Sign-changing solutions for Schrödinger–Kirchhoff-type fourth-order equation with potential vanishing at infinity

The purpose of this paper is to study the existence of sign-changing solution to the following fourth-order equation: 0.1Δ2u−(a+b∫RN|∇u|2dx)Δu+V(x)u=K(x)f(u)in RN,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ \Delta ^{2}u- \biggl(a+ b \int _{\mathbb{R}^{N}} \vert \nabla u \vert ^{2}\,dx \biggr) \Delta u+V(x)u=K(x)f(u) \quad\text{in } \mathbb{R}^{N}, $$\end{document} where 5≤N≤7\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$5\leq N\leq 7$\end{document}, Δ2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\Delta ^{2}$\end{document} denotes the biharmonic operator, K(x),V(x)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$K(x), V(x)$\end{document} are positive continuous functions which vanish at infinity, and f(u)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$f(u)$\end{document} is only a continuous function. We prove that the equation has a least energy sign-changing solution by the minimization argument on the sign-changing Nehari manifold. If, additionally, f is an odd function, we obtain that equation has infinitely many nontrivial solutions.


Introduction and main results
This article is concerned with the following fourth-order Kirchhoff-type equation: where 5 ≤ N ≤ 7, 2 denotes the biharmonic operator, and a, b are positive constants. When a = 1, b = 0, equation (1.1) becomes the following fourth-order equation (replace R N with ): Problem (1.1) stems from the following Kirchhoff equation: (1.3) where ⊂ R N is a bounded domain, a, b > 0.
Problem (1.3) comes from the following equation: Because equation (1.4) is regarded as a good approximation for describing nonlinear vibrations of beams or plates, it is used to describe some phenomena that appear in different physical, engineering, and other sciences [5,9].
It is noticed that in the past decades many mathematicians have paid much of their attention to nonlocal problems. The appearance of nonlocal terms in the equations not only marks their importance in many physical applications but also causes some difficulties and challenges from a mathematical point of view. Certainly, this fact makes the study of nonlocal problems particularly interesting. In addition to the equations of Kirchhoff type, there are also some nonlocal problems, such as Schrödinger-Poisson systems, equations with the fractional Laplacian operator, and so on. Especially, these days there is a good trend of existence of solutions for fractional-order differential equations which are definitely the generalized study [1, 2, 15-20, 22, 32, 33].
Throughout this paper, as in [3], we say that (V , K) ∈ K if continuous functions V , K : R N → R satisfy the following conditions: (VK 0 ) V (x), K(x) > 0 for all x ∈ R N and K ∈ L ∞ (R N ).
(VK 1 ) If {A n } n ⊂ R N is a sequence of Borel sets such that |A n | ≤ R for all n ∈ N and some R > 0, then One of the following conditions occurs: N-4 is the critical Sobolev exponent. As for the function f , we assume f ∈ C(R, R) and satisfies the following conditions: (f 3 ) f has a "quasicritical growth", that is, It follows from (V , K) ∈ K that the space B given by In this paper, we discuss our problem on the space So, it is easy to see that E is a Hilbert space. Furthermore, For problem (1.1), the energy functional is given by It follows from the conditions of this paper that I b (u) belongs to C 1 and The solution of problem (1.1) is the critical point of the functional I b (u). Furthermore, we say that u is a sign-changing solution if u ∈ E is a solution to problem (1.1) and u ± = 0, where u + = max{u(x), 0}, u -= min{u(x), 0}.
As pointed out in the article, since the nonlocal term ( R N |∇u| 2 dx) u is involved, there is an essential difference between problem (1.1) and problem (1.2) when we discussed the existence of sign-changing solutions, see [6-8, 10, 25, 37, 61].
Therefore, to study sign-changing solutions for problem (1.1), as in [6,7,10], we first obtain a minimizer of I b over the constraint The rest is to prove that the minimizer is a sign-changing solution of problem (1.1). It is noticed that there are some interesting results, for example, [11, 14, 23, 24, 30, 31, 34, 35, 38-43, 47, 56, 60], which considered sign-changing solutions for other nonlocal problems.
The main result can be stated as follows.
which is also used to characterize one problem as zero mass.
The rest of this paper proceeds as follows. Sections 2 and 3 are devoted to our variational setting, and necessary lemmas are shown and proved, which shall be used in the proof of our main results in Sect. 4.

