Existence and concentration of positive ground states for a 1-Laplacian problem in RN

Applied Mathematics Letters 100 (2020) 106045

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Applied Mathematics Letters www.elsevier.com/locate/aml

Existence and concentration of positive ground states for a 1-Laplacian problem in RN ✩ Guofeng Che a ,∗, Hongxia Shi b , Zewei Wang c a

School of Applied Mathematics, Guangdong University of Technology, Guangzhou 510006, Guangdong, PR China b School of Mathematics and Computational Science, Hunan First Normal University, Changsha 410205, Hunan, PR China c School of Earth Sciences and Engineering, Sun Yat-sen University, Guangzhou 510275, Guangdong, PR China

article

info

abstract

Article history: Received 18 July 2019 Received in revised form 5 September 2019 Accepted 5 September 2019 Available online 12 September 2019

This paper is concerned with the following quasilinear elliptic problem:

Keywords: 1-Laplacian operator Bounded variation functions Concentration-compactness principle

where 0 < p < N 1−1 , N ≥ 2 and ε > 0 is a small parameter. Under some mild conditions on the nonnegative functions V (x) and K(x), we establish the existence of a ground state solution uε for the above problem. Moreover, uε concentrates on the intersection set of global minimum points of V (x) and maximum points of K(x). The methods are based on the Nehari manifold technique and the Concentration-Compactness Principle of Lions. © 2019 Elsevier Ltd. All rights reserved.

{

−ε∆1 u + V (x) u ∈ BV

u = K(x)|u|p−1 u, |u|

in RN ,

(RN ),

1. Introduction The 1-Laplace problem, which arises from image denoising and restoration, is of crucial importance for many mathematical and physical fields. For instance, solutions of 1-Laplacian equation, i.e., the bounded variation solutions, allow us to construct explicit solutions of the total variation formulation for denoising problems [1,2]. For more details about the applications, see [3] and the references therein. Recently, with the developments of elliptic equations, more and more attention has been paid to p-Laplace operator p−2 △p u = div(|Du| )Du, p ∈ (1, +∞). However, researches on the 1-Laplacian problem have been rarely reported, partially because the compactness of sequences as p → 1 occurs in weak norms, and partially because the associated energy functional is no longer smooth and strictly convex. These rule out a number ✩ This work was supported by the National Natural Science Foundation of China (Grant No. 11801160). ∗ Corresponding author. E-mail address: [email protected] (G. Che).

https://doi.org/10.1016/j.aml.2019.106045 0893-9659/© 2019 Elsevier Ltd. All rights reserved.

