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Intelligent Control and Automation, 2010, 1, 28-35 doi:10.4236/ica.201.11004 Published Online August 2010 (http://www.SciRP.org/journal/ica) Copyright © 2010 SciRes. ICA A Nonmonotone Line Search Method for Symmetric Nonlinear Equations* Gonglin Yuan1, Laisheng Yu2 1College of Mathematics and Information Science, Guangxi University, Nanning, China 2Student Affairs Office, Guangxi University, Nanning, China E-mail: glyuan@gxu.edu.cn Received February 7, 2010; revised March 21, 2010; accepted July 7, 2010 Abstract In this paper, we propose a new method which based on the nonmonotone line search technique for solving symmetric nonlinear equations. The method can ensure that the search direction is descent for the norm function. Under suitable conditions, the global convergence of the method is proved. Numerical results show that the presented method is practicable for the test problems. Keywords: Nonmonotone Line Search, Symmetric Equations, Global Convergence 1. Introduction Consider the following nonlinear equations: 0, n g xxR (1) where :nn g RR be continuously differentiable and its Jacobian () g x is symmetric for all n x R. This problem can come from unconstrained optimization problems, a saddle point problem, and equality con- strained problems [the detail see [1]]. Let () x be the norm function defined by 2 1 () () 2 x gx . Then the nonlinear equation problem (1) is equivalent to the fol- lowing global optimization problem min, n x xR (2) The following iterative formula is often used to solve (1) and (2): 1kkkk x xd where k is a steplength, and k d is one search direc- tion. To begin with, we briefly review some methods for (1) and (2) by line search technique. First, we give some techniques for k . Li and Fukashima [1] proposed an approximate monotone line search technique to obtain the step-size k satisfying: 222 12 kkk kkkkk k xd x dgg (3) where 10 and 20 are positive constants,k i kr , (0,1), k ri is the smallest nonnegative integer i such that (3), and k satisfies 0 k k . Combining the line search (3) with one special BFGS update formula, they got some better results (see [1]). Inspired by their idea, Wei [2] and Yuan [3] made a further study. Brown and Saad [4] proposed the following line search method to obtain the stepsize T kkkkk kk x dx xd (4) where (0,1). Based on this technique, Zhu [5] gave the nonmonotone line search technique: () T kkklkkkk x dx xd (5) where () 0() max, 0,1,2,, lkk j jnk k ()min{ , } nkMk, and 0M is an integer constant. From these two tech- niques (4) and (5), it is easy to see that the Jacobian ma- tric () g x must be computed at every iteration, which will increase the workload especially for large-scale problems or this matric is expensive. Considering these points, we [6] presented a new backtracking inexact technique to obtain the stepsize k *This work is supported by China NSF grands 10761001, the Scientific Research Foundation of Guangxi University (Grant No. X081082), and Guangxi SF grands 0991028. G. L. YUAN ET AL. Copyright © 2010 SciRes. ICA 29 22 2 k T kkk kkk g xdgxgx d (6) where (0,1) Second, we present some techniques for k d. One of the most effective methods is Newton method. It normally requires a fewest number of function evaluations, and it is very good at handling ill-condi- tioning. However, its efficiency largely depends on the possibility to efficiently solve a linear system which arises when computing the