DSpace at VNU: PARALLEL HYBRID METHODS FOR A FINITE FAMILY OF RELATIVELY NONEXPANSIVE MAPPINGS

This article was downloaded by: [The University Of Melbourne Libraries] On: 15 September 2014, At: 18:28 Publisher: Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK Numerical Functional Analysis and Optimization Publication details, including instructions for authors and subscription information: http://www.tandfonline.com/loi/lnfa20 Parallel Hybrid Methods for a Finite Family of Relatively Nonexpansive Mappings a Pham Ky Anh & Cao Van Chung a a Department of Mathematics , Vietnam National University , Hanoi , Vietnam Accepted author version posted online: 06 Aug 2013.Published online: 01 Apr 2014 To cite this article: Pham Ky Anh & Cao Van Chung (2014) Parallel Hybrid Methods for a Finite Family of Relatively Nonexpansive Mappings, Numerical Functional Analysis and Optimization, 35:6, 649-664, DOI: 10.1080/01630563.2013.830127 To link to this article: 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by numerical experiments Thanks to the parallel computation, we can reduce the overall computational effort under widely used assumptions on mappings and spaces Keywords Common fixed point; Hybrid method; Parallel computation; Relatively nonexpansive mapping Mathematics Subject Classification 47H09; 47J25; 65J15; 65Y05 INTRODUCTION Various problems of science and engineering, such as the convex feasibility problems with applications in optimization theory, image processing, radiation therapy treatment planning, etc (see [1]), can be reduced to a problem of finding a common fixed point of a family of nonexpansive mappings In 2005, Matsushita and Takahashi [2] proposed the following hybrid method, called a CQ algorithm, for finding a fixed point of a relatively nonexpansive mapping:   x0 ∈ C chosen arbitrarily,   −1  yn := J ( n Jxn + (1 − n )JTxn ), (1.1) Cn := z ∈ C : (z, yn ) ≤ (z, xn ) ,   Qn := z ∈ C : xn − z, Jx0 − Jxn ≥ ,    xn+1 := Cn ∩Qn (x0 ) Received 24 March 2012; Revised 21 July 2013; Accepted 25 July 2013 Address correspondence to Pham Ky Anh, Department of Mathematics, Vietnam National University, 334 Nguyen Trai, Thanh Xuan, Hanoi, Vietnam; E-mail: anhpk@vnu.edu.vn 649 Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 650 P K Anh and C V Chung Several attempts to generalize the CQ method (1.1) for finding a common fixed point of a finite or infinite family of (relatively) nonexpansive mappings have recently made by Takahashi and Zembayashi [3], Plubtieng and Ungchittrakool [4, 5], Reich and Sabach [6, 7], Su, Wang, and Xu [8], Cholamjiak and Suantai [9], etc Very recently, Liu [10] proposed the following cyclic CQ method for a finite family of N relatively nonexpansive mappings:   x0 ∈ C chosen arbitrarily,     −1  y n Jxn + (1 − n )JT[n] xn ,  n := J (1.2) Cn := z ∈ C : (z, yn ) ≤ n (z, x0 ) + (1 − n ) (z, xn ) ,   Qn := z ∈ C : xn − z, Jx0 − Jxn ≥ ,    x where T[n] := Tn(modN ) n+1 := Cn ∩Qn (x0 ), Clearly, Liu’s algorithm is inherently sequential, hence it will be costly on a single processor when the number of operators N is large In this article we introduce some parallel CQ methods which may be regarded as a counterpart of the cyclic one Our idea consists of determining synchronously the vectors yni for each operator Ti , i = 1, 2, N by N parallel processors Then we find the farthest element yn := ynin from the previous approximation xn and construct the corresponding sets Cn and Qn The next approximation is defined as the generalized projection of x0 onto the set Cn ∩ Qn The benifit of our approach is clear Based on the parallel computation we can reduce the overall computational effort under widely used conditions on the space and mappings (cf [2–5, 8, 9]) Moreover, the additional computation cost of our method is essentially negligible Further, for mappings acting in a real Hilbert space, we modify the algorithm (1.2) by choosing the sets Cn , Qn as halfspaces, hence the generalized projection of x0 onto the set Cn ∩ Qn can be computed effectively (see [11]) Finally, we perform both parallel and sequential CQ methods for finding a common fixed point of two nonlinear nonexpansive integral operators and compare the obtained results The remainder of this article is organized as follows In section 2, we recall some notations and results needed for our further study Section deals with the convergence analysis of the parallel CQ algorithm, while in section we provide a modified algorithm for the Hilbert space case A numerical experiment is considered in the final section 651 Parallel Hybrid Methods PRELIMINARIES Let E be a real Banach space and E ∗ be its dual space For each f ∈ E ∗ and x ∈ E , we denote x, f := f (x) Let J be the normalized duality mapping defined by Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 J (x) = f ∈ E ∗ : x, f = x E = f E∗ We first recall some facts about the geometry of Banach spaces (see [12, 13, 16] for details) i) E (E ∗ ) is uniformly convex if and only if E ∗ (E ) is uniformly smooth ii) if E is uniformly convex, then it is reflexive and strictly convex and it satisfies the Kadec-Klee (or Efimov-Stechkin) property iii) if E is uniformly smooth and uniformly convex then J and J −1 = J ∗ are single-valued and uniformly norm-to-norm continuous on bounded subsets of E and E ∗ , respectively Now let E be a smooth, strictly convex and reflexive real Banach space For all x, y ∈ E , we consider the functional : E × E → + defined by (x, y) = x − x, J (y) + y ∀(x, y) ∈ E × E (2.1) Clearly, ( x − y )2 ≤ (x, y) ≤ ( x + y )2 Let C ⊂ E be a nonempty, convex and closed set The generalized metric projection from E onto C is defined as follows C (x) := arg (z, x) ∀ x ∈ E z∈C It is proved that the minimizer C (x) exists and is unique Besides, in a Hilbert space H , (2.1) reduces to (x, y) = x − y and C coincides with the well-known metric projection PC We have the following properties of the functional and the generalized projection C (see [4, 14–18]) Lemma 2.1 Let E be a reflexive, strictly convex and smooth real Banach space Then for all x, y ∈ E , (x, y) = if and only if x = y Lemma 2.2 Let E be a uniformly convex and smooth real Banach space, xn and yn be two sequences in E If (xn , yn ) → and either xn or yn is bounded, then xn − yn → as n → ∞ 652 P