Nonlinear Analysis 182 (2019) 113–142
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Nonlinear Analysis www.elsevier.com/locate/na
Infinitely many sign-changing solutions for system of p-Laplace equations in RN ✩ Junfang Zhao a ,∗, Xiangqing Liu b , Jiaquan Liu c a b c
School of Science, China University of Geosciences, Beijing, 100083, PR China Department of Mathematics, Yunnan Normal University, Kunming 650500, PR China LMAM, School of Mathematics, Peking University, Beijing, 100871, PR China
article
info
Article history: Received 3 July 2018 Accepted 2 December 2018 Communicated by Enzo Mitidieri MSC: 35B05 35B45 58J70 Keywords: System of p-Laplacian equations Sign-changing solution Truncation method Concentration analysis
abstract In this paper we consider the system of p-Laplacian equations in RN
⎧ ⎨
−∆p uj + µj (x)|uj |p−2 uj =
k ∑
βij |ui |q |uj |q−2 uj
in RN ,
i=1
⎩
uj (x) → 0
as |x| → ∞, j = 1, . . . , k,
p , βij are constants, i, j = where 1 < p < N , max{2, p} < 2q < p∗ = NN−p 1, . . . , k, βjj > 0, j = 1, 2, . . . , k, βij ≤ 0, i ̸= j, i, j = 1, . . . , k; µj , j = 1, . . . , k are the potential functions and satisfy suitable decay assumptions. The existence of infinitely many sign-changing solutions is proved by the truncation method and by the concentration-compactness analysis on the approximating solutions. © 2018 Elsevier Ltd. All rights reserved.
1. Introduction In this paper we consider the following system of p-Laplacian equations in RN ⎧ k ∑ ⎪ ⎨ − ∆ u + µ (x)|u |p−2 u = βij |ui |q |uj |q−2 uj in RN , p j j j j i=1 ⎪ ⎩ uj (x) → 0 as |x| → ∞, j = 1, . . . , k,
(P)
where ∆p is the p-Laplacian operator, ∆p u = div(|∇u|p−2 ∇u). We assume (N) 1 < p < N , max{2, p} < 2q < p∗ . ✩ The first author was supported by the NSFC, China 11601493 and Fundamental Research Funds for the Central Universities, China 2652015194; the second author was supported by NSFC, China 11761082 and Yunnan Province, Young Academic and Technical Leaders Program, China 2015HB028; the third author was supported by NSFC, China 11271331. ∗ Corresponding author. E-mail address:
[email protected] (J. Zhao).
https://doi.org/10.1016/j.na.2018.12.005 0362-546X/© 2018 Elsevier Ltd. All rights reserved.
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
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(B) βij , i, j = 1, . . . , k are constants, βjj > 0, j = 1, . . . , k, βij ≤ 0, i ̸= j, i, j = 1, . . . , k. (Λ) µj ∈ C 1 (RN , R), j = 1, . . . , k are the potential functions satisfying (Λ1 ) There exist µ, µ such that 0 < µ ≤ µj (x) ≤ µ, x ∈ RN , j = 1, . . . , k. (Λ2 ) Let
∂ µj (x), ∂r x x ∂ = · ∇ is the derivative along the direction , where ∂r |x| |x| m(x) = min
1≤j≤k
lim m(x)eν|x| = +∞ for any ν > 0.
|x|→∞
(Λ3 ) There exists a constant c > 1 such that |∇µj (x)| ≤ c
∂ µj (x), x ∈ RN , |x| ≥ c, j = 1, . . . , k. ∂r
Theorem 1.1 (The Existence Theorem). Assume (N ), (B) and (Λ). Then the problem (P) has infinitely many sign-changing solutions. We call a solution U = (u1 , . . . , uk ) sign-changing, if all the components uj , j = 1, . . . , k are sign-changing. The system (P ), in particular with N = 2, 3, p = 2, q = 2, has applications in many physical problems such as in nonlinear optics and in multi-species Bose–Einstein condensates [9,12,22,24,29]. Physically, βjj and βij (i ̸= j) are the intraspecies and interspecies scattering lengths respectively. The sign of the scattering length determines whether the interactions of states are repulsive or attractive. In the repulsive case (βij < 0) the components tend to segregate with each other leading to phase separations. These phenomena have been documented in experiments as well as in numeric simulations [9,21,27]. Mathematical work has been done extensively in recent years [2,4,5,10,14,17,20,23,28,31]. In this paper, we consider the problem (P ) with general p. The problem (P) has a variational structure, given by the functional I(U ) =
1 p
∫ RN
∫ k k ∑ ∑ 1 (|∇uj |p + µj (x)|uj |p ) dx − βij |ui |q |uj |q dx 2q N R j=1 i,j=1
(1.1)
for U = (u1 , . . . , uk ), defined on W = W 1,p (RN ) × · · · × W 1,p (RN ), the k- fold product of W 1,p (RN ). The embedding W 1,p (RN ) ↪→ L2q (RN ), p < 2q < p∗ is continuous, but not compact due to the shifts, therefore the Palais–Smale condition is not satisfied by the functional I, and the problem lacks necessary compactness. Since the pioneering work [6] of H. Brezis and L. Nirenberg, significant progress has been made for this kind of problems without compactness in recent decades. In particular, the authors of [11] deal with the following equation with critical growth { ∗ − ∆u = |u|2 −2 u + λu, in Ω , (Pα ) u = 0 on ∂Ω , where Ω is an open regular domain in RN (N ≥ 3) and 2∗ = N2N −2 is the critical exponent, and prove the existence of infinitely many solutions. The solutions are found as limits of solutions of approximated problems with subcritical growth. The lack of compactness due to the dilations does not allow to deduce that a sequence of approximating solutions must have a convergent subsequence but the fact that they solve
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the approximated problems gives, with the use of a local Pohoˇzaev identity, some extra estimates which lead to a proof of the desired compactness. Later this argument was adapted to subcritical problems on the whole space RN [8] and to quasilinear problems with critical growth [7,13]. As for the existence of multiple solutions of the original problem, we need to check that the multiple solutions of the approximated problems do not converge all to the same solution of the limit problem. In both [8] and [11] some estimates on the Morse index are employed, which has been used as one of the possible devices to distinguish the limits of the multiple approximating solutions by their original variational characterization. For the p-Laplacian equations with p ̸= 2 we have no information on the Morse index, therefore the approaches in [8,11] cannot be extended in a straightforward way to problems involving the p-Laplacian operator with p ̸= 2. In this paper, along the lines of [19], we are going to use the truncation technique. We first consider the truncated problem, solutions of which will be used as approximating solutions. By a concentration analysis as that in [7,8,11] in particular with the use of local Pohoˇzaev identity, theorem of convergence and uniform bound on approximating solutions are proved. The truncated method has the advantage over other approaches that if U = (u1 , u2 , . . . , uk ) is a solution of the approximated problem (Pλ ) (see the problem (Pλ ) below) and satisfies the estimate 1 −λ√1+|x|2 e , x ∈ RN , j = 1, . . . , k, (1.2) λ then U will be a solution of the original problem (P ) too, that is the original problem (P ) and the approximated problem (Pλ ), λ > 0 may share common solutions. It turns out that as λ → 0, more and more solutions of the problem (Pλ ), λ > 0 satisfy the above estimate (1.2). In this way we obtain infinitely many solutions of the original problem. The truncation technique is not new. People use this technique to deal with various problems, because of ∗ ∗ various reasons. For example, in equation (Pa ) with 0 < λ < λ1 , one replaces the term |u|2 −2 by |u|2+ −1 , in order to obtain a positive solution. And when one considers the quasi-linear equation |uj (x)| ≤
1 ∆u + u∆u2 − V (x)u + |u|2 u 2
in RN , p ∈ (4,
4N ) N −2
in [18], one replaces the quasi-linear term u∆u2 by uM ∆(uuM ) if |u| ≤ M , uM = M if u ≥ M . See also [25,26] for Hamiltonian systems and for nonlinear wave equations. But up to the authors’ knowledge, the application of the truncation technique to a problem on the unbounded domains by introducing a decay √ 1 −λ 1+|x|2 in order to overcome the difficulty of lack of compactness seems to be new. function, such as λ e Let us describe the truncation method in more detail. Let ψ ∈ C0∞ (R, [0, 1]) be such that ψ(s) = 1 for |s| ≤ 1; ψ(s) = 0 for |s| ≥ 2, ψ is even and decreasing in [1, 2]. For λ > 0, (x, s) ∈ RN × R define √ 2 bλ (x, s) = ψ(λeλ 1+|x| s), ∫ s mλ (x, s) = bλ (x, τ )dτ, 0 ( )pα 1 s gλ (x, s) = |s|p , p mλ (x, s) where α ∈ (0, 1) is to be chosen (see (3.5)). For x = 0, we simply use the notations ∫ s bλ (s) = bλ (0, s) = ψ(λeλ s), mλ (s) = mλ (0, s) = bλ (τ )dτ, 0
and define fλ (s) =
1 2q
(
s mλ (s)
)2qα
|s|2q .
116
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
We consider the truncated problem ⎧ k ∑ ⎪ d ⎪ ⎨ − ∆p uj + µj (x) ∂ gλ (x, uj ) = βij |ui |q |uj |q−2 uj + βjj fλ (uj ), ∂s ds i=1,i̸=j ⎪ ⎪ ⎩ uj (x) → 0 as |x| → ∞, j = 1, . . . , k .
in RN ,
(Pλ )
The corresponding functional is 1 Iλ (U ) = p
k ∑
∫
|∇uj | dx +
RN j=1
µj (x)gλ (x, uj ) dx
RN j=1 k ∑
∫
1 − 2q
k ∑
∫
p
q
q
∫
βij |ui | |uj | −
RN i,j=1,i̸=j
k ∑
(1.3) βjj fλ (uj )dx,
RN j=1
for U = (u1 , . . . , uk ), defined on the space Wλ = Wλ1,p (RN ) × · · · × Wλ1,p (RN ), where the space Wλ1,p (RN ) is defined as ∫ √ 2 Wλ1,p (RN ) = {u|u ∈ W 1,p (RN ), (λeλ 1+|x| )pα |u|p(1+α) dx < +∞}. RN
Comparing the two functionals I and Iλ , we find that we have replaced the terms |uj |p and |uj |2q by gλ (x, uj ) and fλ (uj ), respectively. Notice that ( )pα 1 s c ( λ√1+|x|2 )pα p(1+α) gλ (x, s) = |s|p ∼ λe |s| , as |s| → ∞. p mλ (x, s) p √ 2 The system (Pλ ) has a coercive potential function eλpα 1+|x| , hence the functional Iλ , λ > 0 satisfies the Palais–Smale condition (see Lemma 3.3), while the functional I does not. On the other hand, the nonlinear term ( )2qα 1 s cλ 2q(1+α) fλ (s) = |s|2q ∼ |s| , as |s| → ∞. 2q mλ (s) 2q k
cλ ∑ βjj |uj |2q(1+α) , as |s| → ∞. This fact makes the 2q j=1 functional Iλ a “pure” super p-linear one in some sense, which is very useful when we construct the pseudogradient vector field (see Lemma 3.7, in particular the formulas (3.18) and (3.19)). We use ∥ · ∥λ to denote the norm of Wλ . For U = (u1 , . . . , uk ) ∈ Wλ , The term of highest order of the functional Iλ is
∥U ∥λ = ∥U ∥ + |U |p,λ , where
⎞ p1 k ∑ (|∇uj |p + |uj |p )dx⎠ ,
⎛ ∫ ∥U ∥ = ⎝
|U |p,λ
RN j=1
⎛ ∫ =⎝
RN
k ∑
√ (λeλ
⎞ 1+|x|2 pα
) |uj |p(1+α) dx⎠
1 p(1+α)
.
j=1
Notice that the space Wλ1,p , and hence Wλ is a Banach space with a weighted Lp -norm. We also use similar notations for a scalar function u: ∥u∥λ = ∥u∥W 1,p (RN ) = ∥u∥ + |u|p,λ , λ
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where (∫ ∥u∥ = ∥u∥W 1,p (RN ) = √
(∫ |u|p,λ =
(λe
λ
p
p
) p1
(|∇u| + |u| )dx
,
RN
1+|x|2 pα
) |u|
p(1+α)
) 1 p(1+α) dx
RN
∂ If U = (u1 , . . . , uk ) is a solution of (Pλ ), λ > 0 and satisfies the estimate (1.2), then gλ (x, uj ) = ∂s d |uj |p−2 uj , fλ (uj ) = |uj |2q−2 uj and U will be a solution of the original problem (P ) too, as we already ds mentioned. Theorem 1.2 (The Uniform Bound Theorem).Assume (N ), (B) and (Λ) hold. Then given M > 0 there exists ν = ν(M ) such that if U = (u1 , . . . , uk ) ∈ Wλ is a solution of the problem (Pλ ), λ > 0 and ∥U ∥ ≤ M , then it holds √ 1 2 |uj (x)| ≤ e−ν 1+|x| , x ∈ RN , j = 1, . . . , k. ν If, in addition, λ < ν, then U is a solution of the problem (P).
The paper is organized as follows. In Section 2 we develop necessary concentration-compactness analysis and prove Theorem 1.2. In Section 3 we construct a sequence of critical values of the functional Iλ , λ > 0 by using the method of multiple invariant sets of descending flow. Please refer to [15,16]. In Section 4, using the results from Sections 2 and 3, we finally prove Theorem 1.1. 2. Uniform bounds First we collect the properties of the auxiliary functions in the following lemma. Lemma 2.1.
For (x, s) ∈ RN × R, it holds that
mλ (x, s) ≤ 1, s √ √ √ 1 2 2 2 2 mλ (x, s) = s for |s| ≤ λ1 e−λ 1+|x| and min{|s|, e−λ 1+|x| } ≤ |mλ (x, s)| ≤ min{|s|, e−λ 1+|x| } . λ λ x ∇x mλ (x, s) = −λ √ (mλ (x, s) − sbλ (x, s)). 1 + |x|2 ∂2 p(p − 1)gλ (x, s) ≤ s2 2 gλ (x, s) ≤ cgλ (x, s), c > 0 is a constant. ∂s Let s1 , s2 ∈ R, if 1 < p < p(1 + α) < 2, then
(1) 0 ≤ bλ (x, s) ≤ (2) (3) (4) (5)
) ⎧( ∂ ∂ ⎪ ⎪ g (x, s ) − g (x, s ) (s2 − s1 ) ⎪ λ 2 λ 1 ⎪ ∂s ∂s ⎪ ⎪ ⎪ ⎪ √ ⎪ (s2 − s1 )2 (s2 − s1 )2 ⎪ λ 1+|x|2 pα ⎨ ≥c + c(λe ) , |s1 |2−p + |s2 |2−p |s1 |2−p(1+α) + |s2 |p(1+α) ⏐ ⏐ ⎪⏐∂ ⏐ ⎪ ⎪ ⏐ gλ (x, s2 ) − ∂ gλ (x, s1 )⏐ ⎪ ⎪ ⏐ ⏐ ⎪ ⎪ ∂s ∂s ⎪ ⎪ √ ⎪ 2 ⎩ ≤ c|s2 − s1 |p−1 + c(λeλ 1+|x| )pα |s2 − s1 |p(1+α)−1 ,
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If 2 < p < p(1 + α), then ) ⎧( ∂ ∂ ⎪ ⎪ gλ (x, s2 ) − gλ (x, s1 ) (s2 − s1 ) ⎪ ⎪ ∂s ∂s ⎪ ⎪ √ ⎪ ⎪ 2 ⎨ ≥ c|s2 − s1 |p + c(λeλ 1+|x| )pα |s2 − s1 |p(1+α) , ⏐ ⏐ ⏐∂ ⏐ ⎪ ⎪ ⏐ gλ (x, s2 ) − ∂ gλ (x, s1 )⏐ ⎪ ⎪ ⏐ ⏐ ⎪ ∂s ∂s ⎪ ⎪ √ ⎪ ⎩ 2 ≤ c(|s1 |p−2 + |s2 |p−2 )|s2 − s1 | + c(λeλ 1+|x| )pα (|s1 |p(1+α)−1 + |s2 |p(1+α)−1 )|s2 − s1 |. Proof . (1), (2) and (3) are easy to prove, we omit the proof. We consider (4) and (5). (4) By straightforward calculus we have ( )pα ( ) s sbλ (x, s) ∂ gλ (x, s) = |s|p−2 s 1 + α − α , ∂s mλ (x, s) mλ (x, s) ( )pα s ∂2 g (x, s) = (p − 1) |s|p−2 λ ∂s2 mλ (x, s) )pα ( )( ) ( sbλ (x, s) sbλ (x, s) s p−2 |s| 1− 2p − 2 + (pα + 1)(1 − ) +α mλ (x, s) mλ (x, s) mλ (x, s) ( )pα s2 ∂ bλ (x, s) s −α |s|p−2 ∂s , mλ (x, s) mλ (x, s) ( )( ( )) 2 sbλ (x, s) sbλ (x, s) 2 ∂ s gλ (x, s) = pαgλ (x, s) 1 − 2p − 2 + (pα + 1) 1 − ∂s2 mλ (x, s) mλ (x, s) ∂ s2 ∂s bλ (x, s) + p(p − 1)gλ (x, s) − pαgλ (x, s) . mλ (x, s) √ 2 Let σ = λeλ 1+|x| s, then √ √ √ 2 2 2 ∂ ∂ s2 ∂s s2 ∂s bλ (x, s) ψ(λeλ 1+|x| s) (λeλ 1+|x| s)2 ψ ′ (λe 1+|x| s) σ 2 ψ ′ (σ) ∫σ √ √ = ∫s = = . λ 1+|x|2 mλ (x, s) ψ(τ )dτ ∫ λeλ 1+|x|2 s τ )dτ ψ(λe 0 0 ψ(τ )dτ 0 Notice that ψ ′ (σ) = 0 for |σ| ≤ 1 or |σ| ≥ 2 and σψ ′ (σ) ≤ 0, we have 0≤−
∂ bλ (x, s) s2 ∂s ≤ c. mλ (x, s)
sbλ (x, s) Also we have 0 ≤ 1 − ≤ 1. Now (4) follows from (2.1). mλ (x, s) ( ) ∂ ∂ (5) We estimate gλ (x, s2 ) − gλ (x, s1 ) (s2 − s1 ). ∂s ∂s ( ) ∂ ∂ gλ (x, s2 ) − gλ (x, s1 ) (s2 − s1 ) ∂s ∂s ∫ 1 d ∂ = gλ (x, τ s2 + (1 − τ )s1 )(s2 − s1 )dτ dτ ∂s 0 ∫ 1 2 ∂ = g (x, τ s2 + (1 − τ )s1 )(s2 − s1 )2 dτ 2 λ ∂s 0 ∫ 1 gλ (x, τ s2 + (1 − τ )s1 ) ≥c (s2 − s1 )2 dτ 2 |τ s + (1 − τ )s | 2 1 0 ∫ 1( √ ) 2 ≥c |τ s2 + (1 − τ )s1 |p−2 + (λeλ 1+|x| )pα |τ s2 + (1 − τ )s1 |p(1−α)−2 (s2 − s1 )2 dτ 0
(2.1)
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∫
119
1
d (|τ s2 + (1 − τ )s1 |p−2 (τ s2 + (1 − τ )s1 )(s2 − s1 ))dτ 0 dτ ∫ 1 √ d 2 +c2 (λeλ 1+|x| )pα (|τ s2 + (1 − τ )s1 |p(1+α)−2 (τ s2 + (1 − τ )s1 )(s2 − s1 ))dτ dτ 0 √ ( ) ( p−2 ) 2 p(1+α)−2 s1 (s2 − s1 ). = c1 |s2 | s2 − |s1 |p−2 s1 (s2 − s1 ) + c2 (λeλ 1+|x| )pα |s2 |p(1+α)−2 s2 − s1
= c1
Now we apply the following elementary inequalities to complete the proof. For 1 < p < 2 and ξ, η ∈ RN ⎧ |ξ − η|2 ⎨ (|ξ|p−2 ξ − |η|p−2 η, ξ − η) ≥ c , |ξ|2−p + |η|2−p (2.2) ⏐ ⎩ ⏐⏐ p−2 p−1 p−2 ⏐ |ξ| ξ − |η| η ≤ c|ξ − η| . For p ≥ 2 and ξ, η ∈ RN {
(|ξ|p−2 ξ − |η|p−2 η, ξ − η) ≥ c|ξ − η|p ,
(2.3)
||ξ|p−2 ξ − |η|p−2 η| ≤ c(|ξ|p−2 + |η|p−2 )|ξ − η|. ⏐∂ ⏐ ∂ ⏐ ⏐ In a similar way we can estimate ⏐ gλ (x, s2 ) − gλ (x, s1 )⏐. ■ ∂s ∂s In this section we always assume λn → 0, Un ∈ Wλn , DIλn (Un ) = 0 and ∥Un ∥λn ≤ M.
