Classification of solutions to Hardy-Sobolev Doubly Critical Systems

This work deals with a family of Hardy-Sobolev doubly critical system defined in $\mathbb{R}^n$. 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.


Introduction
We are concerned with the study of the doubly critical system: Along this article we address the analysis of the classification of positive solutions to (S * ).
In the case γ = ν = 0, system (S * ) reduces to the classical Sobolev critical equation (1.1) It is well-known that the so-called Aubin-Talenti bubbles solve (1.1), where λ > 0 and it is centered at x 0 ∈ R n .Moreover, they realize the equality in the sharp Sobolev inequality in R n .In [29], the authors proved that any positive solution u to (1.1), such that u(x) = O (1/|x| m ) , at infinity for m > 0, is radially symmetric and decreasing about some point in 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 u ∈ H 1 loc (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 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 γ = 0 and ν = 0, system (S * ) becomes the so-called Hardy-Sobolev doubly critical equation 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) , where (1.5) , and µ > 0 is a scaling factor.Obviously, U µ = V µ,0 if γ = 0.
The case of 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 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 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) where V is the system potential and 1 < p ≤ 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 V and ν.
Concerning the critical case p = 2 * 2 , if V is non-zero constant, then system (1.6) admits only the trivial solution (0, 0).This result follows from a proper application of the Pohozaevtype identity.For 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. ν < 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 V = − γ |x| 2 .Under that choice, the problem (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 ν > 0, the existence of ground and bound states are obtained depending on α, β, n and a potential function 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 (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: with U µ introduced in (1.4), µ 0 > 0 and c 1 , c 2 are any positive constants satisfying the system Remark 1.2.This result holds (and is new) also in the case γ = 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.
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 α < 2, β < 2 or both.In such a case, in fact, the coupling term is non locally Lipschitz continuous.
Note that this necessarily occurs if 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 (S * ) are radially symmetric about the origin.Actually we shall provide a more general result and, in particular, holds without requiring global energy information.
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 x 1 -direction move the Hardy potential to 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 for the right choice of τ > 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.where t := log r with r > 0 and δ = n − 2 2 .We prove that there exists C > 0 such that y u = Cy v where C is a zero of the function The number of roots of 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 f has been treated in the literature.Precisely, the hardest part is the case when the function f (•) has more than one zero, a possible issue, see Remark 1.4.When the proportionality of components (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 (S * γ 1,2 ) is radially symmetric about the origin.We shall therefore consider positive continuous (far from the origin) solutions To prove Theorem 1.5, we need to fix some notations.For any real number λ we set which is the reflection through the hyperplane T λ := {x 1 = λ}.Moreover, given any function w, we will set namely the reflected function.A crucial ingredient in our proof is the use of a translation argument.We fix x 0 = (0, x ′ 0 ) ∈ R n \ {0} and we assume, by translation, that (u, v) solves This will allows us to take full advantage of the Kelvin transformation.In fact we set K : R n \ {0} −→ R n \ {0} defined by Such a transformation is a well-known tool and, given any (u, v) solution to (S * x 0 ), its Kelvin transform is defined as By direct computation it follows that (û, v) weakly satisfies Having in mind the last definition, we prove the following key lemma that will be used in the proof of the symmetry result.
Proof.It is immediate to have that for every x ∈ R n .We observe that thanks to our assumptions, the term x, x 0 does not depend on x 1 .Hence, thanks to Schwarz inequality we deduce that for any (2.5) Analogously, we get We shall prove a symmetry result regarding any solution to ( Ŝ * x 0 ), that translates into a symmetry result for the original problem (S * γ 1,2 ).In particular, the couple (û, v) satisfies 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 (û, v) be a solution to ( Ŝ * x 0 ).Then, there exists R > 0 and µ > 0 such that Proof.First of all, we note that (û, v) satisfies (2.6).Borrowing an idea contained in [23], we point out that, for every ε > 0, we can find a function We note that there exists R > 0 such that x 0 ∈ B R(0).Since û > 0 in B R(0) \ {0}, there exists µ > 0 such that û > µ > 0. In the same way v > µ > 0 on ∂B R(0).Hence, setting 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 Using the Young's inequality, we are able to get that (2.11) Finally, we have that (2.12) Passing to the limit for ε that goes to 0 we get the thesis for û.Arguing in a similar fashion, we obtain the same result for v.
The translation argument that we introduced allow us to deduce the following: Proof.Since x 0 = (0, x ′ 0 ) ∈ R n is fixed, then there exists δ > 0 such that x 0 ∈ B δ (0).By our assumptions u, v ∈ C(R n \ {x 0 }), and hence we deduce that u, v ∈ C(B δ (0)).We fix R > 1 δ 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 ( Ŝ * We recall that we are working with the weak formulations (2.6) and (2.7).By Lemma 2.3 we deduce that |û(x)| ≤ C u |x| 2−n and |v(x)| ≤ C v |x| 2−n and for every x ∈ R n such that |x| ≥ R, where C u , C v and R are positive constants (depending on u and v).In particular, for every λ < − R < 0, we have In order to complete the proof of our result, we split the proof into three steps.
Step 1: There exists M > 0 large such that û ≤ ûλ and v ≤ vλ in Σ λ \ R λ ({0, x 0 }), for all λ < −M.We immediately see that ).We point out that, for every ε > 0, we can find a function Therefore, by a standard density argument, we can plug Φ and Ψ as test functions in (2.6) and (2.7) respectively, so that, subtracting we get and (2.17) Exploiting also Young's inequality, recalling that 0 ≤ ξ + λ ≤ û and 0 ≤ ζ + λ ≤ v, we get that (2.18) Similarly, we obtain where c(n) is a positive constant depending only on the dimension n and by the absolute continuity of the Lebesgue integral.Analogously, we deduce that Let us now estimate I 3 and E 3 .Here, we recall the monotonicity property of the function f x 0 stated in Lemma 2.1.Moreover, for λ < 0 sufficiently large we have that . Therefore where (2.23) By Hardy's and Young's inequality we also deduce that (2.24) Employing the argument to estimate I 3 , we get Since û(x), ûλ (x), v(x), vλ (x) > 0, by the convexity of t → t 2 * −1 , for t > 0, we obtain for every x ∈ Σ λ .Thus, Therefore, using Hölder's inequality with exponents 2 * 2 , n 2 , we have where in the last two steps we applied Sobolev and Young's inequalities respectively.Arguing in the same way, we deduce The evaluation of I 5 and E 5 is a delicate issue within this argument.Note that, in particular, we may have that either α < 2, β < 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 Having in mind this argument, one obtains that where in the last inequality we used the cooperativity of our system and the fact that we are working in Σ λ ∩ supp[(u − u λ ) + ].Making use of Lemma 2.3 we deduce that (2.29) By Remark 2.4, we get 3. Then (2.29) rewrites as where the constant C has been relabelled and we have applied Young's inequality.
Similarly, we can obtain an analogous estimate for E 5 , i.e.
As we argued in (2.26) for I 4 in, we get that Analogously, we deduce that (2.33) Collecting all the previous estimates for I k , E k with k = 1, 2, 3, 4, 5, we deduce that where We obtain an analogous estimate for ζ + λ , i.e. where Summing both the contributions (2.34) and (2.35) we get . (2.36) Now, recalling (2.23) and using the absolute continuity of the Lebesgue integral, we fix M > 0 sufficiently large such that for each λ < −M.Finally, passing to the limit for R → +∞ and ε → 0 + , we obtain that (2.37) for each λ < −M.The last inequalities immediately imply the thesis of the first step.
The case (ii) is not possible ,because, since û has a sign at x 0 , it holds that û < ûλ and v < vλ in Σλ \ Rλ({0, x 0 }).Now, we push further the hyperplane Tλ and consider the hyperplane Tλ +ε for some ε > 0. We claim that, for ε > 0 small, we have that where R > 0 is given by Lemma 2.3 and δ > 0 is small.For the reader convenience we set Arguing by contradiction, let us assume that (2.39) does not hold.Hence, if we define we deduce that ∃ P m ∈ G τm , with {τ m } m∈N (τ m → 0) such that û(P m ) > ûλ +τm (P m ) or v(P m ) > vλ +τm (P m ).
Without loss of generality, we assume that û(P m ) > ûλ +τm (P m ).Up to subsequence, as m → +∞ we have that Finally we have that or P ∈ Tλ or P / ∈ Tλ.
• If P ∈ Tλ, then we have ∂ û ∂x 1 ( P ) ≤ 0. But, by the Hopf boundary lemma, we know that ∂ û ∂x 1 ( 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 as test functions respectively in (2.6) and in (2.7) so that, subtracting we get (2.40) (2.41) Now, we can define the following sets where, in the set 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 û and v are bounded away from zero near the origin and near the translation point 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) Finally, collecting (2.39), (2.42) and (2.43), we deduce supp[(û − ûλ +ε ) + ] = ∅, supp[(v − vλ +ε ) + ] = ∅ and hence û < ûλ +ε and v < vλ +ε in Σλ +ε , which is in contradiction with the definition of λ.So, λ must be 0.
Step 3: Conclusion.The symmetry of the Kelvin transform (û, v) in the x 1 -direction follows now performing the moving plane method in the opposite direction.By the definition of (û, v) given in (2.2), the symmetry of û and v w.r.t. the hyperplane {x 1 = 0} implies the same symmetry of the solution (u, v).The symmetry of any solution of (S * γ 1,2 ) follows now recalling that the translation point x 0 is arbitrary.We can repeat the same argument with respect to any fixed direction ν ∈ R n .The last one implies that (u, v) must be radially symmetric about the origin.

