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# Lyapunov stability for nonlinear systems

We discuss the Lyapunov stability for nonlinear systems. Indeed, having a reference solution, a stationary solution, one wonders if the ODE solution is closed to this reference when the time is very large.

Throughout this post, we suppose that $F:\Omega\to \mathbb{R}^d$ is a continuously differentiable function, $C^1(\Omega)$, where $\Omega$ is an open set of $\mathbb{R}^d$. Moreover, we consider the Cauchy problem \begin{align*}\tag{CP}\dot{u}(t)=F(u(t)),\quad t\ge 0,\quad u(0)=x_0.\end{align*}

According to Peano’s, see also Cauchy-Lipschitz theorem, the Cauchy problem $(CP)$ admits a maximal solution.

## The definition of exponentially stable equilibrium

We say that $x^\ast\in \Omega$ is an equilibrium, or critical, point of the application $F$ if it satisfies $F(x^\ast)=0$. Of course, a such point is not necessarily unique.

If $F$ is a linear transformation then one of the equilibrium points of $F$ is $0$.

If we define the constant function $u^\ast(t)=x^\ast$ for any $t$. Then we have $\dot{u}^\ast(t)=0=F(x^\ast)=F(u^\ast(t))$. This shows that $u^\ast$ is a solution of the differential equation $\dot{u}^\ast(t)=F(u^\ast(t))$. This solution is called the stationary solution, reference solution, or target solution.

Definition: We say that an equilibrium point $x^\ast\in \Omega$ is exponentially stable if There exist constants $omega, M>0$ and there exists $\delta>0$ such that if the initial state $u(0)=x_0$ satisfies $\|x_0-x^\ast\|\le \delta,$ then the maximal solution $u$ is golable, this means that $u(t)$ defined for any $t\in [0,+\infty)$, and we satisfies the following estimates \begin{align*}\|u(t)-x^\ast\|\le Me^{-\omega t}\|x_0\|,\qquad \forall t\ge 0.\end{align*}

This means that if the initial state $x_0$ is closed to the critical point, the solution is closed exponentially to this equilibrium.

## Lyapunov’s stability theorem on nonlinear systems

We notice that without losing generalities, we can assume that the point of equilibrium of $F$ is $x^\ast$. This is because, if $x^\ast$ is an equilibrium point, then by replacing $u(t)$ by $v(t)=u(t)-x^\ast$ and $F(x)$ by $G(x)=F(x+x^\ast),$ then we have $G(0)=0$.

Theorem: Let $F$ be a $C^1(\Omega)$ class function such that $F(0)=0$, $0$ is a critical point of $F$. We denote by $\sigma(F'(0))$ the spectrum of the matrix $F'(0)$, the set of eigenvalues. We assume that the spectral bound \begin{align*} s(F'(0)):=\sup{{\rm Re}\lambda: \lambda\in \sigma(F'(0))} < 0. \end{align*} Then $0$ is exponentially stable.

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