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For a linear acoustic perturbation of a uniform fluid,is the sum of kinetic and compressional energy densities. Together with the acoustic energy flux $\mathbf{I}=p^{\prime }\mathbf{u}$, it satisfies $E_t+\nabla \cdot \mathbf{I}=0$.

A membrane of mass per area $m$ and tension $T$ separating two identical acoustic half-spaces has evanescent modes withThe second term is the added mass of an evanescent fluid layer contributed by both sides. Hence the phase speed is below the vacuum membrane speed $\sqrt{T/m}$, and for $k\to 0$ it obeys $c\sim \sqrt{Tk/(2{\rho }_0)}\to 0$. Pressure and normal velocity are in quadrature, so the mean normal acoustic energy flux vanishes.

For a mode $u=a{e}^{i\omega (p_{\alpha }{x}_{\alpha }+q{x}_3-t)}$ with $x_3$ positive downwards, the reduced traction is $(Tq+R)a$. Its mean vertical acoustic energy flux is the displayed expression. Propagating modes are downgoing for positive flux and upgoing for negative flux; phase-slowness sign need not agree in anisotropy. Depth-decaying evanescent modes have $\mathrm{Im}q>0$. At grazing or repeated roots use limiting radiation modes.

An elastic plate coupled to one compressible fluid half-space has an outgoing Fourier transform kernel $D$, where ${\gamma }^2=k^2-{\omega }^2/c_0^2$. A harmonic line force $F_0$ produces displacement transform $F_0\gamma /D$. Propagating acoustic components have $\gamma =-i\sqrt{{\omega }^2/c_0^2-k^2}$, whereas guided components decay normally to the plate. The outgoing prescription separates acoustic energy flux from guided-wave energy transport.

The normal acoustic impedance of a progressive acoustic wave at a planar boundary is the pressure amplitude divided by the normal-velocity amplitude. For propagation angle $\theta$ from the normal,The distinction from the acoustic impedance $\rho c$ matters at oblique incidence: interface reflection depends on the ratio of normal impedances. For an evanescent outgoing wave $k_n$ is imaginary and the normal impedance is reactive, giving no mean normal acoustic energy flux.

With pressure amplitude $P=ZV$ and velocity $V$ directed into a boundary, the mean absorbed acoustic energy flux is $\frac{1}{2}\mathrm{Re}(P{V}^{\ast })=\frac{1}{2}\mathrm{Re}Z|V{|}^2$. A passive boundary cannot supply net energy, so $\mathrm{Re}Z\ge 0$ at a real frequency. Its reflection from a propagating acoustic plane wave has magnitude at most one. A pole continued to a complex incidence angle or frequency must be interpreted separately from real-frequency passive scattering.

Let $\mu =\pi /H$. The end-wall forcing has transverse factor $\cos (\mu y)$, which already satisfies both fixed-wall Neumann boundary conditions. By linearity and separation of variables, seek its periodic response asSubstitution into the wave equation and the moving-wall condition givesThe acoustic waveguide cutoff is thereforeIt is the lowest nonzero transverse-mode frequency. The uniform transverse mode has no cutoff, but is not excited here because ${\int }_0^H\cos (\pi y/H)dy=0$.

For $0<\omega <{\omega }_c$, put $\alpha =\sqrt{{\mu }^2-{\omega }^2/c^2}>0$. The axial equation has solutions $e^{\pm \alpha x}$. Boundedness removes the growing exponential, and ${\Phi }^{\prime }(0)=ϵ\omega$ fixes the coefficient. ThusThis evanescent wave oscillates in time but decays with penetration length $1/\alpha$. Its axial velocity is proportional to $\cos (\omega t)$, while its excess pressure $p^{\prime }=-{\rho }_0{\varphi }_t$ is proportional to $\sin (\omega t)$. They are in phase quadrature, so their mean product, the axial acoustic energy flux, is zero.

For $\omega >{\omega }_c$, put $q=\sqrt{{\omega }^2/c^2-{\mu }^2}>0$. The axial solutions are $e^{\pm iqx}$. The outgoing condition selects $e^{iqx}$ for the chosen time convention, andTaking the real part givesIts axial velocity is $ϵ\omega \cos (\pi y/H)\cos (qx-\omega t)$, so the moving-wall condition is satisfied with the correct sign. This is a traveling acoustic waveguide mode carrying energy toward positive $x$. Its dispersion relation isThe larger axial phase velocity does not imply faster energy transport: the group velocity controls that transport.

