The existence of least energy nodal solutions for some class of Kirchhoff equations and Choquard equations
aa r X i v : . [ m a t h . A P ] J a n The existence of least energy nodal solutions forsome class of Kirchhoff equations and Choquardequations
Hongyu Ye ∗ College of Science, Wuhan University of Science and Technology,Wuhan 430065, P. R. China
Abstract
In this paper, we study the existence of least energy nodal solutions forsome class of Kirchhoff type problems. Since Kirchhoff equation is a nonlocalone, the variational setting to look for sign-changing solutions is different fromthe local cases. By using constrained minimization on the sign-changing Neharimanifold, we prove the Kirchhoff problem has a least energy nodal solution withits energy exceeding twice the least energy. As a co-product of our approaches,we obtain the existence of least energy sign-changing solution for Choquardequations and show that the sign-changing solution has an energy strictly largerthan the least energy and less than twice the least energy.
Keywords:
Least energy sign-changing solutions; Nonlocal problems; Ekelandvariational principle; Nodal solutions
Mathematics Subject Classification(2000):
In the past years, the following nonlinear Kirchhoff equation − (cid:18) a + b Z R N |∇ u | (cid:19) ∆ u + V ( x ) u = f ( u ) , x ∈ R N , (1.1)has attracted considerable attention, where N = 2 , a, b > V : R N → R . Equation (1.1) is a nonlocal one as the appearance of the term R R N |∇ u | implies that (1.1) is no longer a pointwise identity. This causes somemathematical difficulties which make the study of (1.1) particularly interesting. For ∗ a: Partially supported by NSFC NO: 11371159. E-mail address: [email protected] V ( x ) satisfiesone of the following two cases:( V ) V ( x ) is a constant or V ( x ) ∈ C ( R N , R ) such that lim | x |→ + ∞ V ( x ) = + ∞ . Moreover, the nonlinearity f ∈ C ( R , R ) satisfies the following conditions:( f ) lim s → f ( s ) | s | = 0;( f ) There exists 3 < q < ∗ − | s |→ + ∞ f ( s ) | s | q = 0 , where 2 ∗ = + ∞ if N = 2 and 2 ∗ = 6 if N = 3;( f ) lim | s |→ + ∞ F ( s ) s = + ∞ , where F ( s ) = R s f ( t ) dt ;( f ) The function f ( s ) | s | is nondecreasing on R \{ } .To state our main result, for a > V ) holds, we define the Sobolevspace H = H r ( R N ) = { u ∈ H ( R N ) | u ( x ) = u ( | x | ) } when V ( x ) is a constantor H = (cid:26) u ∈ D , ( R N ) | Z R N V ( x ) u < + ∞ (cid:27) with the norm given as k u k = (cid:18)Z R N ( a |∇ u | + V ( x ) u ) (cid:19) , ∀ u ∈ H, (1.2)which is induced by the corresponding inner product on H . By ( V ), the embedding H ֒ → L p ( R N ) (2 < p < ∗ ) is compact, see e.g. [5]. Weak solutions to problem (1.1)are critical points of the following functional I : H → R defined by I ( u ) = 12 Z R N ( a |∇ u | + V ( x ) u ) + b (cid:18)Z R N |∇ u | (cid:19) − Z R N F ( u ) . (1.3)It easily sees that I ∈ C ( H, R ) and for any ϕ ∈ H , h I ′ ( u ) , ϕ i = Z R N ( a ∇ u ∇ ϕ + V ( x ) uϕ ) + b Z R N |∇ u | Z R N ∇ u ∇ ϕ − Z R N f ( u ) ϕ. (1.4)2e call u a least energy sign-changing solution to (1.1) if u is a solution of (1.1) with u ± = 0 and I ( u ) = inf { I ( v ) | v ± = 0 , I ′ ( v ) = 0 } , where u + ( x ) = max { u ( x ) , } and u − ( x ) = min { u ( x ) , } . When b ≡ a ≡ − ∆ u + V ( x ) u = f ( u ) , x ∈ R N . (1.5)There are several ways in the literature to obtain sign-changing solutions for (1.5), seee.g. [4, 12, 6, 8, 29, 37]. In [4], Bartsch, Liu and Weth used minimax arguments in thepresence of invariant sets of a descending flow to prove that (1.5) has a nodal solutionif conditions imposed on V ( x ) ensure the compact embeddings and f satisfies theclassical ( AR ) and monotonicity condition. Under similar assumptions on f , Furtado,Maia and Medeiros in [12] obtained nodal solutions for (1.5) by seeking minimizerson the sign-changing Nehari manifold when V may change sign and satisfy mildintegral conditions. Nodal solutions of (1.5) in bounded domains are proved to existby Noussair and Wei in [29] via the Ekeland variational principle and the implicitfunction theorem, and by Bartsch and Weth in [6] based on the variational methodtogether with the