2x1/2
y c1x2 c2x3 x4 x2 ln x
(a) y c1 cos vx c2 sin vx A cos ax B sin ax,
v a, y c1 cos vx c2 sin vx Ax cos vx
Bx sin vx, v a
(b) y c1evx c2evx Aeax, v a,
y c1evx c2evx Axevx, v a
(a) y c1 cosh x c2 sinh x c3x cosh x c4x sinh x
(b) yp Ax2 cosh x Bx2 sinh x
y exp cos x
y 134 ex 54 ex x 12 sin x
y x2 4
x c1et 32 c2e2t 52
y c1et c2e2t 3
x c1et c2e5t tet
y c1et 3c2e5t tet 2et
14.4 lb
47. 0 m 2
1
(a) q(t) 150
sin 100t 751 sin 50t
(b) i(t) 23 cos 100t 23 cos 50t
np
(c) t , n 0, 1, 2, …
50
d 2x
m 2 kx 0
dt
x
y(x) 2 cos x 2 5 sin x e0 sin (x 2 t) tan t dt
2 cos x 2 4 sin x 2 cos x ln Zsec x tan xZ
36 2
25 x
1. x c1et c2tet
y (c1 c2)et c2tet
x c1 cos t c2 sin t t 1
y c1 sin t c2 cos t t 1
x 12 c1 sin t 12 c2 cos t 2c3 sin !6t 2c4 cos !6t
y c1 sin t c2 cos t c3 sin !6t c4 cos !6t
x c1e2t c2e2t c3 sin 2t c4 cos 2t 15 et
y c1e2t c2e2t c3 sin 2t c4 cos 2t 15 et
3t
x c1 c2 cos t c3 sin t 17
15 e
4 3t
y c1 c2 sin t c3 cos t 15 e
x c1et c2et/2 cos 12 !3t c3et/2 sin 12 !3t
y (32 c2 2 12 !3c3)e t>2 cos 12 !3t
(12 !3c2 2 32 c3)e t>2 sin 12 !3t
x c1e4t 43 et
y 34 c1e4t c2 5et
x c1 c2t c3et c4et 12 t 2
y (c1 c2 2) (c2 1)t c4et 12 t 2
x c1et c2et/2 sin 12 !3t c3et/2 cos 12 !3t
y c1et ( 12 c2 12 !3c3) et/2 sin 12 !3t
( 12 !3c2 12 c3) et/2 cos 12 !3t
z c1et ( 12 c2 12 !3c3) et/2 sin 12 !3t
( 12 !3c2 12 c3) et/2 cos 12 !3t
x 6c1et 3c2e2t 2c3e3t
y c1et c2e2t c3e3t
z 5c1et c2e2t c3e3t
x e3t3 te3t3
y e3t3 2te3t3
mx 0
my mg;
x c1t c2
y 12 gt 2 c3t c4
"7
"7
2 x c3 sin 2 x)
4 3
c3 sin !11
2 x) 5 x
17. y c1e x>3 e 3x>2 (c2 cos
Exercises 3.12, Page 201
Chapter 3 in Review, Page 203
1. y 0
3. false
5. 8 ft
3
7. 4
9. (q, q); (0, q)
11. y c1e3x c2e5x c3xe5x c4ex c5 xex c6 x2 ex,
y c1x3 c2x5 c3 x5 ln x c4 x c5 x ln x
c6 x(ln x)2
ANS-8
13. y c1e (1 "3)x c2e (1 2 "3)x
15. y c1 c2e5x c3 xe5x
46
125
57. (a) u(t) v0 "l>g sin "g>l t
59. (a) x (v0 cos u)t, y 12gt 2 (v0 sin u)t;
y 12
g
v 20 cos 2
Exercises 4.1, Page 217
2 s
1
e 2
1.
s
s
1 e ps
5.
s2 1
1
1
1
2 2 2 e s
9.
s
s
s
1
13.
(s 2 4)2
s2 2 1
17.
(s 2 1)2
10
4
21. 2 2
s
s
6
6
3
1
25. 4 3 2 s
s
s
s
u
x2 sin u
x
cos u
3.
7.
11.
15.
19.
23.
27.
1
1
2 2 e s
s2
s
1 s
1
e 2 e s
s
s
e7
s21
1
2
s 2s 2
48
s5
2
6
3
22
s
s3
s
1
1
s
s24
Exercises 4.2, Page 225
1. 12 t 2
3. t 2t 4
3 2
1 3
5. 1 3t 2 t 6 t
7. t 1 e2t
9. 14 et/4
11. 57 sin 7t
t
13. cos
15. 2 cos 3t 2 sin 3t
2
1
1 3t
17. 3 3 e
19. 34 e3t 14 et
21. 0.3e0.1t 0.6e0.2t
23. 12 e2t e3t 12 e6t
1
1
25. 5 5 cos "5t
27. 4 3et cos t 3 sin t
a sin bt 2 b sin at
29. 13 sin t 16 sin 2t
31.
ab(a 2 2 b 2)
33. y 1 et
1 4t
6t
35. y 10
e 19
37. y 43 et 13 e4t
10 e
39. y 10 cos t 2 sin t !2 sin !2t
5 t
41. y 89 et/2 19 e2t 18
e 12 et
43. y sin h t 2 sin t
10
45. y(t) 10
21 cos 2t 2 21 cos 5t
1 t
1 3t
47. y 4 e 4 e cos 2t 14 e3t sin 2t
Exercises 4.3, Page 234
1
6
1.
3.
2
(s 2 10)
(s 2)4
1
2
1
5.
(s 2 2)2
(s 2 3)2
(s 2 4)2
3
7.
(s 2 1)2 9
s21
s4
s
2
3
9.
2
2
s 25
(s 2 1) 25
(s 4)2 25
1 2 2t
3t
11. 2 t e
13. e sin t
2t
2t
15. e cos t 2e sin t
17. et tet
3 2 t
t
t
19. 5 t 5e 4te 2 t e
21. y te4t 2 e4t
23. y et 2tet
10 3t
1
2
2 3t
25. y 9 t 27 27 e 9 te
27. y 32 e3t sin 2t
29. y 12 12 et cos t 12 et sin t
31. y (e 1) tet (e 1) et
33. x (t) 32 e 7t>2 cos
37.
e s
s2
"15
2
t2
39.
7 "15 7t>2
10 e
2s
e
s2
sin "15
2 t
e 2s
2
s
41.
45.
47.
49.
53.
55.
57.
59.
61.
63.
s
e ps
43. 12 (t 2 2)2 8(t 2 2)
s2 4
sin t ᐁ(t p)
ᐁ(t 1) e(t1) ᐁ(t 1)
(c)
51. (f )
(a)
4
2
f (t) 2 4 ᐁ(t 3); +{ f (t)} 2 e3s
s
s
s
e
e s
e s
f (t) t 2 ᐁ(t 1); +{ f (t)} 2 3 2 2 s
s
s
2s
2s
1
e
e
f (t) t t ᐁ(t 2); +{ f (t)} 2 2 2 2 2
s
s
s
e bs
e as
2
f (t) ᐁ(t a) ᐁ(t b); +{ f (t)} s
s
y [5 5e(t1)] ᐁ(t 1)
65. y 14 12 t 14 e2t 1
4
ᐁ(t 1)
(t 1) ᐁ(t 1) 14 e2(t1) ᐁ(t 1)
1
2
67. y cos 2t 1
6
sin 2(t 2p) ᐁ(t 2p)
1
3
sin(t 2p) ᐁ(t 2p)
69. y sin t [1 cos(t p)] ᐁ(t p)
[1 cos(t 2p)] ᐁ(t 2p)
71. x(t) 54 t 73.
75.
77.
79.
81.
5
16
25
4
5
16
sin 4t 54 (t 5) ᐁ(t 5)
sin 4(t 5) ᐁ(t 5) 254 ᐁ(t 5)
cos 4(t 5) ᐁ(t 5)
q(t) 25 ᐁ(t 3) 25 e5(t3) ᐁ(t 3)
10
1 10t
1
(a) i(t) 101
e
101
cos t 101
sin t
1 10(t3p/2)
ᐁ(t 2 3p>2)
101 e
10
cos (t 2 3p>2) ᐁ(t 2 3p>2)
101
1
sin (t 2 3p>2)
101
(b) imax 艐 0.1 at t 艐 1.6, imin 艐 0.1 at t 艐 4.7
w0 L2 2
w0 L 3
y(x) x 2
x
16EI
12EI
w0 4
w0
x 2
(x 2 12 L)4 8(x 2 12 L)
24EI
24EI
w0 L2 2
w0 L 3
y(x) x 2
x
48EI
24EI
w0 5 4
[ Lx 2 x 5 (x 2 12 L)5 8(x 2 12 L)]
60EI 2
dT
(a)
k (T 70 57.5t (230 57.5t) ᐁ(t 4))
dt
Exercises 4.4, Page 245
1
s2 2 4
1.
3.
2
(s 10)
(s 2 4)2
6s 2 2
12s 2 24
5.
7.
(s 2 2 1)3
[(s 2 2)2 36] 2
9. y 12 et 12 cos t 12 t cos t 12 t sin t
11. y 2 cos 3t 53 sin 3t 16 t sin 3t
13. y 14 sin 4t 18 t sin 4t
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 4
2
1
1
8
15
31. 3 2 2
s
s22
s24
s
s 9
e kt 2 e kt
33. Use sinh k t to show that
2
k
+{sinh k t} 2
.
s 2 k2
1
1
2
35.
2
37. 2
2(s 2 2)
2s
s 16
4 cos 5 (sin 5)s
!p
39.
43. 1>2
s 2 16
s
3!p
45.
4s 5>2
29.
18 (t p) sin 4(t p) ᐁ(t p)
Answers to Selected Odd-Numbered Problems
ANS-9
17. y 23 t 3 c1t 2
25.
29.
33.
37.
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 4
43.
45.
47.
49.
51.
19.
24
s5
1
6
23. 5
s 21
s
1
27.
s(s 2 1)
1
31. 2
s (s 2 1)
21.
2
s21
(s 1)[(s 2 1)2 1]
s1
s[(s 1)2 1]
3s 2 1
35. et 1
s 2(s 2 1)2
et 12 t 2 t 1
41. f (t) sin t
3 t
1 t
1 t
f (t) 8 e 8 e 4 te 14 t 2 et
f (t) et
f (t) 38 e2t 18 e2t 12 cos 2t 14 sin 2t
y(t) sin t 12 t sin t
i(t) 100[e10(t1) e20(t1)] ᐁ(t 1)
100[e10(t2) e20(t2)] ᐁ(t 2)
55.
59.
#e
t2
sin 3(t 2 t) dt
0
as
12e
s(1 e as)
coth (ps>2)
57.
s2 1
1
61. i(t) (1 eRt/L)
R
2 q
a (1)n (1 eR(tn)/L) ᐁ(t n)
R n1
63. x(t) 2(1 et cos 3t 13 et sin 3t)
4 a (1)n [1 e(tnp) cos 3(t np)
q
n1
13 e(tnp) sin 3(t np)] ᐁ(t np)
Exercises 4.5, Page 251
1. y e3(t2) ᐁ(t 2)
3. y sin t sin t ᐁ(t 2p)
5. y cos t ᐁ(t 2 p>2) cos t ᐁ(t 2 3p>2)
7. y 12 12 e2t [ 12 12 e2(t1)]ᐁ(t 1)
9. y e2(t2p) sin t ᐁ(t 2p)
2 !6
2
2
15 sin !6t 5 cos t 2 5 cos !6t
4
1
x2 25 sin t 2 !6
15 sin !6t 5 cos t 5 cos !6t
100
100 900t
(b) i2 9 9 e
i3 809 809 e900t
(c) i1 20 20e900t
375 15t
85
2t
i2 20
1469
e
145
13 e
113 cos t 113 sin t
250 15t
810
2t
i3 30
1469
e
280
13 e
113 cos t 113 sin t
100t
i1 65 65 e100t cosh 50 !2t 9 !2
sinh 50 !2t
5 e
6 !2 100t
6
6 100t
i2 5 5 e
cosh 50 !2t 5 e
sinh 50 !2t
1 2
(a) x(t) (v0 cos u)t, y(t) 2gt (v0 sin u)t
13. x1 15.
17.
19.
21.
a 1
1
a 2 bs
b
s bs
e 21
3
4
3
4
!2 sin !2t
y
t !2 sin !2t
2 3
1 4
9. x 8 t t
11. x 12 t 2 t 1 et
3!
4!
2
1 4
y t3 t
y 13 13 et 13 tet
3!
4!
12
t
53. y(t) 1
5
sin t (b) y(x) g
2v 20 cos 2 u
x 2 (tan u)x is a quadratic
function, for a fixed value of u its graph is a parabola.
(c) Solve y(x) 0 to find the range R. To prove the
complementary-angle property, show that R(u) R(p/2 u).
(d) Solve y(x) 0 and find the corresponding value
of y(x).
(e) For u 38: range is 2728.96 ft, max. height is
533.02 ft
For u 52: range is 2728.96 ft, max. height is
873.23 ft
(f) For u 38: time to hit the ground is t < 11.5437 s,
max. height occurs at t < 5.7718 s
For u 52: time to hit the ground is t < 14.7752 s.
max. height occurs at t < 7.3876 s
(g) y
q = 52°
800
600
q = 38°
400
11. y e2t cos 3t 23 e2t sin 3t
13 e2(tp) sin 3(t p)ᐁ(t p)
13 e2(t3p) sin 3(t 3p)ᐁ(t 3p)
13. y sin t sin t a (1)k 8(t 2 kp)
q
200
500
1000
1500
2000
2500
k1
w0 1 2 1 3
( Lx 2 6 x ),
EI 4
15. y(x) μ
w0 L2 1
( x 2 121 L),
4EI 2
0 # x , L>2
L>2 # x # L
Exercises 4.6, Page 254
3. x cos 3t 53 sin 3t
1. x 13 e2t 13 et
y 2 cos 3t 73 sin 3t
y 13 e2t 23 et
5. x 2e3t 52 e2t 1
2
y 83 e3t 52 e2t 16
ANS-10
7. x 12 t Answers to Selected Odd-Numbered Problems
Chapter 4 in Review, Page 257
1
2
1. 2 2 2 e s
3. false
s
s
1
5. true
7.
s7
2
4s
9. 2
11.
2
s 4
(s 4)2
1 5
1 2 5t
13. 6 t
15. 2 t e
5 5t
5t
17. e cos 2t 2 e sin 2t
19. cos p(t 1) ᐁ(t 1) sin p(t 1) ᐁ(t 1)
x
21. 5
23. ek(sa) F(s a)
25. f (t) ᐁ(t t0)
27. f (t t0) ᐁ(t t0)
29. f (t) t (t 1)ᐁ(t 1) ᐁ(t 4);
31.
33.
37.
39.
41.
43.
45.
47.
Exercises 5.1, Page 270
1. R 12 , [ 12 , 12 )
5. x 2
3
1
2
3
x 7. 1 x2 2
15
5
24
3. R 10, (5, 15)
5
x x4 9. a (k 2 2)ck 2 2 x k
4
315
61
720
x p
x6 p , (p/2, p/2)
7
q
k3
11. 2c1 a f2 (k 1)ck1 6ck1]x k
15. 5; 4
q
k1
3 3
32
x x6
23
2356
33
x9 p d
235689
3 4
32
y2(x) c1 cx x x7
34
3467
33
x 10 p d
3 4 6 7 9 10
17. y1(x) c0 c1 3 4
21 6
x 2
x 2 pd
4!
6!
5 5
45 7
x x pd
5!
7!
42 6
72 42 9
x 2
x pd
6!
9!
52 22 7
x
7!
82 52 22 10
x pd
10!
q
1
23. y1(x) c0; y2(x) c1 a x n
n
n1
2
1 2
1
1
x x3 x4 pd
2
6
6
1
1
1
y2(x) c1 cx x 2 x 3 x 4 p d
2
2
4
1
7
23 7 6
y1(x) c0 c1 x 2 2
x4 x 2 pd
4
4 4!
8 6!
1
14 5
34 14 7
y2(x) c1 cx 2 x 3 x x 2 pd
6
2 5!
4 7!
1 2
1 3
1 4
y(x) 2 c1 x x x p d 6x
2!
3!
4!
8x 2 2e x
y(x) 3 2 12x 2 4x 4
1
1 5
y1(x) c0 c1 2 x 3 x pd
6
120
1 4
1 6
y2(x) c1 cx 2
x x pd
12
180
25. y1(x) c0 c1 27.
29.
31.
33.
Exercises 5.2, Page 278
1. x 0, irregular singular point
3. x 3, regular singular point; x 3, irregular singular
point
5. x 0, 2i, 2i, regular singular points
7. x 3, 2, regular singular points
9. x 0, irregular singular point; x 5, 5, 2, regular
singular points
x(x 2 1)2
x1
5(x 1)
for x 1: p(x) , q(x) x2 x
x21
13. r1 13 , r2 1
15. r1 32 , r2 0
11. for x 1: p(x) 5, q(x) y(x) C1 x 3>2 c1 2
2
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 5
1
1
1
2 2 es e4s;
s
s2
s
1
1
+{etf (t)} 2
e(s1)
2
(s 2 1)
(s 2 1)2
1
e4(s1)
s21
f (t) 2 (t 2)ᐁ(t 2);
1
2
+{ f (t)} 2 e2s;
s
s
2
1
t
+{e f (t)} e2(s1)
s21
(s 2 1)2
e s 2 1
+5 f(t)6 35. y 5te t 12 t 2 et
s(1 e s)
5t
y 256 15 t 32 et 13
254 ᐁ(t 2)
50 e
15 (t 2) ᐁ(t 2) 14 e(t2) ᐁ(t 2)
9
100
e5(t2) ᐁ(t 2)
y 1 t 12 t 2
x 14 98 e2t 18 e2t
y t 94 e2t 14 e2t
i(t) 9 2t 9et/5
w0
y(x) f15 x 5 12 L x 4 2 12 L2 x 3 14 L3 x 2
12EIL
15 (x 2 12 L)5 8(x 2 12 L)g
w0 sinh p2
y(x) sin x sinh x
2EI sinh p
w0 cosh p2
(sin x cosh x 2 cos x sinh x)
4EI sinh p
w0
p
p
csinax 2 bcoshax 2 b
4EI
2
2
p
p
p
2 cosax 2 bsinhax 2 b d 8ax 2 b
2
2
2
+{ f (t)} 1 2
x 2
2!
1 3
y2(x) c1 cx x 3!
1 3
21. y1(x) c0 c1 2
x 3!
22 4
y2(x) c1 cx 2
x 4!
19. y1(x) c0 c1 2
2
22
x
x2
5
752
23
x3 pd
9 7 5 3!
C2 c1 2x 2 2x 2 23 3
x 2 pd
3 3!
Answers to Selected Odd-Numbered Problems
ANS-11
Exercises 5.3, Page 290
17. r1 78 , r2 0
y(x) C1 x 7>8 c1 2
1. y c1 J1/3(x) c2 J1/3(x)
2
22
x
x2
15
23 15 2
3. y c1 J5/2(x) c2 J5/2(x)
3
2
x3 pd
31 23 15 3!
2
2
2
x2
92
C2 c1 2 2x ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 5
19. r1 13 , r2 0
y(x) C1 x 1>3 c 1 1
1
1
x3 pd
x 2 x2 3
3
3 3!
3 2
23. r1 2
3 , r2
y(x) C1 x 2>3 c1 2
25. r1 0, r2 1
y(x) C1x C2 [x ln x 1 12 x2
29. r1 r2 0
1
y(x) C1y(x) C2 cy1(x)ln x y1(x)ax x 2
4
1
1
2
x3 x4 2 p≤ d
3 3!
4 4!
q
1 n
where y1(x) a
x ex
n 0 n!
(1)n
y2(t) t 1 a
n0
(c) y c1x sina
y c1x 1>2I1>3(23ax 3>2) c2x 1>2K1>3(23ax 3>2)
39. (a) j1(x) n
sin x
cos x
2
2
x
x
j2(x) a
j3(x) a
3
1
3 cos x
2 b sin x 2
3
x
x
x2
15
6
15
1
2 2 b sin x 2 a 3 2 b cos x
4
x
x
x
x
51. P2(x), P3(x), P4(x), and P5(x) are given in the text,
P6(x) 1
16
(231x 6 2 315x 4 105x 2 2 5),
P7(x) 161 (429x 7 2 693x 5 315x 3 2 35x)
53. l1 2, l2 12, l3 30
Chapter 5 in Review, Page 294
1. false
3. [12, 12]
121 x3 721 x4 p ]
q
25. y x1>2 [c1 J1>2(18 x 2) c2J1>2(18 x 2)]
1
1
7 3
x x2 2
x pd
2
5
120
q
q
1
1
y(x) C1 a
x 2n C2 x 1 a
x 2n
n 0 (2n 1)!
n 0 (2n)!
q
q
1
1
C1 x 1 a
x 2n 1 C2 x 1 a
x 2n
n 0 (2n 1)!
n 0 (2n)!
1
[C1 sinh x C2 cosh x]
x
27. r1 1, r2 0
q
C1 sin x C2 cos x
35. y c1x 1>2J1>3(23ax 3>2) c2x 1>2Y1>3(23ax 3>2);
1
5 2
1 3
x
x 2
x pd
2
28
21
C2 x 1>3 c1 2
23. y x 1>2 [c1J1>2(x) c2J1>2(x)]
2 cos x
sin xb
a
x
Å px
2 4
x3 pd
11 9 7
1
3
19. y x1 [c1J1>2 (12x 2) c2J1>2(12x 2)]
31. (b) J3>2(x) 3
17. y x 1>2 [c1J3>2(x) c2Y3>2(x)]
C1 x3>2 sin(18 x 2) C2 x3>2 cos(18 x 2)
22
22 3 2
c1 x
x
7
97
("lt)2n 33. (b) y1(t) a
(2n
1)!
n0
ANS-12
15. y x[c1J1(x) c2Y1(x)]
21. r1 52 , r2 0
y(x) C1 x
9. y c1I2>3(4x) c2K2>3(4x)
13. y x1>2 [c1J1(4x 1>2) c2Y1(4x 1>2)]
1
1 2
1
C2 c1 x x x3 pd
2
52
852
5>2
7. y c1J2(3x) c2Y2(3x)
11. y c1x1>2J1>2(ax) c2x1>2J1>2(ax)
23
x3 pd
17 9 3!
2
5. y c1 J0(x) c2Y0(x)
sin( "lt)
"lt
(1)
cos("lt)
("lt)2n (2n)!
t
"l
"l
b c2x cosa
b
x
x
Answers to Selected Odd-Numbered Problems
7. x 2(x 2 1)y0 y9 y 0
9. r1 12 , r2 0
1
y1(x) C1x1/2 [1 13 x 301 x2 630
x3 p ]
y2(x) C2 [1 x 16 x2 901 x3 p ]
x2 12 x3 58 x4 p ]
y2(x) c1 [x 12 x3 14 x4 p ]
13. r1 3, r2 0
1
y1(x) C1x 3 [1 14 x 201 x 2 120
x 3 p]
y2(x) C2 [1 x 12x 2]
1 6
15. y(x) 3[1 x2 13 x4 15
x p ] 2[x 12 x3
18 x5 481 x7 p ]
11. y1(x) c0 [1 17.
1
6
p
3
2
19. x 0 is an ordinary point
21. y(x) c0 c1 2
1 3
1
1
x6 2 3
x9 pd
x 2
3
3 2!
3 3!
c1 cx 2
2
1 4
1 7
x x
4
47
1
5
1
x 10 p d c x 2 2 x 3
4 7 10
2
3
2
(22 2 a2)a2 4
a2 2
x 2
x
2!
4!
(42 2 a2)(22 2 a2)a2 6
2
x 2p
6!
(32 2 a2)(1 2 a2) 5
1 2 a2 3
y2(x) x x x
3!
5!
(52 2 a2)(32 2 a2)(1 2 a2) 7
x p
7!
y1(x) 1 2
Exercises 6.1, Page 301
1. for h 0.1, y5 2.0801; for h 0.05, y10 2.0592
3. for h 0.1, y5 0.5470; for h 0.05, y10 0.5465
5. for h 0.1, y5 0.4053; for h 0.05, y10 0.4054
7. for h 0.1, y5 0.5503; for h 0.05, y10 0.5495
9. for h 0.1, y5 1.3260; for h 0.05, y10 1.3315
11. for h 0.1, y5 3.8254; for h 0.05, y10 3.8840;
at x 0.5 the actual value is y(0.5) 3.9082
13. (a) y1 1.2
(b) y0(c) 12 h 2 4e 2c 12 (0.1)2 0.02e 2c # 0.02e 0.2
0.0244
(c) Actual value is y(0.1) 1.2214. Error is 0.0214.
(d) If h 0.05, y2 1.21.
(e) Error with h 0.1 is 0.0214. Error with h 0.05
is 0.0114.
15. (a) y1 0.8
(b) y (c) 12 h 2 5e–2c 12 (0.1)2 0.025e–2c 0.025
for 0 c 0.1
(c) Actual value is y(0.1) 0.8234. Error is 0.0234.
(d) If h 0.05, y2 0.8125.
(e) Error with h 0.1 is 0.0234. Error with h 0.05
is 0.0109.
17. (a) Error is 19h2 e–3(c–1).
1 2
2h
2
19(0.1) (1) 0.19
(b) y (c)
(c) If h 0.1, y5 1.8207. If h 0.05, y10 1.9424.
(d) Error with h 0.1 is 0.2325. Error with h 0.05
is 0.1109.
1
1 2
2h .
(c 1)2
(b) Zy0(c) 12 h 2 Z # (1) 12 (0.1)2 0.005
(c) If h 0.1, y5 0.4198. If h 0.05, y10 0.4124.
(d) Error with h 0.1 is 0.0143. Error with h 0.05
is 0.0069.
Exercises 6.2, Page 305
1. y5 3.9078; actual value is y(0.5) 3.9082
3. y5 2.0533
5. y5 0.5463
7. y5 0.4055
9. y5 0.5493
11. y5 1.3333
13. (a) 35.7678
mg
kg
tanh
t;
Ä k
Ä m
15. (a) h 0.1, y4 903.0282;
h 0.05, y8 1.1 1015
17. (a) y1 0.82341667
(c) v(t) (b) y(5)(c)
v(5) 35.7678
(0.1)5
h5
h5
40e–2c
40e2(0)
5!
5!
5!
3.333 10–6
(c) Actual value is y(0.1) 0.8234134413. Error is
3.225 10–6 3.333 10–6.
(d) If h 0.05, y2 0.82341363.
(e) Error with h 0.1 is 3.225 10–6. Error with
h 0.05 is 1.854 10–7.
19. (a) y(5)(c)
h5
24
h5
5!
(c 1)5 5!
(0.1)5
24
h5
# 24
2.0000 10–6
5
5!
(c 1) 5!
(c) From calculation with h 0.1, y5 0.40546517.
From calculation with h 0.05, y10 0.40546511.
(b)
Exercises 6.3, Page 309
1. y(x) x ex ; actual values are y(0.2) 1.0214,
y(0.4) 1.0918, y(0.6) 1.2221, y(0.8) 1.4255;
approximations are given in Example 1
3. y4 0.7232
5. for h 0.2, y5 1.5569; for h 0.1, y10 1.5576
7. for h 0.2, y5 0.2385; for h 0.1, y10 0.2384
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 6
1
1
x6 2 3
x9 pd
3 2!
3 3!
23. (a) y c1 J3/2(4x) c2Y3/2(4x)
(b) y c1 I3(6x) c2K3(6x)
29. y(x) c0y1(x) c1y2(x), where
19. (a) Error is
Exercises 6.4, Page 313
1. y(x) 2e2x 5xe2x; y(0.2) 1.4918, y2 1.6800
3. y1 1.4928, y2 1.4919
5. y1 1.4640, y2 1.4640
7. x1 8.3055, y1 3.4199; x2 8.3055, y2 3.4199
9. x1 3.9123, y1 4.2857; x2 3.9123, y2 4.2857
11. x1 0.4179, y1 2.1824; x2 0.4173, y2 2.1821
Answers to Selected Odd-Numbered Problems
ANS-13
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 7
Exercises 6.5, Page 316
1. y1 5.6774, y2 2.5807, y3 6.3226
3. y 1 0.2259, y 2 0.3356, y 3 0.3308,
y4 0.2167
5. y1 3.3751, y2 3.6306, y3 3.6448, y4 3.2355,
y5 2.1411
7. y1 3.8842, y2 2.9640, y3 2.2064, y4 1.5826,
y5 1.0681, y6 0.6430, y7 0.2913
9. y1 0.2660, y2 0.5097, y3 0.7357, y4 0.9471,
y5 1.1465, y6 1.3353, y7 1.5149, y8 1.6855,
y9 1.8474
11. y1 0.3492, y2 0.7202, y3 1.1363, y4 1.6233,
y5 2.2118, y6 2.9386, y7 3.8490
13. (c) y0 2.2755, y1 2.0755, y2 1.8589,
y3 1.6126, y4 1.3275
Chapter 6 in Review, Page 317
1. Comparison of Numerical Methods with h 0.1
xn
Euler
Improved
Euler
RK4
1.10
1.20
1.30
1.40
1.50
2.1386
2.3097
2.5136
2.7504
3.0201
2.1549
2.3439
2.5672
2.8246
3.1157
2.1556
2.3454
2.5695
2.8278
3.1197
Comparison of Numerical Methods with h 0.05
xn
Euler
Improved
Euler
RK4
1.10
1.20
1.30
1.40
1.50
2.1469
2.3272
2.5410
2.7883
3.0690
2.1554
2.3450
2.5689
2.8269
3.1187
2.1556
2.3454
2.5695
2.8278
3.1197
5. –9i 6j; 3i 9j; 3i 5j; 3 !10; !34
7. –6i 27j; 0; 4i 18j; 0; 2 !85
9. 6, 14; 2, 4
15. 2i 5j
17. 2i 2j
y
y
P1P2
P1P2
x
x
19. (1, 18)
21. (a), (b), (c), (e), (f)
23. 6, 15
1
25. k !2,
1
l;
!2
1
1
k !2
, !2
l
5 12
29. k13
, 13 l
27. 0, 1; 0, 1
31.
35.
6
!58
i
3b – a
14
!58
33. k3, 15
2l
j
37. (a b)
3b
a
1
b 2 12 c
43. !2 (i j)
45. (b) approximately 31
qQ
1
47. F i
4pe0 L "L2 a 2
41. a 5
2
Exercises 7.2, Page 331
1. 5
z
(1, 1, 5)
3. Comparison of Numerical Methods with h 0.1
xn
Euler
Improved
Euler
0.60
0.70
0.80
0.90
1.00
0.6000
0.7095
0.8283
0.9559
1.0921
0.6048
0.7191
0.8427
0.9752
1.1163
RK4
0.6049
0.7194
0.8431
0.9757
1.1169
Comparison of Numerical Methods with h 0.05
xn
Euler
Improved
Euler
RK4
0.60
0.70
0.80
0.90
1.00
0.6024
0.7144
0.8356
0.9657
1.1044
0.6049
0.7194
0.8431
0.9757
1.1170
0.6049
0.7194
0.8431
0.9757
1.1169
y
(3, 4, 0)
(6, –2, 0) x
7. The set {(x, y, 5)|x, y real numbers} is a plane perpen9.
11.
13.
15.
17.
19.
5. h 0.2: y(0.2) 3.2; h 0.1: y(0.2) 3.23
7. x(0.2) 1.62, y(0.2) 1.84
ANS-14
11. 10i 12j; 12i 17j
13. 20, 52; 2, 0
Exercises 7.1, Page 325
1. 6i 12j; i 8j; 3i; !65; 3
3. 12, 0; 4, 5; 4, 5; !41; !41
Answers to Selected Odd-Numbered Problems
21.
27.
31.
35.
dicular to the z-axis, 5 units above the xy-plane.
The set {(2, 3, z)|z a real number} is a line perpendicular
to the xy-plane at (2, 3, 0).
(0, 0, 0), (2, 0, 0), (2, 5, 0), (0, 5, 0), (0, 0, 8), (2, 0, 8),
(2, 5, 8), (0, 5, 8)
(–2, 5, 0), (–2, 0, 4), (0, 5, 4); (–2, 5, 2); (3, 5, 4)
the union of the coordinate planes
the point (1, 2, 3)
the union of the planes z 5 and z 5
!70
23. 7; 5
25. right triangle
isosceles
29. d(P1, P2) d(P1, P3) d(P2, P3)
6 or 2
33. (4, 12 , 32 )
P1(–4, 11, 10) 37. –3, 6, 1
39. 2, 1, 1
41. 2, 4, 12
43. –11, 41, 49 45. !139
49. – 23 , 13 , 23 47. 6
a
z
53.
