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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. –  GEdgar Jun 21 '11 at 17:33
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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 $u$. –  Willie Wong Jun 24 '11 at 15:35
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3 Answers

Hint number 2: Forget about the implicit function theorem and do it the pedestrian way. The system $F=0$, $G=0$ is linear in $x^2$ and $y^2$, so go and solve it for these auxiliary variables.

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It is my understanding that (0,0,0,0) is the only solution. If you vary u or v at all away from (0,0) the system is no longer satisfied so in this way it is impossible to define x or y as functions of u and v in any open set around a point (u,v) because such an open set has some variation in it. I guess you would need at least an open ball of solutions to have any chance of such a function existing. –  user9352 Jun 24 '11 at 14:22
    
So your hint is good, you don't need to look at derivatives at all to see there is no chance of a solution but the claim above is still sort of open as another possible way to say that there is not an implicit function possible. –  user9352 Jun 24 '11 at 14:30
    
@user9352: Note that the implicit function theorem has two assumptions: (a) You are given a point $(u_0,v_0,x_0,y_0)$ that satisfies the equations, and (b) a certain determinant evaluated at that point should be $\ne0$. Now in the case at hand there is only one point satisfying the equations, namely $(0,0,0,0)$, and the determinant in question is $=0$ there. –  Christian Blatter Jun 24 '11 at 16:26
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To say $(F,G) = (0,0)$ is to say that $y^2 - 3x^2 = u^2 + 4uv + v^2$ and $y^2 - x^2 = 2uv$. By some algebra, this is equivalent to $x^2 = -{1 \over 2}(u + v)^2$ and $y^2 = -{1 \over 2}(u - v)^2$. So you are requiring the nonnegative quantities on the left to be equal to the nonpositive quantities on the right. Hence the solution set is only $(x,y,u,v) = (0,0,0,0)$, where everything is zero.

Suppose on the other hand you had equations $x^2 = {1 \over 2}(u + v)^2$ and $y^2 = {1 \over 2}(u - v)^2$. Then you could solve them, but there is no uniqueness now; you could take $(x,y) = (\pm {1 \over \sqrt{2}}(u + v),\pm {1 \over \sqrt{2}}(u - v))$ obtaining four distinct smooth solutions that come together at $(0,0,0,0)$.

So these are good examples showing that if the determinant is zero at $(0,0)$ you don't have to have existence of solutions, nor uniqueness when you do have existence.

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Hint: Determine the set of points $(x,y,u,v)$ where the function $F-G$ vanishes.

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I had looked at F-G you get $2x^2+(u+v)^2$ which can only $=0$ when $x,u,v=0 $ but then F implies that $y=0$ So it would vanish only at (0,0,0,0)? –  user9352 Jun 21 '11 at 18:05
    
Not too fast :-) $(u+v)^2=0$ whenever $u=-v$. –  Jyrki Lahtonen Jun 21 '11 at 18:10
    
Ok so $(x,y)=(0,0)$ is necessary to satisfy the equations but in any n-hood of (0,0) there is some (u,-u) (-u,u) combo. –  user9352 Jun 21 '11 at 18:41
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I get a vibe that the contradiction comes from combining what you have already shown with the assumption that the pair $(u,v)$ is supposed to range over a (presumably non-empty) open subset $U$ of the $uv$-plane. Note: Any open subset $U$ - not just those containing the origin of the $uv$-plane. –  Jyrki Lahtonen Jun 21 '11 at 18:50
    
Aren't we supposed to show that it is impossible to find $x=x(u,v)$, $y=y(u,v)$ such that $F(x(u,v),y(u,v),u,v)=G(x(u,v),y(u,v),u,v)=0$ for all $(u,v)$ in some open set? –  Jyrki Lahtonen Jun 21 '11 at 19:41
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