0 & 1 & 1
\end{pmatrix}
\widetilde \Psi^{-1}(y;\rho)\widetilde \Psi(x;\rho)
\begin{pmatrix}
1
\\
0
\\
0
\end{pmatrix}, \qquad x,y \in \mathbb{R}.$$

Differential equations for $\Psi$
---------------------------------

For later use, we need the following linear differential equations for $\Psi$ with respect to $z$ and $\rho$.

\[prop:ODE\] Let $\Psi=\Psi(z;\rho)$ be the unique solution of the RH problem \[rhp: Pearcey\]. We then have $$\label{eq: ODE1}
\frac{\partial \Psi}{\partial z}=
\begin{pmatrix}
0 & 1 & 0
\\
0 & 0 & 1
\\
z & \rho & 0
\end{pmatrix}\Psi,$$ and $$\label{eq:ODE2}
\frac{\partial \Psi}{\partial \rho}=\frac12
\begin{pmatrix}
0 & 0 & 1
\\
z & \rho & 0
\\
1 & z & \rho
\end{pmatrix}\Psi.$$

The differential equation follows directly from , and . To show , we obtain from , and direct calculations that, for $j=1,\ldots,5$, $$\frac{\partial p_j}{\partial \rho}=-\frac12 \int_{\Gamma_j}s^2e^{-\frac14 s^4-\frac{\rho}{2}s^2+is z}{\,\mathrm{d}}s=\frac12 p_j'',$$ $$\frac{\partial p_j'}{\partial \rho}=-\frac i 2 \int_{\Gamma_j}s^3 e^{-\frac14 s^4-\frac{\rho}{2}s^2+is z}{\,\mathrm{d}}s= \frac12 p_j'''=\frac12 (zp_j+\rho p_j'),$$ and $$\frac{\partial p_j''}{\partial \rho}= \frac 1 2 \int_{\Gamma_j}s^4 e^{-\frac14 s^4-\frac{\rho}{2}s^2+is z}{\,\mathrm{d}}s
= \frac12 p_j''''
=\frac12 (zp_j+\rho p_j')'=\frac12 (p_j+zp_j'+\rho p_j''),$$ which leads to .

This completes the proof of Proposition \[prop:ODE\].

Differential identities for $F$
-------------------------------

By , it is readily seen that $$\label{eq:tildeKdef}
K^{\mathrm{Pe}}(x,y;\rho) = \frac{{\mathbf}{f}^t(x){\mathbf}{h}(y)}{x-y},$$ where $$\label{def:fh}
{\mathbf}{f}(x)=\begin{pmatrix}
f_1
\\
