Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 88 / 2

Use polar coordinates with $-\pi <\theta <\pi$; the two faces of the screen are $\theta =\pm \pi$. Take $0<{\theta }_0<\pi$, so the incident wave approaches from the upper half-plane, and retain the specified time dependence $e^{i\omega t}$. Outgoing cylindrical waves then have radial factor $e^{-i{k}_{0}r}$. Define the normalized Fresnel integralThe exact Neumann Sommerfeld half-plane diffraction solution for the total spatial field isOne way to verify the Helmholtz equation is to use parabolic coordinates $\xi =\sqrt{2r}\cos (\alpha /2)$, $\eta =\sqrt{2r}\sin (\alpha /2)$. Each term has the form $e^{i{k}_0({\xi }^2-{\eta }^2)/2}\mathcal{F}(\pm \sqrt{k_0}\xi )$. The operator is $({\xi }^2+{\eta }^2{)}^{-1}({\partial }_{\xi }^2+{\partial }_{\eta }^2)$, and ${\mathcal{F}}^{\prime \prime }(s)+2is{\mathcal{F}}^{\prime }(s)=0$ makes its Helmholtz residual zero. At $\theta =\pm \pi$, the two Fresnel arguments coincide, while the derivatives of both the arguments and plane-wave phases cancel in pairs. Hence ${\partial }_{\theta }\Phi =0$, which is the required normal boundary condition.

Away from the Fresnel transition regions, integration by parts gives the Fresnel tail expansionwhere $H$ is the Heaviside step function. The same expansion follows from the complementary-error-function asymptotics in NIST DLMF. Because $s_{\pm }^2=k_{0}r[1+\cos (\theta \pm {\theta }_0)]$, the tails in both terms have outgoing phase $e^{-i{k}_{0}r}$. The scattered far field is consequentlyThe first term cancels the incident geometrical optics field in its shadow, the second is its reflected plane wave and the last is the diffracted edge wave. The expansion is nonuniform at $\theta ={\theta }_0-\pi$ and $\theta =\pi -{\theta }_0$; the exact Fresnel expression resolves those regions. Incidence from below follows by reflecting $\theta$ and ${\theta }_0$.

Let $G_1=\varphi (\mathbf{r}|{\mathbf{r}}_1)$ and $G_2=\varphi (\mathbf{r}|{\mathbf{r}}_2)$. Green's second identity givesThe common Robin boundary condition makes the surface integrand vanish. Where $\beta \ne 0$, substitute ${\partial }_n{G}_i=-\alpha G_i/\beta$; a Dirichlet portion also contributes zero. The delta sources therefore give the Helmholtz Green-function reciprocityThis is a bilinear identity, with no complex conjugation. Assume the boundary-value problem has a unique Green kernel; in an exterior problem use the same outgoing radiation condition for both kernels.

For the weighted operator, assume smooth positive $f$ and set $\mathbf{b}=\nabla \log f$. ThenIts unweighted transpose, obtained by integration by parts, isThe associated boundary identity isTo cancel it when $v$ satisfies the given Robin condition, the required adjoint kernel satisfiesUsing these two delta sources in the boundary identity proves $\psi ({\mathbf{r}}_2|{\mathbf{r}}_1)=\tilde{\psi }({\mathbf{r}}_1|{\mathbf{r}}_2)$, the adjoint reciprocity for a weighted Helmholtz operator. As a check, $L^{\mathrm{T}}(fv)=fLv$, soThis is equivalent to weighted acoustic Green-function reciprocity, $f({\mathbf{r}}_2)\psi ({\mathbf{r}}_2|{\mathbf{r}}_1)=f({\mathbf{r}}_1)\psi ({\mathbf{r}}_1|{\mathbf{r}}_2)$. Unit delta normalization is essential here. A Hermitian adjoint would instead conjugate coefficients and change the radiation convention; the requested equality uses the transpose.