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.