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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.

For the upper half-space and the sign convention $(\Delta +k^2)G_D={\delta }_{r^{\prime }}$, reflect the source $r^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$ to $r^{\prime \ast }=(x^{\prime },y^{\prime },-z^{\prime })$ and use the displayed difference. This is the negative of the conventional outgoing kernel, whose source sign is $-\delta$. The method of images makes the Dirichlet boundary condition zero on the plane. With $R=|r^{\prime }-r|$ and $\cos \theta =z^{\prime }/R$, its outward normal derivative there isThe outward unit normal is $-e_z$. Differentiating $e^{ikR}/R$ gives $e^{ikR}(ik/R-1/R^2)$; the real and reflected source contributions add in the normal derivative. Green's second identity therefore expresses a solution of the Helmholtz equation as $\varphi (r^{\prime })={\int }_{z=0}\varphi (r){\partial }_n{G}_D(r,r^{\prime })dS$ when the boundary term at infinity vanishes. Suitable decay or compatible outgoing Sommerfeld radiation conditions are needed; vanishing at infinity without further control is not a general uniqueness condition for a real-frequency Helmholtz problem.

The Hankel functions are the combinations $H_{\nu }^{(1)}=J_{\nu }+i{Y}_{\nu }$ and $H_{\nu }^{(2)}=J_{\nu }-i{Y}_{\nu }$ of the Bessel function of the first kind and Bessel function of the second kind. They form a standard independent pair for the Bessel differential equation.

For real $k,r>0$ and time dependence $e^{-i\omega t}$, their leading radial factors are proportional to $e^{ikr}$ and $e^{-ikr}$. Constant wave phase $kr-\omega t$ moves outward with increasing time, whereas $-kr-\omega t$ moves inward. Reversing the time-sign convention reverses these labels. This makes the combinations useful when imposing a Sommerfeld radiation condition, rather than regularity at the origin.^[ref]

For complex arguments, the Hankel asymptotic sectors and the chosen branch of a multivalued function are part of the approximation. A formula continued around zero must not be treated as one global exponential on every argument sheet.

Let $(\Delta +k^2)u=0$ for $x_3>0$, with real $k>0$, $u=O(r^{-1})$ and $r({\partial }_{r}u-iku)\to 0$ uniformly in direction. With sufficient boundary regularity, Green's second identity on an expanding half-ball gives the title formula. The plane integral is interpreted as the limit over expanding disks when it is not absolutely convergent. Its sign uses the upward derivative ${\partial }_{3}u$; the outward normal derivative is $-{\partial }_{3}u$.

For $G=e^{ik|x-y|}/(4\pi |x-y|)$, $G=O(r^{-1})$ and $({\partial }_r-ik)G=O(r^{-2})$ at large $|y|=r$. The hemispherical integrand can be written $u({\partial }_r-ik)G-G({\partial }_r-ik)u$, whose integral tends to zero by the Sommerfeld radiation condition. Add the boundary identities for a source point $x$ and its reflected point $x^{\ast }=(x_1,x_2,-x_3)$. On the plane $G(x,y)=G(x^{\ast },y)$ and their normal derivatives are opposite. The terms involving boundary values of $u$ cancel, leaving twice $G{\partial }_{3}u$ and hence the title formula. In particular, zero Neumann boundary data imply $u=0$ in this radiating class.

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 $r$, $G_k=e^{ikr}{e}^{-ik\hat{\mathbf{r}}\cdot {\mathbf{r}}^{\prime }}/(4\pi r)+O(r^{-2})$.

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.

Write $r=|\mathbf{x}|$, $\theta =\mathbf{x}/r$, and choose $R$ so that the sphere $r=R$ strictly encloses the scatterer. Normalize the spherical harmonics to be an orthonormal basis of $L^2(S^2)$, with surface measure on the unit sphere. The outgoing spherical Hankel function is denoted here by $h_n^{(1)}$, corresponding to the printed $H_n^{(1)}$. For fixed $n$, its large-argument behavior isConsequently the coefficient of $Y_n^m$ in the outgoing expansion is asymptotic to $a_n^m{e}^{ikr}/r$: the factors satisfy $i^{n+1}(-i{)}^{n+1}=1$. With the far-field pattern defined by $u_s(r\theta )=e^{ikr}{u}_{\infty }(\theta )/r+O(r^{-2})$, we obtainOne can first project onto each spherical harmonic and then take the large-$r$ limit. This avoids assuming that the fixed-order asymptotic is uniform in $n$. The Sommerfeld radiation condition selects the outgoing spherical Hankel functions rather than their incoming counterparts.

