# Ground State Solutions for Schrödinger Problems with Magnetic Fields and Hardy-Sobolev Critical Exponents

Authors: Min Liu; Xiaorui Yue
Journal: Journal of Function Spaces
 (2018)

---

## Abstract

A Schrödinger equation and system with magnetic fields and Hardy-Sobolev critical exponents are investigated in this paper, and, under proper conditions, the existence of ground state solutions to these two problems is given.


---

## Body

## 1. Introduction

In the present paper, we consider the following semilinear stationary Schrödinger equation and system with magnetic fields and Hardy-Sobolev critical exponents:(1)-i∇+A2u=λup-2uxs+μu2∗s1-2uxs1,u∈HA1Ω,u=0,on∂Ω,(2)-i∇+A2u=λ1up-2uxs+μ1u2∗s1-2uxs1+αγ2∗s2uα-2uvβxs2,-i∇+B2v=λ2vp-2vxs+μ2v2∗s1-2vxs1+βγ2∗s2uαvβ-2vxs2,u∈HA1Ω,v∈HB1Ω,u=v=0,on∂Ω,where u,v:RN→C,N≥4, A=(A1,⋯,AN),B=(B1,⋯,BN):RN→RN are magnetic potentials, 0≤s<2,0<s1,s2<2,λ,λ1,λ2,μ,μ1,μ2,γ>0, α,β>1 with α+β=2∗(s2)≔2(N-s2)/(N-2), and Ω is a smooth bounded domain containing the origin as a interior point. Set -ΔA≔(-i∇+A)2,∇A≔∇+iA, and (3)HA1Ω≔u∈L2Ω:∇Au∈L2Ω.Then -ΔAu=-Δu-iudivA-2iA·∇u+|A|2u; HA1(Ω) is a Hilbert space obtained by the closure of Cc∞(Ω,C) with respect to scaler product: (4)Re∫Ω∇Au·∇Av¯,where the bar denotes complex conjugation. Here and in the following, ∫· means ∫·dx. We regard the range of function as C, except for emphasizing, that is, R. Lp(Ω,dx/|x|s) denotes the space of Lp-integrable functions with respect to the measure dx/|x|s, endowed with norm (5)up,s≔∫Ωupxs1/p.For Ω=RN, denote the Lp(Ω,dx/|x|s) norm by (6)up,s,RN≔∫RNupxs1/p.Write |u|p≔|u|p,0 and |u|p,RN≔|u|p,0,RN for simplicity.The first existence results for this kind of problems with a nonsingular magnetic potential were established in [1]. Motivated by the seminal work [1], some papers dealt with the Schrödinger equations with nonsingular magnetic field, for example, [2–11] and references therein. For the singular magnetic potential, we refer to [12, 13]. There are a few works about the nonsingular magnetic problems with critical exponents, such as [9, 14, 15] for Sobolev critical exponent and [16] for Hardy critical exponent.As far as we have known, there is no result about the problems of this type with Hardy-Sobolev critical exponents, especially for the system case. The current paper is mainly motivated by [15] and applies the least energy solution obtained in [17, 18] to show the existence of the ground state solution to the critical system (2) under proper conditions. Precisely, we get the following results.Theorem 1.
Assume thatA∈LlocN(RN,RN). If curlA≡0 or A is continuous at 0, where the curlA is the N×N skew-symmetric matrix with entries ajk=∂jAk-∂kAj, then (1) has a nontrivial ground state solution.Theorem 2.
Assume thatA,B∈LlocN(RN,RN) and α,β<2. If curlA≡0≡curlB or A and B are both continuous at 0, then (2) has a nontrivial ground state solution.The corresponding energy functionalE:HA,B→R of (2) is (7)Eu,v=12u,vHA,B2-1pλ1up,sp+λ2vp,sp-12∗s1μ1u2∗s1,s12∗s1+μ2v2∗s1,s12∗s1-γ2∗s2∫Ωuαvβxs2,where HA,B≔HA1(Ω)×HB1(Ω), endowed with norm (8)u,vHA,B2≔∇Au22+∇Bv22.Define (9)M≔u,v∈HA,B∖0,0:u,vHA,B2=λ1up,sp+λ2vp,sp+μ1u2∗s1,s12∗s1+μ2v2∗s1,s12∗s1+γ∫Ωuαvβxs2and M≔inf(u,v)∈ME(u,v). By nontrivial solutions (u,v)∈HA,B of (2), we mean u≠0,v≠0. A solution of (2) is called a ground state solution if (u,v)∈M and E(u,v)=M. A ground state solution is semitrivial if it is of type (u,0) or (0,v). Similar definitions are verified for single equation (1).Remark 3.
Although it has been studied that the existence of ground state solutions is heavily dependent on the type of equations without magnetic fields, the symmetric and decaying information about ground state solutions of equations with magnetic fields is not known. In this paper, we find that the latter equation and system also indeed have the ground state solutions provided suitable conditions.


