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<p>TRNG I HC HI PHNGKHOA TON - TINTRN THANH HIPHNG TRNH LAPLACE V HM IU HALUN VN TT NGHIPHI PHNG - 2012TRNG I HC HI PHNGKHOA TON - TINTRN THANH HIMSV :0851510030PHNG TRNH LAPLACE V HM IU HAChuyn Ngnh: Phng Trnh o Hm RingLUN VN TT NGHIPNgi hng dn : ThS. THU HOIHI PHNG - 2012Mc lc1 Phng trnh Laplace v hm iu ha 11.1 Phng trnh Laplace v hm iu ha . . . . . . . . . . . . . . . . . . . . 11.1.1 Nghim c bn ca phng trnh Laplace . . . . . . . . . . . . . . . 21.1.2 Cng thc Green i vi ton t Laplace. . . . . . . . . . . . . . . 31.1.3 Biu din tch phn ca hm bt k . . . . . . . . . . . . . . . . . . 41.1.4 Biu din tch phn ca hm iu ho. . . . . . . . . . . . . . . . . 61.2 Cc tnh cht c bn ca hm iu ho. . . . . . . . . . . . . . . . . . . . 61.2.1 nh l gi tr trung bnh . . . . . . . . . . . . . . . . . . . . . . . 71.2.2 Nguyn l cc tr ca hm iu ho . . . . . . . . . . . . . . . . . . 81.2.3 nh l trung bnh o. . . . . . . . . . . . . . . . . . . . . . . . . 91.2.4 nh l Harnack . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2.5 nh l Liouville . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.3 Hm iu ha trong mt phng v hm gii tch bin phc . . . . . . . . . 121.4 Hm iu ha trn, hm iu ha di . . . . . . . . . . . . . . . . . . . . 132 Nghim ca mt s bi ton i vi phng trnh Laplace 172.1 S tn ti v nh ngha nghim. . . . . . . . . . . . . . . . . . . . . . . . 172.1.1 Bi ton Dirichlet trong . . . . . . . . . . . . . . . . . . . . . . . . 172.1.2 Bi ton Dirichlet ngoi . . . . . . . . . . . . . . . . . . . . . . . . 182.1.3 Bi ton Newmann . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Mt s bi ton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2.1 Gii bi ton Diricle . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2.2 Bi Tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24Ti liu tham kho 26Ch dn 27iChng 1Phng trnh Laplace v hm iuha1.1 Phng trnh Laplace v hm iu haTrong bn lun vn ny, nu khng c gii thch g thm, ta lun xt hm iu hatrn tp con m ca khng gian Euclid thc vi n s thc dng ln hn 1 v l tpcon m, cha trongRn.nh ngha 1.1.1.Gi s min b chn Rn. Phng trnh Laplace trong min lphng trnhu = 0 trong : =n</p> <p>i=12x2i.