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

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