Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 72 (Direct and Inverse Scattering of Waves)

Choose the time dependence $e^{-i\omega t}$ and measure distances in background-wavenumber units. Thus $L_0=\Delta +1$, the incident field satisfies $L_0{\psi }_i=0$, and the scattering potential convention is $V=n^2-1$, giving $L_0\psi =-V\psi$. To match the minus sign in the paper's integral, take the outgoing Green function to satisfy $L_{0}G=\delta$. This is the negative of the frequently used outgoing Helmholtz equation fundamental solution satisfying $L_{0}G=-\delta$. In physical coordinates one restores $k_0^2$ in the scattering potential, or absorbs it in the integral kernel; the sign convention must remain consistent.

On a region where the incident and total fields are nonzero, introduce the logarithmic wave perturbationChoose a continuous logarithm branch connected to the unperturbed field. Substitution into the Helmholtz equation givesThe quadratic term uses the complex bilinear dot product, not the squared modulus of the gradient. Write $V=\epsilon V_1$ and $\chi =\epsilon {\chi }^{[1]}+\cdots$. The first-order Rytov approximation discards that quadratic term. Since $L_0({\psi }_i\chi )={\psi }_i{\mathcal{L}}_i\chi$, the outgoing first correction isand henceThe validity of the first Rytov approximation concerns the omitted logarithmic correction. Its next contribution satisfies ${\mathcal{L}}_i{\chi }_2=-\nabla {\chi }_1\cdot \nabla {\chi }_1$, so a useful explicit criterion is that the outgoing solution ${\chi }_2$ be small in the region of interest. Weak refractive index contrast and small wave phase gradients on a wavelength scale provide the usual perturbative regime, with weak amplitude fluctuations and no strong focusing or zeros that destroy the logarithm. A sufficient local source comparison is $|\nabla {\chi }_1\cdot \nabla {\chi }_1|\ll |V|$ where the scattering potential is nonzero, together with control of propagation of that error. This is not a universal pointwise test at zeros of $V$.

Small accumulated wave phase is not required in the same way as in a linear field approximation: the exponential retains that wave phase accumulation. Large gradients, strong multiple-scattering amplitude effects or field zeros can still invalidate the Rytov approximation. The integrals also require a finite scattering region or appropriate convergence conditions.

Let $\delta {\psi }_B={\psi }_1^{(B)}-{\psi }_i$ be the first-order scattered correction in the Born approximation for scalar wave scattering, using the paper's Green function convention. Comparing its outgoing integral with the preceding logarithmic correction givesTherefore the exact algebraic relation between the two first-order approximations isIf $V=O(\epsilon )$, then ${\chi }_1=O(\epsilon )$ on a controlled fixed region, andThus Born and Rytov agree to first order in the scattering potential, but differ as finite approximations. The Rytov approximation exponentiates the first logarithmic correction; the Born approximation adds the first field correction. The exponential's higher powers are not a calculation of all higher multiple-scattering terms in the Born series. This distinction explains why a smooth, appreciable wave phase accumulation can be represented more naturally by the Rytov approximation even when the corresponding linear field expansion is inaccurate.

Assume a real refractive index and write the scattering potential as $V=⟨V⟩+W$, so $⟨W⟩=0$ by construction. With the real kernel components in the question, the logarithmic Rytov approximation separates intoInclude the incident wave phase and the mean-potential wave phase in the deterministic reference ${\varphi }_0$: at each observation point it is $\arg {\psi }_i-\int b⟨V⟩$. It need not be spatially constant. The amplitude is $|{\psi }_i|\exp [-\int aV]$. The phase covariance in the first Rytov approximation therefore starts from the fluctuating wave phaseUnder the integrability assumptions needed to interchange the expectation and integral,For a complex absorbing potential, the corresponding phase fluctuation is $-\int [b\mathrm{Re}W+a\mathrm{Im}W]$; the displayed scalar formula is the real-index case. The zero mean comes from centering $V$, not from setting the mean of $V$ equal to zero. In particular the printed $⟨n⟩=0$ does not imply $⟨V⟩=0$. A physical positive refractive index usually has a nonzero background mean; a zero-mean assumption normally refers to its fluctuation. The algebra above remains meaningful for a signed real random field and explicitly retains its mean scattering potential.