Brouwer degree theory. However, all these mentioned methodsheavily reply on the following two decompositions: I ( u ) = I ( u + ) + I ( u − ) , (1.6) h I ′ ( u ) , u + i = h I ′ ( u + ) , u + i and h I ′ ( u ) , u − i = h I ′ ( u − ) , u − i , (1.7)where I ( u ) = Z R N ( |∇ u | + V ( x ) u ) − Z R N F ( u ) . Furthermore, (1.7) and (1.6) imply that the energy of any sign-changing solution to(1.5) is larger than two times the least energy in H . However, for the case b > I no longer possesses the samedecompositions as (1.6) (1.7). Indeed, we have I ( u ) = I ( u + ) + I ( u − ) + b Z R N |∇ u + | Z R N |∇ u − | , h I ′ ( u ) , u + i = h I ′ ( u + ) , u + i + b Z R N |∇ u + | Z R N |∇ u − | , (1.8) h I ′ ( u ) , u − i = h I ′ ( u − ) , u − i + b Z R N |∇ u + | Z R N |∇ u − | . (1.9)So the methods to obtaining sign-changing solutions of the local problem (1.5) andestimating the energy of sign-changing solutions seem not suitable for our nonlocalone (1.1). 3ecently, in [34], Wang and Zhou studied the existence of sign-changing solutionsfor a Schr¨odinger-Poisson system with pure power nonlinearity | u | p − u , which is anonlocal one. By using constrained minimization on the sign-changing Nehari mani-fold and the Brouwer degree theory, they proved that under the assumption ( V ), theSchr¨odinger-Poisson system has a sign-changing solution for all 3 < p < V ( x )is a constant, the energy of any sign-changing solution is strictly larger than the leastenergy. However, their method strongly depends on the fact that the nonlinearity ishomogeneous and the nonlocal term is positive, which makes it could not be appliedto our problem (1.1), which deals with a more general nonlinearity.Recall that under the conditions ( V ) and ( f ) − ( f ), it has been similarly provedas [15, 35] that (1.1) possesses at least one ground state solution w ∈ H satisfyingthat I ( w ) = inf v ∈N I ( v ) := c, (1.10)where N is the corresponding Nehari manifold of (1.1) given as N = { u ∈ H \{ }| h I ′ ( u ) , u i = 0 } . Our main result is as follows:
Theorem 1.1.
Assume that ( V ) and ( f ) − ( f ) hold, then problem (1.1) has atleast one least energy sign-changing solution u ∈ H with I ( u ) > c , moreover, u hasprecisely two nodal domains. Now we give our main idea for the proof of Theorem 1.1. To get least energysign-changing solutions, we define the following constrained manifold: M = { u ∈ H | u ± = 0 , h I ′ ( u ) , u ± i = 0 } . (1.11)Obviously, M contains all sign-changing solutions of (1.1). Therefore, it is enoughto prove that m := inf u ∈M I ( u ) has a minimizer and the minimizer is just a criticalpoint of I on H . However, since it follows from (1.8) and (1.9) that h I ′ ( u ) , u ± i > h I ′ ( u ± ) , u ± i if u ± = 0, it is difficult to show that M 6 = ∅ in a usual way. Moreover,the nonlinearity in (1.1) is more general than that in [34], which makes that themethod used in [34] cannot be applied here. To overcome this difficulty, it needsmore analysis and some new ideas. By using a totally different approach from [34],we succeeded in proving that for each u ∈ H with u ± = 0, there exists a uniquepair ( t, s ) := ( t ( u ) , s ( u )) ∈ R + × R + such that tu + + su − ∈ M . Our result is moredelicate. We furthermore conclude that if h I ′ ( u ) , u ± i < , then the unique pair ( t, s )such that tu + + su − ∈ M must satisfy t ∈ (0 ,
1) and s ∈ (0 , . (1.12)With the help of such conclusion it can be showed that m is attained and the leastenergy sign-changing solution of (1.1) has precisely two nodal domains. If m is4chieved by u ∈ M , then there exist two Lagrange multipliers µ , µ ∈ R such that I ′ ( u ) + µ J ( u + ) + µ J ( u − ) = 0 , where J ( u ± ) = h I ′ ( u ) , u ± i . By using the assumptions f ∈ C and ( f ), it can be proved that µ = µ = 0, i.e. u is a critical point of I . Toestimate the energy of the least energy sign-changing solution u , although we do nothave u ± ∈ N , we do easily see that h I ′ ( u ± ) , u ± i < . (1.13)Then ( f ) − ( f ) show that there exists k, l ∈ (0 ,
1) such that ku + , lu − ∈ N , whichimplies that 2 c ≤ I ( ku + ) + I ( lu − ) < I ( u ) . Therefore we complete the proof ofTheorem 1.1.
Remark 1.2.
