# Use Grönwall's proof to prove boundedness of ODE's solution

We have a function $$F:\mathbb{R}\rightarrow \mathbb{R}$$ continuous, positive and increasing. Let $$u:(a,b)\rightarrow \mathbb{R}$$ differentiable and $$|u'(t)|\leq F(|u(t)|), \hspace{1.0cm} t\in(a,b)$$ If $$\int_{1}^{+\infty}\frac{1}{F(s)}ds=+\infty$$ Prove that $$u$$ is bounded on $$(a,b)$$.

My idea is to get inspiration from this proof of Grönwall's inequality:

Grönwall's lemma

Let $$\omega\in C^1(a,b)$$, if $$\exists \epsilon>0, Q>0:\forall t\in (a,b)$$ is $$|\omega'(t)| \leq \epsilon+Q(|\omega(t)|$$ so $$|\omega(t)|\leq \left(\frac{\epsilon}{Q}+|\omega(t_0)|\right)e^{Q|t-t_0|}$$

Proof

Let $$z(t)=\sqrt{\omega^2(t)+\sigma^2}\geq \omega(t)$$ $$|z'(t)|=|\frac{2\omega(t)\omega'(t)}{2\sqrt{\omega^2(t)+\sigma^2}}|\leq |\omega'(t)|\leq \epsilon+Q|\omega(t)|\leq \epsilon +Qz(t)$$ $$z'(t)\leq \epsilon +Qz(t)$$ $$\frac{z'(t)}{\epsilon+Qz(t)}\leq 1$$ Notice that $$\frac{d}{dt}ln(\epsilon+Qz(t))=\frac{Qz'(t)}{\epsilon+Qz(t)}$$ $$\implies \frac{z'(t)}{\epsilon+Qz(t)}=\frac{1}{Q}\frac{d}{dt}ln(\epsilon+Qz(t))\leq 1 \implies \frac{d}{dt}ln(\epsilon+Qz(t))\leq Q$$ Integrating from $$t_0$$ to $$t$$: $$\int_{t_0}^{t}\frac{d}{dt}ln(\epsilon+Qz(t))dt=\int_{t_0}^{t}Q$$ $$ln\left(\frac{\epsilon+Qz(t)}{\epsilon+Qz(t_0)}\right)\leq Q(t-t_0)$$ $$z(t)\leq \left(\frac{\epsilon}{Q}+z(t_0)\right)e^{Q(t-t_0)}$$ And because $$z(t)\geq \omega(t)$$, $$|\omega(t)|\leq \left(\frac{\epsilon}{Q}+|\omega(t_0)|\right)e^{Q(t-t_0)}$$

But I don't know how to use the improper integral given to me in this verification.

• It should be Gronwall. Commented Dec 26, 2019 at 12:04
• @Dmitry: and even Grönwall. Commented Dec 26, 2019 at 13:47

Set $$G(x)=\int_1^x\frac1{F(s)}ds$$. If $$|u(t)|>1$$ for $$t\in(t_0,t_1)$$, then inside this interval $$\frac{d}{dt} G(|u(t)|)=G'(|u(t)|)\frac{u(t)\cdot u'(t)}{|u(t)|}=\frac{u(t)\cdot u'(t)}{|u(t)|\,F(|u(t)|)}$$ which is smaller than one in absolute value, $$G(|u(t)|)\le G(|u(t_0)|)+|t-t_0|$$. This now means that this expression can not reach infinity in finite time, which translates via the inverse of $$G$$ to $$|u(t)|$$. So the solution is bounded over all finite intervals.
Or perhaps simpler to follow, let $$v(t)$$ be the solution of $$v'(t)=F(v(t))$$ with $$v(0)=1+|u(0)|$$. Then at any point $$t>0$$ you have $$|u(t)|-v(t)\le |u(0)|-v(0)+\int_0^t[|u'(s)|-v'(s)]ds\le -1+\int_0^t[F(|u(s)|)-F(v(s))]$$ Assuming that there exists a point with $$|u(t)|=v(t)$$ leads to a contradiction, as the inequality above gives $$|u(t)|+1\le v(t)$$ for the smallest such $$t$$. Now the boundedness of $$v$$ follows from the separation-of-variables method and the given property of $$F$$.
A similar argument holds for the other direction in comparing $$|u(-t)|$$ and $$v(t)$$ for $$t>0$$.
• after proving that $G(|u(t)|)$ is bounded on a finite time, how can I conclude that $|u(t)|$ is bounded? Commented Dec 26, 2019 at 15:10
• $G$ is positive and monotonously increasing to infinity, which translates to its inverse function. So if $G(|u(t)|)$ is bounded by $M=G(x)$, then $|u(t)|<x$. Commented Dec 26, 2019 at 15:56
• How do we use the fact that $\int_{1}^{+\infty}\frac{1}{F(s)}ds=+\infty$? Commented Dec 27, 2019 at 8:16
• For instance to ensure the existence of a growing sequence $x_n$ with $\int_1^{x_n}\frac{ds}{F(s)}=2^n$. Then if $G(|u(t)|)\le 2^n$, one can conclude that $|u(t)|\le x_n$. Commented Dec 27, 2019 at 8:49
• Why the fact that $|u(t)|\leq x_n$ implies the boundedness of $|u(t)|$, if $x_n$ is a growing sequence? Commented Dec 27, 2019 at 9:11