(Related to [this Math.SE question](https://math.stackexchange.com/q/3012078/8157).)


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.



The blow-up rate of a nonlinear oscillator
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.


