Fixed-frequency inverse scattering and Born linearization — Examples and Counterexamples

By [F2], $M=\mathrm{\mid }D\mathrm{\mid }=1\mathrm{/}64$. By [F3], every distance in $D$ is at most $\sqrt{3}\mathrm{/}4$, and opposite corners attain this value, so $d:=\mathrm{diam}(D)=\sqrt{3}\mathrm{/}4<1\mathrm{/}2$. The indicator $1_D$ is a nonzero $L^2$ class.

For $0\le r\le d$, the curve $\gamma (s)=e^{irs}$ on $[0,1]$ has derivative ${\gamma }^{\mathrm{\prime }}(s)=ir{e}^{irs}$, whose modulus is $r$ by [F4]. Fact [F5] gives $\mathrm{\mid }{e}^{ir}-1\mathrm{\mid }\le r$, and hence $\mathrm{Re}(e^{ir})\ge 1-r\ge 1-d>1\mathrm{/}2$.

For almost every $x\in D$, the kernel integral in [F1] represents $K_{q,1}{1}_D(x)$. For such an $x$, almost every $y\in D$ has $r=\mathrm{\mid }x-y\mathrm{\mid }$ in $(0,d]$ by step 1.1, and [F1] gives $\mathrm{Re}{G}_1(x-y)=\frac{\mathrm{Re}(e^{ir})}{4\pi r}>\frac{1}{8\pi d}\mathrm{.}$ The measurable function $y\mapsto \mathrm{Re}{G}_1(x-y)-1\mathrm{/}(8\pi d)$ is positive almost everywhere on the positive-measure set $D$, so some set where it exceeds $1\mathrm{/}n$ has positive measure and its integral is strictly positive. Integrating the pointwise lower bound against $q=512\pi >0$ therefore yields $\mathrm{Re}(K_{q,1}{1}_D)(x)>qM\mathrm{/}(8\pi d)=1\mathrm{/}d=4\mathrm{/}\sqrt{3}>2$ for almost every $x\in D$; the last inequality follows by squaring the positive sides of $\sqrt{3}<2$.

By step 3.1, $g:=\mathrm{\mid }{K}_{q,1}{1}_D{\mathrm{\mid }}^2-4$ is positive almost everywhere on $D$; [F1] and [A1] put $K_{q,1}{1}_D$ in $L^2(D)$, so $g$ is integrable. Since $M>0$ by step 1.1, the sets $\{x\in D:g(x)>1\mathrm{/}n\}$ cannot all be null, as their countable union contains the positive-measure set where $g>0$. Thus for some $n$, the set $\{x\in D:g(x)>1\mathrm{/}n\}$ has positive measure, so ${\int }_{D}g\ge \mathrm{\mid }\{g>1\mathrm{/}n\}\mathrm{\mid }\mathrm{/}n>0$; hence $\mathrm{\parallel }{K}_{q,1}{1}_D{\mathrm{\parallel }}_2^2>4M=4\mathrm{\parallel }{1}_D{\mathrm{\parallel }}_2^2$. The unit-ball supremum in [F7] gives $\mathrm{\parallel }{K}_{q,1}\mathrm{\parallel }>2$. By [F9] the contrast lies outside ${\mathcal{U}}_{D,1}$, so the small-norm model branch does not apply; no nonexistence or nonuniqueness conclusion follows.

By [F4], $M=\mathrm{\mid }D\mathrm{\mid }={\mathrm{\ell }}^3$. The coordinate formula [F5] shows that the largest distance in the cube is attained by opposite corners and is $\mathrm{diam}(D)=\sqrt{3}\mathrm{\ell }$. Also $\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}=1$.

