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
 
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