Math 257/316 Assignment 5 Solutions
1. For the “triangle wave” function
f (x) =
x
0≤x≤1
2−x 1≤x≤2
defined on [0, 2]:
(a) compute its Fourier sine series
Doing some integration by parts,
Z 1
Z 2
Z
2 2
f (x) sin(kπx/2)dx =
x sin(kπx/2)dx +
(2 − x) sin(kπx/2)dx
bk =
2 0
0
1
Z 1
2
2
1
= −x
cos(kπx/2)|0 +
cos(kπx/2)dx
kπ
kπ 0
Z 2
2
2
− (2 − x)
cos(kπx/2)|21 −
cos(kπx/2)dx
kπ
kπ 1
4
2
4
2
cos(kπ/2) + 2 2 sin(kπx/2)|10 +
cos(kπ/2) − 2 2 sin(kπx/2)|21
=−
kπ
k π kπ
k π
8
8
0
k even
= 2 2 sin(kπ/2) = 2 2
(−1)(k−1)/2 k odd
k π
k π
so
F.S.S. of f (x)
8
= 2
π
∞
X
k=1,k
odd
((−1)(k−1)/2
sin
k2
kπx
2
πx 8
8
= 2 sin
− 2 sin
π
2
9π
(b) computes its Fourier cosine series
Begin with
2
a0 =
2
Z
2
1
Z
f (x)dx =
0
Z
0
1
2
(2 − x)dx =
xdx +
1
1 1
+ =1
2 2
3πx
+· · · .
2
(twice the average value of f on [0, 2]). For k ≥ 1, integrate by parts:
ak =
=
=
=
Z
Z 1
Z 2
2 2
f (x) cos(kπx/2)dx =
x cos(kπx/2)dx +
(2 − x) sin(kπx/2)dx
2 0
0
1
Z 1
Z 2
2
2
2
2
sin(kπx/2)dx +
sin(kπx/2)dx
x sin(kπx/2)|10 −
(2 − x) sin(kπx/2)|21 +
kπ
kπ 0
kπ
kπ 1
2
4
2
4
sin(kπ/2) + 2 2 cos(kπx/2)|10 −
sin(kπ/2) − 2 2 cos(kπx/2)|21
kπ
k π
kπ
k
π
16
4
− k2 π2 k = 4m + 2, m ∈ Z
[cos(kπ/2) − 1 − cos(kπ) + cos(kπ/2)] =
,
0
else
k2 π2
so
∞
1
1 4
4
1 4 X
cos((2m+1)πx) = − 2 cos(πx)− 2 cos(3πx)−· · ·
F.C.S. of f (x) = − 2
2
2 π
(2m + 1)
2 π
9π
m=0
(c) by evaluating f (1), use each of your results from (a) and (b) in turn, to find the
value of the sum of the squares of the reciprocals of the odd integers:
1+
1
1
1
1
+ 2 + 2 + 2 + ··· .
2
3
5
7
9
By the Fourier convergence theorem, the F.S.S of f converges to f (1) = 1 at
x = 1 (since f is continuous there), so
1=
8
π2
∞
X
k=1,k
odd
1
sin2
k2
kπ
2
=
8
π2
∞
X
k=1,k
odd
1
k2
∞
X
=⇒
k=1,k
odd
1
π2
=
.
2
k
8
Also by the Fourier convergence theorem, the F.C.S of f converges to f (1) = 1
at x = 1 (since f is continuous there), so
∞
∞
1 4 X cos((2m + 1)π)
1 4 X
1
1= − 2
=
+
2 π
(2m + 1)2
2 π2
(2m + 1)2
m=0
=⇒
m=0
2. Solve the heat conduction problem:
ut = 2uxx ,
0 < x < 2, t > 0,
u(0, t) = 0 = u(2, t)
x
0≤x≤1
u(x, 0) =
,
2−x 1≤x≤2
2
0 ≤ x ≤ 2.
∞
X
m=0
1
π2
=
.