51. 4i 4j 4k
1
2
(a + b)
b
y
x
17. 49 , 13 , 1
7. 29
21. 1.11 radians or 63.43
23. 1.89 radians or 108.43
25. cos a 1/ !14, cos b 2/ !14, cos g 3/ !14;
a 74.5, b 57.69, g 36.7
27. cos a 1
2,
cos b 0, cos g !3/2; a 60,
b 90, g 150
29. 0.955 radian or 54.74; 0.616 radian or 35.26
31. a 58.19, b 42.45, g 65.06
33.
5
7
31.
35.
39.
43.
35. –6 !11/11
37. 72 !109/109
47.
41.
43.
96
72
25 , 25 55.
47. 0; 150 N-m
59.
28
5
12 6 4
– 7 , 7 , 7 39. – 21
5,
45. 1000 ft-lb
49. approximately 1.80 angstroms
61.
Exercises 7.4, Page 343
1. –5i 5j 3k 3. –12, 2, 6
5. –5i 5k
7. –3, 2, 3
9. 0
11. 6i 14j 4k 13. –3i 2j 5k
17. –i j k
19. 2k
21. i 2j
23. –24k
25. 5i 5j k
27. 0
29. !41
31. –j
33. 0
35. 6
37. 12i 9j 18k
39. –4i 3j 6k
41. –21i 16j 22k
43. –10
45. 14 square units
47.
1
2
65. x 2 t, y 12 t, z t
63. (c) and (d)
12 t, y 12 32 t, z t
69. (–5, 5, 9)
71. (1, 2, 5)
73. x 5 t, y 6 3t, z 12 t
75. 3x y 2z 10
67. x 1
2
77.
z
y
square unit
x
y
x
z
81.
Exercises 7.5, Page 350
1. x, y, z 1, 2, 1 t 2, 3, 3
3. x, y, z 12 , 12 , 1 t –2, 3, 32 5. x, y, z 1, 1, 1 t 5, 0, 0
7. x 2 4t, y 3 4t, z 5 3t
9. x 1 2t, y 2t, z 7t
11. x 4 10t, y 1
2
34 t, z 13 16 t
z
79.
7
2
square units
51. 10 cubic units
53. coplanar
55. 32; in the xy-plane, 30 from the positive x-axis in the
direction of the negative y-axis; 16 !3i 16j
57. A i k, B j k, C 2k
49.
51.
15
2
), (10, 5, 0)
(2, 3, 5)
33. Lines do not intersect.
40.37
37. x 4 6t, y 1 3t, z 6 3t
2x 3y 4z 19
41. 5x 3z 51
6x 8y 4z 11
45. 5x 3y z 2
3x 4y z 0
49. The points are collinear.
x y 4z 25
53. z 12
–3x y 10z 18
57. 9x 7y 5z 17
6x 2y z 12
orthogonal: (a) and (d), (b) and (c), (d) and (f), (b) and (e);
parallel: (a) and (f), (c) and (e)
29. (0, 5, 15), (5, 0,
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 7
Exercises 7.3, Page 336
1. 12
3. –16
5. 48
2 4
9. 25
11. – 5 , 5 , 2 13. 25!2
15. (a) and (f), (c) and (d), (b) and (e)
y24
x21
z9
9
10
7
x7
z25
15.
,y 2
11
4
y 2 10
z2
17. x 5,
9
12
19. x 4 3t, y 6 12 t, z 7 32 t;
x24
2z 14
2y 2 12 3
3
y
x
z
21. x 5t, y 9t, z 4t; 5
9
4
23. x 6 2t, y 4 3t, z 2 6t
25. x 2 t, y 2, z 15
27. Both lines pass through the origin and have parallel
direction vectors.
13.
y
x
Exercises 7.6, Page 357
1. not a vector space, axiom (vi) is not satisfied
3. not a vector space, axiom (x) is not satisfied
5. vector space
Answers to Selected Odd-Numbered Problems
ANS-15
7. not a vector space, axiom (ii) is not satisfied
9. vector space
11. a subspace
13. not a subspace
15. a subspace
17. a subspace
19. not a subspace
23. (b) a 7u1 12u2 8u3
25. linearly dependent
15. a
17. a
a
27. linearly independent
29. f is discontinuous at x 1 and at x 3.
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 8
31. 2"23 p2>3, "p
Exercises 7.7, Page 363
4
1. u 58
13 w1 2 13 w2, where
w1 3. u 5
k12
13 , 13 l,
32
w2 k135 ,
v1 7v2 2
23
2
12
13 l
v3, where
v1 k1, 0, 1l, v2 k0, 1, 0l, v3 k1, 0, 1l
3
2
2
3
l, k !13
, !13
l6
!13
1
1
1
1
5 k !2, !2 l, k !2, !2 l6
1
1
1
1
5 k !2
, !2
l, k !2
, !2
l6
5. (a) B0 5 k !13,
(b) B0 7. (a) B0 (b) B0 5 k1, 0l, k0, 1l6
1
1
1
1
4
, 0l, k3 !2
, 3 !2
, 3 !2
l, k23, 23, 13 l6
!2
1
1
2
1
7
1
B0 5 k !6
, !6
, !6
l, k !3
, 5 !3
, 5 !3
l,
1
3
4
k !2, 5 !2, 5 !2 l6
1
5
2
13
1
4
B0 5 k !30
, !30
, !30
l, k !186
, !186
, !186
l6
7
9
3
1
1
1 1
1
B0 5 k2, 2, 2, 2 l, k2 !35, 2 !35, 2 !35, 2 !35 l6
B9 51, x, 12 (1 3x 2)6
9. B0 5 k !2,
11.
13.
15.
17.
1
19. B0 5 !2,
5
x, 2 !10
(3x 2 2 1)6
41
1
q (x) 3!6 q2(x) !3
q3(x), where
!15 1
1
3
5
!2, q2(x) !6 x, q2(x) 2 !10 (3x 2 2 1)
21. p(x) q1(x)
3
!6
Chapter 7 in Review, Page 364
1. true
3. false
5. true
7. true
9. true
11. 9i 2j 2k 13. 5i
15. 14
17. –6i j 7k 19. (4, 7, 5)
21. (5, 6, 3)
23. 36!2
25. 12, 8, and 6
27. 3!10>2
33. 2
43. 14x 5y 3z 0
45. 30 !2 N-m
7
9
20
9
i j k
37. sphere; plane
y
2
3
x27
z5
39.
4
2
6
41. The direction vectors are orthogonal and the point of
intersection is (3, 3, 0).
35.
47. approximately 153 lb
49. not a vector space
51. a subspace; 1, x
Exercises 8.1, Page 374
1. 2 4
3. 3 3
5. 3 4
7. not equal
9. not equal
11. x 2, y 4
13. c23 9, c12 12
ANS-16
Answers to Selected Odd-Numbered Problems
11
6
b; a
1
14
11
17
9
3
18
19
b; a
31
3
24
3
b; a
8
6
4
21. 180; ° 8
10
23. a
27. a
7
10
1
2
b; a
19
12
6
32
b; a
22
4
19
30
8
16
20
38
7
b; a
75
10
38
b
2
27
b;
1
6
b
22
28
b
12
8
0
b; a
16
0
0
4
b; a
0
8
38
b
75
14
b
1
10
6
20 ¢ ; ° 12 ¢
25
5
25. a
5
b
10
29. 4 5
1 2
2 4
b, B a
b
0 0
1
2
AB is not necessarily the same as BA.
a11x1 a12x2 b1
a21x1 a22x2 b2
b k1, 1l
47. b k2, 0l
1
0
M a
b
0 1
cos b 0 sin b
(b) MR ° 0
1
0
¢,
sin b 0
cos b
37. A a
39.
41.
45.
49.
51.
1
MP ° 0
0
(c) xS 29. 2 units
31. (i j 3k)/ !11
26
9
19. a
2
2
0
cos a
sin a
0
sin a ¢
cos a
3!2 2!3 2 !6 6
1.4072,
8
yS 3!2 2!3 2 3!6 2
0.2948,
8
zS !2 !6
0.9659
8
Exercises 8.2, Page 387
1. x1 4, x2 7
3. x1 23 , x2 13
5. x1 0, x2 4, x3 1 7. x1 t, x2 t, x3 0
9. inconsistent
11. x1 0, x2 0, x3 0
13. x1 2, x2 2, x3 4 15. x1 1, x2 2 t, x3 t
17. x1 0, x2 1, x3 1, x4 0
19. inconsistent
21. x1 0.3, x2 0.12, x3 4.1
23. 2Na 2H2O S 2NaOH H2
25. Fe3O4 4C S 3Fe 4CO
27. 3Cu 8NHO3 S 3Cu(NO3)2 4H2O 2NO
29. i1 35
9 , i2
389 , i3 13
1
31. ° 5
8
1
2
1
1
x1
0
2 ¢ ° x2 ¢ ° 0 ¢
5
x3
0
15. a
3
1
x1
12
1 1 ¢ ° x2 ¢ ° 1 ¢
4 1 1
x3
10
35. Interchange row 1 and row 2 in I3.
37. Multiply the second row of I3 by c and add to the
third row.
2
33. ° 1
a22
a12
a32
a11
41. EA °
a21
ca21 a31
a23
a13 ¢
a33
a12
a22
ca22 a32
Exercises 8.3, Page 392
1. 2
3. 1
7. 2
9. 3
13. linearly independent
17. rank(A) 2
a13
a23
¢
ca23 a33
5. 3
11. linearly independent
15. 5
Exercises 8.4, Page 398
1. 9
3. 1
5. 2
7. 10
11. 17
13. l2 3l 4
17. 62
19. 0
23. –x 2y z
25. –104
29. l1 5, l2 7
9. –7
21. –85
3.
7.
12
° 18
3
8
7. Theorem 8.5.3
11. –5
13. –5
19. –105
27. 0
33. 0
35. 16
1
3
0
11. ° 0
1
6
0
13. ±
2
9
1
27
10
27
4
27
5.
1
2
1
4
1
4
0
1
2
38 ¢
1
8
0
0¢
12
7
9
1
27
17
27
274
13
2
9
29
1
9
19. singular matrix
1
3
12
5 6 3
23. ° 2 2 1 ¢
1 1 1
27. a
13
1
1
3
10 b
3
29. a
25. §
2
3
9.
43
19
≤
179
4
9
7
15
° 151
2
15
1
4
1b
12
5
12
0
21. ° 0
1
0
12
3
b
4
2
3
13
23
23
1
3
13
1
1
3
2
3 ¢
0
16
1
3
1
3
1
2
7
6
43
1¥
3
1
2
31. x 5
47. x1 34 , x2 12
45. x1 6, x2 2
49. x1 2, x2 4, x3 6 51. x1 21, x2 1, x3 11
9
10 ,
x2 13
20 ; x1 6, x2 16; x1 2, x2 7
55. System has only the trivial solution.
57. System has nontrivial solutions.
R3E2 R3E1 R2E1 2 R2E3
59. (c) i1 ,
R3R1 R3R2 R1R2
R3E2 2 R3E1 2 R1E3 R1E2
i2 ,
R3R1 R3R2 R1R2
R2E1 R1E3 R2E3 2 R1E2
i3 R3R1 R3R2 R1R2
53. x1 Exercises 8.7, Page 417
1. x1 35 , x2 65
5. x 4, y 7
27. 48
3. Theorem 8.5.7
1
a 61
4
14
15. –48
Exercises 8.6, Page 413
1
9
5b
9
17. a
9. u 4, v 32 , w 1
Exercises 8.5, Page 404
1. Theorem 8.5.4
5. Theorem 8.5.5
9. Theorem 8.5.1
15. 5
17. 80
29. –15
31. –9
1
a 94
9
0
1
12
1b
4
13
30
152
7
30
7. x1 4, x2 4, x3 5
11. k 6
5
13. T1 450.8 lb, T2 423 lb
Exercises 8.8, Page 425
1. K3, l 1
3. K3, l 0
5. K2, l 3; K3, l 1
2
7
1
1
7. l1 6, l2 1, K1 a b , K2 a b ; nonsingular
0
1b
2
3. x1 0.1, x2 0.3
158
1
15 ¢
2
15
9. l1 l2 4, K1 a
11. l1 3i, l2 3i
1
b ; nonsingular
4
1 2 3i
1 3i
b , K2 a
b ; nonsingular
5
5
13. l1 4, l2 5,
1
8
K1 a b , K2 a b ; nonsingular
0
9
15. l1 0, l2 4, l3 4,
9
1
1
K1 ° 45 ¢ , K2 ° 1 ¢ , K3 ° 9 ¢ ; singular
25
1
1
K1 a
Answers to Selected Odd-Numbered Problems
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 8
a21
39. EA ° a11
a31
1
6
ANS-17
17. l1 l2 l3 2,
2
0
K1 ° 1 ¢ , K2 ° 0 ¢ ; nonsingular
0
1
19. l1 1, l2 i, l3 i,
1
1
1
K1 ° 1 ¢ , K2 ° i ¢ , K3 ° i ¢ ; nonsingular
1
i
i
21. l1 1, l2 5, l3 7,
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 8
1
1
1
K1 ° 0 ¢ , K2 ° 2 ¢ , K3 ° 2 ¢ ; nonsingular
0
0
4
23. eigenvalues of A are l 1 6, l 2 4; eigenvalues of
A1 are l1 16, l2 14; corresponding eigenvectors
1
1
for both A and A1 are K 1 a b, K 2 a b.
1
1
25. eigenvalues of A are l1 5, l2 4, l3 3, eigenvalues of A1 are l1 15, l2 14, l3 13; corresponding
3
eigenvectors for both A and A1 are K 1 ° 1 ¢ ,
1
1
2
K2 ° 0 ¢ , K3 ° 1 ¢ .
0
0
Exercises 8.9, Page 429
3.
1
m m 1
7 f3(1) 2
a
2
7 f(2)m
11 57
a
b
38 106
1
5.
7.
a3
2m2 1 f5m (1)mg
1
m
m
3 f10 2 (2) g
83328 41680
a
b
33344 16640
1
°0
0
2m 2 1
1
m 1
(1)mg
3 f2
1
m
m
3 f2 2 (1) g
a
22528
18432
b
18432 14336
(c)
2(3)m2 1
1
° 6 f9(2)m 2 4(3)mg
1
m
m
6 f9(2) 8(3) g
15. a
3
10
101
2m2 2 f5m 2 (1)mg
b;
1 m2 1 m
f5 5(1)mg
32
2m 2 1
2
m
m
3 f2 2 (1) g ¢ ;
1
m
m
3 f2 2(1) g
3m2 1
3m2 1
1
m
m
6 f3(2) 2 2(3) g
1
m
m
6 f3(2) 4(3) g
1
m
m
6 f3(2) 2 2(3) g
1
m
m
6 f3(2) 4(3) g
2
5
1b
5
17. (b), (c), (d), (e), (f)
Exercises 8.10, Page 436
1. (b) l1 4, l2 1, l3 16
3. (b) l1 18, l2 l3 8 5. orthogonal
7. orthogonal
9. not orthogonal
1
!2
1
!2
1
!2
11. a
15. £
0
1
!2
3
!11
1
17. £ !11
1
!11
1
!2 b
1
!2
1
!2
13. a
0
1≥
0
0
1
!2
1
!66
4
!66
7
!66
19. a 45 , b 3
5
2 (2)mg 0
1
m
m m 1
g 0¢;
3 f4 (1) 2
1
m
m
f4
2
(2)
g 1
3
3m4m2 1
b;
m4m2 1 (1 2 m)4m
Answers to Selected Odd-Numbered Problems
1
!10 b
3
!10
1
!6
2
≥
!6
1
!6
(c) P 1
!2
1
£ !2
0
1
!6
1
!6
2
!6
Exercises 8.11, Page 443
1
1
1. 2, a b
7. 7 and 2
1
!3
1
!3
1
!3
≥
1
1
3
3. 14, a b
9. 4, 3, and 1
11. approximately 0.2087
3 2 1
1
°2 4 2¢
4
1 2 3
(d) 0.59
(e) approximately 9.44EI/L2
13. (c)
1
m
3 f4
3
!10
1
!10
21. (b) l1 l2 2, l3 4
5
3
1 1023 1023
° 0 683 682 ¢
0 341 342
1
2m 1
(2)mg
3 f2
9. ° 13 f22m 1 2 (1)m2m 1g
1
2m 1
(2)m 2 3g
3 f2
699392 349184 0
° 698368 350208 0 ¢
699391 349184 1
7m4m 1 (1 2 m)4m
11. a
3m4m2 1
ANS-18
37 f(2)m 2 5mg
1
m
m b;
7 f(2) 2 6(5) g
5mg
2 5mg
1 1
b, m 1
3 3
(b) Am 0, m 1
13. (a) 4m a
Exercises 8.12, Page 451
3 1
1
1. P a
b, D a
1 1
0
3. not diagonalizable
13 1
7
5. P a
b, D a
2 1
0
1
1 1
7. P a
b, D a 3
1 1
0
0
b
5
0
b
4
0
2b
3
5. 10, ° 1 ¢
0.5
¢
1 1
i 0
b, D a
b
i i
0 i
1 0 1
1
0
11. P ° 0 1 1 ¢ , D ° 0 1
0 0 1
0
0
1 1 1
0 0
13. P ° 0 1 0 ¢ , D ° 0 1
1 1 1
0 0
15. not diagonalizable
9. P a
2
0
D ±
0
0
0
2
0
0
1
!2
1
!2
!10
a !14
2
!14
1
!2
1
° !2
21. P a
0
27. P 2
3
° 32
1
3
0
29. P ° 1
0
0
2
3
1
3
23
1
1
3
2
3 ¢ ,
2
3
1
1
!2
!2
0
1
!2
1
a !2
1
!2
3. LU a
1
7
1
9. LU ° 0
2
1
11. LU a 1
3
1
13. LU a 1
4
1
15. LU ° 3
1
0
b
2
1
° 0
0
3
D °0
0
0
6
0
1
0 ¢ , D °0
1
0
!2
1
0
1
0
0
0¢
9
0
6
0
y
17. LU °
0
0¢
1
1 bX9
!5
we get X 2 /4 Y 2 /4 1.
1
2
3
4
b
10
0
1
7
0
1
8
0
1
0¢ °0
1
0
0
1
10
0 3
ba
1 0
0
1
25
2
1
0
2
1
0
2
1
0
0
1
0¢ °0
1
0
0 4
ba
1
0
0
1
1
0
4
0¢ °0
1
0
2
b
52
0
1
0¢ °0
1
0
1
2
0
0
4
0¢ °0
1
0
2
25
0
1
3¢
21
7
8 ¢
109
1
2¢
21
9
b
8
21. x1 13, x2 56
0
0¢
8
1
1 ¢
1
12
10 ¢
16
0
1 2 1 0
0
0 1 0 1
≤ ±
≤
0
0
0 3 7
1
0
0 0 26
23. x1 1, x2 7
2
25. x1 1, x2 1, x3 5
27. x1 28, x2 5, x3 13
x
y
0
31. X ° 5 ¢
3
35. 2
X
X a
11
b
10
2
b
3
1
0
0
2
1
0
19. LU ±
1 4
1
5 11 2
Y
33. Hyperbola; using
21
22
29. x1 25, x2 4, x3 19
1
!2 bX9
1
!2
2
!5
1 0 1
ba
2 1
0
7. LU ° 2
we get X 2 /4 Y 2 /6 1.
1
!5
2
!5
0 2
ba
1 0
2
0
0
0
0
≤
1
0
0 1
31. Ellipse; using
X
1
1. LU a 1
1
5. LU ° 1
3
0
b
10
39. A5 a
Exercises 8.13, Page 458
1
0
≤,
1
0
1
0
!2 b, D a
1
0
!2
!10
3
!35 b, D a
5
0
!35
1
0
!2
1
0¢, D !2
1
b
1
4
2
X
Y
x
1
41. LU ° 3
1
2
43. LLT ° 1
6
33. X 39. 78
37. 1600
0
2
2
0
5
2
0
1
0¢ °0
1
0
0
2
0¢ °0
4
0
1
4
° 32 ¢
54
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 8
3 1 1
0
1
0
19. P ±
3
0
0
1
0
1
25. P 0
0¢
2
0
0¢
2
1 !5 1 2 !5
2
2
¢,
0
0
0
0
!5
0¢
0 !5
0
17. P ° 0
1
1
D °0
0
23. P 35. A a
1
1
0
1
5
0
1
1
2¢
1
6
2 ¢
4
Answers to Selected Odd-Numbered Problems
ANS-19
Exercises 8.14, Page 462
35 15 38 36
1. (a) a
27 10 26 20
48
3. (a) a
32
31
5. (a) ° 24
1
64
40
120
75
44
29
15
107
67
15
15
15
Chapter 8 in Review, Page 476
2 3 4
3 4 5
3 4
1. ±
≤
3. a
b, (11)
4 5 6
6 8
5 6 7
0
b
0
40
b
25
61
47
0
50
35
15
49
31
5
7. STUDY_HARD
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 9
22
22
26
20
18
26
8
23
14
23
25
23
6
2
16
21
23
26
22
25 ¢
12
Exercises 8.15, Page 467
1. (0 1 1 0)
3. (0 0 0 1 1)
5. (1 0 1 0 1 0 0 1)
7. (1 0 0)
9. parity error
15. (0 1 0 0 1 0 1)
19. code word; (0 0 0 0)
21. (0 0 0 1)
23. code word; (1 1 1 1)
23. (a) a
27. (1 0 1 0)
31. trivial solution only
3. y 1.1x 0.3
99.6
9. f(x) 0.75x 2 2.45x 2.75
Exercises 8.17, Page 475
0.8 0.4
90
1. (a) T a
b , X0 a b
0.2 0.6
60
(c) X̂ a
0
100
0 ¢ , X0 ° 0 ¢
1
0
2
3
1
b
1
1
1
2
K 1 ° 2 ¢ , K 2 ° 0 ¢ , K 3 ° 1 ¢
0
1
2
45. l1 l2 3, l3 5,
1
!6
2
£ !6
≥
1
!6
49. hyperbola
204 13 208 55 124 120 105 214 50 6 138 19 210
b
185 12 188 50 112 108 96 194 45 6 126 18 189
53. HELP_IS_ON_THE_WAY
55. (a) (1 1 0 0 1) (b) parity error
57. x1 3, x2 2, x3 1
59. f(x) 45 x 2 32
a
2
0.2 0.5
3. (a) T ° 0.3 0.1
0.5 0.4
20
19
(b) ° 30 ¢ , ° 9 ¢
50
72
25. x1 12 , x2 7, x3 12
y X sin u Y cos u
39. x1 7, x2 5, x3 23
51.
5. y 1.3571x 1.9286
96
98.4
b, a
b
54
51.6
1
b
1
37. x X cos u Y sin u,
47.
Exercises 8.16, Page 471
116.4,
13. true
19. false
2
3
1
K1 ° 1 ¢ , K2 ° 0 ¢ , K3 ° 2 ¢
0
1
1
(b) 2 16
(c) (0 0 0 0 0 0 0), (0 1 0 0 1 0 1),
(0 1 1 0 0 1 1), (0 1 0 1 0 1 0),
(0 1 1 1 1 0 0), (0 0 1 0 1 1 0),
(0 0 1 1 0 0 1), (0 0 0 1 1 1 1),
(1 0 0 0 0 1 1), (1 1 0 0 1 1 0),
(1 0 1 0 1 0 1), (1 0 0 1 1 0 0),
(1 1 1 0 0 0 0), (1 1 0 1 0 0 1),
(1 0 1 1 0 1 0), (1 1 1 1 1 1 1)
7. v 0.84T 234,
1
1
35. x1 12 , x2 14 , x3 4
1. y 0.4x 0.6
11. false
43. l1 l2 1, l3 8,
17. (1 1 0 0 1 1 0)
29. (a) 2 128
9. 0
17. true
1
2
13. (0 0 1 0 1 1 0)
7
5
15. false
41. l1 5, l2 1, K1 a b , K2 a
11. (1 0 0 1 1)
25. (1 0 0 1)
5
8,
33. I2 10HNO3 S 2HIO3 10NO2 4H2O
11. DAD_I_NEED_MONEY_TODAY
15
13. (a) B9 ° 10
3
7.
29. 240
9. MATH_IS_IMPORTANT_
(b) a
ANS-20
41
21 ¢
19
5. false
Exercises 9.1, Page 485
z
1.
z
3.
y
100
b
50
0
(c) X̂ ° 0 ¢
100
Answers to Selected Odd-Numbered Problems
x
y
x
5.
y
z
7.
y
x
x
z
9.
11.
5. Speed is "5.
z
7. Speed is "14.
z
a(1)
v(2)
C
y
z
a(2)
v(1)
y
x
y
x
x
r(t) = ti + tj +
y
2t 2k
x
z
13.
v(0) 2 j 5 k, a(0) 2 i 2 k,
v(5) 10 i 73 j 5 k, a(5) 2 i 30 j 2 k
11. (a) r (t) (16t2 240t) j 240 "3 t i and
y
x(t) 240 "3 t, y(t) 16t 2 240t
(b) 900 ft (c) 6235 ft (d) 480 ft/s
13. 72.11 ft/s
15. 97.98 ft/s
17. (a) 4300 ft, approximately 7052.15 ft, approximately
576.89 ft/s
y
(b)
x
r(t) = 3 cos ti + 3 sin tj + 9 sin2tk
15. 2i 32j
17. (1/t)i (1/t 2)j; (1/t 2)i (2/t 3)j
19. e2t (2t 1), 3t 2, 8t 1; 4e2t (t 1), 6t, 8
z
y
21.
23.
4000
y
x
x
q = 60°
3000
q = 30°
2000
1000
35.
37.
39.
x 2 t, y 2 2t, z 4t
r(t) r (t)
r(t) [r (t) r (t)]
2 r1(2t) (1/t2) r2(1/t)
3
2 i 9 j 15 k
et (t 1) i 12 e2t j 12 et2 k c
(6t 1) i (3t 2 2) j (t 3 1) k
(2t 3 6t 6) i (7t 4t 3/2 3) j (t 2 2t) k
41.
43.
45.
47.
2"a 2 c 2p
6(e3p 1)
a cos(s/a) i a sin(s/a) j
Differentiate r (t) r (t) c2.
25.
27.
29.
31.
33.
2000
8
3
x
the target at t 0. Then rp rt when t x0 /(v0 cos u) y0 /(v0 sin u). This implies tan u y0 /x0. In other words,
aim directly at the target at t 0.
25. 191.33 lb
1
3
2
27. r(t) k1e 2t i 2
j (k3e t 2 1) k
2t k2
29. (b) Since F is directed along r, we must have F cr for some constant c. Hence T r (cr) c(r r) 0. If T 0, then dL/dt 0. This implies
that L is a constant.
Exercises 9.3, Page 495
y
v(0)
a(1)
x
8000
21. Assume that (x0, y0) are the coordinates of the center of
1. T ( "5 /5)(sin t i cos t j 2 k)
3. Speed is 2.
v(1)
6000
19. approximately 175.62 ft/s
Exercises 9.2, Page 488
1. Speed is "5.
4000
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 9
9. (0, 0, 0) and (25, 115, 0);
C
a(0)
3. T (a2 c2)1/2 (a sin t i a cos t j c k),
y
x
N cos t i sin t j,
B (a2 c2)1/2 (c sin t i c cos t j a k),
k a/(a2 c2)
5. 3 "2x 3 "2y 4z 3p
7. 4t> "1 4t 2,
2> "1 4t 2
Answers to Selected Odd-Numbered Problems
ANS-21
9. 2"6, 0, t . 0 11. 2t> "1 t 2,
15. "3e t, 0
13. 0, 5
43. 0w /0t 3u(u2 v2)1/2 et sin u
2> "1 t 2
3v(u2 v2)1/2 et cos u,
0w /0u 3u(u2 v2)1/2 et cos u
"b 2c 2 sin2 t a 2c 2 cos 2 t a 2b 2
17. k (a 2 sin2 t b 2 cos 2 t c 2 )3>2
23. k 2, r 1
2;
3v(u2 v2)1/2 et sin u
2
k 2/ "125 0.18,
2
Exercises 9.4, Page 500
y
y
3.
2 2
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 9
x
x
2
2 2
0R/0v 2s2t 4uvev 2rst 4 eu 8rs2t 3u2veu v
y cosh rs
0w
xu
47.
2
,
2
1>2
0t
(x y ) (rs tu)
u(x 2 y 2)1>2
sty cosh rs
0w
xs
2
,
0r
(x y 2)1>2(rs tu)
u(x 2 y 2)1>2
ty cosh rs
0w
xt
2 2
2
2
1>2
0u
(x y ) (rs tu)
u (x y 2)1>2
r "125 /2 5.59; the curve is sharper at (0, 0).
1.
2
45. 0R/0u s2t 4 ev 4rst 4uveu 8rs2t 3uv2 eu v ,
49. dz /dt (4ut 4vt3)/(u2 v2)
y
5.
Exercises 9.5, Page 505
1. (2x 3x2y2) i (2x3y 4y3) j
3. ( y2 /z3) i (2xy/z3) j (3xy2 /z4) k
x
7. elliptical cylinders
9. ellipsoids
z
z
11.
57. 5.31 cm2 /s
51. dw/dt|tp 2
5. 4 i 32 j
z
9. !3x y
13.
y
x
y
y
x
c>0
19. 1
17. 3!2
21. 12> !17
31. 31 !17
33. u 35 i 45 j; u 45 i 35 j; u 45 i 35 j
17. 0z /0x 2x1/2 /(3y2 1),
35. Du f (9x2 3y2 18xy2 6x2y)/ "10;
2
Du F (6x2 54y2 54x 6y 72xy)/10
37. (2, 5), (2, 5)
39. 16 i 4 j
41. x 3e4t, y 4e2t
2 2
3
15. 98> !5
29. 38 i 2 12j 2 23 k, !83,281>24
0z /0y 15x4y2 6x2y5 4
2
2)
27. 8!p>6 i 2 8!p>6 j, 8!p>3
15. 0z /0x 20x3y3 2xy6 30x4,
0z /0y 24 "xy /(3y 1)
12 !10
25. 2 i 2 j 4 k, 2!6
c<0
13. 0z /0x 2x y2, 0z/0y 2xy 20y4
2
15
2 ( !3
11.
23. !2 i (!2>2) j, !5>2
x
c=0
7. 2!3 i 2 8 j 2 4!3 k
19. 0z /0x 3x (x y ) ,
0z /0y 2y(x3 y2)2
21. 0z /0x 10 cos 5x sin 5x,
0z /0y 10 sin 5y cos 5y
43. One possible function is f (x, y) x3 23 y3 xy3 e xy.
3
3
23. fx e x y(3x 3y 1), fy x 4e x y
Exercises 9.6, Page 510
25. fx 7y/(x 2y)2, fy 7x/(x 2y)2
2
3
2
2
3
27. gu 8u/(4u 5v ), gv 15v /(4u 5v )
1.
y
y
3.
29. wx x1/2y, wy 2 "x ( y/z)ey/z ey/z,
x
wz ( y2 /z2)ey/z
2
3
2
x
2
31. Fu 2uw v vwt sin(ut ),
Fv 3uv2 w cos(ut2), Fx 128x7t4,
5.
y
7.
Ft 2uvwt sin(ut 2) 64x8t 3
2
2
39. 0z /0x 3x2v2 euv 2uveuv , 0z/0y 4yuveuv
2
41. 0z /0u 16u3 40y(2u v),
0z /0v 96v2 20y(2u v)
ANS-22
Answers to Selected Odd-Numbered Problems
y
x
x
z
9.
35. 2 i (1 8y) j 8z k
z
11.
45. div F 1
y
0. If there existed a vector field G such that
F curl G, then necessarily div F div curl G 0.
y
x
x
Exercises 9.8, Page 523
1. 125/3 !2; 250( !2 4)/12; 125
2
5. 1; (p 2)/2; p2 /8; !2p2 /8
3. 3; 6; 3 !5
13. (4, 1, 17)
7. 21
15. 2x 2y z 9
9. 30
17.
19. 6x 8y z 50
23. 0
25.
21. 2x y !2z (4 5p)/4
31. e
33. 4
23. !2x !2y z 2
25. (1/ !2, !2, 3/ !2), (1/ !2, !2, 3/ !2)
27. (2, 0, 5), (2, 0, 3)
33. x 1 2t, y 1 4t, z 1 2t
35. (x 12 )/4 ( y 13 )/6 (z 3)
Exercises 9.7, Page 514
y
1.
