# Classification of solutions to Hardy-Sobolev Doubly Critical Systems

Francesco Esposito∗, Rafael López-Soriano#, Berardino Sciunzi∗ * Dipartimento
di Matematica e Informatica, UNICAL, Ponte Pietro Bucci 31B, 87036 Arcavacata
di Rende, Cosenza, Italy<EMAIL_ADDRESS><EMAIL_ADDRESS>$\displaystyle\\#$ Departamento de Análisis Matemático, Universidad de
Granada, Campus Fuentenueva, 18071 Granada, Spain<EMAIL_ADDRESS>
(Date: August 27, 2024; Date: August 27, 2024)

###### Abstract.

This work deals with a family of Hardy-Sobolev doubly critical system defined
in $\displaystyle\mathbb{R}^{n}$. More precisely, we provide a classification
of the positive solutions, whose expressions comprise multiplies of solutions
of the decoupled scalar equation. Our strategy is based on the symmetry of the
solutions, deduced via a suitable version of the moving planes technique for
cooperative singular systems, joint with the study of the asymptotic behavior
by using the Moser’s iteration scheme.

###### Key words and phrases:

Semilinear elliptic systems, Hardy potentials, qualitative properties,
critical nonlinearities

###### 2000 Mathematics Subject Classification:

35J61; 35B50; 35B06; 35J47;

F. Esposito and B. Sciunzi are partially supported by PRIN project 2017JPCAPN
(Italy): Qualitative and quantitative aspects of nonlinear PDEs. Moreover they
are members of INdAM. R. López-Soriano is currently supported by the grant
Juan de la Cierva Incorporación fellowship (JC2020-046123-I), funded by
MCIN/AEI/10.13039/501100011033, and by the European Union Next Generation
EU/PRTR. He is also partially supported by Grant PID2021-122122NB-I00 funded
by MCIN/AEI/ 10.13039/501100011033 and by “ERDF A way of making Europe”.

August 27, 2024

##  1\. Introduction

We are concerned with the study of the doubly critical system:

($\displaystyle\mathcal{S}^{*}$) $\begin{cases}\displaystyle-\Delta
u\,=\gamma\frac{u}{|x|^{2}}+u^{2^{*}-1}+\nu\alpha
u^{\alpha-1}v^{\beta}&\text{in}\quad{\mathbb{R}}^{n}\vspace{0.2cm}\\\
\displaystyle-\Delta v\,=\gamma\frac{v}{|x|^{2}}+v^{2^{*}-1}+\nu\beta
u^{\alpha}v^{\beta-1}&\text{in}\quad{\mathbb{R}}^{n}\vspace{0.2cm}\\\
u,v>0&\text{in}\quad{\mathbb{R}}^{n}\setminus\\{0\\},\end{cases}$

where $\displaystyle\gamma\in[0,\Lambda_{n})$ with
$\displaystyle\Lambda_{n}=\left(\frac{n-2}{2}\right)^{2}$ the best constant in
the Hardy’s inequality for $\displaystyle n\geq 3$, $\displaystyle
2^{*}=\dfrac{2n}{n-2}$ is the critical exponent in the Sobolev embedding,
$\displaystyle\nu>0$ is the coupling parameter and
$\displaystyle\alpha,\beta>1$ are real parameters satisfying

$\displaystyle\alpha+\beta=2^{*}.$

Along this article we address the analysis of the classification of positive
solutions to ($\displaystyle\mathcal{S}^{*}$).
In the case $\displaystyle\gamma=\nu=0$, system
($\displaystyle\mathcal{S}^{*}$) reduces to the classical Sobolev critical
equation

(1.1) $\begin{cases}\displaystyle-\Delta
u\,=u^{2^{*}-1}&\text{in}\quad{\mathbb{R}}^{n}\\\
u>0&\text{in}\quad{\mathbb{R}}^{n}.\end{cases}$

It is well-known that the so-called Aubin-Talenti bubbles

(1.2)
$\mathcal{V}_{\lambda,x_{0}}(x):=\left[\frac{\lambda\sqrt{n(n-2)}}{\lambda^{2}+|x-x_{0}|^{2}}\right]^{\frac{n-2}{2}}$

solve (1.1), where $\displaystyle\lambda>0$ and it is centered at
$\displaystyle x_{0}\in{\mathbb{R}}^{n}$. Moreover, they realize the equality
in the sharp Sobolev inequality in $\displaystyle{\mathbb{R}}^{n}$. In [29],
the authors proved that any positive solution $\displaystyle u$ to (1.1), such
that $\displaystyle u(x)=O\left(1/|x|^{m}\right),$ at infinity for
$\displaystyle m>0$, is radially symmetric and decreasing about some point in
$\displaystyle{\mathbb{R}}^{n}$. In the proof the authors used a refinement of
the celebrated moving plane procedure developed by themselves in a previous
paper. Later on, Caffarelli, Gidas and Spruck in [10] classified all the
solutions to (1.1). Making use of the Kelvin transform, the authors showed
that the moving plane procedure can start. Finally, they showed that all the
solutions $\displaystyle u\in H^{1}_{loc}({\mathbb{R}}^{n})$ to (1.1) are
given by the Aubin-Talenti bubbles (1.2), hence solutions are unique up to
rescaling. Recently, these results were also generalized in the case of
critical equations involving the $\displaystyle p$-Laplacian and the Finsler
anisotropic operator, where the use of the Kelvin transform is not-allowed,
see e.g. [12, 15, 37, 40, 48, 49].

When $\displaystyle\gamma\neq 0$ and $\displaystyle\nu=0$, system
($\displaystyle\mathcal{S}^{*}$) becomes the so-called Hardy-Sobolev doubly
critical equation

(1.3) $\begin{cases}-\Delta
u\,=\gamma\frac{u}{|x|^{2}}+u^{2^{*}-1}\quad&\text{in}\quad{\mathbb{R}}^{n}\\\
u>0&\text{in}\quad{\mathbb{R}}^{n}\setminus\\{0\\}.\end{cases}$

Terracini in [46], by means of variational arguments and the concentration
compactness principle, showed the existence of solutions to (1.3), actually a
more general one. Moreover, by using the Kelvin transformation and a fine use
of the moving plane method, she proved that any solution to (1.3) is radially
symmetric about the origin. Finally, thanks to a detailed ODE’s analysis, she
gave a complete classification of the solutions to (1.3), given by

(1.4)
$\mathcal{U}_{\mu}(x)=\mu^{\frac{2-n}{2}}\mathcal{U}\left(\frac{x}{\mu}\right)\qquad\mbox{
with
}\qquad\mathcal{U}(x)=\dfrac{A(n,\gamma)}{|x|^{\tau}\left(1+|x|^{2-\frac{4\tau}{n-2}}\right)^{\frac{n-2}{2}}},$

where

(1.5) $\tau:=\frac{n-2}{2}-\sqrt{\left(\frac{n-2}{2}\right)^{2}-\gamma},\qquad
A(n,\gamma):=\left(\frac{n(n-2-2\tau)^{2}}{n-2}\right)^{\frac{n-2}{4}},$

and $\displaystyle\mu>0$ is a scaling factor. Obviously,
$\displaystyle\mathcal{U}_{\mu}=\mathcal{V}_{\mu,0}$ if
$\displaystyle\gamma=0$.

The case of $\displaystyle p$-Lapalce operator was firstly treated in [2],
where the authors carried out a very fine ODE’s analysis. The radial symmetry
of the solutions there was an assumption. Later on, in [36] the authors showed
that all the positive solutions to $\displaystyle p$-Laplace doubly critical
equation are radial (and radially decreasing) about the origin. Once the
radial symmetry of the solution is proved it is easy to derive the associated
ordinary differential equation fulfilled by the solution $\displaystyle
u=u(r)$ and, hence, to apply the results in [2].

¿From now on we focus our attention to the case of systems. Nonlinear
Schrödinger problems, like the Gross–Pitaevskii type systems, have a strong
connection with some physical phenomena. Such problem appears in the study of
Hartree–Fock theory for double condensates, that is a binary mixture
Bose–Einstein condensates in two different hyperfine states which overlap in
space, see [24, 27] for further details. That type of systems arises also in
nonlinear optics. Actually, it allows one to study the propagation of pulses
in birefringent optical fibers and the beam in Kerr-like photorefractive
media, see [5, 31] and references therein. In particular, solitary-wave
solutions to the coupled Gross–Pitaevski equations satisfies the problem

(1.6) $\left\\{\begin{array}[]{ll}\displaystyle{-\Delta u+V(x)u=\mu
u^{2p-1}+\nu u^{p-1}v^{p}}&\text{in }\mathbb{R}^{n}\vspace{.3cm}\\\ -\Delta
v+V(x)v=\mu v^{2p-1}+\nu u^{p}v^{p-1}&\text{in
}\mathbb{R}^{n},\end{array}\right.$

where $\displaystyle V$ is the system potential and $\displaystyle
1<p\leq\frac{2^{*}}{2}$. This problem is typically known as the Bose–Einstein
condensate system. For the subcritical regime, we refer to [6, 7, 33, 43, 45]
for some results concerning existence and multiplicity of solutions under
different assumptions on $\displaystyle V$ and $\displaystyle\nu$.

Concerning the critical case $\displaystyle p=\frac{2^{*}}{2}$, if
$\displaystyle V$ is non-zero constant, then system (1.6) admits only the
trivial solution $\displaystyle(0,0)$. This result follows from a proper
application of the Pohozaev-type identity. For $\displaystyle V=0$, in [38]
the authors showed the uniqueness of the ground states, under suitable
assumptions on the parameters of the generalized system; whereas in the paper
[16] the competitive setting is considered, i.e. $\displaystyle\nu<0$,
deducing that the system admits infinitely many fully nontrivial solutions,
which are not conformally equivalent.

As a non-constant potential, from now on we shall consider the aforementioned
Hardy-type one $\displaystyle V=-\frac{\gamma}{|x|^{2}}$. Under that choice,
the problem ($\displaystyle\mathcal{S}^{*}$) can be also seen as an extension
of (1.6). The mathematical interest in such system lies in their double
criticality, since both the exponent of the nonlinearities and the
singularities share the same order of homogeneity as the Laplacian. Moreover,
inverse square potentials arise in some physical prototypes, such as
nonrelativistic quantum mechanics, molecular physics or combustion models.

Doubly critical problems has attracted attention in recent years. In the
pioneer work [3], a general Hardy-Sobolev type system is studied by means of
variational techniques. In the cooperative regime $\displaystyle\nu>0$, the
existence of ground and bound states are obtained depending on
$\displaystyle\alpha,\beta$, $\displaystyle n$ and a potential function
$\displaystyle h$ arising in the coupling term. Recently, that kind of results
were extended in [17] by using similar strategies. Such doubly critical system
is widely analyzed in [14]. Actually, the non-existence of ground states for
the competitive case is indeed proved.

The aim of this paper is to classify all the positive solutions to problem
($\displaystyle\mathcal{S}^{*}$), via the study of symmetry and monotonicity
properties in the same spirit of the aforementioned papers [10, 46].

Hence, we state one of the main results of our paper:

######  Theorem 1.1.

Let $\displaystyle(u,v)\in D^{1,2}(\mathbb{R}^{n})\times
D^{1,2}(\mathbb{R}^{n})$ be a solution to ($\displaystyle\mathcal{S}^{*}$).
Then,

(1.7)
$(u,v)=\left(c_{1}\,\mathcal{U}_{\mu_{0}},c_{2}\,\mathcal{U}_{\mu_{0}}\right),$

with $\displaystyle\mathcal{U}_{\mu}$ introduced in (1.4),
$\displaystyle\mu_{0}>0$ and $\displaystyle c_{1},c_{2}$ are any positive
constants satisfying the system

(1.8) $\left\\{\begin{array}[]{ll}\displaystyle{c_{1}^{2^{*}-2}+\nu\alpha
c_{1}^{\alpha-2}c_{2}^{\beta}}=1&\vspace{.3cm}\\\
\displaystyle{c_{2}^{2^{*}-2}+\nu\beta
c_{1}^{\alpha}c_{2}^{\beta-2}=1}.&\end{array}\right.$

######  Remark 1.2.

This result holds (and is new) also in the case $\displaystyle\gamma=0$. Under
that assumption the explicit solutions in (1.4) reduce to those of (1.2). See
[13] for a result related to such a case.

######  Remark 1.3.

As a consequence of Theorem 1.1, we can give an explicit expression of the
solution. Observe that the system (1.8) might admit several solutions
depending on the values $\displaystyle n,\nu,\alpha,\beta$. For instance, if
$\displaystyle n=3$, $\displaystyle\nu=1$, $\displaystyle\alpha=3=\beta$,
there exist three synchronized solutions; whereas if $\displaystyle n=4$,
$\displaystyle\nu=1$, $\displaystyle\alpha=2=\beta$ there exists one
synchronized solution. Concerning the uniqueness of synchronized solutions we
refer [38] for more details.

######  Remark 1.4.

From a pure mathematical point of view, let us emphasize that, throughout the
paper, a very deep and crucial issue is represented by those cases in which
either $\displaystyle\alpha<2$, $\displaystyle\beta<2$ or both. In such a
case, in fact, the coupling term is non locally Lipschitz continuous. Note
that this necessarily occurs if $\displaystyle n>4$. In low dimension,
however, all the situations are possible.

A first important step in our strategy is the proof of the fact that the
solutions to ($\displaystyle\mathcal{S}^{*}$) are radially symmetric about the
origin. Actually we shall provide a more general result and, in particular,
holds without requiring global energy information.

######  Theorem 1.5.

Let $\displaystyle(u,v)\in\left(H^{1}_{loc}(\mathbb{R}^{n})\cap
C({\mathbb{R}}^{n}\setminus\\{0\\})\right)\times\left(H^{1}_{loc}(\mathbb{R}^{n})\cap
C({\mathbb{R}}^{n}\setminus\\{0\\})\right)$ be solution to

($\displaystyle\mathcal{S}^{*}_{\gamma_{1,2}}$)
$\begin{cases}\displaystyle-\Delta
u\,=\gamma_{1}\frac{u}{|x|^{2}}+u^{2^{*}-1}+\nu\alpha
u^{\alpha-1}v^{\beta}&\text{in}\quad{\mathbb{R}}^{n}\vspace{0.2cm}\\\
\displaystyle-\Delta v\,=\gamma_{2}\frac{v}{|x|^{2}}+v^{2^{*}-1}+\nu\beta
u^{\alpha}v^{\beta-1}&\text{in}\quad{\mathbb{R}}^{n}\\\
u,v>0&\text{in}\quad{\mathbb{R}}^{n}\setminus\\{0\\},\end{cases}$

where $\displaystyle\gamma_{1},\gamma_{2}\in[0,\Lambda_{n})$. Then
$\displaystyle(u,v)$ is radially symmetric about the origin.

The proof of the radial symmetry of the solutions is based on a fine
adaptation of the moving plane procedure. The technique goes back to the
seminal works of Alexandrov and Serrin [4, 42] and the well-known
contributions by Berestycki-Nirenberg [8] and Gidas-Ni-Nirenberg [28].
Originally, the technique was introduce to be performed in general domains
providing partial monotonicity results near the boundary and symmetry for
convex and symmetric domains. Regarding elliptic systems, the moving plane
technique was adapted by Troy in [47], where the cooperative case is analyzed.
The procedure was also applied for semilinear systems in the half-space in
[20] and in the whole space by Busca and Sirakov in the work [9]. We refer the
reader to [18, 19, 21, 22, 25, 26, 34, 39, 44] for other interesting
contribution about elliptic systems in bounded or unbounded domains.

For the reader’s convenience we describe the strategy of our proofs that turn
out to be tricky somehow.

###  1.1. The symmetry result

The proof of Theorem 1.5, as recalled above, is based on the moving plane
technique. Unfortunately the adaptation of the technique is not
straightforward since we work in unbounded domains and the coupling term, in
general, is not locally Lipschitz continuous. We overcame such a difficulty
studying a suitable translated problem. To obtain symmetry in the
$\displaystyle x_{1}$-direction move the Hardy potential to

$\displaystyle x_{0}=(0,x^{\prime}_{0})\in{\mathbb{R}}^{n}\setminus\\{0\\}.$

Then we apply the Kelvin transformation. The translation argument allow us to
deal with the presence of the Hardy potential. The problem is not invariant
under this procedure but the equation that arises is not so bad and we succeed
in the adaptation of the moving plane procedure.

###  1.2. The asymptotic analysis

Once we know that the solutions are radially symmetric, we use a suitable
transformation

$(u_{\tau}(x),v_{\tau}(x)):=\left(|x|^{\tau}u(x),|x|^{\tau}v(x)\right)\,,$

for the right choice of $\displaystyle\tau>0$. The Moser iteration scheme,
applyed to the transformed equation provides a first asymptotic information.
Then the study of the associated ODE (together with the Kelvin transform)
allows us to deduce the precise asymptotic information at zero and at
infinity.

###  1.3. The classification result

Once we know the exact behavior of the solutions, we exploit the standard
change of variable

$\begin{split}y_{u}(t):=r^{\delta}u(r)\quad&\text{and}\quad
y_{v}(t):=r^{\delta}v(r),\\\ \end{split}$

where $\displaystyle t:=\log r$ with $\displaystyle r>0$ and
$\displaystyle\delta=\frac{n-2}{2}$. We prove that

there exists $\displaystyle C>0$ such that $\displaystyle y_{u}=Cy_{v}$

where $\displaystyle C$ is a zero of the function

$f(s)=s^{2^{*}-2}+\nu\alpha s^{\alpha-2}-1-\nu\beta s^{\alpha}.$

The number of roots of $\displaystyle f$ is equivalent to the solutions of the
system (1.8), then such quantity gives us the number of synchronized
solutions. Although in this final issue we are reduced to deal with an ODE
analysis, the proof is no longer standard. To the best of our knowledge, only
the case of a single root of $\displaystyle f$ has been treated in the
literature. Precisely, the hardest part is the case when the function
$\displaystyle f(\cdot)$ has more than one zero, a possible issue, see Remark
1.4. When the proportionality of components $\displaystyle(y_{u},y_{v})$ is
guaranteed, one can conclude the classification result by direct computation.

##  2\. Radial symmetry of the solutions, proof of Theorem 1.5

The aim of this section is to prove that any solution to
($\displaystyle\mathcal{S}^{*}_{\gamma_{1,2}}$) is radially symmetric about
the origin. We shall therefore consider positive continuous (far from the
origin) solutions $\displaystyle(u,v)\in H^{1}_{loc}({\mathbb{R}}^{n})\times
H^{1}_{loc}({\mathbb{R}}^{n})$ such that

(2.1) $\begin{split}\int_{\mathbb{R}^{n}}\langle\nabla
u,\nabla\Phi\rangle\,dx\,=\,\gamma_{1}\int_{{\mathbb{R}}^{n}}\frac{u}{|x|^{2}}\Phi\,dx+\int_{\mathbb{R}^{n}}{u}^{2^{*}-1}\Phi\,dx+\nu\alpha\int_{\mathbb{R}^{n}}{u}^{\alpha-1}{v}^{\beta}\Phi\,dx,&\vspace{0.2cm}\\\
\displaystyle\int_{\mathbb{R}^{n}}\langle\nabla{v},\nabla\Psi\rangle\,dx\,=\gamma_{2}\int_{{\mathbb{R}}^{n}}\frac{v}{|x|^{2}}\Psi\,dx+\,\int_{\mathbb{R}^{n}}{v}^{2^{*}-1}\Psi\,dx+\nu\beta\int_{\mathbb{R}^{n}}{u}^{\alpha}{v}^{\beta-1}\Psi\,dx,&\end{split}$

for any $\displaystyle\Phi,\Psi\in C^{1}_{c}(\mathbb{R}^{n})$ and
$\displaystyle\gamma_{1},\gamma_{2}\in[0,\Lambda_{n})$. To prove Theorem 1.5,
we need to fix some notations. For any real number $\displaystyle\lambda$ we
set

$\Sigma_{\lambda}=\\{x\in{\mathbb{R}}^{n}:x_{1}<\lambda\\}$
$x_{\lambda}=R_{\lambda}(x)=(2\lambda-x_{1},x_{2},\ldots,x_{n})$

which is the reflection through the hyperplane $\displaystyle
T_{\lambda}:=\\{x_{1}=\lambda\\}$. Moreover, given any function $\displaystyle
w$, we will set

$w_{\lambda}:=w\circ R_{\lambda},$

namely the reflected function. A crucial ingredient in our proof is the use of
a translation argument. We fix

$\displaystyle x_{0}=(0,x^{\prime}_{0})\in{\mathbb{R}}^{n}\setminus\\{0\\}$

and we assume, by translation, that $\displaystyle(u,v)$ solves

($\displaystyle\mathcal{S}^{*}_{x_{0}}$) $\begin{cases}\displaystyle-\Delta
u\,=\gamma_{1}\frac{u}{|x-x_{0}|^{2}}+u^{2^{*}-1}+\nu\alpha
u^{\alpha-1}v^{\beta}&\text{in}\quad{\mathbb{R}}^{n}\vspace{0.2cm}\\\
\displaystyle-\Delta
v\,=\gamma_{2}\frac{v}{|x-x_{0}|^{2}}+v^{2^{*}-1}+\nu\beta
u^{\alpha}v^{\beta-1}&\text{in}\quad{\mathbb{R}}^{n}\\\
u,v>0&\text{in}\quad{\mathbb{R}}^{n}\setminus\\{x_{0}\\}.\end{cases}$

This will allows us to take full advantage of the Kelvin transformation. In
fact we set $\displaystyle
K:{\mathbb{R}}^{n}\setminus\\{0\\}\longrightarrow{\mathbb{R}}^{n}\setminus\\{0\\}$
defined by

$\displaystyle K=K(x):=\frac{x}{|x|^{2}}.$

Such a transformation is a well-known tool and, given any $\displaystyle(u,v)$
solution to ($\displaystyle\mathcal{S}^{*}_{x_{0}}$), its Kelvin transform is
defined as

(2.2)
$\left(\hat{u}(x),\hat{v}(x)\right):=\left(\frac{1}{|x|^{n-2}}u\left(\frac{x}{|x|^{2}}\right),\frac{1}{|x|^{n-2}}v\left(\frac{x}{|x|^{2}}\right)\right)\quad
x\in{\mathbb{R}}^{n}\setminus\\{0,x_{0}\\}.$

By direct computation it follows that $\displaystyle(\hat{u},\hat{v})$ weakly
satisfies

($\displaystyle\hat{\mathcal{S}}^{*}_{x_{0}}$)
$\begin{cases}\displaystyle-\Delta\hat{u}\,=\gamma_{1}\frac{\hat{u}}{f_{x_{0}}(x)}+\hat{u}^{2^{*}-1}+\nu\alpha\hat{u}^{\alpha-1}\hat{v}^{\beta}&\text{in}\quad{\mathbb{R}}^{n}\vspace{0.2cm}\\\
\displaystyle-\Delta\hat{v}\,=\gamma_{2}\frac{\hat{v}}{f_{x_{0}}(x)}+\hat{v}^{2^{*}-1}+\nu\beta\hat{u}^{\alpha}\hat{v}^{\beta-1}&\text{in}\quad{\mathbb{R}}^{n}\vspace{0.2cm}\\\
\hat{u},\hat{v}>0&\text{in}\quad{\mathbb{R}}^{n}\setminus\\{0,x_{0}\\},\end{cases}$

where

(2.3) $f_{x_{0}}(x):=\big{|}x-x_{0}|x|^{2}\big{|}^{2}.$

Having in mind the last definition, we prove the following key lemma that will
be used in the proof of the symmetry result.

