Conductivity uniqueness on a Euclidean CTA domain with simple transversal

By [F1], all boundary-normal jets of ${\gamma }_1$ and ${\gamma }_2$ agree, so the boundary values and first normal derivatives of ${\rho }_j=\sqrt{{\gamma }_j}$ agree. Put $q_j={\mathrm{\Delta }}_g{\rho }_j\mathrm{/}{\rho }_j$. Their Liouville boundary relations in [F2], together with ${\mathrm{\Lambda }}_{{\gamma }_1}={\mathrm{\Lambda }}_{{\gamma }_2}$, now give ${\mathrm{\Lambda }}_{q_1}g={\mathrm{\Lambda }}_{q_2}g$ for every $g\in H^{1\mathrm{/}2}(\mathrm{\partial }\mathrm{\Omega })$: substitute $f=g\mathrm{/}{\rho }_b$ into the physical-flux relation, using the common positive boundary value ${\rho }_b$ and common ${\mathrm{\partial }}_{\nu }{\rho }_j$. Multiplication and division by smooth positive ${\rho }_b$ preserve $H^{1\mathrm{/}2}(\mathrm{\partial }\mathrm{\Omega })$.

Each $q_j$ is nonresonant. If $w\in H_0^1(\mathrm{\Omega })$ solves $(-{\mathrm{\Delta }}_g+q_j)w=0$, then $v=w\mathrm{/}{\rho }_j\in H_0^1(\mathrm{\Omega })$ solves $-\mathrm{div}({\gamma }_j\mathrm{\nabla }v)=0$ by [F2]. Testing weakly with $\bar{v}$ gives ${\int }_{\mathrm{\Omega }}{\gamma }_j\mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}^{2}dx=0$. Positivity of ${\gamma }_j$ implies $v=0$, hence $w=0$.

The domain is bounded and smooth in ${\mathbb{R}}^3$, the potentials are real and smooth, both operators are nonresonant by step 1.2, and their full Schrödinger DN maps agree by step 1.1. Thus [F3] applies and yields $q_1=q_2=:q$ in $\mathrm{\Omega }$.

Each ${\rho }_j$ solves $(-{\mathrm{\Delta }}_g+q){\rho }_j=0$ and the two functions have the same boundary value by step 1.1. Their difference is a zero-trace solution of the nonresonant operator from step 1.2, hence vanishes. Therefore ${\rho }_1={\rho }_2$ and ${\gamma }_1={\rho }_1^2={\rho }_2^2={\gamma }_2$ in $\mathrm{\Omega }$; continuity extends this equality to $\bar{\mathrm{\Omega }}$.