Understanding Green's Identity: Solving Laplace Equation for Harmonic Functions

[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 .}

Why is it equal zero if u is harmonic function? Why is:

\int_{\Omega} \nabla^2 u \; dx\;dy =0 \hbox { if } \nabla^2 u =0

Or more basic question:

What is \int_{\Gamma} 0 dxdy? Is it not zero?

 

[jackmell]

The integral of zero is zero. I don't see a problem here.

 

[yungman]

The integral of zero is zero. I don't see a problem here.

Thanks

So you mean:

\int_{\Omega} \nabla^2 u \; dx\;dy =0 \hbox { if } \nabla^2 u =0?

 

[jackmell]

Thanks

So you mean:

\int_{\Omega} \nabla^2 u \; dx\;dy =0 \hbox { if } \nabla^2 u =0?

Youngman, I don't want to get over my head with this or nothing, but let me try to explain since I jumped in: if the function u is harmonic, then the Laplacian is zero, not a limit which approaches zero but identically zero and surely the integral of zero is just zero.

 

[yungman]

Youngman, I don't want to get over my head with this or nothing, but let me try to explain since I jumped in: if the function u is harmonic, then the Laplacian is zero, not a limit which approaches zero but identically zero and surely the integral of zero is just zero.

I was just thinking if you differentiate x twice you get zero, if integration is the reverse of differentiation, should integrating zero be something! I dug up books and I did not see anything about this, that's why I posted.

Thanks for the help.