SIGN-CHANGING SOLUTIONS FOR FOURTH ORDER ELLIPTIC EQUATIONS WITH KIRCHHOFF-TYPE

. In this paper, we study the following fourth-order elliptic equation with Kirchhoﬀ-type where the constants a > 0 ,b ≥ 0. By constraint variational method and quantitative deformation lemma, we obtain that the problem possesses one least energy sign-changing solution u b . Moreover, we also prove that the energy of u b is strictly larger than two times the ground state energy. Finally, we give a convergence property of u b when b as a parameter and b → 0.

1. Introduction and main results. This paper is concerned with the existence of sign-changing solutions to the following fourth order elliptic equations of Kirchhoff type where a, b are positive constants, ∆ 2 is the biharmonic operator. Problem (1.1) arises in the study of travelling waves in suspension bridge and the study of the static deflection of an elastic plate in a fluid, see [23]. For the case b = 0, there are many results focused on the existence, multiplicity and concentration of solutions, see for instance [17,18,34,35,36,39,40,41,42]. For the case b = 0, problem (1.1) is a nonlocal one as the appearance of the term R N |∇u| 2 dx implies that (1.1) is not a pointwise identity. This causes some mathematical difficulties which make the study of the problem particularly interesting and has received considerable attention in mathematical analysis in the last years, see [8,9,10,13,15,24,25,28,31,32,33,38] and the references therein.
From the discussion in these literatures, we know that problem (1.1) has deep mathematical and physical background. In fact, if we set V (x) = 0, replace R N by a bounded smooth domain Ω ⊂ R N and set u = ∆u = 0 on ∂Ω, then problem (1.1) is reduced to the following equation which is related to the following stationary analogue of the equation of Kirchhoff type (1.3) this problem is used to describe some phenomena appeared in different physical, engineering and other sciences because it is regarded as a good approximation for describing nonlinear vibrations of beams or plates, see [2,3]. Later, in [19], In [29], by inserting a parameter λ, the authors considered the existence of nontrivial solutions for (1.2) with λ. Recently, Xu and Chen [31,32,33] obtained infinitely many large energy solutions and negative energy solutions to (1.1), respectively. However, up to our knowledge, there seem no any results on the existence of sign-changing solutions to (1.1). In this paper, the first aim is to establish the existence result of sign-changing solutions to problem (1.1). In what follows, we make the following assumptions: where meas(·) denotes the Lebesgue measure in R N ; (F 1 ) f ∈ C 1 (R, R) and f (s) = o(|s|) as s → 0; (F 2 ) for some constant p ∈ (2, 2 * ), lim s→∞ f (s) |s| 3 is a non-decreasing function of s ∈ R \ {0}. To state our main results, for a > 0 fixed, define the Sobolev space with the inner product as When (V 1 ) or (V 2 ) holds, we know that the embedding E → L p (R N ) for 2 < p < 2 * is compact and is continuous for 2 ≤ p ≤ 2 * (see [4,7]), then (1.5) Weak solutions to problem (1.1) are critical points of the following functional Obviously, the functional Φ b is well-defined for every u ∈ E and belongs to C 2 (E, R). Moreover, for any u, v ∈ E, we have We call u a least energy sign-changing solution to (1.1) if u is a solution of (1.1) with u ± = 0 and When b = 0, problem (1.1) turns to be the following one As we know, in the past decades, there are some powerful methods developed to study the existence and multiplicity of sign-changing solutions for nonlinear elliptic problems, see [22,16,8,37,12,5,1,26] and the references therein. In [16,11], the authors used the descended flow methods, in [12], the authors used super and sub solution combining with truncation techniques. From discussions on the existence of sign-changing solutions to (1.8), we know, methods of finding sign-changing solutions to (1.8) heavily rely on the following decompositions: (1.10) However, for the case b > 0, since the effect of the nonlocal term, the functional Φ b no longer satisfies (1.9) and (1.10). Indeed, we have (1.13) In this paper, the nonlocal term ( R N |∇u| 2 dx)∆u is involved in the equation, we borrow some ideas from [5,1,26] and first try to seek a minimizer of the energy functional Φ b over the following constraint: (1.14) Our main results are as follows.
