Need to verify some proof involving Green's function.

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Discussion Overview

The discussion revolves around verifying a proof involving Green's function, specifically the symmetry property G(\vec{x},\vec{x_0}) = G(\vec{x_0},\vec{x}) within the context of mathematical physics. The focus is on the mathematical formulation and steps involved in the proof, including the use of limits and integrals over specific surfaces.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant outlines a proof involving Green's function, detailing the setup with regions and surfaces, and the need to verify a specific limit involving integrals over these surfaces.
  • The proof includes breaking down the limit into parts and evaluating integrals, with claims about the harmonic nature of functions involved leading to certain results.
  • Another participant suggests that a simpler proof of the symmetry exists in a specific textbook, indicating an alternative approach to the problem.
  • There is mention of difficulty in finding appropriate forums for discussion on this advanced topic, with a request for advice on where to post.

Areas of Agreement / Disagreement

Participants express differing views on the complexity of the proof and the appropriate resources for understanding Green's function. There is no consensus on the correctness of the proof presented, nor on the best forum for discussion.

Contextual Notes

Participants note the challenges of discussing advanced topics in physics and mathematics, highlighting the potential for varying levels of understanding and expertise among readers.

Who May Find This Useful

Readers interested in mathematical physics, specifically those studying Green's functions, boundary value problems, or advanced electromagnetism may find this discussion relevant.

yungman
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This is not homework. This is actually a subset of proofing [itex]G(\vec{x},\vec{x_0}) = G(\vec{x_0},\vec{x})[/itex] where G is the Green's function. I don't want to present the whole thing, just the part I have question.

Let D be an open solid region with surface S. Let [itex]P \;=\; G(\vec{x},\vec{a}) \;\hbox{ and } P \;=\; G(\vec{x},\vec{b}) \;[/itex] where both are green function at point a and b resp. inside D. This means Q is defined at point a ( harmonic at point a ) and P is defined at point b. Both P and Q are defined in D except at a and b resp. Both equal to zero on surface S.

Green function defined:

[tex]G(\vec{x},\vec{x_0}) \;=\; v + H \;\hbox { where } \;v=\; \frac{-1}{4\pi|\vec{x}-\vec{x_0|}} \;\hbox{ and }\; H \;\hbox { is a harmonic function in D and on S where }\; G(\vec{x},\vec{x_0}) \;=\; 0 \;\hbox { on D}.[/tex]

In this proof, I need to make two spherical cutout each with radius =[itex]\epsilon[/itex] with center at a and b. I call the spherical region of this two sphere A and B resp and the surface [itex]S_a \;&\; S_b[/itex] resp. Then I let [itex]D_{\epsilon} = D -A-B[/itex] so both P and Q are defined and harmonic in [itex]D_{\epsilon}[/itex].



Now come to the step I need to verify:

I want to prove:

[tex]^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} P\frac{\partial Q}{\partial n} \;-\; Q\frac{\partial P}{\partial n} \;dS \;=\; ^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} v\frac{1}{4\pi\epsilon^2} \;dS[/tex]



This is my work:

[tex]^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} P\frac{\partial Q}{\partial n} \;-\; Q\frac{\partial P}{\partial n} \;dS \;=\; ^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} (-\frac{1}{4\pi r} + H)\frac{\partial Q}{\partial n} \;-\; Q\frac{\partial }{\partial n}(-\frac{1}{4\pi r} + H) \;dS[/tex] (1)

Where:

[tex]^{lim}_{\epsilon\rightarrow 0}\; v\; =\; \frac{-1}{4\pi |\vec{x}-\vec{a}|} \;=\; ^{lim}_{\epsilon\rightarrow 0} \;\frac{-1}{4\pi r} \;[/tex]. in sphere region A.

[tex]^{lim}_{\epsilon\rightarrow 0}( P=v+H )\;=\; ^{lim}_{\epsilon\rightarrow 0} (\frac{-1}{4\pi r } + H)[/tex]


Form (1) I break into 3 parts:

[tex]^{lim}_{\epsilon\rightarrow 0} [ \int\int_{S_a} -\frac{1}{4\pi r}\frac{\partial Q}{\partial n} dS + \int\int_{S_a} (H\frac{\partial Q }{\partial n} \;-\; Q\frac{\partial H}{\partial n}) dS + \int\int_{S_a} Q \frac{\partial}{\partial n}(-\frac{1}{4\pi r}) \;dS][/tex]

[tex]^{lim}_{\epsilon\rightarrow 0} [ \int\int_{S_a} -\frac{1}{4\pi r}\frac{\partial Q}{\partial n} dS \;=\; -\frac{1}{4\pi \epsilon} \int\int_{S_a} \frac{\partial Q}{\partial n} dS \;=\; 0[/tex]

Because Q is harmonic and [itex]\int\int_{S_a} \frac{\partial Q}{\partial n} dS \;=\; 0[/itex]

From second identity:

[tex]\int\int_{S_a} (H\frac{\partial Q }{\partial n} \;-\; Q\frac{\partial H}{\partial n}) dS \;= \int\int\int_A (H\nabla^2 Q - Q\nabla^2 H) dV =0[/tex]

because both H and Q are harmonic in A and on surface [itex]S_A[/itex].

Therefore.

[tex]^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} P\frac{\partial Q}{\partial n} \;-\; Q\frac{\partial P}{\partial n} \;dS \;=\; ^{lim}_{\epsilon\rightarrow 0}\int\int_{S_a} Q \frac{\partial}{\partial n}(-\frac{1}{4\pi r}) \;dS = \frac{1}{4\pi \epsilon^2} \int\int_{S_a} Q dS[/tex]

The proof of the Strauss's book is very funky to put it politely. This is the way I proof it and please bare with the long explanation and tell me whether I am correct or not.

Thanks
Alan
 
Last edited:
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Anyone please?

Am I even posting in the correct sub-forum? I tried Green's function both in the Differential equation sub-forum and also in Advance Applied math in another forum with no response except a Math PHD advice to go to Electro-Dynamics type of section because PDE barely touch this.

Please advice.
Thanks
Alan
 
If all you need is a proof of the symmetry, there is a simple one, using the definition of the GF, in "Classical Eletromagnetism" by Franklin
 
Meir Achuz said:
If all you need is a proof of the symmetry, there is a simple one, using the definition of the GF, in "Classical Eletromagnetism" by Franklin

Thanks, I just bought it on Amazon. It is a really new book, used ones are just as expensive, cost me $80 big dollars!

Yes I can use one in between normal EM and Jackson book. It is getting hard to get help in these advanced topics.
 

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