K Anh and C V Chung Lemma 2.3 Let E be a reflexive, strictly convex and smooth real Banach space, C be any nonempty closed convex set in E Then for all x ∈ E and for all z ∈ C , x∗ = C (x) (z, if and only if x ∗ − z, J (x) − J (z) ≥ 0; C (x)) + ( C (x), x) ≤ (z, x) (2.2) (2.3) Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 Let T : C → C be a mapping with a nonempty set of fixed points F (T ) := u ∈ C : u = T (u) A point p in C is said to be an asymptotic fixed point of T if C contains a sequence xn such that xn p and T (xn ) − xn → as n → ∞ We denote the set of all asymptotic fixed point of T by F (T ) The operator T is said to be relatively nonexpansive if Fˆ (T ) ≡ F (T ) and (p, T (x)) ≤ (p, x) for all p ∈ F (T ), x ∈ C The structure of the fixed point set of a relatively nonexpansive mapping is described in the following theorem Lemma 2.4 ([15, 19]) Let E be a strictly convex and smooth real Banach space, C be a closed convex subset of E , and T be a relatively nonexpansive mapping from C into itself Then F (T ) is closed and convex STRONG CONVERGENCE OF A PARALLEL HYBRID ALGORITHM Throughout this section we assume that C is a nonempty, closed and convex subset of a uniformly convex and uniformly smooth real Banach space E , and Ti : C → C , i = 1, 2, N , is a finite family of relatively nonexpansive mappings Moreover, suppose that N F := F (Ti ) = ∅ (3.1) i=1 For finding an element x ∈ F , we propose the following parallel CQ method Algorithm Let x0 ∈ C be an arbitrarily chosen initial approximation and k ⊂ (0, 1) be a vanishing numerical sequence For k ≥ 0, assuming xk is known, we • Calculate yki := J −1 ( k J (xk ) + (1 − k )J (Ti (xk ))), i = 1, 2, ,N (3.2) 653 Parallel Hybrid Methods • Find ik := arg max yki − xk (3.3) i=1,2, ,N • Define Ck := v ∈ C : (v, ykk ) ≤ (v, xk ), i and Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 Qk := u ∈ C : xk − u, J (x0 ) − J (xk ) ≥ • Compute xk+1 := Ck ∩Qk (x0 ) (3.4) • If xk+1 = xk then stop Else, set k := k + and repeat The following lemma shows that the Algorithm is well defined Lemma 3.1 If Algorithm reaches a step k ≥ 0, then F ⊂ Ck ∩ Qk and xk+1 is well defined Proof Obviously, Qk is closed and convex for all k ≥ Further, the relation (v, ykk ) ≤ (v, xk ) i is equivalent to v, J (xk ) − J (ykk ) ≤ i xk − ykk i , which shows that Ck is also convex and closed Thus, both sets Ck and Qk are closed and convex Noting that the function : C → , (x) := x is convex, for all p ∈ F ⊂ F (Tik ), we have (p, ykk ) = (p, J −1 ( k J (xk ) + (1 − i = p + ≤ p + − p, kJ kJ (xk ) + (1 − (xk ) + (1 − −2 k xk k k )J k )J (Tik (xk )))) k )J (Tik (xk )) (Tik (xk )) p, J (xk ) − 2(1 − + (1 − k) Tik (xk ) k) p, J (Tik (xk )) ≤ k (p, xk ) + (1 − k) (p, Tik (xk )) ≤ k (p, xk ) + (1 − k) (p, xk ) = (p, xk ) 654 P K Anh and C V Chung Hence, p ∈ Ck Thus, F ⊂ Ck for all k ≥ Clearly, F ⊂ Q0 ≡ C Assume that F ⊂ Qk−1 for some k ≥ 1, we will show that F ⊂ Qk Indeed, since xk = Ck−1 ∩Qk−1 (x0 ), property (2.2) implies that xk − z, J (x0 ) − J (xk ) ≥ for all z ∈ Ck−1 ∩ Qk−1 Therefore, from F ⊂ Ck−1 ∩ Qk−1 we have xk − z, J (x0 ) − J (xk ) ≥ for all z ∈ F Hence, the definition