Our approach to prove Theorem 1.1 can be used to deal with theexistence of nodal solutions for the following Schrodinger-Poisson system (cid:26) − ∆ u + V ( x ) u + φ ( x ) u = f ( u ) , x ∈ R , − ∆ φ = u , x ∈ R (1.14) under the assumptions ( V ) and ( f ) − ( f ) , while [34] is a special case of (1.14) with f ( u ) = | u | p − u . Another aim of this paper is to study the existence of sign-changing solutions forthe following Choquard equation: − ∆ u + V ( x ) u = ( I α ∗ | u | p ) | u | p − u, x ∈ R N , (1.15)where N ≥ α ∈ (0 , N ), N + αN < p < N + αN − . The potential V : R N → R satisfies ( V )given above. I α : R N → R is the Riesz potential [32] defined as I α ( x ) = Γ( N − α )Γ( α ) π N α | x | N − α , ∀ x ∈ R N \{ } . Problem (1.15) is also a nonlocal one due to the existence of the nonlocal nonlinearity.It arises in various fields of mathematical physics, such as quantum mechanics, physicsof laser beams, the physics of multiple-particle systems, etc. When N = 3, V ( x ) ≡ α = p = 2, (1.15) turns to be the well-known Choquard-Pekar equation: − ∆ u + u = ( I ∗ | u | ) u, x ∈ R , which was proposed as early as in 1954 by Pekar [30], and by a work of Choquard1976 in a certain approximation to Hartree-Fock theory for one-component plasma,see [20, 22]. (1.15) is also known as the nonlinear stationary Hartree equation sinceif u solves (1.15) then ψ = e it u ( x ) is a solitary wave of the following time-dependentHartree equation iψ t = − ∆ ψ − ( I α ∗ | ψ | p ) | ψ | p − ψ in R + × R N , R N with somesymmetries when 2 ≤ p < N + αN − and V ( x ) is continuous and radially symmetric andinf x ∈ R N V ( x ) > V ( x ) → V ∞ > | x | → + ∞ and V ( x ) ≤ Ce − k | x | for some C, k > V ) in the literature.Clearly, since V ( x ) satisfies condition ( V ), our working space is still H and itsnorm is also given as (1.2) with a ≡
1. Weak solutions to (1.15) correspond to criticalpoints to the following functional Ψ : H → R :Ψ( u ) = 12 Z R N ( |∇ u | + V ( x ) u ) − p Z R N ( I α ∗ | u | p ) | u | p . (1.16)Recall the well-known Hardy-Littlewood-Sobolev inequality, we have for some C > Z R N ( I α ∗ | u | p ) | u | p ≤ C (cid:18)Z R N | u | NpN + α (cid:19) N + αN . Then Ψ ∈ C ( H, R ) and for any ϕ ∈ H , h Ψ ′ ( u ) , ϕ i = Z R N ( ∇ u ∇ ϕ + V ( x ) uϕ ) − Z R N ( I α ∗ | u | p ) | u | p − uϕ. Similarly, we call u a least energy sign-changing solution of (1.15) if u is a solutionof (1.15) with u ± = 0 andΨ( u ) = inf { Ψ( v ) | v ± = 0 , Ψ ′ ( v ) = 0 } . Recall it has been shown that (1.15) has at least one ground state solution w ∈ H such that Ψ( w ) = inf v ∈N Ψ( v ) := ¯ c, where N = { u ∈ H \{ }| h Ψ ′ ( u ) , u i = 0 } is the associated Nehari manifold, see e.g.[27, 28]. Moreover, each ground state solution has constant sign.Our main result is as follows: Theorem 1.3.
Assume that N ≥ , α ∈ (( N − + , N ) , here ( N − + = N − if N ≥ , ( N − + = 0 if N = 3 ; ≤ p < N + αN − and V ( x ) satisfies ( V ) . Then problem (1.15) has at least one least energy sign-changing solution u ∈ H with ¯ c < Ψ( u ) < c. Due to the effect of the nonlocal nonlinearity, the functional Ψ does not have thesame decompositions like (1.6)(1.7). In fact, for any u ∈ H with u ± = 0,Ψ( u ) = Ψ( u + ) + Ψ( u − ) − p Z R N ( I α ∗ | u + | p ) | u − | p (1.17)6nd h Ψ ′ ( u ) , u + i = h Ψ ′ ( u + ) , u + i − Z R N ( I α ∗ | u − | p ) | u + | p < h Ψ ′ ( u + ) , u + i , (1.18) h Ψ ′ ( u ) , u − i = h Ψ ′ ( u − ) , u − i − Z R N ( I α ∗ | u + | p ) | u − | p < h Ψ ′ ( u − ) , u − i . (1.19)Then the methods to deal with local equations cannot be applied here. Moreover, themethod used in [34] do not work since their method heavily depends on the associatedfunctional has a positive nonlocal term. Motivated by Theorem 1.1, we try to lookfor a minimizer of Ψ constrained on the following manifold: M = { u ∈ H | u ± = 0 , h Ψ ′ ( u ) , u ± i = 0 } (1.20)and show that the minimizer is a critical point of Ψ on H . However, as the nonlocalterm in Ψ is negative while that in I is positive, there must be some differences.The main difference and difficulty is that Ψ does not have properties like (1.12) and(1.13), which makes that the arguments used in Theorem 1.1 to prove the existenceof minimizers and to estimate the energy of sign-changing solutions cannot work. Itneeds some improvement. We succeeded in overcoming the difficulty by using Ekelandvariational principle and the implicit function theorem to construct a Palais-Smalesequence and with the help of the positive ground state solution of (1.15).Throughout this paper, we use standard notations. For simplicity, we write R Ω h to mean the Lebesgue integral of h ( x ) over a domain Ω ⊂ R N . L p := L p ( R N ) (1 ≤ p < + ∞ ) is the usual Lebesgue space with the standard norm | · | p . We use “ → ”and “ ⇀ ” to denote the strong and weak convergence in the related function spacerespectively. C will denote a positive constant unless specified. We use “ := ” todenote definitions and B r ( x ) := { y ∈ R N | | x − y | < r } . We denote a subsequence ofa sequence { u n } as { u n } to simplify the notation unless specified.The paper is organized as follows. In §
2, we prove Theorem 1.1. In §
3, we proveTheorem 1.3.
