Assignment 4
Due date: August 04, 2015
1. Solve the following problem:







ut = uxx ,
t > 0,
BC: u(0, t) = 20, u(100, t) = 0,
IC:
u(x, 0) = f (x),
where
f (x) =
Answer.
0 < x < 100,
t > 0,
0 ≤ x ≤ 100,
2x,
0 ≤ x ≤ 50,
200 − 2x, 50 ≤ x ≤ 100.
Let w(x, t) = ax + b such that
w(0, t) = 20,
and w(100, t) = 0,
t > 0.
Then
b = 20,
So we get a = −
1
and b = 20, that is,
5
and
100a + b = 0.
1
w(x, t) = − x + 20.
5
Let v(x, t) = u(x, t) − w(x, t), then v satisfies

vt = vxx , 0 < x < 100, t > 0,




BC: v(0, t) = v(100, t) = 0,
t > 0,



 IC: v(x, 0) = f (x) + 1 x − 20 = f˜(x),
5
(
We know that the “ general” solution to
vt = vxx ,
(1)
0 ≤ x ≤ 100,
0 < x < 100,
BC: v(0, t) = v(100, t) = 0,
v(x, t) =
∞
X
n2 π 2
an e− 1002 t sin
n=1
t > 0,
t > 0,
is given by:
nπ x .
100
Notice that the Four sine series of f˜(x) is:
∞
X −40πn + 800 sin
1
f˜(x) = f (x) + x − 20 =
5
n2 π 2
n=1
nπ
2
sin
nπ x .
100
So the solution to (1) is given by:
v(x, t) =
∞
X
−40πn + 800 sin
nπ
2
n2 π 2
n=1
n2 π 2
e− 1002 t sin
nπ x .
100
So the solution to our problem is given by:
∞
nπ
2
X −40πn + 800 sin
1
u(x, t) = w(x, t) + v(x, t) = − x + 20 +
5
n2 π 2
n2 π 2
e− 1002 t sin
n=1
nπ x .
100
2. Solve the following problem:







Answer.
ut = uxx ,
0 < x < 30,
t > 0,
BC: u(0, t) = 40, ux (30, t) = 0,
u(x, 0) = 30 − x,
IC:
t > 0,
0 ≤ x ≤ 30.
Let w(x, t) = ax + b such that
w(0, t) = 40,
and wx (30, t) = 0,
t > 0.
Then b = 40 and a = 0, that is,
w(x, t) = 40.
Let v(x, t) = u(x, t) − w(x, t), then v satisfies

vt = vxx , 0 < x < 30, t > 0,



BC: v(0, t) = vx (30, t) = 0,
t > 0,



IC: v(x, 0) = 30 − x − 40 = −x − 10 =: f (x),
(
We know that the “ general” solution to
v(x, t) =
vt = vxx ,
(2)
0 ≤ x ≤ 30,
0 < x < 30,
t > 0,
BC: v(0, t) = vx (30, t) = 0,
∞
X
−
an e
(2n+1)2 π 2
t
2·302
sin
n=1
Let
f (x) = −x − 10 =
∞
X
an sin
n=1
t > 0,
(2n − 1)π
x .
2 · 30
(2n − 1)π
x ,
2 · 30
then
an =
=
Z 30
2
(2n − 1)π
(−x − 10) sin
x dx
30 0
2 · 30
40[−(2n − 1)π + 6(−x)n ]
.
π 2 n2
Page 2
is given by:
So the solution to (2) is given by:
v(x, t) =
∞
X
40[−(2n − 1)π + 6(−x)n ]
π 2 n2
n=1
e
−
(2n+1)2 π 2
t
2·302
sin
(2n − 1)π
x .
2 · 30
So the solution to our problem is given by:
u(x, t) = w(x, t) + v(x, t) = 40 +
∞
X
40[−(2n − 1)π + 6(−x)n ]
π 2 n2
n=1
3. Let T1 , T2 be two constants, solve the following problem:

2x


, 0 < x < L,
ut = uxx +


L
(2n+1)2 π 2
t
2·302
sin
(2n − 1)π
x .
2 · 30
t > 0,
BC: u(0, t) = T1 , u(L, t) = T2 ,




IC: u(x, 0) = f (x), 0 ≤ x ≤ L.
Answer.
e−
t > 0,
Let w(x, t) = ax + b such that
w(0, t) = T1 ,
w(L, t) = T2 ,
t > 0.
Then
b = T1 ,
That is, a =
and aL + b = T2 .
T2 − T1
and b = T1 , that is,
L
w(x, t) =
T2 − T1
x + T1 .
L
Let v(x, t) = u(x, t) − w(x, t), then v satisfies:






2x
, 0 < x < L, t > 0,
L
BC: v(0, t) = u(L, t) = 0,
t > 0,




 IC:
vt = vxx +
v(x, 0) = f (x) −
(
Recall the “general” solution to
vt = vxx ,
T2 − T1
x − T1 ,
L
0 < x < L,
BC: u(0, t) = u(L, t) = 0,
v(x, t) =
∞
X
an e−
n2 π 2
t
L2
n=1
Page 3
sin
(3)
0 ≤ x ≤ L.
t > 0,
t > 0.
nπ x .
L
is:
Let
∞
nπ X
T2 − T1
f (x) −
x − T1 =
fn sin
x ,
L
L
n=1
then
an =
2
L
Z
L
f (x) −
0
nπ T2 − T1
x − T1 sin
x dx,
L
L
Let
n ≥ 1.
∞
nπ 2x X
=
bn sin
x ,
L
L
n=1
then
2
bn =
L
Z
0
L
nπ 2x
4(−1)n+1
sin
x dx =
,
L
L
nπ
Let
v(x, t) =
∞
X
vn (t) sin
n=1
n ≥ 1.
nπ x ,
L
then
∞
X
n=1
∞
X
n=1
vn0 (t) sin
∞ ∞
nπ X
nπ nπ X
nπ 2
4(−1)n+1
x
= −
sin
x +
sin
x
L
L
L
nπ
L
n=1
n=1
∞
nπ nπ X
x
=
fn sin
x .
vn (0) sin
L
L
n=1
Then for any n ≥ 1, we have


vn0 (t) = −
nπ 2
 v (0) = f . L
n
n
Then
So we get
vn (t) +
4(−1)n+1
nπ
4(−1)n L2 − n2 π2 2 t 4(−1)n+1 L2
vn (t) = fn +
e L +
,
n3 π 3
n3 π 3
n ≥ 1.
∞ nπ X
4(−1)n L2 − n2 π2 2 t 4(−1)n+1 L2
L
v(x, t) =
fn +
e
+
sin
x .
n3 π 3
n3 π 3
L
n=1
Therefore, the solution to our problem is:
∞
X
T2 − T1
u(x, t) = w(x, t) + v(x, t) =
x + T1 +
L
n=1
nπ 4(−1)n L2 − n2 π2 2 t 4(−1)n+1 L2
L
fn +
x .
e
+
sin
n3 π 3
n3 π 3
L
Page 4
4. Solve the following problem:







Answer.
ut = uxx ,
0 < x < 1,
t > 0,
BC: u(0, t) = 2e−t , u(1, t) = 0,
IC:
t > 0,
0 ≤ x ≤ 1.
u(x, 0) = x,
Let w(x, t) = ax + b such that w(0, t) = 2e−t and w(1, t) = 0, then
b = 2e−t ,
and a + b = 0.
So we get
a = −2e−t ,
and b = 2e−t .
That is, we have
w(x, t) = −2e−t x + 2e−t .
Let v(x, t) = u(x, t) − w(x, t), then v satisfies

vt − vxx = ut − wt − uxx + wxx = −2e−t x + 2e−t ,



BC: v(0, t) = v(1, t) = 0,
t > 0,



IC: v(x, 0) = u(x, 0) − w(x, 0) = x + 2x − 2 = 3x − 2,
(
Recall that the “general” solution to the problem
v(x, t) =
vt − vxx = 0,
0 < x < 1,
(4)
0 ≤ x ≤ 1.
0 < x < 1,
BC: v(0, t) = v(1, t) = 0,
∞
X
2 π2 t
e−n
sin(nπx).
n=1
Let
−2e−t x + 2e−t =
∞
X
σn (t) sin(nπx),
n=1
then
σn (t) =
=
=
Z
2 1
[−2e−t x + 2e−t ] sin(nπx) dx
1 0
4e−t (−1)n 4e−t [1 − (−1)n ]
+
nπ
nπ
−t
4e
, n ≥ 1.
nπ
Let
3x − 2 =
∞
X
fn sin(nπx),
n=1
Page 5
t > 0,
t > 0,
t > 0.
is:
then
2
1
fn =
= −
1
Z
(3x − 2) sin(nπx) dx
0
2[2 + (−1)n ]
,
πn
Let
v(x, t) =
∞
X
n ≥ 1.
vn (t) sin(nπx)
n=1
be the solution to (4), then
∞
X
vn0 (t) sin(nπx)
+
n=1
∞
X
2 2
n π vn (t) sin(nπx) =
n=1
∞
X
∞
X
4e−t
nπ
n=1
∞
X
vn (0) sin(nπx) = −
n=1
n=1
sin(nπx)
2[2 + (−1)n ]
sin(nπx).
πn
So for all n ≥ 1, we have

