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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.

Using $V=k_0^2(n^2-1)$, the scalar Lippmann-Schwinger equation is $\psi ={\psi }_i+{\int }_D{G}_{k_0}V\psi$. The Born approximation replaces the total field inside the integral by the known incident field:For a unit plane wave, its far-field pattern is $(4\pi {)}^{-1}{\int }_{D}V({\mathbf{r}}^{\prime })e^{i{k}_0({\hat{\mathbf{r}}}_0-\hat{\mathbf{r}})\cdot {\mathbf{r}}^{\prime }}d{\mathbf{r}}^{\prime }$. Thus each incident direction samples a shifted sphere in the Fourier transform of the potential.

If a unit incident plane wave has scattered field ${\psi }_s=e^{ikR}{f}_{\infty }(\hat{\mathbf{r}},{\hat{\mathbf{r}}}_0)/R+O(R^{-2})$, the differential wave scattering cross-section isIt is the far-field limit of $R^2$ times the scattered-to-incident energy flux ratio. The common physical amplitude normalization cancels when the incident and scattered waves are in the same background medium. The formula assumes that the $1/(4\pi )$ Green-function factor, if present, is already included in the far-field pattern.

For a bounded source region and fixed positive wavenumber, expand $|R\mathbf{\theta }-{\mathbf{r}}^{\prime }|=R-\mathbf{\theta }\cdot {\mathbf{r}}^{\prime }+O(R^{-1})$. In the outgoing Green function for the three-dimensional Helmholtz equation, this gives $G=e^{ikR}{e}^{-ik\mathbf{\theta }\cdot {\mathbf{r}}^{\prime }}/(4\pi R)+O(R^{-2})$. Volume integration gives the outgoing field's far-field pattern with an $O(R^{-2})$ remainder. The estimate is an asymptotic statement for fixed source size and frequency.

A source is nonradiating at frequency $k$ if its outgoing field has zero far-field pattern. For every $\chi \in C_c^{\infty }(A)$, the source $Q=(\Delta +k^2)\chi$ has outgoing field $\psi =\chi$ and is invisible outside $A$. Integration by parts shows that its Fourier transform vanishes on $|\mathbf{\xi }|=k$, so it belongs to the kernel of a linear map defined by the source-to-far-field operator at fixed frequency. This family proves nonuniqueness independently of the instability of the inverse.

Pairing a far-field pattern with the conjugate density of a Herglotz wave function converts the angular integral into boundary data:This follows by substituting the surface representation of an obstacle far-field pattern and interchanging the integrals.

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.

Write the observation point as $r\hat{r}$, put $q=k_0(\hat{r}-{\hat{r}}_0)$, and take ${\psi }_0(r\hat{r})=e^{i{k}_{0}r{\hat{r}}_0\cdot \hat{r}}$. For a bounded medium, the far-field approximation for an outgoing source usesFor finite-distance accuracy require $r\gg \ell$ and $k_0{\ell }^2/r\ll 1$ as well as being in the radiation region. Define the Fourier transform sample $S(q)={\int }_{D}V(r^{\prime })e^{-iq\cdot r^{\prime }}{d}^3{r}^{\prime }$ and the known factorThen the required total Born approximation for scalar wave scattering and total Rytov approximation areThe first Born approximation far-field pattern is $S(q)/(4\pi )$. In both formulas the incident field is retained; an outgoing scattered term alone would not be a total-field answer.

For $\mathbf{r}=R\hat{\mathbf{r}}$ with the observation distance much larger than the scatterer, expand $|\mathbf{r}-{\mathbf{r}}^{\prime }|=R-\hat{\mathbf{r}}\cdot {\mathbf{r}}^{\prime }+O(R^{-1})$. The far-field pattern in the Born approximation for scalar wave scattering is consequentlyThis convention includes the factor $1/(4\pi )$ in $f_{\infty }$; moving that factor into the outgoing-wave definition would change the pattern's normalization.