At $\omega ={\omega }_c$, the axial equation becomes ${\Phi }^{\prime \prime }=0$. The imposed nonzero derivative forces $\Phi =ϵ{\omega }_{c}x+$ a constant, which is unbounded. Thus there is no bounded periodic response at exact cutoff for this infinite, undamped, velocity-driven model. The divergent coefficients as $q\to 0$ or $\alpha \to 0$ also show why the assumed small perturbations fail sufficiently close to cutoff at fixed forcing. Initial conditions can add free transients; the displayed formulas give the selected periodic forced response.

Use the usual minor and major symmetries of the elastic stiffness tensor, positive density, and strong ellipticity in elasticity. For the plane elastic wave $u=a{e}^{ik(n\cdot x-vt)}$, substitution into momentum balance gives the acoustic tensor problemThe major symmetry makes $\Gamma$ a real symmetric matrix. Moreover $a^{\mathsf{T}}\Gamma a=c_{ijpq}{a}_i{n}_j{a}_p{n}_q>0$ for nonzero $a$ under strong ellipticity in elasticity. Hence its three eigenvalues are positive and the three positive wave speeds are real. Symmetry alone proves only reality of $v^2$: an unstable isotropic tensor with negative shear modulus has negative transverse $v^2$. The requested reality therefore uses stability as part of the physical elastic-solid assumption.

The elastic slowness surface consists of three sheets, with $s=n/v_{\alpha }(n)$, satisfyingIt is centrosymmetric. Sheets can meet at degenerate polarizations; they need not be spherical or convex in an anisotropic solid. The group velocity is normal to the corresponding smooth sheet: for a sheet tangent $ds$, differentiation of $a^{\mathsf{T}}[c_{ijpq}{s}_j{s}_q-\rho I]a=0$ gives $a_i{c}_{ijpq}{a}_p{s}_{q}d{s}_j=0$. The coefficient vector is proportional to the group velocity, so its scalar product with every tangent vanishes.

Fix horizontal slowness $p=(n_1/c,n_2/c)$ and put $q=s_3$. DefineThe symbol $p$ on a tensor index is an index, whereas $p_{\alpha }$ is a horizontal slowness component. The vertical slowness equation is the vertical slowness sextic of an anisotropic solidThe leading coefficient is $\det T>0$, so it is a degree-six polynomial with real coefficients. Nonreal roots occur in conjugate pairs, leaving between zero and six real roots counted with multiplicity. Generically the counts are $0,2,4,6$; a grazing repeated root can give an odd count of distinct real roots. Geometrically these are the intersections of a vertical line of fixed horizontal slowness with the three sheets.

All four generic counts are actually possible in a stable material. For example, take $\rho =1$ and orthotropic Voigt stiffnessesThe stiffness matrix is positive definite. For horizontal slowness $(p,0)$, the three diagonal acoustic equations areThe choices $p=0.1,0.3,0.8,1.2$ give respectively six, four, two and zero real vertical roots.

For the general depth-dependent amplitude, substitution of the horizontal harmonic factor yieldsFor six simple roots $q_{\ell }$, choose nonzero null vectors $a_{\ell }$ of the corresponding matrix. The complete solution isComplex vertical roots give evanescent waves. At defective repeated roots the usual polynomial-in-depth factors multiplying exponentials must be included, or the solution can be defined by a limiting procedure.

To classify the modes, take $x_3$ positive downwards and $\omega >0$. The reduced traction for a mode isMechanical energy crosses a plane at rate $-{\sigma }_{i3}{u}_{i,t}$. Averaging the real harmonic fields uses $⟨\mathrm{Re}(f{e}^{-i\omega t})\mathrm{Re}(g{e}^{-i\omega t})⟩=\mathrm{Re}(f{g}^{\ast })/2$. With velocity amplitude $-i\omega a_{\ell }$ and traction amplitude $i\omega b_{\ell }$, the time-averaged vertical acoustic energy flux isFor a real propagating mode, write its physical wavevector as $k=\omega (p_1,p_2,q)$. Differentiate $a^{\mathsf{T}}\Gamma (k)a=\rho \omega (k{)}^2{a}^{\mathsf{T}}a$ in $k_3$. The derivatives of $a$ cancel by the eigenvalue equation, while $a^{\mathsf{T}}({\partial }_{k_3}\Gamma )a=2\omega a^{\mathsf{T}}(Tq+R)a$. Thus its vertical group velocity and flux areThus the elastic mode classification by vertical energy flux is upgoing if $J_3<0$, downgoing if $J_3>0$. In anisotropy, the sign of vertical phase slowness alone need not classify energy propagation. For evanescent modes, select $\mathrm{Im}q>0$ for decay into a downward half-space and $\mathrm{Im}q<0$ for decay into an upward half-space. Away from grazing and degeneracy, continuation from zero horizontal slowness, with a small radiation damping if needed, separates three outgoing modes on each side. Superpose only the appropriate three to construct the most general upgoing or downgoing field; limiting modes handle exceptional cases.