−t

 vn0 (t) + n2 π 2 vn (t) = 4e ,
nπ
n

 vn (0) = − 2[2 + (−1) ] .
πn
So we get
2[2 + (−1)n ]
4
4
2 2
vn (t) = −
+
e−n π t +
,
2
2
2
πn
nπ(n π − 1)
nπ(n π 2 − 1)
Hence
v(x, t) =
n ≥ 1.
∞ X
4
4
2[2 + (−1)n ]
−n2 π 2 t
+
e
+
sin(nπx).
−
πn
nπ(n2 π 2 − 1)
nπ(n2 π 2 − 1)
n=1
Therefore, the solution to our problem is:
u(x, t) = w(x, t) + v(x, t)
−t
= −2e x + 2e
−t
∞ X
4
4
2[2 + (−1)n ]
−n2 π 2 t
+
−
+
e
+
sin(nπx).
πn
nπ(n2 π 2 − 1)
nπ(n2 π 2 − 1)
n=1
5. Consider the following heat conduction problem:

ut = uxx + sin(3πx), 0 < x < 1,



BC: u(0, t) = u(1, t) = 0,
t > 0,



IC: u(x, 0) = sin(πx), 0 ≤ x ≤ 1.
Page 6
t > 0,
(a) Determined the steady-state temperature.
Answer.
The steady-state temperature u(x, t) = w(x), then
w00 (x) + sin(3πx) = 0,
w(0) = w(1) = 0.
So we get
w(x) =
1
sin(3πx).
9π 2
(b) Find the temperature u(x, t).
Let v(x, t) = u(x, t) − w(x, t), then v satisfies:

vt = vxx , 0 < x < 1, t > 0,




BC: v(0, t) = v(1, t) = 0,
t > 0,



 IC: v(x, 0) = sin(πx) − 1 sin(3πx), 0 ≤ x ≤ 1.
9π 2
(
vt = vxx , 0 < x < 1, t > 0,
Recall the ”general” solution to
is:
BC: v(0, t) = v(1, t) = 0,
t > 0.
Answer.
v(x, t) =
∞
X
2 π2 t
an e−n
sin(nπx).
n=1
Sine v(x, 0) = sin(πx) −
1
sin(3πx), then we have
9π 2
2
v(x, t) = e−π t sin(πx) −
1 −9π2 t
e
sin(3πx).
9π 2
Therefore, the solution to our problem is:
u(x, t) = w(x, t) + v(x, t) =
1
1
2
2
sin(3πx) + e−π t sin(πx) − 2 e−9π t sin(3πx).
2
9π
9π
6. Solve utt = c2 uxx , u(x, 0) = ex , ut (x, 0) = sin(x) for x ∈ R and t ≥ 0.
Answer.
By D’Alembert’s solution, we have
u(x, t) =
=
=
Z
1 x+ct
ex+ct + ex−ct
+
sin(τ ) dτ
2
2c x−ct
ex+ct + ex−ct cos(x + ct) − cos(x − ct)
−
2
2c
x+ct
x−ct
e
+e
1
+ sin(x) sin(ct).
2
c
Page 7
7. Let α > 0 and γ > 0 be two constants, use separation of varaibels to solve the following problem:

