Given the following differential equations:

\begin{equation} \begin{aligned} \dot{x}_1 &= f_1(x_1,\ldots,x_n) \\ \vdots \\ \dot{x}_n &= f_n(x_1,\ldots,x_n) \end{aligned} \end{equation}

In a compact way: $$\dot{\hat{x}} = \hat{F}(\hat{x})$$

Let the group $\Psi\subset S_n$, where $S_n$ is a symmetric group.

Suppose the following key property holds: $$\hat{F}(P_\sigma \hat{x})=P_\sigma \hat{F}(\hat{x}), \ \ \ \ \ \forall \sigma\in\Psi$$ where $P_\sigma$ is the permutation matrix corresponding to $\sigma\in \Psi$.

For example: \begin{equation} \begin{aligned} \dot{x}_1 &= x_1(x_1-1)(x_1+1) +x_2 \\ \dot{x}_2 &= x_2(x_2-1)(x_2+1) +x_1 + x_3 \\ \dot{x}_3 &= x_3(x_3-1)(x_3+1) +x_2\end{aligned} \end{equation}

In this case, we can choose $\sigma = (13)\in \Psi = \{(),(13)\}$, and the corresponding permutation matrix: $$P_\sigma = \begin{bmatrix}0 & 0 & 1\\0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$ It is easy to see that $\hat{F}(P_\sigma \hat{x})=P_\sigma \hat{F}(\hat{x})$

My question:

If $\hat{x}^*$ is a stable equilibrium point for $\dot{\hat{x}} = \hat{F}(\hat{x})$, can I say $P_\sigma \hat{x}^*$ is also a stable equilibrium point, for all $\sigma\in \Psi$?

Note: Easily to see that $P_\sigma \hat{x}^*$ is also an equilibrium point.

  • 1
    $\begingroup$ Yes. The linear map that implements the symmetry conjugated the Jacobean at $x$ to that at $P_\sigma(x)$, so the eigenvalues are the same. $\endgroup$ Aug 5, 2018 at 10:14
  • $\begingroup$ @AnthonyQuas Kindly ask: The eigenvalues you mention is the eigenvalue of the matrix after linearization? Also what do you mean "symmetry conjugated"? Thanks! $\endgroup$ Aug 5, 2018 at 10:28

1 Answer 1


I think that the question is more suitable for the Mathematics StackExchange, but I'll try to (reasonably fully) answer it here.

In order not to be burdened with unnecessary technicalities, let us assume that $F$ is so regular that Picard's theorem holds. Also, let $\lVert \cdot \rVert$ stand for the Euclidean metric.

Let $\varphi \colon (\alpha, \beta) \to \mathbb{R}^n$ be a nonextendible solution of $\dot{x} = F(x)$ (I deleted hats). That is, one has $\dot{\varphi}(t) = F(\varphi(t))$ for all $t \in (\alpha, \beta)$. We claim that, for any $\sigma \in \Psi$, the mapping $$ (\alpha, \beta) \ni t \mapsto P_{\sigma}\varphi(t) \in \mathbb{R}^n $$ is a solution. Indeed, one has $$ \dot{(P_{\sigma}\varphi)}(t) = P_{\sigma} \dot{\varphi}(t) = P_{\sigma} F(\varphi(t)) = F(P_{\sigma} \varphi(t)) \qquad \forall \, t \in (\alpha, \beta). $$ In the theory of continuous-time dynamical systems, a mapping taking (in our setting) the solutions of one system of ODEs onto the solutions of another system of ODEs is called conjugacy.

Recall the definition of (Lyapunov) stability: an equilibrium $x^*$ is stable if for each $\varepsilon > 0$ there exists $\delta > 0$ such that for any $\tilde{x} \in \mathbb{R}^n$, if $\lVert \tilde{x} - x^* \rVert < \delta$ then the solution $\varphi(\cdot;\tilde{x})$ of the IVP $\dot{x} = F(x)$, $x(0) = \tilde{x}$, is defined at least on $[0, \infty)$ and the inequality $$ \lVert \varphi(t;\tilde{x}) - x^* \lVert < \epsilon $$ holds for all $t \ge 0$.

We have $$ \lVert P_{\sigma} x - P_{\sigma} y \lVert = \lVert x - y \rVert $$ for any $\sigma \in \Psi$ and any $x, y \in \mathbb{R}^n$. This should be enough to conclude the proof.

A proof for asymptotic stability goes in an analogous way.


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