So this property of Brownian Motion is so elegant, in my opinion, that when I was recently asked what my ‘favourite theorem’ was, I suggested this. With this result, we can use this probabilistic structure to specify solutions to an important PDE, with boundary conditions, over a large class of domains.
Given a domain D, Laplace’s equation is: on D, and
on the boundary dD, where f is any continuous function defined there. This PDE arises wherever the notion of potentials is defined, for example electromagnetism, fluids and thermodynamics.
Theorem: Given suitable regularity conditions on D to be discussed later, Laplace’s equation has a unique solution, given by:
Notation: First, what does this mean? Define , to be the time at which a Brownian Motion leaves the domain D. This is a stopping time, and so will be suitable for application of the Strong Markov Property.
means that we are taking expectation with respect to a BM started at x. So informally, we are defining u(x) as: start a BM at x; see where it hits the boundary of D; record the value of f at that point. Then set u(x) to be the expected value of this process.
Existence: First, we are going to check that the solution conjectured is a solution. We will need a lemma:
Lemma: A locally-bounded function u satisfies on a domain D if and only if it has the property that for every closed ball
we have:
where is the surface area measure on the boundary S(x,r) of the ball radius r centred on x. Essentially, this says that u(x) is equal to the average value of u on a ball around x.
Proof of Theorem: First, existence. Set u as specified in the statement of the theorem. Given a Brownian Motion started at x, we have stopping times corresponding to the hitting times of the ball radius r around x and the boundary dD. The domination condition holds by continuity provided B(x,r) is contained within D. So we may apply the Strong Markov Property:
By definition, the left hand expression is u(x). But also, because the distribution of is uniform on S(x,r), the right hand side is equal to:
and so by the lemma, this guarantees that the function u is harmonic on the interior of D.
The lemma can also be used to show uniqueness. Continue reading