What is the Fourier Transform for the following:
$$ \mathscr{F} \left\{\frac{\partial^2 (x^2p(x,t))}{\partial x^2} \right\} = ? $$
Here is my problem:
Suppose we are given the diffusion $$ dX(t)= \mu dt + \sigma dW(t) \hspace{10mm} (1) \\ X(0)=x_0 $$
And a function $$ k(x)= \theta x^2 \hspace{10mm} (2) $$
Then by Ito Lemma $$ dk[X(t)]=(2 \mu \theta x + \theta \sigma ^2)dt + (2 \sigma \theta x)dW(t) \hspace{10mm} (3) \\ k[X(0)]=0 $$
And the Forward Kolmogorov to find the transition density
$$ \frac{\partial p(x,t)}{\partial t}=- \frac{\partial ((2 \mu \theta x+ \theta \sigma^2 )p(x,t))}{\partial x} + \frac{1}{2} \frac{\partial^2 ((2 \sigma \theta x)^2p(x,t))}{\partial x^2} \hspace{10mm} (4) \\ p(x,0)= \delta (x-x_0) $$
So far so good
Now, I need to convert the 2nd order pde to ordinary so I can solve it. This is why I need the Fourier Transform.
List of properties:
$ (a) \ \mathscr{F} \left\{ p(x,t) \right\}= \overline{p}(\xi,t)= \intop\nolimits_{-\infty}^{\infty} p(x,t) e^{-i2 \pi x \xi} dx $
$ (b) \ \mathscr{F}^{-1} \{ \overline{p} (\xi,t) \}= p(x,t)= \intop\nolimits_{-\infty}^{\infty} \overline{p} (\xi,t) e^{i2 \pi x \xi} d\xi $
$ (c) \ \mathscr{F} \{ xp(x,t) \}= \frac{i}{-2 \pi} \frac{\partial \overline{p} (\xi,t)}{\partial \xi} $
$ (d) \ \mathscr{F} \{ \frac{\partial^n p(x,t)}{\partial x^n} \}=(i2 \pi \xi)^n \overline{p} (\xi,t) $
$ (e) \ \mathscr{F} \{ \delta (x-x_0) \}= \intop\nolimits_{-\infty}^{\infty} \delta (x-x_0) e^{-i2 \pi x \xi}dx = e^{-i2 \pi x_0 \xi } $
So now I can start to re-write (4) term by term:
$$ \ \mathscr{F} \{ \frac{\partial p(x,t)}{\partial t} \}= \frac{\partial \overline{p} (\xi,t)}{\partial t} \hspace{10mm} (4-1) $$
By $(f+g)'=f'+g'$
$$ \frac{\partial ((2 \mu \theta x + \theta \sigma^2) p(x,t) )}{\partial x}= 2 \mu \theta \frac{\partial (xp(x,t))}{\partial x} + \theta \sigma^2 \frac{\partial p(x,t)}{\partial x} $$
$$ \ \mathscr{F} \left\{ 2 \mu \theta \frac{\partial (xp(x,t))}{\partial x} \right\} = (2 \mu \theta) (i2 \pi \xi) \frac{i}{-2 \pi } \frac{\partial p(\xi,t)}{\partial \xi} \hspace{10mm} (4-2) $$
$$ \ \mathscr{F} \left\{ \theta \sigma^2 \frac{\partial p(x,t)}{\partial x} \right\} = (\theta \sigma^2) (i2 \pi \xi) \overline{p} (\xi,t) \hspace{10mm} (4-3) $$
And I get stuck on the 2nd term of (4)
$$ \ \mathscr{F} \left\{\frac{\partial^2 ((x)^2p(x,t))}{\partial x^2} \right\} =? \hspace{10mm} (4-4) $$
Any help/hint if I can re-write (4-4) in terms of the properties would be tremendous. Thank you!
Additional stuff - 2nd derivative for $xp(x,t)$
$$ \ \mathscr{F} \left\{ \frac{\partial^2 (xp(x,t))}{\partial x^2} \right\} = (i2 \pi \xi)^2 \frac{i}{-2 \pi } \frac{\partial p(\xi,t)}{\partial \xi} $$