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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.