# Kolmogorov Backward Equation Boundary Value Problem

I need to solve the backward equation $u_t - \gamma x u_x + \frac{1}{2} b^2 u_{xx}$ subject to the final condition $u(x,T) = (x-a)^2$. Here a and b and $\gamma$ are constants. I am given a strong hint to use the ansatz (assumed form of the solution) $u(x,t) = B(t)(x - A(t) )^2 + C(t)$, which seems to only make sense.

Upon differentiation and substituting the ansatz into the PDE, I get $$B'(t) (x - A(t) )^2 + C'(t) + B(t)\left[2(x-A(t))A'(t) - 2\gamma x (x-A(t)) + b^2 \right] = 0$$ subject to the final conditions $A(T) = a, B(T) = 1$, and $C(T) = 0$. But I'm unable to make sense of this, as I want to take B(t) = 1 for all t and C(t) = 0 for all t, but then I'm left with the last term in square brackets which doesn't seem nice.

Any insight would be greatly appreciated as I seem to be pretty bad at solving random PDEs. Thanks in advance.

-
You should remove the part depending on $x$ and $x^2$ by a proper choice of $A$ and $B$. Then you are done. – Jon Dec 11 '11 at 18:48
@doraemonpaul, maybe that's enough editing for today. – Gerry Myerson Aug 27 '12 at 3:18

Just do this. Expand your computation starting from

$$B'(t) (x - A(t) )^2 + C'(t) + B(t)\left[2(x-A(t))A'(t) - 2\gamma x (x-A(t)) + b^2 \right] = 0.$$

You will get

$$B'(t)x^2+B'(t)A^2(t)-2xB'(t)A(t)+C'(t)+2B(t)A'(t)x-2B(t)A(t)A'(t)$$ $$-2\gamma B(t)x^2-2\gamma xB(t)A(t)+b^2B(t)=0.$$

Then, just collect $x$ and $x^2$ terms and set them to zero to get

$$B'(t)-2\gamma B(t)=0$$

$$B(t)A'(t)-B'(t)A(t)-2\gamma A(t)B(t)=0$$

that give you $A$ and $B$. Then, you are left with the equation for $C$

$$C'(t)+B'(t)A^2(t)-2B(t)A(t)A'(t)+b^2B(t)=0.$$