By Theorem 3.3 in [1] (see also [30]) we have the following profile decomposition for a renumbered subsequence ∑ Un = Vl (· − xn,l ) + Rn , (2.4) l∈Λ N
where Vl , Rn ∈ W, xn,l ∈ R , Λ ⊂ N an index set, satisfying (1) xn,1 = 0, |xn,l − xn,m | → +∞ as n → ∞, l ̸= m. (2) V1 = U = w − lim Un , Vl = w − lim Un (· + xn,l ) in W . n→+∞
n→+∞
(3) Rn → 0 in Ls (RN ) as n → ∞, p < s < p∗ . ∑ (4) l∈Λ |Vl |ss ≤ limn→∞ |Un |ss , p < s < p∗ , where | · |s is the norm in Ls (RN ). ˜n = Lemma 2.2. Assume λn ≥ 0, λn → 0, xn ∈ RN , Un ∈ Wλn , DIλn (Un ) = 0, ∥Un ∥λn ≤ M, U Un (· + xn ) ⇀ V ̸= 0 in W . Then √ 2 (1) limn→∞ λn eλn 1+|xn | < +∞. 1,p N ˜n → V in W (R ). (2) U loc (3) Let V = (v1 , . . . , vk ), zj = vj , j = 1, . . . , k. Then zj satisfies the differential inequality ∫ ∫ (|∇z|p−2 ∇z∇φ + µz p−1 φ) dx ≤ β z 2q−1 φ dx, for φ ≥ 0, φ ∈ W 1,p (RN ), (2.5) RN
RN
where µj (x) ≥ µ, x ∈ RN , j = 1, . . . , k, and β = max{β11 , . . . , βkk }. Proof . For any R > 0 we have √ { } 2 2(x, xn ) + |x|2 λn eλn 1+|x+xn | √ √ = exp λn √ → 1, 2 1 + |xn |2 + 1 + |x + xn |2 λn eλn 1+|xn |
as
n→∞
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120
∫ uniformly in x ∈ BR = BR (0). Choose L > 0 such that B (0) |V |p(1+α) dx > 0. By the assumption L ∫ √ 2 p(1+α) c ≥ ∥Un ∥λn ≥ (λn eλn 1+|x| )pα |Un |p(1+α) dx N R ∫ ∫ √ √ 2 2 ≥ (λn eλn 1+|x| )pα |Un |p(1+α) dx = (λn eλn 1+|x+xn | )pα |Un (x + xn )|p(1+α) dx BL (xn ) BL (0) ∫ √ λn 1+|xn |2 pα p(1+α) ) ∼ (λn e |V | dx. √
BL (0) 2
Therefore limn→∞ λn eλn 1+|xn | < +∞. ˜n = (˜ ˜n satisfies the system (2) Let U un,1 , . . . , u ˜n,k ), U ∫ ∫ ∂ µj (x + xn ) gλn (x + xn , u |∇˜ un,j |p−2 ∇˜ un,j ∇φdx + ˜n,j )φdx ∂s RN RN ∫ ∫ k ∑ d q q−2 βij |˜ un,i | |˜ un,j | u ˜n,j φ dx + βjj fλn (˜ = unj )φ dx, ds N N R R
(2.6)
i=1,i̸=j
j = 1, . . . , k, φ ∈ Wλ1,p (RN ). n ˜n converges in Ls (RN ), 1 ≤ s < p∗ , we have Let φ ∈ C0∞ (RN ), φ ≥ 0, suppφ ⊂ BR (0). Since U loc ∫ (|∇˜ un,j |p−2 ∇˜ un,j − |∇˜ um,j |p−2 ∇˜ um,j , ∇˜ un,j − ∇˜ um,j )φ dx RN ∫ =− (|∇˜ un,j |p−2 ∇˜ un,j − |∇˜ um,j |p−2 ∇˜ um,j , ∇φ)(˜ un,j − u ˜m,j ) dx RN ( ) ∫ ∂ ∂ − µj (x + xn ) gλn (x + xn , u ˜n,j ) − µj (x + xm ) gλm (x + xm , u ˜m,j ) (˜ un,j − u ˜m,j )φdx ∂s ∂s RN ∫ k ∑ βij (|˜ un,i |q |˜ un,j |q−2 u ˜n,j − |˜ um,i |q |˜ um,j |q−2 u ˜m,j )(˜ un,j − u ˜m,j )φdx + RN i=1,i̸=j
( ) d d + βjj fλn (˜ un,j ) − fλm (˜ um,j ) (˜ un,j − u ˜m,j )φdx ds ds N ∫R ( ) ≤c |∇˜ un,j |p−1 + |∇˜ um,j |p−1 |˜ un,j − u ˜m,j ||∇φ|dx N R∫ ( ) +c |˜ un,j |p−1 + |˜ un,j − u ˜m,j |dx um,j |p−1 |˜ N ∫R √ 2 +c (λn eλn 1+|x+xn | )pα (|˜ un,j |p(1+α)−1 + |˜ um,j |p(1+α)−1 )|˜ un,j − u ˜m,j |φdx N R ∫ +c (|˜ un,j |2q−1 + |˜ um,j |2q−1 )|˜ un,j − u ˜m,j |φdx ∫
RN
∫ +c RN
∫ +c RN
(|˜ un,j |2q−1 + (λn eλn )pα |˜ un,j |2q(1+α)−1 + |˜ um,j |2q−1 )|˜ un,j − u ˜m,j |φdx (λm eλm )pα |˜ um,j |2q(1+α)−1 |˜ un,j − u ˜m,j |φdx
(2.7)
≤ c|˜ un,j − u ˜m,j |Lp (BR ) + c|˜ un,j − u ˜m,j |Lp(1+α) (BR ) + c|˜ un,j − u ˜m,j |L2q (BR ) + c|˜ un,j − u ˜m,j |L2q(1+α) (BR ) → 0, ˜n converges in W 1,p (RN ). as n, m → ∞. Now it follows from (2.2), (2.3) that U loc ˜n,j ˜n satisfies the system (2.6). Let φ ≥ 0, φ ∈ C ∞ (RN ). Take φε = φ √ u (3) U ∈ Wλ1,p (RN ), ε > 0 0 n 2 2 ε +u ˜n,j as the test function in (2.6) and let ε → 0.
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
Notice that
121
∂ gλ (x, s) ≥ sp−1 for s > 0 and βij ≤ 0 for i ̸= j, we obtain ∂s ∫ (∇|˜ un,j |p−2 ∇|˜ un,j |∇φ + µ|˜ un,j |p−1 φ)dx RN ∫ ≤ β(|˜ un,j |2q−1 + (λn eλn )pα |˜ un,j |2q(1+α)−1 )φdx.
(2.8)
RN 1,p ˜n → V in W 1,p (RN ), |˜ Since U un,j | → |vj | = zj in Wloc (RN ). Let n → ∞ in (2.8), we obtain the desired loc ∞ N result for φ ≥ 0, φ ∈ C0 (R ). By a density argument we complete the proof. ■
The index set Λ in the profile decomposition (2.4) is finite.
Corollary 2.1.
Proof . By (2.3) and the Sobolev embedding theorem ∫ ∫ ∫ ( ( ) 2q 2q p p p ≤c zj dx ≤ c (|∇zj | + zj ) dx RN
Therefore
∫ RN
and
∫
RN
RN
RN
|V |2q dx ≤ c
(
∫
|V |2q dx
) 2q zj2q dx p .
) 2q p
RN
2q
|V | dx ≥ m > 0 for some m > 0. By (4) of the profile decomposition (2.4) Λ is finite.
■
Proposition 2.1. Assume z solves the differential inequality (2.5), then z decays exponentially, i.e. there exists ν > 0 such that √ 1 2 |z(x)| ≤ e−ν 1+|x| , x ∈ RN . ν The proof of this proposition is somewhat well-known [8,18,19] and will be given in the Appendix. Remark 2.1. Assume λn ≥ 0, Un ∈ W , DIλn (Un ) = 0 and ∥Un ∥λn ≤ M . Then the components of Un satisfy the differential inequality (2.8). By Proposition 2.1, with 2q replaced by 2q(1 + α), there exists a constant ν > 0 such that √ 1 2 |Un (x)| ≤ e−ν 1+|x| , for x ∈ RN . (2.9) ν Moreover if Un converges in Ls (RN ), p < s < p∗ , then the constant ν in the above estimate is independent of n. In fact in this case all the constants appearing in the proof of Proposition 2.1 are independent of n. (n)
Lemma 2.3. Assume the profile decomposition (2.4) holds. Denote ΩR = RN \ exists a constant ν > 0 such that for Un = (un,1 , . . . , un,j ) k ∑ 1 1 (|∇un,j |p + |un,j |p ) dx ≤ e−νR , |Un (x)| ≤ e−νR (n) ν ν Ω j=1
∫
⋃
l∈Λ
BR (xn,l ). Then there
(n)
for x ∈ ΩR .
R
We have the following result, a Pohozaev-type identity (see [8]). Lemma 2.4.
Let U = (u1 , . . . , uk ) ∈ Wλ satisfy the system ⎧ ⎨ − ∆p uj = ∂ Fλ (x, U ) in RN , ∂uj ⎩ uj (x) → 0, as |x| → ∞, j = 1, . . . , k.
(2.10)
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
122
Then for all t = (t1 , . . . , tk ) ∈ RN , φ ∈ C0∞ (RN ), we have ∫ − (t, ∇x Fλ (x, U ))φdx RN
∫ ∫ k k ∑ ∑ 1 |∇uj |k (t, ∇φ)dx + |∇uj |p−2 (t, ∇uj )(∇uj , ∇φ)dx p RN j=1 N R j=1 ∫ + Fλ (t, U )(t, ∇φ)dx,
=−
(2.11)
RN ∂F ∂F ). For our problem (Pλ ), , . . . , ∂x where ∇x F = ( ∂x n 1 k
Fλ (x, U ) = −
1∑ uj 1 µj (x)( )pα |uj |p + p j=1 mλ (x, uj ) 2q
k ∑
βij |ui |q |uj |q +
i,j=1,i̸=j
k 1 ∑ uj βjj ( )2qα |uj |2q . 2q j=1 mλ (uj )
Proof . Multiplying the equations in the system (2.10) by (t, ∇uj )φ and integrating by parts, we obtain (2.11). In fact, ∫ −∆p uj (t, ∇uj )φ dx RN ∫ ∫ p−2 = |∇uj | ∇uj ∇(t, ∇uj )φ dx + |∇uj |p−2 (t, ∇uj )(∇uj , ∇φ) dx N N R∫ R ∫ (2.12) 1 = (∇(|∇uj |p ), t)φ dx + |∇uj |p−2 (t, ∇uj )(∇uj , ∇φ) dx p RN N ∫ ∫R 1 =− |∇uj |p (t, ∇φ) dx + |∇uj |p−2 (t, ∇uj )(∇uj , ∇φ) dx p RN RN and
k ∑ ∂ Fλ (x, U )(t, ∇uj )φ dx ∂u j j=1
∫ RN
∫ = N R∫
=−
RN
where ∇F = ∇x F +
(2.13)
((t, ∇Fλ (x, U )) − ∇x Fλ (x, U ))φ dx ∫ Fλ (x, U )(t, ∇φ) dx − (t, ∇x Fλ (x, U ))φ dx, RN
k ∑ ∂F ∇uj . Then (2.11) follows from (2.12), (2.13) and the system. ∂u j j=1
■
Proposition 2.2. Let λn ≥ 0, Un ∈ W, DIλn (Un ) = 0, ∥Un ∥λn ≤ M, Un ⇀ U in W . Then Un converges to U in Ls (RN ), p < s < p∗ . Proof . Now assume λn ≥ 0, Un ∈ W, DIλn (Un ) = 0, ∥U ⏐ n ∥λn ≤ M , and the decomposition (2.4) holds. Without loss of generality we assume |xn,2 | = min{|xn,l | ⏐l ∈ Λ, l ≥ 2}. Denote xn = xn,2 . According to [8], we construct a sequence of cones Cn having the vertex 21 xn and generated by the ball Bˆ (xn ): rn Cn = {w ∈ RN |w = where rˆn satisfies
1 1 xn + λ(x − xn ), x ∈ Bˆ (xn ), λ ∈ [0, ∞)}, rn 2 2
γ |xn | |xn | 1 · = rn ≤ rˆn ≤ l0 rn = γ , γ= l0 2 2 5(c + 1)
and c is the constant in the condition (Λ3 ), l0 is the cardinal number of the index set Λ.
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123
The cone Cn has the following property ⋃
∂Cn ∩
B 1 rn (xn,l ) = ∅.
(2.14)
2
l∈Λ
We now apply the identity (2.11). Take U = Un = (un,1 , . . . , un,k ), t = tn = |xxnn | and φ = χφR , where χ, φR ∈ C0∞ (RN ) such that χ(x) = 0 for x ̸∈ Cn , χ(x) = 1 for x ∈ Cn and dist(x, ∂Cn ) ≥ 1; φR (x) = 1 for |x| ≤ R, φR (x) = 0 for |x| ≥ 2R. Let R → ∞ we obtain ∫ ∫ ∑ k 1 − (tn , ∇x Fλ (x, Un ))χ dx = − |∇un,j |p (tn , ∇χ) dx p RN j=1 RN ∫ + RN
k ∑
|∇un,j |p−2 (tn , ∇un,j )(∇un,j , ∇χ) dx
(2.15)
j=1
∫ Fλ (x, Un )(tn , ∇χ) dx.
+ RN
(n)
By (2.14) and the definition of χ, the support of ∇χ is contained in the domain ΩR with R = 21 rn − 1. By Lemma 2.3 the right hand side of (2.13) decays exponentially, say RHS of (2.15) ≤ ce−ν|xn | .
