Dynamical Systems Gradient Method for Solving Nonlinear Equations with Monotone Operators

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1 Acta Appl Math (29) 16: DOI 1.17/s Dynamical Systems Gradient Method for Solving Nonlinear Equations with Monotone Operators N.S. Hoang A.G. Ramm Received: 28 June 28 / Accepted: 26 August 28 / Published online: 18 September 28 Springer Science+Business Media B.V. 28 Abstract A version of the Dynamical Systems Gradient Method for solving ill-posed nonlinear monotone operator equations is studied in this paper. A discrepancy principle is proposed and justified. A numerical experiment was carried out with the new stopping rule. Numerical experiments show that the proposed stopping rule is efficient. Equations with monotone operators are of interest in many applications. Keywords Dynamical systems method (DSM) Nonlinear operator equations Monotone operators Discrepancy principle Mathematics Subject Classification (2) 47J5 47J6 47J35 65R3 1 Introduction In this paper we study a version of the Dynamical Systems Method (DSM) (see [1]) for solving the equation F(u)= f, (1) where F is a nonlinear, twice Fréchet differentiable, monotone operator in a real Hilbert space H,and(1) is assumed solvable, possibly nonuniquely. Monotonicity means that F(u) F(v),u v, u, v H. (2) Equations with monotone operators are important in many applications and were studied extensively, see, for example, [5, 7, 21, 24], and references therein. One encounters many N.S. Hoang A.G. Ramm ( ) Mathematics Department, Kansas State University, Manhattan, KS , USA ramm@math.ksu.edu N.S. Hoang nguyenhs@math.ksu.edu

2 474 N.S. Hoang, A.G. Ramm technical and physical problems with such operators in the cases where dissipation of energy occurs. For example, in [9] and[8], Chap. 3, pp , a wide class of nonlinear dissipative systems is studied, and the basic equations of such systems can be reduced to (1) with monotone operators. Numerous examples of equations with monotone operators can be found in [5] and references mentioned above. In [19] and[2] it is proved that any solvable linear operator equation with a closed, densely defined operator in a Hilbert space H can be reduced to an equation with a monotone operator and solved by a convergent iterative process. In this paper, apparently for the first time, the convergence of the Dynamical Systems Gradient method is proved under natural assumptions and convergence of a corresponding iterative method is established. No special assumptions of smallness of the nonlinearity or other special properties of the nonlinearity are imposed. No source-type assumptions are used. Consequently, our result is quite general and widely applicable. It is well known, that without extra assumptions, usually, source-type assumption about the right-hand side, or some assumption concerning the smoothness of the solution, one cannot get a specific rate of convergence even for linear ill-posed equations (see, for example, [1], where one can find a proof of this statement). On the other hand, such assumptions are often difficult to verify and often they do not hold. By this reason we do not make such assumptions. The result of this paper is useful both because of its many possible applications and because of its general nature. Our novel technique consists of an application of some new inequalities. Our main results are formulated in Theorems 17 and 19, and also in several lemmas, for example, in Lemmas 3, 4, 8, 9, 11, 12. Lemmas 3, 4, 11, 12 may be useful in many other problems. In [23] a stationary equation F(u)= f with a nonlinear monotone operator F was studied. The assumptions A1 A3 on p. 197 in [23] are more restrictive than ours, and the Rule R2 on p.199, formula (4.1) in [23] for the choice of the regularization parameter is quite different from our rule and is more difficult to use computationally: one has to solve a nonlinear equation (equation (4.1) in [23]) in order to find the regularization parameter. To use this equation one has to invert an ill-conditioned linear operator A + αi for small values of α. Assumption A1 in [23] is not verifiable practically, because the solution x is not known. Assumption A3 in [23] requires F to be constant in a ball B r (x ) if F (x ) =. Our method does not require these assumptions, and, in contrast to equation (4.1) in [23], it does not require inversion of ill-conditioned linear operators and solving nonlinear equations for finding the regularization parameter. The stopping time is chosen numerically in our method without extra computational effort by a discrepancy-type principle formulated and justified in Theorem 17, in Sect. 3. We give a convergent iterative process for stable solution of (1.1) and a stopping rule for this process. In [23] the source-type assumption is made, that is, it is assumed that the right-hand side of the equation F(u)= f belongs to the range of a suitable operator. This usually allows one to get some convergence rate. In our paper, as was already mentioned above, such an assumption is not used because, on the one hand, numerically it is difficult to verify such an assumption, and, on the other hand, such an assumption may be not satisfied in many cases, even in linear ill-posed problems, for example, in the case when the solution does not have extra smoothness. We assume the nonlinearity to be twice locally Fréchet differentiable. This assumption, as we mention below, does not restrict the global growth of the nonlinearity. In many practical and theoretical problems the nonlinearities are smooth and given analytically. In these cases one can calculate F analytically. This is the case in the example, considered in Sect. 4. This example is a simple model problem for non-linear Wiener-type filtering (see [18]).

3 Dynamical Systems Gradient Method for Solving Nonlinear Equations 475 If one drops the nonlinear cubic term in the equation Bu + u 3 = f of this example, then the resulting equation Bu= f does not have integrable solutions, in general, even for very smooth f, for example, for f C ([, 1]), asshownin[18]. It is, therefore, of special interest to solve this equation numerically. It is known (see, e.g., [1]), that the set N := {u : F(u)= f } is closed and convex if F is monotone and continuous. A closed and convex set in a Hilbert space has a unique minimal-norm element. This element in N we denote by y, F(y)= f.weassumethat sup F (j) (u) M j (R), j 2, (3) u u R where u H is an element of H, R> is arbitrary, and f = F(y) is not known but f δ, the noisy data, are known, and f δ f δ. Assumption (3) simplifies our arguments and does not restrict the global growth of the nonlinearity. In [12] this assumption is weakened to hemicontinuity in the problems related to the existence of the global solutions of the equations, generated by the DSM. In many applications the nonlinearity F is given analytically, and then one can calculate F (u) analytically. If F (u) is not boundedly invertible then solving (1)foru given noisy data f δ is often (but not always) an ill-posed problem. When F is a linear bounded operator many methods for stable solving of (1) were proposed (see [2, 4 1] and references therein). However, when F is nonlinear then the theory is less complete. DSM consists of finding a nonlinear map (t,u) such that the Cauchy problem u = (t,u), u() = u, has a unique solution for all t, there exists lim t u(t) := u( ),andf(u( )) = f,! u(t) t ; u( ); F(u( )) = f. (4) Various choices of were proposed in [1]for(4) to hold. Each such choice yields a version of the DSM. The DSM for solving equation (1) was extensively studied in [1 17]. In [1], the following version of the DSM was investigated for monotone operators F : u δ = (F (u δ ) + a(t)i) 1 (F (u δ ) + a(t)u δ f δ ), u δ () = u. (5) Here I denotes the identity operator in H. The convergence of this method was justified with some apriorichoice of stopping rule. In [22] a continuous gradient method for solving (1) was studied. A stopping rule of discrepancy type was introduced and justified under the assumption that F satisfies the following condition: F( x) F(x) F (x)( x x) =η F(x) F( x), η<1, (6) for all x, x in some ball B(x,R) H. This very restrictive assumption is not satisfied even for monotone operators. Indeed, if F (x) = forsomex B(x ) then (6) implies F(x)= f for all x B(x,R), provided that B(x,R)contains a solution of (1). In this paper we consider a gradient-type version of the DSM for solving (1): u δ = (F (u δ ) + a(t)i)(f (u δ ) + a(t)u δ f δ ), u δ () = u, (7)

