带有两个奇异项的p(x)-Laplace方程解的存在性多解性研究.pdf

第 43 卷第 2 期2023 年 6 月数学理论与应用MATHEMATICAL THEORY AND APPLICATIONSVol.43No.2Jun.2023Existence and Multiple Solutions ofp(x)Laplace Equation with TwoSingular TermsHu XincunChen Haibo(School of Mathematics and Statistics,Central South University,Changsha 410083,China)AbstractInthispaper,westudytheexistenceandmultiplicityofpositivesolutionsforthefollowingdoublesingularproblem with p(x)Laplace operatorp(x)u+V(x)|u|p(x)2u=|u|s(x)2u|x|s(x)+h(x)u(x)in,u=0on.Due to the presence of singular term u(x)and singular potential|x|s(x)in the equation,it is more difficult to dealwith the existence of positive solutions.By using the decomposition of Nehari manifold and some refined estimates,we show that there admits at least two positive solutions for the double singular problem.Key wordsNehari manifoldp(x)Laplace operatorSingular termsVariational method带有两个奇异项的p(x)Laplace 方程解的存在性多解性研究胡新存陈海波(中南大学数学与统计学院,长沙,410075)摘要本文研究带有两个奇点的 p(x)Laplace 算子p(x)u+V(x)|u|p(x)2u=|u|s(x)2u|x|s(x)+h(x)u(x)in,u=0on的正解的存在性和多解性.由于上述方程中奇异项 u(x)和|x|s(x)的出现,使得其正解存在性的证明更加困难.我们通过使用 Nehari 流形的分解和一些精确的估计,证明上述方程至少有两个正解.关键词Nehari 流形p(x)Laplace 算子奇数项变分法doi:10.3969/j.issn.10068074.2023.02.0051IntroductionIn this paper,we consider the following double singular problem with p(x)Laplace operator(p(x)u+V(x)|u|p(x)2u=|u|s(x)2u|x|s(x)+h(x)u(x)in,u=0on,(P,)This work is supported by the National Natural Science Foundation of China(No.12071486)收稿日期:2022 年 5 月 25 日带有两个奇异项的 p(x)Laplace 方程解的存在性多解性研究69where is a bounded domain in RN(N 3)with C2boundary,and are positive parameters,h(x)C()is the nonnegative potential function with compact support in,(x):(0,1)is continuous,and p(x),s(x)C+()=?q C():q(x)1,x?.For any continuous and bounded function m(x),letm+:=esssupm(x),m:=essinfm(x).The following are two assumptions on functions s(x),(x),p(x)and V(x)in the double singularproblem(P,):(A0)0 1 +1 (x)1 p p(x)p+s s(x)s+maxp(x),N,where p(x)=Np(x)Np(x)(V0)V C(),0 V V+.The p(x)Laplace operator p(x)is defined as p(x)u=div(|u|p(x)2)u,which a naturalgeneralization of the pLaplace operator pu=div(|u|p2)u.However,p(x)possesses morecomplicated nonlinearity than the pLaplace operator due to the fact that p(x)is not homogeneous.Recently,the study of the partial differential equations with variable exponents has received considerableattention.The interest can be justified by many physical applications,such as elastic mechanics,electrorheo logical fluids 22,image restoration 4,dielectric breakdown,electrical resistivity,polycrystal plasticity 3 and continuum