Math 4470/5470 Partial Differential Equations Prof. Carlson
Homework 9
Text section 5.1
2. Let φ(x) = x2 for 0 ≤ x ≤ 1.
a) Calculate the Fourier sine series.
Solution Since l = 1 the Fourier sine series is
∞
X
Am sin(mπx),
m=1
where the Fourier coefficients are
Z 1
x2 sin(mπx) dx.
Am = 2
0
Integration by parts gives
1
(cos mπx)
dx
mπ
mπ
0
0
Z 1
(sin mπx)
(− cos mπ)
(sin mπx) 1
dx
=2
+ 4x
−4
2
mπ
(mπ)
(mπ)2
0
0
(−1)m+1
(−1)m+1
(cos mπx) 1
(−1)m − 1
=2
+4
+4
.
=2
mπ
(mπ)3 0
mπ
(mπ)3
2 (− cos mπx) Z
1
+4
Am = 2x
x
b) Calculate the Fourier cosine series.
Solution The Fourier cosine series series is
∞
X
A0
+
Am cos(mπx),
2
m=1
where the Fourier coefficients are
Z 1
x2 cos(mπx) dx.
Am = 2
0
For m 6= 0 integration by parts gives
Am
Z 1
1
(sin mπx)
(sin
mπx)
x
= 2x2
dx
−4
mπ
mπ
0
0
1
Z 1
(cos mπx)
(−1)m
(cos mπx) 1
dx = 4
.
= 4x
−4
(mπ)2 0
(mπ)2
(mπ)2
0
We also have A0 = 2/3.
5. Given the Fourier sine series of φ(x) = x on (0, l), assume that the
series can be integrated term by term, a fact that will be shown later.
a) Find the Fourier cosine series for x2 /2. Find the constant of integration.
Solution From p. 109 we have
x=
∞
X
(−1)m+1
m=1
2l
sin(mπx/l).
mπ
Integration of both sides gives
Z
x
∞
X
2l2
cos(mπx/l).
t dt = x /2 = C +
(−1)
2 π2
m
m=1
2
m
The constant of integration is the missing coefficient
A0
1
C=
=
2
l
l
Z
0
x2
dx = l2 /6.
2
b) By setting x = 0, find the sum
∞
X
(−1)n+1
.
2
n
n=1
Solution Setting x = 0 gives
∞
X
2l2
0 = l /6 +
(−1)
,
2 π2
m
m=1
2
or
2
π /12 =
∞
X
m
(−1)m+1
m=1
2
1
.
m2
8. A rod has length l = 1 and constant k = 1. Its temperature satisfies
the heat equation with boundary conditions u(t, 0) = 0, u(t, 1) = 1. The
initial temperature is
φ(x) =
n
5x/2,
3 − 2x,
0 < x < 2/3,
2/3 < x < 1.
o
Find the solution, including the coefficients.
Solution The equilibrium solution is U = x. We consider v(t, x) =
u(t, x) − x. The function v satisfies the heat equation and boundary conditions
vt = vxx , v(0) = 0 = v(1).
In addition the initial condition is
v(0, x) =
n
3x/2,
3 − 3x,
o
0 < x < 2/3,
.
2/3 < x < 1.
The solution of the original problem may then be written as
u(t, x) = v(t, x) + x.
The function v(t, x) may be represented as
v(t, x) =
∞
X
2
An e−n
π2 t
sin(nπx),
n=1
where
An = 2
Z
1
v(0, x) sin(nπx) dx.
