**Background:**


I am reading the paper: *Best constant in Sobolev inequality* by Talenti (see [here](https://math.jhu.edu/%7Ejs/Math646/talenti.sobolev.pdf)) and I am trying to understand the following step.


On p. 365, the author is arguing that the solutions to the following equation


$$\left(r^{m-1}\left|u^{\prime}\right|^{\dot{p}-1} \operatorname{sgn} u^{\prime}\right)^{\prime}+C r^{m-1}|u|^{q-1} \operatorname{sgn} u=0 \quad(C=\text { a positive constant })$$


take the form $(a+br^p)^{1-m/p}.$ Here $m$ is the dimension of the space, $q=\frac{2m}{m-2}$ and $1<p<m.$ The above ODE corresponds to the radial solutions for the Euler Lagrange equation associated to the Sobolev inequality.


He begins by considering the case $p=2$. Then we have,


$$\left(r^{m-1}\left|u^{\prime}\right| \operatorname{sgn} u^{\prime}\right)^{\prime}+C r^{m-1}|u|^{q-1} \operatorname{sgn} u=0.\label{1}\tag{28}$$ By considering $u(r)=r^{1-m/2}v(r)$ we obtain the following ODE for $v$,
$$r\left(r v^{\prime}\right)^{\prime}=(1-m / 2)^{2} v-C v^{q-1}.$$
By multiplying the above ODE by $v$ and integrating this is equivalent to solving,
$$\left(r v^{\prime}\right)^{2}=(1-m / 2)^{2} v^{2}-(2 C / q) v^{q}+\text { constant }.\label{2}\tag{$\ast$}$$
From this the author concludes that all the solutions of \eqref{1} that are positive decreasing and satisfy the following decay conditions
$$u(r)=o(r^{1-m/2}),u'(r)=o(r^{-m/2})$$
must be of the form $u(r)=(a+br^2)^{1-m/2}$ for constants $a$ and $b$ related to $C$.


**Question:** I am not sure how looking at the ODE \eqref{2} allows the author to deduce that the solutions must take the form $u(r)=(a+br^2)^{1-m/2}$?



Extremizers of the Sobolev inequality
Take any extremizer $u$ as in the [paper you linked](https://math.jhu.edu/%7Ejs/Math646/talenti.sobolev.pdf).
From the conditions $u(r)=r^{1-m/2}v(r)$, $u(r)=o(r^{1-m/2})$, and $u'(r)=o(r^{-m/2})$ (as $r\to\infty$), we deduce $v(r)=r^{m/2-1}u(r)$, $v(r)=o(1)$, and $v'(r)=(m/2-1)r^{m/2-2}u(r)+r^{m/2-1}u'(r)=o(1/r)$.  

So, it follows from your ODE $(\*)$ that the unnamed constant there is $0$, so that
\begin{equation\*}
(r v'(r))^2=(1-m/2)^2 v(r)^2-(2C/q) v(r)^q. \tag{$\*\*$}
\end{equation\*}
Since $v$ is decreasing, we have $v'\le0$ and hence
\begin{equation\*}
r v'(r)=-\sqrt{(1-m/2)^2 v(r)^2-(2C/q) v(r)^q}. \tag{$\*\*\*$}
\end{equation\*}
Note that $2=p<m$, and hence $m>2$.


Take any real $r\_\*>0$ and then any real $U\_\*>0$ such that $U\_\*^2-4c\_1 r\_\*^2>0$, where  

\begin{equation\*}
 c\_1:=\frac C{m(m-2)}. 
\end{equation\*}
Let then  

\begin{equation\*}
 a:=a(r\_\*):=\frac{U\_\*-\sqrt{U\_\*^2-4c\_1 r\_\*^2}}2\quad\text{and}\quad 
 b:=b(r\_\*):=\frac{U\_\*+\sqrt{U\_\*^2-4c\_1 r\_\*^2}}{2r\_\*^2},
\end{equation\*}
so that $a$ and $b$ are positive real numbers, solving the system of equations $a+br\_\*^2=U\_\*$ and $abm(m-2)=C$.
Moreover, it is now straightforward to check that, for such $a$ and $b$, the formulas
\begin{equation\*}
 v\_\*(r)=r^{m/2-1}u\_\*(r)\quad\text{and}\quad u\_\*(r)=(a+br^2)^{1-m/2} \tag{1}
\end{equation\*}
define a solution $v\_\*$ of ODE $(\*\*\*)$ on $(0,\infty)$ with the "initial" condition $u\_\*(r\_\*)=U\_\*^{1-m/2}$.


Recall that the extremizer $u$ is a solution of $(\*\*\*)$. Moreover, condition (25b) in that paper implies that $u(r)=o(r^{1-m/2})$ as $r\downarrow0$, and hence
$U(r)/r\to\infty$ as $r\downarrow0$, where
\begin{equation\*}
 U(r):=u(r)^{1/(1-m/2)}. 
\end{equation\*}
So, for any small enough real $r\_\*>0$, choosing at this point $U\_\*:=U(r\_\*)$, we see that the condition $U\_\*^2-4c\_1 r\_\*^2>0$ holds.
So, the functions $u$ and $u\_\*$ are both solutions of $(\*\*\*)$ on $(0,\infty)$ satisfying the same "initial" condition: $u\_\*(r\_\*)=U\_\*^{1-m/2}=u(r\_\*)$. So, by the uniqueness, $u=u\_\*$, as desired.


