Extremizers of the Sobolev inequality Background:
I am reading the paper: Best constant in Sobolev inequality by Talenti (see here) 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 $u(r)=(a+br^2)^{1-m/2}$?
 A: Take any extremizer $u$ as in the paper you linked.
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.
