带位势的基尔霍夫方程规范解的存在性.pdf

Chin.Quart.J.of Math.2023,38(2):196209The Existence of Normalized Solution to the KirchhoffEquation with PotentialLIANG Yan-xia(School of Mathematics and Statistics,Guangdong University of Technology,Guangzhou 510520,China)Abstract:In this paper we discuss the following Kirchhoffequation(?a+bRR3|u|2dx?u+V(x)u+u=|u|q2u+|u|p2u in R3,RR3u2dx=c2,where a,b,and c are positive numbers,is unknown and appears as a Lagrangemultiplier,143qp6 and V is a continuous non-positive function vanishing at in-finity.Under some mild assumptions on V,we prove the existence of a mountain passnormalized solution.Here we study the existence of normalized solution to mass super-critical Kirchhoffequation with potential via the minimax principle and Nehari-Pohozaevmanifold.Keywords:Kirchhoffequation;Normalized solutions;Variational methods2000 MR Subject Classification:35A15,35B38,35J60CLC number:O175Document code:AArticle ID:1002-0462(2023)02-0196-14DOI:10.13371/ki.chin.q.j.m.2023.02.0071.IntroductionThis paper concerns the existence of normalized solution to the following Kirchhoffequation?a+bZR3|u|2dx?u+V(x)u+u=|u|q2u+|u|p2u in R3,(1.1)where a,b,are positive real numbers,is unknown and appears as a Lagrange multiplier,143qp6.Problem like Eq.(1.1)is introduced firstly by Kirchhoff6 in 1883 which can be seen asan extension of the classical DAlembert wave equation.It shows the free vibration of elasticstrings and describes the motions of moderately large amplitude.Received date:2023-03-13Biographies:LIANG Yan-xia(1997-),female,native of Guangzhou,Guangdong,postgraduate of GuangdongUniversity of Technology,engages in nonlinear functional analysis.Corresponding author LIANG Yan-xia:.196No.2LIANG Yan-xia:The Existence of Normalized Solution to197From the mathematical point of view,Eq.(1.1)is nonlocal since the appearance of the termRR3|u|2dx.This kind of problem has been paid much attention after the pioneering work ofLions 8,in which an abstract functional analysis framework was introduced.At present,there are two different ways to study Eq.(1.1).One is to regard the frequency as a given constant.In this situation,solutions are critical points of the following actionfunctional F:H1(R3)R,defined byF(u)=a2ZR3|u|2dx+b4(ZR3|u|2dx)2+12ZR3V(x)u2dx+2ZR3|u|2dxqZR3|u|qdx1pZR3|u|pdx.The other one is to consider the case R is unknown.In such a point of view,the mass isprescribed.Physicists call a solution u of Eq.(1.1)withRR3u2dx=c2a normalized solution.Compared with the problem without the normalization condition,it presents more diffficul-ties.A natural approach to such a question is by variational methods.This case seems verymeaningful from the physical point of view.Hence,we focus on this case.As far as we know,many scholars have focused on the following Kirchhoffequation?a+bZR3|u|2dx?u+V(x)u+u=|u|p2u in R3,(1.2)satisfying the normalization constraintRR3u2dx=c2.The corresponding energy functional isFV(u)=a2ZR3|u|2dx+b4(ZR3|u|2dx)2+12ZR3V(x)u2dx1pZR3|u|pdx.If V(x)0,the existence of normalized solutions of Eq.