In many applications, it is highly desirable to know when the solution $x(t)$ to an ODE $\dot x = f(x)$, is uniformly continuous.

However, there seems to be very few words written on conditions on $f$ when this is true.

In general, the uniform continuity of $x(t)$ is given by the Caratheodory existence theorem (https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_existence_theorem). But this is a very general theorem that works for discontinuous functions $f$, and the conditions are a bit difficult to check. In any case, the existence theorem provides conditions on $f$ such that $x(t)$ is absolutely continuous, hence uniformly continuous. Is there a way to conclude uniform continuity without resorting to absolute continuity?

Can we conclude uniform continuity of the solution directly from Cauchy-Lipschitz or Peano's existence theorem?

What are some "nice" conditions on $f$ that allows us to immediately conclude $x(t)$ is uniformly continuous? (i.e. suppose $f$ is Lipschitz, then is $x(t)$ u.c.?)

  • $\begingroup$ What interval do you want solutions on? Any continuous function on a compact set is uniformly continuous. Alternatively, if $f$ is bounded, then $x$ is Lipschitz. $\endgroup$
    – Jason
    Jun 14 '17 at 18:17
  • $\begingroup$ @Jason the maximal interval of existence. So you are saying that if $f$ is Lipschitz, then it is bounded, then $x$ is Lipschitz (by Cauchy-Lipschitz?) and hence uniformly continuous? Sorry I'm not quite following the Lipschitz implies uniformly continuous part. $\endgroup$
    – Olórin
    Jun 14 '17 at 18:19
  • $\begingroup$ Lipschitz functions need not be bounded, take $f(x)=x$ for a simple example. I think boundedness is probably the best you can do. Indeed, the example I gave was not just Lipschitz, but cotinuously differentiable, smooth, even analytic, yet will not lead to a uniformly continuous solution. $\endgroup$
    – Jason
    Jun 14 '17 at 18:39

I will only consider an initial value problem on $[0,\infty)$ for simplicity. Also let's assume that $f$ is continuous and defined on all $\mathbb{R}^n$.

The solution is obviously uniformly continuous on any closed bounded interval on which it exists.

Let $I$ be any interval containing $0$ on which the solution is uniformly continuous, then it may be extended to the closure of the interval and then the solution can be extended to a strictly larger interval.

If the right hand side $f(x(t))$ is uniformly bounded for all $t$ in the maximal interval of existence, then the solution exists for all $t > 0$ and is uniformly continuous there.

But if the solution exists for all $t>0$, it may or may not be uniformly continuous on this set (example $x' = x$).

In short, uniform continuity of the solution on its maximal interval of existence implies global existence but not the other way around. And uniform continuity of the solution on its maximal interval of existence is essentially equivalent to uniform boundedness of the solution.

  • 1
    $\begingroup$ $x'=1$ gives $x(t)=t+x(0)$, which is uniformly continuous but not uniformly bounded. $\endgroup$
    – user254433
    Jul 24 '17 at 23:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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