Spring 2014 – Advanced Differential Equations
Midterm
Problem 1. Find the Fourier series of the 2π-periodic function f (x) whose expression over the interval [−π, π] is f (x) = x2 .
Since the function f is even, then bn = 0 for every n and

2π 2


if n = 0

∫ π
 3
2
2
an =
x cos(nx)dx =

π 0
n


 4(−1)
if n ≥ 1
2
n
The Fourier series of f is therefore
∞
∑
π2
(−1)n
+4
cos(nx)
3
n2
n=1
Problem 2. Consider the function f (x)
[−2, 2] by

if
 x
1
if
f (x) =

−1 if
with period 4 and given on the interval
−1≤x≤1
1≤x<2
− 2 < x ≤ −1
The Fourier series of f is:
)
∞ (
∑
2
nπ
4
nπ
nπx
−
cos
+ 2 2 sin
sin
.
nπ
2
π
n
2
2
n=1
(a) Can you use the Fourier series of f to deduce the Fourier series of f ′ ? Explain.
Since the the function f is not continuous throughout R, then its Fourier series
cannot be differentiated termwise to obtain the Fourier series of f ′
(b) Can you use∫the Fourier series of f to deduce the Fourier series of the antix
derivative F (x) =
f (t)dt.
∫
0
2
The function f is piecewise smooth and
f (x)dx = 0, then its antiderivative
−2
F has period 4 and the Fourier series of F can be obtained from that of f by
termwise integration
(c) Find the Fourier series of F .
We have
)∫ x
∞ (
∑
2
nπ
4
nπ
nπt
F (x) =
−
cos
+ 2 2 sin
sin
dt
nπ
2
π n
2
2
0
n=1 (
)(
)
∞
)
∑
2
nπ
4
nπ
−2 (
nπx
nπx
=
−
cos
+ 2 2 sin
cos
− 1 sin
nπ
2
π
n
2
nπ
2
2
n=1
)(
)
∞ (
∑
2
nπ
4
nπ
−2
nπx
7
−
cos
+ 2 2 sin
cos
= +
6 n=1
nπ
2
π n
2
nπ
2
Problem 3. (a) Write Parseval Identity
1
2
For a 2P -periodic peiecewise continuous function we have
∫ P
∞
1
a2
1∑ 2
f (x)2 dx = 0 +
(a + b2n )
2P −P
4
2 n=1 n
(b) The Fourier series of the 2π-periodic function f (x) given on the interval
[−π, π] by f (x) = x3 − π 2 x is
6π
∞
∑
(−1)n sin(nx)
n3
n=1
Use the Parseval Identity to evaluate
1
We have
2π
∞
∑
1
n6
n=1
∫
π
−π
(x3 − π 2 x)2 dx = 18π 2
∞
∑
1
. From this relation, we get
6
n
n=1
∞
∑
1
4π 4
=
n6
945
n=1
Problem 4. Solve the following boundary value problem
uxx + uyy = 0
ux (0, y) = ux (1, y) = 0
u(x, 0) = sin(4x)
u(x, 0) = 1
u(x, 2) = x
0 < x < 1, 0 < y < 2
0<y<2
0 < x < (π/2)
0<x<1
0<x<1
Hint: The Fourier sine series over the interval [0, 1] of the functions 1 and x are:
for 1 :
∞
∑
1 − (−1)n
sin(nπx);
πn
n=1
for x :
∞
2 ∑ (−1)n+1
sin(nπx) .
π n=1
n
The Fourier cosine series over the interval [0, 1] of the functions 1 and x are:
for 1 : 1 +
∞
∑
0 cos(nπx);
n=1
for x :
∞
1
4 ∑ cos[(2k + 1)πx]
− 2
.
2
π k=0
(2k + 1)2
If u(x, y) = X(x)Y (y) solves the homogeneous part of the BVP, then the functions X(x) and Y (y) solve the ODE problems:
X ′′ (x) + λX(x) = 0 0 < x < 1;
X(0) = X(1)
=0
and
Y ′′ (y) − λY (y) = 0 0 < y < 2
where λ is the separation constant.
The eigenvalues and eigenfunctions of the X-problem are: λ0 = 0, X0 (x) = 1,
and λn = (nπ)2 , Xn (x) = cos(nπx) with n ∈ Z+ .
For λ = 0, the Y -problem has independent solution 1 and y; and for λ = (nπ)2
the independent solutions are cosh(nπy) and sinh(nπy).
The general formal solution of the homogeneous part is
u(x, y) = A0 + B0 y +
∞
∑
n=1
(An cosh(nπy) + Bn sinh(nπy)) cos(nπx)
3
We use the nonhomogeneous conditions to get the constants Ak ’s and Bk ’s.
∞
∑
u(x, 0) = A0 +
An cos(nπx) = 1
n=1
So A0 = 1 and An = 0 for n ≥ 1. Then,
∞
∑
u(x, 2) = 1 + 2B0 +
Bn sinh(2nπ) cos(nπx) = x .
n=1
1
We get B0 = − , B2k = 0, and
4
B2k+1 =
π 2 (2k
1)2
−4
sinh(2π(2k + 1))
+
The solution of the BVP is:
∞
y
4 ∑ sinh((2k + 1)πy) cos((2k + 1)πx)
u(x, y) = 1 − − 2
.
4 π
(2k + 1)2 sinh(2π(2k + 1))
k=0