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Two-step General Linear Methods for Retarded Functional Differential Equations PDF
Preview Two-step General Linear Methods for Retarded Functional Differential Equations
9 Two-step General Linear Methods for 0 0 Retarded Fun tional Di(cid:27)erential Equations 2 n a J 9 2 ∗ Anton Tuzov ] A N . h t Abstra t a m This paper presents a lass of Two-Step General Linear Methods for the numeri al [ solution of Retarded Fun tional Di(cid:27)erential Equations. Expli it methods up to order (cid:28)ve 1 are onstru ted. To avoid order redu tion for mildly sti(cid:27) problems the uniform stage order v of the methods is hosen to be lose to uniform order. 9 0 6 4 1 Two-step General Linear Methods for . 1 0 Ordinary Di(cid:27)erential Equations 9 0 y(t) : For the numeri al approximation of the solution of a system of Ordinary Di(cid:27)erential v Equations i X y′(t) = f(t,y), t [t ,T], 0 r ∈ a y(t ) = y , (1) 0 0 f : R Rd Rd, y Rd 0 where × −→ ∈ , we onsider the lass of General Linear Methods [5℄ s r Y[n] = a hF[n] + u y[n−1], i = 1,...,s, i ij j ij j Xj=1 Xj=1 s r y[n] = b hF[n] + v y[n−1], i = 1,...,r, i ij j ij j (2) Xj=1 Xj=1 F[n] = f(t +c h,Y[n]), i = 1,...,s, i n−1 i i y[n−1],...,y[n−1] n where 1 r (cid:22) input ve tors, available at step number , Y[n],...,Y[n] F[n],...,F[n] a , u , b , v . 1 s 1 s ij ij ij ij (cid:22) stage values, (cid:22) derivative values, (cid:22) oe(cid:30) ients of the method. ∗ Department of Control systems, Siberian State Aerospa e University, Krasnoyarsk, Russia, e-mail: tuzovsibsau.ru. 1 r = s+2 Let us restri t ourselves to two-step GLMs and hoose , y[n−1] y(t ), y[n−1] y(t ), y[n−1] hy′(t +c h), i = 1,...,s. 1 ≈ n−1 2 ≈ n−2 2+i ≈ n−2 i Then y[n−1] = y , y[n−1] = y , y[n−1] = hf(t +c h,Y[n−1]) = hF[n−1], i = 1,...,s. 1 n−1 2 n−2 2+i n−2 i i i and (2) takes the form s s Y[n] = h a F[n] + u y +u y +h u F[n−1], i = 1,...,s, i ij j i1 n−1 i2 n−2 i,2+j j Xj=1 Xj=1 s s y = h b F[n] + v y +v y +h v F[n−1], n 1j j 11 n−1 12 n−2 1,2+j j (3) Xj=1 Xj=1 F[n] = f(t +c h,Y[n]), i = 1,...,s. i n−1 i i y[n−1] = u y(t ) +v hy′(t )+O(h2) i i n−1 i n−1 In the onstru tion of GLMs it is assumed that and 'pre onsistensy onditions' holds Vu = u, Uu = 1. (4) u = 1, v = 0, u = 1, v = 1, u = 0, v = 1, i = 1,...,s 1 1 2 2 2+i 2+i For (3) we have − . It follows from (4) that u = 1 u , i = 1,...,s, i2 i1 − v = 1 v . 12 11 − Let us denote K[n] := F[n] a := u , u := u , u = 1 u , j = 1,...,s, i = 1,...,s, i i ij i,2+j i i1 ⇒ i2 − i b := b , b := v , v := v , v = 1 v, j = 1,...,s, j 1j ej 1,2+j 11 12 ⇒ − e then the method (3) satisfying 'pre onsistensy onditions' (4) takes the form s s y = (1 v)y +vy +h b K[n−1] +h b K[n], n − n−2 n−1 j j j j Xj=1 Xj=1 e K[n] = f(t +c h,Y[n]), i = 1,...,s, i n−1 i i (5) s s Y[n] = (1 u )y +u y +h a K[n−1] +h a K[n], i = 1,...,s. i − i n−2 i n−1 ij j ij j Xj=1 Xj=1 e 2 2 Two-step General Linear Methods for Retarded Fun tional Di(cid:27)erential Equations We begin with notations introdu ed in [1℄. r [0,+ ) [ r,0] Rd Let ∈ ∞ , and C be tφhe=spam eaxof φo(nθt)in,uouφs fun tions − −→ , equipped withRthde maximum(uniform)norm k k θ∈[−r,0]| | ∈ C, where |·| is an arbitrary norm on . u [a r,b) Rd a < b t [a,b) Let be ontinuous fun tion − −→ , where . Then ∀ ∈ shift fun tion is u (θ) = u(t+θ), θ [ r,0] u t t well de(cid:28)ned by ∈ − , and ∈ C. Let us onsider a system of Retarded Fun tional Di(cid:27)erential Equations y′(t) = f(t,y ), t [t ,T], t 0 ∈ y (θ) = φ(θ), θ [ r,0]. (6) t0 ∈ − (t ,φ) Ω, f : Ω Rd, Ω R , Ω 0 where ∈ −→ ⊂ ×C is open set. It is assumed that there exists a unique solution of (6). We introdu e the lass of two-step GLMs for RFDEs on the base of approa h proposed in [1℄. We an reformulate the method (5) for RFDEs (6) as follows s s η[n](αh) = (1 v(α))η[n−1](0) +v(α)η[n−1](h) +h b (α)K[n−1] +h b (α)K[n], − j j j j Xj=1 Xj=1 e α [0,1], ∈ K[n] = f(t +c h,Y[n]i ), i = 1,...,s, i n−1 i cih s s Y[n]i(αh) = (1 u (α))η[n−1](0)+u (α)η[n−1](h)+h a (α)K[n−1] +h a (α)K[n], − i i ij j ij j (7) Xj=1 Xj=1 e α [0,c ], i = 1,...,s, i ∈ η[n](θ) = η[n−1](θ), θ [ r,0], h ∈ − Y[n]i(θ) = η[n−1](θ), θ [ r,0], i = 1,...,s, h ∈ − where η[n−1] : [ r,h] Rd, K[n−1] Rd − −→ i ∈ are available as approximations n 1, omputed in the step − Y[n]i , (cid:22) stage fun tions K[n] , i (cid:22) stage values u ( ), a ( ),a ( ), v( ), b ( ), b ( ) . i ij ij j j · · · · · · (cid:22) oe(cid:30) ients of the method e e 3 η[n](αh) y(t +αh), α [0,1] η[n](θ) y(t +θ), θ [ r,0] n−1 n−1 Thus ≈ ∈ , ≈ ∈ − , hen e η[n] y [ r h,0] t = t +h h ≈ tn on − − , where n n−1 . k 2 Remark 2.1. We hose two-step methods among multi-step methods ( ≥ ) for the following reasons. E(h,t ,y ) • For multistep methods, the lo al error n−1 tn−1 has the required order only if exa t y(t) [t ,t ] n−k n solution is su(cid:30) iently smooth on . This is rather severe assumption for many k = 2 problems (6). For the ase of two-step methods ( ) this ondition imposes the weakest restri tion on stepsize. • Furthermore, in the ase of two-step methods, starting pro edure and stepsize strategy seem to be simplest ones. Let us denote η := η[n], K := K[n], η := η[n−1], K := K[n−1], σ := t , i i i i n−1 then the method (7) an be reformulated in Stefano Maset's notations as follows s s η(αh) = (1 v(α))η(0) +v(α)η(h) +h b (α)K +h b (α)K , α [0,1], j j j j − ∈ Xj=1 Xj=1 e K = f(σ +c h,Yi ), i i cih s s Yi(αh) = (1 u (α))η(0) +u (α)η(h)+h a (α)K +h a (α)K , α [0,c ], i i ij j ij j i − ∈ (8) Xj=1 Xj=1 e η(θ) = η (θ), θ [ r,0], h ∈ − Yi(θ) = η (θ). θ [ r,0], h ∈ − 4 3 Two-step GLMs for RFDEs in Stefano Maset's notations Let us onsider a system of Retarded Fun tional Di(cid:27)erential Equations x′(t) = f(t,x ), t [t ,T], t 0 ∈ x (θ) = φ(θ), θ [ r,0]. (9) t0 ∈ − (t ,φ) Ω, f : Ω Rd, Ω R , Ω 0 where ∈ −→ ⊂ ×C is open set. It is assumed that onditions of existen e and uniqueness theorem for the (9) hold. s When -stage Two-Step General Linear Method for RFDEs (TSGLM) with oe(cid:30) ients (a ( ),b ( ),c ,a ( ),b ( ),u ( ),v( )) h ij j i ij j i i,j=1,...,s · · · · · · isappliedwithstepsize to(9)forthe omputation x(t) [ r,h] y := x(σ+ ) of the solutione , iet yields, as an approximation on − of the shift fun tion · , the fun tion s s η(αh) = (1 v(α))η(0)+v(α)η(h)+h b (α)K +h b (α)K , α [0,1], j j j j − ∈ (10) Xj=1 Xj=1 e η(θ) = η (θ), θ [ r,0], h ∈ − where η y [ r h,0] K , h i fun tion ≈ on − − and stage values are available as approximations omputed in the previous step, K = f(σ+c h,Yi ), i = 1,...,s, i i cih (11) Yi : [ r,c h] Rd i and − −→ is a stage fun tion given by s s Yi(αh) = (1 u (α))η(0)+u (α)η(h)+h a (α)K +h a (α)K , α [0,c ], i i ij j ij j i − ∈ (12) Xj=1 Xj=1 e Yi(θ) = η (θ). θ [ r,0], h ∈ − (a ( ),b ( ),c ,a ( ),b ( ),u ( ),v( )) ij j i ij j i i,j=1,...,s It is assumed that oe(cid:30) ients · · · · · · of TSGLMs satisfy the following onditions: e e a ( ), a ( ), u ( ), [0,c ] R, i, j = 1,...,s. ij ij i i • · · · are polynomial fun tions −→ (13) b ( ), b ( ), v( ), [0,1] R, j = 1,...,s. j ej · · · are polynomial fun tions −→ c R, c 0, i = 1,...,s. i e i • ∈ ≥ (14) a (0) = a (0) = 0, u (0) = 1, i, j = 1,...,s. ij ij i • (15) b (0) = b (0) = 0, v(0) = 1, j = 1,...,s. j ej • (16) e Yi The last two onditions orrespondingly gurantee ontinuity of the stage fun tions cih ∈ C and η h the approximate solution ∈ C provided that approximate solution omputed in the previous η . step is ontinuous fun tion h ∈ C 5 Remark 3.1. If the onditions u ( ) = 1, a ( ) = 0, i, j = 1,...,s, i ij · · (17) v( ) = 1, b ( ) = 0, j = 1,...,s, ej · · e hold, the two-step method (8) be omes the one-step RK method for RFDEs introdu ed in [1℄, φ := η h where initial fun tion . (a ( ),b ( ),c ,a ( ),b ( ),u ( ),v( )) ij j i ij j i i,j=1,...,s De(cid:28)nition3.2. TSGLMwith oe(cid:30) ients · · · · · · is alled a ( ) = 0 j : j i, i,j = 1,...,s. ij expli it if · for all ≥ e e E = η y : [0,h] Rd De(cid:28)nition 3.3. The fun tion − −→ omputed under the assumption that η = y [ r h,0] h on − − is alled the lo al error of TSGLM (8). Ei = Yi y : [0,c h] Rd i De(cid:28)nition 3.4. The fun tion − −→ omputed under the assumption η = y [ r h,0] h that on − − is alled the lo al stage error of TSGLM (8). By analogy with [7℄ we de(cid:28)ne stage order for RFDEs. Ei = Yi y : [0,c h] Rd, i = 1,...,s i De(cid:28)nition 3.5. Let fun tions − −→ , Es+1 = η y : [0,h] Rd η = y −[ r h,0],−→K = ya′r(ee hom+pcuhte)d, unKder=tyh′e(cashs)u,mpjti=on1t,h..a.t,s, eh on − − j − j j j that is s s Ei(αh) = (1 u (α))y( h)+u (α)y(0)+h a (α)y′( h+c h)+h a (α)y′(c h) i i ij j ij j − − − − Xj=1 Xj=1 e e y(αh), α [0,c ], i = 1,...,s, i − ∈ (18) s s Es+1(αh) = (1 v(α))y( h) +v(α)y(0) +h b (α)y′( h+c h) +h b (α)y′(c h) j j j j − − − − Xj=1 Xj=1 e e y(αh), α [0,1]. − ∈ c := 1 p D > 0, H > 0 s+1 i i Denote . If there are positive integers and reals su h that max Ei(αh) D hpei+1, h [0,H],e i = 1,...,s+1, i α∈[0,ci]| | ≤ ∈ (19) e p = min p ,...,p 1 s+1 then positive integer { } is alled uniform stage order of TSGLM (8). e e e 6 4 Order onditions f Cl l Assume that is of lass with respe t to the se ond argument for a su(cid:30) iently large and x(t) Cm m solution of (9) is of pie ewise lass for a su(cid:30) iently large . Γ : [0,1] R Γ : [0,c ] R k ik i We introdu e the polynomial fun tions −→ and −→ given by 1 (1 v(α))( 1)k s s αk Γ (α) = − − + b (α)( (1 c ))k−1 + b (α)ck−1 , k (k 1)!(cid:20) k j − − j j j − k (cid:21) − Xj=1 Xj=1 e α [0,1], ∈ (20) 1 (1 u (α))( 1)k s s αk Γ (α) = − i − + a (α)( (1 c ))k−1 + a (α)ck−1 , ik (k 1)!(cid:20) k ij − − j ij j − k (cid:21) − Xj=1 Xj=1 e α [0,c ], i = 1,...,s. i ∈ Γ , Γ ik k Remark 4.1. If the onditions (17) hold the are the same as for the one-step RK method [1℄. c∗,...,c∗ c∗ < c∗ < < c∗ c∗,...,c∗ = c ,...,c c∗ Let 1 s∗ su h that 1 2 ··· s∗ and { 1 s∗} { 1 s}, i.e. i are c i distin t in in reasing order. p x Cp+1 Lemma 4.2. Let be a positive integerE. Iif= Ois(hopf),piie =ew1i,s.e.. ,lsass and the lo alEstage errors omputed in the previous step are , then the lo al error and Ei the lo al stage errors satisfy s p E(αh) = h b (α)D + y(k)(0)hkΓ (α) +O(hp+1), α [0,1], j j k ∈ (21) Xj=1 Xk=1 s p Ei(αh) = h a (α)D + y(k)(0)hkΓ (α)+O(hp+1), α [0,c ], i = 1,...,s, ij j ik i ∈ (22) Xj=1 Xk=1 where D = f(σ +c h,y +Ei ) f(σ +c h,y ), i = 1,...,s. i i cih cih − i cih (23) *The Lemma 4.2 has been proved, but its proof is omitted here for brevity.* Γ = 0 1 In the following we assume that the TSGLM satis(cid:28)es the onditions and Γ = 0, i = 1,...,s, i1 that is s s v(α) 1 + b (α) + b (α) = α, α [0,1], j j − ∈ Xj=1 Xj=1 e (24) s s u (α) 1+ a (α)+ a (α) = α, α [0,c ], i = 1,...,s. i ij ij i − ∈ Xj=1 Xj=1 e 7 The above ondition is an equivalent form of uniform stage order one ondition. Γ = 0 2 Theorem 4.3. A TSGLM satisfying (24) has uniform order two i(cid:27) . *The theorem 4.3 has been proved, but its proof is omitted here for brevity.* Theorem 4.4. Let TSGLM satisfy (24) and has uniform order two. s Γ = 0 b (α)Γ (β) = 0, α [0,1], β [0,c∗ ], m = 1,...,s∗ If 3 and i i2 ∈ ∈ m ciiP==c1∗m then the method has uniform order three. *The theorem 4.4 has been proved, but its proof is omitted here for brevity.* Theorem 4.5. Let TSGLM satisfy (24) and has uniform order three. Γ = 0, 4 If s b (α)Γ (β) = 0, α [0,1], β [0,c∗ ], m = 1,...,s∗, i i3 ∈ ∈ m Xi=1 ci=c∗m (25) s s ∗ ∗ ∗ b (α)a (β)Γ (γ) = 0, α [0,1], β [0,c ], γ [0,c ], l,m = 1,...,s . i ij j2 ∈ ∈ m ∈ l Xi=1 Xj=1 ci=c∗mcj=c∗l then the method has uniform order four. *The theorem 4.5 has been proved, but its proof is omitted here for brevity.* p Theorem 4.6. TSGLM has uniform stage order i(cid:27) Γ = 0, Γ = 0, i = 1,...,s, k = 1,...,p. ik k e e Ei Proof. Follows by Taylor series expansion of fun tions given by (18). e The following results an be obtained as orollary of theorems (4.6) and (4.4), (4.5). Corollary 4.7. Let TSGLM has uniform stage order two. Γ = 0 3 If then the method has uniform order three. Corollary 4.8. Let TSGLM has uniform stage order three. Γ = 0 4 If then the method has uniform order four. The results of Corollary (4.7) and (4.8) an be easily generalized as follows. p Theorem 4.9. Let TSGLM has uniform stage order . p = p+1 Γpe+1 = 0 It has uniform order i(cid:27) . e e *The theorem 4.9 has been proved, but its proof is omitted here for brevity.* 8 5 Constru tion of expli it two-stage GLMs of uniform stage order four and (cid:28)ve Consider two-stage expli it TSGLM satisfying (24). It's But her tableau is Table 5.1. But her tableau for 2-stage expli it TSGLMs c u (α) a (α) a (α) 1 1 11 12 0 0 c u (α) a (α) a (α) a (α) 2 2 e21 e22 21 0 v(α) b (α) b (α) b (α) b (α) e1 e2 1 2 e e c , i = 1,...,s i A natural hoi e will be to spa e out the abs issae uniformly in the 1 s 2 interval [0,1] so that [6℄ c1 = 0, c2 = s 1,..., cs−1 = s−1, cs = 1. In the ase of s = 2 we − − c = 0, c = 1 1 2 have . c = 0, Γ (α) = 0, α [0,c ], k = 1,2,... Γ (0) = 0, 1 1k 1 1k Sin e onditions ∈ redu e to k = 1,2,... u ( ) = 1, a ( ) = 0, a ( ) = 0 1 11 12 that follows from (15). It also follows that · · · . α For brevity we omit the argument of the method oe(cid:30) ient efun tions. Bey theorem (4.9), the method has uniform order four and uniform stage order three if Γ = 0, k = 1,2,3,4 Γ = 0, k = 1,2,3, k 2k and that is (1 v)+b +b +b +b = α, 1 2 1 2 − − 1 v α2 e e − b +b = , 1 2 2 − 2 1 v e α3 − +b +b = , 1 2 − 3 3 1 v e α4 − b +b = , 1 2 4 − 4 (26) (1 u )+ea +a +a = α, 2 21 22 21 − − 1 u α2 − 2 ea e = , 21 2 − 2 1 u α3 − 2 +ae = , 21 − 3 3 e α [0,1] where ∈ . 9 The oe(cid:30) ients are de(cid:28)ned by u = (2α 1)(α+1)2, 2 − − v = (α 1)2(α+1)2, − a = α2(α+1), 2,1 1 eb = α2(α+1)(5α 7), 1 −12 − (27) ae = α (α+1)2 a , 2,1 2,2 − 1 b = α (2α 3e)(α+1)2 b , 1 2 −3 − − 1 e b = α2(α+1)2, 2 12 4 a , b Γ (1) = = 0 2,2 2 5 where remain free. The relation 15 6 implies that it is impossible to attain e dis retee order (cid:28)ve. The uniform order and the uniform stage order an be in reased by (cid:28)nding a suitable value c c = 0, c = 0 c = 1 2 1 2 2 for . Assume that 6 (in general ase 6 ). By theorem (4.9), the method has uniform order (cid:28)ve and uniform stage order four if Γ = 0, k = 1,2,3,4,5 Γ = 0, k = 1,2,3,4, k 2k and that is (1 v)+b +b +b +b = α, 1 2 1 2 − − 1 v α2 e e − b (1 c )b +c b = , 1 2 2 2 2 2 − − − 2 1 v e e α3 − +b +(1 c )2b +c2b = , − 3 1 − 2 2 2 2 3 1 v e e α4 − b (1 c )3b +c3b = , 4 − 1 − − 2 2 2 2 4 1 v e e α5 − +b +(1 c )4b +c4b = , − 5 1 − 2 2 2 2 5 (28) (1 u )+ea +a e +a = α, 2 21 22 21 − − 1 u α2 − 2 ea e(1 c )a = , 21 2 22 2 − − − 2 1 u α3 − 2 +ae +(1 c )e2a = , 21 2 22 − 3 − 3 1 u α4 − 2 ea (1 c )3ea = , 21 2 22 4 − − − 4 e e α [0,1] α [0,c ] 2 where ∈ in the (cid:28)rst (cid:28)ve equations (28) and ∈ in other ones. 10