Counterexample to the Markus–Yamabe 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 $Df(x)$ be Hurwitz-stable for every $x \in \R^n$, that is, all its eigenvalues have negative real part. Then the equilibrium point $0$ 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.