Geography Reference
In-Depth Information
Dividing (8.53) by (u
−
c) and separating the resulting equation into real and
imaginary parts we obtain
ερ
0
∂
r
∂z
∗
k
2
δ
r
(∂q/∂y)
r
−
∂
2
r
∂y
2
1
ρ
0
∂
∂z
∗
+
−
−
δ
i
∂q/∂y
i
=
0 (8.55)
ερ
0
∂
i
∂z
∗
k
2
δ
r
(∂q/∂y)
i
+
∂
2
i
∂y
2
1
ρ
0
∂
∂z
∗
+
−
−
δ
i
∂q/∂y
r
=
0 (8.56)
where
−
u
c
r
c
i
δ
r
=
δ
i
=
,
and
c
r
)
2
c
i
c
r
)
2
c
i
−
+
−
+
(u
(u
Similarly, dividing (8.54) through by (
c) and separating into real and imaginary
parts gives for the boundary condition at z
∗
=
u
¯
−
0:
∂
r
∂z
∗
+
∂u
∂z
∗
∂
i
∂z
∗
−
∂u
∂z
∗
(δ
i
i
−
=
;
(δ
r
i
+
=
δ
r
r
)
0
δ
i
r
)
0
(8.57)
Multiplying (8.55) by
i
, (8.56) by
r
, and subtracting the latter from the former
yields
ρ
0
i
i
ερ
0
ερ
0
∂
2
r
∂y
2
∂
2
i
∂y
2
∂
∂z
∗
∂
r
∂z
∗
∂
∂z
∗
∂
i
∂z
∗
−
r
+
−
r
ρ
0
δ
i
(∂q/∂y)
i
r
−
+
=
0
(8.58)
Using the chain rule of differentiation, (8.58) can be expressed in the form
i
∂
r
ερ
0
i
∂
r
∂
∂y
r
∂
i
∂y
∂
∂z
∗
r
∂
i
∂z
∗
ρ
0
∂y
−
+
∂z
∗
−
ρ
0
δ
i
(∂q/∂y)
i
r
−
+
=
0
(8.59)
The first term in brackets in (8.59) is a perfect differential in y; the second term
is a perfect differential in z
∗
. Thus, if (8.59) is integrated over the y, z
∗
domain
the result can be expressed as
i
+
L
ερ
0
i
∞
∞
+
L
∂
r
∂y
−
∂
i
∂y
∂
r
∂z
∗
−
∂
i
∂z
∗
ρ
0
dz
∗
+
r
r
dy
−
L
0
0
−
L
+
L
∞
i
r
dydz
∗
∂q
∂y
=
ρ
0
δ
i
+
(8.60)
−
L
0
