in Lemma \[l:limitN\]. Then

-   there exists $\ell\in {{\mathbb N}}$ such that $\gamma=\ell$;

-   for every sequence $\lambda_n\to0^+$, there exist a subsequence $\{\lambda_{n_k}\}_{k\in{{\mathbb N}}}$ and $2M_\ell$ real constants $\beta_{\ell,m},\beta'_{\ell,m}$, $m=1,2,\dots,M_\ell$, such that $\sum_{m=1}^{M_\ell}((\beta_{\ell,m})^2+(\beta'_{\ell,m})^2)=1$ and $$\frac{U(\lambda_{n_k}z)}{\sqrt{H(\lambda_{n_k})}}\to
    |z|^{\ell}
    \sum_{m=1}^{M_\ell}\beta_{\ell,m}Y_{\ell,m}\Big(\frac
    z{|z|}\Big),\quad
    \frac{V(\lambda_{n_k}z)}{\sqrt{H(\lambda_{n_k})}}\to
    |z|^{\ell}
    \sum_{m=1}^{M_\ell}\beta'_{\ell,m}Y_{\ell,m}\Big(\frac
    z{|z|}\Big),$$ weakly in $H^1(B_1^+)$ and strongly in $H^1(B_r^+)$ for all $r\in(0,1)$. See Section \[s:introduction\] for the definition of $M_\ell$ and $Y_{\ell,m}$.

Let us define $$\label{eq:UVtau}
U_\lambda(z)=\frac{U(\lambda z)}{\sqrt{H(\lambda)}},
\quad
V_\lambda(z)=\frac{V(\lambda z)}{\sqrt{H(\lambda)}}.$$ We notice that $$\label{eq:26}
\Delta U_\lambda=\lambda^2V_\lambda\quad\text{and}\quad \int_{S_1^+}(U_{\lambda}^2+V_{\lambda}^2)dS=1.$$ By scaling and we have $$\begin{gathered}
\label{eq:8bis}
  \int_{B_{1}^+} \left(|\nabla U_\lambda(z)|^2 +|\nabla V_\lambda(z)|^2+\lambda^2 U_\lambda(z) V_\lambda(z)\right)\,dz -
  \lambda
\int_{B_1'}h(\lambda
x)U_\lambda(x,0) V_\lambda(x,0)\,dx
  \\={\mathcal N}(\lambda)=O(1)\end{gathered}$$ as $\lambda\to0^+$. On the other hand, Lemmas \[l:hardyball\] and \[l:Poincare\] imply $$\begin{gathered}
  \mathcal N(\lambda)\geq
\left(\int_{B_{1}^+} \left(|\nabla U_\lambda(z)|^2 +|\nabla V_\lambda(z)|^2\right)\,dz
\right)\left(1-\frac{\lambda^2}{2N}-
\frac{\lambda\|h\|_{L^{\infty}(B_{r_0}')}}{2\widetilde \Lambda}\right)\\
-\frac{\lambda^2}{2N}-\frac{\lambda\|h\|_{L^{\infty}(B_{r_0}')}(N-1)}{4\widetilde \Lambda}\end{gathered}$$ so that and Lemma \[l:Poincare\] imply that $$\{U_\lambda\}_{\lambda\in(0,\tilde \lambda)}
\text{ and }\{V_\lambda\}_{\lambda\in(0,\tilde \lambda)}
\text{ are bounded in }H^1(B_1^+)$$ for some $\tilde\lambda>0$.

Therefore, for any given sequence $\lambda_n\to 0^+$, there exists a subsequence $\lambda_{n_k}\to0^+$ such that $U_{\lambda _{n_k}}{\rightharpoonup}\widetilde U$ and $V_{\lambda _{n_k}}{\rightharpoonup}\widetilde V$ weakly in $H^1(B_1^+)$ for some $\widetilde U,\widetilde V\in H^1(B_1^+)$. From compactness of the trace embedding $H^1(B_1^+)\hookrightarrow L^2(S_1^+)$ and from we deduce that $$\label{eq:27}
\int_{S_1^+}(\widetilde U^2+\widetilde V^2)dS=1,$$ hence $(\widetilde U,\widetilde V)\neq(0,0)$, i.e. $\widetilde U$ and $\widetilde 