ORDINARY DIFFERENTIAL
Chapter
J
EQUATIONS
on the study of the different
application to geometry and classical
(Newtonian) physics. The motivating problem throughout is the central
problem of the subject of differential equations : to find a function on the
basis of given information on its derivatives. Observed phenomena in the
sciences seem always to involve rates of change. For example, it is observed
that the rate of acceleration of a falling body is a constant independent of
mass, height, or velocity; the progress of a chemical reaction slows down as
it proceeds, dependent on the quantities of the chemicals involved. These
observations, when made precise, appear as differential equations. In
order to predict (the time it takes for the body to fall a given height, the
amount of new chemicals produced before the reaction stops), the function
described by the differential equation must be found.
The first two sections of the present chapter are devoted to the description
In these next three
ial calculus of
one
chapters
we
shall elaborate
variable and its
of the basic concepts involved ; in the first we shall discuss the differentiation
of vector-valued functions, and the second is devoted to approximation and
Taylor's formula. We also include a brief excursion into the computation of
maxima and minima of functions of several variables subject to constraints
the technique of Lagrange multipliers.
The main theoretical tool in this study is Picard's theorem which gives
conditions under which a differential equation has a solution and only one
by
solution.
essentially tells us what a well-posed problem is,
that well-posed problems are always solvable. The question
This theorem
and asserts
227
228
3
Ordinary Differential Equations
actually producing a formula for the solution, or an algorithm for com
puting approximate values for the solution is another matter altogether.
Several techniques will be exposed in this chapter and Chapter 5 (successive
approximations, series expansions); there are many more very efficient
computational techniques which we shall not develop here.
It will become clear that the subject of ordinary differential equations has a
lot to do with the study of curves (paths of motion). Thus in the next
chapter we shall investigate the geometry of curves and its relation with the
subject of differential equations.
of
3.1
Differentiation
The first
important step
in the
study
of differential
equations
is to consider
vector- valued functions of a real variable as well as real- valued functions.
This is the
appropriate setting for many problems involving differential
particularly relevant when studying equations involving
and is
equations,
derivatives of order greater than one. In the first sections we shall consider
differentiable vector-valued functions of a real variable and introduce a
special technique
Definition 1.
in
a
for
approximating
Let x0 e
of x0
neighborhood
,.
lim
f(*o
+
an
Revalued function defined
f is differentiable at x0 if
f(*o)
t
f-0
exists.
~
Taylor's expansion.
R, and suppose f is
.
0
values:
The limit is called the derivative of f at x0 and is denoted by
an open set U, we say f is differentiable (written f is
If f is defined in
U if
as t
[f(x
-+
+
t)
f(x)]/f
converges for all
x on
U to
a
f'(x0).
C1)
in
continuous function f
0.
That this definition is not
is demonstrated
Proposition 1.
by
the
so
far from the derivative encountered in calculus
following
assertion.
Let f be
an R"-valued function defined in a neighborhood of
(fu ...,/) in coordinates, f is differentiable at x0 if and
only iffi,...,fH are differentiable at x0. Further, f'(x0) (f[(x0),
fn(Xo))-
x0
e
R.
Write f
=
=
.
.
.
,
3.1
Differentiation
229
Proof.
f (Xp +
Q
-
f (Xo)
//l(Xo + Q-/1(Xo)
=
/
The limit
on
the left
as t
I
-*
t
)
0 exists if and
(Proposition 10 in Chapter 2),
Proposition 1 says.
Now if f is
/(x0 + Q-/(x0)\
"'
*
and
only if all the limits on the right exist
equality holds also in the limit. That is all that
differentiable function
interval taking values in R", its
f'(x0) is a vector in R" and points in
the direction of motion of the curve (Figure 3.1). That is, the line through
f(x0) and parallel to f'(x0) is the limiting position of the line through f(x0)
and a nearby point f(x0 + t). For that line is parallel to t~ 1(f(x0 + t)
f(x0)),
0. This line through
and by definition this vector has f'(*o) as limit as t
f(x0) and parallel to f'(x0) is called the tangent line of the curve at f(x0).
From Proposition 1 it easily follows that iff, g are differentiable, so is f + g,
and (f + g)'(xo)
f'(x0) + g'(*o)- The chain rule also follows easily:
image is
a
a curve
in R".
on an
The derivative
-
=
Proposition 2. (Chain Rule I) Let g be a real-valued function defined in a
neighborhood of x0 in R, and differentiable at x. Suppose f is an R"-valued
function which is differentiable at g(x0) (see Figure 3.2). Then fgis differ
entiable at x0 and (f g)'(x0)
g'(x0)f'(g(x0)). (We have written g'(x0)
before f'(g(x0)) as this is the customary way of writing the product of a scalar
and a vector.)
=
Figure
3.1
230
3
Ordinary Differential Equations
i(g(x))
x
g(x)
Figure 3.2
This is of
ordinary
course
true,
just
because it is true in each coordinate, by the
,/), then fg (fig,---,fg),
(fu
Thus if f
chain rule.
=
=
.
.
.
so
(f <?)'
=
=
((/i <?)',. --,(/s0')
(j i0', ,/; 0')
=
0'f
Example
1. Let
f(x)
=
(x, x2, x3), g(t)
=
sin t.
Then
(f g)(t)
=
(sin
t,
sin2f,
sin3 0
(f g)'
o
=
cos
t(l, 2 sin t, 3 sin2f)
Now, there is also
chain rule for
differentiable function
taking a real-valued function of a
(Figure 3.3). Suppose now g is a continuously
defined on an interval / taking values in a domain D
in R".
a
a
vector-valued function
partial
Suppose /
is
real-valued function defined
derivatives continuous.
Then/
g is
a
on
D which has all
real-valued function
on
the
interval /.
For
clarity of exposition, let us take the
as g(x)
(g^(x), g2(x)). Then
coordinates
/(g(*o
case n
=
2.
We
can
write g in
=
+
0) -/(g(*o))
=f(ffi(x0
+
t), g2(x0
+
t)) -f(gi(x0), g2(x0))
=f(gi(x0 0. gi(xo 0) -f(gi(x0), g2(x0
+JXffi(x0), g2(x0 + t)) -f(gi,(x0), g2(x0))
+
+
+
0)
(3.1)
3.1
231
Differentiation
f(s, g2(x0 + 0) is differentiable (it is the restriction of /
g2(xo + 0)- By the mean value theorem, the first difference is
Now the function
to the line y
=
Pi f
-ir
(i> 02(*o + 0)[ffi(*o +
dx
for
some
theorem
between
t,x
we see
0i(*o
for
some
+
gt(x0
+
t)
and
ff i(*o)]
Now
gt(xQ).
applying
the
mean
O-0i(*o)=0iOi)'
^r(Si,02(*o
+
dx
9i(x0)
<
Thus the first difference in
.
O)0'i("i)<
$i
Similarly, the second
dy
~
that
nx between x0 + t and x0
df
0
<
0i(*o
+
<
<
x0 + t
0
x0
0
x0<n2<x0
nx
difference is
(0l(*o) >2)9l(*l2)t
g2(x0)
<
2
<
g2(x0
+
+ t
/(gU))
Figure 3.3
(3.1)
is
value
232
3
Thus,
Ordinary Differential Equations
may rewrite
we
/(g(x0
+
Q
(3.1)
as
/(g(x0))
-
t
=
yx i, 92(x0
Taking the limit as
tinuous), and g2(x0
t
+
-0,
+
we
t), 2
0)ffi(li)
+
have
the
on
both tend to
(9i(x0), kteaOh)
right ^ -*#i(x0) (since
Notice that,
using
<*(/g)
dx
true
(x0)
along
only
in not
Y% (g(*o))0i(*o)
+
=
df(g(x0), g'(x0))
of/ along the
=
exists,
con
so
y (g(*o)) 02'(xo)
the directional derivative
Thus the derivative
derivative
=
gt is
since g2 is continuous. Also
lie between x0 and x0 + t. Since all the
g2(x0)
!, w2 bth tend to x0 since they
derivatives in (3.2) are continuous, the limit
^^ (xo)
(3.2)
(3.3)
notation, (3.3) becomes
<V/(g(x0), g'(x0)>
curve x
=
g(x)
is the
same as
(3.4)
its directional
the tangent direction to the curve (Figure 3.4). This is
R2, but for all R". The derivation is of course the same,
only with the notational complication of many
Figure
3.4
more
variables.
Thus
3.1
Differentiation
233
Proposition 3. (Chain Rule II) Let g be a continuously differentiable
function of a real variable, taking values in a domain D in Rn, and suppose f
is a continuously differentiable real-valued function defined on D.
Thenfo g
is a differentiable function and
(f g)'(0
=
df(g(t), g'(t))
Examples
2. Let
g(t)
=
(sin t,
t),f(x, y)
cos
(a,b))=d-fa
8-fb
dx
dy
df((x, y),
g'(t)
=
+
sin
(cos t,
(f g)'(0
=
=
2t
y2a
+
Then
2xyb
f)
df(g(t), g'(0)
cos
=
xy2.
=
cos2
=
r cos t
+ 2 cos t sin
t(-sin 0
cos t
We can, of course,
sin t cos2 t
verify
this
by
direct substitution, since f
g(t)
=
.
3. Let
g(r)
=
(t, t2, 2t),f(x,
y,
z)
=
xy +
log z.
c
df((x,
g'(t)
(f
=
y,
z), (a, b, c))
=
ya + xb +
-
z
(1, 2t, 2)
g)'(0
=
df((t, t2, 2t), (1, 2t, 2))
=
3t2
+
=
t2
+
2t2
+
-
-
t
4.
mum
Then V/(g(x0)) is
Proposition 3, and/ g has a maxi
orthogonal to g'(t0). For (/ g)'(t0)
df(g(t0), g'(fo))
<V/(g(<o)). g'Oo)>
Suppose/
at r0
.
g
are
given
as
in
=
0,but
(/ g)'Oo)
=
=
234
3
Ordinary Differential Equations
Lagrange Multipliers
This last example serves to provide a method for finding maxima (or
minima) of functions subject to certain constraints. This is the process of
Lagrange multipliers. Suppose/, # are differentiable functions in a certain
domain D in jR". We consider/as the function we are studying and g(x)
0
the constraint. Suppose / has a maximum on g(x)
0 at x0
if
T
Thus,
is a curve in the set {g(x)
0} going through x0 then V/(x0) is orthogonal
to the tangent line to T at x0
For if T is the image of a function <b of a real
and
then
as in Example 4, <V/(x0), 0'(ro)>
variable,
0, and
(b(t0) x0
is
the
line
to
T
at
Now
also
<b
constant, so
g
(p'(t0) spans
tangent
x0
0.
of
on
Thus
at
the
maximum
<V#(x0), <t>'(t0)y
point xQ
/
{g(x) 0},
V/(x0) and Vg(x0) are both orthogonal to all curves through x0 subject to
the constraint g(x)
0. If there are enough such curves, say, so that the
set of tangent vectors fills out a subspace of R" of dimension n
1 then
and
must
be
We
not
here
that
collinear.
will
there
are
V/(x0)
V#(x0)
worry
of
these
but
take
it
for
After
we
are
not
here
curves,
all,
enough
granted.
studying the theory, but only seeking a technique which will provide candi
dates for a maximum point. We can state this principle : if x0 is a maximum
(or minimum) point for /subject to the constraint g(x)
0, then there is a A
=
=
.
=
,
.
=
=
,
.
=
=
=
,
=
such that
V/(x0)
Thus
we can
V/(x)
g(x)
=
Mx0)
find
=
=
possible
x0
by solving the system of equations
Xg(x)
0
(3.5)
for x, X.
Examples
x2
5. We shall find the maximum value of xyz on the unit
1.
+ y2 + z2
1.
Let/(x) xyz, g(x) x2 + y2 + z2
=
V/(x)
Thus
x2
(yz,
+
(yz,
=
xz,
we must
y2
xz,
+
xy)
z2
=
=
xy)
solve
=
1
2X(x, y,z)
Vg(x)
=
=
(2x, 2y, 2z)
-
sphere
3.1
X from
Eliminating
xz
yz
Equations (3.6),
Differentiation
235
obtain
we
xy
(3.7)
xyz
This
z
=
can
0
be written
or
x
=
0
as
or
y
0
=
or
-
=
Thus either
y
of the coordinates is
one
=
-,-
x
zero
(3.8)
-
z
y
x2
or
=
y2
=
z2
Near
point where one of the coordinates is zero, / changes sign, so
these points are disqualified. This leaves any one of the points
l/>/3(+l +1, +!) The value of /at any one of these points is
any
thus 3_3/2 is the maximum.
3_3/2,
+
l)2
6. Find the
point on the curve 2(x
totheorigin. Here#(x,y) 2(x
l)2
=
-
+
+
j2
3y2
4
-
=
4 which is closest
and/(x, y) =x2
+
y2.
Thus
V/= (2x, 2y)
Vg
The
equations
x
=
2X(x
y
=
2(x
-
1), 2y)
1)
Xy
l)2
-
gives
x
values
+
y2
4
=
off at
/
7. Find the
+
is 6.
curve on
1
y2
+
either y
2z2
=
8
=
0
or
(1
Ji)2, (1
are
-
X
=
1.
The second
case
V^O), (2, + ^2). The
+ V2)2 '> and at the s_econd
pair is (1
Clearly, the minimum
(see Figure 3.5).
the first
the maximum is 6
=
equation,
Thus, the candidates
=2.
the value of
x2
(4(x
become
From the second
xyz
=
distance is
|1
the intersection of the two surfaces
-
v
2| and
236
3
Ordinary Differential Equations
Figure
3.5
origin. In this problem we have two constraints,
through the technique. The tangent vector to the
curve is orthogonal to the gradient of both constraining functions, and
at the maximum point V(x2 + y2 + z2) is orthogonal to the curve.
Thus this gradient must be coplanar with the gradients of the
1,
constraining functions. Let f(x) x2 + y2 + z2, g(x) xyz
8. Then V/=2(x, y, z), Vg
h(x) x2 + y2 + 2z2
(yz, xz, xy),
Vn
2(x, y, 2z). We must solve these five equations for x, y, z,X,\i:
which is closest to the
but
we
can
see
=
=
=
=
-
=
2(x, y, z)
xyz
x2
=
+
X(yz,
=
xz,
xy)
+
zfi(x,
2z)
1
y2
+
2z2
8. Let M
V7/x, x
=
We show this
<Tx, x>
=
=
=
for all i and /
8
(a/)
be
a
symmetric
n x n
If T is the transformation
matrix.
on
=
a?
27x
by computation:
(3.9)
j
The kth component of VTx,
with respect to x\ this gives
Ifl*'x' + Zfl7v
j
That
is, af
R" defined by M,
2 fly'xV
".
*
y,
x
is found
by differentiating (3.9)
3.1
But since M is
2(Tx)k.
/OO
The
Then
symmetric, this is the
Vrx, x
=
(Tx, x> must attain a maximum on
Lagrange multiplier procedure tells us
=
V(2xi2-l)|x=Xo
Thus the transformation T has
unit
sphere
2; Ok'x'
+
or
y ak}x>
=
'function
Now, the
the unit sphere, say at x.
that there is a X such that
2Tx
=
2Xx
eigenvector, namely that
an
237
x0
on
the
which maximizes the function (Tx, x>.
continue this idea in order to prove that a transformation given by
symmetric transformation has an orthogonal basis of eigenvectors. For,
We
a
same as
2Tx is established.
=
Vrx,x|x=Xo
Differentiation
can
let x1 be the eigenvector found as in
1
subject to the constraints <x, x>
=
point subject
||x2 1|
to
=
1,
these
constraints,
<x, Xl>
=
0,
we
Example 8. Now maximize (Tx, x>
0. If x2 is the maximum
<x, xx >
have X2, \i2 such that
=
,
2Tx2
=
2X2x2
,
2Tx2
=
/i2Vx, xx
Thus, by the first two equations, x2 is nonzero and orthogonal to xl5 and
by the third, x2 is an eigenvector of T. Now proceed to the constraints
0. The same technique works to produce
1, <x, x^
0, <x, x2>
<x, x>
=
=
third
eigenvector.
eigenvectors.
a
=
We
can
go
on
until
we
have found
n
independent
Examples
9. Let
eigenvectors of M.
eigenvector of M if and only if there is a nonzero vector x
such that (M
XI)x 0. We know the necessary and sufficient
condition for that: det(M
AI) 0. Thus the eigenvalues of M are
0.
