THE DYNAMICAL SYSTEMS METHOD FOR SOLVING NONLINEAR EQUATIONS WITH MONOTONE OPERATORS

Transcription

1 Asian-European Journal of Mathematics Vol. 3, No. 1 (2010) c World Scientific Publishing Company THE DYNAMICAL SYSTEMS METHOD FOR SOLVING NONLINEAR EQUATIONS WITH MONOTONE OPERATORS N. S. Hoang Mathematics Department, Kansas State University, Manhattan, Kansas, KS , USA nguyenhs@math.ksu.edu A. G. Ramm Mathematics Department, Kansas State University, Manhattan, Kansas, KS , USA ramm@math.ksu.edu Communicated by Jörg Koppitz Received January 27, 2009 Revised July 14, 2009 A review of the authors results is given. Several methods are discussed for solving nonlinear equations F (u) =f, wheref is a monotone operator in a Hilbert space, and noisy data are given in place of the exact data. A discrepancy principle for solving the equation is formulated and justified. Various versions of the Dynamical Systems Method (DSM) for solving the equation are formulated. These versions of the DSM include a regularized Newton-type method, a gradient-type method, and a simple iteration method. Apriori and a posteriori choices of stopping rules for these methods are proposed and justified. Convergence of the solutions, obtained by these methods, to the minimal norm solution to the equation F (u) =f is proved. Iterative schemes with a posteriori choices of stopping rule corresponding to the proposed DSM are formulated. Convergence of these iterative schemes to a solution to equation F (u) =f is justified. New nonlinear differential inequalities are derived and applied to a study of large-time behavior of solutions to evolution equations. Discrete versions of these inequalities are established. Keywords: Ill-posed problems; nonlinear operator equations; monotone operators; nonlinear inequalities; dynamical systems method. AMS Subject Classification: 47H05, 47J05, 47N20, 65J20, 65M30 1. Introduction Consider equation F (u) =f, (1.1) 57

2 58 N. S. Hoang & A. G. Ramm where F is an operator in a Hilbert space H. Throughout this paper we assume that F is a monotone continuous operator. Monotonicity is understood as follows: F (u) F (v),u v 0, u, v H. (1.2) We assume that equation (1.1) has a solution, possibly non-unique. Assume that f is not known but f, the noisy data, f f, areknown. There are many practically important problems which are ill-posed in the sense of J. Hadamard. Problem (1.1) is well-posed in the sense of Hadamard if and only if F is injective, surjective, and the inverse operator F 1 is continuous. To solve illposed problem (1.1), one has to use regularization methods rather than the classical Newton s or Newton-Kantorovich s methods. Regularization methods for stable solution of linear ill-posed problems have been studied extensively (see [13], [15], [38] and references therein). Among regularization methods, the variational regularization (VR) is one of the frequently used methods. When F = A, wherea is a linear operator, the VR method consists of minimizing the following functional: Au f 2 + α u 2 min. (1.3) The minimizer u,a of problem (1.3) can be found from the Euler equation: (A A + αi)u,α = A f. Inthe VR method the choiceof the regularizationparameterα is important. Various choices of the regularization parameter have been proposed and justified. Among these, the discrepancy principle (DP) appears to be the most efficient in practice (see [13]). According to the DP one chooses α as the solution to the following equation: Au,α f = C, 1 <C= const. (1.4) When the operator F is nonlinear, the theory is less complete (see [2], [37]). In this case, one may try to minimize the functional F (u) f 2 + α u 2 min (1.5) as in the case of linear operator F. The minimizer to problem (1.5) solves the following Euler equation F (u,α ) F (u,α )+αu,α = F (u,α ) f. (1.6) However, there are several principal difficulties in nonlinear problems: there are no general results concerning the solvability of (1.6), and the notion of minimalnorm solution does not make sense, in general, when F is nonlinear. Other methods for solving (1.1) with nonlinear F have been studied. Convergence proofs of these methods often rely on the source-type assumptions. These assumptions are difficult to verify in practice and they may not hold. Equation (1.1) with a monotone operator F is of interest and importance in many applications. Every solvable linear operator equation Au = f can be reduced to solving operator equation with a monotone operator A A. For equations with a

