具有超临界相位的特殊拉格朗日型方程的Neumann问题.pdf

数学杂志Vol.43(2023)J.of Math.(PRC)No.6THE NEUMANN PROBLEM FOR A SPECIALLAGRANGIAN TYPE EQUATION WITHSUPERCRITICAL PHASEJIA Hao-hao,XU Wen-zhao(School of Mathematics and Statistics,Ningbo University,Ningbo 315211,China)Abstract:In this paper,we explore the Neumann problem for special lagrangian type equa-tions with supercritical phase in Rr.We show the global C2 a priori estimates of the solution andestablish the existence of classical solutions by the methods of continuity.Keywords:Neumann problem;special Lagrangian type equation;supercritical phase2010 MR Subject Classification:35J60;35B45Document code:A1 Introduction and main resultsWe consider the Neumann problem of a special Lagrangian equationwherearctanu l n -D?u =:a r c t a n n i +a r c t a n n 2 +.+a r c t a n n n.Denote n:=(ni,n2,:,nn)which are the eigenvalues of the matrix uln-D2u in 1 withwhere 入=(Ai,2,*.:,An)are the eigenvalues of the Hessian matrix D?u.Here()isusually studied under three different types of two boundary value conditions:the phase,the critical phase,supereritical phase.More preisely,(a)e(-,),e=Pr,pr (a)n his papr,we conider h speia agrangian equation(.)with2supercritical phase,that is the third type.The first boundary value problem(Dirichlet problem)for elliptic partial differentialequations has been intensively studied many years.For the Laplace equation,results canbe found in Gilbarg-Trudinger 2.The Dirichlet problem for Monge-Ampere equations*Received date:2022-09-04Foundation item:Supported by National Natural Science Foundation of China(12171260).Biography:Jia Haohao(1995-),female,born at Handan,Hebei,postgraduate,major in partialdifferential equations.E-mail:Article ID:0255-7797(2023)06-0471-16arctanu In -D u)=O(c),in 2 C R,n=Z入,Vi=1,2,.,n,kiAccepted date:2022-12-05(1.1)2，22472was investigated in Caffarelli-Nirenberg-Spruck 3 and Krylov 4.They showed the globalregularity of solutions.Caffarelli-Nirenberg-Spruck 5 studied the existence of admissiblesolutions and the global regularity of k-Hessian equations.The Hessian quotient equationswhich have different structure conditions were studied in Trudinger 6.To the best of myknowledge,the special Lagrangian equationJournal of MathematicsVol.43was introduced in Harvey-Lawson 7 firstly and is a constant called the phase angle.In their study,the graph a (,Du(a)defines a calibrated,minimal submanifold of R2n.Collins-Picard-Wu 8 considered the Dirichlet problem to Lagrangian phase operator in boththe real and complex setting.They solved the concavity of Lagrangian phase operator,theessential condition,to obtain the existence theorem by using the classical methods.Recently,Zhu 1 established the global C2 estimates and showed the existence theorem of the Dirichletproblem to(1.1).For the Neumann and oblique derivative problem of elliptic equations,there are manyresearch results.A priori estimates and the existence theorem of the Laplace equation canbe found in 2.And,we can see more results about the