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For one acoustic half-space and vacuum on the other side, a harmonic point force gives displacement amplitude $F\gamma /D$ and fluid acoustic velocity potential amplitude $i\omega F{e}^{-\gamma y}/D$. For time dependence $e^{-i\omega t}$, the outgoing acoustic square-root branch has $\gamma >0$ for $|k|>k_0$ and $\gamma =-i\sqrt{k_0^2-k^2}$ for $|k|<k_0$. The causal limit $\omega \mapsto \omega +i0$ places the positive guided-mode pole above the inversion contour and the negative pole below. The saddle point generates cylindrical radiation, whereas guided residues decay normally to the membrane.

Take $\omega >0$, with time convention $e^{-i\omega t}$, and use the Fourier transform pair $\hat{h}(k)=\int h(x)e^{-ikx}dx$, $h(x)=(2\pi {)}^{-1}\int \hat{h}(k)e^{ikx}dk$. For an acoustic velocity potential, $\mathbf{u}=\nabla \varphi$ and $p=-{\rho }_0{\varphi }_t$. Its harmonic amplitude solves the Helmholtz equation. The outgoing Fourier component has the formThe outgoing acoustic square-root branch is fixed by decay for evanescent components and upward radiation for propagating components:A precise causal prescription is to replace $\omega$ by $\omega +i\sigma$, $\sigma >0$, select $\mathrm{Re}\gamma >0$ for real $k$, invert on the real $k$ axis, and take $\sigma \downarrow 0$. It corresponds to switching on the forcing from the remote past. The limiting inversion contour $C$ passes below the positive outgoing surface-wave pole and above the negative one. The branch points are approached with the same causal prescription; analytic continuation during contour deformation must remain on this sheet. A principal real square root with positive imaginary part inside the acoustic interval would produce incoming sound and the wrong solution.

The elastic membrane's normal velocity equals the fluid velocity at the boundary, so $-\gamma A=-i\omega \hat{\eta }$. ConsequentlyThe elastic membrane balance therefore givesWriting $D=(T{k}^2-m{\omega }^2)\gamma -{\rho }_0{\omega }^2$, the one-sided fluid-loaded membrane radiation formulas areThe sign of the fluid term describes positive added mass of an evanescent fluid layer, not negative inertia.

For $0<\theta <\pi$, the acoustic saddle point is $k_s=k_0\cos \theta$, with ${\gamma }_s=-i{k}_0\sin \theta$. Applying the stated method of steepest descent result givesSince $p=i{\rho }_0\omega \varphi$ in amplitude notation, the angular acoustic directivity of the pressure, apart from angle-independent factors, isThe intensity directivity is proportional to ${\mathcal{D}}^2$. This cylindrical radiating field has amplitude of order $r^{-1/2}$. The formula is for fixed interior angles; grazing-angle limits require a separate uniform approximation.

For real $k>k_0$, $\gamma$ is positive, $D\to -{\rho }_0{\omega }^2$ as $k\downarrow k_0$, and $D\to +\infty$ as $k\to \infty$. A root exists. A root also requires $T{k}^2-m{\omega }^2>0$. In that range,so the root is unique. Evenness supplies the second root, $-k_{\ast }$. There are no real roots inside $|k|<k_0$: the fluid-loading term is real and nonzero while the other term is purely imaginary. Thus there are exactly two real physical-sheet roots, $\pm k_{\ast }$, for nonzero forcing frequency, positive elastic-sheet tension and positive fluid mass density. The zero-frequency static limit is not a two-distinct-root statement.

These roots are counterpropagating evanescent acoustic surface waves coupled to the membrane:Their normal profiles decay as $e^{-{\gamma }_{\ast }y}$, and their phase speed is below both $c_0$ and the vacuum membrane speed. They matter for the elastic membrane motion and for observations near its surface, but are not automatically part of the leading far-field radiation. A residue is picked up only when the causal contour deformation crosses its pole; at fixed $y/r>0$ its magnitude is exponentially small, $e^{-{\gamma }_{\ast }r\sin \theta }$.

For example, in the right half-plane put $k=k_0\cos \zeta$, $\gamma =-i{k}_0\sin \zeta$ and $k_{\ast }=k_0\cosh a$. The pole is at $\zeta =ia$ and the saddle at $\zeta =\theta$. The steepest-descent path has $\mathrm{Re}\cos (\zeta -\theta )=1$. At the pole height its left branch has real coordinate $\theta -\arccos (\mathrm{sech}a)$; comparing it with the original contour just to the right of the pole shows that its residue is crossed forThe analogous sector lies near $\theta =\pi$ for the negative pole. This is why the existence of real dispersion zeros alone does not make their residues relevant in every observation direction.

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.

Define $k_0=\omega /c_0$ and use the outgoing acoustic square-root branchFor positive real frequency reached from below, $\gamma =i\sqrt{k_0^2-k^2}$ on the propagating interval and is positive real for $|k|>k_0$. The outgoing field in the lower fluid has pressure amplitude $B{e}^{\gamma y}$, since $y<0$. If $\eta =\hat{\eta }{e}^{i\omega t-ikx}$, the shared normal velocity is $i\omega \hat{\eta }$. The linear homentropic acoustic equations therefore giveThis lower-fluid wave is outgoing; no additional incoming sound is included in defining the impedance seen by the upper fluid.