(y l mt biu din n gin ca lp phng trnh eliptic ).Dng khng thun nht ca phng trnh Laplace : u = f(x) c gi l phngtrnh Poisson. Nghim ca phng trnh Poisson trong min l hmu(x) C2() saochou = f(x) ,x ( nghim ny cn c gi l nghim c in ).Gi su C2(),hmu(x)thomnphngtrnhLaplacec gilhmiuho trong .Hmu(x, y) c gi l hm iuha tiim(x0, y0) nunc o hm cp2lin tc ti im v tha mn phng trnhu=0. Hmu(x, y) iu ha timi im(x, y) c gi l iu ha trong min gii ni .Trng hp khng gii ni, hm u(x, y) c gi l hm iu ha trong nu niu ha trong mi min con gii ni1 ca v tha mn nh gi ti v cng :|u(x, y)| C khi |(x, y)| , C l hng s.1Chng1.Phngtrnh Laplacevhmiuha. V d :1. u(x, y) =xx2+ y2iu ho trongR2\ {(0, 0)}.2. u(x, y, z) = x2+ y22z2iu ho trong mi min gii ni trongR3.3. u(x, y) = sin(xy) khng phi l hm iu ha trongRn.1.1.1 Nghim c bn ca phng trnh LaplaceDng 1.Ta tm nghim ca phng trnh Laplaceu = 0viu C2(Rn\ {0}) di dngu(x) = v(r),x Rn.Chng minh. y ta chor = |x| = (x21 + ... + x2n)12v chnv sao chou = 0 .Ta c =n</p> <p>i=12rx2ivirxi=12_x21 + ... + x2n_122xi=xir, vi (x = 0)V thuxi= v(r)xir , uxixi= v(r)x2ir2+ v(r)_1r x2ir3_, i = 1, ..., nKhi :u = v(r) +n 1rv(r)Nh vyu = 0 khi v ch khiv(r) +n 1rv(r) = 0nuv = 0, ta thy[ ln(v)]=vv=1 nr v(r) =arn1via l hng s no . Suy ra, nur &gt; 0 ta nhn cv(r) =___brn2+ c, vi n 3b ln r + c, vi n = 2vib vc l hng s .Ta thay i vai tr cau(x) vv(r) khi hm s tng ng viu(x) = u(|x|) =___1n(2 n)n.|x|2n, vi n 312. ln|x|, vi n = 2()2Chng1.Phngtrnh Laplacevhmiuhavix Rn,x = 0 hm s (*) c gi l nghim c bn ca phng trnh Laplace.T ta c nhy v a vo nghim c bn ca phng trnh Laplace.(x y) = ( |x y| ) =___1n(2 n)n. | x y |2nvi n &gt; 212.ln | x y | vi n 2n l th tch hnh cu n v trongRn.Dng 2.i khi ta cn nghin cu phng trnh Laplace trong h to cc hay to cu.Phng trnh Laplace trong mt phng :u =2ux2+2uy2= 0vi php th bin___x = r cos y= r sin vit c di dng :u =2ur2+1r ur+1r2 2u2= 0Phng trnh Laplace trong khng gian :u =2ux2+2uy2+2uz2= 0vi php th bin___x = r cos . sin y= r sin . sin z= r cos vit c di dngu =1r2 r_r2ur_+1r2sin _sin u_ 2ur2+1r2sin2 2u2=1r r_rur_+1r2 2u2= 0.1.1.2 Cng thc Green i vi ton t Laplace.Ch 1.Gi n l php tuyn trong mt mt S v, , l cc gc hp bi n vi cctrc. Ta c :vn=vx. cos +vy. cos +vz. cos (1)3Chng1.Phngtrnh LaplacevhmiuhaCh 2.Cng thc Ostrogradski :Gi s l min gii ni, gii hn bi mt binStrn tng mnh .P, Q, R l cc hm lin tc trong Sv c cc o hm ring lin tc trong .Ta c cng thc Ostrogradski :____Px+Qy+Rz_dxdydz +__S(P. cos + Q. cos + R. cos ) dS= 0. (2)trong , , l nhng gc hp bi php tuyn trong ca mt S v cc trc ta .Gi s u(x, y, z) ,v(x, y, z) l cc hm bt k c cc o hm ring cp 2 lin tc trong, cnu, v v cc o hm ring cp 1 ca chng lin tc trong min ng S.Ta p dng cng thc(2) cho cc hm :P= u.vx ,Q = u.vy ,R = u.vzPx=ux.vx+ u.2vx2Rz=uz.vz+ u.2vz2Suy ra :___u.v+____ux.vx+uy.vy+uz.vz_.dV ++__Su._vx. cos +vy. cos +vz. cos _dS= 0T(1) ta c :___u.v+____ux.vx+uy.vy+uz.vz_.dV+__Su.vndS= 0 (3)y l cng thc Green I.Trong cng thc ny ta i vai tr ca sau tr cho nhau ta c cng thc Green II.