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Existence of solution to Poisson's equation

  1. Jan 5, 2007 #1
    I'm reteaching electrodynamics to myself on a more rigorous footing, and I'm trying to prove to myself that setting the divergence of the vector potential is justified using a gauge shift. I could use the Helmholtz theorem to do this, but the problem with this from my perspective is that I haven't actually justified the full version of the theorem, only the weaker version which requires that a vector function decay to zero faster than 1/r at infinity. This isn't a problem for the field quantities (since all physical fields decay like 1/r^2); nevertheless, they do pose a problem for the potential quantities (which in general will not even decay). Basically, given a vector potential [tex]\vec{A}[/tex], I want to show that fixing the divergence of the gauge-shifted potential [tex]\vec{A}\prime[/tex] to some scalar function [tex]D[/tex] is equivalent to adding the gradient of some some scalar function [tex]\phi[/tex].

    [tex]\nabla \cdot \vec{A}' = D[/tex]
    [tex]\nabla \cdot (\vec{A} + \nabla \phi) = D[/tex]
    [tex]{\nabla}^2 \phi = D - \nabla \cdot \vec{A}[/tex]

    Since the right-hand side is just some function of position, proving that the divergence can be adjusted by adding the gradient of a scalar amounts to proving that Poisson's equation has a solution for an arbitrary source term. No boundary conditions are specified, so I would expect that there are actually an infinite number of solutions; however, I cannot prove this. Does anyone have any insights? Thanks.
     
    Last edited: Jan 5, 2007
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  3. Jan 6, 2007 #2

    Hurkyl

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    Can we solve this system of equations?

    [tex]\nabla \cdot \mathbf{X} = \theta[/tex]
    [tex]\nabla \times \mathbf{X} = 0[/tex]

    If we know [itex]\nabla \times \mathbf{V} = 0[/itex], can we solve this equation?

    [tex]\nabla \varphi = \mathbf{V}[/tex]

    If so, then we can chain these results to solve the Poisson equation.

    [tex]\nabla^2 \varphi = \theta[/tex]

    If you know a counterexample to one of the above questions, I suspect you could use it to construct a Poisson equation without a solution.
     
  4. Jan 7, 2007 #3
    Thanks for the response, Hurkyl. I've been thinking about it some more, and I've been able to informally construct a solution to the equation, but I'm still not quite satisfied. Since there are no boundary conditions specified, and I'm just trying to construct any solution, I arbitrarily specified that

    [tex]\phi (x=0) = \frac{\partial \phi}{\partial x} = 0 [/tex]

    It would then seem to me that if you numerically integrated with a small enough dx, you could construct phi as follows:

    [tex]\phi (x+dx,y,z) = \phi (x,y,z) + \frac{\partial \phi}{\partial x} (x,y,z) \cdot dx[/tex]

    [tex]\frac{\partial \phi}{\partial x} (x+dx,y,z) = \frac{\partial \phi}{\partial x} (x,y,z) + \frac{{\partial}^2\phi}{\partial {x}^2} (x,y,z) \cdot dx [/tex]

    [tex]= \frac{\partial \phi}{\partial x} (x,y,z) + (f(x,y,z) - \frac{{\partial}^2\phi}{\partial {y}^2}- \frac{{\partial}^2\phi}{\partial {z}^2}) \cdot dx[/tex]

    (f(x,y,z) is the source term.) Obviously, this method fails for some source terms, like the pathological sin(1/x) or any function which blows up, but other than those cases, it seems to be pretty solid. I've tried to "un-discretize" the construction to form integrals, but the expression gets pretty messy quickly. Could anyone explain how to to it a little more formally, since I'm still not convinced?
     
    Last edited: Jan 7, 2007
  5. Jan 7, 2007 #4

    Hurkyl

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  6. Jan 7, 2007 #5
    Yeah, but AFAIK, the problem with that solution is that the integral will only converge if the source term decays faster than 1/r^2, which is not in general the case when dealing with potentials.
     
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