Infinitely many solutions for Schrödinger – Kirchhoff-type equations involving indefinite potential

In this paper, we study the multiplicity of solutions for the following Schrödinger–Kirchhoff-type equation { − ( a + b ∫ RN |∇u| 2dx ) 4u + V(x)u = f (x, u) + g(x, u), x ∈ RN , u ∈ H1(RN), where N ≥ 3, a, b > 0 are constants and the potential V may be unbounded from below. Under some mild conditions on the nonlinearities f and g, we obtain the existence of infinitely many solutions for this problem. Recent results from the literature are generalized and significantly improved.


Introduction and main results
In this paper, we consider the following Schrödinger-Kirchhoff-type equation where N ≥ 3 and a, b > 0 are constants.If in (1.1), we set V(x) ≡ 0 and replace R N by a smooth bounded domain Ω, then (1.1) reduces to the following Dirichlet problem Corresponding author.Email: qingyezhang@gmail.comProblem (1.2) is related to the stationary analogue of the Kirchhoff equation which was presented by Kirchhoff in 1883 [8] as a generalization of the classical D'Alembert's wave equation for free vibrations of elastic strings.Kirchhoff's model takes into account the changes in length of the string produced by transverse vibrations.In [11], Lions first introduced an abstract functional analysis framework to this model.After that, problems like type (1.2) have been studied by many authors, see [3,4,15,16,20,25,27] and the references therein.
More recently, with the aid of variational methods, the existence and multiplicity of various solutions for equations of type (1.1) have also been extensively investigated in the literature, see, for instance, [1, 2, 5, 6, 9, 10, 12, 21-24, 26, 29] and the references therein.Here we emphasize that almost in all these mentioned papers the conditions imposed on the potential V always imply that V is bounded from below, which is crucial for the corresponding results.
In the present paper, different from the references mentioned above, we are going to study the existence of infinitely many solutions for (1.1) in the case where the potential V may be unbounded from below.Specifically, we first assume that V satisfies This type of assumptions on the potential V has already been introduced in [13] to study Schrödinger equations (see also [28]), which ensures that the Schrödinger operator S := −a∆ + V, defined as a form sum, is self-adjoint and semibounded on L 2 (R N ) (see Theorem A.2.7 in [19]).We denote by σ(S ) ⊂ R the spectrum, σ ess (S) the essential spectrum and σ pp (S) the pure point spectrum of S respectively.Consider the nondecreasing sequence of min-max values defined by where U k is the family of all k-dimensional subspaces of [17,18] for details).Then we make the further assumption on V.
For the nonlinearities, we present the following assumptions.
(S 4 ) There exist an x 0 ∈ R N and a constant r 0 > 0 such that lim inf where B r 0 (x 0 ) is the ball in R N centered at x 0 with radius r 0 and Our main result reads as follows.
Theorem 1.1.Suppose that (S 1 )-(S 5 ) are satisfied.Then (1.1) possesses a sequence of nontrivial Remark 1.2.In Theorem 1.1, the potential V satisfying (S 1 ) and (S 2 ) may not be coercive or bounded from below.Moreover, the nonlinear term f satisfying (S 3 ) and (S 4 ) may be partially oscillatory near the origin.This is in sharp contrast with the aforementioned references.To the best of our knowledge, there is little literature concerning infinitely many solutions for (1.1) in this situation.In fact, it is easy to see that conditions (S 1 ) and (S 2 ) are rather weaker than the usual one in the existing literature that the potential Remark 1.3.Theorem 1.1 also essentially improves some related results in the existing literature.Compared to Theorem 6 in [26], our conditions (S 1 ) and (S 2 ) on the potential V are weaker than (V 1 ) there, and our conditions (S 3 ) and (S 4 ) on the nonlinear term f are much weaker than (f 5 ) there if we just take g = 0 in (1.1).In fact, there are many functions V and f which satisfy our conditions (S 1 )-(S 4 ) but do not satisfy the condition (V 1 ) and (f 5 ) in [26].
For instance, let where V 0 ∈ L q (R N ) for some q ≥ 2 is a given non-positive function and unbounded from below.Then it is evident that V satisfies (S 1 ) and (S 2 ) if the positive constant V is chosen to be large enough.Moreover, V is also unbounded from below.In addition, let be the primitive function of f with respect to u, where > 0 is small enough and α ∈ (1 + , 2).Then it is easy to check that f satisfies conditions (S 3 ) and (S 4 ) with ν = α − and ξ(x) = (α + )e −|x| 2 .

