Liouville transform and conductivity DN conjugacy

Let $v\in H^1(M)$ be a weak conductivity solution and put $u=\rho v$. Since $\rho$ is smooth, the distributional Leibniz rule in charts gives ${\mathrm{div}}_g({\rho }^2{\mathrm{\nabla }}_{g}v)={\rho }^2{\mathrm{\Delta }}_{g}v+2\rho g({\mathrm{\nabla }}_g\rho ,{\mathrm{\nabla }}_{g}v),{\mathrm{\Delta }}_g(\rho v)=\rho {\mathrm{\Delta }}_{g}v+2g({\mathrm{\nabla }}_g\rho ,{\mathrm{\nabla }}_{g}v)+v{\mathrm{\Delta }}_g\rho \mathrm{.}$ The first expression is zero distributionally by the weak equation. Dividing by the smooth positive $\rho$ in that identity and substituting into the second gives ${\mathrm{\Delta }}_g(\rho v)=v{\mathrm{\Delta }}_g\rho =q(\rho v)$, hence $u\in H^1(M)$ solves $(-{\mathrm{\Delta }}_g+q)u=0$ weakly. Equivalently, the same identity reads ${\mathrm{div}}_g({\rho }^2{\mathrm{\nabla }}_g(u\mathrm{/}\rho ))=\rho ({\mathrm{\Delta }}_{g}u-qu)$, so the transformation also sends every weak Schrödinger solution in $H^1$ back to a weak conductivity solution. These distributional identities use only the divergence convention in [F4], and are independent of orientability.

If $w\in H_0^1(M)$ solves $(-{\mathrm{\Delta }}_g+q)w=0$, then $v=w\mathrm{/}\rho$ lies in $H_0^1(M)$ because multiplication by the smooth functions $\rho$ and ${\rho }^{-1}$ preserves this space. The distributional identities of step 1.1 in reverse give ${\mathrm{div}}_g(\gamma {\mathrm{\nabla }}_{g}v)=0$; by density this is the weak equation on all $H_0^1$ tests. Testing with $v$ yields ${\int }_M\gamma \mathrm{\mid }{\mathrm{\nabla }}_{g}v{\mathrm{\mid }}_g^{2}d{V}_g=0$. Positivity of $\gamma$ implies ${\mathrm{\nabla }}_{g}v=0$ almost everywhere; the zero trace then gives $v=0$, hence $w=0$. Therefore $q$ is nonresonant and ${\mathrm{\Lambda }}_q$ is defined.

Let $f,h\in H^{1\mathrm{/}2}(\mathrm{\partial }M)$, let $v_f$ be the conductivity solution, and put $u=\rho v_f$. Let $U_h$ be the Schrödinger solution with trace ${\rho }_{b}h$; it exists by step 2.1 and the nonresonant solvability supplier. Then $U_h\mathrm{/}\rho$ is the conductivity solution with trace $h$ by step 1.1. Use these solutions as extensions in the weak DN pairings from [F1]. The weak product calculation, valid almost everywhere for $H^1$ functions and smooth $\rho$, is ${\rho }^{2}g\text{\hspace{0.17em}⁣}({\mathrm{\nabla }}_g(u\mathrm{/}\rho ),{\mathrm{\nabla }}_g(\overline{U_h}\mathrm{/}\rho ))=g({\mathrm{\nabla }}_{g}u,{\mathrm{\nabla }}_g\overline{U_h})+qu\overline{U_h}-{\mathrm{div}}_g\text{\hspace{0.17em}⁣}(u\overline{U_h}{\mathrm{\nabla }}_g\log \rho )\mathrm{.}$ Here $q={\mathrm{\Delta }}_g\log \rho +\mathrm{\mid }{\mathrm{\nabla }}_g\log \rho {\mathrm{\mid }}_g^2$. The product $u\overline{U_h}$ lies in $W^{1,1}(M)$. Approximate $u,U_h$ in $H^1$ by smooth functions using the completion defining $H^1$; their products converge in $W^{1,1}$, and their boundary products converge in $L^1$ by the trace bound in [F1]. Thus the smooth Gauss--Green formula passes to this product, whose trace is $({\rho }_{b}f)\overline{({\rho }_{b}h)}$. Integrating therefore gives $⟨{\mathrm{\Lambda }}_{\gamma }f,h⟩=⟨{\mathrm{\Lambda }}_q({\rho }_{b}f),{\rho }_{b}h⟩-{\int }_{\mathrm{\partial }M}{\rho }_b({\mathrm{\partial }}_{{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}}\rho )f\bar{h}d{S}_g$. By duality, multiplication of an $H^{-1\mathrm{/}2}$ distribution by ${\rho }_b$ is characterized by pairing against ${\rho }_{b}h$, so this is exactly the asserted operator identity for arbitrary $H^{1\mathrm{/}2}$ data.