4 476 N.S. Hoang, A.G. Ramm where F is a monotone operator and A denotes the adjoint to a linear operator A. IfF is monotone then F ( ) := A. If a bounded linear operator A is defined on all of the complex Hilbert space H and A, i.e., Au, u, u H,thenA = A,soA is selfadjoint. In a real Hilbert space H a bounded linear operator defined on all of H and satisfying the inequality Au, u, u H is not necessary selfadjoint. Example: H = R 2, A= ( ) 2 1, 2 Au, u =2u u 1u 2 + u 2 2, but A = ( ) 2 A. 1 2 The convergence of the method (7) for any initial value u is proved for a stopping rule based on a discrepancy principle. This a posteriori choice of stopping time t δ is justified provided that a(t) is suitably chosen. The advantage of method (7), a modified version of the gradient method, over the Gauss- Newton method and the version (5) of the DSM is the following: no inversion of matrices is needed in (7). Although the convergence rate of the DSM (7) maybe slower than that of the DSM (5), the DSM (7) might be faster than the DSM (5) for large-scale systems due to its lower computation cost at each iteration. In this paper we investigate a stopping rule based on a discrepancy principle (DP) for the DSM (7). The main results of this paper are Theorem 17 and Theorem 19 in which a DP is formulated, the existence of a stopping time t δ is proved, and the convergence of the DSM with the proposed DP is justified under some natural assumptions. 2 Auxiliary Results The inner product in H is denoted u, v. Let us consider the following equation F(V δ ) + av δ f δ =, a >, (8) where a = const. It is known (see, e.g., [1, 25]) that equation (8) with monotone continuous operator F has a unique solution for any f δ H. Let us recall the following result from [1]: Lemma 1 Assume that (1) is solvable, y is its minimal-norm solution, assumption (2) holds, and F is continuous. Then where V a solves (8) with δ =. lim V a y =, a Of course, under our assumption (3), F is continuous. Lemma 2 If (2) holds and F is continuous, then V δ =O( 1 ) as a, and a lim a F(V δ) f δ = F() f δ. (9)

5 Dynamical Systems Gradient Method for Solving Nonlinear Equations 477 Proof Rewrite (8) as F(V δ ) F() + av δ + F() f δ =. Multiply this equation by V δ, use inequality F(V δ ) F(), V δ and get: a V δ 2 f δ F() V δ. Therefore, V δ =O( 1 ). This and the continuity of F imply (9). a Let a = a(t) be a strictly monotonically decaying continuous positive function on [, ), <a(t), and assume a C 1 [, ). These assumptions hold throughout the paper and often are not repeated. Then the solution V δ of (8) is a function of t, V δ = V δ (t). Fromthe triangle inequality one gets: F(V δ ()) f δ F() f δ F(V δ ()) F(). From Lemma 2 it follows that for large a() one has: F(V δ ()) F() M 1 V δ () =O ( ) 1. a() Therefore, if F() f δ >Cδ,then F(V δ ()) f δ (C ɛ)δ, whereɛ>issufficiently small and a()> is sufficiently large. Below the words decreasing and increasing mean strictly decreasing and strictly increasing. Lemma 3 Assume F() f δ >. Let <a(t), and F be monotone. Denote ψ(t):= V δ (t), φ(t):= a(t)ψ(t)= F(V δ (t)) f δ, where V δ (t) solves (8) with a = a(t). Then φ(t) is decreasing, and ψ(t) is increasing. Proof Since F() f δ >, one has ψ(t), t. Indeed, if ψ(t) t=τ =, then V δ (τ) =, and (8) implies F() f δ =, which is a contradiction. Note that φ(t) = a(t) V δ (t). One has Thus, F(V δ (t 1 )) F(V δ (t 2 )), V δ (t 1 ) V δ (t 2 ) = a(t 1 )V δ (t 1 ) + a(t 2 )V δ (t 2 ), V δ (t 1 ) V δ (t 2 ) = (a(t 1 ) + a(t 2 )) V δ (t 1 ), V δ (t 2 ) a(t 1 ) V δ (t 1 ) 2 a(t 2 ) V δ (t 2 ) 2. (1) (a(t 1 ) + a(t 2 )) V δ (t 1 ) V δ (t 2 ) a(t 1 ) V δ (t 1 ) 2 a(t 2 ) V δ (t 2 ) 2 = (a(t 1 ) V δ (t 1 ) a(t 2 ) V δ (t 2 ) )( V δ (t 2 ) V δ (t 1 ) ) = (φ(t 1 ) φ(t 2 ))(ψ(t 2 ) ψ(t 1 )). (11) If ψ(t 2 )>ψ(t 1 ) then (11) implies φ(t 1 ) φ(t 2 ),so a(t 1 )ψ(t 1 ) a(t 2 )ψ(t 2 )>a(t 2 )ψ(t 1 ).