utt = α2 uxx − α2 γ 2 u, 0 < x < L, t > 0,



BC: u(0, t) = 0, u(L, t) = 0,
t > 0,



IC: u(x, 0) = f (x), ut (x, 0) = 0, 0 ≤ x ≤ L.
Answer.
Let u(x, t) = X(x)T (t) be a non-zero separated solution to
(
utt = α2 uxx − α2 γ 2 u, 0 < x < L, t > 0,
BC: u(0, t) = 0, u(L, t) = 0,
t > 0.
,
(5)
then XT 00 = α2 X 00 T − α2 γ 2 XT , that is,
−
T 00 + α2 γ 2 T
X 00
=−
= λ.
X
α2 T
It’s easy to see that λx = λt = 0, then λ is a constant. Since u(0, t) = u(L, t) = 0, then X(0) = X(L) = 0,
that is, X satisfies
00
X + λX = 0,
X(0) = X(L) = 0.
nπ 2
nπ We know that the eigenvalues are λ =
x ,
and the corresponding eigenfunctions are X = C sin
L
L
with n ∈ N.
nπ 2
For each λ =
, since T 00 + α2 γ 2 T + α2 λT = 0, then
L
!
!
r
r
2π2
2π2
n
n
T (t) = C1 cos α γ 2 + 2 t + C2 sin α γ 2 + 2 t .
L
L
So we find a non-zero solution to (5)):
"
r
un (x, t) = C1 cos α
γ2
!
!#
r
nπ n2 π 2
n2 π 2
2
+ 2 t + C2 sin α γ + 2 t
sin
x .
L
L
L
Then the “general” solution to (5) is given by:
"
!
!#
r
r
∞
nπ 2π2
2π2
X
n
n
u(x, t) =
an cos α γ 2 + 2 t + bn sin α γ 2 + 2 t
sin
x .
L
L
L
n=1
Since ut (x, 0) = 0, then
0 = ut (x, 0) =
∞
X
r
α
γ2 +
n=1
Page 8
nπ n2 π 2
b
sin
x ,
n
L2
L
which implies that
n ≥ 1.
bn = 0,
Since
f (x) = u(x, 0) =
∞
X
an sin
n=1
then
2
an =
L
Z
L
f (x) sin
0
nπ x ,
L
nπ x ,
L
n ≥ 1.
Therefore, the solution to our problem is:
u(x, t) =
∞
X
!
nπ 2π2
n
γ 2 + 2 t sin
x .
L
L
r
an cos α
n=1
where
an =
2
L
Z
L
f (x) sin
0
nπ x ,
L
n ≥ 1.
8. Use the D’Alembert’s formula to solve the following problem:

x

utt = uxx + , 0 < x < π, t > 0,


π


t2
BC:
u(0,
t)
=
0,
u(π,
t)
=
,
t > 0,


2



IC: u(x, 0) = sin(x), ut (x, 0) = sin(3x), 0 ≤ x ≤ π.
Answer.
Let w(x, t) = ax + b such that w(0, t) = 0 and w(π, t) =
b = 0,
t2
,
2
and aπ + b =
then
a=
t2
, that is,
2
t2
.
2π
That is,
w(x, t) =
t2
x.
2π
Let v(x, t) = u(x, t) − w(x, t), then v satisfies

x x

vtt − vxx = utt − wtt − uxx + wxx = − = 0, 0 < x < π, t > 0,



π π
BC:
v(0,
t)
=
v(π,
t)
=
0,
t
>
0,



 IC: v(x, 0) = u(x, 0) − w(x, 0) = sin(x) = f (x), v (x, 0) = u (x, 0) − w (x, 0) = sin(3x) = g(x),
t
Page 9
t
t
0 ≤ x ≤ π.
Since sin(x) is odd with period 2π, sin(3x) is odd with period 2π
3 , then the odd 2π-periodic extension of f (x)
and g(x) on [0, π] are sin(x) and sin(3x), respectively. By D’Alembert’s solution, we have
Z
f0 (x + t) + f0 (x − t) 1 x+t
g0 (τ )τ
u(x, t) =
+
2
2 x−t
Z
sin(x + t) + sin(x − t) 1 x+t
=
sin(3τ ) dτ
+
2
2 x−t
x+t
1 1
= sin(x) cos(t) − · cos(3τ )
2 3
x−t
cos(3x + 3t) − cos(3x − 3t)
6
sin(3x) sin(3t)
= sin(x) cos(t) +
.
3
= sin(x) cos(t) −
So we get
u(x, t) = w(x, t) + v(x, t) =
t2
sin(3x) sin(3t)
x + sin(x) cos(t) +
.
2π
3
9. Use the method of eigenfunction expansion to solve the following problem:

x

utt = uxx + sin(x) + , 0 < x < π, t > 0,


π


t2
t > 0,
BC: u(0, t) = 0, u(π, t) = ,


2



IC: u(x, 0) = sin(x), ut (x, 0) = sin(3x), 0 ≤ x ≤ π.
Answer.
Let w(x, t) = ax + b such that w(0, t) = 0 and w(π, t) =
b = 0,
and aπ + b =
then
t2
, that is,
2
t2
,
2
t2
.
2π
a=
That is,
w(x, t) =
t2
x.
2π
Let v(x, t) = u(x, t) − w(x, t), then v satisfies

