0$. By analogous proof, one can see that
if $P$ be non-positive then $Q$ will be non-positive as well.
Next, assume the maximum $P$ is attained at a point $x_0 \in D_{i_0}$.
Then, we get
\begin{align*}
0< P&= \widehat{u}_{i_0}(x_0) - \widehat{v}_{i_0}(x_0)\\
&= (u_{i_0}(x_0) - v_{i_0}(x_0)) + \sum_{j\neq {i_0}}( v_{j} (x_0) - u_{j}(x_0))\\
&= \sum_{j\neq {i_0}}( v_{j} (x_0) - u_{j}(x_0)).
\end{align*}
This shows that
\[
\sum_{j\neq i_0} v_{j} (x_0)= \sum_{j\neq i_0} u_{j}(x_0)+P> 0.
\]
Since $(v_1, \dots ,v_m)\in S, $ then there exists $j_0\neq i_0$ such that
$v_{j_0}(x_0)> 0$. This implies
\[
0< P= \widehat{u}_{i_0}(x_0) - \widehat{v}_{i_0}(x_0)
= v_{j_0}(x_0)- \sum_{j\neq i_0}u_{j}(x_0)
\le \widehat{v}_{j_0}(x_0)-\widehat{u}_{j_0}(x_0) \le Q.
\]
The same argument shows that $Q \le P$ which yields $P=Q$. Hence, we can write
\[
P=v_{j_0}(x_0)-\sum_{j\neq i_0} u_{j}(x_0)
= \widehat{v}_{j_0}(x_0)-\widehat{u}_{j_0}(x_0)=Q.
\]
This gives us $2 \sum_{j\neq j_0}u_{j}(x_0)= 0$, and therefore
\[
u_{j} (x_0)=0, \quad \forall j\neq j_0,
\]
which completes the last statement of the proof.
\end{proof}
We are ready to prove the uniqueness of a limiting configuration.
\begin{theorem} \label{thm3.3}
There exists a unique vector $(u_1,\dots,u_m) \in S $, which satisfies
the limiting solution of \eqref{s1}.
\end{theorem}
\begin{proof}
To show the uniqueness of the limiting configuration, we assume that two
m-tuples $(u_1,\dots,u_m)$ and $(v_1,\dots,v_m )$ are the solutions of
system \eqref{s1} as $ \varepsilon$ tends to zero. These two solutions
belong to the class $S$. For them we set $P$ and $Q$ as above.
Then, we consider two cases $P\le 0 $ and $P> 0$.
If we assume that $ P \le 0$ then Lemma \ref{s21} implies that $Q \le 0$.
This leads to
\[
0 \le -Q\le \widehat{u}_i(x) - \widehat{v}_i(x) \le P\le 0,
\]
for every $1\le i\le m$, and $x\in\Omega$.
This provides that
\[
\widehat{u}_i(x) =\widehat{v}_i(x) \quad i= 1,\dots ,m,
\]
which in turn implies
\[
u_i(x) =v_i(x).
\]
Now, suppose $P > 0$. We show that this case leads to a contradiction.
Let the value $P$ is attained for some $i_0$, then due to Lemma \ref{s21}
there exist $ x_0 \in \Omega $ and
$ j_0\neq i_0 $ such that:
\[
0 < P= Q = \widehat{u}_{i_0}(x_0) - \widehat{v}_{i_0}(x_0)
= \max_{\{ u_{i_0}=v_{i_0}=0 \}} (\widehat{u}_{i_0}(x)
- \widehat{v}_{i_0}(x))= v_{j_0}(x_0)- u_{j_0}(x_0).
\]
Let $\Gamma$ be a fixed curve starting at $x_0$ and ending on the boundary
of $\Omega$. Since $\Omega$ is connected, then one can always choose such
a curve belonging to $\overline{\Omega}$. By the disjointness and smoothness
of $v_{j_0}$ and $u_{j_0}$ there exists a ball centered at $x_0$, and
with radius $r_0$ ($ r_0 $ depends on $x_0$) which we denote
$B_{r_0}(x_0)$, such that
\[
{v}_{j_0}(x)-u_{j_0}(x)>0 \quad \text{in } B_{r_0}(x_0).