mechanics 2.Theexistenceandqualitativepropertiesofnontrivialsolutionsforp(x)Laplacianproblemshavebeenextensively investigated,such as 1,7,8,15,19,21,23,24,25,27,28 and the references therein.Someinteresting papers on the application of the Nehari manifold method in variable exponent problems haverecently been published,see,for example,1,19,23,24,25.Let us briefly review some results relatedto our work.In 19,Mashiyev,Ogras,Yucedag and Avci investigated the existence and multiplicity ofsolutions for the p(x)Laplacian Dirichlet problem(p(x)u=a(x)uq(x)2+b(x)|u|h(x)2uin,u=0on,whereisaboundeddomainwithsmoothboundaryinRN(N 2),isapositiveparameter,a(x),b(x)C()are nonnegative potential functions with compact support in,p(x),q(x),h(x)C(),satisfying1 q(x)p(x)h(x)0in,|u|p(x)2 uv=b(x)|u|q(x)2uon,70数学理论与应用where is a bounded domain in RN(N 2)with C2boundary,is a positive parameter,a(x),b(x)C()are nonnegative potential functions with compact support in,p(x),q(x),(x)C(),such that0 1 +1 (x)1 p p(x)p+q q(x)q+0,0 0,such that Eq.(P,)has at least two positive solutions for all (0,0),(0,0).Throughout this paper,we use ci(i=1,2,)to denotes the general nonnegative or positiveconstant.The remainder of this paper is organized as follows.In section 2 we give some preliminaryresults and in Section 3 we give the proof of Theorem 1.1.2PreliminariesHere we recall some results and basic properties on the variable exponent Lebesgue and Sobolevspaces,see,for example,5,10,11 and the references therein for more details.Denote by S()the set ofall measurable realvalued functions defined on.Note that two measurable functions are considered asthe same element of S()when they equal almost everywhere.LetLp(x)()=?u S():Z|u(x)|p(x)dx 0:Z?u(x)?p(x)dx 1).For any u Lp(x)()and v Lp(x)(),where1p(x)+1p(x)=1,we have the following Hlderinequality:Z|uv|dx?1p+1(p)?|u|p(x)|v|p(x).(2.1)带有两个奇异项的 p(x)Laplace 方程解的存在性多解性研究71Now,we introduce the modular of the LebesgueSobolev space Lp(x)()as the mapping p(x):Lp(x)()R defined byp(x)(u)=Z|u|p(x)dx,u Lp(x)().Define the variable exponent Sobolev space W1,p(x)()byW1,p(x)()=nu Lp(x)():|u|Lp(x)()o,equipped with the normuW1,p(x)()=|u|p(x)+|u|p(x).Let W1,p(x)0()be the closure of C0()in W1,p(x)()with respect to the above norm.Proposition 2.1(9)If u Lp(x)()and un Lp(x)(),then the following properties holdtrue:(i)|u|p(x)1|u|p+p(x)p(x)(u)|u|pp(x)(ii)|u|p(x)1|u|pp(x)p(x)(u)|u|p+p(x)(iii)limn|un|p(x)=0 limnp(x)(un)=0(iv)limn|un|p(x)=limnp(x)(un)=.Proposition 2.2(9)(i)The spaces Lp(x)(),W1,p(x)()and W1,p(x)0()are separable andreflexive Banach spaces.