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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 $p$.

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Perhaps you can take the laplace transform and then use the Final value theorem to find $f(\infty)$ –  Navin Oct 29 '12 at 1:54
    
That's true, but I don't see how to usefully take the Laplace transform to begin with. –  djws Oct 29 '12 at 8:42

1 Answer 1

Hint:

Let $\begin{cases}v=x-\dfrac{t^2}{2}\\w=t\end{cases}$ ,

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial v}\dfrac{\partial v}{\partial x}+\dfrac{\partial u}{\partial w}\dfrac{\partial w}{\partial x}=\dfrac{\partial u}{\partial v}$

$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial v}\right)=\dfrac{\partial}{\partial v}\left(\dfrac{\partial u}{\partial v}\right)\dfrac{\partial v}{\partial x}+\dfrac{\partial}{\partial w}\left(\dfrac{\partial u}{\partial v}\right)\dfrac{\partial w}{\partial x}=\dfrac{\partial^2u}{\partial v^2}$

$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial v}\dfrac{\partial v}{\partial t}+\dfrac{\partial u}{\partial w}\dfrac{\partial w}{\partial t}=-t\dfrac{\partial u}{\partial v}+\dfrac{\partial u}{\partial w}$

$\therefore-t\dfrac{\partial u}{\partial v}+\dfrac{\partial u}{\partial w}=-t\dfrac{\partial u}{\partial v}+\dfrac{1}{2}\dfrac{\partial^2u}{\partial v^2}$

$\dfrac{\partial u}{\partial w}=\dfrac{1}{2}\dfrac{\partial^2u}{\partial v^2}$

Let $u(v,w)=V(v)W(w)$ ,

Then $V(v)W'(w)=\dfrac{1}{2}V''(v)W(w)$

$\dfrac{2W'(w)}{W(w)}=\dfrac{V''(v)}{V(v)}=s^2$

$\begin{cases}\dfrac{W'(w)}{W(w)}=\dfrac{s^2}{2}\\V''(v)-s^2V(v)=0\end{cases}$

$\begin{cases}W(w)=c_3(s)e^{\frac{ws^2}{2}}\\V(v)=\begin{cases}c_1(s)e^{vs}+c_2(s)e^{-vs}&\text{when}~s\neq0\\c_1v+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(v,w)=\int_{-\infty}^\infty C_1(s)e^{\frac{ws^2}{2}+vs}~ds+\int_{-\infty}^\infty C_2(s)e^{\frac{ws^2}{2}-vs}~ds$

$u(x,t)=\int_{-\infty}^\infty C_1(s)e^{\frac{ts^2}{2}+\left(x-\frac{t^2}{2}\right)s}~ds+\int_{-\infty}^\infty C_2(s)e^{\frac{ts^2}{2}-\left(x-\frac{t^2}{2}\right)s}~ds$

$u(x,t)=\int_{-\infty}^\infty C_1(s)e^{\frac{t}{2}\left(s^2+\left(\frac{2x}{t}-t\right)s\right)}~ds+\int_{-\infty}^\infty C_2(s)e^{\frac{t}{2}\left(s^2-\left(\frac{2x}{t}-t\right)s\right)}~ds$

$u(x,t)=\int_{-\infty}^\infty C_1(s)e^{\frac{t}{2}\left(s^2+\left(\frac{2x}{t}-t\right)s+\left(\frac{x}{t}-\frac{t}{2}\right)^2-\left(\frac{x}{t}-\frac{t}{2}\right)^2\right)}~ds+\int_{-\infty}^\infty C_2(s)e^{\frac{t}{2}\left(s^2-\left(\frac{2x}{t}-t\right)s+\left(\frac{x}{t}-\frac{t}{2}\right)^2-\left(\frac{x}{t}-\frac{t}{2}\right)^2\right)}~ds$

$u(x,t)=e^{-\frac{t}{2}\left(\frac{x}{t}-\frac{t}{2}\right)^2}\int_{-\infty}^\infty C_1(s)e^{\frac{t}{2}\left(s+\frac{x}{t}-\frac{t}{2}\right)^2}~ds+e^{-\frac{t}{2}\left(\frac{x}{t}-\frac{t}{2}\right)^2}\int_{-\infty}^\infty C_2(s)e^{\frac{t}{2}\left(s-\frac{x}{t}+\frac{t}{2}\right)^2}~ds$

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Yes, you've removed the drift term from the PDE, but now the location of the boundary, $x=0$, is $v=-t^2/2$, so unfortunately I don't see how this helps. –  djws Dec 17 '12 at 20:39

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