Another aim of this paper is to show that the energy of any sign-changing solution of (1.1) is strictly larger than two times the ground state energy. This is trivial for the typical equation (1.8).
is the least energy sign-changing solution obtained in Theorem 1.1. In particular, c b will be achieved either by a positive or a negative function.
In the following, we will show a convergence property of u b as b → 0, this result can be stated in the following theorem. Theorem 1.3. If the assumptions of Theorem 1.1 hold, for any sequence {b n } with b n → 0 as n → ∞, there exists a subsequence, still denoted by {b n }, such that u bn convergent to u 0 strongly in E as n → ∞, where u 0 is a least energy sign-changing solution of the problem (1.8), which changes sign only once.
This paper is organized as follows. In section 2, we prove some crucial lemmas. In section 3, we obtain that the minimizer of the constrained problem is a sign-changing solution by quantitative deformation lemma and degree theory. Furthermore, we use some energy estimations and comparisons to prove Theorem 1.2 and 1.3.

2.
Prelimanaries. We first show that the set M b is nonempty in E and to seek a critical point of Φ b by constraint minimization on M b .
For fixed u ∈ E with u ± = 0, we denote B : Hence, the problem su + + tu − ∈ M b is reduced to verify that there is only one solution (s, t) ∈ (R + × R + ) of system (2.1). In order to solve this problem, we consider the solvability of the following system with a parameter µ ∈ [0, 1] Proof. Since g 0 (s, t) is independent of t and h 0 (s, t) is independent of s, without loss of generality, we need only to prove that there is a unique t > 0 such that h 0 (s, t) = 0. Since u − = 0, from (F 1 )-(F 4 ), we get that h 0 (s, 0) = 0, h 0 (s, t) > 0 for t > 0 small and h 0 (s, t) < 0 for t large. Thus, there exists t 1 > 0 such that h 0 (s, t 1 ) = 0. Next, we prove the uniqueness of t 1 . Arguing by contradiction, suppose that there exist t 1 , t 2 such that 0 < t 1 < t 2 and h 0 (s, Similarly, we have dx. which is absurd in view of (F 4 ) and 0 < t 1 < t 2 . The proof of Lemma 2.1 is completed.
Therefore, we have Then, from the implicit function theorem, we can find open neighborhoods U 0 ⊂ U of µ 0 , V 0 ⊂ V of (s,t) such that system (2.2) is uniquely solvable in U 0 × V 0 , that is, we can find a unique function ϕ : Extend U 0 as much as possible, we regard this extended domain asŨ 0 , setŨ 0 = (δ 1 , δ 2 ). If δ 2 < +∞, then there exists {µ n } ⊂Ũ 0 such that From the above proof, for (s n , t n ) = ϕ(µ n ), we have s n →s, t n →t as n → ∞, and g µn (s n , t n ) = 0, h µn (s n , t n ) = 0. It follows from the continuity of g µ and h µ that This contradicts to the fact that δ 2 ∈Ũ 0 . Thus δ 2 = +∞. Similarly, we have δ 1 = +∞. Hence, we know the system (2.2) is solvable, particularly, when µ = 1, that is, the system (2.1) is solvable.
Next, we prove the uniqueness of the solution.
Proof. Set (s 0 , t 0 ) be the solution of (2.1), without loss of generality, we assume that 0 < s 0 ≤ t 0 , then and Combining (2.11) with (2.12), we have 4 dx, From this inequality, we can conclude that t 0 ≤ 1. In fact, arguing by contradiction, assume that t 0 > 1, the left side of the above inequality is negative, while the right side is non-negative. This is absurd, thus, t 0 ≤ 1. Similarly, we can prove s 0 ≥ 1. Since 0 < s 0 ≤ t 0 , we have s 0 = t 0 = 1.
