# solution orbit connects fixed point

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 $$0, then $$(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, $$\frac{1}{2}x^2-\frac{1}{3}x^3+\frac{1}{2}y^2$$.

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.
• I could choose $(0,0)$,$(1,0)$,$(1,\frac{-c-\sqrt{c^2-4}}{2})$ as the trapping region, Oct 26, 2022 at 6:58
• Sorry, I meant $(1,-2/c)$, not $(0,-2/c)$. Oct 26, 2022 at 11:03
• 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. Oct 26, 2022 at 11:07