# PDE with problematic but natural boundary conditions.

I am looking for the solution of the PDE $$\frac{\partial p}{\partial t} = \frac{\partial^2}{\partial x^2} \left[ x p \right]$$ where $p(x,t)$ is a probability density. With the ansatz $p(x,t) = T(t) \cdot W(x)$ we can solve this equation with separation of variables. Setting each side equal to the constant $c_1$, the equation for the time variable is then straight forward $T(t) = A e^{c_1 t}$. It seems natural to assume that the time part is decaying rather than growing, and so we can assume that $c_1 < 0$. Let me thus replace $c_1$ by $- c_1^2$ and assume $c_1 > 0$. For the spatial variable we then have $$\frac{\partial^2}{\partial x^2} \left[ x W \right] = - c_1^2 W.$$ Solving this equation (e.g. with Mathematica) yields the solution $$W(x) = c_2 \frac{I_1\left(2^{3/2} c_1 \sqrt{x}\right)}{\sqrt{x}}+ c_3 \frac{K_1\left(2^{3/2} c_1 \sqrt{x}\right)}{\sqrt{x}}.$$ where $I_\nu$ is the modified Bessel function of first and $K_\nu$ the modified Bessel function of second kind. So far so good. My problem are now the boundary and initial conditions that I want to impose:
1. $\left. \frac{dp}{dx} \right|_{x=0} > 0$ (reflecting boundary)
2. $p(x,t=0) = \delta(x-x_0)$ (well-known initial position)
together with some additional obvious constraints such as $p > 0$ and $\lim_{x \to \infty} p(x,t) = 0$ which ensure that $p$ is a well-defined probability density. This is where I am stuck. How can I impose condition number 2 on my solution? This does not seem to work... Did I restrict myself somewhere or is this really the most general solution?
(Btw: I am actually quite convinced that there must be a solution to this equation together with these boundary conditions, since the PDE above is nothing but the Fokker-Planck equation arising from the stochastic process $dX_t = \sqrt{X_t} dW_t$. For comparison: I have already solved the PDE corresponding to $dX_t = X_t dW_t$ with the same boundary conditions successfully. )

Alternatively, it might be helpful to take a look at the integral representations of $I_\nu$ and $K_\nu$.
• I am not quite sure how this helps me. I can use the asymptotic expansion to check that my solution decays sufficiently fast, but this is currently the least of my problems (worst case, I introduce a cut-off). But how will the integral representation help me with my $\delta$-initial condition? Commented Jun 1, 2015 at 9:16