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I would like to have an Itô Diffusion $$ X_t = \int_0^t b(s) \mathrm{d}s + \int_0^t \sigma(s) \mathrm{d}B_s.$$ where the (vector- and matrix-valued, respectively) functions $b$ and $\sigma$ have lower regularity than the usual requirements that they be Lipschitz continuous.

My question is: Can such a process be well-defined?

Let me briefly explain how this came up: For the Brownian motion on a Riemannian manifold, locally the functions $U$ and $\sigma$ should locally be given by $$\sigma_{ij} = g^{ij}, \quad b^k = \sum_{ij}g^{ij}\Gamma_{ij}^k,$$ $g^{ij}$ being the coefficients of the (inverse) metric and $\Gamma_{ij}^k$ the Christoffel symbols of the Levi-Civita-connection, which contain derivatives of $g^{ij}$.

Now, for example, I would like consider Brownian motion on the suspension of a circle (http://upload.wikimedia.org/wikipedia/commons/c/c3/Suspension.svg) embedded in $\mathbb{R}^3$.

This is not a Riemannian manifold in the usual sense, as the metric is not smooth: In case of the suspension, if you give it the smooth structure obtained by projecting to the $2$-sphere, but take the metric induced from $\mathbb{R}^3$, this metric will not be smooth: at the fold in the middle, the coefficient funcions will be only Lipschitz, so the Christoffel symbols will not be continuous there (even though bounded). Near the cusps at the top and bottom, the metric is not even bounded.

What to do here?

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You can obtain a Brownian motion on certain metric spaces going via Dirichlet forms: projecteuclid.org/euclid.kjm/1250281992 –  Bati Jul 1 '13 at 13:40
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2 Answers

up vote 2 down vote accepted

You can have weak solutions with quite low regularity. In the compact case, continuity of drift and diffusion is sufficient for existence of a weak solution, and in the non-compact case you will need to impose some growth conditions on the coefficient. This is a theorem of Skorokhod(1965), and for its proofs see a recent paper and references to several standard books therein. Strong solutions may fail to exist.

In 1D I remember that it is sufficient that the coefficients are bounded and measurable, and I think that this is a theorem of Veretennikov, but I failed to find a reference in a reasonably short time.

Depending on which examples you want to consider, you may go along a variety of ways. First, the suspension of a circle seems to be only a topological space homeomorphic to a sphere. Is there a canonical metric on a suspension space? The picture that you borrowed from Wikipedia may be misleading. Anyway, suppose you have a space and a metric on it. You can

(1) Mollify it in some way and take the limit of the resulting diffusions. You will need to prove convergence and that the limiting process is a truly Markov process with some infinitesimal properties.

(2) Define the process in different ways on smooth part of the surface and on singularities. You can define a Markov process in several ways. One is to use infinitesimal generators. With a little experience you can write down the generator or a pre-generator right away, but then you will have to prove that it does define a Markov semi-group.

(3) Use some trick like introducing an extra variable if your process allows for some symmetries. For example, the circle suspension is a finite length cylinder $S^1\times [-1,1]$ with two boundary circles contracted into points. You can define the second component of the candidate process to be a Browinian motion on $[-1,1]$ with reflections at the endpoints, and the first component to be a Brownian motion on the circle with random jumps at each time when the second component is reflected. You will have to explain though what it means because the second component will be reflected uncountably many times, but it is still possible.

It's hard to tell what is the shortest path to understand all these notions. One useful book is Ethier&Kurtz on Markov processes.

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Thanks for your answer. Do you think that your suggestion 1) will work? –  Kofi Jul 6 '13 at 20:55
    
@Kofi: Well, it is still not clear to me what a typical example of a nonsmooth coefficient you want to consider, but yes, it should work in most reasonable situations. Weak solutions can be recast as solutions of appropriate martingale problems and probably that is the way to go, see the theory in Ethier&Kurtz. Also, smoothening is one step in the proof of Skorokhod's theorem. –  Yuri Bakhtin Jul 6 '13 at 23:00
    
For a start, I would like to have the Coefficients $\sigma$ to be Lipschitz and the coefficients $b$ to be derivatives of Lipschitz functions, as in my example above. –  Kofi Jul 7 '13 at 11:10
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A weak solution (is it good enough for you?) can be obtained by a Girsanov transformation that gets rid of the $b$, so the main issue is the diffusion coefficient, and for that the martingale problem initiated by Stroock and Varadhan in this context is the way to go. Their 1979 book is an excellent starting point (and there are the original articles of course).

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What do you mean with "weak solution"? And can you elaborate how Girsanov transformation can help here? –  Kofi Jul 2 '13 at 11:04
    
1Basically, strong solutions are pathwise, ie given a Brownian motion you can construct the solution as a measurable map of it. In weak solutions you construct the bm together with the solution, and the solution may use additional randomization. The above mentioned book of stroock- varadhan, as well as karatzas-shreve's book, are good places to read about that and about girsanov's theorem. –  ofer zeitouni Jul 2 '13 at 11:20
    
Just an extra comment on girsanov transformation: once you solve the equation with the given sigma and b=0, you can get a weak solution of your original equation by changing the measure using girsanov ( and checking the appropriate integrability condition). This will give a weak, not strong, solution. –  ofer zeitouni Jul 2 '13 at 11:31
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