With the help of the monotonicity trick, a nonautonomous Choquard equations with general nonlinearity is studied and a nontrivial solution is obtained

Ding Ling

Liu Jiu

Yuan Yan-Xiang

English

University of Szeged

Electronic Journal of Qualitative Theory of Differential Equations2022, No. 66, 1–11; https://doi.org/10.14232/ejqtde.2022.1.66 www.math.u-szeged.hu/ejqtde/A nontrivial solution for a nonautonomousChoquard equation with general nonlinearityLing Ding1, Jiu LiuB 2, 3 and Yan-Xiang Yuan2, 31School of Mathematics and Statistics, Hubei University of Arts and Science,Xiangyang, 441053, People’s Republic of China2School of Mathematics and Statistics, Qiannan Normal University for Nationalities,Duyun, 558000, People’s Republic of China3Key Laboratory of Complex Systems and Intelligent Optimization of Guizhou Province,Duyun, 558000, People’s Republic of ChinaReceived 12 September 2022, appeared 30 December 2022Communicated by Patrizia PucciAbstract. With the help of the monotonicity trick, a nonautonomous Choquard equa-tions with general nonlinearity is studied and a nontrivial solution is obtained.Keywords: Choquard equations, monotonicity trick, Pohožaev identity.2020 Mathematics Subject Classification: 35J20, 35J60.1 Introduction and main resultIn the paper, we explore nontrivial solutions for the following nonlocal problem−∆u + V(x)u =(1|x| ∗ u2)u + g(u) in R3, (1.1)where 1|x| ∗ u2 =∫R3u2(y)|x−y|dy, the nonlinearity g satisfies general subcritical growth conditions(g1) g ∈ C(R, R) is odd;(g2) −∞ < lims→0+g(s)s = −m < 0;(g3) lims→+∞g(s)s5 = 0; and the potential function V verifies(V1) V ∈ C(R3, (−m, 0]) and lim|x|→∞ V(x) = 0;(V2) (∇V, x) ∈ L32 (R3) and|(∇V, x)| 32:=( ∫R3|(∇V, x)| 32 dx) 23< 2S := 2 inf0 ̸=u∈D1,2(R3)∫R3|∇u|2dx(∫R3u6dx)13.BCorresponding author. Email: jiuliu2011@163.com2 L. Ding, J. Liu and Y.-X. YuanWhen V ≡ 0 and g(s) = −s, Eq. (1.1) is simplified to the classical Choquard equation−∆u + u =(1|x| ∗ u2)u in R3. (1.2)Eq. (1.2) appeared at least as early as in 1954, in a work by S. I. Pekar describing the quantummechanics of a polaron at rest [11]. In 1976, P. Choquard used Eq. (1.2) to describe an electrontrapped in its own hole in a certain approximation to Hartree–Fock theory of one componentplasma [4]. For more details in the physics aspects, please refer to [7]. Therefore, manyscholars have carried out in-depth research on Choquard equations and related problems. Forrecent results, we refer the readers to [6, 8, 9, 12, 14] and references therein. See also [10] for abroad survey of Choquard equations.It is important to point out that Liu et al. in [6] considered the following special case ofEq. (1.1)−∆u =(1|x| ∗ u2)u + g(u) in R3. (1.3)Under the assumptions (g1)-(g3), they investigated ground states of Eq. (1.3) by using thePohožaev manifold method. In the present paper, we study Eq. (1.1) which can be regardedas the perturbation equation of Eq. (1.3). By using the monotonicity trick we obtain thefollowing main result.Theorem 1.1. Suppose that (V1)–(V2) and (g1)–(g3) hold. Then Eq. (1.1) possesses a nontrivialsolution.Set K(x) = V(x) +m and f (s) = g(s)+ms. Then Eq. (1.1) equals to the following equation−∆u + K(x)u =(1|x| ∗ u2)u + f (u) in R3, (1.4)where f satisfies( f1) f ∈ C(R, R) is odd;( f2) lims→0+f (s)s = 0;( f3) lims→+∞f (s)s5 = 0;and K verifies(K1) K ∈ C(R3, (0, m]) and lim|x|→∞ K(x) = m;(K2) (∇K, x) ∈ L32 (R3) and |(∇K, x)| 32< 2S.Then we convert to consider the followingTheorem 1.2. Suppose that (K1)–(K2) and ( f1)–( f3) hold. Then Eq. (1.4) has a nontrivial solution.Remark 1.3. If K(x) ≡ m, then Theorem 1.2 was proved in [6]. Thus we assume that K(x) ̸≡ m.Nonautonomous Choquard equations 3For the rest of this paper, we make the following marks. H := H1(R3) is the usual Sobolevspace endowed with the standard norm ∥ · ∥. Ls(R3), 2 ≤ s ≤ 6, denotes the usual Lebesguespace with the norm | · |s. C, C1, C2, . . . denote different positive constants whose exact valueis inessential. For any u ∈ H, we define ut(·) := u(t−1·) for t > 0.It is widely known that the solutions of Eq. (1.4) correspond to the critical points of thefunctional defined byI(u) =12∫R3(|∇u|2 + K(x)u2)dx − 14∫R3∫R3u2(x)u2(y)|x − y| dxdy −∫R3F(u)dx, u ∈ H,where F(s) =∫ s0 f (t)dt. Using the Hardy–Littlewood–Sobolev inequality [5], one has∫R3∫R3u2(x)u2(y)|x − y| dxdy ≤ C|u|4125≤ C1∥u∥4.Combining with ( f1)–( f3) and (K1) we know that I is well defined and I is of C1. But it ishard to obtain a bounded (PS) sequence for the functional I under the assumptions ( f1)–( f3).In addition, another difficulty we face is the lack of space compactness.From (K1) we know that there exits a > 0 such that a∥u∥2 ≤∫R3(|∇u|2 + K(x)u2)dx.2 PreliminariesIn order to prove Theorem 1.2, we cannot directly apply the mountain pass theorem [1]. In-stead we use an indirect approach which dated to Struwe [13] and was developed by Jeanjeanin [2]. Exactly, we apply the followingProposition 2.1. Let X be a Banach space equipped with a norm ∥ · ∥X and let J ⊂ R+ be an interval.We consider a family {Φµ}µ∈J of C1-functionals on X of the formΦµ(u) = A(u)− µB(u), ∀µ ∈ J,where B(u) ≥ 0 for all u ∈ X and such that either A(u) → +∞ or B(u) → +∞ as ∥u∥X → +∞.We assume that there are two points v1, v2 in X such thatΓ = {γ ∈ C([0, 1], X) : γ(0) = v1, γ(1) = v2},there hold, ∀µ ∈ J,cµ = infγ∈Γmaxt∈[0,1]Φµ(γ(t)) > max{Φµ(v1), Φµ(v2)}.Then for almost every µ ∈ J, there is a sequence {un} ⊂ X such that(i) {un} is bounded in X,(ii) Φµ(un) → cµ and(iii) Φ′µ(un) → 0 in the dual X∗ of X.Moreover, the map µ → cµ is non-increasing and continuous from the left.4 L. Ding, J. Liu and Y.-X. YuanDefine f±(s) = max{± f (s), 0}, F1(s) =∫ s0 f+(t)dt and F2(s) =∫ s0 f−(t)dt, from ( f1)–( f3)one haslims→0f±(s)s= 0 and lims→∞f±(s)s5= 0. (2.1)SetX = H, ∥ · ∥X = ∥ · ∥, Φµ = Iµ, J = [2−1, 1],B(u) =14∫R3∫R3u2(x)u2(y)|x − y| dxdy +∫R3F1(u)dxandA(u) =12∫R3(|∇u|2 + K(x)u2)dx +∫R3F2(u)dx.Then A(u) → +∞ as ∥u∥ → +∞ andIµ(u) = A(u)− µB(u)=12∫R3(|∇u|2 + K(x)u2)dx +∫R3F2(u)dx− µ4∫R3∫R3u2(x)u2(y)|x − y| dxdy − µ∫R3F1(u)dx.Specially, I1(u) = I(u). The following limit equations−∆u + mu = µ(1|x| ∗ u2)u + lµ(u) in R3, (2.2)will play an important role, where lµ(s) = µ f+(s)− f−(s). The energy functional of Eq. (2.2)is defined byI∞µ (u) =12∫R3(|∇u|2 + mu2)dx − µ4∫R3∫R3u2(x)u2(y)|x − y| dxdy −∫R3Lµ(u)dx,where Lµ(s) =∫ s0 lµ(t)dt. Setc∞µ = inf{I∞µ (u) : 0 ̸= u ∈ H, (I∞µ )′(u) = 0}.Let ω be a positive ground state solution of Eq. (2.2) with µ = 1. By the proof of [6,Lemma 3.3], one has c∞1 = maxt>0 I∞1 (ω(t−1x)).3 The proof of Theorem 1.2The following lemma is to verify the assumptions of Proposition 2.1.Lemma 3.1. Suppose that (K1) and ( f1)–( f3) hold. Then there exist v1, v2 ∈ H such that for anyµ ∈ J, cµ > max{Iµ(v1), Iµ(v2)}.Proof. From ( f1)–( f3) it follows thatF(s) ≤ a4|s|2 + C|s|6 for all s ∈ R.Nonautonomous Choquard equations 5Combining with the Hardy–Littlewood–Sobolev and Sobolev inequality, for any u ∈ H andµ ∈ J one hasIµ(u) ≥ I(u)≥ a4∥u∥2 − C14∥u∥4 − C∥u∥6which implies that there exist α, ρ > 0 such that Iµ(u) ≥ α for all µ ∈ J and ∥u∥ = ρ. Letω ∈ H be a positive ground state solution of Eq. (2.2) with µ = 1. For any µ ∈ J, one hasIµ(ωt) ≤ I∞12(ωt)=t2∫R3|∇ω|2dx + mt32∫R3ω2dx − t58∫R3∫R3ω2(x)ω2(y)|x − y| dxdy −t32∫R3L 12(ω)dx.Combining with∥ωt∥2 = t∫R3|∇ω|2dx + t3∫R3|ω|2dx,there exists t0 > 0 such that ∥ωt0∥ > ρ and Iµ(ωt0) < 0. Set v1 = 0 and v2 = ωt0 . Thus for anyγ ∈ Γ, maxt∈[0,1] Iµ(γ(t)) ≥ α > 0. So cµ ≥ α > max{Iµ(v1), Iµ(v2)}.Lemma 3.2. Suppose that (K1) and ( f1)–( f3) hold. Then there exists δ ∈ [ 12 , 