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yields
n
2
b
X n (
x
) −
X n (
x
)=
0
.
Its general solution is
a n e n π x
n
π
x
b n e −
X n
(
x
)=
+
,
b
b
where the a n and the b n are undetermined constants. Thus
n = 1 a n e n π x
b sin n
∞
π
y
b .
n
π
x
b n e −
u 1 (
x
,
y
)=
+
(7.3)
b
By applying boundary conditions u 1
(
0
,
y
)= ϕ
(
y
)
and u 1
(
a
,
y
)= ϕ
(
y
)
, we obtain
1
2
⎧
⎨
b
0 ϕ 1 (
2
b
sin n
π
y
a n +
b n =
y
)
d y
,
b
(7.4)
b
0 ϕ 2 (
⎩
2
b
sin n
π
y
e n π b a n +
n π b b n =
e −
y
)
d y
,
b
so that the a n and the b n can be determined. Similarly, we can also obtain u 2 (
x
,
y
)
.
Example 2. Solve the PDS in a cube
⎧
⎨
⎩
Δ
u
=
0
,
0
<
x
,
y
,
z
<
a
,
(
,
,
)=
(
,
,
)=
,
u
0
y
z
u
a
y
z
0
(7.5)
u
(
x
,
0
,
z
)=
u
(
x
,
a
,
z
)=
0
,
u
(
x
,
y
,
0
)= ϕ (
x
,
y
) ,
u
(
x
,
y
,
a
)=
0
.
Solution. Based on the given boundary conditions, consider
∞
m , n = 1 Z n ( z ) sin m π x
sin n
π
y
(
,
,
)=
u
x
y
z
.
(7.6)
a
a
Substituting it into the equation in PDS (7.5) and using the orthogonality of
sin m
in 0
π
x
sin n
π
y
<
x
,
y
<
a yield
a
a
m
2
n
2
a
a
Z mn (
) − λ
(
)=
, λ
=
+
,
(
)=
.
(7.7)
Its two linearly-independent particular solutions are e √ λ mn z ,e − √ λ mn z . Thus its gen-
eral solution can be written as
z
mn Z mn
z
0
Z mn
a
0
mn
a mn ch
b mn sh
Z mn (
z
)=
λ mn (
a
−
z
)+
λ mn (
a
−
z
) ,
where the a mn and the b mn are constants. Applying Z mn (
a
)=
0 yields a mn =
0.
Therefore,
m , n = 1 b mn sh λ mn ( a − z ) · sin m π x
∞
sin n
π
y
a .
u
(
x
,
y
,
z
)=
a
 
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