The variational framework and preliminary results
Remark 2.1 Since E is continuously embedded in B, it is easy to see that Proposition 2.1 and Proposition 2.2 also hold if B is replaced with E.

Lemma 2.2
Assume that (V , K) ∈ K and (f 1 )-(f 5 ) hold. Then, for any u ∈ E\{0}, Proof The proof is similar to that of Lemma 2.4 in [24], so we omit it here.
Similarly, we have the following results. The following results are very important, because they allow us to overcome the nondifferentiability of N (see Lemma 2.5(iii)).
On the other hand, thanks to F(s) ≥ 0, ∀s ∈ R, we get where D ⊂ suppu is a measurable set with finite and positive measures. Hence, by combining Fatou's lemma and (f 4 ), one has Therefore, by the continuity of h u and (f 5 ), there is t u > 0 which is a maximum global point of h u with t u u ∈ N .
We assert that t u is the unique critical point of h u . In fact, suppose, by contradiction, that there are t 1 > t 2 > 0 such that h u (t 1 ) = h u (t 2 ) = 0. Then which is absurd. Next, we prove (ii). For any u ∈ S, according to (i), there exists t u > 0 such that By (2.1), we have that So, there exists τ > 0, independent of u, such that t u ≥ τ .
On the other hand, let Q ⊂ S be compact. Suppose that there exist {u n } ⊂ Q, u ∈ Q such that t n := t u n → ∞, u n → u in E. So, it follows from (2.6) that (2.10) According to (f 5 ), we obtain that is increasing when t > 0 and decreasing when t < 0. Hence we have, for each u ∈ N , that Thanks to {t u n u n } ⊂ N , replaced u with t u n u n in (2.12), from (2.10) we have a contradiction. Therefore, (ii) holds.
Finally, we prove (iii). We assert that m, m, m -1 are well defined.
Hence, m is bijective and m -1 is continuous.
In what follows, we prove m : E\{0} → N is continuous. Suppose {u n } ⊂ E\{0} and u ∈ E\{0} such that u n → u in E. According to (ii), there is t 0 > 0 such that u n t u n = t ( un un ) → t 0 . So, we have t u n → t 0 u =: t 0 . Thanks to t u n u n ∈ N , one has that From the above equality, let n → ∞, one has that which indicates that (t 0 / u )u ∈ N and t u = t 0 / u . Therefore, m(u n ) → m(u). So, the proof is completed.
Define : E → R and : S → R by (u) = I b m(u) and := | S . By Lemma 2.5 and the result from [37], one has the following.

Technical lemmas
Proof It is easy to see that where g u (s, t) = s 2 u + 2 + bs 4 Hence, item (i) is obvious.
In the following, we prove (ii). Firstly, we assert that M = ∅. By (i), we only prove the existence of a critical point of ϕ u (s, t). Let u ∈ E with u ± = 0 and t 0 ≥ 0 fixed, it follows from (3.1) that Together with Lemma 2.4 and Lemma 2.2, one gets g u (s, t 0 ) > 0 for s small enough; g u (s, t 0 ) < 0 for s large enough.
Since g u (s, t 0 ) is continuous, there exists s 0 > 0 such that g u (s 0 , t 0 ) = 0. We assert that s 0 is unique. In fact, supposing by contradiction, there exist 0 < s 1 < s 2 such that g u (s 1 , t 0 ) = g u (s 2 , t 0 ) = 0, and then we have Therefore, it follows from (f 5 ) and 0 < s 1 < s 2 that we have a contradiction. That is, there exists unique s 0 > 0 such that g u (s 0 , t 0 ) = 0.
Let φ 1 (t) := s(t), where s(t) satisfies the properties just mentioned previously, with t in the place of t 0 . Then the map φ 1 : R + → (0, ∞) is well defined.
(1) φ 1 (t) is continuous. To this end, let t n → t 0 as n → ∞ and suppose, by contradiction, that there is a subsequence, still denoted by t n , such that φ 1 (t n ) → ∞.
Obviously, φ 1 (t n ) ≥ t n for n large enough. According to (3.3), one has that In view of Lemma 2.2, we have a contradiction. So φ 1 (t n ) is bounded. Therefore, there exists s 0 ≥ 0 such that, passing to a subsequence,
(3) φ 1 (t) < t for t large. Indeed, if there exists a sequence {t n } with t n → ∞ such that φ 1 (t n ) ≥ t n for all n ∈ N, then arguing as in (3.4), we have a contradiction. Thus, φ 1 (t) < t for t large.

