\mathcal{H} (\mathbb{C}\setminus [-1,\, 1]$ with $\sqrt{x^2-1}_\pm=\pm i \sqrt{1-x^2}$, $x \in (-1,\,1)$. We introduce the matrix function $S$ as follows $$S(z)=\left\{\begin{array}{c l l}
T(z) & \mbox{when} & z\in \mathbb{C} \setminus (\gamma \cup \Omega) \\ 
T(z) \left(\begin{array}{c c}
1 & 0 \\ 
-i\sqrt{z^2-1}\, e^{-2 n \mathcal{A}(z)}& 1
\end{array} \right) & \mbox{when} & z\in \Omega
\end{array} 
\right..$$ The matrix function $S$ is the solution of the following Riemann-Hilbert problem (S-RHP):

1.  $S \in \mathcal{H}(\mathbb{C} \setminus (\gamma \cup [-1,\,1]))$,

2.  $\displaystyle S_+(x)=S_-(x) \left(\begin{array}{c c}
    0 & \displaystyle (1-x^2)^{-1/2} \\ 
    - (1-x^2)^{1/2} &0 
    \end{array}
    \right)$, when $x \in (-1,\,1)$ and

3.  $\displaystyle S_+(\zeta)=S_-(\zeta) \left(\begin{array}{c c}
    1 & 0 \\ 
    -i\sqrt{z^2-1}\, e^{-2 n \mathcal{A}(z)}& 1
    \end{array} \right),$ when $\zeta \in \gamma.$

4.  $S(z)=\mathbb{I}+\mathcal{O}\left(1/z\right)$ as $z \to \infty$,

5.  $\displaystyle S(z)=\mathcal{O}\left(\begin{array}{c c}
    1 & |z\pm 1|^{-1/2} \\
    1 & |z\pm 1|^{-1/2}
    \end{array}\right)$ as $z \to \mp 1$.

The jump matrix on $\gamma$ approaches uniformly the identity matrix $I$. However it does not happen in $[-1,\, 1]$. We fix this problem in the interval following the steps in [@Ku]. Consider the matrix $$\label{Ndef}
N(z)=\left(\begin{array}{l l}
\displaystyle \frac{\displaystyle a(z)+a^{-1}(z)}{\displaystyle 2} \frac{D(\infty)}{D(z)}& \displaystyle \frac{\displaystyle a(z)-a^{-1}(z)}{\displaystyle 2i} D(\infty) D(z) \\ 
\displaystyle \frac{\displaystyle a(z)-a^{-1}(z)}{\displaystyle -2i}\frac{1}{D(\infty)D(z)} & \displaystyle \frac{\displaystyle a(z)+a^{-1}(z)}{\displaystyle 2}\frac{D(z)}{D(\infty)} 
\end{array}
\right),$$ 