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How to prove the uniqueness theorem in an unbounded domain?

  1. Apr 7, 2010 #1
    I read a lot of books on the uniqueness theorem of Poisson equation,but all of them are confined to a bounded domain [tex]\Omega[/tex] ,i.e.
    "Dirichlet boundary condition: [tex]\varphi[/tex] is well defined at all of the boundary surfaces.
    Neumann boundary condition: [tex]\nabla\varphi[/tex]is 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 [tex]\varphi \to 0{\rm{ }}(r \to \infty )[/tex],
    I ended up with[tex]\int_S {\phi \frac{{\partial \phi }}{{\partial n}}} dS = {\int_V {\left( {\nabla \phi } \right)} ^2}dV[/tex]in which [tex]\phi[/tex] is the difference between two possible solution of the electric potential.

    Let S be the surface of an infinite sphere,we have
    [tex]4\pi \int_{r \to \infty } {{r^2}\phi \frac{{\partial \phi }}{{\partial r}}} dr = {\int_V {\left( {\nabla \phi } \right)} ^2}dV[/tex]
    We have[tex]\phi \to 0(r \to \infty )[/tex],but it doesn't indicate [tex]{r^2}\phi \frac{{\partial \phi }}{{\partial r}} \to 0(r \to \infty )[/tex] So we can't conclude that [tex]\nabla \phi \equiv 0[/tex] 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 [tex]{r^2}\frac{{\partial \phi }}{{\partial r}}[/tex] is bounded?
     
  2. jcsd
  3. Apr 7, 2010 #2

    fluidistic

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    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.
     
  4. Apr 7, 2010 #3
    I do find a "proof" of uniqueness theorem in an exterior domain,but it just assumes that [tex]E = O\left( {\frac{1}{{{r^2}}}} \right)(r \to \infty )[/tex] which is the same as what I want to prove,that is [tex]{r^2}\frac{{\partial \phi }}{{\partial r}} = O\left( 1 \right)(r \to \infty )[/tex]
     
  5. Apr 8, 2010 #4

    Meir Achuz

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    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.
     
  6. Apr 8, 2010 #5
    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 [tex]{r^2}\frac{{\partial \phi }}{{\partial r}} = O(1)(r \to \infty )[/tex]?
     
  7. Apr 8, 2010 #6

    Meir Achuz

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    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.
     
  8. Apr 8, 2010 #7
    I don't think it is obvious that the total charge being zero makes that BC hold.You should give me a derivation.
     
  9. Apr 9, 2010 #8

    clem

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    Just use Gauss's law.
     
  10. Apr 9, 2010 #9
    You are right!

    But I found a glitch in my previous "demonstration"
    It should have been
    [tex]{\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 [/tex]

    So here we cannot separate the two variables[tex]\phi [/tex]and[tex]{{r^2}\frac{{\partial \phi }}{{\partial r}}}[/tex] into two integrals

    How to go on with it then?(We still have [tex]\phi \to 0\left( {r \to \infty } \right)[/tex]
     
    Last edited: Apr 9, 2010
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