If I have the following systems of PDE \begin{align} u_t+x^2u_{xx}-\dfrac{h_1(t)}{h_0(t)}e^{-(v-u)}-\dfrac{h_0'(t)}{h_0(t)}=0\\ v_t-\dfrac{h_0(t)}{h_1(t)}e^{-(u-v)}-\dfrac{h_1'(t)}{h_1(t)}=0, \end{align} where $x\in[-L,L]$ and $t\in (0,T)$, $(h_0(t),h_1(t))$ are solutions to the system ${\bf h}'(t)={\bf M}{\bf h}(t)$. I have the explicit expressions for $h_0(t)$ and $h_1(t)$.
I want to determine the system under the transformation $\tau =T-t$. I know that $u_t=-u_\tau$ and $v_t=-v_\tau$. I am not sure how the rest of the terms should be.
If you employ the explicit representation of $\,h_0(t)\,$ and $\,h_1(t)\,$ then under the transformation $τ=T−t\,$ your system will take the form \begin{align} -\tilde{u}_\tau+x^2\tilde{u}_{xx}-\dfrac{h_1(T-\tau)}{h_0(T-\tau)} e^{-(\widetilde{v}-\widetilde{u})}- \dfrac{h_0'(T-\tau)}{h_0(T-\tau)}=0,\\ -\widetilde{v}_\tau-\dfrac{h_0(T-\tau)}{h_1(T-\tau)}e^{-(\widetilde{u}-\widetilde{v})}-\dfrac{h_1'(T-\tau)}{h_1(T-\tau)}=0 \end{align} with the new unknowns $\,\tilde{u}(s)=u(T-s)\,$ and $\,\widetilde{v}(s)=v(T-s)\,$. There's nothing else to it.