given in , and , respectively, and $$\label{X1-formula}
        \mathsf{X}_1 = \Psi_1 + \Psi_0^{-1} \mathsf{Y}_1 \Psi_0$$ with $\Psi_1$ and $\mathsf{Y}_1$ given in and .

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

We only need to show that $X$ does not have a jump over $(-s,s)$, while the other claims follow directly from and the RH problem \[rhp: Pearcey\] for $\Psi$.

By , we have, for $-s<x<s$, $$I-2\pi i {\mathbf}{f}(x){\mathbf}{h}^t(x)=\widetilde\Psi(x)
\begin{pmatrix}
1 & -1 & -1
\\
0 & 1  & 0
\\
0 & 0 & 1
\end{pmatrix}\widetilde\Psi(x)^{-1}.$$ This, together with and , implies that for $x\in(-s,s)$, $$\begin{gathered}
X_+(x)=Y_+(x)\widetilde \Psi(x)=Y_-(x)(I-2\pi i {\mathbf}{f}(x){\mathbf}{h}^t(x))\widetilde \Psi(x)
\\
=Y_-(x)\widetilde \Psi(x)\begin{pmatrix}
1 & -1 & -1
\\
0 & 1  & 0
\\
0 & 0 & 1
\end{pmatrix}=X_-(x),\end{gathered}$$ as desired.

This completes the proof of Proposition \[rhp:X\].

The connections between the above RH problem and the partial derivatives of $F(s;\rho)$ are revealed in the following proposition.

\[prop:derivativeandX\] With $F$ defined in , we have $$\begin{aligned}
    \frac{\partial}{\partial \rho} F(s;\rho) &=
\frac{{\,\mathrm{d}}}{{\,\mathrm{d}}\rho} \ln \det\left(I-K_{s,\rho}^{\mathrm{Pe}}\right)
= -\frac{1}{2} \left[ (\mathsf{X}_1)_{12} + (\mathsf{X}_1)_{23} \right] + \frac{\rho^3}{54},  \label{eq:derivativein-rho-X}\end{aligned}$$ where $\mathsf{X}_1$ is given in and $(M)_{ij}$ stands for the $(i,j)$th entry of a matrix $M$, 