Substituting the values from step 1.1 into [F2] gives ${\rho }_1(q)\le C_D=\sqrt{\frac{{\mathrm{\ell }}^3\sqrt{3}\mathrm{\ell }}{4\pi }}=\sqrt{\frac{\sqrt{3}{\mathrm{\ell }}^4}{4\pi }}=\frac{1}{2\sqrt{2}}<1.$ The final strict inequality follows from $\sqrt{2}>1$. Hence $q\in {\mathcal{U}}_{D,1}$ by [F1].

Apply [F3] with $M={\mathrm{\ell }}^3$, $\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}=1$, and $C_D=1\mathrm{/}(2\sqrt{2})$. Since $C_D<1$, it gives the uniform error bound ${\sup }_{\theta ,d\in S^2}\mathrm{\mid }{\mathcal{S}}_1(q)(\theta ,d)-{\mathcal{B}}_{1}q(\theta ,d)\mathrm{\mid }\le \frac{{\mathrm{\ell }}^3}{4\pi }\frac{C_D}{1-C_D}=\frac{{\mathrm{\ell }}^3}{4\pi (2\sqrt{2}-1)}\mathrm{.}$ This verifies the claimed Born-regime example.

Put $N=m+1$ and partition $D$ into the $N$ measurable slabs $E_n=[(n-1)\mathrm{/}N,n\mathrm{/}N)\times [0,1{]}^2(1\le n<N),E_N=[(N-1)\mathrm{/}N,1]\times [0,1{]}^2\mathrm{.}$ They are pairwise disjoint, cover $D$, and have measure $1\mathrm{/}N$ by [F3]. For each pair $({\theta }_b,d_a)$ set ${\xi }_{ba}=k({\theta }_b-d_a)\mathrm{/}(2\pi )$ and define $F\in {\mathbb{C}}^{m\times N}$ by $F_{(b,a),n}=\frac{1}{4\pi }{\int }_{E_n}{e}^{-2\pi ix\cdot {\xi }_{ba}}dx$.

Regard $F$ as the linear map ${\mathbb{C}}^{m+1}\to {\mathbb{C}}^m$. Its image is a subspace of dimension at most $m$ by [F5], so [F4] gives $\dim \ker F=(m+1)-\dim \mathrm{im}F\ge 1$. Choose a nonzero $c=(c_n{)}_{n=1}^{m+1}\in \ker F$.

Define $h:={\sum }_{n=1}^{m+1}{c}_n{1}_{E_n}$. Since at least one $c_n\mathrm{\ne }0$ and each $E_n$ has positive measure, $h$ is a nonzero element of $L^{\mathrm{\infty }}(D;\mathbb{C})$. By [F1] and [F2], ${\mathcal{B}}_{k}h({\theta }_b,d_a)=(Fc{)}_{(b,a)}=0$ for every $1\le b\le r$, $1\le a\le s$. Thus the finite Born sample map has a nontrivial kernel.

For any scalar $t$, [F6] gives ${\rho }_k(th)\le C_D\mathrm{\parallel }th{\mathrm{\parallel }}_{\mathrm{\infty }}=C_D\mathrm{\mid }t\mathrm{\mid }\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}$. The right-hand side tends to zero as $\mathrm{\mid }t\mathrm{\mid }\to 0$, so choose $t\mathrm{\ne }0$ small enough that $th\in {\mathcal{U}}_{D,k}$. Set $q:=th$. Linearity of ${\mathcal{B}}_k$ gives ${\mathcal{B}}_{k}q({\theta }_b,d_a)=0$ for every acquisition pair while $q\mathrm{\ne }0$. The vector $c$ is a nonzero coefficient representation of a contrast whose sampled first Born data are exactly those of zero.

For input $y=0$, any filter application in the algorithm gives $\widehat{c}=0$ by [F7]; the reconstruction is zero and $R=\mathrm{\parallel }F(0)-0{\mathrm{\parallel }}_2=0\le \delta$, so the status is discrepancy for every $\delta \ge 0$ and positive budget. This status cannot distinguish the zero contrast from the nonzero contrast in step 3.1 within the finite Born model, and no statement about the exact nonlinear data was used.