Let $g(\theta )=u_s(R\theta )$. Its coefficients are $g_n^m=d_n{a}_n^m$, where $d_n=k{i}^{n+1}{h}_n^{(1)}(kR)$. Since the field is smooth on a sphere outside the scatterer, $g\in L^2(S^2)$; Parseval's identity givesThe required coefficient condition uses fixed-argument growth of spherical Hankel functions, a different limit from the far-field limit. For fixed $t>0$,For completeness, the series of the ordinary Bessel function gives $J_{-n-1/2}(t)\sim (t/2{)}^{-n-1/2}/\Gamma (1/2-n)$. The identities $y_n(t)=(-1{)}^{n+1}\sqrt{\pi /(2t)}{J}_{-n-1/2}(t)$ and $\Gamma (1/2-n)=(-4{)}^{n}n!\sqrt{\pi }/(2n)!$ give the first formula; the corresponding positive-order series gives the estimate for $j_n$. The successive series corrections at fixed $t$ are $O(1/n)$. Applying Stirling's approximation to the factorial ratio yieldsIn particular there are positive constants giving upper and lower bounds by the last weight for all sufficiently large $n$. The finite trace norm is therefore equivalent to the spherical far-field range conditionwhere the $n=0$ weight is defined to be one. The same conclusion holds at every larger sphere on which the outgoing field is defined.

The printed comparison with $\infty$ must mean a finite sum: a literal non-strict inequality allows a divergent sum. Also, the supplied one-sided big-$O$ bound alone does not imply this necessary condition from a finite trace norm; a lower bound is needed in that direction. The two-sided asymptotic above supplies it. Conversely, its upper bound makes the weighted condition sufficient for an $L^2$ trace at the chosen sphere. This is a condition on exterior-field continuation, not a sufficient condition that the coefficients come from a particular Dirichlet obstacle.

Use the Fourier transform convention $\hat{p}(k,y)={\int }_{\mathbb{R}}{e}^{ikx}p(x,y)dx$. Let $\gamma (k{)}^2=k^2-k_0^2$, choosing $\mathrm{Re}\gamma >0$ on the real contour with $\mathrm{Im}\omega <0$. For time dependence $e^{i\omega t}$, this is the decaying continuation of the outgoing Sommerfeld radiation condition.

The transformed Helmholtz equation has upper and lower solutions $P(k)e^{-\gamma y}$ and $Q(k)e^{\gamma y}$. The two kinematic traces give $-\gamma P={\rho }_0{\omega }^2\hat{\eta }=\gamma Q$, henceThe elastic membrane equation gives $[\hat{p}]=(m{\omega }^2-T{k}^2)\hat{\eta }$. Therefore the acoustic wave on a tensioned massive membrane has dispersion relationEquivalently, $T{k}^2={\omega }^2(m+2{\rho }_0/\gamma )$. The fluid on both sides supplies a positive added mass of an evanescent fluid layer, which is a useful independent check on the sign. In particular, the incident pressure isThe symbol $m$ here denotes elastic membrane mass per area, rather than the fluctuating Mach number used in Question 1.

Choose the time convention $e^{-i\omega t}$ and the outgoing Sommerfeld radiation condition. Define the outgoing Green function for the three-dimensional Helmholtz equation byAway from ${\mathbf{r}}^{\prime }$ it solves the homogeneous Helmholtz equation. Integrating its radial derivative over a small sphere gives $-1$ in the zero-radius limit, verifying the sign of the Dirac delta distribution. At infinity it satisfies $({\partial }_R-ik)G_k=o(R^{-1})$.

For the positive source on the right-hand side, the outgoing field isApplying the Helmholtz equation operator under this source representation gives $Q$. The minus sign is essential with the displayed Green function convention; defining the Green function to satisfy $LG=+\delta$ instead would absorb it. Here $A$ is the solid ball of radius $r_0$, as required by the volume source equation. Its finite volume and $Q\in L^2(A)$ make $Q$ integrable. The Sommerfeld radiation condition specifies the outgoing solution; without it one could add a homogeneous Helmholtz equation solution.

Use the physical refractive index convention $k=k_{0}n$, with $k_0$ the background wavenumber and time dependence $e^{-i\omega t}$. Treat $B$ as a bounded region in ${\mathbb{R}}^3$; the printed $B\in {\mathbb{R}}^3$ should be $B\subset {\mathbb{R}}^3$. Define the contrast $q=n^2-1=2n_{ϵ}+n_{ϵ}^2$ and the outgoing Green functionThe Sommerfeld radiation condition selects this sign of the outgoing wave phase. Since $(\Delta +k_0^2)\psi =-k_0^{2}q\psi$, the Lippmann-Schwinger equation isThe positive sign in this integral follows from the minus sign in the defining Green function equation. One may instead define the scattering potential with the opposite sign, provided both equations change consistently.

If $\Vert T_q\Vert <1$ on the chosen field space inside $B$, its Neumann series converges. The Born series for the scattered field isFor observation points outside $B$, restrict the intermediate factors to $B$ and use the same outgoing integral for the final factor. For example, the first two terms areThe $m$th term describes $m$ successive scattering interactions. The first Born approximation for scalar wave scattering keeps one interaction and replaces the field inside the medium by the incident field. Higher terms describe multiple scattering and the resulting feedback on the internal field.

