Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 335 (Direct and Inverse Scattering of Waves)

DefineIf $A$ and $B$ commute, the Helmholtz equation operator factorizes asWhen $n$ varies with $x$, the exact product also contains the commutator $[A,B]$; neglecting it assumes longitudinal changes are slow. The forward-propagating factor isbecause it admits $e^{i{k}_{0}x}$ in a uniform medium.

Put $\psi =E{e}^{i{k}_{0}x}$. The one-way equation becomesForthe Taylor expansion $(1+Q{)}^{1/2}=1+Q/2+O(Q^2)$ givesThus the parabolic wave equation is

The expansion is accurate under the paraxial approximation: transverse wavenumbers satisfy $|k_z|/k_0\ll 1$, the envelope varies slowly on the carrier scale, the refractive-index contrast is weak enough for $Q^2$ to be negligible, and $n$ varies slowly in $x$ so $[A,B]$ is small. The one-way factor discards backward propagation and reflection; the square-root expansion additionally discards large-angle and higher-order diffraction, and it does not accurately represent strongly evanescent components or abrupt longitudinal interfaces.

Withthe parabolic wave equation is $E_x=(D+S)E$. Freeze $S$ at $x_0$, or preferably at the step midpoint. Over a short distance $\Delta x$, Lie-Trotter splitting givesThe reversed ordering has the same first-order accuracy, while symmetric half-steps in $S$ give the more accurate Strang form.

The commutator can be displayed explicitly. If $q=n^2-1$, thenThe splitting assumption requires $\Delta x^2[D,S]E$ to be small relative to $E$. It is favored by a short range step, a transversely smooth refractive index, and a field without unresolved large transverse wavenumbers. Freezing $S$ also requires $\Delta x{S}_x$ to be small. These conditions supplement the one-way and paraxial approximation already used in part i.

The phase-screen substep is pointwise:DefineThen solve the free-diffraction initial-value problemto $x_0+\Delta x$. In transverse Fourier transform variables, this substep is simplyThis is the split-step Fourier method.

Partition the range into planes $x_j=x_0+j\Delta x$. At each step, evaluate the refractive-index phase screen at $x_j$ or the midpoint, multiply the current envelope by $e^{\Delta x{S}_j}$, transform in $z$, multiply each transverse Fourier component by the diffraction phase $e^{-i{k}_z^2\Delta x/(2k_0)}$, and apply the inverse Fourier transform. Repeating this split-step Fourier method $m$ times produces $E(x_m,z)$ from $E(x_0,z)$. The step size must continue to resolve both longitudinal medium variation and the Lie-Trotter splitting error.

Use the sign convention in the question,Then the total field satisfiesThe outgoing free-space Green function iswith $({\nabla }^2+k_0^2)G_0=-\delta$. Hence the Lippmann-Schwinger equation isReplacing the unknown interior total field by the incident field gives the first Born approximation

For the Rytov approximation, put $\psi ={\psi }_i{e}^{\varphi }$. After division by $\psi$, the wave equation givesNeglecting the quadratic term $(\nabla \varphi {)}^2$ makes ${\psi }_i\varphi$ obey the same inhomogeneous equation as the first Born scattered field. ThusandIts power series beginsso the two approximations agree through first order in the scattering potential.

Let the incident plane wave beand define the scattering vectorFor $r$ much larger than the diameter of $D$, the far-field pattern follows fromWritingthe Born result is

The corresponding Rytov logarithmic perturbation isTherefore

Since $n=1+W$ with small variance,To leading order, $V$ is therefore a centered stationary Gaussian random field. Let its autocorrelation function of a random field beThe $W^2$ contribution gives higher-order mean and non-Gaussian corrections and is consistently omitted at this order.

The far-field Rytov approximation from part ii has unit incident intensity andThe real random variable $X={\varphi }_1+{\varphi }_1^{\ast }$ is centered Gaussian. Its moment-generating function givesDefineThenand henceThis expression depends only on the two-point autocorrelation of the scattering potential. In the weak-fluctuation expansion it becomes

Let $({\sigma }_j,u_j,v_j)$ be a singular system of a compact operator $A:X\to Y$, soThe Moore–Penrose inverse of an operator has domainand acts bywith the orthogonal component of $y$ sent to zero. Equivalently, its domain consists of the data satisfying the Picard criterion. It obeys

If $Ax=y$ is exactly solvable, every solution is $x^{†}+z$ with $z\in \ker A$, andis the unique minimum-norm least-squares solution. It recovers the component of the original $x$ orthogonal to the null space; no data can determine the null-space component.

The normal equation for a linear inverse problem isLandweber iteration is the stationary iterationA sufficient step-size condition isIf $y\in \mathcal{D}(A^{†})$ and $x_0\in (\ker A{)}^{\perp }$, thenFor an arbitrary initial iterate, its null-space component is unchanged and the limit is

Forthe Frechet derivative in direction $h$ isThus the Hilbert-space gradient isThe gradient descent update with step size $\gamma$ is consequentlywhich is exactly Landweber iteration. Its stationary points solve the normal equation and minimize the convex quadratic functional $J$.

SetStarting from $x_0=0$, repeated substitution in Landweber iteration givesThis is the finite partial sum of a Neumann series. Whenever $A^{\ast }A$ is boundedly invertible on the relevant subspace and $\Vert I-\gamma A^{\ast }A\Vert <1$,so

For a genuinely compact operator on an infinite-dimensional space, nonzero singular values can accumulate at zero, so this inverse is generally unbounded and the Neumann series need not converge in operator norm. Under the condition in part ii it nevertheless converges componentwise on admissible data to the Moore–Penrose inverse of an operator; stopping after finitely many terms suppresses poorly determined small-singular-value components and acts as a regularization of an inverse problem.