5
$\begingroup$

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.

  • 1
    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

2 Answers 2

3

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.

  • 0
    @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.2011-06-24
1

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.