# On Nonlinear Schrödinger-Bopp-Podolsky System with Asymptotically Periodic Potentials

Authors: Heng Yang; Yanxiang Yuan; Jiu Liu
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In the paper, we study the existence of ground states for a nonlinear Schrödinger-Bopp-Podolsky system with asymptotically periodic potentials. As a consequence, we also prove ground states for the nonlinear Schrödinger-Bopp-Podolsky system with periodic potentials.


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## Body

## 1. Introduction and Main Result

In [8], d’Avenia and Siciliano investigated the following nonlinear Schrödinger-Bopp-Podolsky system
(1)−Δu+ωu+λϕu=up−2uinℝ3,−Δϕ+a2Δ2ϕ=4πu2inℝ3,wherea>0,ω>0,λ>0,2<p<6, which has been considered for the first time in the mathematical literature. Such a system appears when we couple a Schrödinger field ψ=ψt,x with its electromagnetic field in the Bopp-Podolsky electromagnetic theory, and, in particular, in the electrostatic case for standing waves ψt,x=eiωtux, see ([8], Section 2]) for more details. For more physical details, we refer the reader to the recent papers [1, 3, 4, 7], etc. In addition, we point out that the operator −Δ+Δ2 appears also in other different interesting mathematical and physical situations (see [2, 10] and their references).Since [8], the existence of nontrivial solutions for the nonlinear Schrödinger-Bopp-Podolsky system has begun to receive much attention. We refer the reader to [5, 6, 11, 15, 17, 18], etc.For the subcritical case, we can see from [8] that there is a difference in the result depending on the range where p varies. Moreover, major treatment methods are different if p∈2,3, p∈3,4, or p∈4,6.In the present paper, we are mainly concerned with the positive solutions of the following nonlinear Schrödinger-Bopp-Podolsky system with subcritical exponent(2)−Δu+Vxu+λϕu=Qxup−2uinℝ3,−Δϕ+a2Δ2ϕ=4πu2inℝ3,where a>0,λ>0,4≤p<6, and the potentials V,Q satisfyVV∈Cℝ3,ℝ, 0≤Vx≤V∞x∈L∞ℝ3, for all x∈ℝ3 and Vx−V∞x∈FQQ∈Cℝ3,ℝ, Q∞x≤Qx∈L∞ℝ3, for all x∈ℝ3 and Qx−Q∞x∈FHere,(3)F=kx:foranyε>0,mx∈B1y:kx≥ε⟶0asy⟶∞.B1y=x∈ℝ3:x−y<1, m denotes the Lebesgue measure in ℝ3, and V∞,Q∞ satisfyV∞V∞∈Cℝ3,ℝ, infx∈ℝ3V∞x>0, and V∞·+z=V∞· for all z∈ℤ3Q∞Q∞∈Cℝ3,ℝ, infx∈ℝ3Q∞x>0, and Q∞·+z=Q∞· for all z∈ℤ3ConditionV was introduced by Liu et al. in [13] and unifies the following two types of asymptotic conditions:V1V∈Cℝ3,ℝ, 0<infx∈ℝ3Vx≤supx∈ℝ3Vx<+∞, and limx⟶∞Vx=supx∈ℝ3VxV2V∈Cℝ3,ℝ, 0<infx∈ℝ3Vx≤supx∈ℝ3Vx<+∞, and Vx−supx∈ℝ3Vx∈L3/2ℝ3As is well known, conditionV1 was given by Rabinowitz in [16] to study the nonlinear Schrödinger equations and has been widely used in the study of various elliptic problems.Although Chen et al. [5] also considered nonlinear Schrödinger-Bopp-Podolsky system under condition V1, they required that the nonlinearity is autonomous, and the function V satisfies more assumptions. Thus, to the best of our knowledge, under V and Q, there is no result for systems (2).Before stating our results, few preliminaries are in order.(i)
H≔H1ℝ3 is the usual Sobolev space endowed with the norm(4)·=∫ℝ3∇·2+·2dx1/2.(ii)
Lsℝ3 is the Lebesgue space endowed with the norm(5)·s=∫ℝ3·sdx1/sforalls∈1,∞.(iii)
D=ϕ∈D1,2ℝ3:Δϕ∈L2ℝ3 is the Sobolev space endowed with the norm(6)·D2=a2Δ·22+∇·22.From [8], we know that for every u∈H, there is a unique solution ϕu∈D such that
(7)−Δϕ+a2Δ2ϕ=4πu2.Moreover, it turns out that(8)ϕux=1−e−x/ax∗u2x=∫ℝ31−e−x−y/ax−yu2ydy,14πϕuD2=∫ℝ3ϕuu2dx=∫ℝ3∫ℝ31−e−x−y/ax−yu2xu2ydxdy≥0.By using the classical reduction argument, one is led to study, equivalently, the single equation(9)−Δu+Vxu+λϕuu=Qxup−2uinℝ3.This means that we just need to find the solutions of Eq. (9).Now, we introduce the main result of the present paper.Theorem 1.
Leta>0,λ>0 and 4≤p<6. Assume that V and Q hold. Then, Eq. (9) has a ground state solution.
IfVx≡V∞x and Qx≡Q∞x, then Eq. (9) reduces to the following periodic case
(10)−Δu+V∞xu+λϕuu=Q∞xup−2uinℝ3.
From Theorem1, we have the following corollary.Corollary 2.
Leta>0,λ>0 and 4≤p<6. Assume that V∞ and Q∞ hold. Then, Eq. (10) has a ground state solution.Remark 3.
In fact, the ground state solution we find is positive.


