Complex Analysis/Harmonic function

Let ${\displaystyle U\subseteq \mathbb {C} }$ be an open set. A function ${\displaystyle u\colon U\to \mathbb {R} }$ is called harmonic if it is twice differentiable and satisfies

${\displaystyle \Delta u:={\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}=0}$

have. The real part of a holomorphic function is harmonic, as follows from the Cauchy-Riemann-Differential equation. Interestingly, the converse also holds: every harmonic function is the real part of a holomorphic function.

Let ${\displaystyle U\subseteq \mathbb {C} }$ be simply connected. For ${\displaystyle u\in C^{2}(U,\mathbb {R} )}$, the following are equivalent:

1. ${\displaystyle \Delta u=0}$
2. There exists ${\displaystyle v\in C^{2}(U,\mathbb {R} )}$ such that ${\displaystyle u+iv}$ is holomorphic.

(2). ${\displaystyle \implies }$ (1).By the Cauchy-Riemann-Differential equation, we have:

${\displaystyle {\begin{array}{rl}{\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}&={\frac {\partial }{\partial x}}\left({\frac {\partial u}{\partial x}}\right)+{\frac {\partial }{\partial y}}\left({\frac {\partial u}{\partial y}}\right)\\&={\frac {\partial }{\partial x}}\left({\frac {\partial v}{\partial y}}\right)+{\frac {\partial }{\partial y}}\left(-{\frac {\partial v}{\partial x}}\right)\\&={\frac {\partial ^{2}v}{\partial x\partial y}}-{\frac {\partial ^{2}v}{\partial y\partial x}}\\&=0.\end{array}}}$

since partial derivatives commute. 1. ${\displaystyle \implies }$ 2.Define the function ${\displaystyle \textstyle g:={\frac {\partial u}{\partial x}}-i{\frac {\partial u}{\partial y}}}$. By the Cauchy-Riemann-Differential equation, ${\displaystyle g}$ is holomorphic.since${\displaystyle U}$ is simply connected, there exists a primitive ${\displaystyle f\colon U\to \mathbb {C} }$ from ${\displaystyle g}$, assume (by Adding a constant) ,that ${\displaystyle u(z_{0})=f(z_{0})}$ for a ${\displaystyle z_{0}\in U}$ applies. write ${\displaystyle f=u_{1}+iv_{1}}$. it is

${\displaystyle {\frac {\partial u_{1}}{\partial x}}-i{\frac {\partial u_{1}}{\partial y}}=f'=g={\frac {\partial u}{\partial x}}-i{\frac {\partial u}{\partial y}}}$

so is ${\displaystyle u_{1}-u}$ constant. because ${\displaystyle u_{1}(z_{0})=f(z_{0})=u(z_{0})}$ ist ${\displaystyle u_{1}=u}$ und ${\displaystyle v:=v_{1}}$ does what is desired.

This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:

• Source: Harmonische Funktion - URL:

https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Harmonische_Funktion

• Date: 01/08/2024