5
$\begingroup$

I have two functions $f,g:\mathbb{R}\longrightarrow{\mathbb{R}}$, both continuous and with $g$ bounded. I also have the following Cauchy problem $$ \begin{cases}x'=f(t)g(x)\\x(t_0)=x_0 \end{cases} $$ If $\phi$ is the solution of the system defined in $(-\infty, b)$, I have $$ \lim_{t \to{+}b}{\phi'(t)=f(t)g(\phi(t))=f(b)k} $$ with $k \geq g(\phi(t))$ since g is bounded. And now the question:
in my notes it claimed that $\phi(t)$ goes to $\infty$ in $b$ since is not defined there. Is it true?
I don't see why $\phi \to \infty$ in $b$. I think it is simply not defined but I don't see why it should go to infinity. Showing this I would get a contradiction since $\phi(t)'$ is bounded, hence proving that $\phi$ is defined in $\mathbb{R}$

Regards.

Edit -------

Original problem:

Let $f,g:\mathbb{R}\longrightarrow{\mathbb{R}}$, both continuous and with $g$ bounded. Prove that for all $(t_0,x_0)\in \mathbb{R^2}$, the maximal solutions of the Cauchy problem $$ \begin{cases}x'=f(t)g(x)\\x(t_0)=x_0 \end{cases} $$ are defined for all $\mathbb{R}$.

The attempt of solution is done by contradiction; supposing $\phi : (-\infty,b) \to \mathbb{R}$ is a solution, we want to show the derivative of $\phi$ is bounded (as I noted at the begining of this post) while $\phi$ goes to $\infty$ in $b$. This last part is what I don't get. I don't see why is going to $\infty$.

$\endgroup$
  • $\begingroup$ Hi, without more information on $f$ and $g$, it is difficult to find the behavior of $x$ as $t$ gets near $b$. Perhaps you could see if there is further information in your notes. $\endgroup$ – Daniele Tampieri Oct 27 '18 at 16:06
  • 1
    $\begingroup$ Hi. I am going to add the original problem. $\endgroup$ – Odestheory12 Oct 27 '18 at 16:11
5
$\begingroup$

A direct argument, not a proof by contradiction, for proving the existence of a global in time solution for every initial data is the following one. Assuming $k$ is an upper bound for $g:\mathbb{R}\to\mathbb{R}$, i. e. $|g(x)|\le k$ for all $x\in\mathbb{R}$, we also know that $f:\mathbb{R}\to\mathbb{R}$ is continuous: this implies that, in any (time) interval we have $$ |f(x)|\le \max_{x\in I} |f(x)|<\infty. $$ Considering $I=[t_0,t_1]$ or $[t_1,t_0]$ (we must also consider the behavior of the solution backward in time) and defining $\max_{t\in I} |f(t)|\triangleq M^{t_1}_{t_0}<\infty$, we have that $$ |\phi^\prime(t)|\le k M^{t_1}_{t_0}<\infty\quad\forall t\in I\label{1}\tag{1} $$ Note that $M^{t_1}_{t_0}$ depends in general on both $t_1$ and $t_2$. Equation \eqref{1} implies $$ |x(t)-x_0|=\Bigg|\int\limits_{t_0}^{t}\phi^\prime(s)\mathrm{d}s\Bigg|\le \begin{cases} \displaystyle\int^{t_0}_tk M^{t_1}_{t_0} \mathrm{d}s &t_1<t_0\\ \\ \displaystyle\int_{t_0}^tk M^{t_1}_{t_0} \mathrm{d}s &t_1>t_0 \end{cases} \le k M^{t_1}_{t_0} |t_1-t_0| $$ i. e. $$ |x(t)|\le k M^{t_1}_{t_0} |t_1-t_0|+|x_0|<\infty\quad \forall t\in I,\;\forall(t_0,x_0)\in\mathbb{R}\label{2}\tag{2} $$ The arbitrariness of $t_1$ and formula \eqref{2} imply that the solution $x(t)$ of the posed Cauchy problem exists finite for each $(t_0,x_0)\in\mathbb{R}$ and each time $t$.

$\endgroup$
  • 2
    $\begingroup$ Hello. Good answer! I got it all. But I am still worried about various facts on my original question. 1. Since $\phi$ is defined in $(-\infty, b)$, then $\lim_{x\to +b}$ shouldn't exist since $\phi$ is not defined as we approach from the right. 2. $\phi(t)$ goes to $\infty$ in $b$ because "is not defined". I rechecked my notes and coulldn't find nothing related to. If a solution is not defined at some point it doesnt mean it should necessary explode (i.e: that it should go to infty or -infty), right? Regards. $\endgroup$ – Odestheory12 Oct 28 '18 at 16:27
  • $\begingroup$ One thing, I don't see why $\int^{t_0}_tk M^{t_1}_{t_0} \mathrm{d}s = k M^{t_1}_{t_0} |t_1-t_0|$ for $t_1<t_0$. Where the $t_1$ come from? What I see is that the integral is equal to $k M^{t_1}_{t_0} |t-t_0|$ or $\leq k M^{t_1}_{t_0} |t_1-t_0|$ Can you elabore a bit more please? Thanks. $\endgroup$ – Odestheory12 Oct 28 '18 at 16:59
  • $\begingroup$ @Odestheory12 It is a consequence of the fact that we have chosen an interval $I=[t_0,t_1]$ (or $I=[t_1,t_0]$, if we consider backward times) with arbitrary endpoint (start point) $t_1$: there $\max_If$ exists and is the constant $M_{t_0}^{t_1}$. So, when integrating the by using estimate (1), it can be treated as a constant. $\endgroup$ – Daniele Tampieri Oct 28 '18 at 17:08
  • $\begingroup$ @DanieleTampieri Yeah I got that part, sorry for being unclear. I mean I don't see why do we have $|t_1 - t_0|$ instead of $|t-t_0|$ in the result of the integral, i.e:$k M^{t_1}_{t_0} |t_1-t_0|$ , since we are integrating over $(t,t_0)$.. $\endgroup$ – Topologicalife Oct 28 '18 at 17:45
  • 2
    $\begingroup$ @DanieleTampieri The issue I see with that is $\phi$ is not even defined in $b$, so it doesn't make sense to me to talk about $\displaystyle \lim_{x\to b^+} \phi(x)$ since is not defined. If we could prove that $\phi$ goes to $\infty$ in $b$ and if $\lim_{t \to{+}b}{\phi'(t)=f(t)g(\phi(t))=f(b)k}$ then it would be a contradiction, since we got a bounded derivative with a function that diverges, which is not possible by MVT. $\endgroup$ – Topologicalife Oct 29 '18 at 4:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.