Schwarz and Poisson formulas

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 . 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 .

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 SchwarzAndPoissonFormulas 2013-03-22 16:05:58 2013-03-22 16:05:58 perucho (2192) perucho (2192) 12 perucho (2192) Theorem msc 30D10 Schwarz formula Poisson formula