are constant matrices $$\begin{aligned}
    \label{def:Lpm}
    L_{+}=
    \begin{pmatrix}
    -\omega & \omega^2 & 1 \\ -1&1&1 \\ -\omega^2 & \omega & 1
    \end{pmatrix},
    \qquad
    L_{-}=
    \begin{pmatrix}
    \omega^2 & \omega & 1 \\ 1&1&1 \\ \omega & \omega^2 & 1
    \end{pmatrix},\end{aligned}$$ with $\omega=e^{2\pi i/3}$, and $\Theta(z)$ is given by $$\begin{aligned}
    \label{def:Theta}
    \Theta(z)=\Theta(z;\rho)&= \begin{cases}
    \operatorname{diag}(\theta_1(z;\rho),\theta_2(z;\rho),\theta_3(z;\rho)), & \text{${\mathrm{Im}\,}z >0$,} \\
    \operatorname{diag}(\theta_2(z;\rho),\theta_1(z;\rho),\theta_3(z;\rho)), & \text{${\mathrm{Im}\,}z <0$,} \\
    \end{cases}\end{aligned}$$ with $$\label{eq: theta-k-def}
    \theta_k(z;\rho)=\frac34 \omega^{2k}z^{\frac43}+\frac{\rho}{2}\omega^kz^{\frac23}, \qquad k=1,2,3.$$

-   $\Psi(z)$ is bounded near the origin.

It is shown in [@BK3 Section 8.1] that the above RH problem has a unique solution expressed in terms of solutions of the Pearcey differential equation . Indeed, note that admits the following solutions: $$\label{def:pj}
p_j(z)=p_j(z;\rho)=\int_{\Gamma_j}e^{-\frac14 s^4-\frac{\rho}{2}s^2+is z}{\,\mathrm{d}}s, \qquad j=0,1,\ldots,5,$$ where $$\begin{aligned}
\Gamma_0&=(-\infty,+\infty), \quad && \Gamma_1=(i\infty, 0]\cup[0,\infty),
\\
\Gamma_2&=(i\infty,0]\cup[0,-\infty), \quad && \Gamma_3=(-i\infty, 0]\cup[0,-\infty),
\\
\Gamma_4&=(-i\infty,0]\cup[0,\infty), \quad && \Gamma_5=(-i\infty, i\infty).\end{aligned}$$ We then have $$\label{def:soltopsi}
\Psi(z)=\left\{
        \begin{array}{ll}
          \begin{pmatrix} -p_2(z) & p_1(z) & p_5(z)\\ -p_2'(z) & p_1'(z) & p_5'(z) \\-p_2''(z) & p_1''(z) & p_5''(z) \end{pmatrix}, & \quad \hbox{$z\in \Theta_0$,} \\
          \begin{pmatrix}  p_0(z) & p_1(z) & p_4(z)\\ p_0'(z) & p_1'(z) & p_4'(z) \\ p_0''(z) & p_1''(z) & p_4''(z) \end{pmatrix}, & \quad \hbox{$z\in \Theta_1$,} \\
          \begin{pmatrix} -p_3(z) & -p_5(z) & p_4(z) \\ -p_3'(z) & -p_5'(z) & p_4'(z) \\-p_3''(z) & -p_5''(z) 