---
tag: 8452SY027Z
kind: counterexample
name: Counterexample to the Markus–Yamabe conjecture
msc: [34D23, 34D20]
uses: [0N8NGKDZDZ, 047NDTEPCV, 0BQAA2BK3Q, 0NB689782W, 9A3TBBHRFA]
sources:
  - url: https://en.wikipedia.org/wiki/Markus%E2%80%93Yamabe_conjecture
---

## Counterexample

The Markus–Yamabe conjecture asked the following.
Let $f \colon \R^n \to \R^n$ be $C^1$ with $f(0) = 0$.
Let the [[Jacobian matrix|047NDTEPCV]] $Df(x)$ be [[Hurwitz-stable|9A3TBBHRFA]] for every $x \in \R^n$, that is, all its [[eigenvalues|0BQAA2BK3Q]] have negative real part.
Then the [[equilibrium point|0N8NGKDZDZ]] $0$ of $\dot x = f(x)$ is globally asymptotically stable.

It fails for $n = 3$.
Let $f \colon \R^3 \to \R^3$ be
$$f(x, y, z) = \bigl(-x + z (x + y z)^2,\ -y - (x + y z)^2,\ -z\bigr).$$
Write $w = x + y z$.
The Jacobian matrix is
$$Df = \begin{pmatrix} -1 + 2 z w & 2 z^2 w & w^2 + 2 y z w \\ -2 w & -1 - 2 z w & -2 y w \\ 0 & 0 & -1 \end{pmatrix}.$$
The last row shows that $-1$ is an eigenvalue, and the other two are the eigenvalues of the upper left $2 \times 2$ block.
That block has trace $-2$ and determinant $(-1 + 2 z w)(-1 - 2 z w) + 4 z^2 w^2 = 1$.
So its characteristic polynomial is $(\lambda + 1)^2$, and all eigenvalues of $Df(x, y, z)$ equal $-1$ at every point.
Also $f(0, 0, 0) = 0$.

But the curve $x(t) = 18 e^t$, $y(t) = -12 e^{2t}$, $z(t) = e^{-t}$ is a solution.
Indeed $w = 18 e^t - 12 e^t = 6 e^t$, so the first component of $f$ is $-18 e^t + 36 e^t = 18 e^t = \dot x$, the second is $12 e^{2t} - 36 e^{2t} = -24 e^{2t} = \dot y$, and the third is $-e^{-t} = \dot z$.
This solution is unbounded, so $0$ is not globally asymptotically stable.

## Remarks

The example is due to Cima, van den Essen, Gasull, Hubbers and Mañosas (1997).
Adding the equations $\dot u_i = -u_i$ for extra coordinates gives counterexamples in every dimension $n \ge 3$.
For $n = 1$ the conjecture is easy, and for $n = 2$ it is a theorem, proved in the 1990s.
The condition at every point implies [[asymptotic stability|0NB689782W]] of $0$, by linearisation; only the global part fails.

## Explanations

### L1 human

If a system pulls inward at every single point when you look very closely, one might expect every orbit to fall to the rest point.
This example shows that in three dimensions it is false.
Each local picture is a contraction, but the nonlinear terms carry one orbit off to infinity.