ODE system requires a trick

Question

Let's consider a system of non-linear ODE: $$ x' = 1-x^{2}-y^{2} \\ y ' = 2xy$$

The final goal is to draw a phase portrait (of course, manually). The trouble is that this system, as i can suggest right know, requires a trick, taking into account that it cannot be transformed in something, which is more convenient to work with only by applying routine tranformations. It's likely to use polar coordinates, but this particular approach does not work as well.

Could anybody suggest some trick (i suppose that it should be something well-known )? Any help would be much appreciated.

Answer 1

I'd start with $$(x+y)' = 1-(x-y)^2\\(x-y)'=1-(x+y)^2.$$ After that make an obvious change of variables and then deduce a phase portrait.

Answer 2

Converting this system of ODEs to the first order DE we get

$$ \frac{dy}{dx} = \frac{2xy}{1-x^2-y^2}.$$

It also could be written this way

$$ 2xy \, dx + (x^2+y^2-1) \, dy = 0 .$$

It's easy to see that this ODE is exact: there exists function $F(x,y)$ such that $\frac{\partial F}{\partial x} = 2xy$ and $\frac{\partial F}{\partial y} = x^2+y^2-1$. From this we find that $F(x,y) = x^2y + \frac{y^3}{3} - y$. It's well known fact that integral curves of exact equation lie on level sets $F(x,y) = {\rm const}$. The most important fact that you obtain from this is that trajectories are closed in neighbourhoods of some equilibria. By the way, you could obtain all integral curves just plotting $x^2y + \frac{y^3}{3} - y = C$ for different values of $C$.