Environmental Engineering Reference
In-Depth Information
Therefore,
t
=
0
=
t
=
0
+
∂
t
=
0
∂
W
ϕ
t
=
0
+
∂
u
W
ϕ
∂
W
ϕ
1
∂
A
2
B
2
Δ
Δ
∂
t
t
t
B
2
B
2
B
2
=
Δϕ
+
ϕ
1
(
M
)=
Δϕ
−
Δϕ
=
0
.
Theorem 2.
Let
u
=
W
ψ
(
M
,
t
)
be the solution of the well-posed PDS
⎧
⎨
u
t
τ
0
+
B
2
∂
∂
A
2
u
tt
=
Δ
u
+
t
Δ
u
,
Ω
×
(
0
,
+
∞
)
,
(6.8)
⎩
L
(
u
,
u
n
)
|
∂Ω
=
0
,
u
(
M
,
0
)=
0
,
u
t
(
M
,
0
)=
ψ
(
M
)
.
The solution of the well-posed PDS
⎧
⎨
u
t
τ
B
2
∂
∂
A
2
0
+
u
tt
=
Δ
u
+
t
Δ
u
+
f
(
M
,
t
)
,
Ω
×
(
0
,
+
∞
)
,
(6.9)
⎩
L
(
u
,
u
n
)
|
∂Ω
=
0
,
(
,
)=
,
(
,
)=
u
M
0
0
u
t
M
0
0
is
t
u
=
W
f
τ
(
M
,
t
−
τ
)
d
τ
,
(6.10)
0
where
f
τ
=
f
(
M
,
τ
)
.
Proof
. By the definition of
W
f
τ
(
M
,
t
−
τ
)
,wehave
⎧
⎨
2
W
f
τ
∂
τ
0
∂
1
W
f
τ
∂
+
∂
A
2
B
2
∂
t
Δ
−
Δ
W
f
τ
−
W
f
τ
=
0
,
t
t
2
L
W
f
τ
,
∂
∂Ω
=
W
f
τ
∂
,
(6.11)
0
⎩
n
t
=
τ
=
t
=
τ
=
∂
∂
W
f
τ
(
M
,
t
−
τ
)
0
,
t
W
f
τ
(
M
,
t
−
τ
)
f
(
M
,
τ
)
.
Satisfaction of Equation
By substituting Eq. (6.10) into the equation of PDS (6.9) and applying Eq. (6.11),
we obtain
u
t
τ
0
+
B
2
∂
∂
A
2
−
−
u
tt
Δ
u
t
Δ
u
Search WWH ::
Custom Search