V½ dö 2.3.1. Ta x²t B i to¡n (2.1) vîi Ci ⊂Rn v Qj ⊂ Rm ÷ñc x¡c ành bði
Ci= {x ∈RN : haCi , xi ≤ bCi }, Qj ={x∈RM : haQj, xi ≤bQj },
trong â aCi ∈RN, aQj ∈RM v bCi , bQj ∈ R vîi måi i = 1,2, ..., N, j = 1,2, ..., M
v Tk l ph²p chi¸u m¶tric tø RN l¶n Sk vîi
Sk ={x∈Rn : kx−Ikk2 ≤ R2k},
vîi måik = 1,2, ..., K v A l mët to¡n tû tuy¸n t½nh bà ch°n tø RN l¶n RM vîi ma trªn câ c¡c ph¦n tû ÷ñc sinh ng¨u nhi¶n trong o¤n [2,4].
Ti¸p theo, ta l§y ng¨u nhi¶n gi¡ trà c¡c tåa ë cõa aCi , aQj trong o¤n [1,3]
v bCi , bQj trong o¤n [2,4], tåa ë t¥mIk trong o¤n [−1,1] v b¡n k½nh Rk cõa h¼nh c¦u Sk trong o¤n [2,10], t÷ìng ùng.
D¹ th§y S = N \ i=1 Ci \ M \ j=1 A−1(Qj) \ K \ k=1 F(Tk) 6=∅, v¼ 0∈S. B¥y gií, ta kiºm tra sü hëi tö cõa Thuªt to¡n 2.1, vîi ph¦n tû ban ¦u
x0 ∈ RN câ c¡c tåa ë ÷ñc sinh ng¨u nhi¶n trong o¤n [−5,5], N = 20,
M = 40, N = 50, M = 100, K = 200 v tn = 1
2kAk2. Sau n«m l¦n thû, ta thu ÷ñc b£ng k¸t qu£ sè d÷îi ¥y.
i·u ki»n døng: TOLn <10−5 i·u ki»n døng: TOLn <10−6
No. TOLn n No. TOLn n
1 9.73191e−006 525 1 9.82257e−007 2692 2 9.72380e−006 382 2 9.88394e−007 1084 3 9.74093e−006 594 3 9.99178e−007 1878 4 9.81788e−006 793 4 9.82163e−007 1922 5 9.77395e−006 250 5 9.98486e−007 1644 B£ng 2.1: B£ng k¸t qu£ sè cho V½ dö 2.3.1
Chó þ 2.3.2. Trong v½ dö tr¶n, h m sè TOL ÷ñc x¡c ành bði TOLn = 1 N N X i=1 kxn−PCixnk2+ 1 M M X j=1 kAxn−PQjAxnk2+ 1 K K X k=1 kxn−Tkxnk2,
vîi måi n ≥1. Chó þ r¬ng, n¸u t¤i b÷îc l°p thùn, TOLn = 0 th¼ xn ∈ S, tùc l
xn l mët nghi»m cõa b i to¡n.
V½ dö 2.3.3. Ta l§y E =F = L2([0,1]) vîi t½ch væ h÷îng hf, gi= Z 1 0 f(t)g(t)dt v chu©n x¡c ành bði kfk= Z 1 0 f2(t)dt !1/2 , vîi måi f, g ∈L2([0,1]). °t Ci= {x ∈L2([0,1]) : hai, xi=bi}, trong â ai(t) =ti−1, bi = 1 i+ 1 vîi måi i = 1,2, . . . , N v t ∈[0,1], Qj ={x∈ L2([0,1]) : hcj, xi ≥dj}, trong â cj(t) =t+j, dj = 1 4 vîi måi j = 1,2, . . . , M v t∈[0,1], Tk =PSk,
ð ¥y Sk = {x ∈ L2([0,1]) : kx− Ikk ≤ k + 1}, vîi Ik(t) = t + k vîi måi
k = 1,2, . . . , K v t∈[0,1]. Gi£ sû
A: L2([0,1])−→ L2([0,1]), (Ax)(t) = x(t) 2 .
Ta x²t b i to¡n t¼m mët ph¦n tû x† sao cho
x† ∈S = N \ i=1 Ci \ M \ j=1 A−1(Qj) \ K \ k=1 F(Tk) . (2.16) D¹ th§y S 6=∅, v¼ x(t) =t∈ S. Ta câ ΠCi(x) =PCi(x) = bi− hai, xi kaik2 ai+x, PQj(x) = max 0, dj− hcj, xi kcjk2 cj +x,
v Tk(x) = x, n¸u kx−Ikk ≤k+ 1, Ik+ k+ 1 kx−Ikk(x−Ik), trong c¡c tr÷íng hñp kh¡c.
