Applying boundary conditions and simplifying

We now proceed term-by-term in (I.2). The zeroth-order is

$$\psi \ \stackrel{{\mbox{\tiny BCs}}}{\longrightarrow}\ 0,$$ (I.15)

where the arrow implies application of the Dirichlet boundary conditions (BCs), namely $\psi(z{=}0,s) = 0$ for all $s$ (remember that we already chose $z=0$ in order to be able to write the metric). The first order is

$${\mathbf a}\cdot\nabla \psi \ = \ a_n\psi^n + a_t\psi^t \ \stackrel{{\mbox{\tiny BCs}}}{\longrightarrow}\ a_n\psi^n ,$$ (I.16)

where the result $\psi^t = 0$ was used. In will use in future the fact that all higher $t$-derivatives $\psi^{tt\cdots t}$ vanish upon application of the BCs. The second order requires application of the rules (I.10):

$$\frac{1}{2!}({\mathbf a}\cdot\nabla)({\mathbf a}\cdot\nabla) \psi = \frac{1}{2} (a_n\partial_n + a_t\partial_t)(a_n\psi^n + a_t\psi^t)$$

$$= \frac{1}{2} \left[ a_n^2 \psi^{nn} + 2 a_n a_t \psi^{nt} + a_t^2 \psi^{tt} + \alpha a_t(a_t\psi^n - a_n\psi^t)\right]$$

$$\stackrel{{\mbox{\tiny BCs}}}{\longrightarrow} a_n a_t \psi^{nt} + \frac{\alpha}{2}(a_t^2 - a_n^2) \psi^n .$$ (I.17)

The Helmholtz equation and the BCs together imply $\psi^{nn} = -\alpha \psi^n$, a relation used both above and below. In general our task is to reduce the number of $n$-derivatives; it is always possible using such manipulations to leave only terms containing a single $n$-derivative. This will be desirable for manipulation of boundary integrals by parts later.

The third-order result (included below) requires use of the following. $\psi^{nnn} = \partial_n(-\psi^{tt}-\alpha \psi^n - k_\mu^2 \psi)$ simplifies to $-\psi^{ntt} + (2\alpha^2 - k_\mu^2) \psi^n$ when the BCs are applied. This required the normal derivative of curvature $\partial_n \alpha = -\alpha^2$ since $\alpha$ now needs to be regarded as a scalar field with $z$-dependence. The tangential derivative is given the name $\partial_t \alpha = \alpha'$. Also $\psi^{nnt} = \partial_t(-\psi^{tt}-\alpha \psi^n - k_\mu^2 \psi)$ simplifies to $-\alpha' \psi^n - \alpha \psi^{nt}$ when BCs are applied.

Combining everything and finally substituting for ${\mathbf a}$ gives the expansion of a Dirichlet scaling eigenfunction at location $s$ on $\Gamma$:

$$\psi(k_\mu + \delta,s) = \frac{\delta}{k_\mu} r_n \psi^n$$

$$+ \ \frac{\delta^2}{k_\mu^2} \left[ \alpha \frac{r_t^2 - r_n^2}{2} \psi^n \; + \; r_n r_t \psi^{nt} \right]$$

$$- \ \frac{\delta^3}{6k_\mu} r_n^3 \psi^n \; + \; \frac{\delta^3}{... ...3r_t^2) + \alpha' r_t(r_t^2 - 3r_n^2) \right] \psi^n \rule{0in}{0.25in} \right.$$

$$\hspace{1in} \left. \rule{0in}{0.25in} + \; 3\alpha r_t(r_t^2 - 2r_n^2) \psi^{nt} + r_n(3r_t^2 - r_n^2) \psi^{ntt} \right\}$$

$$+ \ O(\delta^4) \cdots$$ (I.18)

The expression is believed to be correct to order $\delta^3$. The complexity increases greatly with each power of $\delta$, and higher derivatives of the curvature $\alpha$ enter. However, in the case of the circle billiard (where $\alpha$ is constant) it is easy to verify the expansion. One general pattern is that the $m^{{\mbox{\tiny th}}}$-order terms all contain $m$ powers of $r \delta$, and for each term the number of spatial derivatives of $\psi$ must equal the power of $1/k_\mu$ minus the power of $\alpha$. It is important to notice that the $O(\delta^2)$ term differs by the presence of a factor of $\alpha$ from the $O(\delta^2)$ term given by VS.