Sự không tồn tại nghiệm dương của phương trình tích phân

Chia sẻ: Tran Cong Dat | Ngày: | Loại File: PDF | Số trang:11

Thêm vào BST

Báo xấu

208
lượt xem 34
download

Download Vui lòng tải xuống để xem tài liệu đầy đủ

Tài liệu tham khảo Sự không tồn tại nghiệm dương của phương trình tích phân

Chủ đề:

Bình luận(0) Đăng nhập để gửi bình luận!

Lưu

Nội dung Text: Sự không tồn tại nghiệm dương của phương trình tích phân

LU{J.nvan tot nghi~p Trang 24 CHUONG 4 st; KHONG TON T~I NGHIEM DUONG ? "'".., CUA PHUONG TRINH TICH PHAN Val (J = N -1, N > 2 Trang phgn nay chung ta xet sv kh6ng t6n t~i nghit%mdu'dng cua phu'dng trlnh tich phan phi tuye'n sau day (4.1) U( x ) = bN g(y,u(y)) dy "dx E IRN , f N-l' IRN Iy - xl trang do bN = 2((N -l)lUN+ltl voi lUN+1 dit%n la tich cua m~t c~u ddn vi trong IRN+I, N > 2 va g: IRN x IR+ ~ IR la ham lien t\!Ccho tru'oc thoa di~u kit%n: T6n t~i cae hftng s6 a,fJ ~ 0, M > 0 sao cho (4.2) g(x,u) ~ MlxlP ua, "dx E IRN, "du ~ 0, va mQt sf) di~u kit%nph\! sau do. Phudng trlnh tich phan (4.1) duQc thanh l~p tu bai loan Neumann phi tuye'n sau dayvoiN=n-l>2: TIm mQt ham v Ia nghit%mcua bai loan Neumann (4.3) ~v=O, xEIR: ={(xl,xn):xl EIRn-l,xn >O}, (4.4) - vxn(Xl ,0) = g(XI, V(XI ,0)), Xl E IRn-l, thoa cae tinh cha't: (8]) VEC2(IR:)nC(IR:), vxn EC(IR:), (82 ) lim SUP I vex) + R. sup ov (x) I = 0, k--HOO ( Ixl=R,xn>O Ixl=R,xn>O fun J d day g: IRn-1x [0,+00)~ [0,+00)cho tru'oc thoa cac di~u kit%nsau: (G]) g la ham lien t\!e, (G2) 3a~0,3M>0: g(xl,v)~Mva, "dv~O, "dxl EIRn-l. va mQt sf) di~u kit%nph\! se d~t sau.
Lu(jn van tot nghi~p Trang 25 Khi do, n€u g 1a ham lien t\lC va nghi~m v bai loan (4.3), (4.4) co cac tinh cha'"t(SI)' (S2)' thi v 1anghi~mcua phudngtrinhtich phan sau day I - 2 g(l , vel ,0))dl I n (4.5) vex ,xn) - (n-2)OJn f I I 2 2 (n-2)/2 ' VeX ,xn) E IRp Rn-I (1 Y -x' I +Xn ) trang do OJn1a di~n tich cua m~t c~u ddn vi trong IRn. Day 1a k€t qua trong ph~n thi€t l~p phudng trinh tich phan (chudng 2, dinh 1y 2.1), trang do co stf thay d6i cac ky hi~u trang cach vi€t bang cach thay (a/,an) va (xl,xn) 1~n1u'