## 2. Preliminaries

Consider(10)-Δu=μ1u2∗s1-2uxs1+αγ2∗s2uα-2uvβxs2,-Δv=μ2v2∗s1-2vxs1+βγ2∗s2uαvβ-2vxs2,u,v∈D1,2RN,and the corresponding energy functional I:D→R(11)Iu,v=12u,vD2-12∗s1μ1u2∗s1,s1,RN2∗s1+μ2v2∗s1,s1,RN2∗s1-γ2∗s2∫RNuαvβxs2,where D≔D1,2(RN)×D1,2(RN), endowed with the norm (12)u,vD2≔∇u2,RN2+∇v2,RN2.Define (13)N≔u,v∈D∖0,0:u,vD2=μ1u2∗s1,s1,RN2∗s1+μ2v2∗s1,s1,RN2∗s1+γ∫RNuαvβxs2and M0≔inf(u,v)∈NI(u,v). Then, by [18], we see that, under the condition α,β<2, (10) has a positive ground state solution (U,V), where U and V are radial symmetric decreasing with the following decay information: (14)Ux+Vx≤C1+x2-N,∇Ux+∇Vx≤C1+x1-N.That is, M0 is attained by (U,V). Define (15)Uεx,Vεx≔ε-N-2/2Uxε,ε-N-2/2Vxε.As proved in [1, 19], for any u∈HA1(Ω), the followingdiamagnetic inequality holds pointwise for almost every x∈Ω:(16)∇u=Re∇uu¯u=Re∇u+iAuu¯u≤∇Au.Then, for u∈HA1(Ω), we see that |u| belongs to the usual Sobolev space H01(Ω). Moreover, we have the following.Lemma 4 (see [20]).
The embeddingHA1Ω↪Lp(Ω,dx/|x|s) is continuous for 1≤p≤2∗(s), and it is compact for 1≤p<2∗(s), where 0≤s<2.Lemma 5.
Ifun⇀0 weakly in HA1(Ω), where A∈LlocN(RN,RN), then ∫Ω|∇Aun|2=∫Ω|∇un|2+o(1).Proof.
Obviously,|un|∈H01(Ω). Lemma 4 ensures that un⇀0 weakly in L2∗(Ω) and un→0 a.e. in Ω. Then, un2⇀0 weakly in L2∗/2(Ω). Since |A|2∈LlocN/2(RN)=Lloc2∗/2RN′, the dual space of Lloc2∗/2(RN), we see that (17)∫ΩA2un2=A2,un2→0,where the duality product is taken with respect to LN/2(Ω) and L2∗/2(Ω). It is easy to see that (18)∫Ω∇Aun2=∫Ω∇un2+∫ΩA2un2+2Re∫ΩiAun∇un¯.By Hölder inequality and (17), we have (19)Re∫ΩiAun∇un¯≤∫Ω∇un21/2∫ΩA2un21/2→0.Thus, the proof is completed by (17), (18), and (19).Remark 6.
It is easy to see thatM is closed and bounded away from 0, for any (u,v)∈HA,B∖{(0,0)}, there exists unique tu,v>0 such that (tu,vu,tu,vv)∈M. M>0 follows from Lemma 4. It is standard to see that any (PS) sequence of E with constraint M is a (PS) sequence of E without constraints and every (PS) sequence of E is bounded in HA,B. Define (20)M=infu,v∈HA,B∖0,0⁡supt>0⁡Etu,tv,M=infh∈Γ⁡maxt∈0,1⁡Eht,where Γ≔{h∈C([0,1],HA,B):h(0)=(0,0),I(h(1))<0}. Then it can be proved that M=M=M.


