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  "body": "\n## Counterexample\n\nThe Markus–Yamabe conjecture asked the following.\nLet $f \\colon \\R^n \\to \\R^n$ be $C^1$ with $f(0) = 0$.\nLet 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.\nThen the [[equilibrium point|0N8NGKDZDZ]] $0$ of $\\dot x = f(x)$ is globally asymptotically stable.\n\nIt fails for $n = 3$.\nLet $f \\colon \\R^3 \\to \\R^3$ be\n$$f(x, y, z) = \\bigl(-x + z (x + y z)^2,\\ -y - (x + y z)^2,\\ -z\\bigr).$$\nWrite $w = x + y z$.\nThe Jacobian matrix is\n$$Df = \\begin{pmatrix} -1 + 2 z w \u0026 2 z^2 w \u0026 w^2 + 2 y z w \\\\ -2 w \u0026 -1 - 2 z w \u0026 -2 y w \\\\ 0 \u0026 0 \u0026 -1 \\end{pmatrix}.$$\nThe last row shows that $-1$ is an eigenvalue, and the other two are the eigenvalues of the upper left $2 \\times 2$ block.\nThat block has trace $-2$ and determinant $(-1 + 2 z w)(-1 - 2 z w) + 4 z^2 w^2 = 1$.\nSo its characteristic polynomial is $(\\lambda + 1)^2$, and all eigenvalues of $Df(x, y, z)$ equal $-1$ at every point.\nAlso $f(0, 0, 0) = 0$.\n\nBut the curve $x(t) = 18 e^t$, $y(t) = -12 e^{2t}$, $z(t) = e^{-t}$ is a solution.\nIndeed $w = 18 e^t - 12 e^t = 6 e^t$, 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 $-e^{-t} = \\dot z$.\nThis solution is unbounded, so $0$ is not globally asymptotically stable.\n\n## Remarks\n\nThe example is due to Cima, van den Essen, Gasull, Hubbers and Mañosas (1997).\nAdding the equations $\\dot u_i = -u_i$ for extra coordinates gives counterexamples in every dimension $n \\ge 3$.\nFor $n = 1$ the conjecture is easy, and for $n = 2$ it is a theorem, proved in the 1990s.\nThe condition at every point implies [[asymptotic stability|0NB689782W]] of $0$, by linearisation; only the global part fails.\n\n## Explanations\n\n### L1 human\n\nIf a system pulls inward at every single point when you look very closely, one might expect every orbit to fall to the rest point.\nThis example shows that in three dimensions it is false.\nEach local picture is a contraction, but the nonlinear terms carry one orbit off to infinity.\n",
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