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 (S * ) that we recall for the reader convenience For any τ > 0, let us define In particular, we set which are solutions to the algebraic equation 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 |x| −τ 1 , while |x| −τ 2 gives us the behavior at infinity.In any case, with this choice of τ , it is easy to check that (u τ , v τ ) weakly solves Such equation, in the weak formulation, is well defined in the weighted Sobolev space D 1,2 τ (R n ) for τ = τ 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 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: Proof.For any η ≥ 1 and T > 0 we define the following Lispchitz function Making easy computations one can check that for any w ∈ D 1,2 τ (R n ) it holds the following inequality in the weak distributional meaning Now we make all the computations on the first equation of system (W * ).We remark that, since Υ is a Lipschitz function, then Υ(u τ ), Υ(v τ ) ∈ D 1,2 τ (R n ).Hence, it holds the Sobolev inequality By (3.5), we deduce where in the last inequality we used the estimate tΥ ′ (t) ≤ ηΥ(t) for every t > 0. Integrating by parts in the left hand side of (3.7), and using (3.6) we deduce Arguing in the same way with the second equation of (W * ), we obtain (3.9) Our aim is to apply the Moser's iteration scheme (see [35]).In order to do this, we take into account that u τ , v τ ∈ D 1,2 τ (R n ).Now, let η := 2 * 2 and m u , m v ∈ 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 g(t) := Υ(t)t α−2 is non-decreasing for every t ≥ 0 and for every α > 1.Hence, taking into account this fact and using the definition of Υ in (3.4), we have Using Hölder's inequality with conjugate exponents Arguing in an analogous way to (3.10) for the second equation of (W * ), we are able to deduce . (3.12) Hence, thanks to (3.11), by (3.8) we obtain Arguing in a similar way with v τ , thanks to (3.12), by (3.9) we deduce Summing both (3.13) and (3.14), we deduce that Hence, from (3.15) we obtain where C depends on m u , m v , ν, n and C S .Moreover, we used that Υ(t) ≤ t η and that Hence, recalling that η = 2 * 2 and taking T → +∞, thanks also to Fatou's lemma one has Finally, we set (3.21) and we obtain the recursive inequality By induction we can easily deduce that log (3.22)where in the last inequality we used that the series +∞ j=2 log C j converges.Hence, thanks to the monotonicity of the integral and by (3.22), for every R > 0, we have (3.23)log Passing to the limit for k → +∞ and thus η k → +∞, we deduce that The thesis follows by the arbitrariness of R.
In the previous section we showed that any solution (u Obviously also u τ , v τ are radial and they satisfy Thanks to to Proposition 3.1, we deduce that 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: ) be any solution to (S * ).Then, we have where u 0 , v 0 ∈ (0, +∞).Moreover, we have where u ∞ , v ∞ ∈ (0, +∞).

