## 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.