3 The DSM for Solving NOE 59 bounded operator A this is a simple fact, and for unbounded, closed, densely defined linear operators A it is proved in [28], [30], [31], [15]. Physical problems with dissipation of energy often can be reduced to solving equations with monotone operators (see, e.g., [35]). For simplicity we present the results for equations in Hilbert spaces, but some results can be generalized to the operators in Banach spaces. When F is monotone then the notion minimal-norm solution makes sense (see, e.g., [15], p. 110). In [36], Tautenhahn studied a discrepancy principle for solving equation (1.1). The discrepancy principle in [36] requires solving for α the following equation: (F (u,α )+αi) 1 (F (u,α ) f ) = C, 1 <C= const, (1.7) where u,α solves the equation: F (u,α )+αu,α = f. For this discrepancy principle optimal rate of convergence is obtained in [36]. However, the convergence of the method is justified under source-type assumptions and other restrictive assumptions. These assumptions often do not hold and some of them cannot be verified, in general. In addition, equation (1.7) is difficult to solve numerically. A continuous analog of the Newton method for solving well-posed operator equations was proposed in [3], in In [1], [4] [33], and in the monograph [15] the Dynamical Systems Method for solving operator equations is studied systematically. The DSM consists of finding a nonlinear map Φ(t, u) such that the Cauchy problem u =Φ(t, u), u(0) = u 0, (1.8) has a unique solution for all t 0, there exists lim t u(t) := u( ), and F (u( )) = f,! u(t) t 0; u( ); F (u( )) = f. (1.9) Various choices of Φ were proposed in [15] for (1.9) to hold. Each such choice yields a version of the DSM. In this paper, several methods developed by the authors for solving stably equation (1.1) with a monotone operator F in a Hilbert space H and noisy data f,given in place of the exact data f, are presented. A discrepancy principle (DP) for solving stably equation (1.1) is formulated and justified. In this DP the only assumptions on F are the continuity and monotonicity. Thus, our result is quite general and can be applied for a wide range of problems. Several versions of the Dynamical Systems Method (DSM) for solving equation (1.1) are formulated. These versions of the DSM are a Newton-type method, a gradient-type method and a simple iterations method. A priori and a posteriori choices of stopping rules for several versions of the DSM and for the corresponding iterative scheme are proposed and justified. Convergence of the solutions of these versions of the DSM to the minimal-norm

4 60 N. S. Hoang & A. G. Ramm solution to the equation F (u) = f is proved. Iterative schemes, corresponding to the proposed versions of the DSM, are formulated. Convergence of these iterative schemes to a solution to equation F (u) =f is established. When one uses these iterative schemes one does not have to solve a nonlinear equation for the regularization parameter. The stopping time is chosen automatically in the course of calculations. Implementation of these methods is illustrated in Sec. 6 by a numerical experiment. In Secs. 2 and 3 basic and auxiliary results are formulated, in Sec. 4 proofs are given, in Sec. 5 ideas of applications of the basic nonlinear inequality (2.87) are outlined. 2. Basic Results 2.1. A discrepancy principle Let us consider the following equation F (V,a )+av,a f =0, a > 0, (2.1) where a = const. It is known (see, e.g., [15], p. 111) that equation (2.1) with a monotone continuous operator F has a unique solution V,a for any f H. Assume that equation (1.1) has a solution. It is known that the set of solutions N := {u : F (u) =f} is convex and closed if F is monotone and continuous (see, e.g., [15], p. 110). A closed and convex set N in H has a unique minimal-norm element. This minimal-norm solution to (1.1) is denoted by y. Theorem 1. Let γ (0, 1] and C > 0 be some constants such that C γ >. Assume that F (0) f >C γ.lety be the minimal-norm solution to equation (1.1). Then there exists a unique a() > 0 such that F (V,a() ) f = C γ, (2.2) where V,a() solves (2.1) with a = a(). If 0 <γ<1 then lim V,a() y =0. (2.3) 0 Instead of using (2.1), one may use the following equation: F (V,a )+a(v,a ū) f =0, a > 0, (2.4) where ū is an element of H. DenoteF 1 (u) :=F (u +ū). Then F 1 is monotone and continuous. Equation (2.4) can be written as: F 1 (U,a )+au,a f =0, U,a := V,a ū, a > 0. (2.5) Applying Theorem 1 with F = F 1 one gets the following result: Corollary 2. Let γ (0, 1] and C > 0 be some constants such that C γ >. Let ū H and z be the solution to (1.1) with minimal distance to ū. Assume that F (ū) f >C γ. Then there exists a unique a() > 0 such that F (Ṽ,a()) f = C γ, (2.6)