x x

vtt − vxx = utt − wtt − uxx + wxx = sin(x) + − = sin(x), 0 < x < π, t > 0,



π π
BC:
v(0,
t)
=
v(π,
t)
=
0,
t
>
0,



 IC: v(x, 0) = u(x, 0) − w(x, 0), v (x, 0) = u (x, 0) − w (x, 0) = sin(3x), 0 ≤ x ≤ π.
t
t
Page 10
t
(6)
(
Recall the “general” solution to
vtt − vxx = 0,
0 < x < π,
BC: v(0, t) = v(π, t) = 0,
v(x, t) =
∞
X
t > 0,
t > 0.
is given by:
2
an e−n t sin(nx).
n=1
Let
v(x, t) =
∞
X
vn (t) sin(nx)
n=1
be the solution of (6), then
∞
X
vn00 (t) sin(nx)
n=1
+
∞
X
n2 vn (t) sin(nx) = sin(x)
n=1
∞
X
n=1
∞
X
vn (0) sin(nx) = sin(x)
vn0 (0) sin(nx) = sin(3x).
n=1
So we get
00
v1 (t) + v1 (t) = 1,
,
v1 (0) = 1, v10 (0) = 0
v300 (t) + 9v3 (t) = 0,
,
v3 (0) = 0, v30 (0) = 1.
and
vn00 (t) + n2 vn (t) = 0,
,
vn (0) = 0, vn0 (0) = 0
n 6= 1, 3
So we get
v1 (t) = 1,
v3 =
1
sin(3t),
3
and vn (t) = 0,
n ≥ 1, 3.
So we get
v(x, t) = sin(x) +
1
sin(3t) sin(3x).
3
Therefore, the solution to our problem is:
u(x, t) = w(x, t) + v(x, t) =
1
t2
x + sin(x) + sin(3t) sin(3x).
2π
3
10. Find the solution u(x, y) of Laplace’s equation in the rectangle 0 < x < a, 0 < y < b, that satisfies the
boundary conditions:
u(0, y) = 0,
u(a, y) = 0, 0 < y < b,
u(x, 0) = h(x), u(x, b) = 0, 0 ≤ x ≤ a.
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Answer.
First, let’s look the problem:

∆u = uxx + uyy = 0, 0 < x < a, 0 < y < b,

BC : u(x, b) = 0, 0 ≤ x ≤ a,

u(0, y) = 0, u(a, y) = 0, 0 ≤ y ≤ b.
(7)
Let u(x, y) = X(x)Y (y) be a non-zero separated solution to (7), then X 00 (x)Y (y) + X(x)Y 00 (y) = 0, that is,
−
X 00 (x)
Y 00 (y)
=
= λ.
X(x)
Y (y)
It’s easy to see that λx = λy = 0, that is, λ is a constant. Since u(0, y) = u(a, y) = 0, then X(0) = X(a) = 0.
Then X(x) satisfies:
00
X + λX = 0,
(8)
X(0) = X(a) = 0.
For the eigenvalue problem (8), we know that the eigenvalues are λ =
nπ functions are X(x) = C sin
x , with n ∈ N.
a
nπ 2
For λ =
, since u(x, b) = 0, then Y (b) = 0. Then Y satisfies
a
(
nπ 2
Y 00 − λY = Y 00 −
Y = 0,
a
Y (b) = 0.
Then we know that
nπ 2
a
, and the corresponding eigen-
(9)
i
h nπ
2nπ
nπ
Y (y) = C e a y − e a b e− a y .
Then for any n ∈ N, we can find a non-zero solution to (8):
i
nπ h nπ
2nπ
nπ
un (x, y) = e a y − e a b e− a y sin
x .
a
Let
u(x, y) =
∞
X
h
an e
nπ
y
a
−e
n=1
2nπ
b
a
e
− nπ
y
a
i
nπ sin
x
a
by our solution, it’s easy to see that u(x, b) = 0 for 0 ≤ x ≤ a, and u(0, y) = u(a, y) = 0 for 0 ≤ y ≤ b. Since
u(x, 0) = h(x) for 0 ≤ x ≤ a, then
h(x) =
∞
X
n=1
So we get
h
i
nπ 2nπ
an 1 − e a b sin
x .
a
h
i 2Z a
nπ 2nπ
b
an 1 − e a
=
h(x) sin
x dx,
a 0
a
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for all n ≥ 1.
So we get
an =
Z
2
h
a 1−e
2nπ
b
a
i
a
h(x) sin
0
nπ x dx,
a
for all n ≥ 1.
Therefore, the solution is given by:
u(x, y) =
∞
X
n=1
where
an =
i
nπ h nπ
nπ
2nπ
x ,
an e a y − e a b e− a y sin
a
Z
2
h
a 1−e
2nπ
b
a
i
a
h(x) sin
0
nπ x dx,
a
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for all n ≥ 1.