Math 285 - Spring 2012 - Review Material - Exam 3
Section 9.2 - Fourier Series and Convergence
• State the definition of a Piecewise Continuous function.
Answer: f is Piecewise Continuous if the following to conditions are satisfied:
1) It is continuous except possibly at some isolated points.
2) The left and right limits f (x+ ) and f (x− ) exist (finite) at the points of discontinuity.
For example, Square-Wave functions are piecewise continuous.
But f (x) = sin 1x for 0 < x < 1, extended to R periodically, is not piecewise continuous,
since the right limit at zero DNE.
Also f (x) = tan x for 0 < x < π2 and for π2 < x < π, extended to R periodically, is not
piecewise continuous, since the right and left limits at π2 DNE (infinite).
• Define Fourier Series for functions for a Piecewise Continuous periodic function with
period 2L.
Answer:
∞
π
π
a0
f (x) ∼ + ∑ an cos n x + bn sin n x
2 n=1
L
L
where for n ≥ 0
1
an =
L
Z L
1
bn =
L
Z L
−L
π
f (x) cos n x dx
L
and for n ≥ 1
−L
π
f (x) sin n x dx
L
• State the definition of a Piecewise Smooth function.
Answer:
f (x) is Piecewise Smooth if both f (x) and f 0 (x) are piecewise continuous.
For example, Square-Wave functions are piecewise smooth.
• State the Convergence Theorem for Fourier Series.
Answer:
If f (x) is periodic and piecewise smooth, then its Fourier Series converges to
1) f (x) at each point x where f is continuous.
2) 12 ( f (x+ ) + f (x− )) at each point where f is NOT continuous.
• Compute the Fourier Series of Square-Wave Function with period 2L:
(
+1 0 < x < L
f (x) =
−1 L < x < 2L
For which values of f (x) at the points of discontinuity the Fourier series is convergent
for all x?
1
2
Answer:
an = 0 for all n ≥ 0, and bn =
2
nπ (1 − cos nπ)
f (x) =
for all n ≥ 1. Thus
π
4
sin n x
L
n=odd nπ
∑
Note that f (x) is discontinuous at x = integer multiples of L at which the average of left
and right limit of f (x) is zero. Thus by Convergence Theorem f (x) must be 0 at those
points.
• Letting x = L2 in the Fourier Series representation of the Square-Wave Function, obtain
the following relation:
1 1 1
π
(−1)k
∑ 2k + 1 = 1 − 3 + 5 − 7 + · · · = 4
k=0
∞
Answer:
Note that if we let x = L2 , then the Fourier series is convergent to f (x) = 1, thus we have
4 ∞ sin(2k + 1) π2
4 ∞ (−1)k
L
= ∑
= ∑
1= f
2
π k=0 2k + 1
π k=0 2k + 1
Hence
∞
(−1)k
1
1
1
π
∑ 2k + 1 = 1 − 3 + 5 − 7 + · · · = 4
k=0
3
Section 9.3 - Fourier Sine and Cosine Series
• Recall f (x) is odd if f (−x) = − f (x), i.e. its graph is symmetric w.r.t y-axis.
Example: x2n+1 , sin nx for all integers n 6= 0, the square-wave functions
(
−1 − L < x < 0
f (x) =
1 0<x<L
• Recall f (x) is even if f (−x) = f (x), i.e. its graph is symmetric w.r.t origin.
Example: x2n , cos nx for all integers n.
• Remarks:
RL
f (x)dx = 0 for any L
1) If f (x) is odd, then −L
R
RL
2) If f (x) is even, then −L f (x)dx = 2 0L f (x)dx for any L
3) If f and g are odd, then f g is even.
4) If f and g are even, then f g is even.
3) If f is odd and g is even, then f g is odd.
• FourierR series of odd functions
with period 2L:
1 L
1 RL
a0 = L −L f (x)dx = 0, an = L −L f (x) cos n Lπ xdx = 0 since f (x) cos n Lπ x is odd.
RL
R
bn = L1 −L
f (x) sin n Lπ xdx = L2 0L f (x) sin n Lπ xdx since f (x) sin n Lπ x is even.
In this case, if f is piecewise smooth, f (x) = ∑ bn sin n Lπ x only involves sine.
Example: f (x) =
x
2
n+1
(−1)
on (−π, π), then f (x) = ∑∞
n=1
n
sin nx
• FourierR series of even functions
with period 2L:
2 RL
1 L
π
an = L −L f (x) cos n L xdx = L 0 f (x) cos n Lπ xdx since f (x) cos n Lπ x is even.
RL
bn = L1 −L
f (x) sin n Lπ xdx = 0 since f (x) sin n Lπ x is odd.
In this case, if f is piecewise smooth, f (x) = a20 + ∑ an cos n Lπ x only involves cosine.
Example: Try f (x) = x2 on (−π, π).
• Even and odd extensions of a function:
Suppose f is a piecewise continuous function defined on interval (0, L).
Even extension of f to the interval (−L, 0) is
(
f (x) 0 < x < L
fE (x) =
f (−x) − L < x < 0
Example: f (x) = x2 + x + 1 on (0, L)
Then
(
x2 + x + 1 0 < x < L
fE (x) = 2
x −x+1 −L < x < 0
Odd extension of f to the interval (−L, 0) is
(
f (x) 0 < x < L
fO (x) =
− f (−x) − L < x < 0
4
Example: f (x) = x2 + x + 1 on (0, L)
Then
(
x2 + x + 1 0 < x < L
fO (x) =
−x2 + x − 1 − L < x < 0
Remark:
1) fE (x) is an even function with Fourier Series of the form
Hence f (x) = a20 + ∑ an cos n Lπ x for x in (0, L).
This is called the Fourier cosine series of f
a0
2
+ ∑ an cos n Lπ x
2) fO (x) is an odd function with Fourier Series of the form ∑ bn sin n Lπ x
Hence f (x) = ∑ bn sin n Lπ x for x in (0, L).
This is called the Fourier sine series of f
• Remark: Note that for x in (0, L), f (x) = fE (x) = fO (x).
In many cases we are not concerned about f (x) on (−L, 0), so the choice between (1)
and (2) depends on our need for representing f by sine or cosine.
• Example: f (t) = 1 on (0, π). Compute the Fourier sine and cosine series and graph the
