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A theorem about identically zero potential function

  1. Jun 2, 2012 #1
    I found this theorem in my book on optics which I cannot prove: if f is a potential function in the plane, which is zero along a curve and such that the normal derivative to the curve is itself zero at any point along the curve, then f is zero in the whole plane. Can you give me a reference on this result or briefly explain how is it so?
     
  2. jcsd
  3. Jun 2, 2012 #2
    I think there are details missing. One of the following might possibly fix your statement.

    Is it true for any curve?

    Is f an analytic function between complex numbers?


    Double check the buildup to the author's statement, details may be hiding.
     
  4. Jun 2, 2012 #3
    I feel the statement as it is is false.

    If we take f(x,y)=e^(-1/x),
    then f(0,y)=0.
    The normal derivative is ∂f/∂y(0,y)=0.
    But f(x,y) is not identically zero.

    But this function is not analytic from the view point of complex analysis, and the condition of analyticity may correct the statement.
     
  5. Jun 2, 2012 #4
    Here is the full passage:

    "A well-known theorem in riemann's theory of functions says that if a two-dimensional potential v vanishes together with its normal derivative along a finite curve segment s, then v vanishes identically in the whole plane."

    Analyticity is not stated, but maybe it is given for granted. In this case, would the theorem be true?
     
  6. Jun 2, 2012 #5
    It sounds like you are talking about a function whose Laplacian is 0:
    $$ \frac{\partial^2v}{\partial x ^2}+\frac{\partial^2v}{\partial y ^2}=0.$$
    This would explain why you are calling it a potential function. So yes, because of that, v is analytic. But what's more important is that there exists another conjugate harmonic function u such that [itex]f(z) = u(x,y)+iv(x,y)[/itex] is complex analytic. Your hypotheses imply that along the curve, f'(z) =0. But f'(z) is complex analytic, and such functions can only have isolated zeros unless they are zero everywhere. Therefore, being zero over a whole curve implies f'(z) is zero everywhere. Which implies that f(z) is constant which implies that f(z) is zero which implies that v is zero everywhere.
     
  7. Jun 3, 2012 #6
    Thank you, Vargo!
     
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