Environmental Engineering Reference
In-Depth Information
⎧
⎨
a
2
u
tt
=
Δ
u
,
Ω
×
(
0
,
+
∞
)
,
L
u
∂ Ω
=
,
∂
u
0
,
(2.2)
⎩
∂
n
(
,
)=
,
(
,
)=
ψ
(
)
,
u
M
0
0
u
t
M
0
M
⎧
⎨
⎩
a
2
u
tt
=
Δ
u
+
f
(
M
,
t
)
,
Ω
×
(
0
,
+
∞
)
,
L
u
∂ Ω
=
,
∂
u
,
0
(2.3)
∂
n
(
,
)=
,
(
,
)=
,
u
M
0
0
u
t
M
0
0
⎧
⎨
a
2
=
+
(
,
)
,
Ω
×
(
,
+
∞
)
,
u
tt
Δ
u
f
M
t
0
∂ Ω
=
L
u
,
∂
u
0
,
(2.4)
⎩
∂
n
u
(
M
,
0
)=
ϕ
(
M
)
,
u
t
(
M
,
0
)=
ψ
(
M
)
,
(
,
)
(
,
,
)
where
M
represents the point
x
,
in one-, two- and three-dimensional
space respectively. For the one-dimensional case,
x
y
and
x
y
z
Δ
u
is defined as
u
xx
.
Theorem.
Suppose that
u
2
=
W
ψ
(
M
,
t
)
is the solution of (2.2), then
1.
u
1
=
∂
∂
t
W
ϕ
(
M
,
t
)
is the solution of (2.1).
t
2.
u
3
=
W
f
τ
(
M
,
t
−
τ
)
d
τ
is the solution of (2.3) with
f
τ
=
f
(
M
,
τ
)
.
0
t
u
3
=
∂
∂
3.
u
=
u
1
+
u
2
+
t
W
ϕ
+
W
ψ
+
W
f
τ
(
M
,
t
−
τ
)
d
τ
is the solution of (2.4).
0
Proof.
1. Since
W
ϕ
(
M
,
t
)
satisfies
⎧
⎨
2
W
ϕ
∂
∂
a
2
=
Δ
W
ϕ
,
Ω
×
(
0
,
+
∞
)
,
t
2
∂ Ω
=
L
W
ϕ
,
∂
W
ϕ
∂
0
,
⎩
n
,
∂
∂
W
ϕ
(
M
,
0
)=
0
t
W
ϕ
(
M
,
0
)=
ϕ
(
M
)
,
thus
∂
2
u
1
∂
2
2
W
ϕ
∂
∂
u
1
=
∂
t
2
∂
W
ϕ
∂
Δ
∂
W
ϕ
∂
t
=
∂
a
2
a
2
a
2
t
2
−
Δ
t
−
−
Δ
W
ϕ
=
0
.
t
2
∂
∂
t
Therefore,
u
1
=
∂
∂
t
W
ϕ
(
M
,
t
)
satisfies the equation of (2.1).
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