L=\left(\begin{array}{c c}
2^{n a} & 0 \\ 
0 & 2^{-n a} 
\end{array}
\right).$ We define the matrix function $T=LYGL^{-1}$. So $T$ is the unique solution of the following Riemann-Hilbert problem (T-RHP)

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

2.  $T_+(x)=T_-(x)M(x)$, $x\in (-1,\,1),$

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

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

where according to (\[roi\]) and (\[casta\]) the jump matrix $\displaystyle M(x)=\left(\begin{array}{c c}
e^{-2 n \wp(x)} & (1-x^2)^{-1/2}  \\
0 & e^{2 n\wp(x)} 
\end{array}
\right),$ $x\in (-1,\,1).$

According to [@DKMc Theorem 1.34] there exists a domain $D$ containing the interval $[-1,\,1]$ where the function $\wp$ in (\[california\]) admits an analytic extension on $D\setminus [-\infty,\,1]$ as $$\label{defwp}
\mathcal{A}(z)=\pi i \int_z^1 \, \mbox{d}\overline{\nu}(\zeta)=\pi i \int_z^1 \, \overline{\nu}^{\prime}(\zeta) \mbox{d} \zeta,$$ where $\displaystyle \overline{\nu}^{\prime}(\zeta)=\frac{\psi(\zeta)}{\sqrt{1-\zeta^2}},$ with $\psi \in \mathcal{H}(D)$ and $\psi(x)>0,$ $x\in (-1,1).$ Observe that $\mathcal{A}_+(x)=\wp(x)=-\mathcal{A}_-(x)$, then we write $\displaystyle M(x)=\left(\begin{array}{c c}
e^{-2 n \mathcal{A}_+(x)} & (1-x^2)^{-1/2}\\
0 & e^{-2 n\mathcal{A}_-(x)} 
\end{array}
\right)$.

We now seek jump conditions as we have in R-RHP. Consider a closed Jordan curve $\gamma \in D$ surrounding $[-1,\,1]$ as we have in R-RHP. Let $\Omega$ denote the bounded connected component of $\mathbb{C}\setminus\gamma$. We consider the function $\sqrt{z^2-1}\in 