(1.2)has been studied in 1215.Inparticular,Ye in 13 studied Eq.(1.2)with V(x)0 and p(2,6).Using some simple energyestimates rather than the concentration-compactness principles,Zeng and Zhang 15 improvedthe results of 13.Other results about normalized solutions of Kirchhoffequation can referto 2,7.When V(x)6=0,normalized solutions of Eq.(1.2)have been studied in 1,9,10.In particular,in 10 the authors studied the existence of normalized solutions when V(x)satisfies appropriateassumptions in the mass super-critical case.Different from 10,we use the minimax methodgiven by N.Ghoussoub 3,Theorem 5.2 to study the problem.To the authors knowledge,there are few results in studying mountain pass type normalizedsolutions to Kirchhoffequations with potential and combined nonlinearities in the Sobolevsubcritical case.In this paper,inspired by 5,we are interested in the existence of normalizedsolution of Eq.(1.3)and our result seems to extend the results of 5,which studied the nonlinearSchr odinger equation with mixed nonlinearities,to a class of Kirchhoffequation:(?a+bRR3|u|2dx?u+V(x)u+u=|u|q2u+|u|p2u in R3,RR3u2dx=c2.(1.3)198CHINESE QUARTERLY JOURNAL OF MATHEMATICSVol.38Throughout this paper we will make the following assumptions on V:(V1)lim|x|+V(x)=supxR3V(x)=0,V(x)=V(|x|),there exists 01mina2,(143(q2)asuch that for any uH1(R3),|ZR3V(x)u2dx|1kuk22.(V2)V(x)exists for a.e.xR3,defineW(x):=12hV(x),xi.There exists 02min3(q2)(a1)4a,2a31,a21 such that for any uH1(R3),|ZR3W(x)u2dx|2kuk22.(V3)W(x)exists for a.e.xR3,defineY(x):=(3(q2)2)W(x)+hW(x),xi.There exists 033(q2)2a2a such that for any uH1(R3),|ZR3|Y(x)u2|dx|3kuk22.(V4)V(x)+W(x)0 a.e.on R3.Our main result is stated as follows:Theorem 1.1.Assume that143qp0,b0,c0and 0,there exists a couple(,u)RH1r(R3)solving Eq.(1.3)and Iq(u)=mq,c.This paper is organized as follows.In section 2,we give some preliminary results.In section3,we prove the main result.Notations:Throughout the paper we use standard notations.For 1p00 such thatkukqCqkukq2kuk1q2,for any uH1(R3),(2.1)where q=3(q2)2qand Uqis the unique positive solution ofU+(1q1)U=2qq|U|q2U.No.2LIANG Yan-xia:The Existence of Normalized Solution to199Hereafter,E denotes the spaceE=uH1r(R3):ZR3V(x)|u|2dx+(2.2)endowed with the normkuk=(ZR3|u|2+V(x)u2+u2dx)12.Moreover,under our assumptions of V in present paper,we deduce that the norm kuk isequivalent to the usual norm kukH.The embedding E,Lp0(R3,R)is continuous with p0(143,6).In what follows,we shall consider the functional Iq:E7R given byIq(u)=a2ZR3|u|2dx+b4(ZR3|u|2dx)2+12ZR3V(x)|u|2dxqZR3|u|qdx1pZR3|u|pdx,(2.3)restricted to the sphere in L2(R3):Sc=uE:ZR3u2dx=c2.We intend to consider the minimization of Iqon a subset of Scin this work.It is easy to seethat if u is a solution to Eq.(1.3),then the following Nehari identity holdsaZR3|u|2dx+b(ZR3|u|2dx)2+ZR3V(x)|u|2dx+ZR3|u|2dxZR3|u|pdxZR3|u|qdx=0.(2.4)Moreover,if uE is a solution to Eq.