# Brouwer FPT and solutions to a system of equations

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

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I would suggest using the Intermediate Value Theorem instead of Brouwer's theorem. –  Jim Belk Jun 1 '11 at 0:53
Hi Jim: the question follows an exercise on proving Brouwer's FPT, so I assume there is an intended method which uses the theorem. Is there any way I can use it? I feel like I'm almost at a solution, barring the one degenerate case where 4 solutions coincide. –  Peter Hewkin Jun 1 '11 at 9:05
@Jim Belk: sorry, I didn't realise I was meant to use your username to notify you of my response re: the above: it isn't obvious to me how to use IVT at any rate, while I feel like I almost have a workable solution with BFPT if I can just iron out the one above issue. Please do let me know if you have any thoughts. –  Peter Hewkin Jun 1 '11 at 13:59
Both of your problems are solved by noting that you assumed that $f,g$ are continuous and positive. Hence $f^2\neq 0 \neq g^2$ and also that the image of your maps $F_{\pm\pm}$ are strictly inside the relevant quadrants.
If you don't assume the strict inequality, the statement is false. Let $f(x,y) = x$ and $g = 1$. Then on the unit circle where $(1-x^2) = y^2$ your equations reduce to the one equation
$$(1-x^2) x^2 = x^2 \iff (1-x^2 - 1)x^2 = 0 \iff - x^4 = 0$$