Edit: I've reformulated the problem in a way that makes it easier to express the boundary and initial conditions. This involved expressing it in terms of a different function $g(x,t)$ (which was related to the previous formulation as $g_x=f$)
I'm trying to find a general solution $g(x,t)$ on $0<x<1$, $t>0$ to the following PDE:
$$\rho(g_{xt}+g_{xxt})=(1-\rho)g_{xx}+g_{xxx}$$
where $\rho>0$. Boundary conditions are:
$$\rho(g_t(0,t)+g(0,t))=g_x(0,t)$$
$$g(1,t)=1$$
And the initial condition is $g(x,0)=1$. We also require that
$$\lim_{\rho\to 0}g(x,t)=1$$
If I set the time derivatives to zero I can obtain a steady state solution:
$$g(x)= \frac{\rho e^{x (\rho-1) }-1}{\rho e^{\rho-1 }-1}$$
I am looking for a general solution in $(x,t)$ that converges to the steady-state solution as $t\to\infty$.
Any ideas? I suspect a change of variables is going to make this simpler, but I haven't worked out a good way to do that. Note that by setting $h=g_{x}+g_{xx}$ the PDE can be re-written as:
$$h-g_{x}=\frac{1}{\rho}h_x-h_t$$
which suggests that a change in variables like $u=x+\rho t$ might be a good approach?