5 The DSM for Solving NOE 61 where Ṽ,a() solves the following equation: If γ (0, 1) then this a() satisfies F (Ṽ,a)+a()(Ṽ,a ū) f =0. (2.7) lim 0 Ṽ,a() z =0. (2.8) The following result is useful for the implementation of our DP. Theorem 3. Let, F, f,andy be as in Theorem 1 and 0 <γ<1. Assume that v H, θ>0 is a constant, α() > 0, and the following inequalities are satisfied: and Then one has: F (v )+α()v f θ, θ > 0, (2.9) C 1 γ F (v ) f C 2 γ, 0 <C 1 <C 2. (2.10) lim v y =0. (2.11) 0 Remark 1. Based on Theorem 3 an algorithm for solving nonlinear equations with monotone Lipschitz continuous operators is outlined in [11]. Remark 2. It is an open problem to choose γ and C 1,C 2 optimal in some sense. Remark 3. Theorem 1 and Theorem 3 do not hold, in general, for γ = 1. Indeed, let Fu = u, p p, p =1,p N(F ):={u H : Fu =0}, f = p, f = p + q, where p, q =0, q =1,Fq =0, q =. For linear operator F we write Fu, rather than F (u). One has Fy = p, wherey = p, is the minimal-norm solution to the equation Fu = p. EquationFV,a + av,a = p + q, has the unique solution V,a = q/a + p/(1 + a). Equation (2.2) is C = q +(ap)/(1 + a). Thisequation yields a = a() =c/(1 c), where c := (C 2 1) 1/2, and we assume c < 1. Thus, lim 0 V,a() = p + c 1 q := v, andfv = p. Therefore v = lim 0 V,a() is not p, i.e., is not the minimal-norm solution to the equation Fu = p. This argument is borrowed from [14], p. 29. If equation (1.1) has a unique solution, then one can prove convergence results (2.3) and (2.11) for γ = The dynamical systems method Let a(t) 0 be a positive and strictly decreasing function. Let V (t) solvethe following equation: F (V (t)) + a(t)v (t) f =0. (2.12) Throughout the paper we assume that equation F (u) =f has a solution in H, possibly nonunique, and y is the minimal-norm solution to this equation. Let f be unknown, but f be given, f f.

6 62 N. S. Hoang & A. G. Ramm The Newton-type DSM In this section we assume that F is a monotone operator, twice Fréchet differentiable, and F (j) (u) M j (R, u 0 ), u B(u 0,R), 0 j 2. (2.13) This assumption is satisfied in many applications. Denote A := F (u (t)), A a := A + ai, (2.14) where I is the identity operator. Let u (t) solve the following Cauchy problem: u = A 1 a(t) [F (u )+a(t)u f ], u (0) = u 0. (2.15) We assume below that F (u 0 ) f >C 1 ζ,wherec 1 > 1andζ (0, 1] are some constants. We also assume without loss of generality that (0, 1). Assume that equation (1.1) has a solution, possibly nonunique, and y is the minimal norm solution to equation (1.1). Recall that we are given the noisy data f, f f. Lemma 4 ([8] Lemma 2.7). Suppose M 1,c 0,andc 1 are positive constants and 0 y H. Then there exist >0 and a function a(t) C 1 [0, ), 0 <a(t) 0, such that the following conditions hold M 1, (2.16) y c 0 a(t) ȧ(t) c 1 a(t) a(t) 2 [ 1 ȧ(t) ], (2.17) 2a(t) a(t) [ 1 ȧ(t) ], (2.18) a(t) F (0) f a2 (0). (2.19) In the proof of Lemma 2.7 in [8] we have demonstrated that conditions (2.17) (2.19) are satisfied for a(t) = d,whereb (0, 1], c, d > 0areconstants,c>6b, (c+t) b and d is sufficiently large. Theorem 5. Assume a(t) = d,whereb (0, 1], c, d > 0 are constants, c> (c+t) b 6b, andd is sufficiently large so that conditions (2.17) (2.19) hold. Assume that F : H H is a twice Fréchet differentiable monotone operator, (2.13) holds, u 0 is an element of H satisfying inequalities u 0 V 0 F (0) f, h(0) = F (u 0 )+a(0)u 0 f 1 a(0) 4 a(0) V (0). (2.20) Then the solution u (t) to problem (2.15) exists on an interval [0,T ], lim 0 T =, and there exists a unique t, t (0,T ) such that lim 0 t = and F (u (t )) f = C 1 ζ, F (u (t)) f >C 1 ζ, t [0,t ), (2.21)