Now
the roots of det(M
XT)
and find the
X is
an
=
-
-
-
=
=
238
3
Ordinary Differential Equations
After
a
det(M
computation
AI)
-
(2
=
we
A)3
-
find that
3(2
-
A)
-
+ 2
=
-(A
We find the
eigenvalues are 1,4.
by solving the equations (M
Thus the
I)x
-
=
1)2(A
-
4)
-
corresponding eigenvectors
0, (M
-
4I)x
=
0 for
nonzero
vectors.
1
eigenvalue
:
/I
1
1\
1
1
1
\l
1
1/
M-I=
corresponding eigenvectors: (1, -1, 0), (0, -1, 1)
(Any two independent vectors such that vx + v2 +
1-2
1
1
-2
M-4I=
\
sum
0 will
do.)
1\
1
1-2/
1
is zero, so
independent, so the
of the three
and second
on
=
4:
eigenvalue
The
v3
are
rows
they are dependent. The first
corresponding eigenvector lies
the line
-2x+
2y
x
Such
(0,
y +
a
+
z
=
0
z
=
0
vector is
1, 1)
with
Here
det(M
of
eigenvalues
AI)
-
1, and
eigenvalue
10. Find the
eigenvalue
1
eigenvalue
5: M
:
=
(2
M + I
51
A)2
-
=
=
-
9 which has the roots -1,5,
(3
3\
,
_|
I
1-3
I
eigenvectors of M are (1,
(1, 1, 1) with eigenvalue 4.
Thus the
(1, 1, 1).
,
kills the vector
3\
(1,
1).
.1 kills the vector (1, 1).
-
1, 0),
3.1
Differentiation
239
EXERCISES
1. Differentiate these functions and
graph the
curve
function
defined by the
(a) f(t) e", c a complex number.
(b) f (f ) (cos f, sin f, f ).
(c) f (t)
(a cos t, b sin t).
(d) t(t)
(t2,t3).
(e) t(t)=(t,t2,t3).
(f) f(f)=(sinf,cosf,0).
2. What is the length of t'(t) in each of Exercises
l(a)-(f ) ?
angle between f '(f) and f "(f)?
=
=
=
=
3. At which
pairs of points
are
and (c) of Exercise 1 parallel?
4. At which pairs of points
the tangent lines to the
are
What is the
curves
(a) (c
the tangent lines to Exercises
=
i),
1(b), (f)
parallel?
5. Find the maximum of xy
6. Find the minimum of
on
the
ax2 +
ellipse
by2
=
1.
+ y on the curve xy
1 in the first quadrant.
7. Find the two points on the curves y
x2 and xy
1 which are
closest.
x
=
=
=
8. Minimize x2 +
y2 + z2 on the ellipsoid ax2 + by2 + cz2
1.
straight lines L1 and L2 in space how would you try to find
the points PieZ,1, p2eL2 which are closest (i.e., minimize
I'p
q|| for
=
9. Given two
peL\qeL2)?
10. Find the eigenvalues and
(a)
(I !)
eigenvectors
of these matrices:
("i ?)
*>
(b)
11
Find the
.
(a)
eigenvalues
and
1
0
0
1
0
\-l
0
3/
eigenvectors of
-1\
.
(b)
these matrices :
/2
1
3\
1
0
3
\3
3
PROBLEMS
1. Let f, g be differentiable /{"-valued functions defined on an interval I.
<f, g> is differentiable and
(a) Show that the inner product h
=
h'
<f ', g> + <f, g'>.
(b) Show that ||f ||
<f, f>1/2 is constant if and only if f (x), V(x)
orthogonal for all x.
(c) Give a condition for f to lie on a straight line.
2. Find the point on the intersection of these two surfaces
a2x2 + b2y2 + c2z2
1
\
x2+y2
which is closest to the origin.
=
=
=
=
are
240
3
Ordinary Differential Equations
rectangular box of maximum volume is to be constructed, with sides
parallel to the coordinate planes, one vertex at the origin and the diagonally
1
Find the volume of that
opposite vertex on the plane ax + by + cz
3. A
=
.
box.
community consumes water at the rate of sin2(27rf/24) gallons per
They wish to build a storage tank of capacity Q with a pump of rate
The
w gallons per hour, so that the community will never run out of water.
Minimize this cost for them.
cost is Q + kw.
5. Show that if /is any differentiable function on R3, there are at least
two points x on the unit sphere at which V/(x) is parallel to x.
4. A
hour.
3.2
Formula
Taylor's
order derivatives appear for vector-valued functions
one- variable calculus.
Higher
just
as
they
do in the usual
Let f be
Definition 2.
an
.Revalued function defined
f is A:-times differentiable
UcJJ.
on
on an open set
U if there exist differentiable function
defined on U such that gt
f, g2 g't ,
, gk
gk
gl5
denote gk by fm. f is Ac-times continuously differentiable
f e Ck(U)) if f(t) is continuous on U.
=
.
.
=
=
.
.
The
following proposition
is
an
.
.
,
obvious extension of
g'k-!
on
We will
U
(written
Proposition
1
by
induction.
Proposition 4. Let f= (ft, ...,/,) be an R"-valued function defined on U.
f is k-times (continuously) differentiable on U iffu ,/ are each k-times
(continuously) differentiable on U. Further, fm (f^k),
,f(k)).
=
.
Knowing
be
a
that
a
given
function is differentiable at
great aid in computing approximations
These considerations in turn lead to
a
better
.
.
particular point can
nearby points.
understanding of the notion of
a
to its values at
differentiability. Suppose that/is a differentiable i?"-valued function
in a neighborhood of 0. By definition the difference quotient,
defined
-[/(0-/(0)]
t
converges to
<t)
=
In other
f'(0).
-
U(t)
-
/(0)]
-
words, the function e(f) defined for
/'(0)
t
# 0
by
3.2
has limit 0
as t
->
0.
e(t)
8(0
+
Thus
0.
=
Formula
241
Rewriting this,
fit) =/(0) +f'(0)t
where lim
Taylor's
a
(3.10)
t
good approximation
to
the value f(t) would be
<->o
/(0)
+ /'(0)' ; how
good depends
of
the function
course on
But since
e(t).
the difference between this approximation and f(t) is e(t) t, it suffices to
know just the maximum of |e(f)|. We give an illustration of how to go
about determining this.
Suppose /is
M
a
C2 function defined in
an
interval \_-R, R].
Let
sup{|/"(x)|:|x|<K}
=
Then
1/(0
0/(0) +/'(0)0l
-
This follows
<
from the
easily
for t
MR\t\
mean
e
(3.11)
[-*, K]
There is
value theorem.
between
a
t and 0 such that
Further, there is
for
a
given
(0
* e
=
=
an n
between , and 0 such that /'OS)
Thus,
-f'(0) =/"(")
[-R, R],
^C(/(0-/(0))]-/'(0)
f'(0
-
/'(0)
=
/"()
r,,Zer_-R,K]
from (3.10) and this inequality.
|(f)| < MR. Inequality (3.1 1) follows
describe the function
to
difficult
adequately
be
very
Now, although it could
In
/" is monotonic
obtain.
to
practice,
easier
much
e(0, the maximum M is
at the end points -R and R to
near 0 so we need only look at its values
Thus
obtain this estimate.
We shall
obtain estimates which
are even more
now
generalize
this argument
in
order to
accurate.
illustration above
we can
assert
special
Rereading Equation (3.10)
of the
a
at
function
a
point shows us how the values
of
that differentiability
a firstof
values
the
function at nearby points can be well approximated by
that the error is sma
order polynomial. (Well approximated here means
this well
and the
relative to the distance between the two points.)
approximability is a criterion for differentiability.
Furthermore,
242
3
Ordinary Differential Equations
Proposition 5. Suppose that fis an R" -valuedfunction defined in a neighbor
of x0eR. f is differentiable at x0 if and only if there exists a linear
s(t) 0
function L: R^R" and a function e defined for small t such that lim
(-.0
hood
=
and
f(xQ
+
t)=f(x0)
+
Furthermore, L(t) =f'(x0)
L(t)
+
e(t)t
t.
Proof. We have seen above that differentiability implies this condition.
versely, suppose this condition is verified. Then
r
hm
/(* + ') -/(*>
=
f
I-.0
for since L is linear,
Now,
L0)
L(t )
fL(l).
=
..
.
,
e(t
)
,.
h lim
,
f
t-*o
=
hm
t-o
r-.o
L(t)
=
f
L(\)
Thus /is differentiable at x0
evaluation of /(f) for t
approximate
an
,lim
near
Con
,
and /'(x0)
0 with
error
=
L(\).
that is
small relative to |r| may not be as good as required. A better approximation
would be one whose error is small as compared to t2, or even better \t\k
for
sufficiently large
We shall
now
derivation follows
higher order derivatives come in.
gives such approximations. The
This is where the
k.
derive
a
theorem which
by induction directly
from the above remarks.
Theorem 3.1. (Taylor's Theorem) Suppose that f is a (k + l)-times con
tinuously differentiable Revalued function defined in an interval I about x0
Then there is a polynomial P (with coefficients in R") of degree k, and a function
s defined for t in I such that
.
(i)
e(t)
is bounded
by max{|/(lt+1)(x) |
: x
between x0
andx0
+
r},
E(i)tk+1
(ii)
/(x0
+
0
F(0
=
+
A^
(3-12)
Furthermore, P is unique and is given by
P(t)
If
we
=
write
/(x0)
x
=
+
/'(*o)'
+
^
t2
x0 + t, (3.12) becomes
of degree k about x0 :
+
+
i^
tk
a more
familiar
x0)(x
xoy+1
expression,
called
Taylor's expansion
fix)
=
I
>=o
rr2 (x
i-
-
x0y
+
b(x
-
-
(3.13)
3.2
Taylor's
1
by induction on k. The case k
1 and
n
above. We now assume the proposition for k
applying the induction hypothesis to /'. For simplicity
is
proof
The
Proof.
=
=
I
=
243
Formula
already discussed
n, by
was
prove it for k
we
take
x0
=
=
0, and
{x: \x\<a}.
By the induction hypothesis
Let tel.
Now let
t}.
e0(t) is bounded by M
integrate (3.14) from 0 to x:
+ 1>
since/'(0=/(,
us
Here
The
/""+1>(0) r
"-1
<*
rfr
/'(f)
integral
we can
=
T1
I
f*
I
*' dt +
-,
max{|/(*+n(x)|
: x
between 0 and
/-,
kn
(3-15)
*o(0" *
is, by the fundamental theorem of calculus, /(x)-/(0).
the left
on
1
=
write
Thus, letting
ix)
we
=
n+ 1 C*
-^r
x
Jo
e0(t)t"dt
obtain from (3.15)
n+l
m =/(o)
which is
just the
+
2 J
771
^T ^TT
eW
We must show that
(3.12).
same as
+
e(x)
is bounded by M.
But,
"W'-SH* ^(Ol^f<^M{;r.f<M
since e0 is bounded by M.
Examples
Taylor expansion
11. Find the
1 +
/(I)
3t4.
t +
=
/"(1)
thus the
f(t)
/'(l)=l
5
=
=
of degree 3 about 1
f"(l)
36
+
=
12'3|,=
5 +
13(t
-
D
+
and
72
Taylor expansion
180
=
i
13
/(4)(0
=
72
is
-
D2
+
120
"
D
,
+
(0
"24
'
4
of
f(t)
=
244
Ordinary Differential Equations
3
where
|e(t)|
<
72.
/(5)(0
Notice that, since
=
> the Taylor expansion of degree 4 is
accurate :
f(t)
5 + 13
=
for all
+
18(1
l)2
-
+
12(f
-
I)3
+
30
-
I)4
t.
12. Find the
(1
1)
-
Taylor expansion of degree
4 about 0 of
t2)-1
+
/(0)
1
=
/'(f)=-2f(l
+
f2)-l
/'(0)
=
0
-2
/"(0)
f2)"1 + 4f2(l + r2)"1
2
/'"(0)
8f3(l + f2)~
f'"(t) 4*1 + t2)'2 +St(l + t2)-1
12
/W(f) 4(1 + *T2 + 8(1 + t2)-1 + t[_- ]
t2 + f + e(t)t5
1
f{t)
f"(t)
-2(1
=
=
+
-
=
=
/(36)
x
f(x)
\x-112
=
=
=
/'"(36)
for
(40)1/2
three decimal
to
places.
y/x about 36.
=
f'"(x)
0
-
13. Calculate
f'(x)
=
=
=
f(x)
/(f)
f"(x)
\x-^2
6
=
/W(x)
/'(36)
=
^x-7'2
f"^
=
^
^ l/WWI^
6 +
+
=
1
Thus
between 36 and 40.
=
\x-3>2
=
1 (x
-L (X
8.6
-
-
36)
36)3
+
^ (x
+
\e(x
6
-
-
36)2
36)4(x
-
36)4
We
expand
=
3.2
where
(40)
e(0
<
15/16.67.
Taylor's Formula
245
Thus
i/2-H+8 <iJ743<L^10~
67
6 16.67
and the desired
approximation
is 6.334.
14. Calculate e4 to three decimal places. We first write down the
have
Taylor expansion f(x) ex about 0. Since f'(x) =/(x), we
Thus the Taylor expansion of ex, degree n is
ex for all x.
=
/<k>(x)
=
xn+1
x'
"
/-,
<c\
(316)
'-S^^oTTiii
e* we now
|e(x)| < max{| e'\ : 0<t<x}. Thus to estimate
the
Taylor expansion
take |e(x)| < e4 < 34. The approximation by
of degree n is bounded by
where
00.xn+1^344l)!
_(n
+
(n+1)!
3
large that this is bounded by 10"
by the following succession of inequalities
We must choose
do,
as we see
4*+ 5
1
3*4"+
n so
(7+7)!
22n+1
zr^
~
~
2~3^ri
n >
41 will
1
~
ir3"1
1
<
^
Thus
we
irj3/10(n-31)
must have
(3/10)(n
Taylor expansion (3.16)
dominated by
In the
-
31)
<
3,
or n >
41.
of ex observe that the remainder
term
is
x"+1
e"
(n
+
1)!
and therefore tends to
zero as
-
a..
Thus, if
we
let
n
-
oo
in
(3.16),
we
246
3
Ordinary Differential Equations
obtain (once
i
again)
=
I-
0
Now, this kind of
happen
differentiable in
fix)
where
M"
+
argument
an
be
can
applied
to know has derivatives of all orders.
1(x)
interval /about x0
/(x0)
=
|e(x)|
an
E
+
(x
max{|/("+1)(f) | :
<
+
-
x0)
+
That is, if
write the
e(x)
between x0 and
(w
x}
/ is infinitely
Taylor expansion
+
(317)
1),
valid for every
Let
n.
If
be thus bound.
limM"
t
we can
to any function which we
1(x)(*,~Xi"+1=0
in
1)
(3.18)
+
as n -+ oo in (3.17) and represent /as a
Taylor Expansion of/aboutx0. In Chapters
5 and 6 we shall return to the consideration of series expansions for functions.
In Section 5.8 we shall construct infinitely differentiable functions which are
For the present we mean only
not represented by these Taylor expansions.
to remark on these approximations of the Taylor expansion as a tool for
then
clearly
series.
take the limit
can
we
This series is called the
approximation.
Examples
15. Consider
f'(x)
=
/(4)(x)
/"OO
cos x
=sin x,
and the
.
.
+
/<4" 4>(x)
cos x
=
sin
/'"OO
x
cos x
Thus
/(4n 2)(x)
about
=
sin
x
/(4"+3)(x)
zero
is thus found to be
7
^
X
+
=
x
T
f(x)
sin
We have
x.
+
=
X
x
sin
itself.
Taylor expansion
=
=
=
.
cycle repeats
/<4"+1>(x)
The
now/(x)
X
-
+ + Remainder term
=
-cos x
3.2
Since all derivatives of sin
bounded by 1,
is bounded by
are
x
one
the remainder for the
so
Taylor's
Formula
of +sinx, +cosx,
Taylor expansion of
247
they are
degree k
1
(k
iy.
+
which tends to
zero as
k
x5
x7
x3
sinx
=
x
__
accurately
compute
+
x2
16. Find sin
bound
on
expansion
\kfk+ 1
(-1/2*
sine
+
to
'"+l2l0r
an
ensure
+
calculating
this bound is
<
<
4/5
to
We need to compute
terms of the Taylor
n
10" 3.
