\tau(1 + \delta) \log \log \tau, \quad \tau \geq \tau^*. \label{eq:y2bdhigh}\\
\intertext{or $\tau_* < \tau^*$ such that}
Y(\tau) &\leq (1 + \delta) \tau \log \tau, \quad \tau \geq \tau_*. \label{eq:y2bdsub}\end{aligned}$$

If we look for $r \geq 0$, in , which solves we have: $$\begin{aligned}
\tau \log \tau [1 + r] & = \tau[\log \tau + \log \log \tau + \log(1 + r)] \label{eq:y2proofbegin}\\
\intertext{or}
r = \varphi(r), &\quad \varphi(X) = \xi + \eta \log (1 + X), \quad r > 0 \label{eq:y2proofbegin.2}\\
\intertext{where}
\xi &= \frac{\log \log \tau}{\log \tau}, \quad \eta = \frac{1}{\log \tau} \label{eq:y2proofbegin.3}\end{aligned}$$ For any ${\displaystyle}0 < \delta\leq \frac{1}{2}$ we can choose $\tau^*$ such that $$0 < \xi \leq \frac{\delta}{2}, \quad 0 < \eta < \frac{\delta}{2} \quad \text{ if }\tau \geq \tau^*.$$ Then the function $\varphi$, $\varphi: [0, \delta] \to [0, \delta]$ is a contraction mapping, and has the unique solution $$r = r(\tau), \quad 0 < r(\tau) \leq \delta. \quad \label{eq:rconfirm}$$ Therefore, $$r = \frac{\log \log \tau}{\log \tau} + \frac{\rho}{\log \tau},\quad  0 < \rho \leq \delta. \label{eq:rform}$$ This implies with $$\frac{\rho}{\log \log \tau} = o(1).$$

\[cor:xtbehave\] The solution $X(t)$ of , $0 <   \beta \leq 1$, goes to $\infty$ when $t \to \infty$ and $$X(t) \leq 2^{1/\beta} 