I have the following exercise:
Consider the Cauchy problem: $$\begin{cases}x' = \log x \\ x(0) = a \end{cases} \quad \text{where } a > 0 .$$ Perform a qualitative study of the solution of the problem.
Qualitative study here means the beahaviour of the maximal solution $x : (\omega_-, \omega_+) \to \mathbb R$ and its limits as $x \to \pm \infty$.
By studing the sign of $\log x$, we have that the maximal solution
- has its graph contained in $\mathbb R \times (1, +\infty)$ and is increasing if $a > 1$
- is constant to 1 if $a = 1$
- has its graph contained in $\mathbb R \times (0, 1)$ and is decreasing if $a < 1$.
Essentially, the cases to be considered are only two.
Case $a > 1$. I am going to see if the maimal solution is bounded by two continuous functions. One of the is the one constant to $1$. As for the other, I observe that, being $x \ge 1$ here, we have $\log x \le x$. Now I consider the solution of the problem $$\begin{cases} u' = u \\ u(0) = a \end{cases}$$ which is $u(t) = ae^t$ and is defined over all $\mathbb R$. Then $$1 \le x(t) \le u(t) \quad \text{for every } t \in (\omega_-, \omega_+) .$$ In the current case, the maximal solution has domain $(\omega_-, \omega_+) = \mathbb R$. Both limits $\ell_- := \lim_{t \to \omega_\_} x(t)$ and $\ell_+ := \lim_{t \to \omega_+} x(t)$ do exist, being $x$ monotonic. In particular the former limit is finite; hence $$0 = \lim_{t \to \omega_\_} x'(t) = \lim_{t \to \omega_-} \log x(t) = \log \ell_- \implies \ell_- = 1 .$$ Now assume $\ell_+ < +\infty$. In that case again $$0 = \lim_{t \to \omega_+} x'(t) = \lim_{t \to \omega_+} \log x(t) = \log \ell_+ \implies \ell_+ = 1$$ which is absurd because $x(t) \ge a > 1$ for $t \ge 0$. We have to conlude that $\ell_+ = +\infty$.
Case $0 < a < 1$. For $t \le 0$, we have $a \le x(t) < 1$: that results in $\omega_- = -\infty$. The computation of $\ell_-$ is done in a similar fashion, and $\ell_- = 1$. Now I suppose $\omega_+ = +\infty$. So $$0 = \lim_{t \to \omega_+} x'(t) = \lim_{t \to \omega_+} \log x(t) = \log \ell_+ \implies \ell_+ = 1$$ which is absurd because $x(t) \le a < 1$ for $t \ge 0$. It must be $\omega_+ < +\infty$.
If my attempt is correct -- is it? --, any idea on how to compute $\ell_+ = \lim_{t \to \omega_+} x(t)$?
