We want to prove the following theorem:

*If *
* is a displacement field whose
divergence *
*
has the following properties:
*

then

We note that the static displacement field of a moment-tensor
source in a homogeneous isotropic full space has the
required properties, as can be analytically derived from the
explicit solution (the use of computer algebra is recommended).
For a LVD source *q* is given by eq. (6). It is a multiple of
the classical quadrupole potential,
,
and therefore a solution of
.
For our present purpose
we need the theorem only for point sources, but may as well
prove it for distributed sources at the same time.

The proof of the theorem is as follows:

Under the above conditions, *q* can be expanded into
a series of spherical harmonics associated with negative
powers of the radius:

(7) |

The origin may be chosen arbitrarily. The expansion will converge
outside the smallest sphere around the origin that encloses the
source. **The condition that spherical surfaces enclose
the source region will from now on be understood.**
The
and
terms are absent for any choice of the
origin due to our assumption on the asymptotic behaviour of *q*
for large radii. The integral or average of *q* over any spherical
surface, which is related to the term of the expansion, is therefore zero. Generally, any solution
of the Laplace equation that decays faster than *r*^{-1} for
must have zero average over any spherical
surface enclosing the source. The same result can be derived without
invoking the series expansion by setting up Green's formula

(8) |

with

Consequently, if we integrate *q* over a spherical shell of any
thickness, the result must be zero.

Since *q* is the divergence of ,
this implies that displaces the same total volume through the inner and outer surfaces
of the shell. The volume displacement through all spherical surfaces
having a common center is thus the same.

We now consider two such sets of surfaces. We select surface *S*_{1}from the first set and *S*_{2} from the second set so that they have
the same radius *r*. The surfaces intersect because they both
enclose the source. Let the volumes displaced by
through
these surfaces be *D*_{1} and *D*_{2}. The difference *D*_{2} - *D*_{1}may be calculated by integrating *q* over the differential volume
enclosed between *S*_{1} and *S*_{2} (one of the two bowl-shaped
halves of this volume enters with a negative sign). By making *r*
sufficiently large, *D*_{2} - *D*_{1} can be made arbitrarily small because
the volume of integration increases with *r*^{2} while the integrand
tends uniformly to zero faster than *r*^{-2}. On the other hand,
the displaced volume is constant within each set, so *D*_{1} must
equal *D*_{2} for any value of *r*. The displaced volume is thus
the same for all surfaces of both sets. This completes our proof.

2003-05-30