Full-data uniqueness for the Euclidean Schrödinger Calderón problem

Set $Q=q_1-q_2$. For exact $H^1$ solutions $u_j$ let $f_j=\mathrm{tr}{u}_j\in H^{1\mathrm{/}2}(\mathrm{\partial }\mathrm{\Omega })$ and define the Hermitian weak forms $a_j(v,w)={\int }_{\mathrm{\Omega }}(\mathrm{\nabla }v\cdot \mathrm{\nabla }\bar{w}+q_{j}v\bar{w})dx$. By the weak DN definition, $⟨{\mathrm{\Lambda }}_{q_1}{f}_1,f_2⟩=a_1(u_1,u_2)$. Let ${\tilde{u}}_1\in H^1(\mathrm{\Omega })$ be the $q_2$ solution with trace $f_1$, which exists by nonresonant weak solvability. Then $⟨{\mathrm{\Lambda }}_{q_2}{f}_1,f_2⟩=a_2({\tilde{u}}_1,u_2)$. Since $u_1-{\tilde{u}}_1\in H_0^1(\mathrm{\Omega })$ and $u_2$ solves the $q_2$ weak equation, $a_2(u_1-{\tilde{u}}_1,u_2)=\overline{a_2(u_2,u_1-{\stackrel{~}{u}}_1)}=0$, where the conjugation uses real-valuedness of $q_2$. Subtracting the two weak pairings gives $⟨({\mathrm{\Lambda }}_{q_1}-{\mathrm{\Lambda }}_{q_2})f_1,f_2⟩={\int }_{\mathrm{\Omega }}Q{u}_1\overline{u_2}dx$. The maps agree on all of $H^{1\mathrm{/}2}(\mathrm{\partial }\mathrm{\Omega })$, so this integral vanishes. This establishes the weak-trace extension of [F1] used here.

Fix $\xi \in {\mathbb{R}}^n$. Since $n\ge 3$, choose orthonormal ${\omega }_1,{\omega }_2\in {\mathbb{R}}^n$ perpendicular to $\xi$; when $\xi =0$, choose any orthonormal pair. For $\tau \ge 1$, set ${\sigma }_{\tau }=({\tau }^2+\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}4{)}^{1\mathrm{/}2}$ and ${\zeta }_1=\tau {\omega }_1+i{\sigma }_{\tau }{\omega }_2+\xi \mathrm{/}2,{\zeta }_2=-\tau {\omega }_1-i{\sigma }_{\tau }{\omega }_2+\xi \mathrm{/}2.$ The orthogonality gives ${\zeta }_j\cdot {\zeta }_j={\tau }^2-{\sigma }_{\tau }^2+\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}4=0$, ${\zeta }_1+{\zeta }_2=\xi$, and $\mathrm{\mid }{\zeta }_j\mathrm{\mid }\to \mathrm{\infty }$. Apply [F2] directly to $q_1$ with phase ${\zeta }_1$ and to $q_2$ with phase $-\overline{{\zeta }_2}$, which is also null. This gives $u_1=e^{i{\zeta }_1\cdot x}(1+r_1),u_2=e^{-i\overline{{\zeta }_2}\cdot x}(1+r_2),\mathrm{\parallel }{r}_j{\mathrm{\parallel }}_{L^2(\mathrm{\Omega })}=O({\tau }^{-1})\mathrm{.}$ Substitute in step 1.1. Since $Q\in L^2(\mathrm{\Omega })\cap L^{\mathrm{\infty }}(\mathrm{\Omega })$, $\mid {\int }_{\mathrm{\Omega }}Q{e}^{i\xi \cdot x}(r_1+\overline{r_2}+r_1\overline{r_2})dx\mid \le \mathrm{\parallel }Q{\mathrm{\parallel }}_2(\mathrm{\parallel }{r}_1{\mathrm{\parallel }}_2+\mathrm{\parallel }{r}_2{\mathrm{\parallel }}_2)+\mathrm{\parallel }Q{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{r}_1{\mathrm{\parallel }}_2\mathrm{\parallel }{r}_2{\mathrm{\parallel }}_2=O({\tau }^{-1})\mathrm{.}$ Taking $\tau \to \mathrm{\infty }$ proves ${\int }_{\mathrm{\Omega }}Q{e}^{i\xi \cdot x}dx=0$ for every $\xi \in {\mathbb{R}}^n$.

Define the finite complex Borel measure $\mu (E)={\int }_{E\cap \mathrm{\Omega }}Q(x)dx$ on ${\mathbb{R}}^n$. It has finite total variation because $\mathrm{\Omega }$ is bounded and $Q$ is bounded. The identity in step 2.1 gives $\widehat{\mu }(\eta )=0$ for every $\eta \in {\mathbb{R}}^n$ (take $\xi =-2\pi \eta$). By [F3], $\mu =0$. Hence ${\int }_E\mathrm{Re}Qdx={\int }_E\mathrm{Im}Qdx=0$ for every Borel $E$; applying this to the sets $\{\mathrm{Re}Q\ge 1\mathrm{/}m\}$, $\{\mathrm{Re}Q\le -1\mathrm{/}m\}$, and their imaginary-part analogues shows $Q=0$ almost everywhere. Continuity of $Q$ gives $Q=0$ throughout $\mathrm{\Omega }$.