(2.16)
We estimate the left hand side of (2.15). According to Lemma 4.2 of [8] it holds that for x ∈ Cn x 1−γ x x (tn , )≥ , τn = tn − (tn , ) , |τn | ≤ γ, |x| 1+γ |x| |x| x 1−γ 1 ) − c|τn | ≥ − cγ ≥ . (tn , |x| 1+γ 2 By Lemma 2.1 and the assumptions (Λ), one has (tn , x) ≥ 0, ( ) x ∂ (tn , ∇µj (x)) = tn , µj (x) + (τn , ∇µj (x)) |x| ∂r ( ) x ∂ ≥ (tn , − c|τn |) µj (x) |x| ∂r 1 ∂ 1 ≥ µj (x) ≥ m(x). 2 ∂r 2 ( )pα k 1∑ uj ∇x Fx (x, U ) = − ∇µj (x) |uj |p p j=1 mλ (x, uj ) )pα ( k ∑ uj ∇x mλ (x, uj ) +α µj (x) |uj |p mλ (x, uj ) mλ (x, uj ) j=1 ( ) k pα 1∑ uj =− ∇µj (x) |uj |p p j=1 mλ (x, uj ) ( )pα ( ) k ∑ αλx uj uj bλ (x, uj ) p −√ µj (x) |uj | 1 − mλ (x, uj ) mλ (x, uj ) 1 + |x|2 j=1 and
k
(−tn , ∇x Fx (x, U )) ≥ k ∑ 1 ≥ m(x) |uj |p . 2p j=1
1∑ ⟨tn , ∇µj (x)⟩ p j=1
(
uj mλ (x, uj )
)pα
(2.17)
(2.18)
|uj |p (2.19)
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J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
∫
|V2 |p dx = β > 0. Then
Choose L > 0 such that BL (0)
∫ (−tn , ∇x Fx (x, U ))χdx
LHS of (2.15) = ∫RN
1 m(x)|Un |p dx 2p BL (xn ) ∫ 1 ≥ inf m(x) |Un |p dx 2p BL (xn ) ∫BL (xn ) 1 ˜n |p dx inf m(x) = |U 2p BL (xn ) ∫BL (0) 1 inf m(x) |V2 |p dx. ∼ 2p BL (xn ) BL (0) ≥
(2.20)
Now it follows from (2.19), (2.20) that 1 β inf m(x) ≤ ce−ν|xn | , 4p BL (xn ) which contradicts the assumption (Λ). Then Λ = {1} and by the property (3) of the decomposition (2.4) we have proved the proposition. ■ Proof of Theorem 1.2. Assume λn ≥ 0, Un = (un,1 , . . . , un,k ) ∈ Wλn , DIλn (Un ) = 0, ∥Un ∥λn ≤ M . Without loss of generality, we assume Un weakly converges in W . By Proposition 2.2, Un strongly converges in Ls (RN ), p < s < p∗ . By Remark 2.1 there exists a constant ν > 0 such that |un,j (x)| ≤
1 −ν √1+|x|2 , e ν
for x ∈ RN , j = 1, . . . , k.
■
3. Solutions of the perturbed systems In this section we consider the perturbed system (Pλ ), λ > 0. The method of multiple invariant sets of descending flow developed in [15,16] is employed to obtain multiple sign-changing solutions of the perturbed problem (Pλ ), which will be used as approximating solutions of the original problem (P ). First we have the following abstract critical point theorem (Theorem 3.1), which is essentially a consequence of known results in a more general setting (see [16, Theorem 4.1]). For readers’ convenience we sketch the proof of Theorem 3.1 in Appendix A. Let X be a Banach space, f be an even C 1 -functional on X. Let Pj , Qj , j = 1, . . . , k be a family of open ⋃k ⋂k convex sets of X, Qj = −Pj . Set M = j=1 (Pj ∪ Qj ), Σ = j=1 (∂Pj ∩ ∂Qj ). Assume (I1 ) f satisfies the Palais–Smale condition. (I2 ) c∗ = inf x∈Σ f (x) > 0. Assume there exists an odd, continuous mapping A : X → X satisfying (A1 ) Given c0 , b0 > 0, there exists b = b(c0 , b0 ) > 0 such that if ∥Df (x)∥ ≥ b0 , |f (x)| ≤ c0 , then ⟨Df (x), x − Ax⟩ ≥ b∥x − Ax∥ > 0. (A2 ) A(∂Pj ) ⊂ Pj , A(∂Qj ) ⊂ Qj , j = 1, . . . , k.
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Define Γj = {E|E ⊂ X, E compact, −E = E, γ(E ∩ σ −1 (Σ )) ≥ j, for σ ∈ Λ}, Λ = {σ|σ ∈ C(X, X), σ odd, σ(Pj ) ⊂ Pj , σ(Qj ) ⊂ Qj , j = 1, . . . , k, σ(x) = x if f (x) < 0}, where γ is the genus of symmetric sets γ(E) = inf{n| there exists an odd map φ : E → Rn \{0}}. Assume (Γ )
Γj is nonempty, j = 1, 2, . . . .
Define cj = inf
sup f (x), j = 1, 2, . . . ,
E∈Γj x∈E\M
(3.1)
Kc = {x|Df (x) = 0, f (x) = c}, Kc∗ = Kc \M. Theorem 3.1 (The Abstract Critical Point Theorem). Assume (I1 ), (I2 ), (A1 ), (A2 ) and (Γ ) hold. Then (1) cj ≥ c∗ , Kc∗j ̸= ∅. (2) cj → +∞ as j → ∞. (3) If cj = cj+1 = · · · = cj+k−1 = c, then γ(Kc∗ ) ≥ k. In the following, we will verify that all the assumptions of Theorem 3.1 are fulfilled by the functional Iλ , λ > 0. First we verify the assumption (I1 ) of Theorem 3.1, that is the Palais–Smale condition. We need Lemmas 3.1 and 3.2. Lemma 3.1.
The embedding from Wλ into Ls (RN ), p < s < p∗ is compact.
Proof . Obviously we need only to consider the scalar case. Let un be a sequence such that ∫ ∫ ( √ ) 2 pα |un |p(1+α) dx ≤ c < +∞, (|∇un |p + |un |p ) dx + λeλ 1+|x| RN
RN
we find a subsequence (still denoted by un ) convergent in Lp(1+α) (BR ) for any R > 0. We have ∫ |un − um |p(1+α) dx N R∫ ∫ = |un − um |p(1+α) dx + |un − um |p(1+α) dx RN \BR BR ∫ √ √ )−pα ∫ ( )pα ( λ 1+R2 λ 1+|x|2 p(1+α) λe ≤ λe |un − um | dx + |un − um |p(1+α) dx RN \BR BR ∫ −λpαR ≤ ce + |un − um |p(1+α) dx → 0 as n, m → ∞. BR
Therefore un converges in Lp(1+α) (RN ). For general s ∈ (p, p∗ ), by interpolation un converges in Ls (RN ), ∗ since un is bounded in both Lp (RN ) and Lp (RN ). ■ We denote by Jλ the principal part of the functional Iλ : ∫ ∑ ∫ ∑ k k ( )pα 1 1 uj p Jλ (U ) = |∇uj | dx + µj (x) |uj |p dx. p RN j=1 p RN j=1 mλ (x, uj )
(3.2)
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126
It holds that for U = (u1 , . . . , uk ), V = (v1 , . . . , vk ) ∈ Wλ
Lemma 3.2.
p(1+α)−1
(1) ∥DJλ (U ) − DJλ (V )∥ ≤ c∥U − V ∥p−1 + c|U − V |p,λ
for 1 < p < p(1 + α) < 2,
∥DJλ (U ) − DJλ (V )∥ ≤c(∥U ∥p−2 + ∥V ∥p−2 )∥U − V ∥ p(1+α)−2
+ c(|U |p,λ
p(1+α)−2
+ |V |p,λ
)|U − V |p,λ for p(1 + α) > p > 2 .
p(1+α)−1
(2) ∥DJλ (U )∥ ≤ c(∥U ∥p−1 + |U |p,λ ). p(1+α) (3) ⟨DJλ (U ) − DJλ (V ), U − V ⟩ ≥ c∥U − V ∥p + c|U − V |p,λ for p(1 + α) > p > 2 , ⟨DJλ (U ) − DJλ (V ), U − V ⟩ ≥c
|U − V |2p,λ ∥U − V ∥2 + c 2−p(1+α) 2−p(1+α) ∥U ∥2−p + ∥V ∥2−p |U |p,λ + |V |p,λ
for 1 < p < p(1 + α) < 2. Proof . Let Φ = (φ1 , . . . , φn ) ∈ Wλ . By Lemma 2.1 ⟨DJλ (U ) − DJλ (V ), Φ⟩ ∫ ∫ ∑ k p−2 p−2 (|∇uj | ∇uj − |∇vj | ∇vj )∇φj dx + =
k ∑
RN j=1
RN j=1
≤
RN j=1
RN j=1
k ∑ (
∫ +c
√ λeλ
RN j=1
≤c
) ⏐ 1+|x|2 pα ⏐
⏐ |uj |p(1+α)−2 uj − |vj |p(1+α)−2 vj ⏐ |φj | dx
k ) p−1 (∫ ∑ ⏐ ⏐ p ⏐ |∇uj |p−2 ∇uj − |∇vj |p−2 ∇vj ⏐ p−1 dx p
(∫
RN j=1
+c
(∫
+c
k ∑
) p1 |∇φj |p dx
RN j=1
k ) ∑ ⏐ ⏐ p ⏐ |uj |p−2 uj − |vj |p−2 vj ⏐ p−1 dx
p−1 p
RN j=1
(∫
) (∂ ∂ gλ (x, uj ) − gλ (x, vj ) φj dx ∂s ∂s
k ∑ ⏐ ⏐ ⏐|uj |p−2 uj − |vj |p−2 vj ⏐ |φj | dx
∫ k ∑ ⏐ ⏐ ⏐|∇uj |p−2 ∇uj − |∇vj |p−2 ∇vj ⏐ |∇φj | dx + c
∫
µj (x)
k ∑ (
√ λeλ
1
(∫
k ∑
)p |φj | dx p
RN j=1 p(1+α)
1+|x|2
RN j=1
) p(1+α)−1 ⏐ p(1+α) )pα ⏐ ⏐ |uj |p(1+α)−2 uj − |vj |p(1+α)−2 vj ⏐ p(1+α)−1 dx 1
·
(∫
k √ ) p(1+α) ∑ λ 1+|x|2 pα p(1+α) (λe ) |φj | dx .
RN j=1
Now we apply the inequalities (2.2) and (2.3) to complete the proof (1). Inequality (2) follows from (1) by setting V = 0. Finally, by Lemma 2.1 ⟨DJλ (U ) − DJλ (V ), U − V ⟩ ∫ ∑ k = (|∇uj |p−2 ∇uj − |∇vj |p−2 ∇vj )(∇uj − ∇vj ) dx RN j=1
∫ +
k ∑
RN j=1
µj (x)
(∂ ) ∂ gλ (x, uj ) − gλ (x, vj ) (uj − vj ) dx ∂s ∂s
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
127
k ∑ (|∇uj |p−2 ∇uj − |∇vj |p−2 ∇vj )(∇uj − ∇vj ) dx
∫ ≥
RN j=1 k ∑
∫ +c RN
j=1 k ∑
∫ +c
(|uj |p−2 uj − |vj |p−2 vj )(uj − vj ) dx √ 2 (λeλ 1+|x| )pα (|uj |p(1+α)−2 uj − |vj |p(1+α)−2 vj )(uj − vj ) dx.
RN j=1
Again we apply the inequalities (2.2) and (2.3) to get the proof of (3). Corollary 3.1.
■
Let Un be a bounded sequence of Wλ and ⟨DJλ (Un ) − DJλ (Um ), Un − Um ⟩ → 0
as n, m → ∞.
Then Un is a convergent sequence of Wλ . Proof . By Lemma 3.2 (3), we have ∥Un − Um ∥ → 0, |Un − Um |p,λ → 0 as n, m → ∞, and ∥Un − Um ∥λ → 0 as n, m → ∞. Un is a Cauchy sequence, hence a convergent sequence, since Wλ is Banach space. Lemma 3.3.
■ The functional Iλ , λ > 0 is a C 1 -functional and satisfies the Palais–Smale condition.