0
Given the piecewise definition of v(0, x) we compute
αn = 2
Z
0
2/3
3x
sin(nπx) dx = 3
2
Z
2/3
x sin(nπx) dx
0
Z 2/3
cos(nπx)
− cos(nπx) 2/3
= 3x
dx
+3
nπ
nπ
0
0
3
cos(2nπ/3)
sin(nπx) 2/3
sin(2nπ/3)
cos(2nπ/3)
=
−2
+3
+
3
= −2
nπ
n2 π 2 0
nπ
n2 π 2
The second piece is
βn = 2
Z
1
(3 − 3x) sin(nπx) dx = 6
Z
1
Z
1
(1 − x) sin(nπx) dx
2/3
2/3
cos(nπx) 1
= −6(1 − x)
−6
nπ
2/3
=2
2/3
cos(nπx)
dx
nπ
cos(2nπ/3)
sin(2nπ/3)
dx
+6
nπ
n2 π 2
Finally,
An = αn + βn
= −2
cos(2nπ/3)
sin(2nπ/3)
sin(2nπ/3)
cos(2nπ/3)
+
2
+3
+
6
nπ
n2 π 2
nπ
n2 π 2
=9
sin(2nπ/3)
.
n2 π 2
9. Solve utt = c2 uxx for 0 < x < π with the boundary conditions
ux (t, 0) = ux (t, π) = 0, and the initial conditions u(0, x) = 0, ut (0, x) =
cos2 (x).
Solution The general solution has the form (see p. 89 (7))
∞
A0
B0 t X
u(t, x) =
+
+
[An cos(nct) + Bn sin(nct)] cos(nx).
2
2
n=1
Since u(0, x) = 0, we have coefficients An = 0 for all n. Use the trig identity
cos2 (x) =
1 1
+ cos(2x).
2 2
Now equate both forms of the initial condition
∞
1 1
B0 X
ut (0, x) = + cos(2x) =
+
ncBn cos(nx).
2 2
2
n=1
4
We find that
B0 = 1,
B2 =
1
.
4c
Otherwise Bn = 0. Thus
u(t, x) =
1
t
+
sin(2ct) cos(2x).
2 4c
Text section 5.2
3. Prove that even functions f (x) and odd functions g(x) satisfy
Z
l
g(x) dx = 0,
−l
Z
l
f (x) dx = 2
l
Z
f (x) dx = 0.
0
−l
Solution By definition an even function satisfies f (x) = f (−x) while
an odd function satisfies g(−x) = −g(x).
For an odd function g(x),
Z
l
g(x) dx =
=
0
g(x) dx +
l
g(−t) dt +
0
Z
l
Z
g(x) dx
0
−l
−l
Z
Z
l
g(x) dx ==
Z
l
(−g(x) + g(x)) dx = 0.
0
0
Here we made a substitution t = −x in the integral.
For an even function f (x),
Z
l
f (x) dx =
=
l
f (−t) dt +
0
0
f (x) dx +
Z
l
f (x) dx =
0
Z
Z
l
f (x) dx
0
−l
−l
Z
Z
l
(f (x) + f (x)) dx = 2
0
Z
l
f (x) dx.
0
4. a) Use property (5) to prove that if φ(x) is an odd function, then
its full Fourier series on (−l, l) has only sine terms.
Solution The Fourier cosine coefficients are
Z
1 l
An =
f (x) cos(nπx/l) dx.
l −l
5
Since cos(nπx/l) is even and the product of an even function and an odd
function is odd, the integrand is odd. Thus An = 0 for all n.
b) Also, if φ(x) is even, then its full Fourier series on (−l, l) has only
cosine terms.
Solution The Fourier sine coefficients are
1
Bn =
l
Z
l
f (x) sin(nπx/l) dx.
−l
Since sin(nπx/l) is odd and the product of an even function and an odd
function is odd, the integrand is odd. Thus Bn = 0 for all n.
7. Show how the full Fourier series on (−l, l) can be derived from the
full series on (−π, π) by changing variables x′ = πx/l.
Solution Suppose we have a function f (x) defined on (−l, l) whose
Fourier coefficients are desired. Assume we know the Fourier coefficients A′n
and Bn′ of g(x′ ) = f (lx′ /π) for −π < x′ < π. Then making the indicated
substitution we have
1
An =
l
1
=
l
Z
Z
l
f (x) cos(nπx/l) dx
−l
π
1
f (lx /π) cos(nx )l/π dx =
π
−π
′
′
′
6
Z
π
g(x′ ) cos(nx′ ) dx′ = A′n .
−π