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1、 PHNG PHP SAI PHN HU HN 1. Pht biu bi ton: Cho min l hnh vung c dng = (0,l) x (0,l), S l bin ca . Phng trnh elliptic c dng: uxx + uyy = f (x,y) trn u = g(x,y) trn S (1)2. Phn r bi ton : Cho N, ta chia min thnh (N+1)2 hnh vung cnh l vi cc im chia ( xm , yn ) = ( (m-1)h, (n-1)h ). p dng cng thc sai ph
2、n trung tm xp x (uxx + uyy) : t Umn = u( xn , ym ), fmn = f( xn , ym ) , ta c cc phng trnh sai phn hay Ti cc im trn bin, ta c: khi m = 1 hoc n = 1 Nh vy ta c c mt h gm ( N-1)2 phng trnh v ( N-1)2 n cn tm l Umn , vi 2 m,n N . nh ch s li theo cch : ( i , j ) ( N+1) ( i-1) + j ta vit li U dng vect v c
3、h phng trnh : AU = B (2)* H (2 ) c cc tnh cht sau :1/- S chiu ca vect U bng ( N-1)2 2/- Vect B xc nh bi cc iu kin bin v cc s hng khng ph thuc U nh fm,n , h .3/- Ma trn A c cp ( N-1)2 ( N-1)2 . Cp ca A l mt hm gim theo h . Do , mc d cc phn t ca A khng ph thuc h nhng cc tr ring ca A, cng nh tr ring ca
4、 cc ma trn dn xut t A, l hm ph thuc h. A l mt ma trn khi gm (N-1)(N-1) ma trn con, A c dng:Trong I l ma trn n v cp (N-1) v H l ma trn (N-1) (N-1) c dng4/- Nu bi ton (1) ban u c duy nht nghim v h nh, ma trn A khng suy bin, tc l h (2) c duy nht nghim .3. Phng php lp gii h AU = b Ta hon ton c th gii h
5、AU = b mt cch trc tip (chng hn bng phng php kh Gauss). Tuy nhin, khi ma trn A c kch thc qu ln v nhiu c phn t nhiu phn t bng 0, ngi ta thng dng phng php lp ci t v phng php ny tit kim b nh v cho kt qu s chnh xc hn vic gii trc tip. Phng php lp s to ra mt dy cc nghim Uk xp x nghim chnh xc. Cc phng php l
6、p im gii h (2) gm c: Phng php lp im Jacobi (3) Phng php lp im Gauss Seidel (4) Phng php lp im SOR. (5)* Cc tnh cht ca nhng phng php lp im trn nh sau:1/ Cc iu kin bin c m bo bng cch t ti cc im bin vi mi k = 0, 1, 2, . . . 2/ Trong 3 phng php trn, gi tr khi to c th chn ty v khi k tin v , dy Uk s hi t
7、v nghim chnh xc. 3/ Ta c: Fmn = - h2 fmn4/ Trong phng php Jacobi, th t duyt cc im nt khng quan trng.5/ Trong phng php Gauss-Seidel, th t duyt cc im nt rt quan trng. C th y, cc nt s c duyt theo chiu tng dn ca cc hng, trn mi hng duyt theo chiu tng dn ca cc ct,6/ Phng php Gauss-Seidel khc phng php Jaco
8、bi ch: cc gi tr mi ca U s c cp nht ngay tnh nhng gi tr k tip.7/ Phng php SOR tung t phng php Gauss-Seidel nhng c a thm h s ( relaxation paraneter) y nhanh tc hi t. Vi = 1, ta c tr li phng php Gauss-Seidel. 4. S hi t ca phng php lp im:Vit ma trn A nh sau: A = - L + D U , trong L , D , U ln lt l cc ma
9、 trn tam gic trn ngt , ma trn cho v ma trn tam gic di ngt . Ta gi s thm det (D) khc 0. Cc phng php (2), (3), (4) c biu din di dng ma trn nh sau : Phng php lp im Jacobi Uk+1 = TJ Uk + CJ (5)( Trong TJ = D-1 (L+U) , CJ = D-1B ) Phng php lp im Gauss-Seidel : Uk+1 = TG Uk + CG (6)(Trong TG = ( D-L)-1 U
10、, CG = ( D-L)-1 B ) Phng php lp im SOR: (7) (Trong )Nh vy c 3 phng php lp u c dng tng qut l: Uk+1 = TUk + CXt nh x: Ta thy: Uk+1 = F (Uk) . Do F lin tc nn nu U* l gii hn ca dy Uk th ta c U* = F( U*) , tc l U* = TU* + C. Nh vy, trong c 3 phng php (5), (6), (7), nu U* l gii hn ca dy Uk th r rng AU* = b , ngha l U* l nghim ca phng trnh ban u. iu kin dy Uk hi t c cp trong cc nh l sau: nh l 1: Dy Uk xc nh bi Uk+1 = TUk + C , vi U0 ty , hi t v mt vect duy nht U* ( U* c lp vi U0 ) khi v ch khi r(T) 1( trong r(T) l bn knh ph ca ma trn T, tc l r(T) = max | l tr ring (phc) ca ma trn T Nhn