We first give some technical lemmas, which are important in the proof of Theorem1.1.By f ∈ C ( R ) satisfies ( f ) − ( f ), we see that f ′ ( s ) s − f ( s ) ≥ s ; (2.1) e F ( s ) := f ( s ) s − F ( s ) is increasing on R and 14 f ( s ) s ≥ F ( s ) ≥ s . By ( f ) and ( f ), for any ε > , there exists C ε > | f ( s ) | ≤ ε | s | + C ε | s | q , | F ( s ) | ≤ ε | s | + C ε q + 1 | s | q +1 , ∀ s ∈ R . (2.2)7n this section, for any u ∈ H with u ± = 0, we denote for simplicity as follows: α := k u + k , β := b (cid:18)Z R N |∇ u + | (cid:19) , A := b Z R N |∇ u + | Z R N |∇ u − | ,α := k u − k , β := b (cid:18)Z R N |∇ u − | (cid:19) . (2.3)Then α i , β i ( i = 1 , , A > Lemma 2.1.
Let M be defined in (1.11) . Assume that f satisfies ( f ) − ( f ) , thenfor any u ∈ H with u ± = 0 , there exists a unique pair ( t, s ) := ( t ( u ) , s ( u )) ∈ R + × R + such that tu + + su − ∈ M .Proof. For any u ∈ H with u ± = 0, to show that there exists a unique ( t, s ) ∈ R + × R + such that tu + + su − ∈ M is equivalent to show that the following system t α + t β + t s A − Z R N f ( tu + ) tu + = 0 ,s α + s β + t s A − Z R N f ( su − ) su − = 0 ,t, s > g ( t ) : R + → R as follows: g ( t ) = 1 A (cid:18)Z R N f ( tu + ) tu + t − t β − α (cid:19) , ∀ t > , then by ( f ), ( f ), we see that lim t → + g ( t ) = − αA < t → + ∞ g ( t ) = + ∞ . By directcomputation, we have g ′ ( t ) = 1 A (cid:18)Z R N f ′ ( tu + )( tu + ) − f ( tu + ) tu + t − tβ (cid:19) , which and (2.1)(2.2) imply that g ( t ) has a unique zero point t ∗ > g ( t ) < t < t ∗ ; g ( t ) > t > t ∗ . Moreover, g ′ ( t ) > t ≥ t ∗ .By the first equation of (2.4), we see that s = g ( t ) , t ∗ < t < + ∞ . So to prove this lemma, it is enough to prove that h ( t ) := α + g ( t ) β + t A − Z R N f ( p g ( t ) u − ) u − p g ( t ) = 0 , t ∗ < t < + ∞ has a unique solution. By the definition of g ( t ) and ( f ), ( f ), we have thatlim t → t + ∗ h ( t ) = α + t ∗ A > , lim t → + ∞ h ( t ) = −∞ . h ′ ( t ) = g ′ ( t ) β − Z R N f ′ ( p g ( t ) u − )( p g ( t ) u − ) − f ( p g ( t ) u − ) p g ( t ) u − g ( t ) ! + 2 tA. By the definition of g ′ ( t ) and (2.1), we see that the continuous function h ′ ( t ) has aunique zero point on [ t ∗ , + ∞ ) and h ′ ( t ∗ ) = g ′ ( t ∗ ) β + 2 At ∗ > h ′ ( t ) → −∞ as t → + ∞ . Therefore, there exists a unique t := t ( u ) ∈ ( t ∗ , + ∞ ) such that h ( t ) = 0. Let s = g ( t ), then ( t , s ) is a unique pair of solution of system (2.4), i.e. t u + + s u − ∈M . Lemma 2.2.
Let ( f ) − ( f ) hold. (1) If h I ′ ( u ) , u ± i < , then there exists a unique pair ( t, s ) ∈ (0 , × (0 , suchthat tu + + su − ∈ M . (2) If h I ′ ( u ) , u ± i > , then there exists a unique pair ( t, s ) ∈ (1 , + ∞ ) × (1 , + ∞ ) such that tu + + su − ∈ M .Proof. (1) Since h I ′ ( u ) , u ± i <
0, we see that u ± = 0. Then Lemma 2.1 implies thatthere exists a unique pair ( t, s ) ∈ R + × R + such that tu + + su − ∈ M . Let us nextshow that t, s ∈ (0 , h I ′ ( u ) , u ± i <
0, we have α + β + A < Z R N f ( u + ) u + , α + β + A < Z R N f ( u − ) u − , where α i , β i ( i = 1 , , A > tu + + su − ∈ M and t, s >
0, we have α (cid:18) − t (cid:19) + (cid:18) − s t (cid:19) A < Z R N (cid:18) f ( u + )( u + ) − f ( tu + )( tu + ) (cid:19) ( u + ) (2.5)and α (cid:18) − s (cid:19) + (cid:18) − t s (cid:19) A < Z R N (cid:18) f ( u − )( u − ) − f ( su − )( su − ) (cid:19) ( u − ) . (2.6)By contradiction, we just suppose that t >
1. We have to discuss the followingtwo cases:
Case 1. s ≥ t .Then s > ts ≤
1, which and ( f ) imply that α (cid:18) − s (cid:19) + (cid:18) − t s (cid:19) A > Z R N (cid:18) f ( u − )( u − ) − f ( su − )( su − ) (cid:19) ( u − ) ≤ . It contradicts (2.6). So Case 1 is impossible.
