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(Related to this Math.SE question.)

For $p>1$, let $u$ be a solution to $$\tag{1}\frac{d^2 u}{dt^2} + u = |u|^{p-1}u$$ that blows up at $T>0$, that is $$\lim_{t\nearrow T}u(t)=+\infty.$$

Remark. If $u$ solves (1), then the following quantity is independent on $t$; $$\tag{2} E=\frac12\left(\frac{du}{dt}\right)^2 +\frac{u^2}{2} -\frac{|u|^{p+1}}{p+1}.$$

Question. Is it true that $$\lim_{t\nearrow T} \frac{u(t)}{C(T-t)^{-\alpha}}=1,\qquad \text{where }C=\left(2\frac{p+1}{(p-1)^2}\right)^\frac1{p-1}\text{ and }\ \alpha=\frac{2}{p-1}? $$

I believe that the answer is affirmative. My motivation is the heuristic that $u_0:=C(T-t)^{-\alpha}$ should solve both $$\tag{3}\frac{d^2 u_0}{dt^2} =|u_0|^{p-1}u_0, $$ and $$\tag{4}\left(\frac{du_0}{dt}\right)^2 = \frac{2}{p+1}|u_0|^{p+1}, $$ and that is indeed the case.

The equations (3) and (4) are obtained from (1) and (2) by considering the highest power of $u$ only.

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Your heuristics is correct. Indeed, suppose that $u(t)\to\infty$ as $t\uparrow T$. Then for some real $h>0$ and all $t\in[T-h,T)$ we have $u(t)>0$ and, by the "conservation of energy" stated in your post,
\begin{equation} u'(t)^2=2E-u(t)^2+\frac{2u(t)^{p+1}}{p+1}\sim \frac{2u(t)^{p+1}}{p+1}, \end{equation} because $p>1$ and hence $p+1>2$; everywhere here, the asymptotic relations are given for $t\uparrow T$. Because $u''$ exists and $u(t)\to\infty$, we see that \begin{equation} u'(t)\sim \sqrt{\frac{2}{p+1}} \,u(t)^{(p+1)/2},\quad -\frac d{dt}\,\big(u(t)^{-(p-1)/2}\big)\sim c_p:=\sqrt{\frac{2}{p+1}}\,\frac{p-1}2, \end{equation} \begin{equation} u(t)^{-(p-1)/2}=\int_t^T \Big(-\frac d{ds}\,\big(u(s)^{-(p-1)/2}\big)\Big)\, ds \sim c_p(T-t), \end{equation} and thus \begin{equation} u(t)\sim [c_p(T-t)]^{-2/(p-1)}=C(T-t)^{-\alpha} \end{equation} for $C=\big(2\frac{p+1}{(p-1)^2}\big)^\frac1{p-1}$ and $\alpha=\frac{2}{p-1}$, as desired.

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