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Volume integrals of scalar quantities
By taking the divergence of all the types of terms appearing in relevant
vector identities, the following matrix relation is reached,

(H.10) 
The scalar coefficient matrix is formed by finding
the linear amounts of each of the 6 scalar functions which result
from performing the divergences on each vector term.
Deducing a few of the entries requires Einstein summation form,
and substitution of the Helmholtz equation, but all are simple.
It is convenient that 6 inputs gives 6 outputs since the matrix is
then possibly invertible. Its determinant is found to be
,
implying that it is invertible for .
The relation (H.10) can be written

(H.11) 
where the sum over is implied. Care should be taken not to
confuse spatial coordinates (boldface vectors) with coefficient
indices
(Greek letters).
Integrating over the domain , applying Gauss' theorem and
premultiplying by the inverse of gives

(H.12) 
Thus we have closedform expressions for the volume integrals of any of the
6 scalar functions, in terms of boundary integrals.
That is, we have `pushed' the problem onto the boundary, as desired.
The symbolic inverse can be found
(e.g. using the Mathematica software package),

(H.13) 
where
is the `energy difference'.
Rows of this matrix give the desired expressions for volume integrals.
For instance, (H.5) results from row 1.
When rows 2 and 3 of
become identical, so the matrix is
singular and the above inversion fails.
However, it is still possible to extract solutions to this singular
problem by finding a vector such that
for a desired unit vector .
This is only possible if the unit vector lies entirely
within the row space
of [188].
For instance, the first unit vector
gives
,
corresponding to
This is a manifestly symmetric form of (H.7).
Its derivation is much more routine than that of Boasman[33] because
the matrix method automatically handles the coefficients.
Next: Volume integrals of vector
Up: Matrix trick for pushing
Previous: Matrix trick for pushing
Alex Barnett
20011003