search k d at each iteration kk k g xd gx (7) Moreover, the exact solution of the system (7) could be too burdensome, or it is not necessary when k d is far from a solution [7]. Inexact Newton methods [5,7] represent the basic approach underlying most of the Newton-type large-scale algorithms. At each iteration, the current estimate of the solution is updated by ap- proximately solving the linear system (7) using an itera- tive algorithm. The inner iteration is typically “trun- cated” before the solution to the linear system is obtained. Griewank [8] firstly proposed the Broyden’s rank one method for nonlinear equations and obtained the global convergence. At present, a lot of algorithms have been proposed for solving these two problems (1) and (2) (see [9-15]). The famous BFGS formula is one of the most effective quasi-Newton methods, where the k d is the solution of the system of linear equations 0 kk k Bd g (8) where k B is generated by the following BFGS update formula 1 TT kkkkk k kk TT kkkkk BssByy BB s Bss y (9) where 1kk k s xx and 1 ()(). kk k ygx gx Recently, there are some results on nonlinear equations can be found at [6,17-20]. Zhang and Hager [21] present a nonmonotone line search technique for unconstrained optimization problem min( ) n xR f x , where the nonmonotone line search technique is defined by , T kkk k kkk TT kkkk kk f xdC fxd fxddfx d where 000 11 1 01, ,1, 1, kkkkkk kkk k k CfxQ QCf xd QQC Q min max [, ], k and min max 01 . They proved the global convergence for nonconvex, smooth functions, and R-linear convergence for strongly convex functions. Numerical results show that this method is more effec- tive than other similar methods. Motivated by their tech- nique, we propose a new nonmonotone line search tech- nique which can ensure the descent search direction on the norm function for solving symmetric nonlinear Equa- tions (1) and prove the global convergence of our method. The numerical results are reported too. Here and throughout this paper, ||.|| denote the Euclidian norm of vectors or its induced matrix norm. This paper is organized as follows. In the next section, we will give our algorithm for (2). The global conver- gence and the numerical result are established in Section 3 and in Section 4, respectively. 2. The Algorithm Precisely, our algorithm is stated as follows. Algorithm 1. Step 0: Choose an initial point 0, n x R an initial sy- mmetric positive definite matrix 0 nn BR , and constants 2 12100 0 (0,1),01,01,, 1rJgE , and 0;k Step 1: If 0; k g then stop; Otherwise, solving the following linear Equations (10) to obtain k dand go to step 2; 0 kk k Bd g (10) Step 2: Let k i be the smallest nonnegative integer i such that 22 2 k T kkkkkk g xdgxgx d (11) holds for. i r Let k i kr ; Step 3: Let 1, kkkk x xd 1kk k s xx and 1. kkk y gx gx If 0, k T k ys update k B to generate 1k B by the BFGS formula (9). Otherwise, let 1; kk BB Step 4: Choose [0,1], k and set 2 1 11 1 1, kkk k kkkk k EJg x EEJE (12) Step 5: Let k: = k + 1, go to Step 1. Remark 1: 1) By the technique of the step 3 in the algorithm [see [1]], we deduce that 1k B can inherits the positive and symmetric properties ofk B. Then, it is not difficult to get 0. k T k dg G. L. YUAN ET AL. Copyright © 2010 SciRes. ICA 30 2) It is easy to know that 1k J is a convex combina- tion of k J and 2 1 (). k gx By 2 00 ,Jgit follows that k J is a convex combination of function values 22 2 01 ,,,. k gg g The choice of k controls the degree of nonmonotonicity. If 0 k for each k, then the line search is the usual monotone line search. If 1 k for each k, then 2 0 1 1 k ki i Vg k , kk J V where is the average function value. 