Exterior Boundary Value Problem: Scattering

Let us apply the properties of the Bessel function to solve the following exterior boundary value (``scattering'') problem:

Find that solution to the Helmholtz equation $(\nabla^2+k^2)\psi =0$ in the Euclidean plane which satisfies

1. the Dirichlet boundary condition on the circular boundary $r=a$ and
2. the condition that its asymptotic form, as $r\to\infty$ , is that of a plane wave propagating into the $x$ -direction,
   $$\psi_{inc} = e^{ikr\cos\theta}$$
   plus only outgoing waves, if any; i.e. no incoming waves.

Mathematically the second condition is a type of boundary condition at infinity. It is evident that this boundary condition states that the solution consists of ``plane wave + outgoing wave''. The physical meaning of this condition is that it represents a scattering process.

Figure 5.13: Scattering by a cylinder. An incoming plane wave $\psi _{incident}$ in the presence of a cylindrical boundary gives rise to a circular scattered wave $\psi _{scattered}$ which at large radii propagates away from the cylindrical boundary.

If the circular boundary were absent, then there would have been no scattering. The Dirichlet boundary condition 1. would have been replaced by the regularity requirement that $\psi =$ finite at $r=0$ while the second boundary condition 2. at $r=\infty$ would have remained the same. In that case the resulting ``no scattering'' solution is immediate, namely,

$$\psi = e^{ikr\cos\theta} =\sum^\infty_{m=-\infty} J_m(kr)i^m e^{im\theta}\,.$$

By contrast, if the circular boundary is present as stipulated by the problem, then this solution must be augmented so that the Dirichlet boundary conditions are satisfied,

$$\psi =e^{ikr\cos\theta} +\psi_{\textrm{scatt}}~~.$$

This augmentation can be implemented with Hankel functions of the first kind, or of the second kind, or with a combination of the two. The boundary condition that the solution represent a plane wave plus a scattered wave, outgoing only, demands that the augmentation have the form

$$\psi_{\textrm{scatt}} = \sum^\infty_{m=-\infty} a_m H^{(1)}_m (kr)i^m e^{im\theta} ~~.$$

It expresses the requisite outgoing wave condition for $e^{-i\omega t}$ time dependence. This is because as $r\to\infty$ ,

$$H^{(1)}_m(kr)\simeq\sqrt{\frac{2}{\pi kr}} e^{i\left[ kr-\left( m+\frac{1}{2} \right) \pi /2\right]}\,.$$

By contrast, at $r=a$ , the Dirichlet boundary condition demands that

$$= \psi (r=a,\theta )$$ 0

$$= \sum^\infty_{m=-\infty} J_m(ka)i^m e^{im\theta} +a_m H^{(1)}_m (ka)i^m e^{im\theta}\,.$$

The fact that this holds for all angles $\theta$ implies that each term (= ``partial wave amplitude'') in the sum must vanish. Consequently,

$$a_m =- \frac{J_m(ka)}{H^{(1)}_m(ka)}~~\qquad~~\qquad~~ m=0,\pm 1,\dots\,.$$

It follows that the hard cylinder scattering process yields the modified plane wave

$$\psi = e^{ikr\cos\theta} +\sum^\infty_{m=-\infty}(-)\frac{J_m(ka)} {H^{(1)}_m(ka)}i^m H^{(1)}_m(kr)e^{im\theta}$$

$$= \sum^\infty_{m=-\infty} \left\{ J_m(kr)-\frac{J_m(ka)}{H^{(1)}_m (ka)} H^{(1)}_m (kr)\right\} i^m e^{im\theta}\,.$$

This solution to the Helmholtz equation represents the incident plane wave plus the scattered outgoing cylinder waves.