This tiny post is about stability in the sense of Lyapunov of stationary points of autonomous ordinary differential equation (ODE) in $\mathbb{R}^n$ given by $x'(t)=f(x(t)),\quad x(0)=x_0$ where $f:O\subset\mathbb{R}^n\to\mathbb{R}^n$ is a $\mathcal{C}^1$ vector field. This includes two important examples:

• Gradient flows : $f=-\nabla\mathcal{E}$ where $\mathcal{E}:O\to\mathbb{R}$ is $\mathcal{C}^2$
Geometrically, the system descents down the levelsets of $\mathcal{E}$
• Hamiltonian systems : $f=\Omega\nabla H$ with $\Omega$ skew-symmetric and $H:O\to\mathbb{R}$ is $\mathcal{C}^2$
Geometrically, the system remains in the levelset of $H$ where it started

The $\mathcal{C}^1$ regularity of $f$ ensures, via the Cauchy-Lipschitz or Picard–Lindelöf theorem, the local existence and uniqueness of solutions for all $x_0\in O$. As a consequence, the solution that takes the value $x$ at initial time $0$, denoted $t\mapsto\varphi_t(x)$, is uniquely defined on a maximal time interval ending up at $T_\max(x)$, and blows as $t\to T_\max(x)$ if $T_\max(x)<\infty$. We say that $\varphi_t:O\to O$ is the flow of the ODE. The phase space $O$ is partitioned into disjoint integral curves associated to the solutions. We are interested in the long time behavior of solutions near stationary points.

Stability of stationary points. Suppose that $x_0$ is stationary : $f(x_0)=0$. Then $x(t)=x_0$ for all $t$. We say that this stationary point is stable in the sense of Lyapunov when for all $\varepsilon>0$, there exists $\eta>0$ such that for all $x\in\overline{B}(x_0,\eta)$,
$T_\max(x)=+\infty \quad\text{and}\quad \|\varphi_t(x)-x_0\|\leq\varepsilon\text{ for all t\geq0},$ the solution remains as close to $x_0$ as we want and for all time provided that we start close enough. We say that the stationary point $x_0$ is unstable otherwise. Moreover, we say that a stable stationary point $x_0$ is asymptotically stable if for some $\eta_0>0$ and all $x\in\overline{B}(x_0,\eta_0)$,
$\lim_{t\to\infty}\varphi_t(x)=x_0.$ Furthermore, we say that it is exponentially stable if there exist positive constants $\alpha,c,C>0$ such that forall $x\in\overline{B}(x_0,\eta_0)$ and all $t\geq0$,
$\|\varphi_t(x)-x_0\|\leq C\mathrm{e}^{-\alpha t}\|x-x_0\|.$

Lyapunov function. It is a $\mathcal{C}^2$ function $\Phi:O\to\mathbb{R}$ such that for all $x\in O$,
$\langle\nabla \Phi(x),f(x)\rangle\leq0.$ This is equivalent to say that $t\mapsto\Phi(\varphi_t(x))$ is non-increasing for all $x$, since $\varphi_0(x)=x$ and
$\partial_t\Phi(\varphi_t(x))=\langle\nabla\Phi(\varphi_t(x),\varphi_t'(x)\rangle =\langle\nabla\Phi(\varphi_t(x)),f(\varphi_t(x))\rangle\leq0.$ For a gradient flow $f=-\nabla\Phi$, the function $\Phi$ is a Lyapunov function and in this case
$\partial_t\Phi(\varphi_t(x))=-\|\nabla\Phi(\varphi_t(x))\|^2.$

Prime integral. It is a $\mathcal{C}^2$ function $H:O\to\mathbb{R}$ such that for all $x\in O$,
$\langle\nabla H(x),f(x)\rangle=0.$ This is equivalent to say that both $H$ and $-H$ are Lyapunov functions, and also to say that $t\mapsto H(\varphi_t(x))$ is constant for all $t$ and $x$. For a Hamiltonian system $f=\Omega\nabla H$, the Hamiltonian $H$ is a prime integral, indeed thanks to the skew-symmetry of $\Omega$,
$\partial_t H(\varphi_t(x)) =\langle\nabla H(\varphi_t(x)),f(\varphi_t(x))\rangle =\langle\nabla H(\varphi_t(x),\Omega\nabla H(\varphi_t(x))\rangle =0.$

Criterion for stability.. If $x_0$ is a strict local minimum of a Lyapunov function $\Phi$ then $x_0$ is stationary and stable. Actually the proof shows that we only need to know that $\Phi$ is locally a Lyapunov function, namely Lyapunov on a small enough ball centered at $x_0$ included in $O$, this modification of the open set on which we define the ODE has potentially an impact on $T_\max$, but at the end, we conclude that $T_\max=+\infty$ and it does not matter.