(1.3),u satisfies the following Pohozaev identitya2ZR3|u|2dx+b2(ZR3|u|2dx)2+32ZR3V(x)|u|2dx+12ZR3hV(x),xi|u|2dx+32Z|u|2dx3pZR3|u|pdx3qZR3|u|qdx=0.(2.5)Hence,by(2.4)and(2.5),u satisfiesPq(u)=0,(2.6)wherePq(u)=aZR3|u|2dx+b(ZR3|u|2dx)212ZR3hV(x),xi|u|2dxqZR3|u|qdxpZR3|u|pdx.To find the normalized solutions of Eq.(1.3),we introduce the following Pohozaev constrainedset Pq,c=uE:Pq(u)=0Sc.Obviously,the set Pq,ccontains all of the normalized solutionsto Eq.(1.3).Hence,this paper mainly studies the following minimization problemmq,c=infuPq,cIq(u).200CHINESE QUARTERLY JOURNAL OF MATHEMATICSVol.38For the limit problem to Eq.(1.3),namely(?a+bRR3|u|2dx?u+u=|u|q2u+|u|p2u in R3,RR3u2dx=c2.(2.7)Define the energy functional of Eq.(2.7)Iq(u)=a2ZR3|u|2dx+b4(ZR3|u|2dx)2qZR3|u|qdx1pZR3|u|pdx,and the Pohozaev constrained set and the minimization problem are defined asPq,c=uH1(R3):Pq(u)=0Sc,mq,c=infuPq,cIq(u).Recall that the limit problem:Eq.(2.7),was studied by authors in 7.They got the followingtheorem.Theorem 2.1.Let143qp0,then Eq.(2.7)has a positive radial nor-malized solution u satisfying Iq(u)=mq,c.For any uSc,let(su)(x):=e3s2u(esx),then suSc.Next we give the property of theground state energy of the limit problem 4,Lemma 4.5.Lemma 2.2.Let143qp0.Then c7mq,cwith c0 is nonincreasing.Now we setu(s):=Iq(su)=ae2s2ZR3|u|2dx+be4s4(ZR3|u|2dx)2+12ZR3V(esx)|u|2dxe3(p21)spZR3|u|pdxqe3(q21)sZR3|u|qdx,(2.8)andPq(su)=ae2sZR3|u|2dx+be4s(ZR3|u|2dx)2ZR3W(esx)|u|2dxpe3(p21)sZR3|u|pdxqe3(q21)sZR3|u|qdx.(2.9)Obviously,we know that 0u(s)=Pq(su),which implies that for any uSc,sR is a criticalpoint for u(s)if and only if suPq,c.Lemma 2.3.Assume(V2)holds and143qp0 suchthat kuk2.Proof.For any uPq,c,by(V2)and Gagliardo-Nirenberg inequality(2.1),we obtain thataZR3|u|2dx+b(ZR3|u|2dx)2=12ZR3hV(x),xi|u|2dx+pZR3|u|pdx+qZR3|u|qdx2kuk22+pCppcp(1p)kukpp2+qCqqcq(1q)kukqq2,(2.10)No.2LIANG Yan-xia:The Existence of Normalized Solution to201which implies that(a2)kuk22+bkuk42pCppcp(1p)kukpp2+qCqcq(1q)kukqq2.Since143qp4.Therefore we havebkuk42pCppcp(1p)kukpp2+qCqcq(1q)kukqq2.It follows thatbpCppcp(1p)kukpp42+qCqcq(1q)kukqq42.(2.11)Hence,for any 0,one obtains that there exists 0 satisfying kuk2.Lemma 2.4.Assume that(V1)and(V2)hold.For any143qp0.Proof.By(V2)and Pq=0,one infers that(a+2)kuk22+bkuk42aZR3|u|2dx+b(ZR3|u|2dx)2ZR3W(x)|u|2dx=qZR3|u|qdx+pZR3|u|pdxqZR3|u|qdx+qZR3|u|pdx.(2.12)Then,Iq(u)=a2ZR3|u|2dx+12ZR3V(x)|u|2dx+b4(ZR3|u|2dx)21pZR3|u|pdxqZR3|u|qdx12(a1)kuk22+b4kuk421qZR3|u|pdxqZR3|u|qdx12(a1)kuk22+b4kuk421qq(a+2)kuk22+bkuk42=12(a1)1qq(a+2)kuk22+(b4bqq)kuk42.(2.13)From q143and(V2),we deduce that mq,c0.Consider the decomposition of Pq,cinto the disjoint unionPq,c=(Pq,c)+(Pq,c)0(Pq,c),where(Pq,c)+(resp.0,)=uPq,c:00u(0)(resp.=,)0.Lemma 2.5.Assume that(V2)and(V3)hold and143qp6.Then Pq,c=(Pq,c).202CHINESE QUARTERLY JOURNAL OF MATHEMATICSVol.38Proof.For any uPq,c,then Pq(u)=0,namelyaZR3|u|2dx+b(ZR3|u|2dx)2ZR3W(x)|u|2dxqZR3|u|qdxpZR3|u|pdx=0.(2.14)And00u(0)=2aZR3|u|2dx+4b(ZR3|u|2dx)2+ZR3hW(x),xi|u|2dxp3(p2)2ZR3|u|pdxq3(q2)2ZR3|u|qdx.