2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ It should be Gronwall. $\endgroup$
    – Dmitry
    Commented Dec 26, 2019 at 12:04
  • $\begingroup$ @Dmitry: and even Grönwall. $\endgroup$
    – Bernard
    Commented Dec 26, 2019 at 13:47

1 Answer 1

1
$\begingroup$

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 $t>0$.

$\endgroup$
8
  • $\begingroup$ after proving that $G(|u(t)|)$ is bounded on a finite time, how can I conclude that $|u(t)|$ is bounded? $\endgroup$
    – JBond007
    Commented Dec 26, 2019 at 15:10
  • $\begingroup$ $G$ 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 $|u(t)|<x$. $\endgroup$ Commented Dec 26, 2019 at 15:56
  • $\begingroup$ How do we use the fact that $\int_{1}^{+\infty}\frac{1}{F(s)}ds=+\infty$? $\endgroup$
    – JBond007
    Commented Dec 27, 2019 at 8:16
  • $\begingroup$ 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 $|u(t)|\le x_n$. $\endgroup$ Commented Dec 27, 2019 at 8:49
  • $\begingroup$ Why the fact that $|u(t)|\leq x_n$ implies the boundedness of $|u(t)|$, if $x_n$ is a growing sequence? $\endgroup$
    – JBond007
    Commented Dec 27, 2019 at 9:11

You must log in to answer this question.

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