# Does stability of equilibrium point preserved by permutation matrix (symmetry)?

Given the following differential equations:

\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}

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{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}

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.

• 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. Aug 5 '18 at 10:14
• @AnthonyQuas Kindly ask: The eigenvalues you mention is the eigenvalue of the matrix after linearization? Also what do you mean "symmetry conjugated"? Thanks! Aug 5 '18 at 10:28

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.