(2.15)Then by(2.14),(V2)and(V3),we deduce that00u(0)=00u(0)3(q2)2Pq(u)=(2a3(q2)2a)ZR3|u|2dx+(4b3(q2)2b)kuk42+ZR3hW(x),xi|u|2dx+3(q2)4ZR3hV(x),xi|u|2dx+(3(q2)23(p2)2)ZR3|u|pdx(2a3(q2)2a)ZR3|u|2dx+(4b3(q2)2b)kuk42+ZR3hW(x),xi+3(q2)2W(x)|u|2dx(2a3(q2)2a)ZR3|u|2dx+ZR3Y(x)u2dx(2a3(q2)2a+3)ZR3|u|2dx0.The last inequality means Pq,c=(Pq,c).Lemma 2.6.Let(V2)and(V3)hold and143qp0Iq(su).Proof.Since 0u(s)=Pq(su),then we only prove that 0u(s)has a unique root in R.By(V2),for any uSc,one has0u(s)=ae2skuk22+be4skuk42ZR3W(esx)|u|2dxpe3(p21)ZR3|u|pdxe3(q21)qZR3|u|qdx(a2)e2skuk22pe3(p21)ZR3|u|pdxe3(q21)qZR3|u|qdx.(2.16)When143qp0 as s.Hence there exists s0R such thatu(s)is increasing on(,s0).In addition,u(s)as s+.Thus there is s1Rwith s1s0such thatu(s1)=maxs0u(s).(2.17)No.2LIANG Yan-xia:The Existence of Normalized Solution to203Therefore 0u(s1)=0,i.e.,s1uPq,c.Assume that there exists s2R such that s2uPq,c.Without loss of generality,assume that s2s1.From Lemma 2.5 we have 00u(s2)0 and 0u(s3)=0,which implies that s3uPq,cand s3u(Pq,c)+(Pq,c)0.This is a contradiction.Hence setting su=s1,we know that suR is a uniquenumber satisfying suuPq,c.And it follows from(2.17)that Iq(suu)=maxs0Iq(su).Furthermore,0u(s)su.(2.19)Hence 0u(0)=Pq(u)0 if and only if su0.Lemma 2.7.Suppose that(V1)and(V2)hold.For143qp0small enough such that0supAkIq0,(2.20)where Ak=uSc:kuk220 for any uAkwith k0 small enough.Similarly by(V2)and(2.1),we can alsoprove that Pq(u)0 for any uAkwith k0 small enough.Moreover,choosing k sufficientlysmall,from Lemma 2.4 we haveIq(u)a2|u|2dx+b4(ZR3|u|2dx)2mq,c.So we complete the proof.Lemma 2.8.Assume that(V1)holds and143qp6.Then mq,cmq,c.Proof.Let Sc,r:=ScH1r(R3).By Theorem 2.1,we know there exists uSc,rsuch thatIq(u)=mq,c.Hence uPq,c.From Lemma 2.6,there exists a unique suR such that suuPq,c.By(V1)we know that supxR3V(x)=0,hencemq,cIq(su)=Iq(su)+12ZR3V(esx)|u|2dxIq(suu)Iq(u),where 7,Theorem 5.2 is applied in the last inequality.Thus mq,cmq,c.Lemma 2.9.If uPq,cis a critical point for Iq|Pq,cwith143qp6 and(Pq,c)0=,thenu is a critical point for I|Sc.Proof.The proof of this lemma is similar to that of 5,Lemma 2.9,we omit it here.204CHINESE QUARTERLY JOURNAL OF MATHEMATICSVol.383.Proof of Theorem 1.1Now we prove the existence of normalized solution for Eq.(1.3)with143qp6.Proof of Theorem 1.1.For any143qp0 since(0)Akby Lemma 2.7.Now we claim that Pq?(1)=Pq(1)0 for s and 0u(s)0 ass.Hence from Lemma 2.6 we obtain that(1)(s)0 for ss(1).Moreover,(1)(0)=Iq(1)0 because(1)I0q,which implies that s(1)sup(AkI0q)Sc,rIq=sup(0,Ak)(0,I0q)(RSc,r)Iq.Using the terminology in 3,Section 5,we can prove that(0,1):is a homotopy stablefamily of RSc,rwith extended closed boundary(0,Ak)(0,I0q)and that the superlevel setIqq,c is a dual set for.Therefore,applying 3,Theorem 5.2,taking any minimizingsequence n=(n,n)nforIq|RSc,rat the level q,cwith the property n=0 and n0a.e.in R3for every 0,1,there exists a sequence(sn,wn)RSc,r(0,Ak)(0,I0q)such that as n,(i)Iq(sn,wn)q,c.(ii)Iq0|RSc,r(sn,wn)0.