This may be impossible to answer, but I was curious as to general theorems that let one guarantee that an ODE system ($ f(x) \in \mathbb{R}^{n}$) \begin{align*} \dot{x} &= f(x) \end{align*}

has at least one equilibrium solution? For example, I think for planar systems, if the solution remains bounded, it must contain at least one equilibrium point (a corollary of Poincare Bendixson).

Can this be extended to higher dimensions? What about solutions in the positive orthant, say that are bounded; can we be sure it contains a positive equilibrium point? Are there any general theorems or techniques to attack such a question, based on the dynamics of the system.

  • $\begingroup$ You are asking about solutions to the non-linear system of equations $f(x)=0$. In general, a root-free region has index 0, so a non-zero index implies a root. The index is the generalization of the winding number, telling you how many unit spheres the normalized vector field covers. $\endgroup$ Nov 10, 2023 at 7:04
  • $\begingroup$ If the system has a compact convex forward-invariant set $K$, then $K$ contains an equilibrium, see here mathoverflow.net/questions/68174/… $\endgroup$
    – Gerd
    Nov 11, 2023 at 14:26

1 Answer 1


There is no such extension to higher dimensions: The following simple example in $\mathbb{R}^3$ has some bounded solutions but no equilibrium: $$ x_1'=-x_2, \quad x_2'=x_1 \quad x_3'=1-(x_1^2+x_2^2). $$ A bounded solution is $(x(t),y(t),z(t))=(\cos(t),\sin(t),0)$, for example. Of course this system also has unbounded solutions. The question if there is a dynamical system $x'=f(x)$ in $\mathbb{R}^3$ with no equilibrium but ${\bf all}$ solutions bounded is much harder, but answered by "yes" in a paper of Jones & Yorke: The Existence and Nonexistence of Critical Points in Bounded Flows. A way to show the existence of an equilibrium based on the dynamics of the system is to look for invariant compact convex sets:

Assume that $f:\mathbb{R}^n \to \mathbb{R}^n$ is locally Lipschitz and that there is a nonempty compact convex set $C \subseteq \mathbb{R}^n$ such that for each $c \in C$ the solution of $x'=f(x)$, $x(0)=c$ stays in $C$ (that is $x(t,c) \in C$ $(t\ge 0)$), then $C$ contains an equilibrium.

To outline a proof first note that $f$ is Lipschitz continuous on $C$ (since $C$ is compact), that is for some $L\ge 0$ we have $\|f(x)-f(y)\| \le L\|x-y\|$ $(x,y \in C)$. Choose $P > 0$ such that $LP < 6$ and consider the Poincaré map $g:C \to C$, $g(c)=x(P,c)$. Since $g$ is continuous it has a fixed point $c_0 \in C$ by Brouwers FPT. Hence $t \mapsto x(t,c_0)$ is a periodic solution of $x'=f(x)$ in $C$. By the Busenberg-Fisher-Martelli inequality we have $$ \int_0^P \int_0^P \|x(t,c_0)-x(s,c_0)\| dsdt \le\frac{P}{6} \int_0^P \int_0^P \|x'(t,c_0)-x'(s,c_0)\|dsdt $$ $$ \le \frac{LP}{6} \int_0^P \int_0^P \|x(t,c_0)-x(s,c_0)\|dsdt, $$ with $LP/6<1$ hence $$ \int_0^P \int_0^P \|x(t,c_0)-x(s,c_0)\| dsdt =0. $$ Thus $x(t,c_0)=c_0$ $(t \in \mathbb{R})$, which yields $f(c_0)=0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .