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A pinned semi-infinite membrane scatters an incoming guided mode into reflected guided modes and outgoing sound. For unit incident displacement the scattered endpoint value is $-1$. Its Half-range Fourier transform therefore has nonzero endpoint terms even though the total displacement vanishes. The Wiener-Hopf equation kernel is $2/(m{\omega }^2-T{k}^2)+\gamma /({\rho }_0{\omega }^2)$ with the outgoing branch ${\gamma }^2=k^2-{\omega }^2/c_0^2$.

Suppress the common time factor $e^{i\omega t}$, take $\omega >0$, and fix the spatial Fourier transform conventionA plus transform is supported on $x>0$ and analytic above $\Gamma$; a minus transform is supported on $x<0$ and analytic below it. For the outgoing radiation condition with this time convention, initially take $k_0=\omega /c_0-i\epsilon$, $\epsilon >0$, and pass to the limit at the end. Thus $+k_0$ lies below the contour and $-k_0$ above it.

Choose $\gamma =(k^2-k_0^2{)}^{1/2}$ to have positive real part on the real transform line. Its branch cuts run from $+k_0$ into the lower half-plane and from $-k_0$ into the upper half-plane, without crossing $\Gamma$; downward and upward vertical rays are suitable. In the zero-absorption limit, $\gamma >0$ for real $|k|>k_0$ and $\gamma =+i\sqrt{k_0^2-k^2}$ between the branch points. This ensures that $e^{-\gamma |y|}$ represents decay or outgoing radiation, rather than an incoming exterior field.

Evenness in $y$ and the Helmholtz equation give the transformed fieldsAt $y=b$, the one-sided normal derivatives of the scattered field agree for $x<0$, because that part of the interface is open. For $x>0$ they both vanish by rigidity. Their common trace is therefore a minus function, denoted $V^{-}(k)={\partial }_y{\Phi }^{-}{|}_{y=b}$. ConsequentlyThe jump in the total scattered transform isFor $x<0$, continuity of the total mass density requires the scattered jump to cancel the incident jump, so $\varphi (b^{+})-\varphi (b^{-})=e^{i{k}_{0}x}$. Its minus transform is $-i/(k+k_0)$. The unknown plate-side jump is the plus function $J^{+}=[{\Phi }^{+}{]}_{b^{-}}^{b^{+}}$. Thus $J=J^{+}-i/(k+k_0)$, and the Wiener-Hopf equation follows:

Use the Wiener-Hopf factorization $L=L^{+}{L}^{-}$, with factors analytic and nonzero in their designated half-planes. Their analytic continuations allocate outgoing modal zeros to $L^{+}$ below the contour, and the opposite zeros to $L^{-}$ above it. Set $C=L^{+}(-k_0)$. Multiplication by $L^{+}$ and pole subtraction giveThe pole in the upper expression is removable by its numerator. The two expressions analytically continue to the common entire function, which is zero under the stipulated edge/growth assumption. Henceand the transformed fields areA constant reciprocal rescaling of the factors does not alter $C{L}^{-}$ or the physical field.

Inside the guide, $\cosh (\gamma y)$ and $\gamma \sinh (\gamma b)$ are even entire functions of $\gamma$, while $L^{-}$ is analytic in the lower half-plane. Thus continuation across the lower branch cut changes none of the interior transform: that cut is removable. The exterior expression retains the cut, corresponding to radiation into the open exterior.

For $x>0$, the inverse-transform contour closes downwards, clockwise. Away from modal cutoffs, its enclosed singularities are the outgoing simple poleswith positive real part for propagating modes and negative imaginary part for decaying modes; $k_0$ is the $n=0$ root. The opposite roots lie above the contour or cancel against zeros of $L^{-}$. Symmetry excludes odd transverse modes. At $n\ge 1$, $\gamma =in\pi /b$, and differentiation of $D(k)=\gamma \sinh (\gamma b)$ givesEach inverse-transform contribution is $-i$ times its residue. Therefore the outgoing modes of an open rigid acoustic waveguide areThe original time factor multiplies this expression. Thus every cut-on mode propagates in the positive $x$ direction; cut-off modes are evanescent waves decaying into the guide, not additional backward waves. Strictly, a complete mode sum includes these evanescent modes as well as propagating ones. The displayed simple-pole amplitudes apply away from exact cutoffs; cutoff values use the outgoing limiting-absorption continuation before taking the limit. The reflected plane-wave amplitude relative to the unit incident wave is $A_0$ above. It has the expected negative sign in the long-wavelength leading kernel approximation $L\simeq b(k^2-k_0^2)$.

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.