6 478 N.S. Hoang, A.G. Ramm Thus, if ψ(t 2 )>ψ(t 1 ) then a(t 2 )<a(t 1 ) and, therefore, t 2 >t 1, because a(t) is strictly decreasing. Similarly, if ψ(t 2 )<ψ(t 1 ) then φ(t 1 ) φ(t 2 ). This implies a(t 2 )>a(t 1 ),sot 2 <t 1. Suppose ψ(t 1 ) = ψ(t 2 ), i.e., V δ (t 1 ) = V δ (t 2 ).From(1), one has V δ (t 1 ) 2 V δ (t 1 ), V δ (t 2 ) V δ (t 1 ) V δ (t 2 ) = V δ (t 1 ) 2. This implies V δ (t 1 ) = V δ (t 2 ), and then equation (8) implies a(t 1 ) = a(t 2 ). Hence, t 1 = t 2, because a(t) is strictly decreasing. Therefore φ(t) is decreasing and ψ(t) is increasing. Lemma 4 Suppose that F() f δ >Cδ, C>1, and a() is sufficiently large. Then, there exists a unique t 1 > such that F(V δ (t 1 )) f δ =Cδ. Proof The uniqueness of t 1 follows from Lemma 3 because F(V δ (t)) f δ =φ(t),andφ is decreasing. We have F(y)= f,and = F(V δ ) + av δ f δ,f(v δ ) f δ = F(V δ ) f δ 2 + a V δ y,f(v δ ) f δ +a y,f(v δ ) f δ = F(V δ ) f δ 2 + a V δ y,f(v δ ) F(y) +a V δ y,f f δ +a y,f(v δ ) f δ F(V δ ) f δ 2 + a V δ y,f f δ +a y,f(v δ ) f δ. Here the inequality V δ y,f(v δ ) F(y) was used. Therefore On the other hand, we have F(V δ ) f δ 2 a V δ y,f f δ a y,f(v δ ) f δ a V δ y f f δ +a y F(V δ ) f δ aδ V δ y +a y F(V δ ) f δ. (12) = F(V δ ) F(y)+ av δ + f f δ,v δ y = F(V δ ) F(y),V δ y +a V δ y 2 + a y,v δ y + f f δ,v δ y a V δ y 2 + a y,v δ y + f f δ,v δ y. where the inequality V δ y,f(v δ ) F(y) was used. Therefore, This implies a V δ y 2 a y V δ y +δ V δ y. a V δ y a y +δ. (13) From (12) and(13), and an elementary inequality ab ɛa 2 + b2, ɛ >, one gets: 4ɛ F(V δ ) f δ 2 δ 2 + a y δ + a y F(V δ ) f δ δ 2 + a y δ + ɛ F(V δ ) f δ ɛ a2 y 2, (14)

7 Dynamical Systems Gradient Method for Solving Nonlinear Equations 479 where ɛ> is fixed, independent of t, and can be chosen arbitrary small. Let t and a = a(t). Then (14) implies lim t (1 ɛ) F(V δ ) f δ 2 δ 2. This, the continuity of F, the continuity of V δ (t) on [, ), and the assumption F() f δ >Cδ imply that equation F(V δ (t)) f δ =Cδ must have a solution t 1 >. The uniqueness of this solution has already established. Remark 5 From the proof of Lemma 4 one obtains the following claim: If t n then there exists a unique n 1 > such that F(V n1 +1) f δ Cδ < F(V n1 ) f δ, V n := V δ (t n ). Remark 6 From Lemmas 2 and 3 one concludes that a n V n = F(V n ) f δ F() f δ, a n := a(t n ), n. Remark 7 Let V := V δ (t) δ=,so F(V)+ a(t)v f =. Let y be the minimal-norm solution to (1). We claim that Indeed, from (8) one gets V δ V δ a. (15) F(V δ ) F(V)+ a(v δ V)= f f δ. Multiply this equality with (V δ V)and use the monotonicity of F to get a V δ V 2 δ V δ V. This implies (15). Similarly, multiplying the equation by V y one derives the inequality: F(V)+ av F(y)=, Similar arguments one can find in [1]. From (15) and(16), one gets the following estimate: V y. (16) V δ V + δ a y + δ a. (17) Lemma 8 Suppose a(t) = constants. Then d 2 ( 1 2b ) t 2 c θ d 2 d, ϕ(t) = t a 2 (s) (c+t) b ds where b (, 1 ], d and c are positive 2 4 e ϕ(s) eϕ(t) ds <, t >, θ= 1 2b>. (18) (s + c) 3b (c + t) b

8 48 N.S. Hoang, A.G. Ramm Proof We have t d 2 ϕ(t) = 2(c + s) ds = d 2 ) ((c + t) 1 2b c 1 2b = p(c + t) θ C 2b 3, (19) 2(1 2b) where θ := 1 2b, p := d2 2θ,C 3 := pc θ. One has d e p(c+t)θ dt (c + t) = pθep(c+t)θ b (c + t) = ep(c+t)θ (c + t) b (c + t) b+1 b+1 θ bep(c+t)θ ( ep(c+t)θ (c + t) b d 2 2(c + t) 2b d 2 2(c + t) b 2b c + t ) ( 1 2b c θ d 2 ). Therefore, d 2 ( 1 2b ) t 2 c θ d 2 e p(c+s)θ t ds (s + c) 3b d e p(c+s)θ ds (c + s) ds b ep(c+t)θ (c + t) epcθ b c b ep(c+t)θ (c + t) b. Multiplying this inequality by e C 3 and using (19), one obtains (18). Lemma 8 is proved. Lemma 9 Let a(t) = d and ϕ(t) := t a 2 (s) (c+t) b ds where d,c >, b (, ] and c1 2b d 2 6b. One has t e ϕ(t) e ϕ(s) ȧ(s) V δ (s) ds 1 2 a(t) V δ(t), t. (2) Proof From Lemma 8, one has ( 1 1 2b ) t e ϕ(s) d 3 2 c θ d 2 (s + c) ds < d 3b eϕ(t), (c + t) b c,b, θ= 1 2b>. (21) Since c 1 2b d 2 6b or This implies a 3 ( (s) 1 2b ) = 2 c θ d 2 6b 1, one has c θ c b c θ d 2 4b c θ d 2 d 3 ( 1 2b ) 2(c + s) 3b c θ d 2 4b (c + s) 1 2b d 2, s. 4db = 2 ȧ(s), 2(c + s) b+1 s. (22) Multiplying (21) by V δ (t), using inequality (22) and the fact that V δ (t) is increasing, then for all t one gets t ( e ϕ(t) a(t) V δ (t) > e ϕ(s) V δ (t) a3 (s) 1 2b ) t ds 2 e ϕ(s) ȧ(s) V 2 c θ d 2 δ (s) ds.

