# Application of Gronwall Inequality to existence of solutions

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,

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

Therefore,

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

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.

• 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. Dec 19, 2018 at 19:41
• 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. Dec 19, 2018 at 20:20
• 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. Dec 19, 2018 at 20:37
• 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. Dec 19, 2018 at 22:16

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.