 Solution

For the second-order rough-surface scattered field, the Dirichlet boundary condition expanded at the mean plane isAt normal incidence, ${\psi }^{[0]}=e^{-ikz}-e^{ikz}$, so its second normal derivative vanishes at zero. Define the Dirichlet-to-Neumann map for a Helmholtz half-space through the Fourier multiplier $\mathcal{B}$:The first-order trace is $2ikh$, hence ${\partial }_z{\psi }_s^{[1]}(x,0)=-2k\mathcal{B}h(x)$. It follows thatwhere $h=O(\epsilon )$ with fixed regular profile. In integral notation the quadratic contribution isThe second-order field above the mean plane is $\mathcal{E}[2kh\mathcal{B}h]$ added to $-e^{ikz}+\mathcal{E}[2ikh]$. No local replacement of $\beta (q)$ by $k$ has been made; such a replacement would be an additional long-spatial-scale approximation.

The physical surface trace and reference-plane trace are distinct. At the actual rough boundary $z=h(x)$, the condition itself says ${\psi }_s(x,h(x))=-e^{-ikh(x)}$, soThe first boxed expression is the reference-plane trace needed in part (d), at $z=0$, using the perturbative continuation where that plane lies below the actual boundary; the second answers the literal “at the surface” wording if it means the physical boundary. Taylor-expanding the first expression and its normal derivatives from $z=0$ to $z=h$ reproduces the second, so there is no contradiction between them.