3) By (9), we have 111kk kkkkk Bs y gggs 1, k T k g s this means that 1k B approximate to 1k g along . k s 3. The Global Convergence Analysis of Algorithm 1 In this section, we establish global convergence for Al- gorithm 1. The level set is defined by 0 |() (). n x Rgxgx Assumption A. The Jaconbian of g is symmetric and there exists a constant 0M holds kk g xgx Mxx (13) for .x. Since k B approximates k g along direction k s , we can give the following assumption. Assumption B. k B is a good approximation to k g , i.e., kkk k g xBd g (14) where (0,1) is a small quantity. Assumption C. There exist positive constants 1 b and 2 b satisfy 2 1 T kk k g dbg (15) and 2kk dbg (16) for all sufficiently large k.. By (10) and Assumption C, we have 12kk k bgdb g (17) Lemma 3.1. Let Assumption B hold and 1 {,, kkk dx , 1}. k gbe generated by Algorithm 1. Then k d is descent direction for () x at k x , i.e., 0 T kk xd (18) Proof. By (10), we have kk kk TTT kkkkkkk kk TT kkk kk x dggdg gdBdg ggdBdgg (19) Using (14) and taking the norm in the right-hand-side of (19), we get 2 2 1 k TT kkkkkkk k x dggdBd g g (20) Therefore, for (0,1), we get the lemma. By the above lemma, we know that the norm function () x is descent along k d, then 1kk g g holds. Lemma 3.2. Let Assumption B hold and 1 {,, kkk dx , 1} k g be generated by Algorithm 1. Then {} . k x Moreover, {} k g converges. Proof. By Lemma 3.1, we get 1kk g g . Then, we conclude that {} k g converges. Moreover, we have for all k 10 . kk g gg Which means that {}. k x The next lemma will show that for any choice of [0,1), kk J lies between 2 k g and . k V Lemma 3.3. Let 11 {,, ,} kkkk dx g be generated by Algorithm 1, we have 2 1 , kkkkk g JVJJ for each k. Proof. We will prove the lower bound for k J by in- duction. For k = 0, by the initialization 2 00 ,Jg this holds. Now we assume that 2 ii J g holds for all 0.ik By (2.3) and 22 1, ii gg we have 222 11 1 11 22 2 11 1 1 iiiiii ii i ii ii ii i i EJgE gg JEE Eg gg E (21) where 11 iii EE . Now we prove that 1kk J J is true. By (12) again, and using 22 1kk gg , we obtain 2 11 1 11 . kkk kkkk k k kk EJ gEJ J JEE G. L. YUAN ET AL. Copyright © 2010 SciRes. ICA 31 Which means that 1kk J J for all k is satisfied. Then we have 2 1kkk g JJ (22) Let : k LR Rbe defined by 2 1, 1 kk k tJ g Lt t we can get 2 1 2. 1 kk k Jg Lt t By 2 1, kk Jg we obtain '( )0 k Lt for all 0t. Then, k L is nondecreasing, and 2(0)() kk k g LLt for all 0t. Now we prove the upper bound kk J V by induction. For k = 0, by the initialization 2 00 J g, this holds. Now assume that j j J V hold for all 0jk. By using 01E, (12), and [0,1] k , we obtain 1 00 12 ji jjp ip Ej (23) Denote that k L is monotone nondecreasing, (23) im- plies that 11 1 kkkkk k J LELE Lk (24) Using the induction step, we have 22 11 11 kk kk kk kJgkV g Lk V kk (25) Combining (24) and (25) implies the upper bound of k J in this lemma. Therefore, we get the result of this lemma. The following lemma implies that the line search tech- nique is well-defined. Lemma 3.4. Let Assumption A, B and C hold. Then Algorithm 1 will produce an iterate 1kkkk x xd in a finite number of backtracking steps. Proof. From Lemma 3.8 in [4] we have that in a finite number of backtracking steps, k must satisfy 22 T kkkkkk kk g xdgxgx gd (26) where (0,1) . By (20) and (15), we get 2 2 1 1 1 11 T kkkk kk T T kk kkkkk T kk gx gdg gd g gd b gd (27) Using (0,1) k , we obtain 1 2 1 1 1 1 1 TT kkkk kkk T kkk g xgd gd b gd b (28) So let 1 1 1 0, min1,1b . By Lemma 3.3, we know 2. kk g J Therefore, we get the line search (11). The proof is complete. Lemma 3.5. Let Assumption A, B and C hold. Then we have the following estimate for k , when k suffi- ciently large: 00 kb (29) Proof. Assuming the step-size k such that (11). Then 1 k k does not satisfy (11), i.e., 2 2 1 T kkkkk kk g xdJ gxd By 2 kk g J , we get 22 2 2 1 kkk k T kkkkk kk gx dg g xdJ gxd Which implies that 22 2 1 T kkkkkk k g xd gxdg (30) By Taylor formula, (19), (20), and (17), we get 22 222 22 22 2 2 2 2 1 1 1 T kkk kkkkk kk kkkk kkkk gxdgg gd Od gOd dO d b (31) Using (15), (17), (30), and (31) we obtain 2 2 1 2 1 2 22 22 1 2 1 2 2 22 1 2 2 22 22 2 2 2 11 21 11 21 1 21 1 1 kk kkkk T kkkkk kkk k kkkk d b b dd b b dgd b gx dg dO d b (32) G. L. YUAN ET AL. Copyright © 2010 SciRes. ICA 32 which implies when k sufficiently large, 1 2 121 1. 21 k b bb Let 1 02 121 1 0, . 21 b bbb The proof is complete. In the following, we give the global convergence theo- rem. Theorem 3.1. Let 11 {,,,} kkkk dx g be generated by Algorithm 1, Assumption A, B, and C hold, and 2 k g be bounded from below. Then lim inf0 k kg (33) Moreover, if 21, then lim 0 k kg (34) Therefore, every convergent subsequence approaches to a point *, x where * ()0.gx Proof. By (11), (15), (16), and (19), we have 22 11 11 2 011 T kkkkkkkk kk g JdgJbg Jbbg (35) Let 101 .bb Combining (12) and the upper bound of (35), we get 22 1 1 11 2 1 1 kkkkkkkkk k kk k k k EJgEJ Jg JEE g JE (36) Since 2 k g is bounded from below and 2 kk g J for all k, we can conclude that k J is bounded from be- low. Then, using (36), we obtain 2 01 k kk g E (37) By (23), we get 12 k Ek (38) If 2 k g were bounded away from 0, then (37) would violate (38). Hence, (33) holds. If 21 , by (23), we have 1 12 00 0 2 02 11 1 1 j kk j kki jj i k j E (39) Then, (37) implies (34). The proof is complete. 4. Numerical Results In this section, we report the results of some numerical experiments with the proposed method. Problem 1. The discretized two-point boundary value problem is the same to the problem in [22] 2 1 1 g xAxFx n , where A is the nn tridiagonal matrix given by 41 141 1 14 A and 12 ,, T n F xFxFxFx, with i F x sin1, i1,2, i x n . Problem 2. Unconstrained optimization problem min( ),n f xx R, with Engval function [23] defined by 2 22 11 2 43 n ii i i fxxxx The related symmetric nonlinear equation is 14 g xfx where 12 ,,, T n g xgxgx gx with 22 1112 222 11 22 1 1 21,2,3,,1 iiiii nnnn gxxx x g xxx xxin gx xxx In the experiments, the parameters in Algorithm 1 were chosen as 10 0.1,0.001,0.8, k rB is unit matrix. The program was coded in MATLAB 6.1. We stopped the iteration when the condition 26 () 10gx was satisfied. Tables 1 and 2 show the performance of the method need to solve the Problem 1. Tables 3 and 4 show the performance of the method need to solve the Problem 2. The columns of the tables have the following meaning: Dim: the dimension of the problem. NI: the total number of iterations. NG: the number of the function evaluations. GG: the function evaluations. From the above tabulars, we can see that the numerical G. L. YUAN ET AL. Copyright © 2010 SciRes. ICA 33 Test Result for the Problem 1 Table 1. Small-scale. 