 Solution

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.