Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 335 (Direct and Inverse Scattering of Waves)

Use the time convention $e^{-i\omega t}$. The free-space Helmholtz equation, with $\psi =e^{ikx}E$, gives the exact reduced equationFor a forward plane wave at angle $\alpha$, $E_x=O(k{\alpha }^{2}E)$ and $E_{zz}=O(k^2{\alpha }^{2}E)$, whereas $E_{xx}=O(k^2{\alpha }^{4}E)$. Thus the paraxial approximation neglects $E_{xx}$ relative to $2ik{E}_x$. The reduced field satisfies the parabolic wave equationWith the Fourier transform convention used below and its inverse, this becomes an ordinary differential equation for each transverse wavenumber:Consequently the initial-value solution isEquivalently, evaluating the oscillatory Gaussian integral gives the one-dimensional transverse Fresnel propagation formulaThe square-root branch has $\sqrt{1/i}=e^{-i\pi /4}$. The formula is an oscillatory integral for general data, or the Fresnel propagator acting on square-integrable functions. It approaches the initial field as $x\downarrow 0$. A plane wave has transverse wavenumber $\nu =k\sin \alpha$; the reduced axial phase $-{\nu }^{2}x/(2k)$ agrees with $k(\cos \alpha -1)x$ through order ${\alpha }^2$. This also checks the sign. The paraxial approximation applies to $|\nu |\ll k$.

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.

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.

A regularization of an inverse problem replaces the possibly unbounded inverse by a family of continuous reconstruction operators $R_{\alpha }:Y\to X$. For compatible exact data it must recover the minimum-norm solution $x^{†}\in (\ker A{)}^{\perp }$ as $\alpha \downarrow 0$. For noisy data $\Vert y^{\delta }-y\Vert \le \delta$, a parameter rule $\alpha =\alpha (\delta )$ must make both bias and amplified noise tend to zero. The target is the minimum-norm solution because data cannot determine a component in the kernel of a linear map.

Tikhonov regularization minimizesThe quadratic norm penalty makes this functional strictly convex and a coercive function. Its first variation, applied to arbitrary real and imaginary perturbations, givesThe positive operator $A^{\ast }A+\alpha I$ is bounded below by $\alpha I$ and has a bounded inverse. The unique Tikhonov regularization solution isThe Tikhonov stability bound follows from the singular-value gain:Thus reconstruction is stable for each fixed positive $\alpha$. Stability is not uniform as $\alpha \downarrow 0$: that limit restores the unbounded inverse. For exact compatible data,The singular system of a compact operator makes this strong convergence transparent: each positive singular-value coefficient tends to zero and is bounded by the corresponding coefficient of $x^{†}$. Consequentlywhenever $\alpha (\delta )\downarrow 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$, for example $\alpha (\delta )=\delta$. This is a convergent regularization of an inverse problem, not merely a bounded formula.

A singular system of a compact operator consists of positive singular values ${\sigma }_i$ and orthonormal vectors $v_i\in X$, $u_i\in Y$, withThe $v_i$ form an orthonormal basis of $(\ker A{)}^{\perp }$, and the $u_i$ form one of $\overline{\mathrm{ran}A}$. There are finitely many positive singular values for finite-rank $A$; otherwise ${\sigma }_i\to 0$. Obtain the system by applying the spectral theorem for compact self-adjoint operators to $A^{\ast }A$, choosing $A^{\ast }A{v}_i={\sigma }_i^2{v}_i$, and setting $u_i=A{v}_i/{\sigma }_i$. The inner product identities verify orthonormality and the displayed relations.

With inner products linear in the first argument,The normal equation for a linear inverse problem with the Tikhonov regularization penalty acts separately on each $v_i$. Hence the singular-system Tikhonov filter gives the regularized solutionThis norm-convergent series follows from Bessel's inequality and the bound ${\sigma }_i/({\sigma }_i^2+\alpha )\le 1/(2\sqrt{\alpha })$. Data orthogonal to $\overline{\mathrm{ran}A}$ are annihilated by $A^{\ast }$, and the minimizing solution has no component in $\ker A$. The same formula with $y^{\delta }$ reconstructs noisy data. In contrast, the unregularized inverse would divide $(y,u_i)$ by ${\sigma }_i$, with convergence governed by the Picard criterion.

Use the recurrence exactly as printed. Set $B=A^{\ast }A$ and $R=I-B/\alpha$. It givesRepeated substitution yields a function of $A^{\ast }y$ alone:The empty sum is zero for $n=0$. This is Landweber iteration with a Tikhonov initial value. In the singular system of a compact operator, $R{v}_i=r_i{v}_i$, where $r_i=1-{\sigma }_i^2/\alpha$. ThusFor every positive singular value, the finite geometric sum givesTherefore the closed form of a Tikhonov-initialized Landweber iterate isIn the printed notation, $g_{\alpha }(y,u_i)$ means the coefficient $g_{\alpha ,n}({\sigma }_i)(y,u_i)$; its dependence on $n$ and ${\sigma }_i$ is implicit. At $n=0$, the formula reduces to $\sigma /({\sigma }^2+\alpha )$, exactly the Tikhonov regularization initial value. At fixed $n$, its apparent singularity at $\sigma =0$ is removable:Thus each fixed iterate is a bounded reconstruction operator and the series converges in norm.

The terminology in the question requires care. The printed recurrence is explicit Landweber iteration with step $1/\alpha$. Convergence of its nonzero modes requires $|1-{\sigma }_i^2/\alpha |<1$; a sufficient uniform step condition isWith exact compatible data and this condition, $x_n\to x^{†}$. At equality the largest singular-value mode can alternate, and smaller $\alpha$ can produce growing modes. With noisy data, early stopping of Landweber iteration is needed to prevent recovery of arbitrarily unstable small-singular-value components.

Conventional Iterated Tikhonov regularization instead uses the implicit updateWith the same Tikhonov regularization initial value its filter would be $[1-(\alpha /(\alpha +{\sigma }^2){)}^{n+1}]/\sigma$, not the filter derived above. The solution here retains the authoritative printed update rather than silently changing it.