2

G. Che, H. Shi and Z. Wang / Applied Mathematics Letters 100 (2020) 106045

of standard tricks from the calculus of variations and lead to a lot of surprising effects, see [4–6] and the references therein. In this paper, we deal with the existence and concentration of solutions to the following quasilinear elliptic equation: { u p−1 = K(x)|u| u, in RN , −ε∆1 u + V (x) |u| (1.1) u ∈ BV (RN ), where 0 < p < N 1−1 , N ≥ 2, ∆1 u = div(Du/|Du|) is the well known 1-Laplacian operator, ε > 0 is a small parameter and BV (RN ) is a space of functions of bounded variation, which will be defined in Section 2. We assume that functions V (x) and K(x) satisfy the following hypotheses: (H1 ) V ∈ C(RN , R+ ) and V∞ := lim inf |x|→∞ V (x) > V0 = inf x∈RN V (x) > 0. (H2 ) K ∈ C(RN , R+ ), lim|x|→∞ K(x) = K∞ ∈ (0, ∞) and K(x) ≥ K∞ for x ∈ RN . } { } { (H3 ) Λ ∩ Λ1 ̸= ∅, where Λ = x ∈ RN : V (x) = V0 and Λ1 = x ∈ RN : K(x) = K0 := maxx∈RN K(x) . It is obvious that K(x) is a bounded continuous function. Similar hypotheses have been introduced by Liu and Guo [7] and Shao and Chen [8] in their studies of nonlinear Kirchhoff equations and quasilinear Schr¨ odinger equations. Without loss of generality, we may assume that 0 ∈ Λ ∩ Λ1 . Now we state our main result. Theorem 1.1. Suppose that conditions (H1 ) − (H3 ) hold. Then there exists ε0 > 0 such that for each ε ∈ (0, ε0 ), problem (1.1) has a positive solution uε . Moreover, uε has a global maximum point xε ∈ R3 such that limε→0+ V (xε ) = V0 and limε→0+ K(xε ) = K0 . Remark 1.1. The space BV (RN ) is neither reflexive nor uniformly convex. This is the reason why it is so difficult to prove that the functional defined in this space satisfies compactness conditions like the Palais–Smale one. Notation. Throughout this paper, we shall denote by | · |r the Lr -norm, where 1 ≤ r ≤ +∞, and C various positive constants, which may vary from line to line. We use on (1) to denote any quantity which tends to zero as n → ∞. 2. Preliminaries Making the change of variable εz = x, we can rewrite problem (1.1) as: { u p−1 = K(εx)|u| u, in RN , −∆1 u + V (εx) |u| (2.1) u ∈ BV (RN ). { } Let BV (RN ) = u ∈ L1 (RN ); Du ∈ M(RN , RN ) , where M(RN , RN ) is the space of the vectorial random measures. As is pointed in [6], u ∈ BV (RN ) is equivalent to u ∈ L1 (RN ) and {∫ } ∫ 1 N N |Du|dx := sup udivϕdx; ϕ ∈ Cc (R , R ), s.t. |ϕ|∞ ≤ 1 < +∞. RN

RN

We say that {un } ⊂ BV (RN ) converges to u ∈ BV (RN ) in the sense of the intermediate convergence { ∫ ∫ if un → u in L1 (RN ) and RN |Dun |dx → RN |Du|dx as n → ∞. Let us define Xε = u ∈ ⏐ } ∫ ∫ BV (RN )⏐ RN V (εx)|u|dx < ∞ endowed with the norm ∥u∥ε := RN (|Du|+V (εx)|u|)dx. From [5], we know that the functional Jε : Xε → R, given by Jε (u) = ∥u∥ε , is a convex functional and is Lipschitz continuous in its domain. Although non-smooth, the functional Jε admits some directional derivatives. More specifically,

G. Che, H. Shi and Z. Wang / Applied Mathematics Letters 100 (2020) 106045

3

as is shown in [9], given u ∈ BV (RN ), for all v ∈ BV (RN ) such that (Dv)s is absolutely continuous w.r.t. (Du)s and such that v is equal to 0 a.e. in the set where u vanishes, it follows that ∫ ∫ ∫ Dv Du (Du)a (Dv)a s dx + (x) (x)|(Dv) |dx + Jε′ (u)v = V (εx)sgn(u)vdx. (2.2) |(Du)a | |Dv| RN |Du| RN RN In particular, note that, for all u ∈ Xε , Jε′ (u)u = Jε (u). We define the energy functional Iε : Xε → R as follows: Iε (u) = Jε (u) − Fε (u),

(2.3)

p+1

1 K(εx)|u| dx. Now, let us give a detailed explanation of solutions we are where Fε (u) = p+1 RN considering here. From [6], we know that u ∈ Xε is a solution of problem (2.1) if

∫

Jε (v) − Jε (u) ≥ F ′ ε (u)(v − u), ∀ v ∈ Xε .