two extensions.
1) The Even extension is fE (t) = 1 on (−π, π), period 2π.
Then a0 = 2, an = 0, bn = 0, so the cosine series is just
f (t) =
a0
=1
2
2) The Odd extension is fO (t) = 1 on (0, π) and −1 on (−π, 0), period 2π.
2
Then a0 = an = 0, bn = nπ
(1 − (−1)n ), so the cosine series is
fO (t) =
4
sin nx
n=odd nπ
∑
• Example: f (t) = 1 − t on (0, 1). Compute the Fourier sine and cosine series and graph
to the to extensions.
1) The Even extension is fE (t) = 1 − t on (0, 1) and 1 + t on (−1, 0), period 2L, L = 1.
nπ
, bn = 0, so the cosine series is
Then a0 = 1, an = 2 1−cos
n2 π2
f (t) =
1
4
+ ∑ 2 2 cos nπx
2 n=odd n π
2) The Odd extension is fO (t) = 1 −t on (0, 1) and −1 −t on (−1, 0), period 2L, L = 1.
2
Then an = 0 for all n ≥ 0 and bn = nπ
, so the cosine series is
∞
f (t) =
2
∑ nπ sinnπx
n=1
• Termwise differentiation of a Fourier series
Theorem: Suppose f (x) is Continuous for all x, Periodic with period 2L, and f 0 is
5
Piecewise Smooth for all t. If
∞
π
π
a0
f (x) = + ∑ an cos n x + bn sin n x
2 n=1
L
L
then
f 0 (x) =
∞
nπ
π
nπ
π
∑ (−an L ) sin n L x + (bn L ) cos n L x
n=1
Remarks:
1) RHS it the Fourier series of f 0 (x).
2) It is obtained by termwise differentiation of the RHS for f (x).
3) Note that the constant term in the FS of f 0 (x) is zero as
Z L
f 0 (x)dx = f (L) − f (−L) = 0
−L
4) The Theorem fails if f is not continuous!
For example: consider f (x) = x on (−L, L)
Then
2L
π
f (x) = x = ∑ (−1)n+1 sin n x
nπ
L
if we differentiate
2
π
1 6= ∑ (−1)n+1 cos n x
π
L
For example equality fails at t = 0 or L!
• Example: Verify the above Theorem for f (x) = x on (0, L) and f (x) = −x on (−L, 0).
Answer:
n
bn = 0 as f is even, a0 = L, an = n2L
2 π2 ((−1) − 1). Thus
1
4L
π
f (x) = L + ∑ − 2 2 cos n x
2
L
n=odd n π
Then term wise derivative gives
f 0 (x) =
π
4
sin n x
L
n=odd nπ
∑
On the other hand, directly computing the derivative of f (x) we have f 0 (x) = 1 on (0, L)
2
and f (x) = −1 on (−L, 0). Thus an = 0 for all n and bn = nπ
(1 − cos nπ) which gives
the same Fourier Series as in above.
• Applications to BVP’s
Consider the BVP of the form
ax00 + bx0 + cx = f (t), x(0) = x(L) = 0 or x0 (0) = x0 (L) = 0
• Example:
x00 + 2x = 1, x(0) = x(π) = 0
Here f (t) = 1, restrict to the interval (0, π) as in Boundary Values.
The idea is to find a formal Fourier Series solution of the equation.
Since we want x(0) = x(π) = 0, we prefer to consider the Fourier sine series of x(t) on
the interval (0, π),
x(t) = ∑ bn sin nt
6
substitute in the equation and compare the coefficients with that of the sine Fourier
Series of f (t) = 1 which is
f (t) =
4
sin nt
n=odd nπ
∑
Note that by term wise differentiation,
x00 (t) + 2x(t) = (2 − n2 )bn sin nt
Thus bn = 0 for n even and bn =
is
x(t) =
4
πn(2−n2 )
for n odd. Hence a formal power series of x(t)
4
sin nt
2)
πn(2
−
n
n=odd
∑
• Example:
x00 + 2x = t, x0 (0) = x0 (π) = 0
Here f (t) = t, restrict to the interval (0, π) as in Boundary Values.
Since we want x0 (0) = x0 (π) = 0, we prefer to consider the Fourier Cosine Series of
x(t) on (0, π),
a0
x(t) = + ∑ an cos nt
2
substitute in the equation and compare the coefficients with that of the Cosine FS of
f (t) = t which is
π
4
f (t) = − ∑
cos nt
2 n=odd πn2
Note that by termwise differentiation,
x00 (t) + 2x(t) = a0 + (2 − n2 )an cos nt
4
Thus a0 = π2 , an = 0 for n even and an = − πn2 (2−n
2 ) for n odd. Hence a formal power
series of x(t) is
π
4
x(t) = − ∑
cos nt
2
4 n=odd πn (2 − n2 )
• Termwise Integration of the Fourier Series.
Theorem: Suppose f (t) is Piecewise Continuous (not necessarily piecewise smooth)
periodic with period 2L with FS representation
f (t) ∼
∞
a0
π
π
+ ∑ an cos n t + bn sin n t
2 n=1
L
L
Then we can integrate term by term as
Rt
0
f (x)dx =
=
R t a0
Rt
Rt
π
π
∞
0 2 + ∑n=1 an 0 cos n L xdx + bn 0 sin n L xdx
a0
L
π
L
π
∞
2 t + ∑n=1 an nπ sin n L x − bn nπ (cos n L x − 1)
Note that the RHS is not the Fourier series of LHS unless a0 = 0.
• Example: Consider f (x) = 1 on (0, π) and f (x) = −1 on (−π, 0), then
f (t) =
4
sin nt
n=odd πn
∑
7
Then F(t) =
Rt
0
f (x)dx = t for t ∈ (0, 1) and −t for t ∈ (−π, 0), whose Fourier Series is
1
4
1
cos nt
F(t) = π −
∑
2
π n=odd n2
Term by term integration gives the same thing!
4
1
4
(1 − cos nt) =
∑
2
π n=odd n
π
1
π2
∑ n2 = 8
n=odd
!
−
1
4
cos nt
∑
π n=odd n2
8
Section 9.4 - Applications of Fourier Series
• Finding general solutions of 2nd order linear DE’s with constant coefficients:
Example: x00 + 5x = F(t), where F(t) = 3 on (0, π) and −3 on (−π, 0), is odd with
period 2π.
We obtain a particular solution in the following way:
12
Since L = π and the FS of F(t) is ∑n odd nπ
sin nt,
we may assume x(t) is odd and we consider its Fourier sine series F(t) = ∑∞
n=1 bn sin nt.
Substitute in the equation:
x00 + 5x = (5 − n2 )bn sin nt = F(t) =
12
sin nt
nπ
n odd
∑
Hence, comparing the coefficients of sin nt on both sides we get
12
bn = n(5−n
