Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 78 / 2

Suppress the common time factor $e^{i\omega t}$ and invert the stated Fourier transform with $(2\pi {)}^{-1}\int \Phi (k,y)e^{-ikx}dk$. Use limiting absorption principle $k_0\mapsto k_0-i0$, and choose $\gamma =\sqrt{k^2-k_0^2}$ with positive real part for the evanescent spectrum. For $|k|<k_0$ its outgoing continuation is $\gamma =+i\sqrt{k_0^2-k^2}$. The pole $-k_0$ lies above the inversion contour and $+k_0$ below it. These prescriptions fix both radiation and pole signs.

The even interior scattered transform solves ${\Phi }_{yy}-{\gamma }^2\Phi =0$, so it is $A(k)\cosh (\gamma y)$. The exterior transform is $B(k)e^{-\gamma (|y|-b)}$. Let $D^{-}(k)$ be the transform of their common upward normal derivative on $y=b$. On $x>0$ the rigid plates set both derivatives to zero; on $x<0$ continuity requires them to agree. Hence this derivative is supported on the negative half-line and is analytic below the transform contour. It follows thatThe difference between exterior and interior scattered traces on $y=b$ is $-D^{-}/L$, where $L=\gamma \sinh (\gamma b)e^{-\gamma b}$. At the open negative half-line the total traces must match, so this difference equals the incident wave $e^{i{k}_{0}x}$. Its negative-half transform is $-i/(k+k_0)$. The positive-half trace difference has an unknown upper-half-plane transform $C^{+}$, giving the Wiener-Hopf equationFactor $L=L^{+}{L}^{-}$, divide by $L^{-}$, and subtract the forcing pole using $C_0=L^{+}(-k_0)$:The left side is analytic below and the right side above; the pole in the last quotient is removable. Analytic continuation makes their common value entire. The finite-energy edge condition and algebraic factor growth make this entire remainder vanish: the allowed normal-velocity edge singularity is at most inverse-square-root, so its half-line transform is $O(|k{|}^{-1/2})$, while $L^{\pm }=O(|k{|}^{1/2})$. The same edge condition excludes a growing polynomial remainder. This pole subtraction in a Wiener-Hopf equation yieldsThe corresponding exterior result for the Wiener-Hopf solution for an open parallel-plate waveguide isBoth forms are useful: the first makes the common normal derivative transparent, and the second removes the apparent factor singularity at a grazing saddle.

The final unheaded request has a sign error in its printed exponent. To see this from the derived solution rather than an outgoing-decay assertion alone, simplify the interior transform toNear $k=-k_0$, $C_0/L^{+}(k)=1+O(k+k_0)$ and $e^{-\gamma b}\cosh (\gamma y)=1-b\gamma +O({\gamma }^2)$ for fixed $|y|<b$. ThusFor $x<0$, inversion of the first term gives $-e^{i{k}_{0}x}$, cancelling the continued incident potential. The remaining branch-point term uses $\gamma \sim i\sqrt{2k_0}(k+k_0{)}^{1/2}$ on the specified contour, so it is $b\sqrt{2k_0}(k+k_0{)}^{-1/2}$. Its inverse transform at $x=-r$ is proportional to $e^{-i{k}_{0}r+i\pi /4}/\sqrt{\pi r}$. Therefore the upstream decay of an open parallel-plate waveguide isThe amplitude decays as $(-x{)}^{-1/2}$, not $(-x{)}^{1/2}$. The positive exponent is present in the original PDF, so it cannot be proved under the stated outgoing problem. The incident pole cancellation and the explicit branch expansion give the corrected conclusion.

Put $x=r\cos \theta$, $y=r\sin \theta$, and $s=|\sin \theta |$. From the exterior transform, the inverse integral has the amplitudemultiplying $e^{-ikr\cos \theta -\gamma rs}$. Apply the supplied method of steepest descent formula at $k_s=k_0\cos \theta$. On the outgoing sheet, $\gamma (k_s)=i{k}_{0}s$, so $-i\sinh (i{k}_{0}bs)=\sin (k_{0}bs)$. The far-field scattered potential isRestore $e^{i\omega t}$ for the physical time-harmonic potential. This is cylindrical outgoing spreading, with angular directivity determined by the kernel factor and aperture. The ordinary saddle formula applies for a fixed non-grazing direction, away from a coalescing pole or branch point. Its finite upstream angular limit is consistent with the separate branch-point calculation in the root solution; at grazing directions that separate calculation supplies the appropriate justification.

A plane wave returning down the guide has spatial factor $e^{-i{k}_{0}x}$ and arises from the pole at $k=k_0$, lying below the inversion contour. For $x>0$ closing below is clockwise, so the inverse transform is minus $i$ times the residue. Since $\gamma \sinh (\gamma b)\sim 2b{k}_0(k-k_0)$, the interior residue isHence the reflected plane-wave amplitude, relative to the incident amplitude one, isEquivalently, because $L^{+}$ carries the simple zero at $k_0$, $\mathcal{R}=L^{+}(-k_0)/(2k_0(L^{+}{)}^{\prime }(k_0))$. Other guided modes, if propagating at the chosen frequency, come from the additional kernel zeros; they do not alter which residue gives the plane-mode amplitude. Evanescent terms near the aperture are also separate. A multiplicative renormalization of the factors cancels in the product defining $\mathcal{R}$.