## Functional of surface with boundary

Suppose ${M}$ is a compact surface with boundary ${\partial M}$ and metric ${g_0}$. Denote by ${K_0}$ the Gauss curvature of ${M}$ and by ${k_0}$ the geodesic curvature of ${\partial M}$. Define a functional

$\displaystyle E_0[u]=\frac{1}{2}\int_M |\nabla_0u|^2d\mu_0+\int_M K_0ud\mu_0+\int_{\partial M}k_0uds_0$

Lemma: The first variation of the functional ${E[u]}$

$\displaystyle \langle E'_0[u],v\rangle=\int_M Kvd\mu+\int_{\partial M}kvds$

where ${K}$, ${k}$ correspond to the curvature ${g=e^{2u}g_0}$. Suppose there is another metric ${g_1=e^{2u_1}g_0}$, similarly we can define ${E_1[u]}$ replacing by ${K_1}$ and ${k_1}$. Then

Proposition: ${E'_0[u]=E_1'[u-u_1]}$; ${E_0[u]=E_1[u-u_1]+E_0[u_1]}$

Proof: The first conclusion is easy to see from the lemma.

$\displaystyle E_0[u]=E_0[u-u_1+u_1]=E_0[u_1]+\frac{1}{2}\int_M |\nabla_0(u-u_1)|^2d\mu_0$

$\displaystyle +\int_M K_0(u-u_1)d\mu_0+\int_{\partial M}k_0(u-u_1)ds_0+\int_M\langle \nabla_0 u_1,\nabla_0(u-u_1)\rangle_0 d\mu_0$

$\displaystyle =E_0[u_1]+\langle E'_0[u],u-u_1\rangle-\frac{1}{2}\int_{M}|\nabla_0(u-u_1)|^2_0d\mu_0$

$\displaystyle =E_0[u_1]+\langle E'_1[u-u_1],u-u_1\rangle-\frac{1}{2}\int_{M}|\nabla_0(u-u_1)|^2_0d\mu_0$

$\displaystyle =E_0[u_1]+E_1[u-u_1]+\frac{1}{2}\int_{M}|\nabla_1(u-u_1)|^2_1d\mu_1-\frac{1}{2}\int_{M}|\nabla_0(u-u_1)|^2_0d\mu_0$

So we need to prove

$\displaystyle |\nabla_1(u-u_1)|^2_1d\mu_1=|\nabla_0(u-u_1)|^2_0d\mu_0$

By using ${\langle \nabla_0 u,X\rangle_0=X(u)}$, we get

$\displaystyle \nabla_1(u-u_1)=e^{-2u_1}\nabla_0(u-u_1)$

$\displaystyle |\nabla_1(u-u_1)|^2_1=e^{2u_1}|\nabla_0(u-u_1)|^2_0$

which exactly means the former identity. $\Box$