## 3. Ground States

In this section, we firstly suppose that Theorem1 holds, and then we show Theorem 2 below. In fact, the proof of Theorem 1 is similar to and easier than that of Theorem 2 by the application of the ground state solutions to -Δu=|u|2∗(s1)-2u/|x|s1 (cf. [17]), so we shall show Theorem 1 briefly as a remark in the end of this section.By Theorem1, we see that (21)-ΔAu=λ1up-2uxs+μ1u2∗s1-2uxs1,u∈H01Ωand (22)-ΔBu=λ2up-2uxs+μ2u2∗s1-2uxs1,u∈H01Ωhave ground state solutions uμ1 and uμ2 with energies (23)Mμ1≔12∫Ω∇Auμ12-λ1p∫Ωuμ1pxs-μ12∗s1∫Ωuμ12∗s1xs1and (24)Mμ2≔12∫Ω∇Buμ22-λ2p∫Ωuμ2pxs-μ22∗s1∫Ωuμ22∗s1xs1Then, we have the following.Lemma 7.
Assume thatA,B∈LlocN(RN,RN) and α,β<2. If curlA≡0≡curlB or A and B are continuous at 0, then M<min{Mμ1,Mμ2,M0}.Proof.
Firstly, we prove thatM<M0. There exists ρ>0 satisfying B2ρ≔{x:|x|≤2ρ}⊂Ω.
(i) IfcurlA≡0≡curlB holds, then by [21], there exist φ,ψ∈Wloc1,N(RN,R) such that ∇φ=A,∇ψ=B. Let (25)uεx,vεx=ϕUεxe-iφx,ϕVεxe-iψx,where (Uε,Vε) is defined by (15) and ϕ∈C01(B2ρ) is a cut-off function with 0≤ϕ≤1 and ϕ≡1 for |x|≤ρ. By direct computation, we have (26)∫Ω∇Auε2=∫Ω∇ϕUε2≤∫RN∇U2+OεN-2,(27)∫Ωuε2∗s1xs1≥∫RNU2∗s1xs1+OεN-s1,(28)∫Ωuεαvεβxs2≥∫RNUαVβxs2+OεN-s2,(29)∫Ωuεpxs≥CεN-N-2p/2-s+OεN-2p/2,(30)∫Ω∇Bvε2=∫Ω∇ϕVε2≤∫RN∇V2+OεN-2,(31)∫Ωvε2∗s1xs1≥∫RNV2∗s1xs1+OεN-s1,(32)∫Ωvεpxs≥CεN-N-2p/2-s+OεN-2p/2.By (26)–(32), for any t>0, we deduce that (33)Etuε,tvε=t22uε,vεHA,B-tppλ1uεp,sp+λ2vεp,sp-t2∗s12∗s1μ1uε2∗s1,s12∗s1+μ2vε2∗s1,s12∗s1-t2∗s2γ2∗s2∫Ωuεαvεβxs2≤t22Aε-tppBε-t2∗s12∗s1Cε-t2∗s22∗s2Dε≤tε22Aε-tεppBε-tε2∗s12∗s1Cε-tε2∗s22∗s2Dε,where (34)Aε=U,VD2+OεN-2,Bε=dεN-N-2p/2-s+OεN-2p/2,Cε=μ1U2∗s1,s1,RN2∗s1+μ2V2∗s1,s1,RN2∗s1+OεN-s1,Dε=γ∫RNUαVβxs2+OεN-s2,d is a positive constant, and t(ε) solves (35)Aε-tεp-2Bε-tε2∗s1-2Cε-tε2∗s2-2Dε=0.Then, (35) yields that t′(ε)=P(ε)/Q(ε), where (36)Pε=OεN-3-tp-2dN-N-2p2-sεN-N-2p/2-s-1+OεN-2p/2-1-t2∗s1-2OεN-s1-1-t2∗s2-2OεN-s2-1,Qε=p-2tp-3Bε+2∗s1-2t2∗s1-3Cε+2∗s2-2t2∗s2-3Dε.Since t(0)=1, we see that as ε→0, (37)Pε=-dN-N-2p2-sεN-N-2p/2-s-11+o1,Qε=2∗s1-2μ1U2∗s1,s1,RN2∗s1+μ2V2∗s1,s1,RN2∗s1+2∗s2-2γ∫RNUαVβxs21+o1.Setting κ=N-(N-2)p/2-s and (38)Λ=d2∗s1-2μ1U2∗s1,s1,RN2∗s1+μ2V2∗s1,s1,RN2∗s1+2∗s2-2γ∫RNUαVβ/xs2,we have (39)t′ε=-Λκεκ-11+o1,tε=1-Λεκ1+o1,tε2=1-2Λεκ1+o1,tεp=1-pΛεκ1+o1,tε2∗s1=1-2∗s1Λεκ1+o1,tε2∗s2=1-2∗s2Λεκ1+o1.Then, by (33), we deduce that (40)Etuε,tvε≤1-2Λεκ1+o12Aε-1-pΛεκ1+o1pBε-1-2∗s1Λεκ1+o12∗s1Cε-1-2∗s2Λεκ1+o12∗s2Dε=12U,VD2-12∗s1μ1U2∗s1,s1,RN2∗s1+μ2V2∗s1,s1,RN2∗s1-γ2∗s2∫RNUαVβxs2-dεκp1+o1<M0,which implies that (41)M≤supt>0⁡Ituε,tvε<M0.
(ii) IfA and B are continuous at 0, then setting (42)θx≔-∑j=1NAj0xj,ϑx≔-∑j=1NBj0xj,we have (43)∇θx=-A10,⋯,-AN0=-A0,∇ϑx=-B10,⋯,-BN0=-B0,which imply that (∇θ+A)(0)=0 and (∇ϑ+B)(0)=0. Then, there exists δ>0 such that (44)∇θ+Ax2≤m,∇ϑ+Bx2≤m,∀x<δ,where m is a positive number. Let 2r≔min{δ,2ρ} and (45)uεx,vεx=ϕxUεxeiθx,ϕxVεxeiϑx,where ϕ∈C01(B2r) is a cut-off function such that ϕ(x)=1 in Br and (Uε,Vε) is defined by (15). By (44), we deduce that (46)∫Ω∇Auε2+∇Bvε2=∫Ω∇ϕUε2+ϕ2Uε2∇θ+A2+∇ϕVε2+ϕ2Vε2∇ϑ+B2≤∫RN∇U2+∇V2+OεN-2+m∫B2rϕ2Uε2+Vε2≤∫RN∇U2+∇V2+Oε2.Hence, replacing Aε with (U,V)D2+O(ε2) and following the previous proof for case curlA≡0≡curlB, we may prove that M<M0 with minor modifications.
Next to proveM<Mμ1, recalling (9), choose t(ϵ)≔tuμ1,ϵuμ1 satisfying (47)∇Auμ12+ϵ2∇Buμ12=tϵp-2λ1+λ2ϵpuμ1p,sp+tϵ2∗s1-2μ1+μ2ϵ2∗s1uμ12∗s1,s12∗s1+tϵ2∗s2-2γuμ12∗s2,s22∗s2ϵβ.Then, (t(ϵ)uμ1,t(ϵ)ϵuμ1)∈M. Noting that |∇Auμ1|2=λ1|uμ1|p,sp+μ1|uμ1|2∗(s),s2∗(s) and t(0)=1, we deduce that (48)t′ϵ=-Λ1βϵβ-2ϵ1+o1,tϵ=1-Λ1ϵβ1+o1,tϵp=1-pΛ1ϵβ1+o1,tϵ2∗s1=1-2∗s1Λ1ϵβ1+o1,tϵ2∗s2=1-2∗s2Λ1ϵβ1+o1,where (49)Λ1=γuμ12∗s2,s22∗s2p-2λ1uμ1p,sp+2∗s1-2μ1uμ12∗s1,s12∗s1.Then, (50)M≤Itϵuμ1,tϵϵuμ1=12-1pλ1+λ2ϵpuμ1p,sptϵp+12-12∗s1μ1+μ2ϵ2∗s1uμ12∗s1,s12∗s1tϵ2∗s1+12-12∗s2γϵβuμ12∗s2,s22∗s2tϵ2∗s2=12-1pλ1uμ1p,sp+12-12∗s1μ1uμ12∗s1,s12∗s1-12-1pλ1uμ1p,sppΛ1+12-12∗s2μ1uμ12∗s1,s12∗s12∗s1Λ1-12-12∗s2γuμ12∗s2,s22∗s2ϵβ+oϵβ<Mμ1-12+12∗s2γuμ12∗s2,s22∗s2ϵβ+oϵβ<Mμ1forϵsmall  enough.Similarly, M<Mμ2.Proof of Theorem2.
