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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,

\nabla \cdot \left( \begin{array}{c}
{\mathbf r} a^* b \\ a^...
...cdot ({\mathbf r}\cdot\nabla) \nabla a^*
\end{array} \right) .
\end{displaymath} (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 $(k_a^2 - k_b^2)^2$, implying that it is invertible for $k_a \neq k_b$. The relation (H.10) can be written
\nabla \cdot {\mathbf v}_\alpha \; = \; {\mathcal{M}}_{\alpha \beta} \,s_\beta ,
\end{displaymath} (H.11)

where the sum over $\beta$ is implied. Care should be taken not to confuse spatial coordinates (boldface vectors) with coefficient indices $\alpha,\beta = 1\cdots6$ (Greek letters). Integrating over the domain ${\mathcal{D}}$, applying Gauss' theorem and premultiplying by the inverse of $\mathcal{M}$ gives
\int_{\mathcal{D}} \!\! d{\mathbf r} \, s_\alpha \; = \;
...Gamma \!\! d{\mathbf s} \,{\mathbf n}\cdot {\mathbf v}_\beta .
\end{displaymath} (H.12)

Thus we have closed-form 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),

{\mathcal{M}}^{-1}\; \; = \; \;
\frac{1}{\varepsilon } \left...
...\varepsilon & k_a^2 &
-k_a^2 & -k_b^2\\
\end{array} \right) ,
\end{displaymath} (H.13)

where $\varepsilon \equiv k_a^2 - k_b^2$ is the `energy difference'. Rows of this matrix give the desired expressions for volume integrals. For instance, (H.5) results from row 1.

When $k_a = k_b = k$ rows 2 and 3 of $\mathcal{M}$ 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 ${\mathbf \xi}$ such that ${\mathcal{M}}^T {\mathbf \xi} = {\mathbf e}$ for a desired unit vector ${\mathbf e}$. This is only possible if the unit vector lies entirely within the row space of $M$ [188]. For instance, the first unit vector $e_\alpha = \delta_{\alpha 1}$ gives ${\mathbf \xi} = (k^2, d/2, d/2, -1, 1, 1)^T$, corresponding to

$\displaystyle \left\langle a \left\vert b\right.\right\rangle _{\mathcal{D}} \;$ $\textstyle =\;$ $\displaystyle -\frac{1}{2k^2} \oint_\Gamma \!\! d{\mathbf s} \,
\left[ \mbox{\s...
...\mathbf r})
( k^2 a^* b - \nabla a^* \cdot \nabla b )
\rule{0in}{0.2in} \right.$  
    $\displaystyle \hspace{1in} \left. \rule{0in}{0.2in}
+ \; a^* {\mathbf n}\cdot({\mathbf r}\cdot\nabla) \nabla b
+ a^*\!{\leftrightarrow}b\right] .$ (H.14)

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 up previous
Next: Volume integrals of vector Up: Matrix trick for pushing Previous: Matrix trick for pushing
Alex Barnett 2001-10-03