I've been struggling for a while with the following system of PDEs in the context of an optimal control problem ($u$ and $v$ are value functions).
$$\begin{cases} \displaystyle \frac{\partial u}{\partial t} + \frac{f(x)}{t+1}\frac{\partial^2u}{\partial x^2} &= \displaystyle a - be^{v-u} \\ \\ \displaystyle\frac{\partial v}{\partial t} + \frac{f(x)}{t+1}\frac{\partial ^2v}{\partial x^2} &= \displaystyle a - be^{u-v} \end{cases}$$
where $u=u(x,t)$, $v=v(x,t)$ and the domain is $x\in[-1,1]$ and $t\in[0,T]$, with $T$ being a positive real number. Furthermore, $a$ and $b$ are positive real constants, and $f(x)$ is a function of $x$ such that $f(0)=1$ and $f(-1)=f(1)=0$, and it decays smoothly.
The initial conditions are $$u(x,T)=\;\,\,\, Dx$$ $$v(x,T)=-Dx$$
where $D$ is a positive real constant, and the boundary conditions are $$ \frac{\partial^2u}{\partial x^2}(1,t) = \frac{\partial^2v}{\partial x^2}(1,t) = \frac{\partial^2u}{\partial x^2}(-1,t) = \frac{\partial^2v}{\partial x^2}(-1,t) = 0$$
I've attempted to solve the equations numerically using a forward finite differences approach, but given the initial conditions, the exponential term in the right hand side (which is the one that couples both equations) becomes very problematic for $D\gg1$. This is because at $t=T$, $e^{v-u}=e^{-2Dx}$, so as $x$ approaches $-1$ that term becomes $e^{2D}$. Also, by exploring the solution to the related system of ODEs:
$$\begin{cases} \displaystyle \frac{d u}{d t} &= \displaystyle a - be^{v-u} \\ \\ \displaystyle \frac{d v}{d t} &= \displaystyle a - be^{u-v} \end{cases}$$
with the same initial conditions (considering $x$ now as a parameter), which turns out to be
$$ u,v(x,t) = a(t-T) + \log[\cosh(Dx) \pm \sinh(Dx)e^{2b(t-T)}] $$
I get the intuition that the second derivative with respect to $x$ will blow up with $D$ as well. At this point I'm pretty lost, help would be most welcome.