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 $\mathbb{R}$


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 $\infty$.

  • $\begingroup$ 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
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    $\begingroup$ Hi. I am going to add the original problem. $\endgroup$ – Odestheory12 Oct 27 '18 at 16:11

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 $t$.

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    $\begingroup$ 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
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    $\begingroup$ @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

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