Neumann and the oblique derivativeproblems of linear and quasilinear elliptic equations in Lieberman 9.The Neumann problemof Monge-Ampere equations was solved in Lions-Trudinger-Urbas 10.Ma-Qiu 11 studiedthe Neumann problem of k-Hessian equations in uniformly convex domain.And,Chen-Zhang 12 solved the Neumann problem of Hessian quotient equations,the general formsof k-Hessian equations.For the special Lagrangian equation with supercritical phase instrictly convex domain,Chen-Ma-Wei established the global C2 estimates and obtained theexistence theorem by the method of continuity in 13 recently.It is worth mentioning that the key to solving of the existence and uniqueness of classicalsolutions for elliptic partial differential equations is to establish the global a priori estimatesand the method of continuity in above works.To our best knowledge,the existence theoremof the classical Neumann problem to(1.1)with supercritical phase has not been studiedbefore.In this paper,we apply the method used in 9,10 and show the existence theoremof the Neumann problem of special Lagrangian equation following the classical idea(see forexample 14 or 15).More precisely,we get our theorem.Theorem 1.1 Suppose 2 C Rn is a C4 strictly convex domain and v is outer unitnormal vector of 00.Let p E C8(0n)and e(a)e C(2)with(n-)e(an)0.We need to establish a priori estimate of ue which is independent ofe,and the strict convexity of 2 plays an important role.By taking the limit on e and theperturbation argument,we can obtain the existence of a solution of(1.2).2 PreliminariesIn this section,we show some properties of the special Lagrangian equation with super-critical phase.Property 2.1 Let R beadomain and(a)ith Pr(a)0,In-1l nn,InilCo,maxewhere Co=max(tan(-)-min O),an(2The proofs are analogous to Property 2.1 and Lemma 2.1 in 1,13,16,17 and are omit-ted.The following property is Property 2.2 in 1 and we give the proof here for convenience.Property 2.21 Suppose 2 C R is a domain and e(a)e C2(2)with(n-2)0(c)0and o(m)C()with pr 0.So u attains its maximum at some boundary pointCo E a2.Then we have(3.2)andeu eu(ro)p(ro)maxlpl.We can assume O e 2,and denote B=2(a-1 tan(maxen+oo.Then we havearctan n(Du)=max =arctan(D(Blal2).Using the comparison principle,we get u-Blcj2 to attain its minimum at a boundary pointyo E a2.Therefore,(3.5)Then,we haveeu e(u-Blal2)e(u(yo)-Blal)-2Bdiam(2)-max -Bdiam(2)2.The Neumann problem for a special lagrangian type equation with supercritical phase4750 ur(co)=-u(co)+(co),(3.3)8(3.4)0 (u-Blal)(yo)=uv(yo)-2Byo V-u(yo)-(yo)-2Bdiam(2).(3.6)3.2 Global C1 estimateIn this subsection,we prove the Ci estimate of solutions for the special Lagrangianequation(1.3)with supercritical phase.We show the following theorem.Theorem 3.3Suppose 2 c Rn is a C3 uniformly convex domain and E C2(02).Let e(a)=Ci()with(2(a)0,then we havesup|DulMi,where Mi depends only on n,2,max O,min O,Mo,Olc1 and plc2.Proof We just have to provewhere =(S1,:,Sn),Isl=1.Choose(3.7)Deu(a)Mi,V(a,t)e Sn-1,(3.8)w(a,E)=Dgu()-(v,s)(-eu+)+e?u?