___(u.v v.u) dV +__S_u.vn v.un_dS= 0 (4)1.1.3 Biu din tch phn ca hm bt k .Ly imx0 bt k .GiB(x0, ) l hnh cu tmx0, bn knh nm trong . K hiuS= B(x0, )= \B(x0, ) = S4Chng1.Phngtrnh LaplacevhmiuhaGi_ x x0_ =1r=1|x x0|l nghim c bn ca phng trnh Laplace ( xt trng hp 3 chiu ).p dng cng thc Green II trn cho hai hmu, ___(u.v v.u)dV+___u.vn v.un_dS= 0u kh vi lin tc n cp 2 :u C1() C0()___.udxdydz +___un u.n_dS= 0Xt__S_un un_dS=__SundS. .I1+__SundS. .I2TrnS ch php tuyn trong n hng theo phng bn knh ca hnh cuS :n=r=n_1r_ = 1r2Thay vo ta cI2= __Su.ndS=__Su. 1r2dS=12__SudSTheo nh l gi tr trung bnh :I2=12.u(p)__SdS=12.u(p). 42= 4.u(p) vi p STng t viI1 :I1=__S1r.undS=1__SundS=1.un(p)__SdS= 4.un(p)I1 0 khi 0I2 4.u(x0) khi 0Vy__S_un u.n_dS 4.u(x0) khi 05Chng1.Phngtrnh Laplacevhmiuhavi 0 ta c___.u.dxdydz +____un u.n_dS + 4.u(x0) = 0Hayu(x0) =14____u.n un_dS 14___.u (5).l cng thc biu din tch phn ca hmu bt k,x0 .1.1.4 Biu din tch phn ca hm iu ho.Khiu l hm iu ho th(5) c dng :u(x0) =14____u.n un_dS (6)( biu din tch phn ca hm iu ho )1.2 Cc tnh cht c bn ca hm iu honh l 1.2.1. Nuu l hm iu ho trong th __undS= 0( p dng cng thc Green I viu l hm iu ho,v 1 ).nh l 1.2.2. Hm iu ha trong Rnc o hm ring mi cp trong min .Chng minh. Gi su l hm iu ha trong Rnv mt imx0bt k trong .Ly nm gn trong chax0.Khi u kh vi lin tc n cp hai trong. Xt hnh cu B(x0, ) tm x0 bn knh nh sao choB . Theo cng thc biu din tch phn hm iu ha trong min ta c :u(x, y) =12__uv_ln1r_ ln1ruv_dSQ.TrongQ(, ) l mtim btk, r =rPQiunyc nghavimiimP V v vi mi Q , hm di du tch phn ca u(x, y) l hm lin tc v c o hmmi cp i vi bin P. Vy theo nh l v tch phn ph thuc tham bin th hm u co hm mi cp trong min .6Chng1.Phngtrnh Laplacevhmiuha1.2.1 nh l gi tr trung bnhnh l1.2.3. Gi sulhm iu ho tronghnh cuB(x0, R);SRl bin hnh cukhigitr utitmx0cahnhcusbnggitrtrungbnhcautrnhnhcuSR. Tc l :u(x0) =1| SR|._SRu.dS( gi tr trung bnh ca u trn mt cu )Chng minh. T cng thc biu din tch phn ca hm iu ha :u(x0) =14___u.n un_dSTa t =1rn= r= r_1r_ =1r2u(x0) =14__SR_u.n un_dS=14__SR_u._ 1r2_1r.un_dS=14R2__SRu.dS 14R__SRun.dS. .0u(x0) =14R2__SRu.dS=1| SR|._SRu.dS.Nhn xt. Ta c th vit cng thc di dng :u(x0) =1nrn1_B(x0,r)u() dSB(x0, r) nh l 1.2.4. Gi su l hm iu ho trong hnh cuB(x0, R) thu(x0) =1| B(x0, R) |.