1
21. 83
29. 19
8
27. 70
35. 0
37. On each curve the line integral has the value
Exercises 9.9, Page 533
1. 16
3. 14
3
9. 1096
13. not a conservative field
17. 3 e1
19. 63
23. 16
25. p 4
y
9.
5. 3
7. 330
4 3
11. f x y 3x y
15. f 14 x4 xy 14 y4
21. 8 2e3
27. f (Gm1m2)/|r|
x
ln 5
2
5.
y
11.
y
3.
208
3 .
41. x 32 , y 2>p
Exercises 9.10, Page 540
2
1. 24y 20ey
3. x 2e 3x 2 x 2e x
7. 2 sin y
x
2
19.
13. 1
643
x
x
1
x
2
13.
1
21
15.
19. 2 ln 2 1
21.
25. 18
27. 2p
5.
41. x 39. p/8
45. x x
51.
17
21 ,
1
105
53. 4k/9
Exercises 9.11, Page 545
7. (x y) i (x y) j; 2z
9. 0; 4y 8z
2
11. (4y 6xz ) i (2z 3x ) k; 6xy
13. (3ez 8yz) i xez j; ez 4z2 3yez
15. (xy2 ey 2xyey x3yzez x3yez) i y2 ey j
(3x2yzez xex) k; xyex yex x3zez
1. 27p/2
15.
19.
23.
256
21
3
4
7
57.
941
10
3
67. a !3> 3
3. (4p 3 !3)/6 5. 25p/3
7 )
9. 54
x 13>3p, y 13>3p 13. x 125, y 3!3>2
x (4 3p)>6, y 43 17. pa4 k/4
(ka/12)(15 !3 4p)
21. pa4 k/2
4k
25. 9p
27. (p/4)(e 1)
7. (2p/3)(15
11.
55.
65. 16 !2k/3
63. ka /6
3
y2
sin 8
43. x 3, y 32
61. ab p/4; a bp/4; b/2; a/2
4
2
8
3,
2
3
37.
55
147
59. a!10>5
3
29. 4
23. (c), 16p
31. 30 ln 6
y
47. x 0, y 49. x (3e 4 1)> f4(e 4 2 1)g,
y 16(e 5 2 1)> f25(e 4 2 1)g
1
1
17. 96
35. (23/2 1)/18
33. 15p/4
y
25
84
14
3
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 9
15. 460
17. 6x 2y 9z 5
11. 1
26
9
123
2
3/2
3/2
Answers to Selected Odd-Numbered Problems
ANS-23
29. 3p/8
31. 250
33. approximately 1450 m3
Exercises 9.12, Page 550
1. 3
3. 0
9.
56
3
11.
21. 16!2
1
5. 75p
2
3
13.
7. 48p
1
8
19. 3a p/8
23. 45p/2
25. p
27. 27p/2
29. 3p/2
33. 3p
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 9
3!29
3.
25p/6
9.
2pa(c2 c1)
972p
23. 9(173/2 1)
29. 18
31.
3
37. 8pa
39.
10p/3
5. (p/6)(173/2 1)
2
2 a (p 2) 11. 8a2
15. 26
17. 0
3
5/2
7/2
21. (3 2 1)/15
25. 12!14
27. k!3>12
28p
33. 8p
35. 5p/2
2
4pkq
41. (1, 3 , 2)
Exercises 9.14, Page 563
1. 40p
3.
9. 3p/2
45
2
5.
3
2
7. 3
13. 152p
11. p
17. Take the surface to be z 0; 81p/4.
15. 112
Exercises 9.15, Page 572
1. 48
3. 36
7. 14 e2 12 e
9. 50
4
11.
2 2 (x>2)
5. p 2
4
1 "1 2 x 2 2y 2
39.
43.
47.
51.
55.
8
4
# # # dz dy dx
0 x3 0
4 2 8
(c)
59.
61.
63.
67.
71.
75.
79.
Exercises 9.16, Page 579
1.
3
2
3. 12a5p/5
9. 4p(b 2 a)
5. 256p
11. 128
7. 62p/5
13. p>2
(4, 2)
x
R
(6, –4)
###
0
8
(b)
4
7. 2v
9. 13 u2
R
3
###
0
y
5.
!y
x
dx dz dy
0
0
11. y
dy dx dz
x3
0
z
15.
37. (!3/2, 32 , 4)
0
(z 2 x)>2
0 0
2
13. (a)
33. k/30
x 2y
0 x 0
4 z (z 2 x)>2
0
(x y 4) dz dy dx
( !2, p/4, 9)
41. (2 !2, 2p/3, 2)
2
2
r z 25
45. r2 z2 1
2
2
zx y
49. x 5
(2p/3)(64 123/2)
53. 625p/2
(0, 0, 3a/8)
57. 8pk/3
1
2
( !3/3, 3 , 0); ( 3 , p/6, 0)
(4, 4, 4 !2); (4 !2, 3p/4, 4 !2)
(5 !2, p/2, 5p/4)
65. (!2, p/4, p/6)
r8
69. f p/6, f 5p/6
2
2
2
x y z 100
73. z 2
9p(2 !2)
77. 2p/9
(0, 0, 76 )
81. pk
0 2y 0
4 z>2 z 2 2y
0 0
4 4
27. x 0, y 2, z 0
82y
3. y
## #
# # # F(x, y, z) dx dz dy;
# # # F(x, y, z) dx dy dz;
# # # F(x, y, z) dy dz dx;
# # # F(x, y, z) dy dx dz
z 2 2y
#
8
3
Exercises 9.17, Page 585
1. (0, 0), (2, 8), (16, 20), (14, 28)
F(x, y, z) dz dy dx;
0 0
2 4
##
z
35. (10/ !2, 10/ !2, 5)
Exercises 9.13, Page 557
1.
7.
13.
19.
29.
"1 2 x 2
32
7,
31. 2560k/3; !80>9
15. (b a) (area of region bounded by C)
2
23. 16p
25. x 45, y R
x
(0, 0) is the image of every point on the boundary u 0.
15. 12
17. 14 (b 2 a) (d 2 c)
315
1
19. 2 (1 2 ln 2)
21. 4
23. 14 (e 2 e1)
d
25. 126
27. 52 (b 2 a) ln
29. 15p/2
c
y
13. 16
x
z
17.
z
19.
y
y
x
ANS-24
x
Answers to Selected Odd-Numbered Problems
Chapter 9 in Review, Page 586
1. true
3. true
5. false
9. false
11. false
13. true
15. true
17. true
7. true
y
x
i2 2
j
2 3>2
(x y )
(x y 2)3>2
21. v (1) 6 i j 2 k, v (4) 6 i j 8 k, a (t) 2 k
for all t
23. i 4 j (3p/4) k
19. =f 2
z
25.
Exercises 10.2, Page 609
1
1 t
1. X c1 a b e5t c2 a
be
2
1
2
1
2
5
3. X c1 a b e3t c2 a b et
5
2
1
4
5. X c1 a b e8t c2 a b e10t
y
1
x
31. 4px 3y 12z 4p 6 "3
1
##
##
0
y>2
"1 2 x 2 dy dx;
2
"1 2 x dx dy 35. 41k/1512
39. 6xy
37. 8p
41. 0
2
##
1
1
y>2
2
"1 2 x dx dy;
1
3
56 "2 p3/3
45. 12
2 2/3p
49. p2 /2
3/2
3/2
(ln 3)(17 5 )/12
4pc
55. 0
125p
59. 3p
5
61. 3
63. 0
65. p
43.
47.
51.
53.
57.
Exercises 10.1, Page 597
3 5
x
1. X a
b X, where X a b
4
8
y
3
4 9
x
3. X ° 6 1
0 ¢ X, where X ° y ¢
10
4
3
z
1 1
1
x
5. X ° 2
1 1 ¢ X, where X ° y ¢
1
1
1
z
dx
4x 2y et
7.
dt
dy
x 3y et
dt
dx
9.
x y 2z et 3t
dt
dy
3x 4y z 2et t
dt
dz
2x 5y 6z 2et t
dt
17. Yes; W(X1, X2) 2e8t 0 implies that X1 and X2
are linearly independent on (q, q).
19. No; W(X 1, X 2, X 3) 0 for every t. The solution
vectors are linearly dependent on (q, q). Note that
X3 2X1 X2.
11. X c1 °
1
2
1
1
1
0 ¢ et c2 ° 4 ¢ e3t c3 ° 1 ¢ e2t
1
3
3
9. X c1 °
2x
0 x
1 y
1
4
12
0 ¢ et c2 ° 6 ¢ et/2
1
5
4
c3 ° 2 ¢ e3t/2
1
1
1
0
1
13. X 3 a b et/2 2 a b et/2
15. (a) X9 a
3
100
1
50
1
100
bX
501
25 1 t>100
35 1 t>25
a be
a be
3
1
3 2
(d) approximately 34.3 minutes
(b) X 1
3
1
3
21. X c1 a b c2 c a bt a
1
1
1
1
1
1
1
4
bd
14
23. X c1 a b e2t c2 c a b te 2t a
13 2t
be d
0
1
25. X c1 ° 1 ¢ et c2 ° 1 ¢ e2t c3 ° 0 ¢ e2t
1
0
4
2
0 ¢ e5t
1
27. X c1 ° 5 ¢ c2 °
2
1
2
12
c3 £ ° 0 ¢ te 5t ° 12 ¢ e 5t §
1
1
0
0
0
t
t
29. X c1 ° 1 ¢ e c2 £ ° 1 ¢ te ° 1 ¢ e t §
1
1
0
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 10
0
27. (6x2 2y2 8xy)/ "40 29. 2; 2/ "2; 4
33.
2
7. X c1 ° 0 ¢ et c2 ° 3 ¢ e2t c3 ° 0 ¢ et
1
0
0
2
t2 t
t
c3 £ ° 1 ¢ e ° 1 ¢ te ° 0 ¢ e t §
2
1
0
0
2
1
31. X 7 a b e4t 13 a
2t 1 4t
be
t1
Answers to Selected Odd-Numbered Problems
ANS-25
33. Corresponding to the eigenvalue l1 ⫽ 2 of multiplicity
five, eigenvectors are
1
0
0
0
0
0
K1 ⫽ • 0μ , K2 ⫽ • 1μ , K3 ⫽ • 0μ .
0
0
1
0
0
0
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 10
35. X ⫽ c1 a
37. X ⫽ c1 a
39. X ⫽ c1 a
cos t
sin t
b e4t ⫹ c2 a
b e4t
2 cos t ⫹ sin t
2 sin t 2 cos t
cos t
sin t
b e4t ⫹ c2 a
b e4t
⫺ cos t 2 sin t
⫺ sin t ⫹ cos t
5 cos 3t
5 sin 3t
b ⫹ c2 a
b
4 cos 3t ⫹ 3 sin 3t
4 sin 3t 2 3 cos 3t
1
41. X ⫽ c1 ° 0 ¢ ⫹ c2 °
0
0
⫺ cos t
⫺ sin t
cos t ¢ ⫹ c3 ° ⫺ sin t ¢
sin t
⫺ cos t
sin t
3. X ⫽ a
⫺c1e⫺4t ⫹ c2e 2t
5. X ⫽ ° c1e⫺4t ⫹ c2e 2t ¢
c2e 2t ⫹ c3e 6t
c1e⫺t 2 c2e 2t 2 c3e 2t
7. X ⫽ °
c1e⫺t ⫹ c2e 2t
¢
⫺t
2t
c1e ⫹ c3e
c1e t ⫹ 2c2e 2t ⫹ 3c3e 3t
9. X ⫽ ° c1e t ⫹ 2c2e 2t ⫹ 4c3e 3t ¢
c1e t ⫹ 2c2e 2t ⫹ 5c3e 3t
m1 0
k ⫹ k2 ⫺k2
b and K ⫽ a 1
b
0 m2
⫺k2
⫺k2
Since M is a diagonal matrix with m1 and m2
nonzero, it has an inverse.
11. (a) M ⫽ a
k1 ⫹ k2
m1
(b) B ⫽ M⫺1K ⫽ ±
k2
⫺
m2
⫺ cos t
43. X ⫽ c1 ° 2 ¢ et ⫹ c2 ° cos t ¢ et ⫹ c3 ° ⫺ sin t ¢ et
1
cos t
⫺ sin t
25
cos 5t 2 5 sin 5t
t
47. X ⫽ ⫺ ° ⫺7 ¢ e ⫺ °
cos 5t
¢
6
cos 5t
5 cos 5t ⫹ sin 5t
sin 5t
¢
sin 5t
dx1
1
1
49. (a)
⫽ ⫺ x1 ⫹
x
dt
20
10 3
dx2
1
1
⫽
x1 2
x
dt
20
20 2
dx3
1
1
⫽
x2 2
x
dt
20
10 3
⫺cos 201 t ⫹ sin 201 t
(b) X ⫽ ⫺6 °
⫺sin 201 t
¢ e⫺t>10
1
cos 20 t
⫹ 6°
Exercises 10.4, Page 619
⫺1 ⫺t
⫺3
⫺1
1. X ⫽ c1 a
b e ⫹ c2 a b e t ⫹ a b
1
1
3
3. X ⫽ c1 a
1 ⫺2t
1
⫺1
b e ⫹ c2 a b e 4t ⫹ a 43 b t 2
⫺1
1
4
5. X ⫽ c1 a
55
1 3t
1
b e ⫹ c2 a b e 7t ⫹ a 36
b et
⫺3
9
⫺194
⫹ a
7.
9.
11.
⫺cos 201 t 2 sin 201 t
2
1
⫺t>10
2 2°
cos 20 t
¢e
⫹ 11 ° 2 ¢
sin 201 t
1
Exercises 10.3, Page 613
3c1e 7t 2 2c2e ⫺4t
1. X ⫽ a
b
3c1e 7t ⫹ 3c2e ⫺4t
ANS-26
Answers to Selected Odd-Numbered Problems
k2
m1
≤
k2
⫺
m2
⫺
1
1
(c) X ⫽ c1 a b cos t ⫹ c2 a b sin t
2
2
⫺2
⫺2
⫹ c3 a b cos "6t ⫹ c4 a b sin "6t
1
1
28
5 cos 3t
2t
45. X ⫽ ° ⫺5 ¢ e ⫹ c2 ° ⫺4 cos 3t 2 3 sin 3t ¢ e⫺2t
25
0
5 sin 3t
⫹ c3 ° ⫺4 sin 3t ⫹ 3 cos 3t ¢ e⫺2t
0
⫺2c1e t>2 ⫹ c2e 3t>2
b
⫺2c1e t>2 ⫹ 2c2e 3t>2
13.
1
4
bt
⫺14
⫹ a
⫺2
3b
4
3
1
1
1
2
t
2t
5t
X ⫽ c1 ° 0 ¢ e ⫹ c2 ° 1 ¢ e ⫹ c3 ° 2 ¢ e 2 ° 72 ¢ e 4t
0
0
2
2
1
⫺4
⫺9
X ⫽ 13 a b e t ⫹ 2 a b e 2t ⫹ a b
⫺1
6
6
3
1
⫺
0
100
(a) X9 ⫽ a 1001
1 bX ⫹ a b
⫺
1
50
25
80 1 ⫺t>50
10
70 ⫺1
a be
⫹ a b
(b) X ⫽ ⫺ a be⫺t>20 ⫹
3
2
3 1
30
(c) 10; 30; as t S q the total amount of salt in the system
of mixing tanks approaches a constant 40 lb
1
3
11
15
X ⫽ c1 a b ⫹ c2 a b e t 2 a b t 2 a b
1
2
11
10
2
1
15. X ⫽ c1 a b e t>2 ⫹ c2 a
13
15
10 3t>2
b e 2 a 132 b te t>2 2 a 29 b e t>2
3
4
4
2
1
1
1
3
3
4
2
17. X c1 a b e t c2 a b e 2t a b e t a b te t
4
1
19. X c1 a b e 3t c2 a
4
2 3t
12
be a
b t 2 a 34 b
1
0
3
1
2
1 t
t
21. X c1 a
b e c2 a 1
b e t a b e t
1
2
2 2 t
cos t
sin t
cos t
b c2 a
b a
bt
sin t
cos t
sin t
25. X c1 a
cos t t
sin t t
cos t
b e c2 a
be a
b te t
sin t
cos t
sin t
a
sin t
b ln Z cos tZ
cos t
cos t
sin t
cos t
27. X c1 a
b c2 a
b a
bt
sin t
cos t
sin t
a
sin t
sin t
b 2 a
b ln Z cos tZ
sin t tan t
cos t
a
cos t
2 cos t t
b e t ln Z sin tZ a
b e ln Z cos tZ
12 sin t
sin t
1
0
11. X c1 a
13. X °
15. e At a
1
3
i1
i2
6
3
4 19
a be 12t 2
a b cos t
29 1
29 42
35. a b 2 a be 2t 1
3
4 83
a b sin t
29 69
2
7
37. X c1 a bet c2 a be2t a
39. X c1 a
20
b
53
1
1
1
1
b c2 a be 10t a bt 2
1
1
2 1
1 41
41 1
a
bt 2
a b
10
100 1
39
Exercises 10.5, Page 625
et 0
et
At
1. eAt a
a
2t b ; e
0 e
0
t1
3. e ° t
2t
At
1
0
t
t1
2t
0
1
0
b
e2t
t
t
¢
2t 1
5. X c1 a be t c2 a be 2t
3 2t
3 2t
4e 2 4e
b;
1 2t
2 2e 32e2t
e e
3 2t
3 2t
e 2 1e2t
e 2 3e2t
X c1 a 2 2t 2 2t b c2 a 4 1 2t 4 3 2t b or
e e
2e 2e
X c3 a
3 2t
1
be c4 a be2t
2
2
X c1 a
1 3t 2t
9t
be c2 a
be 2t
t
1 2 3t
e 2t 3te 2t
te 2t
19. e At a 2
1 2t
2
2
be a bte 4t a be 4t
1
2
0
1b
2
cosh t
sinh t
1
b c2 a
b 2 a b
sinh t
cosh t
1
3 2t
1 2t
2e 2 2e
2t
2t
1
1
0
2t
31. X c1 ° 1 ¢ c2 ° 1 ¢ e c3 ° 0 ¢ e 3t
0
0
1
2
2
3
t1
t
t
t ¢ 2 4°t 1¢ 6°
t
¢
2t
2t
2t 1
17. e At a
33. X a bte 2t a
0
1
9. X c3 a b et c4 a b e2t a
2 sin t t
2 cos t t
3 sin t
29. X c1 a
b e c2 a
b e a3
b te t
cos t
sin t
2 cos t
14e 2t 12te 2t
° e t 14e 2t 12te 2t ¢
1 2 3t
2t e
t1
t
t
t ¢ c2 ° t 1 ¢ c3 °
t
¢
2t
2t
2t 1
4 t
5e
t
5e
15e 6t
2 25e 6t
4 t
9te 2t
b;
e 2t 2 3te 2t
2 t
5e
1 t
5e
2 25e 6t
b;
45e 6t
2 t
2 6t
e 15e 6t
5e 2 5e
b
c
a
2
2 6t
4 6t b or
1 t
t
5e 2 5e
5e 5e
X c1 a 25
2
1
X c3 a be t c4 a be 6t
1
2
21. e At a
e 3t
0
2et 2e 3t
b;
et
1
2et 2e 3t
X c1 a be 3t c2 a
b or
0
et
X c2 a
2 t
1
be c4 a be 3t
1
0
3 3t
2e
3t
2e
25. X c1 a 3
2 12e 5t
12e 3t 12e 5t
b or
3 5t b c2 a 1 3t
2 2e
2e 32e 5t
1
1
X c3 a b e 3t c4 a b e 5t
1
3
Chapter 10 in Review, Page 626
1
1
0
1. k 13 5. X c1 a b e t c2 c a b te t a b e t d
1
1
1
cos 2t t
sin 2t t
7. X c1 a
be c2 a
be
sin 2t
cos 2t
2
0
7
9. X c1 ° 3 ¢ e 2t c2 ° 1 ¢ e 4t c3 ° 12 ¢ e3t
1
1
16
1 2t
4 4t
16
11
11. X c1 a be c2 a be a
bt a b
0
1
4
1
Answers to Selected Odd-Numbered Problems
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 10
23. X c1 a
7. X c1 °
ANS-27
cos t sin t
sin t 2 cos t
b c2 a
b
2 cos t
2 sin t
1
sin t
2 a b a
b ln Z csc t 2 cot tZ
1
sin t cos t
1
1
1
15. (b) X c1 ° 1 ¢ c2 ° 0 ¢ c3 ° 1 ¢ e 3t
0
1
1
13. X c1 a
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 11
Exercises 11.1, Page 635
1. x y
y 9 sin x; critical points at (np, 0)
3. x y
y x 2 y(x 3 1); critical point at (0, 0)
5. x y
y Px3 x;
1
1
critical points at (0, 0), a
, 0b, a
, 0b
!P
!P
7. (0, 0) and (1, 1)
9. (0, 0) and (43 , 43 )
11. (0, 0), (10, 0), (0, 16), and (4, 12)
13. (0, y), y arbitrary
15. (0, 0), (0, 1), (0, 1), (1, 0), (1, 0)
17. (a) x c1e5t c2et
y 2c1e5t c2et
(b) x 2et
y 2et
19. (a) x c1(4 cos 3t 3 sin 3t) c2(4 sin 3t 3 cos 3t)
y c1(5 cos 3t) c2(5 sin 3t)
(b) x 4 cos 3t 3 sin 3t
y 5 cos 3t
21. (a) x c1(sin t cos t)e4t c2(sin t cos t)e4t
y 2c1(cos t)e4t 2c2(sin t)e4t
(b) x (sin t cos t)e4t
y 2(cos t)e4t
1
1
23. r 4
, u t c2; r 4 4
, u t;
!4t c1
!1024t 1
the solution spirals toward the origin as t increases.
25. r 1
"1 c1e2t
, u t c2; r 1, u t (or x cos t
and y sin t) is the solution that satisfies X(0) (1, 0);
1
r
, u t is the solution that satisfies
"1 2 34e2t
X(0) (2, 0). This solution spirals toward the circle
r 1 as t increases.
27. There are no critical points and therefore no periodic
solutions.
29. There appears to be a periodic solution enclosing the
critical point (0, 0).
Exercises 11.2, Page 642
1. (a) If X(0) X0 lies on the line y 2x, then X(t) approaches (0, 0) along this line. For all other initial
conditions, X(t) approaches (0, 0) from the direction determined by the line y x/2.
ANS-28
Answers to Selected Odd-Numbered Problems
3. (a) All solutions are unstable spirals that become un-
bounded as t increases.
5. (a) All solutions approach (0, 0) from the direction
specified by the line y x.
7. (a) If X(0) X0 lies on the line y 3x, then X(t) ap-
9.
13.
17.
19.
23.
25.
proaches (0, 0) along this line. For all other initial
conditions, X(t) becomes unbounded and y x
serves as the asymptote.
saddle point
11. saddle point
degenerate stable node 15. stable spiral
|µ| 1
µ 1 for a saddle point; 1 µ 3 for an unstable
spiral point
(a) (3, 4)
(b) unstable node or saddle point
(c) (0, 0) is a saddle point.
(a) (12 , 2)
(b) unstable spiral point
(c) (0, 0) is an unstable spiral point.
Exercises 11.3, Page 650
1. r r0e at
3. x 0 is unstable; x n 1 is asymptotically stable.
5. T T0 is unstable.
7. x a is unstable; x b is asymptotically stable.
9. P a/b is asymptotically stable; P c is unstable.
11. (12 , 1) is a stable spiral point.
13. ( !2, 0) and (!2, 0) are saddle points; (12 , 74 ) is a
stable spiral point.
15. (1, 1) is a stable node; (1, 1) is a saddle point; (2, 2)
is a saddle point; (2, 2) is an unstable spiral point.
17. (0, 1) is a saddle point; (0, 0) is unclassified; (0, 1)
is stable but we are unable to classify further.
19. (0, 0) is an unstable node; (10, 0) is a saddle point;
(0, 16) is a saddle point; (4, 12) is a stable node.
21. u 0 is a saddle point. It is not possible to classify
either u p/3 or u p/3.
23. It is not possible to classify x 0.
25. It is not possible to classify x 0, but x 1/ !e and
x 1/ !e are each saddle points.
29. (a) (0, 0) is a stable spiral point.
33. (a) (1, 0), (1, 0)
35. |v0| 12 !2
37. If b 0, (0, 0) is the only critical point and is stable.
If b 0, (0, 0), (x̂, 0), and (x̂, 0), where x̂2 a/b,
are critical points. (0, 0) is stable, while (x̂, 0) and
(x̂, 0) are each saddle points.
39. (b) (5p/6, 0) is a saddle point.
(c) (p/6, 0) is a center.
Exercises 11.4, Page 657
1. |v0| , !3g>L
1 x2
b.
1 x 02
9. (a) The new critical point is (d/c P2/c, a/b P1/b).
(b) yes
11. (0, 0) is an unstable node, (0, 100) is a stable node,
(50, 0) is a stable node, and (20, 40) is a saddle point.
17. (a) (0, 0) is the only critical point.
5. (a) First show that y2 v 20 g ln a
0(dQ)
0(dP)
r
.
0x
0y
Kx
15. If n (2x, 2y), show that V n 2(x y)2 2y4.
17. Yes; the sole critical point (0, 0) lies outside the invariant
region 161 x2 y2 1, and so Theorem 11.5.5(ii) applies.
19. V n 2y 2 (1 x 2 )
2y2 (1 r2) and P/ x 2
Q/ y x 1 0. The sole critical point is (0, 0)
and this critical point is a stable spiral point. Therefore,
Theorem 11.5.6(ii) applies.
0Q
0P
21. (a)
2xy 1 x2 2x 1 x2
0x
0y
(x 1)2 0
(b) limtS X(t) (32 , 29 ), a stable spiral point
Chapter 11 in Review, Page 667
1. true
3. a center or a saddle point
5. false
7. false
9. true
1
11. r 3
, u t; the solution curve spirals toward
!3t 1
the origin.
13. center; degenerate stable node
15. stable node for µ 2; stable spiral point for
2 µ 0; unstable spiral point for 0 µ 2;
unstable node for µ 2
17. Show that y2 (1 x02 x2)2 1.
0Q
0P
19.
1
0x
0y
21. (a) Hint: Use the Bendixson negative criterion.
(d) In (b), (0, 0) is a stable spiral point when b 2ml
!g>l 2 v2 . In (c), (x̂, 0) and (x̂, 0) are stable
spiral points when b 2ml "v2 2 g 2>(v2l 2).
Exercises 12.1, Page 676
7. !p>2
9. !p>2
p
np
xg 11. i1i !p; g cos
p
Ä 2
21. T 1
23. T 2p
25. T 2p
Exercises 12.2, Page 681
1
1 q 1 2 (1)n
sin nx;
1. f (x) p na
n
2
1
converges to 12 at x 0
3. f (x) q (1)n 2 1
3
1
sin npxf ;
ae
cos npx 2
np
4 n1
n2p2
converges to 12 at x 0
5. f (x) q
2(1)n
p2
ae
cos nx
6
n2
n1
a
(1)n 1p
2
f(1)n 2 1g b sin nxr
n
pn3
7. f (x) p 2 a
q
n1
9. f (x) (1)n 1
sin nx
n
1
1 q (1)n 1
1
cos nx
sin x a
p
p n 2 1 2 n2
2
11. f (x) 1
1 q
1
np
np
a b sin
cos
x
pn1
n
4
2
2
3
np
np
a1 2 cos
b sin
xr;
n
2
2
converges to 1 at x 1, 12 at x 0,
and 12 at x 1
13. f (x) q
(1)n 2 1
9
np
5a b
cos
x
2 2
4
5
np
n1
15. f (x) (1)n 1
np
sin
xr
np
5
q (1)n
2 sinh p 1
(cos nx 2 n sin nx)d
c a
2
p
2
n1 1 n
Exercises 12.3, Page 687
1. odd
3. neither even nor odd
5. even
7. odd
9. neither even nor odd
11. f (x) 2 a
q
n1
13. f (x) (1)n 2 1
sin npx
n
p
2 q (1)n 2 1
cos nx
p na
2
n2
1
15. f (x) 1
4 q (1)n
cos npx
2 a
3
p n 1 n2
17. f (x) q (1)n 1
2p2
4a
cos nx
3
n2
n1
19. f (x) 2 q 1 2 (1)n(1 p)
sin nx
p na
n
1
3
4 q
21. f (x) 2 a
4
p n1
23. f (x) ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 12
Exercises 11.5, Page 665
1. The system has no critical points.
0Q
0P
3.
2 , 0
0x
0y
0Q
0P
5.
µ 9y2 0 if µ 0
0x
0y
7. The single critical point (0, 0) is a saddle point.
9. d(x, y) ey/2
0Q
0P
11.
4(1 x2 3y2) 0 for x2 3y2 1
0x
0y
13. Use d(x, y) 1/(xy) and show that
cos
np
21
2
np
cos
x
2
2
n
2
2 q 1 (1)n
cos nx
p
p na
1 2 n2
2
Answers to Selected Odd-Numbered Problems
ANS-29
1
2 q
25. f (x) p na
2
1
1 2 cos
2 q
f (x) p na
1
27. f (x) f (x) ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 12
29. f (x) f (x) 31. f (x) np
2
cos npx
n
sin
np
2
n
sin npx
2
4 q (1)n
cos 2nx
2
p
p na
1 1 2 4n
8 q
n
sin 2nx
2
p na
4n
21
1
p
2
p na
4
1
q
np
2
sin
4
p na
1
q
2 cos
n2
3
4
2 a
4
p n1
(1)
f (x) 4 a e
np
n1
n1
np
21
2
np
cos
x
2
n2
n
(1) 2 1
f sin npx
n3p3
4p
1
p
35. f (x) 4 a e 2 cos nx 2 sin nx f
n
3
n
n1
2
37. f (x) q
3
1 q 1
sin 2npx
2
p na
2
1 n
10 q 1 2 (1)
sin nt
2
p na
1 n(10 2 n )
n
43. xp(t) 45. xp(t) q
p2
1
cos nt
16 a 2 2
18
n
(n
2 48)
n1
47. (a) x(t) 1
10 q 1 2 (1)n 1
e sin nt 2
sin!10tf
2
p na
n
!10
1 10 2 n
49. (b) y(x) 2w0 L4 q (1)n 1
np
sin
x
5
L
EIp5 na
n
1
Exercises 12.4, Page 691
1. f (x) 3. f (x) a
q
n q, n20
1
4
5. f (x) p ANS-30
1 2 (1)n inpx>2
e
npi
a
1 2 e inp>2 2inpx
e
2npi
a
i inx
e
n
q
n q, n20
q
n q, n20
exL m(x) L n(x) dx 0, m 2 n
11. (a) ln 16n2, yn sin (4n tan1 x), n 1, 2, 3, . . .
5
2 q 3(1)n 2 1
2 a
cos npx
6
p n1
n2
q
q
0
q
4
np
2
np
f (x) a e 2 2 sin
2
(1)n f sin x
np
2
2
n1 n p
33. f (x) #
#
sin nx
cos
q
np
2 (1)n 2 1
2
cos nx
n2
Exercises 12.5, Page 697
1. y cos an x; a defined by cot a a;
l1 0.7402, l2 11.7349,
l3 41.4388, l4 90.8082
y1 cos 0.8603x, y2 cos 3.4256x,
y3 cos 6.4373x, y4 cos 9.5293x
5. 12 f1 sin2ang
np 2
np
7. (a) ln a
b , yn sin a
ln xb, n 1, 2, 3, . . .
ln 5
ln 5
d
l
(b)
fxy9g y 0
x
dx
5
1
mp
np
(c)
sin a
ln xb sin a
ln xb dx 0, m 2 n
x
ln 5
ln 5
1
d
9.
fxe x y9g nexy 0;
dx
Answers to Selected Odd-Numbered Problems
(b)
#
1
1
sin (4m tan1x) sin (4n tan1x) dx 0,
2
0 1 x
mn
Exercises 12.6, Page 703
1. a1 1.277, a2 2.339, a3 3.391, a4 4.441
q
1
3. f (x) a
J0(ai x)
i 1 ai J1(2ai)
J0(ai x)
5. f (x) 4 a
2
2
i 1 (4ai 1) J 0 (2ai)
q
ai J1(2ai)
7. f (x) 20 a
q
ai J2 (4ai)
J1(ai x)
2
2
(2a
i 1
i 1) J 1 (4ai)
q
J2 (3ai)
9
J0(ai x)
9. f (x) 2 4 a 2 2
2
i 1 ai J 0 (3ai)
5
15. f (x) 14 P0(x) 12 P1(x) 16
P2(x) 323 P4(x) …
3
21. f (x) 12 P0(x) 58 P2(x) 16
P4(x) …,
f (x) |x| on (1, 1)
Chapter 12 in Review, Page 704
1. true
3. cosine
5. false
7. 5.5, 1, 0
9. true
1
2 q
1
13. f (x) a e 2 f(1)n 2 1g cos npx
pn1 n p
2
2
(1)n sin npxr
n
q 1 2 (1)ne1
15. f (x) 1 2 e1 2 a
cos npx;
1 n2p2
n1
f (x) a
q
n1
2npf1 2 (1)ne1g
sin npx
1 n2p2
(2n 2 1)2p2
, n 1, 2, 3, p ,
36
2n 2 1
yn cos a
p ln xb
2
17. ln 19. p(x) 1
2
, the interval (1, 1),
"1 2 x
1
Tm(x) Tn(x) dx 0, m 2 n
2
1 "1 2 x
1 q J1(2ai)
J (a x)
21. f (x) a
4 i 1 ai J 21(4ai) 0 i
#
1
0u
2
0, u(x, 2) 0, 0 , x , 4
0y y 0
Exercises 13.3, Page 718
np
1 k(n2p2>L2)t
2 q cos
np
2
¢e
x
1. u(x, t) °
sin
a
p n1
L
n
2
c4 sinh "1 2 a y).