Proof. We use a reasoning by continuity. Since $x_0$ is a strict minimum, for $\varepsilon>0$ small enough, $\overline{B}(x_0,\varepsilon)\subset O$ and $\Phi(x_0)<\Phi(x)$ for all $x\in\overline{B}(x_0,\varepsilon)\setminus\{x_0\}$. Since the sphere $\overline{S}(x_0,\varepsilon)$ is compact because the dimension is finie, and $\Phi$ is continuous, there exists $x_1\in S(x_0,\varepsilon)$ such that $m:=\min_{S(x_0,\varepsilon)}\Phi=\Phi(x_1)$, and since $\Phi(x_1)>\Phi(x_0)$, we get $m>\Phi(x_0)$. On the other hand, since $\Phi$ is continuous at $x_0$ and $\Phi(x_0)< m$, there exists $0<\eta<\varepsilon$ such that the maximum $M$ of $\Phi$ on $\overline{B}(x_0,\eta)$ is $< m$. Now, for all $x\in\overline{B}(x_0,\eta)$, since $\varphi_0(x)=x$ and $t\mapsto\Phi(\varphi_t(x))$ is non-increasing because $\Phi$ is a Lyapunov function, the function $\Phi$ remains $\leq M< m$ along the trajectory, hence the trajectory cannot reach the sphere $S(x_0,\varepsilon)$, thus it remains in $B(x_0,\varepsilon)$. Therefore $x_0$ is stable. We have proved stability before stationarity. To show that it is stationary, it suffices to linearize the flow as $\varphi_t(x)-x_0=tf(x)+o_{t\to0}(t)$ and to take $\eta=t\eta'$. Finally, since the trajectory is bounded and the dimension is finite, we have necessarily $T_{\max}(x)=+\infty$.

Criterion for being exponentially stable.. If $x_0$ is stationary and
$\max\Re\mathrm{spec}(\mathrm{Jac}(f)(x_0))<0,$then $x_0$ is exponentially stable. This is a spectral criterion by linearization of the ODE.