######  Lemma 2.1.

Let $\displaystyle x_{0}\in{\mathbb{R}}^{n}\setminus\\{0\\}$ be any fixed
point such that $\displaystyle
x_{0}=(0,x^{\prime}_{0})=(0,x_{0,2},\dots,x_{0,n})$. Then the function
$\displaystyle f_{x_{0}}$ is increasing in the $\displaystyle x_{1}$-direction
for any $\displaystyle x\in{\mathbb{R}}^{n}\setminus\Sigma_{0}$, decreasing in
in the $\displaystyle x_{1}$-direction for any $\displaystyle x\in\Sigma_{0}$,
and symmetric with respect to $\displaystyle T_{0}$.

###### Proof.

It is immediate to have that

(2.4) $\begin{split}f_{x_{0}}(x)&=\big{|}x-x_{0}|x|^{2}\big{|}^{2}=\langle
x-x_{0}|x|^{2},x-x_{0}|x|^{2}\rangle\\\ &=|x|^{2}-2|x|^{2}\langle
x,x_{0}\rangle+|x_{0}|^{2}|x|^{4}\\\ &=|x|^{2}\left(1-2\langle
x,x_{0}\rangle+|x_{0}|^{2}|x|^{2}\right),\end{split}$

for every $\displaystyle x\in{\mathbb{R}}^{n}$. We observe that thanks to our
assumptions, the term $\displaystyle\langle x,x_{0}\rangle$ does not depend on
$\displaystyle x_{1}$. Hence, thanks to Schwarz inequality we deduce that for
any $\displaystyle x\in{\mathbb{R}}^{n}\setminus\Sigma_{0}$ it holds

(2.5) $\begin{split}\frac{\partial f_{x_{0}}}{\partial
x_{1}}(x)&=2x_{1}\left(1-2\langle x,x_{0}\rangle+2|x_{0}|^{2}|x|^{2}\right)\\\
&\geq 2x_{1}\left(1-2|x|\cdot|x_{0}|+2|x_{0}|^{2}|x|^{2}\right)\\\
&=2x_{1}\left[\left(1-|x_{0}|\cdot|x|\right)^{2}+|x_{0}|^{2}\cdot|x|^{2}\right]\geq
0.\end{split}$

Analogously, we get $\displaystyle\frac{\partial f_{x_{0}}}{\partial
x_{1}}(x)\leq 0$ for any $\displaystyle x\in\Sigma_{0}$.

∎

We shall prove a symmetry result regarding any solution to
($\displaystyle\hat{\mathcal{S}}^{*}_{x_{0}}$), that translates into a
symmetry result for the original problem
($\displaystyle\mathcal{S}^{*}_{\gamma_{1,2}}$). In particular, the couple
$\displaystyle(\hat{u},\hat{v})$ satisfies

(2.6)
$\begin{array}[]{lr}\displaystyle\int_{\mathbb{R}^{n}}\langle\nabla\hat{u},\nabla\Phi\rangle\,dx\,=\,\gamma_{1}\int_{{\mathbb{R}}^{n}}\frac{\hat{u}}{f_{x_{0}}(x)}\Phi\,dx+\int_{\mathbb{R}^{n}}\hat{u}^{2^{*}-1}\Phi\,dx+\nu\alpha\int_{\mathbb{R}^{n}}\hat{u}^{\alpha-1}\hat{v}^{\beta}\Phi\,dx,\\\
\\\
\displaystyle\int_{\mathbb{R}^{n}}\langle\nabla\hat{v},\nabla\Psi\rangle\,dx\,=\,\gamma_{2}\int_{{\mathbb{R}}^{n}}\frac{\hat{v}}{f_{x_{0}}(x)}\Psi\,dx+\int_{\mathbb{R}^{n}}\hat{v}^{2^{*}-1}\Psi\,dx+\nu\beta\int_{\mathbb{R}^{n}}\hat{u}^{\alpha}\hat{v}^{\beta-1}\Psi\,dx,\end{array}$

for any $\displaystyle\Phi,\Psi\in
C^{1}_{c}(\mathbb{R}^{n}\setminus\\{0,x_{0}\\})$, while reflecting through
$\displaystyle T_{\lambda}$ we deduce that
$\displaystyle(\hat{u}_{\lambda},\hat{v}_{\lambda})$ satisfies

(2.7)
$\begin{array}[]{lr}\displaystyle\int_{\mathbb{R}^{n}}\langle\nabla\hat{u}_{\lambda},\nabla\Phi\rangle\,dx\,=\,\gamma_{1}\int_{{\mathbb{R}}^{n}}\frac{\hat{u}_{\lambda}}{f_{x_{0}}(x_{\lambda})}\Phi\,dx+\int_{\mathbb{R}^{n}}\hat{u}_{\lambda}^{2^{*}-1}\Phi\,dx+\nu\alpha\int_{\mathbb{R}^{n}}\hat{u}_{\lambda}^{\alpha-1}\hat{v}_{\lambda}^{\beta}\Phi\,dx,\\\
\\\
\displaystyle\int_{\mathbb{R}^{n}}\langle\nabla\hat{v}_{\lambda},\nabla\Psi\rangle\,dx\,=\gamma_{2}\int_{{\mathbb{R}}^{n}}\frac{\hat{v}_{\lambda}}{f_{x_{0}}(x_{\lambda})}\Psi\,dx+\,\int_{\mathbb{R}^{n}}\hat{v}_{\lambda}^{2^{*}-1}\Psi\,dx+\nu\beta\int_{\mathbb{R}^{n}}\hat{u}_{\lambda}^{\alpha}\hat{v}_{\lambda}^{\beta-1}\Psi\,dx,\end{array}$

for any $\displaystyle\Phi,\Psi\in C^{1}_{c}(\mathbb{R}^{n}\setminus
R_{\lambda}(\\{0,x_{0}\\}))$.

A crucial point in all the paper, as recalled in the introduction, is
represented by the fact that the coupling term may be not Lipschitz continuous
at zero. To face this difficulty, we will also use the following:

######  Lemma 2.2.

Let $\displaystyle(\hat{u},\hat{v})$ be a solution to
($\displaystyle\hat{\mathcal{S}}^{*}_{x_{0}}$). Then, there exists
$\displaystyle\bar{R}>0$ and $\displaystyle\mu>0$ such that

(2.8) $\hat{u}(x)\geq\mu>0\quad\text{and}\quad\hat{v}(x)\geq\mu>0\quad\text{in
}B_{\bar{R}}.$

###### Proof.

First of all, we note that $\displaystyle(\hat{u},\hat{v})$ satisfies (2.6).
Borrowing an idea contained in [23], we point out that, for every
$\displaystyle\varepsilon>0$, we can find a function
$\displaystyle\psi_{\varepsilon}\in C^{0,1}({\mathbb{R}}^{N},[0,1])$ such that

$\displaystyle\int_{\Sigma_{\lambda}}|\nabla\psi_{\varepsilon}|^{2}<\varepsilon$

and $\displaystyle\psi_{\varepsilon}=0$ in an open neighborhood
$\displaystyle\mathcal{B_{\varepsilon}}$ of $\displaystyle 0$, with
$\displaystyle\mathcal{B_{\varepsilon}}\subset{\mathbb{R}}^{n}$. We note that
there exists $\displaystyle\bar{R}>0$ such that $\displaystyle x_{0}\not\in
B_{\bar{R}}(0)$. Since $\displaystyle\hat{u}>0$ in $\displaystyle
B_{\bar{R}}(0)\setminus\\{0\\}$, there exists $\displaystyle\mu>0$ such that
$\displaystyle\hat{u}>\mu>0$. In the same way $\displaystyle\hat{v}>\mu>0$ on
$\displaystyle\partial B_{\bar{R}}(0)$. Hence, setting

(2.9)
$\Phi:=(\mu-\hat{u})^{+}\psi_{\varepsilon}^{2}\chi_{B_{\bar{R}}(0)}\quad\text{and}\quad\Psi:=(\mu-\hat{v})^{+}\psi_{\varepsilon}^{2}\chi_{B_{\bar{R}}(0)},$

one can check that $\displaystyle\Phi,\Psi\in D^{1,2}({\mathbb{R}}^{n})$, and
by density arguments we can choose these as test functions in (2.6). For the
reader convenience we make the computations just for the first equation of
(2.6). Hence. we are able to deduce that

(2.10)
$\begin{split}\displaystyle-\int_{B_{\bar{R}}(0)}|\nabla(\mu-\hat{u})^{+}|^{2}\psi_{\varepsilon}^{2}\,dx\,+2\int_{B_{\bar{R}}(0)}\langle\nabla\hat{u},\nabla\psi_{\varepsilon}\rangle(\mu-\hat{u})^{+}\psi_{\varepsilon}\geq
0.\end{split}$

Using the Young’s inequality, we are able to get that

(2.11)
$\begin{split}\displaystyle\int_{B_{\bar{R}}(0)}|\nabla(\mu-\hat{u})^{+}|^{2}\psi_{\varepsilon}^{2}\,dx&\leq
2\int_{B_{\bar{R}}(0)}\langle\nabla\hat{u},\nabla\psi_{\varepsilon}\rangle(\mu-\hat{u})^{+}\psi_{\varepsilon}\,dx\\\
&\leq\frac{1}{2}\int_{B_{\bar{R}}(0)}|\nabla(\mu-\hat{u})^{+}|^{2}\psi_{\varepsilon}^{2}\,dx+2\int_{B_{\bar{R}}(0)}|\nabla\psi_{\varepsilon}|^{2}[(\mu-\hat{u})^{+}]^{2}\,dx.\end{split}$

Finally, we have that

(2.12)
$\int_{B_{\bar{R}}(0)}|\nabla(\mu-\hat{u})^{+}|^{2}\psi_{\varepsilon}^{2}\,dx\leq
4\mathcal{C}\int_{B_{\bar{R}}(0)}|\nabla\psi_{\varepsilon}|^{2}\,dx\leq
4\mathcal{C}\varepsilon.$

Passing to the limit for $\displaystyle\varepsilon$ that goes to
$\displaystyle 0$ we get the thesis for $\displaystyle\hat{u}$. Arguing in a
similar fashion, we obtain the same result for $\displaystyle\hat{v}$.

∎

The translation argument that we introduced allow us to deduce the following:

######  Lemma 2.3.

Let $\displaystyle(\hat{u},\hat{v})$ be a solution to
($\displaystyle\hat{\mathcal{S}}^{*}_{x_{0}}$). Then, there exist
$\displaystyle c_{\hat{u}},C_{\hat{u}}>0$, $\displaystyle
c_{\hat{v}},C_{\hat{v}}>0$ and $\displaystyle\tilde{R}:=\tilde{R}(x_{0})>0$
such that

(2.13)
$\frac{c_{u}}{|x|^{n-2}}\leq\hat{u}(x)\leq\frac{C_{u}}{|x|^{n-2}}\quad\text{and}\quad\frac{c_{v}}{|x|^{n-2}}\leq\hat{v}(x)\leq\frac{C_{v}}{|x|^{n-2}}\quad\text{in
}{\mathbb{R}}^{n}\setminus B_{\tilde{R}}(0).$

###### Proof.

Since $\displaystyle x_{0}=(0,x_{0}^{\prime})\in{\mathbb{R}}^{n}$ is fixed,
then there exists $\displaystyle\delta>0$ such that $\displaystyle
x_{0}\not\in B_{\delta}(0)$. By our assumptions $\displaystyle u,v\in
C({\mathbb{R}}^{n}\setminus\\{x_{0}\\})$, and hence we deduce that
$\displaystyle u,v\in C(B_{\delta}(0))$. We fix
$\displaystyle\tilde{R}>\frac{1}{\delta}$ in such a way that by the definition
of Kelvin transformation (2.2) we easily deduce (2.13).

∎

Now, we are ready to prove that any positive solution to
($\displaystyle\hat{\mathcal{S}}^{*}_{x_{0}}$) is symmetric in the
$\displaystyle x_{1}$ direction.

###### Proof of Theorem 1.5.

Let us consider

$\begin{split}\xi_{\lambda}(x):=\hat{u}(x)-\hat{u}_{\lambda}(x)=\hat{u}(x)-\hat{u}(x_{\lambda}),\\\
\zeta_{\lambda}(x):=\hat{v}(x)-\hat{v}_{\lambda}(x)=\hat{v}(x)-\hat{v}(x_{\lambda}).\end{split}$

We recall that we are working with the weak formulations (2.6) and (2.7). By
Lemma 2.3 we deduce that $\displaystyle|\hat{u}(x)|\leq C_{u}|x|^{2-n}$ and
$\displaystyle|\hat{v}(x)|\leq C_{v}|x|^{2-n}$ and for every $\displaystyle
x\in{\mathbb{R}}^{n}$ such that $\displaystyle|x|\geq\tilde{R}$, where
$\displaystyle C_{u},C_{v}$ and $\displaystyle\tilde{R}$ are positive
constants (depending on $\displaystyle u$ and $\displaystyle v$). In
particular, for every $\displaystyle\lambda<-\tilde{R}<0$, we have

$\hat{u},\hat{v}\in L^{2^{*}}(\Sigma_{\lambda})\cap
L^{\infty}(\Sigma_{\lambda})\cap C^{0}(\overline{\Sigma_{\lambda}})\,.$

In order to complete the proof of our result, we split the proof into three
steps.

_Step 1: There exists $\displaystyle M>0$_ large such that
$\displaystyle\hat{u}\leq\hat{u}_{\lambda}$ and
$\displaystyle\hat{v}\leq\hat{v}_{\lambda}$ in
$\displaystyle\Sigma_{\lambda}\setminus R_{\lambda}(\\{0,x_{0}\\})$, for all
$\displaystyle\lambda<-M$. We immediately see that
$\displaystyle\xi_{\lambda}^{+},\zeta_{\lambda}^{+}\in
L^{2^{*}}(\Sigma_{\lambda}),$ since

$\displaystyle 0\leq\xi_{\lambda}^{+}\leq\hat{u}\in
L^{2^{*}}(\Sigma_{\lambda})\ \ \text{ and }\ \
0\leq\zeta_{\lambda}^{+}\leq\hat{v}\in L^{2^{*}}(\Sigma_{\lambda}).$

We point out that, for every $\displaystyle\varepsilon>0$, we can find a
function $\displaystyle\psi_{\varepsilon}\in C^{0,1}({\mathbb{R}}^{N},[0,1])$
such that

$\displaystyle\int_{\Sigma_{\lambda}}|\nabla\psi_{\varepsilon}|^{2}<\varepsilon$

and $\displaystyle\psi_{\varepsilon}=0$ in an open neighborhood
$\displaystyle\mathcal{B_{\varepsilon}}$ of $\displaystyle
R_{\lambda}(\\{0,x_{0}\\})$, with
$\displaystyle\mathcal{B_{\varepsilon}}\subset\Sigma_{\lambda}$.

Fix $\displaystyle R_{0}>0$ such that $\displaystyle
R_{\lambda}(\\{0,x_{0}\\})\in B_{R_{0}}$ and, for every $\displaystyle
R>R_{0}$, let $\displaystyle\eta_{R}$ be a standard cut off function such that
$\displaystyle 0\leq\eta_{R}\leq 1$ on $\displaystyle{\mathbb{R}}^{n}$,
$\displaystyle\eta_{R}=1$ in $\displaystyle B_{R}$, $\displaystyle\eta_{R}=0$
outside $\displaystyle B_{2R}$ with $\displaystyle|\nabla\eta_{R}|\leq 2/R,$
and consider

$\displaystyle\Phi\,:=\begin{cases}\,\xi_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,&\text{in}\quad\Sigma_{\lambda},\\\
0&\text{in}\quad{\mathbb{R}}^{n}\setminus\Sigma_{\lambda}\end{cases}\quad\text{and}\quad\Psi\,:=\begin{cases}\,\zeta_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,&\text{in}\quad\Sigma_{\lambda},\\\
0&\text{in}\quad{\mathbb{R}}^{n}\setminus\Sigma_{\lambda}.\end{cases}$

Now, it is easy to infer that $\displaystyle\Phi,\Psi\in
C^{0,1}_{c}({\mathbb{R}}^{n})$ with $\displaystyle\text{supp}(\Phi)$ and
$\displaystyle\text{supp}(\Psi)$ contained in
$\displaystyle\overline{\Sigma_{\lambda}\cap B_{2R}}\setminus
R_{\lambda}(\\{0,x_{0}\\})$ and

(2.14)
$\nabla\Phi=\psi_{\varepsilon}^{2}\eta_{R}^{2}\nabla\xi^{+}_{\lambda}+2\xi_{\lambda}^{+}(\psi_{\varepsilon}^{2}\eta_{R}\nabla\eta_{R}+\psi_{\varepsilon}\eta_{R}^{2}\nabla\psi_{\varepsilon}),$
(2.15)
$\nabla\Psi=\psi_{\varepsilon}^{2}\eta_{R}^{2}\nabla\zeta^{+}_{\lambda}+2\zeta_{\lambda}^{+}(\psi_{\varepsilon}^{2}\eta_{R}\nabla\eta_{R}+\psi_{\varepsilon}\eta_{R}^{2}\nabla\psi_{\varepsilon}).$

Therefore, by a standard density argument, we can plug $\displaystyle\Phi$ and
$\displaystyle\Psi$ as test functions in (2.6) and (2.7) respectively, so
that, subtracting we get

(2.16)
$\begin{split}\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx=&-2\int_{\Sigma_{\lambda}}\langle\nabla\xi^{+}_{\lambda},\nabla\psi_{\varepsilon}\rangle\xi_{\lambda}^{+}\psi_{\varepsilon}\eta_{R}^{2}\,dx-2\int_{\Sigma_{\lambda}}\langle\nabla\xi^{+}_{\lambda},\nabla\eta_{R}\rangle\xi_{\lambda}^{+}\eta_{R}\psi_{\varepsilon}^{2}\,dx\\\
&+\gamma_{1}\int_{\Sigma_{\lambda}}\left(\frac{\hat{u}}{f_{x_{0}}(x)}-\frac{\hat{u}_{\lambda}}{f_{x_{0}}(x_{\lambda})}\right)\xi_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,\,\\\
&+\int_{\Sigma_{\lambda}}(\hat{u}^{2^{*}-1}-\hat{u}_{\lambda}^{2^{*}-1})\xi_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,\,\\\
&+\nu\alpha\int_{\Sigma_{\lambda}}(\hat{u}^{\alpha-1}\hat{v}^{\beta}-\hat{u}^{\alpha-1}_{\lambda}\hat{v}_{\lambda}^{\beta})\xi_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,\,\\\
=:&\,\mathcal{I}_{1}+\mathcal{I}_{2}+\mathcal{I}_{3}+\mathcal{I}_{4}+\mathcal{I}_{5}\,\end{split}$

and

(2.17)
$\begin{split}\int_{\Sigma_{\lambda}}|\nabla\zeta^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx=&-2\int_{\Sigma_{\lambda}}\langle\nabla\zeta^{+}_{\lambda},\nabla\psi_{\varepsilon}\rangle\zeta_{\lambda}^{+}\psi_{\varepsilon}\eta_{R}^{2}\,dx-2\int_{\Sigma_{\lambda}}\langle\nabla\zeta^{+}_{\lambda},\nabla\eta_{R}\rangle\zeta_{\lambda}^{+}\eta_{R}\psi_{\varepsilon}^{2}\,dx\\\
&+\gamma_{2}\int_{\Sigma_{\lambda}}\left(\frac{\hat{v}}{f_{x_{0}}(x)}-\frac{\hat{v}_{\lambda}}{f_{x_{0}}(x_{\lambda})}\right)\zeta_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,\,\\\
&+\int_{\Sigma_{\lambda}}(\hat{v}^{2^{*}-1}-\hat{v}_{\lambda}^{2^{*}-1})\zeta_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,\,\\\
&+\nu\beta\int_{\Sigma_{\lambda}}(\hat{u}^{\alpha}\hat{v}^{\beta-1}-\hat{u}^{\alpha}_{\lambda}\hat{v}_{\lambda}^{\beta-1})\zeta_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,\\\
=:&\,\mathcal{E}_{1}+\mathcal{E}_{2}+\mathcal{E}_{3}+\mathcal{E}_{4}+\mathcal{E}_{5}\,.\end{split}$

