Remarkable fact about Brownian Motion #4: The Dirichlet Problem

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: \Delta u=0 on D, and u=f 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 T_D:=\inf\{t:B_t\not\in D\}, 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. \mathbb{E}_x 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 \Delta u=0 on a domain D if and only if it has the property that for every closed ball \bar{B(x,r)}\subset D we have:


where \sigma_{x,r} 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 T_r<T_D 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 B_{T_r} 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. It suffices to show uniqueness when the boundary function is zero everywhere. Then an obvious solution is the function u  which is identically zero. Suppose there is another such solution v. Then suppose v attains a maximum M at x in the interior of D. But by the local average property, this means v must equal M almost everywhere on every radius r ring around x, and by continuity, in fact everywhere. This property therefore lifts to balls around x, and so by continuity, taking a ball around x that just touches the boundary, must have M=0. A similar result obviously applies to minima, completing the proof.

Proof of Lemma: I’m going to give a proof in the direction needed for the proof of existence of the Dirichlet solution. The converse is similar but requires a bit more detail, without being much more interesting.

Assume that u is harmonic and we have a ball B(x,r) in D. First we can show that u is infinitely differentiable which is not hugely surprising, as in some sense the definition of harmonic could be interpreted as ‘u is its own regularisation’. See the Stochastic Calculus course notes for a fuller explanation. But now can expand as a Taylor series. For ease of notation, assume 0 is in the domain, and we show that \Delta u=0 at 0. Consider the ball radius r around 0, and point x on that ball. Then:

u(x)=u(0)+\nabla u\cdot x+\frac12 \sum_{i,j}\frac{\partial^2 u}{\partial x^i\partial x^j}|_0 x_ix_j+o(r^2)

Now integrate over S(x,r), and discount by the appropriate measure. Clearly \int_S u(0)dx=u(0) remains unchanged, while \int_S \nabla u|_0\cdot xdx=\nabla u|_0\cdot \int_S xdx=0. Similarly, the i\neq j integral disappears, and so the definition of u harmonic becomes:

u(0)=u(0)+\frac12 \sum \int_S\frac{\partial^2 u}{\partial x^i\partial x^i}x_i^2dx+o(r^2)=u(0)+\left(\sum \frac{\partial^2 u}{\partial x^i\partial x^i}\right)\int_S x_i^2dx

\Rightarrow 0=c\Delta u|_0 + o(r^2)

where c is some non-zero constant. And so we are done.

Technical Point: One step is missing. We need u to be continuous on \bar{D}, and this is where we need to impose some condition on D. For this we define the Poincare Cone Condition for a domain. D satisfies this condition if whenever we take a boundary point x\in\partial D, if you can insert an open cone into the complement of D locally with apex at x. An example of a domain without this condition might be a ‘peach-shape’ where at the critical point, the tangent rotates through 180 degrees. Alternatively, a a disc missing a small slit, as discussed in the construction of SLE.

For a formal proof of the continuity property from the cone condition see the Advanced Probability notes. Informally, take a fixed ball B around a boundary point, and observe that as the starting point of a Brownian Motion moves closer to the boundary point, the probability of hitting the boundary first dominates the probability of hitting the fixed ball first. In particular, can bound the boundary hitting time by the hitting time of the external cone at x, which is much more tractable. Also, the cone condition is invariant under scaling at the apex, so can exploit this property to get an exponential relation which is strong enough to give the continuity condition.


Karatzas, Shreve – Brownian Motion and Stochastic Calculus

Part III Stochastic Calculus Notes –

Part III Advanced Probability Notes –


One thought on “Remarkable fact about Brownian Motion #4: The Dirichlet Problem

  1. Pingback: Advanced Probability Revision Summary | Eventually Almost Everywhere

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