# Does $\nabla f\cdot\nu=g\cdot \tau$ Imply $\nabla f\cdot\tau=-g\cdot\nu$?

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?

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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
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.