7 The DSM for Solving NOE 63 where C 1 > 1 and 0 <ζ 1. Ifζ (0, 1) and t satisfies (2.21), then lim u (t ) y =0. (2.22) 0 Remark 4. One can choose u 0 satisfying inequalities (2.20). Indeed, if u 0 is a sufficiently close approximation to V (0) the solution to equation (2.12), then inequalities (2.20) are satisfied. Note that the second inequality in (2.20) is a sufficient condition for the following inequality (see also (4.55)) e t 1 2 h(0) 4 a(t) V (t), t 0, (2.23) to hold. In our proof inequality (2.23) (or inequality (4.55)) is used at t = t.the stopping time t is often sufficiently large for the quantity e t 2 a(t ) to be large. Note that V (t) is a strictly increasing function of t (0, ) (see Lemma 20). In this case inequality (2.23) with t = t is satisfied for a wide range of u 0. Condition c > 6b is used in the proof of Lemma 27 (see below) The dynamical system gradient method In this section we assume that F is a monotone operator, twice Fréchet differentiable, and estimates (2.13) hold. Denote A := F (u (t)), A a := A + ai, a = a(t), where I is the identity operator. Let u (t) solve the following Cauchy problem: u = A a(t) [F (u )+a(t)u f ], u (0) = u 0. (2.24) Let us recall the following result: Lemma 6 ([9] Lemma 11). Suppose M 1,c 0,andc 1 are positive constants and 0 y H. Then there exist >0 andafunctiona(t) C 1 [0, ), 0 <a(t) 0, such that ȧ(t) a3 (t) 4, (2.25) and the following conditions hold M 1 y, (2.26) [ a 2 (t) 2 ȧ(t) ], (2.27) a(t) c 0 (M 1 + a(t)) 2a 2 (t) [ ȧ(t) c 1 a(t) a2 (t) a 2 (t) 2 ȧ(t) ], (2.28) 2 a(t) a 2 g(0) < 1. (2.29) (0)

8 64 N. S. Hoang & A. G. Ramm We have demonstrated in the proof of Lemma 11 in [9] that conditions (2.25) (2.29) are satisfied with a(t) = d (c+t), where b (0, 1 b 4 ], c 1, and d > 0 are constants, and d is sufficiently large. Theorem 7. Let a(t) = d, where b (0, 1 (c+t) b 4 ], c 1, and d > 0 are constants, and d is sufficiently large so that conditions (2.25) (2.29) hold. Assume that F : H H is a twice Fréchet differentiable monotone operator, (2.13) holds, u 0 is an element of H, satisfying inequalities F (u 0 ) f > C 1 ζ and h(0) = F (u 0 ) + a(0)u 0 f 1 4 a(0) V (0). (2.30) Then the solution u (t) to problem (2.24) exists on an interval [0, T ], lim 0 T =, and there exists t, t (0, T ), not necessarily unique, such that F (u (t )) f = C 1 ζ, lim t =, (2.31) 0 where C 1 > 1 and 0 < ζ 1 are constants. If ζ (0, 1) and t satisfies (2.31), then lim u (t ) y = 0. (2.32) 0 Remark 5. One can easily choose u 0 satisfying inequality (2.30). Note that inequality (2.30) is a sufficient condition for the following inequality (cf. (4.95)) e ϕ(t) h(0) 1 4 a(t) V (t), t 0, (2.33) to hold. In our proof inequality (2.33) (see also (4.95)) is used at t = t. The stopping time t is often sufficiently large for the quantity e ϕ(t ) a(t ) to be large. In this case inequality (2.33) (cf. (4.95)) with t = t is satisfied for a wide range of u 0. The parameter ζ is not fixed in (2.31). While we could fix it, for example, by setting ζ = 0.9, it is an interesting open problem to propose an optimal in some sense criterion for choosing ζ The simple iteration DSM In this section we assume that F is a monotone operator, Fréchet differentiable, and sup F (u) M 1 (u 0, R). (2.34) u B(u 0,R) Let us consider a version of the DSM for solving equation (1.1): u = ( F (u ) + a(t)u f ), u (0) = u 0, (2.35) where F is a monotone operator. The advantage of this version compared with (2.15) is the absence of the inverse operator in the algorithm, which makes the algorithm (2.35) less expensive than (2.15). On the other hand, algorithm (2.15) converges faster than (2.35) in many

