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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.

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    @doraemonpaul, maybe that's enough editing for today.2012-08-27

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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.

This solves your problem.

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    Thanks Jon, that did the trick2011-12-13