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# Schwarz and Poisson formulas

# Introduction

Fundamental boundary-value problems of potential theory, i.e., the so-called Dirichlet and Neumann problems occur in many branches of applied mathematics such as hydrodynamics, elasticity theory and electrodynamics. While solving the two-dimensional problem for special types of boundaries is likely to present serious computational difficulties, it is possible to write down formulas for a circular 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 $u$. ^{1}^{1}Since $u+iv$ 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

$\displaystyle v(x,y)=\int_{{z_{0}}}^{z}\frac{\partial v}{\partial x}dx+\frac{% \partial v}{\partial y}dy=\int_{{z_{0}}}^{z}-\frac{\partial u}{\partial y}dx+% \frac{\partial u}{\partial x}dy\>,$ |

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

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 equivalent 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 ^{2}^{2}From Taylor’s formula

$\displaystyle w(z)=w(0)+w^{{\prime}}(0)z+\frac{1}{2!}w^{{\prime\prime}}(0)z^{2% }+O(z^{3}).$ |

$\displaystyle\overline{w}(\overline{\zeta})=\overline{w}(0)+\overline{w}^{{% \prime}}(0)\frac{1}{\zeta}+\frac{1}{2!}\overline{w}^{{\prime\prime}}(0)\frac{1% }{\zeta^{2}}+O\bigg(\frac{1}{\zeta^{3}}\bigg)$ |

$\displaystyle\frac{1}{2\pi i}\int_{\gamma}\frac{d\zeta}{\zeta^{n}(\zeta-z)}=% \left\{\begin{array}[]{ll}1,&if\;\;n=0,\\ 0,&otherwise.\end{array}\right.$ |

$\displaystyle w(z)=\frac{1}{\pi i}\int_{\gamma}\frac{f(\theta)}{\zeta-z}d\zeta% -a+ib.$ | (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.^{3}^{3}It 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 represents the solution under the assumption that $f(\theta)$ is a piecewise continuous function.^{4}^{4}See [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.^{5}^{5}For 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.

## Mathematics Subject Classification

30D10*no label found*

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## Corrections

two formulas by CWoo ✓

reword and suppress link by Mathprof ✓

link suppress by Mathprof ✓

## Comments

## Schwarz and Poisson formulas corrections

Hello all,

I have opened my entry so that any PM people feels free to make the corrections that consider advisable. I'll specially thank CWoo(*) and Mathprof(**) to include their suggestions. Thanks in advance!

(*)>Hi, please add the two terms "Schwarz formula" and "Poisson formula" in the also defines box, in case people want to search only one or the other term.

Could you arrange that? I don't understand well your idea. (If you will do, please accept the correction when you save the changes)

(**)>Also can you add a definition for "bidimensional problem"?

I think that you are not refering to change ``bidimensional'' by ``two-dimensional''(I did it though!), but to give a ``two-dimensional problem'' definition. Unfortunately that question is not so easy to make it in my entry. Why? Because physical considerations it could be different (and in fact, they it are!) for electrodynamics, hydrodynamics or elastostatic, even in the latter ``plane strain'' and ``plane stress'' differs! (Perhaps I'll write soon the entries ``Airy's stress function'' and ``general solution of biharmonic equation'', and you will realize of which I am saying).

perucho

## Re: Schwarz and Poisson formulas corrections

I took the liberty of taking care of the correction that CWoo filed, and I also fixed some grammar and linking issues that I found. Also, the formulas provided seem to be fully justified within the article, so I checked the "contains own proof" box. We have too many things on the "unproven" list as is. :-)

Speaking of which, shouldn't we try to go through the "unproven" list and clean it out by proving some stuff? I've tried to do that as much as possible, but the number of items there keeps growing.

Warren