\]
This yields
\[
\Delta (\widehat{v}_{j_0}(x)-\widehat{u}_{j_0}(x))\ge 0\quad \text{in }
B_{r_0}(x_0).
\]
The maximum principle implies that
$$
\max_{\overline{B_{r_0}(x_0)}} (\widehat{v}_{j_0}(x)-\widehat{u}_{j_0}(x))
= \max_{\partial B_{r_0}(x_0)}\;(\widehat{v}_{j_0}(x)-\widehat{u}_{j_0}(x))\leq P.
$$
One the other hand, in view of Lemma \ref{s21} we have
$$
\widehat{v}_{j_0}(x_0)-\widehat{u}_{j_0}(x_0)=v_{j_0}(x_0)- u_{j_0}(x_0)=P,
$$
which implies that $P$ is attained at the interior point $x_0\in B_{r_0}(x_0)$. Thus,
$$
\widehat{v}_{j_0}(x)-\widehat{u}_{j_0}(x)\equiv P>0\quad\text{in }
\overline{B_{r_0}(x_0)}.
$$
Next let $x_1\in \Gamma\cap \partial B_{r_0}(x_0)$. We get
$\widehat{v}_{j_0}(x_1)-\widehat{u}_{j_0}(x_1)=P>0$,
which leads to ${v}_{j_0}(x_1)~\ge ~ {u}_{j_0}(x_1)$. We proceed as follows:
If ${v}_{j_0}(x_1)> {u}_{j_0}(x_1)$, then as above ${v}_{j_0}(x)> {u}_{j_0}(x)$
in $B_{r_1}(x_1)$. This in turn implies
\[
\Delta (\widehat{v}_{j_0}(x)-\widehat{u}_{j_0}(x))\ge 0\quad \text{in }
B_{r_1}(x_1).
\]
Again following the maximum principle and recalling that
$\widehat{v}_{j_0}(x_1)-\widehat{u}_{j_0}(x_1)=P$ we conclude that
$$
\widehat{v}_{j_0}(x)-\widehat{u}_{j_0}(x)= P>0\quad \text{in }
\overline{B_{r_1}(x_1)}.
$$
If ${v}_{j_0}(x_1)={u}_{j_0}(x_1)$, then clearly the only possibility is
${v}_{j_0}(x_1)={u}_{j_0}(x_1)=0$. Thus,
$$
00\quad \text{in }
\overline{B_{r_1}(x_1)}.
$$
Then we take $x_2\in \Gamma\cap \partial B_{r_1}(x_1)$ such that $x_1$
stands between the points $x_0$ and $x_2$ along the given curve $\Gamma$.
According to the previous arguments for the point $x_2$ we will find an
index $l_0\in{\{1,\dots,m}\}$ and corresponding ball $ B_{r_{2}}(x_2)$, such that
$$
|\widehat{u}_{l_0}(x)-\widehat{v}_{l_0}(x)|=P \quad\text{in }
\overline{B_{r_{2}}(x_2)}.
$$
We continue this way and obtain a sequence of points $x_n$ along the curve
$\Gamma$, which are getting closer to the boundary of $\Omega$.
Since for all $j=1, \dots ,m$ and $x\in\partial\Omega$ we have
$$
\widehat{u}_{j}(x)-\widehat{v}_{j}(x)=\widehat{v}_{j}(x)-\widehat{u}_{j}(x)=0,
$$
then obviously after finite steps $N$ we find the point $x_N$, which will
be very close to the $\partial\Omega$ and for all $j=1,\dots,m $
$$
|\widehat{u}_{j}(x_N)-\widehat{v}_{j}(x_N)|