H^1(B_R^+)$ be a weak solution to such that $(U,V)\not=(0,0)$ (i.e. $U$ and $V$ are not both identically null). Let $D=D(r)$ and $H=H(r)$ be the functions defined in and . Then there exists $r_0\in(0,R)$ such that $H(r)>0$ for any $r\in (0,r_0)$.

Suppose by contradiction that for any $r_0>0$ there exists $r\in (0,r_0)$ such that $H(r)=0$. Then there exists a sequence $r_n\to 0^+$ such that $H(r_n)=0$, i.e. $U=V=0$ on $S_{r_n}^+$. From it follows that $$\label{eq:2}
\int_{B_{r_n}^+} \left(|\nabla U|^2+|\nabla
V|^2+UV\right)dz=\int_{B_{r_n'}}
h(x)u(x)v(x)\, dx.$$ From , Lemma \[l:Poincare\], and Lemma \[l:hardyball\] it follows that $$\begin{aligned}
\left(1-\frac{r_n^2}{2N}\right)& \int_{B_{r_n}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz{\leqslant}\int_{B_{r_n}^+} \left(|\nabla U|^2+|\nabla
V|^2+UV\right)dz\\
&=\int_{B_{r_n'}}h(x)u(x)v(x)\, dx{\leqslant}{\rm
  const\,}r_n\left(\int_{B_{r_n}'}\frac{u^2}{|x|}\,dx+
\int_{B_{r_n}'}\frac{v^2}{|x|}\,dx\right)
\\
&{\leqslant}{\rm
  const\,}r_n
\int_{B_{r_n}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz.\end{aligned}$$ Since $r_n\to0^+$ as $n\to+\infty$, the above inequality implies that $\int_{B_{r_n}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz=0$ for $n$ sufficiently large. Hence, in view of Lemma \[l:Poincare\], $U\equiv V\equiv 0$ in $B_{r_n}^+$. Classical unique continuation principles then imply that $U\equiv V\equiv 0$ in $B_R^+$ giving rise to a contradiction.

\[l:Dgeq\] Letting $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be as in Lemma \[l:Hpos\] and $D,H$ as in –, there holds $$\begin{aligned}
\label{eq:5}
&D(r){\geqslant}r^{1-N} \left(\int_{B_{r}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz\right)(1+O(r))-H(r)O(r),\\
\label{eq:6}
&D(r){\geqslant}Nr^{-1-N} \left(\int_{B_{r}^+} (U^2+V^2)dz\right)(1+O(r))-H(r)O(1),\end{aligned}$$ as $r\to0^+$.

From Lemma \[l:Poincare\] we have that $$\label{eq:4}
\int_{B_r^+}(U^2+V^2)\,dz{\leqslant}\frac{r^{1+N}}{N}H(r)+\frac{r^2}{N}\int_{B_{r}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz.$$ From it follows that $$\label{eq:1}
