Environmental Engineering Reference
In-Depth Information
Example
Find the solution of
⎧
⎨
∂
∂
a
2
u
tt
=
x
[
w
(
x
)
u
x
]+
f
(
x
,
t
)
,
(
0
,
l
)
×
(
0
,
+
∞
)
,
(2.28)
⎩
u
(
0
,
t
)=
u
(
l
,
t
)=
0
,
u
(
x
,
0
)=
u
t
(
x
,
0
)=
0
.
where
a
>
0 is constant and
w
(
x
)
>
0 is a differentiable function.
Solution
1. Homogenization of the Equation
t
The solution of PDS (2.28) can be expressed by
u
(
x
,
t
)=
v
(
x
,
t
,
τ
)
d
τ
,where
0
v
(
x
,
t
,
τ
)
is the solution of
⎧
⎨
∂
∂
a
2
v
tt
=
x
[
w
(
x
)
v
x
]
,
(
0
,
l
)
×
(
0
,
+
∞
)
,
⎩
(
,
,
τ
)=
(
,
,
τ
)=
,
|
t
=
τ
=
,
|
t
=
τ
=
(
,
τ
)
.
v
0
t
v
l
t
0
v
0
v
t
f
x
Considering the solution of type
v
=
X
(
x
)
T
(
t
)
, we arrive at an eigenvalue problem
d
d
x
[
w
(
x
)
X
x
]+
λ
X
=
0
,
X
(
0
)=
X
(
l
)=
0
,
(2.29)
and
a
2
T
T
tt
+
λ
=
0
,
T
(
τ
)=
0
,
(2.30)
where
is the separation constant.
For a PDS with the
λ
δ
-function as the nonhomogeneous term
⎧
⎨
d
d
x
[
w
(
x
)
G
x
]=
δ
(
x
−
s
)
,
(2.31)
⎩
s
−
,
s
+
,
G
(
0
,
s
)=
G
(
l
,
s
)=
0
,
G
(
s
)=
G
(
s
)
,
the solution is the Green function
a
1
+
b
1
+(
a
2
+
b
2
)
u
,
x
≤
s
,
G
(
x
,
s
)=
a
1
−
b
1
+(
a
2
−
b
2
)
u
,
x
≥
s
.
)=
d
x
w
s
−
,
s
+
,
Here
G
(
s
)=
lim
s
−
0
G
(
x
,
s
)
,
G
(
s
)=
lim
s
+
0
G
(
x
,
s
)
,
u
(
x
. By applying
(
x
)
x
→
x
→
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