9 The DSM for Solving NOE 65 cases. The algorithm (2.35) is cheaper than the DSM gradient algorithm proposed in (2.24). The advantage of method (2.35), a modified version of the simple iteration method, over the Gauss-Newton method and the version (2.15) of the DSM is the following: neither inversion of matrices nor evaluation of F is needed in a discretized version of (2.35). Although the convergence rate of the DSM (2.35) maybe slower than that of the DSM (2.15), the DSM (2.35) might be faster than the DSM (2.15) for large-scale systems due to its lower computation cost. In this section we investigate a stopping rule based on a discrepancy principle (DP) for the DSM (2.35). The main results of this section is Theorem 9 in which a DP is formulated, the existence of a stopping time t is proved, and the convergence of the DSM with the proposed DP is justified under some natural assumptions. Lemma 8 ([10] Lemma 11). Suppose M 1 and c 1 are positive constants and 0 y H. Then there exist a number > 0 and a function a(t) C 1 [0, ), 0 < a(t) 0, such that and the following conditions hold ȧ(t) a2 (t) 2, (2.36) M 1, (2.37) y 0 [ a(t) ȧ(t) ], (2.38) 2a(t) a(t) ȧ(t) c 1 a(t) a(t) [ a(t) ȧ(t) ], (2.39) 2 a(t) g(0) < 1. (2.40) a(0) It is shown in the proof of Lemma 11 in [10] that conditions (2.36) (2.40) hold for the function a(t) = d, where b (0, 1 (c+t) b 2 ], c 1 and d > 0 are constants, and d is sufficiently large. Theorem 9. Let a(t) = d (c+t), where b (0, 1 b 2 ], c 1 and d > 0 are constants, and d is sufficiently large so that conditions (2.36) (2.40) hold. Assume that F : H H is a Fréchet differentiable monotone operator, condition (2.34) holds, and u 0 is an element of H, satisfying inequalities F (u 0 ) f > C 1 ζ and h(0) = F (u 0 ) + a(0)u 0 f 1 4 a(0) V (0). (2.41) Assume that equation F (u) = f has a solution y B(u 0, R), possibly nonunique, and y is the minimal-norm solution to this equation. Then the solution u (t) to problem (2.35) exists on an interval [0, T ], lim 0 T =, and there exists t, t (0, T ), not necessarily unique, such that F (u (t )) f = C 1 ζ, lim t =, (2.42) 0

10 66 N. S. Hoang & A. G. Ramm where C 1 > 1 and 0 <ζ 1 are constants. If ζ (0, 1) and t satisfies (2.42), then lim u (t ) y =0. (2.43) 0 Remark 6. One can easily choose u 0 satisfying inequality (2.41). Again, inequality (2.41) is a sufficient condition for (2.33) (cf. (4.131)) to hold. In our proof inequality (2.33) is used at t = t. The stopping time t is often sufficiently large for the quantity e ϕ(t) a(t ) to be large. In this case inequality (2.33) with t = t is satisfied for a wide range of u Iterative schemes Let 0 <a n 0 be a positive strictly decreasing sequence. Denote V n := V n, where V n, solves the following equation: Note that if a n := a(t n )thenv n, = V (t n ). F (V n, )+a n V n, f =0. (2.44) Iterative scheme of Newton-type In this section we assume that F is a monotone operator, twice Fréchet differentiable, and F (j) (u) M j (R, u 0 ), u B(u 0,R), 0 j 2. (2.45) Consider the following iterative scheme: u n+1 = u n A 1 n [F (u n)+a n u n f ], A n := F (u n )+a n I, u 0 = u 0, (2.46) where u 0 is chosen so that inequality (2.52) holds. Note that F (u n ) 0sinceF is monotone. Thus, A 1 n 1 a n. Lemma 10 ([7] Lemma 2.5). Suppose M 1,c 0,andc 1 are positive constants and 0 y H. Then there exist >0 and a sequence 0 < (a n ) n=0 0 such that the following conditions hold a n 2a n+1, (2.47) f F (0) a2 0, (2.48) M 1 y, (2.49) a n a n+1 a 2 1 n+1 2c 1, (2.50) a n c a n a n+1 c 1 a n+1 a n+1. (2.51) It is shown in the proof of Lemma 2.5 in [7] that conditions (2.47) (2.51) hold for the sequence a n = d,wherec 1, 0 <b 1, and d is sufficiently large. (c+n) b

