Say that an autonomous system $\dot{u} = f(u)$ in $\mathbb{R}^{m}$ has the property that for any two solutions $x(t), y(t)$ corresponding to initial conditions $x(0)$ and $y(0)$ the trajectories are converging towards ech other:

$d(x(t),y(t)) \leq e^{-\alpha t}d(x(0),y(0))$

for some $\alpha > 0$. Must there be an equilibrium point of this system?

In the discrete time setting this is what the contraction mapping theorem is saying: If we define the trajectories of the system by $x_{n+1} = T^{n+1}(x)$, then the condition $d(x_{n},y_{n}) \leq e^{-\alpha n}d(x,y)$ is sufficient to prove the existence of (and convergence to) an equillibrium point. I am just wondering when this criteria can be carried over to the continuous setting.

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    $\begingroup$ You can consider the time-1 map of the ODE. This map will be a contraction map in $\mathbb R^m$. $\endgroup$ Jan 7, 2014 at 21:35

1 Answer 1


As observed by P. Grover, the flow of the system $g^t:\mathbb{R}^m\to\mathbb{R}^m$ is a contraction for any $t>0$. In particular, for any $k\in\mathbb{N}_ +$ there is a fixed point of $g^{1/k}$, which is, of course, a fixed point of the $k$ th compositional iterate, $g^1$. By uniqueness, these fixed points coincides, and are therefore a fixed point of the system (rmk: this is a general argument to derive continuous versions of some results about discrete DS. An example is the Grossman-Hartman Thm).

Also note that the weaker requirement $\|x(t)-y(t)\|=o(1)$ as $t\to+\infty$ for any pair of slutions, would not guarantee the result (for instance: $\dot u=\operatorname{sech} (u)$ in $\mathbb{R}^1$ has no equilibrium point, though all solutions are $\operatorname {arcsinh}(t-c)$ and their distance is infinitesimal for $|t|\to+\infty$).


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