# Mean curvature and the solution of $\Delta u = F(u)$

Let M be a compact manifold with boundary. Suppose $H$ denotes the mean curvature of $\partial M$, and if $u$ is a non-negative solution of $$\Delta u =F(u) \\ u|_{\partial M} =0$$ By a suitable choice of orthonormal frame $e_1,..., e_n$ such that $e_n=\frac{\partial}{\partial v}$, outer normal vector, and $e_\alpha$ are tangential to $\partial M$ for $\alpha <n$. Then how to get $$(n-1)Hu_v =\sum_{\alpha < n } u_{\alpha\alpha}$$

This question is from theorem 2 of Li and Yau's Estimates of eigenvalues of a compact Riemannian manifold

• Is $F$ some specific function?
– mcd
Commented Jan 5, 2017 at 10:02
• @mcd $F$ is a smooth function. Commented Jan 5, 2017 at 11:28

The fact that the desired expression does not depend on $F$ should be a hint that this fact is independent of the equation $\Delta u = F(u)$. This equation holds for any Riemannian submanifold. Let $\nabla$ denote the connection of $M$ and $D$ that of $\partial M$, and $$A(X,Y)\nu = \langle \nabla_X \nu, Y \rangle \nu = D_X Y - \nabla_X Y$$ the second fundamental form. Then the second covariant derivatives $D^2u, \nabla^2u$ differ by
$$\nabla^2 u(X,Y) - D^2 u(X,Y) =(X(Yu) - (\nabla_X Y)u) - (X(Yu) - (D_XY)u) = A(X,Y)\partial_\nu u.$$
Since $\{e_\alpha : \alpha < n\}$ forms an orthonormal basis for $T\partial M$ and $D^2u = 0$ (from the boundary condition), we thus have
$$\sum_{\alpha<n} u_{\alpha\alpha} = \sum_{\alpha<n} \nabla^2 u(e_\alpha, e_\alpha) = {\rm tr}_{T\partial M} (A) \partial_\nu u = (n-1)H \partial_\nu u.$$