11 The DSM for Solving NOE 67 Remark 7. In Lemmas 10 14, one can choose a 0 and so that a0 is uniformly bounded as 0evenifM 1 (R) as R at an arbitrary fast rate. Choices of a 0 and, satisfying this condition, are discussed in [7], [9] and [10]. Let a n and satisfy conditions (2.47) (2.51). Assume that equation F (u) =f has a solution y B(u 0,R), possibly nonunique, and y is the minimal-norm solution to this equation. Let f be unknown but f be given, and f f. Wehavethe following result: d (c+n) b Theorem 11. Assume a n = where c 1, 0 <b 1, andd is sufficiently large so that conditions (2.47) (2.51) hold. Let u n be defined by (2.46). Assume that u 0 is chosen so that F (u 0 ) f >C 1 ζ γ >ζ and g 0 := u 0 V 0 F (0) f. (2.52) a 0 Then there exists a unique n, depending on C 1 and γ (see below), such that F (u n ) f C 1 γ, C 1 γ < F (u n ) f, n <n, (2.53) where C 1 > 1, 0 <γ 1. Let 0 < ( m ) m=1 be a sequence such that m 0. IfN is a cluster point of the sequence n m satisfying (2.53), then lim u n m m = u, (2.54) where u is a solution to the equation F (u) =f. If lim n m =, (2.55) m and γ (0, 1), then lim u n m m y =0. (2.56) Note that by Remark 9, inequality (2.52) is satisfied with u 0 = An iterative scheme of gradient-type In this section we assume that F is a monotone operator, twice Fréchet differentiable, and estimates (2.45) hold. Consider the following iterative scheme: u n+1 = u n α n A n[f (u n )+a n u n f ], A n := F (u n )+a n I, u 0 = u 0, (2.57) where u 0 is chosen so that inequality (2.65) holds, and {α n } n=1 is a positive sequence such that 2 0 < α α n a 2 n +(M 1 + a n ) 2, A n M 1 + a n. (2.58) It follows from this condition that 1 α n A a n A an = sup a 2 n (M1+an)2 1 α n 1 α n a 2 n. (2.59)

12 68 N. S. Hoang & A. G. Ramm Note that F (u n ) 0 since F is monotone. Lemma 12 ([9] Lemma 12). Suppose M 1, c 0, c 1 and α are positive constants and 0 y H. Then there exist > 0 and a sequence 0 < (a n ) n=0 0 such that the following conditions hold a n a n+1 2, (2.60) f F (0) a3 0, (2.61) M 1 y, (2.62) c 0 (M 1 + a 0 ) 1 2, (2.63) a 2 n αa4 n 2 + a n a n+1 c 1 a2 n+1 a n+1. (2.64) It is shown in the proof of Lemma 12 in [9] that the sequence (a n ) n=0 satisfying conditions (2.60) (2.64) can be chosen of the form a n = d (c+n), where c 1, 0 < b b 1 4, and d is sufficiently large. Assume that equation F (u) = f has a solution in B(u 0, R), possibly nonunique, and y is the minimal-norm solution to this equation. Let f be unknown but f be given, and f f. We prove the following result: d (c+n) b where c 1, 0 < b 1 4, and d is sufficiently Theorem 13. Assume a n = large so that conditions (2.60) (2.64) hold. Let u n be defined by (2.57). Assume that u 0 is chosen so that F (u 0 ) f > C 1 ζ > and Then there exists a unique n such that g 0 := u 0 V 0 F (0) f a 0. (2.65) F (u n ) f C 1 ζ, C 1 ζ < F (u n ) f, n < n, (2.66) where C 1 > 1, 0 < ζ 1. Let 0 < ( m ) m=1 be a sequence such that m 0. If the sequence {n m := n m } m=1 is bounded, and {n mj } j=1 is a convergent subsequence, then where ũ is a solution to the equation F (u) = f. If lim u n mj = ũ, (2.67) j and ζ (0, 1), then lim n m =, (2.68) m lim u n m y = 0. (2.69) m It is pointed out in Remark 9 that inequality (2.65) is satisfied with u 0 = 0.

13 The DSM for Solving NOE A simple iteration method In this section we assume that F is a monotone operator, Fréchet differentiable. Consider the following iterative scheme: u n+1 = u n α n [F (u n )+a n u n f ], u 0 = u 0, (2.70) where u 0 is chosen so that inequality (2.77) holds, and {α n } n=1 is a positive sequence such that 2 0 < α α n a n +(M 1 + a n ), M 1(u 0,R)= sup F (u). (2.71) u B(u 0,R) It follows from this condition that 1 α n (J n + a n ) = sup 1 α n 1 α n a n. (2.72) a n M 1+a n Here, J n is an operator in H such that J n = Jn 0and J n M 1, u B(u 0,R). A specific choice of J n is made in formula (4.186) below. Lemma 14 ([10] Lemma 12). Suppose M 1, c 1 and α are positive constants and 0 y H. Then there exist a number >0 andasequence0 < (a n ) n=0 0 such that the following conditions hold a n a n+1 2, (2.73) f F (0) a2 0, (2.74) M 1 y, (2.75) a n αa2 n + a n a n+1 c 1 a n+1 a n+1. (2.76) It is shown in the proof of Lemma 12 in [10] that conditions (2.73) (2.76) hold for the sequence a n = d,wherec 1, 0 <b 1 (c+n) b 2,andd is sufficiently large. Let a n and satisfy conditions (2.73) (2.76). Assume that equation F (u) =f has a solution y B(u 0,R), possibly nonunique, and y is the minimal-norm solution to this equation. Let f be unknown but f be given, and f f. Weprove the following result: Theorem 15. Assume that F is a Fréchet differentiable monotone operator and F is selfadjoint. Assume a n = d where c 1, 0 <b 1 (c+n) b 2,andd is sufficiently large so that conditions (2.73) (2.76) hold. Let u n be defined by (2.70). Assume that u 0 is chosen so that F (u 0 ) f >C 1 ζ >and g 0 := u 0 V 0 F (0) f. (2.77) a 0 Then there exists a unique n such that F (u n ) f C 1 ζ, C 1 ζ < F (u n ) f, n <n, (2.78)

