Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 70 / 4

Choose the scattering-potential convention $V(\mathbf{r})=k^2[n(\mathbf{r}{)}^2-1]$. Let $G_k(\mathbf{r},{\mathbf{r}}^{\prime })=e^{ik|\mathbf{r}-{\mathbf{r}}^{\prime }|}/(4\pi |\mathbf{r}-{\mathbf{r}}^{\prime }|)$, so $(\Delta +k^2)G_k=-\delta$. Since $(\Delta +k^2)\psi =-V\psi$, the outgoing Green representation is the Lippmann-Schwinger equationIterating it produces the Born seriesFor example its first two terms areThe first term describes one scattering event driven by the incident field. The second propagates its first scattered wave to another interaction before reaching the observer. Higher terms contain more successive interactions, including repeated visits to the same region; they account for multiple scattering rather than additional incident waves.

A sufficient convergence condition on an appropriate field norm is $\Vert T\Vert =\rho <1$. Then the first-Born remainder is bounded by ${\rho }^2\Vert {\psi }_i\Vert /(1-\rho )$, while $\Vert T{\psi }_i\Vert \le \rho \Vert {\psi }_i\Vert$. A simple concrete sufficient condition on the bounded scatterer, using the supremum norm, isThis makes the field inside the inhomogeneity close to the incident field. Physically weak index contrast with small accumulated phase shift, such as $k\ell |n-1|\ll 1$ for a smooth weak scatterer of thickness $\ell$, is a common first-Born regime, provided resonant enhancement and multiple reflections are negligible. Weak local contrast alone is not sufficient for an arbitrarily thick scatterer. If the opposite sign convention for $V$ is used, the integral operator changes sign and the same consistent iteration applies.

In first Born order the unknown potential enters linearly. Let $M_i$ multiply by the known incident field, $\mathcal{G}$ be the outgoing Green integral over the known support $D$, and $\mathcal{R}$ restrict the resulting field to the measurement set. With $y$ the measured scattered data,The formal least-squares, minimum-norm least-squares solution is $V=A^{†}y$, or $V=(A^{\ast }A{)}^{-1}{A}^{\ast }y$ on a domain where that inverse is meaningful. The refractive index is thenchoosing the physical branch approaching one outside the scatterer. At weak contrast $n-1\simeq V/(2k^2)$.

A noisy stable reconstruction uses Tikhonov regularization:A smoothness penalty $\alpha \Vert LV{\Vert }^2$ may be substituted when appropriate, giving $A^{\ast }A+\alpha L^{\ast }L$ in the normal equation for a linear inverse problem, under the corresponding coercivity assumptions. The parameter balances noise amplification and smoothing bias.

The data geometry matters: knowing a scattered field on a measurement surface does not automatically mean knowing it throughout the support. A single incident wave and fixed-frequency far field sample only a shifted sphere of the potential's Fourier transform, and in general do not determine an arbitrary three-dimensional potential uniquely. Tikhonov regularization stabilizes a selected minimum-norm least-squares solution; it cannot create missing information or guarantee uniqueness from inadequate data. If the total field and its derivatives were known throughout $D$, the first-Born equation could instead be inverted formally as $V\simeq -(\Delta +k^2){\psi }_s/{\psi }_i$ where ${\psi }_i\ne 0$, but differentiation still amplifies measurement noise.

The adjoint operators, orthonormal singular vectors and squared-norm normal equation for a linear inverse problem conditions require Hilbert spaces. For arbitrary normed spaces the Banach-space adjoint takes values in the dual spaces and need not produce this formula. Interpret the given singular value system in Hilbert spaces, with $u_j$ in the unknown space and $v_j$ in the data space, as in the printed identities. Positive singular values tending to zero provide the compact ill-posed case. A general bounded linear operator need not have this property; the identity operator is a well-conditioned counterexample. Components in the kernel must be fixed by a minimum-norm convention, and the unregularized inverse is defined only for data satisfying the corresponding range condition.

Using the singular value system with $A{u}_j={\sigma }_j{v}_j$ and $A^{\ast }{v}_j={\sigma }_j{u}_j$, the formal minimum-norm inverse isIt defines an element of the unknown Hilbert space only if ${\sum }_j|(y,v_j){|}^2/{\sigma }_j^2<\infty$, the Picard criterion. A null-space component is undetermined by the data.

Adding an error along a single left singular vector changes precisely one coefficient:Although this perturbation has data norm $\delta$, its amplification factor is $1/{\sigma }_j$. If ${\sigma }_j\to 0$, these factors are unbounded. More explicitly choose $y^{(j)}={\sigma }_j{v}_j$, tending to zero, while the minimum-norm inverse is $u_j$, of norm one. Thus the inverse is discontinuous at zero. It may also fail to exist for data outside its nonclosed range, so merely minimizing a residual does not guarantee that the formal inverse exists. This is ill-posedness by unbounded small-singular-value amplification, not a claim that every finite-dimensional or bounded forward map has an unstable inverse.

For $\alpha >0$, minimize $J_{\alpha }(x)=\Vert Ax-y{\Vert }^2+\alpha \Vert x{\Vert }^2$. Its first variation in every direction givesThe self-adjoint operator on the left has lower bound $\alpha I$, so it is invertible and the minimizer is unique. The regularized solution has the singular expansionSince ${\sigma }^2+\alpha \ge 2\sigma \sqrt{\alpha }$, its gain is at most $1/(2\sqrt{\alpha })$. ConsequentlyThis proves existence, uniqueness and continuous data dependence at fixed positive parameter. For the particular perturbation from part (i), the error is $\delta {\sigma }_j/({\sigma }_j^2+\alpha )$ instead of $\delta /{\sigma }_j$. To recover a minimum-norm exact solution as noise tends to zero, choose $\alpha \to 0$ while $\delta /\sqrt{\alpha }\to 0$; dominated convergence of the exact-data singular coefficients removes the bias.

To connect this with the provided inverse expansion, put $B=A{A}^{\ast }$ and $C=A^{\ast }A$. The exact intertwining identity isIt follows by multiplying $(C+\alpha I)A^{\ast }=A^{\ast }(B+\alpha I)$ on either side by the appropriate resolvent. A useful exact inverse identity isExpanding $(I+{\alpha }^{-1}C{)}^{-1}$ to two terms gives the printed approximation, namelyThis is the truncated Tikhonov resolvent. Multiplication gives $(\alpha I+B)R_{\alpha ,2}=I+{\alpha }^{-3}{B}^3$, so the expression is not an exact inverse. For example $A=1$ and $\alpha =1$ give the approximate value $1$ instead of the exact $1/2$. Its exact remainder isIn the regime $\Vert C\Vert /\alpha <1$ it is a controlled Neumann expansion. Using it to reconstruct givesFor each fixed $\alpha >0$ this is a bounded polynomial in bounded operators, withThus the supplied approximation also has continuous data dependence in its domain of use. It must not be extrapolated to $\alpha \downarrow 0$ with $A$ fixed: its gain on any nonzero singular mode diverges, so it would not define a convergent regularization family there. The exact Tikhonov filter supplies the well-posed regularized problem; the truncated formula is only a controlled computational approximation.