1) such that for anyµ ∈ [δ, 1], cµ < c∞µ .Proof. According to the proof of Lemma 3.1, for any µ ∈ J, there exists tµ ∈ (0, t0) such thatIµ(ω(t−1µ x)) = maxt∈(0,1] Iµ(ω((t0t)−1x)) ≥ cµ. Set θ = infµ∈J tµ. We claim θ > 0. Otherwise,there exists µn ∈ J such that tµn → 0 and thenc1 ≤ cµn ≤ Iµn(ω(t−1µn x)) → 0.It is a contradiction. Note that K(x) ≤ m and K(x) ̸≡ m. Defineδ = max12, 1 −θ3 mins∈[θ,t0]∫R3[m − K(sx)]ω2dx2t30∫R3F1(ω)dx +t502∫R3∫R3ω2(x)ω2(y)|x−y| dxdy .Then for any µ ∈ [δ, 1], we getc∞µ ≥ c∞1≥ I∞1 (ω(t−1µ x))= Iµ(ω(t−1µ x)) +t3µ2∫R3[m − K(tµx)]ω2dx − (1 − µ)t3µ∫R3F1(ω)dx−(14− µ4)t5µ∫R3∫R3ω2(x)ω2(y)|x − y| dxdy> cµ +θ32mins∈[θ,t0]∫R3[m − K(sx)]ω2dx − (1 − µ)t30∫R3F1(ω)dx−(14− µ4)t50∫R3∫R3ω2(x)ω2(y)|x − y| dxdy≥ cµ.6 L. Ding, J. Liu and Y.-X. YuanLemma 3.3. Fix µ ∈ [δ, 1]. Suppose that (K1) and ( f1)–( f3) hold and that {un} ⊂ H is a bounded(PS)cµ sequence for Iµ. Then there exists u ∈ H, k ∈ N, vi ∈ H\{0}, yn,i ∈ R3 for 1 ≤ i ≤ k suchthat up to a subsequence,(i) |yn,i| → ∞, |yn,i − yn,j| → ∞, i ̸= j, for 1 ≤ i, j ≤ k,(ii) (Iµ)′(u) = 0 and (I∞µ )′(vi) = 0 for 1 ≤ i ≤ k,(iii) un − u − ∑ki=1 vi(· − yn,i) → 0 in H,(iv) cµ = Iµ(u) + ∑ki=1 I∞µ (vi) + o(1),where we agree that in the case k = 0 the above holds without vi, yn,i.Proof. The proof is in the spirit of [3]. Obviously, there exists u ∈ H such that up to asubsequence un ⇀ u in H, un → u in Lploc(R3) with 2 ≤ p < 6 and un(x) → u(x) a.e. in R3.For any φ ∈ C∞0 (R3), one has0 = ⟨I′µ(un), φ⟩+ o(1) = ⟨I′µ(u), φ⟩.Set un,1 = un − u. If un → 0 in H, we are done. So we can assume that {un,1} does notconverge strongly to 0 in H. Thus up to a subsequence un,1 ⇀ 0 in H, un,1 → 0 in Lploc(R3)and un,1(x) → 0 a.e. in R3. Then we have∥un,1∥2 = ∥un∥2 − ∥u∥2 + o(1),∫R3K(x)u2n,1dx =∫R3K(x)u2ndx −∫R3K(x)u2dx + o(1),∫R3∫R3u2n,1(x)u2n,1(y)|x − y| dxdy =∫R3∫R3u2n(x)u2n(y)|x − y| dxdy −∫R3∫R3u2(x)u2(y)|x − y| dxdy + o(1)and ∫R3Lµ(un,1)dx =∫R3Lµ(un)dx −∫R3Lµ(u)dx + o(1).Therefore,Iµ(un,1) = Iµ(un)− Iµ(u) + o(1).Defineβ1 = lim supn→∞supz∈R3∫B1(z)u2n,1dx.If β1 = 0, one sees un,1 → 0 in Lp(R3) with 2 < p < 6 from Lion’s lemma [15]. Then∫R3∫R3u2n(x)u2n(y)|x − y| dxdy →∫R3∫R3u2(x)u2(y)|x − y| dxdyand ∫R3lµ(un)undx →∫R3lµ(u)udx.Therefore,0 = ⟨I′µ(un), un⟩+ o(1) ≥ ⟨I′µ(u), u⟩ = 0,Nonautonomous Choquard equations 7which infers un → u in H. It is a contradiction. If β1 > 0, we may assume the existence ofyn,1 ∈ R3 such that∫B1(yn,1)u2n,1dx >β12 . Set wn,1 = un(·+ yn,1), there exists v1 ∈ H such that upto a subsequence wn,1 ⇀ v1 in H, wn,1 → v1 in Lploc(R3) with 