Exploiting also Young’s inequality, recalling that $\displaystyle
0\leq\xi_{\lambda}^{+}\leq\hat{u}$ and $\displaystyle
0\leq\zeta_{\lambda}^{+}\leq\hat{v}$, we get that

(2.18)
$\begin{split}|\mathcal{I}_{1}|&\leq\frac{1}{4}\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+4\int_{\Sigma_{\lambda}}|\nabla\psi_{\varepsilon}|^{2}(\xi_{\lambda}^{+})^{2}\eta_{R}^{2}\,dx\\\
&\leq\frac{1}{4}\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+4\varepsilon\|\hat{u}\|^{2}_{L^{\infty}(\Sigma_{\lambda})}.\\\
\end{split}$

Similarly, we obtain

(2.19)
$\begin{split}|\mathcal{E}_{1}|&\leq\frac{1}{4}\int_{\Sigma_{\lambda}}|\nabla\zeta^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+4\varepsilon\|\hat{v}\|^{2}_{L^{\infty}(\Sigma_{\lambda})}.\\\
\end{split}$

Furthermore, using Hölder’s inequality with exponents
$\displaystyle\left(\frac{2^{*}}{2},\frac{n}{2}\right)$, one has that

(2.20)
$\begin{split}|\mathcal{I}_{2}|\leq&\frac{1}{4}\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+4\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}|\nabla\eta_{R}|^{2}(\xi_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\,dx\\\
\leq&\frac{1}{4}\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\\\
&+4\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}|\nabla\eta_{R}|^{n}\,dx\right)^{\frac{2}{n}}\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}\hat{u}^{2^{*}}\,dx\right)^{\frac{n-2}{n}}\\\
\leq&\frac{1}{4}\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,+\,c(n)\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}\hat{u}^{2^{*}}\,dx\right)^{\frac{n-2}{n}}.\end{split}$

where $\displaystyle c(n)$ is a positive constant depending only on the
dimension $\displaystyle n$ and

$\displaystyle\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}\hat{u}^{2^{*}}\,dx\right)^{\frac{n-2}{n}}\rightarrow 0\qquad$ as
$\displaystyle R\rightarrow+\infty$

by the absolute continuity of the Lebesgue integral. Analogously, we deduce
that

(2.21)
$\begin{split}|\mathcal{E}_{2}|\leq&\frac{1}{4}\int_{\Sigma_{\lambda}}|\nabla\zeta^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,+\,c(n)\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}\hat{v}^{2^{*}}\,dx\right)^{\frac{n-2}{n}}.\end{split}$

Let us now estimate $\displaystyle\mathcal{I}_{3}$ and
$\displaystyle\mathcal{E}_{3}$. Here, we recall the monotonicity property of
the function $\displaystyle f_{x_{0}}$ stated in Lemma 2.1. Moreover, for
$\displaystyle\lambda<0$ sufficiently large we have that

$\displaystyle\frac{1}{f_{x_{0}}(x_{\lambda})}\geq\frac{1}{f_{x_{0}}(x)}.$

Therefore

(2.22)
$\begin{split}|\mathcal{I}_{3}|&\leq\gamma_{1}\int_{\Sigma_{\lambda}}\frac{1}{f_{x_{0}}(x)}\left|\hat{u}-\hat{u}_{\lambda}\right|\xi_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,\,\\\
&\leq\gamma_{1}C\int_{\Sigma_{\lambda}}\frac{(\xi_{\lambda}^{+})^{2}}{|x|^{4}}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\leq\gamma_{1}CC_{\lambda}\int_{\Sigma_{\lambda}}\frac{(\xi_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}}{|x|^{2}}\,dx,\end{split}$

where

(2.23) $C_{\lambda}:=\sup_{x\in\Sigma_{\lambda}}\frac{1}{|x|^{2}}\ \text{ and
}\ C_{\lambda}\rightarrow 0\ \text{ as }\ \lambda\rightarrow-\infty.$

By Hardy’s and Young’s inequality we also deduce that

(2.24)
$\begin{split}|\mathcal{I}_{3}|\leq&\gamma_{1}C_{H}^{u}C_{\lambda}\int_{\Sigma_{\lambda}}|\nabla(\xi_{\lambda}^{+}\psi_{\varepsilon}\eta_{R})|^{2}\,dx\\\
=&\gamma_{1}C_{H}^{u}C_{\lambda}\int_{\Sigma_{\lambda}}|\psi_{\varepsilon}\eta_{R}\nabla\xi_{\lambda}^{+}+\xi_{\lambda}^{+}\psi_{\varepsilon}\nabla\eta_{R}+\xi_{\lambda}^{+}\eta_{R}\nabla\psi_{\varepsilon}|^{2}\,dx\\\
\leq&3\gamma_{1}C_{H}^{u}C_{\lambda}\int_{\Sigma_{\lambda}}|\nabla\xi_{\lambda}^{+}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+3\gamma_{1}C_{H}^{u}C_{\lambda}\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}|\nabla\eta_{R}|^{2}(\xi_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\,dx\\\
&+3\gamma_{1}C_{H}^{u}C_{\lambda}\int_{\Sigma_{\lambda}}|\nabla\psi_{\varepsilon}|^{2}(\xi_{\lambda}^{+})^{2}\eta_{R}^{2}\,dx\\\
\leq&3\gamma_{1}C_{H}^{u}C_{\lambda}\int_{\Sigma_{\lambda}}|\nabla\xi_{\lambda}^{+}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+3\gamma_{1}C_{H}^{u}C_{\lambda}c(n)\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}\hat{u}^{2^{*}}\,dx\right)^{\frac{n-2}{n}}\\\
&+3\gamma_{1}C_{H}^{u}C_{\lambda}\varepsilon\|\hat{u}\|^{2}_{L^{\infty}(\Sigma_{\lambda})}.\end{split}$

Employing the argument to estimate $\displaystyle\mathcal{I}_{3}$, we get

(2.25)
$\begin{split}|\mathcal{E}_{3}|\leq&3\gamma_{2}C_{H}^{v}C_{\lambda}\int_{\Sigma_{\lambda}}|\nabla\zeta_{\lambda}^{+}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+3\gamma_{2}C_{H}^{v}C_{\lambda}c(n)\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}\hat{v}^{2^{*}}\,dx\right)^{\frac{n-2}{n}}\\\
&+3\gamma_{2}C_{H}^{v}C_{\lambda}\varepsilon\|\hat{v}\|^{2}_{L^{\infty}(\Sigma_{\lambda})}.\end{split}$

Since
$\displaystyle\hat{u}(x),\hat{u}_{\lambda}(x),\hat{v}(x),\hat{v}_{\lambda}(x)>0$,
by the convexity of $\displaystyle t\mapsto t^{2^{*}-1},$ for $\displaystyle
t>0$, we obtain

$\displaystyle\hat{u}^{2^{*}-1}(x)-\hat{u}_{\lambda}^{2^{*}-1}(x)\leq\frac{n+2}{n-2}\hat{u}^{2^{*}-2}(x)(\hat{u}(x)-\hat{u}_{\lambda}(x))$

and

$\displaystyle\hat{v}^{2^{*}-1}(x)-\hat{v}_{\lambda}^{2^{*}-1}(x)\leq\frac{n+2}{n-2}\hat{v}^{2^{*}-2}(x)(\hat{v}(x)-\hat{v}_{\lambda}(x)),$

for every $\displaystyle x\in\Sigma_{\lambda}$. Thus,

$\displaystyle(\hat{u}^{2^{*}-1}-\hat{u}_{\lambda}^{2^{*}-1})\xi_{\lambda}^{+}\leq\frac{n+2}{n-2}\hat{u}^{2^{*}-2}(\hat{u}-\hat{u}_{\lambda})\xi_{\lambda}^{+}\leq\frac{n+2}{n-2}\hat{u}^{2^{*}-2}(\xi_{\lambda}^{+})^{2}$

and

$\displaystyle(\hat{v}^{2^{*}-1}-\hat{v}_{\lambda}^{2^{*}-1})\zeta_{\lambda}^{+}\leq\frac{n+2}{n-2}\hat{v}^{2^{*}-2}(\hat{v}-\hat{v}_{\lambda})\zeta_{\lambda}^{+}\leq\frac{n+2}{n-2}\hat{v}^{2^{*}-2}(\zeta_{\lambda}^{+})^{2}.$

Therefore, using Hölder’s inequality with exponents
$\displaystyle\left(\frac{2^{*}}{2},\frac{n}{2}\right)$, we have

(2.26)
$\begin{split}|\mathcal{I}_{4}|&\leq\frac{n+2}{n-2}\int_{\Sigma_{\lambda}}\hat{u}^{2^{*}-2}(\xi_{\lambda}^{+}\psi_{\varepsilon}\eta_{R})^{2}\,dx\\\
&\leq\frac{n+2}{n-2}\left(\int_{\Sigma_{\lambda}}\hat{u}^{2^{*}}\,dx\right)^{\frac{2}{n}}\left(\int_{\Sigma_{\lambda}}(\xi_{\lambda}^{+}\psi_{\varepsilon}\eta_{R})^{2^{*}}\,dx\right)^{\frac{n-2}{n}}\\\
&\leq\frac{n+2}{n-2}\mathcal{C}_{S}^{u}\left(\int_{\Sigma_{\lambda}}\hat{u}^{2^{*}}\,dx\right)^{\frac{2}{n}}\int_{\Sigma_{\lambda}}|\nabla(\xi_{\lambda}^{+}\psi_{\varepsilon}\eta_{R})|^{2}\,dx\\\
&\leq
3\frac{n+2}{n-2}\mathcal{C}_{S}^{u}\left(\int_{\Sigma_{\lambda}}\hat{u}^{2^{*}}\,dx\right)^{\frac{2}{n}}\left[\int_{\Sigma_{\lambda}}|\nabla\eta_{R}|^{2}(\xi_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\,dx+\int_{\Sigma_{\lambda}}|\nabla\psi_{\varepsilon}|^{2}(\xi_{\lambda}^{+})^{2}\eta_{R}^{2}\,dx\right]\\\
&\quad+3\frac{n+2}{n-2}\mathcal{C}_{S}^{u}\left(\int_{\Sigma_{\lambda}}\hat{u}^{2^{*}}\,dx\right)^{\frac{2}{n}}\int_{\Sigma_{\lambda}}|\nabla\xi_{\lambda}^{+}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx,\end{split}$

where in the last two steps we applied Sobolev and Young’s inequalities
respectively. Arguing in the same way, we deduce

(2.27) $\begin{split}|\mathcal{E}_{4}|&\leq
3\frac{n+2}{n-2}\mathcal{C}_{S}^{v}\left(\int_{\Sigma_{\lambda}}\hat{v}^{2^{*}}\,dx\right)^{\frac{2}{n}}\left[\int_{\Sigma_{\lambda}}|\nabla\eta_{R}|^{2}(\zeta_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\,dx+\int_{\Sigma_{\lambda}}|\nabla\psi_{\varepsilon}|^{2}(\zeta_{\lambda}^{+})^{2}\eta_{R}^{2}\,dx\right]\\\
&\quad+3\frac{n+2}{n-2}\mathcal{C}_{S}^{v}\left(\int_{\Sigma_{\lambda}}\hat{v}^{2^{*}}\,dx\right)^{\frac{2}{n}}\int_{\Sigma_{\lambda}}|\nabla\zeta_{\lambda}^{+}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx.\end{split}$

The evaluation of $\displaystyle\mathcal{I}_{5}$ and
$\displaystyle\mathcal{E}_{5}$ is a delicate issue within this argument. Note
that, in particular, we may have that either $\displaystyle\alpha<2$,
$\displaystyle\beta<2$ or both. In all this cases we have to face a nonlinear
term which is not Lipschitz continuous at zero. We shall exploit the following

######  Remark 2.4.

Let us consider $\displaystyle h(t)=t^{s}$ with $\displaystyle s>0$ and
$\displaystyle b\in[\theta_{1}\,,\,\theta_{2}]$ with
$\displaystyle\theta_{i}>0$. Then, for $\displaystyle 0<a\leq b$

$\displaystyle\frac{h(b)-h(a)}{b-a}\leq C(s,\theta_{1},\theta_{2})\,.$

To prove this fact it is sufficient to exploit the Mean Value Theorem for
$\displaystyle a\in[b/2,b]$ and the Weierstrass Theorem for $\displaystyle
a\in(0,b/2]$ .

Having in mind this argument, one obtains that

(2.28)
$\begin{split}\mathcal{I}_{5}&=\nu\alpha\int_{\Sigma_{\lambda}}(\hat{u}^{\alpha-1}\hat{v}^{\beta}-\hat{u}^{\alpha-1}_{\lambda}\hat{v}^{\beta})\xi_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+\nu\alpha\int_{\Sigma_{\lambda}}(\hat{u}^{\alpha-1}_{\lambda}\hat{v}^{\beta}-\hat{u}^{\alpha-1}_{\lambda}\hat{v}_{\lambda}^{\beta})\xi_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\\\
&\leq\nu\alpha\int_{\Sigma_{\lambda}}(\hat{u}^{\alpha-1}-\hat{u}^{\alpha-1}_{\lambda})\hat{v}^{\beta}\xi_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+\nu\alpha\int_{\Sigma_{\lambda}\cap\\{\hat{v}>\hat{v}_{\lambda}\\}}\hat{u}^{\alpha-1}(\hat{v}^{\beta}-\hat{v}_{\lambda}^{\beta})\xi_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx,\end{split}$

where in the last inequality we used the cooperativity of our system and the
fact that we are working in
$\displaystyle\Sigma_{\lambda}\cap\text{supp}[(u-u_{\lambda})^{+}]$. Making
use of Lemma 2.3 we deduce that

(2.29)
$\begin{split}\mathcal{I}_{5}&\leq\nu\alpha\int_{\Sigma_{\lambda}}|\hat{u}^{\alpha-1}-\hat{u}^{\alpha-1}_{\lambda}|\hat{v}^{\beta}\xi_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+\nu\alpha\int_{\Sigma_{\lambda}\cap\\{\hat{v}>\hat{v}_{\lambda}\\}}\hat{u}^{\alpha-1}|\hat{v}^{\beta}-\hat{v}_{\lambda}^{\beta}|\xi_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\\\
&\leq\nu\alpha\mathcal{C}\int_{\Sigma_{\lambda}}|x|^{-\beta(n-2)}|x|^{-(\alpha-1)(n-2)}|(|x|^{n-2}\hat{u})^{\alpha-1}-(|x|^{n-2}\hat{u}_{\lambda})^{\alpha-1}|\xi_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\\\
&\quad+\nu\alpha\mathcal{C}\int_{\Sigma_{\lambda}\cap\\{\hat{v}>\hat{v}_{\lambda}\\}}|x|^{-(\alpha-1)(n-2)}|x|^{-(\beta-1)(n-2)}\xi_{\lambda}^{+}\zeta_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\\\
&\leq\nu\alpha\mathcal{C}\int_{\Sigma_{\lambda}}|x|^{-(2^{*}-1)(n-2)}\frac{|(|x|^{n-2}\hat{u})^{\alpha-1}-(|x|^{n-2}\hat{u}_{\lambda})^{\alpha-1}|}{|x|^{n-2}\hat{u}-|x|^{n-2}\hat{u}_{\lambda}}|x|^{n-2}(\xi_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\\\
&\quad+\nu\alpha\mathcal{C}\int_{\Sigma_{\lambda}}|x|^{-(2^{*}-2)(n-2)}\xi_{\lambda}^{+}\zeta_{\lambda}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx.\\\
\end{split}$

By Remark 2.4, we get

$\displaystyle\frac{(|x|^{n-2}\hat{u})^{\alpha-1}-(|x|^{n-2}\hat{u}_{\lambda})^{\alpha-1}}{|x|^{n-2}\hat{u}-|x|^{n-2}\hat{u}_{\lambda}}\leq\overline{C},$

whereas $\displaystyle c_{u}\leq|x|^{n-2}\hat{u}\leq C_{u}$ and $\displaystyle
c_{v}\leq|x|^{n-2}\hat{v}\leq C_{v}$ by Lemma 2.3. Then (2.29) rewrites as

(2.30)
$\begin{split}|\mathcal{I}_{5}|&\leq\nu\alpha\mathcal{C}\int_{\Sigma_{\lambda}}\hat{u}^{2^{*}-2}(\xi_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,+\nu\alpha\mathcal{C}\int_{\Sigma_{\lambda}}\hat{u}^{\frac{2^{*}-2}{2}}\xi_{\lambda}^{+}\psi_{\varepsilon}\eta_{R}\hat{v}^{\frac{2^{*}-2}{2}}\zeta_{\lambda}^{+}\psi_{\varepsilon}\eta_{R}\,dx\\\
&\leq\mathcal{C}_{1}\int_{\Sigma_{\lambda}}\hat{u}^{2^{*}-2}(\xi_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,+\mathcal{C}_{2}\int_{\Sigma_{\lambda}}\hat{v}^{2^{*}-2}(\zeta_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx,\end{split}$

where the constant $\displaystyle\mathcal{C}$ has been relabelled and we have
applied Young’s inequality.

Similarly, we can obtain an analogous estimate for
$\displaystyle\mathcal{E}_{5}$, i.e.

(2.31)
$\begin{split}|\mathcal{E}_{5}|\leq\tilde{\mathcal{C}}_{1}\int_{\Sigma_{\lambda}}\hat{u}^{2^{*}-2}(\xi_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,+\tilde{\mathcal{C}}_{2}\int_{\Sigma_{\lambda}}\hat{v}^{2^{*}-2}(\zeta_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx.\end{split}$

As we argued in (2.26) for $\displaystyle\mathcal{I}_{4}$ in, we get that

(2.32) $\begin{split}|\mathcal{I}_{5}|&\leq
3\frac{n+2}{n-2}\mathcal{C}_{S}^{u}\mathcal{C}_{1}\left(\int_{\Sigma_{\lambda}}\hat{u}^{2^{*}}\,dx\right)^{\frac{2}{n}}\left[\int_{\Sigma_{\lambda}}|\nabla\eta_{R}|^{2}(\xi_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\,dx+\int_{\Sigma_{\lambda}}|\nabla\psi_{\varepsilon}|^{2}(\xi_{\lambda}^{+})^{2}\eta_{R}^{2}\,dx\right]\\\
&\quad+3\frac{n+2}{n-2}\mathcal{C}_{S}^{v}\mathcal{C}_{2}\left(\int_{\Sigma_{\lambda}}\hat{v}^{2^{*}}\,dx\right)^{\frac{2}{n}}\left[\int_{\Sigma_{\lambda}}|\nabla\eta_{R}|^{2}(\zeta_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\,dx+\int_{\Sigma_{\lambda}}|\nabla\psi_{\varepsilon}|^{2}(\zeta_{\lambda}^{+})^{2}\eta_{R}^{2}\,dx\right]\\\
&\quad+3\frac{n+2}{n-2}\mathcal{C}_{S}^{u}\mathcal{C}_{1}\left(\int_{\Sigma_{\lambda}}\hat{u}^{2^{*}}\,dx\right)^{\frac{2}{n}}\int_{\Sigma_{\lambda}}|\nabla\xi_{\lambda}^{+}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\\\
&\quad+3\frac{n+2}{n-2}\mathcal{C}_{S}^{v}\mathcal{C}_{2}\left(\int_{\Sigma_{\lambda}}\hat{v}^{2^{*}}\,dx\right)^{\frac{2}{n}}\int_{\Sigma_{\lambda}}|\nabla\zeta_{\lambda}^{+}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx,\end{split}$

Analogously, we deduce that

(2.33) $\begin{split}|\mathcal{E}_{5}|&\leq
3\frac{n+2}{n-2}\mathcal{C}_{S}^{u}\tilde{\mathcal{C}}_{1}\left(\int_{\Sigma_{\lambda}}\hat{u}^{2^{*}}\,dx\right)^{\frac{2}{n}}\left[\int_{\Sigma_{\lambda}}|\nabla\eta_{R}|^{2}(\xi_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\,dx+\int_{\Sigma_{\lambda}}|\nabla\psi_{\varepsilon}|^{2}(\xi_{\lambda}^{+})^{2}\eta_{R}^{2}\,dx\right]\\\
&\quad+3\frac{n+2}{n-2}\mathcal{C}_{S}^{v}\tilde{\mathcal{C}}_{2}\left(\int_{\Sigma_{\lambda}}\hat{v}^{2^{*}}\,dx\right)^{\frac{2}{n}}\left[\int_{\Sigma_{\lambda}}|\nabla\eta_{R}|^{2}(\zeta_{\lambda}^{+})^{2}\psi_{\varepsilon}^{2}\,dx+\int_{\Sigma_{\lambda}}|\nabla\psi_{\varepsilon}|^{2}(\zeta_{\lambda}^{+})^{2}\eta_{R}^{2}\,dx\right]\\\
&\quad+3\frac{n+2}{n-2}\mathcal{C}_{S}^{u}\tilde{\mathcal{C}}_{1}\left(\int_{\Sigma_{\lambda}}\hat{u}^{2^{*}}\,dx\right)^{\frac{2}{n}}\int_{\Sigma_{\lambda}}|\nabla\xi_{\lambda}^{+}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\\\
&\quad+3\frac{n+2}{n-2}\mathcal{C}_{S}^{v}\tilde{\mathcal{C}}_{2}\left(\int_{\Sigma_{\lambda}}\hat{v}^{2^{*}}\,dx\right)^{\frac{2}{n}}\int_{\Sigma_{\lambda}}|\nabla\zeta_{\lambda}^{+}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx.\end{split}$

