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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.

Use an acoustic velocity potential $\phi$ with $\mathbf{u}=\nabla \phi$ and $p^{\prime }=-{\rho }_0{\phi }_t$. It obeys the wave equation ${\phi }_{tt}=c_0^2{\nabla }^2\phi$. The rigid walls require zero normal velocity, hence Neumann boundary conditions on the transverse variables. Separation givesDefine ${\kappa }_{mn}^2=m^2/{\alpha }^2+n^2/{\beta }^2$. The dispersion relation, phase velocity and group velocity for positive $k,\omega$ areFor a dispersive mode, $\omega (0)=c_0{\kappa }_{mn}>0$, $c$ decreases from infinity to $c_0$, and $c_g$ increases from zero to $c_0$. The $(0,0)$ mode has $\omega =c_{0}k$ and $c=c_g=c_0$: it propagates without dispersion.

Let $\mathcal{A}={\pi }^2\alpha \beta$ be the cross-sectional area. The appropriate time and cross-sectional average at fixed $x$ isWrite $F_{mn}(y,z)$ for the product of cosines and $Q={\mathcal{A}}^{-1}\int F_{mn}^{2}dydz$. Its transverse eigenvalue and wall conditions give $\int |{\nabla }_{\perp }{F}_{mn}{|}^2={\kappa }_{mn}^2\int F_{mn}^2$ by integration by parts. The kinetic energy density and compressive acoustic energy density are ${\rho }_0|\mathbf{u}{|}^2/2$ and $(p^{\prime }{)}^2/(2{\rho }_0{c}_0^2)$. Time averaging each squared real oscillation contributes a factor $1/2$, soThis holds for every dispersive mode, and also for the zero mode.

For localized smooth initial data, a dispersive mode contains integrals of the form $\int a(k)e^{it[kV-\omega (k)]}dk$ along $x=Vt$. When $0<V<c_0$, a stationary point exists at $k_{\ast }={\kappa }_{mn}V/\sqrt{c_0^2-V^2}$. It is nondegenerate because ${\omega }^{\prime \prime }(k)=c_0{\kappa }_{mn}^2/(k^2+{\kappa }_{mn}^2{)}^{3/2}>0$. The stationary phase method therefore gives an oscillatory amplitude $O(t^{-1/2})$ for each excited dispersive mode, and for their sum under sufficient mode summability.

At $V=c_0$, no dispersive mode has a finite stationary wavenumber; with smooth localized data these contributions decay, while the nondispersive right-moving pulse retains its original amplitude. Thus a general disturbance with nonzero zero-mode component has an undiminished $O(1)$ pulse on the sound-speed ray. If that component is absent, this surviving contribution is absent too. For $V>c_0$, finite propagation speed puts a compactly supported initial disturbance outside this ray for all sufficiently large $t$, so the field is eventually zero. Rapidly localized, rather than strictly compact, data give a correspondingly negligible tail instead.

Write the density and pressure as ${\rho }_0+{\rho }^{\prime }$ and $p_0+p^{\prime }$, and let the small fluid velocity be the gradient of an acoustic velocity potential, $\mathbf{u}=\nabla \varphi$. The linear homentropic acoustic equations arewhere the adiabatic sound speed of a perfect gas is $c_0^2=\gamma p_0/{\rho }_0$. Absorbing a function of time into $\varphi$, the momentum equation givesEliminating $p^{\prime }$ and ${\rho }^{\prime }$ gives the acoustic wave equation

Let the shell displacement be $a(t)-a_0=\mathrm{Re}(\eta e^{-i\omega t})$ and put $k=\omega /c_0$, $\theta =k{a}_0$. The spherically symmetric solution regular at the centre isThe kinematic boundary condition ${\varphi }_r(a_0,t)=\dot{a}(t)$ givesSince $p^{\prime }=-{\rho }_0{\varphi }_t$, the pressure amplitude on the shell is thereforeThe radial Newton's second law per unit area isso its complex amplitude satisfiesFor a nonzero mode, cancellation of $\eta$ and multiplication by $a_0^2/(m{c}_0^2)$ yield the frequency equation for spherically symmetric vibration of a gas-filled elastic shell:The spectrum is discrete and contains infinitely many positive solutions. Indeed, for sufficiently large $n$, $\theta \cot \theta -1$ runs from $+\infty$ to $-\infty$ on $(n\pi ,(n+1)\pi )$, forcing at least one root there. These high modes are predominantly standing acoustic modes of the gas, while the lowest mode is predominantly the shell's radial breathing mode; the coupling shifts both families.

Use the acoustic velocity potential conventionPutFor a spherically symmetric harmonic solution, writeThe wave equation andgive the radial Helmholtz equation

The rigid inner sphere requiresA form satisfying this condition automatically isIndeed, if the bracket is $h(r)$, then $h(R)=1$ and $h^{\prime }(R)=1/R$, so $(h/r{)}^{\prime }=0$ at $R$.

At the outer sphere,soSincewe obtainThereforeThe assumption $\alpha \le \pi /2$ keeps the displayed denominator positive.

Use the acoustic velocity potential conventionFor a spherically symmetric outgoing wave with time dependence $e^{i\omega t}$ and $k=\omega /c_0$, writeLinearizing the boundary condition at the mean surface $r=a$ givesSincewe findThus the outgoing acoustic field of a pulsating sphere is