Let the stationary scattering potential fluctuation have autocorrelation function of a random fieldBecause $W$ is centered, this is also its covariance function. Multiplying the two real wave phase integrals and taking expectations givesThe two minus signs cancel. This is the wave phase variance, since the wave phase mean is zero. More generally the phase covariance in the first Rytov approximation replaces the first kernel by $b({\mathbf{r}}_1,{\mathbf{r}}^{\prime })$ and the second by $b({\mathbf{r}}_2,{\mathbf{r}}^{\prime \prime })$. Finite observation/scattering windows, or suitable weighted-integrability hypotheses, make these double integrals well-defined in a stationary infinite-medium model.

The correlation needed here is that of the scattering potential fluctuation. With the printed $V=n^2-1$, the covariance of a squared random field isIt involves a fourth moment of $n$, so its value is not generally determined by the ordinary two-point correlation $C_n=⟨n({\mathbf{r}}^{\prime })n({\mathbf{r}}^{\prime }+\mathbf{\rho })⟩$ alone. If one additionally assumes a zero-mean Gaussian random field, Isserlis theorem yields $C_W=2C_n^2$. That assumption is not printed and must not be inserted silently. Alternatively, for a physical weak fluctuation $n_{\mathrm{phys}}=1+\eta$, $W\simeq 2\eta$ gives $C_W\simeq 4C_{\eta }$. The general answer uses $C_W$; either reduction to a refractive index two-point correlation requires an extra assumption.

Work above the surface, with $e^{-i\omega t}$ time dependence. Define the scattered field by $\psi ={\psi }_i+{\psi }_s$, so it includes the reflection from a flat surface. For small-height Dirichlet scattering, write $h=\epsilon \eta$ and expandThe zeroth-order total field ${\psi }^{[0]}={\psi }_i+{\psi }_s^{[0]}$ satisfies the Dirichlet boundary condition ${\psi }^{[0]}(x,0)=0$. Taylor expansion at the perturbed boundary givesThus the first-order rough-surface scattered field has mean-plane boundary data $g_1(x)=-h(x){\partial }_z{\psi }^{[0]}(x,0)$.

Use the outgoing angular spectrum to solve this boundary-value problem. For $\hat{g}(q)=\int g(x)e^{-iqx}dx$, letThe branch ensures upward propagation or upward evanescent decay. Each component solves the Helmholtz equation, and $\mathcal{E}g$ has trace $g$ at $z=0$. ConsequentlyThis gives the scattered field through first order by adding the two contributions. If one reserves “rough scattered field” for the non-specular correction, it is ${\psi }_s^{[1]}$ alone; the convention here keeps the flat reflection as well.

The expansion is in height for a fixed sufficiently regular profile. The condition $|kh|\ll 1$ controls the incident wave's height expansion, but very short spatial scales can create large evanescent normal derivatives. The surface regularity and relevant spectral moments must also control the subsequent boundary expansions; small amplitude alone is not a uniform guarantee for arbitrarily fine roughness.

For the incident acoustic plane wave, set $q_0=k\sin \theta$ and ${\beta }_0=k\cos \theta >0$. The flat reflected field and total field areTherefore ${\partial }_z{\psi }^{[0]}(x,0)=-2i{\beta }_0{e}^{i{q}_{0}x}$ and the first-order rough-surface scattered field isIt is linear in the height. Since $⟨h(x)⟩=0$, its mean vanishes, with the expectation interpreted through finite windows or stationary spectral distributions when needed. HenceThe coherent first-order reflection is the flat-surface reflection. If the field symbol is instead used only for the rough correction, its first-order mean is zero. Stationarity ensures the coherent reflection retains the incident horizontal wavenumber, but zero mean height already explains the vanishing linear correction. The nonzero root mean square height does not enter this mean at first order; it does enter the fluctuating reflected field and its intensity.