(ii) For a2 1,
u (c1 cosh ax c2 sinh ax)(c3 cos "a2 2 1y
c4 sin "a2 2 1y).
(iii) For a2 1,
u (c1 cosh x c2 sinh x)(c3y c4).
The results for the case l a2 are similar.
For l 0 :
u (c1x c2) (c3 cosh y c 4 sinh y)
17. elliptic
19. parabolic
21. hyperbolic
23. parabolic
25. hyperbolic
Exercises 13.2, Page 716
2
0u
0u
, 0 , x , L, t . 0
0t
0x 2
0u
u(0, t) 0,
2
0, t . 0
0x x L
1. k
u(x, 0) f (x), 0 x L
0 2u
0u
3. k 2 , 0 , x , L, t . 0
0t
0x
0u
u(0, t) 100,
2
hu (L, t), t . 0
0x x L
u(x, 0) f (x), 0 x L
0 2u
0u
5. k 2 2 hu , 0 , x , L, t . 0, h a constant
0t
0x
u(0, t) sin(pt/L), u(L, t) 0, t 0
u(x, 0) f (x), 0 x L
L
# f (x) dx
1
3. u(x, t) L
2 q
a
L na
1
0
L
# f (x) cos
0
5. u(x, t) eht c
2 q
a
L na
1
1
L
np
np
2 2 2
x dxbek(n p >L )t cos
x
L
L
L
# f (x) dx
0
L
# f (x) cos
0
np
np
2 2 2
x dxb ek(n p >L )t cos
xd
L
L
7. u(x, t) A0 a ek(np>L) t aAn cos
q
2
k1
where A0 An 1
L
#
1
2L
#
L
f (x) dx,
L
L
f (x) cos
L
np
np
x Bn sin xb ,
L
L
np
1
dx, Bn L
L
#
L
f (x) sin
L
np
dx
L
Exercises 13.4, Page 722
L2 q 1 2 (1)n
npa
np
1. u(x, t) 3 a
cos
t sin
x
3
L
L
p n1
n
1
3. u(x, t) sin at sin x
a q
1 2 (1)n
4
5. u(x, t) 3 a c
cos npat
p n1
n3
1 2 (1)n
sin npatd sin npx
n4pa
np
sin
8h q
2
npa
np
7. u(x, t) 2 a
cos
t sin
x
2
L
L
p n1 n
Answers to Selected Odd-Numbered Problems
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 13
Exercises 13.1, Page 711
1. The possible cases can be summarized in one form
u c1e c2(x y), where c1 and c2 are constants.
3. u c1e y c2(x 2 y)
5. u c1(xy)c2
7. not separable
2
2
9. u et(A1e ka t cosh ax B1e ka t sinh ax)
2
2
u et(A2eka t cos ax B2eka t sin ax)
u et(A3x B3)
11. u (c1 cosh ax c2 sinh ax)(c3 cosh aat c4 sinh aat)
u (c5 cos ax c6 sin ax)(c7 cos aat c8 sin aat)
u (c9x c10)(c11t c12)
13. u (c1 cosh ax c2 sinh ax)(c3 cos ay c4 sin ay)
u (c5 cos ax c6 sin ax)(c7 cosh ay c8 sinh ay)
u (c9x c10)(c11y c12)
15. For l a2 0 there are three possibilities:
(i) For 0 a2 1,
u (c1 cosh ax c2 sinh ax) (c3 cosh"1 2 a2y
0 2u
0 2u
2 , 0 , x , L, t . 0
2
0x
0t
u(0, t) 0, u(L, t) 0, t 0
0u
u(x, 0) x (L 2 x),
2
0, 0 , x , L
0t t 0
0 2u
0u
0 2u
9. a 2 2 2 2b
2 , 0 , x , L, t . 0
0t
0x
0t
u(0, t) 0, u(L, t) sin pt, t . 0
0u
u(x, 0) f (x),
2
0, 0 , x , L
0t t 0
0 2u
0 2u
11.
0, 0 , x , 4, 0 , y , 2
0x 2
0y 2
0u
2
0, u(4, y) f (y), 0 , y , 2
0x x 0
7. a 2
ANS-31
6h q 1 2 (1)n
np
npa
np
9. u(x, t) 2 a
sin
cos
t sin
x
2
3
L
L
p n1
n
L
2L q (1)n 2 1
npa
np
11. u(x, t) 2 a
cos
t cos
x
2
2
L
L
p n1
n
13. u(x, y) a aAn cosh
y Bn sinh
yb sin
x,
a
a
a
n1
q
where An 13. u(L>2, t) 0 for t $ 0
Bn 15. u(x, t) e bt a An e cos qnt sin qnt f sin nx,
qn
n1
q
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 13
#
b
p
2
where An f (x) sin nx dx and qn "n2 2 b2
p 0
19. u(x, t) t sin x cos at
1
21. u(x, t) sin 2x sin 2at
2a
q
n2p2
23. u(x, t) a aAn cos
at
L2
n1
2
L
L
# f (x) sin L
np
#
np
g(x) sin
x dx
L
2
°
a na
1
q
#
a
0
np
f (x) sin
x dx ¢
a
np
np
y sin
x
a
a
1
a
# f (x) sin
np
x dx ¢
a
2 q f1 2 (1)ng
p na
n
1
n cosh nx 1 sinh nx
3
sin ny
n cosh np 1 sinh np
9. u(x, y) a (An cosh npy Bn sinh npy) sin npx,
q
n1
1 2 (1)n
np
f1 2 (1)ng f2 2 cosh npg
Bn 200
np
sinh np
where An 200
ANS-32
0
#
np
x dx
a
15. u u1 u2 where
u1(x, y) u2(x, y) 2 q 1 2 (1)n
sinh ny sin nx
p na
1 n sinh np
2 q 1 2 (1)n
p na
n
1
3
sinh nx sinh n(p 2 x)
sin ny
sinh np
200 q (1)n 2 1 kn2p2t
e
sin npx
p na
n
1
q
u0
r
r
3 3d
x (x 2 1) 2 a c
np
2k
kn
p
n1
2
3 f(1)n 2 1g ekn p t sin npx
5. u(x, t) c(x) a Anekn p t sin npx,
q
2
2
n1
A
where c(x) 2 febx (eb 2 1) x 1g
kb
1
np 0
sinh
b
a
np
np
3 sinh
(b 2 y) sin
x
a
a
1
2 q 1 2 (1)n
5. u(x, y) x 2 a 2
sinh npx cos npy
2
p n 1 n sinh np
2 q
a
p na
1
# g(x) sin
np
b≤
a
2 An cosh
3. u(x, t) u0 1
np
sinh
b
a
3 sinh
11. u(x, y) a
2
2 q
1. u(x, y) a °
a n1
7. u(x, y) np
x dx
a
2
1
a
a
np
sinh
b
a
1. u(x, t) 100 Exercises 13.5, Page 728
3. u(x, y) 0
np
Exercises 13.6, Page 736
L
0
# f (x) sin
np
17. max temperature is u 1
x dx
0
2L
Bn 2 2
npa
a
n2p2
np
at≤ sin
x,
L
L2
Bn sin
where An 2
a
np
p
f (x) sin nx dxb eny sin nx
0
Answers to Selected Odd-Numbered Problems
An 2
7. c(x) u0 c1 2
9. u(x, t) # f f (x) 2 c(x)g sin npx dx
0
sinh !h>k x
sinh !h>k
A
(x x3)
6a 2
d
2A q (1)n
cos npat sin npx
a 2p3 na
n3
1
11. u(x, y) (u0 u1) y u1
2 q u0(1)n 2 u1 npx
e
sin npy
p na
n
1
13. u(x, t) (1 2 x) sin t
2 q n2p2e n p t 2 n2p2 cos t 2 sin t
d sin npx
ac
p n51
n(n4p4 1)
2
1
2
b
15. u(x, t) x sin t 2 a c a 2 2 2 2 2 2
np
n p (n p 2 1)
n51
q
3 sin npt 2
(1)n
(1)n
(1)n
sin td sin npx
np(n2p2 2 1)
17. u(x, t) 2 a
e 3t sin nx
2
n(n
2
3)
n51
Chapter 13 in Review, Page 744
(1)n11
q
1 2a
1. u c1e (c2x y>c2)
n
q
(1)
2
e n t sin nx
2
n51 n(n 2 3)
c 2 2 (1)n
19. u(x, t) a
np
np
n51
q
n2p2 cos t sin t
d
n4p4 1
1
2
3 sin npx a c
q
n51
3. c(x) u0 n
4 2 2(1)
2np
d
2 (1)n 4 4
3 3
np
np 1
sin npx
Exercises 13.7, Page 740
q
sin an
2
1. u(x, t) 2h a
ekan t cos anx, where
2
n 1 (h sin an)
the an are the consecutive positive roots of cot a a/h
3. u(x, y) a An sinh any sin anx, where
q
a
# f (x) sin a x dx
n
0
and the an are the consecutive positive roots of
tan aa a/h
5. u(x, t) a Anek(2n 2 1) p t>4L sin a
q
2
2
2
n1
2
where An L
7. u(x, y) #
L
0
f (x) sin a
4u0 q
p na
1
3 cosh a
2n 2 1
b px,
2L
1. u(x, y, t) a a Amne k(m
2n 2 1
bp
2
2
n2)t
sin mx sin ny,
4u0
[1 (1)m][1 (1)n]
mnp2
16
[(1)m 1][(1)n 1]
m n p2
sin "n2 1tg sin nx
15. u(x, t) u0 12(u1 2 u0)x
q
cos an
2
ea ntsin anx
2(u1 2 u0) a
2
n 1 an(1 cos an)
17. u(x, y) x 2 px q
q
mp
2
np
b
##
0
a
0
sin h n(p 2 y) sin h ny
bsin nx
sin h np
q0a 4b 4
py
px
sin
sin
4
2
2 2
a
b
p D(a b )
Exercises 14.1, Page 751
u0 q 1 2 (1)n n
u0
a
1. u(r, u) r sin nu
p n1
n
2
5. u(r, u) q
2p2
1
2 4 a 2 r n cos nu
3
n1 n
u0
2u0 q 1
np r n
sina
b a b cos nu
p na
2
2
2
1n
7. u(r, u) a Anr 2n sin 2nu,
n51
where
An 2
where vmn "(mp>a) (np>b) and
4
ab sinh (cvmn)
4 q f(1)n 2 1g
a
p n51
n3
q
3 3
5. u(x, y, z) a a Amn sinh vmnz sin
y,
x sin
a
b
m 1 n 1
Amn
n1
q
m 1 n 1
where Amn 13. u(x, t) e(x t) a An f "n2 1 cos "n2 1t
3. u(r, u) 3. u(x, y, t) a a Amn sin mx sin ny cos a "m2 n2t,
q
100 q 1 2 (1)n nx
e sin ny
p na
n
1
11. (a) u(x, t) et sin x
19. w(x, y) m 1 n 1
where Amn 9. u(x, y) 100 q 1 2 (1)n
sinh nx sin ny
p na
1 n sinh np
1
2n 2 1
2n 2 1
b px sin a
bpy
2
2
q
7. u(x, y) 3 a
Exercises 13.8, Page 744
q
np
3np
2 cos
4
4
¢
n2
3 sin npat sin npx
2n 2 1
b px dx
2L
(2n 2 1) cosh a
cos
q
n1
2h
An sinh anb (ah cos 2ana)
2h
°
p2a na
1
q
mp
np
y dx dy
f (x, y) sin
x sin
a
b
7. Use a b c 1 with f (x, y) u0 in Problem 5 and
f (x, y) u0 in Problem 6. Add the two solutions.
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 14
3e
n2p2t
5. u(x, t) (u1 2 u0)
x
1p
2
pc n
#
p>2
f (u) sin 2nu du
0
9. u(r, u) a Anr np>bsin
u,
b
n51
q
where
An np
2
bc np>b
b
# f (u) sin b u du
np
0
Answers to Selected Odd-Numbered Problems
ANS-33
9. u(r, t) a An J0(anr)ekant ,
q
r
b
11. u(r, u) A0 ln a b
b
r
a c a b 2 a b d (An cos nu Bn sin nu),
r
b
n1
n
q
where
n
n
1
a
A0 ln a b b
2p
n
b
a
1
c a b 2 a b d An p
a
b
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 14
b n
a n
1
c a b 2 a b d Bn p
a
b
13. u(r, u) A0 a
#
2p
f (u) du
0
2p
#
f (u) cos nu du
0
2p
#
f (u) sin nu du
0
(r 2n a 2n)
(An cos nu Bn sin nu),
rn
n1
q
A0 where
1
2p
#
2p
f (u) du
0
2p
(b 2n a 2n)
1
An n
p
b
#
f (u) cos nu du
(b 2n a 2n)
1
Bn p
bn
#
f (u) sin nu du
0
2n
2n
n
17. u(r, u) A0ln r a An(r 2n 2 r 2n) cos 2nu,
q
n51
where
An 2
A0 pln 2
#
p>2
#
p>2
f (u) du,
0
4
p(2 2 22n)
2n
where An f (u) cos 2nu du
0
n1
2p
1
2p
n
#
c
An p
#
c
Bn p
#
f (u) du
0
2p
f (u) cos nu du
0
n 2p
0
0
n
n1
1
2a2n
rJ0(anr )f (r)dr
(a2n h 2)J 20(an) 0
q
J1(an)
2
J0(anr)ean t
13. u(r, t) 100 50 a
2
a
J
(2a
)
n1 n 1
n
#
where An 15. (b) u(x, t) a An cos(an !gt) J0(2an !x),
q
n1
#
!L
2
vJ0(2anv) f (v 2)dv
2
L J 1(2an !L) 0
q
sinh anz
17. u(r, z) 16 a 3
J1(anr)
n51 anJ2(an)sinh an
where An 1
3 r
P0( cos u) a b P1( cos u)
2
4 c
7 r 3
11 r 5
a b P3( cos u) a b P ( cos u) p R
16 c
32 c 5
r
3. u(r, u) cos u
c
1. u(r, u) 50 c
5. u (r, u) a An
q
where
b
f (u) sin nu du
2 a 2n 1
An
b 2n 1a n 1
2a
(1)n 2 1
I0 (npr) cos npz
2 2
n 1 n p I0 (np)
2n 1
2
p
# f (u)P ( cos u) sin u du
n
0
7. u(r, u) a A2nr 2nP2n( cos u),
q
n0
4n 1
where A2n c 2n
9. u(r, t) 100 0
Exercises 14.2, Page 758
2 q sin anat
J0(anr)
1. u(r, t) 2
ac na
1 an J1(anc)
q
sinh an(4 2 z)
3. u(r, z) u0 a
J0(anr)
n 1 an sinh 4an J1(2an)
b 2n 1 2 r 2n 1
Pn( cos u),
b 2n 1r n 1
2n 1
11. u(r, t) 1
2
# rJ (a r) f (r)dr
2
n1
19. u(r, u) A0 a r n(An cos nu Bn sin nu),
where A0 c
2
c 2J 21(anc)
11. u(r, t) a An J0(anr)ekan t,
q
q
5. u (r, z) #
p>2
0
f (u)P2n( cos u) sin u du
200 q (1)n n2p2t
e
sin npr
pr na
n
1
1 q
npa
npa
np
aAn cos
t Bn sin
tb sin
r,
r na
c
c
c
1
where An Bn #
c
2
np
r dr,
rf (r) sin
c 0
c
#
c
2
np
r dr
rg(r) sin
npa 0
c
q
2n 2 1
prb
4u0
2
2n 2 1
7. u (r, z) sin
pz
p na
2
2n
2
1
1
(2n 2 1)I0 a
pb
2
q
ANS-34
n1
Exercises 14.3, Page 761
4 q 1 2 (1) r 2 b
a
a b sin nu
a
3
2n
2n
pn1
r
n
a 2b
n
15. u(r, u) 0
2p
2
I0 a
Answers to Selected Odd-Numbered Problems
Chapter 14 in Review, Page 763
2u0 q 1 2 (1)n r n
a b sin nu
1. u(r, u) p na
n
c
1
3. u (r, u) 4u0 q 1 2 (1)n n
r sin nu
p na
n3
1
5. u (r, u) A0 a Anr n cos nu,
q
13. u(x, t) u0 c1 2 e erfc a
n1
where A0 1
p
p
# f (u) du
2 e x terfc a !t 0
2
An pc n
x
15. u(x, t) 2!p
p
# f (u) cos nu du
0
2u0 q r 4n r 4n 1 2 (1)n
sin 4nu
4n
p na
n
24n
1 2
7. u(r, u) q
1
2
J (a r)e an t
an J1(an) 0 n
n1
cosh anz
13. u(r, z) 50 a
J0(anr)
n 1 an cosh 4an J1(2an)
q
3
7
15. u (r, u) 100 c rP1( cos u) 2 r 3P3( cos u)
2
8
11 5
r P5( cos u) p R
16
21. u (r, z) 100 200 a
J0(anr)
n 1 an cosh an J1(an)
q
eanz
q
1. (a) Let t u2 in the integral erf ( !t).
9. y(t) ept erf( !pt)
b
11. Use the property
a
b
# # # #
0
2
0
0
a
1
x 2
x
g at 2 b A sin vat 2 b d
a
a
2
x
1
3 8at 2 b 2 gt 2
a
2
q
F0
2nL L 2 x
7. u(x, t) a
(1)n e at 2
b
a
a
E n0
2 at 2
2nL L 2 x
b
a
2nL L x
2nL L x
b 8at 2
br
a
a
9. u(x, t) (t x) sinh (t x) 8(t x)
xex cosh t ext sinh t
x
11. u(x, t) u1 (u0 u1) erfc a
b
2!t
x
b ᐁ(t 2)
2!t 2 2
erfc a
12x
bd
2!t
erfc a
2n 1 x
bd
2!kt
12x
b
2!t
p
xb
L
q
2n 1 2 x
23. u(x, t) u0 u0 a (1)n cerfc a
b
2!kt
n0
21. u(x, t) u0 u0e(p >L )t sin a
2
25. u(x, t) u0e Gt>Cerf a
2
x
RC
b
2Ä t
100
r21
erfc a
b
r
2!t
29. u(x, t) u0 erfc a
x
2"kt
b; u0
Exercises 15.3, Page 781
3.
q
sin a cos ax 3(1 2 cos a) sin ax
#
a
1
f (x) # fA(a) cos ax B(a) sin axg da,
p
1. f (x) 1
p
da
0
q
0
x
x
3. u(x, t) f at 2 b 8at 2 b
a
a
3 8at 2
t
0
2
e x >4tdt
0
Exercises 15.2, Page 774
apt
px
1. u(x, t) A cos
sin
L
L
5. u(x, t) c
3>2
19. u(x, t) 100 ce 1 2 x terfc a !t 27. u (r, t) Exercises 15.1, Page 769
f (t 2 t)
17. u(x, t) 60 40 erfc a
cosh anz
23. u(r, z) 200 a
J0(anr)
n 1 anJ1(an)
#
t
x
brd
2!t
3a sin 3a cos 3a 2 1
a2
sin 3a 2 3a cos 3a
B(a) a2
where A(a) 7.
q
#
10
f (x) p #
2
f (x) #
p
4
f (x) #
p
2k
f (x) p #
2
f (x) #
p
1
5. f (x) p
0
q
0
q
9.
0
q
11.
0
13.
q
0
q
0
cos ax a sin ax
da
1 a2
(1 2 cos a) sin ax
da
a
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 15
11. u(r, t) 2e ht a
x
b
2!t
(pa sin pa cos pa 2 1) cos ax
da
a2
a sin ax
da
4 a4
cos ax
da
k 2 a2
a sin ax
da
k 2 a2
Answers to Selected Odd-Numbered Problems
ANS-35
15. f (x) 2
p
f (x) 8
p
17. f (x) #
q
#
q
Exercises 15.5, Page 797
(4 2 a2) cos ax
da
(4 a2)2
0
3. 1
7.
a sin ax
da
(4 a2)2
1
1
1
2
1
, x.0
p 1 x2
1
"2
"2
2 1i 2
0
19. Let x 2 in (7). Use a trigonometric identity and
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 16
replace a by x. In part (b) make the change of
variable 2x kt.
Exercises 15.4, Page 786
1
p
1. u(x, t) #
q
2
5. u(x, t) p
2
7. u(x, t) p
#
q
#
#
q
#
0
1 2 eka t
sin ax da
a
1 2 cos a ka2t
sin ax da
e
a
sin a ka2t
e
cos ax da
a
0
¢ F(a) cos aat G(a)
#
11. u(x, y) 2
p
13. u(x, y) 100
p
15. u(x, y) 17. u(x, y) 2
p
2
p
q
sinh a(p 2 x)
cos ay da
(1 a2) sinh ap
0
#
q
sin a ay
e cos ax da
a
0
#
q
#
q
F(a)
0
0
sin aat iax
da
≤e
aa
sinh a(2 2 y)
sin ax da
sinh 2a
a
feax sin ay eay sin axg da
1 a2
1
2
19. u(x, t) ex >(1 4kt)
!1 4kt
1
2!p
1
2!p
2u0
25. u(r, z) p
"2
"2
2 1i 2
1
"2
1 "2
2 i 2
"2
"2
2 i 2
1
#
q
0
#
q
q
#
q
q
1
1
i
i
1
"2
"2
2 i 2
1
1
1
1
i
1
1
i
1
1
i
i
1
"2
"2
2 i 2
1
"2
"2
2 1i 2
i
1
"2
"2
2 i 2
i
"2
i "2
2 i 2
1
1
2
e a >4 cosh ay iax
e
da
cosh a
2
e a >4 cosh ay
cos ax da
cosh a
I0(ar)
sin a cos az da
aI0(a)
Answers to Selected Odd-Numbered Problems
2
p
1
i
i
"2
"2
2 1i 2
q
#
# erfc a 2!t b dt
u
sin a(p 2 x) sin ax
u(x, t) e
a
2p #
7.
x
0
q
0
ka2t
da
q
9. u(x, y) #
q
0
a
1 2 cos a
b fe ax sin ay 2eay sin axg da
a
q
#
1
u (x, t) 2p #
11. u (x, y) 2
p
0
q
a
B cosh ay
A
b sin ax da
2
a
(1 a ) sinh ap
cos ax a sin ax ka2t
e
da
1 a2
q
17. u (x, t) 1 e4t sin 2x
13.
19. u(x, y) 2
p
#
q
F(a) sinh"a2 h y
sinh"a2 h p
0
where F(a) #
sin ax da,
q
f (x) sin ax dx
0
x
t x2>4t
e
2 x erfc a
b
p
Å
2!t
21. u (x, t) 2
or
1
u(x, t) !p
t x2>4(t2 t)
# !t 2 t dt
e
0
Exercises 16.1, Page 806
11
π
"2
i "2
2 1i 2
"2
"2
"2
i "2
2 i 2 1 2 1i 2
t
5. u(x, t) 100
p
q
q
"2
1 "2
2 1i 2 i
1
1
"2
"2
2 i 2
sinh ay
cos ax da
a(1
a2) cosh ap
0
x
3. u(x, t) u0ehterf a
b
2!t
9. (a) u(x, t) 1
2p
1
1. u(x, y) 2
q
q
i
i
1
Chapter 15 in Review, Page 798
cos ax ka2t
e
da
1 a2
q
2u0
p
3. u(x, t) eka t iax
e
da
1 a2
q
#
1
"2
"2
2 1i 2
2
q
1
p
21. u(x, y) ANS-36
F8 ©
1
14
1. u11 15 , u21 15
3. u11 u21 !3/16, u22 u12 3 !3/16
5. u21 u12 12.50, u31 u13 18.75, u32 u23 37.50,
u11 6.25, u22 25.00, u33 56.25
7. (b) u 14 u 41 0.5427, u 24 u 42 0.6707,
u34 u43 0.6402, u33 0.9451, u44 0.4451
Exercises 16.2, Page 811
The tables in this section give a selection of the total number
of approximations.
1.
Time x ⴝ 0.25 x ⴝ 0.50 x ⴝ 0.75 x ⴝ 1.00 x ⴝ 1.25 x ⴝ 1.50 x ⴝ 1.75
1.0000
0.3728
0.2248
0.1530
0.1115
0.0841
0.0645
0.0499
0.0387
0.0301
0.0234
1.0000
0.6288
0.3942
0.2752
0.2034
0.1545
0.1189
0.0921
0.0715
0.0555
0.0432
1.0000
0.6800
0.4708
0.3448
0.2607
0.2002
0.1548
0.1201
0.0933
0.0725
0.0564
1.0000
0.5904
0.4562
0.3545
0.2757
0.2144
0.1668
0.1297
0.1009
0.0785
0.0610
0.0000
0.3840
0.3699
0.3101
0.2488
0.1961
0.1534
0.1196
0.0931
0.0725
0.0564
0.0000
0.2176
0.2517
0.2262
0.1865
0.1487
0.1169
0.0914
0.0712
0.0554
0.0431
0.0000
0.0768
0.1239
0.1183
0.0996
0.0800
0.0631
0.0494
0.0385
0.0300
0.0233
3.
Time x ⴝ 0.25 x ⴝ 0.50 x ⴝ 0.75 x ⴝ 1.00 x ⴝ 1.25 x ⴝ 1.50 x ⴝ 1.75
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.700
0.800
0.900
1.000
1.0000
0.4015
0.2430
0.1643
0.1187
0.0891
0.0683
0.0530
0.0413
0.0323
0.0253
1.0000
0.6577
0.4198
0.2924
0.2150
0.1630
0.1256
0.0976
0.0762
0.0596
0.0466
1.0000
0.7084
0.4921
0.3604
0.2725
0.2097
0.1628
0.1270
0.0993
0.0778
0.0609
1.0000
0.5837
0.4617
0.3626
0.2843
0.2228
0.1746
0.1369
0.1073
0.0841
0.0659
0.0000
0.3753
0.3622
0.3097
0.2528
0.2020
0.1598
0.1259
0.0989
0.0776
0.0608
0.0000
0.1871
0.2362
0.2208
0.1871
0.1521
0.1214
0.0959
0.0755
0.0593
0.0465
0.0000
0.0684
0.1132
0.1136
0.0989
0.0814
0.0653
0.0518
0.0408
0.0321
0.0252
5.
Time x ⴝ 0.25 x ⴝ 0.50 x ⴝ 0.75 x ⴝ 1.00 x ⴝ 1.25 x ⴝ 1.50 x ⴝ 1.75
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.0000
0.3972
0.2409
0.1631
0.1181
0.0888
0.0681
0.0528
0.0412
0.0322
0.0252
1.0000
0.6551
0.4171
0.2908
0.2141
0.1625
0.1253
0.0974
0.0760
0.0594
0.0465
1.0000
0.7043
0.4901
0.3592
0.2718
0.2092
0.1625
0.1268
0.0991
0.0776
0.0608
1.0000
0.5883
0.4620
0.3624
0.2840
0.2226
0.1744
0.1366
0.1071
0.0839
0.0657
0.0000
0.3723
0.3636
0.3105
0.2530
0.2020
0.1597
0.1257
0.0987
0.0774
0.0607
0.0000
0.1955
0.2385
0.2220
0.1876
0.1523
0.1214
0.0959
0.0754
0.0592
0.0464
Absolute errors are approximately 1.8 10–2, 3.7 10–2, 1.3 10–2.
Absolute errors are approximately 2.2 10–2, 3.7 10–2, 1.3 10–2.