9 Dynamical Systems Gradient Method for Solving Nonlinear Equations 481 This implies inequality (2). Lemma 9 is proved. Let us recall the following lemma, which is basic in our proofs. Lemma 1 ([1], p. 97) Let α(t), β(t), γ(t) be continuous nonnegative functions on [t, ), t is a fixed number. If there exists a function μ C 1 [t, ), μ >, lim t μ(t) =, such that α(t) μ 2 [ γ μ(t) μ(t) β(t) 1 [ γ μ(t) 2μ μ(t) ], μ := dμ dt, (23) ], (24) μ()g()<1, (25) and g(t) satisfies the inequality then g(t) exists on [t, ) and ġ(t) γ(t)g(t)+ α(t)g 2 (t) + β(t), t t, (26) g(t) < 1, as t. (27) μ(t) If inequalities (23) (25) hold on an interval [t,t), then g(t) exists on this interval and inequality (27) holds on [t,t). Lemma 11 Suppose M 1,c, and c 1 are positive constants and y H. Then there exist > and a function a(t) C 1 [, ), <a(t), such that and the following conditions hold ȧ(t) a3 (t) 4, M 1, (28) y c (M 1 + a(t)) 2a 2 (t) ȧ(t) c 1 a(t) a2 (t) 2 [ a 2 (t) 2 ȧ(t) a(t) [ a 2 (t) 2 ȧ(t) a(t) ], (29) ], (3) g()<1. (31) a 2 () Proof Take a(t) = d (c + t) b, <b 1 4, 4b c1 2b d 2,c 1. (32)

10 482 N.S. Hoang, A.G. Ramm Note that ȧ = ȧ. Wehave ȧ a 3 = Hence, a 2 (t) 2 Thus, inequality (29) is satisfied if b d 2 (c + t) 1 2b b d 2 c 1 2b 1 4. a 2 (t) 2 ȧ(t) a(t). (33) c (M 1 + a()) 4. (34) Take Then (28) is satisfied and ( max 8c M 1, M ) 1. (35) y For any given g(), choose a() sufficiently large so that Then inequality (31) is satisfied. Choose κ 1 such that Define c M 1 8. (36) a 2 () g()<1. Using inequalities (36), (37)and(38), one gets ( 4c1 b κ>max, 8c ) a(), 1. (37) d 4 ν(t) := κa(t) κ := κ 2. (38) c (M 1 + ν()) 8 + c ν() κ 8 + κ 8 = κ 4. Thus, (34) holds for a(t) = ν(t), = κ. Consequently, (29) holds for a(t) = ν(t), = κ since (33) holds as well under this transformation, i.e., ν 2 (t) 2 ν 2 (t) 2 ν(t) ν(t). (39) Using the inequalities (37)andc 1 and the definition (38), one obtains ν(t) 4 κ c 1 ν 5 (t) = 4c b 1 κ 2 d 4 (c + t) 4c b 1 4b 1 κ 2 d 1. 4

11 Dynamical Systems Gradient Method for Solving Nonlinear Equations 483 This implies [ ν c 1 ν(t) ν4 (t) ν2 (t) ν 2 2 ν ]. 4 κ 2 κ ν Thus, one can replace the function a(t) by ν(t) = κa(t)and by κ = κ 2 in the inequalities (28) (31). Lemma 12 Suppose M 1,c,c 1 and α are positive constants and y H. Then there exist > and a sequence <(a n ) n= such that the following conditions hold a n 2, (4) a n+1 f δ F() a3, (41) M 1 y, (42) c (M 1 + a ) 1 2, (43) an 2 αa4 n 2 + a n a n+1 c 1 a2 n+1 a n+1. (44) Proof Let us show that if a > is sufficiently large, then the following sequence a n = a (1 + n) b, b= 1 4, (45) satisfies conditions (41) (44) if ( ) M1 max y, 4c M 1. (46) Condition (4) is satisfied by the sequence (45). Inequality (42) is satisfied since (46)holds. Choose a() so that a 3 f δ F(), (47) then (41) is satisfied. Assume that (a n ) n= and satisfy (4), (41)and(42). Choose κ 1 such that ( 4c a 4 κ max, αa 22 2, c 1 αa 4 ). (48) It follows from (48) that Define 4 κ 2 a 22 2 α, c 1 α. (49) κ 2 a 4 (b n ) n= := (κa n) n=, κ := κ 2. (5)

12 484 N.S. Hoang, A.G. Ramm Using inequalities (46), (48) and the definitions (5), one gets c (M 1 + b ) κ c a κ = 1 2. Thus, inequality (43) holds for a replaced by b = κa and replaced by κ = κ 2,where κ satisfies (48). For all n one has a 2 n a2 n+1 a 4 n = a4 n a4 n+1 an 4(a2 n + a2 n+1 ) a4 n a4 n+1 = 2an+1 2 a4 n Since a n is decreasing, one has a 4 n+1 a4 n+2 2 a2 n+2 a 4 n+1 = 1 a 22 n a (51) a n a n+1 a 4 n a n+1 = a 4 n a4 n+1 a 4 n a n+1(a n + a n+1 )(a 2 n + a2 n+1 ) a4 n a4 n+1 4a 4 n a4 n+1 Using inequalities (51)and(49), one gets = a 4 n+1 a4 n+2 4 a4 a 4 n+2 n+1 Similarly, using inequalities (52) and(49), one gets Inequalities (53) and(54) imply 1, n. (52) 4a 4 4(an 2 a2 n+1 ) 4 κ 2 an 4 κ 2 a 22 α. (53) 2 4(a n a n+1 )c 1 κ 2 a 4 n a n+1 c 1 α. (54) κ 2 a 4 bn 2 b2 n+1 + b n b n+1 c 1 = a2 n a2 n+1 + a n a n+1 c 1 κ b n+1 a n+1 = κ2 a 4 n 4 4(a 2 n a2 n+1 ) κ 2 a 4 n + κ2 a 4 n 4 κ2 an 4 4 α + κ2 an 4 4 α = κ2 an 4 α 2 = αb4 n. 2 κ 4(a n a n+1 )c 1 κ 2 a 4 n a n+1 Thus, inequality (44) holds for a n replaced by b n = κa n and replaced by κ = κ 2,whereκ satisfies (48). Inequalities (4) (42) hold as well under this transformation. Thus, the choices a n = b n and := κ max( M 1 y, 4c M 1 ),whereκ satisfies (48), satisfy all the conditions of Lemma 12. Remark 13 The constant c and c 1 used in Lemma 11 and 12 will be used in Theorems 17 and 19. These constants are defined in (67). The constant α, usedinlemma12, is the one from Theorem 19. This constant is defined in (94).