For a free surface at $x_3=0$ with solid occupying $x_3>0$, let $(q_0,a_0)$ be a unit-amplitude incident upgoing mode. The reflected field consists of the three downgoing propagating or depth-decaying modes:Frequency and horizontal slowness are common to all modes, by translation invariance of the surface. Zero surface traction imposes the three equationsHence the free-surface anisotropic reflection amplitudes arewhenever $B$ is invertible. This specifies mode conversion as well as reflection. If $B$ is singular, a nonzero outgoing homogeneous surface mode exists; these are surface-wave poles, including Rayleigh waves in the isotropic case. At such parameters one must use the appropriate limiting radiation solution and check solvability instead of assuming a unique finite inverse.

Fluid-loaded plate. Use complex amplitudes with time dependence $e^{-i\omega t}$ and take the velocity potential in the fluid above the elastic plate to be $\Phi$. Let ${\rho }_0$ and $c_0$ be the undisturbed mass density and speed of sound. The linearized Euler equations give $p=i{\rho }_0\omega \Phi$; the kinematic boundary condition is ${\Phi }_y(x,0)=-i\omega \eta (x)$. We model loading from one fluid half-space, with no fluid loading on the other face. Loading by identical fluids on both faces instead doubles the fluid term below.

Define the Fourier transform and its inverse byThe Sommerfeld radiation condition selects $\hat{\Phi }(k,y)=\hat{\Phi }(k,0)e^{-\gamma y}$, with ${\gamma }^2=k^2-k_0^2$ and $k_0=\omega /c_0$. Here and in the diffraction calculation below, the outgoing square root is specified by first taking $\mathrm{Im}\omega >0$. Draw a vertical cut upwards from $+k_0$ and another downwards from $-k_0$, and putThus ${\gamma }_u$ is analytic in the upper half-plane and ${\gamma }_l$ in the lower half-plane. In the real-frequency limit,The Fourier transform contour $C$ follows the real axis with the limiting-absorption pole prescriptions. In particular, an outgoing propagating component has positive vertical wavenumber.

The kinematic boundary condition gives $\hat{\Phi }(0)=i\omega \hat{\eta }/\gamma$, so $\hat{p}(0)=-{\rho }_0{\omega }^2\hat{\eta }/\gamma$. Eliminating the fluid pressure from the bending balance gives the line-forced fluid-loaded bending plate kernelFor harmonic line force amplitude $F_0$ at the origin, the plate displacement and fluid velocity potential are thereforeAn arbitrary localized loading replaces $F_0$ by its Fourier transform. This representation both calculates the response and identifies its different wave components through poles and branch points.

For an observer $x=r\cos \theta$, $y=r\sin \theta$, with $0<\theta <\pi$, the method of steepest descent uses the saddle point $k_s=k_0\cos \theta$. At fixed angle away from grazing, the acoustic far field isThe $r^{-1/2}$ spreading is that of cylindrical acoustic waves. The phase velocity parallel to the plate is $\omega /|k|$: components with $|k|<k_0$ are supersonic in that sense and radiate, whereas those with $|k|>k_0$ are subsonic and exponentially decay normally to the plate. The latter have reactive fluid loading and zero mean normal acoustic energy flux.

There is also a real guided root $\kappa >k_0$. Indeed, $K(k_0+)=-\infty$, $K(+\infty )=+\infty$, andThe residue theorem then gives an outgoing displacement contribution $i{F}_0{e}^{i\kappa |x|}/K^{\prime }(\kappa )$. It transports energy along the plate while its fluid field decays as $e^{-\gamma (\kappa )y}$. Other zeros on continued sheets can describe leaky or resonant waves; branch point contributions describe near and grazing fields. A fixed-angle acoustic far field expansion is not uniform near the plate, where these components can matter.

