(bilinearity), a(u, v) M u V v V (continuity), a(v, v) m v 2 V (coercivity).

Transcription

1 Precoditioed fiite elemets method Let V be a Hilbert space, (, ) V a ier product o V ad V the correspodig iduced orm. Let a be a coercive, cotiuous, biliear form o V, that is, a : V V R ad there exist m, M, 0 < m M such that for all u, v, w V, α, β R, a(αu + βv, w) = αa(u, w) + βa(v, w), a(u, αv + βw) = αa(u, v) + βa(u, w) (biliearity), a(u, v) M u V v V (cotiuity), a(v, v) m v 2 V (coercivity). Let V be the set of all cotiuous, liear forms o V. The the followig Lax-Milgra results hold: (LM1) For ay F V, there exists a uique elemet u = u F a(u, v) = F(v), v V. V such that (LM2) If, moreover, V h is a fiite-dimesioal subspace of V, the to F we ca also associate a elemet u h V h such that a(u h, v h ) = F(v h ), v h V h, which is uiquely defied too. Approximatig u by u h Ituitively, such elemet u h ca be used as a approximatio of u if V h belogs to a family of subspaces {V h } h 0 of icreasig dimesio, such that the closure of h 0 V h coicides with V. I fact, it ca be show that a hypothesis of cosistecy o {V h } h 0 (implyig the latter property) yields the result: h 0 u u h V 0. (cov) Cosistecy of {V h } h 0 i V. {V h } h 0, V h V, is said to be cosistet i V if there exist V V dese i V (with respect to V ) ad a operator R h : V V h such that for ay v V, R h (v) v V 0 as h 0 (R h might be a iterpolatio operator). Let us show that the cosistecy of {V h } h 0 implies (cov). First we prove that the error u u h V is proportioal to the miimal error we ca have with V h. Note that a(u, v h ) = F (v h ), a(u h, v h ) = F (v h ) a(u u h, v h ) = 0 v h V h ad this remark implies m u u h 2 V a(u u h, u u h ) = a(u u h, u v h ) M u u h V u v h V, u u h V M m if u v h V. v h V h (cea) Now we ca prove (cov). Let v V be such that v u V < ε. By (cea) ad the cosistecy hypothesis, if h < h ε (h is suitably small), the u u h V M m u R h(v) V M m ( u v V + v R h (v) V ) < M m 2ε. 1

2 How to compute u h Let N h be the dimesio of V h, ad ϕ i, i = 1,..., N h, a basis for V h. The u h = N h (u h) j ϕ j ad the coditio a(u h, v h ) = F (v h ), v h V h, ca be rewritte as follows: N h (u h ) j a(ϕ j, ϕ i ) = F (ϕ i ), i = 1,..., N h. So the (u h ) j defiig u h ca be obtaied by solvig a liear system Ax = b, beig a ij = a(ϕ j, ϕ i ), b i = F (ϕ i ), 1 i, j N h. It is importat to otice that the symmetric part of the coefficiet matrix A is positive defiite, that is z T Az > 0, z R z 0. I fact, by the coercivity of a, we have z T Az = ij z i a(ϕ j, ϕ i )z j = a( z j ϕ j, z i ϕ i ) m z i ϕ i 2 V > 0 uless the z i are all ull (the ϕ i are assumed liearly idepedet). Example: a differetial problem solved by the fiite elemet method Assume that u : R is the uique solutio of the