2 ≤ p < 6 and wn,1(x) → v1(x)a.e. in R3. From∫B1(0)v21dx = limn→∞∫B1(0)w2n,1dx = limn→∞∫B1(yn,1)u2ndx = limn→∞∫B1(yn,1)(u2n,1 + u2)dx ≥ β12,we know v1 ̸= 0. Since un,1 ⇀ 0 in H, {yn,1} is unbounded in R3 and, up to a subsequence,we can assume that |yn,1| → ∞. Thus0 = ⟨I′µ(un), φ(· − yn,1)⟩+ o(1)= ⟨(I∞µ )′(v1), φ⟩.Set un,2 = un − u − v1(· − yn,1). If un,2 → 0 in H, we are done. So we can assume that {un,2}does not converge strongly to 0 in H. Thus up to a subsequence un,2 ⇀ 0 in H, un,2 → 0 inLploc(R3) and un,2(x) → 0 a.e. in R3. Thus we have∥un,2∥2 = ∥un∥2 − ∥u∥2 − ∥v1∥2 + o(1),∫R3K(x)u2n,2dx =∫R3K(x)u2ndx −∫R3K(x)u2dx −∫R3mv21dx + o(1),∫R3∫R3u2n,2(x)u2n,2(y)|x − y| dxdy =∫R3∫R3u2n(x)u2n(y)|x − y| dxdy −∫R3∫R3u2(x)u2(y)|x − y| dxdy−∫R3∫R3v21(x)v21(y)|x − y| dxdy + o(1)and ∫R3Lµ(un,2)dx =∫R3Lµ(un)dx −∫R3Lµ(u)dx −∫R3Lµ(v1)dx + o(1).Therefore,Iµ(un,2) = Iµ(un)− Iµ(u)− I∞µ (v1) + o(1).Defineβ2 = lim supn→∞supz∈R3∫B1(z)u2n,2dx.We replaced un,1 by un,2 and repeat the above arguments. If β2 = 0, then un,2 → 0. It is acontradiction. If β2 > 0, we may assume the existence of yn,2 ∈ R3 such that∫B1(yn,2)u2n,2dx >β22 . Set wn,2 = un(·+ yn,2), there exists v2 ∈ H such that up to a subsequence wn,2 ⇀ v2 in H,wn,2 → v2 in Lploc(R3) with 2 ≤ p < 6 and wn,2(x) → v2(x) a.e. in R3. From∫B1(0)v22dx = limn→∞∫B1(0)w2n,2dx = limn→∞∫B1(yn,2)u2ndx = limn→∞∫B1(yn,2)u2n,2dx ≥β22,we know v2 ̸= 0. Since un,2 ⇀ 0 in H, {yn,2} is unbounded in R3 and, up to a subsequence,we can assume that |yn,2| → ∞ and |yn,2 − yn,1| → ∞. Similarly, (I∞µ )′(v2) = 0. Setun,3 = un − u − v1(· − yn,1)− v2(· − yn,2).8 L. Ding, J. Liu and Y.-X. YuanAgain we repeat the above arguments, then there exists k ∈ N, vi ∈ H\{0}, yn,i ∈ R3 for1 ≤ i ≤ k such that up to a subsequence, |yn,i| → ∞, |yn,i − yn,j| → ∞, i ̸= j, for 1 ≤ i, j ≤ k,(I∞µ )′(vi) = 0 for 1 ≤ i ≤ k,un,k+1 = un − u −k∑i=1vi(· − yn,i)andcµ = Iµ(un,k+1) + Iµ(u) +k∑i=1I∞µ (vi) + o(1).Note that there exists α > 0 such that ∥v∥ ≥ α for any v ∈ {v ∈ H : v ̸= 0 and (I∞µ )′(v) = 0}.The iterations must stop after steps because {un} is bounded in H.For almost every µ ∈ [δ, 1], by Proposition 2.1 there is a sequence {un} ⊂ H such that(i) {un} is bounded in H,(ii) Iµ(un) → cµ,(iii) I′µ(un) → 0 in the dual H∗ of H.