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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