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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.

Linearity permits subtraction of the incident acoustic membrane wave. The scattered pressure solves the homogeneous Helmholtz equation, and the kinematic relation defines its line displacement ${\eta }^{\prime }$ on both halves. On $x<0$, subtracting the incident elastic membrane equation gives the scattered dynamic equation. On $x>0$, there is no elastic membrane, so the total pressure jump must vanish, giving $[p^{\prime }]=-[p_I]$. The pinned endpoint has $\eta (0)=0$, hence ${\eta }^{\prime }(0)=-1$, rather than zero.

Use full and Half-range Fourier transforms with the same $e^{ikx}$ convention as part (a), and putWrite $P^{-}$ for the left transform of the upper scattered pressure, $P^{+}$ for its right transform, and ${\eta }^{\pm }$ for the corresponding transforms of ${\eta }^{\prime }$. Outgoing antisymmetry in $y$ impliesThe second relation comes from the prescribed pressure cancellation on $x>0$.

The endpoint terms in the left Fourier transform of ${\eta }_{xx}^{\prime }$ are crucial:Thus $2P^{-}=a{\eta }^{-}+TA+iT(k+k_I)$. Eliminating ${\eta }^{-}$ gives the Wiener-Hopf equationThere is a defect in the printed right side: with the stated incident exponential, pressure jump, and pinned total displacement, it omits the incident endpoint term and has the opposite sign on the incident-pressure term. The displayed corrected equation follows directly from both boundary traces. The given right side cannot be derived with these definitions. Using $a(k_I)=-2\beta /{\gamma }_I$, an especially useful equivalent form isThe correction is necessary for the reconstructed pressure to satisfy the original physical boundary conditions.

For Wiener-Hopf factorization, take $K=K^{+}{K}^{-}$ with the plus factor analytic and nonzero in the upper half-plane and the minus factor analytic and nonzero in the lower half-plane. Singularities listed next refer to continuation out of each factor's own analytic half-plane. Let $b=\omega \sqrt{m/T}$, so $a=-T(k-b)(k+b)$ and $b$ lies below the real axis. Generically $K^{+}$ inherits the pole at $b$ and the square-root branch point at $k_0$, while $K^{-}$ inherits the pole at $-b$ and the branch point at $-k_0$. The kernel is finite but nonanalytic at the acoustic branch points: its local behavior is a constant plus a square-root term, not an inverse-square-root divergence. Coincident branch points and poles require a limiting treatment.

Zeros of $K$ are the fluid-loaded dispersion relation roots. Lower-half-plane roots, including the incoming guided root $k_I$, belong to the continued $K^{+}$; upper-half-plane roots, including the reflected root $-k_I$ for the even kernel, belong to the continued $K^{-}$. The roots represent membrane-guided modes; the branch cuts represent radiated acoustic waves. Only roots on the selected outgoing sheet are included, not spurious roots created by squaring the dispersion relation. No explicit factorization is required.

Here is an explicit solution in terms of those factors. DefineThe pole of $B$ at $-b$ has residue $C$; the pole at $b$ is canceled by the pole of $K^{+}$. Thus $B^{\pm }$ have the required respective analyticity. Divide the corrected Wiener-Hopf equation by $K^{+}$ and split its right side as $F^{+}+F^{-}$, whereThe difference quotient in $F^{-}$ is removable at $k_I$ and is analytic below. Moving the plus and minus terms to opposite sides yields the common entire function. With the assumed $E(k)=0$, the solution isCombining $P^{-}$ with the known $P^{+}$ gives the full scattered pressure transformThe requested pressure integral, for $y\ne 0$, is consequentlyThe real contour $\mathcal{C}$ uses the causal continuation $\mathrm{Im}\omega <0$; its undamped limit retains the induced pole and branch-cut prescriptions. The residue theorem shows that upper-half-plane zeros of $K^{-}$ yield left-going scattered elastic membrane modes. On the right, the lower incident-pole residue of the scattered integral cancels $p_I$ at the open line, as required.

Finally reconstruct the left scattered displacement:For a generic unspecified $A$, this has a lower-half-plane pole at the bare elastic membrane wavenumber $b$. Such a pole is incompatible with analyticity of an outgoing left-supported scattered displacement: it would represent an additional right-going incoming elastic membrane contribution. It must be removed. Its residue is $-[2P^{-}(b)-TA-iT(b+k_I)]/(2Tb)$, so the incoming-pole cancellation at a pinned membrane edge condition isSince $P^{-}$ is linear in $A$, this fixes the endpoint slope generically. Explicitly it isThe physical lower incident pole of the total displacement is already prescribed by ${\eta }_I$; it must not be confused with this removable spurious pole of the scattered field.

For identical unit-density, unit-sound-speed fluids above and below a sheet on $x<0$, use time $e^{i\omega t}$ and transform $\int e^{ikx}\varphi dx$. The outgoing acoustic square-root branch obeys ${\gamma }^2=k^2-{\omega }^2$. If $J$ is the pressure jump upper minus lower, $s={\eta }^{\prime }(0)$ and $d^{+}$ the right Half-range Fourier transform of the scattered normal derivative, the Wiener-Hopf equation isThe total displacement is pinned, so its transformed second derivative is $s-k^2{\eta }^{-}$. The bare-sheet pole $b=\omega \sqrt{m/T}$ in reconstructed displacement must cancel: $J(b)=Ts$. This is an incoming-pole cancellation at a pinned membrane edge condition, distinct from the outgoing coupled-mode pole at the zero of $K$.