2 )π if n is odd and zero otherwise.
12
Thus a particular solution is x(t) = ∑n odd n(5−n
2 )π sin nt
Definition: We call this a steady periodic solution, denoted by xsp (t).
Thus, if x1 (t), x2 (t) are the solutions of the associated homogeneous equation, then the
general solution is
x(t) = c1 x1 (t)√
+ c2 x2 (t) + xsp
√(t)
12
= c1 cos( 5t) + c2 sin( 5t) + ∑n odd n(5−n
2 )π sin nt
• Remark: Consider the equation x00 + 9x = F(t). Then when trying to find a particular
solution we get
x00 + 9x = (9 − n2 )bn sin nt = F(t) =
12
sin nt
n odd nπ
∑
We can not find b3 as 9 − n2 = 0 for n = 3.
In this case we need to use the method of undetermined coefficients to find a function
y(t) such that
12
y00 + 9y =
sin 3t
3π
2
Take y = At sin 3t + Bt cos 3t, and substitute to find A = 0 and B = − 3π
.
Therefore, the general solution is
x(t) = c1 cos(3t) + c2 sin(3t) +
12
2
sin nt − t cos 3t
2
n(9 − n )π
3π
n odd,n6=3
∑
Definition: We say in this case a pure resonance occurs.
Remark: To determine the occurrence of pure resonance, just check if for some n,
sin n Lπ t is a solution of the associated homogeneous equation.
9
• Application: Forced Mass-Spring Systems
Let m be the mass, c be the damping constant, and k the constant of spring. Then
mx00 + cx0 + kx = F(t)
Consider the case that the external force F(t) is odd or even periodic function.
Remark: If F(t) is periodic for t ≥ 0, it can be arranged to be odd or even by passing
to odd or even extension for values of time t.
Case 1) Undamped Forced Mass-Spring Systems: c = 0
mx00 + kx = F(t)
Let ω0 =
q
k
m
be the natural frequency of the system, then we can write
1
F(t)
m
Assume F(t) is periodic odd function with period 2L.
Then
∞
π
F(t) = ∑ Fn sin n t
L
n=1
x00 + ω20 x =
Consider the odd extension of x(t), so that
∞
x(t) =
π
∑ bn sin n L t
n=1
substituting in the equation we get
∞
nπ π
1
1
π
x00 + ω20 x = ω20 − ( )2 bn sin n t = F(t) = ∑ Fn sin n t
L
L
m
L
n=1 m
Thus
1
nπ ω20 − ( )2 bn = Fn
L
m
q
2
If for all n ≥ 1, ω20 − ( nπ
L ) 6= 0, or equivalently
can solve for bn as
1
m Fn
bn = 2
2
ω0 − ( nπ
L)
Hence we have a steady periodic solution
1
m Fn
nπ 2
2
n=1 ω0 − ( L )
∞
xsp (t) =
∑
k π
m·L
is not a positive integer, then we
π
sin n t
L
Example: If m = 1, k = 5, L = π,
x00 + 5x = F(t)
2
2
then ω20 − ( nπ
L ) = 5 − n 6= 0 for all positive integers n ≥ 1.
q
On the other hand, if n0 = mk · Lπ is a positive integer, then a particular solution is
1
m Fn
x(t) =
nπ 2
2
n=1,n6=n0 ω0 − ( L )
∞
∑
π
Fn0
sin n t −
t cos ω0t
L
2mω0
10
Example: m = 1, k = 9, L = π,
x00 + 9x = F(t)
2
2
then ω20 − ( nπ
L ) = 9 − n = 0 for n = 3. There will be a pure resonance.
Example: m = 1, k = 9, L = 1,
x00 + 9x = F(t)
2
2
then ω20 − ( nπ
L ) = 9 − (nπ) 6= 0 for all positive integers n. We have a steady periodic
solution.
Case 2) Damped Forced Mass-Spring Systems: c 6= 0
mx00 + cx0 + kx = F(t)
Assume F(t) is periodic odd function with period 2L.
Then
∞
π
F(t) = ∑ Fn sin n t
L
n=1
For each n ≥ 1, we seek a function xn (t) such that
π
mxn00 + cxn0 + kxn = Fn sin n t = Fn sin ωnt
L
π
where ωn = n L . Note that there will never be a duplicate solution as c 6= 0.
Hence using the method of undetermined coefficients, we can show
Fn
xn (t) = p
sin(ωnt − αn )
(k − mω2n )2 + (cωn )2
where
cωn
−1
0≤α≤π
αn = tan
k − mω2n
Therefore, we have a steady periodic solution
∞
xsp (t) =
∞
∑ xn(t) =
n=1
Fn
∑ p(k − mω2)2 + (cω )2 sin(ωnt − αn)
n=1
n
n
Example: m = 3, c = 1, k = 30, F(t) = t − t 2 for 0 ≤ t ≤ 1 is odd and periodic with
L = 1. Compute the first few terms of the steady periodic solution.
3x00 + x0 + 30x = F(t) =
∞
π
∑ Fn sin n L t = ∑
n=1
n=odd
8
n3 π3
sin nπt
So, we have
xsp (t) =
Fn
(k−mω2n )2 +(cωn )2
√
∑∞
n=1
sin(ωnt − αn )
8
n3 π3
= ∑n= odd √
sin(nπt − αn )
(30−3n2 π2 )2 +n2 π2
nπ
−1
αn = tan
0≤α≤π
30 − n2 π2
The fist two terms are
0..0815 sin(πt − 1.44692) + 0.00004 sin(3π3 − 3.10176) + · · ·
11
Section 9.5 - Heat Conduction and Separation of Variables
• Until now, we studied ODE’s - which involved single variable functions.
In this section will consider some special PDE’s - Differential Equations of several variable functions involving their Partial Derivatives - and we will apply Fourier Series
method to solve them.
• Heat Equations:
Let u(x,t) denote the temperature at pint x and time t in an ideal heated rod that extends
along x-axis. then u satisfies the following equation:
ut = kuxx
where k is a constant - thermal diffiusivity of the material - that depends on the material
of the rod.
Boundary Conditions:
Suppose the rod has a finite length L, then 0 ≤ x ≤ L.
1) Assume the temperature of the rod at time t = 0 at every point x is given. Then
we are given a function f (x) such that u(x, 0) = f (x) for all 0 ≤ x ≤ L.
2) Assume the temperature at the two ends of the rod is fixed (zero) all the time - say by
putting two ice cubes! Then u(0,t) = u(L,t) = 0 for all t ≥ 0.
Thus we obtain a BVP,