Assume that (s 1 , t 1 ) and (s 2 , t 2 ) are solutions of system (2.1), then From the above proof and the fact that The uniqueness of solution is obtained, hence, the proof of this lemma is completed.
Proof. Without loss of generality, we assume 0 < t u ≤ s u , since s u u + + t u u − ∈ M b , then (2.13) The assumption g 1 (1, 1) ≤ 0 gives that (2.14) Combining (2.13) with (2.14), we get If s u > 1, then the left side of this inequality is negative. But from (F 4 ), the right is positive, hence, we must have s u ≤ 1. Then, the proof is completed.
Lemma 2.6. For fixed u ∈ E with u ± = 0, then the vector (s u , t u ) which obtained in Lemma 2.3 is the unique maximum point of the functionφ : (R + × R + ) → R defined asφ(s, t) = Φ b (su + + tu − ).
Proof. From the proof of Lemma 2.1, 2.2 and 2.3, we know that (s u , t u ) is the unique critical point ofφ in (R + × R + ). From the assumption (F 3 ), we havẽ φ(s, t) → −∞ uniformly as |(s, t)| → ∞, thus, it is sufficient to check that a maximum point cannot be achieved on the boundary of (R + × R + ). Without loss of generality, we may assume that (0,t) is a maximum point ofφ. By calculation, we havẽ

From the above equation, we knowφ(s,t) is an increasing function with respect to
s if s is small enough, the pair (0,t) is not a maximum point ofφ in (R + × R + ).
Proof. For every u ∈ M b , we have Φ b (u), u = 0. Then from (F 1 ), (F 2 ), (1.5) and Sobolev embedding theorem, we get thus, there exists a constant α > 0 such that Similar as the proof of (2.16), there exists a constant β > 0 such that u ± n 2 ≥ β for all n ∈ N. Since {u n } ⊂ M b , thus It follows from the boundedness of u n that there is C 1 > 0 such that Choosing ε = β 2C1 , we get From the above inequality and the compactness of the embedding E → L p (R N ) for 2 < p < 2 * , we have thus, u ± b = 0. Combining (F 1 )-(F 2 ) with the compactness lemma of Strauss [27], we get (2.20) By the weak semicontinuity of norm, we have (2.21) Then from (2.20), we get From the above inequalities and Lemma 2.5, there exists ( Since condition (F 4 ) implies that G(s) := sf (s) − 4F (s) is a non-negative function, increasing in |s|, we have (2.23) 3. Proof of main results. The main aim of this section is to prove our main results. We first prove that the minimizer u b for the minimization problem (2.15) is indeed a sign-changing solution of (1.1), that is Φ (u b ) = 0.
Lemma 2.4 and the degree theory now yields deg(Ψ 1 , D, 0) = deg(Ψ 0 , D, 0) = 1. Therefore, Ψ 1 (s 0 , t 0 ) = 0 for some (s 0 , t 0 ) ∈ D, so that η(1, g(s 0 , t 0 )) = h(s 0 , t 0 ) ∈ M b , which is a contradiction. Thus, u b is a critical point of Φ b , and so, a signchanging solution for problem (2.1). Now, we show that u b has exactly two nodal domains, arguing by contradiction, we assume that u has at least three nodal domains Ω 1 , Ω 2 , Ω 3 . Without loss of generality, we may assume that u > 0 a.e. in Ω 1 and u < 0 a.e. in Ω 2 . Set then u i ∈ E and u i = 0 and Φ b (u b ), u i = 0 for i = 1, 2, 3. Setting v := u 1 + u 2 , we see that v + = u 1 and v − = u 2 , i.e., v ± = 0. Then from Lemma 2.4, there is a unique pair (s v , t v ) ∈ D of positive numbers such that Moreover, combining with the fact Φ b (u b ), u i = 0 that Φ b (v), v ± < 0. From Lemma 2.5, we have that On the other hand, Thus, similar as (2.23), we can get which is a contradiction. Hence, u 3 = 0 and u b has exactly two nodal domains. From Theorem 1.1, we know that the problem (1.1) has a least energy signchanging solution u b which changes sign only once. Now, we show that the energy of u b is strictly larger than two times the ground state energy.
