Diff. Eqns.
Problem Set 7
Solutions
1. Let
(
x
0≤x≤1
2
1<x≤2
f (x) =
(1)
Find S(x), the Fourier sine series of f (x), and C(x), the Fourier cosine series of
f (x), both for L = 2. Sketch the graphs of S(x) and C(x) for −6 ≤ x ≤ 6.
The Fourier sine series is given by
f (x) =
∞
X
bn sin
n=1
nπx L
,
(2)
with
Z
nπx 2 L
f (x) sin
dx
bn =
L 0
L
Z 1
Z 2
nπx nπx =
x sin
dx +
2 sin
dx
2
2
0
1
nπx 1 Z 1 2 nπx nπx 2
2
2
−
−
+2 −
=x −
cos
cos
cos
nπ
2 0
nπ
2
nπ
2 1
0
nπ 2 2
nπ 1
nπ 2
4 =−
cos
+
sin
(−1)n − cos
−
nπ
2
nπ
2 nπ
2
0
nπ nπ 2 2
4
2
cos
+
sin
−
(−1)n
=
nπ
2
nπ
2
nπ
(3)
(4)
(5)
(6)
(7)
So the Fourier sine series is
f (x) =
∞
X
n=1
nπ 2
cos
+
nπ
2
2
nπ
2
sin
nπ 2
4
−
(−1)n
nπ
!
sin
nπx 2
(8)
The Fourier cosine series is given by
∞
f (x) =
nπx a0 X
+
bn cos
,
2
L
n=1
(9)
with
2
a0 =
L
Z
L
1
Z
2dx =
xdx +
f (x)dx =
1
0
0
2
Z
5
2
(10)
and
Z
nπx 2 L
an =
dx
f (x) cos
L 0
L
Z 2
Z 1
nπx nπx 2 cos
x cos
dx +
dx
=
2
2
1
0
nπx 1 Z 1 2 nπx nπx 2
2
2
−
sin
sin
sin
=x
dx + 2
nπ
2 0
nπ
2
nπ
2 1
0
nπ nπ 2 2 2
cos
sin
+
−1 .
=−
nπ
2
nπ
2
(11)
(12)
(13)
(14)
Then the Fourier cosine series is
∞
nπ 5 X
2
f (x) = +
−
sin
+
4
nπ
2
n=1
2
nπ
2 cos
nπ 2
!
nπx −1
cos
2
(15)
2. Find the solution of the heat equation
0 ≤ x ≤ 3,
ut = 5uxx ,
t≥0
for the initial condition and boundary conditions given by
πx πx u (x, 0) = 2 sin
− sin
,
u(0, t) = 0, ux (3, t) = 0.
6
2
(16)
(17)
Assume we have a separable solutions, u(x, t) = X(x)T (t). Then
XT 0 = 5X 00 T
T0
(18)
X 00
1
=
,
5T
X
(19)
which must equal some constant, call this −λ. Then we get the ordinary differential equations
X 00 + λX = 0,
0
T + 5λT = 0.
X(0) = 0,
X 0 (3) = 0
(20)
(21)
First we will solve (20). The eigenvalues and eigenfunctions of this differential equation
were found in Problem Set 6, #2(a):
2n − 1 2
λn =
π
6
2n − 1
Xn = sin
πx .
6
(22)
(23)
Next we solve (21). We use the values of λ that give us nontrivial solutions to (20), so (21)
becomes
2n − 1 2
0
T +5
π T = 0.
(24)
6
2n − 1 2
0
T = −5
π T
(25)
6
#
" 2n − 1 2
(26)
π t
T ∝ exp −5
6
So we have
"
un (x, t) = exp −5
2n − 1
π
6
2 #
2n − 1
t sin
πx
6
(27)
and
u (x, t) =
∞
X
"
cn exp −5
n=1
2n − 1
π
6
2 #
2n − 1
t sin
πx .
6
(28)
Applying the initial condition, we have
2 sin
πx 6
− sin
πx 2
= u(x, 0) =
∞
X
cn sin
n=1
2n − 1
πx .
6
(29)
Matching the left- and right-hand sides, we get
c1 = 2,
c2 = −1,
cn = 0 for n = 3, 4, ...
(30)
So our solution is
π π π 2 π 2
t sin
x − exp −5
t sin
x
u (x, t) = 2 exp −5
6
6
2
2
(31)
3. Find the solution of the wave equation
utt = 4uxx ,
0 ≤ x ≤ 1,
t≥0
(32)
for the initial conditions and boundary conditions given by
u (x, 0) = 3 sin (πx) ,
ut (x, 0) = 0,
u (0, t) = 0,
u (1, t) = 0.
(33)
Use a trigonometric identity to show that the solution can be written in the
form
1
[F (x − 2t) + F (x + 2t)] .
2
u (x, t) =
(34)
Find F . (This form is known as D’Alembert’s solution.)
The solution to this problem is derived in Section 10.7, Elastic String with Nonzero Initial
Displacement, and is given by equations (20) and (22):
u (x, t) =
∞
X
cn sin
nπx n=1
L
cos
nπat
L
.
(35)
Applying the initial condition, we have
3 sin (πx) = u(x, 0) =
∞
X
cn sin (nπx) .
(36)
n=1
Matching the left- and right-hand sides, we get
c1 = 3,
cn = 0 for n = 2, 3, ...
(37)
So our solution is
u (x, t) = 3 sin (πx) cos (2πt) .
(38)
Using the trigonometric identity
sin θ cos φ =
1
(sin (θ − φ) + sin (θ + φ)) ,
2
(39)
we can rewrite (38) as
3
(sin (πx − 2πt) + sin (πx + 2πt))
2
1
= (3 sin (π (x − 2t)) + 3 sin (π (x + 2t)))
2
1
= (F (x − 2t) + F (x + 2t)) ,
2
u (x, t) =
(40)
(41)
(42)
where
F (z) = 3 sin (πz) .
(43)