of Qk ensures that F ⊂ Qk Thus, Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 F ⊂ Ck ∩ Qk for all k ≥ Since F = ∅, the next approximation xk+1 is well defined We have the following convergent results of Algorithm Lemma 3.2 If Algorithm finishes at a finite iteration k < ∞, then xk is a common fixed point of Ti , i = 1, N Proof For all x ∈ E , C (x) is uniquely defined From stopping rule of Algorithm 1, we see that if it finishes at step k < ∞, then xk = xk+1 := i (xk , ykk ) ≤ (xk , xk ) = Ck ∩Qk (x0 ) ∈ Ck Using the definition of Ck , we have ik i Hence, (xk , yk ) = From Lemma 2.1, it follows xk = ykk By the definition of ik , we have xk = yki for i = 1, N Taking into account yki = xk and (3.2), we have J (xk ) = kJ (xk ) + (1 − k )J (Ti (xk )) Hence, J (xk ) = J (Ti (xk )), i = 1, N Since E is reflexive, uniformly convex and uniformly smooth, J and J −1 is uniformly norm-to-norm continuous Therefore, xk = Ti (xk ), i = 1, N Thus, xk ∈ F Otherwise, we are able to prove the main convergence theorem Theorem 3.1 Let xk be the (infinite) sequence generated by Algorithm and Ti be continuous for i = 1, N Then xk → x † := F (x0 ) as k → ∞ Proof Since Ti is relatively nonexpansive, F (Ti ) is convex and closed for i = 1, N Therefore, F is a convex, closed set and there exists a unique element x † = F (x0 ) Using (2.2) and the definition of Qk , we get xk = (xk , x0 ) ≤ Qk (x0 ) From xk+1 = Ck ∩Qk (x0 ) ∈ Qk and Lemma 2.3, it follows (xk , x0 ) is nondecreasing Taking (xk+1 , x0 ) Therefore, the sequence into account the relations x † ∈ F ⊂ Qk and xk = Qk (x0 ) and using (2.3), we find (xk , x0 ) ≤ (x † , x0 ) − (x † , xk ) ≤ (x † , x0 ), ∀k ≥ (3.5) 655 Parallel Hybrid Methods The last inequalities ensure the boundedness of xk and (xk , x0 ) Hence, there exists a finite limit limk→∞ (xk , x0 ) From xk+1 = Ck ∩Qk (x0 ) ∈ Qk and xk = Qk (x0 ), we have Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 (xk+1 , xk ) ≤ (xk+1 , x0 ) − (xk , x0 ) The last relation implies that limk→∞ (xk+1 , xk ) = 0, hence, by Lemma 2.2, limk→∞ xk+1 − xk = Using the definition of Ck and the inclusion xk+1 ∈ Ck , we also have (xk+1 , ykk ) ≤ (xk+1 , xk ) i Due to limk→∞ (xk+1 , xk ) = 0, from the last inequality we have (xk+1 , ykk ) → i as k→∞ Again, Lemma 2.2 ensures that limk→∞ xk+1 − ykk = i i Using this relation and xk − ykk ≤ xk+1 − ykk + xk − xk+1 , we find i xk − ykk → as k → ∞ i By the definition of ik , we also have xk − yki → as k → ∞ for i = 1, 2, , N Hence, yki is also bounded for i = 1, 2, , N From (3.2) it follows that J (Ti (xk )) − J (yki ) = J (Ti (xk )) − = n nJ (xk ) + (1 − J (xk ) − J (Ti (xk )) , n )J (Ti (xk )) i = 1, 2, ,N The relative nonexpansiveness of Ti and the boundedness of xk yield the boundedness of Ti (xk ) for all i = 1, N Since E is uniformly smooth, J is uniformly continuous on every bounded subsets Therefore, J (xk ) − J (Ti (xk )) is bounded Using limk→∞ k = 0, we find J (Ti (xk )) − J (yki ) → as k → ∞, (i = 1, N ) Since E is reflexive, uniformly convex, J −1 is uniformly norm-to-norm continuous on every bounded subsets Hence, Ti (xk ) − yki → as k → ∞, (i = 1, N ) Using this