Conditional uniqueness in the transversely anisotropic Calderon problem

For exact $H^1$ solutions $u_j$ let $f_j=\mathrm{tr}{u}_j\in H^{1\mathrm{/}2}(\mathrm{\partial }M)$ and define $a_j(v,w)={\int }_M(g({\mathrm{\nabla }}_{g}v,{\mathrm{\nabla }}_g\bar{w})+q_{j}v\bar{w})d{V}_g$. By the weak DN definition, $⟨{\mathrm{\Lambda }}_{q_1}{f}_1,f_2⟩=a_1(u_1,u_2)$. Let ${\tilde{u}}_1\in H^1(M)$ 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 $u_1-{\tilde{u}}_1\in H_0^1(M)$ 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 weak pairings and using equality of the DN maps on all $H^{1\mathrm{/}2}(\mathrm{\partial }M)$ gives ${\int }_{M}Q(x_1,x^{\mathrm{\prime }})u_1(x_1,x^{\mathrm{\prime }})\overline{u_2(x_1,x^{\mathrm{\prime }})}d{V}_g=0$, where $Q=q_1-q_2$. This derives the needed weak-trace extension of [F2]. Because $q_j\in C_c^{\mathrm{\infty }}(M^{\circ })$ and $M⋐\mathbb{R}\times M_0^{\circ }$, extension of $Q$ by zero gives a smooth compactly supported function on $\mathbb{R}\times M_0$.

Fix $\lambda \in \mathbb{R}$, put $s=\tau +i\lambda$, and choose the beam $v_s$ of [F6] with residual $O({\tau }^{-K})$. Use the signed real-weight solver [F5] to obtain exact solutions $u_1=e^{-s{x}_1}(v_s+r_1),u_2=e^{s{x}_1}(v_s+r_2)$ for the respective potentials, with $r_j\in H^1(M)$ and $\mathrm{\parallel }{r}_j{\mathrm{\parallel }}_2=O({\tau }^{-1})$. Indeed the conjugated residual is $(-{\mathrm{\Delta }}_{g_0}-s^2)v_s+q_j{v}_s$, of bounded $L^2$ norm on a fixed product cylinder; multiplication by $e^{\pm i\lambda x_1}$ is bounded there. Correcting with the real-weight solver gives precisely this bound. Their product in step 1.1 is $e^{-2i\lambda x_1}(v_s+r_1)(\overline{v_s}+\overline{r_2})$. Every error integral tends to zero by Cauchy--Schwarz, the boundedness of $Q$, $\mathrm{\parallel }{v}_s{\mathrm{\parallel }}_2=O(1)$, and $\mathrm{\parallel }{r}_j{\mathrm{\parallel }}_2=o(1)$. Integrating first in $x_1$ gives the smooth compactly supported transversal test function ${\widehat{Q}}_{\lambda }$ below. Applying [F6] proves the attenuated ray identity for every $\lambda$ and every nontangential $\gamma$. This argument works on the base manifold; it does not assume injectivity of a ray transform on its orientation cover.

Step 2.1 supplies ${\int }_{\gamma }{\widehat{Q}}_{\lambda }(\gamma (t))e^{-2\lambda t}dt=0\text{for every real }\lambda \text{ and every nontangential geodesic }\gamma ,$ where ${\widehat{Q}}_{\lambda }(x^{\mathrm{\prime }})={\int }_{\mathbb{R}}{e}^{-2i\lambda x_1}Q(x_1,x^{\mathrm{\prime }})d{x}_1$. At $\lambda =0$, [F3] gives ${\widehat{Q}}_0=0$. Assume inductively that ${\mathrm{\partial }}_{\lambda }^j{\widehat{Q}}_0=0$ for every $j<k$. Differentiate the ray identity $k$ times at $\lambda =0$. Differentiation under the finite geodesic integral is valid because $Q$ is smooth with compact $x_1$ support. Leibniz' rule leaves the unweighted ray integral of ${\mathrm{\partial }}_{\lambda }^k{\widehat{Q}}_0$; all other terms contain lower derivatives and vanish by the induction hypothesis. Thus [F3], applied to real and imaginary parts, gives ${\mathrm{\partial }}_{\lambda }^k{\widehat{Q}}_0=0$. Induction proves that every Taylor coefficient of $\lambda \mapsto {\widehat{Q}}_{\lambda }(x^{\mathrm{\prime }})$ vanishes for each fixed $x^{\mathrm{\prime }}$.

For each fixed $x^{\mathrm{\prime }}$, compact $x_1$ support makes ${\widehat{Q}}_{\lambda }(x^{\mathrm{\prime }})$ entire in $\lambda$, so its zero Taylor series implies it vanishes for all real $\lambda$. This is the Fourier transform of the $L^1(\mathbb{R})$ function $Q(\cdot ,x^{\mathrm{\prime }})$ at every frequency $2\lambda$; [F4] gives $Q(\cdot ,x^{\mathrm{\prime }})=0$ almost everywhere. Smoothness of $Q$ then yields $Q=0$ on $M$.