Millersville University Name Answer Key Department of Mathematics ...

Millersville University Name Answer Key
Department of Mathematics
MATH 467, Partial Differential Equations, Test 1
February 26, 2006
Please answer the following questions. Show all work and write neatly. Answers without
justifying work will receive no credit. Partial credit will be given as appropriate, do not leave
any problem blank. The point values of problems are indicated in parentheses.
1. Consider the following initial boundary value problem for the heat equation.
ut = α 2 uxx + βx 2
u(0, t) = 0 for t > 0
ux(1, t) = 0 for t > 0
u(x, 0) = x for 0 < x < 1
for 0 < x < 1, t > 0
(a) (7 points) Find the steady-state solution to this heat equation.
If v ≡ v(x) is the steady-state solution then
Since v ′ (1) = 0 then C = β
3α 2.
Since v(0) = 0 then D = 0, so
0 = α 2 v ′′ (x) + βx 2
v ′′ (x) = − β
α 2x2
v ′ (x) = − β
3α 2x3 + C
v ′ (x) = β β
−
3α2 3α2x3 v(x) = β β
3α2x −
12α2x4 + D
v(x) = β β
3α2x −
12α2x4 (b) (10 points) Suppose that α = 1 and β = 0 and find the eigenfunctions and
eigenvalues of the boundary value problem.
If we assume u(x, t) = X(x)T(t) then after separating variables we have the
following ODE for X.
X ′′ + λ 2 X = 0

If λ = 0 or if λ = iγ where i = √ −1 and γ ∈ R there are no solutions which
satisfy the boundary conditions. Thus we may assume λ > 0. There are two
linearly independent solutions to this ODE:
X(x) = sin λx and X(x) = cosλx.
Since cos(λ0) = 0 then we may ignore this solution.
Since X ′ (1) = 0 then λ =
for n ∈ N.
X ′ (x) = d
sin λx
dx
= λ cosλx
X ′ (1) = λ cosλ
(2n − 1)π
2
where n ∈ N. Thus we have
Eigenvalues : λ 2 n = (2n − 1)2 π 2
Eigenfunctions : Xn(x) = sin
4
(2n − 1)πx
2
(c) (13 points) Suppose that α = 1 and β = 0 and find the Fourier Series solution to
the heat equation.
The series form of the solution is
where for n ∈ N,
u(x, t) =
∞
n=1
Ane −(2n−1)2π2t (2n − 1)πx
4 sin ,
2
1
An = 2
(2n − 1)πx
x sin dx
0 2
=
8
(2n − 1) 2 (2n − 1)π
sin
π2 2
= 8(−1)n−1
(2n − 1) 2 π 2.
2. (5 points each) Consider the function
sin x if −π ≤ x < 0,
f(x) =
1 − x/π if 0 ≤ x ≤ π.
(a) Sketch the graph the sum of the Fourier series for f(x) for three complete periods.

1.0
0.5
5 5
0.5
1.0
(b) Sketch the graph the sum of the Fourier sine series for f(x) for three complete
periods.
1.0
0.5
y
5 5
0.5
1.0
(c) Sketch the graph the sum of the Fourier cosine series for f(x) for three complete
periods.
1.0
0.8
0.6
0.4
0.2
y
y
5 5
x
x
x

3. (15 points) Consider the following initial boundary value problem for the wave equation
with damping.
utt + 2γut = c 2 uxx for 0 < x < L, t > 0
u(0, t) = 0 = u(L, t) for t > 0
u(x, 0) = f(x) for 0 < x < L
ut(x, 0) = g(x) for 0 < x < L
Assuming the solution u(x, t) has the form u(x, t) = X(x) T(t) use separation of variables
to find the ordinary differential equations which X(x) and T(t) must satisfy. You
do not need to find the solutions to these ordinary differential equations.
If u(x, t) = X(x) T(t) then upon differentiating this solution and substituting into the
damped wave equation we obtain
X T ′′ + 2γX T ′ = c 2 X ′′ T
X T ′′ + 2γX T ′
c2X T
= c2X ′′ T
c2 T
X T
′′ + 2γT ′
c 2 T
= X′′
X
Since the left-hand side of this equation depends only on t and the right-hand side
depends only on x they are both constant. If the constant is written as −λ 2 then the
following two ODEs are implied.
4. (10 points) The Fourier Series for
is
X ′′ + λ 2 X = 0
T ′′ + 2γT ′ + λc 2 T = 0
f(x) =
f(x) ∼ 1
2 +
0 for −π < x < 0,
1 for 0 ≤ x ≤ π,
∞
k=1
2
sin ((2k − 1)x) .
(2k − 1)π
Use these facts to find the sum of the following infinite series.
∞
k=1
Note that f(x) is continuous at x = 1, thus
f(1) = 1
2 +
∞
k=1
sin(2k − 1)
2k − 1
2
sin ((2k − 1))
(2k − 1)π

1 = 1 2
+
2 π
k=1
∞
k=1
1
=
2
∞ 2 sin(2k − 1)
π (2k − 1)
k=1
π
4 =
∞ sin(2k − 1)
(2k − 1) .
1
sin ((2k − 1))
(2k − 1)
5. Consider the following initial boundary value problem for the wave equation.
utt = uxx for 0 < x < 1, t > 0
u(0, t) = 0 = u(1, t) for t > 0
u(x, 0) = f(x) for 0 < x < 1
ut(x, 0) = g(x) for 0 < x < 1
(a) (10 points) Suppose f(x) = sin(2πx) and g(x) = 0. Find the solution to this
initial boundary value problem (hint: d’Alembert’s solution).
Functions f and g are both odd periodic functions of period 2 so d’Alembert’s
solution is of the form
x−t
u(x, t) = 1
1
[f(x + t) + f(x − t)] + g(u) du
2 2 x−t
= 1
[sin(2π(x + t)) + sin(2π(x − t))]
2
= 1
[sin 2πx cos 2πt + cos 2πx sin 2πt + sin 2πx cos 2πt − cos 2πx sin 2πt]
2
= sin(2πx) cos(2πt)
(b) (5 points) Sketch the solution to the wave equation found above for t = 1/3 on
the axes provided below.
u
1.0
0.5
0.5
1.0
0.2 0.4 0.6 0.8 1.0 x

(c) (15 points) Suppose f(x) = 0 and g(x) = sin(πx). Find the solution to this initial
boundary value problem (hint: d’Alembert’s solution).
Functions f and g are both odd periodic functions of period 2 so d’Alembert’s
solution is of the form
u(x, t) = 1
1
[f(x + t) + f(x − t)] +
2 2
= 1
2
x+t
x−t
= − 1
2π
sin πu du
cosπu| x+t
x−t
x−t
x−t
= − 1
[cos(π(x + t)) − cos(π(x − t))]
2π
= 1
[cos(π(x − t)) + cos(π(x + t))]
2π
g(u) du
= 1
[cosπx cosπt + sin πx sin πt + cosπx cos πt − sin πx sin πt]
2π
= 1
cos(πx) cos(πt)
π

Useful Formulas
sin u sin v = 1
[cos(u − v) − cos(u + v)]
2
sin u cos v = 1
[sin(u + v) + sin(u − v)]
2
cos u cos v = 1
[cos(u + v) + cos(u − v)]
2