Lu(in win tot nghifp Trang 26 Ch6 thich 4.1, K€t qua nay m~nh hdn k€t qua tfong [2], [8]. Th~t v~y, vOi CY =N -1, d cling phu'dng trlnh rich phan (4.7), cae gia thi€t sau day dii sa dt,mg trong cae bai baa [2], [8] ma trong ehu'dngnay khong e~n d€n: (G3) g(x,u) la ham khong giam d6i vdi bi€n u, i.e., (g(x,u)-g(x,v))(u-v)~O VxEIRN, Vu~O, Vv~O. (G4) Tich phan J ( g (1,0; ~-I 1/1' 1+ x ) t6n t~i va du'dng. Tru'de h€t ta e~n mQt sO'ba't d&ng thue sau day: B6 d~ 4.1. Vai mQi q ~ 0, X E IRN, fa dijt: (4.8) A[q](x):=A[(1+lylrq](x)= J(1+lylr:_,dy. lRN Iy - x I Khi an (4.9) A[q](x) = +00, ne'u q:::; 1, (4.10) A[q](x) hQifl;l va A[q](x)~ (q-I)2 OJNN-I (1+ 111 -I' x)q ne'u q>1. Chung minh b6 d~ 4.1. a) Gia sa q :::; . Chti 1 Y d€n ba't d&ng thue tam giae (4.11 ) Iy - xl :::;Iyl + Ix! vdi mQi x, y E IRN , ta suy fa tu eong thue (4.8) ding (4.12) A[q](x) = J (1+lyl )-:-~y [RN Iy - x I - (1 + Iyl rq d = +oo (1+ rrqNlr d J d:S > J J 1/' 1III ( Y + x ) - N1Y II ) - lyl=r 0 ( r + x r' trong d6 J dSr la rich phan m~t tren m~t e~u, tam 0, ban kinh r trong IRN. Iyl=r Tich phan n~y ehinh la dit%nrich eua m~t tren m~t e~u Iyl= r, tue la: N-l (4.13) J dSr = r OJN' Iyl=r
LucJn van tot nghifp Trang 27 Do do, ta suy tu (4.12), (4.13) ding (4.14) +00 N-} dr J A[q](x)~wN I( r:'xl)N 1 (1+r)q =wN q' +00 N-I d Tich philo Jq = f (r+ rll 0 x) N-I (1+r)q r philo ky khi q ~ 1 va hQi t1;1 q > 1. khi Do do, rich philo (4.15) A[q](x) philo ky khi q ~ 1. a) Gia sa q > 1. i) Xet t~i x = 0, ta co - (1 + Iylrq dy - (4.16) A[q](O)- f m~ I y N-I I -wN + f oo(1+ rrq 0 r, N-I rl-Ndr = w f + oo~ N 0 (I+r)q . / / A +00 dr A' , Do do, hch Phan f hOI tu VI q > 1. 0 (1+r)q . . V~y, rich philo (4.17) A [q](0) hQi t1;1khi q > 1. ii) Xet t~i x =F chQn R > 31xJ O. Ta vie't l~i A[q](x) thanh t6ng hai tich philo 0, > (4.18) A[q](x)= f (1+IYI)~q_~y+ f (1+IYI)~q_~y =J~I>CX)+J~2)(X). IY-Xl$/?Iy - xl Jy-xl"/? Iy - xl (1+lylrqdy U)Banhgia J~I)(X)= f N 1 . IY-Xl$/? IY - xl - Ta co: J Ii ()X = f ~ (l) (1+lylrqdy< -q (4.19) f N-I - sup (I + II) Y N-I IY-XI$R Iy - xl ly-xl:>R ly-xl:SR - xl Iy = sup (1+ !ylrq IY-XI$R f Izl:SRIzi d :-1 = sup (1 + !ylrq wN ly-xl:SR r R N-Id 0 r N-/ = sup (1 + Iylrq wNR < +00. ly-xl:SR
Lugn wln tot nghi~p Trang 28 (1+lyl)-qdy OJ) Danhgia J~2)(X)= f N I . ly-4~1I IY - xl - Ta co: (21 = (1+lylrqdy < (1+lylrqdy < (1+lylrqdy (4.20) JII (x) f NI - f NI - f NI ly-xl~R Iy-xl - lyl~R-lxl Iy-xl - IYI~R-Ixillyl-Ixil - +00 (1 +r ) -'1 r N-I dr +00 r N-I dr = OJN f N 1 = OJN f N I- . II-Ixl Ir-Ixll - R-Ixllr-Ixll - (1+r)q Chu y rang, do R>3Ixl>O,ta colr-lxll=r-lxl:=::R-2Ixl>lxl>O, voi mQi r:=::R-Ixl. +00 ' ' A r N-] dr A' ~. D d h h h Q1 tl,l 0 0, tIc p an f N I 'I VOl q> 1. R-Ixl I r -Ixll - (1+ r) V~y, tich phan (4.21 ) J~2) (x) hQi W khi q > 1. T6 h 1, ta vie"t +00 N-l +00 d r N-I d r J = (4.23) f r r :=:: f q o(r+lxl)N-I(1+r)q Ixl(r+lxl)N-I(1+r)q +00 rN-Idr 1 +00 dr :=:: J( r+r )N-I(1+r)q = 2N-I J(1+r)q = 1 1 \Ix E JRN (q-l)2N-l (1+lxl)q-l . Do do b6 d~ 4.1 du