14 70 N. S. Hoang & A. G. Ramm where C 1 > 1, 0 <ζ 1. Let 0 < ( m ) m=1 be a sequence such that m 0. If the sequence {n m := n m } m=1 is bounded, and {n m j } j=1 is a convergent subsequence, then lim u n mj =ũ, (2.79) j where ũ is a solution to the equation F (u) =f. If and ζ (0, 1), then lim n m =, (2.80) m lim u n m y =0. (2.81) m Remark 8. If H is a complex Hilbert space, then a bounded non-negative-definite operator A = F,theFréchet derivative of a monotone operator F,isselfadjoint;if A is a bounded linear operator defined on all of H and Au, u 0 for all u H, then A is selfadjoint. This is not ( true, ) in general, if H is a real Hilbert space. 11 Example: H = R 2, A is matrix.thena is not selfadjoint, but Au, u = 01 u u 1 u 2 + u for all u 1,u 2 R. Remark 9. In Theorems we choose u 0 H such that g 0 := u 0 V 0 F (0) f. (2.82) a 0 It is easy to choose u 0 satisfying this condition. Indeed, if, for example, u 0 =0, then by Lemma 20 in Sec. 3.2 (see below) one gets g 0 = V 0 = a 0 V 0 a 0 If (2.82) and either (2.48) or (2.74) hold then F (0) f a 0. (2.83) g 0 a 0. (2.84) This inequality is used in the proof of Theorems 11 and 15. If (2.82) and (2.61) hold, then g 0 a2 0. (2.85) This inequality is used in the proof of Theorem Nonlinear inequalities A nonlinear differential inequality In [15] the following differential inequality ġ(t) γ(t)g(t)+α(t)g 2 (t)+β(t), t τ 0, (2.86)

15 The DSM for Solving NOE 71 was studied and applied to various evolution problems. In (2.86) α(t),β(t),γ(t) and g(t) are continuous non-negative functions on [τ 0, ) whereτ 0 is a fixed number. In [15], an upper bound for g(t) is obtained under some conditions on α, β, γ. In [12] the following generalization of (2.86): ġ(t) γ(t)g(t)+α(t)g p (t)+β(t), t τ 0, p > 1, (2.87) is studied. We have the following result: Theorem 16 ([12] Theorem 1). Let α(t),β(t) and γ(t) be continuous functions on [τ 0, ) and α(t) > 0, t τ 0. Suppose there exists a function μ(t) > 0, μ C 1 [τ 0, ), such that α(t) μ p (t) + β(t) 1 [ γ(t) μ(t) ]. (2.88) μ(t) μ(t) Let g(t) 0 beasolutiontoinequality(2.87) such that μ(τ 0 )g(τ 0 ) < 1. (2.89) Then g(t) exists globally and the following estimate holds: 0 g(t) < 1 μ(t), t τ 0. (2.90) Consequently, if lim t μ(t) =, then lim g(t) =0. (2.91) t Theorem 16 remains valid if the sign < in (2.89) and (2.90) is replaced by the sign (see Theorem 2 in [12]). When p = 2 we have the following corollary: Corollary 17 ([15] p. 97). Suppose there exists a monotonically growing function μ(t), such that μ C 1 [τ 0, ), μ > 0, lim t μ(t) =, 0 α(t) μ(t) [ 2 β(t) 1 2μ(t) γ(t) μ(t) ], u := du μ(t) dt, (2.92) [ γ(t) μ(t) ], (2.93) μ(t) where α(t),β(t) and γ(t) are continuous non-negative functions on [τ 0, ), τ 0 0. Let g(t) 0 beasolutiontoinequality(2.87) such that Then g(t) exists globally and the following estimate holds: μ(τ 0 )g(τ 0 ) < 1. (2.94) 0 g(t) < 1 μ(t), t τ 0. (2.95)

