Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 335 (Direct and Inverse Scattering of Waves)

Putand let $q_1=nk\cos {\theta }_T$ be the downward vertical wavenumber in the lower half-space. The given form of Snell's law saysWith the $e^{-i\omega t}$ convention, write the incident, reflected, and transmitted plane waves asThe two interface conditions at $z=0$ giveSolving this linear system gives the reflection and transmission coefficients at a scalar-wave interfaceThus the exact fields are

After separating the conserved horizontal factor as $\psi =e^{ipx}f(z)$, the Helmholtz equation becomeswhere $H$ is the Heaviside step function. The outgoing Green function for $d^2/d{z}^2+q_0^2$ isThe Born approximation at first order replaces $f$ on the right-hand side by the incident profile $f_i(z^{\prime })=e^{-i{q}_0{z}^{\prime }}$. For $z>0$ this givesThe integral is understood with the usual outgoing-wave convergence factor. Sincewe obtainTherefore the Born reflected field is

The Rytov approximation at first order writes the total profile as $f_R=f_i{e}^{{\chi }_1}$ and identifies the first logarithmic perturbation with the Born relative field:Hence, in the upper half-space,Expanding the exponential function to first order shows that its reflected component isThe exponentiated expression is the Rytov approximation to the total field; only its term linear in $\alpha$ is a single specular reflected plane wave.

The Taylor expansion of the square root isSubstitution into the exact reflection coefficient givesConsequentlyThis is exactly the reflected field furnished by both the Born approximation and the term linear in $\alpha$ in the Rytov approximation. Thus the exact, Born, and Rytov fields agree through first order. If the Rytov exponential is retained without re-expansion, its higher powers generate spatial harmonics $e^{i(px+(2m-1)q_{0}z)}$; those terms are part of the approximation and should not be confused with the exact interface's single reflected wave.

For a slowly varying envelope $E=\psi e^{-ikx}$, the paraxial approximation to the Helmholtz equation isBecause $n=1+\mu W$ and ${\mu }^2\ll 1$, passage through a sufficiently thin phase screen producesIts modulus is one at the screen exit. Beyond the screen, $n=1$, so the parabolic wave equation is $E_x=i{E}_{zz}/(2k)$. A Taylor expansion in propagation distance givesSincewe findIt follows thator, equivalently, $|E{|}^2=1-x{\varphi }^{\prime \prime }/k+O(x^2)$. Thus random phase curvature produces local focusing and defocusing: free-space diffraction converts phase fluctuations into amplitude fluctuations immediately after the screen.

To first order in the weak fluctuation, $n^2-1=2\mu W+O({\mu }^2)$, soForintroduce the signs $s=(1,-1,1,-1)$. The product rule gives

Assume that $W$ is a zero-mean stationary Gaussian random field, that its longitudinal correlation length is short compared with the envelope's evolution scale, and that the propagation distance is long compared with that correlation length. The forward Markov approximation then neglects diffraction during one correlation length. Applying the Furutsu–Novikov formula closes the last average at second order in $\mu$. Define the integrated longitudinal autocorrelation function of a random fieldThenwhereWriting $C_{ij}=C(z_i-z_j)$ and using the evenness of the covariance gives the equivalent expressionThe derivation also assumes paraxial propagation, weak scattering, sufficient regularity to interchange differentiation and expectation, and statistical homogeneity in both coordinates. Without the short-correlation approximation, the Gaussian identity produces a nonlocal longitudinal memory integral rather than this local closed equation.

Substitute $m_4=e^{\Psi }$ into the given equation. The chain rule givesThe incident reduced field is the constant one, so $m_4(-\Delta x)=1$ and $\Psi (-\Delta x)=0$. During a thin-screen crossing, $\Psi =O(\Delta x)$. Its mixed-derivative contribution integrates to $O(\Delta x)$. More precisely, its contribution to $\Psi$ is $O((\Delta x{)}^2)$, while the product of its first derivatives contributes only at still higher order. Thereforeand hence

In free space $Q=0$, and the fourth moment obeysFor the two-dimensional Fourier transformthe Fourier transform of a derivative turns this equation into the ordinary differential equationThusUsing the screen-exit value from part (a), the inverse transform givesthrough first order in $\Delta x$. Equivalently, the free-space fourth-moment propagator has kernel

Because $A$ is a real symmetric positive-definite matrix, the finite-dimensional spectral theorem supplies an orthonormal eigenbasis $u_1,\dots ,u_n$ withIts eigendecomposition is also its singular value decomposition. If $e=y^{(\delta )}-y$, thensoThe operator norms satisfy $\Vert A{\Vert }_2={\lambda }_n$ and $\Vert A^{-1}{\Vert }_2=1/{\lambda }_1$. Therefore the worst-case relative perturbation bound isThe ratio ${\kappa }_2(A)$ is the spectral condition number of a positive-definite matrix. A large ratio means that data noise aligned with an eigenvector for the smallest eigenvalue is strongly amplified, so the inverse problem is ill conditioned.

Let $({\sigma }_j,u_j,v_j)$ be a singular system of a compact operator, withThe Moore–Penrose inverse of an operator is the generally unbounded mapdefined when the Picard criterion holds, with the component in $\ker A^{\ast }$ sent to zero. It is the minimum-norm least-squares solution of $Af=g$.

A regularization of an inverse problem consists of bounded maps $R_{\alpha }:Y\to X$ and a parameter rule $\alpha =\alpha (\delta ,g^{(\delta )})$ such thatwhenever $\Vert g^{(\delta )}-g\Vert \le \delta$ and $g$ is in the domain of $A^{†}$.

For Tikhonov regularization, minimizinggivesThe scalar spectral filter satisfiesand consequentlyFor exact data, each filter factor ${\sigma }_j^2/({\sigma }_j^2+\alpha )$ tends to one, so $R_{\alpha }g\to A^{†}g$. Choosingtherefore makes both the approximation error and propagated data error vanish. For example, $\alpha (\delta )=\delta$ is an admissible a priori regularization parameter choice.

For the Volterra integration operatorthe adjoint operator isSince $1/{\sigma }_n=(2n-1)\pi /2$, direct integration givesSimilarly,because $\cos ((2n-1)\pi /2)=0$. The half-integer sine and cosine families are orthonormal bases of $L^2([0,1])$, so $({\sigma }_n,u_n,v_n)$ is a singular system of a compact operator.

The Tikhonov solution for noisy data $g^{(\delta )}$ is thereforeWriting out the inner product and the singular functions makes this explicit:The factors ${\sigma }_n/({\sigma }_n^2+\alpha )$ suppress the unstable reciprocal growth $1/{\sigma }_n$ at high index.