2 Weighted Residual Methods

Transcription

1 2 Weighed Residua Mehods Fundamena equaions Consider he probem governed by he differenia equaion: Γu = f in. (83) The above differenia equaion is soved by using he boundary condiions given as foows: u = ū on u (84) Bu = on = \. (85) where he boundary consiss of u and. The boundary condiions given in eqs.(84) and (85) are caed rigid and naura boundary condiions, or iriche and Neumann boundary condiions, respecivey. In mos engineering probems, Γ and B are differenia operaors in he forms of Γ = Σ and B = n Σ, where Σ is anoher differenia operaor and n is he norma vecor on. For he Lapace probem, he governing equaion is given by 2 u = f in, (86) and he Neumann boundary condiion on is given as = n u =. Then he differenia operaors Γ, B and Σ are defined by Γ = 2, B = n and Σ =, respecivey. Tria funcions uppose ha u is approximaed by ria funcion ũ of ũ(x) = N J (x) (87) where are coefficiens and N J (x) are ineary independen funcions caed ria bases, es funcions or shape funcions. Weighed residua ince he ria funcion ũ does no saisfy he governing equaion and he boundary condiion rigorousy, he residuas defined in he foowing are nonzero R (x) = {Γũ(x) + f(x) = in (88) R (x) = ũ ū(x) on u, or Bũ(x) (x) on. (89) The probem is o find ũ o make zero he residuas weighed by he appropriae funcion w(x) as foows: w(x) R (x)dv + w(x) R (x)d = (9) where w(x) is caed he weigh funcion. Eq.(9) is aso caed a weak form o he origina differenira equaion (83), caed a srong form, because he souion ha saisfies eq.(9) does no aways saisfy he origina differenia equaion (83) and he boundary condiion (85). ubsiuion of eq.(87) ino eq.(9) yieds he foowing equaion: w I R dv + w I R d = w I (x) ( ΓN J (x) f(x))dv + w I (x) (N J (x) ū(x))d + w I (x) (BN J (x) (x))d u { = w I (x) ΓN J (x)dv + w I (x)n J (x)d + w I (x) BN J (x)d u {{ { w I (x) f(x)dv + w I (x)ū(x)d + w I (x) (x)d u {{ = (91) 13

2 or =. (92) where M independen weigh funcions w I (I = 1, 2,..., M) are chosen. This equaion can be soved for. As shown in he seque, here are severa residua mehods depending on he seecion of weigh funcions. I is no difficu o find ria funcions ũ saisfying he boundary condiion ũ = ū on u. In such a case, no residua exiss on u and he inegras on u disappear in eq.(91). In he foowing secions, i is assumed ha ria funcions ũ saisfy ũ = ū on u. 2.1 Poin coocaion mehod In he poin coocaion mehod, we seec a eas he same number of coocaion poins as he unknown parameers and deermine he parameers such ha he residua is zero a he seeced poins. The coocaion mehod is equivaen o he residua mehod in which he weigh funcion is chosen as w I = 1δ(x x I ) (93) where 1 denoes he uni marix and δ(x) is he irac dea funcion saisfying he equaions δ(x x I )R(x)dV = R(x I ), for x I (94) or δ(x x I )R(x)d = R(x I ), for x I (95) In his case, he marix and he vecor in eq.(92) can be expressed by 2.2 Leas squares mehod = ΓN J (x I ) or BN J (x I ), = f(x I ) or (x I ) (96) In he eas squares mehod, he sum of he squares of he residuas defined by I = R 2 dv + R 2 d (97) is minimized. The necessary condiion o minimize I is yieding he sysem of equaions: or I =, I = 1, 2,... (98) I û R R dv + R R d = (99) û I û I [ ] ΓN I (x) ΓN J (x)dv + BN I (x) BN J (x)d {{ [ ] ΓN I (x) f(x)dv + BN I (x) (x)d {{ = (1) As shown in he above equaion, he eas squares mehod is considered as a weighed residua mehod wih he weigh funcions ΓN I (x) in he domain and BN I (x) on he boundary. 14

