Environmental Engineering Reference
In-Depth Information
z
=
x
+
i
y
in
D
and must satisfy the Cauchy-Riemann equation
u
x
=
v
y
,
u
y
=
−
v
x
(7.82)
Thus
Δ
u
=
0,
Δ
v
=
0, so both
u
and
v
are harmonic functions.
v
(
x
,
y
)
is called the
conjugate harmonic function of
u
(
x
,
y
)
. Note that
u
(
x
,
y
)
is not necessarily the con-
jugate harmonic function of
v
(
x
,
y
)
.Once
u
is available, we can use Eq. (7.82) to find
the harmonic function
v
i
v
.
This can help us to find the Poisson formula of internal problems of two-dimensional
harmonic equations.
(
x
,
y
)
and consequently the analytical function
f
(
z
)=
u
+
Example
. Find the solution of
Δ
u
(
r
,
θ
)=
0
,
0
<
r
<
R
,
(7.83)
u
(
R
,
θ
)=
f
(
θ
)
.
Solution.
Let
v
be the conjugate harmonic function of
u
,then
g
(
z
)=
u
+
i
v
is an
analytical function. By the Cauchy integral formula we have
1
2
(
ξ
)
ξ
−
g
g
(
z
)=
z
d
ξ
,
π
i
|
ξ
|
=
R
1
2
(
ξ
)
ξ
−
g
0
=
d
ξ
,
z
π
i
|
ξ
|
=
R
R
2
r
r
e
iθ
R
.
z
=
e
iθ
=
<
<
where
z
is any point in a circle 0
r
is the symmetric point
R
e
iθ
. A subtraction
z
|
=
R
2
.Let
of
z
with respect to the circle
r
=
R
so that
|
z
||
ξ
=
of the latter from the former yields
z
1
z
−
g
(
z
)=
u
+
i
v
=
g
(
ξ
)
d
ξ
z
)
2
π
i
(
ξ
−
z
)(
ξ
−
|
ξ
|
=
R
2π
R
2
r
r
e
iθ
−
e
iθ
1
R
ie
iθ
d
θ
=
(
+
)
u
i
v
2
π
i
R
2
r
R
e
iθ
r
e
iθ
)(
R
e
iθ
e
iθ
)
(
−
−
0
2π
R
r
2
R
2
r
(
−
)
1
2
θ
=
(
u
+
i
v
)
d
π
R
2
r
R
e
i
(
θ
−
θ
)
−
e
i
(
θ
−
θ
)
)
(
r
)(
R
−
0
2π
r
2
R
2
1
2
−
θ
=
(
u
+
i
v
)
d
π
Rr
e
i
(
θ
−
θ
)
−
R
r
e
i
(
θ
−
θ
)
)
(
r
2
)(
1
−
0
2π
R
2
r
2
1
2
−
θ
.
=
(
u
+
i
v
)
d
R
2
r
2
(
θ
−
θ
)
π
+
−
2
Rr
cos
0
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