Note that the functionalE has a mountain pass structure. It follows from the mountain pass theorem (see [22]) that there exists {(un,vn)}⊂HA,B such that (51)limn→∞⁡Eun,vn=M,limn→∞⁡E′un,vn=0.By Remark 6, we see that {(un,vn)} is bounded in HA,B. By Lemma 4, we may assume that(52)un⇀uweakly  inHA1Ω,vn⇀vweakly  inHB1Ω,un→u,vn→vstrongly  inL2Ω,un⇀u,vn⇀vweakly  inL2∗s1Ω,dxxs1,unq-1un⇀uq-1u,vnq-1vn⇀vq-1vweakly  inL2∗s1/qΩ,dxxs1,unα-2unvnβ⇀uα-2uvβ,unαvnβ-2vn⇀uαvβ-2vweakly  inL2∗s2/2∗s2-1Ω,dxxs2,where q∈(1,2∗(s1)). From (53)∫Ω∇Aun∇Aφ¯+∇Bvn∇Bψ¯-λ1∫Ωunp-2unφ-xs-λ2∫Ωvnp-2vnψ-xs-μ1∫Ωun2∗s1-2unφ-xs1-μ2∫Ωvn2∗s1-2vnψ-xs1-αγ2∗s2∫Ωunα-2unφ-vnβxs2-βγ2∗s2∫Ωunαvnβ-2vnψ-xs2=o1,∀φ,ψ∈H,we get that (54)∫Ω∇Au∇Aφ¯+∇Bv∇Bψ¯-λ1∫Ωup-2uφ-xs-λ2∫Ωvp-2vψ-xs-μ1∫Ωu2∗s1-2uφ-xs1-μ2∫Ωv2∗s1-2vψ-xs1-αγ2∗s2∫Ωuα-2uφ-vβxs2-βγ2∗s2∫Ωuαvβ-2vψ-xs2=0,i.e., E′(u,v)=0 in HA,B′, the dual space of HA,B. Setting ωn=un-u and σn=vn-v, then ωn⇀0 in HB1(Ω) and σn⇀0 in HB1(Ω). Thus, ωn→0,σn→0 strongly in L2(Ω), ωn→0,σn→0 a.e. in Ω, up to a subsequence. Brezis-Lieb Lemma guarantees that (55)un2∗s1,s12∗s1=u2∗s1,s12∗s1+ωn2∗s1,s12∗s1+o1,vn2∗s1,s12∗s1=v2∗s1,s12∗s1+σn2∗s1,s12∗s1+o1,∫Ωunαvnβxs2=∫Ωuαvβxs2+∫Ωωnασnβxs2+o1.Noting that E′(un,vn)(un,0)=o(1),E′(un,vn)(0,vn)=o(1), and E′(u,v)(u,v)=0, by (55) and Lemma 5, we deduce that(56)∫Ω∇ωn2-∫Ωμ1ωn2∗s1xs1+αγ2∗s2ωnασnβxs2=o1,∫Ω∇σn2-∫Ωμ2σn2∗s1xs1+βγ2∗s2ωnασnβxs2=o1,and (57)Iωn,σn=Eun,vn-Eu,v+o1.Assume that (58)limn→∞⁡∫Ωμ1ωn2∗s1xs1=a1,limn→∞⁡∫Ωμ2σn2∗s1xs1=a2,limn→∞⁡∫Ωγωnασnβxs2=b,up to a subsequence. By (56), we see that (59)Iωn,σn=12-12∗s1a1+a2+12-12∗s1b+o1.Letting n→∞, we have (60)M=limn→∞⁡Eun,vn=12-12∗s1a1+a2+12-12∗s1b+Eu,v≥Eu,v≥0.
