Non-autonomous double phase eigenvalue problems with indefinite weight and lack of compactness

In this paper


INTRODUCTION
In this paper, we investigate eigenvalues to the following double phase problem with unbalanced growth and indefinite weight, where N ≥ 2, 1 < p, q < N , p = q, a ∈ C 0,1 (R N , [0, +∞)), a ≡ 0 and m : R N → R is an indefinite sign weight which may admit nontrivial positive and negative parts.Here ∆ q is the q-Laplacian operator and ∆ a p is the weighted p-Laplace operator defined by ∆ a p u := div(a(x)|∇u| p−2 ∇u).Throughout of this paper, we shall always assume that the weight function m : R N → R satisfies the following assumption, E-mail addresses: tianxiang.gou@xjtu.edu.cn,radulescu@inf.ucv.ro.Date: January 9, 2024.Acknowledgments.T. Gou was supported by the National Natural Science Foundation of China (No. 12101483) and the Postdoctoral Science Foundation of China (No. 2021M702620).V.D. Rȃdulescu was supported by a grant of the Romanian Ministry of Research, Innovation, and Digitization, CNCS/CCCDI-UEFISCDI (No. PCE 137/2021), within PNCDI III.The authors would like to thank warmly the anonymous referees for his/her very precise reading of our manuscript and for giving constructive comments and suggestions.
Problems like (1.1) arise when one looks for the stationary solutions of reaction-diffusion systems of the form u t = div [D(x, ∇u)∇u] + g(x, u) (x, t) ∈ R N × (0, ∞), where D(x, ∇u) = a(x)|∇u| p−2 + |∇u| q−2 .This system has a wide range of applications in physics and related fields, such as biophysics, plasma physics, and chemical reaction design (see [7,26]).In such applications, the function u is a state variable and describes density or concentration of multi-component substances, div [D(x, ∇u)∇u] corresponds to the diffusion with a diffusion coefficient D(x, ∇u), and g(x, u) is the reaction and relates to source and loss processes.Typically, in chemical and biological applications, the reaction term g(x, u) has a polynomial form with respect to the unknown concentration denoted by u.
The analysis of the double phase eigenvalue problem (1.1) is closely associated with the following single phase quasilinear eigenvalue problem, The first part of the paper is devoted to the study of (1.2).The main results we establish regarding (1.2) are upcoming Theorem 3.1 and Proposition 3.1, which reveal that there exist a sequence of eigenvalues to (1.2) and the first eigenvalue is simple.In the case of bounded domains and r = 2, this problem is related to the Riesz-Fredholm theory of self-adjoint and compact operators.The anisotropic linear case (if r = 2 and m(•) is non-constant) was first considered in the pioneering papers of Bocher [6], Hess and Kato [17] and Pleijel [25].An important contribution in the case of unbounded domains is due to Allegretto and Huang [1] and Szulkin and Willem [27].In [27], the authors assumed that weight function may have singular points.Equation (1.1) contains the contribution of two differential operators in the left-hand side, so this problem is not homogeneous.In fact, the differential operator u → −∆ a p u − ∆ q u is related to the "double-phase" variational functional defined by The integrand of this functional is the function ξ(x, t) = a(x)t p + t q for all x ∈ R N and t ≥ 0.
