- lostmanthm
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Suppose I have a time homogeneous ito process X,such that the transition probability satisfies Fokker Planck/Kolmogorov forward equation (KFE):and the boundary conditions I can solve the KFE by separation of variables and then apply Sturm Liouville theorem to determine the solution. However, how can I prove that the solution to KFE is unique and should be in this separable form? Many thanks.

QuoteI can solve the KFE by separation of variables and then apply Sturm Liouville theorem to determine the solution. Can you?

- lostmanthm
**Posts:**8**Joined:**

Yes I can, but the proof is a bit long:By separation of variables I have 1. which is easy to solve, and 2. problem 2 can be rewritten as self-adjoint form:which is an eigenvalue / eigenfunction problem, and then we can apply regular Sturm Liouville theorem, the eigenvalues / eigenfunctions are Since probability density has to be positive, of all the eigenfunctions only is positive and never changes sign. So it is the only valid solution to the problem, and its eigenvalue is also the only solution.

Suggest you try all this with f_t = f_xx, x in [0,1], which should answer some of your orig. question and resolve some misconceptions.

- Cuchulainn
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QuoteOriginally posted by: isometryQuoteI can solve the KFE by separation of variables and then apply Sturm Liouville theorem to determine the solution. Can you?AFAIK SOV hits a dead end except for the the simplest PDEs.

- Cuchulainn
**Posts:**62100**Joined:****Location:**Amsterdam-
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QuoteOriginally posted by: isometryQuoteI can solve the KFE by separation of variables and then apply Sturm Liouville theorem to determine the solution. Can you?AFAIK SOV hits a dead end except for the the simplest PDEs. edit: even if it does work, it may not be numerically feasible in general. Even heat flow PDE in an infinite rod begins to look daunting.

Last edited by Cuchulainn on April 2nd, 2013, 10:00 pm, edited 1 time in total.

Intuitively, I think it more likely that with practical problems and the lack of t-dependence in the Ito coefficients, the solution would look more like f(x,t,y,T) = g(y-x,T-t). But I'm well-known to be an amateur with Fokker-Planck....

Last edited by Fermion on April 8th, 2013, 10:00 pm, edited 1 time in total.

a) What you call boundary conditions aren't really. You probably mean f is zero at plus and minus infinity. b) What's your initial condition?c) Dump the x and t, it just makes everything look messy.d) You really want to use transforms. (That's like having a continuum of lambdas.)P

Am I wrong or Linetsky followed this path ? (though not for the Fokker Planck equation)http://www.worldscientific.com/doi/abs/ ... 4904002451

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