Case 1 (u≡0,v≡0). By M>0 and (60), we see that a1+a2+b>0. Then, we may assume that (ωn,σn)≠(0,0) for n large enough. Recalling (56), (57), and N in (13), there exists tn>0 with tn→1 as n→∞ such that (tnωn,tnσn)∈M. Then, by (60), we obtain that (61)M=12-12∗s1a1+a2+12-12∗s1b=limn→∞⁡Iωn,σn=limn→∞⁡Itnωn,tnσn≥M0,which contradicts Lemma 7. Hence, u≡0,v≡0 is impossible.
Case 2 (u≢0,v≡0 or u≡0,v≢0). Without loss of generality, we may assume that u≢0,v≡0. Thus, u is a nontrivial solution of ΔAu+λ1(|u|p-2u/|x|s)+μ1|u|2∗(s1)-2u/|x|s1=0. Then, by (60), we deduce that M≥E(u,0)≥Mμ1, a contradiction to Lemma 7. Therefore, Case 2 is impossible.
Since Cases 1 and 2 are both impossible, we haveu≢0,v≢0. Remark 6 yields that E(u,v)=M=M. It follows from (9) and (20) that there exists a Lagrange multiplier L∈R such that (62)E′u,v+LF′u,v=0,where (63)Fu,v=u,vHA,B2-λ1up,sp+λ2vp,sp-μ1u2∗s1,s12∗s1+μ2v2∗s1,s12∗s1-γ∫Ωuαvβxs2,Since E′u,vu,v=Fu,v=0 and (64)F′u,vu,v=-p-2λ1up,sp+λ2vp,sp-2∗s1-2μ1u2∗s1,s12∗s1+μ2v2∗s1,s12∗s1-2∗s2-2γ∫Ωuαvβxs2<0,we obtain that L=0 and, hence, E′(u,v)=0 in HA,B′. That is, (u,v) is a ground state solution of (2).Remark 8.
It follows from Remark6 that the ground state solution of (2) is also a mountain pass solution.Remark 9.
Note that we only use the result in Theorem1 to construct the values of Mμ1 and Mμ1 which are used in Lemma 7 and then in Case 2 at last of the proof of Theorem 2 to exclude the semitrivial solutions as (u,0) and (0,v) to system (2), that is to say that the part of proof of Theorem 2 before Case 2 is independent of Theorem 1. So we can repeat the proof above with a little modification to prove Theorem 1. Indeed, the existence of a ground state solution to -Δu=|u|2∗(s1)-2u/|x|s1 is ensured in [17], and we only need to exclude the trivial solution of u=0 to (1) which is similar to Case 1 in the proof for the system.



---

*Source: 2382717-2018-08-01.xml*