I've managed to reduce a question in probability to the following simple looking PDE: $ u_t = -t u_x + \frac{1}{2} u_{xx}, {\rm ~for~} x>0, \, t \in \mathbb{R} \;, $ with a limiting initial condition: $ u(x,t) \to 1, {\rm ~as~} t \to -\infty \;, $ a boundary condition at $x=0$: $ u_x(0,t) = \left\{ \begin{array}{lc} -2 a t u(0,t) \;, & t \le 0 \\ 0 \;, & t \ge 0 \end{array} \right. \;, $ where $a > 0$ is a constant, and a suitable boundary condition as $x \to \infty$: either $u \to 1$ or $u$ is bounded.
All I want is an exact expression for $p = \lim_{t \to \infty} u(0,t)$, which represents the particular probability that I am after, but I don't know how to do this or if it's possible. I'm hoping there's some sort of method out there that I don't know about to derive $p$ without having to solve the whole boundary value problem. Any ideas?
A few comments:
1) I've solved it numerically (with finite differences) and $u(x,t)$ looks nice and smooth and only takes values between 0 and 1.
2) I don't expect the piecewise nature of the boundary condition at $x=0$ to be a major issue. Presumably we can solve up to $t=0$, pause, and then solve for $t>0$. For $t \le 0$, the boundary condition is a time-dependent Robin boundary condition, which I've never seen dealt with before.
3) To me the major problem seems to be the explicit time-dependency in the PDE and the boundary condition at $x=0$. I can get rid of the time-dependency in the PDE by defining $ v(x,t) = {\rm e}^{\frac{1}{6} t^3 - t x} u(x,t) \;, $ which gives $ v_t = -x v + \frac{1}{2} v_{xx} \;, $ $ v_x(0,t) = \left\{ \begin{array}{lc} -(1+2a)tv(0,t) \;, & t \le 0 \\ -tv(0,t) \;, & t \ge 0 \end{array} \right. \;. $
4) To the boundary value problem in $v$, I can't get separation of variables, or Laplace transforms in time, or some sort of half Fourier transform in space, to work.
5) Incidentally, $\lim_{t \to \infty} u(x,t)$, should be independent of $x$, that is, pointwise, $u(x,t)$ approaches the constant value $p$.