# Functional relation between u(x,y,z) and v(x,y,z)

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1. Apr 15, 2016

### arpon

1. The problem statement, all variables and given/known data
Let $u$ and $v$ be differentiable functions of $x,~y$ and $z$. Show that a necessary and sufficient condition that $u$ and $v$ are functionally related by the equation $F(u,v)=0$ is that $\vec \nabla u \times \vec \nabla v= \vec 0$

2. Relevant equations
(Not applicable)

3. The attempt at a solution
$\vec \nabla u$ and $\vec \nabla u$ are the normal vectors to the constant $u$-surface and the constant $v$-surface respectively. As, $\vec \nabla u \times \vec \nabla v= \vec 0$, i.e, $\vec \nabla u$ and $\vec \nabla v$ are in the same (or opposite) direction for a particular value of $(x, y, z)$, a constant $u$-surface also represents a constant $v$-surface. Therefore, for a particular value of $u$, there exists a corresponding value of $v$. So, we can conclude that $u$ and $v$ are functionally related.
But, how can I prove it mathematically?

2. Apr 15, 2016

### andrewkirk

By 'functionally related', do you think that means that
- for any given value of $u$ there is a unique value of $v$ such that $F(u,v)=0$ and
- for any given value of $v$ there is a unique value of $u$ such that $F(u,v)=0$?

If so, I'm not convinced by the above argument. I can imagine the set with $u=1$ as being two disconnected spheres, on one of which we have $v=2$ and on the other we have $v=3$. Then there would be no unique $v$ value for $u=1$.

I think a successful argument is going to have to bring the function $F$ into the argument, which the above does not.

3. Apr 15, 2016

### andrewkirk

In fact, I think the proposition may not even be true.

First, note that the 'necessary' part is easy to prove. Just express $u$ as a function of $v$ and then write out $\nabla u(v)\times \nabla v$ in coordinate form and we see that everything cancels.

I think the following may be a counterexample to the 'sufficient' claim though.
Define $P=(1,0,0),Q=(-1,0,0)\in \mathbb{R}^3$ and define $u:\mathbb{R}\to\mathbb{R}$ by
• $u(x)=\max(0,B(1-\|x-P\|))$ if $x^1\geq 0$; and
• $u(x)=-\max(0,B(1-\|x-Q\|))$ if $x^1< 0$
Where $B:\mathbb{R}\to\mathbb{R}$ is a bump function with support $(0,1)$ (to ensure $u$ and $v$ are differentiable).

Then define $v=|u|$, and $F(u,v)=u^2-v^2$.

Then for any $v\in(0,1)$ the set of $(u,v)$ satisfying $F(u,v)=0$ is a pair of congruent, non-intersecting spheres, of radius in $(0,1)$, centred on $P$ and $Q$. The value of $v$ is constant everywhere on both spheres, but the values of $u$ on the two spheres have opposite signs. So $F(u,v)=0$ does not generate a functional relationship between $u$ and $v$.

The best we could do would be to prove something like the Implicit Function Theorem, that requires additional conditions such as continuous differentiability, and only concludes that a functional relationship exists locally, not necessarily globally.

4. Apr 16, 2016

### Samy_A

Thanks, @andrewkirk . I was also puzzled by this exercise.

For the 'necessary' part, it is straightforward if one assumes that one of the partial derivatives of F never vanishes.
Without any assumption on F, no way this can be true (F≡0 as a silly counterexample for the 'necessary' part).

5. Apr 16, 2016

### andrewkirk

In the necessary case one needs to prove that:

(1) $\nabla u\times\nabla v\not\equiv 0\Rightarrow u,v$ are not functionally related by $F$.

For a counterexample to that we would have to affirm the antecedent and deny the consequent, that is, we'd need:

(2) $(\nabla u\times\nabla v\not\equiv 0)$ and ($u,v$ are functionally related by $F$)

The case $F\equiv 0$ affirms the consequent of (1), and hence denies the second conjunct of (2), so (2) will not be true and the case cannot be a counterexample.

I've written out a componentwise proof of the necessary part, but then I realised I mustn't post it, as that would be giving away part of the answer wholesale.