When a ≡ 1, then (1.1) becomes the so-called p & q Laplacian problem, which was investigated by Benouhiba and Belyacine [4,5].A feature of the present paper is that we do not assume that the function a(•) is bounded away from zero, that is, we do not require that essinf x∈R N a(x) > 0. This implies that the integrand ξ(x, t) exhibits unbalanced growth, namely there holds that where C 0 > 0 is a constant.In this scenario, the study is carried out in the framework of Musielak-Orlicz-Sobolev spaces.Such functionals were first investigated by Marcellini [18,19,20] in the context of problems of the calculus of variations and of nonlinear elasticity for strongly anisotropic materials.For such problems, there is no global (that is, up to the boundary) regularity theory.There are only interior regularity results, which are primarily due to Baroni et al. [3] and Marcellini [10,20,21].In fact, most of works dealt with double phase problems having unbalanced growth in bounded domains of R N , we refer the readers to [13,14,15,22,23,24] and references therein.However, there exist relatively few ones treating the problems in R N .The study of eigenvalue problems like (1.1) is open until now.Since (1.1) is set in the whole space R N , then lack of compactness is one of major difficulties we encounter to discuss the eigenvalue problem (1.1) in Musielak-Orlicz-Sobolev spaces and more careful analysis is needed in suitable weighted functions spaces.Indeed, this is mainly because the embedding W 1,ξ (R N ) ֒→ L r (R N ) is only continuous for any q ≤ r ≤ q * (see Lemma 2.3) and the weight function m : R N → R is indefinite, which cause that the verification of the compactness of the underlying (minimizing and Palasi-Smale) sequences becomes difficult.Consequently, we manage to study the problem (1.1) in a new weighted Sobolev space E defined by the completion of C ∞ 0 (R N ) under the norm where • ξ denotes the standard norm in D 1,ξ (R N ).Here W 1,ξ (R N ) and D 1,ξ (R N ) are Musielak-Orlicz-Sobolev spaces defined in Section 2. In the present paper, when p < q, we establish the existence of a continuous family of eigenvalues to (1.1), starting from the principal frequency to (1.2), see Theorems 3.2 and 3.3.While q < p, we prove the existence of a discrete family of positive eigenvalues to (1.1), which diverges to infinity, see Theorem 3.4 and Proposition 3.2.The results we derive reveal new facts of eigenvalues to double phase problems in R N .In both cases, we actually need to assume that q < q * := N q N −q , because of the unbalanced growth property (1.3) with respect to the double phase operator and the dominance is the q-Laplacian term.Thus the problem under consideration is Sobolev subcritical and the energy functional J corresponding to (1.1) is well-defined in the Sobolev space E by Theorem 2.3, where Observe that p q < 1 + 1 N implies that p < q * .When double phase problems are set in bounded domains in R N , then the condition p q < 1 + 1 N can be applied to prove the desired compact embedding results, for example [22,Proposition 4].While double phase problems are set in R N , then the condition p q < 1 + 1 N can no longer be applicable to derive the compact embedding results, which leads to lack of compactness for the study.In the present paper, such a condition is actually used to guarantee the regularity of solutions to (1.1) (see [8,9]), which along with the maximum principle developed in [23,24] can lead to the simplicity of eigenvalues, see Proposition 3.2.

PRELIMINARIES
In the section, we are going to present some preliminary results used to establish our main theorems.To deal with the eigenvalue problem (1.1), we shall work in the corresponding Musielak-Orlicz-Sobolev space.For the convenience of the readers, let us first present a few definitions from [11, Section 2] concerning the main notions and function spaces used in this paper.
) is said to be a N -function if it is continuous and positive on [0, +∞).In addition, it satisfies that whenever s, t ≥ 0 and |x − t| ≥ ǫ max {|s|, |t|}.
With these definitions in hand, we are now ready to introduce the double phase function ξ : It is simple to check that ξ is a generalized N -function.Moreover, ξ is uniformly convex and it satisfies the ∆ 2 -condition.Let us denote by M (R N ) the space consisting of all Lebesgue measurable function u : R N → R.
The Musielak-Orlicz space L ξ (R N ) is defined by where ρ ξ is the modular function given by Here the space L ξ (R N ) is equipped with the Luxemburg norm given by Using the above properties satisfied by ξ, we can easily check that L ξ (R N ) is a Banach space, which is also separable and reflexive.The Musielak-Orlicz-Sobolev space W 1,ξ (R N ) is defined by Here the space W 1,ξ (R N ) is equipped with the norm where Next we are going to show some relations between the norm in L ξ (R N ) and the modular function ρ ξ given by (2.2) and (2.3) respectively, proofs of which can be completed by using the ingredients presented in [16, Section 3.2].Lemma 2.1.Let ξ : R N × [0, +∞) → [0, +∞) be defined by (2.1).Then the following assertions hold.
Note that t q ≤ ξ(x, t) for any x ∈ R N and t ∈ R, by the assertion (ii) of Lemma 2.1, then there holds the following embedding result.
As a consequence of Lemma 2.2 and Sobolev's embeddings in W 1,q (R N ) and D 1,q (R N ) for 1 < q < N , we have the following embedding result.

MAIN RESULTS
In this section, we shall consider the eigenvalue problem (1.1) under the assumption (H).The hypothesis (H) is always assumed to hold in what follows.First we shall present some results related to the following eigenvalue problem, ) and a ≡ 0. Then there exists a sequence of solutions (µ a,r,k , u a,r,k where η(x, t) = a(x)t r for x ∈ R N and t ≥ 0, where the Sobolev space V is the completion of C ∞ 0 (R N ) under the norm Reasoning as the proof of [1, Lemma 1], we are able to show that Ψ(u) restricted on M r satisfies the Palais-Smale condition.Then, by adapting Ljusternik-Schnirelman theory as the proof of forthcoming Theorem 3.4, we can derive the desired conclusion.Thus the proof is completed.
