Geoscience Reference
In-Depth Information
Hence, the series (1-110) converges for
r
1, and the series (1-111) con-
verges for
r ≥
1; furthermore, both series have been found to represent
harmonic functions. Therefore, we see that Dirichlet's problem is solved by
V
i
(
r, ϑ, λ
) for the interior of the sphere
r
=1,andby
V
e
(
r, ϑ, λ
)foritsexte-
rior.
For a sphere of arbitrary radius
r
=
R
, the solution is similar. We expand
the given function
≤
V
(
R, ϑ, λ
)=
∞
Y
n
(
ϑ, λ
)
.
(1-114)
n
=0
The surface spherical harmonics
Y
n
are determined by
Y
n
(
ϑ, λ
)=
2
n
+1
4
π
2
π
π
V
(
R, ϑ
,λ
)
P
n
(cos
ψ
)sin
ϑ
dϑ
dλ
.
(1-115)
λ
=0
ϑ
=0
Then the series
r
R
n
Y
n
(
ϑ, λ
)
V
i
(
r, ϑ, λ
)=
∞
(1-116)
n
=0
solves the first boundary-value problem for the interior, and the series
V
e
(
r, ϑ, λ
)=
∞
R
r
n
+1
Y
n
(
ϑ, λ
)
(1-117)
n
=0
solves it for the exterior of the sphere
r
=
R
.
Thus, we see that Dirichlet's problem can always be solved for the sphere.
It is evident that this is closely related to the possibility of expanding an
arbitrary
function on the sphere into a series of
surface
spherical harmonics
and a
harmonic
function in space into a series of
solid
spherical harmonics.
Dirichlet's boundary-value problem can be solved not only for the sphere
but also for any suciently smooth boundary surface. An example is given
in Sect. 1.16.
The solvability of Dirichlet's problem is also essential to Molodensky's
problem (Sect. 8.3). See also Kellogg (1929: Chap. XI).
Poisson's integral
A more direct solution is obtained as follows. We consider only the exterior
problem, which is of greater interest in geodesy. Substituting
Y
n
(
ϑ, λ
)from
(1-89) into (1-117), we obtain
V
e
(
r, ϑ, λ
)=
∞
R
r
n
+1
2
n
+1
4
π
2
π
π
V
(
R, ϑ
,λ
)
P
n
(cos
ψ
)sin
ϑ
dϑ
dλ
.
λ
=0
ϑ
=0
n
=0
(1-118)
