Environmental Engineering Reference
In-Depth Information
Remark 1.
The boundary conditions in Examples 1-3 are all of the first kind. For
Neumann problems with boundary conditions of the second kind, there exist some
constraints on the boundary values to ensure the existence of solutions. Take the
two-dimensional case as an example. By the Gauss formula,
C
∂
u
Δ
u
d
σ
=
∇
·
∇
u
d
s
=
C
∇
u
·
n
d
s
=
n
d
s
,
∂
D
D
where
C
is the positive-directed boundary curve of plane domain
D
and
n
is the ex-
ternal unit normal of
C
. This shows that
C
∂
u
n
d
s
must vanish when
u
is a solution of
the Laplace equation. Thus the boundary value
u
n
∂
|
C
=
(
)
in a Neumann problem
must satisfy the necessary condition for the existence of solution
g
x
g
(
x
)
d
s
=
0
.
C
In a circular domain, it reduces to
a
r
=
a
∂
u
r
d
s
=
f
(
θ
)
d
θ
=
0
.
∂
r
=
a
Example 4
. Find the solution of the Neumann problem in a circular domain
⎧
⎨
r
∂
2
u
∂θ
1
r
∂
∂
u
r
2
∂
1
Δ
u
(
r
,
θ
)=
+
2
=
0
,<
r
<
a
,
0
<
θ
<
2
π
,
r
∂
r
r
=
a
=
(7.29)
⎩
∂
u
(
θ
)
, |
(
,
θ
)
| <
∞
,
(
,
θ
+
π
)=
(
,
θ
)
.
f
u
0
u
r
2
u
r
∂
r
Solution.
By a separation of variables similar to that in Example 1, we obtain
∞
n
=
1
r
n
u
(
r
,
θ
)=
A
0
+
(
a
n
cos
n
θ
+
b
n
sin
n
θ
)
.
(7.30)
Thus
∞
n
=
1
na
n
−
1
u
r
(
a
,
θ
)=
(
a
n
cos
n
θ
+
b
n
sin
n
θ
)=
f
(
θ
)
.
Using Fourier coefficients, we have
2π
2π
1
na
n
−
1
1
na
n
−
1
a
n
=
f
(
θ
)
cos
n
θ
d
θ
,
b
n
=
f
(
θ
)
sin
n
θ
d
θ
.
π
π
0
0
Since
A
0
can take any value, the solution of the Neumann problem is not unique. It
can be proven that subject to the necessary condition for existence, the solution is
unique up to an arbitrary constant term.
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