Collecting all the previous estimates for
$\displaystyle\mathcal{I}_{k},\mathcal{E}_{k}$ with $\displaystyle
k=1,2,3,4,5$, we deduce that

(2.34)
$\begin{split}\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\leq&C_{1,u}\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+C_{1,v}\int_{\Sigma_{\lambda}}|\nabla\zeta_{\lambda}^{+}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx,\\\
&+\varepsilon
C_{2,u}\|\hat{u}\|^{2}_{L^{\infty}(\Sigma_{\lambda})}+\varepsilon
C_{2,v}\|\hat{v}\|^{2}_{L^{\infty}(\Sigma_{\lambda})}\\\
&+C_{3,u}\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}\hat{u}^{2^{*}}\,dx\right)^{\frac{n-2}{n}}+C_{3,v}\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}\hat{v}^{2^{*}}\,dx\right)^{\frac{n-2}{n}},\end{split}$

where

  * $\displaystyle\displaystyle C_{1,u}:=\frac{1}{2}+3\gamma_{1}C_{H}^{u}C_{\lambda}+3\frac{n+2}{n-2}\mathcal{C}_{S}^{u}(1+\mathcal{C}_{1})\|\hat{u}\|_{L^{2^{*}}(\Sigma_{\lambda})}^{\frac{4}{n-2}},$

  * $\displaystyle C_{1,v}:=\displaystyle 3\frac{n+2}{n-2}\mathcal{C}_{S}^{v}\mathcal{C}_{2}\|\hat{v}\|_{L^{2^{*}}(\Sigma_{\lambda})}^{\frac{4}{n-2}},$

  * $\displaystyle\displaystyle C_{2,u}:=4+3\gamma_{1}C_{H}^{u}C_{\lambda}+3\frac{n+2}{n-2}\mathcal{C}_{S}^{u}(1+\mathcal{C}_{1})\|\hat{u}\|_{L^{2^{*}}(\Sigma_{\lambda})}^{\frac{4}{n-2}},$

  * $\displaystyle\displaystyle C_{2,v}:=3\frac{n+2}{n-2}\mathcal{C}_{S}^{v}\mathcal{C}_{2}\|\hat{v}\|_{L^{2^{*}}(\Sigma_{\lambda})}^{\frac{4}{n-2}},$

  * $\displaystyle\displaystyle C_{3,u}:=c(n)+3\gamma_{1}C_{H}^{u}C_{\lambda}c(n)+3\frac{n+2}{n-2}\mathcal{C}_{S}^{u}(1+\mathcal{C}_{1})\|\hat{u}\|_{L^{2^{*}}(\Sigma_{\lambda})}^{\frac{4}{n-2}},$

  * $\displaystyle\displaystyle C_{3,v}:=3\frac{n+2}{n-2}\mathcal{C}_{S}^{v}\mathcal{C}_{2}\|\hat{v}\|_{L^{2^{*}}(\Sigma_{\lambda})}^{\frac{4}{n-2}}.$

We obtain an analogous estimate for $\displaystyle\zeta_{\lambda}^{+}$, i.e.

(2.35)
$\begin{split}\int_{\Sigma_{\lambda}}|\nabla\zeta^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\leq&\tilde{C}_{1,u}\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+\tilde{C}_{1,v}\int_{\Sigma_{\lambda}}|\nabla\zeta_{\lambda}^{+}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx,\\\
&+\varepsilon\tilde{C}_{2,u}\|\hat{u}\|^{2}_{L^{\infty}(\Sigma_{\lambda})}+\varepsilon\tilde{C}_{2,v}\|\hat{v}\|^{2}_{L^{\infty}(\Sigma_{\lambda})}\\\
&+\tilde{C}_{3,u}\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}\hat{u}^{2^{*}}\,dx\right)^{\frac{n-2}{n}}+\tilde{C}_{3,v}\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}\hat{v}^{2^{*}}\,dx\right)^{\frac{n-2}{n}},\end{split}$

where

  * $\displaystyle\displaystyle\tilde{C}_{1,u}:=3\frac{n+2}{n-2}\mathcal{C}_{S}^{u}\tilde{\mathcal{C}}_{1}\|\hat{u}\|_{L^{2^{*}}(\Sigma_{\lambda})}^{\frac{4}{n-2}},$

  * $\displaystyle\displaystyle\tilde{C}_{1,v}:=\frac{1}{2}+3\gamma_{1}C_{H}^{v}C_{\lambda}+3\frac{n+2}{n-2}\mathcal{C}_{S}^{v}(1+\tilde{\mathcal{C}}_{2})\|\hat{v}\|_{L^{2^{*}}(\Sigma_{\lambda})}^{\frac{4}{n-2}},$

  * $\displaystyle\displaystyle\tilde{C}_{2,u}:=3\frac{n+2}{n-2}\mathcal{C}_{S}^{u}\tilde{\mathcal{C}}_{1}\|\hat{u}\|_{L^{2^{*}}(\Sigma_{\lambda})}^{\frac{4}{n-2}},$

  * $\displaystyle\displaystyle\tilde{C}_{2,v}:=4+3\gamma_{1}C_{H}^{v}C_{\lambda}+3\frac{n+2}{n-2}\mathcal{C}_{S}^{v}(1+\tilde{\mathcal{C}}_{2})\|\hat{v}\|_{L^{2^{*}}(\Sigma_{\lambda})}^{\frac{4}{n-2}},$

  * $\displaystyle\displaystyle\tilde{C}_{3,u}:=3\frac{n+2}{n-2}\mathcal{C}_{S}^{u}\tilde{\mathcal{C}}_{1}\|\hat{u}\|_{L^{2^{*}}(\Sigma_{\lambda})}^{\frac{4}{n-2}},$

  * $\displaystyle\displaystyle\tilde{C}_{3,v}:=c(n)+3\gamma_{1}C_{H}^{u}C_{\lambda}c(n)+3\frac{n+2}{n-2}\mathcal{C}_{S}^{v}\tilde{\mathcal{C}}_{2}\|\hat{v}\|_{L^{2^{*}}(\Sigma_{\lambda})}^{\frac{4}{n-2}}.$

Summing both the contributions (2.34) and (2.35) we get

(2.36)
$\begin{split}\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx&+\int_{\Sigma_{\lambda}}|\nabla\zeta^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\\\
\leq&(C_{1,u}+\tilde{C}_{1,u})\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx+(C_{1,v}+\tilde{C}_{1,v})\int_{\Sigma_{\lambda}}|\nabla\zeta_{\lambda}^{+}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\\\
&+\varepsilon(C_{2,u}+\tilde{C}_{2,u})\|\hat{u}\|^{2}_{L^{\infty}(\Sigma_{\lambda})}+\varepsilon(C_{2,v}+C_{2,v})\|\hat{v}\|^{2}_{L^{\infty}(\Sigma_{\lambda})}\\\
&+(C_{3,u}+\tilde{C}_{3,u})\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}\hat{u}^{2^{*}}\,dx\right)^{\frac{n-2}{n}}\\\
&+(C_{3,v}+\tilde{C}_{3,v})\left(\int_{\Sigma_{\lambda}\cap(B_{2R}\setminus
B_{R})}\hat{v}^{2^{*}}\,dx\right)^{\frac{n-2}{n}}.\end{split}$

Now, recalling (2.23) and using the absolute continuity of the Lebesgue
integral, we fix $\displaystyle M>0$ sufficiently large such that

$\displaystyle C_{1,u}+\tilde{C}_{1,u}<1\quad\text{and}\quad
C_{1,v}+\tilde{C}_{1,v}<1$

for each $\displaystyle\lambda<-M$. Finally, passing to the limit for
$\displaystyle R\rightarrow+\infty$ and $\displaystyle\varepsilon\rightarrow
0^{+}$, we obtain that

(2.37)
$\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\,dx+\int_{\Sigma_{\lambda}}|\nabla\zeta^{+}_{\lambda}|^{2}\,dx\leq
0,$

for each $\displaystyle\lambda<-M$, which implies

(2.38) $\int_{\Sigma_{\lambda}}|\nabla\xi^{+}_{\lambda}|^{2}\,dx\leq
0\quad\text{and}\quad\int_{\Sigma_{\lambda}}|\nabla\zeta^{+}_{\lambda}|^{2}\,dx\leq
0,$

for each $\displaystyle\lambda<-M$. The last inequalities immediately imply
the thesis of the first step.

To proceed further we define

$\Lambda=\\{\lambda<0:\hat{u}\leq\hat{u}_{t}\,\,\,\text{and}\,\,\,\hat{v}\leq\hat{v}_{t}\,\,\,\text{in}\,\,\,\Sigma_{t}\setminus
R_{t}(\\{0,x_{0}\\})\,\,\,\text{for all $\displaystyle
t\in(-\infty,\lambda]$}\\}$

and

$\bar{\lambda}=\sup\,\Lambda.$

_Step 2: $\displaystyle\bar{\lambda}=0$._ We argue by contradiction and
suppose that $\displaystyle\bar{\lambda}<0$. By continuity we know that
$\displaystyle\hat{u}\leq\hat{u}_{\bar{\lambda}}$ and
$\displaystyle\hat{v}\leq\hat{v}_{\bar{\lambda}}$ in
$\displaystyle\Sigma_{\bar{\lambda}}\setminus
R_{\bar{\lambda}}(\\{0,x_{0}\\})$. By the strong comparison principle we
deduce that there hold two possibilities:

  * (i)

$\displaystyle\hat{u}<\hat{u}_{\bar{\lambda}}$ and
$\displaystyle\hat{v}<\hat{v}_{\bar{\lambda}}$ in
$\displaystyle\Sigma_{\bar{\lambda}}\setminus
R_{\bar{\lambda}}(\\{0,x_{0}\\})$;

  * (ii)

$\displaystyle\hat{u}\equiv\hat{u}_{\bar{\lambda}}$ and
$\displaystyle\hat{v}\equiv\hat{v}_{\bar{\lambda}}$ in
$\displaystyle\Sigma_{\bar{\lambda}}\setminus
R_{\bar{\lambda}}(\\{0,x_{0}\\})$.

The case (ii) is not possible ,because, since $\displaystyle\hat{u}$ has a
sign at $\displaystyle x_{0}$, it holds that
$\displaystyle\hat{u}<\hat{u}_{\bar{\lambda}}$ and
$\displaystyle\hat{v}<\hat{v}_{\bar{\lambda}}$ in
$\displaystyle\Sigma_{\bar{\lambda}}\setminus
R_{\bar{\lambda}}(\\{0,x_{0}\\})$.

Now, we push further the hyperplane $\displaystyle T_{\bar{\lambda}}$ and
consider the hyperplane $\displaystyle T_{\bar{\lambda}+\varepsilon}$ for some
$\displaystyle\varepsilon>0$. We claim that, for $\displaystyle\varepsilon>0$
small, we have that

(2.39) $\text{supp}[(\hat{u}-\hat{u}_{\bar{\lambda}+\varepsilon})^{+}],\
\text{supp}[(\hat{v}-\hat{v}_{\bar{\lambda}+\varepsilon})^{+}]\subset(\Sigma_{\bar{\lambda}+\varepsilon}\setminus
B_{\tilde{R}}(0))\cup R_{\bar{\lambda}+\varepsilon}(B_{\delta}(0)\cup
B_{\delta}(x_{0})),$

where $\displaystyle\tilde{R}>0$ is given by Lemma 2.3 and
$\displaystyle\delta>0$ is small. For the reader convenience we set

$\displaystyle\Xi_{\bar{\lambda}+\varepsilon}:=(\Sigma_{\bar{\lambda}+\varepsilon}\setminus
B_{\tilde{R}}(0))\cup R_{\bar{\lambda}+\varepsilon}(B_{\delta}(0)\cup
B_{\delta}(x_{0})).$

Arguing by contradiction, let us assume that (2.39) does not hold. Hence, if
we define

$\displaystyle\mathcal{G}_{\varepsilon}:=B_{\tilde{R}}(0)\cap(\Sigma_{\bar{\lambda}+\varepsilon}\setminus\\{R_{\bar{\lambda}+\varepsilon}(B_{\delta}(0)\cup
B_{\delta}(x_{0}))\\}),$

we deduce that

$\begin{split}\\\ \;\exists\ P_{m}\in\mathcal{G}_{\tau_{m}},\text{ with
}\\{\tau_{m}\\}_{m\in{\mathbb{N}}}\ (\tau_{m}\rightarrow 0)\text{ such that
}\\\ \hat{u}(P_{m})>\hat{u}_{\bar{\lambda}+\tau_{m}}(P_{m})\;\text{ or
}\;\hat{v}(P_{m})>\hat{v}_{\bar{\lambda}+\tau_{m}}(P_{m}).\end{split}$

Without loss of generality, we assume that
$\displaystyle\hat{u}(P_{m})>\hat{u}_{\bar{\lambda}+\tau_{m}}(P_{m})$. Up to
subsequence, as $\displaystyle m\rightarrow+\infty$ we have that

$\displaystyle P_{m}\longrightarrow\bar{P}\in\overline{\mathcal{G}_{0}}.$

Finally we have that or $\displaystyle\bar{P}\in T_{\bar{\lambda}}$ or
$\displaystyle\bar{P}\notin T_{\bar{\lambda}}$.

  * •

If $\displaystyle\bar{P}\notin T_{\bar{\lambda}}$, then we have a
contradiction with the fact that
$\displaystyle\hat{u}(\bar{P})<\hat{u}_{\bar{\lambda}}(\bar{P})$ and
$\displaystyle\bar{P}\not\in R_{\bar{\lambda}}(B_{\delta}(0)\cup
B_{\delta}(x_{0}))$.

  * •

If $\displaystyle\bar{P}\in T_{\bar{\lambda}}$, then we have
$\displaystyle\displaystyle\frac{\partial\hat{u}}{\partial x_{1}}(\bar{P})\leq
0$. But, by the Hopf boundary lemma, we know that
$\displaystyle\displaystyle\frac{\partial\hat{u}}{\partial x_{1}}(\bar{P})>0$
providing a contradiction. Note that here it is needed to run over the
argument in [28] that works thanks to the cooperativity condition.

Now, we know that (2.39) holds true and we argue as in Step 1 choosing

$\displaystyle\Phi\,:=\begin{cases}\,\xi_{\bar{\lambda}+\varepsilon}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,&\text{in}\quad\Sigma_{\bar{\lambda}+\varepsilon},\\\
0&\text{in}\quad{\mathbb{R}}^{n}\setminus\Sigma_{\bar{\lambda}+\varepsilon}\end{cases}\quad\text{and}\quad\Psi\,:=\begin{cases}\,\zeta_{\bar{\lambda}+\varepsilon}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,&\text{in}\quad\Sigma_{\bar{\lambda}+\varepsilon},\\\
0&\text{in}\quad{\mathbb{R}}^{n}\setminus\Sigma_{\bar{\lambda}+\varepsilon},\end{cases}$

as test functions respectively in (2.6) and in (2.7) so that, subtracting we
get

(2.40)
$\begin{split}\int_{\Xi_{\bar{\lambda}+\varepsilon}}|\nabla\xi^{+}_{\bar{\lambda}+\varepsilon}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx=&-2\int_{\Xi_{\bar{\lambda}+\varepsilon}}\langle\nabla\xi^{+}_{\bar{\lambda}+\varepsilon},\nabla\psi_{\varepsilon}\rangle\xi_{\bar{\lambda}+\varepsilon}^{+}\psi_{\varepsilon}\eta_{R}^{2}\,dx\\\
&-2\int_{\Xi_{\bar{\lambda}+\varepsilon}}\langle\nabla\xi^{+}_{\bar{\lambda}+\varepsilon},\nabla\eta_{R}\rangle\xi_{\bar{\lambda}+\varepsilon}^{+}\eta_{R}\psi_{\varepsilon}^{2}\,dx\\\
&+\gamma_{1}\int_{\Xi_{\bar{\lambda}+\varepsilon}}\left(\frac{\hat{u}}{f_{x_{0}}(x)}-\frac{\hat{u}_{\bar{\lambda}+\varepsilon}}{f_{x_{0}}(x_{\bar{\lambda}+\varepsilon})}\right)\xi_{\bar{\lambda}+\varepsilon}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,\,\\\
&+\int_{\Xi_{\bar{\lambda}+\varepsilon}}(\hat{u}^{2^{*}-1}-\hat{u}_{\bar{\lambda}+\varepsilon}^{2^{*}-1})\xi_{\bar{\lambda}+\varepsilon}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,\,\\\
&+\nu\alpha\int_{\Xi_{\bar{\lambda}+\varepsilon}}(\hat{u}^{\alpha-1}\hat{v}^{\beta}-\hat{u}^{\alpha-1}_{\bar{\lambda}+\varepsilon}\hat{v}_{\bar{\lambda}+\varepsilon}^{\beta})\xi_{\bar{\lambda}+\varepsilon}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,.\end{split}$
(2.41)
$\begin{split}\int_{\Xi_{\bar{\lambda}+\varepsilon}}|\nabla\zeta^{+}_{\bar{\lambda}+\varepsilon}|^{2}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx=&-2\int_{\Xi_{\bar{\lambda}+\varepsilon}}\langle\nabla\zeta^{+}_{\bar{\lambda}+\varepsilon},\nabla\psi_{\varepsilon}\rangle\zeta_{\bar{\lambda}+\varepsilon}^{+}\psi_{\varepsilon}\eta_{R}^{2}\,dx\\\
&-2\int_{\Xi_{\bar{\lambda}+\varepsilon}}\langle\nabla\zeta^{+}_{\bar{\lambda}+\varepsilon},\nabla\eta_{R}\rangle\zeta_{\bar{\lambda}+\varepsilon}^{+}\eta_{R}\psi_{\varepsilon}^{2}\,dx\\\
&+\gamma_{1}\int_{\Xi_{\bar{\lambda}+\varepsilon}}\left(\frac{\hat{v}}{f_{x_{0}}(x)}-\frac{\hat{v}_{\bar{\lambda}+\varepsilon}}{f_{x_{0}}(x_{\bar{\lambda}+\varepsilon})}\right)\zeta_{\bar{\lambda}+\varepsilon}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,\,\\\
&+\int_{\Xi_{\bar{\lambda}+\varepsilon}}(\hat{v}^{2^{*}-1}-\hat{v}_{\bar{\lambda}+\varepsilon}^{2^{*}-1})\zeta_{\bar{\lambda}+\varepsilon}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,\,\\\
&+\nu\beta\int_{\Xi_{\bar{\lambda}+\varepsilon}}(\hat{u}^{\alpha}\hat{v}^{\beta-1}-\hat{u}^{\alpha}_{\bar{\lambda}+\varepsilon}\hat{v}_{\bar{\lambda}+\varepsilon}^{\beta-1})\zeta_{\bar{\lambda}+\varepsilon}^{+}\psi_{\varepsilon}^{2}\eta_{R}^{2}\,dx\,.\end{split}$

Now, we can define the following sets

$\displaystyle A:=R_{\bar{\lambda}+\varepsilon}(B_{\delta}(0)\cup
B_{\delta}(x_{0}))\quad\text{and}\quad
B:=\Sigma_{\bar{\lambda}+\varepsilon}\setminus B_{\tilde{R}}(0),$

so that
$\displaystyle\text{supp}[(\hat{u}-\hat{u}_{\bar{\lambda}+\varepsilon})^{+}],\
\text{supp}[(\hat{v}-\hat{v}_{\bar{\lambda}+\varepsilon})^{+}]\subset A\cup
B$. Arguing as in Step 1, from (2.18) to (2.35) it is possible to show that

(2.42) $\int_{A\cup B}|\nabla\xi^{+}_{\bar{\lambda}+\varepsilon}|^{2}\,dx\leq
0\quad\text{and}\quad\int_{A\cup
B}|\nabla\zeta^{+}_{\bar{\lambda}+\varepsilon}|^{2}\,dx\leq 0,$

where, in the set $\displaystyle A$, we argue as in the case of bounded
domains, see [8]. Note that Lemma 2.2 ensures that we are working far from
zero since $\displaystyle\hat{u}$ and $\displaystyle\hat{v}$ are bounded away
from zero near the origin and near the translation point $\displaystyle
x_{0}$. This allows us to apply the weak comparison principle in small domains
(see e.g. [22, 23, 41]), and deduce that

(2.43) $\int_{A}|\nabla\xi^{+}_{\bar{\lambda}+\varepsilon}|^{2}\,dx\leq
0\quad\text{and}\quad\int_{A}|\nabla\zeta^{+}_{\bar{\lambda}+\varepsilon}|^{2}\,dx\leq
0.$

Finally, collecting (2.39), (2.42) and (2.43), we deduce
$\displaystyle\text{supp}[(\hat{u}-\hat{u}_{\bar{\lambda}+\varepsilon})^{+}]=\emptyset$,
$\displaystyle\text{supp}[(\hat{v}-\hat{v}_{\bar{\lambda}+\varepsilon})^{+}]=\emptyset$
and hence

$\displaystyle\hat{u}<\hat{u}_{\bar{\lambda}+\varepsilon}\ \text{ and }\
\hat{v}<\hat{v}_{\bar{\lambda}+\varepsilon}\text{ in
}\Sigma_{\bar{\lambda}+\varepsilon},$

which is in contradiction with the definition of $\displaystyle\bar{\lambda}$.
So, $\displaystyle\bar{\lambda}$ must be 0.

