tài liệu tham khảo chỉnh hóa bài toán moment tổng quát
Chuang III: Chlnh h6a bai toaD moment t6ng quat , ? " A?' CHU0NG III: CHINH HOA BAI TOAN MOlVIENT TONG QUAT 3.1 Bai toan moment t6ng quat: TIm mQt ham u tren mi~n D c Rdthoa di~u ki~n : Iu(x)gn (x) dx = ~n voi (3.1) nEN d6, (gn) la day ham cho truac, /.l=(/.ll, /.l2, ) 13.day so 12 hay B3.i roan moment r" thuong khong chlnh rhea nghIa 13 chung khong c6 nghi~m va truong h Vc}yW E X.Lvdi X 1a kh6ng gian sinh bdi cac gj, i = 1,2, ., n Theo tinh chat gian Hilbert, ta co: Lz =X + X.L ma X =Lz X.L ={a} => w =0 => UI nen =Uz Vc}yta co tinh nhat nghi~m (neu co nghi~m) cua bai roan (3.1) truonghQpkh6ng gian da k.i~nla t' Vdi cach d~t roan ta A nhutren ta co m~nh d~ sail: Menh d~: Cho A: H ~ tJ U H Au = ((u, gn)H )n=1.2,,' A 1a anh X? tuyen tinh lien t\1c A 1a don anh Range A chua h~n 100 A-I: Range A ~ H kh6ng lien t\1c Chung minh: A tuyen tinh tinh tuyen tinh cua rich v6 hudng Ta co: IIAu!!t"' supl(u, O)H ::; supllgllllH = =0 g lIullH n I IluliH => A lien t\1C Gia sa Au =«u, gn)H)n=1,2, ,= O Do (gn) 1a co sd cua H nen UE H.L => u =O Vc}yA la don anh Ta chung rninh b~ng phan chung Gia sa Range A = 100 Theo h~ qua cua dinh 1yanh X?md (xem [2] ) thl A la d6ng ph6i tit H vao 100 VI H phan X?, A la d6ng ph6i tuyen tinh, nen 100phan X? (xem [1] ) f)i~u v6ly 20 Chuang III: Chinh hoa bai roan moment t6ng quat V~y Range A chua h:1n r-c nghla la bai roan 3.1 khong luau t6n t~i nghi€$mtrong tntong h nen: ( v, gi,) = ( u, gj,) = ~i (3.6) (1 :::; :::; ) i n (do tinh cha't phep chieu va thoa (3.4)) n V~y v thoa man tinh cha't tli (1) (3) nen ta duQc v = LAiei i=1 M?t khac rhea tinh cheithlnh chieu vuong goc ta co: II vii:::; II till Mall ull=min{11 vII :vEL2(0),(v,gj)=~j}=>11 =II Tli hai ke't qua tren suy : II u II V~y II w 112 = II U 112-II ull :::;11 vii' v II V 112 (dinh 1y Pythagore) =0 =>w=o n Ta lai co u = v + w => u =v =~ Xe II i=1 n (do ta da chung minh duQc v = LAiei i=1 ) V~y ta co (3) Dao lai: Gia sa ta co (3) Cho v E L2 (D) cho U = 1, , = ~j n) Cho Pyla hlnh chieu vuong goc cua v len < gJ, g2, , gn>' 2-1- Chuang III: Chinh hoa bfli loan moment t6ng quat Khi rhea cach d~t tu (3.6) ta c6: (Py, gj) = (v, gj) = Ilj (Py Ia hinh chie'u cua v Ien < gl, g2, gn> ~ Pythoa (1)) n Thea chung minh (1) Q (3) ta du n ~ = I~igi i=l pn(ll) (3.8) M~t khac pn(ll) thoa man di€u ki~n (1) cua dinh 1y 3.1 nen (3.9) (pn(Il), gj) = Ilj n The'(3.8)vao(3.9) : (I i=1 ~ig/!gJ) = J-ij n Q I~I(gl,g)=!lj O=l, ,n) i=1 Day Ia h u = IXi(,u)e; ;=1 ;=\ j=1 I ma thee = ( 3 ) "CII.IJI'"'"JA ~ j=1 Do m~nh d~ 3.1 (3) va ph§n chung minh (2) Q (3) ta co: u=v=P u } Vn Ma pn (p)= U (pn(/l) dong vai tro nhu' u) ~ pn (p) = Pr-.