## 2. Proof of Theorem1

For the sake of simplicity, we assumeλ=1. We also use C,C0,C1,⋯ to denote different positive constants whose exact value is inessential. From ([13], Lemma 5), it follows that there is L>0 such that
(11)LuH2≤∫ℝ3∇u2+Vxu2dx≤max1,V∞∞u2forallu∈H,i.e.,(12)·V≔∫ℝ3∇·2+Vx·2dx1/2is equivalent to the norm ·.The energy functional associated with Eq. (9) is
(13)Iu=12uV2+14∫ℝ3ϕuu2dx−1p∫ℝ3Qxupdx.WhenVx≡V∞x and Qx≡Q∞x, the functional I turns into
(14)I∞u=12∫ℝ3∇u2+V∞xu2dx+14∫ℝ3ϕuu2dx−1p∫ℝ3Q∞xupdx.Obviously,I is of class C1 and for any u,v∈H,
(15)I′u,v=∫ℝ3∇u·∇vdx+∫ℝ3Vxuvdx+∫ℝ3ϕuuvdx−∫ℝ3Qxup−2uvdx.As everyone knows, the weak solutions of Eq. (9) correspond to the critical points of the functional I. For obtaining a ground state solution (also known as least energy solution), we define
(16)M=u∈H0:Ju=0andm=infu∈MIu,where Ju=I′u,u. In addition, for the functional I∞, we also define
(17)M∞=u∈H0:I∞′u,u=0andm∞=infu∈M∞I∞u.To prove the theorem, we introduce a series of lemmas in order.Lemma 4.
Leta>0 and 4≤p<6. Define
(18)N=u∈H:∫ℝ3ϕuu2dx<∫ℝ3Q∞xupdx.
Then,N≠∅.Proof.
From [8], we assume that u is a nontrivial solution of the following equation
(19)−Δu+u+ϕuu=infx∈ℝ3Q∞xup−2uinℝ3.
Then, one has(20)∫ℝ3ϕuu2dx<u2+∫ℝ3ϕuu2dx=infx∈ℝ3Q∞x∫ℝ3updx≤∫ℝ3Q∞xupdx,i.e., N≠∅.From now on, we assume that all the conditions of Theorem1 are always true.Lemma 5.
For eachu∈N, there exists a unique tu>0 such that tuu∈M. Moreover, the function t↦Itu is increasing on 0,tu and decreasing on tu,+∞ and Ituu=maxt>0Itu.Proof.
Ifu∈N, then
(21)∫ℝ3ϕuu2dx<∫ℝ3Q∞xupdx≤∫ℝ3Qxupdx.
By definingft≔Itu, one has
(22)f′t=tuV2+t3∫ℝ3ϕuu2dx−tp−1∫ℝ3Qxupdx.
Because4≤p<6, we have that f′t>0 for t>0 small enough and f′t<0 for t>0 large enough. Then, there exists tu>0 such that f′tu=0. Note that I′tuu,tuu=f′tutu. Thus, tuu∈M.
Suppose that there existst1,t2∈ℝ such that t1>t2>0 and t1u,t2u∈M. Then,
(23)t12−puV2+t14−p∫ℝ3ϕuu2dx=∫ℝ3Qxupdx=t22−puV2+t24−p∫ℝ3ϕuu2dx>t12−puV2+t14−p∫ℝ3ϕuu2dx.It is a contradiction. Thus,tu is unique. It is obvious that the function t↦Itu is increasing on 0,tu and decreasing on tu,+∞ and Ituu=maxt>0ItuLemma 6.
m≤m∞.Proof.
From Lemma5, we know that for any u∈M∞, one has I∞u=maxt>0I∞tu, and there exists tu>0 such that tuu∈M and Ituu=maxt>0Itu. Recall that Vx≤V∞x and Q∞x≤Qx for all x∈ℝ3. Thus, we have
(24)m≤Ituu=maxt>0Itu≤maxt>0I∞tu=I∞u.
According to the arbitrariness ofu, we get that m≤m∞.Lemma 7.
m>0.Proof.
For anyu∈M, one has
(25)Lu2≤uV2+∫ℝ3ϕuu2dx=∫ℝ3Qxupdx≤Cup.
Then, there existsα>0 such that
(26)u≥αforallu∈M.
Thus, for anyu∈M, we have
(27)Iu=Iu−1pJu=12−1puV2+14−1p∫ℝ3ϕuu2dx≥12−1pLu2.