"
n/4
can
we
'"
10"3.
accuracy of
Similarly,
(see Exercise 15),
Now the remainder
by [_(2k + 1 ) !] \nj4)2k +1.
verify that k 3 will work:
after k terms is bounded
fact that
...
sum.
for the cosine
the remainder after
and then
infinite
(-l)kx2k
+
6!
n/4
an
as
x6
x4
Taylor expansion
..+___ +
+
Taylor expansion
a
COSX=2!-4!
a
___
expresses
Thus the
->oo.
We shall
use
the
=
I^7<d/4<_L<10-3
7!\4/ _6.4457_6.54
Thus
an
n
n3
4
6.64
estimate to sin
+
17. The
n/4
to
within
one
thousandth is
120.45
is infinitely differentiable around the point 1.
we
infinite Taylor expansion there ? By computation,
logarithm
Does it have
an
find
log(')(x)
log<">(x)
log('")(x)
log<4>(x)
log()(x)
=
=
=
x"1
log'(D
=
l
log(")(D=-1
2x"3
log('")(l) 2
13.2x"4
log(4)(l) (- 1)3 2
log(n)(l) (-!)"(-!)!
(-l)"(n-l)!x-"
-x"2
=
=
=
-
=
=
(3-19)
248
3
Ordinary Differential Equations
The
Taylor expansion
of
degree
n
about 1 is thus
^'-S^^^fr-tf ^OT
+
Notice that from the first
|e(x)|<(n)!x"<"
+
equation
1
/
x-
1\
x
1
zero
3/2. Thus,
Taylor series
=
is bounded
by
n
->
long
1
oo, so
that the
in the interval
1/2
as
> x >
1/2.
Similarly,
remainder goes to zero if
< x < 3/2, the logarithm has
l)k
(x
00
log(x)
as
(Exercise 18)
< x <
the
(3.20)
J
show
can
1,
\n+1
1
which tends to
we
<
1)
and thus the remainder of
n+
(3.19), if x
of
(3.20,
I(-l)^-X
K
i=l
EXERCISES
12. Find the
Taylor expansion about the origin of degree 5 of
tan x;
(l+*)-\
13. Find sin \
14. Find
V3 accurately
15. Derive the
1 6. Find
(1
+
x)-1
accurately
=
an
to 4 decimal
to 4 decimal
=
places.
Taylor expansion (given after Example 15) of cos
interval about the
origin in which the substitution
l-x
is accurate to three decimal places.
(l+x)-1
places.
l-x + x1?
What about the substitution
x.
of
3.2
17. Find
interval about the
an
x2
e*
l +
=
x
+
T
Taylor's Formula
249
in which the substitution
origin
x3
+
-
is accurate to three decimal
places.
18. Show that the series
(x
1)<
-
k
1=1
represents the logarithm in the interval l/2<x<3/2.
series converges for all
in the interval
x
(0, 2).
Observe that the
Does it converge there to
logx?
PROBLEMS
Suppose/is a A>times differentiable real-valued function defined on the
0 for all k.
Show that / is a polynomial of
Suppose /<k)
1
degree at most k
7. Suppose that/ g are C functions defined on an interval containing 0,
and /(0)
=g-1)(0) =0, but gm(0)^0.
---=/"-1)(0)=0, g(0)
6.
interval /.
=
.
=
=
Prove that
/0)_/""(0)
9m(0)
9(t)
8.
(Taylor's form of
the interval [ R, R].
such that
1
=
9. Let
m
be any
nm
X
fix)
I
=
n=
o
mean
value
theorem) Suppose that /is C on
[R, R], there is a between 0 and f
t e
K\
11
0
the
Show that for
(mn
integer and define the functions f0
,
.
.
.
,
fm-i by
+ I
+
i) !
(
(a) e*=fi(x)+---+fm(x).
l,...,/w-l.
(b) //=/i-i for ;
(c) /o =/m-l(d) The functions /i, ...,/ are all solutions of
equation
=
y(m)
_
y
the differential
250
3
Ordinary Differential Equations
(a) Suppose that /is continuous
10.
0(0=
f
the interval [ R,
R].
Define
te[-R,R]
f(tx)dx
>0
on
and show that g is also continuous.
(b) Suppose that h is C1 on [-R, R]. Prove that there is a con
tinuous function k such that h(t )
h(0) + tk(t ). (Hint: Consider
tx.)
'(T) dr and make the substitution t
=
jo
3.3
=
Differential
Now,
Equations
ordinary differential equation is (roughly speaking) an equation
function/ and some of its deriva
an
involving the variable x, an "unknown"
tives/',/", .,fk\ Thus
..
fix)
=
k(x)
f"+f=0
f'(x)
=
x/(x)
C/(4)(x)]2
e*''w
+
=
|/(3)(x)|
+
log|x
+
1|
examples of differential equations. A solution is a function which makes
equation true. For example, |5 k, sin x, exp(^x2) solve the first three
equations respectively (as for the fourth, we cannot easily exhibit a solution).
We prefer to think about differential equations in this vague sense rather
than to try to attempt a formal definition of such, so we shall do so.
Many equations do not admit solutions and some equations admit many.
are
the
Consider these :
|/|
(y'f
y"
|y-x|
+
+ i
+ y
=
The first has
and
=
=
0
o
0
no
solution y
=f(x),
because
we cannot
have
both/(x)
=
x
/'(x)
supposed
0; the second has no solution because the derivative of the
solution would be imaginary. The third equation has as solutions
sin x,
x,
cos
must be
if
we
=
as
well
discarded
as
any linear combination of these.
The first
equation
the second admits solutions
being self-contradictory;
permit ourselves to consider complex-valued
as
functions.
As
we
shall
3.3
see
this turns out to be
the third
The
Differential Equations
very fruitful course, for it
a
251
permits understanding
well.
as
of calculus derives from the fact that it is
necessary to the
problems (mainly derived from the study of physics and
the natural sciences). These problems usually are stated
mathematically
importance
solution of concrete
as
differential
equations.
Examples
18.
A bank likes to pay its depositors on the
deposited and the length of time they have been
able to use these deposits. Thus every (say) June 30 your bank
would add to your deposit an amount equal to (say) 5 % of that part
of your deposit which they have held for the past year (and if they are
decent about it a reasonable fraction of that 5 % for parts left in for
fractions of that year). Many years ago, that great financial wizard,
L. Waverly Oakes, pointed out that that amount that he kept in his
bank for the first half-year was working for the bank and he should
be paid for it. Furthermore, argued Mr. Oakes, the payment he
Compound
interest.
basis of the amount
should have received
so
also
Finally,
was
also sunk back into the bank's investments
income for the bank, and thus for its depositors.
Mr. Oakes pointed out that there is nothing special in half a
earning
was
"
Over
His very words were
any particular fraction thereof.
a
the
of
how
of
no
matter
small,
particular
time,
earning
any period
balance relative to that balance should be directly proportional to
year,
or
period of time. In order to best approach the interest due its
depositors, our banks should be computing interest as often as
possible." The banks all responded to this profound utterance by
recomputing their interest every month instead of every year. Some
body even suggested that, with an army of secretaries, they could so
that
compute the accrued interest every 30 seconds.
would have rested
were
it not for
an
And there the matter
obscure student of Isaac Newton
who dabbled in the stock market.
at time
Suppose
f(t)
according
t-y ^
t0
f1
^ t
to
n
a sum
of s0
pounds
are
deposited
in the bank.
t0 be the balance accruing from this deposit
the Oakes system. Then, Oakes' assertion is, for all
for all times
Let
t >
,
fih) /Pi)
fih)
-
=
k(t2
-
(3-2D
h)
where k is the earning power
(interest rate)
of money.
The first
252
3
Ordinary Differential Equations
thing this brilliant person remarked is that (3.21) cannot possibly
always hold. Let us illustrate his discussion.
Suppose that 500 pounds are deposited in the bank at a 5% per
/(0) 500 and at the end of one year,
pounds, so/(l) 525. Now, if interest is computed
we
obtain, by (3.21),
half-year,
Then
interest rate.
annum
the interest is 25
every
=
=
^-^ 0.05ft)
K2>
500
or/(l/2)
/(l)
=
512.50.
Then,
over
the second half of the year,
we
obtain
512.50
-
=
512.50
0.05ft)
by this computation /(l) 525.31. As this is closer to the
actual earnings of the initial deposit, this is more like the amount the
depositor should get. Furthermore, this semiannual computation
has neglected the earnings during the last three-quarters of the 6.25
accrued during the first quarter. In fact, when we compute the
interest quarterly we find that the value of/(l) should be no less than
525.504. And so it goes: no matter what period we choose for the
computation of interest, we will be neglecting the interest accrued
by the growing total during that interest. Thus Oakes' formulation
cannot be correct.
However, our student was moved by the basic
justice of Oakes' ideas and after rewriting Oakes' formula as
so
that
f(t2)
:
h
~
=
fih)
:
_
=
.,v
fc/Oi)
h
he asserted that he had found the
precise
statement of the Oakes
Oakes should have said "over any infinitesimally small
interval of time ..." rather than over any period of time, no matter
formula.
"
of change of the balance
time; that is,/'
kf,
proportional
where k is the interest rate (0.05 above). Thus, the problem is to
ky
0 so that
find a solution / for the differential equation y'
how small ..."
at any
time is
Precisely, then:
the rate
to the balance at that
=
=
/Oo)
=
*o
Population explosion. Population tends to grow also according
differential equation. That is, it is assumed that every
individual has the same propensity to reproduce and that propensity
19.
to the above
3.3
is
independent
of time.
Differential Equations
253
Thus
over any infinitesimal period of time
population to the initial population is
proportional to the time elapsed. (You know what mobs are like:
the larger they are the faster they seem to grow.) This assertion is
supposed to be true for brief periods of time ; thus we should more
precisely assert that the rate of change is directly proportional to the
total population; thus if/ is the total population,/'
kfi where the
constant k is called the growth rate.
In some societies the growth rate varies with time; among certain
mammals it peaks at certain times of the year. In these cases the
population as a function / of time satisfies a differential equation :
f'(t) k(t)f(t), where k(t) is the variable growth rate. It may even
happen that the growth rate depends on the total population; in a
well-regulated society (1984) this would be the case. Then the
population function is a solution to a more complicated equation,
/ Ky)y.
the ratio of the increment in
=
=
=
20. Survival
On
of thefetahs.
Pacific there live only two
garibs. These animals are
everpresent undergrowth
to
a
remote volcanic atoll in the
South
species of animals, the fetahs and the
essentially vegetarian and there is an
feed them. However fetahs especially
garibs and garibs find the succulent fetahs hard to resist.
Now each fetah tends to reproduce at the rate of one young each per
7 per year. Conversely
year, and consume garibs at the rate of
eat fetahs at the rate of
and
the garibs have only one young per year
of
fetahs and garibs in a
17 per year. Thus the increment A/ kg
year should be given by
love to eat
A/=/0
-
Hg0
where f0,g0
the Oakes
are
Ag
=
g0-
the initial
reasoning
lf0
(3-22)
populations of these groups. However,
applied to this case; because as the
must be
continually affect the increment. The
solution is, as in the above case, to rewrite (3.22) as a differential
and garibs at
equation. If f(t), g(t) are the populations of fetahs
and
g:
time t, then these equations describe the growth off
population changes,
f'=f-iig
it will
g'=g-7f
less remote island there are n different
on the growth patterns
species of animals, all of which have some effect
of all the others (some feed on others; some house, or protect others).
21. The biotic matrix.
On
a
254
3
Ordinary Differential Equations
This kind of
society
can
be
represented by
a
biotic
nx n
matrix
(i,j)th entry is described as follows: The increment
(a/).
in the rth species in one year which is attributable to each member
of the y'th species is a/. (Thus the effect of one member of the /th
species on the rth species in an interval of time Af years, is a- At .) If
A
The
=
/0)
=
(f1^),
.
.
.
,/"(0) is the population function
equation must be satisfied:
on
this island,
then this differential
(3.23)
f'=Af
We consider
22. Particle motion.
in R".
Let
f(f)
be the location of that
Revalued function of
at a time
a
t0 is the limit
real variable.
as t
->
now
the motion of
particle
at time t
The rate of
a
particle
f is thus
.
change
of
an
position
t0 of
-J_(f(0-f(to))
i0
i
f'Oo), called the velocity of the particle at f0.
change of velocity, f ", is the acceleration of the particle.
thus is
f
'
The rate of
The velocity vector has both magnitude and direction; we can write
(f) v(t)T(t) (at least when f # 0), where v(t) is a positive function of t,
'
=
T(f) is a unit vector. T(f) points out the direction in which the particle
traveling at time t and v(t) is the speed at which it is moving. Also, v(t)
The length of the path that the
can be given the following description.
particle traces out in a certain period of time is the distance traveled by the
particle. \v(t)\ is also the rate of change of that distance at time t. We will
have to await a full discussion of arc length (Section 4.2) before justifying
this; however some heuristic arguments are possible (see Problem 20).
According to this description, the distance s(t) traveled by the particle from
time t0 to t is a solution of the differential equation y'
||f '(t) || with s(t0) 0.
We would hope that there is only one solution, for there is no further way to
determine this function. (Fortunately, by the fundamental theorem of
calculus this problem has a unique solution.)
Consider, for example, a particle moving on the unit circle in the plane.
Let s(t) be the arc length on
Let / be the position function of this particle.
the circle from the point (1, 0) to /(f) at time f. Then (since arc length on
the unit circle is the same as the angle)
and
is
=
f(t)
=
(cos s(t),
sin
s(t))
=
3.3
Differential Equations
255
velocity vector is /'(f) s'(t)(- sin s(t), cos s(t)). Notice that
l/'(OI, giving further weight to our description of speed above.
k'(OI
Notice also that /'(f) is tangent to the circle at the point /(f); this reflects
the fact that the motion is constrained to the circle. Differentiating further,
The
=
=
we
find that the acceleration is
f"(t)
s"(t)(- sin s(t),
=
=
cos
s(t))
+
s'(f)2(- cos s(t),
-sin
s(t))
s"(t)T(t)-[s'(t)-]2f(t)
Thus the acceleration has
to the circle
(in the direc
motion)
magnitude
change
speed, and a
to
the
direction
of
whose
motion,
component perpendicular
magnitude is
to
the
For
if
the
is
equal
speed squared.
example,
particle rotating around
the circle with constant speed, it is accelerating toward the center of the
tion of the
a
component tangent
whose
is the rate of
of
circle.
According to Newton's laws of motion the situation is as follows. Given
particle at time f 0 situated at p0 and having velocity v0 all further motion
is determined uniquely by the forces acting on the object. The motion is
determined by this law: the acceleration is directly proportional to the force
acting on the particle. Thus, in the absence of any forces, if f(f) is the
position of the particle at time t, we have
a
,
fOo)
=
Po
f'0o)
=
vo
f"0)
=
0
allf
uniquely determined by these conditions. We say that f is a solu
tion of
differential equation y" 0 with the initial conditions y(0) p0
y'(0) v0. Newton's laws require that the solution exists and is unique.
Thus, in the
Mathematics bears this out ; the solution is f(r)
p0 + fv0
absence of force, a particle will move with constant velocity, that is, in a
straight line at a constant speed.
Now, in general, the mechanics of motion can be described as follows.
There is a function F defined on R" x R taking values in R". The value
F(x, f) represents the force that will act on a unit mass acting at point x at
time t
The function F is called a force field. A particle of mass m situated
According to
at the point x will experience the force mF(x, t) at time t.
Newton's law it will accelerate in the direction of F. The magnitude of this
acceleration is determined by or according to this announcement of Newton's
and f is
the
=
=
,
=
=
.
law : Force
mF
(a
=
=
=
mass
nza
acceleration).
acceleration,
.
256
3
Ordinary Differential Equations
Suppose
we
place
a
particle
of
mass m
at p0
with
velocity
v0 into this
situation at time t0
Let f be the function describing its subsequent motion
to
Newton's
law. Then at time t it is at f(f) and it experiences a
according
.
force
Thus
F(f(0, 0f"(0
=
have
we
F(f(0,0
Thus f is the solution of the differential
conditions
exist
f(f0)
p0 f'Oo)
vo
In the next section
=
=
uniquely.