) and a ≡ 0. Then the first eigenvalue µ a,r,1 obtained in Theorem 3.1 is simple and the eigenfunction u a,r,1 has constant sign.Moreover, if u ∈ D 1,η (R N ) is a nontrivial solution to (3.1) corresponding to µ > µ a,r,1 , then u is signchanging.
Since the function m is an indefinite sign weight, then proof of Proposition 3.1 is not straightforward.To prove this, we need the following auxiliary result.
Then I(u, v) ≥ 0.Moreover, I(u, v) = 0 if and only if u = kv for some k ∈ R.
Proof.Observe that Then, by the divergence theorem, we see that Using Young's inequality, we have that As a consequence, coming back to (3.2), we can conclude that It then follows that |u∇v − v∇u| = 0.This implies that there exists k ∈ R such that u = kv and the proof is completed.If u ∈ M r satisfies that Ψ(u) = u a,r,1 , then |u| ∈ M r and Ψ(|u|) = u a,r,1 .Therefore, without restriction, we may assume that u a,r,1 is nonnegative.Observe that u a,r,1 ∈ D 1,η (R N ) satisfies the equation By maximum principle, then u a,r,1 > 0. Let u a,r,1 ∈ M r and v a,r,1 ∈ M r be two positive eigenfunctions corresponding to µ a,r,1 , then It is simple to calculate that I(u a,r,1 , v a,r,1 ) = 0.As a result of Lemma 3.1, we have that u a,r,1 = kv a,r,1 for some k ∈ R.This indicate that µ a,r,1 is simple.
Arguing by contradiction, we suppose that u ∈ D 1,η (R N ) is a nonnegative solution to (3.1) corresponding to µ > µ a,r,1 .By the maximum principle, then u > 0. Notice that In addition, we know that if u ∈ D 1,η (R N ) is a solution to (3.1), then ku ∈ D 1,η (R N ) is also a solution to (3.1) for any k ∈ R\{0}.Then, by scaling, we may assume that Let u a,r,1 ∈ M and u a,r,1 > 0 be an eigenfunction to (3.1) corresponding to µ a,r,1 .Then u a,r,1 solves the equation −∆ a r u a,r,1 = µ a,r,1 m(x)|u a,r,1 | r−2 u a,r,1 in R N .As a consequence of Lemma 3.1 and (3.3), we have thta This is impossible, hence u is sign-changing and the proof is completed.
3.1.Case p < q.In this case, to establish the existence of solutions to (1.1), we shall adapt some ideas from [1].Let us first introduce the weight function Let E be the completion of C ∞ 0 (R N ) under the norm .
It is standard to conclude that E is a separable and reflexive Banach space.In order to prove the existence of solutions to (1.1), we shall define the associated energy functional J : E → R by ) and a ≡ 0. Then there exist positive solutions to (1.1) for any λ > µ 1,q,1 .
In this case, we find that J is unbounded from below in E. Indeed, let u 1,q,1 ∈ D 1,q (R N ) be an eigenfunction of (3.1) corresponding to µ 1,q,1 .We observe that Since p < q and λ > µ 1 , then J(tu 1,q,1 ) → −∞ as t → +∞.In this situation, to seek for solutions to (1.1), we introduce the Nehari manifold N := {u ∈ E\{0} : where Then we are able to define the minimization problem Obviously, any minimizer of (3.7) is a solution to (1.1).
Proof of Theorem 3.3.Let {u n } ⊂ N be a minimizing sequence to (3.7).Then m = J(u n ) + o n (1) and I(u n ) = 0. Since I(|u|) ≤ I(u) = 0 for any u ∈ N , then there exists a unique 0 < t |u| ≤ 1 such that I(t |u| |u|) = 0, where Moreover, for any u ∈ N , we see that Therefore, for any u ∈ N , As a consequence, we shall assume that {u n } ⊂ N is a nonnegative minimizing sequence to (3.7

Let us first assume that
(3.9) Since I(u n ) = 0, then there holds that In this case, we set It is easy to see that {v n } ⊂ M q .Since I(u n ) = 0, then In view of (3.9) and (3.11), then It then yields that Invoking Hölder's inequality, Sobolev's inequality and (3.12), we then get that where 0 < θ < 1 and q = θp + (1 − θ)q * .This is a contradiction, because of v n ∈ M q .Let us next assume that there exists some λ 0 > 0 such that Therefore, there holds that ∇v n p p = λ + o n (1).In virtue of [1, Lemma 1], we then get that {v n } is compact in V , where the Sobolev space V is the completion of C ∞ 0 (R N ) under the norm It then infers that v ∈ V is a nonnegative eigenfunction to (3.1) corresponding to λ.By Lemma 3.1, we reach a contradiction, because of λ > µ 1,q,1 .As a consequence, we have that m > 0.