7. (a)
Time
x ⴝ 2.00
x ⴝ 4.00
x ⴝ 6.00
x ⴝ 8.00
x ⴝ 10.00
x ⴝ 12.00
x ⴝ 14.00
x ⴝ 16.00
x ⴝ 18.00
0.00
2.00
4.00
6.00
8.00
10.00
30.0000
27.6450
25.6452
23.9347
22.4612
21.1829
30.0000
29.9037
29.6517
29.2922
28.8606
28.3831
30.0000
29.9970
29.9805
29.9421
29.8782
29.7878
30.0000
29.9999
29.9991
29.9963
29.9898
29.9782
30.0000
30.0000
29.9999
29.9996
29.9986
29.9964
30.0000
29.9999
29.9991
29.9963
29.9898
29.9782
30.0000
29.9970
29.9805
29.9421
29.8782
29.7878
30.0000
29.9037
29.6517
29.2922
28.8606
28.3831
30.0000
27.6450
25.6452
23.9347
22.4612
21.1829
(b)
Time
x ⴝ 5.00
x ⴝ 10.00 x ⴝ 15.00 x ⴝ 20.00 x ⴝ 25.00
x ⴝ 30.00
x ⴝ 35.00
x ⴝ 40.00
x ⴝ 45.00
0.00
2.00
4.00
6.00
8.00
10.00
30.0000
29.5964
29.2036
28.8212
28.4490
28.0864
30.0000
29.9973
29.9893
29.9762
29.9585
29.9363
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
29.9999
29.9997
29.9993
29.9986
30.0000
29.9973
29.9893
29.9762
29.9585
29.9363
30.0000
29.5964
29.2036
28.8213
28.4490
28.0864
30.0000
30.0000
29.9999
29.9997
29.9992
29.9986
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
0.0000
0.0653
0.1145
0.1145
0.0993
0.0816
0.0654
0.0518
0.0408
0.0320
0.0251
(c)
Time
x ⴝ 2.00
x ⴝ 4.00
x ⴝ 6.00
x ⴝ 8.00
x ⴝ 10.00
x ⴝ 12.00
x ⴝ 14.00
x ⴝ 16.00
x ⴝ 18.00
0.00
2.00
4.00
6.00
8.00
10.00
18.0000
15.3312
13.6371
12.3012
11.1659
10.1665
32.0000
28.5348
25.6867
23.2863
21.1877
19.3143
42.0000
38.3465
34.9416
31.8624
29.0757
26.5439
48.0000
44.3067
40.6988
37.2794
34.0984
31.1662
50.0000
46.3001
42.6453
39.1273
35.8202
32.7549
48.0000
44.3067
40.6988
37.2794
34.0984
31.1662
42.0000
38.3465
34.9416
31.8624
29.0757
26.5439
32.0000
28.5348
25.6867
23.2863
21.1877
19.3143
18.0000
15.3312
13.6371
12.3012
11.1659
10.1665
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 16
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.700
0.800
0.900
1.000
(d)
Time
x ⴝ 10.00 x ⴝ 20.00 x ⴝ 30.00 x ⴝ 40.00 x ⴝ 50.00
x ⴝ 60.00
x ⴝ 70.00
x ⴝ 80.00
x ⴝ 90.00
0.00
2.00
4.00
6.00
8.00
10.00
8.0000
8.0000
8.0000
8.0000
8.0000
8.0000
32.0000
31.9918
31.9686
31.9323
31.8844
31.8265
24.0000
23.9999
23.9993
23.9978
23.9950
23.9908
16.0000
16.0000
16.0000
15.9999
15.9998
15.9996
8.0000
8.0000
8.0000
8.0000
8.0000
8.0000
16.0000
16.0000
16.0000
15.9999
15.9998
15.9996
24.0000
23.9999
23.9993
23.9978
23.9950
23.9908
32.0000
31.9918
31.9686
31.9323
31.8844
31.8265
40.0000
39.4932
39.0175
38.5701
38.1483
37.7498
Answers to Selected Odd-Numbered Problems
ANS-37
9. (a)
Time
x ⴝ 2.00
x ⴝ 4.00
x ⴝ 6.00
x ⴝ 8.00
x ⴝ 10.00
x ⴝ 12.00
x ⴝ 14.00
x ⴝ 16.00
x ⴝ 18.00
0.00
2.00
4.00
6.00
8.00
10.00
30.0000
27.6450
25.6452
23.9347
22.4612
21.1829
30.0000
29.9037
29.6517
29.2922
28.8606
28.3831
30.0000
29.9970
29.9805
29.9421
29.8782
29.7878
30.0000
29.9999
29.9991
29.9963
29.9899
29.9783
30.0000
30.0000
30.0000
29.9997
29.9991
29.9976
30.0000
30.0000
29.9997
29.9988
29.9966
29.9927
30.0000
29.9990
29.9935
29.9807
29.9594
29.9293
30.0000
29.9679
29.8839
29.7641
29.6202
29.4610
30.0000
29.2150
28.5484
27.9782
27.4870
27.0610
(b)
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 16
ANS-38
Time
x ⴝ 5.00
x ⴝ 10.00 x ⴝ 15.00 x ⴝ 20.00 x ⴝ 25.00
x ⴝ 30.00
x ⴝ 35.00
x ⴝ 40.00
x ⴝ 45.00
0.00
2.00
4.00
6.00
8.00
10.00
30.0000
29.5964
29.2036
28.8212
28.4490
28.0864
30.0000
29.9973
29.9893
29.9762
29.9585
29.9363
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
29.9999
29.9997
29.9995
30.0000
29.9991
29.9964
29.9921
29.9862
29.9788
30.0000
29.8655
29.7345
29.6071
29.4830
29.3621
30.0000
30.0000
29.9999
29.9997
29.9992
29.9986
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
30.0000
(c)
Time
x ⴝ 2.00
x ⴝ 4.00
x ⴝ 6.00
x ⴝ 8.00
x ⴝ 10.00
x ⴝ 12.00
x ⴝ 14.00
x ⴝ 16.00
x ⴝ 18.00
0.00
2.00
4.00
6.00
8.00
10.00
18.0000
15.3312
13.6381
12.3088
11.1946
10.2377
32.0000
28.5350
25.6913
23.3146
21.2785
19.5150
42.0000
38.3477
34.9606
31.9546
29.3217
27.0178
48.0000
44.3130
40.7728
37.5566
34.7092
32.1929
50.0000
46.3327
42.9127
39.8880
37.2109
34.8117
48.0000
44.4671
41.5716
39.1565
36.9834
34.9710
42.0000
39.0872
37.4340
36.9745
34.5032
33.0338
32.0000
31.5755
31.7086
31.2134
30.4279
29.5224
18.0000
24.6930
25.6986
25.7128
25.4167
25.0019
(d)
Time
x ⴝ 10.00 x ⴝ 20.00 x ⴝ 30.00 x ⴝ 40.00 x ⴝ 50.00
x ⴝ 60.00
x ⴝ 70.00
x ⴝ 80.00
x ⴝ 90.00
0.00
2.00
4.00
6.00
8.00
10.00
8.0000
8.0000
8.0000
8.0000
8.0000
8.0000
32.0000
31.9918
31.9687
31.9324
31.8846
31.8269
24.0000
24.0000
24.0002
24.0005
24.0012
24.0023
16.0000
16.0102
16.0391
16.0845
16.1441
16.2160
8.0000
8.6333
9.2272
9.7846
10.3084
10.8012
16.0000
16.0000
16.0000
15.9999
15.9998
15.9996
24.0000
23.9999
23.9993
23.9978
23.9950
23.9908
32.0000
31.9918
31.9686
31.9323
31.8844
31.8265
40.0000
39.4932
39.0175
38.5701
38.1483
37.7499
1
11. (a) c(x) 2 x 20
(b)
Time
x ⴝ 4.00
x ⴝ 8.00
x ⴝ 12.00
x ⴝ 16.00
0.00
10.00
20.00
30.00
50.00
70.00
90.00
110.00
130.00
150.00
170.00
190.00
210.00
230.00
250.00
270.00
290.00
310.00
330.00
350.00
50.0000
32.7433
29.9946
26.9487
24.1178
22.8995
22.3817
22.1619
22.0687
22.0291
22.0124
22.0052
22.0022
22.0009
22.0004
22.0002
22.0001
22.0000
22.0000
22.0000
50.0000
44.2679
36.2354
32.1409
27.4348
25.4560
24.6176
24.2620
24.1112
24.0472
24.0200
24.0085
24.0036
24.0015
24.0007
24.0003
24.0001
24.0001
24.0000
24.0000
50.0000
45.4228
38.3148
34.0874
29.4296
27.4554
26.6175
26.2620
26.1112
26.0472
26.0200
26.0085
26.0036
26.0015
26.0007
26.0003
26.0001
26.0001
26.0000
26.0000
50.0000
38.2971
35.8160
32.9644
30.1207
28.8998
28.3817
28.1619
28.0687
28.0291
28.0124
28.0052
28.0022
28.0009
28.0004
28.0002
28.0001
28.0000
28.0000
28.0000
Answers to Selected Odd-Numbered Problems
Exercises 16.3, Page 814
1. (a) Time
x ⴝ 0.25
(b)
0.00
0.20
0.40
0.60
0.80
1.00
0.1875
0.1491
0.0556
0.0501
0.1361
0.1802
Time
x ⴝ 0.4
0.00
0.20
0.40
0.60
0.80
1.00
0.0032
0.0652
0.2065
0.3208
0.3094
0.1450
x ⴝ 0.50
0.2500
0.2100
0.0938
0.0682
0.2072
0.2591
x ⴝ 0.8
0.5273
0.4638
0.3035
0.1190
0.0180
0.0768
x ⴝ 0.75
0.1875
0.1491
0.0556
0.0501
0.1361
0.1802
x ⴝ 1.2
0.5273
0.4638
0.3035
0.1190
0.0180
0.0768
x ⴝ 1.6
0.0032
0.0652
0.2065
0.3208
0.3094
0.1450
(c)
x ⴝ 0.1
Time
0.00
0.12
0.24
0.36
0.48
0.60
0.72
0.84
0.96
0.0000
0.0000
0.0071
0.1623
0.1965
0.2194
0.3003
0.2647
0.3012
x ⴝ 0.2
0.0000
0.0000
0.0657
0.3197
0.1410
0.2069
0.6865
0.1633
0.1081
x ⴝ 0.3
0.0000
0.0082
0.2447
0.2458
0.1149
0.3875
0.5097
0.3546
0.1380
x ⴝ 0.4
0.0000
0.1126
0.3159
0.1657
0.1216
0.3411
0.3230
0.3214
0.0487
x ⴝ 0.5
0.0000
0.3411
0.1735
0.0877
0.3593
0.1901
0.1585
0.1763
0.2974
x ⴝ 0.6
0.5000
0.1589
0.2463
0.2853
0.2381
0.1662
0.0156
0.0954
0.3407
x ⴝ 0.7
0.5000
0.3792
0.1266
0.2843
0.1977
0.0666
0.0893
0.1249
0.1250
x ⴝ 0.8
x ⴝ 0.9
0.5000
0.3710
0.3056
0.2104
0.1715
0.1140
0.0874
0.0665
0.1548
0.5000
0.0462
0.0625
0.2887
0.0800
0.0446
0.0384
0.0386
0.0092
x ⴝ 0.2
x ⴝ 0.4
x ⴝ 0.6
x ⴝ 0.8
0.00
0.10
0.20
0.30
0.40
0.50
0.5878
0.5599
0.4788
0.3524
0.1924
0.0142
0.9511
0.9059
0.7748
0.5701
0.3113
0.0230
0.9511
0.9059
0.7748
0.5701
0.3113
0.0230
0.5878
0.5599
0.4788
0.3524
0.1924
0.0142
Time
x ⴝ 0.2
x ⴝ 0.4
x ⴝ 0.6
x ⴝ 0.8
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.5878
0.5808
0.5599
0.5257
0.4790
0.4209
0.3527
0.2761
0.1929
0.1052
0.0149
0.9511
0.9397
0.9060
0.8507
0.7750
0.6810
0.5706
0.4467
0.3122
0.1701
0.0241
0.9511
0.9397
0.9060
0.8507
0.7750
0.6810
0.5706
0.4467
0.3122
0.1701
0.0241
0.5878
0.5808
0.5599
0.5257
0.4790
0.4209
0.3527
0.2761
0.1929
0.1052
0.0149
(b)
(b)
0.00000
0.60134
1.20268
1.80401
2.40535
3.00669
3.60803
4.20936
4.81070
5.41204
6.01338
6.61472
7.21605
7.81739
8.41873
9.02007
9.62140
x ⴝ 0.40
x ⴝ 0.60
x ⴝ 0.80
0.2000
0.2000
0.2000
0.2000
0.2000
0.1961
0.4000
0.4000
0.4000
0.4000
0.3844
0.3609
0.6000
0.6000
0.6000
0.5375
0.4750
0.4203
0.8000
0.8000
0.5500
0.4250
0.3469
0.2922
(c) Yes; the table in part (b) is the table in part (a)
shifted downward.
Exercises 17.1, Page 823
1. 3 3i
3. 1
7. 7 5i
9. 11 10i
7
13. 2i
19.
25.
15. 17 23
64
37 37 i
7
9
130 130 i
x ⴝ 20
x ⴝ 30
x ⴝ 40
x ⴝ 50
0.1000
0.0984
0.0226
0.1271
0.0920
0.0932
0.0284
0.1064
0.1273
0.0625
0.0436
0.0931
0.1436
0.0625
0.0287
0.0654
0.1540
0.2000
0.1688
0.0121
0.1347
0.2292
0.1445
0.0205
0.1555
0.2060
0.1689
0.0086
0.1364
0.2173
0.1644
0.0192
0.1332
0.2189
0.3000
0.1406
0.0085
0.1566
0.2571
0.2018
0.0336
0.1265
0.2612
0.2038
0.0080
0.1578
0.2240
0.2247
0.0085
0.1755
0.2089
0.2000
0.1688
0.0121
0.1347
0.2292
0.1445
0.0205
0.1555
0.2060
0.1689
0.0086
0.1364
0.2173
0.1644
0.0192
0.1332
0.2189
0.1000
0.0984
0.0226
0.1271
0.0920
0.0932
0.0284
0.1064
0.1273
0.0625
0.0436
0.0931
0.1436
0.0625
0.0287
0.0654
0.1540
11
17 i
17. 8 i
102
5
23.
31. "(x 2 1)2 (y 2 3)2
29. 2y 4
7
10 i
1
37. z 30
!2
2
1
!2
2
39. 11 6i
i,2 !2
2 2
3. 3acos
7. 2acos
9.
3p
3p
1 i sin
b
2
2
5p
5p
1 i sin
b
6
6
5. !2 acos
3!2
5p
5p
acos
1 i sin
b
2
4
4
53
5
11. 2 2 i
!2
4
13. 5.5433 2.2961i
!2
4 i
Chapter 16 in Review, Page 815
1. u11 0.8929, u21 3.5714, u31 13.3929
3. (a) x ⴝ 0.20
x ⴝ 0.40
x ⴝ 0.60
17. 30!2 fcos (25p>12) i sin (25p>12)g ;
0.6000
0.6000
0.5375
0.4750
0.4203
0.3734
i
p
p
1 i sin b
4
4
15. 8i;
0.4000
0.4000
0.4000
0.3844
0.3609
0.3346
!2
2
Exercises 17.2, Page 827
1. 2(cos 0 i sin 0) or 2(cos 2p i sin 2p)
Note: Time is expressed in milliseconds.
0.2000
0.2000
0.2000
0.2000
0.1961
0.1883
116
5 i
27. x/(x2 y2)
33. x 5 22, y 5 1 35.
x ⴝ 10
5. 7 13i
11. 5 12i
21. 20i
9
5.
Time
x ⴝ 0.20
2
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 17
3. (a)
Time
40.9808 10.9808i
x ⴝ 0.80
0.8000
0.5500
0.4250
0.3469
0.2922
0.2512
19.
1
2 !2
1
4
fcos (p>4) i sin (p>4)g;
21. 512
23.
1
32 i
2 14 i
25. i
27. w0 2, w1 1 !3i, w2 1 !3i
29. w0 !2
2
!2
2
i, w1 !2
2 2
!2
2
i
Answers to Selected Odd-Numbered Problems
ANS-39
31. w0 33. !2
2
!2
2 (1
35. 32 acos
!6
2
i, w1 !2
2 2
i), !2
2 (1 2 i)
!6
2
Exercises 17.4, Page 834
i
v
1.
13p
13p
i sin
b, 16!3 16i
6
6
v
3.
2
u= v – 4
16
37. cos 2u cos2u sin2u, sin 2u 2 sin u cos u
u≤0
v=0
u
u
Exercises 17.3, Page 830
1.
y
3.
y
x
x=5
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 17
5.
x
y
5.
v
y = –3
7. y
v≥0
u=0
x
u
(4, –3)
x
7. f (z) (6x 5) i(6y 9)
9. domain
9. f (z) (x2 y2 3x) i(2xy 3y 4)
11. domain
y
11. f (z) (x3 3xy2 4x) i(3x2y y3 4y)
y
x
13. f (z) ax x
y
x
b i ay 2 2
b
2
x y
x y2
2
15. 4 i; 3 9i; 1 86i
17. 14 20i; 13 43i; 3 26i
13. domain
15. not a domain
y
y
21. 4i
27. 12z (6 2i) z 5
29. 6z2 14z 4 16i
31. 6z(z2 4i)2
33.
x
x
19. 6 5i
2
8 2 13i
(2z i)2
35. 3i
37. 2i, 2i
41. x(t) c1e2t and y(t) c2e2t; the streamlines lie on lines
through the origin.
43. y cx; the streamlines are lines through the origin.
17. not a domain
19. domain
45.
v
y
y
x
x
u
21. domain
y
Exercises 17.5, Page 839
15. a 1, b 3
21. f (z) ex cos y iex sin y
23. f (z) x i( y C)
x
25. f (z) x2 y2 i(2xy C )
23. the line y x
ANS-40
2
2
25. the hyperbola x y 1
Answers to Selected Odd-Numbered Problems
27. f (z) loge(x2 y2) i a2 tan1
y
Cb
x
29.
y
33. 1, 1
35. pure imaginary numbers
37. f (z) (2y 5) 2xi
v = c2
Exercises 18.1, Page 858
13.
u = c1
31. the x-axis and the circle |z| 1
15. e x
23.
25.
27.
29.
33.
37.
41.
47.
2
!2
2 iR
7. 1.8650 4.0752i
11. 0.9659 0.2588i
3. e 1 Q
!2
2
2 y2
(cos 2xy i sin 2xy)
1.6094 i(p 2np)
1.0397 i(3p/4 2np)
1.0397 i(p/3 2np)
2.1383 (p/4)i
31.
3.4657 (p/3)i
35.
3 i(p/2 2np)
39.
2np
e
(0.2740 0.5837i) 43.
no; no; no
2.5649 2.7468i
1.3863 i(p/2 2np)
e(28n)p
e2
Chapter 17 in Review, Page 851
7
1. 0; 32
3. 25
5. 45
7. false
9. 0.6931 i (p/2 2np) 11. 0.3097 0.8577i
p
i
13. false
15. 3 2
17. 58 4i
2
19. 8 8i
23.
y
x
x
17.
1
2i
19. 0
5pe 5
21. 43 2 53 i
23. 43 2 53 i
25.
12
27. 6!2
31. 11 38i ; 0
33. circulation 0, net flux 4p
35. circulation 0, net flux 0
Exercises 18.2, Page 862
9. 2pi
11. 2pi
13. 0
15. 2pi ; 4pi ; 0 17. 8pi ; 6pi
19. p (1 i) 21. 4pi
23. 6pi
Exercises 18.3, Page 867
3. 48 24i
5. 6 26
3i
7. 0
1
1
9.
2
11. 2
13. 2.3504i
i
p
p
15. 0
17. pi
19. 12 i
21. 11.4928 0.9667i
23. 0.9056 1.7699i
Exercises 18.4, Page 873
1. 8pi
3. 2pi
5. p(20 8i)
7. 2p; 2p
9. 8p
11. 2pe1i
13. 43 pi
15. 5pi; 5pi; 9pi; 0
17. p(3 i); p(3 i)
19. p( 83 12i)
21. 0
23. pi
Chapter 18 in Review, Page 874
1. true
3. true
5. 0
7. p(6p i)
9. true
11. 0 if n 1, 2pi if n 1
88
13. 72
15. 136
17. 0
15 3 i
19. 14.2144 22.9637i
21. 2pi
23. 83 pi
25. 25 pi
27. 2p
29. 2npi
Exercises 19.1, Page 882
1. 5i, 5, 5i, 5, 5i
3. 0, 2, 0, 2, 0
5. converges
7. converges
9. diverges
11. limnS Re(zn) 2 and limnS Im(zn) 32
13. The series converges to 1/(1 2i).
15. divergent
17. convergent, 15 25 i
19. convergent, 95 12
5i
21. |z 2i| !5, R !5
23. |z 1 i| 2, R 2
25. |z i| 1/ !10, R 1/ !10
27. |z 4 3i| 25, R 25
29. The series converges at z 2 i.
167
Exercises 17.8, Page 851
1. np (1)n 1i log e(1 !2)
3. np
5. 2np i loge(2 !3)
7. p/3 2np
9. p/4 np
11. (1)n loge 3 npi
y
15. 0
1. 2i
Exercises 17.7, Page 848
1. 10.0677
3. 1.0911 0.8310i
5. 0.7616i
7. 0.6481
9. 1
11. 0.5876 1.3363i
15. p>2 2np 2 ilog e(2 !3)
17. (p/2 2np)i
19. p/4 np
21. 2np 2i
21.
3
p
2
2
4
5. (2 p)i
11. e 1
22
3i
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 19
Exercises 17.6, Page 845
!3
1. 2 12 i
5. ep
9. 0.2837 0.9589i
13. ey (cos x i sin x)
3. 48 736
3 i
7
1
9. 12 12 i
1. 28 84i
7. pi
x
Exercises 19.2, Page 886
a (1)
q
25. an ellipse with foci (0, 2) and (0, 2)
27. 1.0696 0.2127i, 0.2127 1.0696i,
1.0696 0.2127i, 0.2127 1.0696i
29. 5i
31. the parabola v u2 2u
1.
k1 k
k1
a (1)
q
3.
k1
k21
z ,R 1
k (2z)k 2 1, R 1
2
Answers to Selected Odd-Numbered Problems
ANS-41
a
q
5.
k0
q
7.
a
k0
q
9.
a
k0
q
11.
a
k
k
a (1) (z 2 1) , R 1
k0
q
13.
(1)k
(2z)k, R q
k!
z 2k 1
,R q
(2k 1)!
(1)k z 2k
a b ,R q
(2k)! 2
(1)k
z 4k 2, R q
(2k 1)!
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 19
k0
q
15.
17.
19.
21.
23.
(z 2 2i)k
, R !13
a
k1
k 0 (3 2 2i)
q (z 2 1)k
,R 2
a
2k
k1
!2
!2
p
!2
p 2
2
az 2 b 2
az 2 b
2
2 1!
4
2 2!
4
!2
p 3
az 2 b p , R q
2 3!
4
q (z 2 3i)k
e 3i a
,R q
k!
k0
1 3
2 5
z 3 z 15 z . . .
1
3
7 2
15 3
z
z z p, R 1
2
3
2i
(2i)
(2i)
(2i)4
27. 2!5
25.
29. a (1)k(z 1)k, R 1;
q
k0
q
(1)k
k
a (2 i)k 1 (z 2 i) , R !5
k0
y
x
31. (a) The distance from z0 to the branch cut is 1 unit.
(c) The series converges within the circle
|z 1 i| !2. Although the series converges
in the shaded region, it does not converge to (or
represent) Ln z in this region.
y
x
ANS-42
z
1
z3
z5
2
2
p
z
2!
4!
6!
1
1
1
3. 1 2
2
p
1! z 2
2! z 4
3! z 6
e(z 2 1)
e(z 2 1)2
e
5.
e
p
z21
2!
3!
1
1
z
z2
7. 2 2 2 3 2 4 2 p
3z
3
3
3
(z 2 3)2
1
1
z23
9.
2 2
2
p
3
3(z 2 3)
3
3
34
1
1
z24
1
11. p 2
2
2
3(z
2
4)
12
3(z 2 4)
3 42
2
(z 2 4)
2
p
3 43
1
1
z
z2
1
13. p 2 2 2 2 2 2 2 3 2 p
z
2
z
2
2
1
15.
2 1 2 (z 2 1) 2 (z 2 1)2 2 p
z21
2(z 1)
2(z 1)2
1
2
17.
2 22
2
2p
3(z 1)
3
33
34
1
1
z
z2
1
19. p 2
2 2
2
2p
2
3z
3
3 2
3z
3 22
1
2 3z 4z 2 p
21.
z
1
23.
3 6(z 2) 10(z 2)2 . . .
z22
3
4 4z 4z2 . . .
25.
z
2
2
2
27. p 1 (z 2 1)
z21
(z 2 1)3
(z 2 1)2
Exercises 19.4, Page 897
1. Define f (0) 2.
3. 2 i is a zero of order 2.
5. i and i are zeros of order 1; 0 is a zero of order 2.
7. 2npi, n 0, 1, . . . , are zeros of order 1.
9. order 5
11. order 1
13. 1 2i are simple poles.
15. 2 is a simple pole; i is a pole of order 4.
17. (2n 1)p/2, n 0, 1, . . . , are simple poles.
19. 0 is a pole of order 2.
21. 2npi, n 0, 1, . . . , are simple poles.
23. 0 is a removable singularity; 1 is a simple pole.
25. nonisolated
Exercises 19.5, Page 902
1.
1. 25
3. 3
5. 0
1
7. Res ( f (z), 4i) 2 , Res ( f (z), 4i) 12
1
9. Res ( f (z), 1) 13 , Res ( f (z), 2) 12
,
–1 + i
33. 1.1 0.12i
Exercises 19.3, Page 894
(1)k
2 q
35.
z 2k 1
!p ka
0 (2k 1)k!
Answers to Selected Odd-Numbered Problems
11.
13.
15.
17.
Res ( f (z), 0) 14
Res ( f (z), 1) 6, Res ( f (z), 2) 31,
Res ( f (z), 3) 30
Res ( f (z), 0) 3/p4, Res ( f (z), p) (p2 6)/2p4
Res ( f (z), (2n 1)p/2) (1)n1, n 0, 1, 2, . . .
0; 2pi/9; 0
19. pi; pi; 0
21. p/3
23. 0
25. 2pi cosh 1
p
p
31. i
3
3
Exercises 19.6, Page 908
1. 4p/ !3
9. p/6
17. p/2
25. pe3
27.
5. p/ !3
13. p/16
21. pe1
29. 6i
7. p/4
15. 3p/8
23. pe1
pe!2
(cos !2 sin !2)
2!2
p e 3
a
2 e 1 b
8
3
Chapter 19 in Review, Page 908
1. true
3. false
5. true
7. true
1
9.
11. Zz 2 iZ !5
p
q ( !2)k cos (kp>4)
13. 1 a
zk
k!
k1
i
1
i
1
i
15. 3 2
2 zp
2
3!z
4!
5!
z
2!z
1
1
17. p 2
(z 2 i)
3!(z 2 i)
5!(z 2 i)3
2
8
26 2
19.
z
z p;
3
9
27
2
p 2 1 2 1 2 1 2 1 2 z 2 z 2 p;
3
2
2
z
3
z
z
3
33
2
8
26
3 4 p;
2
z
z
z
(z 2 1)2
1
1
z21
2 2
2
2p
2
z21
2
2
23
404p
2p
21.
i
23.
i
81
!3
25. (p pe 2 cos 2) i
27. pi
9p3 2
29.
i
31. 7p/50
p2
90 2 52!3
b
33. pa
12 2 7!3
Exercises 20.1, Page 915
1.
5.
7.
11.
13.
15.
17.
19.
21.
23.
27.
the line v u
3. the line v 2
open line segment from 0 to pi
the ray u 12 u0
9. the line u 1
the fourth quadrant
the wedge p/4 Arg w p/2
the circle with center w 4i and radius r 1
the strip 1 u 0
the wedge 0 Arg w 3p/4
w i(z i) iz 1
w 2(z 1)
25. w z4
3z/2
we
29. w z i
Exercises 20.2, Page 920
1. conformal at all points except z 1
3. conformal at all points except z pi 2npi
5. conformal at all points outside the interval [1, 1] on
the x-axis
7. The image is the region shown in Figure 20.2.2(b). A
horizontal segment z(t) t ib, 0 t p, is mapped
onto the lower or upper portion of the ellipse
v2
u2
1
2
cosh b
sinh2b
according to whether b 0 or b 0.
9. The image of the region is the wedge 0 Arg w p/4.
The image of the line segment [p/2, p/2] is the union
of the line segments joining eip/4 to 0 and 0 to 1.
11. w cos(pz /2) using H-4
v
R′
B′
A′
1
13. w a
v
u
1 z 1>2
b using H-5 and w z1/4
12z
A′
v=u
R′
B′
u
B′ = eiπ /4
15. w a
v
A′
B′
C′
i
e p>z ep>z 1>2
b using H-6 and w z1/2
e p>z 2 ep>z
R′
u
17. w sin(iLn z p /2); AB is the real interval
19.
21.
23.
25.
(q, 1].
1
4
u Arg (z4) or u (r, u) u
p
p
1 2 x 2 2 y2
1
12z
1
u Arg ai
b tan1 a
b
p
p
1z
2y
1
u fArg (z 2 2 1) 2 Arg (z 2 1)g
p
10
fArg (e pz 2 1) 2 Arg (e pz 1)g
u
p
Exercises 20.3, Page 927
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 20
29. 3. 0
11. p
19. p/ !2
27. 4i
1. T(0) q, T(1) i, T(q) 0; |w| 1 and the line
v 12 ; |w| 1
3. T(0) 1, T(1) q, T(q) 1; the line u 0 and
the circle |w 1| 2; the half-plane u 0
w 2 1
w1
5. S 1(w) ,
w i
w2i
(1 i)z 2 1
S 1(T (z)) 2z i
Answers to Selected Odd-Numbered Problems
ANS-43
w 2
3
w 2 2 1
, S (T (z)) z
w 1
w21
z1
2z
w 2
11. w z22
z 2 1 2 2i
i z21
w
2 z
(1 i) z 1 2 i
w3
(3 5i) z 2 3 2 5i
1
z2
u
log e 2
2. The level curves are the images
log e2
z21
of the circles |w| r, 1 r 2, under the linear fractional transformation T(w) (w 2)/(w 1). Since
the circles do not pass through the pole at w 1, the
images are circles.
Construct the linear fractional transformation that sends
1, i, i to 0, 1, 1.
7. S 1(w) 9.
13.
15.
17.
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 20
19.
a1z b1
b b2
c1z d1
21. Simplify T2(T1(z)) .
a1z b1
c2 a
b d2
c1z d1
a2 a
Exercises 20.4, Page 931
1. first quadrant
3.
v
(1 2 i)z 2 (1 i)
1
b
Arg a
p
12z
1
12z
2 Arg a
b
p
(1 i)z 1 2 i
11. u(0, 0) 13 , u(0.5, 0) 0.5693, u(0.5, 0) 0.1516
#
p
1
u(e it) dt.
2p p
15. u(r, u) r sin u r cos u or u(x, y) x y
13. Show that u(0, 0) y
1.2
0.8
0.4
–1
1
x
0
– 0.8
–1.2
– 0.4
Exercises 20.6, Page 940
1. g(z) e iu0 is analytic everywhere and G(z) e iu0 z
is a complex potential. The equipotential lines are the
lines x cos u0 y sin u0 c.
y
θ0 = π
6
6
ai
x
4
u
0
5. f (z) A(z 1)1/2z1/2(z 1)1/2 for some constant A
7. f (z) A(z 1)1/3z1/3 for some constant A
9. Show that f (z) 1
A
and conclude that
(z 2 1)1>2
2
f (z) cosh z.
11. Show that f (z) S A/z as w1 S q and conclude that
f (z) Ln z.
13. Show that f (z) S A(z 1)1/2z(z 1)1/2 Az /(z2 1)1/2 as u1 S 0.
2
–2
–6 –4
0
3. g(z) 1/z is analytic for z 0 and G(z) Ln z is
analytic except for z x 0. The equipotential lines
are the circles x2 y2 e2c.
y
0.5
1
0.75
0
x
Exercises 20.5, Page 935
1. u 3. u 1
1
z21
z
b 2 Arg a
b
Arg a
p
p
z
z1
5
1
z
b
fp 2 Arg (z 2 1)g Arg a
p
p
z1
2
1
z1
b
Arg a
p
z2
y
y2 2 x 2
x21
x
ctan1 a
b 2 tan1 a b d
5. u e 1 p
y
y
y
7. u ANS-44
9. u (x 2 1)2 y 2
x log e c
dr
x 2 y2
1
z2 2 1
5
b Arg (z 2 1)
Arg a
2
p
p
z
Answers to Selected Odd-Numbered Problems
4
4
4
Arg z or f(r, u) u, and G(z) Ln z is a
p
p
p
complex potential. The equipotential lines are the rays
5. f u
y
p
4
x
c and F a 2
,
b.
p x y2 x 2 y2
4
7. The equipotential lines are the images of the rays
u u0 under the successive transformations z w1/2
and z (z 1)/(z 1). The transformation z w1/2
maps the ray u u0 to the ray u u0/2 in the z-plane,
and z (z 1)/(z 1) maps this ray onto an arc of
a circle that passes through z 1 and z 1.
9. (a) c(x, y) 4xy(x2 y2) or, in polar coordinates,
(c)
c(r, u) r4 sin 4u. Note that c 0 on the boundary of R.
y
y=π
(b) V 4z 3 4(x3 3xy2, y3 3x2y)
(c) y
y = π /2
x
17. (a) f (t) 11. (a) c(x, y) cos x sinh y and c 0 on the boundary
of R.
(b) V cos z (cos x cosh y, sin x sinh y)
(c)
y
1, t , 1
and Re ( f (t)) 0 for 1 t 1.
0, t . 1
Hence, Im (G(z)) c(x, y) 0 on the boundary
of R.
1
(b) x Re c ¢ ((t ic)2 2 1)1>2 cosh1(t ic) ≤ d
p
e
y Im c
for c
(c)
–π
2
π
2
0
1
¢ ((t ic)2 2 1)1>2 cosh1(t ic) ≤ d
p
y
x
2xy
or, in polar coordinates,
(x 2 y 2)2
2
2
c(r, u) (r 1/r ) sin 2u. Note that c 0 on
the boundary of R.
13. (a) c(x, y) 2xy (b) V 2z 2 2>z 3
(c)
x
0
19. z 0 in Example 5; z 1, z 1 in Example 6
21. The streamlines are the branches of the family of
hyperbolas x2 Bxy y2 1 0 that lie in the first
quadrant. Each member of the family passes through
(1, 0).
23. Hint: For z in the upper half-plane,
y
k [Arg (z 1) Arg (z 1)] k Arg a
1
x
1
2
i Arg (t 1) i Arg (t 1)] and so
15. (a) f (t) pi [loge|t 1| loge|t 1|
0,
Im ( f ( t)) c p>2,
p,
t , 1
1 , t , 1
t . 1.
z21
b.
z1
Chapter 20 in Review, Page 942
1. v 4
3. the wedge 0 Arg w 2p/3
5. true
7. 0, 1, q
9. false
11. The image of the first quadrant is the strip 0 v p/2.
Rays u u0 are mapped onto horizontal lines v u0
in the w-plane.
i 2 cos pz
13. w i cos pz
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, CHAPTER 20
x
1 2
1
((t 1)1/2 cosh1 t) ((t 2 1)1/2
p
p
Ln (t (t 2 1)1/2)) and so Im (f (t)) v
A′
Hence, Im (G(z)) c(x, y) 0 on the boundary
of R.
R′
1
u
(b) x 12 [loge|t 1 ic| loge|t 1 ic|]
y p 12 [Arg (t 1 ic) Arg (t 1 ic)],
for c 0
B′
Answers to Selected Odd-Numbered Problems
ANS-45
15. u 2 2y/(x2 y2)
17. (a) Note that a 1 S 0, a 2 S 2p , and a 3 S 0 as
u1 S q.
(b) Hint: Write f (t) 12 A[loge|t 1| loge|t 1| i Arg (t 1) i Arg (t 1)] B.
19. G(z) f 1(z) maps R to the strip 0
v
p, and
U(u, v) v/p is the solution to the transferred boundary problem. Hence, f (x, y) (1/p )Im (G(z)) (1/p)c(x, y), and so the equipotential lines f(x, y) c
are the streamlines c(x, y) cp.
