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How to prove uniqueness (or non-uniqueness) of solution

  1. Feb 26, 2010 #1
    I've only learned differential equations for use in physics, and never took a rigorous math course on all their amazing features. So I'm hoping someone can teach me a bit here, in the context of this question:

    Consider Maxwell's equations in vacuum, units don't matter here so I'll get rid of all constants:
    [tex]\nabla \cdot \vec{E} = 0[/tex]
    [tex]\nabla \cdot \vec{B} = 0[/tex]
    [tex]\nabla \times \vec{E} = - \frac{\partial}{\partial t} \vec{B}[/tex]
    [tex]\nabla \times \vec{B} = \frac{\partial}{\partial t} \vec{E}[/tex]

    Now consider a finite region of space, with the boundary condition that the fields and their derivatives are zero on the boundary at time 0<=t<T. What solutions are there for the fields in the region during this time?

    One obvious solution is: E=0, B=0 everywhere.

    Is this question well posed enough to prove that this solution is unique?
    If so, how? If not, what is missing?
  2. jcsd
  3. Feb 26, 2010 #2
    Okay, I came up with another solution.

    If we define, at t=0, an E field with no divergence, and let B=0. Then I can use Maxwell's equations to evolve the time dependence, right? So the problem is reduced to finding a finite volume E field with no divergence, which I don't see why that is a problem.

    Using cylindrical coordinates, I can define:
    [tex]\mathbf{E}(t_0) = \hat{\phi} f(r)g(z)[/tex]
    This field has no divergence.

    Now looking at the time dependence
    [tex] -\frac{\partial \mathbf{B}}{\partial t} = -\hat{r} f(r) \frac{\partial}{\partial z}g(z) + \hat{z} \frac{1}{r} g(z) \frac{\partial}{\partial}[r f(r)] [/tex]
    So B will have r and z components. But these components only depend r and z. So the curl of B will only have
    [tex] \frac{\partial \mathbf{E}}{\partial t} = \nabla \times \mathbf{B} = \hat{\phi}(\frac{\partial B_r}{\partial z}- \frac{\partial B_z}{\partial r})[/tex]
    So E will remain in the phi direction, and so on for all time.

    This is true for any function f(r) and g(z). So I can just choose a solution initially confined enough that it doesn't have time to propagate to the boundary.

    Does this look correct?
    To do this I'd need f(r) to be non-analytic (since it needs to be identically zero for a region of r). Is that somehow a problem?
    Last edited: Feb 26, 2010
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