,
force fields this is the
equation y" F(y, 0 with the initial
require that the solution
shall show that for smoothly varying
=
Newton's laws
we
case.
PROBLEMS
11. Find
all
complex-valued solutions
of the
differential
equation
(/)2+l=0.
12. Solve the differential
y(0)
=
13.
equation y'
=
y
with the initial condition
0.
long will it take 100 dollars to double at a compound
% per year ?
(b) How long will it take 350 dollars to double at the same rate?
(c) How long will it take 100 dollars to double at a rate of 10%
per year?
14. It is observed that radioactive elements decay into heavy metals. It
is assumed that the probability of any given atom decaying is independent
of the particular atom. Let k be the probability that a given atom of a
particular element will decay within one year. Show that the function /
is governed by the differential equation y'
ky if /(f) is the mass of the
given element after time f.
(a)
How
interest rate of 5
=
15. The time it takes for
the half-life.
If
an
a
radioactive element to halve in
element has
a
mass
is called
half-life of 14 million years, find the
constant k of Problem 14.
Why is Oakes' formulation of the interest problem wrong ? Can you
equation (3.21) so that it holds for a specified period; that is, given
n, find/so that (3.21) holds for ti
k/n, t2=(k+ \)/n, 0^k<nl
11. A weight of mass m is suspended from a rigid support by a spring of
natural length L. According to Hooke's law the spring produces a
restoring force which is proportional to the displacement from its natural
length, and directed toward its natural position (Figure 3.6). Let us denote
this constant of proportionality by k. Let x denote the distance of mass
from the natural position, where the positive direction is upward. Then the
mass has two forces acting on it: a force Fi
kx due to the restoring effort
of the spring, and the force of gravity F2
If the mass is at rest, then
mg.
there is no acceleration, so by Newton's laws Fi + F2= 0, from which we
may conclude that the rest position is at x
k~1mg. Now suppose we
1 6.
solve
=
"
"
=
=
3.3
Differential Equations
257
Figure 3.6
displace the
by an amount ha and let it go. Using Newton's
equation governing the subsequent motion.
mass
the differential
law find
18. A certain insect lays its eggs in the flesh of a mammal. Each insect
Now every time one of these eggs hatches in a
hatches h eggs per year.
horse, it kills the horse.
Assuming the total mammalian population is a
equations governing the growth of
this insect and horse population if we also know the natural death rate (dt)
of the insect and the natural birth and death rates (bH dH) of the horse.
Let I(t), H(t) be the population of the insect, horse, respectively. During
a period of time Af bH
H- Af horses are born, and dH- H At horses die of
natural causes. Now each insect hatches hAt eggs during this interval;
the probability that its host is a horse is HjT. Thus there are hl(H\T) Af
horse deaths attributed to the insect during this time interval. The change
AH in the horse population is thus
constant T,
derive the differential
we can
,
,
AH
=
bHH At
-
dH At
-
hi
(f)"
3
Ordinary Differential Equations
Find the corresponding
change in the insect population and deduce that these
differential equations govern the growth:
hH
H'
V
=
I
(bH-da)H-
=hT-dII
by Galileo that the gravitational attraction of the
small, we may assume the world is flat, thus we
0 is the surface of the
take as a model R3, and assume that the plane z
earth. The gravitational attraction then is a force field F(x, v, z)
(0, 0, g). Suppose a particle of mass m is at p0 and has a velocity v0 at
What is the
time f0
Let f (f ) be the position of this particle at time f.
differential equation governing the motion of the particle? Can you
19. It
was
observed
earth is constant.
In the
=
=
.
solve for f ?
20. Suppose there is a
(c, 0, 0) on our particle,
equation of motion.
21.
Suppose that
on
wind
the
coming
out of the east which exerts
matter what the
no
plane there is
a
position is.
centripetal
a
force
Now find the
force field propor
tional to the distance from the origin. At the time f
0, a particle is
placed at the point z0 and has a velocity v0. What is the equation of
=
motion ?
22. We
ing it by
x
=
x(f)
a
can
try to find
line segment.
v
=
a
formula for the
length of a
curve
by approximat
Let
v(f)
be the equations of a curve, and let (x(t 0), y(t0)) be a point on the curve. For
a very short period of time. Af, the curve can be replaced by its tangent line
(see Figure 3.7). The length of the curve between (x(t0), y(t0)) and
(x(t0 + Af ), y(t0 + Af )) is then approximately equal to ((Ax)2 + (Ay)2)1'2.
V(ax)2
+
(aj)
(jr(f
Figure
3.7
+
A/),y(f + if))
3.4
Then the rate of
Some
change of
arc
259
Techniques for Solving Equations
length
over
the interval Af is
((Ax)2 + (Ay)2)1'2
Af
Letting Af -> 0 deduce that the rate of change of
is the length of the vector (x'(f ), y'(t )).
3.4
for
Techniques
Some
arc
length along the
curve
Solving Equations
The fundamental theorem of calculus is of course the basic existence
on solutions for differential equations, and integration is the primary
theorem
tool. Thus
y'
an
h(x)
=
has the solution
Let
constant.
of the form
equation
f(x)
us
state the
Let h
Definition 3.
the closed interval
\* h(t) dt + c,
=
=
la, b~]
same
unique
but for
a
result for vector-valued functions.
continuous Revalued function on
Define the integral of h over the interval [_a, b~] to be
(hu
.
and this solution is
.
.
.
,
h)
be
a
r>-(jy...j\.)
continuous Revalued function defined
and leta<c<b. Then the differential equation
Theorem 3.2.
interval (a,
y'
b)
=
h(x)
Let h be
h + p0-
and Proposition 1
By the fundamental theorem of calculus
f(x)=
f
the open
(3-24)
y(c)=Po
has the unique solution
Proof.
on
a
,
h+Po
is another solution, then
is differentiable and satisfies the conditions (3.24). If g
derivative and thus is constant.
g' =/' on (a, b) so each coordinate of* -/has zero
Since g(c) =p0 =/(c), this constant is zero, so g =/.
260
3
Ordinary Differential Equations
Separation of Variables
There is a class of differential equations which can be solved simply by
integration, just by recalling the chain rule. This is the class of first-order
equations (only the first derivative of the unknown function y appears) in
which the variables separate ; that is, these are equations of the form
(3.25)
h(y)y'=g(x)
The left-hand side appears to be the result of
we can rewrite (3.25) as
h
we
let H be
[#OyOO)]'
H(yOO)
we can
as a
=
an
indefinite
integral
of h, H
=
j h,
then
(3.25)
becomes
0OO
integrate :
so we can
y
rule;
000
=
dx
If
of the chain
y(x)
d_
Thus, if
application
fg
=
solve
(3.26)
for
(3.26)
y(x),
we
will have the desired
explicit expression
of
function of x.
Examples
23.
yy'
rewritten
=
1.
Let
H(y)
[//(yOO)]'
as
=
=
1,
Jy
or
=
y2/2.
H(y)
=
Then the
y2/2
=
x
constant to be determined
by the initial conditions.
solution of
(2(x
24.
y"2y'
y'
=
yy'
=
x2y2.
x2
Integrate :
-i
1
-y
*3
=
y
=
+
^
1 is y
=
Again,
we
+
write
c))1'2.
equation
+ c,
where
Thus the
can
c
is
be
a
general
'.4
Some
Techniques for Solving Equations
261
so
-3
y'x^Tc
25.
sin y
or
y
y'
=
=
y
=
sin
cos x
+
c
cos
sin(c
arc
A
x).
cos
-
After
x.
integrating
this becomes
particular solution is/(x)
=
x
-
n.
26.
1 +
,
After
y +
It is
x
integration
y2
Y
we
have
x2
=
+
x
now a
+
Y
c
bit difficult to write the solution
x, but it is
possible using
explicitly
function of
as a
the formula for roots of
a
quadratic
polynomial :
y
=
-2(4
+ 8x +
4x2
+
c)1/2
(128)
2
The constant
c
is
presumably
determined by the initial conditions, and
with it the function y. Notice however, that each value of c gives
two candidates for the solution, but they may not both be solutions.
For example, suppose we seek the solution of (3.27) with the initial
condition
x
=
0,
y
=
y(0)
0,
=
we
0.
We
arrive
at
(3.28)
and
upon
substituting
obtain
Q_-2(4 + c)1'2
2
so we must choose c
0 and the positive sign before the radical.
If the initial condition is y(0)
-2, again
This boils down to y
x.
c
0, but we must take the negative root, obtaining y= -(x + 2).
=
=
=
=
262
3
Ordinary Differential Equations
1
Notice also that upon substituting the initial condition y( 1)
this
to
we
find
c
8
both
roots
solutions
and
problem;
(3.28),
give
=
into
=
that is, both functions y
x and y
(x + 2) are solutions with this
initial value. Thus it is not always true that the initial conditions
=
=
uniquely determine the solution of the differential equations. Looking
back at the original equation (3.27) we find what might be a clue to
this bizarre behavior: the function (1 + x)(l + y)"1 is ill-behaved
at y
Uses
=
1.
-
of Exponential
We shall
now
Recall
techniques.
equation
V
has
a
=
study of the exponential function ; because it is
simple differential equation it gives rise to several
from Chapter 2 (Definition 21) that the differential
turn to the
the solution of such
a
c
unique solution, denoted
(ecx)'
=
cecx, (ecx)"
These remarks suggest
A
=1
XO)
cy
a
=
complex
any
...,
(ecx)(s)
method of attack
=
on
csecx
+
ak_iy<k-
++
a,y'
+ aoy
=
(3.30)
another class of
homogeneous constant coefficient equation is
y(k>
(3.29)
Notice that
ecx.
c2ecx,
number
one
equations.
of the form
0
(3.31)
We shall consider this class in greater detail in Section 3.6.
Let us compute
ecx.
By (3.30),
the left-hand side of (3.31) under the substitution y
akecx
+
=
ak-xck~1ecx
(ak
We find that ecx is
in
+
a
+
a^^'1
+
+ a0 ecx
axcecx
+
=
a0)ecx
+ a,c +
solution of (3.3 1) if c is
a
(3.32)
root of the
polynomial appearing
(3.32).
Examples
27. Find solutions for
Substituting
y
=
ecx,
y"
we
y
=
obtain
0.
(c2
-
\)ecx
=
0, thus
We conclude that ex,e~x are solutions.
for any a, b, aex + be~x is also a solution.
c=
+1.
we
must have
Notice also that
3.4
Some
28. Find solution of
y'"
Techniques for Solving Equations
+ y
=
263
0.
Here substitution of y
ecx yields (c3 + l)ecx
0,
cube root of
1
Thus we obtain three solutions :
=
=
so c must
be
a
.
einlix
e~x
e-'"l3x
Of course, all functions of the form ae~x + be'",3x + ce~iK/3x
solutions.
29. Solve the initial value
y'"+v'
=
o
y(0)
Substituting
c
0
=
or c
Let
y(0)
=
can
=
obtain
\
(c3
y"(0)
+
c)ecx
l
=
=
0,
(3.33)
so
we
must have
Thus all functions of the form
if we
can
solve for a, b,
c
by substituting
the
:
0:a + b +
=
c
0
-b-c=\
y"(0)=l:
a
=
=l:ib-ic=l
y'(0)
We
us see
problem
y'(0)
we
i.
=
solutions.
initial conditions
ecx,
=
ce-'x
+
beix
i
y
+
ae0x
are
or c
=
0
=
are
solve this system,
b
1
c
-
=
obtaining
-
=
2
2
Thus the function
/(X)
=
i-!V-^Vfa
will solve
our
problem.
30. Solve the initial value
v" + y
=
0
y(0)
=
l
problem
/(0)
=
0
(3-34)
264
3
Ordinary Differential Equations
Here
have,
we
initial conditions,
a
+ b
1
=
and thus
f(x)
Notice that
we
b
=
i(eix
=
we
ib
ia
a
+
general solution aeix
as
=
=
0
Thus
1/2.
we
obtain
as
solution
e~ix)
already
know from calculus the solution /(x)
shall learn in the next section that the initial value
unique
cos x
\(eix
=
+
=
e~ix)
sinx
e'x
=
=
cis
Now if /is
=
ef,
wards
we
x
=
=
a
cos x
+
/
sin
(3.37)
x
Equations
differentiable function,
how to solve
so
is
ef,
(ef)' =f'ef.
and
an
g
z'
=/.
=
=
new
function
Since
eHg
we can see
are
how
differential
<7(x)
(3.38)
continuous in
are
consider the
since H'
Letting
Thus, working back
equation y' =/'y.
equation of the form
g(x)y
y'+/00y
/
relationships
(3.36)
Namely, exp(J g) is a solution. With a little more ingenuity
to explicitly solve any linear first-order equation.
These
equations of the form
where
a
functions:
obtain the differential
we see
/
exponential
2^(e'je-e-'x)
First-Order Linear
y
and
trigonometric
has
(3.35)
We shall leave to the exercises the verification of these other
between the
We
cos x.
problem (3.34)
interesting equation follows:
Thus this
solution.
Substituting the
ix.
+ be
obtain
z
=
an
interval about
e"y.
Then z'
by (3.38), y' +fy
=
g,
we
=
a.
eHy'
Let
+
have this
H(x)
H'eHy
=
=
\* /
and
eH(y' +fy),
equation in
z:
3.4
which is solvable
Some
Techniques for Solving Equations
265
by integration :
J.XeHg
+
c
a
Finally,
y
=
e~Hz
=
y
e~Hz,
=
thus the general solution of
f eHg
e~H
where H is the indefinite
(3.38)
is found:
ce~H
+
(3.39)
and
integral of/
is to be found
c
by substituting for
the initial condition.
Examples
31.
y'
Here
the
z'
+ xy
we
=
y(0)
x,
take H
=
j" x
0
=
x2/2
=
corresponding equation
y' exp(x2/2)
=
+ yx
in
z
and consider
z
=
y
exp(x2/2).
Thus
is
exp(x2/2)(y'
xy)
+
=
exp(x2/2) x
Thus
z
| exp(x2/2) xdx +
=
c
exp(x2/2)
=
+
c
so
y
z
=
exp(-x2/2)
Substituting
so c
=
1.
-
32.
y'
Here
x"2y.
=
we
c
exp(-x2/2)
0
the initial condition y
1
is
thus
The solution
y
=
=
x,
take H=
Thenz'
=
y(l)
J 2/x
-2x"3y
+
=
In
x
+
c
and
y
=
x2 In
exp(-02/2)
exp(-x2/2).
=
c
+
=
c
+
1,
0.
=
=
-
2 In
x-2y'
x
+
x
=
We obtain
z
-
=
2x"Jy
-
1 +
ex2
and consider
x"2(y'-2x-1y)
nx
z
=
=
ye
=
^"2^=^ '
266
3
Ordinary Differential Equations
EXERCISES
19. Solve these differential
equations:
l
(a) x2exp(x2)y' x3,v(0)
0
(b) y' x sin x + cos x, y(0)
(0, 1 0)
(x(0),
y(0), z(0))
=(t,t2,t
3),
(c) (xm /(f))
1
(d) z(t) e" + ((1 + i)t)2, z(0)
Solve these differential equations:
(a) y'=y2
(b) y' cos x cos y
(c) x2 + y2y' 0
(d) y'=(y2-l)(x2-l)
(e) y"=xy'
(f) y'=(\+x2)y
(g) xy2 + (l-x)/ 0
(h) y' e*+"
(i) y' sin(x + y) + sin(x y)
Solve these differential equations:
(a) y' + xy cos x, y(0) 0
1
(b) y' cos x + y sin x tan x, y(0)
(c) y' + xy x2,y(0)=0
1
(d) e"y' + e*y e-,x, y(0)
l
(e) y'=ye-",y(l)
Solve these differential equations:
(a) y'" 2y" + y' 2y 0, y(0) 0, y'(0) 0, y"(0)
1 y'(0)
0
(b) y" 2/ y 0, y(0)
(c) y'" (1 + 3i)y" + (3i- 2)y' + y 0, y(0) 0, y'(0)
y"(0)=0
=
=
=
=
=
20.
,
=
=
=
=
=
=
-
=
21.
=
=
=
=
=
=
=
=
22.
-
=
-
=
=
=
-
=
=
=
1
=
,
3.5
=
=
-
=
1
,
Existence Theorems
In this section
we
differential
shall state and prove the basic existence theorem for
method is due to Picard and is that of
equations. The
successive approximations. (Recall
solution to the equation y'
cy.)
ordinary
how
we
found, in Section 2.10, the
=
The first theorem is about first-order
the method of successive
equations.
approximations.