3 2.3 Cassica Gaerkin mehod If ony he domain inegra in eq.(91) is aken ino accoun and he shape funcions N I (x) are used as he weigh funcions w I, where are arbirary coefficiens, hen we have { 2.4 Gaerkin mehod { N I (x)γn J (x)dv {{ N I (x) f(x)dv = (11) {{ In he cassica Gaerkin mehod, he boundary condiion is no considered. Therefore, he cassica Gaerkin mehod is no convenien for pracica use. Here he Gaerkin mehod aking accoun of he boundary condiion is formuaed. Consider he weighed residua as foows: w(x)( Γũ(x) f(x))dv + w(x)(bũ(x) (x))d =. (12) Here we assume ha w is a weighed funcion saisfying w = on u and ũ saisfies he boundary condiion ũ = ū on u. ince Γ = Σ, he firs inegra in eq.(12) is wrien as w(x)γũ(x)dv = w(x) Σũ(x)dV = w(x)n Σũ(x)d w(x) Σũ(x)dV = w(x)bũ(x)d w(x) Σũ(x)dV (13) where he divergence heorem is used. Noe ha in he above equaion, he inegra on u vanishes due o w = on u. Then eq.(12) becomes w(x) Σũ(x)dV w(x) f(x)dv w(x) (x)d =. (14) This equaion has he weak form which invoves he Neumann boundary condiion on, which can be soved under he iriche boundary condiion ũ = ū and he consrain condiion w = on u. Now i is assumed ha he ria funcion ũ is expressed by ũ(x) = N J (x) + N (x)ū (15) where N J (x) and N (x) are he shape funcions equa o zero and one on he boundary u, respecivey. Aso he weigh funcion w is seeced as w(x) = N I (x) (16) where N I (x) is he same shape funcion used in eq.(15). Then he condiions u = û and w = on u are saisfied. ubsiuion of eqs.(15) and (16) ino eq.(14) yieds N I (x) ΣN J (x)dv {{ ( N I (x) f(x)dv N I (x) ΣN (x)dv ū + ) N I (x) (x)d =. (17) {{ 15

4 Exampe 2.1 Consider he souion u saisfying he equaion subjeced o he boundary condiions dx 2 + f = x (18) u = a x =, du/dx = f a x =. (19) The exac souion for he above probem is u(x) = f x(2 x/2) and he vaue a x = /2, u(/2), is obained as u(/2) = f /2(2 x/4) = f 7 2 /8. Now he boundary vaue probem menioned above is soved by means of boh he cassica Gaerkin mehod and he Gaerkin mehod assuming ha he souion is approximaed by ũ(x) = N 1 (x)û 1, where N 1 (x) = x(2 x). Noe ha he approximaed funcion is sighy differen from he exac souion. Cassica Gaerkin mehod For N 1 (x) = x(2 x) and ΓN 1 (x) = d 2 N 1 (x)/dx 2 = 2, eq.(11) becomes x(2 x)( 2)dxû 1 + x(2 x) f dx = (11) From eq.(11), we obained û 1 = f /2. The souion ũ(x) is obained by ũ(x) = f x(2 x)/2. A x = /2, ũ(/2) = 3/8 f 2, which is much differen from he exac souion u(/2) = 7/8 f 2. Gaerkin mehod From eq.(17), he foowing equaion based on he Gaerkin mehod is formuaed. ( ) dn 1 dn 1 dx dx dxû1 N 1 (x) f dx + N 1 (1) f = (111) ubsiuion of N 1 (x) = x(2 x) ino he above equaion yieds ( ) 2( x)2( x)dxû 1 (2x x 2 ) f dx + (2 ) f =, (112) from which û 1 = 5 f /4. Hence, ũ(x) = 5 f x(2 x)/4. Aso a x = /2, ũ(/2) = 15/16 f 2. Thus we can concude ha he Gaerkin mehod wih he boundary condiion gives a vaue coser o he exac souion han he cassica Gaerkin mehod, in which no boundary condiion is considered. Exampe 2.2 ove he governing equaion dx = x 1 (113) by using he Gaerkin mehod wih he ria funcion u(x) = 2 N J (x), where N J (x) = x J. The boundary condiions are given by u() = and du/dx(1) = 2. The Gaerkin mehod yieds he foowing equaion for he above menioned probem: ( 1 1 dn I dn J dx dx dx {{ 1 N I (x) fdx dn I dn ) dx dx dxū + N I (1) = (114) {{ where f = 1, ū =, = 2, M = 2, N 1 (x) = x and N 2 (x) = x 2. The cacuaion of he marix and he vecor eads o he equaions [ ] { { 1 1 û /3 û 2 = (115) 3 which can be soved for (J = 1, 2). 16

5 Probem 2.1 Consider he governing equaion dx 2 u = x 1 (116) subjeced o he boundary condiion u() = and du/dx(1) = 2. ove he probem by using he Gaerkin mehod wih he ria funcion u(x) = 2 N J (x), where N J (x) = x J. Probem 2.2 erive he weak form based on he Gaerkin mehod for he mui-dimensiona Lapace probem, for which he governing equaion is given by and he boundary condiions are as foows. u = f in (117) u = ū on u (118) n u = on = \ u (119) Probem 2.3 ove he same probem as Probem using he weak form. 17