13 Dynamical Systems Gradient Method for Solving Nonlinear Equations 485 Remark 14 Using similar arguments one can show that the sequence a n =,where (c+n) b c 1, <b 1, satisfy all conditions of Lemma 4 provided that d is sufficiently large and 4 is chosen so that inequality (46) holds. d Remark 15 In the proof of Lemma 12 and 11 the numbers a and can be chosen so that a2 is uniformly bounded as δ regardless of the rate of growth of the constant M 1 = M 1 (R) from formula (3) whenr, i.e., regardless of the strength of the nonlinearity F(u). To satisfy (46) one can choose = M 1 ( 1 y + 4c ). To satisfy (47) one can choose a = 3 ( f F() + f ) 3 f δ F(), where we have assumed without loss of generality that < f δ f < f. With this choice of a and, the ratio a2 is bounded uniformly with respect to δ (, 1) and does not depend on R. The dependence of a on δ is seen from (47) sincef δ depends on δ. In practice one has f δ f < f. Consequently, 3 fδ F() 3 ( f F() + f ). Thus, we can practically choose a() independent of δ from the following inequality a 3 ( f F() + f ). Indeed, with the above choice one has a2 c( ) c, wherec> is a constant independent of δ, and one can assume that 1 without loss of generality. This Remark is used in the proof of the main result in Sect. 3. Specifically, it is used to prove that an iterative process (93) generates a sequence which stays in the ball B(u,R)for all n n + 1, where the number n is defined by formula (14) (see below), and R>is sufficiently large. An upper bound on R is given in the proof of Theorem 19, below formula (117). Remark 16 One can choose u H such that g := u V F() f δ. (55) a Indeed, if, for example, u =, then by Remark 6 one gets If (41)and(55) hold then g a2. g = V = a V a F() f δ a. 3 Main Results 3.1 Dynamical Systems Gradient Method Assume: ȧ(t) <a(t), lim t a(t) =, ȧ(t) a 3 (t) 1 4. (56)

14 486 N.S. Hoang, A.G. Ramm Denote A := F (u δ (t)), A a := A + ai, a = a(t), where I is the identity operator, and u δ (t) solves the following Cauchy problem: u δ = A a(t) [F(u δ) + a(t)u δ f δ ], u δ () = u. (57) Theorem 17 Assume that F : H H is a monotone operator, twice Fréchet differentiable, sup u B(u,R) F (j) (u) M j (R), j 2, B(u,R):= {u : u u R}, u is an element of H, satisfying inequality (88) (see below). Let a(t) satisfy conditions of Lemma 11. For example, one can choose a(t) = d, where b (, 1 ], c 1, and d>are constants, (c+t) b 4 and d is sufficiently large. Assume that equation F(u)= f has a solution in B(u,R), possibly nonunique, and y is the minimal-norm solution to this equation. Let f be unknown but f δ be given, f δ f δ. Then the solution u δ (t) to problem (57) exists on an interval [,T δ ], lim δ T δ =, and there exists t δ, t δ (,T δ ), not necessarily unique, such that F(u δ (t δ )) f δ =C 1 δ ζ, lim t δ =, (58) δ where C 1 > 1 and <ζ 1 are constants. If ζ (, 1) and t δ satisfies (58), then lim u δ(t δ ) y =. (59) δ Remark 18 One can easily choose u satisfying inequality (88). Note that inequality (88) is a sufficient condition for (91) to hold. In our proof inequality (91) isusedatt = t δ.the stopping time t δ is often sufficiently large for the quantity e ϕ(t δ) h to be small. In this case inequality (91) with t = t δ is satisfied for a wide range of u. The parameter ζ is not fixed in (58). While we could fix it, for example, by setting ζ =.9, it is an interesting open problem to propose an optimal in some sense criterion for choosing ζ. Proof of Theorem 17 Denote Let One has We use Taylor s formula and get: w := u δ V δ, C := C (6) 2 g(t):= w. ẇ = V δ A a(t)[ F(uδ ) F(V δ ) + a(t)w ]. (61) F(u δ ) F(V δ ) + aw = A a w + K, K M 2 2 w 2, (62) where K := F(u δ ) F(V δ ) Aw,andM 2 is the constant from the estimate (3). Multiplying (61) byw and using (62) one gets gġ a 2 g 2 + M 2(M 1 + a) g 3 + V δ g, (63) 2

15 Dynamical Systems Gradient Method for Solving Nonlinear Equations 487 where the estimates: A a A aw,w a 2 g 2 and A a M 1 + a were used. Note that the inequality A a A aw,w a 2 g 2 is true if A. Since F is monotone and differentiable (see (3)), one has A := F (u δ ). Let t > be such that δ a(t ) = 1 y, C >1. (64) C 1 This t exists and is unique since a(t) > monotonically decays to as t.by Lemma 4, there exists t 1 such that F(V δ (t 1 )) f δ =Cδ, F(V δ (t 1 )) + a(t 1 )V δ (t 1 ) f δ =. (65) We claim that t 1 [,t ]. Indeed, from (8)and(17) one gets ( Cδ = a(t 1 ) V δ (t 1 ) a(t 1 ) y + δ ) = a(t 1 ) y +δ, C > 1, a(t 1 ) so δ a(t 1) y C 1. Thus, δ a(t 1 ) y C 1 = δ a(t ). Since a(t), the above inequality implies t 1 t. Differentiating both sides of (8) with respect to t, one obtains A a(t) V δ = ȧv δ. This implies V δ ȧ A 1 a(t) V δ ȧ a V δ ȧ ( y + δ ) ȧ ( a a a y ), t t. C 1 (66) Since g, inequalities (63)and(66)imply ġ a 2 (t)g(t) + c (M 1 + a(t))g 2 + ȧ(t) a(t) c 1, c = M 2 2,c 1 = y Inequality (67) is of the type (26) with ( C 1 γ(t)= a 2 ȧ(t) (t), α(t) = c (M 1 + a(t)), β(t) = c 1 a(t). Let us check assumptions (23) (25). Take μ(t) = a 2 (t), = const. ). (67) By Lemma 11 there exist and a(t) such that conditions (23) (25) hold. Thus, Lemma 1 yields g(t) < a2 (t), t t. (68)

16 488 N.S. Hoang, A.G. Ramm Therefore, F(u δ (t)) f δ F(u δ (t)) F(V δ (t)) + F(V δ (t)) f δ M 1 g(t) + F(V δ (t)) f δ M 1a 2 (t) + F(V δ (t)) f δ, t t. (69) It follows from Lemma 3 that F(V δ (t)) f δ is decreasing. Since t 1 t, one gets F(V δ (t )) f δ F(V δ (t 1 )) f δ =Cδ. (7) This, inequality (69), the inequality M 1 y (see (35)), the relation (64), and the definition C 1 = 2C 1(see(6)) imply F(u δ (t )) f δ M 1a 2 (t ) + Cδ M 1δ(C 1) + Cδ (2C 1)δ = C 1 δ. (71) y We have used the inequality a 2 (t ) a(t ) = δ(c 1) y which is true if δ is sufficiently small, or, equivalently, if t is sufficiently large. Thus, if then there exists t δ (,t ) such that F(u δ ()) f δ C 1 δ ζ, <ζ 1, F(u δ (t δ )) f δ =C 1 δ ζ (72) for any given ζ (, 1],andanyfixedC 1 > 1. Let us prove (59). If this is done, then Theorem 17 is proved. δ First, we prove that lim δ a(t δ ) =. From (69) with t = t δ, and from (17), one gets C 1 δ ζ a 2 (t δ ) M 1 a 2 (t δ ) M 1 Thus, for sufficiently small δ, one gets ( Cδ ζ M1 a() a(t δ ) + a(t δ ) V δ (t δ ) + y a(t δ ) + δ. ) + y, C >, where C <C 1 is a constant. Therefore, δ lim δ a(t δ ) lim δ 1 ζ ( ) M1 a() + y =, δ C <ζ <1. (73)

