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Fluid-loaded plate. Use complex amplitudes with time dependence $e^{-i\omega t}$ and take the velocity potential in the fluid above the elastic plate to be $\Phi$. Let ${\rho }_0$ and $c_0$ be the undisturbed mass density and speed of sound. The linearized Euler equations give $p=i{\rho }_0\omega \Phi$; the kinematic boundary condition is ${\Phi }_y(x,0)=-i\omega \eta (x)$. We model loading from one fluid half-space, with no fluid loading on the other face. Loading by identical fluids on both faces instead doubles the fluid term below.

Define the Fourier transform and its inverse byThe Sommerfeld radiation condition selects $\hat{\Phi }(k,y)=\hat{\Phi }(k,0)e^{-\gamma y}$, with ${\gamma }^2=k^2-k_0^2$ and $k_0=\omega /c_0$. Here and in the diffraction calculation below, the outgoing square root is specified by first taking $\mathrm{Im}\omega >0$. Draw a vertical cut upwards from $+k_0$ and another downwards from $-k_0$, and putThus ${\gamma }_u$ is analytic in the upper half-plane and ${\gamma }_l$ in the lower half-plane. In the real-frequency limit,The Fourier transform contour $C$ follows the real axis with the limiting-absorption pole prescriptions. In particular, an outgoing propagating component has positive vertical wavenumber.

The kinematic boundary condition gives $\hat{\Phi }(0)=i\omega \hat{\eta }/\gamma$, so $\hat{p}(0)=-{\rho }_0{\omega }^2\hat{\eta }/\gamma$. Eliminating the fluid pressure from the bending balance gives the line-forced fluid-loaded bending plate kernelFor harmonic line force amplitude $F_0$ at the origin, the plate displacement and fluid velocity potential are thereforeAn arbitrary localized loading replaces $F_0$ by its Fourier transform. This representation both calculates the response and identifies its different wave components through poles and branch points.

For an observer $x=r\cos \theta$, $y=r\sin \theta$, with $0<\theta <\pi$, the method of steepest descent uses the saddle point $k_s=k_0\cos \theta$. At fixed angle away from grazing, the acoustic far field isThe $r^{-1/2}$ spreading is that of cylindrical acoustic waves. The phase velocity parallel to the plate is $\omega /|k|$: components with $|k|<k_0$ are supersonic in that sense and radiate, whereas those with $|k|>k_0$ are subsonic and exponentially decay normally to the plate. The latter have reactive fluid loading and zero mean normal acoustic energy flux.

There is also a real guided root $\kappa >k_0$. Indeed, $K(k_0+)=-\infty$, $K(+\infty )=+\infty$, andThe residue theorem then gives an outgoing displacement contribution $i{F}_0{e}^{i\kappa |x|}/K^{\prime }(\kappa )$. It transports energy along the plate while its fluid field decays as $e^{-\gamma (\kappa )y}$. Other zeros on continued sheets can describe leaky or resonant waves; branch point contributions describe near and grazing fields. A fixed-angle acoustic far field expansion is not uniform near the plate, where these components can matter.

For a propagating spectral component, write $\ell =(k_0^2-k^2{)}^{1/2}$. Its upward normal velocity is $i\ell \hat{\Phi }$ and its pressure is $i{\rho }_0\omega \hat{\Phi }$. Applying Parseval's identity to the time-averaged acoustic energy flux gives the upward power per unit length of the forcing line:Equivalently, the integrand is $|\hat{\eta }{|}^2/\ell$ with prefactor ${\rho }_0{\omega }^3/(4\pi )$. Integrating the radial acoustic intensity of the far field over a semicircle gives the same expression. The input power $\omega \mathrm{Im}(F_0^{\ast }\eta (0))/2$ also feeds guided waves along the plate, so it must not be identified with $P_y$ alone.

Rigid half-plane diffraction. Suppress the time factor and assume $0<{\theta }_0<\pi$. Write $a=k_0\cos {\theta }_0$, $b=k_0\sin {\theta }_0$, and denote the incident spatial field by $I=e^{-iax-iby}$. The Helmholtz equation, Sommerfeld radiation condition, and square-root branches are as above. Across the open positive axis the scattered field and its normal derivative are continuous. Across the rigid negative axis its normal derivative is also continuous, while its potential can jump. Let $J(x)$ be this jump, supported on $x<0$. The transformed scattered field isThe jump transform is analytic in the upper half-plane. The absence of an even scattered component follows from derivative continuity and the outgoing condition. Its normal derivative on either face is $-\gamma {\hat{J}}_u/2$.

