= Solution
Take $\omega>0$, with time convention $e^{-i\omega t}$, and use the <Fourier transform> pair $\widehat h(k)=\int h(x)e^{-ikx}\,dx$, $h(x)=(2\pi)^{-1}\int\widehat h(k)e^{ikx}\,dk$. For an <acoustic velocity potential>, $\mathbf u=\nabla\phi$ and $p=-\rho_0\phi_t$. Its harmonic amplitude solves the <Helmholtz equation>. The outgoing Fourier component has the form
$$
\widehat\phi(k,y)=A(k)e^{-\gamma y},\qquad \gamma^2=k^2-k_0^2,\qquad k_0=\omega/c_0.
$$
The <outgoing acoustic square-root branch> is fixed by decay for evanescent components and upward radiation for propagating components:
$$
\gamma(k)=\begin{cases}\sqrt{k^2-k_0^2}>0,&|k|>k_0,\\-i\sqrt{k_0^2-k^2},&|k|<k_0.\end{cases}
$$
A precise causal prescription is to replace $\omega$ by $\omega+i\sigma$, $\sigma>0$, select $\operatorname{Re}\gamma>0$ for real $k$, invert on the real $k$ axis, and take $\sigma\downarrow0$. 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 $-\gamma A=-i\omega\widehat\eta$. Consequently
$$
A=\frac{i\omega}{\gamma}\widehat\eta,\qquad \widehat p(k,0)=-\frac{\rho_0\omega^2}{\gamma}\widehat\eta.
$$
The <elastic membrane> balance therefore gives
$$
\left(Tk^2-m\omega^2-\frac{\rho_0\omega^2}{\gamma}\right)\widehat\eta=F.
$$
Writing $D=(Tk^2-m\omega^2)\gamma-\rho_0\omega^2$, the <one-sided fluid-loaded membrane radiation> formulas are
$$
\boxed{\eta(x,t)=\frac F{2\pi}\int_C\frac{\gamma e^{ikx-i\omega t}}{D(k,\omega)}\,dk,\qquad \phi(x,y,t)=\frac{i\omega F}{2\pi}\int_C\frac{e^{ikx-\gamma y-i\omega t}}{D(k,\omega)}\,dk.}
$$
The 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 $\gamma_s=-ik_0\sin\theta$. Applying the stated <method of steepest descent> result gives
$$
\phi_{\rm rad}\sim i\omega F\sqrt{\frac{k_0}{2\pi r}}\frac{\sin\theta}{D_s}e^{ik_0r-i\omega t-i\pi/4},\qquad D_s=-\rho_0\omega^2-ik_0\sin\theta(Tk_0^2\cos^2\theta-m\omega^2).
$$
Since $p=i\rho_0\omega\phi$ in amplitude notation, the angular <acoustic directivity> of the pressure, apart from angle-independent factors, is
$$
\boxed{\mathcal D(\theta)=\frac{\sin\theta}{\sqrt{\rho_0^2+k_0^2\sin^2\theta\,[m-(T/c_0^2)\cos^2\theta]^2}}.}
$$
The 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 $Tk^2-m\omega^2>0$. In that range,
$$
D_k=2Tk\gamma+(Tk^2-m\omega^2)\frac{k}{\gamma}>0,
$$
so the root is unique. Evenness supplies the second root, $-k_*$. 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 \b[there are exactly two real physical-sheet roots, $\pm k_*$, 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:
$$
\omega^2\left(m+\frac{\rho_0}{\gamma_*}\right)=Tk_*^2,\qquad \gamma_*=\sqrt{k_*^2-k_0^2}>0.
$$
Their normal profiles decay as $e^{-\gamma_*y}$, 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, $e^{-\gamma_*r\sin\theta}$.
For example, in the right half-plane put $k=k_0\cos\zeta$, $\gamma=-ik_0\sin\zeta$ and $k_*=k_0\cosh a$. The pole is at $\zeta=ia$ and the saddle at $\zeta=\theta$. The steepest-descent path has $\operatorname{Re}\cos(\zeta-\theta)=1$. At the pole height its left branch has real coordinate $\theta-\arccos(\operatorname{sech}a)$; comparing it with the original contour just to the right of the pole shows that its residue is crossed for
$$
0<\theta<\theta_s,\qquad \theta_s=\arccos(k_0/k_*).
$$
The 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.
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