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.
Best Answer
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 $t>0$.