(2.4)

Therefore, every u ∈ Xε such that (2.4) holds is going to be called a bounded variation solution of problem (2.1). Moreover, by a standard argument, we know that Iε satisfies the mountain pass geometry, then it follows from [4] that there exists a (P S)cε sequence {un } ⊂ Xε such that Iε (un ) → cε > 0, as n → ∞, and

∫ Jε (v) − Jε (un ) ≥

p−1

K(εx)|un |

(2.5)

un (v − un )dx − τn ∥v − un ∥ε , ∀ v ∈ Xε ,

(2.6)

RN

where τn → 0 as n → ∞. The minimax value cε is given by cε = inf γ∈Γε maxt∈[0,1] Iε (γ(t)), where Γε = {γ ∈ C([0, 1], Xε ) : γ(0) = 0, Iε (γ(1)) < 0}. From the work by Figueiredo and Pimenta [4], we } { can define a Nehari set associated to Iε : Nε = u ∈ Xε \ {0} : Iε′ (u)u = 0 . From [10], we know that ∫ ∫ ∫ p+1 1 K∞ |u| dx cε = inf u∈Nε Iε (u). Let us define the functionals: I∞ (u) = RN |Du|dx+ RN V∞ |u|dx− p+1 RN { ∫ ∫ ∫ p+1 1 ′ (u)u = and I0 (u) = RN |Du|dx + RN V0 |u|dx − p+1 K0 |u| dx. Set N∞ = u ∈ BV (RN ) \ {0} : I∞ RN } { } N ′ 0 , c∞ = inf u∈N∞ I∞ (u) and N0 = u ∈ BV (R ) \ {0} : I0 (u)u = 0 , c0 = inf u∈N0 I0 (u). Indeed, it is clear that c∞ , c0 and the Nehari manifolds N∞ , N0 possess similar properties to cε and Nε . Furthermore, similar to the arguments of [6], we know that there exists a critical point ω0 ∈ BV (RN ) of I0 such that I0 (ω0 ) = c0 . 3. Existence of positive ground states Lemma 3.1. limε→0+ cε = c0 . Proof . The proof is analogous to Lemma 4.2 in [7], we omit it here.

□

From Lemma 3.1, we can easily get the following Corollary. Corollary 3.1. There exists ε0 > 0 such that cε < c∞ for all ε ∈ (0, ε0 ). Lemma 3.2. Suppose that conditions (H1 ) − (H3 ) hold. If {un } ⊂ Xε is a (P S)cε sequence for Iε . Then there exists uε ∈ BV (RN ) such that for all ε ∈ (0, ε0 ), where ε0 is given in Corollary 3.1, there holds un → uε in Lr (RN ) for all 1 ≤ r < 1∗ .

G. Che, H. Shi and Z. Wang / Applied Mathematics Letters 100 (2020) 106045

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1 N Proof . Set the function ρn (x) = |u|unn(x)| |1 . It is easy to see that (ρn ) is a bounded sequence in L (R ). Then the Concentration-Compactness Principle implies that one and only one of the following statements holds: (i) (Compactness) there exists {yn } ⊂ RN such that for any ε > 0, there exists R > 0 such that ∫ ρn (x)dx ≥ 1 − ε, n = 1, 2, . . . ; (3.1) BR (yn )

∫ (ii) (Vanishing) for all R > 0, there holds limn→∞ supy∈RN B (y) ρn (x)dx = 0; R (iii) (Dichotomy) there exist α ∈ (0, 1), {yn } ⊂ RN , R1 > 0, Rn → +∞ such that the functions ρ1,n (x) = χBR (yn ) (x)ρn (x) and ρ2,n (x) = χBRc (yn ) (x)ρn (x) satisfy 1

n

∫

∫ ρ1,n dx → α and

RN

ρ2,n dx → 1 − α.

(3.2)

RN

Indeed, we can prove (ρn ) verifies the Compactness condition by excluding the other two possibilities. It is easy to verify that the Vanishing case does not occur. Now we show the Dichotomy case also does not hold. Firstly, it follows from (2.6) that Iε′ (un )un = on (1). As far as the sequence (yn ) is concerned, let us consider the following two possible situations. ∫ |un | dx → α, then • (yn ) is bounded. In this case, the function uε is nontrivial, since B (yn ) |u n |1 R ∫ |un |dx ≥ δ, for n sufficiently large. Thus, by taking R0 > 0 such that BR (yn ) ⊂ BR0 (yn ) for all BR (yn ) ∫ n ∈ N, then B (yn ) |un |dx ≥ δ for n large enough. It follows from the Sobolev embedding inequality that R0

∫ |uε |dx ≥ δ, for n sufficiently large.