It is easy to see that T(s, t) ∈ [0, C] × [0, C] for all (s, t) ∈ [0, C] × [0, C].
Since T is continuous, using the Brouwer fixed point theorem, there exists (s + , t -) ∈ [0, C] × [0, C] such that φ 1 (t -), φ 2 (s + ) = (s + , t -). (3.6) It follows from φ i > 0 that s + , t -> 0. According to the definition, we have We next shall prove the uniqueness of s + , t -. Suppose that ω ∈ M, one has ∇ϕ ω (1, 1) = ∂ϕ ω ∂s (1, 1), ∂ϕ ω ∂t (1, 1) which shows that (1, 1) is a critical point of ϕ ω . Now, we need to prove that (1, 1) is the unique critical point of ϕ ω with positive coordinates. Let (s 0 , t 0 ) be a critical point of ϕ ω such that 0 < s 0 ≤ t 0 . So, one has that and Thanks to 0 < s 0 ≤ t 0 and (3.8), we have that On the other hand, for ω ∈ M, we have Combining (3.9) with (3.10), one has that If t 0 > 1, the left-hand side of the above inequality is negative, which is absurd because the right-hand side is positive by condition (f 5 ). Therefore, we obtain that 0 < s 0 ≤ t 0 ≤ 1. Similarly, by (3.7) and 0 < s 0 ≤ t 0 , we get 1 and from (f 5 ) this is absurd. Therefore, we have s 0 ≥ 1. Consequently, s 0 = t 0 = 1, which indicates that (1, 1) is the unique critical point of ϕ ω with positive coordinates. Let u ∈ E, u ± = 0 and (s 1 , t 1 ), (s 2 , t 2 ) be the critical points of ϕ u with positive coordinates. In view of (i), one has that So, It follows from ω 1 ∈ E and ω ± 1 = 0 that ( s 2 s 1 , t 2 t 1 ) is a critical point of the map ϕ ω 1 with positive coordinates. Thanks to ω 1 ∈ M, one has that Hence, s 1 = s 2 , t 1 = t 2 . Now, we prove that the unique critical point is the unique maximum point of ϕ u . In fact, using Lemma 2.3, we have that Hence, the maximum point of ϕ u (s, t) cannot be achieved on the boundary of (R + × R + ). Without loss of generality, we may assume that (0,t) is a maximum point of ϕ u (s, t). But, according to Lemma 2.4, it is obvious that is an increasing function with respect to s if s is small enough. Hence, (0,t) is not a maximum point of ϕ in R + × R + .