+K|cl2(3.9)476where v is a C2(2)extension of the outer unit normal vector field on 02,e is a small positiveconstant in(1.3)and K is a large positive constant.Note that here E C2()is an extensionwith universal C2 norm.Suppose w(a,s)attains its maximum at(co,So)e Sn-1.Inthe following,we divide(3.8)into two steps.1.We claim that ro E 00.Assume co E 2,and we will prove Fiouw(a,o)la=ao 0to establish a contradiction.For Co E 2,we can assume D2u(co)is diagonal with 入;=uiand 入 入2 .An by rotating the coordinate(ei,*,en),then Fii(co)is diagonal.Then we have(3.10)uijP+i0,Hence we can get from Property 2.1,F11F22.11+%Co0;p+nnFiui=np1+工2P=1where Co=1(n=,)min)1+max(tan2We suppose the maximum of w fixed in some direction So,all the calculations are at coin the following.We get0 EFaw(r,S0)la=ao=Z Fuito-(v,So)a(-eu+p)-2(v,So)(-eui+p)-(v,o)(-euu+pa)+2e(u+wua)+2K)Journal of Mathematicsa arctannFimax,tanVol.43ifi=j,ifi牛j.Fnn.1(3.11)(3.12)(3.13)2+Z F2K-(v,0)a(-u+0)-2(v,S0);pi-(v,0)pia 0o+EFv,0)eui+2e wual+Fi(eui+(v,0),)?+F2K-(v,So)a(-su+)-2(v,So);i-(v,o)Pi-(v,So)n-VOI-2十1+Cwhere Co is defined in(2.5),C1 is a positive constant depending only on n,Mo,Ilc2,Ilolc2.C2 is positive constant depending only on Ci and lc1.diction.Thus ao E a2.np+ZF2K-C1-|D(v,0)1P1(2K-C2),No.62.We now consider the direction so with the following three cases.Case a:So is normal to 2 at o,then we havew(co,So)=2(-su+P)+s2u2(co)+K|col C4.And,Deu(c)=w(c,s)+v,s)(-u+p)-s?u?-K|ro w(co,So)+CsC6.Case b:So is non-tangential but not normal to o2 at co.We can find a tangentialvector T E Sn-1 such that So=T+v,with=o T 0,=So V 0,Q?+2=1 andT V=0.Without loss of generality,we take 0 0 in 2;/Vd|=1 on o2.478And,the matrix dil1i,jn-1 -Kmin,where Kmin is the smallest principal curvature ofthe boundary.Because w(co,S)attains its maximum at direction E=So,it is easy to getul(co)0,u2(ao)=.=un-1(ro)=0,un(co)=-uv(ao)=eu(co)-p(ao).On the other hand,from Vk=-dk,Hence,-DuDiv diiui+Ci1 -Kminw(co,So)+Ci1.Then,we haveSimilar to the(3.14)and(3.15),we haveThe proof is complete.3.3 Global second derivatives estimateWe consider now to the global second derivatives and we can get the following theoremTheorem 3.4 Suppose 2 c Rn is a C4 strictly convex domain and E C3(2).Let 0(m)e C()with Pr e(a)0,then we have(3.16)where M2 depends only on n,2,max O,minO,Mo,Mi,Olc2 and olc3.We firstly reduce global second derivatives to double normal second derivatives onboundary and then we show the estimate of double normal second derivatives on bound-ary following the standard method in Lions-Trudinger-Urbas 10 and Ma-Qiu 11.Lemma 3.5Suppose 2 C IRn is a C4 convex domain and E C3(02).Let 0(a)EC()with-)0(a)0,then we havesup|D?ul C14(1+sup|uvvl),where C14 depends only on n,2,max O,min O,Mo,Mi,Olca and Iplcs.Journal of MathematicsDuDiV=Di(vk)Dkun82dk=1nCdiuk=-Ediu-dinunk=1k=1n-1-Zdik u-C/Dvu(ao)lk=1-d11ui-Cl1w(ao,o)IDel+Cro+Cu.Deu(a)=w(r,5)-u?