(3.3)

BR0 (yn )

Claim 1. Iε′ (uε )uε ≤ 0. Note that, if φ ∈ C0∞ (RN , [0, 1]), φ ≡ 1 in BR (0), φ ≡ 0 in B2R (0)c and φR = φ(·/R), then for all u ∈ BV (RN ), we have ( )s D(φR u) is absolutely continuous w.r.t. (Du)s . (3.4) Taking (3.4) into account, the fact that φR un is equal to 0 a.e. in the set where un vanishes and the fact φR µ µ ′ that |φ µ| = |µ| a.e. in BR (0), then Iε (un )(φR un ) is well defined. Furthermore, it follows from (2.2) that R

Iε′ (un )(φR un ) =

∫

φR |(Dun )a |dx +

RN

RN

∫ + ∫R

∫

N

+ RN

un (Dun )a ∇φR dx |(Dun )a |

Dun φR (Dun )s (x) (x)|φR (Dun )s |dx |Dun | |φR (Dun )s | ∫ p+1 V (εx)φR |un |dx − K(εx)|un | φR dx.

(3.5)

RN

From (3.5), Iε′ (un )(φR un ) = on (1) and the lower semicontinuity of the norm, we obtain ∫ ∫ ∫ ∫ un (Dun )a ∇φR p+1 |Duε |dx + lim inf dx + V (εx)φR |uε |dx ≤ K(εx)|uε | φR dx. a| n→∞ |(Du ) N N N n BR (0) R R R ∫ ∫ ∫ p+1 By doing R → +∞ in the above inequality, we obtain RN |Duε |dx+ RN V (εx)|uε |dx ≤ RN K(εx)|uε | dx, which implies that Iε′ (uε )uε ≤ 0. Claim 1 holds. From Claim 1 and (3.3), we know that there exists tε ∈ (0, 1] such that tε uε ∈ Nε . Note that ∫ 1 p+1 cε + on (1) = Iε (un ) + on (1) = Iε (un ) − Iε′ (un )un = (1 − ) K(εx)|un | dx. (3.6) p + 1 RN

G. Che, H. Shi and Z. Wang / Applied Mathematics Letters 100 (2020) 106045

5

It follows from (3.6) and Fatou’s Lemma that 1 )tp+1 cε ≥ (1 − p+1 ε

∫

p+1

K(εx)|uε | RN

dx = Iε (tε uε ) − Iε′ (tε uε )tε uε = Iε (tε uε ) ≥ cε .

Therefore, tε = 1, Iε (uε ) = cε and by (3.6) ∫

p+1

K(εx)|un |

∫

p+1

dx →

RN

K(εx)|uε |

dx, as n → ∞.

(3.7)

RN

∫ ∫ p+1 p+1 Recalling that ∥un ∥ε = RN K(εx)|un | dx + on (1) and ∥uε ∥ε = RN K(εx)|uε | dx, then together with (3.7), we obtain ∥un ∥ε → ∥uε ∥ε , as n → ∞. It follows from the Sobolev inequality that |un |1 → |uε |1 , as n → ∞.

(3.8)

From (3.8), (yn ) is a bounded sequence and Rn → +∞, we obtain ∫ c (y ) BR n n

|un |dx → 0, as n → +∞.