Put $K_s=T{k}^2-m{\omega }^2$. The sheet's force balance gives $K_s\hat{\eta }=-P+B$, hence $P=({\rho }_0{\omega }^2/\gamma -K_s)\hat{\eta }$. Its prescribed downward velocity amplitude is $V=-i\omega \hat{\eta }$. Thus the tensioned-sheet acoustic impedance isThe first term is the lower fluid's normal acoustic impedance, and the second is the sheet's inertial and elastic-sheet tension response. For a real propagating angle, $Z_f={\rho }_0{c}_0/\sin \theta$.

At fixed nonzero frequency and fixed wavenumber, $m\to \infty$ gives $|Z|\to \infty$, a zero-velocity, in-phase reflecting boundary. At fixed $k\ne 0$, $T\to \infty$ gives the same reflection limit, but there is an important exception: elastic-sheet tension does not resist the spatially uniform mode $k=0$. At normal incidence, the impedance remains ${\rho }_0{c}_0+im\omega$ however large the elastic-sheet tension is. With $m=0$ this mode is transparent. By contrast, arbitrarily large mass resists even a spatially uniform oscillation. These fixed-frequency limits exclude a simultaneously tuned structural resonance.

If $m=T=0$, $Z=Z_f$ and $R=0$. The identical fluids are effectively joined across a massless, untensioned interface: pressure and normal velocity continue without reflection. This is the matched case, rather than the pressure-release case.

With no incoming wave, both fluids obey the outgoing acoustic square-root branch. Their surface pressure amplitudes areSubstitution in the sheet's force balance yields the dispersion relation for an acoustic membrane wave:The equivalent denominator $\Delta =\gamma (T{k}^2-m{\omega }^2)-2{\rho }_0{\omega }^2$ is useful away from the branch points $\gamma =0$. For real guided modes with $|k|>k_0$, this is the familiar added mass of an evanescent fluid layer formThe normal fields decay away from the sheet; continued by analytic continuation roots may describe radiating or leaky modes.

On $D=0$, $T{k}^2-m{\omega }^2=2{\rho }_0{\omega }^2/\gamma$, so the impedance from the preceding solution isThus the divergence of the reflection coefficient corresponds to a reflection pole of a fluid-loaded membrane. The prescribed incoming amplitude is zero, but a homogeneous fluid-sheet mode can have nonzero amplitude. This is a resonance or guided-mode pole of the analytic scattering problem, not arbitrarily large passive reflection at a real propagating angle. In particular, an undamped guided mode has an evanescent normal field and therefore a complex incidence angle in the plane-wave continuation. For real propagating incidence, $Z_f>0$ and the finite-mass, finite-tension sheet has a purely imaginary structural impedance, so its passive reflection remains bounded.

Use the Fourier transform pair $\hat{q}(k)=\int q(x)e^{ikx}dx$, $q(x)=(2\pi {)}^{-1}\int \hat{q}(k)e^{-ikx}dk$. The point force transforms to $F$. With the dispersion relation $D$ defined above, the sheet equation becomesThe point-force radiation from a fluid-loaded sheet is therefore represented exactly byThe causal contour and outgoing acoustic square-root branch are fixed first with $\mathrm{Im}\omega <0$, then continued to the desired real frequency. This prescription fixes how poles and the branch points are passed.

For the acoustic far field $x=r\cos \theta$, $y=r\sin \theta$, take $0<\theta <\pi$ bounded away from grazing and $k_{0}r\gg 1$. The method of steepest descent saddle point is $k_s=k_0\cos \theta$, with ${\gamma }_s=i{k}_0\sin \theta$. The supplied saddle point rule, including its $\sin \theta$ factor, gives, provided the contour deformation crosses no poles,A convenient simplification, free of division by $T$, isEquivalently, when $T\ne 0$,The expression printed in the PDF is missing sound-speed factors for general dimensional $c_0$. It agrees with this result if $c_0=1$ in fully normalized units; when $c_0$ is retained as an arbitrary sound speed, the numerator needs $c_0^{-3/2}$ and the structural term needs $T/c_0^2$ in the last form. These factors arise respectively from cylindrical spreading, the pressure-density relation, and $k_s=\omega \cos \theta /c_0$.

A direct countercheck is the transparent-sheet limit $m=T=0$. The sheet jump condition then gives ${\hat{p}}_{+}(k,0)=F/2$, so the saddle point rule requiresThe printed expression, interpreted continuously after multiplying out its structural factor, instead gives $F(2c_0{)}^{-1}\sqrt{\omega /(2\pi r)}\sin \theta$ times the same phase. It differs by a factor $c_0^{3/2}$; for example it is eight times too large when $c_0=4$. This limit also verifies the normalization of the corrected density field independently of the sheet's elastic-sheet tension.

To decide about poles, track the roots of $\Delta (k,\omega )$ on the chosen square-root sheet and deform the original causal contour to the steepest descent contour. A root contributes a residue exactly when it lies in the region swept out by that deformation; its sign is fixed by the contour orientation. Branch cuts must be retained throughout this comparison. Which roots are crossed can depend on observation angle, producing a change of the modal contribution when a pole meets the deformation boundary. A saddle point approaching a pole or a grazing endpoint requires an approximation uniform in that limit, rather than the isolated saddle point formula above.

The crossed poles are the free fluid-sheet modes of the preceding solution. Real subsonic roots represent evanescent acoustic surface waves carrying energy along the sheet, with normal decay; complex continuations represent leaky or radiating modes. Their residues must be added to the saddle point sound when the causal contour selects them. The specification “no poles contribute” is therefore a substantive condition on the contour, not permission to ignore zeros of the dispersion relation.

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 $K$.