It is standard to show that N is a natural constraint.By the fact that there exists a nonnegative minimizing sequence to (3.7) and applying Ekeland's variational principle, then there exists a Palais-Smale sequence Let us verify that { ∇u n q } is bounded.On the contrary, we may assume that ∇u n q → +∞ as n → ∞.Define v n by (3.10), use the fact that I(u n ) = o n (1) and (3.13), then there holds that ∇v n p p = λ + o n (1).With the help of [1, Lemma 1], we can also reach a contradiction.This implies that { ∇u n q } is bounded.By Hardy's inequality, it then follows that As a result, we get that {u n } is bounded in E. Then there exists u ∈ E such that Therefore, we are able to derive that u ∈ E satisfies the following equation Since the embedding (1).(3.16)This readily indicates that u = 0. Otherwise, there holds that This in turn gives that J(u n ) = o n (1), which is impossible, because of m > 0. Therefore, u is a nontrivial solution to (1.1).Moreover, as a consequence of maximum principle, see [24, Proposition 2.3], we have that u > 0. Thus the proof is completed.
3.2.Case q < p. Next we are going to deal with the case that q < p.In this case, we define Lemma 3.2.Assume that (H) holds, N ≥ 2, 1 < q < p < N , a ∈ C 0,1 (R N , [0, +∞)) and a ≡ 0. Then Φ restricted on S satisfies the Palais-Smale condition at any level c ∈ R.
In addition, we see that Therefore, utilizing the fact that {u n } is bonded in E, we get that It necessarily follows that We then deduce that u n → u in L q loc (R N ) as n → ∞.As a consequence, we have that On the other hand, by Hölder's inequality and Sobolev's inequality, we get that where we also used the facts that where the second fact holds because of m 1 ∈ L N q (R N ) from the assumption (H).Combining (3.23), (3.24) and (3.25), by (3.22), we then obtain that This immediately indicates that u n → u in D 1,ξ (R N ) as n → ∞.Taking advantage of (3.20) and (3.21), we then get that because of λ n = λ + o n (1) and λ = 0.In view of (3.18), then Consequently, we derive that u n → u in E as n → ∞.Thus the proof is completed.Theorem 3.4.Assume that (H) holds, N ≥ 2, 1 < q < p < N , a ∈ C 0,1 (R N , [0, +∞)) and a ≡ 0. Then there exists a sequence of solutions Proof.To establish the existence of a sequence of eigenvalues to (1.1), we shall take into account Ljusternik-Schnirelman theory.Define Σ := {A ⊂ S : A is compact and A = −A} .
For a set A ∈ Σ, the genus of A is defined by If such a minimum does not exist, we set γ(A) = +∞.
Let us now define First we see that, for any k ∈ N + , Σ k = ∅.Indeed, let X k be a k dimensional subspace of E, by Borsuk-Ulam's theorem, then γ(S ∩ X k ) ≥ k.Define λk := inf Since Σ k+1 ⊂ Σ k , then λk ≤ λk+1 for any k ∈ N + .From Lemma 3.2, then λk is a critical points of J restricted on S for any k ∈ N + .Then we derive that , where E ′ denotes the dual space of E. Define X i := span {e i } and Further, there exists u ∈ E such that u k ⇀ u in E as n → ∞.Observe that e ′ i , u = e ′ i , u k + o k (1) = o k (1), because of u k ∈ Z k .Therefore, we have that u = 0 and u k ⇀ 0 in E as k → ∞.This along with the assumption that m 1 ∈ L N q (R N ) from the assumption (H) leads to Since m 2 ≥ 0 from the assumption (H), then which is impossible due to u k ∈ S. Consequently, we get that β k → +∞ as k → ∞.Thanks to λk ≥ β k for any k ∈ N + , then λk → +∞ as k → ∞.Since u k ∈ E is a critical point for Φ restricted on S, then there exists Thus the proof is completed.
Proof.Let us first show that ∇ v q − u q v q−1 = 1 + (q − 1) u v q ∇v − q u v q−1 ∇u.

First we are going to prove that m > 0 .
It follows from (3.8) that m ≥ 0. Let us argue by contradiction that m = 0.Then, by (3.8), we have that R N a(x)|∇u n | p dx = o n (1).