Verify Harmonic function and Laplace equation.

1. Aug 5, 2010

yungman

Harmonic function satisfies Laplace equation and have continuous 1st and 2nd partial derivatives. Laplace equation is $\nabla^2 u=0$.

Using Green's 1st identity:

$$\int_{\Omega} v \nabla^2 u \;+\; \nabla u \;\cdot \; \nabla v \; dx\;dy \;=\; \int_{\Gamma} v\frac{\partial u}{\partial n} \; ds$$

$$v=1 \;\Rightarrow\; \int_{\Omega} \nabla^2 u \; dx\;dy \;=\; \int_{\Gamma} \frac{\partial u}{\partial n} \; ds = 0 \;\hbox { if } \;u \;\hbox{ is a harmonic function .}$$

$$\frac{\partial u}{\partial n} \;=\; \nabla u \cdot \widehat{n} \;\hbox{ where} \; \widehat{n} \;\hbox { is the outward normal of the boundary of the closed region } \Omega$$

$$\frac{\partial u}{\partial n} \;=\; \nabla u \cdot \widehat{n} =0 \;\Rightarrow\; \nabla u \;\hbox{ is tangent to the outward normal which means it is tengent of the boundary } \Gamma\;$$.

This mean for a harmonic function $u,\;\; \nabla u$ on the boundary $\Gamma$ is tangent to the boundary. And in vector calculus term, it is total circulation.

This mean at the points of the boundary $\nabla u = \nabla \;X\; A \;\hbox { where A is some scalar function. }$

Just want to verify with you guys.

Thanks

Alan

2. Aug 6, 2010

3. Aug 7, 2010

yungman

Anyone?

4. Aug 10, 2010

yungman

I think I know what I did wrong.

$$\int_{\Omega} \nabla^2 u \; dx\;dy \;=\; \int_{\Gamma} \frac{\partial u}{\partial n} \; ds = 0 \;\hbox { if } \;u \;\hbox{ is a harmonic function .}$$

$$\int_{\Gamma} \frac{\partial u}{\partial n} \; ds = 0 \hbox { do not mean } \frac{\partial u}{\partial n} = 0$$.

So my logic is wrong to start out.

Thanks