I am trying to solve the following problem:
Let f, g be continuous positive functions $\mathbb{R}^2 \to \mathbb{R}$: show that the system of equations
$(1-x^2)f^2(x,y) = x^2 g^2(x,y)$ $(1-y^2)g^2(x,y) = y^2 f^2(x,y)$
has 4 distinct solutions on the unit circle.
Now my thinking was to use Brouwer's fixed point theorem on the unit disc: rewrite the problem as
$x = \pm \sqrt{\frac{f^2}{f^2 + g^2}},\,y = \pm \sqrt{\frac{g^2}{f^2 + g^2}}$
Now it is clear then that both of these square roots must have modulus at most 1, and furthermore, if x and y satisfy these then $x^2 + y^2 = 1$: thus it comes down to showing that these 4 $\pm$s give us the 4 distinct solutions we require. Defining
$F_{\pm \pm} = (\pm \sqrt{\frac{f^2}{f^2 + g^2}}, \pm \sqrt{\frac{g^2}{f^2 + g^2}})$ in the obvious way, each $F_{\pm \pm}$ is a continuous map to the unit disc, so each has a fixed point: so far so good I think. (Though, do we need to worry about the case where $f^2 = g^2 = 0$?)
Now, since each of the 4 functions $F_{* *}$ maps to a distinct quadrant in the unit disc, it is almost obvious that our 4 fixed points for $F_{* *}$ from Brouwer must be distinct: however, what if they both lie on the boundary where 2 quadrants meet? For example the point (1,0) is in both the "x,y positive, positive" and "x,y positive, negative" quadrant, so if 2 of our 4 fixed points coincide here then we have obviously not solved the problem. Is there a simple way around this? Or am I overlooking something? Many thanks, Peter