On the other hand, from (3.2) and (3.8), we have

∫

(3.9)

|un |dx → (1 − α)|uε |1 > 0, as n → +∞, which

c (y ) BR n n

is a contradiction with (3.9). • (yn ) is unbounded. In this case, we should proceed as the case where (yn ) is bounded, but now dealing with the sequence {˜ un }, where u ˜n = un (·+yn ). Indeed, since ∥un ∥ε = ∥˜ un ∥ε , it follows that {˜ un } is bounded N 1 N and then converges, up to a subsequence, to some function u ˜ ∈ BV (R ) in Lloc (R ), where u ˜ ̸= 0 by (3.2). ′ (˜ u)˜ u ≤ 0. Claim 2. I∞

∫ In order to prove this claim, let us denote, for u ∈ BV (RN ), ∥u∥ε,yn = RN (|Du|dx + V (εx + εyn )|u|)dx ∫ p+1 1 ′ K(εx + εyn )|u| dx. Note that, as before, Iε,y (u)v is well defined for all and Iε,yn (u) = ∥u∥ε,yn − p+1 n RN N s s u, v ∈ BV (R ) such that (Dv) is absolutely continuous (Du) and v is equal to 0 a.e. in the set where u vanishes. By the invariance of translation and Iε′ (un )un = on (1), we get ′ Iε,y (˜ un )(φR u ˜n ) = on (1). n

(3.10)

∫ From (3.10), proceeding as in Claim 1 and taking into account that |yn | → +∞, we have RN |D˜ u|dx + ∫ ∫ p+1 V |˜ u |dx ≤ K |˜ u | dx, which implies that Claim 2 holds. Then it follows from Claim 2 and u ˜ ̸= 0 ∞ RN ∞ RN ˜ ˜ that there exists t ∈ (0, 1] such that tu ˜ ∈ N∞ . Note that cε + on (1) = Iε (un ) + on (1) = Iε (un ) − Iε′ (un )un

= (1 −

1 ) p+1

∫

p+1

K(εx)|˜ un |

dx.

(3.11)

RN

Then it follows from (3.11), (H2 ) and Fatou’s Lemma that ∫ cε ≥

p+1

K(εx)|˜ u| RN

p+1 dx ≥ |t˜|

∫

p+1

K∞ |˜ u| RN

′ ˜ ˜ dx = I∞ (t˜u ˜) − I∞ (tu ˜)tu ˜ = I∞ (t˜u ˜ ) ≥ c∞ ,

which is a contradiction with Corollary 3.1. Therefore, in view of Claims 1 and 2, we know that Dichotomy case does not happen, it follows that Compactness must hold.

G. Che, H. Shi and Z. Wang / Applied Mathematics Letters 100 (2020) 106045

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Claim 3. (yn ) in (3.1) is a bounded sequence in RN . We can prove Claim 3 by contradiction that |yn | → +∞ and then proceed as in the case of Dichotomy, where (yn ) is unbounded, reaching that cε ≥ c∞ , which is a contradiction with Corollary 3.1 when ε ∈ (0, ε0 ). ∫ By Claim 3, for η > 0, it follows from (3.1) that there exists R > 0 such that B c (0) ρn dx < η, ∀ n ∈ N, R that is, ∫ c (0) BR

|un |dx < η|un |1 ≤ Cη, ∀ n ∈ N.

(3.12)

Since uε ∈ L1 (RN ), there exists R0 > 0 such that ∫ c (0) BR

|uε |dx ≤ η.

(3.13)

0

Therefore, for R1 ≥ max{R, R0 }, since un → uε in L1 (BR1 (0)), there exists n0 ∈ N such that ∫ |un − uε |dx ≤ η, ∀ n ≥ n0 .

(3.14)

BR1 (0)

Then by (3.12)–(3.14), it follows that if n ≥ n0 ∫ ∫ ∫ |un − uε |dx ≤ |un − uε |dx + η ≤ c (0) BR

RN

∫

c (0) BR

1

|un |dx +

1

c (0) BR

|uε |dx + η

≤ C1 η.

