Environmental Engineering Reference
In-Depth Information
0
Since
sin
m
is orthogonal in
,
m
=
1
,
π
x
[
,
]
=
0
a
,wehave
b
m
a
100
/
sh
π
,
m
=
1
.
Finally, we obtain the solution
100
sh
sh
π
y
a
sin
π
x
a
.
T
(
x
,
y
)=
π
Example 4.
Solve
⎧
⎨
Δ
u
=
0
,
0
<
x
<
a
,
0
<
y
<
+
∞
,
A
1
x
a
(7.9)
u
(
x
,
0
)=
−
,
u
(
x
,
+
∞
)=
0
,
⎩
u
(
0
,
y
)=
u
(
a
,
y
)=
0
.
Solution.
The
u
(
x
,
y
)
that satisfies
u
(
0
,
y
)=
u
(
a
,
y
)=
0 can be written as
m
=
1
a
m
e
m
π
y
a
sin
m
∞
x
a
.
π
m
π
y
b
m
e
−
u
(
x
,
y
)=
+
a
Applying
u
(
x
,
+
∞
)=
0 yields
a
m
=
0. Substituting the
u
(
x
,
y
)
into the equation in
PDS (7.9) leads to
A
1
a
∞
m
=
1
b
m
sin
m
π
x
x
=
−
,
a
so that the
b
m
can be determined by the integration by parts,
a
A
1
sin
m
2
a
x
a
π
x
2
A
m
b
m
=
−
d
x
=
π
.
a
0
Finally, the solution of PDS (7.9) is
m
π
y
a
∞
m
=
1
e
−
2
A
π
sin
m
x
a
.
π
u
(
x
,
y
)=
(7.10)
m
m
π
y
a
The presence of e
−
in the general term of
u
(
x
,
y
)
indicates that the series general
term decays quickly as
m
→
∞
. In calculation, we can take first a few terms as an
approximation of
u
(
x
,
y
)
. If the first three terms are taken, for example, we have
e
−
π
a
sin
π
2
π
y
3
π
y
e
−
e
−
2
A
π
x
a
+
sin
2
π
x
sin
3
π
x
a
a
(
,
)
≈
u
x
y
+
.
2
a
3
a
Remark.
Examples 1-4 show applications of eigenvalues and eigenfunctions in
Row 1 in Table 2.1. We can follow a similar approach to find solutions subject
to the other boundary conditions simply by using corresponding eigenvalues and
eigenfunctions.
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