How to prove the uniqueness theorem in an unbounded domain?

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Discussion Overview

The discussion centers on the uniqueness theorem of the Poisson equation in unbounded domains, particularly in the context of electrostatics and boundary conditions. Participants explore the implications of applying the theorem to scenarios involving infinite domains, such as the electric field produced by a point charge near an infinitely large grounded conductor plate.

Discussion Character

  • Debate/contested
  • Exploratory
  • Technical explanation

Main Points Raised

  • Some participants note that existing literature on the uniqueness theorem primarily addresses bounded domains, raising questions about its applicability in unbounded domains.
  • There is a suggestion that infinity cannot be treated as a boundary surface due to its infinite area, complicating the proof of the uniqueness theorem in such contexts.
  • One participant attempts to prove the uniqueness theorem using an integral approach but encounters difficulties with the behavior of the potential at infinity.
  • Another participant mentions finding a proof for uniqueness in an exterior domain, which assumes a specific decay of the electric field at infinity.
  • Some participants argue that any boundary condition leading to a zero surface integral is suitable for proving uniqueness, but they question the validity of specific boundary conditions in the context of unbounded domains.
  • There is a discussion about the implications of total charge being zero on the validity of boundary conditions, with some participants expressing skepticism about the assumptions made.
  • Participants reference Gauss's law as a method to support their arguments regarding boundary conditions and field behavior at infinity.
  • A participant acknowledges a mistake in their previous reasoning regarding the separation of variables in their integral demonstration.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the uniqueness theorem in unbounded domains, with no consensus reached on the validity of specific boundary conditions or the proof methods discussed.

Contextual Notes

Limitations include unresolved assumptions about the behavior of potentials at infinity and the specific conditions under which boundary conditions hold in unbounded domains.

netheril96
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I read a lot of books on the uniqueness theorem of Poisson equation,but all of them are confined to a bounded domain \Omega ,i.e.
"Dirichlet boundary condition: \varphi is well defined at all of the boundary surfaces.
Neumann boundary condition: \nabla\varphiis well defined at all of the boundary surfaces.
Mixed boundary conditions (a combination of Dirichlet, Neumann, and modified Neumann boundary conditions): the uniqueness theorem will still hold.
"

However,in the method of mirror image,the domain is usually unbounded.For instance,consider the electric field induced by a point charge with a infinely large grounded conductor plate.In all of the textbooks,it is stated that "because of the uniqueness theorem...",but NO book has ever proved it in such a domain!

Some may say that we can regard the infinity as a special surface,but we CAN'T since this "surface" has a infinite area.I tried to prove it using the same way as that in a bounded domain,i.e. with the electric potential known in a bounded surface and \varphi \to 0{\rm{ }}(r \to \infty ),
I ended up with\int_S {\phi \frac{{\partial \phi }}{{\partial n}}} dS = {\int_V {\left( {\nabla \phi } \right)} ^2}dVin which \phi is the difference between two possible solution of the electric potential.

Let S be the surface of an infinite sphere,we have
4\pi \int_{r \to \infty } {{r^2}\phi \frac{{\partial \phi }}{{\partial r}}} dr = {\int_V {\left( {\nabla \phi } \right)} ^2}dV
We have\phi \to 0(r \to \infty ),but it doesn't indicate {r^2}\phi \frac{{\partial \phi }}{{\partial r}} \to 0(r \to \infty ) So we can't conclude that \nabla \phi \equiv 0 so that the uniqueness theorem doesn't hold (or we cannot prove it with the same way proving uniqueness theorem in a bounded domain)
Or can anybody here prove that {r^2}\frac{{\partial \phi }}{{\partial r}} is bounded?
 
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netheril96 said:
I read a lot of books on the uniqueness theorem of Poisson equation,but all of them are confined to a bounded domain \Omega ,i.e.
"Dirichlet boundary condition: \varphi is well defined at all of the boundary surfaces.
Neumann boundary condition: \nabla\varphiis well defined at all of the boundary surfaces.
Mixed boundary conditions (a combination of Dirichlet, Neumann, and modified Neumann boundary conditions): the uniqueness theorem will still hold.
"

However,in the method of mirror image,the domain is usually unbounded.For instance,consider the electric field induced by a point charge with a infinely large grounded conductor plate.In all of the textbooks,it is stated that "because of the uniqueness theorem...",but NO book has ever proved it in such a domain!