Uniqueness and classification of solutions
The aim of this section is to classify solutions to (S * ) up to the associated rescaling.Given (u, v) a solution to the problem (R * S ), consider y u (t) := r δ u(r) and y v (t) := r δ v(r), (4.1) where t := log r with r > 0 and δ = n − 2 2 .By direct computation, it is possible to deduce Next, we will prove that y u and 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 y u and y v .Moreover, there exist cu , Cu , c u , C u , cv , Cv , c v , C v , M positive constants such that c u e (δ−τ 1 )t u τ 1 (e t ) ≤ y u (t) ≤ C u e (δ−τ 1 )t u τ 1 (e t ) for t ≤ −M, (4.2) cu e (δ−τ 2 )t u τ 2 (e t ) ≤ y u (t) ≤ Cu e (δ−τ 2 )t u τ 2 (e t ) for t ≥ M, (4.3) c v e (δ−τ 1 )t v τ 1 (e t ) ≤ y v (t) ≤ C v e (δ−τ 1 )t v τ 1 (e t ) for t ≤ −M, (4.4) c u e (δ−τ 2 )t v τ 2 (e t ) ≤ y v (t) ≤ C v e (δ−τ 2 )t v τ 2 (e t ) for t ≥ M, (4.5) where τ 1 and τ 2 were introduced (3.2).Furthermore, we have that The proof follows by Proposition 3.1 and 3.2 .

1. 3 .
The classification result.Once we know the exact behavior of the solutions, we exploit the standard change of variable y u (t) := r δ u(r) and y v (t) := r δ v(r),

2 .
are able to apply the Moser's iteration scheme, in order to prove our result.For any k ≥ 1, let us define the sequence {η k } by 2η k+1 + 2 * − 2 = 2 * η k , and we set η 1 := 2 * The choice of η 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

Lemma 4 . 1 .
Let y u , y v be the functions defined in (4.1), they satisfy lim

Case 1 :
f (L) = 0. Without loss of generality we assume that (4.25) f (L) < 0.By (4. Thus f y u (t) y v (t) has a sign at −∞.Hence, we may run over the arguments of Case 1 recovering (4.29).Therefore we have thaty u (t) = Cy v (t) for any t ∈ R,with 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 (u, v) ∈ D 1,2 (R n ) × D 1,2 (R n ) bea solution to (S * ).By Proposition 4.3, one gets y u = Cy v .