Case 2. s < t .Then st < t >
1, which and ( f ) would similarly lead to a contradictionwith (2.5). So Case 2 is impossible.Since Cases 1 and 2 are impossible, we must have 0 < t <
1. Similarly, we canprove that 0 < s < . (2) The proof of (2) is similar to that of (1). Remark 2.3.
We conclude from Lemma 2.1 and (2.5) , (2.6) that(1) if h I ′ ( u ) , u + i = 0 , h I ′ ( u ) , u − i < and u + = 0 , then there exists a unique s ∈ (0 , such that u + + su − ∈ M ;(2) if h I ′ ( u ) , u + i < , h I ′ ( u ) , u − i = 0 and u − = 0 , then there exists a unique t ∈ (0 , such that tu + + u − ∈ M . Lemma 2.4. I ( u ) is bounded from below and coercive on M . Moreover, thereexists a constant C > such that k u ± k > C > for all u ∈ M . Proof.
For any u ∈ M , we have h I ′ ( u ) , u i = 0, then I ( u ) = I ( u ) − h I ′ ( u ) , u i ≥ k u k ≥ , which implies that I ( u ) is bounded from below and coercive on M .Since u ± = 0 and h I ′ ( u ) , u ± i = 0, by (2.2) and the Sobolev embedding theoremwe see that k u ± k < Z R N f ( u ± ) u ± ≤ ε Z R N | u ± | + C ε Z R N | u ± | q +1 ≤ ε k u ± k + C ε k u ± k q +1 . If we take ε ∈ (0 , ), then we have k u ± k > C > C > u .By Lemmas 2.1 and 2.4, define m := inf u ∈M I ( u ) , then m > Lemma 2.5. If m is achieved by u ∈ M , then u is a critical point of I . roof. Assume that u ∈ M such that I ( u ) = m , then u is a critical point of I constained on M . Hence there exist two Lagrange multipliers µ , µ ∈ R such that I ′ ( u ) + µ J ′ ( u + ) + µ J ′ ( u − ) = 0 , where J ( u ± ) := h I ′ ( u ) , u ± i . So, we have h I ′ ( u ) + µ J ′ ( u + ) + µ J ′ ( u − ) , u ± i = 0 , i.e. µ (cid:18)Z R N [3 f ( u + ) u + − f ′ ( u + )( u + ) ] − α − A (cid:19) + µ A = 0 ,µ A + µ (cid:18)Z R N [3 f ( u − ) u − − f ′ ( u − )( u − ) ] − α − A (cid:19) = 0 , (2.7)where α i ( i = 1 , , A > Lemma 2.6. If m is attained, then m = inf { I ( v ) | v ± = 0 , I ′ ( v ) = 0 } . Proof.
Suppose that there exists u ∈ M such that I ( u ) = m , then Lemma 2.5shows that u is a critical point of I with u ± = 0. Hence m = I ( u ) ≥ inf { I ( v ) | v ± = 0 , I ′ ( v ) = 0 } . (2.8)On the other hand, since M 6 = ∅ and { v ∈ H | v ± = 0 , I ′ ( v ) = 0 } ⊂ M , we haveinf { I ( v ) | v ± = 0 , I ′ ( v ) = 0 } ≤ inf u ∈M I ( u ) = m, which and (2.8) imply the lemma. Proof of Theorem 1.1
Proof.
We complete the proof in three steps.
Step 1.
Let { u n } ⊂ M be a minimizing sequence of m , then Lemma 2.4 showsthat the sequences { u ± n } are respectively uniformly bounded in H . Up to a subse-quence, there are u ± ∈ H such that u ± n ⇀ u ± in H (2.9)as n → + ∞ . By the compactness of sobolev embedding H ֒ → L p ( R N ), 2 < p < ∗ ,we see that u + ≥ , u − ≤ u + · u − = 0 a.e. in R N . Moreover, by (2.2) and Lemma 2.4, we have C ≤ lim n → + ∞ k u + n k ≤ lim n → + ∞ Z R N f ( u ± n ) u ± n = Z R N f ( u ± ) u ± , u ± = 0.Set u = u + + u − , by u n ∈ M and (2.9), we see that k u + k + b Z R N |∇ u | Z R N |∇ u + | ≤ Z R N f ( u + ) u + and k u − k + b Z R N |∇ u | Z R N |∇ u − | ≤ Z R N f ( u − ) u − , i.e. h I ′ ( u ) , u ± i ≤
0. Then by Lemma 2.2 and Remark 2.3, there exists a unique pair( t, s ) ∈ (0 , × (0 ,
1] such that tu + + su − ∈ M , hence by (2.9) again and ( f ), wehave m ≤ I ( tu + + su − )= I ( tu + + su − ) − h I ′ ( tu + + su − ) , tu + + su − i = t k u + k + s k u − k Z R N (cid:18) f ( tu + ) tu + − F ( tu + ) + 14 f ( su − ) su − − F ( su − ) (cid:19) ≤ k u k Z R N (cid:18) f ( u ) u − F ( u ) (cid:19) ≤ lim inf n → + ∞ (cid:20) k u n k Z R N (cid:18) f ( u n ) u n − F ( u n ) (cid:19)(cid:21) = lim inf n → + ∞ (cid:18) I ( u n ) − h I ′ ( u n ) , u n i (cid:19) = m, which implies that t = s = 1 and I ( u ) = m , i.e. m is achieved by u ∈ M . Then weconclude from Lemma 2.5 that u is a critical point of I . Step 2.