LucJn van tot nghifp Trang 29 Ta vie't phuong trlnh tich phan (4.7) voi bN = 1 theo d~ng (4.24) u(x) = Tu(x) = A [g(y,u(y))](x), \/x E IRN, trong do (4.25) A [w(y)](x) = J w(y) d~-I' X E IRN. iii' I y- x I Ta chung mint b~ng phan chung. Gia su u Ia nghi~m lien t\lCva duong cua (4.24). Khi do t6n t~i XoE IRN sao cho u(xo) > o. VI u lien t\lc nen t6n t~i ro > 0 sao cho: (4.26) u(x»~u(xo)=L \/xEIRN, Ix-xol:::;;ro. 2 Ta suy tu gia thie't (G2),(4.24)-(4.26) r~ng (4.27) u(x) = A[g(y,u(y))](x) ~ MA[ua(y)](x) 2::MLa J dy N-l' \/x E IRN. Iy-xol:s:ro I y-x I Su d\lng ba't d~ng thuc sau (4.28) I y - x I :::;;Iyl + Ixl :::;; 1 + Ixl)(1 + Iyl) ( = (1+ Ixl)(1+ Iyl- Xo + xo) :::;; + Ixl)( 1+ jxo I+ Iy - Xo (1 I ) :::;;(1+lxl)(1+lxol+ro)' \/x,YEIRN, Iy-xo I:::;;ro' ta suy tu (4.27), (4.28) dng (4.29) u(x) 2:: MLa J Iy-xol:s:ro ~- x yI I N-l > MLa 1 J dy -(1+lxol+ro)N-lx(1+lx l )N-l Iy-xol:s:ro N OJ = MLa X 1 NrO (l+lxol+ro)N-l (1+lxl)N-l N ' \/xEIRN. Ta vie't l~i (4.30) u(x) 2::u1(x) = m](1 + Ixlrq), \/x E IRN, trong do
Lugn win tot nghifp Trang30 a N M L ())NrO (4.31 ) ql = N -1, m] = N(1+lxo!+ro)N-I' Sa dl;lng ffiQtl~n nii'a d&ng thuc (4.24), ta sur tITghl thi~t (G2), (4.27) r[tng (4.32) u(x) 2 MA[ua (y)](x) 2 M4[u~ (y)](x) = Mm~A[(1 + Iylraq, ](x) \::IxE IRN. Bay gid ta xet cac tru'dng hQpkhac nhau cua gia tti a. 1 Truong hQ'p1: O::;a::;-. N-1 Ta sur ra tU (4.9), (4.32) voi q = a ql = a(N -1)::; 1, dng (4.33) u(x) = +00 \::IxE IRN. D6 la di~u vo 19. Truong hdp 2: . ~ N-1 < a 1,ta sur ra tIT(4.32) r[tng: (4.34) u(x) 2 Mm~A[(1+ Iylraq, ](x) = Mm~A[a ql ](x) ()) 2 Mmla N N-I(1+lx!)I-aq" \::IxEIRN. (aql -1)2 hay (4.35) U(X)2u2(x)=m2(1+lxlrq2, \::IxEIRN, trong d6 m2 - M()) N ma (4.36) q2 =aq ] -1 , - I 2N-l q2 . Gia sa dng (4.37) u(x) 2 Uk-I (X) =mk-I(1+!X!rqk-l, \::IXEIRN. N€u aqk-I > 1, khi d6 ta dung ba"td&ng thuc (4.10) voi q = aqk-I > 1, ta thu du'Qc tITgia thi€t (G2), (4.24), (4.37), r[tng (4.38) u(X) 2 M4[ua (y)](x) 2 M m:_]A[ (1 + Iylraqk-' ](x)