We shall first illustrate
Example
33. Successive approximations.
of
y>
=ex + y
y(0)
=
0
There is
one
and
only
one
solution
3.5
if/(x)
Now
/(x)=
solves this
equation,
ffV)dt= \\e'
then
by integration we
we
fV
=
see
that
+
Tf
0(0]
dt
f;
that
=
Newton's method
sequence
/0
sequence.
T/0
=
T2f0
T3f0
T%
=
=
=
we
is, / is
...,
=
T(T/0)
f(2e<
=
T(T2f0)
=
of T.
point
T"f0
,
...
Let
.
for/0
,
According
to
the limit of the
as
us
say f0
compute
=
0.
this
Then
ex-l
=
j\3e<
dt
1)
-
2
-
-
=
t)
2ex
dt
2
-
=
3ex
x
-
-
3
-
2x
-
%-
4ex-4-3x-2X--%x2
T5/0
fixed
a
We may choose any function
fe' dt
that
continuous functions by
on
should be able to find T
TfQ, T(Tf0),
,
see
267
f(ty]dt
+
Thus if T is the transformation defined
Tg(x)
Existence Theorems
x4
x3
5^-5-4x-3y-2--x2
T%
=
nex-n-(n- l)x -(n-2)x""1
xj
--(n-j)
77-
(n-1)!
y!
but if
We can't tell yet that this sequence of functions converges,
better
a
picture:
replace ex by its Taylor expansion we can get
co
T"fo
=
Y-*
n
1
we
y/
ln--Y(n-j)Jj=o
j!
j=o
-l
=
I
X1j
XJ
In
-
(n -y)]
-
+n
-
n-1
=
"^
X
XJ
x1
^ (jyj + ;!
(3.40)
268
3
As
Ordinary Differential Equations
lim
the last
oo
->
n
T%
-^
=
y
n->oo
in
sum
=
o
(3.40) tends
to zero, and we obtain
xex
=
1)!
U
given problem! Now we would like
unique. This is easy, because it is easy
Indeed xex solves the
to show
that the solution is
to
that T is
a
Tf(x)
Tg(x)
so
-
contraction
fV
=
+
'o
in the interval
|x|
verify
:
<
f(t) -e'- g(t))
\,
dt
=
f(/(f)
-
Jo
g(t)) dt
say
\Tf-T9\<>\\f-9\
Thus if
T/
=
/
and
g ||, which is
||/
Now, the
most
Tg
=
g,
we
obtain
possible only if/= g
general
differential
=
i||/- g\\
>
\\Tf- Tg\\
=
0.
equation of first
order that
we
shall
consider is
v'
=
where F is
F(x, y)
a
(3.41)
real-valued function defined in
in the
a
(a, b)
plane. A solution is a function
neighborhood of a with these properties
f(a)
If/is
a
=
6
/'(x)
solution, it is
Tg(x)
=
a
=
fixed
neighborhood of the point
=/(x) defined for x in a
y
F(x,f(x))
point
of the transformation
fV(f, ^(O) dt + b
(3.42)
"a
The fixed
point will be found by the method of successive approximations :
/0 anything, fx
T/0 f2
Tflt and in general /
T/n_x. In order to
that
this
a
limit
has
and
the
fixed
guarantee
sequence
point is unique, we
must guarantee the hypothesis of the fixed point theorem.
More precisely,
we must know enough about the function F in order to
guarantee that the
transformation defined by (3.42) is a contraction on the space of continuous
=
=
=
,
=
3.5
functions
shows)
on
269
suitable interval about a. It suffices
(as the proof below
condition is satisfied.
a
if the
Existence Theorems
following
Definition 4.
D in
Let F be a function of two variables
x, y in the domain
R"+m (x ranges in R" and y in Rm). F is Lipschitz in
y if there is a
constant M such that
\F(x, yt)
-
F(x, y2)\
for all yl,y2 such that
Notice that since
<
(x, yt)
(1
+
M\yt
and
y2\
-
(x, y2)
x)(l +y)_1
are
in D.
is not
Lipschitz near y= -1,
apply
Example 26.
of the generality we need for Picard's theorem. Notice that if n
m
F will be Lipschitz if the partial derivative dF/dy exists and is bounded.
by the mean value theorem (along the line x constant)
we
Picard's theorem; and in fact it does not hold as we saw in
We have allowed x, y to range through many variables because
cannot
=
=
1,
For
=
dF
F(x, yt)
and thus
we can
Now let
order k is
(x, OCfi
=
dy
-
y2)
yt
<
<
y2
take the M of Definition 4 to be the bound of
higher
order
equations.
dF/dy.
equation
A differential
a
of
in the form
F(x,y,y',y",...,yW)
=
where F is
F(x, y2)
turn to
us
given
/k)
R*+1.
-
function defined in
A solution is
a
a
(3.43)
neighborhood
of
(a, b0,
=/(x)
A:- times differentiable function y
..
.,
/Jt-i) in
with these
properties
f(a)
=
fm(x)
b0 ,f'(a)
=
=
bu... jC-'Xa)
F(x,/(x),/'(x),
.
.
.
=
bk.it
,f(k^\x))
(3.43) with the given initial conditions by successive
but the method is not transparent. However, the problem
does reduce to the first-order case by means of a great idea. First, we
We would like to solve
approximations,
illustrate.
270
3
Ordinary Differential Equations
Example
34.
y"=2y'-y,y(0)
We introduce
Then the
y'
z'
y(0)
2z
=
z(0)
y
-
=
0
=
1
Thus, what
which is first order.
of the vector differential
(y, z)'
(z,
=
2z
z
and
require
that
y'
=
z.
seek is the vector-valued solution
we
equation
(y(0), z(0))
y)
-
0,y'(0)=l.
is reduced to the system
given equation
z
=
=
unknown function
a new
(0, 1)
=
by integration and thus solve by successive
approximations. Precisely, the solution is the fixed point of the
transformation (defined on pairs of functions) :
This
we
T(f, g)(x)
Let
rewrite
can
ffo(0, 2g(t)
=
compute
us
(/o,</0)
(fi,9i)
(f2 92)
=
=
=
,
(A g3)
=
,
ih,9d
T(f0,g0)
=
T(fu gt)
=
T(f2 92)
\3!
/
1
+
and that
approximations.
(x,2x+l)
=
(x2 + x, f x2 + 2x + 1)
(i*3 + x2 + x, f x3 + \x2
*3
+
*2
2
5x4
+ *'
x""1
7T7 + 7
\(n-l)!
^T, +
(n-2)!
lim/
This then is the
first order.
of the successive
~4\
2
+
=
"
"
+
x
+ X,
3
+
general
+ 2x +
1)
2
%1
+ 2X + X
form of (/
,
\
)
gn) is
\
,
-
X*
3
not hard to surmise that the
x"
-,
(0, 1)
T{f3,g3)
/x4
now
dt +
(0,l)
=
It is
some
,
=
f(t))
-
'o
.
.
.
/
xex.
typical means of reducing the higher order equation to
Equation (3.43), we introduce new unknown functions
Given the
3.5
y0
,
yi,
yk-
,
y'o
=
J^
y\
=
y2
1
and
Existence Theorems
271
replace (3.43) by the first-order system
=F{x,y0,yl, ...y^)
yk-i
.
y0(a)
=
b0, yi(a)
=
bl,..., yk-i(a)
=
bk.Y
Now the
general existence theorem for &th-order equations falls directly
first-order equations for systems. The
beauty of this
trick is that Picard's theorem is no harder for systems and consists
merely of
verifying that the appropriate transformation defined by an integral on
vector-valued functions is a contraction, so the fixed point theorem
applies.
Here, then, are the fundamental existence and uniqueness theorems for
ordinary differential equations.
out of the theorem for
Theorem
3.3. (Fundamental Existence and Uniqueness Theorem) Let
(a, b) be a point in R x R", and F an Revalued Lipschitz function defined in a
neighborhood of (a, b). There is an s > 0 and a unique continuously differenti
able R" valued function f defined on (a- e,a + e) such that ((a)
b and
f'(x) F(x, f(x))for all x in (a s, a + s).
=
=
-
Proof. The idea behind the proof is to change the given problem to a problem
involving integration. In fact, by the fundamental theorem of calculus, our
desired function is that function f such that
f(x)=b+
for all
[C((a
points
e,
a
x near a.
+
7T(x)
f F(f, f (f)) rff
e))]" (the
=
b+
That is,
space of
we
seek
n-tuples
a
fixed
point of the
function T defined
of continuous functions
on
(a
e,
a
+
on
e))
\\(t,f(t))dt
Because F is Lipschitz, we can choose e so that T is a contraction. We shall of
refer to the distance between functions introduced in Chapter 2.
First, since F is Lipschitz in a neighborhood of (a, b), there is an M and some
course
rectangle
B centered at
(a, b) such that
|F(x,y),F(x',y')l<M|y-y'|
272
3
Ordinary Differential Equations
(x, y), (x\ y') in that rectangle. In particular, F is bounded on that rectangle
byii:= |F(a, b)| + Me0. Let e < fi0tf"\ M~l/2, e0. Let X be the set of n-tuples
b || < e0
If
of continuous functions f on the interval (a
e, a + e) such that || f
feX, then for all f e (a
e, a + e), (t, f (f)) is in B and F is defined on B, so the
for all
.
transformation
7T(x)
=
f
b+
F(t,f(t))dt
"a
is well defined
on
We
X.
verify
now
that it is
a
contraction
on
Let f e X.
X.
Then
H7T(x)-b||<
f
Jfl
|F(f, f (f)) <7f
<K\x-a\ <Ke<e0
Thus
||rf-b||<e0,sorf6^ralso.
Let f, g
e
X.
\\Tf(x)
7g(x)||
-
<
f
J a
|F(f, f(f))
<m\
-
F(f, g(f)| dt
llf-gll*
<M|x-a| ||f-g||<Me!|f-g||<i||f-g||
Thus T is
a
contraction,
so
by
the fixed
point theorem
it has
a
unique fixed point
f.
We have
f(x)
so
=
b+
f
F(f,f(f))rff
by the fundamental theorem of calculus f is continuously differentiable (because
b, f '(x)
F(x, f (x)) for all x e (a e, a + e).
right-hand side is so), and f (a)
the
=
Certain remarks
on
=
this theorem
are
necessary.
First of all, the
differential equation of first order is of the form F(y', y, x)
0, noty'
The question arises: when can we rewrite the relation F(y', y, x)
=
=
=
general
F(y, x).
0 in the
theorem, for in this case we will know that solutions exist.
0 for one of its
of explicitly solving an equation H(u, v)
form of Picard's
This
question,
variables,
say
u
=
(so
that there is
a
function
G(v)
such that
H(u, v)
=
0 if and
3.5
only
if
u
=
will be discussed further in
G(v)),
Theorem 2.16 that
we
exists, for each y0
f '(x)
f(x0)
The
=
=
answer
from
(Recall
a
e
R",
for all
y0
for
general,
asserts the existence of local solutions.
/
x
R", / any interval in R,
function defined
a
F(x, f(x))
is in
7.
Chapter
273
condition for functions F of two real variables:
that this is the general condition.)
have
dF/dy # 0. We shall see
Secondly, Picard's theorem only
Supposing that F(x, y) is defined in
if there
Existence Theorems
on
we can
ask
all of /such that
/
x 6
Siven ^o^
For
no.
the function
example,
F(x, y)
=
y2
is
certainly Lipschitz in any rectangle, so local solutions always exist. But we
already know that if y' =y2, y must be of the form (c x)"1 for some
constant
c.
Thus, if
we
impose
an
initial condition
/(x0)
=
c0
,
the
(local)
solution is
/W
=
(I
On any interval
Exercise
interval
+ x0
x)
-
which the solution exists it is given by this formula (see
Thus there is no solution to this initial value problem in any
on
19(a)).
containing the point x0
+
l/c0
.
higher order and the reduction to systems of
Let us represent a point of R1+<*+1>" by coordinates
first order.
R".
y^ where x is a real number and the yt range through
(x, y0
We
now
turn to
equations
of
Revalued
(a0 b0 ,...,bk) e R1+<*+1>" and let F be
bk). There is an
Lipschitz function defined in a neighborhood of (a0,b0,
Revalued
function
e > 0 and a unique (k + l)-times continuously differentiable
that
such
+
defined on(a-e,a e)
Theorem 3.4.
Let
an
,
...,
f(a)
=
b0
fii\a0)^bi
l<i<k
F(x,f(x),f'(x),...,f(k\x))=fk+1)(x)
Proof. Consider
(a, b0,..., bk) by
in
the R(t+1 '"-valued function G defined
G(x, y0,...,y.)= (yt,
>
*->
pix,
Vo
,
.
,
y<d)
a
neighborhood
of
274
3
Ordinary Differential Equations
Clearly, G is Lipschitz wherever
unique function g defined in (a
g(a)
=
By Theorem 3.3, there is
F is.
e,
a
+
e) taking
an e
> 0
and
a
values in ,R<*+1>" such that
(b0,...,bk)
(3.44)
g'(x)
=
G(x, g(x))
gk)'(x) (gi(x),
Writing g =(g0, ...,gk) we have g,(a)=b, and (g0
gk-i(x), F(x, g0,..., gk)). Thus, splitting this into coordinates, gl =g,+i,0<,i<k
and yk(x)
F(x, g0
gk). Thus g6 gi gl g'i gi and in general g0J) gs
=
,
=
=
=
.
.
.
.
,
=
.
.
,
=
.
,
Thus
g0(a)
=
gX\a)
b0
=
b,
l<i<k
and
g0k+lKx)
=
F(x, g0(x), g6(x),
...,
g0k\x))
problem. The uniqueness follows immediately, for if /is a solu
original problem then clearly (/,/', ...,/(") solves (3.44), but the
solution of that is unique.
which solves
tion of
our
our
PROBLEMS
23. Let
h0,
...,
/ and suppose f is
(
=
hk-i be infinitely differentiable functions
+
the interval
0
Show that f also must be infinitely differentiable.
y<t+1)
on
solution of
a
A*-.y< +
kf(h,-i + h',)ya) + h0 y
=
(Hint: Any solution of
0
1=1
solution of the first equation.
{/} is a sequence of bounded functions in C(I) such that
/n-i II < C, where 2 C, < oo, then the sequence {/} converges to a
is also
a
24. Prove: If
ll/n
continuous function.
25. The differential
ponding
y(0)
y(0)
equation y" + y
to the initial conditions
=
l
y'(0)=0
=
0
y'(0)
=
l
=
0 has
unique solutions
corres
3.6
Linear
Differential Equations
275
respectively. Let C, S be these two functions. Prove :
l
(a) C2 + 52
-S
(b) S'
C, C'
(c) S(2x)
2S(x)C(x)
(d) e"
C(x) + iS(x)
Of course, the reader will recognize that C(x)
cos x and Six)
sin x
and thus these equations should follow. However, the intention here is to
verify these equations on the basis only of the defining differential equation.
26. Sometimes it is of value to find a linear differential equation which
has as its space of solutions the vector space spanned by n given functions.
We find an equation of nth order by substituting the n functions in the
0.
+goy
equation y(n) + ^-iy<""1) -I
For example, suppose we want to find the linear equation whose solution
=
=
=
=
=
=
=
=
is
set
the
of
span
y" + gy' + hy
=
and
x
sin
0 and substitute
x
We
x.
and sin
try
a
second-order
equation
x:
g + hx=0
sin
We
x
+g
can
cos x
x
x cos
equations :
x
g(x)
x cos x
sin x
0
=
x
=
Thus the differential
(sin
x
solve these linear
sin
hix)
+ h sin
x)y"
equation
y'x
sin
=
sin
x
-.
sin x
x cos x
is
x
+ y sin
x
=
0
Find the linear differential equation whose solution set is the vector space
spanned by the given set of functions.
(a)
(b)
(c)
(d)
(e)
(f )
3.6
x,
x2, x3
+ nx
ex, e", ea
xex, exp(x2)
sin x, cos x, tan
x sin x, cos x
x,
ex, tan
Linear Differential
x
x
Equations
The most important and best understood class of differential equations
We
and its derivatives.
are those which are linear in the unknown function
now
give
the definition of this class.
276
3
Ordinary Differential Equations
Definition 5.
order k is
on
Let / be
an
interval in R.