On the negative axis this derivative must cancel $-ib{e}^{-iax}$. Since the Half-range Fourier transform of $H(-x)e^{-iax}$ is $i/(k+a)$, the Wiener-Hopf equation has the formwhere $G_l$ is the transform of an unknown function supported on $x>0$ and analytic in the lower half-plane. The incident pole $k=-a$ lies below the Fourier contour. Divide by ${\gamma }_l$ and perform pole subtraction in a Wiener-Hopf equation:The first term is upper analytic and the second lower analytic, since the latter's apparent pole cancels. Equality of the two continued sides gives an entire function. The finite-energy edge condition, with $J(x)=O(\sqrt{-x})$, and the decay of the transforms at infinity force that function to vanish by Liouville's theorem. ConsequentlyThis is the Wiener-Hopf solution of rigid half-plane diffraction with the present transform and time conventions.

Deforming to a steepest descent contour gives a residue from the incident pole precisely when $x+|y|\cot {\theta }_0<0$. At that pole $\gamma (-a)=-ib$; its residue producesThus the total geometrical optics field is $I$ plus a reflected wave in the upper reflection region; in the lower shadow region the scattered residue cancels $I$, and outside that shadow the direct incident wave remains. This gives both the amplitude and the spatial support of the geometrical optics fields rather than merely naming a reflected wave.

For $x=r\cos \theta$, $y=r\sin \theta$, $-\pi <\theta <\pi$, the saddle contribution uses ${\gamma }_l(-a)=-i\sqrt{2k_0}\cos ({\theta }_0/2)$ and ${\gamma }_u(k_s)=\sqrt{2k_0}\cos (\theta /2)$. The diffracted wave isThe total field is $I+{\varphi }_{\mathrm{GO}}+{\varphi }_d$ to this order. This far-field expression applies away from the shadow/reflection boundaries $|\theta |=\pi -{\theta }_0$. There the pole and saddle coalesce, and a uniform Fresnel integral approximation replaces the divergent separate terms by a smooth transition. The exact contour integral remains well defined.

Take $0<{\theta }_0<\pi$, put $a=k_0\cos {\theta }_0$ and $b=k_0\sin {\theta }_0>0$, and suppress the common factor $e^{i\omega t}$. The upper-half-plane incident acoustic plane wave is $e^{iax+iby}$. A rigid zero-thickness screen imposes the same normal-derivative cancellation on both faces; the scattered velocity potential is consequently odd in $y$. Across the open part $x>0$ it is continuous, so its trace there is zero. These complementary half-line data are what make the Wiener-Hopf method applicable.

Use the Fourier transform $\hat{\varphi }(\kappa ,y)={\int }_{\mathbb{R}}\varphi (x,y)e^{i\kappa x}dx$, with inverse exponential $e^{-i\kappa x}$. The outgoing Helmholtz equation solution is $\hat{\varphi }=F^{-}(\kappa )e^{-\gamma y}$ for $y>0$, where $\gamma =\sqrt{{\kappa }^2-k_0^2}$ has $\gamma =i\sqrt{k_0^2-{\kappa }^2}$ on the propagating interval and positive real part on evanescent components. The limiting absorption principle supplies the contour prescription. The transform $F^{-}$ of the trace supported on $x<0$ is analytic below the contour. Let $G^{+}$ be the analytic upper transform of the unknown derivative on $x>0$. The prescribed derivative on $x<0$ has transformSince the upper normal derivative is $-\gamma F^{-}$, the Wiener-Hopf equation isFor the Wiener-Hopf factorization, choose ${\gamma }_{+}=(\kappa -k_0{)}^{1/2}$ analytic above and ${\gamma }_{-}=(\kappa +k_0{)}^{1/2}$ analytic below. Their branch cuts go into the opposite half-planes, with ${\gamma }_{+}(-a)=i\sqrt{k_0+a}$. Dividing by ${\gamma }_{+}$ and performing pole subtraction in a Wiener-Hopf equation givesThe forcing pole is allocated to the lower analytic part; the second quotient has a removable pole and is upper analytic. Subtracting these parts leaves a common entire function. The finite-energy edge condition excludes an inverse-square-root singularity in the potential: its odd scattered trace is $O(\sqrt{|x|})$, so $F^{-}=O(|\kappa {|}^{-3/2})$. The normal derivative on the open side can be $O(x^{-1/2})$, giving $G^{+}=O(|\kappa {|}^{-1/2})$. Both sides of the entire remainder tend to zero. The Liouville theorem sets that remainder to zero, giving the explicit Wiener-Hopf solution of rigid half-plane diffraction:The scattered field on either side of the screen is thereforeIts forcing-pole residue at $\kappa =-a$ is $-i$. Deforming to the outgoing steepest descent contour crosses this pole precisely when $|\theta |>\pi -{\theta }_0$, where $x=r\cos \theta$, $y=r\sin \theta$ and $-\pi <\theta <\pi$. The residue contributes $\mathrm{sgn}(y)e^{iax-ib|y|}$. Thus, away from transition boundaries, the scattered geometrical optics field isAdding the incident acoustic plane wave gives incident plus unit specular reflection in the upper reflected sector, just the incident wave in the remaining illuminated sectors, and zero in the lower shadow sector. The geometrical shadow boundaries also follow directly by tracing a straight ray to $y=0$: the reflected ray hits the screen at $x+y\cot {\theta }_0<0$, and the transmitted incident ray crosses the open part at $x-y\cot {\theta }_0>0$.

