Question: Transforming a SDE to a ODE

I am studying the application file "Maple in Finance"   www.maplesoft.com/applications/view.aspx

I have posted some of the stuff below. I have a hard time following along with the reasoning.

Could anyone please explain what is going one here is simple words ?!   There are a lot of new stuff here.

Fourier transformations etc. The conlussion seem to be that we transform a stochastic differential equation

into ODE which we solve. What can we do with such a solution and what is the interpretation ?!

The process can be expressed in terms of diffusion PDE

restart:
with(PDEtools):
with(inttrans):

eq := diff(u(S, t), t)+r*S*(diff(u(S, t), S))-(1/2)*sigma^2*S^2*(diff(u(S, t), S, S)) = 0;
 

To solve this PDE analytically, we will try to change it into constant-coefficient PDE by change of variables

tr := {S = exp(x)}; pde := simplify(dchange(tr, eq));
           

We want to solve this equation with initial condition u(x, 0) = u[0](x) and u[0](x) = delta[x]

ic := u(x, 0) = Dirac(x);

Since the PDE is now in "desirable format", we apply Fourier transform to solve it:

assume(sigma > 0, t > 0, r > 0); interface(showassumed = 0); fourier(pde, x, w); subs(fourier(u(x, t), x, w) = v(t), %);

We have now removed a spatial variable from the equation ⇒ transformed PDE into ODE, which we can now solve with the given initial condition:

dsolve({%, v(0) = fourier(rhs(ic), x, w)}, v(t)); subs(v(t) = fourier(u(x, t), x, w), %);
              

Since Fourier transform in invertible, we retrieve back the desired probability density function:

invfourier(lhs(%), w, x) = invfourier(rhs(%), w, x); expand(%, exp); subs(a = 'a', t = 't', b = 'b', %); u := unapply(rhs(%), x, t);
         

Finally, we want to check the correctness of our result:

Test1 := simplify(lhs(pde)-rhs(pde));