Verify Harmonic function and Laplace equation.

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Discussion Overview

The discussion revolves around the verification of properties related to harmonic functions and the Laplace equation, specifically focusing on the implications of Green's first identity and the behavior of the gradient of harmonic functions at the boundary of a region.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states that harmonic functions satisfy the Laplace equation and have continuous first and second partial derivatives, presenting the equation as ∇²u = 0.
  • The same participant uses Green's first identity to argue that if u is a harmonic function, then the integral of the Laplacian over a region is zero, implying that the normal derivative on the boundary is also zero.
  • Another participant questions the initial logic, noting that while the integral of the normal derivative over the boundary being zero does not imply that the normal derivative itself is zero at every point on the boundary.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there is a disagreement regarding the interpretation of the implications of the integral of the normal derivative being zero.

Contextual Notes

The discussion highlights potential misunderstandings about the relationship between integrals of derivatives and pointwise values, indicating a need for careful consideration of the conditions under which these statements hold.

yungman
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Harmonic function satisfies Laplace equation and have continuous 1st and 2nd partial derivatives. Laplace equation is [itex]\nabla^2 u=0[/itex].

Using Green's 1st identity:

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

[tex]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 .}[/tex]

[tex]\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[/tex]

[tex]\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\;[/tex].

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

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

Just want to verify with you guys.

Thanks

Alan
 
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Anyone please?
 
Anyone?
 
I think I know what I did wrong.

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

[tex]\int_{\Gamma} \frac{\partial u}{\partial n} \; ds = 0 \hbox { do not mean } \frac{\partial u}{\partial n} = 0[/tex].

So my logic is wrong to start out.

Thanks
 

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