For a propagating spectral component, write $\ell =(k_0^2-k^2{)}^{1/2}$. Its upward normal velocity is $i\ell \hat{\Phi }$ and its pressure is $i{\rho }_0\omega \hat{\Phi }$. Applying Parseval's identity to the time-averaged acoustic energy flux gives the upward power per unit length of the forcing line:Equivalently, the integrand is $|\hat{\eta }{|}^2/\ell$ with prefactor ${\rho }_0{\omega }^3/(4\pi )$. Integrating the radial acoustic intensity of the far field over a semicircle gives the same expression. The input power $\omega \mathrm{Im}(F_0^{\ast }\eta (0))/2$ also feeds guided waves along the plate, so it must not be identified with $P_y$ alone.

Rigid half-plane diffraction. Suppress the time factor and assume $0<{\theta }_0<\pi$. Write $a=k_0\cos {\theta }_0$, $b=k_0\sin {\theta }_0$, and denote the incident spatial field by $I=e^{-iax-iby}$. The Helmholtz equation, Sommerfeld radiation condition, and square-root branches are as above. Across the open positive axis the scattered field and its normal derivative are continuous. Across the rigid negative axis its normal derivative is also continuous, while its potential can jump. Let $J(x)$ be this jump, supported on $x<0$. The transformed scattered field isThe jump transform is analytic in the upper half-plane. The absence of an even scattered component follows from derivative continuity and the outgoing condition. Its normal derivative on either face is $-\gamma {\hat{J}}_u/2$.

On the negative axis this derivative must cancel $-ib{e}^{-iax}$. Since the Half-range Fourier transform of $H(-x)e^{-iax}$ is $i/(k+a)$, the Wiener-Hopf equation has the formwhere $G_l$ is the transform of an unknown function supported on $x>0$ and analytic in the lower half-plane. The incident pole $k=-a$ lies below the Fourier contour. Divide by ${\gamma }_l$ and perform pole subtraction in a Wiener-Hopf equation:The first term is upper analytic and the second lower analytic, since the latter's apparent pole cancels. Equality of the two continued sides gives an entire function. The finite-energy edge condition, with $J(x)=O(\sqrt{-x})$, and the decay of the transforms at infinity force that function to vanish by Liouville's theorem. ConsequentlyThis is the Wiener-Hopf solution of rigid half-plane diffraction with the present transform and time conventions.

Deforming to a steepest descent contour gives a residue from the incident pole precisely when $x+|y|\cot {\theta }_0<0$. At that pole $\gamma (-a)=-ib$; its residue producesThus the total geometrical optics field is $I$ plus a reflected wave in the upper reflection region; in the lower shadow region the scattered residue cancels $I$, and outside that shadow the direct incident wave remains. This gives both the amplitude and the spatial support of the geometrical optics fields rather than merely naming a reflected wave.

For $x=r\cos \theta$, $y=r\sin \theta$, $-\pi <\theta <\pi$, the saddle contribution uses ${\gamma }_l(-a)=-i\sqrt{2k_0}\cos ({\theta }_0/2)$ and ${\gamma }_u(k_s)=\sqrt{2k_0}\cos (\theta /2)$. The diffracted wave isThe total field is $I+{\varphi }_{\mathrm{GO}}+{\varphi }_d$ to this order. This far-field expression applies away from the shadow/reflection boundaries $|\theta |=\pi -{\theta }_0$. There the pole and saddle coalesce, and a uniform Fresnel integral approximation replaces the divergent separate terms by a smooth transition. The exact contour integral remains well defined.

For inviscid barotropic flow, conservation gives ${\rho }_t+\nabla \cdot (\rho u)=0$ and $\rho (u_t+u\cdot \nabla u)=-\nabla p$. Linearizing about uniform rest ${\rho }_0,p_0$ giveswith the derivative adiabatic for ordinary sound.

Taking curl gives ${\partial }_t(\nabla \times u)=0$. Therefore the acoustic velocity is irrotational if the initial vorticity is zero. Uniform rest as the background does not alone forbid a superposed stationary vortical perturbation: for example $u=\epsilon (\sin y,0,0)$, ${\rho }^{\prime }=p^{\prime }=0$ solves the linearized equations but has nonzero curl. Thus the printed irrotational conclusion requires the usual restriction to the longitudinal acoustic mode. In that mode set $u=\nabla \varphi$ and absorb a time-only gauge into $\varphi$, so $p^{\prime }=-{\rho }_0{\varphi }_t$. Continuity givesThis separates the sound wave from the stationary vorticity mode in linear acoustics.