17 Dynamical Systems Gradient Method for Solving Nonlinear Equations 489 Secondly, we prove that Using (57), one obtains: lim δ t δ =. (74) d dt (F (u δ) + au δ f δ ) = A a u δ +ȧu δ = A a A a (F (u δ) + au δ f δ ) +ȧu δ. This and (8) imply: d dt [F(u δ) F(V δ ) + a(u δ V δ )]= A a A a [F(u δ) F(V δ ) + a(u δ V δ )]+ȧu δ. (75) Denote v := F(u δ ) F(V δ ) + a(u δ V δ ), h = v. Multiplying (75) byv and using monotonicity of F, one obtains hḣ = A a A a v,v + v,ȧ(u δ V δ ) +ȧ v,v δ h 2 a 2 + h ȧ u δ V δ + ȧ h V δ, h. (76) Again, we have used the inequality A a A a a2, which holds for A, i.e., monotone operators F. Thus, ḣ ha 2 + ȧ u δ V δ + ȧ V δ. (77) Since F(u δ ) F(V δ ), u δ V δ, one obtains two inequalities and a u δ V δ 2 v,u δ V δ u δ V δ h, (78) F(u δ ) F(V δ ) 2 v,f(u δ ) F(V δ ) h F(u δ ) F(V δ ). (79) Inequalities (78) and(79) imply: Inequalities (77) and(8) imply Since a 2 ȧ a Inequality (82) implies: a u δ V δ h, F(u δ ) F(V δ ) h. (8) ( ḣ h a 2 ȧ ) + ȧ V δ. (81) a 3a2 > a2 by the last inequality in (56), it follows from inequality (81) that 4 2 h(t) h()e t a 2 (s) 2 ds + e t a 2 (s) 2 ds ḣ a2 2 h + ȧ V δ. (82) t e s a 2 (ξ) 2 dξ ȧ(s) V δ (s) ds. (83)

18 49 N.S. Hoang, A.G. Ramm Denote From (83) and(8), one gets t a 2 (s) ϕ(t) := 2 ds. Therefore, t F(u δ (t)) F(V δ (t)) h()e ϕ(t) + e ϕ(t) e ϕ(s) ȧ(s) V δ (s) ds. (84) F(u δ (t)) f δ F(V δ (t)) f δ F(V δ (t)) F(u δ (t)) t a(t) V δ (t) h()e ϕ(t) e ϕ(t) e ϕ(s) ȧ V δ ds. (85) From Lemma 9 it follows that there exists an a(t) such that 1 t 2 a(t) V δ(t) e ϕ(t) e ϕ(s) ȧ V δ (s) ds. (86) For example, one can choose a(t) = c 1 (c + t), b ( b, 1 ],c b, (87) where c 1,c>. Moreover, one can always choose u such that h() = F(u ) + a()u f δ 1 4 a() V δ(), (88) because the equation is solvable. If (88) holds, then F(u ) + a()u f δ = h()e ϕ(t) 1 4 a() V δ() e ϕ(t), t. (89) If (87) holds, c 1and2b c1 2, then it follows that e ϕ(t) a() a(t). (9) Indeed, inequality a() a(t)e ϕ(t) is obviously true for t =, and (a(t)e ϕ(t) ) t, provided that c 1and2b c1 2. Inequalities (89) and(5) imply e ϕ(t) h() 1 4 a(t) V δ() 1 4 a(t) V δ(t), t. (91) where we have used the inequality V δ (t) V δ (t ) for t t, established in Lemma 3. From (72) and(85) (91), one gets Cδ ζ = F(u δ (t δ )) f δ 1 4 a(t δ) V δ (t δ ).

19 Dynamical Systems Gradient Method for Solving Nonlinear Equations 491 Thus, lim a(t δ) V δ (t δ ) lim 4Cδ ζ =. δ δ Since V δ (t) is increasing, this implies lim δ a(t δ ) =. Since <a(t), it follows that (74) holds. From the triangle inequality and inequalities (68) and(15) one obtains: u δ (t δ ) y u δ (t δ ) V δ + V(t δ ) V δ (t δ ) + V(t δ ) y a2 (t δ ) + δ a(t δ ) + V(t δ) y. (92) From (73), (74), inequality (92) and Lemma 1, one obtains (59). Theorem 17 is proved. 3.2 An Iterative Scheme Let V n,δ solve the equation: Denote V n := V n,δ. Consider the following iterative scheme: F(V n,δ ) + a n V n,δ f δ =. u n+1 = u n α n A n [F(u n) + a n u n f δ ], A n := F (u n ) + a n I, u = u, (93) where u is chosen so that inequality (55) holds, and {α n } n=1 is a positive sequence such that 2 < α α n an 2 + (M 1 + a n ), 2 A n M 1 + a n. (94) It follows from this condition that 1 α n A a n A an = sup a 2 n (M 1 +a n) 2 1 α n 1 α n a 2 n. (95) Note that F (u n ) sincef is monotone. Let a n and satisfy conditions (4) (44). Assume that equation F(u)= f has a solution in B(u,R), possibly nonunique, and y is the minimal-norm solution to this equation. Let f be unknown but f δ be given, and f δ f δ. We prove the following result: d (c+n) b where c 1, <b 1, and d is sufficiently large so that 4 Theorem 19 Assume a n = conditions (4) (44) hold. Let u n be defined by (93). Assume that u is chosen so that (55) holds. Then there exists a unique n δ such that F(u nδ ) f δ C 1 δ ζ, C 1 δ ζ < F(u n ) f δ, n<n δ, (96) where C 1 > 1, <ζ 1. Let <(δ m ) m=1 be a sequence such that δ m. If the sequence {n m := n δm } m=1 is bounded, and {n mj } j=1 is a convergent subsequence, then lim u n mj =ũ, (97) j

