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Recover a compactly supported acoustic source $Q$ from measurements of its outgoing field, such as its far-field pattern. The Helmholtz equation forward problem is specified with the Sommerfeld radiation condition. At a single frequency, the source-to-far-field operator at fixed frequency has invisible fixed-frequency nonradiating sources, and its compactness makes recovery of the visible component unstable.

The angular spectrum is the transverse Fourier transform of a monochromatic field on a reference plane. In a uniform medium, each transverse wavevector component propagates with a longitudinal wavenumber satisfying the Helmholtz equation. Components beyond the propagating wavenumber range are evanescent waves. The outgoing angular spectrum chooses the longitudinal branch that propagates or decays away from the plane.

Let ${\Gamma }_9$ have center weight $-10/3$, axial weights $2/3$ and diagonal weights $1/6$, and let $M_h$ have center weight $2/3$ and axial weights $1/12$. Taylor expansion gives $h^{-2}{\Gamma }_{9}u=\Delta u+h^2{\Delta }^{2}u/12+O(h^4)$ and $M_{h}u=u+h^2\Delta u/12+O(h^4)$. On a smooth solution of the constant-coefficient Helmholtz equation, the second-order term becomes $h^2\Delta (\Delta u+\lambda u)/12=0$. Thus the normalized PDE residual is fourth order, while the original unscaled stencil residual is sixth order. This local cancellation alone does not prove uniform global accuracy at or near a resonance.

For the cubic polytropic interior, $\nabla \rho$ vanishes at each vertex, so the locally defined potential $\Phi =C-2K\rho$ predicts zero gravitational acceleration there. But at the vertex $(0,0,0)$, the actual self-gravitational acceleration of the positive mass distribution iswhose three components are strictly positive. The fields cannot match continuously. Thus the interior Helmholtz equation and zero density on the faces do not construct an isolated static cubic star. External stresses or an external gravitational field would be needed to realize such a boundary-value construction.

The interior Helmholtz equation for a polytrope of index one also admits a positive solution vanishing on the faces of a cube:Its central value is ${\rho }_c$ and its average is $\bar{\rho }=8{\rho }_c/{\pi }^3$. Together with $\Phi =C-2K\rho$, it solves the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. A local interior solution need not match the external field of its own mass.

Two Green functions of the same Helmholtz equation with the same homogeneous Robin boundary condition satisfy the displayed reciprocity by Green's second identity. It is a bilinear identity without complex conjugation, so it concerns the transpose rather than the Hermitian adjoint. In an exterior problem both kernels use the same radiation convention.

A Herglotz wave function is a superposition of incident plane waves,It solves the Helmholtz equation everywhere and is real analytic. The finite sphere area makes the density integrable, so each spatial derivative can be taken under the integral.

The incident wave is the prescribed background solution that illuminates a scatterer. It solves the unperturbed Helmholtz equation in scalar time-harmonic models. The scattered contribution is defined by subtracting this field from the total field; reflected waves from a planar boundary are included unless a separate reference reflection is explicitly removed.

The logarithmic wave perturbation measures relative complex amplitude and phase through $\psi ={\psi }_i{e}^{\chi }$. A continuous branch requires nonzero incident and total fields in the region considered. For the Helmholtz equation with scattering potential convention $L_0\psi =-V\psi$, its equation is $(\Delta +2\nabla \log {\psi }_i\cdot \nabla )\chi +\nabla \chi \cdot \nabla \chi =-V$. The dot product is bilinear, not a squared modulus.

The outgoing angular spectrum propagates prescribed scalar boundary data into a half-space. For $e^{-i\omega t}$ time dependence, $\beta (q)=\sqrt{k^2-q^2}$ has nonnegative real part on propagating modes and nonnegative imaginary part on evanescent modes. Its trace at the boundary is $g$, and each mode solves the Helmholtz equation. The spectral representation can be interpreted through distributions for nondecaying plane-wave data.

The outgoing Green function satisfies $(\Delta +k^2)G_k=-{\delta }_{{\mathbf{r}}^{\prime }}$ and the Sommerfeld radiation condition. For time dependence $e^{-i\omega t}$, its radial phase travels outward. Away from its source it solves the homogeneous Helmholtz equation. At large $r$, $G_k=e^{ikr}{e}^{-ik\hat{\mathbf{r}}\cdot {\mathbf{r}}^{\prime }}/(4\pi r)+O(r^{-2})$.

The parabolic wave equation is a one-way, slowly varying envelope approximation to the Helmholtz equation. For carrier wavenumber $k$ and transverse coordinate $z$, a common form is $2ik{E}_x+E_{zz}=0$.

The paraxial approximation describes waves whose propagation directions make small angles with a preferred axis. Factoring out a carrier $e^{ikx}$ and neglecting the envelope's second longitudinal derivative turns the Helmholtz equation into a parabolic wave equation and suppresses backward propagation.