Since u ∈ M and I ( u ) = m , u ± = 0 and h I ′ ( u ± ) , u ± i <
0. Then by( f ) − ( f ), there exist k, l ∈ (0 ,
1) such that ku + ∈ N and lu − ∈ N . Hence by ( f ),we have c ≤ I ( ku + ) = I ( ku + ) − h I ′ ( ku + ) , ku + i = k k u + k Z R N (cid:18) f ( ku + ) ku + − F ( ku + ) (cid:19) < k u + k Z R N (cid:18) f ( u + ) u + − F ( u + ) (cid:19) . Similarly, c < k u − k Z R N (cid:18) f ( u − ) u − − F ( u − ) (cid:19) . c < k u + k + k u − k Z R N (cid:20)(cid:18) f ( u + ) u + − F ( u + ) (cid:19) + (cid:18) f ( u − ) u − − F ( u − ) (cid:19)(cid:21) = k u k Z R N (cid:18) f ( u ) u − F ( u ) (cid:19) = I ( u ) − h I ′ ( u ) , u i = I ( u ) = m. Step 3.
By contradiction, we assume that u has at least three nodal domainsΩ , Ω , Ω . Without loss of generality, we may assume that u > and u < . Set u i := χ Ω i u, i = 1 , , , (2.10)where χ Ω = (cid:26) , x ∈ Ω i , x ∈ R N \ Ω i . (2.11)Then u i ∈ H and u i = 0. Hence it follows from step 1 that h I ′ ( u ) , u i i = 0. Since R R N |∇ u | = 0, we see that h I ′ ( u + u ) , ( u + u ) ± i <
0. By Lemma 2.2, there exists( t, s ) ∈ (0 , × (0 ,
1) such that tu + su ∈ M . Hence by ( f ), we see that m ≤ I ( tu + su )= I ( tu + su ) − h I ′ ( tu + su ) , tu + su i = t k u k + s k u k Z R N (cid:20)(cid:18) f ( tu ) tu − F ( tu ) (cid:19) + (cid:18) f ( su ) su − F ( su ) (cid:19)(cid:21) < k u + u k Z R N (cid:18) f ( u + u )( u + u ) − F ( u + u ) (cid:19) < k u + u k Z R N (cid:18) f ( u + u )( u + u ) − F ( u + u ) (cid:19) + k u k Z R N (cid:18) f ( u ) u − F ( u ) (cid:19) = k u k Z R N (cid:18) f ( u ) u − F ( u ) (cid:19) = I ( u ) − h I ′ ( u ) , u i = I ( u ) = m, which is impossible, so u has precisely two nodal domains.13 Proof of Theorem 1.3
In this section, we consider the existence of sign-changing solutions for the Choquardequation (1.15).
Lemma 3.1. ([21], Theorem 9.8) Let N ≥ and α ∈ (0 , N ) , then for any f, g ∈ L NN + α ( R N ) , (cid:18)Z R N ( I α ∗ f ) g (cid:19) ≤ Z R N ( I α ∗ f ) f Z R N ( I α ∗ g ) g, with equality for g if and only if f = Cg for some constant C. For simplicity, for any u ∈ H with u ± = 0, we use the following notations in whatfollows: A := k u + k , B := Z R N ( I α ∗ | u + | p ) | u + | p , B := Z R N ( I α ∗ | u − | p ) | u + | p ,A := k u − k , B := Z R N ( I α ∗ | u − | p ) | u − | p , (3.1)where N ≥ α ∈ (( N − + , N ) and 2 ≤ p < N + αN − . Then A i , B i ( i = 1 , , B > B < B B . (3.2) Lemma 3.2.
Let M be defined by (1.20) . If ≤ p < N + αN − , then for any u ∈ H with u ± = 0 , there exists a unique pair ( t, s ) := ( t ( u ) , s ( u )) ∈ R + × R + such that tu + + su − ∈ M .Proof. For any u ∈ H with u ± = 0, tu + + su − ∈ M for some t, s > A t − B t p − Bt p s p = 0 , A s − B s p − Bs p t p = 0 , t, s > . (3.3)If p = 2, then (3.3) turns to be a linear system about ( t , s ): B t + Bs = A ,Bt + B s = A ,t, s > . Hence (3.2) implies that the linear system has a unique solution in (0 , + ∞ ) × (0 , + ∞ ).For the case 2 < p < N + αN − , by (3.3) we have s p = t − p A − t p B B >
0, which impliesthat 0 < t < ( A B ) p − . Then (3.3) is equivalent to h ( t ) := A (cid:18) A Bt p − − B B (cid:19) − pp − B − B B B t p − − B A B = 0 , < t < (cid:18) A B (cid:19) p − . p > t → + h ( t ) = − B A B < , lim t → (cid:18) ( A B ) p − (cid:19) − h ( t ) = + ∞ . (3.4)Moreover, by (3.2), we have for any t ∈ (0 , ( A B ) p − ), h ′ ( t ) = 2(2 − p )(1 − p ) A A pB (cid:18) A Bt p − − B B (cid:19) − pp t − p − p − B − B B B t p − > , which and (3.4) imply that there exists a unique 0 < t < ( A B ) p − such that h ( t ) = 0.Let s = ( A t − p − B t p B ) p , so there exists a unique pair ( t , s ) ∈ R + × R + such that t u + + s u − ∈ M . Lemma 3.3. Ψ( u ) is bounded from below and coercive on M .Proof. For any u ∈ M , we have h Ψ ′ ( u ) , u i = 0, thenΨ( u ) = Ψ( u ) − p h Ψ ′ ( u ) , u i = (cid:18) − p (cid:19) k u k ≥ , which implies that Ψ( u ) is bounded from below and coercive on M .By Lemmas 3.1 and 3.3, define m := inf { Ψ( u ) | u ∈ M} , (3.5)then m > Lemma 3.4. m < c. Proof.
