with boundary terms in $N+1$-dimensional half-balls.

\[l:sobolev\] For any $r>0$ and any $U\in H^1(B_r^+)$ we have $$\widetilde S \left(\int_{B_r'}
|u|^{\frac{2N}{N-1}}dx\right)^{\!\!\frac{N-1}{N}} {\leqslant}\int_{B_r^+}
 |\nabla U|^2 dz+\frac{N-1}{2r} \int_{S_r^+}  U^2 dS$$ where $u={\rm Tr\, }U$ and $\widetilde S$ is a positive constant depending only on $N$.

\[l:hardyball\] For any $r>0$ and any $U\in
  H^1(B_r^+)$ we have that $$\widetilde\Lambda \int_{B_r'} \frac{u^2}{|x|} \,
dx{\leqslant}\int_{B_r^+} |\nabla U|^2 dz+\frac{N-1}{2r}
\int_{S_r^+} U^2 dS$$ where $u={\rm Tr\,} U$ and $\widetilde \Lambda$ is a positive constant depending only on $N$.

The following Poincaré type inequality on half-balls will be useful in the sequel.

\[l:Poincare\] For every $r>0$ and $W\in H^1(B_r^+)$ we have that $$\frac{N}{r^2}\int_{B_r^+}W^2(z)\,dz{\leqslant}\frac{1}{r}\int_{S_r^+}W^2(z)\,dS+\int_{B_r^+}|\nabla
W(z)|^2\,dz.$$

From the Divergence Theorem we have that $$\begin{aligned}
(N+1)\int_{B_r^+}W^2(z)\,dz&=
\int_{B_r^+}\left(\dive(W^2z)-2W\nabla W\cdot z\right)\,dz\\
&=r\int_{S_r^+}W^2(z)\,dS-2 \int_{B_r^+}W\nabla W\cdot z\,dz\\
&{\leqslant}r\int_{S_r^+}W^2(z)\,dS+\int_{B_r^+}W^2(z)\,dz+r^2 \int_{B_r^+}|\nabla W|^2\,dz\end{aligned}$$ thus yielding the stated inequality.

The following lemma contains a Pohozaev type identity for solutions to system .

Let $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be a weak solution to . Then for a.e. $r\in(0,R)$ $$\label{Pohozaev-1}
\int_{B_r^+} \left(|\nabla U|^2+|\nabla
V|^2+UV\right)dz=\int_{S_r^+} \left(\frac{\partial U}{\partial
\nu} U+\frac{\partial V}{\partial \nu}V\right)dS  +\int_{B_r'}
h(x)u(x)v(x)\, dx$$ and $$\begin{aligned}
 \label{Pohozaev-2}
-\frac{N-1}2 & \int_{B_r^+}  \left(|\nabla U|^2+|\nabla
V|^2\right)dz+\int_{B_r^+}  V(z\cdot \nabla U)\, dz +\frac r2
\int_{S_r^+}  \left(|\nabla U|^2+|\nabla V|^2\right)dS \\
\notag & =\int_{B_r'} h(x)u(x)(x\cdot \nabla_x v)\,
dx+ r\int_{S_r^+}  \left(\left|\frac{\partial
U}{\partial\nu}\right|^2+\left|\frac{\partial
V}{\partial\nu}\right|^2\right)dS\end{aligned}$$ 