_Step 3: Conclusion._ The symmetry of the Kelvin transform
$\displaystyle(\hat{u},\hat{v})$ in the $\displaystyle x_{1}$-direction
follows now performing the moving plane method in the opposite direction. By
the definition of $\displaystyle(\hat{u},\hat{v})$ given in (2.2), the
symmetry of $\displaystyle\hat{u}$ and $\displaystyle\hat{v}$ w.r.t. the
hyperplane $\displaystyle\\{x_{1}=0\\}$ implies the same symmetry of the
solution $\displaystyle(u,v)$. The symmetry of any solution of
($\displaystyle\mathcal{S}^{*}_{\gamma_{1,2}}$) follows now recalling that the
translation point $\displaystyle x_{0}$ is arbitrary. We can repeat the same
argument with respect to any fixed direction
$\displaystyle\nu\in{\mathbb{R}}^{n}$. The last one implies that
$\displaystyle(u,v)$ must be radially symmetric about the origin.

∎

##  3\. Asymptotic behavior of solutions

The main contribution of the paper is to show that solutions are unique modulo
rescaling, in the same spirit of Teraccini’s paper, [46]. In order to achieve
our goal, we need to study the asymptotic behavior of the solutions to
($\displaystyle\mathcal{S}^{*}$) that we recall for the reader convenience

($\displaystyle\mathcal{S}^{*}$) $\begin{cases}\displaystyle-\Delta
u\,=\gamma\frac{u}{|x|^{2}}+u^{2^{*}-1}+\nu\alpha
u^{\alpha-1}v^{\beta}&\text{in}\quad{\mathbb{R}}^{n}\vspace{0.2cm}\\\
\displaystyle-\Delta v\,=\gamma\frac{v}{|x|^{2}}+v^{2^{*}-1}+\nu\beta
u^{\alpha}v^{\beta-1}&\text{in}\quad{\mathbb{R}}^{n}\\\
u,v>0&\text{in}\quad{\mathbb{R}}^{n}\setminus\\{0\\}.\end{cases}$

For any $\displaystyle\tau>0$, let us define

(3.1) $(u_{\tau}(x),v_{\tau}(x)):=\left(|x|^{\tau}u(x),|x|^{\tau}v(x)\right).$

In particular, we set

(3.2)
$\tau_{1}=\frac{n-2}{2}-\sqrt{\left(\frac{n-2}{2}\right)^{2}-\gamma}\quad\text{or}\quad\tau_{2}=\frac{n-2}{2}+\sqrt{\left(\frac{n-2}{2}\right)^{2}-\gamma},$

which are solutions to the algebraic equation

(3.3) $\tau^{2}-(n-2)\tau+\gamma=0.$

This parameters are well-known in the case of a single equation and appear
describing the behavior of the solutions at zero and at infinity. An heuristic
argument shows that the behavior at zero should be driven by the weight
$\displaystyle|x|^{-\tau_{1}}$, while $\displaystyle|x|^{-\tau_{2}}$ gives us
the behavior at infinity. In any case, with this choice of
$\displaystyle\tau$, it is easy to check that
$\displaystyle(u_{\tau},v_{\tau})$ weakly solves

($\displaystyle\mathcal{W}^{*}$)
$\begin{cases}\displaystyle-\Delta_{\tau}u_{\tau}:=-\operatorname{div}\left(\frac{\nabla
u_{\tau}}{|x|^{2\tau}}\right)=\frac{u_{\tau}^{2^{*}-1}}{|x|^{2^{*}\tau}}+\nu\frac{\alpha}{|x|^{2^{*}\tau}}u_{\tau}^{\alpha-1}v_{\tau}^{\beta}&\text{in}\quad{\mathbb{R}}^{n}\vspace{0.2cm}\\\
\displaystyle-\Delta_{\tau}v_{\tau}:=-\operatorname{div}\left(\frac{\nabla
v_{\tau}}{|x|^{2\tau}}\right)=\frac{v_{\tau}^{2^{*}-1}}{|x|^{2^{*}\tau}}+\nu\frac{\beta}{|x|^{2^{*}\tau}}u_{\tau}^{\alpha}v_{\tau}^{\beta-1}&\text{in}\quad{\mathbb{R}}^{n}\\\
u_{\tau},v_{\tau}>0&\text{in}\quad{\mathbb{R}}^{n}\setminus\\{0\\}.\end{cases}$

Such equation, in the weak formulation, is well defined in the weighted
Sobolev space $\displaystyle D^{1,2}_{\tau}({\mathbb{R}}^{n})$ for
$\displaystyle\tau=\tau_{1}$ as above. This space has been already exploited
in the literature and, after defining it in the classical way [30], we may
look at it as the completion (clousure) of $\displaystyle
C^{\infty}_{c}({\mathbb{R}}^{n})$ under the norm

$\displaystyle\|u\|_{D^{1,2}_{\tau}({\mathbb{R}}^{n})}=\left(\int_{\mathbb{R}^{n}}\frac{|\nabla
u|^{2}}{|x|^{2\tau}}\,dx\right)^{1/2}.$

The weighted Sobolev inequality in this space is related to the Caffarelli-
Kohn-Nireberg inequality [11]; we refer also to [32, 3].

With such a transformation at hand, we shall exploit the Moser iteration
scheme to prove the following:

######  Proposition 3.1.

Let $\displaystyle\tau=\tau_{1}$ and $\displaystyle(u_{\tau},v_{\tau})\in
D^{1,2}_{\tau}({\mathbb{R}}^{n})\times D^{1,2}_{\tau}({\mathbb{R}}^{n})$ be
the solution to ($\displaystyle\mathcal{W}^{*}$). Then
$\displaystyle(u_{\tau},v_{\tau})\in L^{\infty}({\mathbb{R}}^{n})\times
L^{\infty}({\mathbb{R}}^{n})$.

###### Proof.

For any $\displaystyle\eta\geq 1$ and $\displaystyle T>0$ we define the
following Lispchitz function

(3.4) $\Upsilon(t)=\Upsilon_{\eta,T}(t):=\begin{cases}t^{\eta},&\text{if
}0\leq t\leq T\\\ \eta T^{\eta-1}(t-T)+T^{\eta},&\text{if }t>T.\end{cases}$

Making easy computations one can check that for any $\displaystyle w\in
D^{1,2}_{\tau}({\mathbb{R}}^{n})$ it holds the following inequality in the
weak distributional meaning

(3.5) $-\Delta_{\tau}\Upsilon(w)\leq-\Upsilon^{\prime}(w)\Delta_{\tau}w.$

Now we make all the computations on the first equation of system
($\displaystyle\mathcal{W}^{*}$). We remark that, since
$\displaystyle\Upsilon$ is a Lipschitz function, then
$\displaystyle\Upsilon(u_{\tau}),\Upsilon(v_{\tau})\in
D^{1,2}_{\tau}({\mathbb{R}}^{n})$. Hence, it holds the Sobolev inequality

(3.6)
$\left(\int_{{\mathbb{R}}^{n}}\frac{|\Upsilon(w)|^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{1}{2^{*}}}\leq\mathcal{C}_{S}\left(\int_{{\mathbb{R}}^{n}}\frac{|\nabla\Upsilon(w)|^{2}}{|x|^{2\tau}}\,dx\right)^{\frac{1}{2}}\qquad\forall
w\in D^{1,2}_{\tau}({\mathbb{R}}^{n}).$

By (3.5), we deduce

(3.7)
$\begin{split}\int_{{\mathbb{R}}^{n}}\Upsilon(u_{\tau})(-\Delta_{\tau}\Upsilon(u_{\tau}))\,dx&\leq\int_{{\mathbb{R}}^{n}}\Upsilon(u_{\tau})\Upsilon^{\prime}(u_{\tau})(-\Delta_{\tau}u_{\tau})\,dx\\\
&=\int_{{\mathbb{R}}^{n}}\Upsilon(u_{\tau})\Upsilon^{\prime}(u_{\tau})\left(\frac{u_{\tau}^{2^{*}-1}}{|x|^{2^{*}\tau}}+\nu\frac{\alpha}{|x|^{2^{*}\tau}}u_{\tau}^{\alpha-1}v_{\tau}^{\beta}\right)\,dx\\\
&\leq\eta\int_{{\mathbb{R}}^{n}}\Upsilon^{2}(u_{\tau})\left(\frac{u_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}+\nu\frac{\alpha}{|x|^{2^{*}\tau}}u_{\tau}^{\alpha-2}v_{\tau}^{\beta}\right)\,dx,\end{split}$

where in the last inequality we used the estimate $\displaystyle
t\Upsilon^{\prime}(t)\leq\eta\Upsilon(t)$ for every $\displaystyle t>0$.
Integrating by parts in the left hand side of (3.7), and using (3.6) we deduce

(3.8)
$\left(\int_{{\mathbb{R}}^{n}}\frac{|\Upsilon(u_{\tau})|^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\leq\eta\mathcal{C}_{S}\int_{{\mathbb{R}}^{n}}\Upsilon^{2}(u_{\tau})\left(\frac{u_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}+\nu\frac{\alpha}{|x|^{2^{*}\tau}}u_{\tau}^{\alpha-2}v_{\tau}^{\beta}\right)\,dx.$

Arguing in the same way with the second equation of
($\displaystyle\mathcal{W}^{*}$), we obtain

(3.9)
$\left(\int_{{\mathbb{R}}^{n}}\frac{|\Upsilon(v_{\tau})|^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\leq\eta\mathcal{C}_{S}\int_{{\mathbb{R}}^{n}}\Upsilon^{2}(v_{\tau})\left(\frac{v_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}+\nu\frac{\beta}{|x|^{2^{*}\tau}}u_{\tau}^{\alpha}v_{\tau}^{\beta-2}\right)\,dx.$

Our aim is to apply the Moser’s iteration scheme (see [35]). In order to do
this, we take into account that $\displaystyle u_{\tau},v_{\tau}\in
D^{1,2}_{\tau}({\mathbb{R}}^{n})$. Now, let

$\displaystyle\eta:=\frac{2^{*}}{2}$

and $\displaystyle m_{u},m_{v}\in{\mathbb{R}}^{+}$ to be chosen later. We
claim to give an estimate of the right hand side of (3.8). We point out that
it is easy to check that the function $\displaystyle
g(t):=\Upsilon(t)t^{\alpha-2}$ is non-decreasing for every $\displaystyle
t\geq 0$ and for every $\displaystyle\alpha>1$. Hence, taking into account
this fact and using the definition of $\displaystyle\Upsilon$ in (3.4), we
have

(3.10)
$\begin{split}\eta\mathcal{C}_{S}\int_{{\mathbb{R}}^{n}}\Upsilon^{2}(u_{\tau})&\frac{u_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\,+\,\eta\mathcal{C}_{S}\nu\alpha\int_{{\mathbb{R}}^{n}}\frac{\Upsilon^{2}(u_{\tau})u_{\tau}^{\alpha-2}v_{\tau}^{\beta}}{|x|^{2^{*}\tau}}\,dx\\\
=&\eta\mathcal{C}_{S}\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\leq
m_{u}\\}}\Upsilon^{2}(u_{\tau})\frac{u_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx+\eta\mathcal{C}_{S}\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}>m_{u}\\}}\frac{\Upsilon^{2}(u_{\tau})}{|x|^{2\tau}}\frac{u_{\tau}^{2^{*}-2}}{|x|^{(2^{*}-2)\tau}}\,dx\\\
&+\eta\mathcal{C}_{S}\nu\alpha\left[\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\leq
v_{\tau}\\}}\frac{\Upsilon^{2}(u_{\tau})u_{\tau}^{\alpha-2}v_{\tau}^{\beta}}{|x|^{2^{*}\tau}}+\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}>v_{\tau}\\}}\frac{\Upsilon^{2}(u_{\tau})u_{\tau}^{\alpha-2}v_{\tau}^{\beta}}{|x|^{2^{*}\tau}}\,dx\right]\\\
\leq&\eta\mathcal{C}_{S}\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\leq
m_{u}\\}}\Upsilon^{2}(u_{\tau})\frac{u_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx+\eta\mathcal{C}_{S}\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}>m_{u}\\}}\frac{\Upsilon^{2}(u_{\tau})}{|x|^{2\tau}}\frac{u_{\tau}^{2^{*}-2}}{|x|^{(2^{*}-2)\tau}}\,dx\\\
&+\eta\mathcal{C}_{S}\nu\alpha\left[\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\leq
v_{\tau}\\}}\frac{\Upsilon^{2}(v_{\tau})v_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}+\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}>v_{\tau}\\}}\frac{\Upsilon^{2}(u_{\tau})u_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\right]\\\
\leq&\eta\mathcal{C}_{S}\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\leq
m_{u}\\}}\Upsilon^{2}(u_{\tau})\frac{u_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx+\eta\mathcal{C}_{S}\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}>m_{u}\\}}\frac{\Upsilon^{2}(u_{\tau})}{|x|^{2\tau}}\frac{u_{\tau}^{2^{*}-2}}{|x|^{(2^{*}-2)\tau}}\,dx\\\
&+\eta\mathcal{C}_{S}\nu\alpha\left[\int_{{\mathbb{R}}^{n}}\frac{\Upsilon^{2}(v_{\tau})v_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}+\int_{{\mathbb{R}}^{n}}\frac{\Upsilon^{2}(u_{\tau})u_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\right]\,.\end{split}$

Using Hölder’s inequality with conjugate exponents
$\displaystyle\left(\frac{2^{*}}{2},\frac{2^{*}}{2^{*}-2}\right)$, we get

(3.11)
$\begin{split}\eta\mathcal{C}_{S}\int_{{\mathbb{R}}^{n}}\Upsilon^{2}(u_{\tau})&\frac{u_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\,+\,\eta\mathcal{C}_{S}\nu\alpha\int_{{\mathbb{R}}^{n}}\frac{\Upsilon^{2}(u_{\tau})u_{\tau}^{\alpha-2}v_{\tau}^{\beta}}{|x|^{2^{*}\tau}}\,dx\\\
\leq&\eta\mathcal{C}_{S}(1+\nu\alpha)\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\leq
m_{u}\\}}\Upsilon^{2}(u_{\tau})\frac{m_{u}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\\\
&+\eta\mathcal{C}_{S}(1+\nu\alpha)\left(\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\geq
m_{u}\\}}\frac{\Upsilon^{2^{*}}(u_{\tau})}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\left(\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\geq
m_{u}\\}}\frac{u_{\tau}^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2^{*}-2}{2^{*}}}\\\
&+\eta\mathcal{C}_{S}\nu\alpha\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\leq
m_{v}\\}}\Upsilon^{2}(v_{\tau})\frac{m_{v}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\\\
&+\eta\mathcal{C}_{S}\nu\alpha\left(\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\geq
m_{v}\\}}\frac{\Upsilon^{2^{*}}(v_{\tau})}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\left(\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\geq
m_{v}\\}}\frac{v_{\tau}^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2^{*}-2}{2^{*}}}.\end{split}$

Arguing in an analogous way to (3.10) for the second equation of
($\displaystyle\mathcal{W}^{*}$), we are able to deduce

(3.12)
$\begin{split}\eta\mathcal{C}_{S}\int_{{\mathbb{R}}^{n}}\Upsilon^{2}(v_{\tau})&\frac{v_{\tau}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\,+\,\eta\mathcal{C}_{S}\nu\beta\int_{{\mathbb{R}}^{n}}\frac{\Upsilon^{2}(v_{\tau})u_{\tau}^{\alpha}v_{\tau}^{\beta-2}}{|x|^{2^{*}\tau}}\,dx\\\
\leq&\eta\mathcal{C}_{S}(1+\nu\beta)\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\leq
m_{v}\\}}\Upsilon^{2}(v_{\tau})\frac{m_{v}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\\\
&+\eta\mathcal{C}_{S}(1+\nu\beta)\left(\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\geq
m_{v}\\}}\frac{\Upsilon^{2^{*}}(v_{\tau})}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\left(\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\geq
m_{v}\\}}\frac{v_{\tau}^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2^{*}-2}{2^{*}}}\\\
&+\eta\mathcal{C}_{S}\nu\beta\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\leq
m_{u}\\}}\Upsilon^{2}(u_{\tau})\frac{m_{u}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\\\
&+\eta\mathcal{C}_{S}\nu\beta\left(\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\geq
m_{u}\\}}\frac{\Upsilon^{2^{*}}(u_{\tau})}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\left(\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\geq
m_{u}\\}}\frac{u_{\tau}^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2^{*}-2}{2^{*}}}.\end{split}$

Hence, thanks to (3.11), by (3.8) we obtain

(3.13)
$\begin{split}&\left(\int_{{\mathbb{R}}^{n}}\frac{|\Upsilon(u_{\tau})|^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\leq\eta\mathcal{C}_{S}(1+\nu\alpha)\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\leq
m_{u}\\}}\Upsilon^{2}(u_{\tau})\frac{m_{u}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\\\
&\qquad+\eta\mathcal{C}_{S}(1+\nu\alpha)\left(\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\geq
m_{u}\\}}\frac{\Upsilon^{2^{*}}(u_{\tau})}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\left(\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\geq
m_{u}\\}}\frac{u_{\tau}^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2^{*}-2}{2^{*}}}\\\
&\qquad+\eta\mathcal{C}_{S}\nu\alpha\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\leq
m_{v}\\}}\Upsilon^{2}(v_{\tau})\frac{m_{v}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\\\
&\qquad+\eta\mathcal{C}_{S}\nu\alpha\left(\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\geq
m_{v}\\}}\frac{\Upsilon^{2^{*}}(v_{\tau})}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\left(\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\geq
m_{v}\\}}\frac{v_{\tau}^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2^{*}-2}{2^{*}}}.\end{split}$

Arguing in a similar way with $\displaystyle v_{\tau}$, thanks to (3.12), by
(3.9) we deduce

(3.14)
$\begin{split}&\left(\int_{{\mathbb{R}}^{n}}\frac{|\Upsilon(v_{\tau})|^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\leq\eta\mathcal{C}_{S}(1+\nu\beta)\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\leq
m_{v}\\}}\Upsilon^{2}(v_{\tau})\frac{m_{v}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\\\
&\qquad+\eta\mathcal{C}_{S}(1+\nu\beta)\left(\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\geq
m_{v}\\}}\frac{\Upsilon^{2^{*}}(v_{\tau})}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\left(\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\geq
m_{v}\\}}\frac{v_{\tau}^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2^{*}-2}{2^{*}}}\\\
&\qquad+\eta\mathcal{C}_{S}\nu\beta\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\leq
m_{u}\\}}\Upsilon^{2}(u_{\tau})\frac{m_{u}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\\\
&\qquad+\eta\mathcal{C}_{S}\nu\beta\left(\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\geq
m_{u}\\}}\frac{\Upsilon^{2^{*}}(u_{\tau})}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\left(\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\geq
m_{u}\\}}\frac{u_{\tau}^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2^{*}-2}{2^{*}}}.\end{split}$

Summing both (3.13) and (3.14), we deduce that

(3.15)
$\begin{split}&\left(\int_{{\mathbb{R}}^{n}}\frac{|\Upsilon(u_{\tau})|^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}+\left(\int_{{\mathbb{R}}^{n}}\frac{|\Upsilon(v_{\tau})|^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\\\
&\ \
\leq\eta\mathcal{C}_{S}(1+2^{*}\nu)\left[\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\leq
m_{u}\\}}\Upsilon^{2}(u_{\tau})\frac{m_{u}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx+\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\leq
m_{v}\\}}\Upsilon^{2}(v_{\tau})\frac{m_{v}^{2^{*}-2}}{|x|^{2^{*}\tau}}\,dx\right]\\\
&\qquad+\eta\mathcal{C}_{S}(1+2^{*}\nu)\left(\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\geq
m_{u}\\}}\frac{\Upsilon^{2^{*}}(u_{\tau})}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\left(\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\geq
m_{u}\\}}\frac{u_{\tau}^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2^{*}-2}{2^{*}}}\\\
&\qquad+\eta\mathcal{C}_{S}(1+2^{*}\nu)\left(\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\geq
m_{v}\\}}\frac{\Upsilon^{2^{*}}(v_{\tau})}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\left(\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\geq
m_{v}\\}}\frac{v_{\tau}^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2^{*}-2}{2^{*}}}.\end{split}$

Since $\displaystyle u_{\tau},v_{\tau}\in D^{1,2}_{\tau}({\mathbb{R}}^{n})$,
we can fix $\displaystyle m_{u},m_{v}$ such that

(3.16) $\left(\int_{{\mathbb{R}}^{n}\cap\\{u_{\tau}\geq
m_{u}\\}}\frac{u_{\tau}^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2^{*}-2}{2^{*}}},\left(\int_{{\mathbb{R}}^{n}\cap\\{v_{\tau}\geq
m_{v}\\}}\frac{v_{\tau}^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2^{*}-2}{2^{*}}}\leq\frac{1}{2\eta\mathcal{C}_{S}(1+2^{*}\nu)}$

