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ACM 95/100c Problem Set 7 Solutions
Lei Zhang
May 22, 2006
Each Problem is worth 10 points. Each part of a multipart problem is weighted equally.
Problem 1
(No collaboration) Solve
∂
2
u
∂x
2
+
∂
2
u
∂y
2
= 0 for 0
< x < L,
y >
0
subject to the following boundary conditions. If there is a solvability condition, state it and explain it
physically.
(a)
∂u
∂x
(0
,y
) =
g
1
(
y
)
,
∂u
∂x
(
L,y
) =
g
2
(
y
)
,
u
(
x,
0) =
f
(
x
)
(b)
∂u
∂x
(0
,y
) = 0
,
∂u
∂x
(
L,y
) = 0
,
∂u
∂y
(
x,
0) =
f
(
x
)
Solution 1
(a) Let
u
≡
v
+
w
. Here,
v
solves the system
∂
2
v
∂x
2
+
∂
2
v
∂y
2
= 0
(1)
with
∂v
∂x
(0
,y
) =
g
1
(
y
),
∂v
∂x
(
L,y
) =
g
2
(
y
) and
v
(
x,
0) = 0.
w
solves the system
∂
2
w
∂x
2
+
∂
2
w
∂y
2
= 0
(2)
with
∂w
∂x
(0
,y
) = 0,
∂w
∂x
(
L,y
) = 0 and
w
(
x,
0) =
f
(
x
). For
w
, the boundary conditions at
x
= 0 and
x
=
L
imply that we should use a Fourier cosine series expansion, i.e.,
w
(
x,y
) =
∞
X
n
=0
A
n
(
y
)cos
‡
nπx
L
·
.
(3)
Whether we plug in this series directly into the PDE or use the method of ﬁnite transforms, by
orthogonality we get

‡
nπ
L
·
2
+
d
2
A
n
dy
2
= 0
.
(4)
1
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View Full Document Therefore,
A
n
(
y
) =
(
a
0
+
b
0
y,
n
= 0
a
n
exp
±

nπy
L
)
+
b
n
exp
±
nπy
L
)
,
n
= 1
,
2
,
3
,...
.
(5)
Since
w
is regular as
y
→ ∞
,
b
n
= 0, so we may write
w
(
x,y
) =
a
0
+
∞
X
n
=1
a
n
exp
‡

nπy
L
·
cos
‡
nπx
L
·
.
(6)
Finally, applying
w
(
x,
0) =
f
(
x
) yields
w
(
x,
0) =
f
(
x
) =
a
0
+
∞
X
n
=1
a
n
cos
‡
nπx
L
·
,
(7)
so by orthogonality
a
0
=
R
L
0
f
(
x
)
dx
R
L
0
1
2
dx
=
1
L
Z
L
0
f
(
x
)
dx
(8)
and
a
n
=
R
L
0
f
(
x
)cos
±
nπx
L
)
dx
R
L
0
cos
2
±
nπx
L
)
dx
=
2
L
Z
L
0
f
(
x
)cos
‡
nπx
L
·
dx,
n
= 1
,
2
,
3
,...
.
(9)
For
v
, we assume that
v
(
x,y
)
→
0,
g
1
(
y
)
→
0 and
g
2
(
y
)
→
0 as
y
→ ∞
. Since we have the condition
v
(
x,
0) = 0, we Fourier sine transform the PDE for
v
. Using the convention
F
s
[
v
(
x,y
)]
≡
r
2
π
Z
∞
0
v
(
x,y
)sin(
ωy
)
dy
≡
V
(
x,ω
)
,
(10)
we have
∂
2
v
∂x
2
+
∂
2
v
∂y
2
= 0
→
∂
2
V
∂x
2
+
r
2
π
v
(
x,
0)

ω
2
V
=
∂
2
V
∂x
2

ω
2
V
= 0
.
(11)
Thus,
V
(
x,ω
) =
c
exp(
ωx
) +
d
exp(

ωx
) =
c
0
cosh(
ωx
) +
d
0
cosh[
ω
(
L

x
)]
.
(12)
Now,
∂v
∂x
(0
,y
) =
g
1
(
y
)
→
∂V
∂x
(0
,ω
) =
G
1
(
ω
)
(13)
and
2
∂x
(
L,y
) =
g
2
(
y
)
→
∂V
∂x
(
L,ω
) =
G
2
(
ω
)
,
(14)
where
G
1
(
ω
)
≡
r
2
π
Z
∞
0
g
1
(
y
)sin(
ωy
)
dy
(15)
and
G
2
(
ω
)
≡
r
2
π
Z
∞
0
g
2
(
y
)sin(
ωy
)
dy
.
(16)
So,
∂V
∂x
(0
,ω
) =
G
1
(
ω
)
=
c
0
ω
sinh(
ω
0)

d
0
ω
sinh[
ω
(
L

0)]
=

d
0
ω
sinh(
ωL
)
,
(17)
implying that
d
0
=

G
1
(
ω
)
ω
sinh(
ωL
)
.
(18)
Also,
∂V
∂x
(
L,ω
) =
G
2
(
ω
)
=
c
0
ω
sinh(
ωL
)

d
0
ω
sinh[
ω
(
L

L
)]
=
c
0
ω
sinh(
ωL
)
,
(19)
implying that
c
0
=
G
2
(
ω
)
ω
sinh(
ωL
)
.
(20)
Thus,
V
(
x,ω
) =
G
2
(
ω
)
ω
sinh(
ωL
)
cosh(
ωx
)

G
1
(
ω
)
ω
sinh(
ωL
)
cosh[
ω
(
L

x
)]
,
(21)
so
v
(
x,y
) =
r
2
π
Z
∞
0
V
(
x,ω
)sin(
ωy
)
dω
=
r
2
π
Z
∞
0
‰
G
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This note was uploaded on 01/08/2011 for the course ACM 95c taught by Professor Nilesa.pierce during the Spring '09 term at Caltech.
 Spring '09
 NilesA.Pierce

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