Define the quadratic acoustic energy density and acoustic energy flux byMultiplying momentum by $u$ and the pressure form of continuity by $p^{\prime }/({\rho }_0{c}_0^2)$ gives

For any progressive plane wave, not necessarily sinusoidal, let $\varphi =F(\xi )$ with $\xi =\hat{k}\cdot x-c_{0}t$. Then $u=F^{\prime }(\xi )\hat{k}$ and $p^{\prime }={\rho }_0{c}_0{F}^{\prime }(\xi )$. Its kinetic and compressional energy densities both equal ${\rho }_0(F^{\prime }{)}^2/2$, soThe energy is transported at the sound speed in the propagation direction. A superposition of oppositely traveling waves can form a standing wave and need not have this pointwise equality; the statement concerns a single progressive wave.

For a spherical outgoing wave, $\varphi =F(r-c_{0}t)/r$. Its radial velocity and pressure areThereforeThey are generally unequal because of the additional near-field velocity term; their difference is ${\rho }_0[F^2/r^4-2F{F}^{\prime }/r^3]/2$. They agree only to leading far-field order, or for special profile values. This is the near-field energy imbalance in a spherical acoustic wave.

Linearized continuity and Euler equations about a uniform resting inviscid fluid are ${\rho }_t^{\prime }+{\rho }_0\nabla \cdot \mathbf{u}=0$ and ${\rho }_0{\mathbf{u}}_t=-\nabla p^{\prime }$. For a velocity potential $\mathbf{u}=\nabla \varphi$ and sound-speed relation $p^{\prime }=c^2{\rho }^{\prime }$, integration of Euler's equation fixes the temporal gauge so $p^{\prime }=-{\rho }_0{\varphi }_t$. Substitution into continuity givesUse complex amplitudes with time factor $e^{i\omega t}$. The outgoing wave in fluid two is ${\Phi }_2(x)=A{e}^{-i{k}_{2}x}$, with $k_i=\omega /c_i$. Write ${\Phi }_1=C\cos (k_{1}x)+D\sin (k_{1}x)$. Continuity of pressure and normal velocity at zero givesThe piston velocity is $i\omega ϵ$, so ${\Phi }_1^{\prime }(-L)=i\omega ϵ$. With $\theta =\omega L/c_1$, this givesand hence the requested amplitude isFor complex pressure and velocity amplitudes, the mean acoustic energy flux is $\frac{1}{2}\mathrm{Re}(P{U}^{\ast })$. Here $P=-i\omega {\rho }_2{\Phi }_2$ and $U=-i{k}_2{\Phi }_2$, soFor ${\rho }_2\ll {\rho }_1$ and comparable sound speeds, weak transmission loads the finite segment. There are narrow large-flux resonances near $\theta =(n+1/2)\pi$, separated by $\Delta L=\pi c_1/\omega$. The off-resonance scale is proportional to ${\rho }_2$, while peak flux is proportional to ${\rho }_1^2/{\rho }_2$; the angular width is of order $({\rho }_2/{\rho }_1)(c_2/c_1)$. These amplified solutions remain within linear acoustics only when the resulting wave amplitudes are small.

Put $\beta =k\xi \mu$. The random phase screen producesInterpret the normality assumption as a jointly Gaussian process, not merely Gaussian one-point marginal distributions. Write its normalized autocorrelation as $\rho (\zeta )=⟨W(z+\zeta )W(z)⟩$. The difference $W(z+\zeta )-W(z)$ is a zero-mean Gaussian random variable with variance $2(1-\rho (\zeta ))$. Its characteristic function therefore gives the Gaussian phase-screen correlationWithout joint Gaussianity the autocorrelation alone does not determine this expectation.

For an acoustic pressure amplitude $\psi$, with background mass density ${\rho }_0$, the time-averaged acoustic energy flux isAn acoustic potential convention changes the dimensional prefactor. The physical field includes the carrier $e^{ikx}$: a transverse Fourier transform mode has axial wavenumber ${\kappa }_{\nu }=k-{\nu }^2/(2k)$ in the parabolic wave equation. Thus its spectral acoustic flux is proportional to ${\kappa }_{\nu }|\hat{E}(x,\nu ){|}^2$, or to $k|\hat{E}(x,\nu ){|}^2$ at leading paraxial order. Differentiating the reduced field alone would omit the carrier contribution.