ut = kuxx
u(x, 0) = f (x) for all 0 ≤ x ≤ L

u(0,t) = u(L,t) = 0 for all t ≥ 0
Remark: Other possible boundary conditions are, insulating the endpoints of the rod,
so that there is no heat flow. This means
ux (0,t) = ux (L,t) = 0 for all t ≥ 0
Remark: Geometric Interpretation of the BVP.
We would like to find a function u(x,t) such that on the boundary of the infinite strip
t ≥ 0 and 0 ≤ x ≤ L satisfies the conditions u = 0 and u = f (x).
Remark: If f (x) is ”piecewise smooth”, then the solution of the BVP is unique.
• Important observations:
1) Superposition of solutions: If u1 , u2 , . . . satisfy the equation ut = kuxx , then so does
any linear combination of the ui ’s. In other words, the equation ut = kuxx is linear!
2) Same is true about the boundary condition u(0,t) = u(L,t) = 0 for all t ≥ 0. We
say this is a linear or homogeneous condition.
3) The condition u(x, 0) = f (x) for all 0 ≤ x ≤ L is not homogenous, or not linear!
12
• General Strategy: Find solutions that satisfy the linear conditions and then take a
suitable linear condition that satisfies the non-linear conditions.
2
• Example: Verify that un (x,t) = e−n t sin nx is a solution of ut = uxx (here k = 1) for any
positive integer n. For example, u1 (x,t) = e−t sin x and u2 (x,t) = e−4t sin 2x.
• Example: Use the above example to construct a solution of the following BVP.