Proof . Let U, Φ ∈ Wλ , U = (u1 , . . . , uk ), Φ = (φ1 , . . . , φk ), we have ⟨DIλ (U ), Φ⟩ ∫ ∑ ∫ k p−2 = |∇uj | ∇uj ∇φj dx + RN
j=1 k ∑
∫ − RN
= RN
q
q−2
βij |ui | |uj |
µj (x)
j=1
∫ uj φj dx − RN
i,j=1,i̸=j
k ∑
∫
RN
k ∑
∂ gλ (x, uj )φj dx ∂s
k ∑ j=1
βjj
d fλ (x, uj )φj dx ds
|∇uj |p−2 ∇uj ∇φj dx
j=1
∫ + RN
∫ −
k ∑
( )pα uj uj bλ (x, uj ) )( µj (x) 1 + α − α |uj |p−2 uj φj dx m (x, u ) m (x, u ) λ j λ j j=1 k ∑
βij |ui |q |uj |q−2 uj φj dx
RN i,j=1,i̸=j
∫ − RN
k ∑
( )2qα uj bλ (uj ) )( uj βjj 1 + α − α |uj |2q−2 uj φj dx. m (u ) m (x, u ) λ j λ j j=1
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
128
By Lemma 2.1, 1 ⟨DIλ (U ), U ⟩ 2q ∫ ∑ k (1 1) = − |∇uj |p dx p 2q RN j=1
Iλ (U ) −
k ∑
∫
)pα (1 1 + α uj α uj bλ (x, uj ) )( |uj |p dx − + µj (x) p 2q 2q m (x, u ) m (x, u ) λ j λ j j=1
+ RN
+
α 2q
k ∑
∫
( uj bλ (uj ) )( uj )2qα |uj |2q dx βjj 1 − m (u ) m (u ) λ j λ j j=1
RN k ∑
∫ ≥c
|∇uj |p dx + c
RN j=1 k ∑
∫ ≥c
∫
k ( ∑
RN j=1
|∇uj |p dx + c
∫
(3.3)
)pα uj |uj |p dx mλ (x, uj )
k √ ∑ 2 (|uj |p + (λeλ 1+|x| )pα |uj |p(1+α) ) dx.
RN j=1
RN j=1
Here and in the following we always assume p(1 + α) < 2q < 2q(1 + α) < p∗ ,
(3.4)
and p(1 + α) < 2,
if
1 < p < 2.
(3.5)
By (3.3) a Palais–Smale sequence of Iλ is bounded in Wλ . Let Un = (un,1 , . . . , un,k ) ∈ Wλ be a Palais– Smale sequence of Iλ . By Lemma 3.1 we assume that Un weakly converges in Wλ and strongly in Ls (RN ), p < s < p∗ . We have ⟨DJλ (Un ) − DJλ (Um ), Un − Um ⟩ = ⟨DIλ (Un ) − DIλ (Um ), Un − Um ⟩ ∫ k ∑ βij (|un,i |q |un,j |q−2 un,j − |um,i |q |um,j |q−2 um,j )(un,j − um,j ) dx + RN i,j=1,i̸=j
∫ +
k ∑
RN j=1
βjj
(d ) d fλ (un,j ) − fλ (um,j ) (un,j − um,j ) dx ds ds
∫
≤ o(1) + c (|Un |2q−1 + |Um |2q−1 )|Un − Um | dx N R ∫ +c (|Un |2q(1+α)−1 + |Um |2q(1+α)−1 )|Un − Um | dx RN
≤ o(1) + c|Un − Um |2q + c|Un − Um |2q(1+α) → 0 By Corollary 3.1, Un is a Cauchy sequence in Wλ .
as n, m → ∞.
■
Let P be the positive cone in W 1,p (RN ) and Q = −P : P = {u|u ∈ W 1,p (RN ), u(x) ≥ 0, a.e. x ∈ RN }. For δ > 0, j = 1, . . . , k define open convex subsets of Wλ Pj = {U |U = (u1 , . . . , uk ) ∈ Wλ , dist(uj , P ) < δ}, Qj = −Pj = {U |U = (u1 , . . . , uk ) ∈ Wλ , dist(uj , Q) < δ},
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
129
where dist(u, P ) = inf{∥u − v∥W 1,p (RN ) |v ∈ P }. Denote also k ⋃
M=
k ⋂
(Pj ∪ Qj ), O =
j=1
k ⋂
(Pj ∩ Qj ), Σ =
j=1
(∂Pj ∩ ∂Qj ).
j=1
Next we verify the assumption (I2 ) of Theorem 3.1. Lemma 3.4.
There exist 0 < δ0 (≤ 1) and c0 > 0 such that for 0 < δ < δ0 it holds that Iλ (U ) ≥ c0 δ p , Iλ (U ) ≥ 0,
U ∈ Σ,
for
U ∈ O.
for
− Proof . For U = (u1 , . . . , uk ) ∈ ∂Pj , ∥u− j ∥W 1,p (RN ) ≥ dist(uj , P ) = δ and |uj |s = distLs (uj , P ) ≤ cdist(uj , P ) = cδ, p ≤ s ≤ p∗ . Similar estimates hold for u+ j . Hence we have for U ∈ Σ
Iλ (U ) =
1 p −
1 ≥ p
k ∑
∫
|∇uj |p dx +
RN j=1
1 2q ∫
k ∑
∫
1 p
∫
k ( ∑
)pα uj |uj |p dx mλ (x, uj )
RN j=1
βij |ui |q |uj |q dx −
RN i,j=1,i̸=j
1 2q
∫ k ∑ p p (|∇uj | + |uj | ) dx − c
≥ c1 δ p − c2 (δ 2q + δ 2q(1+α) ) ≥ 2q(1+α)
βjj
(
RN j=1 2q
|U |
uj )2qα |uj |2q dx mλ (uj )
∫
|U |2q(1+α) dx
dx − c
RN
RN j=1
provided 21 c1 δ0p ≥ c2 (δ02q + δ0
k ∑
∫
RN
1 p c1 δ , 2
). Similarly Iλ (U ) ≥ 0 for U ∈ O.
■
Now we are going to define the operator A : Wλ → Wλ , which will be used as a pseudo-gradient vector field to construct the descending flow. At this point we need to distinguish the case 1 < q < 2 and the case q ≥ 2. Given U = (u1 , . . . , uk ) ∈ Wλ , define AU = V = (v1 , . . . , vk ) ∈ Wλ by the following system 1 1 ⟨DJλ (V ), Φ⟩ + ⟨DJλ (V − U ) + DJλ (U ), Φ⟩ − 2 2 ∫ = RN
k ∑
d βjj fλ (uj )φj dx, ds j=1
∫
k ∑
βij |ui |q |zj |q−2 vj φj dx
RN i,j=1,i̸=j
(3.6)
for Φ = (φ1 , . . . , φk ) ∈ Wλ ,
where zj = uj for q ≥ 2 and zj = vj for 1 ≤ q < 2 and Jλ is as defined in (3.2). Lemma 3.5.
The operator A is well-defined and continuous.
Proof . We consider the case 1 < q < 2. The other case q ≥ 2 is similar. V = AU can be obtained by solving the following minimization problem, inf{G(V )|V ∈ Wλ }
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
130
where 1 1 1 1 G(V ) = Jλ (V ) + Jλ (V − U ) + ⟨DJλ (U ), V ⟩ − 2 2 2 q ∫ − RN
k ∑
βjj
j=1
k ∑
∫
βij |ui |q |vj |q dx
RN i,j=1,i̸=j
d fλ (uj )vj dx. ds
Notice that βij ≤ 0, i ̸= j, so we have ∫ ∑ k 1 d 1 Jλ (V ) + ⟨DJλ (U ), V ⟩ − βjj fλ (uj )vj dx 2 2 ds RN j=1 ∫ ∑ k ( ( ) ) pα vj |vj |p dx − c0 ∥V ∥λ ≥c |∇vj |p + mλ (x, vj ) RN j=1 ∫ ∫ ( λ√1+|x|2 )pα ≥c (|∇vj |p + vjp ) dx + c λe |vj |p(1+α) dx − c0 ∥V ∥λ
G(V ) ≥
RN
≥ c1 ∥V ∥pλ − c0 ∥V ∥λ ,
RN
hence G is coercive. Let Vn be a minimizing sequence for the functional G, Vn ⇀ V in Wλ . By the lower semicontinuity G(V ) ≤ lim G(Vn ) = inf{G(V )|V ∈ Wλ }, n→∞
and so V is a solution of the system (3.6). Assume V1 = (v11 , . . . , v1k ), V2 = (v21 , . . . , v2k ) are solutions of (3.6). Then, using V1 , V2 as the test function, we have 1 1 ⟨DJλ (V1 ) − DJλ (V2 ), V1 − V2 ⟩ + ⟨DJλ (V1 − U ) − DJλ (V2 − U ), V1 − V2 ⟩ 2 ∫ 2 k ∑ q q−2 βij |ui | (|v1j | v1j − |v2j |q−2 v2j )(v1j − v2j ) dx = 0. −
(3.7)
RN i,j=1,i̸=j
All the terms of (3.7) are nonnegative. Hence ⟨DJλ (V1 ) − DJλ (V2 ), V1 − V2 ⟩ = 0. By Lemma 2.1 (and Lemma 2.2) V1 = V2 . We have proved that the system (3.6) has a unique solution V = AU . We will show that the map A is continuous. First of all A maps bounded sets into bounded sets. In fact, let V = AU , then by using V as the test function in (3.6), we have p(1+α)
c min{∥V ∥pλ , ∥V ∥λ
}≤
1 ⟨DJλ (V ), V ⟩ 2
1 1 ⟨DJλ (V ), V ⟩ + (⟨DJλ (V − U ), V ⟩ + ⟨DJλ (U ), V ⟩) − 2 2 ∫ ∑ k d = βjj fλ (uj )vj dx ds RN j=1 ∫ ∑ k ≤c (|uj |2q−1 + |uj |2q(1+α)−1 )|vj | dx ≤
≤
RN j=1 c(∥U ∥λ2q−1
2q(1+α)−1
+ ∥U ∥λ
)∥V ∥λ .
∫
k ∑
RN i,j=1,i̸=j
βij |ui |q |vj |q dx
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
131
Let Vi = AUi , Ui = (ui1 , . . . , uik ), Vi = (vi1 , . . . , vik ), i = 1, 2. By (3.6) one has 1 1 ⟨DJλ (V1 ) − DJλ (V2 ), V1 − V2 ⟩ + ⟨DJλ (V1 − U1 ) − DJλ (V2 − U2 ), (V1 − U1 ) − (V2 − U2 )⟩ 2 2 1 1 = − ⟨DJλ (V1 − U1 ) − DJλ (V2 − U2 ), U1 − U2 ⟩ − ⟨DJλ (U1 ) − DJλ (U2 ), V1 − V2 ⟩ 2 2 ∫ k ∑ q q q q−2 + βij (|u1i | |v1j | v1j − |u2i | |v2j | v2j )(v1j − v2j ) dx RN i,j=1,i̸=j k ( ∑
∫
) d d fλ (u1j ) − fλ (u2j ) (v1j − v2j ) dx ds ds RN j=1 1 1 ≤ − ⟨DJλ (V1 − U1 ) − DJλ (V2 − U2 ), U1 − U2 ⟩ − ⟨DJλ (U1 ) − DJλ (U2 ), V1 − V2 ⟩ 2 2 ∫ k ∑ + βij (|u1i |q − |u2i |q )|v1j |q−2 v1j (v1j − v2j ) dx βjj
+
RN i,j=1,i̸=j k ( ∑
∫
βjj
+
RN j=1
(3.8)
) d d fλ (u1j ) − fλ (u2j ) (v1j − v2j ) dx. ds ds
The right hand side of (3.8) tends to zero as U2 → U1 . The left hand side of (3.8) is greater than 1 ■ 2 ⟨DJλ (V1 ) − DJλ (V2 ), V1 − V2 ⟩. By Lemma 3.2, we have ∥V1 − V2 ∥λ = o(1) as U2 → U1 in Wλ . In the following lemma we verify the assumption (A2 ) of Theorem 3.1. Lemma 3.6.
There exists δ0 > 0 such that if 0 < δ < δ0 , then A(∂Pj ) ⊂ Pj ,
A(∂Qj ) ⊂ Qj , j = 1, . . . , k.