A linear differential
operator of
transformation from the space of A>times differentiable functions
/to the space of continuous functions on /of the form
a
L(/)
=
/(k)+I1"i/(0
(3.45)
;=o
where
h0
,
.
.
.
,
hk_i
are
given continuous functions
Notice that the coefficient of the
highest order
on
/.
term is 1
More
.
generally,
it could be any function hk
In this case, if hk is never zero on /, we could
divide by hk and obtain the form (3.45). If hk sometimes has the value zero,
.
then the
theory
to
be
presented
A transformation of the type
L (/ + g )
=
L(f)
+
here will fail
(3.45)
L(g)
(see
Problem
is linear, in the
L(cf)
=
31).
sense
that
cL(f)
It follows that the collection of functions which get mapped into zero by L,
{/ L(f) 0}, the kernel of L, is a vector space of functions. We shall
K(L)
=
now
=
show that this is
a
fc-dimensional vector space.
First of all, the equation /(/)
g, for
defined on the interval / has a solution /
given continuous function g
the whole interval, which is
determined
initial
conditions
uniquely
by given
f(a) f>0, ...,/(*"1)(fl)
h-iIn other words, in this case, Picard's theorem is more than local ; it gives
a solution on the whole interval.
We shall verify this fact below (in Pro
position 9). Thus, we can state :
a
=
on
=
=
Proposition 6. Let I be an interval in R, a el, and L a linear differential
operator of order k defined on I.
(i) if g is continuous on I and b0, ...,bk_l are any real numbers, there is a
unique Ck function f defined on I such that
f(a)
(ii)
=
b0 f'(a)
The space
,
=
bu... J^Xa)
K(L) of solution
on
I
=
ofLf=
6k_i
0 is
a vector
space
of dimension k.
Proof.
(i) will follow immediately from Proposition 9 below according to the same
procedure as in the preceding section for reducing a th-order equation to a firstorder system.
3.6
(ii)
Linear
Let Ea be the transformation from
E*(f)
=
Kit)
Differential Equations
to Rk defined
277
by evaluation
at
a:
(f(a),f'(a),...,f-lXa))
By the existence and uniqueness theorem, E is one-to-one and
Thus
onto.
K(L)
also has dimension k.
Let
us
reconsider
L(/)
=
briefly
case
of constant coefficient linear operators :
/W+Z1i/(
=
(
We associate to L the
PL(X)
the
=
(3.46)
0
polynomial
Xk+t^aiXi
i=0
(called the characteristic polynomial of L).
stitution of f(x)
erx, that if PL(r)
0, then
already seen, by sub
K(L). Now if PL has k
distinct roots rx,...,rk, then all of the functions exp(rxx),
exp(yvO
Since K(L) has
are in K(L), as well as all linear combinations of these.
dimension k, these exponential functions form a basis for K(L) and every
solution L(f)
0 is of the form
=
=
We have
e is in
.
.
.
,
=
At exp(rtx)
+
+
Ak exr>(rkx)
by the initial conditions. In case PL
2X + 1), the
X2
does not have k distinct roots (for example, PL(X)
in the
this
discussion
We
shall
situation is more complicated.
complete
of
the
discuss
next chapter, where we shall also
question factoring polynomials.
where the
Aj
are
to be determined
=
-
Examples
35. Solve
ym
y'(0) =1, y"(0)
has the roots
+
=
3y"
1.
conditions:
A+B+C=0
4B + C
=
=
2y'
0,-2,-1.
A + Be~2x + Ce~x.
-2B-C
+
1
l
=
0 with the initial conditions
The characteristic
Thus the
polynomial, X3
general
We solve for A, B, C
+
y(0)
3X2
=
0,
+ 2X
solution is of the form
by substituting the initial
278
3
Ordinary Differential Equations
Solving, we
f(x) 2 + e~2x
=
Linear
We
Systems
now
find
A
2, B
=
=
1,
C
3,
=
the
so
solution is
3e~x.
-
with Constant
Coefficients
turn to the solution of
First, let
with constant coefficients.
systems of linear differential equations
us try to see an example through to the
end.
36. Consider the system
x[
=
x'2
=
xx + x2
xt(0)
x2
x2(0)
Xj
=
a
=
b
(3.47)
According to the fundamental theorem we can approach
by successive approximations using the transformation
T(xi(0, x2(f))
=
f (xj(0 + x2(0, xx(f)
It is convenient to
(3.48)
use
matrix notation.
-l)x
*()
Equation (3.48)
becomes
=
Tx(0=j^{ _jWr)dT
Now,
=
x1
=
X2
=
we
*o
+ x0
successively approximate
Xn
|L _1jx0dT + x0
=
L
_1j<x0
/0[(l -l)(l _l)0 X0]dT
+
dt +
(a, b)
Thus, writing
becomes
x=\l
Xn
x2(f))
-
Jo
+
+ X0
^
a
solution
(3.48)
3.6
Linear
=G -!) yx+(! _!)*<>+
i\3 13
ii
-lj
=
ll
x
n2 12
ii
+
-1.)
ll
3!
279
x
a
+
2!
Differential Equations
n
(l -l)'
+
/*
=
to the fundamental theorem the series converges to the
According
solution
k
tk-\
>=['+!. C -H,
(3.49)
c0
The formula
(3.49) represents the solution in the sense that it describes
computing approximations to the pair of functions xx(t),
x2(0- (The question of measuring the accuracy of those approxima
tions is important; we shall return to those questions in Chapter 5.)
However, we have not obtained formulas for the functions individually.
That is not really surprising since the functions are given by an
interdependent relation (3.47).
By analogy with the series for ex, we defined the exponential of a
a
way of
matrix
as
exp(M)
=
eM
=
/ +
f^
Then
x(f)
We
now
we can
=
expfL
state a
(3.50)
k\
fc=i
write the solution to
(3.47)
as
_1jx0
proposition
which summarizes this discussion for
general
linear first-order systems.
Proposition 7. Consider
unknown functions :
x'(t)
=
Mx(0
the linear
x(0)
=
x0
first-order system of
n
equations
in
n
3
280
where
x
Ordinary Differential Equations
=
(xu
x(t)
=
7x(f )
.
.
,
x) and
M is
an n x n
matrix.
The solution is given
by
eMtx0
find, by successive approximations, the fixed point of
We
Proof.
.
f
=
Jo
Mx(f )
dt + x0
We obtain
Xo
=
Xi
=
Xo
MfXo +
Xo
(Mt)2
-
=
x2
Xo
+ MfXo +
x0
(Mt)-1
/(Mt)"
A
By the fundamental theorem the sequence of vector-valued functions
to the
solution of the
x(0
=
Although
[
we
know there is
{x}
1+
I
x converges
But the limit of x is
(Mt)"'
given by
(3.51)
T-
questioned the convergence of the series (3.50), we
problem. For, by the fundamental theorem the sequence
so the series in (3.51) must converge.
Finally, eM is just
have not
no
converges,
1.
eM'atf
given differential equations.
=
exponential of a matrix is not an easy thing to do; ordinarily
just work with the series and approximate solutions. However,
cases we can obtain explicit formulas for the solution.
Finding
the
it is best to
in certain
Examples (Eigenvectors)
37.
Suppose
(di
\
the matrix M is
diagonal.
Then
3.6
and the
xi
equations
dxxu x'2
=
=
d2x2,...,x'
in
Thus,
(dx
not
The solutions
expOAOXifO),
=
Xi
Differential Equations
281
are
However, this system is
equations.
Linear
.
particular,
.
.
,
xn
/expOA)
'dj
\
dx
system
a
at
all, but just
n
independent
are
=
exp040xn(0)
that
we see
0\
=
0 \
'exp(4,)/
0
38.
Suppose that the vector of initial conditions x0 is an eigenvector
Mx0 Xx0 for some X. Then M2x0 X2x0, ...,M"x0
X"x0 so we can compute the solution explicitly,
of M:
=
=
=
,
x(f)
eM'x0
=
(/\
=
Ii
+
n
=
oo
^)x0
/
n
=
!
n
f)n
_i_ V
x0 + 2j
r x0
~i n!
_
x0 +
e
=
-}!
i n
M"x0
tx
x0
This
computation leads us to speculate as to the existence and quantity of
eigenvectors of the (n x n)-matrix M. In general, this is a difficult quest and
still does not lead to a complete explicit solution of the differential equation
x'
Mx. However, if there is a basis of R" of eigenvectors of M, then we
can give a complete explicit solution.
=
Proposition 8. Suppose Vj
v are independent eigenvectors of the
x n)-matrix M, with
Xn, respectively. Then the equation
eigenvalues Xlt
x'
as follows. Write x0
can
be
solved
Mx, x(0)
c1v1 +
explicitly
x0
The solution is
c"v
(n
...,
=
=
=
.
x(t)
Proof.
=
c1 exp(A1f)v1
+ c"
-I
We compute the series
Mxo
=
M 2x0
=
M'xo
=
IVKc'vi -I
Mfc^v, +
c'A/vi -|
(3.51) directly:
+ Cv)
+
exp(A f)v
=
c'AtVt -|
c"Av)
h c"Av
=
c'A,2?! +
h
c"Av
+ c"A 2v
282
3
Ordinary Differential Equations
Thus
Mktk\
/
tk
I
"
\
"("-(,+.?1Tr)^-J?,4+,?^M^)
I
y=i
tkXk
I
n
=
"
\
+ I
Vy
4*0
-rfk!
\
/
fc=i
^
I
=
exp(l, OVy
y=i
Examples
39. Consider the system of differential
equations
*-(:Si)" >-(,J)
We find the
eigenvalues
of equations
as
and
eigenvectors corresponding to this system
in Section 1 .7.
Let
=(-l i)
Then
X
=
det(M
-
1, 2 of this
XT) X2 -3X
polynomial.
=
+ 2.
The
The eigenvalues are the roots
eigenvalue 1 has an eigenspace the
kernel of
m-.=(:26 J)
(1,2) spans the kernel. Similarly, the vector (1, 3)
eigenvector of M with eigenvalue 2 since it is in the kernel of
The vector
is
an
--G \)
The
general
solution of the
given
differential
This vector has the initial conditions
Our initial conditions
are
(4, 12),
x(0)
=
equation
cx + c2,
so we can
is
y(0)
=
2ct
+
3c2
.
solve for ci,c2:c1= 0,
3.6
c2
4.
=
x(f)
=
Thus,
4e2t
we
y(0
obtain the
=
Linear
283
Differential Equations
explicit solution
12e2'
40.
*'=(_i/4 !) x()=(S)
eigenvalues are 1/2, 3/2, and they have eigenvectors (2, 1),
(2, 1), respectively. Thus the general solution is
The
x(0
=
cie"2(_J) c2e3"2(^
+
Substituting
in
initial conditions,
our
we
obtain these
equations
for
c-i, c2:
3
=
3
=
2ct
+
2c2
-Ci + c2
The solutions
given
3
x
are
ct
=
-3/4,
c2
Thus the solution of the
=4/9.
system is
=
"
3,/2/2\
t,i( 2W4
S
+
4
l-l)
9
ll)
=
/-3/2e"2
I 3/4e"2
+
+
8/9e3"2\
4/9e3'/3)
41.
*-(_! !)" *-(!)
,3H)
X)2 + 1 0, so X 1 + /'.
(1
The root 1=1 + / gives
eigenvector (1, -i), and for the root
Now our initial con
X
1
/ we obtain the eigenvector (1, +0thus we obtain the
and
+
ditions are (1,0)
(1, -i)/2 (1, +i)/2,
The
equation
for X here turns out to be
the
=
-
=
solution
-
=
=
284
3
Ordinary Differential Equations
There is
easier way to solve this equation, and that consists in
that the matrix is of the form
an
recognizing
r>)
and represents
replace
z'O)
(1
=
complex number: in our case
system (3.53) by the single equation
our
a
i)z(t)
-
z(0)
by substituting z(f)
z(f)
=
y(0
same as
=
Re
z(0
=
Im
z(t)
=
-
=
42. Find the
x'=
x(t)
+
(e{
-.
(3.53),
l
"
'>'
+
(e(1-f)'
-
Thus,
we can
1
iy(t).
This has the solution
/I
-3
3\
3
-5
3
\6
-6
4/
thus M has the
eigenvalue
M-/II=
Thus the
of course,
e( 1 + ')
e(1+i)')
general solution
of
x
Now, after computation,
Two
/.
e(1-'"
which is the
x(f)
=
=
1
eigenvalues
we
find
det(M
-
XI)
=
(-2
-
A)2(4
X),
-
2, 4.
2:
(3
-3
3\
3
-3
3
\6
-6
6/
corresponding eigenvectors lie in the plane x y
independent vectors in this plane are (1, 2, 1), (0, 1, 1).
+
z
=
0.
3.6
eigenvalue
M-AI
The
x
-
Linear
285
4:
=
corresponding eigenvectors
0, which is spanned by ( 1
y
Differential Equations
lie
1
=
,
,
on
2).
the line -x-y + z
0,
Thus, the general solution is
=
43.
Hi)
This matrix is
it has
spanning set of eigenvectors. We
(1, 1, 0), (0, 1, 1) have the
already
Example
has
the
1,
1,
eigenvalue
(1,
1)
eigenvalue a- The general solution is
symmetric,
so
found them in
a
9:
The initial condition is
(-M-M-i)*(i)
Thus the solution is
XjO)
=
2e' + e2'
x2(t)
=
-
4e' + e2t
x3(t)
=
2e' + e2'
44.
Hi !)* x(0)=(?)
&54>
286
3
Ordinary Differential Equations
X)2 0, so we obtain only
equation for the eigenvalues is (1
has
the
X=\.
This
eigenvector (1,0). Thus we
eigenvalue,
know one solution of the general equation
The
=
-
one
X(o
=*<(;)
However, this does
satisfy the given conditions.
equation without further study of
not
to
solve this
and that is
generally
proceed
first
row
x'(t)
and
=
=
we
after
y,
y(0)
=
by observing
This has the solution
1.
the
matrix,
In the present case we can
that the second row of (3.54) is
difficult search.
a
avoid such difficulties
just y'
We cannot
y(t)
=
e'.
Then the
is
x(t)
+ e'
x(0)
=
0
know how to solve this
equation : x(0
=
te'.
Thus
our
sought-
pair is (te1, e').
example the solutions are not linear combinations
exponentials, but admit polynomial factors. Only when there is a basis
of eigenvectors are the solutions linear combinations of exponentials ; when
there are too few eigenvectors, we must expect more complicated coeffic
Notice that in the last
of
ients.
There is
a
theorem that any solution of a first-order linear system
a combination of exponentials with polynomial
with constant coefficients is
factors.
This theorem follows from the Jordan canonical form of
shall not go into it here.
We conclude this section with the
a
matrix;
we
theorem from which
Proposition
6
proof
was
of the
global version
of Picard's
obtained.
Proposition 9. (Global Version of Picard's Theorem) Let I be an interval
Suppose F is a continuous R"-valued function defined on I x R" which
satisfies this strong Lipschitz property : there is a constant K>0 such that
for allyuy2eR"
in R.
sup{| Fix, yi)
Then the system
y'
has
a
=
-
F(x, y2) | :
x e
1}
<
K\\7l
-
y2
1|
of n equations
F(x, y)
y(c)
unique solution for
=
a
any initial condition
a at c e
I.
(3.55)
Linear
3.6
We cannot
Proof.
simply
use
the fixed
287
Differential Equations
point theorem, for the transformation
T
defined by
7T(x)
is not
a
=
+
a
contraction
| F(f, fit))
on
theless the successive
continuous functions
fo(x)
d(x)
=
=
dt
the space of functions continuous
works.
on
the interval /.
Define
approximations procedure
a
Never
sequence
{f} of
/ by induction :
on
a
a
j Fit, f0(f ))
+
f(x)=a+
f
dt
F(f,f-i(r))rff
The sequence {f} converges in C(I).
besides (3.55) we also have ||F(x, a) ||,
By making K larger we may assume that
We prove by induction that
< K.
|x)-.-.(x)|^-^-|x-c|"
(i)
=
1
I fi(x)
(ii)
-
fo(x)|
=
f F(f, a)
\
dt < K
dt
=
K \x
-
c\
nl=>n
lx)-t-i(x)|
=
f
J
[F(f, f_i(f))
-
F(f, f-2(r))]
*
c
<Kf\f-i(t)-f-i(t)\dt
(3.56)
From
(3.56),
we
obtain
[K(b-a)]"
|f-f-ilU<
n\
288
3
Ordinary Differential Equations
Since the series
2 [Kib
a)f/n ! converges, it follows that {f} is a Cauchy sequence
C(/) such that f -> f. Since T is continuous on C(7),
Tf->Tf. But rf
f+i, so f^Tf also, thus Tf f. Since f is a fixed point of T
we conclude as in Picard's theorem that f solves our problem.
