# Determine value of harmonic function at origin given boundary values on a regular hexagon

Let $E$ be a regular hexagon centered at the origin of $\mathbb{R}^2$. Let $f$ be the harmonic function in $E$ with boundary value 1 on one of the sides of $E$ and boundary value $0$ on each of the remaining sides. What is the value of $f$ at the origin?

This question has shown up on an old PDE qual I am studying. This problem is causing me a lot of concern, because $f$ seems to be discontinuous on the boundary of $E$. But, given that $f$ is harmonic (and thus continuous) in $E$, shouldn't the boundary values of $f$ also define a continuous function?

Hints or explanations are greatly appreciated!

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I have undone the last edit. See Old John's answer below. –  Giuseppe Negro Jan 19 '13 at 21:19

The solution is harmonic and thus analytic in the interior of $E$ and continuous up to the boundary, except at the end points of the segment where it is equal to 1.

When you rotate the segments, you'll get six functions of this form which are rotational images of each other (by uniqueness) and thus all have the same value at the origin. Adding these six functions, you get a harmonic function with boundary values equal to 1, hence it is 1 everywhere.

Therefore, $f(0) = 1/6$.

This is also the probability that a Brownian motion exits through one of the side of the hexagon assuming it starts at the origin (which could be turned into another proof).

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For example, you might consider the harmonic measure. A good book on this is by Garnett: "Harmonic Measure" (Cambridge), and on page 1 he shows that the harmonic measure of an interval on the $x$-axis gives a harmonic function on the upper half-plane, but has boundary values which are $\pi$ inside the interval and zero elsewhere.