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Define $$F(x,y,u,v)= 3x^2-y^2+u^2+4uv+v^2$$ $$G(x,y,u,v)=x^2-y^2+2uv$$

Show that there is no open set in the $(u,v)$ plane such that $(F,G)=(0,0)$ defines $x$ and $y$ in terms of $u$ and $v$.

If (F,G) is equal to say (9,-3) you can just apply the Implicit function theorem and show that in a neighborhood of (1,1) $x$ and $y$ are defined in terms of $u$ and $v$. But this question seems to imply that some part of the assumptions must be necessary for such functions to exist?

I believe that since the partials exist and are continuous the determinant of $$\pmatrix{ \frac{\partial F}{\partial x}&\frac{\partial F}{\partial y}\cr \frac{\partial G}{\partial x}&\frac{\partial G}{\partial y} }$$ must be non-zero in order for x and y to be implicitly defined on an open set near any point (u,v) but since the above conditions require x=y=0 the determinant of the above matrix is =0.

I have not found this in an analysis text but this paper http://www.u.arizona.edu/~nlazzati/Courses/Math519/Notes/Note%203.pdf claims it is necessary.

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    Maybe it means: for every open set there is some $(u,v)$ in that open set such that more than one pair $(x,y)$ satisfies the equations.2011-06-21
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    Just a note: while a certain non-singular condition on the matrix of determinants must be impose for the implicit function theorem to hold, the failure of those hypothesis *does not* automatically imply that you cannot "find an open set..." For example, consider the case $F(x,u) = x^3 - u$. $\partial_xF = 0$ when $x = 0$, so implicit function theorem doesn't hold there. But the set $F(x,u) = 0$, in a neighborhood of $u = 0$, still describes $x$ as a function of $u$.2011-06-24

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