The infinite stationary illumination has a spectral measure of a stationary random field, rather than a finite ordinary value of $⟨|\hat{E}{|}^2⟩$ at each $\nu$. To specify the normalization, truncate the screen to an aperture of length $L$, propagate this truncated initial field, and setThe propagation multiplier has modulus one. Using the Gaussian phase-screen correlation and the overlap of the two aperture intervals givesMultiplication by ${\kappa }_{\nu }/(2{\rho }_0\omega )$ gives the ensemble-averaged pressure-normalized modal acoustic energy flux for this finite-aperture convention.

The infinite-aperture spectral normalization consistent with the specified Fourier transform isThe integral and limit are understood as spectral measures of a stationary random field when necessary. By Parseval's identity, this is power per unit transverse length: ${\int }_{\mathbb{R}}{S}_E(d\nu )=C_E(0)=1$. The ensemble-averaged axial spectral acoustic flux is independent of $x$:Here the notation includes Dirac delta distributions in the spectral measure of a stationary random field; the leading-order total flux is $I_{\mathrm{inc}}$. The ${\kappa }_{\nu }$ correction is meaningful within $|\nu |\ll k$, the range of the paraxial approximation.

If $\rho (\zeta )\to 0$ and $e^{{\beta }^2\rho (\zeta )}-1$ is integrable, the stationary phase-screen power spectrum separates into coherent and diffuse wave fields:The coherent fraction is $e^{-{\beta }^2}$ and the diffuse fraction is $1-e^{-{\beta }^2}$. If additionally $|\beta |\ll 1$, the stationary phase-screen power spectrum is $(1-{\beta }^2)\delta +{\beta }^2\hat{\rho }+O({\beta }^4)$ with the same Fourier transform convention. Small $\mu$ alone does not imply small accumulated phase $k\xi \mu$.

At each point the Gaussian process has a standard normal marginal distribution. Its characteristic function gives the coherent screen fieldIt is constant in $z$, so its Fourier transform is $m\delta (\nu )$. Applying the Fresnel propagator to this zero-transverse-wavenumber component gives the ensemble-averaged reduced spectrumThus the reduced mean is independent of $x$. The physical mean $⟨\psi (x,z)⟩=e^{ikx}{e}^{-{\beta }^2/2}$ still has its carrier phase. The coherent attenuation by a Gaussian phase screen takes place at the screen, with no further coherent attenuation in homogeneous space.

The full spectral acoustic flux depends on the second moment, not the square of this mean. For a stationary process, the generalized cross-spectral correlation isIndeed its propagation factors cancel on $\nu ={\nu }^{\prime }$. This explains why the ensemble-averaged spectral acoustic flux in (b) is also independent of $x$, even for nonzero transverse wavenumbers whose individual complex amplitudes change phase. The coherent atom has weight $|m{|}^2=e^{-{\beta }^2}$; the full stationary phase-screen power spectrum also contains fluctuations. Under the decay assumptions in (b), their integrated weight is $1-e^{-{\beta }^2}$. One must use these weights in a spectral measure of a stationary random field, rather than square a Dirac delta distribution. A phase-only screen preserves the leading-order total acoustic energy flux, although it reduces its coherent part.

Write the density and pressure as $\rho ={\rho }_0+{\rho }^{\prime }$ and $p=p_0+p^{\prime }$, and let $\mathbf{u}$ be the small fluid velocity. The linear homentropic acoustic equations arewhere the adiabatic sound speed isIf the flow is irrotational, write $\mathbf{u}=\nabla \varphi$. The momentum equation integrates spatially toa purely time-dependent integration term having been absorbed into $\varphi$. Substitution into continuity gives the wave equationso the wave speed is $c_0$.

To obtain energy conservation, dot the momentum equation with $\mathbf{u}$ and use the product rule:The continuity and pressure-density relations giveTherefore the acoustic energy conservation law iswith acoustic energy density and intensity

The normal acoustic energy flux isIn either half-space, the pressure fields above are proportional to $\cos (kx-\omega t)$, whereas the normal velocities are proportional to $\sin (kx-\omega t)$. They are therefore in temporal quadrature. Averaging over one period givesThus the field is evanescent and stores reactive acoustic energy flux near the membrane without transporting net acoustic energy away from it in the normal direction; the wave's mean energy transport is along the membrane.