ut = uxx
u(0,t) = u(π,t) = 0 for all t ≥ 0

u(x, 0) = 2 sin x + 3 sin 2x for all 0 ≤ x ≤ π
Answer:
2
Here L = π. Note that un (x,t) = e−n t sin nx also satisfy the linear condition u(0,t) =
u(π,t) = 0. Thus it is enough to take u(x,t) to be a linear combination of un ’s that
satisfies the non homogenous condition u(x, 0) = 2 sin x + 3 sin 2x. Since un (x, 0) =
sin nx, we take
u(x,t) = 2u1 + 3u2 = 2e−t sin x + 3e−4t sin 2x
so that
u(x, 0) = 2e0 sin x + 3e0 sin 2x = 2 sin x + 3 sin 2x
Remark: The above method in the example for ut = uxx works whenever f (x) is a finite
linear combination of sin x, sin 2x, . . .
• Example: Use the above example to construct a solution of the following BVP.


ut = uxx
u(0,t) = u(π,t) = 0 for all t ≥ 0

u(x, 0) = sin 4x cos x for all 0 ≤ x ≤ L
Solution: Again we have L = π. Note that
f (x) = sin 4x cos x =
1
1
1
1
sin(4x + x) + sin(4x − x) = sin 5x + sin 3x
2
2
2
2
Thus we take
1
1
1
1
u(x,t) = u3 + u5 = e−9t sin 3x + e−25t sin 5x
2
2
2
2
• Remark: When f (x) is a not a finite linear combination of the sine functions, then represent it as an infinite sum using Fourier sine series.
• Example: Construct a solution of the following BVP.


ut = uxx
u(0,t) = u(π,t) = 0 for all t ≥ 0

u(x, 0) = 1 for all 0 ≤ x ≤ π
Answer: Note that f (x) = 1 and L = π, so represent f (x) as a Fourier sine series with
period 2L = 2π
4
f (x) = ∑
sin nx
n= odd nπ
13
Thus we take
u(x,t) =
4 n2t
4
un (x,t) = ∑
e sin nx
n= odd nπ
n= odd nπ
∑
This is a formal series solution of the BVP, one needs to check the convergence, ...
We only take finitely many terms for many applications.
• In general, to solve the following BVP (k is anything, not necessarily 1, and L is anything, not just π)


ut = kuxx
u(0,t) = u(π,t) = 0 for all t ≥ 0

u(x, 0) = f (x) for all 0 ≤ x ≤ L
we observe that
nπ 2
π
un (x,t) = e−k( L ) t sin n x
L
satisfies the equation ut = kuxx and the homogenous boundary condition u(0,t) = u(L,t) =
0 for all t ≥ 0. Thus, represent f (x) as a Fourier sine series with period 2L,
∞
f (x) =
π
∑ bn sin n L x
n=1
Then
u(x,t) :=
∞
∞
n=1
n=1
nπ 2
π
∑ bnun(x,t) = ∑ bne−k( L ) t sin n L x
satisfies the non homogenous condition u(x, 0) = f (x) for all 0 ≤ x ≤ L.
• Case of a rod with insulated endpoints:
Consider the BVP corresponding to a heated rod with insulated endpoint,


ut = kuxx
ux (0,t) = ux (π,t) = 0 for all t ≥ 0

u(x, 0) = f (x) for all 0 ≤ x ≤ L
we observe that for n ≥ 0,
nπ 2
π
un (x,t) = e−k( L ) t cos n x
L
satisfies the equation ut = kuxx and the homogenous boundary condition ux (0,t) =
ux (L,t) = 0 for all t ≥ 0.
Remark: for n = 0, we get u0 = 1 which satisfies the linear conditions!
Thus, represent f (x) as a Fourier cosine series with period 2L,
f (x) =
∞
a0
π
+ ∑ an cos n x
2 n=1
L
Then
∞
∞
nπ 2
a0
a0
π
u(x,t) := + ∑ an un (x,t) = + ∑ an e−k( L ) t cos n x
2 n=1
2 n=1
L
satisfies the non homogenous condition u(x, 0) = f (x) for all 0 ≤ x ≤ L.
14
• Remarks:
1) In the BVP for heated rod with zero temperature in the endpoints, we have
∞
nπ 2
π
∑ bne−k( L ) t sin n L x = 0
t→∞
lim u(x,t) = lim
t→∞
n=1
in other words, heat goes away with no insulation, thus temperature is zero at the end!
2) In the BVP for heated rod with insulated endpoints, we have
∞
nπ 2
π
a0
a0
+ ∑ bn e−k( L ) t cos n x =
t→∞ 2
L
2
n=1
lim u(x,t) = lim
t→∞
which means, with insulation heat distributes evenly throughout the rod, which is the
average of the initial temperature as
Z
a0 1 L
=
f (x)dx
2
L 0
15
Section 9.6 - Vibrating Strings and the One-Dimensional Wave Equation
• Consider a uniform flexible string of length L with fixed endpoints,
stretched along x-axis in the xy-plane from x = 0 to x = L.
Let y(x,t) denote the displacement of the points x on the string at time t
(we assume the points move parallel to y-axis).
Then y satisfies the One-dimensional wave equation:
ytt = a2 yxx
where a is a constant that depend on the material of the string and the tension!
Boundary Conditions:
1) Since the endpoints are fixed, y(0,t) = y(L,t) = 0.
2) The initial position of the string y(x, 0) at each x is given as a function y(x, 0) = f (x).
3) The solution also depends on the initial velocity yt (x, 0) of the string at each x, given
as a function yt (x, 0) = g(x).
Thus we obtain the following BVP

ytt = a2 yxx



y(0,t) = y(L,t) = 0 for all t ≥ 0

y(x, 0) = f (x) for all 0 ≤ x ≤ L



yt (x, 0) = g(x) for all 0 ≤ x ≤ L
• Important Observations:
1) ytt = a2 yxx is a linear equation. Thus the superposition of solutions applies.
2) Condition y(0,t) = y(L,t) = 0 is linear.
3) Conditions y(x, 0) = f (x) and yt (x, 0) = g(x) are not linear.
• General Strategy: Split the BVP into two problems,