Proof. Let $A:=\mathrm{Jac}(f)(x_0)$ be the Jacobian matrix of $f$ at $x_0$, and its Jordan reduction
$A=P^{-1}(D+E)P, \quad D:=\mathrm{Diag}(\lambda_1,\ldots,\lambda_n),\quad P\in\mathrm{GL}_n(\mathbb{C}),\quad E\in\mathcal{M}_n(\mathbb{R}).$ The vector $x_0$ (and $x$ below) and the matrix $A$ are real but $P$ and $D$ are complex ($E$ is real and $DE=ED$ but we do not use that). We can always assume that $\|E\|$ is arbitrary small since the Jordan blocs can be conjugated by diagonal matrices, at the price of making $\|P\|$ bigger:
$\begin{bmatrix} \lambda & 1 & 0 & \cdots & 0 \\ 0 & \lambda & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & 0 & \lambda \end{bmatrix} = \mathrm{Diag}(1,\varepsilon,\varepsilon^2,\ldots) \begin{bmatrix} \lambda & \varepsilon & 0 & \cdots & 0 \\ 0 & \lambda & \varepsilon & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \lambda & \varepsilon \\ 0 & 0 & 0 & 0 & \lambda \end{bmatrix} \mathrm{Diag}(1,\varepsilon^{-1},\varepsilon^{-2},\ldots).$ We have $\alpha:=\max\Re\mathrm{spec}(\mathrm{Jac}(f)(x_0))=\max_{1\leq k\leq n}\Re\lambda_k<0$. Now
$\Phi(x):=\|P(x-x_0)\|^2=\langle \overline{P}^\top P(x-x_0),x-x_0\rangle,$ hence, using the fact that $x-x_0$ is real for the first equality,
$\nabla\Phi(x)=2\Re(\overline{P}^\top P)(x-x_0) \quad\text{and in particular}\quad \left\|\nabla\Phi(x)\right\|=O(\|x-x_0\|).$ On the other hand, $f(x_0)=0$ since $x_0$ is stationary and thus
$f(x)=A(x-x_0)+o(\|x-x_0\|).$ We have then, using the fact that $A$ and $x-x_0$ are real for the first equality,
\begin{align*}
\langle\nabla\Phi(x),f(x)\rangle
&=2\Re\langle\overline{P}^\top P(x-x_0),A(x-x_0)\rangle+o(\|x-x_0\|^2)\\
&=2\Re\langle P(x-x_0),(D+E)P(x-x_0)\rangle+o(\|x-x_0\|^2)\\
&=2\Re\langle z,Dz\rangle+2\Re\langle z,Ez\rangle+o(\|x-x_0\|^2)
\end{align*} where $z:=P(x-x_0)$. But now by the Cauchy–Schwarz inequality
$2\Re\langle z,Ez\rangle\leq2\|E\|\|z\|^2,$ while
$\langle z,Dz\rangle =\langle z,(\Re D)z\rangle-\mathrm{i}\langle z,(\Im D)z\rangle =\sum_k|z_k|^2\Re D_{k,k}-\mathrm{i}\sum_k|z_k|^2\Im D_{k,k}$ gives
$2\Re\langle z,Dz\rangle\leq-2\alpha\|z\|^2.$ Thus, for $\beta<2\alpha$, $\|E\|$ small enough, and $x\in\overline{B}(x_0,\eta_0)$, with $\eta_0>0$ small enough, we get
\begin{align*}
\langle\nabla\Phi(x),f(x)\rangle
&\leq-2\alpha\|P(x-x_0)\|^2+2\|E\|\|P(x-x_0)\|^2+o(\|x-x_0\|^2)\\
&\leq-\beta\|P(x-x_0)\|^2=-\beta\Phi(x).
\end{align*} In particular $\Phi$ is a Lyapunov function, on the restricted set $O_0:=B(x_0,\eta_0)\subset O$, with a strict minimum at $x_0$ because $P$ is invertible, hence $x_0$ is stable. Moreover $\partial_t\Phi(\varphi_t(x))\leq-\beta\Phi(\varphi_t(x))$ for $x\in\overline{B}(x_0,\eta_0)$, and the Grönwall lemma below gives
$\Phi(\varphi_t(x))\leq\Phi(\varphi_0(x))\mathrm{e}^{-\beta t}=\Phi(x)\mathrm{e}^{-\beta t}.$ Next, since $P$ is invertible, there exists a positive constant $C>0$ such that
$C^{-1}\|x-x_0\|^2\leq\Phi(x)\leq C\|x-x_0\|^2,$ hence for $x\in\overline{B}(x_0,\eta_0)$ and all $t\geq0$,
$\|\varphi_t(x)-x_0\|\leq C^2\mathrm{e}^{-\tfrac{\beta}{2} t}\|x-x_0\|$ which means that $x_0$ is exponentially stable.

Grönwall lemma.. If $u:[a,b]\to\mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and if
$u'(t)\leq\alpha(t)u(t)$ for all $t\in(a,b)$, with $\alpha:[a,b]\to\mathbb{R}$ continuous, then for all $t\in[a,b]$,
$u(t)\leq u(a)\exp\bigr(\int_a^t\alpha(s)\mathrm{d}s\bigr).$ The equality case $u'(t)=\alpha(t)u(t)$ is a one-dimensional non-autonomous linear ODE. To see it, we observe that the solution of the equality case is positive : $v(t):=\mathrm{e}^{\int_a^t\alpha(s)\mathrm{d}s}>0$ for all $t\in[a,b]$, and $v'(t)=\alpha(t)v(t)$ for all $t\in(a,b)$, while $v(a)=1$, hence
$\Bigr(\frac{u(t)}{v(t)}\Bigr)’ =\frac{u'(t)v(t)-u(t)v'(t)}{v(t)^2} =\frac{(u'(t)-\alpha(t)u(t))v(t)}{v(t)^2} \leq0$ for all $t\in(a,b)$, therefore $u(t)/v(t)\leq u(a)/v(a)=u(a)$ for all $t\in[a,b]$.

It seems that it was already known to Alexandre Lyapunov (1857 – 1918) as well as to Henri Poincaré (1854 – 1912) that the solution of a linear non-autonomous ODE $x'(t)=A(t)x(t)$ in dimension $n\geq2$ can perfectly diverge as $t\to\infty$ while $A(t)$ has all eigenvalues $<0$ for all $t$.

Note. This tiny post is taken from a course on topology and differential calculus, that I have the pleasure of teaching, succeeding to my former colleague Dmitry Chelkak.