February 7th, 2020, 3:45 pm

Let me switch to (mostly) book notation for a moment. My main goal in Appendix 1.2 is to make the standard PDE-SDE probability connection for parabolic PDE's on the real line (so no bc). So, for each PDE treated, one finds the (formal) probabilistic solution: run an SDE and take an expectation. Effectively, this would be a Monte Carlo solution.

For example, to formally solve [$]u_t = \frac{1}{2} a^2(x) u_{xx} + b(x) u_x[$] with [$]u(x,0) = f(x)[$], you run the SDE

[$] dX_t = b(X_t) \, dt + a(X_t) \, dB_t [$], with [$]X_0 = x[$].

Then the probabilistic solution is [$]u(x,t) = E[f(X_t)][$], where [$]E[\cdots][$] is a time-0 expectation, referencing the time-t value of all possible [$]\{X_t\}[$] paths that start at [$]x[$].

As a by-product, you also get the solution to the hyperbolic problem [$]u_t = b(x) u_x[$] with [$]u(x,0) = f(x)[$] -- by taking [$]a = 0[$]. Doing so, the SDE reduces to the ODE problem for the characteristic, namely:

(*) [$]dX_t = b(X_t) \, dt[$], with [$]X_0 = x[$].

Then, the probabilistic solution reduces to the deterministic solution [$]u(x,t) = f(X_t)[$], by simply dropping the expectation. There's no longer any randomness in the SDE. For each starting value [$]x[$], there's a unique path solution. The ODE (*) is easily solved: [$]X_t = \xi(x,t)[$], where, as I've posted several times

(**) [$]\int_x^{\xi} \frac{dy}{b(y)} = t[$].

You fix x and t, and solve (**) for [$]\xi = \xi(x,t)[$].

This is all for general problems. Finally, for the specific problem at hand, take [$]b(y) = (1 - y)^2[$], and solve (**) for [$]\xi[$], taking in this case [$]x \in (0,1)[$] and so [$]\xi \in (0,1)[$]. This is a slight 'cheat' because I started out saying the problems were on the whole real axis and now restrict the domain. But, I think it is legitimate for this case. Go ahead and do the integral in (**), which is a very easy one, and you'll get

[$]\xi(x,t) = \frac{x + t (1-x)}{1 + t (1-x)}[$].

So, that's how I got the answer I posted.