Therefore,m≥1/2−1/pLα2.Lemma 8.
For anyu∈M, J′u≠0.Proof.
From (26), we know that for any u∈M,
(28)J′u,u=J′u,u−pJu=2−puV2+4−p∫ℝ3ϕuu2dx≤2−pLα2<0,which implies that for any u∈M, J′u≠0.Lemma 9.
There exists a bounded sequenceun⊂M such that Iun⟶m and I′un⟶0 in H∗, where H∗,·∗ is the dual space of H,·.Proof.
By using the Ekeland variational principle [9], we get that there exists un⊂M and λn⊂ℝ such that Iun⟶m and I′un−λnJ′un⟶0 in H∗. From (27), we have
(29)Iun=Iun−1pJun≥12−1pLun2,which deduces that un is bounded in H. Thus,
(30)0=I′un,un=λnJ′un,un+o1.Coupling with (28), we obtain λn⟶0. Since un is bounded in H, by using the Hölder inequality, we get
(31)J′un∗=supφ=1,φ∈HJ′un,φ≤C.Therefore,(32)I′un∗≤I′un−λnJ′un∗+λnJ′un∗=o1.In order to prove thatm is achieved, we need the following lemmas.Lemma 10 (see [13], Lemma 8).
Ifun is bounded in H and zn⟶+∞, then for any v∈C0∞ℝ3, the following identities
(33)∫ℝ3V∞x−Vxunv·−zndx=o1,∫ℝ3Q∞x−Qxunp−2unv·−zndx=o1hold.Lemma 11.
Ifun⟶0 in H, then the following identities
(34)∫ℝ3Vx−V∞xun2dx=o1,∫ℝ3Qx−Q∞xunpdx=o1hold.Proof.
Suppose thatkx∈F. Then, from ([13], Lemma 10), we know that for any ε>0, there exists Rε>0 such that
(35)∫kx≥εu2dx≤C0∫BRε+1u2dx+C1ε2/3u2forallu∈H,where C0,C1 are independent on ε. Thus,
(36)∫ℝ3kxun2dx=∫kx≥εkxun2dx+∫kx<εkxun2dx≤C0∫BRε+1un2dx+C1ε2/3un2+ε∫ℝ3un2dx≤C2ε2/3+C3ε+o1.
According to the arbitrariness ofε, we obtain that
(37)∫ℝ3kxun2dx=o1.
By using the Hölder inequality and the Sobolev inequality, we have also(38)∫ℝ3kxunpdx=∫ℝ3kx6−p/4un6−p/2kxp−2/4un3p−6/2dx≤∫ℝ3kxun2dx6−p/4∫ℝ3kxun6dxp−2/4≤C∫ℝ3kxun2dx6−p/4=o1.
BecauseVx−V∞x∈F and Qx−Q∞x∈F, the conclusions hold.Lemma 12 (see [8]).
Ifun⟶u in H, then for any v∈C0∞ℝ3, the formula
(39)∫ℝ3ϕununvdx=∫ℝ3ϕuuvdx+o1holds.
Now, we prove thatm is achieved.Lemma 13.
There existsu∈M such that Iu=m.Proof.
It follows from Lemma9 that there exists a bounded sequence un satisfying un⊂M, Iun⟶m, and I′un⟶0 in H∗. Up to a subsequence, there exists u∈H such that un⟶u in H, un⟶u in Llocsℝ3 for each s∈1,6, and unx⟶ux, a.e., in ℝ3. Since I′un⟶0 in H∗, for any v∈C0∞ℝ3 using Lemma 12, one has
(40)0=I′un,v+o1=∫ℝ3∇u·∇vdx+∫ℝ3Vxuvdx+∫ℝ3ϕuuvdx−∫ℝ3Qxup−2uvdx,i.e., u is a solution of Eq. (9). Thus, u∈M.
We divide two steps to complete the proof.Step 1.
Suppose thatu≠0. Then, Iu≥m. By the Fatou lemma and the weakly lower semicontinuity of norm, we have
(41)m=liminfn⟶∞Iun−14I′un,un=liminfn⟶∞14unV2+14−1p∫ℝ3Qxunpdx≥14uV2+14−1p∫ℝ3Qxupdx=Iu−14I′u,u=Iu≥m,which deduces Iu=m.Step 2.
Suppose thatu=0. Define
(42)β=limsupn⟶∞supz∈ℝ3∫B1zun2dx.
Ifβ=0, the Lions lemma [12] implies un⟶0 in Lsℝ3 for each s∈2,6. We have
