Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A colleague asked me a question and I got stumped. Suppose that $f:\mathbb{R}^2\rightarrow \mathbb{R}$, $g:\mathbb{R}^2\rightarrow\mathbb{R}^2$, and that $\Gamma$ is a closed $C^\infty$ curve in $\mathbb{R}^2$. Further let $\tau_x$ be a unit tangent at each point $x\in\Gamma$ and let $\nu_x$ be a unit normal. If we have

$$ \nabla f\cdot\nu_x=g\cdot \tau_x $$

at each point of $\Gamma$, does that imply that we have

$$ \nabla f\cdot\tau_x=-g\cdot\nu_x? $$

I did an example with the circle $x^2+y^2=1$. If we work in polar, a unit normal is $\langle \cos\theta,\sin\theta\rangle$ and a unit tangent is $\langle\sin\theta,-\cos\theta\rangle$. The examples I tried all worked out.

What I'd like to say is that, given an $x\in\Gamma$, we can change coordinates to make the curve at $x$ tangent to some circle and then use the above argument. Clearly we can't do this for all $x\in\Gamma$ simultaneously, which sort of bothers me.

What I want to know is, is the above true? Can my argument be made more rigorous? If either of those questions turn out negative, what is a counterexample? Or, what is a valid argument?

share|improve this question
2  
I don't see how the first equation can force $g$ to be anything, since it only involves the tangential component of $g$. (The normal component of $g$ can be anything; given any function $g$ which satisfies the first equation, the function $g+h\nu$ also does so, for any function $h:\mathbb{R}^2\rightarrow\mathbb{R}^2$.) –  Hans Lundmark Nov 15 '11 at 12:52
    
@Hans Lundmark: Good point. I will remove the comment from my question. –  Joe Johnson 126 Nov 15 '11 at 13:09
    
OK. But doesn't that settle the question (in the negative), since it means that the right-hand side of the second equation can be anything? –  Hans Lundmark Nov 15 '11 at 13:14
    
@Hans Lundmark: Agreed. Did you want to post that as an answer, so I can give you credit and close this silly question? –  Joe Johnson 126 Nov 15 '11 at 14:08
add comment

1 Answer

up vote 1 down vote accepted

Given $f$, the first equation only determines the tangential component of $g$. The normal component (the right-hand side of the second equation) can be anything, and therefore the second equation need not hold.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.