After separating the conserved horizontal factor as $\psi=e^{ipx}f(z)$, the <Helmholtz equation> becomes
$$
f''+q_0^2f=-\alpha q_0^2H(-z)f,
$$
where $H$ is the <Heaviside step function>. The outgoing <Green function> for $d^2/dz^2+q_0^2$ is
$$
G(z,z')=\frac{e^{iq_0|z-z'|}}{2iq_0}.
$$
The <Born approximation> at first order replaces $f$ on the right-hand side by the incident profile $f_i(z')=e^{-iq_0z'}$. For $z>0$ this gives
$$
f_{s,B}(z)
=-\alpha q_0^2\int_{-\infty}^{0}
\frac{e^{iq_0(z-z')}}{2iq_0}e^{-iq_0z'}\,dz'.
$$
The integral is understood with the usual outgoing-wave convergence factor. Since
$$
\int_{-\infty}^{0}e^{-2iq_0z'}\,dz'=-\frac{1}{2iq_0},
$$
we obtain
$$
f_{s,B}(z)=-\frac{\alpha}{4}e^{iq_0z}.
$$
Therefore the Born reflected field is
$$
\boxed{\psi_{r,B}(x,z)=-\frac{\alpha}{4}e^{i(px+q_0z)}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 335 / 1 / ii / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-335/1/ii/a/solution