Hence, from (3.15) we obtain

(3.17)
$\begin{split}&\left(\int_{{\mathbb{R}}^{n}}\frac{|\Upsilon(u_{\tau})|^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}+\left(\int_{{\mathbb{R}}^{n}}\frac{|\Upsilon(v_{\tau})|^{2^{*}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\\\
&\ \
\leq\eta\mathcal{C}_{S}(1+2^{*}\nu)\left[m_{u}^{2^{*}-2}\int_{{\mathbb{R}}^{n}}\frac{\Upsilon^{2}(u_{\tau})}{|x|^{2^{*}\tau}}\,dx+m_{v}^{2^{*}-2}\int_{{\mathbb{R}}^{n}}\frac{\Upsilon^{2}(v_{\tau})}{|x|^{2^{*}\tau}}\,dx\right]\\\
&\ \ \leq\eta
C\left[\int_{{\mathbb{R}}^{n}}\frac{u_{\tau}^{2\eta}}{|x|^{2^{*}\tau}}\,dx+\int_{{\mathbb{R}}^{n}}\frac{v_{\tau}^{2\eta}}{|x|^{2^{*}\tau}}\,dx\right],\end{split}$

where $\displaystyle C$ depends on $\displaystyle m_{u},m_{v},\nu,n$ and
$\displaystyle\mathcal{C}_{S}$. Moreover, we used that
$\displaystyle\Upsilon(t)\leq t^{\eta}$ and that $\displaystyle
u_{\tau},v_{\tau}\in D^{1,2}_{\tau}({\mathbb{R}}^{n})$. Hence, recalling that
$\displaystyle\eta=\frac{2^{*}}{2}$ and taking $\displaystyle
T\rightarrow+\infty$, thanks also to Fatou’s lemma one has

(3.18)
$\left(\int_{{\mathbb{R}}^{n}}\frac{u_{\tau}^{2^{*}\eta}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}+\left(\int_{{\mathbb{R}}^{n}}\frac{v_{\tau}^{2^{*}\eta}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}<+\infty.$

Moreover, since $\displaystyle\frac{2}{2^{*}}<1$, we also deduce

(3.19)
$\left(\int_{{\mathbb{R}}^{n}}\frac{u_{\tau}^{2^{*}\eta}}{|x|^{2^{*}\tau}}\,dx+\int_{{\mathbb{R}}^{n}}\frac{v_{\tau}^{2^{*}\eta}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}\leq\left(\int_{{\mathbb{R}}^{n}}\frac{u_{\tau}^{2^{*}\eta}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}+\left(\int_{{\mathbb{R}}^{n}}\frac{v_{\tau}^{2^{*}\eta}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{2}{2^{*}}}<+\infty.$

Now, we are able to apply the Moser’s iteration scheme, in order to prove our
result. For any $\displaystyle k\geq 1$, let us define the sequence
$\displaystyle\\{\eta_{k}\\}$ by

$\displaystyle 2\eta_{k+1}+2^{*}-2=2^{*}\eta_{k},$

and we set $\displaystyle\eta_{1}:=\frac{2^{*}}{2}$. The choice of
$\displaystyle\eta_{k}$ at each step is clear in the first term of the right
hand side of (3.10). Then starting again by (3.10) and (3.12), iterating we
deduce that

(3.20)
$\begin{split}&\left(\int_{{\mathbb{R}}^{n}}\frac{u_{\tau}^{2^{*}\eta_{k+1}}}{|x|^{2^{*}\tau}}\,dx+\int_{{\mathbb{R}}^{n}}\frac{v_{\tau}^{2^{*}\eta_{k+1}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{1}{2^{*}(\eta_{k+1}-1)}}\\\
&\qquad\leq(C\eta_{k+1})^{\frac{1}{2(\eta_{k+1}-1)}}\left(\int_{{\mathbb{R}}^{n}}\frac{u_{\tau}^{2^{*}\eta_{k}}}{|x|^{2^{*}\tau}}\,dx+\int_{{\mathbb{R}}^{n}}\frac{v_{\tau}^{2^{*}\eta_{k}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{1}{2^{*}(\eta_{k}-1)}}.\end{split}$

Finally, we set

(3.21)
$\mathcal{I}_{k}:=\left(\int_{{\mathbb{R}}^{n}}\frac{u_{\tau}^{2^{*}\eta_{k}}}{|x|^{2^{*}\tau}}\,dx+\int_{{\mathbb{R}}^{n}}\frac{v_{\tau}^{2^{*}\eta_{k}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{1}{2^{*}(\eta_{k}-1)}},\
\ \mathcal{C}_{k}:=(C\eta_{k+1})^{\frac{1}{2(\eta_{k+1}-1)}},$

and we obtain the recursive inequality

$\displaystyle\mathcal{I}_{k+1}\leq\mathcal{C}_{k}\mathcal{I}_{k},\ \ k\geq
1.$

By induction we can easily deduce that

(3.22)
$\begin{split}\log\mathcal{I}_{k+1}&\leq\sum_{j=2}^{k+1}\log\mathcal{C}_{j}+\log\mathcal{I}_{1}\leq\sum_{j=2}^{+\infty}\log\mathcal{C}_{j}+\log\mathcal{I}_{1}<+\infty,\end{split}$

where in the last inequality we used that the series
$\displaystyle\sum_{j=2}^{+\infty}\log\mathcal{C}_{j}$ converges. Hence,
thanks to the monotonicity of the integral and by (3.22), for every
$\displaystyle R>0$, we have

(3.23)
$\log\left(\left(\int_{B_{R}}\frac{u_{\tau}^{2^{*}\eta_{k}}}{|x|^{2^{*}\tau}}\,dx+\int_{B_{R}}\frac{v_{\tau}^{2^{*}\eta_{k}}}{|x|^{2^{*}\tau}}\,dx\right)^{\frac{1}{2^{*}(\eta_{k}-1)}}\right)\leq
C.$

Since $\displaystyle|x|<R$, we have

(3.24)
$\begin{split}\frac{\tau}{\eta_{k}-1}\log\left(\frac{1}{R}\right)&+\log\left(\left(\int_{B_{R}}u_{\tau}^{2^{*}\eta_{k}}\,dx\right)^{\frac{1}{2^{*}(\eta_{k}-1)}}\right)\\\
&\leq\frac{\tau}{\eta_{k}-1}\log\left(\frac{1}{R}\right)+\log\left(\left(\int_{B_{R}}u_{\tau}^{2^{*}\eta_{k}}\,dx+\int_{B_{R}}v_{\tau}^{2^{*}\eta_{k}}\,dx\right)^{\frac{1}{2^{*}(\eta_{k}-1)}}\right)\leq
C.\end{split}$

Passing to the limit for $\displaystyle k\rightarrow+\infty$ and thus
$\displaystyle\eta_{k}\rightarrow+\infty$, we deduce that

$\displaystyle(u_{\tau},v_{\tau})\in L^{\infty}(B_{R})\times
L^{\infty}(B_{R}).$

The thesis follows by the arbitrariness of $\displaystyle R$.

∎

In the previous section we showed that any solution $\displaystyle(u,v)\in
H^{1}_{loc}({\mathbb{R}}^{n})\times H^{1}_{loc}({\mathbb{R}}^{n})$ to
($\displaystyle\mathcal{S}^{*}$) is radial, and, hence,
($\displaystyle\mathcal{S}^{*}$) becomes

($\displaystyle\mathcal{R}_{\mathcal{S}}^{*}$)
$\begin{cases}\displaystyle(r^{n-1}u^{\prime})^{\prime}+r^{n-1}\left(\gamma\frac{u}{r^{2}}+u^{2^{*}-1}+\nu\alpha
u^{\alpha-1}v^{\beta}\right)=0&\text{in}\quad(0,+\infty)\\\
\displaystyle\displaystyle(r^{n-1}v^{\prime})^{\prime}+r^{n-1}\left(\gamma\frac{v}{r^{2}}+v^{2^{*}-1}+\nu\beta
u^{\alpha}v^{\beta-1}\right)=0&\text{in}\quad(0,+\infty)\\\
u,v>0&\text{in}\quad(0,+\infty).\end{cases}$

Obviously also $\displaystyle u_{\tau},v_{\tau}$ are radial and they satisfy

($\displaystyle\mathcal{R}_{\mathcal{W}}^{*}$)
$\begin{cases}\displaystyle(r^{n-1-2\tau}u_{\tau}^{\prime})^{\prime}+r^{n-1-2^{*}\tau}\left(u_{\tau}^{2^{*}-1}+\nu\alpha
u_{\tau}^{\alpha-1}v_{\tau}^{\beta}\right)=0&\text{in}\quad(0,+\infty)\\\
\displaystyle\displaystyle(r^{n-1-2\tau}v_{\tau}^{\prime})^{\prime}+r^{n-1-2^{*}\tau}\left(v_{\tau}^{2^{*}-1}+\nu\beta
u_{\tau}^{\alpha}v_{\tau}^{\beta-1}\right)=0&\text{in}\quad(0,+\infty)\\\
u,v>0&\text{in}\quad(0,+\infty).\end{cases}$

Thanks to to Proposition 3.1, we deduce that

$\displaystyle|x|^{\tau_{1}}\cdot u(x)$ and $\displaystyle|x|^{\tau_{1}}\cdot
v(x)$ are bounded functions in $\displaystyle{\mathbb{R}}^{n}$ .

This is a first information regarding the behavior of the solutions at zero
(the behavior that we get at infinity is not sharp). In any case we are in
position to get a complete picture of the asymptotic behavior of the solutions
by proving the following:

######  Proposition 3.2.

Let $\displaystyle(u,v)\in D^{1,2}({\mathbb{R}}^{n})\times
D^{1,2}({\mathbb{R}}^{n})$ be any solution to
($\displaystyle\mathcal{S}^{*}$). Then, we have

(3.25) $\lim_{|x|\rightarrow 0^{+}}|x|^{\tau_{1}}\cdot
u(x)=:u_{0},\qquad\lim_{|x|\rightarrow 0^{+}}|x|^{\tau_{1}}\cdot v(x)=:v_{0},$

where $\displaystyle u_{0},v_{0}\in(0,+\infty)$. Moreover, we have

(3.26) $\lim_{|x|\rightarrow+\infty}|x|^{\tau_{2}}\cdot
u(x)=:u_{\infty},\qquad\lim_{|x|\rightarrow 0^{+}}|x|^{\tau_{2}}\cdot
v(x)=:v_{\infty},$

where $\displaystyle u_{\infty},v_{\infty}\in(0,+\infty)$.

###### Proof.

Let $\displaystyle r:=|x|$. By Problem
($\displaystyle\mathcal{R}_{\mathcal{W}}^{*}$) we deduce that both
$\displaystyle r^{n-1-2\tau}u_{\tau}^{\prime}(r)$ and $\displaystyle
r^{n-1-2\tau}v_{\tau}^{\prime}(r)$ are strictly decreasing functions.
Therefore, we get $\displaystyle r^{n-1-2\tau}u_{\tau_{1}}^{\prime}(r)$ and
$\displaystyle r^{n-1-2\tau}v_{\tau_{1}}^{\prime}(r)$ have a sign near zero,
and this immediately implies that $\displaystyle
u_{\tau_{1}}^{\prime}(r),v_{\tau_{1}}^{\prime}(r)$ have a sign near zero.
Hence $\displaystyle u_{\tau_{1}},v_{\tau_{1}}$ are monotone bounded functions
and then they must admit limit $\displaystyle u_{0},v_{0}\in[0,+\infty)$ as
$\displaystyle r\rightarrow 0^{+}$. By the strong maximum principle (see e.g.
[1]) for the system

$\displaystyle\begin{cases}-\Delta_{\tau_{1}}u_{\tau_{1}}\geq 0&\text{in
}B_{\delta}(0)\\\ -\Delta_{\tau_{1}}v_{\tau_{1}}\geq 0&\text{in
}B_{\delta}(0)\\\ u_{\tau_{1}},v_{\tau_{1}}\geq 0&\text{in
}B_{\delta}(0),\end{cases}$

since $\displaystyle u_{\tau_{1}},v_{\tau_{1}}>0$ in $\displaystyle
B_{\delta}(0)\setminus\\{0\\}$, we get $\displaystyle
u_{\tau_{1}},v_{\tau_{1}}>0$ in $\displaystyle B_{\delta}(0)$ and hence
$\displaystyle u_{0},v_{0}>0$. Now, we have in force the estimate as
$\displaystyle r\rightarrow 0^{+}$, i.e. (3.25) holds true. Since
$\displaystyle(u,v)\in D^{1,2}({\mathbb{R}}^{n})\times
D^{1,2}({\mathbb{R}}^{n})$, it is standard to prove that the Kelvin
transformations $\displaystyle(\hat{u},\hat{v})\in
D^{1,2}({\mathbb{R}}^{n})\times D^{1,2}({\mathbb{R}}^{n})$. Hence, we get

(3.27) $\lim_{|x|\rightarrow
0^{+}}|x|^{\tau_{1}}\cdot\hat{u}\left(\frac{x}{|x|^{2}}\right)=\hat{u}_{0},\qquad\lim_{|x|\rightarrow
0^{+}}|x|^{\tau_{1}}\cdot\hat{v}\left(\frac{x}{|x|^{2}}\right)=\hat{v}_{0},$

which implies (3.25). By the last one, if we change variable $\displaystyle
y:=x/|x|^{2}$ and use the identity $\displaystyle\tau_{1}+\tau_{2}=n-2$, we
deduce

(3.28) $\begin{split}\hat{u}_{0}&=\lim_{|x|\rightarrow
0^{+}}|x|^{\tau_{1}}\cdot\hat{u}(x)=\lim_{|y|\rightarrow+\infty}|y|^{-\tau_{1}}\cdot|y|^{n-2}u(y)=\lim_{|y|\rightarrow+\infty}|y|^{\tau_{2}}\cdot
u(y),\\\ \hat{v}_{0}&=\lim_{|x|\rightarrow
0^{+}}|x|^{\tau_{1}}\cdot\hat{v}(x)=\lim_{|y|\rightarrow+\infty}|y|^{-\tau_{1}}\cdot|y|^{n-2}v(y)=\lim_{|y|\rightarrow+\infty}|y|^{\tau_{2}}\cdot
v(y),\end{split}$

which implies (3.26).

∎

##  4\. Uniqueness and classification of solutions

The aim of this section is to classify solutions to
($\displaystyle\mathcal{S}^{*}$) up to the associated rescaling. Given
$\displaystyle(u,v)$ a solution to the problem
($\displaystyle\mathcal{R}_{\mathcal{S}}^{*}$), consider

(4.1) $\begin{split}y_{u}(t):=r^{\delta}u(r)\quad&\text{and}\quad
y_{v}(t):=r^{\delta}v(r),\\\ \end{split}$

where $\displaystyle t:=\log r$ with $\displaystyle r>0$ and
$\displaystyle\delta=\frac{n-2}{2}$. By direct computation, it is possible to
deduce that system ($\displaystyle\mathcal{R}_{\mathcal{S}}^{*}$) is
equivalent to

($\displaystyle\mathcal{R}_{*}^{\prime}$)
$\begin{cases}y^{\prime\prime}_{u}-(\delta^{2}-\gamma)y_{u}+y_{u}^{2^{*}-1}+\nu\alpha
y_{u}^{\alpha-1}y_{v}^{\beta}=0&\text{in}\quad{\mathbb{R}}\\\
y^{\prime\prime}_{v}-(\delta^{2}-\gamma)y_{v}+y_{v}^{2^{*}-1}+\nu\beta
y_{u}^{\alpha}y_{v}^{\beta-1}=0&\text{in}\quad{\mathbb{R}}\\\
y_{u},y_{v}>0&\text{in}\quad{\mathbb{R}}.\end{cases}$

Next, we will prove that $\displaystyle y_{u}$ and $\displaystyle y_{v}$ reach
a unique maximum point, which actually is the same for both components. To
show this fact, we will use the asymptotic behavior of $\displaystyle y_{u}$
and $\displaystyle y_{v}$.

######  Lemma 4.1.

Let $\displaystyle y_{u},y_{v}$ be the functions defined in (4.1), they
satisfy

$\displaystyle\lim_{t\rightarrow-\infty}y_{u}(t)=0,\quad\lim_{t\rightarrow+\infty}y_{u}(t)=0,\quad\lim_{t\rightarrow-\infty}y_{v}(t)=0,\quad\lim_{t\rightarrow+\infty}y_{v}(t)=0.$

Moreover, there exist
$\displaystyle\bar{c}_{u},\bar{C}_{u},\underline{c}_{u},\underline{C}_{u},\bar{c}_{v},\bar{C}_{v},\underline{c}_{v},\underline{C}_{v},M$
positive constants such that

(4.2)
$\displaystyle\displaystyle\underline{c}_{u}e^{(\delta-\tau_{1})t}u_{\tau_{1}}(e^{t})\leq
y_{u}(t)\leq\underline{C}_{u}e^{(\delta-\tau_{1})t}u_{\tau_{1}}(e^{t})\text{
for }t\leq-M,$ (4.3)
$\displaystyle\displaystyle\bar{c}_{u}e^{(\delta-\tau_{2})t}u_{\tau_{2}}(e^{t})\leq
y_{u}(t)\leq\bar{C}_{u}e^{(\delta-\tau_{2})t}u_{\tau_{2}}(e^{t})\text{ for
}t\geq M,$ (4.4)
$\displaystyle\displaystyle\underline{c}_{v}e^{(\delta-\tau_{1})t}v_{\tau_{1}}(e^{t})\leq
y_{v}(t)\leq\underline{C}_{v}e^{(\delta-\tau_{1})t}v_{\tau_{1}}(e^{t})\text{
for }t\leq-M,$ (4.5) $\displaystyle\displaystyle
c_{u}e^{(\delta-\tau_{2})t}v_{\tau_{2}}(e^{t})\leq y_{v}(t)\leq
C_{v}e^{(\delta-\tau_{2})t}v_{\tau_{2}}(e^{t})\text{ for }t\geq M,$

where $\displaystyle\tau_{1}$ and $\displaystyle\tau_{2}$ were introduced
(3.2). Furthermore, we have that

(4.6)
$\lim_{t\rightarrow-\infty}\frac{y_{u}(t)}{y_{v}(t)}=L_{1}\in(0,+\infty)\
\text{ and }\
\lim_{t\rightarrow+\infty}\frac{y_{u}(t)}{y_{v}(t)}=L_{2}\in(0,+\infty).$

The proof follows by Proposition 3.1 and 3.2 .

Let us point out that, unfortunately, one could not apply the ODE approaches
of Abdellaoui-Felli-Peral [2] and Terracini [46] in a straightforward way. On
the contrary, we shall prove the following result.

######  Lemma 4.2.

The functions $\displaystyle y_{u}$ and $\displaystyle y_{v}$ admits only a
simultaneous global maximum at a point $\displaystyle\tilde{t}\in\mathbb{R}$.