(iii)dist(sn,wn),(0,n()0.Let un:=snwn=e3sn2wn(esnx).It follows from(i)thatlimnIq(un)=limnIq(snwn)=limnIq(sn,wn)=q,c=mq,c.(3.5)Moreover,from(ii)and(3.1)we know thatsIq(sn,wn)=hI0q(sn,wn),(1,0)i0 as n,wheresIq(sn,wn)=ae2snZR3|wn|2dx+be4s(ZR3|wn|2dx)2ZR3W(esnx)|wn|2dxpe3(p21)snZR3|wn|pdxqe3(q21)snZR3|wn|qdx.Hence we havePq(un)=Pq(snwn)=sIq(sn,wn)0 as n.(3.6)And setting hnTun:=zE:hz,uniL2=RR3zundx=0,by simple calculation,we havehI0q(un),hni=ZR3aunhn+V(x)unhndx+bZR3|un|2dxZR3unhndxZR3|un|p2unhndxZR3|un|q2unhndx=ae2snZR3wne5sn2hn(esnx)dx+be4sZR3|wn|2dxZR3wne5sn2hn(esnx)dx+ZR3V(esnx)wne32snhn(esnx)dxe3(p21)snZR3|wn|p2wne3sn2hn(esnx)dx206CHINESE QUARTERLY JOURNAL OF MATHEMATICSVol.38e3(q21)snZR3|wn|q2wne3sn2hn(esnx)dx=ae2snZR3wnhndx+be4snZR3|wn|2dxZR3wnhndx+ZR3V(esnx)wnhndxe3(p21)snZR3|wn|p2wnhndxe3(q21)snZR3|wn|q2wnhndx,which implies thathI0q(un),hni=hIq0(sn,wn),(0,hn)i,(3.7)wherehn(x)=e3sn2hn(esnx).Then it is standard as the proof of 5 that we can find aPalais-Smale sequence unSc,rfor Iq|Sc,rat the level q,c=mq,csatisfyingIq(un)mq,c,I0q|Sc,r(un)0 and Pq(un)0 as n.(3.8)We can easily deduce that un is bounded in E.By the Lagrange multipliers rule,there existsnR such thatnc2=akunk22bkunk42ZR3V(x)u2dx+kunkpp+kunkqq+o(1).(3.9)Then similar to 5 we can show that n is bounded in R.So there exists R satisfyingn as n.Therefore(,u)RH1r(R3)satisfies(a+bA)u+V(x)u+u=|u|q2u+|u|p2u in R3,where A:=limnkunk22.Hence we haveakuk22+bAkuk22=12ZR3hV(x),xiu2dx+pkukpp+qkukqq=limn(12ZR3hV(x),xiu2ndx+pkunkpp+qkunkqq)=limn(akunk22+bkunk42)=aA+bA2.Since(a+bA)(kuk22A)=0.By a0,b0,A0,we have that kuk22=A.Hence,(,u)satisfies(a+bZR3|u|2dx)u+V(x)u+u=|u|q2u+|u|p2u in R3.Now we prove that 0.Since Pq(un)0 as n,i.e.,aZR3|un|2dx+b(ZR3|un|2dx)2=12ZR3hV(x),xi|un|2dx+qZR3|un|qdx+pZR3|un|pdx+o(1),No.2LIANG Yan-xia:The Existence of Normalized Solution to207which and(3.9)imply thatnc2=(1q)kukqq+(1p)kukppZR3(V(x)+W(x)|un|2dx+o(1),(3.10)where p,q0.Next we claim that u6=0.Assume that u=0.Then it follows from thatmq,c=Iq(un)=Iq(un)+o(1),andPq(un)=Pq(un)+o(1),which shows that Pq(un)0 as n.Hence,for any nN,there exists a unique tn0 suchthat tnunPq,c.Thenmq,cIq(tnun)=mq,c+o(1),which contradicts with Lemma 2.8.Hence u6=0.Step 2.we prove that uSc.Let kuk2=d(0,c,then uPq,d.We can prove Iq(u)0 asLemma 2.4.If d6=c,setting e2=c2d2and vn:=unu*0 in E as n,then kvnk2=e0.By Brezis-Lieb type splitting Lemma and the fact that Pq(u)=0,one infersIq(un)=Iq(vn)+Iq(u)+b2kvnk22kuk22+o(1)=Iq(vn)+Iq(u)+b2kvnk22kuk22+o(1).(3.11)Similar to 2,Lemma 3.13,we can prove that there exists a uniquetn0 such thattnvnPq,e.Hence from(3.11)and Iq(u)0,one hasmq,eIq(tnvn)=Iq(vn)+o(1)=mq,cIq(u)b2kvnk22kuk22+o(1),which and Lemma 2.8 show that mq,emq,cmq,c.This contradicts Lemma 2.2.Hence d=c,i.e.,uSc.Step 3.we show that Iq(u)=mq,c.Since un*u in L2(R3,R)as n and uSc,r,one hasas n,unu in L2(R3,R).208CHINESE QUARTERLY JOURNAL OF MATHEMATICSVol.38Then by Gagliardo-Nirenberg inequality(2.1),we have as n,unu in Lt(R3,R)with t(2,6).Since uPq,c,we have Iq(u)mq,c.Hence we havemq,cIq(u)=Iq(