Notations and preliminaries
• H 1 (R N ) is the usual Sobolev space equipped with the standard norm and , denotes a Lebesgue space, and the norm in L p (Ω) is denoted by u p, Ω when Ω is a proper subset of R N , by u p when Ω = R N .
• For any R > 0, B R denotes the ball in R N centered at 0 with radius R.
In what follows it will always be assumed that (S 1 ) and (S 2 ) are satisfied.As pointed out in [13], the form domain of the Schrödinger operator S is which becomes a Hilbert space if it is equipped with the inner product where l 0 > − inf σ(S ) = −λ 1 is a fixed positive constant.We denote by • 0 the associated norm.
Lemma 2.1.E is continuously embedded into H 1 (R N ), that is, for some c 0 > 0.
Proof.Arguing indirectly, we assume that there exists a sequence {u n } n∈N ⊂ E such that for some c 1 > 0. By (2.2) and (2.3), we get , where V − and q are given in (S 1 ).Combining (2.1), (2.4), Hölder's inequality and the Gagliardo-Nirenberg inequality, we have where c 2 > 0 is a constant depending on q.This together with (2.2) and (2.4) yields which contradicts (2.1).The proof is completed.
For later use, we introduce the new inner product in E as follows.Choose d ∈ (d, λ ∞ ) such that d = λ k for all k ∈ N, where d is the constant given in (S 5 ).Denote by λ k 0 the first eigenvalue of the Schrödinger operator S greater than d.Let E − be the subspace of E spanned by the eigenfunctions with corresponding eigenvalues less than d.Note the fact that If there is no eigenvalue of the Schrödinger operator S greater than d, then we set λ k 0 = λ ∞ and E − is empty in this case.Let E + be the orthogonal complement of E − in E with respect to the inner product (•, •) 0 .Then E possesses the orthogonal decomposition E = E − ⊕ E + .By definition, it holds that Now we can define the new inner product (•, •) and the induced norm Note the fact that E − and E + are also orthogonal with respect to the usual inner product in L 2 (R N ).Then it is evident that E possesses the same orthogonal decomposition E = E − ⊕ E + with respect to the new inner product (•, •).Moreover, we have Lemma 2.2.The norms • and • 0 are equivalent in E.
Proof.It suffices to show that • and • 0 are equivalent in E + since E − is finite dimensional.On the one hand, by (2.8), there holds (2.9) On the other hand, invoking (2.5) and (2.8), we get where Hereafter, we always use the inner product (•, •) and the induced norm • in E.Moreover, we write E * for the dual space of E, and •, • : E * × E → R for the dual pairing.From Lemma 2.1 and Lemma 2.2, we immediately know that E is continuously embedded into H 1 (R N ).Furthermore, using the Sobolev embedding theorem, we also get the following lemma.
Lemma 2.3.E is continuously embedded into D 1,2 (R N ) and L p (R N ) for all p ∈ [2, 2 * ], and hence there exist constants c 4 , τ p > 0 such that Moreover, for any bounded domain

Variational setting and proof of the main result
In this section, we will first introduce the variational setting for (1.1).To this end, we define functionals Ψ i (i = 1, 2, 3) and Φ on E by Here d is the constant in (2.8) and G(x, u) := u 0 g(x, t)dt is the primitive function of g(x, u) with respect to u. Proposition 3.1.Assume that (S 1 )-(S 3 ) and (S 5 ) are satisfied.Then ) Proof.First, we show that Evidently, For notational simplicity, we set Then for any u ∈ E, by (2.12), (3.7) and Hölder's inequality, we have where τ µ * is the constant given in (2.12).Thus Ψ 2 is well defined.For any given u ∈ E, define an associated linear operator J (u) : E → R as follows: By (S 3 ), (2.12) and Hölder's inequality, there holds which shows that J (u) is well defined and bounded.On the other hand, it follows from (S 3 ) that Then for any u, v ∈ E, combining (3.9)-(3.12), the mean value theorem and Lebesgue's dominated convergence theorem, we have where θ(x) ∈ [0, 1] depends on u, v, t.This shows that Ψ 2 is Gâteaux differentiable on E and the Gâteaux derivative of Ψ 2 at u is J (u).
Finally, combining (2.6) and (3.1)-(3.4),we immediately know that Φ ∈ C 1 (E, R) and (3.5) holds.In addition, it is known that any critical point u ∈ E ⊆ H 1 (R N ) of the functional Φ is a solution of (1.1).The proof is completed.
We will use the following variant symmetric mountain pass lemma due to [7] (see also [14]) to prove that (1.1) possesses a sequence of weak solutions.Before stating this theorem, we first recall the notion of genus.
Let E be a Banach space and A a subset of E. A is said to be symmetric if u ∈ A implies −u ∈ A. Denote by Γ the family of all closed symmetric subset of E which does not contain 0. For any A ∈ Γ, define the genus γ(A) of A by the smallest integer k such that there exists an odd continuous mapping from A to R k \ {0}.If there does not exist such a k, define
(Φ 2 ) For each k ∈ N, there exists an A k ∈ Γ k such that sup u∈A k Φ(u) < 0.