__SRudxdydz( gi tr trung bnh ca u trong hnh cu )7Chng1.Phngtrnh Laplacevhmiuha1.2.2 Nguyn l cc tr ca hm iu honhl1.2.5. Numthmiuhotrong Rn,tgitrccihaycctiuca n ti mt im trong ca th hm ch c th l mt hng s.Chng minh. iu ny c ngha l nu hm iu ha lin tc trong min ng khngphi l hng s th gi tr cc i hay cc tiu ca n ch c th t c ti cc im trn bin ca min .Gi su l hm iu ho trong ,u t gi tr cc i ti mt im trongx0 .Ta chng minh rngu const trong.tM= u(x0),M= { x , u(x) = M }.Ta s chng minhM= :M = dox0 M .Ml tp ng trong vu l hm lin tc trongM( dou l hm iu ho ) .Ml tp m trong .Lyz M z hnh cuB(z, ) , &gt; 0 nh. Ta s chng minhB(z, ) M.p dngnhltrungbnhchohm iuhow=u MtrongB(z, ) .Gi S=B(z, )u(z) M. .0=1| S|__S(u M) dS 0suy ra __S(u M) dS = 0 .Mu(x) M. .hmiuha 0 x u MtrnS.Cho 0 thu MtrnB(z, ) B(z, ) M MmVyM= , suy ra iu phi chng minh .nh l 1.2.6.( Nguyn l cc i ) Gi su C2() C2() l hm iu ha trnvi l tp m b chn trongRn.Khi max= maxHn na , nu l tp lin thng v tn tix0 sao chou(x0) = maxuKhi u l hm hng trn.Chng minh. Gi s tn tix0 ta tM= u(x0) = maxuXt tp {x , u(x) = M} l tp ng dou lin tc v khc rng .(1)Khi 0 &lt; r &lt; dist(x0, ).p dng cng thc gi tr chnh cho phng trnh laplace ta c :u(x0) =_B(x0,r)udy M8Chng1.Phngtrnh Laplacevhmiuhang thc xy ra khi v ch khiu MtrongB(x0, r) .Suy rau(y) = Mvi miy B(x, r) hay tp {x , u(x) = M} l tp m .T (1) v (2) suy ra tp {x , u(x) = M} va m va ng tng i trong,Vy n trng vi .nhl1.2.7. GisminbchnRn,ulhmiuhotrong, u C2() C0(). Khi :infu u(x) supu, x Chng minh.infu infu u(x) supu supuH qu1: Gi su, v C2() C0() vu=vtrong,u=vtrn .Khiu = v trong .Chng minh. t= u v. Khi = 0 trong v= 0 trn.Theo nh l (1.2.7) ta c= 0 trong. Suy rau = v .H qu 2: Gi su C2() C0() vu 0 ( 0) trong. Khi :supu=supu_infu=infu_1.2.3 nh l trung bnh onh l 1.2.8. Gi s l mt min gii ni vu(x) l hm lin tc trong..Nu ivi bt k hnh cuB(x0, r) nm gn trong. Hmu(x) u tha mn ng thc gi trtrung bnh :u(x) =14r2__udSthu(x) l mt hm iu ha trong .Chng minh. Tht vy, gi s trong , u(x) tha mn ng thc v gi tr trung bnh.Tahy xt hnh cu B(x0, r) nm gn trong . Xt hm iu ha v(x) trong hnh cu A saocho trn bin ca hnh cu trng vi gi tru(x) .v(x)|=u(x)|Hm iu ha v(x) ny tn ti, c th n c cho bi cng thc Poatxong, Hm v(x) lhm iu ha nn n tha mn ng thc v gi tr trung bnh.Hm u(x) theo gi thit cng tha mn v gi tr trung bnh.Do , hiuW(x)=u(x) v(x) cng tha mn ng thc v gi tr trung bnh .9Chng1.Phngtrnh LaplacevhmiuhaV vy, i viW(x) c th p dng c nguyn l cc i.Trn hnh cu bin , ta c :W(y)| =u(y)| v(y)| = 0Do , theo h qu ca nguyn l cc i :W(x) 0 trongATc lu(x) v(x) trongA.Nh vyu(x) trng vi hm iu hav(x) va xy dng .Vyu(x) iu ha trongA. V hnh cuA l