(u) Sl,l'hQi t1,lcua pn(~) d€n u L2 (Q ) theo tinh cha't cua khong gian Hilbert :f; Th~t v~y voi u E H ta co: U = CX) I n=\ n n=\ :f; = Lakek Ta co: U = Lanen + Iakek k=1 k=,,+1 n :f; Iakek k=1 = PYn(U)= pn(~) V~y pn(~) a"e" voi an = (u,en) ~ U = Lakek k=1 u L2 (Q ) ~ Chung minh U E L2 en ): CX) Ta co: (u, ej) i = ( I (2:Cjjll)ej,ej i=1 ) j=l i I Cijllj j=1 = ex) => IIuI12= ; 2: (2: Cijpj)2< ;=1 (vi (ej) la h~ trl,l'cchucfn) 00 ( gia thuy€t (3.11)) j=1 Trong (3.10) the' Pvn = pn (11) ta dliqc: U IIpn(Il) - till ~ cllull~:(Q)n 'v'n 28 Chuang III: Chlnh h6a bai toan moment tong quat Dinh 1y sail chI r~ng tru'ong hc;5P kic$nkh6ng chinh xac, nghic$m du cua bai roan moment (3.4) 1a xap Xl 6n oinh nghic$m Clla bai roan (3.1) Ky hi~u II Il II ro = sup {Illi I; i EN} 3.2.4 Dinh IV 3.4: Cho Un E L2 (D ) 1a nghic$m Clla (3.1) tu'ang ling vdi: I' 0 = ( 1'1,1'2 , ) vOl VOl < £ < i I gl I:'\~" " 11 ham tang ng~t tu [ 1,00) 1en [II g&l, co) rho rt- ta ui;n: n(£ ) IICnl1 =[[ (£ -112 " ' )] VOl f1 a ~ fen) \j n 2: Khi t5n t~i ham s6'1 (£ ) rho lim17(E)= va d6i vdi nhung day Il thoa: i:~1) 111l-IlOlloo t:}v = 0) Tli c6ng thuc Green thu hai : (f6g)dx-1 (g6f)dx = !of ~CY - L g ~ dCY u la phap vec to don vi ngoai cua rn;rtta ; f, g la hai ham du trail Ap dl:lllg c6ng thuc tren voi f = u , g =bm ta duQc: (,0.u , bm ) =( u , ,0.bm ) (do f = u = tren ao ) (3.23) M;rttkhac rhea gia thuyet - ,0.u= a u suy bm = - ambm V~y (u , ,0.bm ) = ( u, - am bm ) = - ameli, bm) Tli (3.23) (3.24) ta co: (,0.u , bm ) = (3.24) (u,,0 bm) = - ameli, bm) vdi u 1aham du trail va tri~t tieu tren bien a O Thay u = ,0.j voi j ~ 1- vao ket qua tren ta duQC: v (,0.1 bm) = v, - am (6]-1 V, bm)= = (-1)1 a~(vJbm) , , -) _1 (3.25) Chuang III: Chlnh h6a bai toan moment t6ng quat un Vt).y ex) IIIV 1112 = I a~ (v, bm)2 m=! ex) "\' a21 ( V, b m ) L m = (vi k=2l) m=! ex) = I(a~(v,bm))2 m=! = = (do 3.25) II(~/v,bmr m=! II ~lv 112 a2 = { (~+ ax! a2 ! +~) axel V } dx (vi d la s6 chi~u cua Q va dinh nghla cua ~ ) = 1(LDav)2dx aeI, (d~ohamca'p21=1 0.1) ~ dl I(Dav)2 aeJI dx (co d1 s6 h~ng) (ap d\lilg ba't d~ng thlic Cauchy - Schwartz: (al.bl+ +anbn)2~(aI2+ I = d1 ae!, ~ d21 2: .+an\(bI2 Ca (DaV)2 dx J (DaV)2dx lal=21=k[2 = d21 L f (DClV)2dx aEl:Q 34 + + bn2) vdin=d1;aj=1) Chuang III: Chlnb boa bfli roan moment t6ng qmit LIIDav)r =dk ( chuffn L (0) ) aEl2 ?: = dk Ivl ~,Q L (DaV)2 1a t6ng Trong d6: co du'Qc khai tri~n mQt cach hint thli'c aEII , ? ? bi~u thli'c (~-2 + I + :2)1 v ma khong nit gQn cac sO' h(lIo).b,< 1=11+\ a (do an la day tang => an+1 < an+2 < an+3 => IIwl12:::;a~~IIII(ulJII12 (do d~nh nghla => ~ anTI ' ~ voi i ~ 1) chu.1n 111.111) (3.30) M~t khac : ? un