Environmental Engineering Reference
In-Depth Information
Remark 1.
For the unit analysis of PDS (3.33) and (3.34), note that
1
L
=
Θ
,
[
δ
(
x
−
ξ
)] = [
1
·
δ
(
x
−
ξ
)] =
L
Θ
·
1
T
=
T
,
[
δ
(
x
−
ξ
,
t
−
τ
)] = [
1
·
δ
(
x
−
ξ
,
t
−
τ
)] =
L
Θ
·
L
·
1
L
=
Θ
.
[
V
]=[
1
·
V
]=
L
Θ
·
satisfies
the conditions for a Fourier transformation. Therefore, Eq. (3.27) is only a nominal
solution. Here we show that it is indeed the solution of PDS (3.25) at
f
Remark 2.
In developing Eq. (3.27) we did not consider whether
ϕ
(
x
)
=
0 [due to
ϕ
(
x
and is bounded.
By taking the derivative of the fundamental solution (3.29), we obtain
)
]if
ϕ
(
x
)
∈
C
(
−
∞
,
+
∞
)
a
2
V
xx
.
V
t
=
(3.35)
Note that the integrand in Eq. (3.27) contains a decaying factor of exponential order,
which also exists in all derivatives of
u
with respect to
x
or
t
. This ensures the
uniform convergence of integrals after differential operations inside the integration.
Therefore, the integral in Eq. (3.27) satisfies
+
∞
V
t
a
2
V
xx
ϕ
(
ξ
)
a
2
u
xx
−
=
−
ξ
=
,
u
t
d
0
−
∞
where we have used Eq. (3.35). This shows that Eq. (3.27) is a solution of the ho-
mogeneous equation in PDS (3.25) (
f
0).
Also, the integral in Eq. (3.27) is uniformly convergent with respect to
t
. We can
take the limit inside the integration,
=
+
∞
+
∞
2
a
√
t
1
√
π
2
e
−
η
(
,
)=
(
,
ξ
,
)
ϕ
(
ξ
)
ξ
=
ϕ
(
+
η
)
lim
t
→
+
0
u
x
t
lim
t
→
+
0
V
x
t
d
x
d
η
−
∞
−
∞
+
∞
1
√
π
2
e
−
η
=
ϕ
(
x
)
d
η
=
ϕ
(
x
)
.
−
∞
Therefore, Eq. (3.27) also satisfies the initial condition
u
(
x
,
0
)=
ϕ
(
x
)
.
Remark 3.
Equation (3.35) shows that
V
satisfies the equation of PDS (3.30). Here
we show that it also satisfies the initial condition. Note that
t
=
0 means
t
→
+
0. By
Eq. (3.29),
∞
,
x
=
ξ
,
V
|
t
=
0
=
0
,
x
=
ξ
.
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