I also want to know how to define the unknown function p(x,t) in Mathematica so I can take the FT of its derivative w.r.t to x and t? - Then solve it as a regular ODE with Dsolve including the initial condition.

• All the examples in Wolfram Documentation are FTs of know functions.

Mr. Krause, I was wondering if you were able to figure out anything from the book above. I will follow up the thread with the "Characteristic Method," but I have not had much time lately. Thank you for all your help though.

Dear Udo,

Thank you so much for your reply again! You are awesome.

So to my understanding, at the last part - you have shown that the equation in my first post is indeed the solution to the PDE with the delta initial condition.

However, the solution (joint density in (t,x)) is unknown and what we are solving for. To demonstrate my point let me purposely do it wrong below. I am going to slightly change the notation to match the steps in my second post.

First lets define deltafunction as in k0 :

f[t_, x_] := 1/Sqrt[2 \[Pi] t] Exp[-(x - x0)^2/(2 *t)]

Take the Fourier transform and suppress the output because it pretty long (perhaps an indication that is very wrong) :

fhat[t_, s_] := FourierTransform[f[t, x], t, s]

Then lets try to use DSolve to get the solution in transform space: Note: I am trying to use the same steps as above.

DSolve[{D[uhat[s], {x, 1}] == (a^2) D[uhat[s], {x, 2}], 
   uhat[0] == fhat[t, s]}, uhat, s] // Simplify

And, as one may guess, we do not get anything that makes sense - and most importantly if I input a zero in fhat[0,s] I get the infinite expression which emphasizes the problem on how to define the DiracDelta. Perhaps, your reply already has the answer but I am unable to see it. I am sorry.

I'm afraid there is no shorter learning path than the lecture of Vladimirov's book.

I'm afraid I'm working in greyed software producing chain gangs too long already ...

The initial condition u[0,x]=DiracDelta[x-x0] means that in the limit t->0 the solution becomes a delta function. The delta function can itself be defined as a limit. Why not take an appropriate limit definition and fit it into the solution?

Start with k0

Clear[k0, b]
k0[\[Tau]_, \[Xi]_] := 1/Sqrt[2 \[Pi] \[Tau]] Exp[-(\[Xi] - \[Xi]0)^2/(2 \[Tau])]

In[28]:= D[k0[t, x], t] - b^2/2 D[k0[t, x], {x, 2}] // Simplify
Out[28]= ((-1 + b^2) E^(-((x - \[Xi]0)^2/(2 t))) (t - (x - \[Xi]0)^2))/(2 Sqrt[2 \[Pi]] t^(5/2))

there is a problem with the b, check the coefficient needed

Clear[k1, b, n]
k1[\[Tau]_, \[Xi]_] := 1/Sqrt[2 \[Pi] \[Tau]] Exp[- (\[Xi] - \[Xi]0)^2 b^n/(2 \[Tau])]

In[19]:= D[k1[t, x], t] - b^2/2 D[k1[t, x], {x, 2}] // Simplify
Out[19]= -(((-1 + b^(2 + n)) E^(-((b^n (x - \[Xi]0)^2)/(2 t))) (-t + b^n (x - \[Xi]0)^2))/(2 Sqrt[2 \[Pi]] t^(5/2)))

this gives 2 + n == 0. Now take the convection term a != 0 into consideration

In[31]:= Clear[k2, b, a]
k2[\[Tau]_, \[Xi]_] := 1/Sqrt[2 \[Pi] \[Tau]] Exp[- ((\[Xi] - \[Xi]0)/b)^2/(2 \[Tau])]

In[33]:= D[k2[t, x], t] + a D[k2[t, x], x] - b^2/2 D[k2[t, x], {x, 2}] // Simplify
Out[33]= (a E^(-((x - \[Xi]0)^2/(2 b^2 t))) (-x + \[Xi]0))/(b^2 Sqrt[2 \[Pi]] t^(3/2)) 

that did not work, of course. How to modify the evolving guess k3? For a = 0 the function k2 should result. Any term a f[?] would do. Should f depend on both - time and space - variables or on time variable alone? There is already a dependency on space variable in the term - just do something simple:

Clear[k3, b, a, f]
k3[\[Tau]_, \[Xi]_] := 1/Sqrt[2 \[Pi] \[Tau]] Exp[- ((\[Xi] - \[Xi]0 + a f[\[Tau]])/ b)^2/(2 \[Tau])]

In[77]:= D[k3[t, x], t] + a D[k3[t, x], x] - b^2/2 D[k3[t, x], {x, 2}] // Simplify
Out[77]= -((a E^(-((x - \[Xi]0 + a f[t])^2/(2 b^2 t))) (x - \[Xi]0 + a f[t]) (1 + Derivative[1][f][t]))/(b^2 Sqrt[2 \[Pi]] t^(3/2)))

okay, we are done if one could solve 1 + f'[t] == 0 and that's possible :-)

In[81]:=DSolve[D[f[t], t] == -1 , f , t]
Out[81]= {{f -> Function[{t}, -t + C[1]]}}

with

In[82]:= Clear[k4, b, a]
k4[\[Tau]_, \[Xi]_] := 1/Sqrt[2 \[Pi] \[Tau]] Exp[- ((\[Xi] - \[Xi]0 - a \[Tau])/b)^2/(2 \[Tau])]

In[84]:= D[k4[t, x], t] + a D[k4[t, x], x] - b^2/2 D[k4[t, x], {x, 2}] // Simplify
Out[84]= 0

the fundamental solution k4 has flippered out of the initial condition.

P.S. In eqn. (10) convolution integrates over y, not x.

This Forward Kolmogorov Equation is in physics known as heat conduction equation. The DiracDelta as initial condition lets one find a so-called fundamental solution. It's called fundamental solution because one gets a solution for other initial conditions - roughly speaking - by convolution of the fundamental solution with the current initial condition (with other words, you must not solve over and over again for different initial conditions). All this is described e.g. in Generalized Functions in Mathematical Physics by V. S. Vladimirov.

I'm afraid there is no shorter learning path than the lecture of Vladimirov's book.