(ii)If r C+()and r(x)0 such that|u|p(x)C|u|p(x),u W1,p(x)().We see that up(x):=|u|p(x)and W1,p(x)()are equivalent norms in W1,p(x)0().We canalso define the modular function 0:W1,p(x)0()R given by0(u)=Z|u|p(x)dx.Proposition 2.3(9)Let u W1,p(x)0()and un W1,p(x)0().Then,the same conclusionsof Proposition 2.1 remain ture for up(x)and 0(u)instead of|u|p(x)and p(x)(u).Proposition 2.4(19)Let p(x)and q(x)be two measurable functions withp(x)L()and1 p(x)q(x)for a.e.x .If u Lq(x)()with u=0,then we haveminn|u|p+p(x)q(x),|u|pp(x)q(x)o|u|p(x)|q(x)maxn|u|pp(x)q(x),|u|p+p(x)q(x)o.72数学理论与应用In the following we consider the norm(noticing the assumption(V0)u=inf(0:Z?u?p(x)+V(x)?u?p(x)!dx 1).It is easy to see that u is an equivalent norm of up(x)in W1,p(x)0().Consequently,throughout thispaper,we shall simply denote the W1,p(x)0()norm by.Proposition 2.5The functional J:W1,p(x)0()R defined byJ(u)=Z?|u|p(x)+V(x)|u|p(x)?dxhas the following properties:(i)u 1 up+J(u)up.(ii)u 1 up J(u)up+.In particular,if u=1,then J(u)=1 Moreover,un 0 if and only if J(un)0.The following classical HardyLittlewoodSobolev inequality will be frequently used.Theorem 2.1(15 HardyLittlewoodSobolev inequality)Suppose that the exponent functionsp,q C()satisfy1 p p(x)p+0 such that the(p(),s()Hardy inequalityZ|u|s(x)|x|s(x)dx 1RZ|u|p(x)dx+Cs+us+,u W1,p(x)0()holds,whereR:=max?Rp,Rs,Rs+?,is a positive constant in the(p(),s(.)Hardy inequality,withRp:=?N pp?p,Rs+:=?N s+s+?s+,Rs:=?N ss?s,and C is the best constant of the embedding W1,p(x)0()into Ls+().Next,we recall the following strong maximum principle.Theorem 2.2(26)Suppose that for some 0 1,u,v C1,(),we have 0 u,0 v,andp(x)u+V(x)|u|p(x)2u u(x)=g(x)q(x)=p(x)v+V(x)|v|p(x)2v v(x)with u=v=0 on,and g,q L(),0 q 0andvn 0on,where n is the inward unit normal on.Then u v in.带有两个奇异项的 p(x)Laplace 方程解的存在性多解性研究73Theorem 2.3(25)Suppose that the domain has the cone property and function p C().Assume that h L(x)(),h(x)0 for x ,C+().If C()and0 1 (x)0 suchthatZh(x)|u|1(x)dx(c1u1ifu 1,c2u1+ifu 0,hu(t)=0 if and only if tu N,.In particular,hu(1)=0 if and only if u N,.Thus,it is natural to split N,into three parts corresponding to localminima,local maxima and points of inflection,defined as follows:N+,=u N,|hu(1)0=ntu W1,p(x)0()0|hu(t)=0,hu(t)0o,N0,=u N,|hu(1)=0=ntu W1,p(x)0()0|hu(t)=0,hu(t)=0o,N,=u N,|hu(1)0=ntu W1,p(x)0()0|h,u(t)=0,hu(t)1.By(2.2),(2.3),and Proposition 2.1,we haveE,(u)=Z|u|p(x)p(x)+V(x)p(x)|u|p(x)!dx Z|u|s(x)s(x)|x|s(x)dx Zh(x)1 (x)|u|1(x)dx1p+upsZ|u|s(x)|x|s(x)dx 1 +Zh(x)|u|1(x)dx=1p+up1s?Z|u|p(x)+V(x)|u|p(x)dx Zh(x)|u|1(x)dx?1 +Zh(x)|u|1(x)dx?1p+1s?up c1?11 +1s?u1.Since 0 1 p p+s,it is seen that E,as u .This implies that E,iscoercive and bounded below on N,.The proof is complete.Lemma 2.2Suppose that u is a local minimizer for E,on subsets N+,or N,of N,such thatu/N0,.Then u is a critical point of E,.Proof Since u is a local minimizer for E,under the constraint,(u)=?E,(u),u?