3
$\begingroup$

Let $U$ be a connected and bounded Domain, w.l.o.g. we choose $[0,1]^2$ and let $f \in \mathcal{C}^2((0,1)^2)$ with $\Delta f(x)=0$ for $x \in (0,1)^2$ and having normal derivative of $0$ almost everywhere on the boundary with respect to the surface measure. Furthermore define $X_t$ to be a process which behaves like the standard Brownian Motion in $(0,1)^2$ and has normal reflection on the boundary (the classical reflected brownian motion)

$\endgroup$

2 Answers 2

2
$\begingroup$

I might eb misunderstanding the problem, but it looks like you have "Neumann boundary conditions" for the Laplacian. What if you extend the function to all of the plane by reflecting in each face of the square and repeating (so $f$ extends to a periodic function with period 2 in both variables)? And replace the Brownian motion by the usual Brownian motion that sees no walls.

$\endgroup$
2
  • $\begingroup$ @ Sigurd Angenent I have given your argument some thought and think that it won't work in my caase. Eventually I want to prove the existence of probabilistic representation of $f$ for a stochastic process which is reflected on one part of the boundary and absorbed at the other - therefore extention in the sense as you are suggesting it probably won't work. $\endgroup$
    – Boldwing
    Jun 17, 2012 at 6:32
  • $\begingroup$ but it is nice idea and I really appreciate the advice - thanks :) $\endgroup$
    – Boldwing
    Jun 18, 2012 at 7:56
0
$\begingroup$

Hi I think that I might have found the solution:

I could extend the derivatives to continuous fuctions on $[0,1]^2$ by defining them via the one sided derivativs along the normal vector.

Now the classical proof of Itô's Lemma (one shows that the result holds true for any polynomial function by using integration by parts formula and kunita-watanabe) should still work, for kunita Watanabe works for continuous functions. Thus Itô's Lemma would be applicable and the problem would be solved.

Could somebody be so kind to give me feedback on this idea (please) :)

Hmm... I might get problems with the 4 corner points - those evil things :(

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.