 Solution

Work above the surface, with $e^{-i\omega t}$ time dependence. Define the scattered field by $\psi ={\psi }_i+{\psi }_s$, so it includes the reflection from a flat surface. For small-height Dirichlet scattering, write $h=\epsilon \eta$ and expandThe zeroth-order total field ${\psi }^{[0]}={\psi }_i+{\psi }_s^{[0]}$ satisfies the Dirichlet boundary condition ${\psi }^{[0]}(x,0)=0$. Taylor expansion at the perturbed boundary givesThus the first-order rough-surface scattered field has mean-plane boundary data $g_1(x)=-h(x){\partial }_z{\psi }^{[0]}(x,0)$.

Use the outgoing angular spectrum to solve this boundary-value problem. For $\hat{g}(q)=\int g(x)e^{-iqx}dx$, letThe branch ensures upward propagation or upward evanescent decay. Each component solves the Helmholtz equation, and $\mathcal{E}g$ has trace $g$ at $z=0$. ConsequentlyThis gives the scattered field through first order by adding the two contributions. If one reserves “rough scattered field” for the non-specular correction, it is ${\psi }_s^{[1]}$ alone; the convention here keeps the flat reflection as well.

The expansion is in height for a fixed sufficiently regular profile. The condition $|kh|\ll 1$ controls the incident wave's height expansion, but very short spatial scales can create large evanescent normal derivatives. The surface regularity and relevant spectral moments must also control the subsequent boundary expansions; small amplitude alone is not a uniform guarantee for arbitrarily fine roughness.