=0,then,applying the theory of Lagrange multiplier,we get an R such thatE,(u)=,(u).Thus,we have?E,(u),u?=?,(u),u?=hu(1)=0.Moreover,u/N0,so hu(1)=0,therefore,=0.Consequently,E,(u)=0,which completes theproof.带有两个奇异项的 p(x)Laplace 方程解的存在性多解性研究75Lemma 2.3There exist 0,0,such that for every 0 0,0 1.By the definition of N0,and(2.3),wehave0=?,(u),u?=Zp(x)?|u|p(x)+V(x)|u|p(x)?dx Zs(x)|u|s(x)|x|s(x)dx Z(1 (x)h(x)|u|1(x)dxpZ?|u|p(x)+V(x)|u|p(x)?dx s+Z|u|s(x)|x|s(x)dx (1 )Zh(x)|u|1(x)dx=(p 1+)Z?|u|p(x)+V(x)|u|p(x)?dx (s+1+)Z|u|s(x)|x|s(x)dx,together with Theorem 2.1,we obtain?,(u),u?(p 1+)up(s+1+)?Cs+us+1RZ|u|p(x)?p 1+1R(1 s+)?up(s+1+)Cs+us+.Therefore,(s+1+)Cs+us+?p 1+1R(1 s+)?up,which implies thatu?p 1+1R(1 s+)(s+1+)Cs+?1s+p.(2.4)Let be sufficiently small,such thatp 1+1R(1 s+)0,that is,0 R(p 1+)s+1+=0 1,we obtain(p+s)up+(s 1+)c1u1 0.So,we haveu?c1(s+1)s p+?1p+1.(2.5)With(2.4)and(2.5),we have?p 1+1R(1 s+)(s+1+)Cs+?1s+p u?c1(s+1)s p+?1p+1.Now taking sufficiently small,such that hp1+1R(1s+)(s+1+)Cs+ihc1(s+1)sp+ip+1s+p=0,we obtain u 1,which is impossible.Therefore,N0,=.The proof is comlpete.3The minimizer on submanifold N+,and N,In this section,we prove the existence of minimizers for the energy functional E,in N+,and N,.By lemma 2.3,for all (0,0),(0,0),N,=N+,N,and by lemma 2.1,we have E,isbounded below on N,and hence also on N+,and N,so we can set+=infuN+,E,(u)and=infuN,E,(u).Lemma 3.1For all (0,0),(0,0),we have E,(u)0,which givesp+Z?|u|p(x)+V(x)|u|p(x)?dxsZ|u|s(x)|x|s(x)dx(1+)Zh(x)|u|1(x)dx 0,(3.1)andE,(u)1pZ?|u|p(x)+V(x)|u|p(x)?dx s+Z|u|s(x)|x|s(x)dx 1 Zh(x)|u|1(x)dx.(3.2)Now,multiplying(2.3)by(1 +)yields(1 +)?Z?|u|p(x)+V(x)|u|p(x)?dx?=(1 +)Z|u|s(x)|x|s(x)dx(1 +)Zh(x)|u|1(x)dx.(3.3)With(3.1)and(3.3),we obtainZ|u|s(x)|x|s(x)dx 1.By(2.3),(3.2)and(3.4),we haveE,(u)?1p11?Z?|u|p(x)+V(x)|u|p(x)?dx?1s+11?Z|u|s(x)|x|s(x)dx?1p11?1s+11?p+1+1 +s?up(p 1+)(s+p)s+(1 +)pup 0.Therefore,+=infuN+,E,(u)0.The proof is complete.Theorem 3.1For all (0,0),(0,0),there exists a minimizer of E,(u)on N+,.Proof Since E,is bounded below on N,and hence also on N+,there exists a sequence un N+,satisfying E,(un)infuN+,E,(u)=+0.Since E,is coercive,un is bounded inW1,p(x)0().Therefore,we can assume that un*u0weakly in W1,p(x)0().By the compact embeddingin Theorem 2.3,we obtain un u0in L1(x)h(x)().Next,we prove un u0in W1,p(x)0().It sufficesto prove that un u0strongly in W1,p(x)0().If not,suppose un u0in W1,p(x)0().ThenZ|u0|p(x)dx 1 ,for u0 1,we obtain+=limninfE,(un)limninf?1p+1s?Z?