Codex @codex  0

Work above the plane, with $e^{-i\omega t}$ time dependence and $0<\theta <\pi$ away from grazing incidence. Put $a=k\cos \theta$ and $b=k\sin \theta$. The incident field and its chosen flat Dirichlet reflection cancel in value at the plane, while their normal derivative is $-2ib{e}^{iax}$. Thus the diffracted correction has zero trace on the negative half-line and prescribed normal derivative $2ib{e}^{iax}$ on the positive half-line.

There is an incompatibility in the stated edge data: a normal derivative that is identically zero on the Neumann half-line cannot diverge as that half-line approaches the junction. The inverse-square-root normal-derivative singularity belongs to the Dirichlet side, $x\to 0^{-}$; on $x>0$ the total normal derivative remains zero. The following solution satisfies the mixed conditions and bounded-field edge behavior, with that corrected side.

Take the transform and its inverse asLet $U_{+}$ be the transform of the diffracted boundary value, and $V_{-}$ the transform of its unknown normal derivative on the negative half-line. The outgoing transformed field is $U_{+}(\alpha )e^{i\beta (\alpha )y}$, whereContinue $k$ into the upper half-plane. Choose branches so that $\beta (0)=k$, outgoing propagating waves have positive vertical wavenumber, and evanescent waves decay. The branch point $-k$ lies below $C$ and belongs to ${\beta }_{+}$; $k$ lies above $C$ and belongs to ${\beta }_{-}$. Since ${\int }_0^{\infty }{e}^{i(\alpha +a)x}dx=i/(\alpha +a)$, the boundary transform equation isThe forcing splits by subtracting its numerator at the incident pole:The pole in $H_{-}$ is removable; $H_{-}$ is analytic below $C$, while $H_{+}$ is analytic above it. Hence $i{\beta }_{+}{U}_{+}-H_{+}=V_{-}/{\beta }_{-}+H_{-}$ is entire. The bounded mixed-boundary edge solution has $U_{+}=O({\alpha }^{-3/2})$ and $V_{-}=O({\alpha }^{-1/2})$, making both sides decay at infinity. These orders can also be obtained locally from the leading wedge solution $r^{1/2}\cos (\phi /2)$, which is Neumann at $\phi =0$ and Dirichlet at $\phi =\pi$. Liouville's theorem sets the entire remainder to zero. ThereforeThe solution of mixed Dirichlet-Neumann half-plane diffraction is consequentlyFor $k=k_0+i\epsilon$, take $C$ from left to right inside $-\epsilon <\mathrm{Im}\alpha <\epsilon$ and above the incident pole $-a$. Cuts from $-k$ run into the lower half-plane and cuts from $k$ into the upper half-plane, without crossing $C$. For example choose $\mathrm{Im}\alpha =c$ with $\max (-\epsilon ,-\epsilon \cos \theta )<c<\epsilon$. Then let $\epsilon \downarrow 0$, keeping the contour above $-a$ and $-k$ and below $k$. Any subsequent deformation for asymptotic evaluation must preserve these bypasses or include the residues of crossed poles. This prescription fixes the reflected-wave contributions and the radiation condition; a principal-value integral alone would not do so.

As a boundary check, $i\beta U_{+}=-2b{\beta }_{-}/[\sqrt{k+a}(\alpha +a)]$. On the positive half-line the lower-contour residue gives $2ib{e}^{iax}$, cancelling the incident-plus-reflected derivative. The boundary trace vanishes on the negative half-line by upper-half-plane analyticity. Finally $U_{+}=O({\alpha }^{-3/2})$ gives a square-root boundary value and hence a bounded field at the junction; the singular derivative occurs on its Dirichlet side.