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS, APPENDIX II
ANS-46
Answers to Selected Odd-Numbered Problems
Appendix II Exercises, Page APP-5
1. 24; 720; 4 !p/3; 8 !p/15
3. 0.297
1
1
1
for x
5. G(x) . t x 2 1e tdt . e 1 t x 2 1dt xe
0
0
As x S 0, 1/x S q.
#
#
0.
Differentiation Rules
1. Constant:
3. Sum:
d
c⫽0
dx
2. Constant Multiple:
d
ff(x) ⫾ g(x)g ⫽ f 9(x) ⫾ g9(x)
dx
5. Quotient:
7. Power:
4. Product:
g(x)f 9(x) 2 f(x)g9(x)
d f(x)
⫽
dx g(x)
fg(x)g 2
d n
x ⫽ nx n 2 1
dx
d
cf (x) ⫽ c f 9(x)
dx
d
f (x)g(x) ⫽ f (x)g9(x) ⫹ g(x) f 9(x)
dx
6. Chain:
d
f (g(x)) ⫽ f 9(g(x))g9(x)
dx
8. Power:
d
fg(x)g n ⫽ nfg(x)g n 2 1g9(x)
dx
Derivatives of Functions
Trigonometric:
9.
12.
d
sin x ⫽ cos x
dx
10.
d
cos x ⫽ ⫺sin x
dx
11.
d
tan x ⫽ sec 2 x
dx
d
cot x ⫽ ⫺csc 2 x
dx
13.
d
sec x ⫽ sec x tan x
dx
14.
d
csc x ⫽ ⫺csc x cot x
dx
17.
d
1
tan⫺1 x ⫽
dx
1 ⫹ x2
d
1
sec ⫺1 x ⫽
dx
ZxZ"x 2 2 1
20.
d
1
csc ⫺1 x ⫽ ⫺
dx
ZxZ"x 2 2 1
Inverse trigonometric:
15.
18.
1
d
sin⫺1 x ⫽
dx
"1 2 x 2
d
1
cot ⫺1 x ⫽ ⫺
dx
1 ⫹ x2
16.
19.
d
1
cos ⫺1 x ⫽ ⫺
dx
"1 2 x 2
Hyperbolic:
21.
d
sinh x ⫽ cosh x
dx
22.
d
cosh x ⫽ sinh x
dx
23.
d
tanh x ⫽ sech2 x
dx
24.
d
coth x ⫽ ⫺csch2 x
dx
25.
d
sech x ⫽ ⫺sech x tanh x
dx
26.
d
csch x ⫽ ⫺csch x coth x
dx
29.
d
1
tanh⫺1 x ⫽
, ZxZ , 1
dx
1 2 x2
32.
d
1
csch⫺1 x ⫽ ⫺
dx
ZxZ"x 2 ⫹ 1
Inverse hyperbolic:
27.
30.
d
1
sinh⫺1 x ⫽
2
dx
"x ⫹ 1
d
1
coth⫺1 x ⫽
, ZxZ . 1
dx
1 2 x2
28.
31.
d
1
cosh⫺1 x ⫽
2
dx
"x 2 1
d
1
sech⫺1 x ⫽ ⫺
dx
x"1 2 x 2
Exponential:
33.
d x
e ⫽ ex
dx
34.
d x
b ⫽ b x(ln b)
dx
36.
d
1
log b x ⫽
dx
x(ln b)
38.
d
dx
Logarithmic:
35.
1
d
lnZxZ ⫽
x
dx
Of an integral:
37.
#
x
d
g(t) dt ⫽ g(x)
dx a
#
b
a
b
g(x, t) dt ⫽
# 0x g(x, t) dt
a
0
Integration Formulas
u n⫹1
⫹ C, n 2 ⫺1
n⫹1
1.
#
2.
# u du ⫽ lnZuZ ⫹ C
3.
#e du ⫽ e
4.
#b du ⫽ ln b b
5.
#sin u du ⫽ ⫺cos u ⫹ C
6.
#cos u du ⫽ sin u ⫹ C
7.
#sec
8.
#csc
9.
#sec u tan u du ⫽ sec u ⫹ C
10.
#csc u cot u du ⫽ ⫺csc u ⫹ C
11.
#tan u du ⫽ ⫺lnZcos uZ ⫹ C
12.
#cot u du ⫽ lnZsin uZ ⫹ C
13.
#sec u du ⫽ lnZsec u ⫹ tan uZ ⫹ C
14.
#csc u du ⫽ lnZcsc u 2 cot uZ ⫹ C
15.
#u sin u du ⫽ sin u 2 u cos u ⫹ C
16.
#u cos u du ⫽ cos u ⫹ u sin u ⫹ C
17.
#sin u du ⫽
18.
#cos
19.
#sin au sin bu du ⫽
20.
#cos au cos bu du ⫽
21.
#e
22.
#e
23.
#sinh u du ⫽ cosh u ⫹ C
24.
#cosh u du ⫽ sinh u ⫹ C
25.
#sech u du ⫽ tanh u ⫹ C
26.
#csch u du ⫽ ⫺coth u ⫹ C
27.
#tanh u du ⫽ ln(cosh u) ⫹ C
28.
#coth u du ⫽ lnZsinh uZ ⫹ C
29.
#ln u du ⫽ u ln u 2 u ⫹ C
30.
#u ln u du ⫽
1 2
2 u ln
# "a
du ⫽ ln P u ⫹ "a 2 ⫹ u 2 P ⫹ C
31.
33.
35.
37.
u n du ⫽
u
2
u
u du ⫽ tan u ⫹ C
2
au
⫹C
1
2u
sin bu du ⫽
2 14 sin 2u ⫹ C
sin(a 2 b)u
sin(a ⫹ b)u
2
⫹C
2(a 2 b)
2(a ⫹ b)
e au
(a sin bu 2 b cos bu) ⫹ C
a ⫹ b2
2
2
# "a
# "a
#a
2
1
2
du ⫽ sin 2 1
2
2u
2
2 u 2 du ⫽
u
⫹C
a
u
a2
u
"a 2 2 u 2 ⫹
sin 2 1 ⫹ C
a
2
2
1
1 a⫹u
du ⫽ ln P
P ⫹C
2
a a2u
2u
# "u
1
2
2 a2
du ⫽ ln P u ⫹ "u 2 2 a 2 P ⫹ C
32.
34.
1
1
u
au
2
u
⫹C
u du ⫽ ⫺cot u ⫹ C
2
u du ⫽ 12 u ⫹ 14 sin 2u ⫹ C
cos bu du ⫽
sin(a 2 b)u
sin(a ⫹ b)u
⫹
⫹C
2(a 2 b)
2(a ⫹ b)
e au
(a cos bu ⫹ b sin bu) ⫹ C
a ⫹ b2
2
2
# "a
36.
#a
38.
#
2
1
2
⫹ u2
2
⫹ u 2du ⫽
u 2 14 u 2 ⫹ C
u
a2
"a 2 ⫹ u 2 ⫹
ln P u ⫹ "a 2 ⫹ u 2 P ⫹ C
2
2
1
1
u
du ⫽ tan⫺1 ⫹ C
2
a
a
⫹u
"u 2 2 a 2du ⫽
u
a2
"u 2 2 a 2 2
ln P u ⫹ "u 2 2 a 2 P ⫹ C
2
2
Table of Laplace Transforms
+{ f (t)} ⴝ F(s)
1
s
f (t)
+{ f (t)} ⴝ F(s)
20. eat sinh kt
1
s2
k
(s 2 a)2 2 k 2
21. eat cosh kt
s2a
(s 2 a)2 2 k 2
22. t sin kt
2ks
(s ⫹ k 2)2
23. t cos kt
s2 2 k2
(s 2 ⫹ k 2)2
24. sin kt 1 kt cos kt
2ks 2
(s 2 ⫹ k 2)2
G(a ⫹ 1)
, a . ⫺1
s a⫹ 1
25. sin kt 2 kt cos kt
2k 3
(s ⫹ k 2)2
7. sin kt
k
2
s 1 k2
26. t sinh kt
2ks
(s 2 k 2)2
8. cos kt
s
2
s 1 k2
27. t cosh kt
s2 ⫹ k2
(s 2 2 k 2)2
f (t)
1. 1
2. t
3. t n
4. t21/2
5. t1/2
6. t
a
9. sin2kt
10. cos2kt
n!
s
n11
, n positive integer
p
Ä s
!p
2s3/2
2
2k
s(s 2 ⫹ 4k 2)
2
2
s ⫹ 2k
s(s 2 ⫹ 4k 2)
2
2
2
28.
eat 2 ebt
a2b
1
(s 2 a)(s 2 b)
29.
aeat 2 bebt
a2b
s
(s 2 a)(s 2 b)
30. 1 2 cos kt
k2
s(s 2 ⫹ k 2)
31. kt 2 sin kt
k
s 2 k2
k3
s 2(s 2 ⫹ k 2)
32. a sin bt 2 b sin at
13. cosh kt
s
s 2 k2
ab(a 2 2 b 2)
(s ⫹ a 2)(s 2 ⫹ b 2)
33. cos at 2 cos bt
14. sinh 2kt
2k 2
s(s 2 4k 2)
s(b 2 2 a 2)
(s 2 ⫹ a 2)(s 2 ⫹ b 2)
34. sin kt sinh kt
2k2s
s4 1 4k4
15. cosh 2kt
s 2 2 2k 2
s(s 2 2 4k 2)
35. sin kt cosh kt
k(s 2 ⫹ 2k 2)
s 4 ⫹ 4k 4
16. eatt
1
(s 2 a)2
36. cos kt sinh kt
k(s 2 2 2k 2)
s 4 ⫹ 4k 4
17. eatt n
n!
, n a positive integer
(s 2 a)n ⫹ 1
37. sin kt cosh kt ⫹ cos kt sinh kt
2ks 2
s 4 ⫹ 4k 4
18. eat sin kt
k
(s 2 a)2 ⫹ k 2
38. sin kt cosh kt 2 cos kt sinh kt
4k 3
s ⫹ 4k 4
19. eat cos kt
s2a
(s 2 a)2 ⫹ k 2
39. cos kt cosh kt
s3
s 4 ⫹ 4k 4
11. eat
12. sinh kt
1
s2a
2
2
2
2
4
40. sinh kt 2 sin kt
2k 3
s 2 k4
41. cosh kt 2 cos kt
2k 2s
s 2 k4
4
4
1
42. J0(kt)
2
"s 1 k2
s2a
ln
s2b
43.
ebt 2 eat
t
44.
2(1 2 cos at)
t
ln
s2 1 a2
s2
45.
2(1 2 cosh at)
t
ln
s2 2 a2
s2
46.
sin at
t
47.
sin at cos bt
t
a
arctan a b
s
1
1
a1b
a2b
1 arctan
arctan
s
s
2
2
e2a!s
!s
1 2a2/4t
e
!pt
a
2
49.
e2a /4t
3
2"pt
48.
50. erfc a
e2a!s
e2a!s
s
a
b
2!t
t ⫺a2>4t
a
e
2 a erfc a
b
p
Ä
2!t
51. 2
2
52. eabeb t erfc ab!t 1
2
a
b
2!t
53. 2eabeb t erfc ab!t 1
54. e atf (t)
a
a
b 1 erfc a
b
2!t
2!t
e2a!s
s !s
e ⫺a!s
!s(!s ⫹ b)
be ⫺!s
s( !s ⫹ b)
F(s 2 a)
55. 8(t 2 a)
e2as
s
56. f (t 2 a)8(t 2 a)
e⫺asF(s)
57. g(t)8 (t 2 a)
e⫺as+ 5g(t ⫹ a)6
58. f (n)(t)
59. t nf (t)
t
60.
# f (t)g(t 2 t) dt
s nF(s) 2 s n 2 1f (0) 2 p 2 f (n 2 1)(0)
dn
(⫺1)n n F(s)
ds
F(s)G(s)
0
61. d(t)
1
62. d(t 2 a)
e2as
Index
A
Absolute convergence:
of a complex series, 880
definition of, 262
of a power series, 262, 881
Absolute error, 72
Absolute value of a complex number, 822
Absolutely integrable, 778, 783
Acceleration:
centripetal, 487
due to gravity, 24, 487
normal component of, 492
tangential component of, 492
as a vector function, 486
Adams–Bashforth–Moulton method, 307
Adams–Bashforth predictor, 307
Adams–Moulton corrector, 307
Adaptive numerical method, 305
Addition:
of matrices, 369
of power series, 263
of vectors, 322, 323, 329
Adjoint matrix:
definition of, 406
use in finding an inverse, 407
Age of a fossil, 75
Aging spring, 155, 284
Agnew, Ralph Palmer, 29
Air resistance:
nonlinear, 27, 43, 90
projectile motion with, 202, 256
projectile motion with no, 202, 255
proportional to square of velocity, 27, 43, 90
proportional to velocity, 24, 43, 81–82
Airy, George Biddell, 267
Airy’s differential equation:
definition of, 155, 267
solution as power series, 267
solution in terms of Bessel functions, 291
various forms of, 267
Algebraic equations, 376
Aliasing, 792
Allee, Warder Clyde, 89
Allee effect, 89
Alternative form of second translation
theorem, 231
Ambient temperature, 21, 76
Amperes (A), 23
Amplitude:
damped, 158
of free vibrations, 153
time varying, 721, 755
Analytic function:
criterion for, 837
definition of, 834
derivatives of, 870
Analytic part of a Laurent series, 888
Analyticity, vector fields and, 936
Analyticity and path independence, 864
Analyticity at a point:
criterion for, 837
definition of, 263, 834
Angle between two vectors, 334
Angle preserving mappings, 916
Anharmonic overtones, 760
Annular domain, 888
Annulus in the complex plane, 802
Anticommute, 476
Antiderivative:
of a complex function, 865
definition of, 865
existence of, 866
Applications of differential equations:
aging springs, 155, 284
air exchange, 83
air resistance, 24, 27, 43, 81–82, 90, 255–256
Allee effect, 89
bacterial growth, 74
ballistic pendulum, 205
bending of a circular plate, 146–147
buckling of a tapered column, 279
buckling of a thin vertical column, 171, 175
cantilever beam, 168
carbon-dating, 75
I-1
INDEX
I-2
Applications of differential
equations:—(Cont.)
caught pendulum, 194
chemical reactions, 21–22, 87
column bending under it’s own
weight, 292
competing species of animals,
95–96, 656
continuous compound interest, 21, 80
cooling fin, 291–292
cooling/warming, 21, 76
coupled spring/mass system, 196–197
cycloid, 101
damped motion, 176
deflection of a beam, 167–169, 171
double pendulum, 253–254
double spring systems, 155
draining a tank, 23, 26
electrical networks, 96, 98, 252, 255
electrical series circuits, 23, 78,
161, 241
emigration, 87
evaporating raindrop, 29, 82
evaporation, 91
falling bodies with air resistance, 24,
27, 43, 81–82
falling bodies with no air resistance,
23–24, 82
falling chain, 65
floating barrel, 27
fluctuating population, 29
forgetfulness, 28
growth of microorganisms in a
chemostat, 659
hard spring, 188
harvesting, 86
heart pacemaker, 58, 83
hitting bottom, 92
hole drilled through the Earth, 28
immigration, 87, 92
infusion of a drug, 28
leaking tanks, 90
linear spring, 188
lifting a heavy rope, 31
logistic population growth, 84–87
marine toads, invasion of, 103
memorization, 28
mixtures, 22, 77
networks, 96
nonlinear springs, 187–188
nonlinear pendulum, 189, 193,
652–653
orthogonal families of curves, 102
oscillating chain, 759
Ötzi (the iceman), 101
Paris guns, 206–209
pendulum of varying length, 292–293
population dynamics, 20, 25
potassium-argon dating, 76, 97
potassium-40 decay, 97
predator-prey, 94–95, 654–655
projectile motion, 202, 206–209,
255–256
Index
pursuit curves, 194–195
radioactive decay, 21, 74–75, 79–80, 97
reflecting surface, 28–29
restocking, 86
rocket motion, 28, 82–83, 191
rope pulled upward by a constant
force, 31, 192–193
rotating fluid, shape of a, 29
rotating pendulum, 668
rotating rod, sliding bead on a, 204
rotating shaft, 176
rotating string, 171–172
sawing wood, 91
series circuit, 78, 161–163
Shroud of Turin, 80
sinking in water, 27
skydiving, 92
sliding bead, 204, 653
sliding box on an inclined plane, 83
snowplow problem, 29
soft spring, 188
solar collector, 90
spread of a disease, 21
spring coupled pendulums, 259
spring/mass systems, 27, 152–161,
204–205, 251–252
spring pendulum, 205–206
streamlines, 65, 861
suspended cables and telephone wires,
24–25, 190–191
temperature in an annular cooling fin,
291–292
temperature in an annular plate,
176, 751
temperature between concentric
cylinders, 146, 762
temperature between concentric
spheres, 175
temperature in a circular plate, 748
temperature in a cylinder,
756–757, 758
temperature in a quarter-circular
plate, 751
temperature in a ring, 176
temperature in a semiannular plate, 752
temperature in a semicircular plate, 750
temperature in a sphere, 175,
760–761, 762
temperature in a wedge-shaped
plate, 751
terminal velocity, 43, 81–82, 90
time of death, 81
tractrix, 28
tsunami, 90
variable mass, 27–28, 191–192
vibrating beam, 724, 741
vibrating string, 719–722
water clock, 102
Aquatic food chain, 475
Arc, 632
Arc length as a parameter, 484
Archimedes’ principle, 27
Area as a double integral, 534, 545
Area:
of a parallelogram, 342
of a surface, 553
of a triangle, 342
Argument of a complex number:
definition of, 824
principal, 824
properties of, 824
Arithmetic modulo 2, 463
Arithmetic of power series, 263
Associated homogeneous equation, 108
Associated homogeneous system, 597
Associated Legendre differential
equation, 293
Associated Legendre functions, 293
Associative laws:
of complex numbers, 821
of matrix addition, 370
of matrix multiplication, 371
Asymptotically stable critical point, 39,
644, 646
Attractor, 39, 601, 642
Augmented matrix:
definition of, 380
elementary row operations on,
380–381
in reduced row-echelon form, 381
in row-echelon form, 381
row equivalent, 381
Autonomous differential equation:
critical points for, 36
definition of, 36, 149
direction field for, 39
first-order, 36
second-order, 150, 652
translation property for, 40
Autonomous system of differential
equations, 630
Auxiliary equation:
for a Cauchy–Euler equation, 142
for a linear equation with constant
coefficients, 120
rational roots of, 124
Axis of symmetry of a beam, 168
B
Back substitution, 379
Backward difference, 314
Bacterial growth, 20, 74
Balancing chemical equations, 384–385
Ballistic pendulum, 205
Banded matrix, 804
Band-limited signals, 793
Basis of a vector space:
definition of, 355
standard, 355–356
BC, 107
Beams:
axis of symmetry, 168
cantilever, 168
clamped, 168
deflection curve of, 168
elastic curve of, 168
Boundary points, 527, 802, 829
Boundary-value problem (BVP):
deflection of a beam, 167–168
eigenfunctions for, 170, 692
eigenvalues for, 170, 692, 693
the Euler load, 171
homogeneous, 169
nonhomogeneous, 169, 730
nontrivial solutions of, 169
numerical methods for ODEs,
313–314, 316
numerical methods for PDEs, 802,
807, 812
for an ordinary differential equation,
107, 167–172, 693
for a partial differential equation, 711,
715, 716, 719, 725, 730, 747, 767
periodic, 176, 695
regular, 676
rotating string, 171–172
second-order, 107, 169
singular, 695
Bounding theorem for contour
integrals, 856
Boxcar function, 233
Branch cut, 844
Branch of the complex logarithm, 843
Branch point, 887
Branch point of an electrical
network, 383
Buckling modes, 171
Buckling of a tapered column, 279
Buckling of a thin vertical column,
171, 175
Buoyant force, 27
BVP, 107
C
Calculation of order h n, 299
Cambridge half-life of C-14, 75
Cantilever beam, 168
Capacitance, 23
Carbon dating, 75–76
Carrying capacity, 84
Cartesian coordinates, 328
Cartesian equation of a plane, 347
Catenary, 191
Cauchy, Augustin-Louis, 141
Cauchy–Euler differential equation:
auxiliary equation for, 142
definition of, 141
general solutions of, 142–143
method of solution, 142
reduction to constant coefficients, 145
Cauchy–Goursat theorem, 860
Cauchy–Goursat theorem for multiply
connected domains, 861
Cauchy principal value of an
integral, 904
Cauchy–Riemann equations, 835
Cauchy–Schwarz inequality, 338
Cauchy’s inequality, 872
Cauchy’s integral formula, 868
Cauchy’s integral formula for
derivatives, 870
Cauchy’s residue theorem, 900
Cauchy’s theorem, 859
Caught pendulum, 194
Cavalieri, Bonaventura, 206
Cayley, Arthur, 367
Cayley–Hamilton theorem, 426
Center:
as a critical point, 640
of curvature, 495
of mass, 538, 566
Central difference:
approximation for derivatives, 314
definition of, 314
Central force, 490
Centripetal acceleration, 487
Centroid, 538
Chain Rule, 833, APP-2
Chain Rule of partial derivatives,
498–499
Chain rule for vector functions, 483
Change of scale theorem, 218
Change of variables:
in a definite integral, 580–581
in a double integral, 544, 581, 583
in a triple integral, 585
Characteristic equation of a matrix, 419
Characteristic values of a matrix, 418
Characteristic vectors of a matrix, 418
Chebyshev, Pafnuty, 295
Chebyshev polynomials, 295
Chebyshev’s differential equation,
295, 704
Chemical equations, balancing of,
384–385
Chemical reactions:
first-order, 21–22
second-order, 22, 87–88
Chemostat, 659
Cholesky, Andre-Louis, 458
Cholesky’s method, 458
Circle:
in complex plane, 828
of convergence, 881
of curvature, 495
Circle-preserving property, 923–924
Circuits, differential equations of, 23, 78,
161–162, 252, 841
Circular helix, 480
Circulation, 857
Circulation of a vector field, 522
Clamped end conditions of a beam, 168
Classification of ordinary differential
equations:
by linearity, 4, 6
by order, 4, 5
by type, 4
Classification of linear partial differential
equations by type, 710
Classifying critical points, 39,
638–641, 648
Clepsydra, 102
Index
INDEX
embedded, 168
free, 168
simply supported, 168
static deflection of a homogeneous
beam, 167–168
use of the Laplace transform, 232–233
Beats, 166
Bell curve, 768
Bending of a thin column, 175, 285
Bendixson negative criterion, 660
Bernoulli, Jacob, 67
Bernoulli’s differential equation:
definition of, 67
solution of, 67
Bessel, Friedrich Wilhelm, 280
Bessel function(s):
aging spring and, 284
differential equations solvable in terms
of, 283–284
differential recurrence relations for,
285–286
of the first kind, 281
graphs of, 281, 282, 283, 287
of half-integral order, 286
modified of the first kind, 283
modified of the second kind, 283
numerical values of, 285
of order n, 281
of order 12, 286, 287
of order ⫺12, 287
orthogonal set of, 696
properties of, 284
recurrence relation for, 291
of the second kind, 282
spherical, 287
zeros of, 285
Bessel series, 698
Bessel’s differential equation:
general solution of, 281, 282
modified of order n, 283
of order n, 280
of order n ⫽ 0, 247
parametric form of, 282
parametric form of modified
equation, 283
series solution of, 280–281
Biharmonic function, 922
Binary string of length n, 463
Binormal, 492
Bits, 461
Boundary conditions (BC):
homogeneous, 183, 693
mixed, 693
nonhomogeneous, 693
for an ordinary differential equation,
107, 169, 693
periodic, 176
for a partial differential equation,
714–715
separated, 693
time dependent, 732
time independent, 730
Boundary of a set, 829
I-3
INDEX
I-4
Clockwise direction, 547
Closed curve, 516, 528
Closed region in the complex plane, 829
Closure axioms of a vector space, 354
Cn[a, b] vector space, 354
Code, 463
Code word, 463
Coefficient matrix, 385
Coefficients of variables in a linear
system, 377
Cofactor, 395
Cofactor expansion of a determinant,
394–397
Column bending under its own
weight, 292
Column vector, 368, 389
Commutative laws of complex
numbers, 821
Compartmental analysis, 472
Compartmental models, 472
Compatibility condition, 729
Competition models, 95–96, 656
Competitive interaction, 656
Complementary error function, 55, 768
Complementary function:
for a linear differential equation, 114
for a system of linear differential
equations, 597, 614
Complete set of functions, 675
Completing the square, 227
Complex eigenvalues of a
matrix, 422
Complex form of Fourier series,
688–689
Complex function:
analytic, 834, 840, 844, 846
continuous, 832, 833
definition of, 830
derivative of, 833
differentiable, 833
domain of, 830
entire, 834, 840, 846
exponential, 840
hyperbolic, 847
inverse hyperbolic, 850
inverse trigonometric, 849
limit of, 832
logarithmic, 842
as a mapping, 830
periodic, 841, 848
polynomial, 832
power, 844
range of, 830
rational, 832
as a source of harmonic functions, 838
as a transformation, 830
trigonometric, 846
as a two-dimensional fluid flow, 831
Complex impedance, 842
Complex line integrals:
definition of, 854
evaluation of, 854–855, 865, 867
properties of, 856
Index
Complex number(s):
absolute value of, 822
addition of, 820
argument of, 824
associative laws for, 821
commutative laws for, 821
complex powers of, 844
conjugate of a, 821
definition of, 820
distributive law for, 821
division of, 820, 824
equality of, 820
geometric interpretation of, 822
imaginary part of, 820
imaginary unit, 820
integer powers of, 825
logarithm of, 842–843
modulus of, 822
multiplication of, 820, 824
polar form of, 823–824
principal argument of, 824
principal nth root of, 826
pure imaginary, 820
real part of, 820
roots of a, 825–826
subtraction of, 820
triangle inequality for, 822
vector interpretation, 822
Complex plane:
definition of, 822
imaginary axis of, 822
real axis of, 822
sets in, 828–829
Complex potential, 937
Complex powers:
of a complex number, 844
principal value of, 844
Complex sequence, 878
Complex series, 878–879
Complex vector space, 353
Complex velocity potential, 938
Components of a vector, 323, 325, 329
Component of a vector on another
vector, 335
Conformal mapping, 916
Conformal mapping and the Dirichlet
problem, 918–919
Conformal mappings, table of, APP-9
Conjugate complex roots, 120, 121–122,
143–144, 422, 606–607
Conjugate of a complex number,
430, 821
Connected region, 527, 829
Conservation of energy, 533
Conservative force field, 533
Conservative vector field:
definition of, 525, 937
potential function for, 525, 527,
530, 937
test for, 529, 531, 562
Consistent system of linear
equations, 377
Constant Rules, 833, APP-2
Constants of a linear system, 377
Constructing an orthogonal basis:
for R2, 360
for R3, 361
for Rn, 362
Continuing numerical method, 307
Continuity equation, 578–579
Continuity of a complex function, 832
Continuity of a vector function, 481
Continuous compound interest, 10
Contour:
definition of, 854
indented, 906
Contour integral:
bounding theorem for, 856
definition of, 854
evaluation of, 855, 860, 861
fundamental theorem for, 865
independent of the path, 863–864, 865
for the inverse Laplace transform, 782
properties of, 856
Contourplot, 65
Convergence:
of a complex sequence, 878
of a complex series, 878–879
of an improper integral, 212
of a Fourier integral, 778
of a Fourier series, 679
of a Fourier-Bessel series, 700
of a Fourier-Legendre series, 702
of an improper integral, 212
of a complex geometric series, 879
of a power series, 262, 881
Convolution integral, 238, 787
Convolution theorem:
for the Fourier transform, 787
inverse form, 240
for the Laplace transform, 239
Cooling of a cake, 76
Cooling fin, temperature in a, 291–292
Cooling and warming, Newton’s law of,
21, 76
Coordinate planes, 328
Coordinates of a midpoint, 329
Coordinates of a vector relative to a
basis, 356
Coordinates of a vector relative to an
orthonormal basis, 359
Coplanar vectors, 343
Coplanar vectors, criterion for, 343
Cosine series, 683
Cosine series in two variables, 743
Coulomb (C), 23
Coulomb’s law, 558
Counterclockwise direction, 547
Coupled pendulums, 259
Coupled spring/mass system,
196–197
Coupled systems, 611
Cover-up method, 224
Cramer’s rule, 415–416
Crank–Nicholson method, 809–810
Critical loads, 171
Curvilinear motion in the plane, 487
Cycle of a plane autonomous system, 632
Cycloid, 101
Cylindrical coordinates:
conversion to rectangular
coordinates, 568
definition of, 568
Laplacian in, 755
triple integrals in, 569
Cylindrical functions, 758
Cylindrical wedge, 569
D
D’Alembert’s solution, 724
Da Vinci, Leonardo, 19
Damped amplitude, 158
Damped motion, 24, 156, 158–159
Damping constant, 156
Damping factor, 156
Daughter isotope, 97
DE, 4
Decay, radioactive, 21
Decay constant, 74
Decoding a message, 463–464
Definite integral, definition of, 516
Deflation, method of, 441
Deflection curve of a beam, 168
Deflection of a beam, 166–167, 175, 232
Deformation of contours, 869
Degenerate nodes:
stable, 639–640
unstable, 639–640
Del operator, 501–502
DeMoivre’s formula, 825
Density-dependent hypothesis, 84
Dependent variables, 496
Derivative of a complex function:
of complex exponential function, 840
of complex hyperbolic functions, 847
of complex inverse hyperbolic
functions, 850
of complex inverse trigonometric
functions, 850
of the complex logarithm
function, 844
of complex trigonometric
functions, 846
definition of, 833
of integer powers of z, 833
rules for, 833
Derivative of a definite integral, 11, 31
Derivative and integral formulas,
APP-2, APP-3
Derivative of a Laplace transform, 237
Derivative of real function, notation for, 5
Derivative of vector function, definition
of, 482
Determinant(s):
of a 3 3 3 matrix, 394
of a 2 3 2 matrix, 394
cofactors of, 395
definition of, 393
evaluating by row reduction, 402
expansion by cofactors, 397
of a matrix product, 401
minor of, 395
of order n, 394
of a transpose, 399
of a triangular matrix, 401
properties of, 399
Diagonal matrix, 373, 425
Diagonalizability:
criterion for, 446, 448
sufficient condition for, 445, 446
Diagonalizable matrix:
definition of, 445
orthogonally, 448
Diagonalization, solution of a linear
system of DEs by, 611–612
Difference equation replacement:
for heat equation, 807, 809
for Laplace’s equation, 802
for a second-order ODE, 314
for wave equation, 812–813
Difference quotients, 314
Differentiable at a point, 833
Differential:
of arc length, 517, 518
of a function of several variables, 59
nth order operator, 108
operator, 108
recurrence relations, 285–286
of surface area, 554
Differential equation (ordinary):
Airy’s, 267, 270, 284, 291
associated Legendre’s, 293
autonomous, 36, 150, 646
Bernoulli’s, 67
Bessel’s, 247, 280
Cauchy–Euler, 141
Chebyshev’s, 295
with constant coefficients, 120
definitions and terminology, 4
differential form of, 5
Duffing’s, 193
exact, 59
explicit solution of, 8
families of solutions of, 9
first-order, 6,
first-order with homogeneous
coefficients, 66
general form of, 5
general solution of, 11, 53, 112, 113,
121–122, 142–144
Gompertz, 87
Hermite’s, 295, 698
higher-order, 123, 139
homogeneous, 51, 66, 108, 120, 142
implicit solution of, 8
Laguerre’s, 248, 697
Legendre’s, 280
linear, 6, 108
as a mathematical model, 19–20
modified Bessel’s, 283
nonhomogeneous, 51, 108, 113
nonlinear, 6, 84, 147, 187
Index
INDEX
Critical points of an autonomous firstorder differential equation:
asymptotically stable, 39
attractor, 39
definition of, 36
isolated, 42
repeller, 39
semi-stable, 39
unstable, 39
Critical points for autonomous linear
systems:
attractor, 601
center, 640
classifying, 641
definition of, 632
degenerate stable node, 639–640
degenerate unstable node, 639–640
locally stable, 636
repeller, 601
saddle point, 638
stable node, 638
stable spiral point, 640
stability criteria for, 642
unstable, 636
unstable node, 638
unstable spiral point, 640
Critical points for plane autonomous
systems:
asymptotically stable, 644
classifying, 648
stability criteria for, 647
stable, 644
unstable, 644
Critical speeds, 176
Critically damped electrical circuit, 162
Critically damped spring/mass
system, 156
Cross product:
component for of, 338
as a determinant, 339
magnitude of, 341
properties of, 339–340
test for parallel vectors, 341
Cross ratio, 925
Crout, Preston D., 458
Crout’s method, 458
Cryptography, 459
Curl of a vector field:
definition of, 512
as a matrix product, 375
physical interpretation of, 514, 562
Curvature, 491, 494
Curve integral, 516
Curves:
closed, 516
defined by an explicit function, 518
of intersection, 481
parallel, 572
parametric, 480
piecewise smooth, 516
positive direction on, 516
simple closed, 516
smooth, 516
I-5
INDEX
I-6
Differential equation (ordinary):—(Cont.)
