Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 70 / 1

Use a Fourier transform parallel to the boundary. Splitting its traces into right- and left-supported parts gives Half-range Fourier transforms $F_{+}$ and $F_{-}$, analytic respectively above and below a common inversion contour. A small positive imaginary part of the wavenumber implements the limiting absorption principle and separates the incident poles and outgoing branch points. Solve the transformed differential equation normal to the boundary using its outgoing or decaying solution. The boundary conditions then yield a Wiener-Hopf equation of the formFactor its Wiener-Hopf kernel as $K=K_{+}{K}_{-}$, with each factor analytic and nonzero in its designated half-plane. Divide by $K_{-}$ and split $H/K_{-}=H_{+}+H_{-}$, allocating poles according to the incident-wave prescription. ThenThe two sides continue to a common entire function. Their growth at infinity, inherited from the field's edge behavior, determines that function, usually a constant or a polynomial; in a decaying case it is zero. Solve for the unknown transforms and apply Fourier inversion, preserving the outgoing contour prescription.

The principal analytic results used are as follows. The Paley–Wiener theorem and its half-line Laplace-transform version relate support on a half-line to holomorphy in a half-plane, with growth or square-integrability bounds. The Cauchy integral formula and Sokhotski–Plemelj theorem provide additive splitting: a Cauchy integral of a sufficiently decaying jump has upper and lower boundary values with that prescribed difference. Applying the same construction to a continuous logarithm of a nonvanishing scalar Wiener-Hopf kernel gives Wiener-Hopf factorization when its winding number is zero; a nonzero index requires an explicit index factor. The identity theorem gives analytic continuation across the common strip. Finally Liouville's theorem makes a bounded entire function constant; its polynomial-growth extension bounds the possible degree. The physical edge and radiation conditions select the remaining constants. Zeros, branch cuts, nonzero index and insufficient decay must be accounted for rather than discarded during splitting.

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.