Mean curvature of sphere cap

Suppose one has the disc {B_1=\{x\in \mathbb{R}^n||x|\leq 1\}}, {n\geq 3}, prescribe the ball with metric

\displaystyle g_{ij}=4u^{\frac{4}{n-2}}\delta_{ij}, \quad u=\left(\frac{\epsilon}{\epsilon^2+|x|^2}\right)^{(n-2)/2}

What is the mean curvature of the boundary? As we all know that under the Euclidean metric, the boundary of unit ball has mean curvature {h=1}. We want to use the formula of mean curvature under the conformal transmformation. Namely, suppose the {(M,g_0)} has mean curvature {h_0}, then under metric {g=v^{\frac{4}{n-2}}g_0}, the mean curvature of {(M,g)} will be

\displaystyle h_g=\frac{2}{n-2}v^{-\frac{n}{n-2}}\left(\frac{\partial v}{\partial \eta}+\frac{n-2}{2}h_0 v\right)

where {\eta} is the normal outer unit vector under {g_0}. Using the above principle, let {v=2^{\frac{n-2}{2}}u}, {g_0} be the Euclidean flat metric, then {h_0=1}.

\displaystyle \frac{\partial v}{\partial \eta}=(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n}{2}}(2-n)

\displaystyle \frac{n-2}{2}h_0 v=\frac{n-2}{2}(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n-2}{2}}

\displaystyle h_g=\frac{2}{n-2}\frac{(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n}{2}}(2-n)+\frac{n-2}{2}(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n-2}{2}}}{(2\epsilon)^{-\frac{n}{2}}(\epsilon^2+1)^{-\frac{n}{2}}}=\frac{\epsilon^2-1}{2\epsilon}

Remark: Escobar. Conformal Defromation of a Riemannnian metric to a constant scalar curvature metric with constant mean curvature on the boundary. Indiana University Mathematics Journal 1996.

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