Proof . Take Φ = (0, . . . , 0, vj+ , 0, . . . , 0) as test function in (3.6). Notice that βij ≤ 0 for i ̸= j, we have 1 2
∫ ∫R
N
|∇vj |p−2 ∇vj ∇vj+ dx +
≤ RN
1 2
∫ µj (x) RN
∂ gλ (x, vj )vj+ dx ∂s
d βjj fλ (uj )vj+ dx. ds
By Lemma 2.1 ∫ RN
(|∇vj+ |p
+
(vj+ )p ) dx
∫
(
≤c RN
) 2q−1 2q(1+α)−1 + (u+ + (u+ vj dx j ) j )
and distp (vj , Q) ≤ c0 (dist2q−1 (uj , Q) + dist2q(1+α)−1 (uj , Q))dist(vj , Q). 1
1 2q(1+α)−p p−1 )
Choose c0p−1 (δ02q−p + δ0
≤ 21 . Then for 0 < δ < δ0 , U ∈ ∂Qj we have dist(vj , Q) ≤
and A(∂Qj ) ⊂ Qj . Similarly A(∂Pj ) ⊂ Pj .
1 1 dist(uj , Q) = δ 2 2
■
As we have mentioned, the operator A will be used to construct the pseudo-gradient vector field. In the following lemma we show that the vector field −(U − AU ) inherits properties of a pseudo-gradient vector field.
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
132
Lemma 3.7.
Let V = AU, U = (u1 , . . . , uk ), V = (v1 , . . . , vk ). Then ⟨DIλ (U ), U − AU ⟩ 1 1 = ⟨DJλ (U − V ), U − V ⟩ + ⟨DJλ (U ) − DJλ (V ), U − V ⟩ 2 2 ∫ k ∑ − βij |ui |q (|uj |q−2 uj − |zj |q−2 vj )(uj − vj ) dx.
(3.9)
RN i,j=1,i̸=j
Consequently 1 ⟨DJλ (U − AU ), U − AU ⟩ 2 p(1+α) ≥ c min{∥U − AU ∥pλ , ∥U − AU ∥λ }.
⟨DIλ (U ), U − AU ⟩ ≥
(3.10)
Moreover there exist positive constant c = cλ and β, γ, δ such that ∥DIλ (U )∥ ≤ c∥U − AU ∥βλ (1 + |Iλ (U )|γ + ∥U − AU ∥δλ ),
(3.11)
where the constants β, γ, δ depend on p and q. Proof . Let U = (u1 , . . . , uk ), V = (v1 , . . . , vk ), V = AU, Φ = (φ1 , . . . , φk ). By (3.6) ⟨DIλ (U ), Φ⟩ =
1 1 ⟨DJλ (U − V ), Φ⟩ + ⟨DJλ (U ) − DJλ (V ), Φ⟩ 2 2 ∫ k ∑ βij |ui |q (|uj |q−2 uj − |zj |q−2 vj )φj dx. −
(3.12)
RN i,j=1,i̸=j
Take Φ = U − V , we obtain (3.9). Each of the three terms on the right hand side of (3.9) is nonnegative, hence is less than the left hand side of (3.9). We obtain (3.10), since by (3.9) ⟨DIλ (U ), U − V ⟩ ≥
1 p(1+α) ⟨DJλ (U − V ), U − V ⟩ ≥ c min{∥U − V ∥pλ , ∥U − V ∥λ }. 2
Also we have k ∑
∫ ⟨DIλ (U ), U − V ⟩ ≥ −
βij |ui |q (|uj |q−2 uj − |zj |q−2 vj )(uj − vj ) dx.
(3.13)
RN i,j=1,i̸=j
By (3.12) we have ∥DIλ (U )∥ ≤
1 1 ∥DJλ (U − V )∥ + ∥DJλ (U ) − DJλ (V )∥ 2 2 ∫ k ∑ + sup (−βij )|ui |q (|uj |q−2 uj − |zj |q−2 vj )φj dx .
(3.14)
RN i,j=1,i̸=j
∥Φ∥λ =1
The first two terms namely ∥DJλ (U − V )∥ and ∥DJλ (U ) − DJλ (V )∥ have been estimated in Lemma 3.2. To estimate the third term, we distinguish 1 < q < 2 and q ≥ 2. If 1 < q < 2, then zj = vj in the formulas (3.13) and (3.14). ∫ ⟨DIλ (U ), U − V ⟩ ≥ −
k ∑
RN i,j=1,i̸=j
βij |ui |q (|uj |q−2 uj − |vj |q−2 vj )(uj − vj )dx
(3.15)
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
133
and k ∑
⏐∫ ⏐ ⏐
RN i,j=1,i̸=j
⏐ ⏐ (−βij )|ui |q (|uj |q−2 uj − |vj |q−2 vj )φj dx⏐
k ∑
( ∫ ≤ −
)1− 1q βij |ui |q (|uj |q−2 uj − |vj |q−2 vj )(uj − vj ) dx
RN i,j=1,i̸=j k ∑
( ∫ · −
βij |ui |q
RN i,j=1,i̸=j
) | |uj |q−2 uj − |vj |q−2 vj | |φj |q dx q−1 |uj − vj |
(3.16)
1 q
1
≤ c⟨DIλ (U ), U − V ⟩1− q |U |2q |Φ|2q . For q ≥ 2, zj = uj in the formulas (3.13) and (3.14). k ∑
∫ ⟨DIλ (U ), U − V ⟩ ≥ −
βij |ui |q |uj |q−2 (uj − vj )2 dx
(3.17)
RN i,j=1,i̸=j
and k ∑
⏐∫ ⏐ ⏐
RN i,j=1,i̸=j
⏐ ⏐ (−βij )|ui |q |uj |q−2 (uj − vj )φj dx⏐
k ∑
( ∫ ≤ − RN
q
q−2
βij |ui | |uj |
) 21 ( ∫ (uj − vj ) dx − 2
RN
i,j=1,i̸=j
≤ c⟨DIλ (U ), U − V ⟩
1 2
k ∑
) 12 βij |ui |q |uj |q−2 φ2j dx
(3.18)
i,j=1,i̸=j
|U |q−1 2q |φ|2q .
By (3.14), Lemma 3.2 and (3.16), (3.18) we obtain the following somewhat “complicated” estimate {
p(1+α)−1
c(∥U − V ∥p−1 + |U − V |p,λ ) for 1 < p < 2 p(1+α)−1 p(1+α)−2 c(∥U − V ∥p−1 + ∥U ∥p−2 ∥U − V ∥ + |U − V |p,λ + |U |p,λ |U − V |p,λ ) for p ≥ 2 { 1 c⟨DIλ (U ), U − V ⟩1− q |U |2q for 1 < q < 2 + 1 for q ≥ 2. c⟨DIλ (U ), U − V ⟩ 2 |U |q−1 2q (3.19) We need to estimate ∥U ∥, |U |p,λ ( for p ≥ 2) and |U |2q . By the definition of V = AU we have ∥DIλ (U )∥ ≤
) 1 (1 1 ⟨DJλ (V − U ), U ⟩ + ⟨DJλ (V ) − DJλ (U ), U ⟩ 2q 2 2 ) 1 1 (1 1 = (Iλ (U ) − ⟨DJλ (U ), U ⟩) + ⟨DJλ (V − U ), U ⟩ + ⟨DJλ (V ) + DJλ (U ), U ⟩ 2q 2q 2 2 ∫ ∑ ∫ ∑ k k (1 ) ( )pα 1 1 1+α α uj bλ (x, uj ) )( uj = − |∇uj |p dx + µj (x) − + |uj |p dx p 2q RN j=1 p 2q 2q mλ (x, uj ) mλ (x, uj ) RN j=1
Iλ (U ) +
1 − 2q
∫
α + 2q
∫
k ∑
βij |ui |q (|uj |q−2 uj − |zj |q−2 vj )uj dx
RN i,j=1,i̸=j
RN
k ∑
( uj bλ (uj ) )( uj )2qα |uj |2q dx. βjj 1 − m (u ) m (u ) λ j λ j j=1 (3.20)
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
134
Notice that bλ (uj ) = 0 if |uj | ≥ λ2 , we have ∫ ∑ ∫ ∑ k ∫ k k ∑ ( λ√1+|x|2 )pα |uj |p(1+α) dx + (|∇uj |p + |uj |p ) dx + λe RN
RN
j=1
2 |uj |≥ λ
j=1
j=1
|uj |2q dx
≤ c(|Iλ (U )| + |⟨DJλ (V − U ), U ⟩| + |⟨DJλ (V ) − DJλ (U ), U ⟩|) k ⏐ ∫ ⏐ ∑ ⏐ ⏐ + c⏐ − βij |ui |q (|uj |q−2 uj − |zj |q−2 vj )uj dx⏐ .
(3.21)
RN i,j=1,i̸=j
By Lemma 3.1 |⟨DJλ (V ) − DJλ (U ), U ⟩| ∫ k ⏐∫ ∑ ⏐ p−2 p−2 (|∇uj | ∇uj − |∇vj | ∇vj )∇uj dx + =⏐ RN j=1
k ∑
RN
⏐ ) (∂ ∂ ⏐ gλ (x, uj ) − gλ (x, vj ) uj dx⏐ µj (x) ∂s ∂s j=1 k ∑ ⏐ ⏐ ⏐ |uj |p−2 uj − |vj |p−2 vj ⏐ |uj | dx
∫ k ∑ ⏐ ⏐ ⏐ |∇uj |p−2 ∇uj − |∇vj |p−2 ∇vj ⏐ |∇uj | dx + c
∫ ≤
RN j=1
∫
√
(
+c
λeλ
RN
RN j=1
) ⏐ 1+|x|2 pα ⏐
⏐ |uj |p(1+α)−2 uj − |vj |p(1+α)−2 vj ⏐ |uj | dx . (3.22)
For any ε > 0 there exists Cε > 0 such that for ξ, η ∈ RN ⏐ p−2 ⏐ ⏐ |ξ| ξ − |η|p−2 η ⏐ |ξ| ≤ C(|ξ|p−1 |ξ − η| + |ξ| |ξ − η|p−1 )
(3.23)
≤ Cε |ξ − η|p + ε|ξ|p . By (3.22), (3.23) we obtain ⏐ ⏐ ⏐⟨DJλ (U ) − DJλ (V ), U ⟩⏐ ∫ k (∫ ∑ p p (|∇(uj − vj )| + |uj − vj | ) dx + ≤ Cε k ∑
√ ) 2 )pα λeλ 1+|x| |uj − vj |p(1+α) dx
RN j=1
RN j=1
(∫ +ε
k ∑ (
p
∫
p
(|∇uj | + |uj | ) dx +
k ∑
(
√ λe
λ
(3.24)
) 1+|x|2 pα
p(1+α)
|uj |
) dx
RN j=1
RN j=1 p(1+α)
= Cε (∥U − V ∥p + |U − V |p,λ
p(1+α)
) + ε(∥U ∥p + |U |p,λ
).
Similarly p(1+α)
⟨DJλ (V − U ), U ⟩ ≤ Cε (∥U − V ∥p + |U − V |p,λ
p(1+α)
) + ε(∥U ∥p + |U |p,λ
).
(3.25)
By (3.24), (3.25) and Lemma 3.2(3) p(1+α)
|⟨DJλ (V − U ), U ⟩| + |⟨DJλ (V ) − DJλ (U ), U ⟩| ≤ Cε ⟨DIλ (U ), U − V ⟩ + ε(∥U ∥p + |U |p,λ
).
By (3.16), (3.18), we have k ⏐ ∫ ⏐ ∑ ⏐ ⏐ βij |ui |q (|uj |q−2 uj − |zj |q−2 vj )uj dx⏐ ≤ Cε ⟨DIλ (U ), U − V ⟩ + ε|U |2q ⏐− 2q .
(3.26)
(3.27)
RN i,j=1,i̸=j
Substituting (3.26), (3.27) into (3.21), we obtain ∫ ∑ ∫ ∑ k ∫ k k ∑ ( λ√1+|x|2 )pα (|∇uj |p + |uj |p ) dx + λe |uj |p(1+α) dx + RN
RN
j=1
j=1
j=1
≤ Cε (|Iλ (U )| + ⟨DIλ (U ), U − V ⟩) ∫ k (∫ ∑ +ε (|∇uj |p + |uj |p ) dx + RN j=1
2 |uj |≥ λ
|uj |2q dx (3.28)
k ∑
RN j=1
√ (
λeλ
) 1+|x|2 pα
|uj |p(1+α) dx +
∫
k ∑
RN j=1
)
|uj |2q dx .
