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

  • #1
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Homework Statement


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##

Homework Equations


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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?
 

Answers and Replies

  • #2
andrewkirk
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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
andrewkirk
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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
Samy_A
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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
andrewkirk
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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).
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.
 

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