For the second-order rough-surface scattered field, the Dirichlet boundary condition expanded at the mean plane isAt normal incidence, ${\psi }^{[0]}=e^{-ikz}-e^{ikz}$, so its second normal derivative vanishes at zero. Define the Dirichlet-to-Neumann map for a Helmholtz half-space through the Fourier multiplier $\mathcal{B}$:The first-order trace is $2ikh$, hence ${\partial }_z{\psi }_s^{[1]}(x,0)=-2k\mathcal{B}h(x)$. It follows thatwhere $h=O(\epsilon )$ with fixed regular profile. In integral notation the quadratic contribution isThe second-order field above the mean plane is $\mathcal{E}[2kh\mathcal{B}h]$ added to $-e^{ikz}+\mathcal{E}[2ikh]$. No local replacement of $\beta (q)$ by $k$ has been made; such a replacement would be an additional long-spatial-scale approximation.

The physical surface trace and reference-plane trace are distinct. At the actual rough boundary $z=h(x)$, the condition itself says ${\psi }_s(x,h(x))=-e^{-ikh(x)}$, soThe first boxed expression is the reference-plane trace needed in part (d), at $z=0$, using the perturbative continuation where that plane lies below the actual boundary; the second answers the literal “at the surface” wording if it means the physical boundary. Taylor-expanding the first expression and its normal derivatives from $z=0$ to $z=h$ reproduces the second, so there is no contradiction between them.

Let the stationary height have covariance function $C_h(\xi )=⟨h(x)h(x+\xi )⟩$, with $C_h(0)={\sigma }^2$. Define the power spectrum of surface height byFor real stationary heights this spectrum is even and nonnegative. The Fourier multiplier identity in part (c) givesThus coherent reflection from a stationary rough surface at normal incidence isRequire the corresponding weighted spectral moment to exist. If the stationary process has a spectral measure rather than a density, the same formula uses that measure with the matching normalization.

Splitting the propagating and evanescent parts makes the effect clear:through second order. The positive real correction reduces the magnitude of the initially negative unit coherent reflection at this order, as some reflection becomes diffuse. The evanescent part produces a coherent wave phase correction. In contrast, the first-order mean was exactly the flat reflected wave.

The height-correlation dependence of coherent reflection cannot generally be determined from $\sigma$ alone: the quadratic term weights the whole spectrum by $\beta (q)$, while ${\sigma }^2=(2\pi {)}^{-1}\int S_h(q)dq$ is unweighted. If the roughness varies only on scales much longer than the wavelength, so its spectrum is concentrated at $|q|\ll k$, then $\beta \simeq k$ andThis is a useful limiting formula, not the general second-order answer under only small-height assumptions. Also the mean field sampled at the moving physical boundary is $-1+k^2{\sigma }^2/2$ through second order, from part (c), and is a different observable.

For a bounded linear operator $A:X\to Y$ between Hilbert spaces, let $N=\ker A$ and $R=\mathrm{ran}A$. The Moore–Penrose inverse of an operator is defined onFor $y=Ax+z$, with $z\in R^{\perp }$, defineThis is independent of the chosen preimage $x$, since two preimages differ by a vector in $N$. Equivalently, it is the inverse of the restriction of $A$ to $N^{\perp }$, applied to the component of the data in $R$, and is zero on $R^{\perp }$.

The generalized solution $x^{†}=A^{†}y$ is the unique minimum-norm least-squares solution. Its residual is orthogonal to the range, giving the operator normal equationThe operator normal equation alone leaves an arbitrary null-space component; the second condition fixes the minimum-norm representative. The identities $A^{†}A=P_{N^{\perp }}$ and $A{A}^{†}=P_{\overline{R}}$ hold on their appropriate domains.

For an infinite-rank compact operator, the range need not be closed, and $A^{†}$ is generally unbounded. Data outside $R\oplus R^{\perp }$ need not have any least-squares solution at all, even though they can be approximated by range elements. This domain qualification is crucial in part (d); the Moore–Penrose notation does not turn an ill-posed inverse into an everywhere-defined bounded operator.

