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This question arises out of confusion about some aspects of the Poisson integral in $R^2$, simplified here to a function f on the boundary of the unit circle $ \partial C$ and on C,

$$u(r,\phi) = \frac{1}{2\pi} \int_0^{2\pi} \frac{ f (\theta) (1-r^2) d\theta}{1-2r \cos(\theta-\phi)+r^2}$$

The intuition behind the derivation of the integral is described nicely here and based on this sort of intuition I wondered whether there might be an approach that did not involve the use of "inversive" geometry.

The article cited mentions that we would typically be considering harmonic functions (temperature, e.g.) and that given an arbitrary piecewise continuous assignment of values to the boundary $\partial C$, "there always exists a harmonic function in R that takes on these values as the boundary is approached."

Well, suppose $f(\phi) = \phi^2 $? There is a discontinuity at $(r = 1, \phi = 0, 2 \pi )$. According to the article, if we are approaching $ \mathit{continuous }$ boundary points this is not an issue.

So for the question(s). If I guess--naively-- that the influence of a point on the boundary on a point inside the circle is inversely proportional to the square of the distance between the two points, I might approximate the integral as

$$ u \approx \frac{\sum_{i=1}^n \frac{f(\phi_i) }{(d_i)^2}}{\sum_{i=1}^n \frac{1}{(d_i)^2}}$$

This approximation is not great, but it yields reasonable values near the center of the circle.

  1. Is there a straightforward Riemann-sum approximation of the integral along these lines? Or are we stuck with the geometry of the article? Is the partial agreement I get for my guess fortuitous?

  2. We could assign pretty outrageous piecewise-continuous values to the boundary--do every one of these correspond to a theoretically possible (for example) heat distribution?

Thanks for any insights, edits.

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up vote 3 down vote accepted

The Poisson kernel $$P(r,\theta,\phi)=\frac{ 1-r^2 }{1-2r \cos(\theta-\phi)+r^2}$$ can be written in the more intuitive form $$P(z,\zeta)=\frac{1-|z|^2}{|z-\zeta|^2}$$ where $z=re^{i\theta}$ and $\zeta=e^{i\phi}$. This form clearly shows both the inverse square relation (in the denominator) and penalty for being close to the boundary (in the numerator). Notice that $1-|z|^2=(1+|z|)(1-|z|)$ where the first factor does not change much: $1\le 1+|z|\le 2$ for all $z$ in the disk. Thus, the penalty for proximity to the boundary is essentially linear, $(1-|z|)$.

Your sum with squared distances ignores the numerator, which of course leads to poor results near the boundary. If you want a Riemann sum approximation, why not just take the Riemann sum of the original Poisson integral itself: that is, split $[0,2\pi]$ into subintervals and pick a point in each?

Concerning (2): Anything piecewise continuous will work as you expect as long as the boundary values are integrable. For example, the boundary values $1/\theta$ for $\theta\in (0,2\pi]$ will not give you any harmonic function inside in the disk: the integral will simply diverge everywhere. But for any integrable boundary values you get a harmonic function, which can be thought of as a stationary heat distribution (under the assumption that the temperature of the boundary is somehow maintained by an outside device).

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The inverse square relation is clear. I wonder if you could characterize in a phrase the effect of the numerator on the behavior of the sum/integral? Near the boundary...? I don't think we can simply say "the contribution of f is weaker near the boundary." But maybe that is the case? My approximation gave inflated numbers near the boundary for the f I chose. – daniel Jun 29 '12 at 20:33
The contribution of $f(\zeta)$ at $z$ gets weaker if the point $z$ approaches a boundary point other than $\zeta$. This is because $z$ is strongly influenced by the nearest boundary points, which do not include $\zeta$. You can consider the ratio $(1-|z|)/|z-\zeta|$ (which is always between 0 and 1) as a measure of how relevant the point $\zeta$ is for the point $z$. Or just use some software to draw the level curves of the Poisson kernel: I believe they are circles internally tangent at $\zeta$. – user31373 Jun 29 '12 at 20:43

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