2y
y
=
a
ytt = a2 yxx


tt
xx




y(0,t) = y(L,t) = 0 for all t ≥ 0
y(0,t) = y(L,t) = 0 for all t ≥ 0
(A)
(B)


y(x, 0) = f (x) for all 0 ≤ x ≤ L
y(x, 0) = 0 for all 0 ≤ x ≤ L






yt (x, 0) = 0 for all 0 ≤ x ≤ L
yt (x, 0) = g(x) for all 0 ≤ x ≤ L
If yA (x,t) and yB (x,t) are the respective solutions, then
y(x,t) = yA (x,t) + yB (x,t)
satisfies the original BVP as
y(x, 0) = yA (x, 0) + yB (x, 0) = f (x) + 0 = f (x)
and
yt (x, 0) = (yA )t (x, 0) + (yB )t (x, 0) = 0 + g(x) = g(x)
• Solving a BVP of type (A)

2

ytt = a yxx

y(0,t) = y(L,t) = 0 for all t ≥ 0
(A)

y(x, 0) = f (x) for all 0 ≤ x ≤ L



yt (x, 0) = 0 for all 0 ≤ x ≤ L
16
Verify directly that for all positive integers n, the function
π
π
yn (x,t) = cos n at · sin n x
L
L
satisfies the equation ytt = a2 yxx and the linear conditions y(0,t) = y(L,t) = 0 and
yt (x, 0) = 0. Thus, by superposition law, we need to find coefficients bn such that
∞
y(x,t) =
∞
π
π
∑ bnyn(x,t) = ∑ bn cos n L at · sin n L x
n=1
n=1
satisfies y(x, 0) = f (x). Note that
∞
f (x) = y(x, 0) =
∞
π
∑ bnyn(x, 0) = ∑ bn sin n L x
n=1
n=1
Thus bn ’s are the Fourier sine coefficients of f (x).
• Example: Triangle initial position((pulled from the midpoint) with zero initial velocity
x if 0 < x < π2
Assume a = 1, L = π and f (x) =
. Then the BVP is type (A)
π − x if π2 < x < π


ytt = yxx





y(0,t) = y(π,t) =(0 for all t ≥ 0
(A)
x if 0 < x < π2

y(x,
0)
=
f
(x)
=


π − x if π2 < x < π



y (x, 0) = 0 for all 0 ≤ x ≤ π
t
Fourier sine series of f (x) is
4 sin nπ
4(−1)
∑ πn2 2 sin nx = ∑ πn2
n=1
n=odd
∞
n−1
2
sin nx
Thus, since a = 1 and L = π,
y(x,t) = ∑∞
n Lπ at · sin n Lπ x
n=1 bn cos
nπ
4 sin 2
= ∑∞
n=1 πn2 · cos nt · sin nx
n−1
=
=
=
4(−1) 2
∑n=odd πn2 · cos nt · sin nx
n−1
4(−1) 2
∑n=odd πn2 · cos nt · sin nx
4
4
4
π cost sin x − 9π cos 3t sin 3x + 25π cos 5t sin 5x + · · ·
• Remark: Using the identity
2 sin A cos B = sin(A + B) + sin(A − B)
we can write the solution as
17
n−1
2
y(x,t) = ∑n=odd 4(−1)
πn2
· cos nt · sin nx
n−1
4(−1) 2
πn2
n−1
4(−1) 2
2
πn
n−1
4(−1) 2
πn2
=
1
2 ∑n=odd
=
1
2 ∑n=odd
=
=
1
· sin n(x + t) + 12 ∑n=odd 4(−1)
2 ∑n=odd
πn2
1
1
f
(x
+
t)
+
f
(x
−
t)
2 O
2 O
· 2 cos nt · sin nx
· (sin(nx + nt) + sin(nx − nt))
n−1
2
· sin n(x − t)
• Solving a BVP of type (B)

ytt = a2 yxx



y(0,t) = y(L,t) = 0 for all t ≥ 0
(A)

y(x, 0) = 0 for all 0 ≤ x ≤ L



yt (x, 0) = g(x) for all 0 ≤ x ≤ L
Verify directly that for all positive integers n, the function
π
π
yn (x,t) = sin n at · sin n x
L
L
satisfies the equation ytt = a2 yxx and the linear conditions y(0,t) = y(L,t) = 0 and
y(x, 0) = 0. Thus, by superposition law, we need to find coefficients cn such that
∞
y(x,t) =
∞
∑ cnyn(x,t) =
n=1
π
π
∑ cn sin n L at · sin n L x
n=1
satisfies yt (x, 0) = g(x). Note that by termwise differentiation with respect to variable t
we have
∞
π
π
π
yt (x,t) = ∑ cn (n a) cos n at · sin n x
L
L
L
n=1
Thus
∞
g(x) = yt (x, 0) =
π
π
∑ cn(n L a) sin n L x
n=1
π
Then cn (n L a)’s are the Fourier sine coefficients of g(x).
π
Hence cn = nLa
· n-th Fourier sine coefficients of g(x)
• Example: Assume a = 1, L = π and g(x) = 1. Then the following BVP is type (B)

