Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 335 (Direct and Inverse Scattering of Waves)

The phase screen adds the phase accumulated across its thickness, so the reduced field just after the screen isFor a zero-mean random variable $w$ with a normal distribution and variance ${\sigma }^2$, its characteristic function givesThe same expression is approximately valid for a non-Gaussian weak fluctuation: the cumulant expansion begins with $-k^2{\xi }^2{\sigma }^2/2$, while higher cumulants give higher-order corrections.

For $x>0$, every realization obeys the parabolic wave equationLinearity permits ensemble averaging, soThe initial mean is independent of $z$, hence diffraction does not change it and $⟨E(x,z)⟩=\exp (-k^2{\xi }^2{\sigma }^2/2)$ for every $x\ge 0$.

LetApplying the parabolic wave equation to each factor givesIf $C(s)=⟨w(z)w(z+s)⟩$, Gaussian averaging at the screen givesStationarity makes this a function only of $s=z_1-z_2$. Since ${\partial }_{z_1}^2+{\partial }_{z_2}^2=2{\partial }_s^2$ on such functions,Writing ${\hat{M}}_0(q)$ for the Fourier transform of the screen value, the solution at arbitrary range is

Choose the overall phase so that the coherent mean $m=⟨E⟩$ is real. With $E_j=m+A_j+i{B}_j$,Because the diffuse field has zero mean, keeping terms through second order givesand, at coincident points,The normalized spatial intensity correlation is thereforeThus weak-scattering intensity fluctuations measure the normalized correlation of the in-phase part of the diffuse field.

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.

UseThe adjoint operator is consequentlyWith a unitary Fourier normalization, this is simply ${\mathcal{F}}^{\ast }={\mathcal{F}}^{-1}$.

At fixed incident direction and wavenumber, defineTaking the complex conjugate of the kernel givesThe least-squares minimizer of $\Vert TV-f_{\infty }{\Vert }^2$ satisfies the normal equation for a linear inverse problem $T^{\ast }TV=T^{\ast }{f}_{\infty }$. Fourier inversion on the measured transfer-vector set is exactly the corresponding Moore--Penrose reconstruction $T^{†}{f}_{\infty }$; hence the formal solution in part (i) is the minimum-norm least-squares solution when the data are incomplete or inconsistent.

The inverse problem is ill posed for several related reasons. At one frequency and one incident direction, the data sample $\hat{V}$ only on the two-dimensional Ewald surface $\mathbf{q}=k_0(\hat{\mathbf{r}}-{\hat{\mathbf{x}}}_0)$, so three-dimensional reconstruction is nonunique without more illuminations or prior information. Finite aperture and a bounded frequency band omit additional Fourier components and limit resolution. The forward map is compact, so its singular values tend to zero and inversion strongly amplifies measurement noise. Finally, the Born approximation neglects multiple scattering; model error becomes significant when the contrast or support is too large, while uncertainty in the incident field, measurement geometry, and domain $D$ creates further instability.

A regularization of an inverse problem consists of bounded operators $R_{\alpha }:Y\to X$ and a parameter rule $\alpha =\alpha (\delta ,y^{\delta })$ such that, whenever $\Vert y^{\delta }-y\Vert \le \delta$ and $y$ lies in the domain of $A^{†}$,as $\delta \to 0$. It is needed because a compact operator on an infinite-dimensional space has singular values tending to zero, so direct inversion divides noisy data by arbitrarily small numbers and is generally discontinuous.

Tikhonov regularization definesand its normal equation givesFor every fixed $\alpha >0$, $A^{\ast }A+\alpha I$ is bounded below by $\alpha I$, and the data-to-solution operator is bounded. Small changes in $y^{\delta }$ therefore produce small changes in $x_{\alpha }^{\delta }$.

A singular system of a compact operator is a family $({\sigma }_i,u_i,v_i)$ with ${\sigma }_i>0$, ${\sigma }_i\to 0$, orthonormal $u_i\in Y$ and $v_i\in X$, andOn each singular vector, $A^{\ast }A+\alpha I$ acts by multiplication by ${\sigma }_i^2+\alpha$. HenceThe bounded filter $\sigma /({\sigma }^2+\alpha )$ replaces the unstable inverse factor $1/\sigma$.

WriteThe stated iteration is $x_{n+1}=B{x}_n+{\alpha }^{-1}{A}^{\ast }y$, so repeated substitution and the geometric series identity giveFor the singular component $v_i$, put $r_i=1-{\sigma }_i^2/\alpha$. Since the initial Tikhonov solution has coefficient ${\sigma }_i(y,u_i)/({\sigma }_i^2+\alpha )$, one obtainsTherefore the requested spectral filter isThus equation (3) reads $x_n={\sum }_i{g}_{\alpha ,n}({\sigma }_i)(y,u_i)v_i$. Convergence of this stationary iteration requires $|1-{\sigma }_i^2/\alpha |<1$ on the nonzero spectrum, for example $\alpha >\Vert A{\Vert }^2/2$.