measure $\overline{\nu}$ is the so called (see [@ST Theorem I.1.3]) equilibrium measure under the influence of the external field $ (1-a) V^{\sigma}(z)$. From (\[equilibriummeasure\]) we have the following equilibrium condition $$\label{equilibriumcondition}
V^{\overline{\nu}}(t)-(1-a) V^{\sigma}(t)=a V^{\lambda_0}(t)=a \log 2, \quad t\in [-1,\,1].$$

Observe the conditions $(3)$ in both Riemann Hilbert problems. $Y$ requires a normalization at infinity to get to $R$’s behavior at infinity. We modify $Y$ to obtain a Riemann-Hilbert problem whose solution is defined on the same set as $Y$, which approaches $\mathbb{I}$ as $n \rightarrow \infty$. Let us introduce the function $g(z,\overline{\nu})$, which is the analytic potential corresponding to the measure $\overline{\nu}$ described in (\[equilibriummeasure\]) $$\label{defgn}
g(z,\overline{\nu})=-\int \log\, (z-t)\, \mbox{d} \overline{\nu}(t)=V^{\overline{\nu}}(z)-\,i\,\int \arg (z-t) \, \mbox{d}\overline{\nu}(t),$$ with $\arg$ denoting the principal argument $g(z,\overline{\nu}) \in \mathcal{H}(K\setminus (-\infty,\,1])$. Substituting $g(z,\overline{\nu})$ in (\[equilibriumcondition\]) we obtain $$\label{roi}
g_+(x,\overline{\nu})+g_-(x,\overline{\nu})-2a \log 2-2(1-a)g(x,\sigma)=0, \quad x \in [-1,\,1].$$ and $$\label{casta}
g_-(x,\overline{\nu})-g_+(x,\overline{\nu})=\left\{\begin{array}{l l l}
0 & \mbox{if} & x\geq 1 \\
2\pi i &\mbox{if} & x\leq 1 \\
2 \wp(x) &\mbox{if} & x\in (-1,\,1),
\end{array}
\right.$$ with $$\label{california}
\wp(x)=\displaystyle \pi i \int^1_{x} \mbox{d} \overline{\nu}(t)=\pi i \left[(1-a) \displaystyle  \int^1_{x} \mbox{d} \widetilde{\sigma}(t)-\frac{a}{\pi}\arccos(x)  \right].$$

Consider the matrices $\displaystyle G(z)=\left(\begin{array}{c c}
e^{n g(z,\overline{\nu})} & 0 \\ 
0 & e^{-n g(z,\overline{\nu})} 
\end{array}
\right)$ and $\displaystyle 