# Divergence of the solution.

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 $$\infty$$.

• 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. – Daniele Tampieri Oct 27 '18 at 16:06
• Hi. I am going to add the original problem. – Odestheory12 Oct 27 '18 at 16:11

## 1 Answer

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 $$|x(t)-x_0|=\Bigg|\int\limits_{t_0}^{t}\phi^\prime(s)\mathrm{d}s\Bigg|\le \begin{cases} \displaystyle\int^{t_0}_tk M^{t_1}_{t_0} \mathrm{d}s &t_1t_0 \end{cases} \le k M^{t_1}_{t_0} |t_1-t_0|$$ 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 $$t$$.

• 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. – Odestheory12 Oct 28 '18 at 16:27
• 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. – Odestheory12 Oct 28 '18 at 16:59
• @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. – Daniele Tampieri Oct 28 '18 at 17:08
• @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 $(t,t_0)$.. – Topologicalife Oct 28 '18 at 17:45
• @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. – Topologicalife Oct 29 '18 at 4:07