ytt = yxx



y(0,t) = y(π,t) = 0 for all t ≥ 0
(B)

y(x, 0) = 0



yt (x, 0) = g(x) = 1 for all 0 ≤ x ≤ π
Fourier sine series of g(x) = 1 with L = π is
4
sin nx
n=odd πn
∑
18
Thus, since a = 1 and L = π
π
π
y(x,t) = ∑∞
n=1 cn sin n L at · sin n L x
π
4
= ∑n=odd nπ(1)
· sin nt · sin nx
· πn
4
= ∑n=odd πn2 · sin nt · sin nx
19
Section 9.7 - Steady-State Temperature and Laplace’s Equation
• Consider the temperature in a 2-dimensional uniform thin plate in xy-plane
bounded by a piecewise smooth curve C.
Let u(x, y,t) denote the temperature of the point (x, y) at time t.
Then the 2-dimensional eat equation states that
ut = k(uxx + uyy )
where k is a constant that depends on the material of the plate.
If we let ∇2 u = uxx + uyy , which is called the laplacian of u, then we can write
ut = k∇2 u
Remark: The 2-dimensional wave equation is
ztt = a2 (zxx + zyy ) = a2 ∇2 z
where z(x, y,t) is the position of the point (x, y) in a vibration elastic surface at time t.
• Case of the steady state temperature: i.e. we consider a 2-dimensional heat equation
in which the temperature does not change in time (The assumption is after a while
temperature becomes steady). Thus
ut = 0
Therefore, ∇2 u = uxx + uyy = 0.
This equation is called the 2-dimensional Laplace equation.
• Boundary problems and Laplace equation:
If we know the temperature on the boundary C of a plate, as a function f (x, y), can we
determine the temp. at every point inside the plate?
In other word, can we solve the BVP
(
uxx + uyy = 0
u(x, y) = f (x, y) on the boundary of the plate
This BVP is called a Dirichlet Problem.
• Remark: If the Boundary C is Piecewise Smooth and the function f (x, y) is Nice!, then
Dirichlet Problem has a unique solution.
We will consider the cases that C is rectangular or circular!
• Case of Rectangular Plates:
Suppose the plate is a rectangle positioned in xy-plane with vertices (0, 0), (0, b), (a, b), (a, 0).
Assume we are given the temp. at each side of the rectangle.
Then we have the following type BVP.
20

uxx + uyy = 0





u(x, 0) = f1 (x, y), 0 < x < a
u(x, b) = f2 (x, y), 0 < x < a


u(0, y) = g1 (x, y), 0 < y < b



u(a, y) = g1 (x, y), 0 < y < b
• General Strategy:
The main equation uxx + uyy = 0 is linear, but all of the boundary conditions are nonlinear, the idea is to split this BVP into 4 simpler BVP denotes by A, B, C, D, in which
only one of the boundary conditions in non-linear and apply Fourier series method there,
and and the end,
u(x, y) = uA (x, y) + uB (x, y) + uC (x, y) + uD (x, y)
• Solving a type BVP of type (A)
For 0 < x < a and 0 < y < b,