From Corollary 2.9 in [14], we know that the critical points of the functional Φ b on N b are critical points of Φ b in E, we can conclude that Φ b (v b ) = 0. Thus, v b is a ground state solution of (1.1). From Theorem 1.1, we know that the problem (1.1) has a least energy sign-changing solution u b which changes sign only once. Suppose As the proof of Lemma 2.1, there is unique t u + b > 0 such that ∈ (0, 1). By the similar arguments, we can prove that there is unique t u − b ∈ (0, 1) such that Then from Lemma 2.6, we have which implies that c b > 0 can not be achieved by a signchanging function. This completes the proof.
In the following, we will analyze the convergence property of u b as b > 0 by regarding b > 0 as a parameter in problem (1.1).
Proof of Theorem 1.3. For any b > 0, let u b ∈ E be the least energy signchanging solution of (1.1) obtained in Theorem 1.1, which changes sign only once. Next, by proving the following Lemma 3.1, 3.2 and 3.3 to complete the proof of Theorem 1.3. Proof. Choose a nonzero function ϕ ∈ C ∞ 0 (R N ) with ϕ ± = 0. From (2.17) and the assumption (F 3 ), for any b ∈ [0, 1], there exists a pair (θ 1 , θ 2 ) of positive numbers, which does not depend on b, such that From Lemma 2.4 and Lemma 2.5, for any b ∈ [0, 1], there exists a unique pair ((s ϕ (b), t ϕ (b))) ∈ (0, 1] × (0, 1] such that (3.7) Thus, for any b > 0, we have where C 0 does not depend on b. For n large enough, we have Hence, {u bn } is bounded in E.
From Lemma 3.1, there exists a subsequence of b n , still denoted by b n , such that u bn u 0 weakly in E. Then, u 0 is a weak solution of (1.8).
Lemma 3.2. u 0 is a least energy sign-changing solution of (1.8) which changes sign only once.
Proof. Suppose that v 0 is a least energy sign-changing solution of (1.8). From [6], the existence of v 0 can be proved. By Lemma 2.4, for each b n > 0, there is a unique pair (s bn , t bn ) of positive numbers such that We can get {s bn }, {t bn } are bounded. In fact, arguing by contradiction, assume that {s bn } is unbounded, that is, there exists a subsequence of {s bn }, still denoted by {s bn }, such that s bn → +∞ as n → ∞. From (3.10), we have (3.13) Observe that from (3.13), the assumption s bn → +∞ as n → ∞ and the assumption (F 4 ), we have b n t bn s bn 2 → +∞, as n → ∞, thus, t bn → +∞ and s bn t bn 2 → 0 + , as n → ∞. (3.14) Combining (3.14) with (3.11), we have The left side of (3.15) goes to zero as n → ∞, while the right side goes to negative infinity as n → ∞. This is absurd. Thus, {s bn } is bounded. Similarly, we can get {t bn } is bounded. By selecting subsequences of {s bn } and {t bn }, still denoted by {s bn } and {t bn }, such that (s bn , t bn ) → (s 0 , t 0 ) as n → ∞. Next, we will show that s 0 = t 0 = 1. In fact, from (3.10), we have Similarly, from (3.11), we have If s 0 > 1, the left side of (3.18) is negative, but the right side is positive, this is a contradiction. If s 0 < 1, the left side of (3.18) is positive, but the right side is negative, this is also a contradiction. Therefore, s 0 = 1. Similarly, we can get t 0 = 1, that is, (s bn , t bn ) → (1, 1), as n → ∞. Now, we can prove u 0 is a least energy sign-changing solution of (1.8) which changes sign only once. From (3.19) and Lemma 2.6, we have (3.20) Combining Lemma 3.1, 3.2 with 3.3, we complete the proof of Theorem 1.3.