Then either (i) or (ii) below holds.
(i) There exists a critical point sequence {u k } such that Φ(u k ) < 0 and lim k→∞ u k = 0.
In order to apply Theorem 3.2, we will show in the following lemmas that the functional Φ defined in (3.1) satisfies conditions (Φ 1 ) and (Φ 2 ) in Theorem 3.2.The proof of these lemmas is partially motivated by [21] and [7].Lemma 3.3.Let (S 1 )-(S 3 ) and (S 5 ) be satisfied.Then Φ is coercive and bounded from below.
Proof.We first prove that Φ is coercive.Arguing indirectly, we assume that for some sequence Note that E − is finite dimensional.Thus, passing to a subsequence if necessary, we can assume by Lemma 2.3 that and since d is chosen to be greater than d in Section 2. Combining (3.1), (3.9) and (3.24), we have Multiplying both sides of (3.26) by u n −2 , we get Then it follows from the weak lower semi-continuity of the norm (3.28) Combining (3.1), (3.9) and (3.24), we have where µ * and τ µ * are the constants given in (3.8) and (2.12) respectively.Multiplying both sides of (3.29) by u n −4 and letting n → ∞, we get which contradicts (3.28).Therefore, Φ is coercive.
Next, we show that Φ is bounded from below.Combining (2.11), ( , we have (3.34)By virtue of (3.32) and Lemma 2.3, we have u n k u 0 in D 1,2 (R N ).Then it follows that Moreover, from (3.32) and Proposition 3.1, we know that for i = 2, 3. Combining (3.33)-(3.37),we obtain This together with (3.33) shows that u n k → u 0 in E. Therefore, Φ satisfies (PS) condition.The proof is completed.Lemma 3.5.Let (S 1 )-(S 5 ) be satisfied.Then for each k ∈ N, there exists an A k ⊆ E with genus γ(A k ) = k such that sup u∈A k Φ(u) < 0.
Proof.We follow the idea of the geometric construction introduced in [7].By coordinate translation, we can assume x 0 = 0 in (S 4 ).Let C denote the cube where r 0 is the positive constant given in (S 4 ).Evidently, C ⊆ B r 0 .By (S 4 ), there exist constants δ, > 0 and two sequences of positive numbers For any fixed k ∈ N, let m ∈ N be the smallest positive integer satisfying m N ≥ k.We divide the cube C equally into m N small cubes by planes parallel to each face of C and denote them by C i with 1 ≤ i ≤ m N .Then the edge of each C i has the length of l := r 0 /m.For each 1 ≤ i ≤ k, we make a cube D i in C i such that D i has the same center as that of C i , the faces of D i and C i are parallel and the edge of D i has the length of l/2.

Choose a function
For each 1 ≤ i ≤ k, let y i ∈ R N be the center of both C i and D i , and define Then it is easy to see that suppϕ and Evidently, V k is homeomorphic to the unit sphere in R k by an odd mapping.Thus γ(V k ) = k.
If we define the mapping H where M n is the constant given in (3.39).Here we use the fact that |δ n s i u ϕ i u (x)| ≡ δ n for all x ∈ D i u and the volume of cube D i u in R N is (l/2) N .Since M n → ∞ as n → ∞, we can choose n 0 ∈ N large enough such that the right-hand side of (3.46) is negative.Define The proof is completed.Now we are in a position to give the proof of our main result.
Taking into account Proposition 3.1 again and the fact that E is continuously embedding into H 1 (R N ), we know that {u k } k∈N is a sequence of nontrivial solutions of (1.1) with u k → 0 in H 1 (R N ) as k → ∞.This ends the proof.
continuous embedding in Lemma 2.3.Since Ψ 1 = Ψ 0 • ι, we immediately know by (3.6) that Ψ 1 ∈ C 1 (E, R) and (3.2) holds.Next, we verify (3.3) by definition and prove that Ψ 2 ∈ C 1 (E, R) with Ψ 2 : E → E * being completely continuous.By (S 3 ), there holds .30) where c 4 and τ 2 are the constants given in(2.11)and(2.12)respectively.This implies that Φ maps bounded sets in E into bounded sets in R. Then it follows from the coercivity that Φ is bounded from below.The proof is completed.Assume that (S 1 )-(S 3 ) and (S 5 ) are satisfied.Then Φ satisfies (PS) condition.Proof.Let {u n } n∈N ⊂ E be a (PS)-sequence, i.e.,|Φ(un )| ≤ D 1 and Φ (u n ) → 0 as n → ∞ (3.31)for some D 1 > 0. Note first that Φ is coercive by Lemma 3.3.This together with (3.31) implies that {u n } n∈N is bounded in E. Thus there exists a subsequence {u n k } k∈N such that 3.43)where c 4 and τ 2 are the constants given in (2.11) and (2.12) respectively.By the definition of V k , there exists some integer 1≤ i u ≤ k such that |s i u | = 1.Then it follows that (x, δ n s i ϕ i )dx = D iu F(x, δ n s i u ϕ i u )dx + C iu \D iu F(x, δ n s i u ϕ i u )dx iu \D iu F(x, δ n s i u ϕ i u )dx + ∑Here we use the fact that the volume of cube C in R N is r N 0 .Combining (3.39) and (3.42)-(3.45),we have 2 − k ∑ i=1 C i F(x, δ n s i ϕ i )dx + d + d k ∑ i=1 C i F(x, δ n s i ϕ i )dx, (FC