1
$\begingroup$

I am looking for a general theorem that guarantees the existence of a global solution for an ODE system in $\mathbb{R}^n$

\begin{equation} \left\{ \begin{aligned} x'(t) &= f(t, x(t)), \qquad t \in [a,b] \\ x(a) &= x_0 \end{aligned} \right. \end{equation}

By "global" I mean that the time interval is fixed, i.e. $[a,b]$, but I am not asking the solution to stay in an a priori fixed compact set of $\mathbb{R}^n$ (though the final solution will be absolutely continuous and thus bounded). The setting is that of a possibly discontinuous vector field, described by the Carathéodory conditions, that is

  1. $x \mapsto f(t,x)$ is continuous for a.e. $t$
  2. $t \mapsto f(t,x)$ is measurable for each $x$
  3. $|f(t,x)| \leq m(t)$, $m(t)$ being summable

A classical Carathéodory existence theorem (see e.g. Filippov, "Differential Equations with Discontinuous Right-Hand Side" (1988)) gives a local existence result in a compact set $K \subset \mathbb{R}^n$ under the above Charathéodory conditions.

Another classical Carathéodory theorem gives instead the global existence and uniqueness under a further Lipschitz continuity assumption:

  1. $|f(t,x)-f(t,y)| \leq L(t) |x-y|$, $L(t)$ being summable

Finally, I found a global existence theorem (see Theorem II.3.2 on Reid, "Ordinary Differential Equations" (1971)), under the assumption

  1. $|f(t,x)| \leq M(t)(1+|x|)$, $M(t)$ being summable

This last result require the vector field to have an at most linear growth in the variable $x$. I was wondering if anyone knows more general results for the existence of a global solutions, which can include also more than linear growth, or if the results I quoted are already the best I can get.

Thank you!

C/p from Math.StackExchange.

$\endgroup$
3
  • 1
    $\begingroup$ Just look at the scalar case $\dot{x} = x^{1+\epsilon}$ with positive initial data: If you allow $f$ to grow superlinearly in $x$, you have finite time blow-up solutions. By making your initial data large you can make the blow-up arbitrarily fast. And so summability of some $M(t)$ will not help you there. $\endgroup$ Mar 27, 2021 at 0:39
  • $\begingroup$ OTOH, if you are willing to put in smalleness assumptions on the initial data, then there can be a competition. $\endgroup$ Mar 27, 2021 at 0:40
  • $\begingroup$ Thank you. What about the smallness assumptions on the initial data? Could you provide some examples? $\endgroup$
    – A. Pesare
    Mar 27, 2021 at 11:46

1 Answer 1

3
$\begingroup$

(N.B. In the below I assume $[a,b] = [0,\infty]$, but the precise values don't matter and appropriate substitutions of $a,b$ into the discussion also gives you the same conclusion.)

Once you have a local existence theorem of the form

For every compact set $K$ and compact subset $K_0 \Subset K$, there exists $T > 0$ such that for every $x_0\in K_0$ there exists a solution to the IVP with initial value $x_0$ defined on $[0,T]$ such that the trajectory remains in $K$ for all $T$

Then upgrading to global existence is simply an issue of proving that any solution cannot escape to infinity in finite time. This is evidenced by the 5th theorem you cited: if you have $$ |x'| \leq M(t) (1+|x|) $$ then you have $$ \frac{|x|'}{1+|x|} \leq M(t) $$ and integrating both sides you find $$ \ln (1+|x|) \Big|_{t = 0}^{t= T} \leq C $$ using that $M(t)$ is summable.

Slight refinement.

Let's abstract the argument a bit. Assume you have a bound on $f$, such that you can write

$$ |x'| \leq M(t) g(|x|) $$

for some given function $g:\mathbb{R}\to \mathbb{R}$ that is locally integrable. Denote by $G$ any primitive of the function $1/g$. Then integrating this differential inequality for a solution guarantees

$$ G(|x|) \Big|_{t = 0}^{t = T} \leq C$$

provided that $M(t)$ is summable. And so as long as $G$ is coercive (in other words, it is proper) then the same argument as above will guarantee that $|x|$ does not blow-up in finite time and hence you have global existence.

You can therefore do slightly better then $(1+|x|)$, since there are superlinear $g$s for which $1/g$ is not integrable on the entire real line. For example, you may wish to take

$$ g(|x|) = (1 + |x|) \ln (1+|x|) $$

Then you have a primitive

$$ G(s) = \ln \ln (1+s) $$

which is still proper. In fact you can extend this using the usual tower of natural logs.

Negative examples

But you can certainly not go beyond logarithmic improvements, at least in general. You can see this easily by considering scalar equations of the form

$$ x' = M(t) (1 + |x|)^{1+\epsilon} $$

where $M$ is a positive summable function.

Note that with this assumption the solution is increasing, and so positive initial data will lead to positive solutions. And so if $x(0) = x_0$ is positive, the solution satisfies

$$ \frac{x'}{(1+x)^{1+\epsilon}} = M(t) $$

which you can integrate to find

$$ \frac{1}{(1+x(T))^\epsilon} = \frac{1}{(1+x_0)^\epsilon} - \epsilon \int_0^T M(t) ~dt $$

And so with $M(t)$ considered fixed, for every $T$ there exists some $R$ such that if $x_0 > R$ then $x$ must blow up prior to $t = T$.

Small data regime

The previous discussion however hints at a small-data version of the result. Those that for the previous scalar equation, if

$$ (1 + x_0)^\epsilon \leq \frac{1}{\epsilon \int_0^\infty M(t) ~dt} $$

then we guarantee that $(1 + x(t))^{-1}$ is bounded away from infinity and cannot blow-up in finite time. And hence we have a statement of the form:

There exists a compact set $K_0$ with non-empty interior such that for all $x_0 \in K_0$ there exists a (future-in-time) global solution.

In the general form we gave earlier: let's assume without loss of generality that $g(|x|)$ is positive, and hence $G(s)$ is either unbounded (in which case you have that it is proper and global existence for all data) or that $\lim_{s\to \infty} G(s) = G_\infty$ exists.

In the latter case, let $C = \int_0^\infty |M(t)|$. Then provided the set $$ \{ r : G(r) < G_\infty - C \} $$ is non-empty, any initial data $x_0$ with $|x_0|$ in this set will generate a solution that exists globally.

Additional Structures

Note also that you may have a situation where you have additional structures. For example, it is possible that the best possible absolute value bound $$ |f(t,x)| \leq M(t) (1+|x|)^\kappa $$ has $\kappa > 1$, but still you have global existence for all initial data. This would be the case if you happen to have $$ x\cdot f(t,x) \leq 0 $$ in which case $|x|^2$ is (weakly) a Lyapunov function.

$\endgroup$
1
  • $\begingroup$ Very useful, thanks a lot! $\endgroup$
    – A. Pesare
    Apr 26, 2021 at 11:05

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.