V$ can not both vanish identically. For every $\lambda\in (0,\tilde \lambda)$, the couple $(U_\lambda,V_\lambda)$ satisfies $$\label{eqlam}
\begin{cases}
   \Delta U_\lambda=\lambda^2 V_\lambda,&\text{in }B_1^+,\\
   \Delta V_\lambda=0,&\text{in }B_1^+,\\
\partial_\nu U_\lambda=0, &\text{on }B_1',\\
\partial_\nu V_\lambda=\lambda h(\lambda x)u_\lambda, &\text{on }B_1',
\end{cases}$$ in a weak sense, i.e. $$\label{eq:8tau}
\begin{cases}
\int_{B_1^+}\nabla
U_\lambda\cdot\nabla \varphi\, dz=-\lambda^2
\int_{B_1^+}V_\lambda\varphi\, dz \\
\int_{B_1^+}\nabla
V_\lambda\cdot\nabla \varphi\, dz=\lambda
\int_{B_1'}h(\lambda x)u_\lambda(x)\mathop{\rm Tr}\varphi(x) \,dx,
\end{cases}$$ for all $\varphi \in H^1(B_1^+)$ such that $\varphi=0\text{ on }S_1^+$, where $u_\lambda=\mathop{\rm Tr}U_\lambda$. From the weak convergences $U_{\lambda _{n_k}}{\rightharpoonup}\widetilde U$ and $V_{\lambda _{n_k}}{\rightharpoonup}\widetilde V$ in $H^1(B_1^+)$, we can pass to the limit in to obtain $$\begin{cases}
\int_{B_1^+}\nabla
\widetilde U\cdot\nabla \varphi\,dz=0,\\
\int_{B_1^+}\nabla \widetilde
V\cdot\nabla \varphi\,dz=0,
\end{cases}\quad \text{for all $\varphi \in H^1(B_1^+)$  such that
$\varphi=0\text{ on }S^1_+$},$$ i.e. $(\widetilde U,\widetilde V)$ weakly solves $$\label{eq:extended_limit}
\begin{cases}
   \Delta \widetilde U=0,&\text{in }B_1^+,\\
   \Delta \widetilde V=0,&\text{in }B_1^+,\\
\partial_\nu \widetilde U=0, &\text{on }B_1',\\
\partial_\nu \widetilde V=0, &\text{on }B_1'.
\end{cases}$$ From elliptic regularity under Neumann boundary conditions (see in particular [@Salsa Theorem 8.13]) we conclude that $$\label{eq:30}
  \{U_\lambda\}_{\lambda\in(0,\tilde \lambda)} \text{ and }
\{V_\lambda\}_{\lambda\in(0,\tilde \lambda)} \text{ are bounded in
$H^2(B_r^+)$ for all $r\in(0,1)$},$$ hence, by compactness, up to passing to a subsequence, $$\label{eq:strong_conv}
U_{\lambda _{n_k}}\to \widetilde U\ \text{and}\
V_{\lambda _{n_k}}\to\widetilde V\ \text{
 weakly in $H^2(B_r^+)$ and strongly in $H^1(B_r^+)$ for all 