For the diffracted contribution, the outgoing phase $-i\kappa r\cos \theta -\gamma r|\sin \theta |$ has saddle ${\kappa }_s=k_0\cos \theta$, value $-i{k}_{0}r$, and second derivative $ir/(k_0{\sin }^2\theta )$. Stationary phase therefore supplies $\sqrt{2\pi k_0/r}|\sin \theta |e^{-i{k}_{0}r+i\pi /4}$. Substituting $F^{-}({\kappa }_s)$, and using $\mathrm{sgn}(y)|\sin \theta |/\sqrt{1+\cos \theta }=\sqrt{2}\sin (\theta /2)$, givesThe full asymptotic scattered field is ${\varphi }_{\mathrm{GO}}+{\varphi }_d$. The angular edge-wave expression is not uniform on the geometrical boundaries. There is also a sign inconsistency in the supplied saddle formula: with its printed inverse exponential $e^{+i\kappa x}$, the outgoing saddle is $-k_0\cos \theta$, not $+k_0\cos \theta$. The consistent inverse convention used above gives the requested outgoing field without that ambiguity.

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.

Suppress the common time factor $e^{i\omega t}$ and use the Fourier transform conventionA right Half-range Fourier transform is analytic above its convergence line and a left transform below it. For the limiting absorption principle with this time convention take $\omega ={\omega }_r-i\epsilon$, ${\omega }_r>0$, $\epsilon >0$, before taking $\epsilon \downarrow 0$. SetThis is the outgoing acoustic square-root branch: at a propagating real wavenumber, $\gamma \to i\sqrt{{\omega }_r^2-k^2}$. Put $J=[\tilde{\varphi }{]}_{y=-0}^{y=+0}$ and $d^{+}={\partial }_y{\Phi }^{+}(k,0)$. Across the fluid part of $y=0$, the field and its normal derivative are continuous; on the sheet both sides have the same normal velocity. The outgoing Helmholtz equation solutions therefore have equal normal derivatives and opposite scattered traces:The jump $J$ is supported on $x<0$, and so is a minus transform.

The linear fluid equation $i\omega v_y=-{\partial }_y{\rho }^{\prime }$ and the sheet velocity $v_y=i\omega \eta$ imply ${\partial }_y{\rho }^{\prime }={\omega }^2\eta$ on either face. The incident contribution to the left derivative transform is $\omega \sin {\theta }_0/(k+\omega \cos {\theta }_0)$. Writing ${\eta }^{-}={\int }_{-\infty }^0{e}^{ikx}\eta (x)dx$, it follows thatThe upward net pressure is lower minus upper, hence $P=-J$ in transform space. If $s={\eta }^{\prime }(0)$, the pinned endpoint givesCombining these identities gives the Wiener-Hopf equationThis is the plane-wave forcing of a pinned elastic half-sheet.

Let $b=\omega \sqrt{m/T}$ and assume initially that the distinguished points are distinct. Besides the branch points $\pm \omega$, $K$ has simple poles at $\pm b$. Its physical-sheet zeros $\pm k_s$ solve the fluid-loaded dispersion relationFor real positive frequency there is exactly one positive real root $k_s>\max ({\omega }_r,{\omega }_r\sqrt{m/T})$: the left side increases strictly from zero to infinity on that interval. Equivalently, with $q=\gamma (k_s)>0$, $T{q}^3+(T-m){\omega }_r^{2}q-2{\omega }_r^2=0$. Its other roots have negative real part and do not belong to this outgoing sheet. Continue $k_s$ with small absorption; $+\omega ,+b,+k_s$ lie below the real line and their negative partners above it. The Wiener-Hopf factorization assigns the lower branch cut and pole to $K^{+}$, and the upper branch cut and pole to $K^{-}$. The continued $K^{+}$ has a zero at $+k_s$, and $K^{-}$ at $-k_s$; these are poles of their reciprocals, rather than poles of the factors. Branch cuts can be drawn away from the real line into the corresponding half-planes.

