I have to prove that if $f\in H^{1/2}(\partial\Omega)$ then $(f,g)$ are Cauchy data for the SchrodingerSchrödinger equation if and only if
$g=\gamma^{-1/2} \Lambda_{\gamma}(\gamma^{-1/2} f)+1/2 \gamma^{-1}\frac{\partial\gamma}{\partial\nu}f$$$g=\gamma^{-1/2} \Lambda_{\gamma}(\gamma^{-1/2} f)+1/2 \gamma^{-1}\frac{\partial\gamma}{\partial\nu}f,$$
Wherewhere $g$ satisfies $g(\varphi)=\int_{\Omega}\nabla u\cdot\nabla\overline\phi-\int_{\Omega}\frac{-\Delta\gamma^{1/2}}{\gamma^{1/2}}u\overline\phi$$$g(\varphi)=\int_{\Omega}\nabla u\cdot\nabla\overline\phi-\int_{\Omega}\frac{-\Delta\gamma^{1/2}}{\gamma^{1/2}}u\overline\phi\,.$$
Then my try is considering a solution for the Schrodinger equation as $\omega=\gamma^{1/2}u$ and then using a Dirichlet condition $\gamma^{-1/2} f$. But then using the mapping $\Lambda_{\gamma}(\gamma^{-1/2} f)(\varphi)=\int_{\Omega}\gamma^{1/2}\nabla\omega\nabla\overline\phi$ the term of the $\gamma^{-1/2}$ cancelled and using the gradient of the product it is almost the left hand side, but the problem is with the term $1/2 \gamma^{-1}\frac{\partial\gamma}{\partial\nu}f$ because I do not know how to pass to an integral. I can not apply the Green formula because we do not have the integral of this term in order to obtain the laplacian.
Does someone know how to proceed from this step? Thanks in advance.
