Consider the $N$-dimensional autonomous system of ODEs $$\dot{x}= f(x),$$ where $f(x)$ is defined for any $x \in \mathbb{R}^N$, and satisfies $||f(x)|| \leq \alpha||x||$, where $\alpha$ is a positive scalar constant, and the norm $||x||$ is the usual quadratic norm (the sum of squared components of a vector under the square root). Using Gronwall’s inequality, show that the solution emerging from any point $x_0\in\mathbb{R}^N$ exists for any finite time.

Here is my proposed solution.

We can first write $f(x)$ as an integral equation,

$$x(t) = x_0 + \int_{t_0}^{t} f(x(s)) ds$$

where the integration constant is chosen such that $x(t_0)=x_0$. WLOG, assume that $t_0=0$. Then,

\begin{equation} \begin{split} ||x(t)|| & = ||x_0 + \int_{0}^{t} f(x(s)) ds|| \\ & \leq ||x_0|| + ||\int_{0}^{t} f(x(s)) ds|| \\ & \leq ||x_0|| + \alpha\int_{0}^{t} ||x(s)|| ds \end{split} \end{equation}

Therefore, by the integral form of Gronwall's inequality, we see that

\begin{equation} \begin{split} ||x(t)|| & \leq ||x_0|| + \alpha\int_{0}^{t} ||x(s)|| ds \\ & \leq ||x_0||e^{\alpha(t)} \end{split} \end{equation}

So, if we let $M = ||x_0||$, then $||x(t)||\leq{{M}e^{\alpha(t)}}$. Therefore, the solution is uniformly bounded on $[0,t]$ for $t>0$.

As $t>0$ was arbitrary, the solution is defined for all positive values of $t$.

We can then analyze what happens for negative values of $t$ by reversing time and applying the same argument to $[-t,0]$.

Once again assume that $t_0=0$. Then,

$$x(t) = x_0 + \int_{-t}^{0} f(x(s)) ds$$


\begin{equation} \begin{split} ||x(t)|| & = ||x_0 + \int_{-t}^{0} f(x(s)) ds|| \\ & \leq ||x_0|| + ||\int_{-t}^{0} f(x(s)) ds|| \\ & \leq ||x_0|| + \alpha\int_{-t}^{0} ||x(s)|| ds \\ & \leq ||x_0||e^{\alpha(0+t)} \\ & = {M}e^{\alpha(t)} \end{split} \end{equation}

So, the solution is uniformly bounded on $[-t,0]$ for $t<0$.

Combining these two bounds, we see that the solution emerging from any point $x_0\in\mathbb{R}^N$ exists for any finite time.

Is this approach correct? Please let me know if there are any better alternatives.

  • 1
    $\begingroup$ I do not think there are better alternatives, but your approach is not complete. You have just proved that if a solution is defined on $[t_1,t_2]$ then its is somehow bounded. Now, apply the extension theorem: if a right-nonextendible solution $x(\cdot)$ is defined on some $[t_1,T)$ with $T<\infty$ then for any compact $K\subset\mathbb{R}^N$ there is $\tau<T$ such that $x(t)\not\in K$ for $t\in(\tau,T)$. And this contradicts your estimates. $\endgroup$
    – user539887
    Dec 19, 2018 at 19:41
  • $\begingroup$ Do you mean the Picard–Lindelöf theorem: math.stackexchange.com/questions/2531735/…? I didn't use that theorem because it is for local solutions. The Gronwall inequality is for global solutions. I'm not sure why I need to apply the extension theorem. $\endgroup$
    – Axion004
    Dec 19, 2018 at 20:20
  • $\begingroup$ No, I mean just the result stating what I wrote, see, e.g., Corollary 2.16 on p. 53 of Teschl's Ordinary Differential Equations and Dynamical Systems. $\endgroup$
    – user539887
    Dec 19, 2018 at 20:37
  • $\begingroup$ I see, I only showed that the solution is bounded. I have to argue by lemma 2.14/corollary 2.15/corollary 2.16 that the solution exists (from reading the proof of theorem 2.17, this follows directly from the compactness of the interval). I don't like the wording in corollary 2.16. I am going to spend some more time reading it and see if I understand. $\endgroup$
    – Axion004
    Dec 19, 2018 at 22:16

1 Answer 1


As explained in the comments, the proposed answer only shows that the solution is bounded between some arbitrary interval $[t_1,t_2]$ where $t_1,t_2,\in\mathbb{R}$. We also need to show that we can extend the solution to any interval of finite length.

To do this, consider Lemma $2.14$ on page $52$ of Teschl.

$\textbf{Lemma 2.14:}$ Let $\phi(t)$ be a solution of $(2.10)$ defined on the interval $(t_-,t_+)$. Then there exists an extension to the interval $(t_-,t_+ + \epsilon)$ for some $\epsilon > 0$ if and only if there exists a sequence $t_m\in(t_-,t_+)$ such that

$$\lim_{m\to\infty}(t_m,\phi(t_m))=(t_+,y)\in{U}. $$

The analogous statement holds for an extension to $(t_- - \epsilon,t_+).$

As $||x(t)||\leq{{M}e^{\alpha(t)}}$, it is clear that $x$ lies in a compact ball. Therefore, by Lemma $2.14$ (and the Bolzano–Weierstrass theorem), we can extend the solution to any interval of finite length.

An alternative argument would be using Corollary $2.16$. I don't like the way Corollary $2.16$ is phrased and have decided to directly apply Lemma $2.14$ instead.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .