Environmental Engineering Reference
In-Depth Information
The solution of
⎨
u
t
τ
0
+
B
2
∂
∂
A
2
u
tt
=
Δ
2
u
+
t
Δ
2
u
+
f
(
x
,
y
,
t
)
,
D
×
(
0
,
+
∞
)
,
(6.65)
⎩
L
(
u
,
u
n
)
|
∂
D
=
0
,
u
(
x
,
y
,
0
)=
ϕ
(
x
,
y
)
,
u
t
(
x
,
y
,
0
)=
ψ
(
x
,
y
)
is
1
τ
W
ϕ
(
0
+
∂
B
2
W
(
λ
m
+
λ
n
)
ϕ
(
u
=
x
,
y
,
t
)+
x
,
y
,
t
)
(6.66)
∂
t
t
+
W
ψ
(
x
,
y
,
t
)+
W
f
τ
(
x
,
y
,
t
−
τ
)
d
τ
(6.67)
0
Proof.
By Theorem 2 in Section 6.1, we only need to prove that the solution of
⎧
⎨
u
t
τ
0
+
B
2
∂
∂
A
2
u
tt
=
Δ
2
u
+
t
Δ
2
u
,
D
×
(
0
,
+
∞
)
,
(6.68)
⎩
L
(
u
,
u
n
)
|
∂
D
=
0
,
u
(
x
,
y
,
0
)=
ϕ
(
x
,
y
)
,
u
t
(
x
,
y
,
0
)=
0
is
1
W
ϕ
(
τ
0
+
∂
B
2
W
u
=
x
,
y
,
t
)+
(
λ
m
+
λ
n
)
ϕ
(
x
,
y
,
t
)
.
(6.69)
∂
t
From the given boundary conditions in PDS (6.64), we may obtain, from Table 2.1,
the complete and orthogonal set of eigenfunctions and their corresponding eigen-
values
,
(
)
,
(
)
.
λ
X
m
x
;
λ
Y
n
y
m
n
Consider now the solution of PDS (6.64) of the form
=
m
,
n
T
mn
(
t
)
X
m
(
x
)
Y
n
(
y
)
.
u
(6.70)
Substituting it into the equation in (6.64) leads to the
T
mn
(
t
)
-equation
1
τ
B
2
T
mn
(
T
mn
(
A
2
T
mn
t
)+
0
+(
λ
+
λ
)
t
)+(
λ
+
λ
)
(
t
)=
0
.
m
n
m
n
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