Environmental Engineering Reference
In-Depth Information
Suppose that
u
∗
is a particular solution of the nonhomogeneous equation
Δ
u
=
u
∗
will transform
f
(
x
,
y
)
. A function transformation
u
(
x
,
y
)=
v
(
x
,
y
)+
Δ
u
=
f
(
x
,
y
)
into
Δ
v
=
0.
Example 2.
Solve the internal problem in a circular domain
r
2
sin
Δ
u
(
r
,
θ
)=
−
θ
cos
θ
,
0
<
r
<
a
,
(7.47)
u
(
a
,
θ
)=
0
.
Solution
. In a Cartesian coordinate system, the equation in PDS (7.47) reads
+
=
−
.
u
xx
u
yy
xy
(7.48)
r
4
12
sin
It is clear that
u
∗
=
−
1
12
xy
x
2
y
2
can satisfy Eq. (7.48) so
u
∗
=
−
θ
is a particular solution of the equation in PDS (7.47). Now consider the function
transformation
(
+
)
θ
cos
u
∗
(
(
,
θ
)=
(
,
θ
)+
,
θ
)
.
u
r
v
r
r
The PDS (7.47) is thus transformed into
Δ
v
(
r
,
θ
)=
0
,
0
<
r
<
a
,
(7.49)
a
4
12
sin
v
(
a
,
θ
)=
−
θ
cos
θ
.
The solution of PDS (7.47) thus follows from the Poisson formula of internal prob-
lems in a circular domain (Eq. (7.17))
2π
r
4
24
sin2
a
4
48
a
2
r
2
θ
(
−
)
sin2
θ
.
u
=
−
θ
+
r
2
d
a
2
−
2
ar
cos
(
θ
−
θ
)+
0
7.3.2 Extremum Principle
Laplace equations can be used to describe the steady temperature field in a do-
main
without any heat source or sink inside. Therefore they satisfy the extremum
principle of heat conduction discussed in Section 3.3.2; their solutions take the
minimum and the maximum values only on the boundary
Ω
∂Ω
of the domain. If
¯
¯
u
|
∂Ω
=
C
(constant) and
u
(
M
)
is continuous on
Ω
(
Ω
=
Ω
∪
∂Ω
)
, the principle
¯
concludes that
u
(
M
)=
C
for all
M
∈
Ω
.
Example 1.
Solve
Δ
x
2
y
2
u
=
1
,
+
<
1
,
(7.50)
u
|
x
2
=
4
.
y
2
+
=
1
Solution.
It is clear that the Poisson equation in PDS (7.50) has a particular solution
1
4
(
u
∗
=
x
2
y
2
+
)
.
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