A concrete sufficient condition follows on $L^{\infty }(B)$. If $d_B$ is the diameter of $B$, then for $\mathbf{r}\in B$ the region lies in the ball of radius $d_B$ about $\mathbf{r}$, andThus $\boxed{\rho \ll 1}$ is a conservative sufficient condition for the first Born approximation for scalar wave scattering. When $\rho <1$, the omitted internal-field terms satisfyThe bound ignores cancellation in the oscillatory Green function, so it is sufficient, not necessary. A common physical small-contrast criterion for an extended weak medium is small accumulated extra wave phase, $k_0\ell \Vert n_{ϵ}{\Vert }_{\infty }\ll 1$, together with weak scattering and no resonant internal enhancement. Small local contrast alone is not enough for an arbitrarily large or resonant object.

Use the outgoing Green function for the three-dimensional Helmholtz equationLet ${\mathbf{n}}^{\prime }$ point out of the obstacle $D$ into the exterior. Apply Green second identity to the field and $G_k$ in a large exterior region. Their Sommerfeld radiation conditions make the outer-sphere contribution vanish. The normal on the exterior region's inner boundary is $-{\mathbf{n}}^{\prime }$, so the Kirchhoff–Helmholtz representation isThe first integral is the incident source field ${\psi }_i$. Therefore the scattered field is the surface term. Since ${\psi }_i$ solves the homogeneous Helmholtz equation within the obstacle, its surface term is zero at exterior observation points; the same representation can be written using only scattered-field traces:The unknown boundary condition determines these traces, but the representation itself does not require choosing that condition.

The outgoing Green function for the three-dimensional Helmholtz equation satisfiesIts normalization follows by integrating over a small ball about $x$: $G_F\sim 1/(4\pi |y-x|)$, so its outward derivative has flux $-1$. Apply Green's second identity in the $y$ variables:The ${\omega }^2$ terms cancel. Thus the surface integral equals $-\Psi (x)$ for $x\in V$, and zero for $x$ outside the closed volume. Equivalently, remove a small source ball and evaluate its flux before taking its radius to zero. A source on the boundary would require a different limiting formula and is not part of this calculation.

For the upper half-space, take a half-ball of radius $\rho$. Interpret the first far-field condition as $\Psi =O({\rho }^{-1})$, and the radiation limit in its usual direction-uniform sense, namely the Sommerfeld radiation conditionThis is the finite-bound interpretation of the PDF's first condition. For fixed $x$ and real $\omega$,Indeed $|y-x|=\rho -\hat{y}\cdot x+O({\rho }^{-1})$, and its radial derivative is $1+O({\rho }^{-2})$. On the large hemisphere the integrand can therefore be rewritten asIts area is $2\pi {\rho }^2$, so its integral tends to zero. The planar integral, taken over expanding disks if needed, consequently equals $-\Psi (x)$ for $x_3>0$ and zero for $x_3<0$.

The outward unit normal vector on the bottom plane is $-e_3$. Denote this planar integral byFor $x_3>0$ reflect the source point to $x^{\ast }=(x_1,x_2,-x_3)$. On the plane,Adding $I(x)=-\Psi (x)$ and $I(x^{\ast })=0$ cancels the boundary values of $\Psi$ and gives $2\int G_F{\partial }_3\Psi =-\Psi (x)$. Thus the Neumann representation for a radiating Helmholtz half-space isThe negative sign is consistent with using the upward derivative ${\partial }_3\Psi$ rather than the outward normal derivative. If that upward derivative is zero everywhere on the plane, the integral is zero. Hence $\Psi =0$ throughout the upper half-space in the specified radiating class; taking a difference also gives uniqueness for prescribed Neumann boundary data.

Impose a homogeneous Dirichlet boundary condition on the Green function in its first variable, $G(r,r^{\prime })=0$ for $r\in S$. The second boundary term then vanishes, givingIn a bounded domain, a Dirichlet Green function requires $k^2$ not to be a Dirichlet eigenvalue of $-\Delta$; otherwise a resonance can obstruct its existence. The representation assumes such a Green function exists.

In all space, put $\rho =|r-r^{\prime }|$ and try $G=A{e}^{ik\rho }/\rho$. For $\rho >0$, direct differentiation gives $({\rho }^2{G}^{\prime }{)}^{\prime }+k^2{\rho }^{2}G=0$, so the Helmholtz equation holds away from the source. Integrate its inhomogeneous version over a small source-centred ball. The $k^{2}G$ volume integral tends to zero, while the flux isIt must equal the integrated delta, one. ThereforeThis is the negative of the convention used in the outgoing Green function for the three-dimensional Helmholtz equation, where the source is $-\delta$. The sign here follows the present $+\delta$ convention. The factor $e^{ik\rho }$ chooses the outgoing solution, with time convention $e^{-i\omega t}$ and the Sommerfeld radiation condition; one could add homogeneous solutions without a radiation prescription.