with nonpolynomial coefficients, 269
normal form of, 6
notation for, 5
order of, 5
ordinary, 4
ordinary points of, 264–265
parametric Bessel, 282
parametric modified Bessel, 283
particular solution of, 9, 51, 113
piecewise linear, 54
with polynomial coefficients, 264, 272
Ricatti’s, 69
second-order, 6, 117, 120, 136–137, 141
self-adjoint form of, 695–696
separable, 43–44
singular points of, 264
singular solution of, 10
solution of, 7–8
standard form of a linear, 51, 118, 137
substitutions in, 65
superposition principles for linear,
109, 114
system of, 10, 93, 196, 251
Van der Pol’s, 663, 664
with variable coefficients, 141, 262,
271, 280
Differential equation (partial):
classification of linear second-order,
710
definition of, 4
diffusion, 500, 715
heat, 712–713, 716–718, 753
homogeneous linear second-order, 708
Laplace’s, 500, 713, 725
linear second-order, 708
nonhomogeneous linear second-order,
708
order of, 5
Poisson’s, 737
separable, 708–709
solution of, 708
superposition principle for
homogeneous linear, 709
time dependent, 732
time independent, 730
wave, 500, 711–712, 719–722, 753
Differential form, 5, 59
Differential operator, nth order, 108
Differential recurrence relation,
285–286
Differentiation of vector functions, rules
of, 483
Diffusion equation, 500, 715
Dimension of a vector space, 356
Dirac delta function:
definition of, 249
Laplace transform of, 249
Direction angles, 334
Direction cosines, 334
Direction field, 34
Direction numbers of a line, 345
Direction vector of a line, 345
Index
Directional derivative:
computing, 503
definition of, 502
for functions of three variables, 504
for functions of two variables,
502–503
maximum values of, 504–505
Dirichlet condition, 714
Dirichlet problem:
for a circular plate, 748
for a cylinder, 756–757, 764
definition of, 726–727, 803
exterior, 752
harmonic functions and, 918
for a planar region, 803
for a rectangular region, 726
for a semicircular plate, 750
solving using conformal
mapping, 919
for a sphere, 760
superposition principle for, 727
Disconnected region, 527
Discontinuous coefficients, 54–55
Discrete compartmental models,
472–473
Discrete Fourier transform, 789
Discrete Fourier transform pair, 790
Discrete signal, 789
Discretization error, 299
Distance formula, 328
Distance from a point to a line, 338
Distributions, theory of, 250
Distributive law:
for complex numbers, 821
for matrices, 371
Divergence of a vector field:
definition of, 513
physical interpretation of, 514, 578
Divergence theorem, 575
Division of two complex numbers,
820, 824
Domain:
in the complex plane, 829
of a complex function, 830
of a function, 8
of a function of two variables, 496
of a solution of an ODE, 7
Dominant eigenvalue, 438
Dominant eigenvector, 438
Doolittle, Myrick H., 454
Doolittle’s method, 454–455
Dot notation for differentiation, 5
Dot product:
component form of, 332
definition of, 332, 333
properties of, 332
in terms of matrices, 431
as work, 336
Double cosine series, 743
Double eigenvalues, 750
Double integral:
as area of a region, 534, 545
as area of a surface, 553
change of variables, 544
definition of, 534
evaluation of, 536
as an iterated integral, 535
in polar coordinates, 542
properties of, 535
reversing the order of integration
in, 537
as volume, 535
Double pendulum, 253
Double sine series, 743
Double spring systems, 155
Doubly connected domain, 859
Downward orientation of a surface, 556
Drag, 24
Drag coefficient, 24
Drag force, 206
Draining a tank, 23, 26
Driven motion:
with damping, 158–160
without damping, 160–161
Driving function, 57, 152
Drosophila, 85
Drug dissemination, model for, 82
Drug infusion, 28
Duffing’s differential equation, 193
Dulac negative criterion, 661
Dynamical system, 25, 631
E
Ecosystem, states of, 472
Effective spring constant, 155
Effective weight, 490
Eigenfunctions:
of a boundary-value problem, 170,
of a Sturm-Liouville problem,
693–695
Eigenvalues of a boundary-value
problem, 170–171, 692–693
Eigenvalues of a matrix:
approximation of, 437
complex, 422, 606
definition of, 418, 599
of a diagonal matrix, 425
distinct-real, 599
dominant, 437–438
of an inverse matrix, 424
of multiplicity m, 602
of multiplicity three, 605
of multiplicity two, 603
repeated, 602
of a singular matrix, 423
of a symmetric matrix, 430, 604
of a triangular matrix, 425
Eigenvector(s) of a matrix:
complex, 422
definition of, 418, 599
dominant, 438
of an inverse matrix, 424
orthogonal, 431
Elastic curve, 168
Electrical circuits, 23, 78, 96, 161–163,
241–242
Euler’s constant, 285
Euler’s formula, 121
Euler’s method:
error analysis of, 72, 298–301
for first-order differential equations,
71, 298
for second-order differential
equations, 309
for systems of differential
equations, 312
Evaluation of real integrals by residues,
902–907
Evaporating raindrop, 29, 82
Evaporation, 91
Even function:
definition of, 681
properties of, 682
Exact differential:
definition of, 59
test for, 59
Exact differential equation:
definition of, 59
solution of, 60
Existence and uniqueness of a solution,
16, 106, 594
Existence of Fourier transforms, 783
Existence of Laplace transform, 216
Expansion of a function:
in a complex Fourier series, 689
in a cosine series, 683
in a Fourier series, 677–678
in a Fourier–Bessel series, 700
in a Fourier–Legendre series, 702
half-range, 684
in a Laurent series, 887–889
in a power series, 262–263
in a sine series, 683
in terms of orthogonal functions,
674–675
Explicit finite difference method, 808
Explicit solution, 8
Exponential form of a Fourier
series, 688
Exponential function:
definition of, 840
derivative of, 840
fundamental region of, 841
period of, 841
properties of, 841
Exponential order, 215
Exponents of a singularity, 275
Exterior Dirichlet problem, 752
External force, 158
Extreme displacement, 153
F
Falling bodies, mathematical models of,
23–24, 27
Falling chain, 65
Falling raindrops, 29, 82
Family of solutions, 9
Farads (f), 23
Fast Fourier transform, 788, 791
Fast Fourier transform, computing
with, 795
Fibonacci, Leonardo, 429
Fibonacci sequence, 429
Fick’s law, 101
Filtered signals, 795
Finite difference approximations,
313–314, 802, 807, 812
Finite difference equation, 315
Finite difference method:
explicit, 314, 808
implicit, 809
Finite differences, 314
Finite dimensional vector space, 356
First buckling mode, 171
First harmonic, 722
First moments, 539
First normal mode, 721
First octant, 328
First shifting theorem, 226
First standing wave, 721
First translation theorem:
form of, 226
inverse form of, 226
First-order chemical reaction, 21
First-order differential equations:
applications of, 74, 64, 93
solution of, 44, 52, 60, 66–68
First-order initial-value problem, 14
First-order Runge–Kutta method, 302
First-order system, 592
Five-point approximation for Laplace’s
equation, 802
Flexural rigidity, 168
Flow:
around a corner, 939
around a cylinder, 939
of heat, 712
steady-state fluid, 938
Fluctuating population, 82
Flux and Cauchy’s integral formula, 870
Flux through a surface, 556
Folia of Descartes, 13, 652
Forced electrical vibrations, 161
Forced motion:
with damping, 158
without damping, 160
Forcing function, 115
Forgetfulness, 28
Formula error, 299
Forward difference, 314
Fossil, age of, 75
Fourier coefficients, 678
Fourier cosine transform:
definition of, 783
operational properties of, 784
Fourier integral:
complex form, 780–781
conditions for convergence, 778
cosine form, 779
definition of, 777–778
sine form, 779
Fourier integrals, 905
Index
INDEX
Electrical networks, 96, 252, 255
Electrical vibrations:
critically damped, 162
forced, 161–162
free, 162
overdamped, 162
simple harmonic, 162
underdamped, 162
Elementary functions, 11
Elementary matrix, 388
Elementary operations for solving linear
systems, 378
Elementary row operations on a matrix:
definition of, 380
notation for, 381
Elimination method(s):
for a system of algebraic equations,
378–379, 381
for a system of ordinary differential
equations, 197
Elliptic partial differential
equation, 710
Elliptical helix, 481
Embedded end conditions of a beam,
168, 724
Empirical laws of heat conduction, 712
Encoding a message, 463
Encoding a message in the Hamming
(7, 4) code, 464
Entire function, 834
Entries in a matrix, 368
Epidemics, 21, 86, 99
Equality of complex numbers, 820
Equality of matrices, 369
Equality of vectors, 322, 323, 329
Equation of continuity, 578–579
Equation of motion, 153
Equidimensional equation, 141
Equilibrium point, 36
Equilibrium position of a spring/mass
system, 152
Equilibrium solution, 36, 632
Equipotential curves, 839
Error(s):
absolute, 72
discretization, 299
formula, 299
global truncation, 300
local truncation, 299
percentage relative, 72
relative, 72
round-off, 298
sum of square, 469
Error function, 55, 768
Error-correcting code, 463–467
Error-detecting code, 464, 467
Escape velocity, 194
Essential singularity, 895
Euclidean inner product, 352
Euler, Leonhard, 141
Euler equation, 141
Euler load, 171
Euler–Cauchy equation, 141
I-7
INDEX
I-8
Fourier series:
complex, 688–690
conditions for convergence, 679
cosine, 683
definition of, 678
generalized, 675
sine, 683
in two variables, 741
Fourier sine transform:
definition of, 782
operational properties of, 784
Fourier transform pairs, 783
Fourier transforms:
definitions of, 783
existence of, 783
operational properties of, 783–784
Fourier–Bessel series:
conditions for convergence, 700
definition of, 698–700
Fourier–Legendre series:
conditions for convergence, 702
definition of, 701–702, 703, 704
Fourth-order partial differential equation,
724, 741
Fourth-order Runge–Kutta methods:
for first-order differential equations,
72, 303
for second-order differential
equations, 309
for systems of differential
equations, 311
Free electrical vibrations, 162
Free motion of a spring/mass system:
damped, 155–156
undamped, 152
Free vectors, 322
Free-end conditions of a beam, 168, 723
Frequency of free vibrations, 152
Frequency filtering, 795
Frequency response curve, 166
Frequency spectrum, 690
Fresnel sine integral function, 58, 768
Frobenius, Georg Ferdinand, 273
Frobenius, method of, 273
Frobenius’ theorem, 273
Fubini, Guido, 536
Fubini’s theorem, 536
Fulcrum supported ends of a beam, 168
Full-wave rectification of sine, 247
Function(s):
complementary, 114
complementary error, 55,
of a complex variable, 830
continuous, 832
defined by an integral, 10, 55
differentiable, 833
directional derivative of, 502–503
domain of, 496
driving, 57, 152, 158
error, 55, 768
even, 681
forcing, 115, 152, 158
Fresnel sine integral, 58
Index
generalized, 250
gradient of, 502
graph of, 496
harmonic, 515, 837–838, 918
inner product of, 672
integral defined, 10–11
input, 57, 115
odd, 681
orthogonal, 672
output, 57, 115
partial derivative of, 497–498
periodic, 676
polynomial, 832
potential, 575, 937
power, 844, 914
of a real variable, 830
range of, 496
rational, 832
sine integral, 58, 782
stream, 938
of three variables, 497
as a two-dimensional flow, 831
of two variables, 496
vs. solution, 8
vector, 480
weight, 675
Fundamental angular frequency, 690
Fundamental critical speed, 176
Fundamental frequency, 722
Fundamental matrix:
definition of, 616
matrix exponential as a, 623
Fundamental mode of vibration, 721
Fundamental period, 676, 690
Fundamental region of the complex
exponential function, 841
Fundamental set of solutions:
definition of, 111, 596
existence of, 112, 596
Fundamental theorem:
of algebra, 872–873
of calculus, 11, 526
for contour integrals, 865
for line integrals, 526
G
g, 24, 152
Galileo Galilei, 24, 206
Gamma function, 217, 281, APP-4
Gauss’ law, 580, 937
Gauss’ theorem, 575
Gaussian elimination, 381
Gauss–Jordan elimination, 381
Gauss–Seidel iteration, 387, 805
General form of an ordinary differential
equation, 5
General solution:
of Bessel’s equation, 281, 282
definition of, 11, 53, 112, 113
of a homogeneous linear differential
equation, 112
of a homogeneous second-order linear
differential equation, 121–122
of a homogeneous system of linear
differential equations, 596
of a linear first-order equation, 53
of linear higher-order equations,
123–125
of modified Bessel’s equation, 283
of a nonhomogeneous linear
differential equation, 113
of a nonhomogeneous system of linear
differential equations, 597
of parametric form of Bessel’s
equation, 282
of parametric form of modified
Bessel’s equation, 283
of a second-order Cauchy–Euler
equation, 142–144
Generalized factorial function, APP-4
Generalized Fourier series, 675
Generalized functions, 250
Generalized length, 673
Geometric series, 879
Geometric vectors, 322
George Washington monument, 168
Gibbs phenomenon, 684
Global truncation error, 300
Globally stable critical point, 659
Gompertz differential equation, 87
Goursat, Edouard, 860
Gradient:
of a function of three variables,
501–502
of a function of two variables, 501–502
geometric interpretation of, 507–508
vector field, 511, 525
Gram–Schmidt orthogonalization
process, 359–362, 676
Graphs:
of a function of two variables, 496
of level curves, 496
of level surfaces, 497
of a plane, 348
Great circles, 572
Green, George, 547
Green’s function:
for an initial-value problem, 178
for a boundary-value problem, 184
relationship to Laplace transform,
242–243
for a second-order differential
equation, 178
for a second-order differential
operator, 178
Green’s identities, 580
Green’s theorem in the plane, 547
Green’s theorem in 3-space, 559
Growth and decay, 21, 74–75
Growth constant, 74
Growth rate, relative, 84
H
Half-life:
of carbon-14, 75
definition of, 75
Homogeneous systems of linear
algebraic equations:
definition of, 377, 384
matrix form of, 385
nontrivial solutions of, 384
properties of, 386
trivial solution of, 384
Homogeneous systems of linear
differential equations:
complex eigenvalues, 606–608
definition of, 592
distinct-real eigenvalues, 599
fundamental set of solutions for, 596
general solution of, 596
matrix form of, 592
repeated eigenvalues, 602–605
superposition principle for, 594
Hoëné-Wronski, Jósef Maria, 111
Hooke’s law, 27, 152
Horizontal component of a vector, 325
Hurricane Hugo, 173–174
Huygens, Christiaan, 206
Hydrogen atoms, distance between,
337–338
Hyperbolic functions, complex:
definitions of, 847
derivatives of, 847
zeros of, 848
Hyperbolic partial differential
equation, 710
I
IC, 14
i, j vectors, 325
i, j, k vectors, 330
Iceman (Ötzi), 101
Identity matrix, 373
Identity property of power series, 263
Ill-conditioned system of equations, 417
Image of a point under a
transformation, 581
Images of curves, 912
Imaginary axis, 822
Imaginary part of a complex number, 820
Imaginary unit, 820
Immigration model, 87, 92
Impedance, 163
Implicit finite difference method, 809
Implicit solution, 8
Improper integral:
convergent, 212, 904
divergent, 212, 904
Improved Euler method, 300
Impulse response, 250
Incompressible flow, 514, 938
Incompressible fluid, 514
Inconsistent system of linear
equations, 377
Indefinite integral, 11, 865
Indented contours, 906
Independence of path:
definition of, 526, 864
test for, 527, 528, 529, 864
Independent variables, 496
Indicial equation, 275
Indicial roots, 275
Inductance, 23, 161–162
Infinite-dimensional vector space, 356
Infinite linearly independent set, 357
Infinite series of complex numbers:
absolute convergence, 880
definition of, 878
convergence of, 879
geometric, 879
necessary condition for
convergence, 879
nth term test for divergence, 880
sum of, 879
Initial conditions (IC), 14, 106, 714
Initial-value problem (IVP):
definition of, 14, 106
first-order, 14, 53
nth-order, 14, 106
second-order, 14
for systems of linear differential
equations, 594
Inner partition, 564
Inner product:
of two column matrices, 431
definition of, 332, 333, 672
properties of, 347, 672
space, 358
of two functions, 672
of two vectors, 332, 672
Inner product space, 358
Input function, 57, 115
Insulated boundary, 714
Integers:
modulo 2, 463
modulo 27, 462
Integrable function:
of three variables, 565
of two variables, 534
Integral-defined function, 10–11
Integral equation, 241
Integral transform:
definition of, 212
Fourier, 783
Fourier cosine, 783
Fourier sine, 783
inverse, 782,
kernel of, 782
Laplace, 212, 782
pair, 782
Integral of a vector function, 484
Integrating factor, 52, 62–63
Integration along a curve, 516
Integration by parts, 877
Integrodifferential equation, 241
Interest, compounded
continuously, 80
Interior point, 802
Interior mesh points, 314
Interior point of a set in the complex
plane, 828
Interpolating function, 306
Index
INDEX
of a drug, 21
of plutonium-239, 75
of radium-226, 75
of uranium-238, 75
Half-plane, 828
Half-range expansions, 685
Half-wave rectification of sine, 247
Hamilton, William Rowan, 351
Hamming (7, 4) code, 464
Hamming (8, 4) code, 468
Hamming, Richard W., 464
Hard spring, 188
Harmonic conjugate functions, 838
Harmonic function, 515, 837–838, 918
Harmonic function, transformation
theorem for, 918
Harmonic functions and the Dirichlet
problem, 918
Harvesting, 86
Heart pacemaker, model for, 58, 83
Heat equation:
derivation of one-dimensional
equation, 712
difference equation replacement for,
807, 809
and discrete Fourier series, 791
and discrete Fourier transform,
791–792
one-dimensional, 711–712
in polar coordinates, 753
solution of, 716
two-dimensional, 753
Heaviside, Oliver, 229
Heaviside function, 229
Helmholtz’s partial differential
equation, 763
Helix:
circular, 480
elliptical, 481
pitch of, 481
Henrys (h), 23
Hermite, Charles, 295
Hermite polynomials, 295
Hermite’s differential equation,
295, 698
Higher-order ordinary differential
equations, 105, 123, 141
Hinged end of a beam, 168
Hitting bottom, 92
Hole through the Earth, 28
Homogeneous boundary conditions,
169, 183, 693
Homogeneous boundary-value problem,
169, 715
Homogeneous first-order differential
equation:
definition of, 66
solution of, 66
Homogeneous function, 66
Homogeneous linear differential
equation:
ordinary, 51, 108
partial, 708
I-9
INDEX
Interval:
of convergence, 262
of definition of a solution, 7
of existence and uniqueness, 16
of validity of solution, 7
Invariant region:
definition of, 662
Types I and II, 662
Invasion of the marine toads, 103
Inverse cosine function:
derivative of, 850
as a logarithm, 849
Inverse hyperbolic functions:
definition of, 850
derivatives of, 850
as logarithms, 850
Inverse integral transform:
Fourier, 783
Fourier cosine, 783
Fourier sine, 783
Laplace, 218, 782
Inverse of a matrix:
definition of, 405
by the adjoint method, 406–407
by elementary row operations, 409
properties of, 406
using to solve a system, 411–412
Inverse power method, 443
Inverse sine function:
definition of, 849
derivative of, 850
as a logarithm, 849
Inverse tangent function:
derivative of, 850
as a logarithm, 849
Inverse transform, 218, 782, 783
Inverse transformation, 582
Inverse trigonometric functions:
definitions of, 849
derivatives of, 850
Invertible matrix, 405
Irregular singular point, 272
Irrotational flow, 514, 938
Isocline, 35, 41
Isolated critical point, 42
Isolated singularity:
classification of, 894–895
definition of, 887
Iterated integral, 535
IVP, 14
J
Jacobian determinant, 582
Jacobian matrix, 647
Joukowski airfoil, 915
Joukowski transformation, 915
K
Kepler’s first law of planetary
motion, 490
Kernel of an integral transform, 782
Kinetic friction, 54
Kirchhoff’s first law, 96
I-10
Index
Kirchhoff’s point and loop rules, 383
Kirchhoff’s second law, 23, 96
L
Lagrange’s identity, 344
Laguerre polynomials, 248
Laguerre’s differential equation,
248, 697
Laplace, Pierre-Simon Marquis de, 213
Laplace transform:
behavior as s S q, 223
of Bessel function of order n ⫽ 0, 247
conditions for existence, 216
convolution theorem for, 238–239
definition of, 212, 782
derivatives of a, 237
of derivatives, 220
of differential equations, 221
differentiation of, 237
of Dirac delta function, 249
existence of, 216
inverse of, 218, 782
of an integral, 240
as a linear transform, 214
and the matrix exponential, 623
of a partial derivative, 770
of a periodic function, 244
of systems of ordinary differential
equations, 251
tables of, 215, APP-6
translation theorems for, 226, 230
of unit step function, 230
Laplace’s equation, 486, 500, 501,
711–712, 725, 748, 802
Laplace’s partial differential equation:
in cylindrical coordinates, 756
difference equation replacement
for, 802
in polar coordinates, 748
maximum principle for, 727
solution of, 725
in three dimensions, 744
in two dimensions, 711–712, 725
Laplacian:
in cylindrical coordinates, 755
definition of, 515, 712
in polar coordinates, 748
in rectangular coordinates, 712
in spherical coordinates, 760
in three dimensions, 712
in two dimensions, 712
Lascaux cave paintings, dating of, 80
Latitude, 572
Lattice points, 802
Laurent series, 888
Laurent’s theorem, 889–890
Law of conservation of mechanical
energy, 533
Law of mass action, 87
Law of universal gravitation, 28
Laws of exponents for complex
numbers, 845
Laws of heat conduction, 712
Leaking tank, 89–90
Leaning Tower of Pisa, 24
Learning theory, 28
Least squares, method of, 468–470
Least squares line, 92, 469
Least squares parabola, 470–471
Least squares solution, 470
Legendre, Adrien-Marie, 280
Legendre associated functions, 293
Legendre functions, 288–289, 293,
294–295
Legendre polynomials:
first six, 289
graphs of, 289
properties of, 289
recurrence relation for, 289
Rodriques’ formula for, 290
Legendre’s differential equation:
associated, 293
of order n, 280
series solution of, 288–289
Leibniz notation, 5
Leibniz’s rule, 31
Length of a space curve, 484
Length of a vector:
in 3-space, 333
in n-space, 352
Leonardo da Vinci, 19
Level curves, 46, 496
Level of resolution of a mathematical
model, 19
Level surfaces, 497
L’Hôpital’s rule, 160–161, 216,
282, 901
Liber Abbaci, 429
Libby half-life, 75
Libby, Willard F., 75
Liebman’s method, 806
Limit cycle, 662, 664
Limit of a function of a complex
variable:
definition of, 832
properties of, 832
Limit of a vector function, 481
Line of best fit, 469
Line integrals:
around closed paths, 519, 528,
546–547
as circulation, 522
complex, 854
in the complex plane, 854
definition of, 516–517
evaluation of, 517, 518, 520
fundamental theorem for, 526
independent of the path, 526
in the plane, 517
in space, 520
as work, 521
Line segment, 345
Lineal element, 34
Linear algebraic equations, systems of,
376–377
Linear combination of vectors, 324
nonhomogeneous, 708
solution of, 708
superposition principle for, 709
Linear spring, 188
Linear system:
of algebraic equations, 376
definition of, 377
of differential equations, 93, 592
rank and, 392
Linear transform, 214, 219
Linearity:
of a differential operator, 108
of the inverse Laplace transform, 219
of the Laplace transform, 212
Linearity property, 108
Linearization:
of a function f(x) at a number, 70, 643
of a function f(x, y) at a point, 643
of a nonlinear differential equation, 189
of a nonlinear system of differential
equations, 645–646
Linearly dependent set of functions, 109
Linearly independent set of
functions, 109
Lines of force, 511
Lines in space:
direction numbers for, 345
direction vector for, 345
normal, 509
parametric equations of, 345
symmetric equations of, 346
vector equation for, 345
Liouville’s theorem, 872
Lissajous curve, 202, 256
Local linear approximation, 70, 643, 646
Local truncation error, 299
Locally stable critical point, 636
Logarithm of a complex number:
branch cut for, 844
branch of, 843
definition of,
derivative of, 844
principal branch, 843
principal value of, 843
properties of, 844
Logistic curve, 85
Logistic equation:
definition of, 69, 84–85
modifications of, 86
solution of, 85
Logistic function, 85
Logistic growth, 84–86
Longitude, 572
Loop rule, Kirchhoff’s, 383
Losing a solution, 45
Lotka–Volterra competition model, 96
Lotka–Volterra predator-prey model, 95
Lower bound for the radius of
convergence, 265
Lower triangular matrix, 372
LRC-series circuit:
differential equation of, 23, 161–162
integrodifferential equation of, 241
LR-series circuit, differential equation
of, 78
LU-decomposition of a matrix, 452
LU-factorization of a matrix, 452–455
M
Maclaurin series, 263, 884, 885
Maclaurin series representation:
for the cosine function, 263
for the exponential function, 263
for the sine function, 263
Magnification in the z-plane, 913
Magnitude of a complex number, 430
Magnitude of the cross product, 341, 342
Magnitude of a vector, 324, 333
Main diagonal entries of a matrix, 368
Malthus, Thomas, 20
Malthusian model, 20
Mapping, 581, 912
Mapping, conformal, 916
Marine toad invasion model, 103
Mass:
center of, 538
as a double integral, 538
of a surface, 554
Mass action, law of, 87
Mathematical model, 19–20, 151,
167, 187
Matrix (matrices):
addition of, 369
adjoint, 406
associative law, 370, 371
augmented, 380
banded, 804
characteristic equation of, 419
coefficient, 385
column vector, 368
commutative law, 370
definition of, 368
determinant of, 393–395
diagonal, 373, 425
diagonalizable, 445
difference of, 369
distributive law, 371
dominant eigenvalue of, 437–438
eigenvalues of, 418, 422, 424, 425
eigenvectors of, 418, 422, 424, 425
elementary, 388
elementary row operations on,
380–381
entries (or elements) of, 368
equality of, 369
exponential, 612–622
fundamental, 616
identity, 373
inverse of, 405, 407–408, 409
invertible, 405
Jacobian, 647
lower triangular, 372, 425
LU-factorization of, 452–453
main diagonal entries of, 368
multiplication, 370
multiplicative identity, 373
Index
INDEX
Linear dependence:
of a set of functions, 109–110
of a set of vectors, 355, 357
of solution vectors, 595
Linear donor-controlled hypothesis, 472
Linear equation in n variables, 376
Linear first-order differential equation:
definition of, 6, 50
general solution of, 53
homogeneous, 51
integrating factor for, 52
method of solution, 52
nonhomogeneous, 51
singular points of, 53
standard form of, 51
variation of parameters for, 51
Linear fractional transformation, 923
Linear independence:
of a set of functions, 109–110
of a set of vectors, 355
of solution vectors, 595
of solutions of linear DEs, 110–111
Linear momentum, 490
Linear operator, 108, 197
Linear ordinary differential equations:
applications of, 58, 74–84, 151, 167
associated homogeneous, 108
auxiliary equation for, 120, 142
boundary-value problems for, 107,
167, 183
with constant coefficients, 120
complementary function for, 114
definition, 6,
first-order, 6, 50
general solution of, 53, 112, 113,
121–122, 142–143
higher-order, 106, 108, 123, 141
homogeneous, 51, 108, 120
indicial equation for, 275
initial-value problems for, 14, 74, 106,
149, 151, 177
infinite series solutions for, 265, 273
nonhomogeneous, 51, 127, 136
nth-order initial-value problem for, 106
ordinary points of, 264–265
particular solution for, 51, 113,
127, 136
piecewise, 54
reduction of order, 117–119
second-order, 6, 107, 117–119, 120
singular points of, 53, 264–265,
272, 273
standard forms of, 51, 118, 137,
139, 177
superposition principles for, 109, 114
with variable coefficients, 141, 261,
264, 269, 271–278, 280
Linear partial differential equation, 708
Linear regression, 103
Linear second-order partial differential
equations:
classification of, 710
homogeneous, 708
I-11
INDEX
I-12
Matrix (matrices):—(Cont.)
