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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