The absolute energy flux depends on what physical amplitude $\psi$ represents. For acoustic pressure in a homogeneous fluid of mass density ${\rho }_0$ and wave speed $c_0$, the momentum equation gives the particle-velocity amplitude $\mathbf{v}=\nabla \psi /(i{\rho }_0\omega )$. The time average of harmonic power and acoustic intensity then giveThe radial derivative of the outgoing factor is $(ik-1/R)e^{ikR}/R$. Its imaginary current is therefore $k|f_{\infty }{|}^2/R^2$ to leading order. The incident unit-amplitude plane wave has imaginary current $k{\hat{\mathbf{r}}}_0$. Hence, with $k=\omega /c_0$,The scattered energy flux points along $\hat{\mathbf{r}}$ at leading order, and the incident energy flux along ${\hat{\mathbf{r}}}_0$. If $\psi$ is a velocity-potential amplitude instead, both leading expressions use the common prefactor ${\rho }_0\omega k/2$ in place of $k/(2{\rho }_0\omega )$. A normalized scalar current uses $k$ as the common prefactor. The paper does not specify this amplitude normalization; the ratio in part (c) is independent of that choice.

Use the same physical normalization for both fields. Part (b) has $E_s=E_i|f_{\infty }^{(1)}{|}^2/R^2+O(R^{-3})$, so the common energy flux prefactor cancels. The differential cross-section for scalar wave scattering isThe far-field pattern has dimensions of length in this normalization, so the differential cross-section for scalar wave scattering has dimensions of area per unit solid angle. Only the scattered field is used in this ratio; interference between the incident and scattered fields is a different contribution to the total field's energy flux.

Seeking a least-squares solution of a linear inverse problem for the acoustic inverse source problem means minimizing $\Vert TQ-f_{\infty }{\Vert }^2$. Its normal equation for a linear inverse problem isFor compatible data, the formal minimum-norm solution can be written with the Moore–Penrose inverse of an operator:The inverse in this expression is defined on the range of the restricted operator, not as a bounded everywhere-defined inverse. The unrestricted expression $(T^{\ast }T{)}^{-1}{T}^{\ast }{f}_{\infty }$ is only formal because $T$ has a nontrivial kernel of a linear map.

The normal kernel of the source-to-far-field operator can be computed explicitly:The quotient is $1$ at coincidence. This smooth kernel already exhibits the loss of information.

There are two distinct causes of ill-posedness. First, a fixed-frequency nonradiating source gives exact nonuniqueness. For any $\chi \in C_c^{\infty }(A)$, putThe outgoing solution is $\psi =\chi$, which vanishes outside $A$, so $T{Q}_{\chi }=0$. Equivalently, integration by parts in the source-to-far-field operator at fixed frequency givessince the exponential solves the homogeneous Helmholtz equation and the boundary terms vanish. This provides an infinite-dimensional family of invisible sources. Single-frequency data determine only the spherical restriction of a three-dimensional Fourier transform.

Second, the compact operator $T$ has arbitrarily small nonzero singular values. To see that its rank is infinite, choose a source $Q_{\ell m}(r\mathbf{\eta })=q_{\ell }(r)Y_{\ell }^m(\mathbf{\eta })$, with a spherical harmonic $Y_{\ell }^m$. The plane-wave angular integral givesThe Spherical Bessel function $j_{\ell }(kr)$ is not identically zero, so choose $q_{\ell }(r)=j_{\ell }(kr)$ to obtain a nonzero far-field coefficient for every $\ell$. The mutually orthogonal angular modes prove infinite rank. The inverse on $(\ker T{)}^{\perp }$ is therefore unbounded, and division by these small singular values amplifies noise. Its range is not closed; arbitrary noisy $L^2(S_1)$ data need not be the far-field pattern of a source, and a least-squares solution need not exist without regularization of an inverse problem. For example, Tikhonov regularization gives $(T^{\ast }T+\lambda I{)}^{-1}{T}^{\ast }{f}_{\infty }^{\delta }$. It stabilizes the recovered visible component but cannot identify the missing fixed-frequency nonradiating source component.

The boundary identity gives a numerical route to shape recovery from a Herglotz boundary identity. First recover an admissible density $g$ from measured far-field patterns, enforcing the supplied Herglotz pairing with an obstacle far field for available incident directions:Then evaluate its Herglotz wave function throughout a search region and find a smooth positive radial profile satisfyingThe same $h(\theta )$ must work for every azimuth $\varphi$, expressing the assumed axial symmetry. In practice expand $g$ and $h$ in basis functions, enforce these equations at collocation points, and minimize the complex residual together with the far-field-data residual. The exclusion of an interior Dirichlet eigenvalue ensures uniqueness of the interior Dirichlet problem for the trial boundary data. Use regularization of an inverse problem to control measurement error and unstable density components; neither uniqueness nor numerical stability follows merely from writing the residual equations.

