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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 $xp(x,t)$

$$ \ \mathscr{F} \left\{ \frac{\partial^2 (xp(x,t))}{\partial x^2} \right\} = (i2 \pi \xi)^2 \frac{i}{-2 \pi } \frac{\partial p(\xi,t)}{\partial \xi} $$

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    $\begingroup$ The Fourier transform you seem to use is wrt $x$ so $\mathscr{F}[ dp/dt] = d\hat{p}/dt$ whereas you have the derivative wrt $\xi$ in 4.1? As for 4.4 then using $x^2 e^{-2\pi i\xi x} = \frac{1}{(-2\pi i)^2} \frac{d^2}{d\xi^2} e^{-2\pi i\xi x}$ and then pulling the derivatives outside of the integral should formally give you something like $\mathscr{F}[ (x^2 p)_{xx}] = \xi^2 \frac{d^2}{d\xi^2}\mathscr{F}[p]$. $\endgroup$ – Winther May 21 '17 at 18:15
  • $\begingroup$ @Winther I fixed (4 - 1) - thank you. Can you please show how did you get there for (4 - 4) so i can mimic your answer next time. How come we are still left with a 2nd derivative. I thought the point of Fourier was to reduce the order. Did you leave out any other constants? $\endgroup$ – Edv Beq May 21 '17 at 22:37
  • $\begingroup$ The Fourier transforms replaces $x$-derivatives with multiplication by powers of $\xi$. Likewise the FT of powers of $x$ times a functions becomes derivatives of the FT of that function wrt $\xi$. It's $x^2p$ which leads to second derivatives. I did the following: $$\mathscr{F}[\frac{d^m}{dx^m}[x^np(x)]] = (2\pi i \xi)^m\int x^np(x)e^{-2\pi i \xi x}\,{\rm d}x = (2\pi i \xi)^m\int p(x)\frac{1}{(-2\pi i)^n}\frac{d^n}{d\xi^n}e^{-2\pi i \xi x}\,{\rm d}x\\= (-1)^n(2\pi i)^{m-n}\xi^m \frac{d^n}{d\xi^n}\mathscr{F}[p(x)]$$ For $n=m$ this gives $(-1)^n\xi^n \frac{\partial^n \hat{p}}{\partial \xi^n}$ $\endgroup$ – Winther May 21 '17 at 23:22
  • $\begingroup$ @Winther This is great! Why don't you write it as an answer so i can give you credit? $\endgroup$ – Edv Beq May 21 '17 at 23:27

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