Complex Analysis/Harmonic function

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Revision as of 23:03, 8 January 2025 by imported>Eshaa2024 (New resource with "==Definition== Let <math>U \subseteq \mathbb{C}</math> be an open set. A function <math>u \colon U \to \mathbb{R}</math> is called ''harmonic'' if it is twice differentiable and satisfies <center><math> \Delta u := \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 </math></center> have. The real part of a holomorphic function is harmonic, as follows from the Cauchy-Riemann-Differential equation. Interestingly, the conv...")
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Definition

Let U be an open set. A function u:U is called harmonic if it is twice differentiable and satisfies

Δu:=2ux2+2uy2=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.

Connection to Holomorphic Functions

Let U be simply connected. For uC2(U,), the following are equivalent:

  1. Δu=0
  2. There exists vC2(U,) such that u+iv is holomorphic.

Proof

(2). (1).By the Cauchy-Riemann-Differential equation, we have:

2ux2+2uy2=x(ux)+y(uy)=x(vy)+y(vx)=2vxy2vyx=0.

since partial derivatives commute. 1. 2.Define the function g:=uxiuy. By the Cauchy-Riemann-Differential equation, g is holomorphic.sinceU is simply connected, there exists a primitive f:U from g, assume (by Adding a constant) ,that u(z0)=f(z0) for a z0U applies. write f=u1+iv1. it is

u1xiu1y=f=g=uxiuy

so is u1u constant. because u1(z0)=f(z0)=u(z0) ist u1=u und v:=v1 does what is desired.

Translation and Version Control

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:

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

  • Date: 01/08/2024


de:Kurs:Funktionentheorie/Harmonische Funktion