z & \rho
\end{pmatrix},$$ which gives us $$\frac{\partial {\mathbf}{h}}{\partial \rho}(y)
=
-\frac12
\begin{pmatrix}
0 & y & 1
\\
0 & \rho & y
\\
1 & 0 & \rho
\end{pmatrix}
{\mathbf}{h}(y).$$ This, together with and , implies $$\begin{gathered}
\frac{{\,\mathrm{d}}}{{\,\mathrm{d}}\rho}K^{{\mathrm{Pe}}}=\frac{\frac{\partial {\mathbf}{f}^t}{\partial \rho}(x){\mathbf}{h}(y)+{\mathbf}{f}^t(x)\frac{\partial {\mathbf}{h}}{\partial \rho}(y)}{x-y}
\\
={\mathbf}{f}^t(x)
\begin{pmatrix}
0 & 1/2 & 0
\\
0 & 0 & 1/2
\\
0 & 0 & 0
\end{pmatrix}{\mathbf}{h}(y)=\frac 12 (f_1(x)h_2(y)+f_2(x)h_3(y)).\end{gathered}$$ Hence, we obtain $$\begin{gathered}
\label{eq:lamdaderivative}
\frac{\partial}{\partial \rho} F(s;\rho)=
\frac{{\,\mathrm{d}}}{{\,\mathrm{d}}\rho} \ln \det\left(I-K_{s,\rho}^{\mathrm{Pe}}\right)
=-\textrm{tr}\left(\left(I-K_{s,\rho}^{\mathrm{Pe}}\right)^{-1}
\frac{{\,\mathrm{d}}}{{\,\mathrm{d}}\rho}K_{s,\rho}^{\mathrm{Pe}}\right)
\\
=-\frac 12 \int_{-s}^s (F_1(v)h_2(v)+F_2(v)h_3(v)){\,\mathrm{d}}v.\end{gathered}$$

We next establish the connection between the functions $\frac{\partial}{\partial s} F(s;\rho)$, $\frac{\partial}{\partial \rho} F(s;\rho)$ and an RH problem with constant jumps, which is based on the fact that the resolvent kernel $R(u,v)$ is related to the following RH problem.

\[rhp:Y\] We look for a $3 \times 3$ matrix-valued function $Y(z)$ satisfying the following properties:

1.  $Y(z)$ is defined and analytic in $\mathbb{C}\setminus [-s,s]$, where the orientation is taken from the left to the right.

2.  For $x\in(-s,s)$, we have $$\label{eq:Y-jump}
     Y_+(x)=Y_-(x)(I-2\pi i {\mathbf}{f}(x){\mathbf}{h}^t(x)),$$ where the functions ${\mathbf}{f}$ and ${\mathbf}{h}$ are defined in .

3.  As $z \to \infty$, $$\label{eq:Y-infty}
     Y(z)=I+\frac{\mathsf{Y}_1}{z}+\mathcal O(z^{-2}).$$

4.  As $z \to \pm s$, we have $Y(z) = \mathcal O(\ln(z \mp s))$.

