on }\ \S.\\
    \end{array}
    \right.$$ Problems and are solvable since $\di\int_{\O}fV^\tau d\O=0$.

In all these three cases, we apply the integral formula (\[formula RF 1\]) to the solutions of the PDEs, i.e. $\phi$ satisfying $\bd ^D \phi =f$. By Cauchy-Schwarz inequality, we have $$\begin{array}[]{rll}
&    \di\frac{n-1}{n}\int_{\O}f^2V^\tau - \int_{\O}V^\tau Ric^{D}(\bn^D \phi, \bn^D \phi),\\ 
\ge&\di\int_{\S}V^{\tau}\left[H^D\<\bn^D \phi,\nu\>^2+\left(h-\g u_\nu g\right)(\n^D\phi, \n^D \phi)-2V^{-\g}\<\n^D \phi, \n^D(V^\g \phi_\nu)\>\right]dA.\label{equ extra 1} 
    \end{array}$$ where the right hand side contains only boundary integrations. Next, we will show that in all these cases, the right hand side boundary integrations are nonnegative.

In case (i), the boundary $\S=\emptyset$ and the result follows immediately.

In case (ii), the first boundary integral in (\[equ extra 1\]) is nonnegative since $H^D\ge0$. Recall from Definition \[notation\] (i), $\n^D \phi=V^{\gamma -\alpha}\n \phi$ where $\n $ is the covariant derivative with respect to the induced metric of the boundary $\S$. Thus we have $\n^D\phi \equiv 0$, since $\phi\equiv 0$ on $\S$.

In case (iii), $\phi_\nu\equiv 0$ on $\S$. We observe that $\left<\bar\n^D\phi, \nu\right>=V^{\g-\alpha}\phi_\nu\equiv 0$ and $\n^D(V^\g\phi_\nu)=V^{\g-\alpha}\n(V^\g\phi_{\nu})\equiv0$. Thus the first and last boundary terms in (\[equ extra 1\]) are all zero. Under the condition that $h-\g u_\nu g\ge0$, we conclude that the right hand side boundary terms are all 