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Given the system of differential equations

$$\begin{cases} x'=x(1-x-y) \\ y'=y(1-y-x)\end{cases}$$

I have seen that the phase portrait on the first quadrant looks like this:

phase portrait

Note that every point of the line $y=1-x$ is a critical point.

I want to see if it is possible to have periodic orbits (it certainly looks like it is not). To do so I thought I could apply Hartman's theorem to see if any of the critical points on the line $y=1-x$ was a centre. However, can I even use Hartman's theorem here? Every critical point on the line $y=1-x$ has another critical point on it's neighborhood.

Bounty edit: I want to know if I can apply the Hartman-Grobman theorem on the crtical points in the line $y=1-x$, and why.

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The dynamic system

$$ \begin{cases} x'=x(1-x-y) \\ y'=y(1-y-x) \end{cases} $$

has as integral curves

$$ \frac{y'}{x'} = \frac yx\Rightarrow \frac{dy}{dx} = \frac yx\Rightarrow y = C_0 x $$

hence no closed (or periodic) orbits are allowed.

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@Cesareo gave a good explanation why there is no periodic solutions in this system: straight lines that pass through an origin are integral curves (i.e., they consist of system's trajectories). Since no trajectory leaves a straight line, no periodic solutions are possible.

I'll explain a bit with Grobman-Hartman's theorem. GH can not be applied to equilibria at $ x + y = 1$. As you've noticed, to each of these equilibria there is another equilibrium that is arbitrarily close. Hyperbolic equilibria (for which GH theorem works: eigenvalues $\lambda$ of their Jacobi matrix doesn't intersect $\mathrm{Re}\, \lambda = 0$) are isolated: there is always a small neighbourhood around them that doesn't contain any other equilibrium point. Also, if you check eigenvalues of Jacobi matrix computed at any of these equilibria, you will find that there is one zero eigenvalue. Hence, GH-theorem can't be applied to this equilibrium.

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