Can a dynamical system have an infinite critical points? I have studied the cosmological evolution of dark energy modeled as a scalar field. I want to make an extension to link and I have arrived at a 
system of differential equations on the following form
$$
x_1'=x_1x_2^3+x_2^2+x_3x_4 \\
x_2'=-4x_1^2+x_4x_3 \\
x_3'=x_3^2-x_4^2 \\
x_4'=x_3^2-x_4^2
$$ The critical points means that the derivative with respect to time(') of the coordinates are critical points $$(x_1'=x_2'=x_3'=x_4'=0)$$. We can see that for this system we cannot solve this system of ODE and that in fact we have a dependence of one coordinates. The question is how to deal with this systems and how to establish stability?(saddle/stable/unstable) since we shall have a dependence of one coordinate which is a function.
Much more, what if this system is non-compact and if we want to study the stability of the critical points at infinity by the means of Poincare central projection method in more than 3 dimensions?
Best regards!
 A: *

*A dynamical system can have infinitely many critical points. For instance, the system $x' = \sin(x)$, with $x = n\pi$.

*For your particular system, you say that it cannot be solved, but it can, indeed. From the third and fourth equation, you get $x_3^2 + x_4^2 = 0$ if the point is critical, but as $x_i$ are real numbers, the only solution is $x_3 = x_4 = 0$. Now go to the second equation: $-4x_1^2 + x_4x_3 = 0$. You already now that $x_3 = x_4 = 0$, thus $-4x_1^2 = 0 \Rightarrow x_1 = 0$. Finally go to the first equation, $x_1x_2^3 + x_2^2 + x_3x_4 = 0$. Plugging in what you already know, you get $x_2^2 = 0$. Again, the only possible solution is $x_2 = 0$, so the only critical point of your system is the origin $x_1 = x_2 = x_3 = x_4 = 0$. 

*Regarding stability, if the problem is well behaved, the proceeding consists in linearising the problem, computing the differential matrix and studying the value of the real part of its eigenvalues (if only one of them is positive, the critical point is unstable). However, your system doesn't have a linear term, so it's not that easy. 