1

∗

Therefore, un → uε in L1 (RN ), and since (un ) is bounded in L1 (RN ), it follows from the interpolation inequality that un → uε in Lr (RN ), 1 ≤ r < 1∗ . The proof is complete. □ Proposition 3.1. Under the assumptions of Theorem 1.1, problem (1.1) has at least a positive ground state solution. Proof . It follows from (H1 ) and Lemma 3.2 that ∫ ∫ p+1 p+1 K(εx)|un | dx → K(εx)|uε | dx, as n → +∞. RN

(3.15)

RN

It follows from (2.4), (3.15) and the lower semicontinuity of the norm that ∫ p−1 ∥v∥ε − ∥uε ∥ε ≥ K(εx)|uε | uε (v − uε )dx, ∀ v ∈ BV (RN ). RN

Hence, uε is a nontrivial solution of problem (2.1). Furthermore, by (2.5), we have ∫ 1 p+1 cε ≤ Iε (uε ) − Iε′ (uε )uε ≤ lim inf (1 − ) K(εx)|un | dx = cε . n→∞ p + 1 RN

(3.16)

Thus, uε is a ground state solution of problem (2.1) and then uε (·/ε) is a ground state bounded variation solution of problem (1.1). The proof is complete. □ 4. Concentration of positive ground states In Section 3, we have proved that for each ε ∈ (0, ε0 ), there exists uε ∈ BV (RN ) of problem (2.1) such that Iε (uε ) = cε . Now let us show that the sequence of solutions concentrate on the intersection set of global minimum points of V (x) and maximum points of K(x).

G. Che, H. Shi and Z. Wang / Applied Mathematics Letters 100 (2020) 106045

Lemma 4.1. Denote vε (x) = uε (εx), then there exist (yε )ε>0 ⊂ RN and R, δ > 0 such that ∫ lim inf |vε |dx ≥ δ > 0. ε→0

7

(4.1)

BR (yε )

Moreover, the family {εyε } is bounded in RN . In particular, if x0 is the limit of the sequence {εn yεn } in the family {εyε }, then we have V (x0 ) = V0 and K(x0 ) = K0 , i.e., x0 ∈ Λ ∩ Λ1 . Proof . Suppose by contradiction that (4.1) does not hold. Then there exists a sequence εn → 0 such that for all R > 0, there holds: It follows from [4, Theorem 1.1] that vεn → 0 in Lr (RN ) for all 1 < r < 1∗ as n → +∞. ∫ p+1 Then RN K(εn x)|vεn | dx → 0, as n → ∞. Taking ω = vεn ± tvεn in (2.6) and passing to the limit as ∫ p+1 + t → 0 , we have ∥vεn ∥ε = RN K(εn x)|vεn | dx = on (1), which implies that cεn = Iεn (vεn ) = on (1). This is a contradiction with Lemma 3.1. Then (4.1) holds. Set yn := yεn and vn := vεn . Argue by contradiction that there exists a sequence εn → 0 such that |εn yn | → ∞ as n → ∞. In the following, we proceed as in the vn )(φR v˜n ) = 0, proof of Claim 2 in Lemma 3.2. If φR is as in the proof of such claim, it follows that Iε′ n yn (˜ N 1 N where v˜n = vn (· + yn ). Since vn , v˜n are bounded in BV (R ), then v˜n → v˜ in Lloc (R ). Then, as before, ′ (˜ v )˜ v ≤ 0 and there exists t˜ ∈ (0, 1] such that t˜v˜ ∈ N∞ . Therefore, arguing in the same way of we get I∞ Claim 2 in Lemma 3.2, we will get a contradiction. Thus, {εn yn } is bounded and there exists x0 ∈ RN such that εn yn → x0 as n → ∞. Define a functional Ix0 as follows: ∫ ∫ 1 p+1 K(x0 )|u| dx. Ix0 (u) = (|Du| + V (x0 )|u|)dx − p + 1 N N R R If V (x0 ) > V0 or K(x0 ) < K0 , similarly to the arguments that used in the proof of Claim 2 in Lemma 3.2, we can get a contradiction. Thus, x0 ∈ Λ ∩ Λ1 , i.e., V (x0 ) = V0 and K(x0 ) = K0 . The proof is complete. □ Proof of Theorem 1.1. From Proposition 3.1 and Lemma 4.1, we can easily verify Theorem 1.1.

□

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