Now the fixed point theorem asserts the uniqueness of our fixed point, and
we seem to have lost that.
But we can regain it on /, because locally we have
Picard's
theorem.
uniqueness, by
Suppose g is another solution of the problem; we
f on /.
For this purpose we may assume that the point c at
have to show that g
which the initial condition is given is one of the end points of /. Let
in
C(7)
so
there is
f
an
e
=
=
=
R
=
sup{r
/
f (x)
:
=
g(x) for all
x
<
r}
a
g(c), cis in the set on the right. Also, b is an upper bound for this set,
We have to show that R=b.
If R < b, then
the least upper bound R exists.
the differential equation is defined in a neighborhood of R. By Picard's theorem,
Since f (c)
=
=
so
there is
equation y'
F(x, y) has a unique solution in
f
But both f and g, when con
(R).
e)
y(R)
sidered as functions on (R
s, R + e), are such solutions.
(Notice g(7?) f (R)
e, R + e), so R + e is in the set above, and R
by continuity.) Thus, f g on (7?
is not an upper bound. Thus the assumption R < b is contradicted, so R
b and
the proposition is proved.
(R
an
s, R +
e
> 0
such that the
with initial condition
=
=
=
=
=
EXERCISES
23. Find the
(a)
y2
(b)
(c)
general solution of these systems of equations
2y2
2y2 + 4yi
y'i =4yi
yi
=
=
yi -y2
y2
=
ay i + y2
yi
=yi + y2 + yi
y2
y'i
=
ayi + y2
=
ayi + y3
24. Find the solution of these initial value
problems
1.
(a) The system in 23(a) with initial condition yi(0)
1, y2(0)
(b) The system in 23(c) with initial conditions Vi(0) =y2(0) =0,
=
25.
yi
1
*(0)
1
y2
yi + y2
*2(0)
l
3yi
(d) y'i
y3
yM
1
y'2 yi + 2y2 y3
*(0)
-l
y'i 2yi
2y3
*(0)
Find the general solution of the equation
(c)
=yi
+ y2
=
=
=
=
=
=
=
(b)
the matrix in Example 10.
the matrices in Exercise 10.
(c)
the matrices in Exercise 1 1
(a)
=
=
by:
=
x'
=
Mx, where M is given
3.7
(d)
/
o
-1
2
3
-2
Second-Order Linear Equations
-3\
(g)
/
2
0^
0
/-2300
3
I
11/
(e)
/
7\
(h)
<f)
(J I)
m
4
3
289
0
0
T
2
PROBLEMS
27.
Suppose M
=
(a/)
that the solutions of x'
(Hint:
M"
is
an n x n
Mx
=
are
matrix such that
all
polynomials
a/ =0ifi< j.
of degree at
Show
most
n
0.)
=
28. Show that
exp(M')
=
(eM)'.
29. Show that if M is skew-symmetric (M'
-M), then eM(eM)' =1.
For such a matrix the rows form an orthonormal basis : A matrix A with the
=
property AA'
3.7
=
I is thus called
Second-Order Linear
The most
y"
Thus the
In this
+
cr1(x)y'
we
rotation.
Equations
equation :
+
techniques
section,
a
type of equation arising from physical problems is the
comon
second-order linear
orthogonal, and represents
a0(x)y
for
shall
=
(3.57)
g(x)
solving such equations
assume
that
we
know
have been well
one
developed.
solution of the associated
homogeneous equation
y"
+
ai(x)y'
+
a0(x)y
=
0
(3.58)
and show how to find the general solution of (3.57). The question of finding
this first solution is of course difficult, and further discussion will be postponed
until Chapter 5. The technique involved in finding the general solution
consists in substituting candidates involving the given solution and a new
to reduce the complication in the
unknown
and
function,
given equation.
thereby attempting
290
3
Ordinary Differential Equations
theory of the first-order
equation: y'
h(x)y g(x).
homogeneous equation is easily solved
of
variables
:
by separation
f(x) exp(Jx h) is a solution of y' hy 0.
to
find
the
solution
of the given equation we substitute y
Now,
z/
general
where z is some new unknown function. From/' + hf= 0, we obtain
In order to motivate this
+
discussion, let
us
recall the
The
=
=
=
=
y'
=
9
hy
+
=
z'f+ z(f'
+
hf)
=
z'f
by integration: z J /" V + cNow, the second-order homogeneous equation (3.58) has two independent
Let us try to find another
solutions. By assumption we know one, call itfx
by substituting y zfx. The new equation in z is
Thus
z
=f~1g,
so z
is found
=
.
=
y"
aiy'
+
+ a0y
=
=
This
find
equation
z(x)
c
=
+
2z'f\
z"/1
+
(2/'1+a1/1)z'
is linear in z' and thus
We have
z.
zf'[
z"fx
+
we can
+
ax(z'fx
+
zf\)
+ a0
zfx
0
=
(3.59)
solve for z' and then
integrate
to
result
as a
f f(t)-2
exp
(- Q
dt
Examples
45. The
We
y'
now
=
equation x2y"
z'x + z,
x2y"
xy' y 0 has the
by substituting y(x)
+
=
find another solution
y"
+
xy'
+
(3z')x2
-y
z"x + 2z',
=
=
z"x3
+
solution
=
z(x)x.
y(x)
=
x.
We have
so
2z'x2
+
z'x2
+
xz
-
zx
=
0
or
z"x3
=
0
Dividing by x2 we have z"x + 3z' 0, which we
separation of variables: z Cxx~2 + C2. We
=
=
and thus the second
46.
xy"
-
sin x2 is
y'
+
4x3y
a
solution,
solution of
=
0
y
=
zx
=
x"1,
can
can
is found.
solve for z'
take
z
=
by
x~2,
Second-Order Linear Equations
3.7
We substitute y
z"x sin x2 +
=
z'(4x2
sin x2 and obtain this differential
z
x2
cos
sin
-
x2)
equation
291
for
z
0
=
Thus
_=
-4xcot x2 +-
z'
x
Integrating,
In z'
obtain
csc(x2)
2 In
=
we
x
+ C
again,
we
+ In
or
z'
=
Cxx esc2 x2
Integrating
once
second solution
might
Now that
have
we
can
guessed
have
a
find
be chosen
the
at
z
the homogeneous equation, we
our cue from the first-order case,
as/i,/2
y(x)
If
we
Now,
=
we
Let
consider
zMhix)
+
a
+
=
C2
cos
Thus, the
x2 (which we
.
independent solutions for
general equation (3.57). Taking
technique for finding
homogeneous equation.
Cx cot x2
x2 sin x2
beginning).
return to the
the
=
cot
as
a
we
try
us
refer
two
combination of the solutions of
to
these
two
solutions of (3.58)
function of the form
(3-6)
z2(x)/2(x)
into (3.57)
compute y' and y" and substitute
we
will get
a
totally
un
two unknown functions z z2
intelligible equation of second order in the
of course, a pair of equations.
What we need, to find two unknown functions, is
We notice, first of all, that
From where is the second equation to come?
functions zt, z2 even 1
the
the formula (3.60) does not uniquely determine
so that (3.60)
if
For,
z z2 are found
we know the sought after function y.
subtract
and
gf from z2
to
r
gives the solution y, then we may add gf2
condition
another
seek
thus
We
another
making (3.60) valid.
.
,
pair
obtaining
(preferably involving derivatives) which
functions zx, z2
Differentiating (3.60),
will
we
serve to
uniquely identify
the
obtain
.
y'ix)
=
z,(x)/;(x)
+
z2(x)/2(x)
+
z\ix)hix)
+
z2(x)f2(x)
(3.61)
292
3
Ordinary Differential Equations
Equations (3.60) and (3.61) will give a pair of linear functions in zx(x) and
z2(x) if the sum z'x(x]fx(x) + z2(x)/2(x) vanishes. This pair of equations
(if noncollinear) will then identify zx(x), z2(x) in terms of y(x), y'(x). Thus, if
that condition is satisfied we know that zx, z2 are uniquely determined by the
solution y. Turning the argument around, we impose the condition
*'i/i
+
(3.62)
zi/2=0
hope now that, together with this condition, the given differential equa
explicitly determine zx,z2. (In fact it will do so theoretically, since
Equation (3.57) determines the solution y which in turn determines zx, z2
in the presence of the condition (3.62).) Let us try our idea on Example 45.
and
tion will
Example
x2y" + xy' y x2.
We have the two solutions x,
47. Solve
=
We consider y
z'1x
+
Now let
=
y''
=
and
x"1 of the homogeneous equation.
impose
the condition
z2x"1=0
(3.63)
substitute this information into the
us
the presence of
y'
1
zxx + z2 x"
=
(3.63),
-z2x"2
z'x
z'2x~2
we
given equation.
In
have
zx
2z2x'3
+
Then
x2
=
=
x2z\
x2y"
x2z\
-
z2
Now the
+
xy'
y
z2 +
=
2z2x_1
z2x_1
+ xzx
ZjX
z2x_1
x2
pair
(3.64)
of linear
Equations (3.63), (3.64)
can
be solved
by
Cramer's rule:
X3
1
-X
=
-2x
z2
2
Integrating, we find
general solution is
y
=
zl/l
+
^2/2
=
=
-2x
that zx
\x
-1
=
-
0-x3
=
,
x2
2
x/2
+ cx, z2
+ cxx + c2
X
=
x2/6
+ c2, and
so
the
Second-Order Linear Equations
3.7
Now, it
not an accident that in this case the
was
equations
293
turned out to be
of linear first-order equations: this is always the case. We shall now
describe the technique in general. Supposing \!na\. fx,f2 are two independent
a
pair
solutions of the homogeneous Equation (3.58) we try a function y
We impose the condition
+ z2/2 as solution of (3.57).
zi/i
z'2/2
+
=
=
zxfx
(3.65)
0
Then
y'
y"
Thus
=
=
Zlf'l
+
Z2/2
z'J'x
+
z'2f'2
+
zxn
z2f2
+
becomes
(3.57)
z'i/i
Z2/2
+
+ z2
a
+
1/2
Zl/l
+
+ ziao
z2/2
/1
+
ZlOl/i
+ z2 o
fi
=
9
or
z'i/i
+
(3-66)
z'2fi=g
vanishing since fx,f2 solve (3.57).
linear Equations (3.65), (3.66) by Cramer's rule.
the rest of the terms
-fiQ
,
_
Zl~/i/2-/2/i
and these
can
=
/i/'2-/2/i
in order to find the general solution. One
functions
is the denominator. If it ever vanishes, these
down.
break
will
discussion
whole
In fact, our
may not be
integrable.
Fortunately,
we can
verify
once
and for all that this function is
The function
=
fx(x)f'2(x)
-
f'x(x)f2(x)
is called the Wronskian of the
W
=
hf\
of
integrated
be
apparent problem
W(x)
pair
/iff
,
'
We solve the
-
fih
pair/j f2
,
=
det(g] /?$)
Notice that
nonzero.
294
3
Ordinary Differential Equations
Since fx,f2 solve
W +
(3.57),
easily check
we can
that
axW=0
and thus
W(x)
W(x0)
=
Thus if W is
exp(- fax(t) dt)
nonzero at one
point,
it is
never zero.
W is
the vectors
nonzero at
(fx(x0), f{(x0)) and (/2(x0), /2'(x0)) are independent;
guaranteed if the functions^ and/2 are independent.
x0 if
this is
Examples
48. Solve
to
y"
that
see
+
is
x
Xy'-y
find another solution
is z"x +
(2
Z
+
x2)z'
=
=
x,
y(0)
solution of the
a
by substituting
0.
0,
y'(0) 0. It is
homogeneous equation.
=
y
=
=
zx.
easy
We
The equation for
z
Thus
X
so
z'
Cx"2exp(-^
=
Thus
we
may take
as
the second solution
*2\
y(x)
=
xz(x)
Now let
us
=
zi
+ z2
+
-
refer to the
by substituting
z'xx
j t~2 exp( -) dt
x
X(j)(x)
z'2(x(b'(x)
y
=
+
=
integral by cb(x). We solve the given equation
z2x<b(x); this gives the pair of equations
zxx +
0
<p(x))
=
1
3. 7
Since
</>'(x)
=
x"2
Second-Order Linear Equations
exp(-x2/2),
we
295
find, by Cramer's rule,
-x(p(x)
exp(-x2/2)
exp(-x2/2)
or
zi{x)=f0-texp{-2-)[sy2exp{-2~)d\
z2(x)
The
exp(yj
=
integrals defining
nevertheless define
y(x)
-x
=
+
This
It
we are
zx
are
not
function.
expressible
in closed form, but
dx
dt
xexp(y)J"0r2exp(~T)dT
solving second-order equations is called variation of
applied to higher order linear equations. Suppose
differential equation :
for
can
be
such
y(n)
I atix)yll>
a
+ y
=
(3.67)
g(x)
Suppose we have somehow found n independent solutions fx, ...,/
homogeneous equation. Then we try a solution
y
=
zi/i
+ z
+
.
.
,
z'i/i
zi/i
+
---
+
z;/;
+
---
+
z;/;'
zi/r2)
+
--
+
of the
fn
As in the second-order case, the solution will
zx,
z if we impose the conditions
.
they
Thus the solution is
fj exp(- ^) [jV> exp(- l)
given
+
a
.
technique
parameters.
dt
=
o
=
o
z;/r2)=o
uniquely determine the
functions
296
3
Ordinary Differential Equations
In the presence of these
zi/S,_1)
We
-"
+
+
conditions, (3.67) becomes
z;/i"-1)
solve this system
can
as a
system of linear equations and then find the
as in the second-order case, this
system is
solvable since the determinant
is
fx, ...,/)
The
y
(called
the Wronskian of the
n
functions
never zero.
\x3y"'
49. Solve
=
0
Just
by integration.
zx, ...,z
=
x2y"
-
2xy'
+
2y
-
=
x5(x2
-
9).
(3.68)
homogeneous equation has the solutions x, x2, x3.
z2x2 + z3x3. We impose these conditions:
Thus
we
try
zxx +
z\x
z\
+
+
z'2x2
2z2 x
z'3x3
=
0
3z'3 x2
=
0
+
+
In the presence of these conditions
z2 +
6z3
x5(x2
=
we
compute (3.68)
to
be
9)
-
The matrix of this system is
tx
X2
x3
1
2x
3x2
1
6
\0
\
/
which has the determinant -2x3 + 6x2
rule,
we must
"
,
""
'
=
-2x2(x
-
3).
have
x5(x2-9)-x4
-2x2(x-3)
,
Z2
=
'
-x5(x2~9)-2x3
-2x2(x 3)
-
x5(x2 9) x2
-2x2(x-3)
-
,
23
After
_
~
integration
we can
x9
y(x)
=
zr
3
rx7
express the
general solution
as
3x8
+
^r
x6
+X1
+ c,
I
7+2"H
+
+c2
Thus, by Cramer's
3.7
Second-Order Linear Equations
297
EXERCISES
26. Show that the general solution of
y" + y=f
can
be expressed
y(t)
=
ci cos f
+
as
c2
sin f +
Jo
sin(f
t)/(t)
-
dr
27. Find the general solution of
y"
+
y'
=
x
28. Find the general solution of:
(a) y" 4y 1
=
-
(b)
(c)
y'"-y' x2
y" + 3y' + 2y
(d)
+
y"-^Lry'
1
(e)
x2y"
=
=
sin
x
^y
x
4xy' + 6y
=
=
0
1
x
x3 + x2
29. Find the solution of
x2y"
-
2y
=
2x2
30. Find the
y(0)
=
1
y'(0)
=
1
general solution of
y" + xe'y' -exy=0
31. Find the solution of
l
y(0)=0
y'(0)
30. A differential
equation of
the form
e"*y" + xy'-y
=
1
=
PROBLEMS
akx"yw
+
ak-i
where the a,'s
y
=
x\
xk-1y('"1) ^
are
constants
1-
<*ixy' +
can
be
a0 y
=
0
easily solved.
