# Stability for Nonlinear System

I am trying to assess the (Liapunov) stability of the equilibrium at $(0,0)$ of the system \begin{align*} x_1' &= -4x_2 + x_1^2 \\ x_2' &= 4x_1 + x_2^2. \end{align*} I plotted the phase portrait in Mathematica, and it looks like a stable (but not asymptotically stable) equilibrium, with orbits circling about the origin. However, my professor tells me the origin is actually unstable. I don't know how to prove or disprove this rigorously. The equilibrium is non-hyperbolic, and I haven't been able to find a Liapunov function for the system. Any guidance would be appreciated.

Edit: I tried polar coordinates: I get $$r' = \frac1r(x_1x_1' + x_2x_2') = r^2(\cos^3 \theta + \sin^3 \theta)$$ and $$\theta' = \frac{x_1x_2'-x_1'x_2}{r^2} = 4 + r\cos\theta\sin\theta(\sin\theta-\cos\theta).$$ So there's a half-plane where $r$ is increasing and a half-plane where it's decreasing. The angle $\theta$ is increasing in a neighborhood of the origin. Is it reasonable to say that since $$0 = \int_{\theta_0}^{\theta_0 + 2\pi} \cos^3 \theta + \sin^3 \theta d\theta = \int_{\theta_0}^{\theta_0 + 2\pi} \frac{r'}{r^2} d\theta = \int_{\theta_0}^{\theta_0 + 2\pi} \left(\frac{1}{r}\right)' d\theta,$$ that the orbits are in fact periodic?

• I tried polar coordinates; I can find that $\theta'$ is positive and $r' = r^2(\sin^3 \theta + \cos^3 \theta)$, but I don't see that this helps much. – ec92 Oct 31 '13 at 17:59
• @ec92 Your last line of reasoning cannot be used, since the angle is changing with non-constant speed ($\dot \theta\neq 1$). – Artem Nov 1 '13 at 17:44
• @ec92 Any luck or some other hints from your instructor? I spent some time on this problem and have some reasons to believe that it is the center, but no definite proof. It would be nice to see it if you have any. – Artem Nov 7 '13 at 15:34
• @Artem You're correct; the professor was mistaken. I was able to use the symmetry of the system to prove this. I rotated my coordinates 45 degrees, setting $\tilde{x_1} = 1/\sqrt{2}(x_1 -x_2)$ and $\tilde{x_2} = 1/\sqrt{2}(x_1+x_2)$. The transformed system is invariant under $(t,\tilde{x_2}) \mapsto (-t,-\tilde{x_2})$, (or maybe the same transformation with $\tilde{x_1}$ in place of $\tilde{x_2}$; I don't remember for certain). There's a theorem (e.g. 2.10.6 in Perko's text) that says if such symmetry exists, the origin remains a center for the nonlinear system. – ec92 Nov 8 '13 at 22:18
• @ec92 Thanks. That's really cool, you should post your solution and accept it. I calculated first Lyapunov coefficient and found that it is zero. This is a very strong indication that the point is the center but not the proof. – Artem Nov 8 '13 at 22:22

## Dynamical system

% \begin{align} % \dot{x} &= -4y + x^{2} \\ % \dot{y} &= 4x + y^{2} % \end{align} %

The sole critical point is the origin.

## Phase portrait

The nullclines are the dashed lines. Red for $\dot{x} = 0$, purple for $\dot{y} = 0$.

## Analysis

Polar coordinate transformation produces $$\dot{r} = r^{2} \left( \cos^{3} \theta + \sin^{3} \theta \right)$$ The plot following shows $$f(\theta) = \cos^{3} \theta + \sin^{3} \theta$$

The red zone shows contraction where $\color{red}{\dot{r} < 0}$, and the blue zone dilation, $\color{blue}{\dot{r} > 0}$

Notice that as we approach the origin, the dynamical system % \begin{align} % \dot{x} &= -4y + x^{2} \qquad \Rightarrow \qquad -4y\\ % \dot{y} &= 4x + y^{2} \qquad \Rightarrow \qquad 4x % \end{align} % approaches a periodic system.

• How did you make these pictures? – lhf May 13 '17 at 1:12
• @lhf: Mathematica. Very simple. Happy to share code. Profile contains email address. – dantopa May 13 '17 at 1:31
• Thanks. Consider publishing the code somewhere, github for instace. – lhf May 13 '17 at 1:34

This is not really an answer to the question. But I think that it can be useful for answering it.

If we divide the differential equations we obtain:

$$\dfrac{dx_2}{dx_1}=\dfrac{4x_1+x_2^2}{-4x_2+x_1^2}$$

In order to simplify notation I will replace $x_1 \to t$ and $x_2 \to x$

$$\dfrac{dx}{dt}=\dfrac{4t+x^2}{-4x+t^2}$$.

This differential equation does have a symmetry, with an infinitesimal generator $$X = \dfrac{t-x-4}{t^2-4x}\partial_{x}.$$

By using this symmetry we can use the method of canonical coordinates to obtain the substitution:

$$t(r,s) = r$$ $$x(r,s)=r-4+\exp\left[W\left(-0.25(r^2-4r+16)\exp(0.25s(r)) \right)-0.25s(r) \right],$$

in which $W$ is the Lambert-$W$-Function.

Using this set of substitutions for the differential equation will result in

$$\dfrac{ds(r)}{dr}=1\implies s(r) = r+c_1=t+c_1$$

and

$$x(t)=t-4+\exp\left[W\left(-0.25(t^2-4t+16)\exp(0.25(t+c_1) \right)-0.25(t-c_1) \right].$$

Turning this back into the original coordinates we obtain:

$$x_2=x_1-4+\exp\left[W\left(-0.25(x_1^2-4x_1+16)\exp(0.25(x_1+c_1) \right)-0.25(x_1-c_1) \right]$$