multiplicative inverse, 405
nilpotent, 430, 476, 626
null-space of, 387
nonsingular, 405, 406
order n, 368
orthogonal, 413, 433–434
orthogonally diagonalizable, 448
partitioned, 376
powers of, 426
product of, 370
rank of, 389–390
reduced row-echelon form, 381
rotation, 375
row-echelon form, 381
row equivalent, 381
row reduction of, 381
row space of, 389
row vector, 368
scalar, 373
scalar multiple of, 369
similar, 426
singular, 406
size, 368
skew-symmetric, 405
sparse, 804
square, 368
stochastic, 426
sum of, 369
symmetric, 373, 604
of a system, 380
trace of a, 636
transpose of, 371
triangular, 372, 425
tridiagonal, 810
upper triangular, 372
zero, 372
Matrix addition, properties of, 370
Matrix exponential:
computation of, 622, 623
definition of, 622
derivative of, 623
as a fundamental matrix, 623
as an inverse Laplace transform, 623
Matrix form of a system of linear
algebraic equations, 385
Matrix form of a system of linear
differential equations, 592
Maximum principle, 727
Maxwell, James Clerk, 514
Maxwell’s equations, 515
Meander function, 247
Memorization, mathematical model for, 28
Meridian, 572
Mesh:
points, 314, 802
size, 802
Message, 463
Methane molecule, 337–338
Method of deflation, 441
Method of diagonalization:
for homogeneous systems of linear
DEs, 511
Index
for nonhomogeneous systems of linear
DEs, 619
Method of Frobenius, 273–277
Method of isoclines, 35
Method of least squares, 468–470
Method of separation of variables:
for ordinary differential equations,
43–44
for partial differential equations, 708
Method of undetermined coefficients:
for nonhomogeneous linear DEs, 127
for nonhomogeneous systems of linear
DEs, 614–616
Method of variation of parameters:
for nonhomogeneous linear DEs, 51,
136–137
for nonhomogeneous systems of linear
DEs, 616–619
Midpoint of a line segment in space, 329
Minor determinant, 395
Mises, Richard von, 438
Mixed boundary conditions, 693
Mixed partial derivatives:
definition of, 498
equality of, 498
Mixtures, 22, 77, 94
ML-inequality, 857
Mm, n vector space, 373
Möbius strip, 555
Modeling process, steps in, 20
Modifications of the logistic equation,
86–87
Modified Bessel equation:
of order n, 283
parametric form of, 283
Modified Bessel function:
of the first kind, 283
of the second kind, 283
Modulus of a complex number, 822
Moments of inertia, 539
Moments of inertia, polar, 546
Motion:
on a curve, 486
in a force field, 151
Moving trihedral, 492
Multiplication:
of complex numbers, 820, 824
of matrices, 370
of power series, 263
by scalars, 322, 323, 329
Multiplication rule for undetermined
coefficients, 132
Multiplicative inverse of a matrix, 405
Multiplicity of eigenvalues, 421,
602–605
Multiply connected domain, 859
Multiply connected region, 527
Multistep numerical method, 307
N
n-dimensional vector, 351–352
Negative criteria, 660, 661
Negative direction on a curve, 547
Negative of a vector, 323
Neighborhood, 828
Net flux, 512
Networks, 96, 252
Neumann condition, 714
Neumann problem:
for a circular plate, 753
for a rectangle, 729
Newton, Isaac, 206
Newton’s dot notation, 5
Newton’s law of air resistance, 206
Newton’s law of cooling/warming,
21, 76
Newton’s law of universal gravitation, 28
Newton’s laws of motion:
first, 23
second, 23, 27, 191–192
Nilpotent matrix, 430, 476, 626
Nodal line, 755
Nodes:
of a plane autonomous system,
638–640, 642
of a standing wave, 721
Nonconservative force, 533
Nonelementary integral, 11, 55
Nonhomogeneous boundary
condition, 693
Nonhomogeneous boundary-value
problem, 169, 183, 693, 730
Nonhomogeneous linear differential
equation:
definition of, 108
general solution of, 113
initial-value problem for, 106
ordinary, 51, 108
partial, 708
particular solution of, 113
Nonhomogeneous systems of algebraic
equations, 377
Nonhomogeneous systems of linear
differential equations:
complementary function of, 597
definition of, 592
general solution of, 597
initial-value problem for, 594
matrix form of, 592
normal form of, 592
particular solution of, 596
solution vector of, 593
Nonisolated singular point, 888
Nonlinear mathematical models,
84, 652
Nonlinear ordinary differential
equation, 6, 147
Nonlinear oscillations, 653
Nonlinear pendulum, 189, 652–653
Nonlinear spring, 188
Nonlinear systems of differential
equations, 93, 629
Nonoriented surface, 555
Nonpolynomial coefficients, 269
Nonsingular matrix, 405, 406, 409
Nontrivial solution, 169
predictor-corrector methods, 300, 307
Runge–Kutta methods, 72, 302,
309, 311
shooting method, 316
single-step method, 307
stability of, 308
stable, 308
starting method, 307
unstable, 308
using the tangent line, 71
Numerical solution curve, 73
Numerical solver, 72
Numerical values of Bessel
functions, 285
O
Octants, 328
Odd function:
definition of, 681
properties of, 682
ODE, 4
Ohms (⍀), 23
Ohm’s law, 79
One-dimensional heat equation:
definition of, 711–712
derivation of, 712–713
One-dimensional phase portrait, 37
One-dimensional wave equation:
definition of, 711–712
derivation of, 713
One-parameter family of solutions, 9
One-to-one transformation, 582
Open annulus, 829
Open disk, 828
Open region, 527
Open set, 828
Operational properties of the Laplace
transform, 214, 220, 226, 230, 231,
237, 239, 240, 244, 249
Operator, differential, 108, 197
Order of a differential equation, 4, 5
Order, exponential, 215
Order of a Runge–Kutta method,
302–303
Order of integration, 537, 565–567
Ordered n-tuple, 351–352, 354
Ordered pair, 323, 327, 351, 354
Ordered triple, 328, 351, 354
Ordinary differential equation, 4
Ordinary point of an ordinary differential
equation:
definition of, 264, 265
solution about, 265–269
Orientable surface:
definition of, 555
of a closed, 556
Orientation of a surface:
downward, 556
inward, 556
outward, 556
upward, 556
Orthogonal basis for a vector space, 359,
360, 361, 362
Orthogonal diagonalizability:
criterion for, 448
definition of, 448
Orthogonal eigenvectors, 431–432
Orthogonal family of curves, 102, 839
Orthogonal functions, 672
Orthogonal matrix:
constructing an, 434
definition of, 433
Orthogonal projection of a vector onto a
subspace, 361
Orthogonal series expansion, 674–675
Orthogonal set of functions, 673
Orthogonal with respect to a weight
function, 675
Orthogonal surfaces at a point, 510
Orthogonal trajectories, 102
Orthogonal vectors, 333
Orthogonally diagonalizable matrix, 448
Orthonormal basis:
definition of, 359
for R n, 359
for a vector space, 359
Orthonormal set of functions, 673
Orthonormal set of vectors, 433
Oscillating chain, 759
Osculating plane, 492
Ötzi (the iceman), 101
Output function, 57, 115
Overdamped electrical circuit, 162
Overdamped spring/mass system, 156
Overdamped system, 654
Overdetermined system of linear
algebraic equations, 386
Overtones, 722
INDEX
Norm:
of a column vector (matrix), 431
of a function, 673
of a partition, 516, 534
square, 673
of a vector, 324, 352, 358
Normal component of acceleration, 492
Normal form:
of an ordinary differential equation, 6
of a system of linear first-order
equations, 592
Normal line to a surface, 509
Normal modes, 721
Normal plane, 492
Normal vector to a plane, 347
Normalization of a vector, 324, 352
Normalized eigenvector, 434
Normalized set of orthogonal
functions, 674
Notation for derivatives, 5
n-parameter family of solutions, 9
n-space (R n):
coordinates relative to an orthonormal
basis, 356
dot (or inner product) in, 352
length (or norm) in, 352
orthogonal vectors in, 352
orthonormal basis for, 359
standard basis for, 356
unit vector in, 352
vector in, 352
zero vector in, 352
nth root of a nonzero complex number,
825–826
nth roots of unity, 797
nth term test for divergence, 880
nth-order differential equation expressed
as a system, 310
nth-order differential operator, 108
nth-order initial-value problem, 14, 106
nth-order ordinary differential equation,
5–6, 106, 108
Nullcline, 42
Null-space of a matrix, 387
Number of parameters in a solution of a
linear system of equations, 391
Numerical methods:
absolute error in, 72
Adams–Bashforth–Moulton, 307
adaptive methods, 305
continuing method, 307
Crank–Nicholson method, 809
deflation method, 441
errors in, 72, 298
Euler’s method, 71, 298–300, 305,
309, 312
finite-difference methods, 314, 802,
807, 812
Gauss–Seidel iteration, 805
improved Euler’s method, 300
inverse power method, 443
multistep method, 307
power method, 438
P
Pacemaker, heart, 58, 83
Parabolic partial differential
equation, 710
Parallel vectors:
definition of, 322
criterion for, 341
Parallels, 572
Parametric curve:
closed, 516
definition of, 480
piecewise smooth, 516
positive direction on, 516
simple closed, 516
smooth, 482, 516
in space, 480
Parametric equations for a line in
space, 345
Parametric form of Bessel equation:
of order n, 696
of order n, 282
in self-adjoint form, 696
Parametric form of modified Bessel
equation of order n, 283
Parent isotope, 97
Paris Guns, 206–209
Parity, 463
Index
I-13
INDEX
I-14
Parity check bits, 463
Parity check code, 463
Parity check equations, 465
Parity check matrix, 465
Parity error, 464
Partial derivatives:
Chain Rule for, 498–499
definition of, 497
generalizations of, 499
higher-order, 498
mixed, 498
with respect to x, 497
with respect to y, 497
second-order, 498
symbols for, 498
third-order, 498
tree diagrams for, 499
Partial differential equation, linear
second order:
definition of, 708
elliptic, 710, 802
homogeneous, 708
hyperbolic, 710, 802, 812
linear, 708
nonhomogeneous, 708
parabolic, 710, 802, 807
separable, 708
solution of, 708
Partial fractions, use of, 219, 223–224
Particular solution:
definition of, 9, 113
of Legendre’s equation, 288–289
of a nonhomogeneous system of linear
DEs, 596, 614, 616, 619
by undetermined coefficients, 127–134
by variation of parameters, 136–140
Partitioned matrix, 376
Path independence:
definition of, 526
tests for, 529, 531
Path of integration, 525
Pauli spin matrices, 476
PDE, 4
Pendulum:
ballistic, 205
double, 253–254
free damped, 191
linear, 189
nonlinear, 189
oscillating, 190
physical, 189
rotating, 668
simple, 187–188
spring, 205–206
spring-coupled, 259
of varying length, 292–293
whirling, 190
Percentage relative error, 72
Perihelion, 491
Period:
of the complex exponential
function, 841
of the complex sine and cosine, 848
Index
of the complex hyperbolic sine and
cosine, 848
Period of free vibrations, 152
Periodic boundary conditions, 176, 695
Periodic boundary-value problem, 695
Periodic driving force, 160, 686
Periodic extension, 680
Periodic functions:
definition of, 244, 676, 841
Laplace transform of, 244
Periodic solution of a plane autonomous
system, 632
Phase angle, 153
Phase line, 37
Phase plane, 593, 600, 637
Phase portrait:
for first-order differential
equations, 37
for systems of two linear first-order
differential equations, 600, 637
for systems of two nonlinear firstorder differential equations,
648–650
Phase-plane method, 649
Physical pendulum, 189
Piecewise-continuous function:
definition of, 215
Laplace transform of, 216
Piecewise-defined solution of an ordinary
differential equation, 10, 47
Piecewise-linear differential equation,
54, 211
Piecewise-smooth curve, 516
Pin supported end of a beam, 168
Pitch, 375–376
Pitch of a helix, 481
Planar transformation, 912
Plane(s):
Cartesian equation of, 347
curvilinear motion in, 487
graphs of, 348–349
line of intersection of two, 349
normal vector to, 347
perpendicular to a vector, 347–348
phase, 600, 637
point-normal form of, 347
trace of, 348
vector equation of, 347
Plane autonomous system:
changed to polar coordinates, 633
types of solutions of a, 632
in two variables, 631
Plane autonomous system, solutions of:
arc, 632
constant, 632
periodic (cycle), 632
Plucked string, 714, 720–721, 725
Plutonium-239, half-life of, 75
Poincare–Bendixson theorems, 662,
664–665
Point-normal form of an equation of a
plane, 347
Point rule, Kirchhoff’s, 383
Poisson integral formula:
for unit disk, 934
for upper half-plane, 932–933
Poisson’s partial differential
equation, 737
Polar coordinates, 544–545,
Polar form of a complex number,
823–824
Polar moment of inertia, 546
Polar rectangle, 542
Pole:
definition of, 894–895
of order n, 894–895, 896
residue of, 897–898
simple, 894–895
Polynomial function, 832
Population, mathematical models for, 20,
69, 74, 84, 86–87, 94–96, 654, 656
Position vector, 323, 329
Positive criteria, 662
Positive direction on a curve, 516,
546–547
Potassium-argon dating, 76, 97
Potassium-40 decay, 97
Potential:
complex, 937
complex velocity, 938
energy, 534
function, 525, 937
Power function, 914
Power method, 438
Power rule of differentiation, 833, APP-2
Power series:
absolute convergence of, 262
arithmetic of, 263
center of, 262
circle of convergence, 881
convergence of, 262
defines a function, 263
definition of, 262
differentiation of, 263
identity property of, 263
integration of, 263
interval of convergence, 262
Maclaurin, 263
radius of convergence, 262
ratio test for, 262
represents a continuous function, 263
represents an analytic function, 263
review of, 262–263
shift of summation index, 263
solutions of differential equations,
264–265
Taylor, 263
Powers, complex, 844
Powers, integer, 825
Powers of a matrix, 426–428
Predator-prey model, 94–95, 654–655
Predictor-corrector methods, 300, 307
Prime meridian, 572
Prime notation, 5
Principal argument of a complex
number, 824
Principal axes of a conic, 451
Principal branch of the logarithm, 843
Principal logarithmic function, 843
Principal normal vector, 492
Principal nth root of a complex
number, 826
Principal part of a Laurent series,
888, 894
Principal value:
of a complex power, 844
of an integral, 904
of logarithmic function, 843
Product Rule, 833, APP-2
Projectile motion, 202, 206–209,
255–256
Projection of a vector onto another, 335
p-series, 880
Pure imaginary number, 820
Pure resonance, 160–161
Pursuit curve, 194–195
Pythagorean theorem, 324
Q
R
Radial symmetry, 753
Radial vibrations, 753
Radioactive decay, 21, 74–75
Radioactive decay series, 93
Radiogenic isotope, 97
Radiometric dating methods, 76
Radius of convergence, 262, 881
Radius of curvature, 495
Radius of gyration, 540
Raindrop, velocity of evaporating, 82
Raleigh differential equation, 651
Range of a complex function, 830
Range of a projectile:
with air resistance, 256
with no air resistance, 255
Rank of a matrix:
definition of, 389
by row reduction, 389–390
Ratio test, 262, 880
Rational function, 832
Rational roots of a polynomial
equation, 124
Rayleigh quotient, 439
RC-series circuit, differential equation
of, 78
zero-input, 224
zero-state, 224
Rest point, 642
Rest solution, 177
Restocking a fishery, 86
Reversing the order of integration, 537
Review of power series, 262–263
Riccati’s differential equation, 69
Riemann mapping theorem, 928
Riemann sum, 564
Right-hand rule, 340
RK4 method, 72, 304
RK4 method for systems, 309, 311
RKF45 method, 305
Robin condition, 714
Robins, Benjamin, 205, 207
Rocket motion, 28, 191
Rodrigues’ formula, 290
Roll, 375
Root mean square, 681
Root test, 880
Roots of a complex number,
825–826
Rope pulled upward by a constant
force, 31
Rotational flow, 514
Rotating fluid, shape of, 29
Rotating pendulum, 668
Rotating rod with a sliding bead, 204
Rotating shaft, 176
Rotating string, 171–172
Rotation and translation, 913
Rotation in the z-plane, 913
Round-off error, 298
Row-echelon form, 381
Row equivalent matrices, 381
Row operations, use in finding the
inverse of a nonsingular
matrix, 409
Row reduction, 381
Row space, 389
Row vector, 368, 389
Row vector form of an autonomous
system, 631
R3, 329
R2, 323
Rules of differentiation, 833, APP-2
Runge–Kutta methods:
first-order, 302
fourth-order, 72, 302
second-order, 303
Runge–Kutta–Fehlberg method, 305
Rutherford, Ernest, 76
INDEX
Quadratic form:
definition of, 450
as a matrix product, 450
Qualitative analysis:
of first-order differential equations,
34, 36
of second-order differential equations,
150, 643–644, 652–654
of systems of differential equations,
600–601, 629
Quasi frequency, 158
Quasi period, 158
Quotient Rule, 833, APP-2
Reactance, 163
Reactions, chemical, 21–22, 87
Real axis, 822
Real integrals, evaluation by residues,
902–907
Real part of a complex number, 820
Real power function, 914
Real vector space, 353
Real-valued function, periodic, 676
Reciprocal lattice, 344
Rectangular coordinates, 327–328
Rectangular pulse, 235
Rectified sine wave, 247
Rectifying plane, 492
Recurrence relation:
three term, 268
two term, 266
Reduced row-echelon form of a
matrix, 381
Reduction of order, 117–119
Reflecting surface, 28–29
Region:
closed, 829
in the complex plane, 829
connected, 527
disconnected, 527
with holes, 549
image of, 581
of integration, 534
invariant, 662
multiply connected, 527
open, 527
simply connected, 527
type I (II), 535, 662
Regression line, 92
Regular singular point of an ordinary
differential equation:
definition of, 272
solution about, 273
Regular Sturm–Liouville problem:
definition of, 693
properties of, 693
Relative error:
definition of, 72
percentage, 72
Relative growth rate, 84
Removable singularity, 894, 895
Repeller, 39, 601, 642
Residue(s):
definition of, 897
evaluation of integrals by, 900, 902
at a pole of order n, 898
at a simple pole, 898
Residue theorem, 900
Resistance, 23
Resonance, pure, 160–161
Resonance curve, 166
Resonance frequency, 166
Response:
definition of, 57
impulse, 250
of a series circuit, 78
of a system, 25, 115, 631
S
Saddle point, 638
Sample point, 516, 534, 554, 564
Sampling Theorem, 793
Sawing wood, 91
Sawtooth function, 247
Scalar, 322, 672
Scalar acceleration, 488
Scalar matrix, 373
Index
I-15
INDEX
I-16
Scalar multiple:
of a matrix, 369
of vectors, 322, 323, 329
Scalar triple product, 342
Scaling, 440
Schwartz, Laurent, 250
Schwarz–Christoffel transformations,
928–929
Second derivative of a complex function,
870–871
Second moments, 539
Second-order boundary-value problem,
169, 313, 316, 693
Second-order chemical reaction, 22,
87–88
Second-order DE as a system, 270, 309
Second-order difference equation, 429
Second-order differential operator, 178
Second-order initial-value problem, 14,
106, 309
Second-order Runge–Kutta
method, 303
Second shifting theorem, 230
Second translation theorem:
alternative form of, 231
form of, 230
inverse form of, 230
Self-adjoint form of a linear secondorder DE, 696
Semi-stable critical point, 39
Separable first-order differential
equation:
definition of, 43
solution of, 44
Separable partial differential
equations, 708
Separated boundary conditions, 693
Separation constant, 709
Sequence:
convergent, 878
criterion for convergence, 878
definition of, 878
Sequence of partial sums, 262
Series (infinite):
absolutely convergent, 262, 880
circle of convergence, 881
complex Fourier, 689
of complex numbers, 878
convergent, 262, 879
cosine, 683
definition of, 878
Fourier, 678
Fourier–Bessel, 700
Fourier–Legendre, 702
geometric, 879
interval of convergence, 262
Laurent, 888–889
Maclaurin, 263, 884, 885
necessary condition for
convergence, 879
nth term test for, 880
orthogonal, 674–675
power, 262, 881
Index
radius of convergence, 262, 881
solutions of ordinary differential
equations, 261
sine, 683
Taylor, 263, 882–884
tests for convergence, 262, 880
trigonometric, 677
Series circuits, 23, 78, 161–162
Sets in the complex plane, 828–829
Shaft through the Earth, 28
Shifting summation index, 263
Shooting method, 316
Shroud of Turin, 75, 80
Sifting property, 250
Signal processing, 793
Similar matrices, 416
Simple closed curve, 516
Simple harmonic electrical
vibrations, 162
Simple harmonic motion, 152
Simple pendulum, 189
Simple pole, 894
Simply connected domain, 527, 859
Simply connected region, 527
Simply supported end of a beam, 168
Simply supported end conditions of a
beam, 168
Sine integral function, 58, 768, 782
Sine series, 683
Sine series in two variables, 743
Single-step numerical method, 307
Singular boundary-value problem, 695
Singular matrix, 406, 409
Singular point of a complex function:
definition of, 887
essential, 895
isolated, 887
nonisolated, 888
pole, 894
removable, 894
Singular point of a linear ordinary
differential equation:
definition of, 53, 264, 265
at infinity, 265
irregular, 272
regular, 272
solution about, 271, 273
Singular solution, 10
Singular Sturm–Liouville problem:
definition of, 693
properties of, 693
Sink, 514
Sinking in water, 90
SIR model, 99
Skew-symmetric matrix, 405
Skydiving, 27, 82, 92
Sliding bead, 653
Sliding box on an inclined plane, 83
Slope field, 34
Smooth curve, 516
Smooth function, 482
Smooth surface, 552
Snowplow problem, 29
Soft spring, 188
Solar collector, 90
Solenoidal vector field, 514
Solution curve:
of an autonomous differential
equation, 37
definition of, 8
Solution of a linear second-order partial
differential equation:
definition of, 708
particular, 708–709
Solution of a linear system of DEs,
methods of, 197, 593
Solution of an ordinary differential
equation:
about an ordinary point, 265
about a regular singular point, 273
definition of, 7
domain of, 7
existence and uniqueness of,
15–16, 112
explicit, 8
general, 11, 53, 112, 113
implicit, 8
integral-defined, 10–11, 55, 58
interval of definition, 7
losing a, 45
nontrivial, 169
n-parameter family of, 9
particular, 9,
piecewise-defined, 10
singular, 10
trivial, 7
verification of a, 7, 8
Solution of a system of differential
equations, 10, 197, 593
Solution of a system of linear algebraic
equations:
definition of, 377, 384
number of parameters in a, 391–392
Solution space, 356
Solution vector, 593
Source, 514
Space curve:
definition of, 480
length of, 484
Span, 357
Spanning set, 357
Sparse matrix, 804
Specific growth rate, 84
Speed, 486
Spherical Bessel function:
of the first kind, 287
of the second kind, 287
Spherical coordinates:
conversion to cylindrical coordinates,
570–571
conversion to rectangular coordinates,
570–571
definition of, 570
Laplacian in, 760
triple integrals in, 571
Spherical wedge, 571
Subscript notation, 5
Subspace:
criteria for, 354
definition of, 354
Substitutions:
in differential equations, 65
in integrals, 544, 580
Subtraction of vectors, 323, 329
Successive mappings, 914
Sum of square errors, 469
Sum Rule, 833, APP-2
Summation index, shifting of, 263
Superposition principle:
for BVPs involving the wave
equation, 722
for Dirichlet’s problem for a
rectangular plate, 727–728
for homogeneous linear ODEs, 109
for homogeneous linear PDEs, 709
for homogeneous systems of linear
algebraic equations, 386
for homogeneous systems of linear
DEs, 594
for nonhomogeneous linear ODEs, 114
Surface, orientable, 555–556
Surface area:
differential of, 554
as a double integral, 553
Surface integral:
applications of, 554, 556
definition of, 554
evaluation of, 554
over a piecewise defined surface, 557
Suspended cable, mathematical model of,
24–25, 49, 190–191
Suspension bridge, 24, 49, 190
Sylvester, James Joseph, 351
Symmetric equations for a line, 346
Symmetric matrix:
definition of, 373
eigenvalues for, 430
eigenvectors of, 604
orthogonality of eigenvectors,
431–432
Syndrome, 465
Systematic elimination, 197
Systems of DEs:
higher-order DEs reduced to, 309, 630
numerical solution of, 309, 311
reduced to first-order systems, 310
Systems of linear algebraic equations:
as an augmented matrix, 380
coefficients of, 377
consistent, 377
elementary operations on a, 380
homogeneous, 377, 384
ill-conditioned, 417
inconsistent, 377
Gaussian elimination, 381
Gauss–Jordan elimination, 381
general form of, 377
matrix form of, 385
nonhomogeneous, 377
overdetermined, 386
solution of, 377
superposition principle for, 386
underdetermined, 386
Systems of first-order differential
equations:
autonomous, 630, 631
definition of, 10, 93, 592, 630
linear form of, 592
matrix form of, 592, 631
solution of, 10, 593
Systems of linear algebraic equations,
methods for solving:
using augmented matrices, 380–381
using Cramer’s rule, 415–416
using elementary operations, 378
using elementary row operations,
380–381
using the inverse of a matrix, 411
using LU-factorization, 456
Systems of linear first-order differential
equations, methods for solving:
using diagonalization, 611, 619
using the Laplace transform, 251, 623
using matrices, 598–608
using a matrix exponential, 621–623
using systematic elimination, 197–198
using undetermined coefficients, 614
using variation of parameters, 616–617
T
Tables:
of conformal mappings, APP-9
of derivatives and integrals, APP-2
of Laplace transforms, 215, APP-6
of trial particular solutions, 131
Tangent line, 70
Tangent plane to a surface:
definition of, 508
equation of, 508
vector equation of, 508
Tangent vectors, 482, 491–493
Tangential component of
acceleration, 492
Taylor series, 149, 883
Taylor’s theorem, 884
Telegraph equation, 716
Telephone wires, shape of, 24–25,
190–191
Temperature:
in an annular cooling fin, 291–292
in an annular plate, 176, 751
in a circular plate, 748, 764
in a circular cylinder, 756–757, 764
between concentric cylinders, 146
between concentric spheres, 175, 762
in a cooling/warming body, 21, 76,
80–81
in an infinite cylinder, 758
in an infinite plate, 763
in a one-eighth annular plate, 752
in a quarter-annular plate, 752
in a quarter-circular plate, 751
Index
INDEX
Spiral points:
stable, 640
unstable, 640
Spread of a disease, 21
Spring constant, 152
Spring coupled pendulums, 259
Spring/mass systems, 152–161
Spring pendulum, 205–206
Square errors, sum of, 469
Square matrix, 368
Square norm of a function, 673
Square wave, 247
Stability criteria:
for first-order autonomous
equations, 646
for linear systems, 642
for plane autonomous systems, 647
Stability for explicit finite difference
method, 809, 814
Stability of linear systems, 636
Stable node, 638, 642
Stable numerical method, 308
Stable critical point, 644
Stable spiral point, 640, 642
Staircase function, 235
Standard basis:
for Pn, 355
for R2, 325, 355
for R3, 330, 355
for Rn, 356
Standard Euclidean inner product, 352
Standard form for a linear differential
equation, 51, 118, 137, 264, 272
Standard inner product in Rn, 352
Standing waves, 721, 755
Starting methods, 307
State of a system, 20, 25, 472, 631
State variables, 25
Stationary point, 36
Steady-state current, 79, 162
Steady-state fluid flow, 938
Steady-state solution, 79, 159, 162, 732
Steady-state temperature, 713, 725, 748,
750, 756, 760
Steady-state term, 79, 159
Stefan’s law of radiation, 101
Step size, 71
Stochastic matrix, 426
Stokes, George G., 559
Stokes’ law of air resistance, 207
Stokes’ theorem, 559
Stream function, 938
Streamlines, 65, 861
Streamlining, 939
String falling under its own weight, 771–772
String of length n, 463
Sturm–Liouville problem:
definition of, 693
orthogonality of solutions, 693
properties of, 693
regular, 693
singular, 695
Submatrix, 376
I-17
INDEX
I-18
Temperature:—(Cont.)
in a rectangular parallelepiped, 744
in a rectangular plate, 725
in a rod, 716
in a semiannular plate, 752
in a semicircular plate, 750
in a semi-infinite plate, 729
in a sphere, 760, 762
in a wedge-shaped plate, 751
Terminal velocity of a falling body, 43,
82, 90
Test point, 924
Theory of distributions, 250
Thermal conductivity, 712
Thermal diffusivity, 713
Three-dimensional Laplacian, 712
Three-dimensional vector field, 495–496
3-space (R3), 329
Three-term recurrence relation, 268
Threshold level, 89
Time of death, 81
TNB-frame, 493
Torque, 343
Torricelli, Evangelista, 206
Torricelli’s law, 23
Trace:
of a matrix, 636
of a plane, 348
Tracer, 472
Tractrix, 28, 101
Trajectories, orthogonal, 102
Trajectory, 593, 600, 631
Transfer coefficients, 473
Transfer function, 224
Transfer matrix, 473
Transform pair, 783
Transformation, 581
Transient solution, 159, 162
Transient term, 79, 159
Translation:
and rotation, 913
in the z-plane, 913
Translation on the s-axis, 226
Translation on the t-axis, 230
Translation property for autonomous
DEs, 40
Translation theorems for Laplace
transform, 226, 230
Transpose of a matrix:
definition of, 371
properties of, 372
Transverse vibrations, 713
Traveling waves, 724
Tree diagrams, 499
Triangle inequality, 822
Triangular matrix, 372
Triangular wave, 247
Tridiagonal matrix, 810
Trigonometric functions, complex:
definitions of, 846
derivatives, 846
Trigonometric identities, 846
Trigonometric series, 677
Index
Triple integral:
applications of, 566
in cylindrical coordinates, 569
definition of, 564–565
evaluation of, 565–566
in spherical coordinates, 571
as volume, 566
Triple scalar product, 342
Triple vector product, 342
Triply connected domain, 859
Trivial solution:
defined, 7
for a homogeneous system of linear
equations, 412
Trivial vector space, 353
Truncation error:
for Euler’s method, 299
global, 300
for improved Euler’s method, 301
local, 299
for RK4 method, 305
Tsunami, mathematical model of, 90
Twisted cubic, 494
Twisted shaft, 739
Two-dimensional definite integral, 534
Two-dimensional fluid flow, 511,
514, 831
Two-dimensional heat equation, 741
Two-dimensional Laplace’s
equation, 712
Two-dimensional Laplacian, 712
Two-dimensional vector field, 510–511
Two-dimensional wave equation, 712
Two-point boundary-value problem,
107, 169
2-space (R2), 323
Two-term recurrence relation, 266
Type I (II) invariant region, 662
Type I (II) region, 535
U
Uncoupled linear system, 611
Undamped forced motion, 160
Undamped spring/mass system, 152, 160
Underdamped electrical circuit, 162
Underdamped spring/mass system, 156
Underdamped system, 654
Underdetermined system of linear
algebraic equations, 386
Undetermined coefficients:
for linear differential equations,
127–134
for linear systems, 614
Uniqueness theorems, 16, 106
Unit impulse, 248
Unit step function:
definition of, 229
graph of, 229
Laplace transform of, 230
Unit tangent, 491
Unit vector, 324
Unstable critical point, 39, 644, 646
Unstable node, 638, 642
Unstable numerical method, 308, 809
Unstable spiral point, 640, 642
Unsymmetrical vibrations, 188
Upper triangular matrix, 372
Upward orientation of a surface, 556
USS Missouri, 176
V
Van der Pol’s differential equation,
663, 664
Van der Waal’s equation, 501
Variable mass, 27, 191–192
Variable spring constant, 155
Variables, separable, 43, 66, 68
Variation of parameters:
for linear DEs, 51, 136–140
for systems of linear DEs, 616–617
Vector(s):
acceleration, 486
addition of, 322–323, 329
angle between, 334
binormal, 492
component on another vector, 335
components of a, 323, 329
in a coordinate plane, 323
coplanar, 343
cross product of, 338–339
difference of, 322, 323, 329
differential operator, 501
direction, 345
direction angles of, 334
direction cosines of, 334
dot product of, 332, 333
equality of, 322, 323, 329
equation for a line, 345
equation for a plane, 347
fields, 511
free, 322
function, 480
geometric, 322
horizontal component of, 325
initial point of, 311
inner product, 332, 352
length of, 324, 329, 352
linear combination of, 324
linearly dependent, 355
linearly independent, 355
magnitude of, 324, 329, 333, 352
multiplication by scalars, 322, 323,
324, 329, 352
negative of, 322
norm of, 324, 352
normal to a plane, 347–348
normalization of, 324, 352
in n-space, 352
orthogonal, 333
orthogonal projection onto a
subspace, 361
orthonormal basis, 359
parallel, 322, 341
position, 323, 329
principal normal, 492
projection on another vector, 335
rules of differentiation, 483
smooth, 482
of three variables, 501, 510
of two variables, 501, 510
as velocity, 486
Vector-valued functions, 480
Vector space:
axioms for a, 352
basis for a, 357
closure axioms for a, 353
complex, 353
dimension of, 356
finite dimensional, 356
infinite dimensional, 356
inner product, 357
linear dependence in a, 355, 357
linear independence in a, 355, 357
real, 353
span of vectors in a, 357
subspace of, 354
trivial, 353
zero, 353
Velocity field, 490
Velocity potential, complex, 938
Velocity vector function, 486
Verhulst, P. F., 85
Vertical component of a vector, 325
Vibrating cantilever beam, 741
Vibrating string, 713
Vibrations, spring/mass systems, 152,
196–197
Virga, 29
Viscous damping, 24
Voltage drops, 23
Volterra integral equation, 241
Volterra’s principle, 658
Volume of a parallelepiped, 343
Volume under a surface:
using double integrals, 535
using triple integrals, 566
Von Mises, Richard, 438
Vortex, 942
Vortex point, 642
W
Water clock, 102
Wave equation:
derivation of the one-dimensional
equation, 713
difference equation replacement
for, 812
one-dimensional, 711–712, 812
solution of, 719–720
two-dimensional, 742
Weight function:
of a linear system, 250
orthogonality with respect to, 675
Weighted average, 302
Wire hanging under its own weight,
24, 190–191
Word:
definition of, 463
encoding, 463
Work:
as a dot product, 336
as a line integral, 521
Work done by a constant force, 336
Wronskian:
for a set of functions, 111
for a set of solutions of a
homogeneous linear DE, 111, 137
for a set of solutions of a
homogeneous linear system, 595
Wronskian determinant, 111
X
x-coordinate of a point in 3-space, 328
xy-plane, 328
xz-plane, 328
Y
Yaw, 375–376
y-coordinate of a point in 3-space, 328
Young’s modulus, 168
yz-plane, 328
INDEX
properties of, 324, 332, 339
right-hand rule, 340
scalar multiple of, 322, 323, 329
scalar product, 332
scalar triple product, 342
as a solution of systems of linear
DEs, 593
span of, 357
spanning set for, 357
standard basis for, 325, 330, 356
subtraction of, 323, 329
sum of, 322
tangent to a curve, 486
terminal point of, 311
triple product, 342
in 3-space, 327
in 2-space, 322
unit, 324, 352
unit tangent to a curve, 491
vector triple product, 342
velocity, 486
vertical component of, 325
zero, 322, 329, 352
Vector differential operator, 501
Vector equation for a curve, 480
Vector equation for a line, 345
Vector equation of a plane, 348
Vector fields:
and analyticity, 936
conservative, 525
curl of, 512
definition of, 510–511
divergence of, 512
flux of, 512
gradient, 511
irrotational, 514
plane autonomous system of, 631
rotational, 514
solenoidal, 514
three-dimensional, 510–511
two-dimensional, 510–511
velocity, 511
Vector functions:
as acceleration, 486
continuity of, 481
definition of, 480
derivative of, 482
differentiation of components, 482
higher derivatives of, 483
integrals of, 484
limit of, 481
Z
z-axis in space, 327–328
z-coordinate of a point in 3-space, 328
Zero matrix, 372
Zero vector, 322, 352
Zero vector space, 353
Zero-input response, 224
Zeros:
of Bessel functions, 285
of complex cosine and sine, 846–847
of complex hyperbolic cosine and sine,
848
of a function, 896
of order n, 896
Zero-state response, 224
z-plane, 822
Index
I-19