###### Proof.

To prove our result we will use the moving plane procedure for solutions to
($\displaystyle\mathcal{R}_{*}^{\prime}$). For
$\displaystyle\lambda\in{\mathbb{R}}$, let us define the following functions

$\displaystyle
w_{u,\lambda}(t):=y_{u}(t)-y_{u,\lambda}(t)=y_{u}(t)-y_{u}(2\lambda-t),\quad
w_{v,\lambda}(t):=y_{v}(t)-y_{v,\lambda}(t)=y_{v}(t)-y_{v}(2\lambda-t),$

Notice that $\displaystyle w_{u},w_{v}$ solves the following system

(4.7)
$\begin{cases}-w^{\prime\prime}_{u,\lambda}=-(\delta^{2}-\gamma_{1})w_{u,\lambda}+(y_{u}^{2^{*}-1}-y_{u,\lambda}^{2^{*}-1})+\nu\alpha(y_{u}^{\alpha-1}y_{v}^{\beta}-y_{u,\lambda}^{\alpha-1}y_{v,\lambda}^{\beta})&\text{in}\quad{\mathbb{R}}\\\
-w^{\prime\prime}_{v,\lambda}=-(\delta^{2}-\gamma_{2})w_{v,\lambda}+(y_{v}^{2^{*}-1}-y_{v,\lambda}^{2^{*}-1})+\nu\beta(y_{u}^{\alpha}y_{v}^{\beta-1}-y_{u,\lambda}^{\alpha}y_{v,\lambda}^{\beta-1})&\text{in}\quad{\mathbb{R}}.\end{cases}$

We pass to the weak formulation of (4.7) given by

(4.8)
$\begin{split}\int_{\mathbb{R}}w^{\prime}_{u,\lambda}\varphi^{\prime}\,dt=&-(\delta^{2}-\gamma)\int_{\mathbb{R}}w_{u,\lambda}\varphi\,dt+\int_{\mathbb{R}}(y_{u}^{2^{*}-1}-y_{u,\lambda}^{2^{*}-1})\varphi\,dt\\\
&+\nu\alpha\int_{\mathbb{R}}(y_{u}^{\alpha-1}y_{v}^{\beta}-y_{u,\lambda}^{\alpha-1}y_{v,\lambda}^{\beta})\varphi\,dt\qquad\forall\varphi\in
C^{1}_{c}({\mathbb{R}}),\\\
\int_{\mathbb{R}}w^{\prime}_{v,\lambda}\psi^{\prime}\,dt=&-\int_{\mathbb{R}}(\delta^{2}-\gamma)w_{v,\lambda}\psi\,dt+\int_{\mathbb{R}}(y_{v}^{2^{*}-1}-y_{v,\lambda}^{2^{*}-1})\psi^{\prime}\,dt\\\
&+\nu\beta\int_{\mathbb{R}}(y_{u}^{\alpha}y_{v}^{\beta-1}-y_{u,\lambda}^{\alpha}y_{v,\lambda}^{\beta-1})\psi\,dt\qquad\forall\psi\in
C^{1}_{c}({\mathbb{R}}).\end{split}$

Thanks to Lemma 4.1 and arguing in a similar way to the proof of Theorem 1.5,
by density argument we can choose the following test functions

$\displaystyle\varphi:=w_{u,\lambda}^{+}\qquad\text{and}\qquad\psi:=w_{v,\lambda}^{+},$

Moreover, we point out that $\displaystyle 0\leq w_{u,\lambda}^{+}\leq y_{u}$
and $\displaystyle 0\leq w_{v,\lambda}^{+}\leq y_{v}$. Putting the test
functions into the weak formulations, (4.8) becomes

(4.9)
$\begin{split}\int_{(-\infty,\lambda]}[(w^{+}_{u,\lambda})^{\prime}]^{2}\,dt=&-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}w_{u,\lambda}w^{+}_{u,\lambda}\,dt+\int_{(-\infty,\lambda]}(y_{u}^{2^{*}-1}-y_{u,\lambda}^{2^{*}-1})w^{+}_{u,\lambda}\,dt\\\
&+\nu\alpha\int_{(-\infty,\lambda]}(y_{u}^{\alpha-1}y_{v}^{\beta}-y_{u,\lambda}^{\alpha-1}y_{v,\lambda}^{\beta})w^{+}_{u,\lambda}\,dt,\\\
\int_{(-\infty,\lambda]}[(w^{+}_{v,\lambda})^{\prime}]^{2}\,dt=&-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}w_{v,\lambda}w^{+}_{v,\lambda}\,dt+\int_{(-\infty,\lambda]}(y_{v}^{2^{*}-1}-y_{v,\lambda}^{2^{*}-1})w^{+}_{v,\lambda}\,dt\\\
&+\nu\beta\int_{\mathbb{R}}(y_{u}^{\alpha}y_{v}^{\beta-1}-y_{u,\lambda}^{\alpha}y_{v,\lambda}^{\beta-1})w^{+}_{v,\lambda}\,dt.\end{split}$

Now, we focus our attention on the first equation, and we consider
$\displaystyle\lambda\leq-M$ (where $\displaystyle M$ is given by Lemma 4.1).
In particular, we argue as in the proof of Theorem 1.5. Hence, using the
definition of $\displaystyle w_{u,\lambda}^{+}$, the Mean Value’s Theorem and
Lemma 4.1

(4.10)
$\begin{split}\int_{(-\infty,\lambda]}[(w^{+}_{u,\lambda})^{\prime}]^{2}\,dt\leq&-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}(w^{+}_{u,\lambda})^{2}\,dt+\int_{(-\infty,\lambda]}y_{u}^{2^{*}-2}(w^{+}_{u,\lambda})^{2}\,dt\\\
&+\nu\alpha\int_{(-\infty,\lambda]}y_{v}^{\beta}(y_{u}^{\alpha-1}-y_{u,\lambda}^{\alpha-1})w^{+}_{u,\lambda}\,dt\\\
&+\nu\alpha\int_{(-\infty,\lambda]}y_{u,\lambda}^{\alpha-1}(y_{v}^{\beta}-y_{v,\lambda}^{\beta})w^{+}_{u,\lambda}\,dt.\end{split}$

Arguing as we did for $\displaystyle\mathcal{I}_{5}$, we get

(4.11)
$\begin{split}\int_{(-\infty,\lambda]}[(w^{+}_{u,\lambda})^{\prime}]^{2}\,dt\leq&-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}(w^{+}_{u,\lambda})^{2}\,dt+\int_{(-\infty,\lambda]}y_{u}^{2^{*}-2}(w^{+}_{u,\lambda})^{2}\,dt\\\
&+\nu\alpha\int_{(-\infty,\lambda]}e^{-(\delta-\tau_{1})(\alpha-1)t}y_{v}^{\beta}\left[(e^{(\delta-\tau_{1})t}y_{u})^{\alpha-1}-(e^{(\delta-\tau_{1})t}y_{u,\lambda})^{\alpha-1}\right]w^{+}_{u,\lambda}\,dt\\\
&+\nu\alpha\int_{(-\infty,\lambda]}y_{u,\lambda}^{\alpha-1}(y_{v}^{\beta}-y_{v,\lambda}^{\beta})w^{+}_{u,\lambda}\,dt\\\
=&-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}(w^{+}_{u,\lambda})^{2}\,dt+\int_{(-\infty,\lambda]}y_{u}^{2^{*}-2}(w^{+}_{u,\lambda})^{2}\,dt\\\
&+\nu\alpha\int_{(-\infty,\lambda]}e^{-(\delta-\tau_{1})(\alpha-1)t}y_{v}^{\beta}\frac{(e^{(\delta-\tau_{1})t}y_{u})^{\alpha-1}-(e^{(\delta-\tau_{1})t}y_{u,\lambda})^{\alpha-2}}{e^{(\delta-\tau_{1})t}y_{u}-e^{(\delta-\tau_{1})t}y_{u,\lambda}}w_{u,\lambda}w^{+}_{u,\lambda}\,dt\\\
&+\nu\alpha\int_{(-\infty,\lambda]}y_{u,\lambda}^{\alpha-1}(y_{v}^{\beta}-y_{v,\lambda}^{\beta})w^{+}_{u,\lambda}\,dt.\\\
\end{split}$

Now, using the same idea contained in Remark 2.4, we get

(4.12)
$\begin{split}\int_{(-\infty,\lambda]}[(w^{+}_{u,\lambda})^{\prime}]^{2}\,dt\leq&-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}(w^{+}_{u,\lambda})^{2}\,dt+\int_{(-\infty,\lambda]}y_{u}^{2^{*}-2}(w^{+}_{u,\lambda})^{2}\,dt\\\
&+\nu\alpha\int_{(-\infty,\lambda]}e^{-(\delta-\tau_{1})(\alpha-1)t}y_{v}^{\beta}\left[(e^{(\delta-\tau_{1})t}y_{u})^{\alpha-1}-(e^{(\delta-\tau_{1})t}y_{u,\lambda})^{\alpha-1}\right]w^{+}_{u,\lambda}\,dt\\\
&+\nu\alpha\int_{(-\infty,\lambda]}y_{u,\lambda}^{\alpha-1}(y_{v}^{\beta}-y_{v,\lambda}^{\beta})w^{+}_{u,\lambda}\,dt\\\
\leq&-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}(w^{+}_{u,\lambda})^{2}\,dt+\int_{(-\infty,\lambda]}y_{u}^{2^{*}-2}(w^{+}_{u,\lambda})^{2}\,dt\\\
&+\nu\alpha\int_{(-\infty,\lambda]}e^{-(\delta-\tau_{1})(\alpha-2)t}y_{v}^{\beta}\frac{(e^{(\delta-\tau_{1})t}y_{u})^{\alpha-1}-(e^{(\delta-\tau_{1})t}y_{u,\lambda})^{\alpha-1}}{e^{(\delta-\tau_{1})t}y_{u}-e^{(\delta-\tau_{1})t}y_{u,\lambda}}w_{u,\lambda}w^{+}_{u,\lambda}\,dt\\\
&+\nu\alpha\int_{(-\infty,\lambda]\cap\\{y_{v}>y_{v,\lambda}\\}}y_{u,\lambda}^{\alpha-1}(y_{v}^{\beta}-y_{v,\lambda}^{\beta})w^{+}_{u,\lambda}\,dt.\\\
\leq&-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}(w^{+}_{u,\lambda})^{2}\,dt+\int_{(-\infty,\lambda]}y_{u}^{2^{*}-2}(w^{+}_{u,\lambda})^{2}\,dt\\\
&+\nu\alpha\int_{(-\infty,\lambda]}e^{-(\delta-\tau_{1})(\alpha-1)t}y_{v}^{\beta}h^{\prime}(\sigma)(w^{+}_{u,\lambda})^{2}\,dt\\\
&+\nu\alpha\beta\int_{(-\infty,\lambda]\cap\\{y_{v}>y_{v,\lambda}\\}}y_{u,\lambda}^{\alpha-1}y_{v}^{\beta-1}w^{+}_{v,\lambda}w^{+}_{u,\lambda}\,dt,\\\
\end{split}$

where we always get

$\displaystyle\frac{(e^{(\delta-\tau_{1})t}y_{u})^{\alpha-1}-(e^{(\delta-\tau_{1})t}y_{u,\lambda})^{\alpha-1}}{e^{(\delta-\tau_{1})t}y_{u}-e^{(\delta-\tau_{1})t}y_{u,\lambda}}\leq
C.$

by Remark 2.4 Now, we recall that all the integrals in (4.12) are computed in
$\displaystyle\text{supp}(w_{u,\lambda}^{+})\cap(-\infty,\lambda]$. Hence, we
have $\displaystyle y_{u,\lambda}(t)\leq y_{u}(t)$ for every $\displaystyle
t\in\text{supp}(w_{u,\lambda}^{+})\cap(-\infty,\lambda]$. Moreover, since
$\displaystyle\lambda\leq-M$, by Lemma 4.1 we know that

$\displaystyle
y_{u}(t)\leq\underline{C}_{u}e^{(\delta-\tau_{1})t}u_{\tau_{1}}(e^{t})\leq
C\underline{C}_{u}e^{(\delta-\tau_{1})t},\ \ \text{ and that }\ \
y_{v}(t)\leq\underline{C}_{v}e^{(\delta-\tau_{1})t}v_{\tau_{1}}(e^{t})\leq
C\underline{C}_{v}e^{(\delta-\tau_{1})t},$

for any $\displaystyle t\leq-M$. Collecting all these information (4.12)
becomes

(4.13)
$\begin{split}\int_{(-\infty,\lambda]}[(w^{+}_{u,\lambda})^{\prime}]^{2}\,dt\leq&-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}(w^{+}_{u,\lambda})^{2}\,dt+C\int_{(-\infty,\lambda]}e^{(2^{*}-2)(\delta-\tau_{1})t}(w^{+}_{u,\lambda})^{2}\,dt\\\
&+\nu\alpha(\alpha-1)C\int_{(-\infty,\lambda]}e^{(2^{*}-2)(\delta-\tau_{1})t}(w^{+}_{u,\lambda})^{2}\,dt\\\
&+\nu\alpha\beta
C\int_{(-\infty,\lambda]}e^{(2^{*}-2)(\delta-\tau_{1})t}w_{v,\lambda}^{+}w^{+}_{u,\lambda}\,dt.\end{split}$

Using Young’s inequality in the last term of the right hand side, we deduce

(4.14)
$\begin{split}\int_{(-\infty,\lambda]}[(w^{+}_{u,\lambda})^{\prime}]^{2}\,dt\leq&-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}(w^{+}_{u,\lambda})^{2}\,dt+\bar{C}\int_{(-\infty,\lambda]}e^{(2^{*}-2)(\delta-\tau_{1})t}(w^{+}_{u,\lambda})^{2}\,dt\\\
&+\nu\alpha\beta
C\int_{(-\infty,\lambda]}e^{(2^{*}-2)(\delta-\tau_{1})t}(w_{v,\lambda}^{+})^{2}\,dt,\end{split}$

where $\displaystyle\bar{C}$ is a positive constant depending only on
$\displaystyle\nu,\alpha$ and $\displaystyle\|u_{\tau}\|_{\infty}$. Arguing in
the same way for the second equation of (4.9), we obtain

(4.15)
$\begin{split}\int_{(-\infty,\lambda]}[(w^{+}_{v,\lambda})^{\prime}]^{2}\,dt\leq&-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}(w^{+}_{v,\lambda})^{2}\,dt+\tilde{C}\int_{(-\infty,\lambda]}e^{(2^{*}-2)(\delta-\tau_{1})t}(w^{+}_{v,\lambda})^{2}\,dt\\\
&+\nu\alpha\beta
C\int_{(-\infty,\lambda]}e^{(2^{*}-2)(\delta-\tau_{1})t}(w_{u,\lambda}^{+})^{2}\,dt,\end{split}$

where $\displaystyle\tilde{C}$ is a positive constant depending only on
$\displaystyle\nu,\alpha$ and $\displaystyle\|u_{\tau}\|_{\infty}$. Summing
both (4.14), (4.15), we have

(4.16)
$\begin{split}\int_{(-\infty,\lambda]}[(w^{+}_{u,\lambda})^{\prime}]^{2}\,dt&+\int_{(-\infty,\lambda]}[(w^{+}_{v,\lambda})^{\prime}]^{2}\,dt\\\
&\leq-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}(w^{+}_{u,\lambda})^{2}\,dt-(\delta^{2}-\gamma)\int_{(-\infty,\lambda]}(w^{+}_{v,\lambda})^{2}\,dt\\\
&+\bar{C}\int_{(-\infty,\lambda]}e^{(2^{*}-2)(\delta-\tau_{1})t}(w^{+}_{u,\lambda})^{2}\,dt+\tilde{C}\int_{(-\infty,\lambda]}e^{(2^{*}-2)(\delta-\tau_{1})t}(w^{+}_{v,\lambda})^{2}\,dt\\\
&+\nu\alpha\beta
C\int_{(-\infty,\lambda]}e^{(2^{*}-2)(\delta-\tau_{1})t}(w_{v,\lambda}^{+})^{2}\,dt+\nu\alpha\beta
C\int_{(-\infty,\lambda]}e^{(2^{*}-2)(\delta-\tau_{1})t}(w_{u,\lambda}^{+})^{2}\,dt,\end{split}$

Let us define $\displaystyle
C_{t}:=\sup_{t\in(-\infty,\lambda]}e^{(2^{*}-2)(\delta-\tau_{1})t}<+\infty$.
Rearranging (4.16), we achieve the following estimate

(4.17)
$\begin{split}\int_{(-\infty,\lambda]}[(w^{+}_{u,\lambda})^{\prime}]^{2}\,dt+\int_{(-\infty,\lambda]}[(w^{+}_{v,\lambda})^{\prime}]^{2}\,dt\leq&\left[C_{t}(\bar{C}+\nu\alpha\beta
C)-(\delta^{2}-\gamma)\right]\int_{(-\infty,\lambda]}(w^{+}_{u,\lambda})^{2}\,dt\\\
&+\left[C_{t}(\tilde{C}+\nu\alpha\beta
C)-(\delta^{2}-\gamma)\right]\int_{(-\infty,\lambda]}(w^{+}_{v,\lambda})^{2}\,dt.\end{split}$

Now, if we fix $\displaystyle\lambda<-M$ sufficiently negative such that

$\displaystyle
C_{t}<\min\left\\{\frac{\delta^{2}-\gamma}{\bar{C}+\nu\alpha\beta
C},\frac{\delta^{2}-\gamma}{\tilde{C}+\nu\alpha\beta C}\right\\},$

we obtain

(4.18)
$\int_{(-\infty,\lambda]}[(w^{+}_{u,\lambda})^{\prime}]^{2}\eta_{R}^{2}\,dt+\int_{(-\infty,\lambda]}[(w^{+}_{v,\lambda})^{\prime}]^{2}\eta_{R}^{2}\,dt\leq
0.$

Hence, we just proved that for any $\displaystyle\lambda<-M$ is true that
$\displaystyle y_{u}\leq y_{u,\lambda}$ and $\displaystyle y_{v}\leq
y_{v,\lambda}$. In other words, if we define

$\Lambda=\left\\{\lambda\in R:y_{u}\leq y_{u,s}\,\,\,\text{and}\,\,\,y_{v}\leq
y_{v,s}\,\,\,\text{in}\,\,\,(-\infty,s]\,\,\,\text{for all $\displaystyle
s\in(-\infty,\lambda]$}\right\\}$

we can say $\displaystyle\Lambda\neq 0$.

To complete the proof it is sufficient to prove that
$\displaystyle\bar{\lambda}=\sup\,\Lambda<+\infty$. The rest of the proof is
standard, in particular one can follows a similar argument to the one used in
the proof of Theorem 1.5.

∎

Now, we shall prove that $\displaystyle(y_{u},y_{v})$ is a synchronized
solution, see (1.7). This follows if and only if there exists a constant
$\displaystyle C>0$ such that

$\displaystyle y_{v}=Cy_{u}\,.$

In the following result we shall exploit some ideas from [50]. The main
difference in our setting is the role played by the function

(4.19) $f(s)=s^{2^{*}-2}+\nu\alpha s^{\alpha-2}-1-\nu\beta s^{\alpha},$

whose number of roots depends on $\displaystyle\alpha,n$ and
$\displaystyle\nu$.

######  Proposition 4.3.

Let $\displaystyle(y_{u},y_{v})$ be a solution of
($\displaystyle\mathcal{R}_{*}^{\prime}$), then there exists
$\displaystyle\tilde{C}>0$ such that $\displaystyle y_{u}=\tilde{C}y_{v}$.

###### Proof.

Let $\displaystyle\tilde{C}$ be a positive constant, then the couple

(4.20)
$(\tilde{y}_{u}(t),\tilde{y}_{v}(t)):=\left(\frac{y_{u}}{\tilde{C}}(t),y_{v}(t)\right)$

satisfies the system

(4.21)
$\begin{cases}\tilde{y}^{\prime\prime}_{u}-(\delta^{2}-\gamma)\tilde{y}_{u}+\tilde{C}^{2^{*}-2}\tilde{y}_{u}^{2^{*}-1}+\nu\alpha\tilde{C}^{\alpha-2}\tilde{y}_{u}^{\alpha-1}\tilde{y}_{v}^{\beta}=0&\text{in}\quad{\mathbb{R}}\\\
\tilde{y}^{\prime\prime}_{v}-(\delta^{2}-\gamma)\tilde{y}_{v}+\tilde{y}_{v}^{2^{*}-1}+\nu\beta\tilde{C}^{\alpha}\tilde{y}_{u}^{\alpha}\tilde{y}_{v}^{\beta-1}=0&\text{in}\quad{\mathbb{R}}\\\
\tilde{y}_{u},\tilde{y}_{v}>0&\text{in}\quad{\mathbb{R}}.\end{cases}$

As a first step, multiplying the first equation of (4.21) by $\displaystyle
y_{v}$, one obtains

(4.22)
$(\tilde{y}_{u}^{\prime}\tilde{y}_{v})^{\prime}-\tilde{y}^{\prime}_{u}\tilde{y}^{\prime}_{v}-(\delta^{2}-\gamma)\tilde{y}_{u}\tilde{y}_{v}+\tilde{C}^{2^{*}-2}\tilde{y}_{u}^{2^{*}-1}\tilde{y}_{v}+\nu\alpha\tilde{C}^{\alpha-2}\tilde{y}_{u}^{\alpha-1}\tilde{y}_{v}^{\beta+1}=0.$

Analogously, multiplying the second equation of (4.21) by $\displaystyle
y_{u}$,

(4.23)
$(\tilde{y}_{v}^{\prime}\tilde{y}_{u})^{\prime}-\tilde{y}^{\prime}_{u}\tilde{y}^{\prime}_{v}-(\delta^{2}-\gamma)\tilde{y}_{u}\tilde{y}_{v}+\tilde{y}_{v}^{2^{*}-1}\tilde{y}_{u}+\nu\beta\tilde{C}^{\alpha}\tilde{y}_{u}^{\alpha+1}\tilde{y}_{v}^{\beta-1}=0.$

Subtracting (4.22) by (4.23) gives

(4.24)
$\begin{split}(\tilde{y}_{u}^{\prime}\tilde{y}_{v}-\tilde{y}_{v}^{\prime}\tilde{y}_{u})^{\prime}&+\tilde{C}^{2^{*}-2}\tilde{y}_{u}^{2^{*}-1}\tilde{y}_{v}+\nu\alpha\tilde{C}^{\alpha-2}\tilde{y}_{u}^{\alpha-1}\tilde{y}_{v}^{\beta+1}-\tilde{y}_{v}^{2^{*}-1}\tilde{y}_{u}-\nu\beta\tilde{C}^{\alpha}\tilde{y}_{u}^{\alpha+1}\tilde{y}_{v}^{\beta-1}=\\\
(\tilde{y}_{u}^{\prime}\tilde{y}_{v}-\tilde{y}_{v}^{\prime}\tilde{y}_{u})^{\prime}&+\tilde{y}_{v}^{2^{*}-1}\tilde{y}_{u}f\left(\tilde{C}\frac{\tilde{y}_{u}}{\tilde{y}_{v}}\right)=0,\end{split}$

where the function $\displaystyle f:(0,+\infty)\to\mathbb{R}$ is defined by
(4.19).

By Lemma 4.1, we recall that
$\displaystyle\tilde{y}_{u}(t),\tilde{y}_{v}(t)\to 0$ as $\displaystyle
t\to-\infty$, $\displaystyle\tilde{y}_{u}(t),\tilde{y}_{v}(t)\to 0$ as
$\displaystyle t\to+\infty$, and by Lemma 4.2 we have
$\displaystyle\tilde{y}^{\prime}_{u}(0)=\tilde{y}^{\prime}_{v}(0)=0$, assuming
by translation that $\displaystyle 0$ is the global maximum of
$\displaystyle\tilde{y}_{u},\tilde{y}_{v}$. By integrating (4.24)

$\displaystyle\int_{-\infty}^{0}y_{v}^{2^{*}-1}y_{u}f\left(\tilde{C}\frac{\tilde{y}_{u}}{\tilde{y}_{v}}\right)\,dt=0,\qquad\int_{0}^{+\infty}\tilde{y}_{v}^{2^{*}-1}\tilde{y}_{u}f\left(\tilde{C}\frac{\tilde{y}_{u}}{\tilde{y}_{v}}\right)\,dt=0.$

If we deduce that $\displaystyle
f\left(\tilde{C}\frac{y_{u}}{y_{v}}\right)\leq 0$ or $\displaystyle
f\left(\tilde{C}\frac{y_{u}}{y_{v}}\right)\geq 0$ for every $\displaystyle
t\in(-\infty,0)$, then it would follows $\displaystyle
f\left(\tilde{C}\frac{y_{u}}{y_{v}}\right)=0$ in $\displaystyle(-\infty,0)$.