-Ke/w(ro,So)+C12 C13.2Vol.43k:un-1.Kminsup|D?ul M2,82(3.17)No.6Proof There exists a small constant 0 such that d(a)e C4()and v=-Dd on02 because 2 is a C4 domain.Define d e C4(2)such that d=d in and denoteNote that v is a C3(2)extension of the outer unit normal vector field on 02.We assume 0 e 2 and consider the auxiliary functionwhere u(a,E)=2($v)S-(Dp-eDu-uDul)=auu+b,E=-(S-v)v,al=-2(.v)(s-Dvl)-2(v)e($),b=2(v)($Dp),and Ki 0 is to be determined later.Andwe know that here p E C3()is an extension with universal C3 norm.RecallBy rotating the coordinates,for any a e 2,we assume D2u is diagonal with A;=ui(i=1,2,.:,n)and 1 2 .入n.We know that(3.10)-(3.13)in the proof of Theorem3.3 stll hold.From Property 2.2,for any fixed$e Sn-1,we haven订1provided byKi=D?0+A|De?+|a|De|+Dal+D?alI ul+2b+1.So maxv(a,$)attains its maximum on 02.Therefore,we can assume the max,v(c,)xSnattains its maximum at some point ao E 0Q and some direction So E Sn-1.We consider twocases in the following and all the calculations are at the point ao and =Eo.Case a:So is tangential to 02 at Co.Naturally,So-v=0,u(co,So)=0,and ugoso(co)0.Recall the following formulas inthe book 9,(3.19)n2vD,v=0.i=1where C is a constant depending only on n and 2.As in 9,we still defineThe Neumann problem for a special lagrangian type equation with supercritical phasen=10s-A0s?-aOt-Da|-FDudu+bul+Ki0IDul C,in,nCvD;vj=0,=1,in,i=1ci=ij-vivi,in Z,479V=-Dd,in2.(a,s)=uet-u(a,s)+Kilal2,Fi _ arctann.Quini=1i1(3.18)nFii=1nni1nFii1(3.20)(3.21)480and for a vector C e R,we denote for the vector with i-th component-cicj.Thenwe have(3.22)i,j=1And,uiul=ci+vivijululj=cD;(ulut)-D,uul+vvivlutj=-cieuj+ciDip-cuDjul+vivivluuj.The last equation used boundary conditions.Then,ulipvl=cpa+vPv9utiaul=cpDa(uivl)-uiDau+vPvuigul=cpDa(-ceuj+ci Dip-cauDjul+vvivluti)-cpauiDaul+vPvyluia,hence we obtainnipl=1nip=1nJournal of MathematicsnI(Du)/2=Zci uiuj and Dul2=(Du)1?+uz.-cpauiDaul+vPvayluialVol.43(3.23)(3.24)i1=-Soseci ujg-Dac uj+SosgDa(ci Di)-SsDa(c Djvul)ut-sgugD,ul+SisgDaviuv-Sosg ugDiu-utoto-26ouitoD;vl+C1s+Ci1slDul+C1sluvl.Without loss of generality,we assume So=e,then we can get the bound for ui(co)fori I due to the maximum of at the So direction.Next,we can ssume e(t)=yr.By calculating,(3.26)(3.25)V1+t2du(co,s(t)0一dtd(t)=2ui(odt=2ui2(co)-2v2(D1-su1-uDivl),t=0du(co,E(t)(0)dtt=0No.6thenAnalogously,for all i 1,we haveFrom D;vl 0,we haveOn the other hand,from the Hopf lemma and(3.23),we haveHence,oso(ao)(Ci7+Cis)(1+|uvvl).Since u is the subharmonic function and(3.31),we obtainmax,luge(a)/(n-1)。ma x,u g s(a)2xSn-1Case b:So is non-tangential.In this case,we have So V+0.In fact,we can find a tangential vector e Sn-1 suchthat Eo=T+v,with =So T 0,=So V 0 and?+2=1.Naturally,T V=0.So,uzoso(co)=2utt(ro)+2uvv(ao)+2utv(co)=2u(co)+2uv(co)+2(o )o-(o v)Dp-Du-uDvl,(3.33)then,(co,So)=2(ao,T)+2v(ao,V)2(co,So)+2v(co,V).The Neumann problem for a special lagrangian type equation with supercritical phaseu12(ro)=?(Di-sul-uDivl)/Cis+C1e|Dul.usoSov -Utoso-Divlutoto+Ci7(1+uvvl)-uoSo+Ci7(1+|uuvl).0 uv(aco,So)=uoov-auv-Dvaul-b+2Ki(D)-uo%o+Ci7(1+uvl)+Ci8.