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.

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.

Use time dependence $e^{i\omega t}$ and write the outgoing two-dimensional Helmholtz equation kernel as $G_2\sim C{e}^{-i{k}_{0}r}/\sqrt{k_{0}r}$, where $C$ is independent of the Mach number. The harmonic Lighthill acoustic analogy givesFor a compact acoustic quadrupole, the source integral is $O({\rho }_0{U}^2{\ell }^2)$. In the radiation region each derivative of the outgoing exponential supplies a factor of $k_0$, soKeeping the same advective source-frequency scaling as in (i), $\omega =O(U/\ell )$ and hence $k_0\ell =O(m)$. Therefore the two-dimensional compact quadrupole scaling isThe changed exponent comes from cylindrical spreading $(k_{0}r{)}^{-1/2}$. Compactness $k_0\ell \ll 1$ and the far-field condition $k_{0}r\gg 1$ are separate assumptions. At fixed imposed $\omega$ and fixed $\ell$, with only $U$ varied, the stress instead gives an $O(m^2)$ dependence; monochromaticity alone does not supply the additional source-frequency assumption.

Take $0<{\theta }_0<\pi$, put $a=k_0\cos {\theta }_0$ and $b=k_0\sin {\theta }_0>0$, and suppress the common factor $e^{i\omega t}$. The upper-half-plane incident acoustic plane wave is $e^{iax+iby}$. A rigid zero-thickness screen imposes the same normal-derivative cancellation on both faces; the scattered velocity potential is consequently odd in $y$. Across the open part $x>0$ it is continuous, so its trace there is zero. These complementary half-line data are what make the Wiener-Hopf method applicable.

Use the Fourier transform $\hat{\varphi }(\kappa ,y)={\int }_{\mathbb{R}}\varphi (x,y)e^{i\kappa x}dx$, with inverse exponential $e^{-i\kappa x}$. The outgoing Helmholtz equation solution is $\hat{\varphi }=F^{-}(\kappa )e^{-\gamma y}$ for $y>0$, where $\gamma =\sqrt{{\kappa }^2-k_0^2}$ has $\gamma =i\sqrt{k_0^2-{\kappa }^2}$ on the propagating interval and positive real part on evanescent components. The limiting absorption principle supplies the contour prescription. The transform $F^{-}$ of the trace supported on $x<0$ is analytic below the contour. Let $G^{+}$ be the analytic upper transform of the unknown derivative on $x>0$. The prescribed derivative on $x<0$ has transformSince the upper normal derivative is $-\gamma F^{-}$, the Wiener-Hopf equation isFor the Wiener-Hopf factorization, choose ${\gamma }_{+}=(\kappa -k_0{)}^{1/2}$ analytic above and ${\gamma }_{-}=(\kappa +k_0{)}^{1/2}$ analytic below. Their branch cuts go into the opposite half-planes, with ${\gamma }_{+}(-a)=i\sqrt{k_0+a}$. Dividing by ${\gamma }_{+}$ and performing pole subtraction in a Wiener-Hopf equation givesThe forcing pole is allocated to the lower analytic part; the second quotient has a removable pole and is upper analytic. Subtracting these parts leaves a common entire function. The finite-energy edge condition excludes an inverse-square-root singularity in the potential: its odd scattered trace is $O(\sqrt{|x|})$, so $F^{-}=O(|\kappa {|}^{-3/2})$. The normal derivative on the open side can be $O(x^{-1/2})$, giving $G^{+}=O(|\kappa {|}^{-1/2})$. Both sides of the entire remainder tend to zero. The Liouville theorem sets that remainder to zero, giving the explicit Wiener-Hopf solution of rigid half-plane diffraction:The scattered field on either side of the screen is thereforeIts forcing-pole residue at $\kappa =-a$ is $-i$. Deforming to the outgoing steepest descent contour crosses this pole precisely when $|\theta |>\pi -{\theta }_0$, where $x=r\cos \theta$, $y=r\sin \theta$ and $-\pi <\theta <\pi$. The residue contributes $\mathrm{sgn}(y)e^{iax-ib|y|}$. Thus, away from transition boundaries, the scattered geometrical optics field isAdding the incident acoustic plane wave gives incident plus unit specular reflection in the upper reflected sector, just the incident wave in the remaining illuminated sectors, and zero in the lower shadow sector. The geometrical shadow boundaries also follow directly by tracing a straight ray to $y=0$: the reflected ray hits the screen at $x+y\cot {\theta }_0<0$, and the transmitted incident ray crosses the open part at $x-y\cot {\theta }_0>0$.