20 492 N.S. Hoang, A.G. Ramm where ũ is a solution to the equation F(u)= f. If where ζ (, 1), then Proof Denote lim n m =, (98) m lim u n m y =. (99) m Let C := C (1) 2 z n := u n V n, g n := z n. We use Taylor s formula and get: F(u n ) F(V n ) + a n z n = A n z n + K n, K n M 2 2 z n 2, (11) where K n := F(u n ) F(V n ) F (u n )z n and M 2 is the constant from (3). From (93) and (11) one obtains z n+1 = z n α n A n A nz n α n A n K(z n) (V n+1 V n ). (12) From (12), (11), (95), and the estimate A n M 1 + a n, one gets g n+1 g n 1 α n A n A n + α nm 2 (M 1 + a n ) gn V n+1 V n g n (1 α n an 2 ) + α nm 2 (M 1 + a n ) gn V n+1 V n. (13) Since <a n, for any fixed δ> there exists n such that By (4), one has δ a n +1 > 1 C 1 y δ a n, C >1. (14) an a n+1 2, n. This and (14) imply 2 2δ y > δ > 1 C 1 a n a n +1 C 1 y δ, C >1. (15) a n Thus, 2 C 1 y > δ, n n + 1. (16) a n The number n, satisfying (16), exists and is unique since a n > monotonically decays to asn. By Remark 5, there exists a number n 1 such that F(V n1 +1) f δ Cδ < F(V n1 ) f δ, (17) where V n solves the equation F(V n ) + a n V n f δ =.

21 Dynamical Systems Gradient Method for Solving Nonlinear Equations 493 We claim that n 1 [,n ]. Indeed, one has F(V n1 ) f δ =a n1 V n1,and V n1 y + δ a n1 (cf. (17)), so ( Cδ < a n1 V n1 a n1 y + δ ) = a n1 y +δ, C > 1. (18) a n1 Therefore, δ< a n 1 y C 1. (19) Thus, by (15), δ < y a n1 C 1 < δ. (11) a n +1 Here the last inequality is a consequence of (15). Since a n decreases monotonically, inequality (11) implies n 1 n. One has a n+1 V n V n+1 2 = (a n+1 a n )V n F(V n ) + F(V n+1 ), V n V n+1 (a n+1 a n )V n,v n V n+1 (a n a n+1 ) V n V n V n+1. (111) By (17), V n y + δ a n, and, by (16), ( V n y δ a n 2 y C 1 a n+1 C 1 for all n n + 1. Therefore, ), n n + 1, (112) and, by (111), V n V n+1 a n a n+1 V n a ( n a n+1 y ), a n+1 C 1 n n + 1. (113) Inequalities (13) and(113) imply g n+1 (1 α n an 2 )g n + α n c (M 1 + a n )gn 2 + a n a n+1 c 1, a n+1 n n + 1, (114) where the constants c and c 1 are defined in (67). By Lemma 4 and Remark 14, the sequence (a n ) n=1, satisfies conditions (4) (44), provided that a is sufficiently large and > is chosen so that (46) holds. Let us show by induction that g n < a2 n, n n + 1. (115) Inequality (115) holds for n = by Remark 16. Suppose (115) holds for some n. From (114), (115)and(44), one gets ( g n+1 (1 α n an 2 ) a2 n a 2 ) 2 + α nc (M 1 + a n ) n + a n a n+1 c 1 a n+1 ( ) = a4 n αn c (M 1 + a n ) α n + a2 n + a n a n+1 c 1 a n+1

22 494 N.S. Hoang, A.G. Ramm α na 4 n 2 + a2 n + a n a n+1 a n+1 c 1 a2 n+1. (116) Thus, by induction, inequality (115) holds for all n in the region n n + 1. From (17) one has V n y + δ a n. This and the triangle inequality imply u u n u + z n + V n u + z n + y + δ a n. (117) Inequalities (112), (115), and (117) guarantee that the sequence u n, generated by the iterative process (93), remains in the ball B(u,R) for all n n + 1, where R a + u + y + δ a n. This inequality and the estimate (16) imply that the sequence u n, n n + 1, stays in the ball B(u,R),where R a + u + y + y C + 1 C 1. (118) By Remark 15, one can choose a and so that a is uniformly bounded as δ evenif M 1 (R) as R at an arbitrary fast rate. Thus, the sequence u n stays in the ball B(u,R) for n n + 1whenδ. An upper bound on R is given above. It does not depend on δ as δ. One has: F(u n ) f δ F(u n ) F(V n ) + F(V n ) f δ M 1 g n + F(V n ) f δ M 1an 2 + F(V n ) f δ, n n + 1, (119) where (115) was used and M 1 is the constant from (3). Since F(V n ) f δ is decreasing, by Lemma 3, andn 1 n, one gets F(V n +1) f δ F(V n1 +1) f δ Cδ. (12) From (42), (119), (12), the relation (14), and the definition C 1 = 2C 1(see(1)), one concludes that Thus, if F(u n +1) f δ M 1an Cδ M 1δ(C 1) + Cδ (2C 1)δ = C 1 δ. (121) y F(u ) f δ >C 1 δ ζ, <ζ 1, then one concludes from (121) that there exists n δ, <n δ n + 1, such that F(u nδ ) f δ C 1 δ ζ < F(u n ) f δ, n<n δ, (122)

23 Dynamical Systems Gradient Method for Solving Nonlinear Equations 495 for any given ζ (, 1], and any fixed C 1 > 1. Let us prove (97). If n>isfixed,thenu δ,n is a continuous function of f δ. Denote where lim n m j = N. j From (123) and the continuity of F, one obtains: ũ := ũ N = lim δ u δ,nmj, (123) F(ũ) f δ = lim j F(u n mj ) f δ lim δ C 1 δ ζ =. Thus, ũ is a solution to the equation F(u)= f,and(97) is proved. Let us prove (99) assuming that (98) holds. From (96) and(119) with n = n δ 1, and from (122), one gets C 1 δ ζ M 1 a 2 n δ 1 + a nδ 1 V nδ 1 M 1 a 2 n δ 1 If δ> is sufficiently small, then the above equation implies Cδ ζ a nδ 1( M1 a ) + y, C >, + y a nδ 1 + δ. where C <C 1 is a constant, and the inequality an 2 δ 1 a n δ 1a was used. Therefore, by (4), δ δ δ 1 ζ ( ) M1 a lim lim lim + y =, δ 2a nδ δ a nδ 1 δ C <ζ <1. (124) In particular, for δ = δ m, one gets δ m lim =. δ m a nm (125) From the triangle inequality and inequalities (15) and(115) one obtains: u nm y u nm V nm + V n V nm, + V nm, y a2 n m + δ m a nm + V nm, y. (126) From (98), (125), inequality (126) and Lemma 1, one obtains (99). Theorem 19 is proved. 4 Numerical Experiments Let us do a numerical experiment solving nonlinear equation (1) with F(u):= B(u) + u3 6 := 1 e x y u(y)dy + u3 13, f(x):= 6 6 e x ex e. (127)