(2m + 1)2
8
By the now familiar formula from class/notes/text with L = 2, and α2 = 2,
u(x, t) =
∞
X
bk sin(kπx/2)e−k
2 π 2 t/2
k=1
with bk already computed in question 1(a), so that
∞
X
8
u(x, t) = 2
π
k=1,k
odd
(−1)(k−1)/2
sin
k2
kπx
2
e
−k
2 π2 t
2
πx π2 t
8
8
e− 2 − 2 sin
= 2 sin
π
2
9π
Note: if you solved the original (typo ridden) version of this problem where the 2 − x
in f (x) was mistakenly written as 1 − x, you would have computed a different set of
coefficients bk .
3. Find the solution of the following problem describing the temperature in a wire with
insulated ends:
ut = 3uxx ,
0 < x < 2, t > 0,
ux (0, t) = 0 = ux (2, t)
0 ≤ x ≤ 2.
u(x, 0) = x,
Again from class/notes/text,
∞
u(x, t) =
a0 X
2 2
ak cos(kπx/2)e−3k π t/4
+
2
k=1
where
2
a0 =
2
Z
2
0
1
xdx = 22 = 2,
2
and for k ≥ 1,
Z 2
2
2
2
x cos(kπx/2)dx = x
sin(kπx/2)dx
sin(kπx/2)|0 −
kπ
kπ 0
0
4
4
− π28k2 k odd
= 2 2 cos(kπx/2)|20 = 2 2 [cos(kπ) − 1] =
,
0
k even
k π
k π
2
ak =
2
Z
2
so
u(x, t) = 1 −
8
π2
∞
X
k=1,k
odd
1
cos
k2
kπx
2
3
e−
3k2 π 2 t
4
=1−
πx 3π2 t
8
cos
e− 4 − · · ·
π2
2
3πx
2
e−
9π 2 t
2
+· · ·
4. Solve the following problem describing heat conduction in a closed thin circular wire:

−1 < x < 1, t > 0,
 ut = uxx ,
u(−1, t) = u(1, t),
ux (−1, t) = ux (1, t),

u(x, 0) = |x|,
−1 ≤ x ≤ 1
If you didn’t see this in class/text/notes, then just carry out the separation of variables, and you’ll find the ’X problem’ to be
X 00 = −λ2 X,
X(−1) = X(1), X 0 (−1) = X 0 (1),
(you can check the separation constant must be ≤ 0, so we may write it as −λ2 ), for
which the general solution is X = A cos(λx) + B sin(λx), so X 0 = λ(−A sin(λx) +
B cos(λx)), and when we impose the periodic BCs we find
A cos(λ) − B sin(λ) = A cos(λ) + B sin(λ)
=⇒
λ(A sin(λ) + B cos(λ)) = λ(−A sin(λ) + B cos(λ))
2B sin(λ) = 0,
=⇒
2Aλ sin(λ) = 0
so (unless A = B = 0), sin(λ) = 0, so λ = kπ, k ∈ {0, 1, 2, 3, . . .}, with corresponding
solution Xk (x) = Ak cos(kπx) + Bk sin(kπx) (we include k = 0 as λ = 0 produces the
constant solution). Thus the general solution to the PDE and BC is
u(x, t) =
∞
X
(Ak cos(kπx) + Bk sin(kπx)) e−k
2 π2 t
.
k=0
To satisfy the IC, we need
|x| = u(x, 0) =
∞
X
(Ak cos(kπx) + Bk sin(kπx))
k=0
which of course is the (full range) Fourier series for f (x) = |x| on [−1, 1]. Thus,
since |x| is an even function, Bk = 0 for all k, while
Z
a0
2 1
1
A0 =
=
|x|dx = ,
2
2 0
2
and for k ≥ 1,
2
2
Z
1
1
1
x cos(kπx)dx = x
sin(kπx)|10 −
kπ
kπ
0
1
0
k even
1
= 2 2 cos(kπk)|0 =
,
2
− k2 π2 k odd
k π
Ak =
so
u(x, t) =
1
2
− 2
2 π
∞
X
k=1,k
4
odd
Z
1
sin(kπx)dx
0
1
2 2
cos(kπx)e−k π t .
k2