Consider $$\Psi_{\gamma,\kappa,\rho}(t)=e^{{\gamma}it}e^{(\kappa A+\rho A^*)}$$
for real parameters $\gamma,\kappa,\rho>0$, real variable $t\in(0,1)$ and $A=\frac{1}{\log(t)},$ $A^*=\frac{1}{\log(1-t)}.$
How do I build the differential equation that $\Psi$ solves?
Consider $h_{\kappa}(t)=e^{\kappa A}$ and $j_{\rho}(t)=e^{\rho A^*}$ and $g_{\gamma}(t)=e^{\gamma it}.$
Then we have a solution of the form: $\Psi(t)=h(t)j(t)g(t).$
We know that $h_{\kappa}(t)$ satisfies the linear parabolic PDE:
$$\kappa \frac{\partial ^2\Psi(\kappa,t)}{\partial {\kappa}^2}=-t \frac{\partial \Psi(\kappa,t)}{\partial t} $$
and $j$ likewise satisfies a linear parabolic PDE:
Now the question is: Does multiplying the solutions together also solve a PDE?
Here's an image of $\Psi$ for $\gamma=7,~ \kappa=0.3,~ \rho=0.9.$