Prove that the origin is an unstable equilibrium point for the system $$\begin{align*}\dot{x}&=x^3+yx^2\\\dot{y}&=-y+x^2\end{align*}$$

I've already tried to linearize the system in order to use the Hartman–Grobman theorem, but the eigenvalues of the respective Jacobian matrix are $0$ and $-1$; non of them are positive, then Lyapunov's indirect method doesn't work either. Do you have any idea of another theorem that allow us to conclude this?

  • $\begingroup$ I also got the same eigenvalues as you for fixed points $(0,0)$- so what makes you think that the equation you have is unstable? do you have a phase portrait for this? $\endgroup$
    – Chinny84
    Jun 25, 2022 at 12:50
  • $\begingroup$ Did you try just looking at the phase portrait? You should be able to argue what happens to a solution curve which starts close to the origin in the region where $x<0$ and $x+y<0$. $\endgroup$ Jun 25, 2022 at 14:56
  • $\begingroup$ In a sense the system is separable. If and while $x,y$ are very small, the first equation tells us that $x$ is nearly constant, slowly changing. The second equation tells us that then $y$ moves exponentially in "normal" time spans towards the equilibrium $y=x^2$. On and around that equilibrium curve $\dot x\approx x^3$. As $x$ moves away from the equilibrium, its velocity becomes larger so that the exponential correction in $y$ is no longer fast enough to keep the solution close to the parabola. $\endgroup$ Jun 25, 2022 at 19:22

2 Answers 2


In order to determine that the origin of this dynamical system is unstable one can use Chetaev's instability theorem:

Let $V: D \to \mathbb{R}$ be a continuously differentiable function on a domain $D \subset \mathbb{R}^n$ such that $V(0)=0$ and $V(x_0)>0$ for some $x_0$ with arbitrarily small $\|x_0\|$. Choose $r > 0$ such that the ball $B_r = \left\{x \in \mathbb{R}^n\,|\,\|x\| \leq r\right\}$ is contained in $D$ and let $U = \left\{x \in B_r\,|\,V(x) > 0\right\}$ and suppose that $\dot{V}(x) > 0$ in $U$. Then, $x=0$ is unstable.

When evaluating the eigenvalues of the linearization at the origin the stable mode associated with eigenvalue of -1 is in the $y$-direction. Therefore, the following continuously differentiable function for in Chetaev's theorem is proposed

$$ V(x,y) = \frac12 (x^2 - y^2), \tag{1} $$

for which it can be shown that it has the following time derivative

$$ \dot{V}(x,y) = x^4 + (x - 1)\,y\,x^2 + y^2. \tag{2} $$

The set $\{z\in\mathbb{R}^2|V(z)>0\}$ using $(1)$ is equivalent to $|x|>|y|$, which does contain elements arbitrarily close to the origin.

When evaluating $(2)$ it can be shown that $\dot{V}(x,y)>0$ around the origin, for example the region $B_1$ (except the origin itself).

Combining all this yields that one can conclude that the origin of this dynamical system is unstable. Furthermore, it can also be stated that any trajectory that enters $U$ will leave $U$ at the boundary of $B_1$.

  • 1
    $\begingroup$ Well done! You managed to find the appropriate instability theorem! $\endgroup$
    – user1054388
    Jun 26, 2022 at 18:27

Determining the stable/unstable manifold $y=\phi(x)$ with $\phi(x) = \sum_{k=1}^n a_k x^k$

$$ \dot y = \phi'(x)\dot x $$

For $n = 3$ we obtain $y = \phi(x) = x^2$ hence $\dot x = x^3+\phi(x)x^2 = x^3+x^4$ which establishes an unstable flow at the origin.

In red, over the stream flow, the approximation $y = \phi(x)$

enter image description here


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