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
Choose ε <
( ) 1 λ 2q−p (< 13 ). 3 2 1 3
∫ 2 |uj |≤ λ
135
Then 1 ( λ )2q−p 3 2
|uj |p dx ≥
∫ 2 |uj |≤ λ
|uj |2q dx ≥ ε
∫ 2 |uj |≤ λ
|uj |2q dx .
By (3.28), we obtain ∫
k ∑
(|∇uj |p + |uj |p ) dx +
k ∑ (
∫
RN j=1
∫ √ 2 )pα |uj |p(1+α) dx + λeλ 1+|x|
k ∑
|uj |2q dx
RN j=1
RN j=1
(3.29)
≤ Cλ (|Iλ (U )| + ⟨DIλ (U ), U − V ⟩). Substituting (3.29) into (3.19), we finally obtain the desired result. As an example, we consider the case p ≥ 2 and q ≥ 2. The other three cases are similar. By (3.19) for p ≥ 2 and q ≥ 2 p(1+α)−1
∥DIλ (U )∥ ≤ c(∥U − V ∥p−1 + ∥U ∥p−2 ∥U − V ∥) + c(|U − V |p,λ
p(1+α)−2
+ |U |p,λ
|U − V |p,λ )
1 2
+ c⟨DIλ (U ), U − V ⟩ |U |q−1 2,q 2
≤ c(∥U − V ∥p−1 + (|Iλ (U )| + ⟨DIλ (U ), U − V ⟩)1− p ∥U − V ∥) p(1+α)−1
+ c(|U − V |p,λ
2 1− p(1+α)
+ (|Iλ (U )| + ⟨DIλ (U ), U − V ⟩) 1 2
+ c⟨DIλ (U ), U − V ⟩ (|Iλ (U )| + ⟨DIλ (U ), U − V ⟩)
1− 1 2 2q
(3.30) |U − V |p,λ )
.
2 1 The exponents of DIλ (U ) in the right hand side of (3.30) are 1 − p2 , 1 − p(1+α) , 1 − 2q respectively. Therefore by Holder’s inequality we obtain p
2
∥DIλ (U )∥ ≤ c(∥U − V ∥p−1 + |Iλ (U )|1− p ∥U − V ∥ + ∥U − V ∥λ2 p(1+α)−1
+ c(|U − V |p,λ
p(1+α) −1 2
+ ∥U − V ∥λ
2 1− p(1+α)
+ |Iλ (U )|
−1
|U − V |p,λ
p
∥U − V ∥ 2 ) (3.31)
|U − V |p,λ ) 1
+ c(∥U − V ∥λ |Iλ (U )|1− q + ∥U − V ∥2q−1 ), λ which leads to the desired estimate ∥DIλ (U )∥ ≤ cλ ∥U − V ∥βλ (1 + |Iλ (U )|γ + ∥U − V ∥δλ ) for suitable positive constants cλ and β, γ, δ.
■
Corollary 3.2 (The Assumption A1 ). Given c0 , b0 > 0 there exists b = b(c0 , b0 ) such that if ∥DIλ (U )∥ ≥ b0 , |Iλ (U )| ≤ c0 then ⟨DIλ (U ), U − AU ⟩ ≥ b∥U − AU ∥λ (> 0). Lemma 3.8.
Γj is nonempty.
Proof . Let n = j + k. Denote t ∈ Rnk by t = (t1 , . . . , tk ), tl = (t1l , . . . , tnl ) ∈ Rn for l = 1, . . . , k. Choose nk functions vil ∈ C0∞ (RN ), 1 ≤ i ≤ n, 1 ≤ l ≤ k with disjoint supports. Define φ(j) : B nk → W by (j)
φ
(t) = R(
n ∑ i=1
ti1 vi1 ,
n ∑ i=2
ti2 vi2 , . . . ,
n ∑
tik vik ),
i=1
where B nk is the unit ball of Rnk and R is sufficiently large such that I1 (φ(j) (t)) < 0 for t ∈ ∂B nk . Then E = φ(j) (B nk ) ∈ Γj . See [16, Lemma 5.6] for the proof of this fact. ■
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
136
Theorem 3.2. solutions.
Assume (N ), (B) hold. Then the problem (Pλ ), λ > 0 has infinitely many sign-changing
Proof . Define Γj = {E|E ⊂ Wλ , E compact, −E = E, γ(E ∩ σ −1 (Σ )) ≥ j, for σ ∈ Λ}, Λ = {σ|σ ∈ C(Wλ , Wλ ), σ odd, σ(Pj ) ⊂ Pj , σ(Qj ) ⊂ Qj , j = 1, . . . , k, σ(U ) = U if I1 (U ) < 0}, cj (λ) = inf
sup Iλ (U ), j = 1, 2, . . . , λ ∈ (0, 1].
E∈Γj U ∈E\W
By Corollary 3.2 the condition (A1 ) of Theorem 3.1 holds, that is given c0 , b0 > 0 there exists b = b(c0 , b0 ) > 0 such that if ∥DIλ (U )∥ ≥ b0 , |Iλ (U )| ≤ c0 then ⟨DIλ (U ), U − AU ⟩λ ≥ b∥U − AU ∥λ > 0. By Lemmas 3.3, 3.4, 3.6 and 3.8, the conditions (I1 ), (I2 ), (A2 ) and (Γ ) hold. All the assumptions of Theorem 3.1 are fulfilled. Then by Theorem 3.1, cj (λ)(≥ c∗ ), j = 1, 2, . . . are critical values of Iλ , and there exists Uj (λ) ∈ Wλ \ M with Iλ (Uj (λ)) = cj (λ), DIλ (Uj (λ)) = 0. Moreover cj (λ) → ∞ as j → ∞. ■ 4. The proof of Theorem 1.1 By Theorem 3.2 the functional Iλ , λ > 0 has a sequence of critical values cj (λ), and the corresponding critical points Uj (λ) belong to Wλ \ M . Since Iλ is increasing in λ, we have c∗ ≤ cj (λ) ≤ βj := cj (1) for 0 < λ ≤ 1. Iλ (Uj (λ)) = cj (λ) ≤ βj , DIλ (Uj (λ)) = 0, by formula (3.6) Uj (λ), 0 < λ ≤ 1 is bounded in Wλ . Given k ∈ N , by Theorem 1.2 (and Remark 2.1) there exists νk > 0 such that |Uj (λ)(x)| ≤
1 −νk √1+|x|2 , x ∈ RN , j = 1, . . . , k, 0 < λ ≤ 1. e νk
Choose λ < νk , U1 (λ), . . . , Uk (λ) will be solutions of the original problem (P ). If c1 (λ) < c2 (λ) < · · · < ck (λ), then U1 (λ), . . . , Uk (λ) are different from each others. If cj (λ) = cj+1 (λ), by (3) of Theorem 3.1, we can assume Uj (λ) ̸= Uj+1 (λ). Any way we find k different solutions U1 (λ), . . . , Uk (λ) of the problem (P ). Since k is arbitrary, we obtain infinitely many sign-changing solutions to problem (P ). Appendix A. Theorem 3.1 (The abstract critical point theorem) In this appendix we sketch the proof of Theorem 3.1. The starting point is a deformation lemma. Deformation Lemma. Let X be a Banach space, f be an even C 1 -functional on X. Let Pj , Qj , j = 1, 2, . . . , k be a family of open convex, Qj = −Pj . Denote f c = {x ∈ X, f (x) ≤ c}, Kc = {x|x ∈ X, Df (x) = 0, f (x) = c}. Let N be an open symmetric neighborhood of Kc . Then there exists a positive constant ε0 such that for 0 < ε < ε′ < ε0 there exists a continuous map σ : [0.1] × X → X satisfying (1) (2) (3) (4) (5)
σt : x ↦→ σ(t, x) is a diffeomorphism of X, σ0 = Id. σt (x) = x for t ∈ [0, 1] and |f (x) − c| ≥ ε′ . σt (−x) = −σt (x) for (t, x) ∈ [0, 1] × X. σ1 (f c+ε \ N ) ⊂ f c−ε . σt (Pj ) ⊂ Pj , σt (Qj ) ⊂ Qj , j = 1, . . . , k, t ∈ [0, 1].
Proof . The proof is somewhat standard, maybe except the property (5) and we sketch the proof here.
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137
The operator A defined by system (3.6) is continuous by Lemma 3.5. By Lemma 2.1 in [3] there exists a locally Lipschitz continuous map B which inherits the properties of A. More precisely let K = {x ∈ X|Df (x) = 0}, X0 = X \ K, then B is defined on X0 and satisfies B(∂Pj ) ⊂ Pj , B(∂Qj ) ⊂ Qj , j = 1, . . . , k.
(A.1)
1 ∥x − Bx∥ ≤ ∥x − Ax∥ ≤ 2∥x − Bx∥. (A.2) 2 1 ⟨Df (x), x − Bx⟩ ≥ ⟨Df (x), x − Ax⟩ . (A.3) 2 Consequently the operator B satisfies assumptions (A1 ), (A2 ). For notation simplicity we assume that A is locally Lipschitz continuous on X0 . For δ > 0 sufficiently small Nδ = {x|x ∈ X, dist(x, Kc ) < δ} ⊂ N . Since f satisfies the Palais–Smale condition, there exist ε0 , b0 > 0 such that ∥Df (x)∥ ≥ b0 for x ∈ f −1 ([c − ε0 , c + ε0 ]) \ N 1 δ . 4
By assumption (A1 ) there exists a constant b > 0 such that ⟨Df (x), x − Ax⟩ ≥ b∥x − Ax∥ > 0 Let V (x) =
x−Ax ∥x−Ax∥
for x ∈ f −1 ([c − ε0 , c + ε0 ]) \ N 1 δ . 4
for x ∈ X0 . Consider the ordinary differential equation ⎧ ⎨ dσ = −g(σ)V (σ), dt ⎩ σ(0, x) = x,
(A.4)
where g is a cut-off function such that g(−x) = g(x) and { 0, if |f (x) − c| ≥ ε or U ∈ N 1 δ , 4 g(x) = 1, if |f (x) − c| ≤ ε and U ̸∈ N 1 δ . 2
1 4 bδ.
We assume ε0 < Then the conclusions (1)–(4) of Theorem 3.1 can be verified as usual. The property (5) follows from the assumption (A2 ). We verify (5). Let x ∈ Pj , consider x(t) = σ(t, x). For t > 0 small, x(t) ∈ Pj . Assume there exists t∗ > 0 such that ∗ x = x(t∗ ) ∈ ∂Pj and x(t) ∈ Pj for 0 ≤ t < t∗ . First of all g(x∗ ) > 0 otherwise x(t) ≡ x∗ . Next x∗ − Ax∗ ̸= 0, since x∗ ∈ ∂Pj and Ax∗ ∈ Pj . By the differential equation (A.4), we have x(t) = x∗ − λ(x∗ − Ax∗ )(t − t∗ ) + o(|t − t∗ |)
as t → t∗ .
1 ∗ where λ = g(x∗ ) ∥x∗ −Ax ∗ ∥ > 0. For t < t we have
x∗ =
x(t) + λ(t∗ − t)(Ax∗ + o(1)) . 1 + λ(t∗ − t)
x∗ is a convex combination of x(t) ∈ Pj and Ax∗ +o(1) ∈ Pj , which contradicts the assumption x∗ ∈ ∂Pj .
■
Corollary. Let N be an open symmetric neighborhood of Kc∗ = Kc \ M . Then there exists a constant ε0 > 0 such that for 0 < 2ε < ε0 there exists a diffeomorphism η of X satisfying (1) (2) (3) (4)
η(−x) = −η(x), x ∈ X. η|f c−2ε = Id. η(f c+ε \ (N ∪ M )) ⊂ f c−ε . η(Pj ) ⊂ Pj , η(Qj ) ⊂ Qj , j = 1, . . . , k.