(n-1)max,u(r,s)+C1o)2xSn-=(n-1)u(co,So)+C19(n-1)ut050(co)+2C19C20(1+|uvvl).481(3.27)u1i(co)|C16+C16|Dul.(3.28)(3.29)(3.30)(3.31)2xSn-(3.32)(3.34)482By the definition of(co,So),we knowandSimilarly to(3.32),we complete the proof of the lemma.Lemma 3.66Suppose 2 C IRn is a C3 strictly convex domain and E C3(02).Let(r)e C()with-2 0(a)0,then we havewhere C22 depends on n,2,max,min O,Mo,Mi,JOlc2 and Iplcs.Proof Since 2 is a C3 strictly convex domain,we can find the defining functionp E C3(2)for it such thatwhere ao is a positive constant depending only on 2,and In is the n x n identity matric.And,V=(vl,v?,.:,v)is a C?()extension of the outer unit normal vector field on 02as in Lemma 3.5.By the classical barrier technique in 11,we consider the functionwhere K2=max2+C(D ll+/Dvl+1),%(I Dl D2v1+?0l)and C isthe constantin(2.5).Also note that here E C2()is an extension with universal C2 norm.RecallFor any E 2,we can assume D?u is diagonal with 入;=ui and 1 2 .An.Weknow that(3.10)-(3.13)in the proof of Theorem 3.3 still hold.Hence we can getni=1Journal of Mathematicsu(co,So)u(co,V),uoso(ao)u(co,o)+C21 v(ao,V)+C21 uvl+2C21:min uvv -C22,852p=0 on 2,p0 in 2;IDpl=1 on a2;D?p aoIn,P(a)=uy+eu(c)-p(ac)+K2p,aoFi_ Daretann.uiin=1=D0u+2Fua(vi):+Fu(u)1i=1-|DOlul-Du|-DullD2vl)K2a01-DO/l-Dl-2?ni=10.Vol.43(3.35)(3.36)(3.37)(3.38)nFi pi1i=1n)u+eZFiu-nni=1i1nPlFi+K2aon三2=1121+Co2K2a0n2i=1i-1n一pi+K2i=11i=1=1“pii(3.39)No.6Also,it is easy to know P=O on o2.Hence P attains its maximum on any boundarypoint.Then we can get for any a E a2,(3.40)hence(3.37)holds.In the following,we establish the upper estimate of double normal second derivativeson boundary.Lemma 3.7Suppose 2 C Rn is a C3 strictly convex domain and E C3(2).Let0(a)E C?()with（n-2)元2 0(c)0,then we have(3.41)where C23 depends on n,2,max,min O,Mo,Mi,JOlc2 and lc3.Proof Similar to the proof of Lemma 3.6,we now consider the test functionwhere K=max(2(+(IDll+Dvl+),%(I Dul/D?vl+ID2l)and Co s the constant in(2.5).And,here E C2()is an extension with universal C?norm.DenoteFor any a E 2,we can assume D?u is diagonal with A,=ui and i 2 :.An.Weknow that(3.10)-(3.13)in the proof of Theorem 3.3 still hold.Hence we can getnuau+2u(ul):+u(ul)al+euu-pu)-RF paThe Neumann problem for a special lagrangian type equation with supercritical phasen4830 P,(a)=uu-Z uCudjv+euy-pul+K2pvuyv+DulD?d+Dul+Dol+K2,max uv C23,P(ac)=uy+eu(c)-p(a)-Kp,aoFiuii(3.42)arctan nni=1i=1=ou+22Fi1|DO|lul+Du+Dul|D?ulnKao1|DOul+Dul+21+C2nn=1=1 0.Let ue=e-Ja u,and itis easy to know ve satisfies(4.2)()=-eus-尚 Jae+(a),E0n.By the gradient estimate(3.7),we know e sup|Due|-0.Naturally,there is a constant and a function ue C2(),such that-sus=,-eve 0,-Jaeue and us-uuniformly in C2()as e 0.It is easy to verify that u is a solution of(4.3)Uv=+p(),E a2.Journal of Mathematics0 P(a)=uv-udiu+euu-pu-Kpu5uv-|Dul|D?d-|Dul-Dl-K,Iulc2.a()C,arctanAvIn-D2vs=O(a),E 2,arctanu In -D?u)=O(),E 2,Vol.43(3.44)No.6If there exists another function U1 E C2()and another constant i satisfyingand we can know =i