For the diffracted contribution, the outgoing phase $-i\kappa r\cos \theta -\gamma r|\sin \theta |$ has saddle ${\kappa }_s=k_0\cos \theta$, value $-i{k}_{0}r$, and second derivative $ir/(k_0{\sin }^2\theta )$. Stationary phase therefore supplies $\sqrt{2\pi k_0/r}|\sin \theta |e^{-i{k}_{0}r+i\pi /4}$. Substituting $F^{-}({\kappa }_s)$, and using $\mathrm{sgn}(y)|\sin \theta |/\sqrt{1+\cos \theta }=\sqrt{2}\sin (\theta /2)$, givesThe full asymptotic scattered field is ${\varphi }_{\mathrm{GO}}+{\varphi }_d$. The angular edge-wave expression is not uniform on the geometrical boundaries. There is also a sign inconsistency in the supplied saddle formula: with its printed inverse exponential $e^{+i\kappa x}$, the outgoing saddle is $-k_0\cos \theta$, not $+k_0\cos \theta$. The consistent inverse convention used above gives the requested outgoing field without that ambiguity.

Let $a(X)=k_0(X{)}^{-2}$, $F(y,X)=\cos (n\pi y/R(X))$, and $N(X)={\int }_{-R}^R{F}^{2}dy$. At the next order of the WKB approximation, the transverse operator $L=a{\partial }_y^2+1-a{k}^2$ givesThe moving-wall Neumann boundary condition gives $P_{1,y}(R)=-ik{R}_{X}AF(R)$ and $P_{1,y}(-R)=+ik{R}_{X}AF(-R)$. Multiply by $F$ and integrate across the duct. Since $LF=0$ and $F_y=0$ at both walls, integration by parts givesAlso $N_X=2\int F{F}_{X}dy+2R_X$, because $F^2=1$ at both walls. Substituting these identities in the next-order equation produces the transport equation for the variable-sound-speed WKB duct amplitude:Consequently $A\sqrt{akN}$ is a complex constant, including a constant phase. For $n\ge 1$, $N=R$; for $n=0$, $N=2R$, with the factor two absorbed into the mode's arbitrary constant. ThusEquivalently,There is an independent flux check. The real-coefficient Helmholtz equation implies $\nabla \cdot \mathrm{Im}(p^{\ast }a\nabla p)=0$, and the wall condition gives zero normal flux. Hence the integrated longitudinal current is constant; at leading order it is $-ak|A{|}^{2}N$. This agrees with the transport equation. The factor $k_0$ must be retained: using only $(kR{)}^{-1/2}$ would ignore the variable coefficient in the given divergence-form equation.

Use $X=ϵx$ and the WKB approximationAt leading order, the divergence-form acoustic Helmholtz equation becomes $P_{0,yy}+[k_0(X{)}^2-k(X{)}^2]P_0=0$. The sloping-wall Neumann boundary condition becomes $P_{0,y}=0$ at $y=\pm R(X)$. In the even transverse family this has normal modes $P_0=A(X)\cos (n\pi y/R(X))$, where $n=0,1,\dots$, andFor the right-going propagating branch take real positive $k$. The displayed local expression $e^{-ikx}$ in the question must be read as a frozen-coefficient shorthand: the actual slowly varying phase is $\exp [-i{\int }^{x}k(ϵs)ds]$. Differentiating $e^{-ik(X)x}$ would instead give the erroneous local wavenumber $k+X{k}_X$.

The symmetric duct also has odd transverse normal modes $\sin [(n+1/2)\pi y/R]$. The integer-cosine form selects the even family requested here, rather than representing every duct mode. The approximation requires smooth slow variation and a positive $k$ separated from cutoff; it fails in a cutoff transition.

Use ${\hat{\psi }}_{\mathrm{sc}}(\nu )=\int {\psi }_{\mathrm{sc}}(x,0)e^{-i\nu x}dx$, inverse factor $1/(2\pi )$, and $k=\omega /c$. Fourier transform the homogeneous Helmholtz equation givesAn upward radiating field has no incoming component from $z=+\infty$. For $q(\nu )=\sqrt{k^2-{\nu }^2}$, take the nonnegative real root when $|\nu |<k$ and the positive imaginary root when $|\nu |>k$. Consequently the outgoing angular spectrum isThe propagating terms go upward; the evanescent terms decay with increasing $z$. Grazing values are understood by a radiation limit.

There is a reference-plane qualification in the question. For an arbitrary rough surface the full line $z=0$ need not lie in the medium: already $f(x)=h>0$ puts it in vacuum. In that case the stated physical trace is undefined without continuation. The exact representation is obtained instead on a plane $z=z_{\ast }>\sup f$,One can formally set ${\hat{\psi }}_{\mathrm{sc}}={\hat{\psi }}_{\ast }{e}^{-iq{z}_{\ast }}$, but convergence of that continued spectrum down to the rough boundary is an additional assumption. This is the reference-plane validity of a rough-surface angular spectrum. The first boxed formula is exact in an appropriate homogeneous upper half-space; it is not automatically an exact representation at every point of a completely arbitrary rough domain. Small-height perturbation theory in part (c) uses the mean plane order by order.