(3.1)Moreover, the map µ → cµ is non-increasing and continuous from the left.Lemma 3.4. Fix µ ∈ [δ, 1]. Suppose that (K1) and ( f1)–( f3) hold and that {un} ⊂ H satisfies (3.1).Then there exists u ∈ H such that Iµ(u) = cµ and I ′µ(u) = 0.Proof. We assume k ≥ 1 in Lemma 3.3. Then∥un − u −k∑i=1vi(· − yn,i)∥ → 0andcµ = Iµ(u) +k∑i=1I∞µ (vi) + o(1),where I′µ(u) = 0 and (I∞µ )′(vi) = 0 for 1 ≤ i ≤ k. Because Iµ(u) ≥ 0 and I∞µ (vi) ≥ c∞µ for1 ≤ i ≤ k, we have cµ ≥ c∞µ . It is a contradiction. Thus k = 0 and ∥un − u∥ → 0. Therefore,Iµ(u) = cµ and I′µ(u) = 0.Lemma 3.5. Suppose that (K1)–(K2) and ( f1)–( f3) hold. Then there exists u ∈ H such that I(u) = c1and I ′(u) = 0.Proof. Choosing µn ∈ [δ, 1] and µn ↗ 1, then Lemma 3.4 implies that there exists a sequence{uµn := un} ⊂ H such thatcµn = Iµn(un)=12∫R3(|∇un|2 + K(x)u2n)dx −µn4∫R3∫R3u2n(x)u2n(y)|x − y| dxdy −∫R3Lµn(un)dx(3.2)and the following Pohožaev identity0 =12∫R3|∇un|2dx +32∫R3K(x)u2ndx +12∫R3(∇K(x), x)u2ndx− 5µn4∫R3∫R3u2n(x)u2n(y)|x − y| dxdy − 3∫R3Lµn(un)dx.(3.3)Nonautonomous Choquard equations 9(3.2) × 3 − (3.3) implies3cµn =∫R3|∇un|2dx −12∫R3(∇K(x), x)u2ndx +µn2∫R3∫R3u2n(x)u2n(y)|x − y| dxdy≥(1 −|(∇K, x)| 322S) ∫R3|∇un|2dx +14∫R3∫R3u2n(x)u2n(y)|x − y| dxdywhich implies that∫R3|∇un|2dx,∫R3∫R3u2n(x)u2n(y)|x − y| dxdy ≤ C, ∀ n ∈ N∗.Combining with (2.1) and the Sobolev inequality, we get∫R3|∇un|2dx +∫R3K(x)u2ndx = µn∫R3∫R3u2n(x)u2n(y)|x − y| dxdy +∫R3lµn(un)undx≤ C +minx∈R3K(x)2∫R3u2ndx + C∫R3u6ndxwhich implies ∫R3u2ndx ≤ C ∀ n ∈ N∗.Then {un} is bounded in H. Recall that µn ↗ 1,I(un) = cµn +µn − 14∫R3∫R3u2n(x)u2n(y)|x − y| dxdy + (µn − 1)∫R3F1(un)dxand∥I′(un)∥∗ = sup∥φ∥E=1∣∣∣∣(µn − 1) ∫R3un(x)φ(x)u2n(y)|x − y| dxdy + (µn − 1)∫R3f+(un)φdx∣∣∣∣,where ∥ · ∥∗ denotes the norm in H∗. So I(un) → c1 and ∥I′(un)∥∗ → 0. According toLemma 3.3, we get that there exists u ∈ H such that I(u) = c1 and I′(u) = 0.Hence, we complete the proof.AcknowledgementsThe study was supported by NNSF of China (11861052), Foundation of Research Ability Culti-vation of HBUAS (2020kypytd006), Science and Technology Foundation of Guizhou Province([2019]5672), Foundation of Education of Guizhou Province (KY[2016]316, KY[2019]067) andFoundation of QNUN (qnsy2018017). We would like to thank the reviewer for careful readingand helpful suggestions which led to an improvement of the original manuscript.References[1] A. Ambrosetti, P. H. Rabinowitz, Dual variational methods in critical point theoryand applications, J. 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A nontrivial solution for a nonautonomous Choquard equation with general nonlinearity

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A nontrivial solution for a nonautonomous Choquard equation with general nonlinearity

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