Geoscience Reference
In-Depth Information
We can rearrange this as
2
π
π
V
e
(
r, ϑ, λ
)=
1
4
π
V
(
R, ϑ
,λ
)
·
λ
=0
ϑ
=0
∞
P
n
(cos
ψ
)
sin
ϑ
dϑ
dλ
.
(1-119)
(2
n
+1)
R
r
n
+1
n
=0
The sum in the brackets can be evaluated. We denote the spatial distance
between the points
P
(
r, ϑ, λ
)and
P
(
R, ϑ
,λ
)by
l
. Then, using (1-104),
R
r
n
+1
∞
1
l
1
1
R
r
2
+
R
2
=
=
P
n
(cos
ψ
)
(1-120)
−
2
Rr
cos
ψ
n
=0
results. Differentiating with respect to
r
,weget
∞
(
n
+1)
R
n
+1
r
n
+2
r
−
R
cos
ψ
l
3
1
R
−
=
−
P
n
(cos
ψ
)
.
(1-121)
n
=0
Multiplying this equation by
−
2
Rr
, multiplying the expression for 1
/l
by
−
R
, and then adding the two equations yields
(2
n
+1)
R
r
n
+1
=
∞
R
(
r
2
R
2
)
−
P
n
(cos
ψ
)
.
(1-122)
l
3
n
=0
The right-hand side is the bracketed expression in (1-119). Substituting the
left-hand side, we finally obtain
2
π
π
V
e
(
r, ϑ, λ
)=
R
(
r
2
R
2
)
V
(
R, ϑ
,λ
)
l
3
−
sin
ϑ
dϑ
dλ
,
(1-123)
4
π
λ
=0
ϑ
=0
where
l
=
r
2
+
R
2
2
Rr
cos
ψ.
(1-124)
This is
Poisson's integral
. It is an explicit solution of Dirichlet's problem for
the exterior of the sphere, which has many applications in physical geodesy.
−
1.13
Other boundary-value problems
There are other similar boundary-value problems. In
Neumann's problem
,or
the
second boundary-value problem of potential theory
, the normal derivative
∂V/∂n
is given on the surface
S
, instead of the function
V
itself. The normal
derivative is the derivative along the outward-directed surface normal
n
to
