Find equilibrium points for Hamilton system.

Question

Here is the given system: $$\begin{cases}x'=x^2+y^2-6 \\ y'=y-x^2 \end{cases}$$ Adding both equations I get: $y^2+y-6=0 \Rightarrow (y-2)(y+3)=0 \Rightarrow y_1=2, y_3=-3, \text{ from there } x_{1,2}=\pm \sqrt{2} \text{ and } x_{2,3}=\pm \sqrt{3}i, \text{ but in the solution only two equilibrium points are given } (\sqrt{2},2) \text{ and } (-\sqrt{2},2). \text{ Why is that?}$

Answer 1

Because we are only looking for real solutions. Why? Because by definition, an equilibrium point must lie in the domain of the function, which in this case is $\Bbb{R}^2$.

Let me be more explicit. We are given a function $F: \Bbb{R}^2 \to \Bbb{R}^2$ defined as \begin{align} F(x,y) = (x^2+y^2-6, y-x^2) \end{align} If we write $\xi = (x,y)$, then (being slightly sloppy with notation) we are considering the system of ODE's \begin{align} \xi' &= F(\xi) \end{align} (I didn't do much... I just rewrote your two ODE's as a single equation). Now, BY DEFINITION, an equilibrium point for this system is any point $\xi_0 \in \text{domain}(F) = \Bbb{R}^2$ such that $F(\xi_0) =0$.

So, it is literally by definition of (the domain of) $F$ that we are only looking for equilibrium points in $\Bbb{R}^2$, and hence discard the complex ones.

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Additional Remark:

Strictly speaking, one should always define the domain and codomain of any functions FIRST, and only afterwards, perform calculations. But very often people are lazy, and it is assumed that such issues are "obvious", so we don't always mention it explicitly. But when in doubt, always do things the proper way systematically.