For rich incident-direction data, the conjugated pairing constraints form an adjoint far-field integral equation for $g$. For only one fixed incidence, the displayed pairing is a single complex scalar constraint and cannot by itself determine an arbitrary density. Additional data or a joint constrained shape fit is then needed; the trace condition supplies the actual geometric information.

For unit plane wave incidence ${\psi }_i=e^{i{k}_0{\hat{\mathbf{r}}}_0\cdot \mathbf{r}}$, define the scattering wavevector difference $\mathbf{q}=k_0({\hat{\mathbf{r}}}_0-\hat{\mathbf{r}})$ andThe outgoing Green function for the three-dimensional Helmholtz equation givesThe Born far-field pattern is $A$, a Fourier transform sample of the potential. Expanding the far-field Rytov exponential in $1/r$ gives the same leading outgoing term, while the logarithmic representation preserves its accumulated phase and amplitude correction.

Let the incident plane wave beand define the scattering vectorFor $r$ much larger than the diameter of $D$, the far-field pattern follows fromWritingthe Born result is

The corresponding Rytov logarithmic perturbation isTherefore

With the scattering potential $V(\mathbf{r})=k_0^2(1-n^2(\mathbf{r}))$, the total field obeysThe outgoing Green function is $G_0(\mathbf{r})=e^{i{k}_0|\mathbf{r}|}/(4\pi |\mathbf{r}|)$. The Lippmann-Schwinger equation and the first Born approximation giveFor $r\to \infty$,sowhere $\mathbf{q}=k_0(\hat{\mathbf{r}}-{\hat{\mathbf{x}}}_0)$ is the momentum transfer. Thus the far-field pattern is $-1/(4\pi )$ times the Fourier transform of $V$ at the measured transfer vectors. If sufficiently many incident directions and frequencies supply all $\mathbf{q}$, the formal reconstruction iswith constants adjusted to the chosen Fourier convention.

If measured far-field patterns determine a density $g$ and its Herglotz wave function equals a prescribed spherical wave on the obstacle boundary, that trace identity can locate the boundary. A star-shaped axisymmetric obstacle can be represented by one radial function $h(\theta )$. Numerically, expand $g$ and $h$ in suitable bases, match the far-field constraints, and minimize the complex boundary residual at angular collocation points, using regularization of an inverse problem for noise and small singular values. A single scalar pairing constraint alone does not determine an arbitrary density.

For $(\Delta +k^2)\psi =Q$, the outgoing Green function for the three-dimensional Helmholtz equation yields the far-field patternThe sign corresponds to the Green function satisfying $(\Delta +k^2)G=-\delta$. This Hilbert-Schmidt operator samples the source Fourier transform on the sphere $|\mathbf{\xi }|=k$; for a ball of radius $r_0$, its squared Hilbert-Schmidt norm is $r_0^3/3$.

For an outgoing field outside a sphere of radius $R$, write $u_s=k{\sum }_{n,m}{i}^{n+1}{a}_n^m{h}_n^{(1)}(kr)Y_n^m$. Its far-field pattern is $\sum a_n^m{Y}_n^m$. If its trace on that sphere lies in L2 space, Parseval's identity gives $k^2\sum |h_n^{(1)}(kR){|}^2|a_n^m{|}^2<\infty$. The fixed-argument growth of spherical Hankel functions yields the displayed equivalent range condition, with the low-order weights defined finitely. Square summability of the far-field coefficients alone does not ensure continuation to the chosen sphere.

Continuation of a far-field pattern to a spherical trace multiplies harmonic coefficients by the displayed factor, whose magnitude grows faster than exponentially with order. A data error of size $\delta$ in one normalized high-order harmonic therefore produces near-field error $\delta k|h_n^{(1)}(kR)|$, unbounded as its order increases. Some noisy square-summable far fields fail the range condition entirely. This proves unstable field continuation; bounds on actual obstacle geometry additionally require an admissible shape class and a stable boundary-extraction procedure.