Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 335 / 2

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.

Write $\mathbf{r}=R\mathbf{\theta }$, $|\mathbf{\theta }|=1$. Uniformly for ${\mathbf{r}}^{\prime }$ in the source ball,For fixed $k$ and $r_0$, the outgoing Green function for the three-dimensional Helmholtz equation therefore has the far-field approximation for an outgoing sourceSubstitution into the source integral yieldsThus the source-to-far-field operator at fixed frequency isIt samples the Fourier transform of the source on the sphere of radius $k$. Its square-integrable kernel makes it a Hilbert-Schmidt operator, hence a compact operator. In factUse complex inner products linear in the first argument: $(f,g)=\int f\overline{g}$. Interchanging the volume and surface integrals givesHence the adjoint source-to-far-field operator isThis is a normalized Herglotz wave function; the change in exponential sign comes from the complex conjugation in the adjoint operator.

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.