Lemma 3.2 If {u n } ∈ M and u n u in E, then u
(3.11) Similar to (2.9), we obtain that there is τ > 0 such that Combining u n u in E with Proposition 2.2, we have which shows that u ∈ E ± .
Next, we consider the following minimization problem: We claim In fact, since M ⊂ N , we have m ≥ c b . On the other hand, for any v ∈ M, according to (i) of Lemma 2.5, there exist positive constants s + and tsuch that s + v + , tv -∈ N . Therefore, from (i) and (ii) of Lemma 3.1, we have Proof Let {u n } be a sequence in M such that (3.17) We will show that u n is bounded in E. In fact, suppose that there exists a subsequence that we still call u n such that u n → ∞. (3.18) Now, we define v n := u n / u n for all n ∈ N. So, there exists v ∈ E such that v n v in E. (3.19) From Proposition 2.2, we conclude that, up to a subsequence, Using (i) of Lemma 3.1, it follows from {u n } ⊂ M that s + (v n ) = t -(v n ) = u n . Therefore, using (i) of Lemma 3.1 again, we obtain Let t ≥ 1 in (3.21), we have that Thanks to v n = 1, there exists a constant α 0 > 0 such that So, we have a contradiction. Hence, v = 0. On the other hand, we get (3.25) Thanks to v = 0, by using Lemma 2.3, we get Therefore, by (3.17), (3.18), and (3.26), passing to the limit as n → ∞ in (3.25), we have a contradiction.
Hence, we deduce that {u n } is bounded in E. Therefore, there exists u ∈ E such that u n u, u ± n u ± . From Lemma 3.2, we have that u ∈ E ± . So, according to Lemma 3.1, there exist s + , t -> 0 such that (3.27) We assert that In fact, according to Lemma 2.1, one has that Since u n u in E, combining the continuous embedding E → D 1,2 (R N ) with the weak semicontinuity of the norm u D 1,2 = ( R N |∇u| 2 dx) Thanks to {u n } ⊂ M, using (3.29), (3.32), and weak semicontinuity of the norm in E, we have Suppose that 0 < s + ≤ t -, then from (3.27) we have that By (3.33), we have that Combining (3.34) with (3.35), we have that From the above inequality and (f 5 ), we have that 0 < t -≤ 1. Now, we prove that I b (s + u + + tu -) = m. Denotingū := su + + tu -. So, from (2.11), (3.27), (3.29), (3.30), and Fatou's lemma, we have that Consequently,s =t = 1. Thus,ū = u and I b (u) = m.

The proof of the main results
In this section, we prove Theorem 1.1.
Furthermore, if f is odd, the functional is even. Now we prove that satisfies the (PS) condition. From (2.4) and (3.16), we have that is bounded from below in S. Suppose that {u n } ⊂ S is a (PS) d sequence of , according to (iii) of Lemma 2.3, we know {v n := m(u n )} ⊂ N is a (PS) d sequence of I b on N . Through the standard agrement at the beginning of this section, we know that v n is bounded in E. So, there exists nonzero v ∈ E such that v n v in E, v n → v a.e. in R N .
Therefore, we have that According to Lemma 2.1, we have that We claim that In fact, if (VK 2 ) holds, since where χ is a character function. Hence { √ K(x)f (v n )χ {|v n ≤1|} } is bounded in L 2 (R N ). Similarly, we have √ K(x)v ∈ L 2 (R N ). So, from v n → va.e. in R N , we can get On the other hand, since |Kf (v n )χ {|v n ≥1|} | 2 * 2 * -1 ≤ Cv 2 * n and v n → v a.e. on R N , we get Then, by similar discussion, we have that (4.9) and (4.10) hold. Therefore, from the above discussion, we have that (4.8) holds. Similarly, we have Since v n v in E and E ⊂ D 1,2 (R N ), we get v n v in D 1,2 . Then, by weak semicontinuity of the norm in D 1,2 (R N ), we have that b( R N |∇v n | 2 dx)( R N ∇v n · ∇(v nv) dx) ≥ 0. Therefore, according to (4.7), we get v n → v in E. From Proposition 2.3, {u n := m -1 (v n )} ⊂ S and u n → u = m -1 (v) ∈ S. That is, satisfies the Palais-Smale condition on S. So, from Lemma 2.5, Proposition 2.3, and [37], the functional I b has infinitely many critical points.

Conclusions
In this paper, by the minimization argument on the sign-changing Nehari manifold and the quantitative deformation lemma, we discussed the existence of least energy sign-changing solution for a class of Schrödinger-Kirchhoff-type fourth-order equations with potential vanishing at infinity. Our results improve and generalize some interesting known results.
Since these days there is a good trend of existence of solutions for fractional-order differential equations which are definitely the generalized study, we will discuss some problems about fractional-order differential equations in the follow-up work.