0 x (4, ..., 4) (20, ..., 20) (100, ..., 100) (-4, ..., -4) (-20, ..., -20) (-100, ..., -100) Dim NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG n = 10 16/22/3.714814e-7 16/20/2.650667e-7 16/20/4.014933e-716/22/3.723721e-7 16/20/2.643713e-7 16/20/4.013770e-7 n = 50 44/47/1.388672e-7 44/47/6.929395e 46/49/3.713174e-844/47/1.388793e-744/47/6.929516e-7 46/49/3.726373e-8 n = 100 68/71/5.905592e-7 70/73/8.759459e 72/75/3.125373e-768/71/5.905724e-770/73/8.759500e-7 72/75/3.125382e-7 0 x (4, .0, ...,) (20, 0, ..., 20) (100, .0, ...,) (-4, .0, ...,) (-20, .0, ...,) (-100, .0, ...,) Dim NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG n = 10 21/23/1.297275e-9 21/23/5.577021e-9 1/23/9.029402e-9 1/23/1.208563e-9 1/23/4.707707e-9 21/23/1.061736e-8 n = 50 63/65/8.204623e-7 67/69/9.997988e-7 9/71/4.511023e-7 3/65/8.204744e-7 7/69/9.997996e-7 69/71/4.511007e-7 n = 100 65/67/6.046233e-7 69/71/7.845951e-7 1/73/6.996085e-7 65/67/6.046254e-7 9/71/7.845962e-7 71/73/6.996092e-7 Table 2. Large-scale. 0 x (4, ..., 4) (20, ..., 20) (30, ..., 30) (-4, ..., -4) (-20, ..., -20) (-30, ..., -30) Dim NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG n = 300 70/73/7.844778e-7 76/79/7.741702e-7 8/81/6.759628e-7 0/73/7.844800e-7 6/79/7.741706e-7 78/81/6.759631e-7 n = 500 70/73/8.547195e-7 76/79/8.435874e-7 8/81/7.366072e-7 0/73/8.547204e-7 6/79/8.435876e-7 78/81/7.366073e-7 n = 800 68/70/6.505423e-7 74/76/6.414077e-7 4/76/9.621120e-7 8/70/6.505425e-7 4/76/6.414078e-7 74/76/9.621120e-7 0 x (4, .0, ...,) (20, 0, ..., 20) (30, .0, ...,) (-4, .0, ...,) (-20, .0, ...,) (-30, .0, ...,) Dim NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG n = 300 67/69/5.896038e-7 71/73/9.997625e-7 3/75/8.731533e-767/69/5.896038e-7 71/73/9.997625e-7 73/75/8.731533e-7 n = 500 67/69/7.145027e-7 73/75/7.057076e-7 5/77/6.163480e-77/69/7.145024e-773/75/7.057075e-7 75/77/6.163479e-7 n = 800 69/71/6.188110e-7 75/77/6.115054e-7 5/77/9.172559e-7 9/71/6.188106e-7 5/77/6.115053e-7 75/77/9.172558e-7 Test Result for the Problem 2 Table 3. Small-scale. 0 x (1, ..., 1) (3, ..., 3) (4, ..., 4) (1, 0, ...,) (3, 0, , ...,) (4, 0, ...,) Dim NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG n = 10 20/22/3.007469e-7 38/47/6.088293e-7 44/48/4.898591e-70/23/3.452856e-75/41/5.833715e-7 29/34/4.338894e-7 n = 50 36/38/6.966974e-7 76/88/6.845101e-7 99/114/8.556270e-76/39/7.812438e-79/77/4.466497e-7 67/75/4.681269e-7 n = 100 36/38/7.207203e-7 6/109/4.173166e-7 0/87/7.911692e-76/39/8.220367e-79/92/8.640158e-7 69/76/8.515673e-7 G. L. YUAN ET AL. Copyright © 2010 SciRes. ICA 34 Table 4. Large-scale. 0 x (1, ..., 1) (3, ..., 3) (4, ..., 4) (1, 0, ...,) (3, 0, ...,) (4, 0, ...,) Dim NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG NI/NG/GG n = 300 40/42/4.452904e-7 4/106/6.146797e-7 63/68/4.232021e-70/43/9.348673e-7 6/72/7.638011e-7 92/106/9.095927e- n = 500 44/46/9.611950e-7 9/171/4.445314e-7 18/133/7.054347-7e 2/45/8.280486e-73/79/7.159029e-7 85/98/4.229872e-7 n = 800 41/43/4.510999e-7 7/185/5.274922e-7 74/198/4.239839e-7 1/44/9.502624e-72/77/6.117626e-7 93/106/8.797380e-7 results are quite well for the test Problems with the pro- posed method. The initial points and the dimension don’t influence the performance of the algorithm 1 very much. However, we find the started points will influence the result for the problem 2 a little in our experiment. In one word, the numerical are attractively. The method can be used to the system of nonlinear equations whose Jaco- bian is not symmetric. 5. 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