Verify Harmonic function and Laplace equation.

yungman
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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
 
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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
 
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