To achieve this claim, we note that, by Lemma 4.1 and in particular by (4.6),
we have

$\displaystyle\lim_{t\to-\infty}\frac{y_{u}(t)}{y_{v}(t)}=L\in(0,+\infty).$

We have to study two different cases.

Case 1: $\displaystyle f(L)\neq 0$. Without loss of generality we assume that

(4.25) $f(L)<0.$

By (4.25) there exists $\displaystyle\hat{t}<0$ such that

(4.26) $f\left(\tilde{C}\frac{\tilde{y}_{u}(t)}{\tilde{y}_{v}(t)}\right)<0\ \
\forall t\in(-\infty,\hat{t})\text{ and
}f\left(\tilde{C}\frac{\tilde{y}_{u}(\hat{t})}{\tilde{y}_{v}(\hat{t})}\right)=f\left(\frac{y_{u}(\hat{t})}{y_{v}(\hat{t})}\right)=0.$

¿From now on, we can fix
$\displaystyle\tilde{C}:=\frac{y_{u}(\hat{t})}{y_{v}(\hat{t})}$ so that

$\displaystyle f(\tilde{C})=0$

and $\displaystyle\tilde{y}_{u}(\hat{t})=\tilde{y}_{v}(\hat{t})$ as in (4.20).
By integrating (4.24) in $\displaystyle(-\infty,t)$ for any $\displaystyle
t\leq\hat{t}$, we get

(4.27)
$(\tilde{y}^{\prime}_{u}\tilde{y}_{v}-\tilde{y}^{\prime}_{v}\tilde{y}_{u})(t)=-\int_{-\infty}^{t}\tilde{y}_{v}^{2^{*}-1}\tilde{y}_{u}f\left(\tilde{C}\frac{\tilde{y}_{u}}{\tilde{y}_{v}}\right)\,ds>0.$

Now, let us assume for a while that there exists $\displaystyle
0<t_{1}<\hat{t}$, such that

(4.28)
$(\tilde{y}^{\prime}_{u}\tilde{y}_{v}-\tilde{y}^{\prime}_{v}\tilde{y}_{u})(t_{1})=0.$

This would immediately imply that

$\displaystyle(\tilde{y}^{\prime}_{u}\tilde{y}_{v}-\tilde{y}^{\prime}_{v}\tilde{y}_{u})(t_{1})+\int_{-\infty}^{t_{1}}\tilde{y}_{v}^{2^{*}-1}\tilde{y}_{u}f\left(\tilde{C}\frac{\tilde{y}_{u}}{\tilde{y}_{v}}\right)\,ds=\int_{-\infty}^{t_{1}}\tilde{y}_{v}^{2^{*}-1}\tilde{y}_{u}f\left(\tilde{C}\frac{\tilde{y}_{u}}{\tilde{y}_{v}}\right)\,ds=0.$

Then we could conclude that

$\displaystyle
f\left(\tilde{C}\frac{\tilde{y}_{u}(t)}{\tilde{y}_{v}(t)}\right)=f\left(\frac{y_{u}(t)}{y_{v}(t)}\right)=0\
\ \text{ for any }t\in(-\infty,t_{1}).$

In particular, this fact would immediately give

(4.29) $\frac{y_{u}(t)}{y_{v}(t)}=\hat{C}\ \text{ for }\ t\in(-\infty,t_{1})\
\text{ such that }\ f(\hat{C})=0.$

By uniqueness of the limits at $\displaystyle-\infty$ we deduce that
$\displaystyle\tilde{C}=\hat{C}=L$, which is a contradiction since we assumed
$\displaystyle f(L)\neq 0$ in this case.

The rest of the proof is devoted to show (4.28). By contradiction, assume that
(4.28) is false. Then, by (4.27),

(4.30)
$(\tilde{y}^{\prime}_{u}\tilde{y}_{v}-\tilde{y}^{\prime}_{v}\tilde{y}_{u})(t)>0\qquad\mbox{
for all }t\in(-\infty,\hat{t}).$

Now, multiplying the first equation of (4.21) by
$\displaystyle\tilde{y}^{\prime}_{u}$, one gets,

(4.31)
$\left(\frac{(\tilde{y}^{\prime}_{u})^{2}}{2}\right)^{\prime}-\frac{(\delta^{2}-\gamma)}{2}(\tilde{y}_{u}^{2})^{\prime}+\frac{\tilde{C}^{2^{*}-2}}{2^{*}}(\tilde{y}_{u}^{2^{*}})^{\prime}+\tilde{C}^{\alpha-2}\nu\alpha\tilde{y}_{u}^{\alpha-1}\tilde{y}_{v}^{\beta}\tilde{y}_{u}^{\prime}=0.$

Similarly, multiplying the second equation of (4.21) by
$\displaystyle\tilde{y}^{\prime}_{v}$,

(4.32)
$\left(\frac{(\tilde{y}^{\prime}_{v})^{2}}{2}\right)^{\prime}-\frac{(\delta^{2}-\gamma)}{2}(\tilde{y}_{v}^{2})^{\prime}+\frac{1}{2^{*}}(\tilde{y}_{v}^{2^{*}})^{\prime}+\tilde{C}^{\alpha}\nu\beta\tilde{y}_{u}^{\alpha}\tilde{y}_{v}^{\beta-1}\tilde{y}_{v}^{\prime}=0.$

Now, subtracting (4.31) and (4.32), integrating in
$\displaystyle(-\infty,\hat{t})$, and using the fact that

$\displaystyle\tilde{y}_{u}(\hat{t})=\tilde{y}_{v}(\hat{t}),$

we have

(4.33)
$\begin{split}&\left(\frac{(\tilde{y}_{u}^{\prime})^{2}}{2}-\frac{(\tilde{y}_{v}^{\prime})^{2}}{2}\right)(\hat{t})\\\
&\qquad\ \
\quad+\int_{-\infty}^{\hat{t}}\left(\frac{\tilde{C}^{2^{*}-2}}{2^{*}}(\tilde{y}_{u}^{2^{*}})^{\prime}-\frac{1}{2^{*}}(\tilde{y}_{v}^{2^{*}})^{\prime}+\tilde{C}^{\alpha-2}\nu\alpha\tilde{y}_{u}^{\alpha-1}\tilde{y}_{v}^{\beta}\tilde{y}_{u}^{\prime}-\tilde{C}^{\alpha}\nu\beta\tilde{y}_{u}^{\alpha}\tilde{y}_{v}^{\beta-1}\tilde{y}_{v}^{\prime}\,\right)ds=0.\end{split}$

We note that, by (4.26), it follows

(4.34)
$\int_{-\infty}^{\hat{t}}(\tilde{y}_{u}^{2^{*}})^{\prime}\,ds=\int_{-\infty}^{\hat{t}}(\tilde{y}_{v}^{2^{*}})^{\prime}\,ds=\int_{-\infty}^{\hat{t}}(\tilde{y}_{u}^{\alpha}\tilde{y}_{v}^{\beta})^{\prime}\,ds=\int_{-\infty}^{\hat{t}}\left(\alpha\tilde{y}_{u}^{\alpha-1}\tilde{y}_{v}^{\beta}\tilde{y}_{u}^{\prime}+\beta\tilde{y}_{u}^{\alpha}\tilde{y}_{v}^{\beta-1}\tilde{y}_{v}^{\prime}\right)\,ds.$

Using (4.34) in (4.33), we get

(4.35)
$\begin{split}\left(\frac{(\tilde{y}_{u}^{\prime})^{2}}{2}-\frac{(\tilde{y}_{v}^{\prime})^{2}}{2}\right)(\hat{t})&+\int_{-\infty}^{\hat{t}}\alpha\left(\frac{\tilde{C}^{2^{*}-2}-1}{2^{*}}+\nu\tilde{C}^{\alpha-2}\right)\tilde{y}_{u}^{\alpha-1}\tilde{y}_{v}^{\beta}\tilde{y}_{u}^{\prime}\,ds\\\
&+\int_{-\infty}^{\hat{t}}\beta\left(\frac{\tilde{C}^{2^{*}-2}-1}{2^{*}}-\nu\tilde{C}^{\alpha}\right)\tilde{y}_{u}^{\alpha}\tilde{y}_{v}^{\beta-1}\tilde{y}_{v}^{\prime}\,ds=0.\end{split}$

Recalling that $\displaystyle f(\tilde{C})=0$, we deduce that

$\displaystyle\tilde{C}^{2^{*}-2}-1=-\nu\alpha\tilde{C}^{\alpha-2}+\nu\beta\tilde{C}^{\alpha}.$

Hence, using the fact that $\displaystyle\alpha+\beta=2^{*}$ one can easily
check that

$\displaystyle\alpha\left(\frac{\tilde{C}^{2^{*}-2}-1}{2^{*}}+\nu\tilde{C}^{\alpha-2}\right)=-\beta\left(\frac{\tilde{C}^{2^{*}-2}-1}{2^{*}}-\nu\tilde{C}^{\alpha}\right)=\frac{\nu\alpha\beta}{2^{*}}\tilde{C}^{\alpha-2}(1+\tilde{C}^{2}).$

Thanks to this information, equation (4.35) becomes

(4.36)
$\begin{split}\left(\frac{(\tilde{y}_{u}^{\prime})^{2}}{2}-\frac{(\tilde{y}_{v}^{\prime})^{2}}{2}\right)(\hat{t})+\alpha\left(\frac{\tilde{C}^{2^{*}-2}-1}{2^{*}}+\nu\tilde{C}^{\alpha-2}\right)\int_{-\infty}^{\hat{t}}\tilde{y}_{u}^{\alpha-1}\tilde{y}_{v}^{\beta-1}(\tilde{y}_{u}^{\prime}\tilde{y}_{v}-\tilde{y}_{v}^{\prime}\tilde{y}_{u})\,ds=0.\end{split}$

By (4.30) and the fact that
$\displaystyle\tilde{y}_{u}(\hat{t})=\tilde{y}_{v}(\hat{t})$, we know that

(4.37)
$(\tilde{y}^{\prime}_{u}\tilde{y}_{v}-\tilde{y}^{\prime}_{v}\tilde{y}_{u})(\hat{t})=\tilde{y}_{u}(\hat{t})[\tilde{y}^{\prime}_{u}(\hat{t})-\tilde{y}^{\prime}_{v}(\hat{t})]>0.$

Finally, we arrive at the conclusion that (4.33) gives us a contradiction,
since (4.30) and (4.37) should imply the left hand to be strictly positive.

Case 2: $\displaystyle f(L)=0$. Since
$\displaystyle\lim_{t\to-\infty}\frac{y_{u}(t)}{y_{v}(t)}=L\in(0,+\infty)$,
then we claim that

(4.38) $\exists\ \tilde{t}<0\text{ such that }\frac{y_{u}(t)}{y_{v}(t)}\leq
L\text{ or }\frac{y_{u}(t)}{y_{v}(t)}\geq L\text{ for any
}t\in(-\infty,\tilde{t}].$

To prove this, we argue by contradiction assuming that there exists a sequence
of points
$\displaystyle\\{t_{k}\\}_{k\in{\mathbb{N}}}\subset(-\infty,\tilde{t}]$ such
that

$\displaystyle\frac{y_{u}(t_{k})}{y_{v}(t_{k})}=L\text{ for any
}k\in{\mathbb{N}}\text{ and }f\left(\frac{y_{u}(t)}{y_{v}(t)}\right)\neq
0\text{ for any }t\in(t_{k},t_{k+1}).$

Let us consider any closed interval $\displaystyle[t_{k},t_{k+1}]$ and fix
$\displaystyle\tilde{C}=L$. Hence, recalling (4.20), we have that
$\displaystyle
f\left(\tilde{C}\frac{\tilde{y}_{u}(t)}{\tilde{y}_{v}(t)}\right)=f\left(\frac{y_{u}(t)}{y_{v}(t)}\right)\neq
0$ for any $\displaystyle t\in(t_{k},t_{k+1})$. Let us consider

$\displaystyle w(t)=\tilde{y}_{u}(t)-\tilde{y}_{v}(t).$

Without loss of generality we can suppose that $\displaystyle w(t)>0$ for any
$\displaystyle t\in(t_{k},t_{k+1})$ and $\displaystyle w(t_{k})=w(t_{k+1})=0$.
We point out that $\displaystyle(\tilde{y}_{u},\tilde{y}_{v})$ weakly solves
(4.21), i.e.

(4.39)
$\begin{split}\int_{\mathbb{R}}\tilde{y}^{\prime}_{u}\varphi^{\prime}\,dt&=-(\delta^{2}-\gamma)\int_{\mathbb{R}}\tilde{y}_{u}\varphi\,dt+\int_{\mathbb{R}}\left(\tilde{C}^{2^{*}-2}\tilde{y}_{u}^{2^{*}-1}+\nu\alpha\tilde{C}^{\alpha-2}\tilde{y}_{u}^{\alpha-1}\tilde{y}_{v}^{\beta}\right)\varphi\,dt,\quad\forall\varphi\in
C^{1}_{c}({\mathbb{R}}),\\\ \\\
\int_{\mathbb{R}}\tilde{y}^{\prime}_{v}\varphi^{\prime}\,dt&=-(\delta^{2}-\gamma)\int_{\mathbb{R}}\tilde{y}_{v}\varphi\,dt+\int_{\mathbb{R}}\left(\tilde{y}_{v}^{2^{*}-1}+\nu\beta\tilde{C}^{\alpha}\tilde{y}_{u}^{\alpha}\tilde{y}_{v}^{\beta-1}\right)\varphi\,dt,\quad\forall\varphi\in
C^{1}_{c}({\mathbb{R}}).\end{split}$

We consider

$\displaystyle\varphi:=w(t)\chi_{(t_{k},t_{k+1})}(t)$

as test function in (4.39) so that, subtracting both the equations, we get

(4.40)
$\begin{split}\int_{(t_{k+1},t_{k})}(w^{\prime})^{2}\,dt=&-(\delta^{2}-\gamma)\int_{(t_{k+1},t_{k})}w^{2}\,dt\\\
&+\int_{(t_{k+1},t_{k})}\left(\tilde{C}^{2^{*}-2}\tilde{y}_{u}^{2^{*}-1}+\nu\alpha\tilde{C}^{\alpha-2}\tilde{y}_{u}^{\alpha-1}\tilde{y}_{v}^{\beta}-\tilde{y}_{v}^{2^{*}-1}-\nu\beta\tilde{C}^{\alpha}\tilde{y}_{u}^{\alpha}\tilde{y}_{v}^{\beta-1}\right)w\,dt\\\
=&-(\delta^{2}-\gamma)\int_{(t_{k+1},t_{k})}w^{2}\,dt+\int_{(t_{k+1},t_{k})}\tilde{y}_{v}^{2^{*}-1}\left[g\left(\frac{\tilde{y}_{u}}{\tilde{y}_{v}}\right)-g(1)\right]w\,dt,\end{split}$

where $\displaystyle g:(0,+\infty)\rightarrow{\mathbb{R}}$ is defined as

$\displaystyle
g(s):=\tilde{C}^{2^{*}-2}s^{2^{*}-1}+\nu\alpha\tilde{C}^{\alpha-2}s^{\alpha-1}-\nu\beta\tilde{C}^{\alpha}s^{\alpha}\quad\text{and}\quad
g(1)=C^{2^{*}-2}+\nu\alpha\tilde{C}^{\alpha-2}-\nu\beta\tilde{C}^{\alpha}=1,$

since $\displaystyle 0=f(\tilde{C})=g(1)-1$. By the Mean Value’s Theorem we
deduce that there exists $\displaystyle
l\in\left(1,\frac{\tilde{y}_{u}}{\tilde{y}_{v}}\right)$ such that (4.40)
becomes

(4.41)
$\begin{split}\int_{(t_{k+1},t_{k})}(w^{\prime})^{2}\,dt\leq-(\delta^{2}-\gamma)\int_{(t_{k+1},t_{k})}w^{2}\,dt+\int_{(t_{k+1},t_{k})}\tilde{y}_{v}^{2^{*}-2}|g^{\prime}(l)|w^{2}\,dt.\end{split}$

Now we observe that $\displaystyle|g^{\prime}(l)|\leq C$. Since
$\displaystyle\frac{\tilde{y}_{u}(t)}{\tilde{y}_{v}(t)}\rightarrow 1$ as
$\displaystyle t\rightarrow-\infty$, up to redefine $\displaystyle\tilde{t}$
we can assume that

$\displaystyle\tilde{y}_{v}(t)\leq\left(\frac{\delta^{2}-\gamma}{C}\right)^{\frac{1}{2^{*}-2}}$

so that, by (4.41), we get

(4.42) $\begin{split}\int_{(t_{k+1},t_{k})}(w^{\prime})^{2}\,dt\leq
0,\end{split}$

which implies that $\displaystyle w^{\prime}(t)=0$ for any $\displaystyle
t\in(t_{k+1},t_{k})$ and hence $\displaystyle w(t)=C$. But, since
$\displaystyle w(t_{k+1})=w(t_{k})=0$ we deduce that $\displaystyle w(t)=0$
for any $\displaystyle t\in(t_{k+1},t_{k})$. As consequence we deduce
$\displaystyle\tilde{y}_{u}(t)=\tilde{y}_{v}(t)$ for any $\displaystyle
t\in(t_{k+1},t_{k})$ providing a contradiction.
Thus $\displaystyle f\left(\frac{y_{u}(t)}{y_{v}(t)}\right)$ has a sign at
$\displaystyle-\infty$. Hence, we may run over the arguments of Case 1
recovering (4.29). Therefore we have that

$\displaystyle y_{u}(t)=\tilde{C}y_{v}(t)\text{ for any }t\in{\mathbb{R}},$

with $\displaystyle\tilde{C}=L$ or, if this is not the case, the argument can
be repeated proving the thesis. ∎

At this point, we are able to prove the classification result.

###### Proof of Theorem 1.1.

Let $\displaystyle(u,v)\in D^{1,2}({\mathbb{R}}^{n})\times
D^{1,2}({\mathbb{R}}^{n})$ be a solution to ($\displaystyle\mathcal{S}^{*}$).
By Proposition 4.3, one gets

$\displaystyle y_{u}=\tilde{C}y_{v}.$

This gives that

$\displaystyle u=\tilde{C}v.$

By direct computations, it follows that $\displaystyle(u,v)$ solves

(4.43) $\begin{cases}\displaystyle-\Delta
u\,=\gamma\frac{u}{|x|^{2}}+\left(1+\frac{\nu\alpha}{\tilde{C}^{\beta}}\right)u^{2^{*}-1}&\text{in}\quad{\mathbb{R}}^{n}\vspace{0.2cm}\\\
\displaystyle-\Delta
v\,=\gamma\frac{v}{|x|^{2}}+\left(1+\nu\beta\tilde{C}^{\alpha}\right)v^{2^{*}-1}&\text{in}\quad{\mathbb{R}}^{n}\\\
u,v>0&\text{in}\quad{\mathbb{R}}^{n}\setminus\\{0\\}.\end{cases}$

By dilatation and rescaling, exploiting [46], it follows that

$\displaystyle(u,v)=(c_{1}\mathcal{U}_{\mu_{0}},c_{2}\mathcal{U}_{\tilde{\mu}})$

with $\displaystyle\mathcal{U}_{\mu}$ given by the expression (1.4) and
$\displaystyle\tilde{\mu},\mu_{0}>0$. Moreover, by the proportionality of
$\displaystyle u$ and $\displaystyle v$, one can deduce that
$\displaystyle\tilde{\mu}=\mu_{0}$. Since $\displaystyle\tilde{C}$ satisfies
$\displaystyle f(\tilde{C})=0$, then

$\displaystyle
1+\frac{\nu\alpha}{\tilde{C}^{\beta}}=\frac{1}{\tilde{C}^{2^{*}-2}}+\frac{\nu\beta}{\tilde{C}^{\beta-2}}.$

Introducing such information in (4.43), we can conclude that

$\displaystyle\displaystyle-\Delta\mathcal{U}_{\mu_{0}}\,=\gamma\frac{\mathcal{U}_{\mu_{0}}}{|x|^{2}}+\left(\frac{1}{\tilde{C}^{2^{*}-2}}+\frac{\nu\beta}{\tilde{C}^{\beta-2}}\right)c_{1}^{2^{*}-2}\mathcal{U}_{\mu_{0}}^{2^{*}-1}=\gamma\frac{\mathcal{U}_{\mu_{0}}}{|x|^{2}}+\left(1+\nu\beta\tilde{C}^{\alpha}\right)c_{2}^{2^{*}-2}\mathcal{U}_{\mu_{0}}^{2^{*}-1}\quad\text{in
}{\mathbb{R}}^{n},$

which immediately implies that $\displaystyle\tilde{C}=\frac{c_{1}}{c_{2}}$
and, moreover, $\displaystyle(c_{1},c_{2})$ solution to the system (1.8),
which completes the proof. ∎

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