differetial problem (α u) + β u + γu = f, x, u = ϕ, x Γ D, u c = ψ, x Γ N. Here is a ope set i R d, Γ D ad Γ N are ope subsets of such that = Γ D Γ N, α : R d2, β : R d, γ, f : R. The, for all v, v ΓD = 0, a(u, v) := α u v + β uv + γuv = fv + ψvdσ. Γ N If we set u = u ϕ + w with u ϕ, w : R, u ϕ ΓD = ϕ ad w ΓD = 0, the the latter equatio becomes: a(w, v) = fv + ψvdσ a(u ϕ, v) =: F (v). Γ N So, we have the followig Problem. Fid w, w ΓD = 0 such that a(w, v) = F (v), v, v ΓD = 0. Moreover, the fuctios w, v must be also such that a(w, v) ad F (v) are well defied, that is we also require w, v H 1 () where H 1 () = {v L 2 () : D i v L 2 ()} (D i v L 2 () iff g i L 2 () g iϕ = vd iϕ ϕ C 0 () (D iv := g i )). Briefly, fid w H 1 0,Γ D () a(w, v) = F (v), v H 1 0,Γ D (). Uder suitable coditios o the data, α, β, γ, ϕ, ψ, the space V = H 1 0,Γ D () is a Hilbert space with respect to the ier product (u, v) V = (u, v) 1, = (u, v) L 2 () + i=1...d (D iu, D i v) L 2 (), ad the forms a ad F are well defied ad satisfy the coditios required by the Lax-Milgra results (LM1) ad (LM2) to hold. So, the w of the problem is well defied, ad for ay fiite-dimesioal 2

3 subspace V h of V = H 1 0,Γ D () is well defied a fuctio w h V h such that a(w h, v h ) = F (v h ), v h V h. Defiitio of w h coverget to w I order to yield fuctios w h coverget to w as h 0, we oly have to defie V h such that {V h } h 0 is cosistet i H 1 0,Γ D (). Let us do this i the case d = 2, = polygo, by usig the fiite elemet method. Let τ h be a triagulatio of of diameter h, that is a set of triagles T such that T τ h T = T ad diam (T ) =: h T h := max T τh h T, T τh T =, T 1, T 2 τ h T 1 T 2 is a commo vertex, a commo side, the whole triagle T 1 = T 2 or the empty set. To ay T i the followig we eed to associate also the umber ρ T, the diameter of the circle eclosed i T. Let S h be the set of all fuctios p : R such that p T is a degree-1 polyomial (i x 1 ad x 2 ) ad set Vh = S h C 0 (). Let i = 1, 2,..., Nh be the odes of the triagulatio τ h (the verteces of the triagles of τ h ) ad deote by ϕ i the elemets of Vh satisfyig the idetities ϕ i (j) = δ ij, i, j = 1, 2,..., Nh. Obviously ay elemet v of V h ca be expressed as v = N h i=1 v(i)ϕ i. Choose V h = Vh H1 0,Γ D () = Spa {ϕ 1,..., ϕ Nh } where 1,..., N h are the odes of the triagulatio τ h which are ot o Γ D. We wat to show that {V h } h 0 is cosistet i H0,Γ 1 D (), so that the well defied fuctios w h =...N h (w h ) j ϕ j V h such that a(w h, v h ) = F (v h ), v h V h, strogly coverge to w as h 0, i.e. w w h V 0. First we itroduce the space V = H0,Γ 1 D () H 2 (), cotaied ad dese i V = H0,Γ 1 D () (H 2 () = {v H 1 () : D ij v L 2 ()}, (u, v) 2, = (u, v) 1, + ij (D iju, D ij v) 0,, v 2 2, = i D iiv 2 0, ). Now let v be a elemet of such V. Notice that v C 0 () (...), thus the fuctio Π h v = N h i=1 v(i)ϕ i of V h, iterpolatig v i the odes of the triagulatio, is well defied. Moreover, Π h v is a fuctio of H 1 () ad oe ca measure the iterpolatig error usig the orm of V: v Π h v 1, c e h v 2,, c e costat. The latter iequality holds if the family of triagulatios {τ h } h 0 is chose regular, that is there exists a costat c r such that h T /ρ T c r for all T τ h ad h. Thus we have the operator R h : V V h required by the cosistecy hypothesis, it is the iterpolatig operator Π h. Observe that i case the fuctio w is i H 2 () we ca say more: w w h V 0 at least at the same rate of h. I fact, w w h V M m w Π hw V M m c eh w 2,. Computatio of w h I order to compute w h = N h (w h) j ϕ j oe has to solve the liear system Ax = b, a ij = a(ϕ j, ϕ i ), b i = F (ϕ i ). 3

4 I fact, (w h ) j = (A 1 b) j. More specifically, i our example, if s(g) deotes the set supp (g), the the etries of A ad b are: a ij = s(ϕ j) s(ϕ i) α ϕ j ϕ i + s(ϕ j) s(ϕ i) β ϕ jϕ i + s(ϕ j) s(ϕ i) γϕ jϕ i, b i = s(ϕ i) fϕ i + Γ N s(ϕ i) ψϕ idσ s(ϕ i) s(u ϕ) α u ϕ ϕ i s(ϕ i) s(u ϕ) β u ϕϕ i s(ϕ i) s(u ϕ) γu ϕϕ i, 1 i, j N h. Here the ϕ i are the Lagrage basis of V h (ϕ i (j) = δ ij ). So, the (w h ) j are the values of w h i the odes j ((w h ) j = w h (j)) ad the matrix A is sparse, i fact for ay fixed i, the umber of j such that the measure of s(ϕ j ) s(ϕ i ) is ot zero is smaller tha a costat (with respect to h) depedet upo the regularity parameter c r of the triagulatios (such costat is a boud for the umber of odes j liked directly to i). But these properties are far from to be essetial: i particular, more importat would be to kow that the matrix A is well coditioed. Ufortuately, eve i case the differetial problem is simply the Poisso problem (α = I, β = 0, γ = 0, Γ N = ) the matrix A has a coditio umber growig as (1/h) 2, if the Lagrage fuctios are used to represet w h. (This estimate of the coditio umber holds more i geeral for the covectio-diffusio problem, if the triagulatios are quasi-uiform (i.e. h T c u h, T τ h h) ad regular). So, cosider a arbitrary basis { ϕ i } of V h, ad represet w h i terms of this basis: w h = N h (w h ) j ϕ j. The, (w h ) j = (Ã 1 b) j where ã ij = s( ϕ j) s( ϕ i) ã ij = a( ϕ j, ϕ i ), bi = F ( ϕ i ), α ϕ j ϕ i + β ϕ j ϕ i + s( ϕ j) s( ϕ i) s( ϕ j) s( ϕ i) bi = s( ϕ i) f ϕ i + Γ N s( ϕ i) ψ ϕ idσ s( ϕ i) s(u ϕ) α u ϕ ϕ i s( ϕ i) s(u ϕ) β u ϕ ϕ i s( ϕ i) s(u ϕ) γu ϕ ϕ i. γ ϕ j ϕ i, If the ϕ i are such that µ 2 (Ã) < µ 2(A) (...), the we ca solve the system Ãx = b, better coditioed tha Ax = b, ad the, if eeded, recover w h = (w h (j)) N h = A 1 b solvig the system Sw h = w h = ((w h ) j ) N h = Ã 1 b. (Of course, all this is coveiet if S is a matrix of much lower complexity tha A). Let us prove this assertio i detail. Let v h be a geeric elemet of V h ad let S be the matrix such that ṽ h = Sv h, beig v h = [(v h ) 1 (v h ) Nh ] T ad ṽ h = [(v h ) 1 (v h ) Nh ] T such that v h V h N h N h (v h ) j ϕ j = v h = (v h ) j ϕ j. The we have ad therefore N h ϕ s = [S T ] sj ϕ j, s = 1,..., N h, a ij = a( N h r=1 [ST ] jr ϕ r, N h m=1 [ST ] im ϕ m ) = r,m [ST ] im a( ϕ r, ϕ m )[S] rj = r,m [ST ] im ã mr [S] rj = [S T ÃS] ij, 4

5 N h N h b i = F ( [S T ] ij ϕ j ) = [S T ] ij F ( ϕ j ) = [S T b]i. Thus, the equalities A = S T ÃS ad b = S T b must hold, ad the thesis follows. I the Poisso case, u = f, x, u = ϕ, x, a basis { ϕ i } for V h ca be itroduced yieldig a matrix à whose coditio umber µ 2(Ã) grows like (log 2 (1/h)) 2. (A aalogous result i the covectio-diffusio case (a ot symmetric) i 1995 was ot kow!). We ow see (ot i all details) that this is possible by usig a particular family of triagulatios τ h. Let τ 0 be a rada triagulatio of. Let us defie τ 1. For each triagle T of τ 0 draw the triagle whose verteces are the middle poits of the sides of T. The four triagles you see (similar to T ) are the triagles of τ 1. Note that if h 0 is the diameter of τ 0 (h 0 = max{h T : T τ 0 }), the h 1, the diameter of τ 1, is equal to 2 1 h 0. Note also that the odes of τ 0 are odes of τ 1 ; the ew odes of τ 1 are the middle poits of the sides of τ 0. We ca cotiue i this way, ad defie the triagulatios τ 2, τ 3,..., τ j,... (obviously, τ j is a abbreviatio for τ hj ). The diameter of the geeric τ j is h j = 2 j h 0, ad the family of triagulatios {τ j } + j=0 is regular ad quasi-uiform. To each τ j we ca associate the space V j = Vh j H0 1 () of the fuctios which are cotiuous, ull o, ad degree-1 polyomials i each T τ j. Note that V j V j+1. Let x jk, k I j = {1,..., N j } {1,..., Nj }, deote the geeric ier ode of the triagulatio τ j, ad {ϕ jk : k I j } the Lagrage basis of V j, ϕ jk (x jl ) = δ kl, k, l I j. Obviously, ay v V j ca be represeted as v = k I j v(x jk )ϕ jk, ad, if Π j is the iterpolatig operator, the v C 0 () Π j (v) = k I j v(x jk )ϕ jk. Istead of v(x jk ) we will write shortly v jk. Now that all is defied, cosider a fuctio v V j+1 ad observe that v j+1,k ϕ j+1,k = v = Π j+1 v = Π j v + (v Π j v) = v jk ϕ jk + (v Π j v). k I j+1 k I j Now the questio is: what must we add to V j i order to obtai V j+1? This questio ca be reduced to: what elemets of {ϕ j+1,k : k I j+1 } are eeded to represet v Π j v? Let x be a poit of ad let T be a triagle of τ j icludig x. Let us observe the above quatities ad, i particular, the fuctio v Π j v o T. Call x jk1, x jk2, x jk3 (k 1, k 2, k 3 I j ) the verteces of T. Note that they are odes also of τ j+1, thus x jki = x j+1,ρ(ki), for some ρ(k i ) Ij+1 o = {k I j+1 : x j+1,k is a ode of τ j }. Call x j+1,σ(k1k 2), σ(k 1 k 2 ) Ij+1 = I j+1\ij+1 o, the middle poit of the side x jk1 x jk2 of T, which is a ew ode, a ode of τ j+1, but ot of τ j. Draw the restrictios to T of the fuctios v ad Π j v. The it is clear that, o T, v Π j v = [v j+1,σ(k1k 2) 1 2 (v j,k 1 + v j,k2 )]ϕ j+1,σ(k1k 2) +[v j+1,σ(k2k 3) 1 2 (v j,k 2 + v j,k3 )]ϕ j+1,σ(k2k 3) +[v j+1,σ(k3k 1) 1 2 (v j,k 3 + v j,k1 )]ϕ j+1,σ(k3k 1) = ṽ j,σ(k1k 2)ϕ j+1,σ(k1k 2) + ṽ j,σ(k2k 3)ϕ j+1,σ(k2k 3) + ṽ j,σ(k3k 1)ϕ j+1,σ(k3k 1) = k I ṽ j+1 jk ϕ j+1,k 5

6 where ṽ jk = v j+1,k 1 2 (v j,k + v j,k ), k I j+1, ad k, k I j are such that x jk = x j+1,ρ(k ), x jk = x j+1,ρ(k ) are the extreme poits of the side of τ j havig x j+1,k as middle poit. Thus, if ψ jk := ϕ j+1,k = ϕ j+1,σ(k k ), k I j+1, the v V j+1 v = v j+1,k ϕ j+1,k = v jk ϕ jk + k I j+1 k I j k I j+1 ṽ jk ψ j,k. It follows that V j+1 = V j + W j, W j := Spa {ψ jk : k Ij+1 }, ad the set {ϕ jk : k I j } {ψ j,k : k Ij+1 } is a alterative basis of V j+1. So, the aswer to the questio is: the Lagragia fuctios of V j+1 correspodig to the ew odes. Observe that the v jk, k I j, ad the ṽ jk, k Ij+1, ca be computed from the v j+1,k, k I j+1, via the formulas v jk = v j+1,ρ(k), k I j ṽ j,k = v j+1,k 1 2 [v j+1,ρ(k ) + v j+1,ρ(k )], k Ij+1. Viceversa, the v j+1,k, k I j+1, ca be computed from the v jk, k I j, a from the ṽ jk, k Ij+1, via the formulas: v j+1,ρ(k) = v jk, k I j v j+1,k = ṽ j,k [v j,k + v j,k ], k I j+1. These formulas ca be writte i matrix form: v jk [ ] v j+1,ρ(k) k I j ṽ jk = I Ij 0 k I j B I I k Ij+1 j+1 v j+1,k k Ij+1, v j+1,ρ(k) k I j v j+1,k k I j+1 where each row of B has oly two ozero elemets, both equal to 1 2. = [ I Ij 0 B I I j+1 Now let J be a positive iteger. We are ready to itroduce a basis ϕ J,k, k I J, of the space V J yieldig a matrix Ã, ã r,s = a( ϕ J,s, ϕ J,r ), r, s I J, whose coditio umber i the Poisso case grows like O((log 2 (1/h J )) 2 ) = O((J + log 2 (1/h 0 )) 2 ), ad thus is smaller tha the coditio umber of the matrix A, a r,s = a(ϕ J,s, ϕ J,r ), r, s I J, yielded by the Lagrage basis ϕ J,k, k I J of V J (recall that µ 2 (A) = O((1/h J ) 2 ) = O((2 J /h 0 ) 2 )). Observe that if v V J the k I J v Jk ϕ Jk = v = Π J v = Π 0 v + J 1 j=0 (Π j+1v Π j v) = k I 0 v 0k ϕ 0k + J 1 j=0 ṽ jk ψ j,k, k I j+1 ψ jk = ϕ j+1,k, k I j+1, j = 0,..., J 1. It follows that V J admits the represetatio V J = V J 1 + W J 1 = V J 2 + W J 2 + W J 1 = V 0 + W W J 1 ] v jk k I j ṽ jk k I j+1 6

7 ad the set { ϕ J,k : k I J } := {ϕ 0k : k I 0 } {ψ 0,k : k I 1 } {ψ J 1,k : k I J } is a alterative basis of V J. Remark. Observe that if v J = (v J,k ) k IJ, ṽ J = (ṽ J,k ) k IJ = ((v 0,k ) k I0 (ṽ 0,k ) k I 1 (ṽ J 1,k ) k I J ), the ṽ J = Sv J = E 0 P 0 E 1 P 1 E J 1 P J 1 v J where the P k are permutatio matrices, the E k are matrices of the form I I0 0 I I1 0 [ E 0 = B 0 I I 1, E 1 = B 1 I I I IJ 1 0 2,..., E J 1 = B J 1 I I I I J ] ad the B k, i the defiitio of the E k, have oly two ozero elemets for each row, both equal to 1 2. So, v J ca be computed from ṽ J (as well as ṽ J ca be computed from v J ) with 2( I J I 0 ) divisios by 2. The trasform of v J ito ṽ J is described i detail here below: v J = [ vj,k k I J ] P J 2 E J 1 P J 1 v J = P J 1 v J = v J 1,ρ(k) k I J 2 v J 1,k k I J 1 ṽ J 1,k k I J v J,ρ(k) k I J 1 v J,k k IJ E 0 P 0 E J 1 P J 1 v J = E J 1P J 1 v J = E J 2 P J 2 E J 1 P J 1 v J = v 0,k k I 0 ṽ 0,k k I 1 ṽ J 1,k k I J. v J 1,k k I J 1 ṽ J 1,k k I J Theorem. Let τ j, V j, Π j, j = 0, 1,..., J, be the triagulatios of, the subspaces of V = H0 1 () ad the iterpolatig operators C 0 () V j defied above. For ay v V J set J 1 ˆ vˆ 2 = Π 0 v 2 1, + j=0 k Ij+1 J 1 (Π j+1 v Π j v)(x j+1,k ) 2 = Π 0 v 2 1, + j=0 k Ij+1 v J 2,k k I J 2 ṽ J 2,k k I J 1 ṽ J 1,k k I J ṽ j,k 2. 7

8 The there exist two positive costats c 1, c 2 (depedig oly o the agles of τ 0 ) such that c 1 ˆ vˆ 2 J 2 v 2 1, c 2ˆ vˆ 2. This iequality, ivolvig the coefficiets ṽ J,k of v V J with respect to the hierarchical basis ϕ J,k, k I J, is due to Yseretat. It allows us to evaluate the coditio umber of Ã, ã r,s = a( ϕ J,s, ϕ J,r ), r, s I J, i the Poisso case where a(u, v) = u v. First ote that Π 0 v 2 1, = k I 0 v 0,k ϕ 0,k 2 1, = ( k I 0 v 0,k ϕ 0,k ) ( k I 0 v 0,k ϕ 0,k ) = k,si 0 v 0k v 0s ϕ 0k ϕ 0s, thus the Yseretat iequality ca be rewritte as follows: [ ] N 0 ṽj T ṽ 0 I J [ c 1 ṽj T Mṽ J c 2 ṽj T N 0 0 I J 2 ] ṽ J where N = ( ϕ 0r ϕ 0,s ) r,s I0 ad M = ( ϕ J,r ϕ J,s ) r,s IJ. Note that N ad M are positive defiite matrices. Note also that i case of the Poisso differetial problem u = f, x, u = ϕ, x, the form a is simply a(u, v) = u v, i.e. we have the cotiuous problem w V = H0,Γ 1 D () w v = fv u ϕ v, v V = H0,Γ 1 D () which is reduced first to the discrete problem w J V J w J v J = fv J ad the, via the represetatio w J = k I J u ϕ v J, v J V J (Y) (w J ) k ϕ J,k, to the liear system Ãx = b, ã r,s = ϕ J,r ϕ J,s, br = f ϕ J,r u ϕ ϕ J,r, (w J ) k = (à 1 b)k. Observe that the coefficiet matrix à of this system is exactly the matrix M i 1 (Y). Now we prove that µ 2 (M) = O((log 2 h J ) 2 ). Cosider the Cholesky factorizatio of N, N = L N L T N, ad ote that [ ] [ ] LN N L := LL I T =. I Set z = L T ṽ J, ṽ J = L T z. By (Y), for all vectors z 0 we have 1 c 1 J 2 zt L 1 ML T z z T c 2. z Thus, if λ is ay eigevalue of the positive defiite matrix L 1 ML T, the c 1 J 2 λ c 2, ad this result implies that the coditio umber of L 1 ML T is bouded by c 2 c 1 J 2. Sice L N is a small matrix ad its dimesio does ot deped o J, it follows that µ 2 (M) cj 2 = c(log 2 h 0 h J ) 2 for some costat c. 8