I'm reading the paper by Giorgio Talenti on the best constant for the Sobolev inequality.
The main theorem states that for $u:\mathbb{R}^m\rightarrow \mathbb{R}$ sufficiently smooth (eg. Lipschitz) and fast enough decay at $\infty$ we have
$$
\|u\|_q\leq C\|Du\|_p\quad 1<p<m,\; q=\frac{mp}{m-p}
$$
where
$$
C=\pi^{-1/2}m^{-1/2}(\frac{p-1}{m-p})^{1-1/p} \left(\frac{\Gamma(1+m/2)\Gamma(m)}{\Gamma(m/p)\Gamma(1+m-m/p)}\right)^{1/m}$$
and equality holds if
$$
u(x)=\left(a+b|x|^{\frac{p}{p-1}}\right)^{1-m/p}.
$$
In Lemma 1 it is shown that $u^*$, the radial rearrangement of $u$, satisfies
$$
\frac{\|u^*\|_q}{\|Du^* \|_p}\geq\frac{\|u\|_q}{\|Du\|_p}
$$ and hence we can restrict to radial functions when we look for maximizers of this ratio.
By the Euler-Lagrange Equations we find that extremizers are a two-parameter family of functions:
$$\big\{\Phi(r)=a(1+br^{p'})^{1-m/p}: a,b>0\big\}$$
where $p'=\frac{p}{p-1}$.
The problem of maximizing
$$
\frac{(\int_0^{\infty} r^{m-1}|u|^q dr)^{1/q}}{(\int_0^{\infty} r^{m-1}|u'|^p dr)^{1/p}}
$$ under the conditions
- $u=o(r^{1-m/p})$ as $r\rightarrow 0$ or $\infty$ and
- $u'=o(r^{-m/p})$ as $r\rightarrow 0$ or $\infty$
is equivalent to the problem of maximizing $$ \int_0^{\infty} r^{m-1}|u_1|^q dr $$ under the condition that
- $u_2'=r^{m-1}|u_1'|^p$ satisfies $u_2(0)=0, u_2(\infty)=1$ and that
- $u_1(\infty)=0$.
Using the extremizers of the original problem, we obtain exremizing pairs for the reformulated problem:
$$
\begin{align}
\Phi_1(r) &=a(1+br^{p'})^{1-m/p},\ a,b>0\\
\Phi_2(r) &=\int_0^r t^{m-1}|\Phi_1'(t)|^p dt=r^{m-1}\Phi_1(r)^p f\left(\frac{br^{p'}}{1+br^{p'}}\right)
\end{align}
$$
where
$$
f(\xi)=\frac{1}{p'}\frac{m-p}{p-1}^p\xi^p\int_0^1 (1-t)^{m/p'}(1-\xi t)^{-m}dt,
$$ i.p. $f$ maps $(0,1)$ injectively to $\mathbb{R}$.
Now these pairs of extremizers form a vector field on $\mathbb{R}^3_1=\{r,u_1,u_2\in\mathbb{R}^3: r,u_1,u_2>0\}$.
Every pair $(\Phi_1,\Phi_2)$ gives a path $(0,\infty)\rightarrow \mathbb{R}^3_1$, $r\mapsto \big(r,\Phi_1(r), \Phi_2(r)\big)$ and these paths are trajectories of a smooth vector field $X$ on $\mathbb{R}^3_1$:
$$
\begin{cases}
X_0(r,u_1,u_2)=1, \\
\\
X_1(r,u_1,u_2)=-\dfrac{m-p}{p-1}\dfrac{u_1}{r}\xi,
\end{cases}\qquad X_0(r,u_1,u_2)=r^{m-1}|X_1(r,u_1,u_2)|^p
$$
where $\xi$ is the unique root in $(0,1)$ s.t. $f(\xi)=r^{p-m}u_1^{-p}u_2$.
To show that the extremizing pairs are indeed maximizers, the goal is to construct an exact differential $dW$, s.t. along any path $r\mapsto (r,u_1(r), u_2(r))$ which statisfies $u_2'(r)=r^{m-1}|u_1'(r)|^p$:
$$
\int_0^{\infty}dW\geq\int_0^\infty r^{m-1}|u_1(r)|^q dr.
$$
For any $(r,u_1,u_2)\in\mathbb{R}^3_1$ define a linear function:
$$\Psi_{r,u_1,u_2}(\xi_0,\xi_1\xi_2)=u_2'(r)\xi_0-\nabla W(r,u_1,u_2).(\xi_0,\xi_1\xi_2)^T$$
restricted to the cone of all directions issuing from $(r,u_1,u_2)$ s.t. $\xi_0>0$ and $\xi_0^{p-1}\xi_2=r^{m-1}|\xi_1|^p$.
We want $X(r,u_1,u_2)$ to be a critical point of $\Psi_{r,u_1,u_2}$ (i.e. the component of the gradient which is parallel to the cone vanishes).
This, by the Lagrange multiplier principle, gives
$$
\begin{align}
\partial_rW(r,u_1,u_2) &=r^{m-1}u_1^q+\lambda(p-1)r^{m-1}X_1|^p \\
\partial_{u_1}W(r,u_1,u_2) &=\lambda p r^{m-1}X_1|^{p-1} \\
\partial_{u_2}W(r,u_1,u_2) &=\lambda
\end{align}
$$
where $\lambda$ is a $C^1$ function to be determined.
Questions.
This is done by examining compatibility conditions of the above equations, I guess this means that we want the equations to hold for any $(r,u_1,u_2)$, but I really don't see how to derive the following:
$$
\begin{pmatrix} 1 & 0 & -(p-1)(r^{m-1}|X_1|^p) \cr
0 & 1 & -p(r^{m-1}|X_1|^{p-1}) \cr
p(r^{m-1}|X_1|^{p-1}) & (p-1)(r^{m-1}|X_1|^p) & 0 \cr
\end{pmatrix}
\begin{pmatrix} \partial_r\lambda \cr
\partial_{u_1}\lambda \cr
\partial_{u_2}\lambda \cr
\end{pmatrix} \\ =
\begin{pmatrix}
0\cr
0\cr
q r^{m-1}u_1^{q-1}\cr
\end{pmatrix}
-p\lambda
\begin{pmatrix}
X_1\partial_{u_2}\cr
-\partial_{u_2}\cr
\partial_r+X_1\partial_{u_1}\cr
\end{pmatrix}\big(r^{m-1}|X_1|^{p-1}\big)
$$
claimed compatability conditions.
The other thing which is unclear to me, is how this system for $\lambda$ is solved:
"Since the matrix on the lefthand side of has rank 2, we must impose orthogonality between the right-hand side and the eigenvectors of the transposed matrix, i.e."
$$p\lambda\frac{\partial}{\partial_X}(r^{m-1}|X_1|^{p-1})=qr^{m-1}u_1^{q-1}$$
Sorry for the very long question and the formating.