Let Q be a positive ground state solution of problem (1.15), i.e.Ψ ′ ( Q ) = 0 , Ψ( Q ) = ¯ c and Q ( x ) > x ∈ R N . Assume that ξ ∈ C ∞ ( R N ) is a cut-off function satisfying that supp ξ ∈ B (0), 0 ≤ ξ ≤ ξ ≡ B (0) and |∇ ξ | <
2. Set u R ( x ) := ξ ( xR ) Q ( x ) ≥ , v R,n ( x ) := − ξ ( x − x n R ) Q ( x ) ≤ , where R > x n = (0 , , · · · , , n ). It easily sees that for n large enough,supp u R ∩ supp v R,n = ∅ . n large there exists a unique pair ( t R , s R ) ∈ R + × R + suchthat t R u R + s R v R,n ∈ M , i.e. t R k u R k − t pR Z R N ( I α ∗ | u R | p ) | u R | p = s pR t pR Z R N ( I α ∗ | v R,n | p ) | u R | p ,s R k v R,n k − s pR Z R N ( I α ∗ | v R,n | p ) | v R,n | p = t pR s pR Z R N ( I α ∗ | u R | p ) | v R,n | p . (3.6)By the definition of u R and v R,n , we have u R → Q in H, v
R,n → − Q in H (3.7)as R → + ∞ . If lim R → + ∞ t R = + ∞ , then by p ≥ ≤ s pR t pR Z R N ( I α ∗ | v R,n | p ) | u R | p = 1 t p − R k u R k − Z R N ( I α ∗ | u R | p ) | u R | p → − Z R N ( I α ∗ | Q | p ) | Q | p < , (3.8)which is impossible. So, t R is uniformly bounded. Similarly, if lim R → + ∞ s R = + ∞ , thenby (3.6) again,0 ≤ t pR s pR Z R N ( I α ∗ | u R | p ) | v R,n | p = 1 s p − R k v R,n k − Z R N ( I α ∗ | v R,n | p ) | v R,n | p → − Z R N ( I α ∗ | Q | p ) | Q | p < , (3.9)which is a contradiction. So s R is also uniformly bounded. Up to a subsequence, wemay assume that there exist t , s ∈ [0 , + ∞ ) such that t R → t and s R → s as R → + ∞ . Moreover, by (3.7)-(3.9), we see that if t = 0 or s = 0, then (cid:18)Z R N ( I α ∗ | Q | p ) | Q | p (cid:19) = lim R → + ∞ (cid:18)Z R N ( I α ∗ | u R | p ) | v R,n | p (cid:19) = lim R → + ∞ (cid:18) k u R k t p − R − Z R N ( I α ∗ | u R | p ) | u R | p (cid:19) (cid:18) k v R,n k s p − R − Z R N ( I α ∗ | v R,n | p ) | v R,n | p (cid:19) = + ∞ , t , s ∈ (0 , + ∞ ). Then we conclude from (3.6) and (3.7) that t k Q k − t p Z R N ( I α ∗ | Q | p ) | Q | p = s p t p Z R N ( I α ∗ | Q | p ) | Q | p ,s k Q k − s p Z R N ( I α ∗ | Q | p ) | Q | p = t p s p Z R N ( I α ∗ | Q | p ) | Q | p . Since Ψ ′ ( Q ) = 0, we have 1 t p − − s p t p , s p − − t p s p . So 0 < t < < s < n large, by t R u R + s R v R,n ∈ M , we have m ≤ Ψ( t R u R + s R v R,n )= Ψ( t R u R + s R v R,n ) − p h Ψ ′ ( t R u R + s R v R,n ) , t R u R + s R v R,n i = (cid:18) − p (cid:19) ( k t R u R k + k s R v R,n k )= ( t + s ) (cid:18) − p (cid:19) k Q k + o R (1)= ( t + s )¯ c + o R (1) , where o R (1) → R → + ∞ . Therefore, by taking R → + ∞ , we get m ≤ ( t + s )¯ c < c. Proof of Theorem 1.3
Proof.
