Environmental Engineering Reference
In-Depth Information
The solution (5.32) reduces to
I 0 b A 2
2 f
t
x + A ( t τ )
1
2 A
t τ
2
e
2
u
=
τ 0 d
τ
(
t
τ )
(
x
ξ )
( ξ , τ )
d
ξ .
(5.33)
0
x
A
(
t
τ )
Remark 3. Verification of CDS.
1. Verify that Eq. (5.30) satisfies u
(
x
,
0
)= ϕ (
x
)
and u t (
x
,
0
)= ψ (
x
)
.
Equation (5.30) clearly satisfies u
by integral properties. Taking the
derivative of Eq. (5.30) with respect to t and using
(
x
,
0
)= ϕ (
x
)
I 1 (
x
)
1
2 ,
lim
x
=
x
0
ξ = x ± At =
tI 1
tI 1
0 b
0 b
lim
ξ
x
±
At
2
2
2
2
4
τ
(
At
)
(
x
ξ )
4
τ
(
At
)
(
x
ξ )
tI 1 ( ε )
4
t
=
lim
ε
0 ε =
0 ,
2
2
τ
8
τ
0
we obtain
1
1
2
1
2 A 2 A
u t
(
x
,
0
)=
0 ϕ (
x
)+
0 ϕ (
x
)+
ψ (
x
)= ψ (
x
) .
2
τ
τ
2. Verify that Eq. (5.32) satisfies u
(
x
,
0
)=
0and u t
(
x
,
0
)=
0. Clearly, u
(
x
,
0
)=
0.
Also,
t
e
τ 0 x + A ( t τ )
x
1
2 A
t
τ
2
u t =
τ 0
t
0
A
(
t
τ )
I 0 b A 2
2 f
2
(
t
τ )
(
x
ξ )
( ξ , τ )
d
ξ
d
τ
I 0 b
2 f
1
x
+
(
x
ξ )
( ξ , τ )
d
ξ ·
.
x
Therefore u t (
x
,
0
)=
0.
 
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