- #1
smallphi
- 441
- 2
I am trying to solve numerically the following PDE:
dF(x,t) / dt = some function of x and F(x,t) ONLY
where 0<x<5.
This equation does NOT need boundary conditions at x=0 and x=5 because each point in x evolves independently from the others (the equation doesn't contain spatial derivatives) and it is enough to know the initial condition F(x, t=0) to obtain the full solution F(x,t).
Unfortunately pdepe (the MATLAB numeric pde solver) asks for boundary conditions. I tried 0=0 but got an error. In principle I can evolve the points x=0 and x=5 and get the boundary conditions. I am trying to avoid that since for the actual problem I am trying to solve F is a column vector of a function, f(x,t) and its first and second spatial derivatives. Obtaining those at x=0 is tough since the function goes to zero so I will have to take a complicated limit which also depends on the behavior of f(x,t) around x=0 and f(x,t) is still unknown.
So is there a way to make pdepe forget about the boundary conditions?
Mathematica does it automatically.
dF(x,t) / dt = some function of x and F(x,t) ONLY
where 0<x<5.
This equation does NOT need boundary conditions at x=0 and x=5 because each point in x evolves independently from the others (the equation doesn't contain spatial derivatives) and it is enough to know the initial condition F(x, t=0) to obtain the full solution F(x,t).
Unfortunately pdepe (the MATLAB numeric pde solver) asks for boundary conditions. I tried 0=0 but got an error. In principle I can evolve the points x=0 and x=5 and get the boundary conditions. I am trying to avoid that since for the actual problem I am trying to solve F is a column vector of a function, f(x,t) and its first and second spatial derivatives. Obtaining those at x=0 is tough since the function goes to zero so I will have to take a complicated limit which also depends on the behavior of f(x,t) around x=0 and f(x,t) is still unknown.
So is there a way to make pdepe forget about the boundary conditions?
Mathematica does it automatically.