By Lemma 3.3 and the Ekeland variational principle, there exists a mini-mizing sequence { u n } ⊂ M such thatΨ( u n ) ≤ m + 1 n , (3.10)Ψ( v ) ≥ Ψ( u n ) − n k u n − v k , ∀ v ∈ M . (3.11)Then the sequences { u ± n } are uniformly bounded in H respectively. Passing to asubsequence, there exist u ± ∈ H such that u ± n ⇀ u ± in H. (3.12)17oreover, u + ≥ , u − ≤ u + · u − = 0 a.e. in R N . To prove this theorem, it isenough to show that Ψ ′ ( u n ) → . (3.13)Indeed, if (3.13) holds, set u := u + + u − , then (3.12) implies that Ψ ′ ( u ) = 0 . Moreover,by the compactness of the Sobolev embedding, we have Z R N ( I α ∗ | u n | p ) | u ± n | p → Z R N ( I α ∗ | u | p ) | u ± | p . Then it follows from u n ∈ M that k u ± n k → k u ± k . Hence u ± n → u ± in H. Therefore, u ∈ M and Ψ( u ) = m , i.e. u is a sign-changing solution of problem (1.15)with Ψ( u ) = m. Similarly to the proof of Lemma 2.6, we see thatΨ( u ) = m = inf { Ψ( v ) | v ± = 0 , Ψ ′ ( v ) = 0 } . By M ⊂ N , we must have ¯ c ≤ m = Ψ( u ). Since each ground state solution ofproblem (1.15) has constant sign, it follows that Ψ( u ) > ¯ c. Therefore, by Lemma 3.4,we have m = Ψ( u ) ∈ (¯ c, c ) . For any ϕ ∈ C ∞ ( R N ) and each n , we define the following two C -functions h n , g n : R → R as follows: h n ( δ, k, l ) = k ( u n + δϕ + ku + n + lu − n ) + k − Z R N ( I α ∗| ( u n + δϕ + ku + n + lu − n ) + | p ) | ( u n + δϕ + ku + n + lu − n ) + | p − Z R N ( I α ∗ | ( u n + δϕ + ku + n + lu − n ) − | p ) | ( u n + δϕ + ku + n + lu − n ) + | p ,g n ( δ, k, l ) = k ( u n + δϕ + ku + n + lu − n ) − k − Z R N ( I α ∗| ( u n + δϕ + ku + n + lu − n ) − | p ) | ( u n + δϕ + ku + n + lu − n ) − | p − Z R N ( I α ∗ | ( u n + δϕ + ku + n + lu − n ) + | p ) | ( u n + δϕ + ku + n + lu − n ) − | p . Then h n (0 , ,
0) = g n (0 , ,
0) = 0 . Moreover, ∂h n ∂k (0 , ,
0) = 2(1 − p ) B n, + (2 − p ) B n , ∂g n ∂l (0 , ,
0) = 2(1 − p ) B n, + (2 − p ) B n and ∂h n ∂l (0 , ,
0) = ∂g n ∂k (0 , ,
0) = − pB n , where B n, := Z R N ( I α ∗ | u + n | p ) | u + n | p > , B n, := Z R N ( I α ∗ | u − n | p ) | u − n | p > B n := Z R N ( I α ∗ | u − n | p ) | u + n | p > . Since Lemma 3.1 shows that B n < p B n, B n, ≤ B n, + B n, , we conclude from p ≥ T := (cid:12)(cid:12)(cid:12)(cid:12) ∂h n ∂k (0 , , ∂h n ∂l (0 , , ∂g n ∂k (0 , , ∂g n ∂l (0 , , (cid:12)(cid:12)(cid:12)(cid:12) = 4(1 − p ) B n, B n, + 2(1 − p )(2 − p )( B n, + B n, ) B n + 4(1 − p ) B n ≥ (cid:26) B n, B n, − B n ) > , if p = 2 , − p )(2 − p ) B n > , if p > . By the implicit function theorem, there exists a sequence { δ n } ⊂ R + and two func-tions k n ( δ ) ∈ C ( − δ n , δ n ), l n ( δ ) ∈ C ( − δ n , δ n ) satisfying k n (0) = 0, l n (0) = 0 and h n ( δ, k n ( δ ) , l n ( δ )) = 0 , g n ( δ, k n ( δ ) , l n ( δ )) = 0 , ∀ δ ∈ ( − δ n , δ n ) . (3.14)Set ϕ n,δ := u n + δϕ + k n ( δ ) u + n + l n ( δ ) u − n . Then (3.14) implies that ϕ n,δ ∈ M forall δ ∈ ( − δ n , δ n ). By (3.11) we haveΨ( ϕ n,δ ) − Ψ( u n ) ≥ − n k δϕ + k n ( δ ) u + n + l n ( δ ) u − n k . (3.15)By the Taylor Expansion, we see thatΨ( ϕ n,δ ) = Ψ( u n ) + δ h Ψ ′ ( u n ) , ϕ i + o ( k δϕ + k n ( δ ) u + n + l n ( δ ) u − n k ) , (3.16)where we have used the fact that h Ψ ′ ( u n ) , u ± n i = 0. Since { u n } is uniformly boundedin H and T >
0, we conclude that { k ′ n (0) } and { l ′ n (0) } are respectively uniformlybounded. Then o ( k δϕ + k n ( δ ) u + n + l n ( δ ) u − n k ) δ → δ → , which and (3.15), (3.16) show that |h Ψ ′ ( u n ) , ϕ i| ≤ Cn , as δ →
0, where