Environmental Engineering Reference
In-Depth Information
u
y
)
∂Ω
=
Example 4.
Let the boundary conditions
L
(
u
,
u
x
,
0 in PDS (4.37) be
u
x
(
0
,
y
,
t
)=
u
x
(
l
,
y
,
t
)=
0
,
(4.47)
u
y
(
x
,
0
,
t
)=
u
y
(
x
,
l
,
t
)=
0
.
Find the Green function and write out its first four terms.
Solution
. Based the given boundary conditions (4.47), we use the eigenfunctions in
Row 5 in Table 2.1 to expand
G
both in
x
and
y
,
cos
m
π
x
cos
n
π
y
u
55
(
x
,
y
)=
.
l
l
Therefore,
+
∞
m
,
n
=
0
τ
0
γ
mn
M
mn
cos
m
1
πξ
l
cos
m
π
x
cos
n
πη
l
cos
n
π
y
t
−
τ
2τ
0
e
−
G
=
sin
γ
mn
(
t
−
τ
)
,
l
l
4
τ
0
a
2
m
2
2
n
l
2
4
,
1
2
l
l
where
M
mn
=
M
m
M
n
=
γ
mn
=
+
−
1.
τ
0
The first four terms are
t
−
τ
2τ
0
t
−
τ
2τ
0
l
2
1
τ
0
e
−
1
τ
0
2
τ
0
i
4
l
2
i
e
−
4
t
−
τ
2
t
−
τ
e
−
e
−
G
00
=
τ
0
=
−
,
2
γ
01
l
2
cos
π
l
4
cos
π
y
l
t
−
τ
2
e
−
G
01
=
τ
0
sin
γ
01
(
t
−
τ
)
,
τ
0
τ
0
γ
10
l
2
cos
π
l
4
cos
π
x
t
−
τ
2τ
0
e
−
=
(
−
τ
)
,
G
10
sin
γ
t
10
l
τ
0
γ
11
l
2
cos
π
l
4
cos
π
x
cos
π
l
cos
π
y
l
t
−
τ
2τ
0
e
−
G
11
=
sin
γ
11
(
t
−
τ
)
.
l
It is always useful to perform a unit analysis. Here
τ
0
a
2
=
L
2
and
T
−
1
so
[
γ
mn
]=
L
−
2
.The
u
due to the nonhomogeneous term
f
[
G
]=
(
x
,
y
,
t
)
reads
t
u
=
d
τ
Gf
(
ξ
,
η
,
τ
)
d
σ
,
0
D
whose unit is
[
u
]=[
d
τ
][
G
][
f
][
d
σ
]=
Θ
.
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