A singular value system of a compact operator consists of positive numbers ${\sigma }_j$ and orthonormal families $v_j\in X$, $u_j\in Y$ satisfyingThus $v_j$ are positive-eigenvalue eigenvectors of $A^{\ast }A$ and $u_j$ of $A{A}^{\ast }$, with eigenvalue ${\sigma }_j^2$. The families are complete in $(\ker A{)}^{\perp }$ and $\overline{\mathrm{ran}A}$ respectively. The singular values can be listed nonincreasingly with multiplicities, and tend to zero in the infinite-rank case. For finite rank there are only finitely many positive singular values; zero-kernel directions are handled separately.

Use the convention that $⟨f,g⟩$ is conjugate-linear in its first entry. Then the singular value system givesThe latter series converges precisely on the admissible range component specified by the Picard criterion:with $R^{\perp }$ components annihilated by the inverse. Merely writing a formal singular expansion does not imply it converges in $X$.

Use the orthonormal Fourier series basis $e_n(x)=(2\pi {)}^{-1/2}{e}^{inx}$, indexed by $n\in \mathbb{Z}$, and the Hilbert space inner product $⟨f,g⟩={\int }_0^{2\pi }\overline{f(x)}g(x)dx$. The integral operator is a periodic convolution operator. Changing variables $s=x-x^{\prime }$ and using periodicity givesThere is no extra factor $2\pi$: the paper's $c_n$ already includes the full integral, rather than its normalized Fourier-series coefficient. The adjoint kernel is $\overline{K(x^{\prime }-x)}$, so $A^{\ast }{e}_n=\overline{c_n}{e}_n$.

A singular system of a periodic convolution operator is thereforeIndeed $A{v}_n={\sigma }_n{u}_n$ and $A^{\ast }{u}_n=(c_n/|c_n|)\overline{c_n}{e}_n={\sigma }_n{v}_n$. The unit factor determined by the complex argument of $c_n$ in $u_n$ is necessary when the complex Fourier coefficients are not positive real numbers. Both families are orthonormal and complete, since every $c_n\ne 0$. One may enumerate $\mathbb{Z}$ by $\mathbb{N}$, or order the positive singular values by decreasing magnitude.

The continuous kernel on a finite square makes $A$ a Hilbert-Schmidt operator, hence compact. Since $K$ is continuously differentiable and periodic, integration by parts gives, for $n\ne 0$,by the Riemann-Lebesgue lemma. In particular the singular values tend to zero. Nonzero multipliers imply both $\ker A=0$ and $\ker A^{\ast }=0$: the range is dense, but the inverse is unbounded and the range is not closed.

Let $y_n=⟨e_n,y⟩$. Using the phase of $u_n$ in the singular value system, the Moore–Penrose inverse of an operator becomesThis gives the admissible data for a periodic convolution inverse. Because the range is dense and the kernel is zero here, the domain of the inverse is exactly the range, not all of $L^2$.

For $y(x)=e^{\alpha x}$ and $\alpha \notin i\mathbb{Z}$,The formal expression would consequently beHowever, the high-frequency obstruction for nonperiodic exponential data prevents this from being a Hilbert-space solution. The numerator is nonzero, $|y_n|$ is asymptotic to a nonzero constant divided by $|n|$, and part (c) proved $|c_n|=o(1/|n|)$. Hence $|y_n/c_n|\to \infty$, so the Picard criterion fails. For $\alpha \notin i\mathbb{Z}$, $A^{†}y$ is not defined in the specified Hilbert space. The formal series is not a convergent generalized solution.

If $\alpha =im$ with $m\in \mathbb{Z}$, the data are a single periodic Fourier mode: $y_n=\sqrt{2\pi }{\delta }_{nm}$. Then there is an exact unique solution,In particular, if the intended parameter is real, only $\alpha =0$ is admissible, giving $f^{†}=1/c_0$.

For the inadmissible cases the inverse problem still has approximate solutions $f_N={\sum }_{|n|\le N}(y_n/c_n)e_n$: their images are Fourier projections converging to $y$ in $L^2$, while their norms diverge. Thus the least-squares residual has infimum zero but no minimizer. This distinguishes an undefined exact inverse from a regularized truncated reconstruction; it is the necessary qualification to the question's unrestricted constant $\alpha$.