uxx + uyy = 0





u(x, 0) = f1 (x, y)
u(x, b) = 0



u(0, y) = 0



u(a, y) = 0
In this case, one can show
nπ
nπ
x · sinh (b − y)
a
a
satisfies the linear conditions of the BVP.
Recall that sinh x = 12 (ex − e−x ) and cosh x = 21 (ex + e−x ),
so (sinh x)0 = cosh x and (cosh x)0 = − sinh x
un (x, y) = sin
• Method of separation of variables: To actually find un ’s
Assume u(x, y) = X(x)Y (y).
Then uxx + uyy = 0 implies X 00Y + XY 00 = 0.
00
00
Thus XX = − YY .
Since RHS only depends on y and LHS only depends on x, these fractions must be constant, say −λ.
00
00
Then XX = −λ and YY = λ.
Thus we obtain X and Y are nonzero solutions of X 00 + λX = 0 and Y 00 − λY = 0, such
that Y (b) = 0, X(0) = X(a) = 0
This is an endpoint problem on X,
we seek those values of λ for which there are nonzero solutions X, such that X(0) =
2
X(a) = 0. The eigen values are λn = ( nπ
a )
and the corresponding eigenfunctions are scalar multiples of Xn (x) = sin nπ
a x.
nπ 2
00
Now substitute λn = ( a ) in Y − λY = 0 with Y (b) = 0, and solve for Y (y).
We obtain the solutions are a scalar multiple of Yn (y) = sinh nπ
a (b − y).
nπ
Hence un (x, y) = Xn (x)Yn (y) = sin nπ
x
·
sinh
(b
−
y)
a
a
21
• Back to solving a type BVP of type (A)
We want u(x, y) = ∑ cn un (x, y) such that u(x, 0) = f1 (x).
Hence
nπ
nπ
f1 (x) = u(x, 0) = ∑ cn un (x, 0) = ∑ cn · sin x · sinh (b)
a
a
nπ
Therefore, cn · sinh a (b) is the n-th Fourier sine coefficient bn of f1 (x) over interval
(0, a). Hence
nπb
cn = bn / sinh
a
• Example: Solve the BVP of type (A) if a = b = π and f1 (x) = 1. Compute u(π/2, π/2),
the temp at the center of the rectangle.
Solution: Note that
nπ
4
4
sin x = ∑
sin nx
f (x) = ∑
a
n=odd nπ
n=odd nπ
Thus
u(x, y) =
∑
n=odd
Then
4
nπb
nπ
nπ
/ sinh
· sin x · sinh (b − y)
nπ
a
a
a
4
· sin nx · sinh n(π − y)
u(x, y) = ∑
n=odd nπ sinh nπ
Also note that after computing the first few terms and using sinh 2t = 2 sinht cosht,
4
u(π/2, π/2) = ∑n=odd nπ sinh
sin nπ/2 · sinh nπ/2
nπ ·
= ∑n=odd nπ sinh2 nπ/2 · sin nπ/2
∼ .25
In fact, one can argue by symmetry that is is exactly .25.
• Case of a semi-infinite strip plate!
Assume the plate is an infinite plate in the first quadrant whose vertices are (0, 0) and
(0, b), and u = 0 along the horizontal sides, u(0, y) = g(y) and u(x, y) < ∞ as x → ∞.
Apply the separation of variable method to solve the corresponding Dirichlet problem.
Answer:
Let u(x, y) = X(x)Y (y).
Then uxx + uyy = 0 implies X 00Y + XY 00 = 0.
00
00
Thus XX = − YY .
Since RHS only depends on y and LHS only depends on x, these fractions must be constant, say λ.
00
00
Then XX = λ and YY = −λ.
Thus we obtain X and Y are nonzero solutions of X 00 − λX = 0 and Y 00 + λY = 0, such
that Y (0) = Y (b) = 0, and u(x, y) = X(x)Y (y) in bounded as x → ∞
This is an endpoint problem on Y .
We want those values of λ for which there are nonzero solutions Y , such that Y (0) =
Y (b) = 0.
2
The eigen values are λn = ( nπ
b )
22
and the corresponding eigenfunctions are scalar multiples of Yn (y) = sin nπ
b y.
nπ 2
00
Now substitute λn = ( b ) in X − λX = 0 and solve for X(x).
nπ
nπ
We obtain Xn (x) = An e b x + Bn e− b x
Since un = XnYn and u and Yn = sin nπ
b y are bounded, then Xn must be bounded too.
Hence An = 0. Suppress bn .
nπ
Thus un (x, y) = Xn (x)Yn (y) = e− b x · sin nπ
b y
Note that un (x, 0) = un (x, b) = 0 and un (x, y) < ∞ as x → ∞.
Now, to solve the BVP:
We want u(x, y) = ∑ cn un (x, y) such that u(0, y) = g(y).
Hence
nπ
y
b
Therefore, cn is the n-th Fourier sine coefficient bn of g(y) over interval (0, b).
g(y) = u(0, y) = ∑ cn un (0, y) = ∑ cn · (1) · sin
• Example: If b = 1 and g(y) = 1, then compute u(x, y).
Solution:
g(y) =
4
4
nπ
sin x = ∑
sin nx
b
n=odd nπ
n=odd nπ
∑
Thus
u(x, y) =
4 −nπx
·e
· sin nπy
nπ
n=odd
∑
• Case of a circular disk:
Using polar coordinates (r, θ) to represent the points in a disk, on can transform the
Laplace equation into
1
1
urr + ur + 2 uθθ = 0
r
r
Note that 0 < r < a = the radius of the disk, and 0 < θ < 2π.
It turns out that the solution is of the form
u(r, θ) =
∞
a0
+ ∑ (an cos nθ + bn sin nθ)rn
2 n=1
Boundary conditions: If we know the temp. on the boundary of the disk, want to
determine the temp. inside. To give the temp. on the boundary means to give a function
f (θ) such that
u(a, θ) = f (θ) for 0 < θ < 2π
Thus
n
f (θ) = u(a, θ) = a20 + ∑∞
n=1 (an cos nθ + bn sin nθ)a
n
n
= a20 + ∑∞
n=1 a an cos nθ + a bn sin nθ
Therefore an an and an bn are the n-th Fourier series coefficient of f (θ) over the interval
(0, 2π).
23
• Example: If f (θ) = 1 on (0, π) and −1 on (π, 2π), and radius r = 1, then compute
u(r, θ).
Answer:
4
sin nθ.
We compute the Fourier series of f (θ) = ∑n=odd nπ
Thus an = 0 for all n. Hence
∞
a0
4 n
u(r, θ) = + ∑ (an cos nθ + bn sin nθ)rn = ∑
r sin nθ
2 n=1
n=odd nπ
• Remark: Intuitively, in the above example u(r, 0) = 0, which is consistent with the
solution: If θ = 0 or π,
4 n
u(r, θ) = ∑
r sin nθ = 0
n=odd nπ
• Remark: In the above example, if we change the boundary condition to f (θ) = 1 on
(0, 2π), then intuitively, u(r, θ) = 1 everywhere by symmetry. This is consistent with the
actual solution to BVP: In this case, the Fourier coefficients of f are a0 = 2, an = bn = 0
for all n ≥ 1. Thus u(r, θ) = 1 for all 0 < r < 1 and 0 < θ < 2π.