# Determine if these two ODE systems are topologically equivalent

Consider the systems $$\begin{cases}\dot{x}=-2y-x\\\dot{y}=y+2x+4x^3\end{cases}\text{ and }\begin{cases}\dot{x}=-y\\\dot{y}=x\end{cases}.$$ Determine if the systems are topologically equivalent or not.

I know some "tricks" to determine whether two systems are topologically equivalent or not. For example:

• If one system has less equilibria than the other, they are not topologically equivalent.

• If one system equals the other system multiplied with a everywhere positve scalar function, both are equivalent.

• If one has periodic orbits and the other systems does not, they are not equivalent.

• If one system has only bounded solutions and the other at least one unbounded, they are not equivalent.

I tried to use one of these statements here but it did not help to determine whether the systems are topologically equivalent or not.

Anyway. my observations are:

1. Both systems have equilibrium $(0,0)$.

2. The second system may be rewritten as $\ddot{x}=-x$ to see that it is an harmonic oscillator.

3. For the first system, linearizing at $(0,0)$ gives the linearization matrix $A=\begin{pmatrix}-1 & -2\\ 2 & 1\end{pmatrix}$ which has eigenvalues $\pm\sqrt{3}i$. Hence it is a non-hyperbolic equilibrium.

Maybe you have an idea?

• For every nonnegative $c$, the level sets of the functional $$U(x,y)=x^2+y^2+xy+cx^4$$ partition the $(x,y)$-plane into topological circles around $(0,0)$ and are invariant by the dynamics $$S(c):\qquad x'=-2y-x,\qquad y'=y+2x+4cx^3,$$ hence the systems $S(c)$ for $c\geqslant0$ are all conjugate. Since $S(0)$ is conjugate to $x'=-y$, $y'=x$, you are done. To find $U$, one solves $$x'U_x+y'U_y=0,$$ that is, $$(2y+x)U_x=(y+2x+4cx^3)U_y.$$ The functional $U$ above coresponds to the choice $$U_y=2y+x,\qquad U_x=y+2x+4cx^3.$$ – Did Jan 9 '16 at 15:28
• To be honest, I have problems to understand this. I never saw this technique before. In particular, the notion of "conjugate" is new for me. Maybe you can give me a link where I can try to understand your procedure? – Rhjg Jan 9 '16 at 19:26
• Replace "conjugate" by "topologically equivalent" if you like. What is it that you do not understand in what you call my "procedure"? Please be specific. – Did Jan 9 '16 at 21:15
• Mainly two things: (1.) What's the idea behind the $U(x,y)$. (2.) How to get $S(c)$ and the homeomorphisms that give the topolog. equivalences. – Rhjg Jan 9 '16 at 21:22
• Idea: Find quantity invariant by the dynamics to deduce that trajectories are included in level sets of said quantity. How to get S(c): ?? One starts from the dynamics S(c) and then tries to find U. Topological equivalence: every S(c) is a topological circle, ergo. – Did Jan 10 '16 at 0:12