I'm preparing for my stochastic calculus final and I'm having some problem solving the PDEs that come out of it. The PDE I have is:
$u_t(t,x) + \frac{1}{2}u_{xx}(t,x)=0$
With the boundary condition:
$u(T,x)=e^{-\gamma x^2}$
My professor showed us a way, but its not one I'm hyper comfortable with. I tried to do it in a way that feels natural to me, but it doesn't seem to be working out.
Following the professor's at first, I start with the proposed solution:
$u(t,x)=e^{a(t)x^2+b(t)x+c(t)}$
So, with this:
$u_t=(a'x^2+b'x+c')u$
$u_x=(2ax+b)u$
$u_{xx}=((2ax+b)^2+2a)u = (4a^2x^2+4abx+b^2+2a)u$
To satisfying the PDE, we need $a'x^2+b'x+c'+\frac{1}{2}(4a^2x^2+4abx+b^2+2a)=0$. We can do this by setting the coefficients in front of $x^2$, $x$, and $1$ to zero. Here is where I diverge from my professor's approach. I tried:
$a'+2a^2=0 \Rightarrow \frac{da}{dt}=-2a^2 \Rightarrow a^{-2}da=-2dt \Rightarrow a^{-1} = 2t \Rightarrow a=\frac{1}{2t}$
$b'+2ab=0 \Rightarrow b'+\frac{b}{t}=0 \Rightarrow \frac{db}{dt}=\frac{-b}{t} \Rightarrow -\frac{db}{b}=\frac{dt}{t} \Rightarrow -ln(b)=ln(t) \Rightarrow b=\frac{1}{t}$
Of course, we can also have $b=0$. I supposed I could see this is the correct solution since I need $b(T)=0$. So let's go with that and move on to $c$
$c'+a=0 \Rightarrow c=-\frac{1}{2t} \Rightarrow c =\frac{1}{2}ln(t)$
This seems wrong for two reasons: (1) the solution doesn't leave any room to enforce my boundary conditon; and (b) the solution the professor showed is different.
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