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I am facing a problem of ODE in dynamical system, which aims to prove the solution connecting fixed pionts is in some interval.

Here is the problem. Consider the ODE below.

$x'=y$; $y'=-cy-x(1-x)$

Then we have 2 fixed points. One of them is the stable fixed point $(0,0)$, and the other is the hyperbolic fixed point $(1,0)$.

If $c>2$, then both two points has two different real eigenvalues after the linearization.

If $0<c<2$, then $(0,0)$ has two different complex eigenvalues, but $(1,0)$ keeps the same.

I could choose the unstable manifold of point $(1,0)$, and there is a solution limit to fixed point $(0,0)$. But I don't know how to show that the $x$ value of the solution will stay in $[0,1]$.

Also we have the Lyapunov function of the system, $\frac{1}{2}x^2-\frac{1}{3}x^3+\frac{1}{2}y^2$.

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I believe the claim is only true for $c>2$, so assume this is the case.

Draw a triangle with vertices at $(0,0)$, $(1,0)$, and $(1,-2/c)$. You need to show that this triangle (including its boundary) is a trapping region for this system of ODEs. That is, a solution that starts in the triangle at $t=0$ stays in the triangle for all time $t\geq 0$. This can be done by showing that at every point on the boundary of the triangle, the vector field $(y,-cy-x(1-x))$ points inside the triangle, not outside. Thus, no solution can pass from the inside to the outside.

Now, using the eigenvectors of the linearization about $(1,0)$, you know that the unstable manifold of $(1,0)$ lies tangent to a line with positive slope through $(1,0)$. This line intersects the triangle. Thus, there exists a point on the unstable manifold close to this line such that if $(x(t),y(t))$ intersects this point at $t=0$, then $(x(t),y(t))$ lies in the triangle for all $t\leq 0$ because it remains close to the positive sloped tangent line. Moreover, since the solution $(x(t),y(t))$ starts in the triangle, it also stays in the triangle for all $t\geq 0$. Finally, by the Poincare-Bendixon Theorem, we observe that this solution is the one we seek. Since $0\leq x\leq 1$ for each point in the triangle, we conclude that the same inequality holds for our heteroclinic solution.

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  • $\begingroup$ Sorry, on the boundary from point (0,0) to point (0,-2/c), the direction goes out of the region. All x direction with y<0 is left. $\endgroup$ Oct 26, 2022 at 6:22
  • $\begingroup$ I could choose $(0,0)$,$(1,0)$,$(1,\frac{-c-\sqrt{c^2-4}}{2})$ as the trapping region, $\endgroup$ Oct 26, 2022 at 6:58
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    $\begingroup$ Sorry, I meant $(1,-2/c)$, not $(0,-2/c)$. $\endgroup$
    – Plutoro
    Oct 26, 2022 at 11:03
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    $\begingroup$ It looks like there is actually quite a lot of freedom in choosing the third vertex of the triangle. Even $(1,-1)$ (independent of $c$) works. $\endgroup$
    – Plutoro
    Oct 26, 2022 at 11:07
  • $\begingroup$ Thank you so much! $\endgroup$ Oct 27, 2022 at 12:02

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