Poisson kernel: Difference between revisions

Revision as of 21:16, 3 May 2008

Definition

The Poisson kernel is the kernel for the Poisson integral operator, the integral operator that solves the Dirichlet problem for a disk. This kernel is defined by the following expression for the open unit disk centered at the origin:

$z_{0}=re^{i\theta }\mapsto P_{r}(\theta )={\frac {1}{2\pi }}{\frac {1-r^{2}}{1-2r\cos \theta +r^{2}}}$

If, instead of the unit disk, we're working with a disk of radius $R$ , the Poisson kernel for that is given by:

${\frac {1}{2\pi }}{\frac {R^{2}-r^{2}}{R^{2}-2Rr\cos \theta +r^{2}}}$

As the real part of a holomorphic function

We can also define the Poisson kernel as the real part of the following holomorphic function on the open unit disk:

${\frac {1}{2\pi }}{\frac {1+z}{1-z}}$

here $z=re^{i\theta }$

This also tells us that the harmonic conjugate of the Poisson kernel is given by the function:

$Q_{r}(\theta )={\frac {1}{2\pi }}{\frac {2r\sin \theta }{1-2r\cos \theta +r^{2}}}$

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 The '''Poisson kernel''' is the kernel for the [[Poisson integral operator]], the integral operator that ''solves'' the Dirichlet problem for a disk. This kernel is defined by the following expression for the [[open unit disk]] centered at the origin:
-<math>P_r(\theta) = \frac{1}{2\pi} \frac{1 - r^2}{1 - 2r \cos \theta + r^2}</math>
+<math>z_0 = re^{i\theta} \mapsto P_r(\theta) = \frac{1}{2\pi} \frac{1 - r^2}{1 - 2r \cos \theta + r^2}</math>
 If, instead of the unit disk, we're working with a disk of radius <math>R</math>, the Poisson kernel for that is given by:
 <math>\frac{1}{2\pi} \frac{R^2 - r^2}{R^2 - 2Rr \cos \theta + r^2}</math>
+===As the real part of a holomorphic function===
+We can also define the Poisson kernel as the real part of the following holomorphic function on the open unit disk:
+<math>\frac{1}{2\pi} \frac{1 + z}{1 - z}</math>
+here <math>z = re^{i\theta}</math>
+This also tells us that the harmonic conjugate of the Poisson kernel is given by the function:
+<math>Q_r(\theta) = \frac{1}{2\pi} \frac{2r \sin \theta}{1 - 2r \cos \theta + r^2}</math>