For this standard assignment a common strip of analyticity and nonvanishing isThe strip shrinks to the real contour as absorption is removed. The factors continue meromorphically beyond their designated half-planes; the allocation of the poles and branch points is part of the radiation condition. The incident pole $-\omega \cos {\theta }_0$ belongs above the inversion contour. For incidence with $0<{\theta }_0<\pi /2$ the real contour is a direct choice. Other nongrazing angles are obtained by analytic continuation with the incident pole kept on its prescribed side, deforming the contour locally if necessary. Coincident distinguished points and grazing incidence are treated by the corresponding limits, not by dividing by a coincident pole.

Divide the Wiener-Hopf equation by $K^{+}$ and use pole subtraction in a Wiener-Hopf equation. With $a=\omega \cos {\theta }_0$, the minus part of $F/K^{+}$ isIndeed these are exactly the incident-pole and upper bare-sheet-pole residues. At the lower bare-sheet pole $+b$, the forcing pole is canceled by the pole of $K^{+}$ before splitting. Thus $B^{+}=F/K^{+}-B^{-}$ is analytic on the plus side of the contour. The separated equation isWith the stipulated $E=0$,The scattered field is consequentlyThe full time-dependent density perturbation is the incident field plus $e^{i\omega t}\varphi$. The two traces at $y=0$ follow by limiting from above and below. A constant rescaling of the two factors does not change this field.

To determine the endpoint slope, reconstruct the sheet transform using the fluid condition:At $+b$ the continued $K^{+}$ has residue $-{\omega }^2/(2Tb{K}^{-}(b))$. Therefore the derivative transform $d^{+}$ generically has the lower-plane poleThis pole produces a forbidden incoming bare-sheet wave in the reconstructed displacement: $e^{i\omega t-ibx}$ travels towards the endpoint from $x=-\infty$. A minus transform of the specified outgoing solution must be analytic there. The incoming-pole cancellation at a pinned membrane edge is preciselyFor generic parameters this determinesThe lower pole is removed; the upper zero $-k_s$ of $K^{-}$ can still produce the legitimate outgoing evanescent acoustic surface wave, whose phase travels to negative $x$. This distinction separates the spurious bare-sheet wave from the coupled fluid-loaded mode. No explicit factorization has been used.

For rigid plates $x>0$, $y=\pm b$, an even incident plane wave travelling toward the opening can be scattered using Wiener-Hopf factorization. With time dependence $e^{i\omega t}$ and transform $\Phi (k)=\int \varphi (x)e^{ikx}dx$, take $\gamma =\sqrt{k^2-k_0^2}$ on the outgoing sheet. The common normal derivative on $y=b$ is supported on $x<0$; its transform is $D^{-}=i{L}^{+}(-k_0)L^{-}(k)/(k+k_0)$. The interior transform is $D^{-}\cosh (\gamma y)/(\gamma \sinh (\gamma b))$, while the exterior transform is $-D^{-}{e}^{-\gamma (|y|-b)}/\gamma$. These follow from continuity of normal velocity and the half-line pressure jump, followed by pole subtraction in a Wiener-Hopf equation and a finite-energy edge condition.

For a rigid screen $x<0$, use time dependence $e^{i\omega t}$ and inverse Fourier transform $e^{-i\kappa x}$. The scattered field is odd in $y$. Its upper-face trace is supported on $x<0$, while its normal derivative on $x>0$ is unknown. With the outgoing factor $\gamma =\sqrt{{\kappa }^2-k_0^2}$, these traces satisfy $\gamma {\hat{\varphi }}^{-}+G^{+}=b/(\kappa +a)$, where $a=k_0\cos {\theta }_0$ and $b=k_0\sin {\theta }_0$. Factor $\gamma =\sqrt{\kappa -k_0}\sqrt{\kappa +k_0}$ into upper/lower analytic factors. Pole subtraction in a Wiener-Hopf equation and the finite-energy edge condition give the displayed trace. The forcing-pole residue yields geometrical optics, while stationary phase yields the outgoing edge waveThe angular formula excludes the shadow-boundary transition regions; the exact integral supplies their uniform continuation.