Finite Difference Method for the Black–Scholes Equation Without Boundary Conditions

Umm - without reading it closely, it looks oddly like early 70's binomial methods, which were also mercifully free of the need of any spatial boundaries.

Finite Difference Method for the Black–Scholes Equation Without Boundary Conditions

"Das ist nicht nur nicht richtig; es ist nicht einmal falsch!"

So they use a Dirichlet BC at x = 0 and they do not use any BC at the far field boundary.
and they wrote "Instead, we reduce grid points by one in every time step" ???????

Here a screenshot of that part of the paper:

So they use a Dirichlet BC at x = 0 and they do not use any BC at the far field boundary.
and they wrote "Instead, we reduce grid points by one in every time step" ???????

Here a screenshot of that part of the paper:

Scheme (3) is explicit Euler on non-uniform meshes. This went out of fashion after Richardson's famous article (1910?).
The authors claim 2nd order convergence in the last sentence of section 3.. And it is slow, very slow in general.
I'll leave further comments to FaridMoussaoui

// This is a parabolic PDE  and not a hyperbolic PDE so I think this approach is doomed.

For a double barrier option the method cannot work. You need to use _all_ of both boundaries.

Way back in 2014 we 3 wrote an article on Asian PDE(S,A) on transformed domain (0,1)X(0,1). I do by Fichera. OK.
We all agreed (me, P and A) that no continuous BC are needed.

https://onlinelibrary.wiley.com/doi/epd ... wilm.10366

Some folk truncate S and A. So, A is now in (0, Amax). By Fichera they say no BC needed at A = 0 but at A = Amax a BC is needed and they search for some kind of BC by taking the analytical solution of the PDE.

My claim is that this is wrong on several levels... the 'truncated PDE' gets a BC even though the original PDE does not.

bit of a fallacy here.. "truncated PDE don't have BC". Otherwise it is guesswork.

I can test my claim by modifying the Anchor code.

Finite Difference Method for the Black–Scholes Equation Without Boundary Conditions

Umm - without reading it closely, it looks oddly like early 70's binomial methods, which were also mercifully free of the need of any spatial boundaries.

I can't read this article but it is possible to get rid of boundary conditions. As Bearish said, x-nomial trees do it. I am also doing it with kernel methods.