|un|p(x)+V(x)|un|p(x)?dx?11 +1s?Zh(x)|u0|1(x)dx?1p+1s?u0p c1?11 +1s?u010,which is a contradiction to Lemma 3.1,hence un u0strongly in W1,p(x)0().The proof is complete.78数学理论与应用Lemma 3.2For all (0,0),(0,0),we have E,(u)0 for all u N,.Proof Let u N,.By(2.3),we obtainE,(u)1p+Z?|u|p(x)+V(x)|u|p(x)?dx sZ|u|s(x)|x|s(x)dx 1 +Zh(x)|u|1(x)dx?1p+1s?Z?|u|p(x)+V(x)|u|p(x)?dx+?1s11 +?Zh(x)|u|1(x)dx.For convenience,assume u 1.By the condition 1 p,and theorem 2.3,we haveE,(u)?1p+1s?up+c1?1s11 +?u01?s p+p+s+c11 +ss(1 +)?up.Choosing 0 0.Since N+,N,=,by lemma 3.1,wehave u N,which cpmpletes the proof.Theorem 3.2For all (0,0),(0,0),there exists a minimizer of E,(v)on N,.Proof Since E,is bounded below on N,and hence also on N,there exists a sequence vn N,satisfying E,(vn)infvN,E,(v)=0.Since E,is coercive,vn is bounded inW1,p(x)0().Therefore,we can assume that vn*v0weakly in W1,p(x)0().By the compact embeddingin Theorem 2.3,we obtain vn v0in L1(x)h(x)().Moreover,using the Hlder type inequality(2.1),wededuce that?Z|vn|s(x)2vn(x)(vn(x)v0(x)|x|s(x)dx?2?|vn|s(x)1|x|s(x)?s(.)s(.)1|vn(x)v0(x)|s().Since vn v0strongly in Ls(x)(),we haveR|vn|s(x)|x|s(x)dx R|v0|s(x)|x|s(x)dx.Now,we prove that for each v0 N,there exists t 0 such that tv0 N,and E,(v0)E,(tv0).Indeed,since(v0)=Zp(x)?|v0|p(x)+V(x)|v0|p(x)?dx Zs(x)|v0|s(x)|x|s(x)dxZ(1 (x)h(x)|v0|1(x)dx,(tv0)=Zp(x)?|tv0|p(x)+V(x)|tv0|p(x)?dx Zs(x)|tv0|s(x)|x|s(x)dxZ(1 (x)h(x)|tv0|1(x)dxtp+p+Z?|v0|p(x)+V(x)|v0|p(x)?dx stsZ|v0|s(x)|x|s(x)dx(1 +)t1+Zh(x)|v0|1(x)dx,带有两个奇异项的 p(x)Laplace 方程解的存在性多解性研究79and 0 1 +p+s,we have tv0 N,.We claim that vn v0strongly in W1,p(x)0().If not,suppose vn v0in W1,p(x)0().ThenZ|v0|p(x)dx limninfZ|vn|p(x)dx,which follows thatE,(tv0)tp+pZ?|v0|p(x)+V(x)|v0|p(x)?dx tss+Z|v0|s(x)|x|s(x)dxt1+1 Zh(x)|v0|1(x)dxlimntp+pZ?|vn|p(x)+V(x)|vn|p(x)?dx limntss+Z|vn|s(x)|x|s(x)dxlimnt1+1 Zh(x)|vn|1(x)dxlimnE,(tvn)limnE,(vn)=infvNE,(v)=.So,we get E,(tv0)0.The proof is complete.Now,we complete the proof of Theorem 1.1.ByTheorem3.1andTheorem3.2,wehaveE,(u0)=infuN+,E,(u)=+0.Sinec E,(|u0|)=E,(u0),and likewise,E,(|v0|)=E,(v0),v0 N,we have u0,v0 0.By Lemma 2.2,both u0,v0are critical points of E,and therefore bothare weak solutions to Eq.(P,).By using the strong maximum 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