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.

This is a big improvement in the sense that it has actual code and a more clearly stated question.

Even with FT methods, I'm not sure you will get far unless you can figure out how to evaluate the transform in question. It seems to be difficult at least if tackled as an integral.

The error message from PDF[bd[\[Beta], \[Delta], x0][t], x] is due to there being no constant term inside the square root in bd[...]. If you add a smallish value (I use 1/1000 below) then that goes away.

bd[\[Beta]_, \[Delta]_, x0_] := 
 ItoProcess[\[DifferentialD]x[
     t] == ((\[Beta] - \[Delta])*x[t])*\[DifferentialD]t + 
    Sqrt[1/1000 + (\[Beta] + \[Delta])*x[t]]*\[DifferentialD]w[t], 
  x[t], {x, x0}, t, w \[Distributed] WienerProcess[]]

Alas, this still does not evaluate: PDF[bd[\[Beta], \[Delta], x0][t], x]. So I'm not sure it helps to make this change.

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).

First, with k4 also a k5 with an arbitrary power of b in front is a solution:

  In[1]:= Clear[k5, b, a, m]
    k5[\[Tau]_, \[Xi]_] := 1/Sqrt[2 \[Pi] b^m \[Tau]] Exp[- ((\[Xi] - \[Xi]0 - a \[Tau])/b)^2/(2 \[Tau])]

  In[3]:= D[k5[t, x], t] + a D[k5[t, x], x] - b^2/2 D[k5[t, x], {x, 2}] // Simplify
  Out[3]= 0

  In[4]:= D[k5[t, x - y], t] + a D[k5[t, x - y], x] - b^2/2 D[k5[t, x - y], {x, 2}] // Simplify
  Out[4]= 0

now, what is the solution k[t,x] for a general initial condition IC (see V. S. Vladimirov for specifications)?

k[t,x] = Integrate[k5[t,x-y] IC[y], dy]
D[k[t, x], t] + a D[k[t, x], x] - b^2/2 D[k[t, x], {x, 2}] = Integrate[D[k5[t, x-y], t] + a D[k5[t, x-y], x] - b^2/2 D[k5[t, x-y], {x, 2}] IC[y], dy]
                                                           = Integrate[0 IC[y], dy] = 0 for t > 0

for t = 0 there is

 Integrate[D[k5[t, x-y], t] + a D[k5[t, x-y], x] - b^2/2 D[k5[t, x-y], {x, 2}] IC[y] dy]
 = Integrate[DiracDeltat[x - y] IC[y], dy]
 = IC[x]

initial condition fulfilled.

What's your point with the DiracDelta?

The important property is

  In[11]:= Integrate[f[x - y] DiracDelta[y], {y, -Infinity, Infinity}]
  Out[11]= f[x]

  In[12]:= Integrate[DiracDelta[x - y] f[y], {y, -Infinity, Infinity}]
  Out[12]= ConditionalExpression[f[x], x \[Element] Reals]

You can not use DiracDelta like an orderly function. That's why it belongs to the generalized functions. Dirac invented it by his needs. Von Neumann said "There is no need for a delta function." Then Schwartz won a Fields Medal for making it intelligible to other people than theoretical physicists. So it can be used safely since then. The mean feature is to have some function space where the generalized functions can act on. All identities have to be written with this function space. Strictly speaking it's non-sense to say

operator[f][x] == DiracDelta[x]

that's slang and an abbreviation for the correct thing to say, which is

Integrate[operator[f][x] g[x], x] == Integrate[DiracDelta[x] g[x], x]

where the g[x] are from the above mentioned function space. It's really functional analysis. In a way you should become familiar with it in order not to run in problems people had before Schwartz worked it out.

I think I found something useful in some course notes online, although - this is a simpler example of a pure diffusion. I retyped the equations and I will try to mimic the logic in Mathematica sometime this coming weekend.

Perhaps, If I replace the initial condition f(x) with DiracDelta[x-x0] it may work.

One thing I have noticed with DiracDelta function in Mathematica is that it does not output in the distribution form. This could be a problem.

If someone is interested to give this a head start -> more than welcome. This stuff is fun.

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.