relation together with limk→∞ xk − yki → and xk − Ti (xk ) ≤ xk − yki + Ti (xk ) − yki , we have lim xk − Ti (xk ) = 0, k→∞ i = 1, N 656 P K Anh and C V Chung From the boundedness of xk , there exists a subsequence xkj such that x˜ and limj →∞ xkj − Ti (xkj ) = Since Ti is relatively nonexpansive, xkj we have x˜ ∈ F (Ti ) = F (Ti ), i = 1, N Hence, x˜ ∈ F Using xk+1 = Ck ∩Qk (x0 ) and x † ∈ F ⊂ Ck ∩ Qk , we have (xk+1 , x0 ) ≤ † (x , x0 ) On the other hand, from the weak lower semicontinuity of the norm, we obtain Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 (x, ˜ x0 ) = x˜ − x, ˜ J (x0 ) + x0 ≤ lim inf xkj j →∞ 2 − xkj , J (x0 ) + x0 ≤ lim inf (xkj , x0 ) ≤ lim sup (xkj , x0 ) j →∞ j →∞ ≤ (x , x0 ) † From definition of x † = F (x0 ), it follows x˜ = x † Hence, limj →∞ (xkj , x0 ) = (x † , x0 ) and limj →∞ xkj = x † Using the Kadec-Klee property of E , we obtain that xkj converges strongly to F (x0 ) Since xkj is an arbitrary weakly convergent sequence of xk , we have xk → F (x0 ) as k → ∞ Remark 3.1 At each step k ≥ 0, yki can be computed simultaneously by N parallel processors Remark 3.2 It is easy to find the optimal index ik by (3.3) Therefore, the additional computation cost in Algorithm is essentially negligible Remark 3.3 In general, it is not easy to determine the generalized projection xk+1 by (3.4) However in a Hilbert space we can modify Algorithm 1, so that the sets Ck and Qk are half spaces, hence xk+1 can be effectively computed A PARALLEL HYBRID ALGORITHM IN HILBERT SPACES Now suppose C is a nonempty, closed and convex subset of a real Hilbert space H In this case, for all x, y ∈ H , we have (x, y) ≡ x − y , J ≡ I (identity mapping in H ) and C (x) ≡ PC (x) Obviously, the relations (2.2) and (2.3) still hold Moreover, we have ([22]) PC (x) − PC (y) ≤ x −y − (PC (x) − x) − (PC (y) − y) ≤ x −y (4.1) Let Ti (i = 1, N ) be a family of relatively nonexpansive mappings from C N into itself and assume that the set F := i=1 F (Ti ) is not empty We consider the following modified algorithm 657 Parallel Hybrid Methods Algorithm Let x0 ∈ C be an arbitrarily chosen element and k ⊂ (0, 1) be a vanishing numerical sequence For k >= 0, assuming xk is known, we • Perform zk := PC (xk ) Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 • Calculate yki := k zk + (1 − k )Ti (zk ), i = 1, 2, ,N (4.2) • Find ik := arg max yki − xk i=1,2, ,N • If ykik − xk = then stop Else: • Define Ck := v ∈ H : v − ykk ≤ v − xk i , and (4.3) Qk := u ∈ H : x0 − xk , xk − u ≥ • Compute xk+1 := PCk ∩Qk (x0 ) (4.4) • If xk+1 = xk then stop Else, set k := k + and repeat For each k ≥ 0, it is easy to see that Qk is a halfspace or Qk = H i i Further, the relation v − ykk ≤ v − xk is equivalent to v, xk − ykk ≤ i xk − ykk or i i v − (xk + ykk ), xk − ykk ≤ (4.5) Hence, for all k ≥ 0, Ck is a halfspace in H or Ck = H An explicit formula for PCk ∩Qk (x0 ) can be obtained similarly as in [11] Therefore, if Ck ∩ Qk = ∅ then xk+1 is easily computed by (4.4) We have the following convergence results for Algorithm Lemma 4.1 If Algorithm finishes at a step k < ∞, then xk is a common fixed point of Ti , i = 1, N , i.e., x ∈ F 658 P