(43)unV2+∫ℝ3ϕunun2dx=∫ℝ3Qxunpdx≤Q∞∫ℝ3unpdx=o1,and then m=Iun+o1=0. It is a contradiction. Thus, β>0. Up to a subsequence, there exists R>0 and zn⊂ℤ3 such that
(44)∫BR0un·+zn2dx=∫BRznun2dx>β2.
Definewn=un·+zn. Because wn=un≤C, there exists w∈H such that up to a subsequence, wn⟶w in H, wn⟶w in Lloc2ℝ3, and wnx⟶wx, a.e., in ℝ3. Note that
(45)∫BR0wn2dx=∫BR0un·+zn2dx.
Thus,w≠0. If zn is bounded, there exists R′>0 such that
(46)∫BR′0un2dx≥∫BRznun2dx>β2,which contradicts with un⟶u=0 in Lloc2ℝ3. Thus, zn is unbounded. Up to a subsequence, we may assume zn⟶∞. Coupling Lemma 10, we get that for any φ∈C0∞ℝ3,
(47)0=I′un,v·−zn+o1=∫ℝ3∇un·∇v·−zn+Vxunv·−zndx+∫ℝ3ϕununv·−zndx−∫ℝ3Qxunp−2unv·−zndx+o1=∫ℝ3∇un·∇v·−zn+V∞xunv·−zndx+∫ℝ3ϕununv·−zndx−∫ℝ3Q∞xunp−2unv·−zndx+o1=∫ℝ3∇wn·∇v+V∞xwnvdx+∫ℝ3ϕwnwnvdx−∫ℝ3Q∞xwnp−2wnvdx+o1=∫ℝ3∇w·∇v+V∞xwvdx+∫ℝ3ϕwwvdx−∫ℝ3Q∞xwp−2wvdx,i.e., w is a nontrivial solution of Eq. (10). Using the Fatou lemma, the weakly lower semicontinuity of norm, Lemma 6, and Lemma 11m we imply
(48)m=Iun−14I′un,un+o1=14unV2+14−1p∫ℝ3Qxunpdx+o1=14∫ℝ3∇un2+V∞xun2dx+14−1p∫ℝ3Q∞xunpdx+o1=14∫ℝ3∇wn2+V∞xwn2dx+14−1p∫ℝ3Q∞xwnpdx+o1≥14∫ℝ3∇w2+V∞xw2dx+14−1p∫ℝ3Q∞xwpdx=I∞w−14I∞′w,w=I∞w≥m∞≥m.
Thus,I∞w=m. Because w is a nontrivial solution of Eq. (10), w∈N. By Lemma 5, there exists tw>0 such that tww∈M. Then, one has
(49)m≤Itww≤I∞tww≤I∞w=m.
Therefore,m is achieved.Lemma 14.
Ifm is achieved by u∈H, then u is a solution of Eq. (9).Proof.
We borrow the idea in [14]. Suppose by contradiction that u is not a solution of Eq. (9). Then there exists v∈H such that I′u,v<−1. Let ε∈0,1 small enough such that for all t−1≤ε and σ≤ε, I′tu+σv,v≤−1/2. We define a smooth cut-off function 0≤ζ≤1 satisfying ζt=1 for t−1≤ε/2 and ζt=0 for t−1≥ε. For t>0, we introduce a curve γt=tu for t−1≥ε and γt=tu+εζtv for t−1<ε. Obviously, γt is a continuous curve and when ε small enough, γt>0 for t−1<ε. We claim Iγt<m, for all t>0. Indeed, if t−1≥ε, Iγt=Itu<Iu=m. If t−1<ε, owing to 0,ε∋σ↦Itu+σζtv is of C1, there exists σ¯∈0,ε such that
(50)Itu+εζtv=Itu+I′tu+σ¯ζtv,εζtv≤Itu−ε2ζt<Iu=m.
Recall thatJu=I′u,u, then Jγ1−ε=J1−εu>0 and Jγ1+ε=J1+εu<0. By the continuity of t↦Jγt there exists t′∈1−ε,1+ε such that Jγt′=0. Thus γt′∈M and Iγt′<m, which is a contradiction. Therefore, u is a solution of Eq. (9).Proof of Theorem 1.
From Lemma13 we know that m is achieved by u∈H. Note that the functionals I,J are even. Thus m is achieved by a nonnegative function u∈H. It follows from Lemma 14 that u is a nonnegative solution of Eq. (9). The strong maximum principle implies that u>0.



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