Mr. Lichtblau, thank you for your guidance. I am tracking 100% with your reply above. The poster above (Henrik) shed some light on the FT conversion of the unknown function - even though there is still an issue as I specified in my reply to him.

Nevertheless, I was able to convert my hand derivation with Mathematica and it scaled properly so I am hopeful we are almost there. I am sorry again for not being clear since the beginning.

My ultimate goal, as I have expressed earlier in the thread, is to be able to solve the equation using Mathematica.

• My first example (pictures of notes) is an example of something that Mathematica can solve.Hence I showed the math and the code how to obtain the same result with Mathematica.

• My second example (picture of notes 2 ) is an example that Mathematica cannot solve using the same technique, so its not simply a rehash (Weiner Process vs Birth&Death Diffusion)

If I do not show the mathematics behind it - how can I write a meaningful code?

Nonetheless, If I am posting in the wrong group - I apologize. Please direct me in the right one, or move/ delete the thread.

Dear Edvin,

posting pictures of basic Fourier identities is not conducive to good questions. This forum is first of all about Wolfram Technologies. Trying to improve your Wolfram Language code and asking questions about your Wolfram Language code is the proper approach.

Also your new notes should really contain more than just a rehash of earlier ones.

Sincerely, Moderation Team

Dear Edvin,

I did not follow this thread very closely, but I would like to make a remark concerning Fourier transform:

There is no specific "Mathematica definition" of Fourier transform, but Mathematica follows - as standard setting - just the standard notation of classical physics. If one does not like this (e.g. myself) it can be changed - as it is well described in the documentation ("Details and Options").

I am referring now to the emphasized formulas above "Definitions & Properties Table": According to the definition given in the first line you need to change to the notation of "signal processing" by setting the option

FourierParameters -> {0, -2 \[Pi]}

Then amazing things can be done, e.g.:

In[1]:= FourierTransform[f''[x], x, k, FourierParameters -> {0, -2 Pi}]

Out[1]= -4 k^2 \[Pi]^2 FourierTransform[f[x], x, k]

In[2]:= InverseFourierTransform[-I  g'[k], k, x, 
 FourierParameters -> {0, -2 \[Pi]}]

Out[2]= -2 \[Pi] x InverseFourierTransform[g[k], k, x]

In[3]:= FourierTransform[DiracDelta[x - x0], x, k, 
 FourierParameters -> {0, -2 Pi}]

Out[3]= E^(-2 I k \[Pi] x0)

From this one can see that the third formula is not consistent with the definition (first line). Furthermore the second formula does not make sense anyway ...

Cheers Henrik

So, I am back at working on this and I suspect I am on the right track this time.

The question I raised earlier in the thread was why do we need the delta function as the initial condition? - One answer that made sense to me is the following: - When we solve an ordinary diff eq we end up with a constant for which we apply the initial condition and solve for.

• In the Kolmogorov equation above, p(t,x) is a function of x and t, but we are taking a partial in respect to time. Therefore:
  • If we do the Fourier Transform of each term - and reduce the PDE to an ODE - since p(t,x) is a function of two variable we need a function instead of a constant as an initial condition (Recall the a = b(x) trick if I remember correctly from my diff eq class).

Thank you very much for the reply. I followed your advise and purchased the book from Amazon. I will update the thread if I figure out something more.