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Definition of the billiard problem

Given the isotropic quadratic dispersion relation corresponding to (5.2), we can choose energy units such that $E = (\hbar k)^2/2m$. If the dispersion relation is not isotropic, it can be made so by a re-parametrization of ${\mathbf r}$. However, note that $E$ will not play any further role. The numerical methods described in this thesis are really about finding the eigenwavenumbers $k_\mu$. Therefore the methods are entirely applicable to any other Helmholtz eigenproblem regardless of the dispersion relation, or indeed the existence of an `energy' (for instance $E$ is physically irrelevant in acoustic problems). The only requirement is that the wavenumber be constant (and isotropic) in the interior.

The billiard has $d$-dimensional `volume' $\mathcal{D}$ and $d{-}1$ dimensional `surface area' ${\mathsf{A}}$, giving a typical length scale ${\mathsf{L}}= {\mathsf{V}}/{\mathsf{A}}$. Our eigenproblem can be written

$\displaystyle (\nabla^2 + k^2) \psi{({\mathbf r})}$ $\textstyle =$ $\displaystyle 0
\hspace{.5in} \mbox{for all $r \in \mathcal{D}$}
\hspace{.5in} \mbox{(Helmholtz equation)}$ (5.5)
$\displaystyle f[ \psi ]$ $\textstyle =$ $\displaystyle 0
\hspace{1.5in} \mbox{(boundary conditions)}$ (5.6)
$\displaystyle g[ \psi ]$ $\textstyle =$ $\displaystyle 1
\hspace{1.5in} \mbox{(normalization condition)}$ (5.7)

where the eigenvalue is $k^2$. The functionals $f$ and $g$ return a scalar, and are quadratic in the wavefunction $\psi$.

The BCs have been incorporated as (5.6) rather than the linear condition (5.3) because satisfaction of the BCs by a wavefunction $\psi$ is then revealed by a single number $f[\psi]$. This number measures the 2-norm of some error function, and is therefore a non-negative quantity. The error function (e.g. (5.3)) gives the amount by which the desired BCs fail to be obeyed. Heller[91] named $f$ `tension', and I shall follow suit. The definition of $f$ is

f[\psi] \; \equiv \; \oint_\Gamma
\!\! w{({\mathbf s})}d{\mathbf s} \, \vert ({\mathcal{L}}\psi){({\mathbf s})}\vert^2 ,
\end{displaymath} (5.8)

where $w{({\mathbf s})}\ge 0$ is a weighting function over the boundary. Arbitrary weighting schemes are possible, but I will stick to $w = 1$ which treats all locations on $\Gamma $ equally. The case of Dirichlet BCs is given by $({\mathcal{L}}\psi){({\mathbf s})}\equiv \psi({\mathbf r}{({\mathbf s})})$, put more simply, $f$ measures the boundary integral of the squared value of the wavefunction. Generally my work will be restricted to this case from now on--however the same techniques would apply to other BCs.

Without a further condition, (5.5) and (5.6) admit the useless solution, $\psi{({\mathbf r})}= 0$ for all ${\mathbf r}$. Therefore the quadratic functional

g[\psi] \; \equiv \; \left\langle \psi \left\vert\psi\right...
...mathcal{D}} \!\! d{\mathbf r} \vert \psi{({\mathbf r})}\vert^2
\end{displaymath} (5.9)

which measures the normalization (norm) in the domain is required. Unit norm is fixed by (5.7).

The solution $\psi{({\mathbf r})}$ is now completely determined, when $k$ reaches one of the eigenwavenumbers $k_\mu$. For other $k$, no solution exists. Therefore in order to be able to define a `best' solution for any given guess at $k$, one of the conditions needs to be relaxed. The condition (5.6) will be replaced by the minimization

f[\psi] \; = \; \mbox{min} .
\end{displaymath} (5.10)

A sweep in $k$ will now show the $k$-dependent tension $f(k)$, indicating how well the best $\psi$ at each $k$ can match the BCs. When $k$ approaches an eigenwavenumber $k_\mu$, the tension will drop to zero, indicating all the conditions (5.5), (5.6) and (5.7) are satisfied simultaneously. Such a sweep is shown in Fig. 5.2.

Figure 5.2: Tension (equal to the inverse largest eigenvalue of (5.14)), as a function of $k$. Minima correspond to billiard eigenstates. The finite value $\epsilon_\mu$ of these minima is visible (for some cases) in the magnified plot.

next up previous
Next: Representation by a Helmholtz Up: Chapter 5: Improved sweep Previous: Brief history of the
Alex Barnett 2001-10-03