24 496 N.S. Hoang, A.G. Ramm Such equation is a model nonlinear equation in Wiener-type filtering theory, see [18]. One can check that u(x) 1 solves the equation F(u)= f. The operator B is compact in H = L 2 [, 1]. The operator u u 3 is defined on a dense subset D of L 2 [, 1], for example, on D := C[, 1].Ifu, v D,then Moreover, u 3 v 3,u v = Therefore, B(u v),u v, so e x = 1 π 1 (u 3 v 3 )(u v)dx. e ix d. F(u v),u v, u, v D. Note that D does not contain subsets, open in H = L 2 [, 1], i.e., it does not contain interior points of H. This is a reflection of the fact that the operator G(u) = u3 is unbounded 6 on any open subset of H. For example, in any ball u C, C = const >, where u := u L 2 [,1], there is an element u such that u 3 =. As such an element one can take, for example, u(x) = c 1 x b, 1 3 <b< 1 2.herec 1 > is a constant chosen so that u C. The operator u F(u)= G(u) + B(u) is maximal monotone on D F := {u : u H, F(u) H } (see [1, p.12]), so that (8) is uniquely solvable for any f δ H. The Fréchet derivative of F is: F (u)h = u2 h e x y h(y)dy. (128) If u(x) vanishes on a set of positive Lebesgue s measure, then F (u) is obviously not boundedly invertible. If u C[, 1] vanishes even at one point x,thenf (u) is not boundedly invertible in H. Let us use the iterative process (93): u n+1 = u n α n (F (u n ) + a n I)(F (u n ) + a n u n f δ ), u =. (129) We stop iterations at n := n δ such that the following inequality holds F(u nδ ) f δ <Cδ ζ, F(u n ) f δ Cδ ζ, n<n δ,c>1, ζ (, 1). (13) Integrals of the form 1 e x y h(y)dy in (127) and(128) are computed by using the trapezoidal rule. The noisy function used in the test is f δ (x) = f(x)+ κf noise (x), κ >, κ = κ(δ). The noise level δ and the relative noise level are determined by δ = κ f noise, δ rel := δ f.

25 Dynamical Systems Gradient Method for Solving Nonlinear Equations 497 In the test, κ is computed in such a way that the relative noise level δ rel equals to some desired value, i.e., δ κ = f noise = δ rel f f noise. We have used the relative noise level as an input parameter in the test. The version of DSM, developed in this paper and denoted by DSMG, is compared with the version of DSM in [3], denoted by DSMN. Indeed, the DSMN is the following iterative scheme u n+1 = u n A 1 n (F (u n ) + a n u n f δ ), u = u, n, (131) where a n = a 1+n. This iterative scheme is used with a stopping time n δ defined by (96). The existence of this stopping time and the convergence of the method is proved in [3]. As we have proved, the DSMG converges when a n = a,b (, 1 (1+n) b 4 ], and a is sufficiently large. However, in practice, if we choose a too large then the method will use too many iterations before reaching the stopping time n δ in (13). This means that the computation time is large. Since F(V δ ) f δ =a(t) V δ, and V δ (t δ ) u δ (t δ ) =O(a(t δ )),wehave Cδ ζ = F(u δ (t δ )) f δ a(t δ ). Thus, we choose a = C δ ζ, C >. The parameter a used in the DSMN is also chosen by this formula. In all figures, the x-axis represents the variable x.inallfigures,bydsmg we denote the numerical solutions obtained by the DSMG, by DSMN we denote solutions by the DSMN and by exact we denote the exact solution. In experiments, we found that the DSMG works well with a = C δ ζ, C [.2, 1]. δ Indeed, in the test the DSMG is implemented with a n := C.99, C (n+1).25 =.5 while the δ DSMN is implemented with a n := C.99 (n+1), C = 1. For C > 1 the convergence rate of DSMG is much slower while the DSMN still works well if C [1, 4]. Figure 1 plots the solutions using relative noise levels δ =.1 and δ =.1. The exact solution used in these experiments is u = 1. In the test the DSMG is implemented with α n = 1, C = 1.1, ζ =.99 and α n = 1, n. The number of iterations of the DSMG for δ =.1 and δ =.1 were 49 and 5 while the number of iteration for the DSMN are 9 and 9, respectively. The number of nodal points used in computing integrals in (127) and (128)wasn = 1. The noise function f noise in this experiment is a vector with random entries normally distributed of mean and variant 1. Figure 1 shows that the solutions by the DSMN and DSMG are nearly the same in this figure. Figure 2 presents the numerical results when N = 1 with δ =.1 u(x) = sin(2πx), x [, 1] (left) and with δ =.1, u(x) = sin(πx), x [, 1] (right). In these cases, the DSMN took 11 and 7 iterations to give the numerical solutions while the DSMG took 512 and 94 iterations for u(x) = sin(2πx) and u(x) = sin(πx), respectively. Figure 2 show that the numerical results of the DSMG are better than those of the DSMN. Numerical experiments agree with the theory that the convergence rate of the DSMG is 1 slower than that of the DSMN. It is because the rate of decaying of the sequence { } (1+n) 1 n=1 4

26 498 N.S. Hoang, A.G. Ramm Fig. 1 Plots of solutions obtained by the DSMN and DSMG when N = 1, u = 1, x [, 1], δ rel =.1 (left) andn = 1, u = 1, x [, 1], δ rel =.1 (right) Fig. 2 Plots of solutions obtained by the DSMN and DSMG when N = 1, u(x) = sin(2πx), x [, 1], δ rel =.1 (left)andn = 1, u(x) = sin(πx), x [, 1], δ rel =.1 (right) is much slower than that of the sequence { 1 1+n } n=1. However, if the cost for evaluating F and F are not counted then the cost of computation at one iteration of the DSMG is of O(N 2 ) while that of the DSMN in one iteration of the DSMN is of O(N 3 ). Here, n is the dimension of u. Thus, for large scale problems, the DSMG might be an alternative to the DSMN. Also, as it is showed in Fig. 2, the DSMG might yield solutions with better accuracy. Experiments show that the DSMN still works with a n = a for 1 b 1. So in (1+n) b 4 practice, one might use faster decaying sequence a n to reduce the time of computation. From the numerical results we conclude that the proposed stopping rule yields good results in this problem.

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