Environmental Engineering Reference
In-Depth Information
Boundary Condition of the Second Kind and the Third Kind
Example 2.
Find the solution of
⎧
⎨
a
2
u
xx
u
t
=
,
0
<
x
<
+
∞
,
0
<
t
,
(
u
x
−
hu
)
|
x
=
0
=
0
,
h
>
0
,
(3.37)
⎩
u
(
x
,
0
)=
ϕ
(
x
)
.
Solution.
Let
v
(
x
)
be the initial distribution in
(
−
∞
,
0
)
, which is the continuation of
ϕ
(
x
)
. The solution of PDS (3.37) is thus, by Eq. (3.27),
0
+
∞
u
=
v
(
ξ
)
V
(
x
,
ξ
,
t
)
d
ξ
+
ϕ
(
ξ
)
V
(
x
,
ξ
,
t
)
d
ξ
−
∞
0
+
∞
=
[
ϕ
(
ξ
)
(
,
ξ
,
)+
(
−
ξ
)
(
,−
ξ
,
)]
V
x
t
v
V
x
t
d
ξ
0
d
+
∞
e
−
(
x
−
ξ
)
2
e
−
(
x
+
ξ
)
2
1
2
a
√
π
=
ϕ
(
ξ
)
+
v
(
−
ξ
)
ξ
.
4
a
2
t
4
a
2
t
t
0
Applying the boundary condition leads to
4
a
2
t
ξϕ
(
ξ
)
−
ξ
d
+
∞
t
e
−
ξ
2
1
2
a
√
π
v
(
−
ξ
)
(
u
x
−
hu
)
|
x
=
0
=
−
h
ϕ
(
ξ
)
−
hv
(
−
ξ
)
ξ
=
0
,
2
a
2
t
0
which can be satisfied by
2
a
2
ht
ξϕ
(
ξ
)
−
ξ
v
(
−
ξ
)
(
−
ξ
)=
ξ
−
−
h
ϕ
(
ξ
)
−
hv
(
−
ξ
)=
0or
v
2
a
2
ht
ϕ
(
ξ
)
.
2
a
2
t
ξ
+
Therefore, the solution of PDS (3.37) becomes
u
=
W
ϕ
(
x
,
t
)
V
d
+
∞
2
a
2
ht
)+
ξ
−
=
ϕ
(
ξ
)
(
,
ξ
,
(
,−
ξ
,
)
ξ
.
x
t
2
a
2
ht
V
x
t
(3.38)
ξ
+
0
Remark 2.
Consider
⎧
⎨
a
2
u
xx
u
t
=
+
f
(
x
,
t
)
,
0
<
x
<
+
∞
,
0
<
t
,
(
u
x
−
hu
)
|
x
=
0
=
0
,
(3.39)
⎩
u
(
x
,
0
)=
0
.
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