Try the substitution
You should obtain
xs[akis)is -l)---(s-k) + ak.iis)is
-
1)
(s
-
k+
1) +
h
ais
+ tfo]
=
0
298
3
Ordinary Differential Equations
Thus
we
need only find the k roots of the polynomial in brackets.
general solution of these differential equations:
Find the
(a)
(b)
(c)
id)
x2y"
2xy' + y
x2y"
3xy'
3y
x2y" + 4xy' + 3y
x2y"-xy' + y
0
=
-
-
=
=
0
=
xs
0
31. Solve the second-order 2x2 system of
equations
*-+(-! jm; 0'iHint: Go
3.8
to the first-order 4x4
system by adding the equations y' =z.)
Summary
An Revalued function defined in
neighborhood
a
of x0 in R is called
differentiable at x0 if
lim/(x0 + t0 -/(x0)
f->0
exists.
This limit is denoted
/ in R its
image
is
a curve
f'(x0).
If
/ is differentiable on an interval
through /(x0) spanned by /'(x0)
If h is differentiable in a neighbor
The line
in R".
is the tangent line to the curve at /(x0).
hood of the curve and has a relative maximum
<Vn(x0), /'(x0)>
h,
g
are
We
can
deduce the
differentiable functions defined in
(or minimum)
x
0.
=
of h
subject
for which there exists
g(x)
=
0
a
Vn(x)
on
the
curve
following principle
at
f(x0),
then
from this.
If
domain in R", then the maximum
to the restraint g(x)
0 is attained at those points
a
=
X such that
=
XVg(x)
If h, gx, ...,gk are differentiable in R", and h has a maximum (or minimum)
Xk
subject to the restraints gx(x) 0,
gk(x) 0 at x0, there exists Xx
=
=
.
.
.
.
,
..,
such that
<7,(x0)
=
0,
.
.
.
,
gk(x0)
=
0
V(x0)
=
X.Vg^Xo)
Suppose / is an Revalued function defined on
(fc-times continuously differentiable) if /', ...,/(t)
+
+
XkWgk(x0)
the interval /.
all exist and
are
/
is Ck
contin-
3.8
If/ is
uous.
fix)
where
such
a
/(x0)
=
e(x x0) is
If/has derivatives
function
+
we
'"'/"W,..
1 '! (x
~
i^-
-
'
limM*(x~Xo)*
k\
,.,
+ six
x0);
"'
by Mk
orders, and
of all
299
have Taylor's expansion about
any x0
bounded
-
Summary
=
"
-
x
max{|/(*'(/) I
e
/:
'
,(x-x0)
xj
""'
:
/
(fc+1)!
between x0 and x}.
0
=
k->OD
then/can
be
expanded in
/(x0)+I
=
n=i
A differential
x, y, y ',...,
/("(x),
equation
^-^(x-xo)"
n!
equation of order k is
y(M.
is
a relation
involving a function of
^-times differentiable function / such that this
after the substitution y =/(x), y' =/'(x),
y(,
If there is
relation holds for all
a
x
a
.
solution of the differential
say /
of order k is
we
infinite Taylor expansion:
f(')(r )
to
/(x)
an
a
.
=
.
,
A linear differential
equation.
relation of the form
k- 1
y(t> +'I
X ai(x)y(0
;=i
+
a0{x)y
=
g(x)
(3.69)
where the functions a-, and # are (at least) continuous on an interval /. If
jsO, the equation is called homogeneous. The space of solutions of the
homogeneous equation is a vector space of dimension k. Equation (3.69)
has a solution on / uniquely determined by the initial conditions
/(x0)
=
has
a
=
Fix,
unique
condition
on
=
*i,...,/('"n(xo)=flt-i
(3-7)
of the form
Any equation
y(k)
/'(*o)
o
y,
ylk~u)
y'
solution
subject
to
the initial conditions (3.70) under this
F:
(i) Fis defined and continuous in a neighborhood
(ii) F is Lipschitz: there is an M such that
of
(x0,
a0,
|F(x,y1,y'1,...,yri>)-F(x,y2,y'2,...,yri,)l
<
Mi\)\
-
y2\
+
\y\
-
y'i\
+
+
\y\k~u- ^""D
.
.
.
,
ak-x).
300
3
Ordinary Differential Equations
Techniques for Solution
1
.
Successive
y'
=
/0
{/}
=
y(x0)
F(x, y),
is solvable if F is
sequence
Lipschitz
defined
as
=
a0
near
x0
fi(x)=
equation
The solution
.
can
be
approximated by
a
follows :
any continuous
/i(x)=
The
approximations.
function,
fF(t,f0(t))dt
+ a0
fF(t,fx(t))dt
+ a0
Jx0
Jx0
fn(x)=CF(t,fn-x(t))dt + a0
2.
If
Separation of variables.
\g-1(y)dy \f(x)dx
=
implicitly
determines y
as a
then the
y' =f(x)g(y),
equation
+ C
function of
The
3. First-order linear equations.
x.
homogeneous equation
y'+/y=0
by separation of variables: y
cexp( |/). The equation
y' +fy g can be reduced by the substitution y z exp( J/). The result
ing equation in z is solved by separation of variables.
4. Constant coefficient linear equations. The characteristic polynomial
can
be solved
=
=
=
of the differential
equation
y(t) + ^_1y(fc-1)+---
+
a1y'
+ a0y
=
+ axX
polynomial Xk + ak_x X*"1 +
polynomial, then erx is a solution of (3.71).
is the
in
5. First-order linear systems.
Let A be
n unknown functions y
y):
(yx,
=
.
.
.
,
(3.71)
0
+ a0
an n x n
.
If
r
is
matrix.
a
root of this
The
equation
3.8
y'
Ay
=
y(0)
has the solution
eM
=
y(f)
exp(M)
=
n=i
If y0 is
The exponential of
a
matrix is defined
by
?L
/ +
=
301
y0
eA'y0.
=
Summary
n!
eigenvector with eigenvalue X, then the solution is y(f)
basis yx,
y of eigenvectors of M, with eigenvalues Al5
respectively, then the general solution is
an
cx
In
a
a
.
exp(Xxt)yx
general,
we
must
.
.
,
+
+ c
allow
y"
find
a
equation
y"
+
ax(x)y'
=
+
Suppose fx,f2
ax(x)y'
by the substitution
zi/i
+
a0(x)y
+
y
=
zxfx
=
+
knowing
one
solution.
.
,
Suppose fx
0
second, by substituting
in z'.
.
is
(3.72)
=
y
are
zfx.
This produces a linear first-order
Then we solve
solutions of (3.72).
g(x)
z2f2
(3.73)
.
z2/2=0
In the presence of the condition
(3.74)
Equation (3.73) becomes
zi/i
+
4/2
=
(3.75)
0
The linear Equations (3.74),
found by integration.
(3.75) can
be solved for zx, z2 and then zx, z2
are
FURTHER READING
E. A.
.
Xn
polynomial coefficients.
a0(x)y
+
.
exp(Xn t)y
6. Second-order linear equations,
solution of
we
ex'y0
=
If R" has
Coddington, An Introduction to Ordinary Differential Equations,
Prentice-Hall, Englewood Cliffs, N.J., 1961. An elementary book on
differential equations which goes more deeply into the material of this
chapter.
M. Tennenbaum and H. Pollard, Ordinary Differential Equations, Harper
302
3
Ordinary Differential Equations
and Row, New York, 1963. This is a thorough treatment of the subject of
differential equations. Many special techniques and applications are
exposed.
F. Brauer and J. A.
Nohel, Qualitative theory of Ordinary Differential
York, 1967. This book studies the theory of
New
Equations, Benjamin,
systems of differential equations, and in particular the behavior of
sets of
solutions.
L. Loomis and S.
Sternberg, Advanced Calculus, Addison-Wesley, Reading,
a very modern approach to the subject.
It goes
This is
Mass., 1968.
into the fundamental theorem.
thoroughly
MISCELLANEOUS PROBLEMS
32. Show that if M is
<Mx, x>
=
skew-symmetric matrix (M'
M), then
symmetric and thus has a basis of
Conclude that, considered as a matrix over C, M also has a
0 for all
eigenvectors.
a
=
Show that M2 is
x.
iHint : M2 A
(M + V A)(M
eigenvector of M2 with eigenvalue A, then either x is
basis of eigenvectors.
an
with
eigenvalue V A,
V
=
or
(M
V
A)x
is
an
an
eigenvector of
A).)
Thus if x is
eigenvector of M
M with
eigenvalue
-Vx.
33. Let / be any linear transformation.
Compute the gradient of
1 is attained at
<Tx, Tx>, and show that the maximum of ||7x||2 on |[x||2
=
an
eigenvalue
of T'T.
34. Show that if T is
whose
major
axes are
35. Find the
of
points p0
a symmetric matrix, 7X{||x||2
1}) is an ellipsoid
length equal to the eigenvalues of T.
e {(x, y) e R2 : xy
1}, pi e {(x, y) e R2 : y + x2 0}
=
=
=
which minimize the distance between these two
curves.
36. Minimize and maximize the volume of
a box with given surface area.
point on the ellipse [x2 + iy2
1} which is closest to (|, 0).
Find the furthest point from (, 0).
38. Find the point on the ellipse {x2 + iy2
1} which is closest to the
circle of radius centered at (I, i).
39. Suppose {} is a bounded sequence. Define fix)
2= i a x".
Show that /is infinitely differentiable in the interval (1, 1), and nla
37. Find the
=
=
=
=
fM(0).
40. Let / be a twice continuously differentiable function defined in a
neighborhood N of (0, 0) in R2. Show that there is a function e defined in
N such that lim e(p)
0 and
=
P-.0
fix, y) =/(0)
41.
+
8 (0, 0)x 8 (0, 0)y
+
+ e(x,
y) \\(x, y)\\
Using Taylor's theorem, we can derive the exponential function in
Suppose that / is a function with the property that
yet another way.
3.8
fix) =/(x) for all
Taylor expansion
*
f(x)=
Then
x.
1
I ~X"+t(x)
=on!
/<"(*) =/(*)
for all x,
Summary
so
/
must
303
have the
xk+1
(A:+l)!
for all k.
Because of the estimate on ek it remains bounded as k ->
oo, so
should expect /to be the limit of the
polynomials Pt(x)
2'=o (l/!)x".
We already know, from the theory of
Chapter 2 that the lim Pkix) exists
,
we
=
for all
Noticing
x.
that Pk
=
/,_
prove
that/(x)
=
lim
k~"
have the
42. With
a
little bit of
patience,
you should be able to find
/'(0)
=
Redoes indeed
property/' =/.
0 and /<"(x) + fix)
a
=
and in the
same
function / defined
0 for all x.
way
on
as
in Exercise 41,
R such that
/(0)
=
l'
43. (a) Suppose that /is C on [-R,R] and
/(0) =/'(0)
/('_I)(0) =0. Then there is a continuous function g such that /(f)
fgit), and<7(0) (l//c!)/<(0).
(b) Suppose that/is C on [- R R]. Show that there is acontinuous
=
=
=
=
function g such that
+ tW)
f(t)=2 r-^t'
l\
1
=
0
44. Change the conditions in Problem 18 as follows: The ratio of horse
population to total population is constant and only the eggs hatched in
horses produce mature insects. Derive the differential equations now
governing the population growth.
45. Suppose now we have an insect which has a natural death rate of d,
The
per insect per year and which lays h eggs per insect per year in the air.
egg hatches if it lands on a horse and the hatching causes the horse's death.
Assuming birth and death rates bH dH for the horse and a probability k
that a given egg will land on a given horse, now find the differential equations
of population.
Let a
z represents a force field on the plane.
46. Suppose f(z)
in
case
the
is i,
the
motion
Describe
be
at
1
at
time
0.
velocity
particle
(1 + 0/2, (1
i)/2.
47. We assume that a particle generates a force field directed toward
the particle and of strength equal to the inverse of the square of the distance
0 there are particles at rest at points pi,
to the particle.
At time t
p*
in R3. Let f,(f ) be the position at time t of the particle originally at p,
What is the differential equation the function (fi,
f) must satisfy?
48. Suppose a river deposits water in a lake at the rate of v gal/day. We
Suppose
may assume that v is a periodic function of time with period 365.
Finally,
two pumps pump water out at the constant rates of wt, w2 gal/day.
,
=
-
=
.
.
.
,
.
.
.
.
,
304
3
Ordinary Differential Equations
evaporates out of the lake at a rate of kit ) gal/day/ft2, where k is also
periodic with period 365. We may assume that the area of the lake is
proportional to W213, where Wit ) is the volume of the water in the lake at
water
Write the differential equation W must satisfy.
Suppose a missile A is moving in a straight line with constant velocity
A tracking missile B of constant speed s0 is always pointed toward the
v0
missile A
Find the differential equation of motion of the tracking missile B.
50. Suppose we have the same situation as in Problem 49, but this time
the speed of B is proportional to the distance between A and B.
Find the
equation of motion of B.
51. A falling body actually experiences a drag due to air resistance which
is proportional to its velocity.
Suppose a body of 100 tons is dropped from
a plane 5 miles high ; and this constant of proportionality (which depends
of course on the shape of the body) is 20.
How long will it take for the
body to reach the ground ?
day
f
.
49.
.
.
52. Two chemicals A, B in solution combine to create chemical C accord
C.
ing to the equation 2A + B
Suppose the rate of the formation of C is
proportional to the product of the amounts of A and B present and inversely
proportional to the amount of C present. Find the differential equation
governing the formation of C, assuming initial amounts A0 B0 of chemicals
-
,
A,B.
53.
10, B0
Suppose in the above problem, A0
5, and the proportion
How long will it take for the reaction to complete?
=
=
constant is 1.
54. If two bodies A B of different temperatures come in contact with each
rate of change of temperature is proportional to the difference in
,
other, the
temperature (the proportion constant depends on the bodies).
TA TB are the temperatures of A, B, respectively, we have
Thus if
,
T'A
T'B
=
=
kAiTA
kiTB
TB)
-
TA)
Find the formula for TA
kA
4, kB
kA =2,kB
55. In Problem 54,
(a)
(b)
What is it in
=
=
=
,
TB with these data :
5, 7^(0)
i, TAi0)
as f
->
oo
=
=
100, TB(0)
120, r(0)
(a),
case
-
=
=
=
=
=
=
=
.
-
-
-
0.
=
50.
the bodies tend to
(b), in general ?
56. Solve these differential equations :
(a) y<4)
3y" + 2y 0.
2e".
(b) y" + 3/ + 2y
cos x.
(c)
y' sin y + cos x cos y
(d)
(x2 + l)y'-2xy x2 + l.
(e)
xy' + 3y x~2 sin x.
x' + ax
b sin f.
(f )
(g) y" xey
(h) y<4> y<3> y(2) y' 2y 0.
case
=
-
=
a common
temperature.
3.8
(i)
(j)
(k)
Summary
305
ay" + by' + cy
0.
1 + y\
y'(l + x2)
x' + y'
2x.
x'-y' 3y.
=
=
=
=
M
,-(_? {),.
57. Solve these initial value problems:
(a) y"
3/ + 2y
e3*, y(0) 0, y'(0)
-
(b)
(c)
(*>
(e)
(f )
(g)
(h)
=
xy' + 3y
y<4>
=
5.
x3, y(0)
3y<2> + 2y 0, y(0)
-
=
=
1
.
=
=
=
1
,
y'(0)
=
0, y"(0)
0,
y '"(0)
=
0
r-(] l)y,y(o) (\).
r-(_93 83)y>y(o) Q.
=
=
e*yf + Xy'-y
gx; ^Q)
Q
0> y(())
x2y" + 3xy' + y 0, y(0)
1 y'(0)
1
x2y" + 4xy' + 2y x1 y(0)
1, y'(0)
=
=
=
=
=
=
,
=
=
.
=
,
58. Show that if all the entries of the matrix M
series
n
=
0.
are
less than 1, then the
2M"
=
0
Show that the limit is (I
converges.
M)_1.
59. Use the idea of the preceding problem to
-
two
decimals,
(a)
A
approximate A"1
where
/
=
1
0
(0.07
0.91
\0.14
()
/
A=i
0.08\
0.11
-0.03
1.13/
0.98
0.01
-0.12
-0.13
1.18
0
-0.03 N
-0.1
0.02
-0.02
1.01
0
0.11
-0.11
0.13
1
to within