# Cauchy's problem, $x'(t) = f(t,x(t))$ with $a < f(t,x(t)) < b$, $f$ is continuous and $C^1$, what are the limits of $x(t)$?

If we have a Cauchy's problem such as :

$$\begin{cases} x'(t) = f(t,x(t)) \\ x(t_0) = x_0 \end{cases}$$

We suppose the function $f(t,x(t))$ is continuous, $C^1$ and bounded (upper and lower). It was assumed that $a < x'(t) = f(t,x(t)) < b$ ($a$ and $b$ are finite). Then, the maximal solution $x(t)$ is defined on the whole $\mathbb{R}$, and we suppose now this solution is increasing.

I just would like to know how to calculate, in a such case, the limits in $\boldsymbol{+\infty}$ and $\boldsymbol{-\infty}$ of $\boldsymbol{x(t)}$. We know $x'(t)$ is bounded, but how to determine these limits of $x(t)$ ? Soemone could help me ?

(I hope I haven't forgotten conditions.)

• "so $x(t)$ is bounded also". Take $f=c>0$. Commented Jun 9, 2017 at 18:20
• Yes, you are right !! I edit it Commented Jun 10, 2017 at 13:53
• Have you an idea for the problem ? I suppose there are several possible cases Commented Jun 10, 2017 at 13:54
• If $a=0$, we can construct an ODE that gives a solution like $x(t) = C\arctan(t)+k$, so that the limits could be anything. On the other hand, if $a > 0$, the solution is bounded below by a line of positive slope, so it has the usual infinite limits of such lines. Of course $a<0$ wouldn't result in an always increasing solution. Commented Jun 10, 2017 at 15:19
• Yes, I believe that many cases are possible. As another example, try to make a differential equation that will have $c+arctan$ as a solution, where $c$ depends on $x_0$. I also believe that the only conclusions that can be drawn without any further information are that both of the limits exist and that $a \ge0$. Commented Jun 10, 2017 at 16:50