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138
Proof . N ∪ M is an open symmetric neighborhood of Kc . According to the deformation lemma we choose η = σ(1, ·). ■ Proof of Theorem 3.1. We verify the conclusion (3): if cj = · · · = cj+k−1 = c, then γ(Kc∗ ) ≥ k. Let N be an open symmetric neighborhood of Kc∗ with γ(N ) = γ(Kc∗ ). By the above corollary there exist ε > 0 and a diffeomorphism η such that (1)–(4) hold true. We assume c − 2ε ≥ c∗ − 2ε > 0, cj+k−1 = c, by the definition of cj+k−1 there exists F ∈ Γj+k−1 with F \ W ⊂ f c+ε . Let E = η(F \ N ). Then E \ M = η(F \ N ) \ M ⊆ (η(F \ (N ∪ M )) ∪ η(M )) \ M ⊆ η(F \ (N ∪ M )) ⊆ f c−ε . Hence E \ M ̸∈ Γj , since cj = c. There exists σ ∈ Λ such that j − 1 ≥ γ(E ∩ σ −1 (Σ )) = γ(η(F \ N ) ∩ σ −1 (Σ )) ≥ γ((F \ N ) ∩ (ση)−1 (Σ )) ≥ γ(F ∩ (ση)−1 ) − γ(N ) ≥ j + k − 1 − γ(N ). We obtain γ(Kc∗ ) = γ(N ) ≥ k.
■
Appendix B. The decay estimate In this appendix we prove Proposition 2.1 and Lemma 2.3. Notice that in the following (see (b1 ), for example) the exponent 2q can be replaced by any s, s ∈ (p, p∗ ). Proposition 2.1. Assume z ∈ W 1,p (RN ) is a nonnegative solution to the differential inequality ∫ ∫ (|∇z|p−2 ∇z∇φ dx + µz p−1 φ) dx ≤ β z 2q−1 φ dx for φ ≥ 0, φ ∈ W 1,p (RN ). RN
(B.1)
RN
Then there exists ν > 0 such that z(x) ≤
1 −ν √1+|x|2 e , ν
for x ∈ RN .
Proof . The proof is somewhat well-known and divided to three steps. Step 1. We use Moser’s iteration to prove ∥z∥L∞ (B 1 (x0 )) ≤ c∥z∥L2q (B1 (x0 )) ,
(B.2)
2
hence z(x) → 0 as |x| → ∞. For 21 ≤ ρ < R ≤ 1, let ψ ∈ C0∞ (RN , [0, 1]) be such that ψ(x) = 1 for x − x0 ≤ ρ, ψ(x) = 1 for 2 |x − x0 | ≥ R, |∇ψ| ≤ R−ρ . Let r ≥ 1, take φ = ψ p z p(r−1)+1 as the test function in (B.1), we have ∫
((p(r − 1) + 1)z p(r−1) ψ p |∇z|p + pz p(r−1)+1 ψ p−1 |∇z|p−2 ∇z∇ψ) dx ∫ ∫ +µ z pr ψ p dx ≤ β z p(r−1)+2q ψ p dx RN
RN
RN
(B.3)
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
hence
∫ z
p(r−1)
p
p
∫
ψ |∇z| dx ≤ c(
RN
pr
∫
p
z |∇ψ| dx + RN
139
z p(r−1)+2q ψ p dx).
RN
We have ∫ |∇(z r ψ)|p dx N R ∫ ∫ r p p ≤ c( |∇z | ψ dx + z p |∇ψ|p dx) R∫N RN ∫ z p(r−1) ψ p |∇z|p dx + c ≤ crp z pr |∇ψ|p dx N N R ∫R ∫ p ‘pr p p(r−1)+2q p ≤ cr z |∇ψ| dx + c z ψ dx RN p
cr ≤ (R − ρ)p ≤
RN
(∫
crp ( (R − ρ)p
p
z
rp∗ · p+p∗ −2q
) p+p∗∗−2q (∫ p dx + crp
) 2q·p (∫ p∗ dx
RN
BR (x0 )
∫
z
p∗
∗
z rp
·d
r
p
p∗ · p+p∗ −2q
(z ψ)
) p+p∗∗−2q p dx
RN
) ∗p p d dx ,
BR (x0 )
p where d = < 1. By (B.4) and the Sobolev embedding theorem we have p + p∗ − 2q ∫ ∫ ) ∗p (∫ ) p∗ crp ( p p d rp∗ d r·p∗ r p z dx ≤c |∇(z ψ)| dx ≤ z . dx p (R − ρ) N BR (x0 ) Bρ (x0 ) R Let χ =
(B.4)
(B.5)
1 d0
> 1, rj = χj , j = 0, 1, . . . , and ρj = ρ + 21j (R − ρ). By (B.5) ( ) r1 (∫ ) 1 ∗ (∫ ) 1∗ ∗ crj rj p r p j r j p∗ ≤ z rj−1 p dx j−1 z dx ρ − ρ j−1 j Bρj−1 (x0 ) Bρj (x0 ) ∫ k 1 ) 1∗ ∏ ( c2j χj ) j ( ∗ c p χ ≤ ∥z∥Lp∗ (BR ) . z p dx ≤ d R − ρ BR (x0 ) (R − ρ) 1−d j=1
Let j → ∞, we obtain ∥z∥L∞ (Bρ (x0 )) ≤
c
∥z∥Lp∗ (BR (x0 )) . d (R − ρ) 1−d For R < 1 we show that ∥z∥Lp∗ (BR (x0 )) ≤ c∥z∥L2q (B1 (x0 )) , hence ∥z∥L∞ (Bρ (x0 )) ≤ c∥z∥L2q (B1 (x0 )) . Choose 2 ψ ∈ C0∞ (RN , [0, 1]) such that ψ(x) = 1 for |x − x0 | ≤ R, ψ(x) = 0 for |x − x0 | ≥ 1 and |∇ψ| ≤ 1−R . Take p φ = zψ as the test function in (B.1), we have ∫ (∫ ) p∗ ∗ p z p dx ≤c |∇(zψ)|p dx N BR (x) R ∫ ∫ p p ≤c z |∇ψ| dx + c z 2q ψ p dx RN RN p p (∫ ) 2q (∫ ) 2q c 2q 2q ≤ z dx + c z dx (2 − R)p B1 (x0 ) B1 (x0 ) p (∫ ) 2q ≤c z 2q dx B1 (x0 )
that is ∥z∥Lp∗ (BR (x0 )) ≤ c∥z∥L2q (B1 (x0 )) . Step 2. We use the hole-filling trick to prove the exponential decay for the integrals: ∫ ∫ p p −νR (|∇z| + z ) dx ≤ ce , z 2q dx ≤ ce−νR . RN \BR
RN \BR
(B.6)
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140
Choose ε and R0 such that βz 2q−p (x) ≤ βε2q−p ≤ 12 µ for |x| ≥ R0 . Then for R > R0 we have ∫ ∫ 1 |∇z|p−2 ∇z∇φ dx + µ z p−1 φ dx ≤ 0, for φ ≥ 0, φ ∈ W01,p (RN \ BR ). 2 RN RN
(B.7)
Choose ηR ∈ C ∞ (RN , [0, 1]) such that ηR (x) = 0 for |x| ≤ R, ηR (x) = 1 for |x| ≥ R + 1 and |∇ηR | ≤ 2. 2 Take φ = zηR as the test function in (B.7), we have ∫ ∫ ∫ 1 2 p 2 p−2 z p ηR dx ≤ 0 |∇z| ηR dx + 2 |∇z| ∇z · zηR ∇ηR dx + µ 2 RN RN RN and ∫
p
∫
p
(|∇z| + z ) dx ≤ RN \BR+1
⏐ ≤ c⏐
∫ ∫RN
RN
2 (|∇z|p + z p )ηR dx
⏐ |∇z|p−2 ∇zzηR ∇ηR dx⏐ (|∇z|p + z p ) dx
≤ c0 BR+1 \BR
= c0
(
∫
(|∇z|p + z p ) dx −
∫
RN \BR
) (|∇z|p + z p ) dx .
RN \BR+1
(|∇z|p + z p ) dx, then an+1 ≤ c0 (an − an+1 ) and an ≤ cθn where θ = Let an = RN \B R0 +n implies that ∫ (|∇z|p + z p ) dx ≤ ce−νR ∫
c0 c0 +1
< 1, which
RN \BR
with ν = −logθ > 0 and
∫ z RN \B
2q
∫
z p dx ≤ ce−νR .
dx ≤ c RN \B
R
Step 3. Return to (B.2), we complete the proof.
R
■
Proof of Lemma 2.3. Lemma 2.3 can be proved in a similar way as we prove Proposition 2.1, and is divided into four steps. Step 1. By Proposition 2.1 and the property (4) of the decomposition (2.4), we have ∫ ∫ ∑∫ |Un |2q dx ≤ c |Vl |2q dx + c |Rn |2q dx (n) N N ΩR R (B.8) l∈Λ R \BR ≤ ce−νR + on (1). Step 2. As in the first step of the proof of Proposition 2.1, we use the Moser’s iteration to prove the L∞ -estimate |Un (x)| ≤ c∥Un ∥L2q (B1 (x)) ≤ ce−νR + on (1),
(n)
for x ∈ ΩR .
(B.9)
In particular for any ε > 0 there exist n0 ∈ N and R0 > 0 such that for n ≥ n0 it holds |Un (x)| ≤ ε,
In the following we assume n ≥ n0 and R ≥ R0 . Step 3. We use the hole-filling trick to prove ∫ (|∇Un |p + |Un |p ) dx ≤ ce−νR , (n) ΩR
(n)
for x ∈ ΩR0 .
∫ (n) ΩR
|Un |2q dx ≤ ce−νR .
(B.10)
(B.11)
J. Zhao, X. Liu and J. Liu / Nonlinear Analysis 182 (2019) 113–142
141
Let Un = (un1 , . . . , unk ). We need only to prove (B.12) for the components un1 . Let zn = |un1 |, then for n ≥ n0 ∫ ∫ ∫ |∇zn |p−2 ∇zn ∇φ dx + µ1 znp−1 φ dx ≤ β1 zn2q−1 φ dx, φ ≥ 0, φ ∈ W 1,p (RN ). RN
RN
RN
Choose ε in (B.11) such that β1 ε2q−p ≤ 12 µ1 , then for n ≥ n0 ∫ ∫ 1 p−2 |∇zn | ∇zn ∇φ dx + z p−1 φ dx ≤ 0, 2 RN n RN
(n)
for φ ≥ 0, φ ∈ W01,p (ΩR0 ). (n)
Let R > R0 and ηR ∈ C ∞ (RN , [0, 1]) be such that ηR (x) = 0 for x ̸∈ ΩR 2 |∇ηR | ≤ 2. Take φ = zn ηR as the test function in (B.13), we can obtain ∫ ∫ 2 (|∇zn |p + znp ) dx ≤ (|∇zn |p + znp )ηR dx (n)
(B.12) (n)
and ηR (x) = 1 for x ∈ ΩR+1 and
RN
ΩR+1
⏐ ≤ c⏐
∫ RN
⏐ |∇zn |p−2 ∇zn zn ηR ∇ηR dx⏐
∫ ≤ c0
(n)
(n)
(|∇z|p + z p ) dx
ΩR \ΩR+1
≤ c0
(
∫ (n)
(|∇z|p + z p ) dx −
ΩR
and
∫ (n) ΩR+1
(|∇zn |p + znp ) dx ≤
∫ (n)
(|∇z|p + z p ) dx
)
ΩR+1
c0 c0 + 1
∫ (n) ΩR
(|∇zn |p + znp ) dx,
(B.13)
which implies the desired exponential decay. Step 4. By the argument in Step 2. we obtain the L∞ -estimate |Un (x)| ≤ c∥Un (x)∥L2q (B1 (x)) ≤ ce−νR
(n)
for x ∈ ΩR .
■
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