A theorem about identically zero potential function

[QuArK21343]

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?

 

[algebrat]

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.

 

[algebrat]

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.

 

[QuArK21343]

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?

 

[Vargo]

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 $f(z) = u(x,y)+iv(x,y)$ 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.

 

[QuArK21343]

Thank you, Vargo!