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Functional relation between u(x,y,z) and v(x,y,z)
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[QUOTE="andrewkirk, post: 5444277, member: 265790"] 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 [LIST] [*]##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## [/LIST] Where ##B:\mathbb{R}\to\mathbb{R}## is a [URL='https://en.wikipedia.org/wiki/Bump_function']bump function[/URL] 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 [I]continuous[/I] differentiability, and only concludes that a functional relationship exists [I]locally[/I], not necessarily globally. [/QUOTE]
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Functional relation between u(x,y,z) and v(x,y,z)
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