Luc7nvan tot nghi~p Trang 31 = M m:-lA[a qk-I ](x) 2 M ma k-l ())N (aqk-I -1)2N-l (1+IXI)I-aqk-1 2mk(I+lxlrqk = Uk(X), '\IxEIRN, trong d6 cac dtiy {qk},{mk} duQCxac d~nh bdi cac cong thuc qui n~p sau: a (4.39) 1m = M())N N I mk-I ' k = 2,3,., qk=aqH-' k 2 qk Tli (4.31), (4.39) ta thu duQc N - k, ntu a = 1, (4.40) qk = k I I-a k-I A'" 1 N (N-l)a - - , neu -
Lugn van tot nghifp Trang 32 (1+lyl rN > - f IIII ( y + x Ii \ ) d NIY- - > +'" (1+rrN d f 0 ( r + x ) -II Nlr lyl=r IdS r +"'(1+rrNrN-I 1\I+rrNrN-I = OJv . f 0 (r + II )N-I x f dr ~ OJN I (r + II )N-I dr x Ixl rN-Idr ~OJN [(1+r)N(r+lxj)N-I. Chu y r~ng voi mQi r sao cho 1 ~ r ~ Ix!ta co N r 1, 1 1 (4.45) ( 1+ r ) ~ 2N va r + Ixl ~ 21xJ. V~y, ta co ta (4.45) dug Ixl rN-Idr 1 1 Ixl dr (4.46) ! (1 + r)N ( r + Ixl)N-I ~ 2N ( 21xl)N-2 ! r( r + Ixl) 1 1 1+ Ixl N = 4N-I x Ixl N-I x In( 2)' "Ix E IR , Ix! ~ 1. Ta (4.43), (4.44), (4.46) ta suy ra ding 0, Ixl~1, (4.47) u(x) ~ V2(x) = ~ ~ In 1+ Ixl PZ, Ixl ~ 1, IxIN-I ( 2 ) voi (4.48) PZ = 1, Cz = MOJNm~ 4N-I Gia su r~ng 0, Ixl~1, (4.49) u(x) ~ vk-l (x) = ~ Ck-l In 1+ ixi Pk-l, Ixl ~ 1, IxlN-l ( 2 J trong do Pk-l>Ck-lla cae h~ng s6 dtiong. Su d\lng gia thie't (G2) va (4.49), ta suy ra dug (4.50) u(x) ~ M A[ua(y)](x)
Lwjn van tot nghi~p Trang 33 ~ M A[v:-1(y)](x) = M J V:-J~~I dy RN Iy - xl >M J v:-I(y) d >M J v:-I(y) d - I?' (lyl+lxl)N-1 Y - lyl~1 (lyl+lxl)N-1 Y +W V:-I (y) dSr = M Jdr Iyl=r ( r + Ixl) N I J I a Pk-I I+r In(- ) +w ( 2 ) dr 1 = M OJNC:-1J r(r + Ixl)N I Ta xet tru'ong hcJp Ixl~ I, ta co 1+ r a Pk-l 1+ r a Pk-I +00 In( -) +00 In( -) 2 ) 2 ) (4.51) J ( dr~ J ( dr I r(r+lxl)N-1 Ixl r(r+lxl)N-l 1+ Ixl a Pk-I +00 dr N-l ~ ( In(-)2 J J r (r + Ixl I x I) II a Pk-I +00 d ;, [ In(l: x). J I~ r(r +:)N-I - 1 1+ x a Pk-I (N -1)2N-Ilxt-1 ( In-fl) . Tli (4.50), (4.51), ta suy ra r~ng 0, Ixl~ 1, Pk (4.52) U(X)~Vk(X)=~ Ck 1+lxl - In- , II x:2:1, IxlN-l ( 2 ) trong do Pk>Ck la cae h~ng s6 du'dng xac dinh b~ng cae cong thu qui n(;lpnhu' sau: a MOJ C (4.53 ) Pk =apk-I' Ck = N k I (N -1)2N-I' k = 3,4,... Ta tinh fa cDng thuc hiSn cua Pk>Ck nho vao (4.48), (4.53), nhu'sau
Lu4n van tot nghi~p Trang 34 k-2 l-N N-I ak-2 (4.54) Pk =a , Ck =dN (dN C2) , k=3,4,... trong a6 MOJN (4.55) dN = (N -1)2N-J . Ta vie't I~i (4.52) voi Ixi ~ 1, ta c6 a k-2 I-N 1 N-I 1+ Ixi (4.56) u(x)~vk(x)=dN IX!N-I ( dN C21n(2) J . ChQn Xl saGcho (4.57) dZ-iC21n(I+lxll»I, 2 Do (4.56), ta suy ra rang u(xi) ~ k->+ooVk(Xl) = +00. lim EHy la ai~u va 19. Dinh 194.2 au'