Some may say that we can regard the infinity as a special surface,but we CAN'T since this "surface" has a infinite area.I tried to prove it using the same way as that in a bounded domain,i.e. with the electric potential known in a bounded surface and \varphi \to 0{\rm{ }}(r \to \infty ),
I ended up with\int_S {\phi \frac{{\partial \phi }}{{\partial n}}} dS = {\int_V {\left( {\nabla \phi } \right)} ^2}dVin which \phi is the difference between two possible solution of the electric potential.

Let S be the surface of an infinite sphere,we have
4\pi \int_{r \to \infty } {{r^2}\phi \frac{{\partial \phi }}{{\partial r}}} dr = {\int_V {\left( {\nabla \phi } \right)} ^2}dV
We have\phi \to 0(r \to \infty ),but it doesn't indicate {r^2}\phi \frac{{\partial \phi }}{{\partial r}} \to 0(r \to \infty ) So we can't conclude that \nabla \phi \equiv 0 so that the uniqueness theorem doesn't hold (or we cannot prove it with the same way proving uniqueness theorem in a bounded domain)
Or can anybody here prove that {r^2}\frac{{\partial \phi }}{{\partial r}} is bounded?
Maybe these notes can help (See chapter 4. There are a lot of orthographic errors. Replace "Poison" by "Poisson" and so on) http://www.famaf.unc.edu.ar/~reula/Docencia/Electromagnetismo/part1.pdf
In case it doesn't help, let us know.
 
fluidistic said:
Maybe these notes can help (See chapter 4. There are a lot of orthographic errors. Replace "Poison" by "Poisson" and so on) http://www.famaf.unc.edu.ar/~reula/Docencia/Electromagnetismo/part1.pdf
In case it doesn't help, let us know.

I do find a "proof" of uniqueness theorem in an exterior domain,but it just assumes that E = O\left( {\frac{1}{{{r^2}}}} \right)(r \to \infty ) which is the same as what I want to prove,that is {r^2}\frac{{\partial \phi }}{{\partial r}} = O\left( 1 \right)(r \to \infty )
 
An 'infinite domain' means a bounding surface of radius R in the limit as R-->infty.
Just like any other surface, any BC that makes the surface integral go to zero is a suitable BC for uniqueness. You can't 'prove' your last equation because that is one of the possible BCs.
 
Meir Achuz said:
An 'infinite domain' means a bounding surface of radius R in the limit as R-->infty.
Just like any other surface, any BC that makes the surface integral go to zero is a suitable BC for uniqueness. You can't 'prove' your last equation because that is one of the possible BCs.

But when do this kind of boundary condition hold?
For instance,how do you know that the electric field induced by a point charge with a infinitely large grounded plate satisfies the boundary condition of {r^2}\frac{{\partial \phi }}{{\partial r}} = O(1)(r \to \infty )?
 
You don't 'know' it. That BC does not hold for an accelerating charge.
Imposing that BC gives the static solution.
For your infinite plane case, the total charge is zero so that BC does hold.
 
Meir Achuz said:
You don't 'know' it. That BC does not hold for an accelerating charge.
Imposing that BC gives the static solution.
For your infinite plane case, the total charge is zero so that BC does hold.

I don't think it is obvious that the total charge being zero makes that BC hold.You should give me a derivation.
 
Just use Gauss's law.
 
clem said:
Just use Gauss's law.

You are right!

But I found a glitch in my previous "demonstration"
It should have been
{\int_V {\left| {\nabla \phi } \right|} ^2}dV = \int_S {\phi \nabla \phi \cdot d{\bf{S}}} = \int_{0 \le \theta \le \pi ,0 \le \varphi < 2\pi } {{r^2}\phi \frac{{\partial \phi }}{{\partial r}}} d\theta d\varphi

So here we cannot separate the two variables\phiand{{r^2}\frac{{\partial \phi }}{{\partial r}}} into two integrals

How to go on with it then?(We still have \phi \to 0\left( {r \to \infty } \right)
 
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