I am given a dynamical system

$$\dot x = f(x,y)= x - (1+\theta(x))x^3-y \\ \dot y =g(x,y)= y - 3x^2y + x$$

where $\theta(x)$ is a step function which is equal to $1$ when $x \geq 0$ and $0$ when $x<0$.

Now I am asked to prove whether or not the fixed point at the origin is asymptotically stable.

My thinking so far is as follows.

1) I know that when $x<|\sqrt{\frac{2}{3}}|$ that there can not exist a periodic orbit due to Bendixson's criterion

2) I have found that $\nabla \cdot \pmatrix{f(x,y) \\ g(x,y)} = 2$ when $x<0$ and $\nabla \cdot \pmatrix{f(x,y) \\ g(x,y)} = 2-3x^2$ when $x \geq 0$

3) I know that if there exists a strict Liapounov function around the fixed point then the fixed point is asymptotically stable

4) Not sure if this is relevant but Poincare bendixson states that if there exists a non empty closed and bounded omega limit set then there is either a fixed point or a periodic orbit. Now I know that there can't be a periodic orbit so there must be a fixed point

What I think I need to do

I think I need to find a strict Liapounov function which will then allow me to state the fixed point is asymptotically stable. How I am going to find this function is still up in the air, but maybe it's something to do with the fact that the orbital derviative

$$\frac{dV}{dt} = \int_{\phi(t,D)} d^nx(\nabla \cdot \pmatrix{f(x,y) \\ g(x,y)})$$ where $\phi(t,D)$ is the region obtained by evolving all the points in a set $D$.

  • $\begingroup$ Wait a minute, I might be saying a stupid thing, but the annoying term $(1-\theta(x))$ is the factor of $x^3$. When you linearize around $x=0, y=0$, that term vanishes. Doesn't this mean that it does not influence the local behavior? (It may be that this affects the stability but not the asymptotic stability) $\endgroup$ – Giuseppe Negro May 29 '18 at 21:29
  • $\begingroup$ You would be correct in saying that it vanishes when you linearize at the origin. But I don't know how that would help me show that the point is asymptotically stable $\endgroup$ – Gragbow May 29 '18 at 21:44
  • $\begingroup$ I am sorry, it does not help you. The eigenvalues of the linearized problem are $0$ and $2$. The presence of a vanishing eigenvalue means that you cannot solve by a simple linearization. $\endgroup$ – Giuseppe Negro May 29 '18 at 21:47
  • 1
    $\begingroup$ Yeah it means the FP is non hyperbolic $\endgroup$ – Gragbow May 29 '18 at 21:50
  • $\begingroup$ Principle of Linearized Instability states that if the linearization at an equilibrium of an autonomous $C^1$ system of ODEs has at least one eigenvalue with positive real part then the equilibrium is unstable (this has nothing to do with hyperbolicity!). You need only to check carefully that your system is $C^1$. $\endgroup$ – user539887 May 30 '18 at 7:44

Multiplying the first by $x$ and the second by $y$ we get

$$ x\dot x = x^2-(1-\theta(x))x^4-x y\\ y\dot y = y^2-3 x^2 y^2+x y $$

adding the equations

$$ \frac{1}{2}\frac{d}{dt}(x^2+y^2) = x^2+y^2 -((1+\theta(x))x^4+3 x^2 y^2) $$

and in polar coordinates

$$ \frac{1}{2}\frac{d}{dt}r^2 = r^2-r^4((1+\theta(x))\cos^4(\theta)+3\cos^2(\theta)\sin^2(\theta)) $$


$$ (1+\theta(x))\cos^4(\theta)+3\cos^2(\theta)\sin^2(\theta) \ge 0 $$

Now making $r^2 = u$ we have the differential equation

$$ \frac{1}{2}u'=u-\sigma(t)u^2 $$


$$ 0 \le \sigma(t) \lt 2 $$

This equation seems not to converge asymptotically to the origin.

I hope this helps.

Attached a stream plot near the origin with an orbit associated to $x(0) = 4, y(0) = 4$ in the original system.

enter image description here

  • $\begingroup$ Is your arguement for it not converging asymptotically because the function of $u$ does not have a fixed negative value? $\endgroup$ – Gragbow May 29 '18 at 21:40
  • $\begingroup$ If you solve the differential equation for any fixed $0 \le \sigma(t) < 2$ fixed value, $u(t)$ does not converges to the origin. $\endgroup$ – Cesareo May 29 '18 at 21:46
  • $\begingroup$ Hmmmmm. This seems correct definitely. But the thing is the question says "refer to any theorems which you know of". Now I know what you have done is correct and that's amazing but I was wondering if there was an easier way of doing this. That's why in my original post I talk about Liapounovs stability theorems. This is just a thought btw :D $\endgroup$ – Gragbow May 29 '18 at 21:49
  • $\begingroup$ I am sorry, I did not understand how you can conclude that the solutions to $u'=2u-2\sigma(t)u^2$ do not converge to $0$. Can you please say a couple more words on that? That would be great, thanks $\endgroup$ – Giuseppe Negro May 29 '18 at 21:53
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    $\begingroup$ The solution for $u'=u-\lambda u^2$ with $\lambda \ge 0$ is $u = \frac{e^t}{C_0+\lambda e^t}$ $\endgroup$ – Cesareo May 29 '18 at 22:11

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