After trying some stuff out I got some interesting results. I would like to share with you.
(1) - Here is the main equation again:
DFQ := D[f[t, x],
t] == -((\[Alpha] (t/\[Beta])^\[Alpha])/t)*f[t, x] - \[Mu]*D[f[t, x], x] - \[Sigma]^2/2*D[f[t, x], x, x]
(2) - Take the Fourier Transform of each term:
In[63]:= FourierTransform[-((\[Alpha] (t/\[Beta])^(-1+\[Alpha]))/\[Beta])*f[x,t],x,\[Xi],FourierParameters->{0,-2 Pi}]
Out[63]= FourierTransform[-((\[Alpha] (t/\[Beta])^(-1+\[Alpha]) f[x,t])/\[Beta]),x,\[Xi],FourierParameters->{0,-2 \[Pi]}]
In[64]:= FourierTransform[-\[Mu]*D[f[t,x],x],x,\[Xi],FourierParameters->{0,-2 Pi}]
Out[64]= -2 I \[Pi] \[Mu] \[Xi] FourierTransform[f[t,x],x,\[Xi]]
In[65]:= FourierTransform[-(\[Sigma]^2/2)*D[f[t,x],x,x],x,\[Xi],FourierParameters->{0,-2 Pi}]
Out[65]= 2 \[Pi]^2 \[Xi]^2 \[Sigma]^2 FourierTransform[f[t,x],x,\[Xi]]
(3) - Solve as a regular ODE:
In[66]:= DSolve[y'[t]+(-((\[Alpha] ((t/\[Beta])^(-1+\[Alpha])) )/\[Beta])-2 I \[Pi] \[Mu] \[Xi]+2 \[Pi]^2 \[Xi]^2 \[Sigma]^2)*y[t]==0,y[t],t]
Out[66]= {{y[t]->E^((t/\[Beta])^\[Alpha]-2 \[Pi] t \[Xi] (-I \[Mu]+\[Pi] \[Xi] \[Sigma]^2)) C[1]}}
(4) - I do not know how to apply the IC correctly so assume C=1, and take the Inverse Fourier Transform:
In[67]:= InverseFourierTransform[E^((t/\[Beta])^\[Alpha]-2 \[Pi] t \[Xi] (-I \[Mu]+\[Pi] \[Xi] \[Sigma]^2)),\[Xi],x,FourierParameters->{0,-2 Pi}]
Out[67]= E^((t/\[Beta])^\[Alpha]-(x+t \[Mu])^2/(2 t \[Sigma]^2))/(Sqrt[2 \[Pi]] Sqrt[t \[Sigma]^2])
- Which looks like it has the correct form.
(5) - I plug in values to check the LHS is equal to the RHS, in case this is a solution:
sl[x_,t_]:=E^((t/\[Beta])^\[Alpha]-(x+t \[Mu])^2/(2 t \[Sigma]^2))/(Sqrt[2 \[Pi]] Sqrt[t \[Sigma]^2])
In[55]:= (D[sl[x,t],t])/.{\[Alpha]->1,\[Beta]->5,\[Mu]->0.2,\[Sigma]->0.4,x->2,t->4}
Out[55]= 0.00177528
In[56]:= (-((\[Alpha] (t/\[Beta])^\[Alpha])/t)*sl[x,t]-\[Mu]*D[sl[x,t],x]-\[Sigma]^2/2*D[sl[x,t],x,x])/.{\[Alpha]->1,\[Beta]->5,\[Mu]->0.2,\[Sigma]->0.4,x->2,t->4}
Out[56]= -0.00177528
So somehow I am off by a negative sign (maybe). My suspicion is that it has to do with the initial condition but if C is a constant then it would flip the sign on both sides again.
When I plot the function sl[x,t] it has a very different shape than the one when I use NDSolve.
I have some issues with NDSolve but I will address that later but I would like to know how to extract some values from NDSolve so I can compare with the Analytical solution for a fixed set of values.
PS: Is there anything I can do to improve my post above? Is it possible to plot Manipulate graphs in here?