K Anh and C V Chung Proof Using stopping rule xk = xk+1 , we have xk ∈ Ck From the definition of Ck , it follows xk − ykk ≤ xk − xk = i Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 Applying the definition of ik , we get yki = xk for i = 1, N Thus, xk ∈ C , that is, zk := PC (xk ) = xk Taking into account (4.2), we have xk = Since k k xk + (1 − i = 1, 2, k )Ti (xk ), ,N < we see xk = Ti (xk ) for i = 1, N , or xk ∈ F Theorem 4.1 Let xk be the (infinite) sequence generated by Algorithm 2, Ti be continuous for i = 1, N Then xk → x † := PF (x0 ) as k → ∞ Proof Since Ti is relatively nonexpansive, F (Ti ) is convex, closed for i = 1, N , the set F is convex and closed, and there exists a unique element x † = PF (x0 ) Moreover, for each k ≥ and i = 1, N we have yki ∈ C because x0 , zk , Ti (zk ) ∈ C Using the convexity of the function (x) := x on C and (4.1), we have for all p ∈ F ⊂ C p − ykk i = p− + (1 − k zk = p − p, ≤ p −2 k zk + k k zk k )Tik (zk ) + (1 − k )Tik (zk ) p, zk − 2(1 − + (1 − k) k) Tik (zk ) p − zk + (1 − k) p − Tik (zk ) ≤ k p − zk + (1 − k) p − zk ≤ p − xk + (1 − k )Tik (zk ) 2 k k zk p, Tik (zk ) = = PC (p) − PC (xk ) + 2 Therefore, p ∈ Ck and, hence, F ⊂ Ck for all k ≥ Clearly, F ⊂ Q0 ≡ H Assume that F ⊂ Qk−1 for some k ≥ 1, we will show that F ⊂ Qk Indeed, from xk = PCk−1 ∩Qk−1 (x0 ), it follows x0 − xk , xk − z ≥ for all z ∈ Ck−1 ∩ Qk−1 Therefore, for all z ∈ F ⊂ Ck−1 ∩ Qk−1 we have x0 − xk , xk − z ≥ Hence, by the definition of Qk , we find F ⊂ Qk Thus, F ⊂ Ck ∩ Qk for all k ≥ Acting as in Lemma 3.1, from F = ∅, we conclude that if Algorithm reaches step k ≥ 0, then xk+1 is well defined 659 Parallel Hybrid Methods Using xk = PQk (x0 ), xk+1 = PCk ∩Qk (x0 ) ∈ Qk and F ⊂ Ck ∩ Qk for all k ≥ 0, then arguing similarly as in Theorem 3.1, we come to the relations x0 − xk ≤ x0 − PF (x0 ) , xk − xk+1 → ∀k; as k → ∞; xk − yki → as k → ∞ Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 (4.6) for i = 1, 2, ,N (4.7) Due to (4.6) and (4.7), xk , and yki are bounded for i = 1, N Further, we have zk − yki = PC (xk ) − PC (yki ) ≤ xk − yki Therefore, using (4.7) we have zk − yki → and zk − xk ≤ zk − yki + xk − yki → as k → ∞, i = 1, N (4.8) This follows that zk is bounded From (4.2) it follows Ti (zk ) − yki = Ti (zk ) − = n n zk + (1 − zk − Ti (zk ) , n )Ti (zk ) i = 1, 2, ,N For a fixed p ∈ F , we have zk − Ti (zk ) = PC (xk ) − PC (Ti (zk )) ≤ xk − Ti (zk ) ≤ xk − p + p − Ti (zk ) ≤ x0 − p + p − zk for i = 1, N The last inequality shows that zk − Ti (zk ) is bounded for i = 1, N Therefore, due to limk→∞ k = 0, we get Ti (zk ) − yki → as k → ∞, (i = 1, N ) Taking into account this relation together with limk→∞ zk − yki → and zk − Ti (zk ) ≤ zk − yki + Ti (zk ) − yki , we find lim zk − Ti (zk ) = 0, i = 1, N k→∞ From the boundedness of zk , there exists a subsequence zkj such that z kj z˜ and limj →∞ zkj − Ti (zkj ) = The relative nonexpansiveness of Ti implies that z˜ ∈ F (Ti ) = F (Ti ), i = 1, N Hence, z˜ ∈ F Using (4.6), we find zkj − PF (x0 ) = PC (xkj ) − PC (PF (x0 )) = x kj − x 2 ≤ xkj − x0 − (PF (x0 ) − x0 ) + x0 − PF (x0 ) ≤ x0 − PF (x0 ) = 2( x0 − PF (x0 ) 2 − xkj − x0 , PF (x0 ) − x0 − xkj − x0 , PF (x0 ) − x0 − zkj − x0 , PF (x0 ) − x0 + zkj − xkj , PF (x0 ) − x0 ) 660 P K Anh and C V Chung z˜ it follows Hence, from (4.8) and zkj lim sup zkj − PF (x0 ) j →∞ ≤2 x0 − PF (x0 ) − z˜ − x0 , PF (x0 ) − x0 (4.9) From the inclusion z˜ ∈ F and the convexity and closedness of F , we have Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 x0 − PF (x0 ) − z˜ − x0 , PF (x0 ) − x0 = x0 − PF (x0 ), z˜ − PF (x0 ) ≤ Combining the last inequality with (4.9), we find limj →∞ zkj − PF (x0 ) = or zkj → PF (x0 ) as j → ∞ Moreover, we also have z˜ ≡ PF (x0 ) Thus, PF (x0 ) is the unique weak accumulation point of zk Clearly, every weakly convergent subsequence of zk strongly converges to PF (x0 ), therefore, zk → PF (x0 ) as k → ∞ From limk→∞ zk − xk = 0, we have limk→∞ xk = PF (x0 ) A NUMERICAL EXAMPLE Consider two integral operators in the Hilbert space E = L [0, 1], defined by [Fi (x)](t ) := Ki (t , s)fi (x(s))ds + gi (t ), i = 1, 2, (5.1) √ √ t +s where K1 (t , s) =√ 23 ts, K2 (t , s) = √e2e2 −1 , f1 (x) = exp(−x ), f2 (x) = sin x + 2e t Since the growth of the derivatives cos x, g1 (t ) = − 43t and g2 (t ) = − e+1 (x), i = 1, is at most linear, that is, |f1 (x)| ≤ 2|x| and |f2 (x)| ≤ f√ i 2, according to [21], √ t √ 1both operators Fi (x) are Frechet differentiable and [F1 (x)h](t ) = − tsx(s) exp(−x (s))h(s)ds, [F2 (x)h](t ) = e 22e−1 e s (cos x(s) − sin x(s))h(s)ds for all x, h ∈ E Let C := B[0, 1] be the closed unit ball centered at the origin A simple computation shows that Fi (x)h ≤ x h , for all x, h ∈ E , which implies the nonexpansiveness of Fi (i = 1, 2) on C Clearly, x = is a common fixed point of Fi Besides, Fi maps C into itself because Fi (x) = Fi (x) − Fi (0) ≤ x for all x ∈ C Now, it can be easily proved (see [2]), that Fi (i = 1, 2) are relatively nonexpansive, hence Algorithm can be implemented Observe that, for each k ≥ 0, the metric projection zk of xk onto C can be defined as zk = xk if xk ≤ 1, otherwise zk = xk / xk Further, we have √ √ 3t s exp(−zk (s))ds − yk (t ) = k zk (t ) + (1 − k ) t , √ t √ t (5.2) 2e 2e yk2 (t ) = k zk (t ) + (1 − k ) e s (sin zk (s) + cos zk (s))ds − e −1 e +1 661 Parallel Hybrid Methods According to [11], for determining xk+1 = PCk ∩Qk (x0 ) we consider three cases If xk = PQk (x0 ) ∈ Ck then xk+1 = xk and iterative process finishes Else if PCk (x0 ) ∈ Qk then x0 , xk − ykk − i xk+1 := PCk (x0 ) = x0 − − y ik x Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 xk − k i ykk (xk − ykk ), i where k := k k In here we note that if xk − ykk = then Ck ≡ H and xk ∈ Ck Otherwise, we have xk+1 = x0 + (xk − Tik (zk )) + (x0 − xk ), where , is a unique solution of the following system of linear equations: xk − ykk , x0 − xk + x0 − xk i i xk − ykk + x0 − xk , xk − ykk i i = − x0 − xk , ik = k − x0 , xk − yk (5.3) In all the following experiments, we choose the initial approximation In this simple example, the determination of xk+1 x0 (t ) ≡ and k = k+1 in method (1.2), which is proposed in [10], is almost the same as in (4.4) All the integrals occured in (5.1)–(5.3) and other places are computed using the trapezoidal formula with the stepsize = 10−3 The programs are written in C and executed on an IBM cluster 1350 with computing nodes with total 51.2GFlops Each node contains two Intel Xeon dual core 3.2GHz, 2GB Ram We use the following notations: PCQM CCQM kmax TOL !NS&ED Tp Ts TL The parallel CQ method The cyclic CQ method (1.2) of X F Liu in [10] Total number of iterations Tolerance xk − x ∗ = xk Method is not stable and explosively divergent Time for PCQM’s execution in parallel mode (2CPUs - in seconds) Time for PCQM’s execution in sequential mode (in seconds) Time for CCQM’s execution (in seconds) In the first experiment, for fixed numbers of iterations we compare the accuracy and execution time of two methods The results are shown in the Table Table shows that for a fixed number of iterations, the PCQM is more accurate than CCQM In parallel mode, the PCQM is less time consuming than CCQM In the second experiment, for given tolerances we compare numbers of iterations and execution time of two methods From Table we see that within a given tolerance, the CCQM is more time consuming than PCQM, 662 P K Anh and C V Chung TABLE Results for a fixed number of iterations Downloaded by [The University Of Melbourne Libraries] at 18:28 15 September 2014 PCQM CCQM kmax Tp Ts TOL TL TOL 500 750 1000 2000 5000 0.97 1.31 1.93 3.61 8.95 1.35 1.80 2.55 5.01 12.98 0.00116 0.00098 0.00089 0.00079 0.00072 1.51 1.97 2.95 5.81 13.47 0.00227 0.00148 0.00126 0.00107 0.00090 TABLE Results for a given tolerance PCQM TOL 0.00135 0.00125 0.00100 0.00075 0.00050 CCQM Tp Ts kmax TL kmax 0.75 0.84 1.12 6.23 15.21 1.05 1.13 1.61 9.95 22.72 356 415 693 4051 9225 2.54 3.02 6.83 33.12 !NS&ED 879 1052 2433 12157 !NS&ED in both parallel and sequential mode Further, whenever the tolerance is too small, the CCQM is unstable CONCLUDING REMARKS We presented a strongly convergent parallel hybrid algorithm for a finite family of relatively nonexpansive mappings Our idea of parallelism is quite different from available parallel methods for finding common fixed points of a family of mappings [4, 5, 8–10] In a Hilbert space our modified method is easily implemented, provided the metric projection onto the given convex closed set C can be computed effectively This is the case, when C is the whole space or a halfspace, or a closed ball The effectiveness of our methods compared to the cyclic CQ method is demonstrated by a numerical example Note that the proposed methods can be used for solving simultaneous systems of equations involving 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University Of Melbourne Libraries] at 18:28 15 September 2014 PARALLEL HYBRID METHODS FOR A FINITE FAMILY OF RELATIVELY NONEXPANSIVE MAPPINGS Pham Ky Anh and Cao Van Chung Department of Mathematics,... hemi -relatively nonexpansive mappings Nonlinear Anal 71:5616–5628 P Cholamjiak and S Suantai (2010) An iterative method for equilibrium problems and a finite family of relatively nonexpansive mappings. .. Plubtieng and K Ungchittrakool (2010) Approximation of common fixed points for a countable family of relatively nonexpansive mappings in a Banach space and applications Nonlinear Anal 72:2896–2908