bt k trong, nnu(x) l hm iuha trong ton.1.2.4 nh l Harnacknh l 1.2.9. Gi s {un} l dy hm iu ho trong , gii ni c binStrn tngmnh v gi sun(x) lin tc trong min ng S.Nu dy {un} hi t u trn, khi :1) {un} hi t u trn.2) Gi sunu trong. Khi u l hm iu ho trong.Chng minh. 1) K hiun=un|. Theo gi thit {un} hi t u trn nn : &gt; 0, N: n N, p NTa c | n+p(x) n(x) | &lt; x .Hay | un+p(x) un(x) | &lt; x .Vy theo tiu chun Cauchy ta cunu trong.2) Ta c nh l trung bnh o :Gi su lin tc trong, vi mi hnh cuB(x0, R) mu(x0) =14R2__SRu dS thu l hm iu ho trong.Gi s vi mi hnh cuB(x0, R) , doun l hm iu ho vi min, nn :un(x0) =14R2__SRu dSlimnun(x0) =14R2__SRlimnundS.n(n)_r2_n1un(x0) =14R2__SRu dS uiu ha trong .10Chng1.Phngtrnh Laplacevhmiuhanh l 1.2.10. ( Nguyn l cc tr mnh )Gi su C2() ,u 0 ( 0) trong v tn tix0 sao chou(x0) = supu(u(x0) = infu)Khi hmu l hng s.1.2.5 nh l Liouvillenhl1.2.11. Gishm iuhau:RnRthamn x Rn, |u(x)| MviMl hng s.Khi u l hm hng .Chng minh. Coix0 ty thucRn.Vi i {1, ..., n}, o hm ring tng phn ca u theo bin xi l hm iu ha theo cngthc gi tr trung bnh ca hm iu ha th vi mir &gt; 0 ta c :uxi(x0)=1(n)_r2_n_B(x0,r2)uxi(x)dx=2n(n)rn_B(x0,r2)uvidSTrong :(n) l th tch ca hnh cu n v trong Rn .vi l to thi ca vect php tuyn n v ca mtB_x0,r2_.Do |uvi| M, ta c :uxi(x0)2n(n)rnM_B(x0,r2)dS=2n(n)rnM=2nMr.Chor 0 ta c :uxi(x0) = 0 .Vyu l hm hng trnRn.nh l 1.2.12. ( nh l Liouville m rng )Gi s hm iu hau : Rn R tha mn :x Rn,|u(x)| M+ A|x|Trong M, A, l cc hng s dng v &lt; 1.Khi u l hm hng trnRn.11Chng1.Phngtrnh LaplacevhmiuhaV d : Xt hm su : R3 R c nh ngha nh sau :Vi x = (x, y, z) R3.u(x) =3x2+ 11y 9z + sin8(x418y2+ z2) 23xy(4x2+ 19y2+ 22z2)23 + ez2+ 12 cos2_16xyz z4_ln6x2+ 11y2+ 7z2+ e(x2+2y2) .Khi u c l hm iu ha trnRnhay khng ?Chng minh. Ta cu(0, 0, 0) = 0 v limxu(x, 0, 0) = +Nnu khng l hm hng trnRn.Vi |x| =_x2+ y2+ z2v |x| kh ln , ta c :u(x) 3 |x|2+ 11 |x|2+ 9 |x|2+ 1 + 23 |x|2_4 |x|2_32+ 12ln_11 |x|2+ 1_47423|x|23+ 12._12 |x|2_13 (47 + 12.3) |x|23= 83 |x|23Mtkhcu khnglhm hngtheonhlLiouvillemrngth ukhnglhmiu ha trnRn.1.3 Hm iu ha trong mt phng v hm gii tchbin phcTa nhc li vi kt qu trong gio trnh hm gii tch bin phc .Nh trong gii tch thc, mt hm phc "trn" f(z) c th c o hm ti mt im no trong min xc nh. Khi tn ti o hm :dfdz=limz0fz=limz0f(z + z) f(z)z()Gia sz= x + iy l mt s phc, v vyz 0 theo nhiu phng thc .Gi sf(z)=u(x, y) + iv(x, y) l hm vi bin phcz=x + iytrong u vvlnhng hm thc ca bin(x, y).Trong lp cc hm bin phcf(z) c mt lp hm quantrng c gi l hm gii tch.Khi o hmf(z) tn ti nu v ch nu :f(z) =ux+ ivx=vy iuyiu kin cn v hm bin phcf(z) l gii tch khi v ch khi phn thc v phno tha mn iu kin Cauchy - Riemanm.ng nht phn thc v phn o ca biu thc ta c :ux=vyuy= vx12Chng1.Phngtrnh LaplacevhmiuhaHoc k hiu khcux= vyuy= vx.Nu hm f(z) khng gii tch, th gii hn (*) hoc khng tn ti i vi t nht mtphng thc tin ti khng ca z, hoc ph thuc vo tng phng thc tin ti khngcaz. i vi hm gii tch, th gii hn (* ) tn ti( khng ph thuc vo phng thc tin ti khng caz ).Khi u(x, y) vv(x, y) c o hm mi cp.Phn thcu(x, y) v phn ov(x, y) ca hm gii tch bt kf(z)=u(x, y) + iv(x, y)trong min l nhng hm iu ha trong min .Tht vy, vi phn h hai phng trnh o hm ring ny, u tin theox, sau theoy ta c :2ux2+2uy2= 0;2vx2+2vy2= 0Hoc di dng k hiu khc :uxx + uyy= vxx + vyy= 0.Phn thc v phn o ca mt hm phc kh vi tha mn phng trnh Laplace . Haihm iu hau(x, y) vv(x, y) to thnh phn thc v phn o ca hm gii tchf(z)c gi l hai hm iu ha lin hp.R rng, nu ta bitu(x, y) th ta c th tm c hm iu ha lin hpv(x, y) ca nbng cch ly tch phn biu thc vi phn hon chnh .dv= uydx +uxdy_=vxdx +uydy_iu ny c ngha l :Bt kmthm iuha notrong mtmincngcth coilphnthc(hocphn o ) ca mt hm gii tch no trong min y.nh l 1.3.1. Gi s l tp m trongRnvu l hm s thc iu ha trn.Khi u l hm gii tch trn.1.4 Hm iu ha trn, hm iu ha dinh ngha 1.4.1.Xt Rnv hm su, R . Hm su c gi l hm iuha ditrn nuu lintc trn v tha mn tnhchtgi tr trung bnhtrnnh sau :a ; B(a, ); r (0, ] ; u(a) 14r2_|xa|=ru(x)dSx(1.1)Tng t hm iu ha trn trong khi trong(1.1) thay du thnh du .Hm iu ha l trng hp c bit ca hm iu ha di ( iu ha trn ) .13Chng1.Phngtrnh Laplacevhmiuhanh l 1.4.2. Xt l tp m lin thng ,u l hm iu ha di trn.Gi su nhn gi tr ln nht trn, tc l tn ti mt imx0 tha mn :u(x0) = sup {u(x) : x }Khi u l nhn gi tr ln nht trn.Chng minh. Gi Ml gi tr ln nht ca u(x) trn . Ta xt tp A = {x : u(x) =M}Mu lin tc v khc rng ( theo gi thit ) =&gt; A l tp ng.Coia ty thucA .u l hm iu ha di nn tn ti hnh cu ngB(a, ) tha mn :u(a) 14r2_|xa|=ru(x)dSxGi s tn tix0 B(a, )\ {a} tha mnu(x0) &lt; M .Gir = |x0 a|Do tnh lin tc cau tix0 , tn ti ln cnVcax0 tha mnx V, u(x) 0 tha mnB(x, ) 0K( do0Kl tp m ).Ta cw(a) = v(a), ng thi dov iu ha trn0Knn :w(a) =14r2_|xa|=ru(x)dSx vi r (0, )* Trng hpa/ 0K:Khi w(a) = u(a) Tn tiB(a, ) = tha mn :w(a) = u(a) 14r2_|xa|=ru(x)dSx 14r2_|xa|=rw(x)dSxVyw l hm iu ha di trn .16Chng 2Nghim ca mt s bi ton i viphng trnh Laplace2.1 S tn ti v nh ngha nghimGi s l min b chn trongRn, ta xt mt s bi ton : bi ton Dirichlet trongv bi ton iricle ngoi i vi...</p>

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