Schwarz and Poisson formulas

Introduction

Fundamental boundary-value problems of potential theory, i.e. (http://planetmath.org/Ie), the so-called Dirichlet and Neumann problems occur in many of applied mathematics such as hydrodynamics, elasticity and electrodynamics. While solving the two-dimensional problem for special of boundaries is likely to present serious computational difficulties, it is possible to write down formulas for a circular (http://planetmath.org/Circle) boundary. We shall give Schwarz and Poisson formulas that solve the Dirichlet problem for a circular domain.

Schwarz formula

Without loss of generality, we shall consider the compact disc $\overline{D}:|z|\leq 1$ in the $z-$ plane, its boundary will be denoted by $\gamma$ and any point on this one by $\zeta=e^{i\theta}$ . Let it be required to determine a harmonic function $u(x,y)$ , which on the boundary $\gamma$ assumes the values

$\displaystyle u\big{|}_{\gamma}=f(\theta),$

(1)

where $f(\theta)$ is a continuous single-valued function of $\theta$ . Let $v(x,y)$ be the conjugate harmonic function which is determined to within an arbitrary constant from the knowledge of the function $u$ . ¹¹Since $u+iv$ is an analytic function of $z=x+iy$ ,it is clear from the Cauchy-Riemann equations that the function $v(x,y)$ is determined by $\displaystyle v(x,y)=\int_{z_{0}}^{z}\frac{\partial v}{\partial x}dx+\frac{% \partial v}{\partial y}dy=\int_{z_{0}}^{z}-\frac{\partial u}{\partial y}dx+% \frac{\partial u}{\partial x}dy\>,$ where the integral is evaluated over an arbitrary path joining some point $z_{0}$ with an arbitrary point $z$ belonging to the unitary open disc $D$ . We are concerned to a simply connected domain, so that the function $v(x,y)$ will be single-valued.Then the function

$\displaystyle w(z)=u(x,y)+iv(x,y)$

is an analytic function for all values of $z\in D$ . We shall suppose that $w(z)\in C(\overline{D})$ the class of continuous functions. Therefore, we may write the boundary condition (1) as

$\displaystyle w(\zeta)+\overline{w}(\overline{\zeta})=2f(\theta)\quad on\,\,\gamma.$

(2)

We define here $\overline{w}(\zeta)=\overline{w(\overline{\zeta})}$ and $\overline{w}(\overline{\zeta})=\overline{w(\zeta)}$ . Next, we multiply (2) by $\frac{1}{2\pi i}\frac{d\zeta}{\zeta-z}$ and, by integrating over $\gamma$ , we obtain

$\displaystyle\frac{1}{2\pi i}\int_{\gamma}\frac{w(\zeta)}{\zeta-z}d\zeta+\frac% {1}{2\pi i}\int_{\gamma}\frac{\overline{w}(\overline{\zeta})}{\zeta-z}d\zeta=% \frac{1}{\pi i}\int_{\gamma}\frac{f(\theta)}{\zeta-z}d\zeta\>,$

(3)

which, by Harnack’s theorem, is to (2). Notice that the first integral on the left is equal to $w(z)$ by Cauchy’s integral formula, and for the same reason ²²From Taylor’s formula $\displaystyle w(z)=w(0)+w^{\prime}(0)z+\frac{1}{2!}w^{\prime\prime}(0)z^{2}+O(% z^{3}).$ But on $\gamma$ , $\overline{z}=1/\zeta$ , so $\displaystyle\overline{w}(\overline{\zeta})=\overline{w}(0)+\overline{w}^{% \prime}(0)\frac{1}{\zeta}+\frac{1}{2!}\overline{w}^{\prime\prime}(0)\frac{1}{% \zeta^{2}}+O\bigg{(}\frac{1}{\zeta^{3}}\bigg{)}$ and term-by-term integration gives the desired result recalling that $\displaystyle\frac{1}{2\pi i}\int_{\gamma}\frac{d\zeta}{\zeta^{n}(\zeta-z)}=% \left\{\begin{array}[]{ll}1,&if\;\;n=0,\\ 0,&otherwise.\end{array}\right.$ the second one is equal to $\overline{w}(0)$ . Let $\overline{w}(0)=a-ib$ , thus (3) becomes

$\displaystyle w(z)=\frac{1}{\pi i}\int_{\gamma}\frac{f(\theta)}{\zeta-z}d\zeta% -a+ib.$

(4)

By setting $z=0$ in (4), we get

$\displaystyle a+ib=\frac{1}{\pi i}\int_{\gamma}\frac{f(\theta)}{\zeta}d\zeta-a% +ib,$

whence

$\displaystyle 2a=\frac{1}{\pi i}\int_{\gamma}\frac{f(\theta)}{\zeta}d\zeta=% \frac{1}{\pi i}\int_{0}^{2\pi}f(\theta)d\theta.$

(5)

As one would expect, $b$ is left undetermined because the conjugate harmonic function $v(x,y)$ is determined to within an arbitrary real constant. Finally we substitute $a$ from (5) in (4),

$\displaystyle w(z)=\frac{1}{\pi i}\int_{\gamma}\frac{f(\theta)}{\zeta-z}d\zeta% -\frac{1}{2\pi i}\int_{\gamma}\frac{f(\theta)}{\zeta}d\zeta+ib=\frac{1}{2\pi i% }\int_{\gamma}f(\theta)\frac{\zeta+z}{\zeta-z}\frac{d\zeta}{\zeta}+ib,$

(6)

the aimed Schwarz formula.³³It is possible to prove that, if $f(\theta)$ satisfies Hölder condition, then the function $w(z)$ given by (6) will be continuous in $\overline{D}$ . Such a condition is less restrictive than the requirement of the existence of a bounded derivative.

Poisson formula

If we substitute $z=\rho\>e^{i\phi}$ and $\zeta=e^{i\theta}$ in (6) and separate the real and imaginary parts, we find

$\displaystyle\Re{w(z)}\equiv u(\rho,\phi)=\frac{1}{2\pi}\int_{0}^{2\pi}\!\!% \frac{(1-\rho^{2})f(\theta)}{1-2\rho\cos{(\theta-\phi)}+\rho^{2}}\;d\theta\>.$

(7)

This is the Poisson formula (so-called also Poisson integral), which gives the solution of Dirichlet problem. It is possible to prove that (7) also the solution under the assumption that $f(\theta)$ is a piecewise continuous function.⁴⁴See [1]. It is also possible to generalize the formulas obtained above so as to make them apply to any simply connected region. This is done by introducing a mapping function and the idea of conformal mapping of simply connected domains.⁵⁵For a discussion of Neumann problem, see [2].

References

1 O. D. Kellog, Foundations of Potential Theory, Dover, 1954.
2 G. C. Evans, The Logarithmic Potential, Chap. IV, New York, 1927.

Title	Schwarz and Poisson formulas
Canonical name	SchwarzAndPoissonFormulas
Date of creation	2013-03-22 16:05:58
Last modified on	2013-03-22 16:05:58
Owner	perucho (2192)
Last modified by	perucho (2192)
Numerical id	12
Author	perucho (2192)
Entry type	Theorem
Classification	msc 30D10
Defines	Schwarz formula
Defines	Poisson formula