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