Theorem. Consider a continuous map $f : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n}$ and suppose that the autonomous dynamical system $\dot{x} = f(x)$ has a semiflow $\varphi : {\mathbb{R}}_{\geq{0}} \times {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n}$. Let $K \subseteq {\mathbb{R}}^{n}$. If $K$ is nonempty, compact, convex and forward-invariant, then $K$ contains an equilibrium of the dynamical system, i.e. a zero of the map $f$.

According to a reliable source, the above theorem is a standard result everyone uses in dynamical systems without proof. I propose a proof in "Equilibria Exist in Compact Convex Forward-Invariant Sets" at I am interested in comments on this proof, in references to this or other proofs in the literature, and in new/better proofs.

  • 1
    Shouldn't you require $f$ to be more than continuous (e.g. Lipschitz)? Currently $f$ doesn't (uniquely) define a (semi-)flow, for example when $f(x) = \sqrt{x}$ in a neighborhood of $x \ge 0$. – Jaap Eldering Jun 21 '11 at 13:33
  • Thanks Jaap, for catching and illustrating this insufficiency. I changed the statement. Now I explicitly require the existence of the semiflow. In my intended application, the map $f$ is a polynomial describing the kinetics of a chemical reaction network and time runs from zero to infinity. So I believe it would be too strong to require (global) Lipschitz continuity and too weak to require local Lipschitz continuity. Thanks again. – Gilles Gnacadja Jun 22 '11 at 1:37
  • A colleague showed me an article that essentially has the result: "The Brouwer Fixed Point Theorem Applied to Rumour Transmission", The article is dated 2005/2006. There have to be earlier references. – Gilles Gnacadja Aug 11 '12 at 22:48

This article from 1962 gives an earlier proof: "Axiomatic treatment of chemical reaction systems",

  • 1
    Thank you, user48647, for this note. For others who might be interested, the specific part of interest in the paper is Part II, Section B (page 1579). – Gilles Gnacadja Mar 25 '14 at 3:28

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.