Sû döng Thuªt to¡n 2.1 vîi N = 10, M = 20 v K = 40, ta thu ÷ñc b£ng k¸t qu£ sè d÷îi ¥y.
i·u ki»n døng: kxn+1−xnk<err
tn = 1, x0(t) =t2 tn = 1, x0(t) = exp(t) err kxn+1−xnk n err kxn+1−xnk n 10−2 9.92326e−003 128 10−2 9.06924e−03 125 10−3 9.90940e−004 2159 10−3 9.95338e−004 1091 10−4 9.98327e−005 47840 10−4 9.97943e−005 11352 B£ng 2.2: B£ng k¸t qu£ sè cho V½ dö 2.3.3
D¡ng i»u cõa kxn+1−xnk trong B£ng 2 ÷ñc mæ t£ bði ç thà d÷îi ¥y.
0 500 1000 1500 2000 2500 10−3 10−2 10−1 100 Number of interations ||x n+1 −x n || x0(t)=exp(t) x0(t)=t2
H¼nh 2.1: D¡ng i»u cõa kxn+1−xnkvîi i·u ki»n døng kxn+1 −xnk< 10−3 (adsbygoogle = window.adsbygoogle || []).push({});
D¡ng i»u cõa nghi»m x§p x¿xn(t)trong c£ hai tr÷íng hñpkxn+1−xnk<10−3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 The solution x*(t)=t xn(t) with x0(t)=exp(t) xn(t) with x0(t)=t2
H¼nh 2.2: D¡ng i»u cõa xn(t) vîi i·u ki»n døng kxn+1−xnk<10−3
Ti¸p theo, nh¬m ÷a ra mët so s¡nh ìn gi£n giúa hai ph÷ìng ph¡p l°p (1.22) v (2.1), ta x²t mët tr÷íng hñp °c bi»t cõa B i to¡n (2.16) nh÷ sau:
T¼m mët ph¦n tû x† ∈C ∩A−1(Q)∩F(T), (2.17) trong â C = C2, Q=Q2 v T = T2.
p döng c¡c Ph÷ìng ph¡p l°p (1.22) v (2.1) vîi tn = 1, αn = 1
n vîi måi
n ≥1v u(t) =x0(t) =exp(t2+ 1)vîi måit ∈[0,1], ta nhªn ÷ñc b£ng k¸t qu£ sè d÷îi ¥y.
i·u ki»n døng: kxn+1 −xnk< err
Ph÷ìng ph¡p (1.22) Ph÷ìng ph¡p (2.1)
err kxn+1 −xnk n err kxn+1−xnk n
10−6 9.81429e−07 18 10−6 8.10708e−07 17 10−7 9.750563778e−08 56 10−7 4.17743e−08 41 10−8 9.97665e−09 174 10−8 9.28195e−09 831
D¡ng i»u cõa nghi»m x§p x¿xn(t) cho tr÷íng hñpkxn+1−xnk<10−6 trong B£ng 2.3 ÷ñc mæ t£ trong h¼nh d÷îi ¥y.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.5 1 1.5 2 2.5 3 Algorithm (2.1) Algorithm (1.16)
K¸t luªn
Luªn v«n ¢ tr¼nh b y l¤i mët c¡ch kh¡ chi ti¸t v h» thèng v· c¡c v§n · sau:
• Mët sè t½nh ch§t °c tr÷ng cõa khæng gian khæng gian Banach ph£n x¤, khæng gian Banach lçi ·u, p-lçi ·u, khæng gian Banach trìn ·u, q-trìn ·u, ¡nh x¤ èi ng¨u;
• Kho£ng c¡ch Bregman, ph²p chi¸u Bregman;
• B i to¡n ch§p nhªn t¡ch, to¡n tû Bregman khæng gi¢n m¤nh tr¡i;
• C¡c k¸t qu£ nghi¶n cùu cõa Tuyen T.M. v Ha N.S. trong t i li»u [17] v· ph÷ìng ph¡p chi¸u lai gh²p t¼m mët nghi»m chung cõa b i to¡n ch§p nhªn t¡ch v b i to¡n iºm b§t ëng chung cho c¡c to¡n tû Bregman khæng gi¢n m¤nh tr¡i trong khæng gian Banach.
T i li»u tham kh£o
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