16 72 N. S. Hoang & A. G. Ramm Consequently, if lim t μ(t) =, then lim g(t) =0. t A discrete version of the nonlinear inequality Theorem 18 ([12] Theorem 4). Let α n,γ n and g n be non-negative sequences of numbers, and the following inequality holds: g n+1 g n γ n g n + α n gn p + β n, h n h n > 0, 0 <h n γ n < 1, (2.96) or, equivalently, g n+1 g n (1 h n γ n )+α n h n g p n + h nβ n, h n > 0, 0 <h n γ n < 1. (2.97) If there is a monotonically growing sequence of positive numbers (μ n ) n=1, such that the following conditions hold: then α n μ p + β n 1 n μ n ( γ n μ ) n+1 μ n, (2.98) μ n h n g 0 1 μ 0, (2.99) 0 g n 1 n 0. (2.100) μ n Therefore, if lim n μ n =, thenlim n g n =0. 3. Auxiliary Results 3.1. Auxiliary results from the theory of monotone operators Recall the following result (see e.g., [15], p. 112): Lemma 19. Assume that equation (1.1) is solvable, y is its minimal-norm solution, assumption (1.2) holds, and F is continuous. Then lim V a y =0, (3.1) a 0 where V a := V 0,a solves equation (2.1) with = Auxiliary results for the regularized equation (2.1) Lemma 20 ([11] Lemma 2). Assume F (0) f > 0. Leta>0, andf be monotone. Denote ψ(a) := V,a, φ(a) :=aψ(a) = F (V,a ) f, where V,a solves (2.1). Thenψ(a) is decreasing, and φ(a) is increasing (in the strict sense).

17 The DSM for Solving NOE 73 Lemma 21 ([11] Lemma 3). If F is monotone and continuous, then V,a = O( 1 a ) as a,and lim F (V,a) f = F (0) f. (3.2) a Lemma 22 ([11] Lemma 4). Let C > 0 and γ (0, 1] be constants such that C γ >. Suppose that F (0) f >C γ. Then, there exists a unique a() > 0 such that F (V,a() ) f = C γ. Lemma 23. If F is monotone and a 0, then ( ) max F (u) F (v),a u v F (u) F (v)+a(u v), u, v H. (3.3) Proof. Denote w := F (u) F (v)+a(u v), h := w. (3.4) Since F (u) F (v),u v 0, one obtains from two equations and w, u v = F (u) F (v)+a(u v),u v, (3.5) w, F(u) F (v) = F (u) F (v) 2 + a u v, F(u) F (v), (3.6) the following inequalities: and a u v 2 w, u v u v h, (3.7) F (u) F (v) 2 w, F(u) F (v) h F (u) F (v). (3.8) Inequalities (3.7) and (3.8) imply Lemma 23 is proved. a u v h, F (u) F (v) h. (3.9) Lemma 24. Let t 0 > 0 satisfy a(t 0 ) = 1 y, C 1 C > 1. (3.10) Then, F (V (t 0 )) f C, (3.11) and V ȧ ( a y 1+ 1 ), t t 0. (3.12) C 1

18 74 N. S. Hoang & A. G. Ramm Proof. This t 0 exists and is unique since a(t) > 0 monotonically decays to 0 as t.sincea(t) > 0 monotonically decays, one has: a(t) 1 C 1 y, 0 t t 0. (3.13) By Lemma 22 there exists t 1 > 0 such that F (V (t 1 )) f = C, F(V (t 1 )) + a(t 1 )V (t 1 ) f =0. (3.14) We claim that t 1 [0,t 0 ]. Indeed, from (3.14) and (3.30) one gets ( C = a(t 1 ) V (t 1 ) a(t 1 ) y + ) = a(t 1 ) y +, C > 1, a(t 1 ) so a(t 1) y C 1. Thus, a(t 1 ) y C 1 = a(t 0 ). Since a(t) 0, one has t 1 t 0. It follows from the inequality t 1 t 0, Lemma 20 and the first equality in (3.14) that F (V (t 0 )) f F(V (t 1 )) f = C. Differentiating both sides of (2.12) with respect to t, oneobtains A a(t) V = ȧv. This and the relations A a := F (u)+ai, F (u) :=A 0, imply V ȧ A 1 a(t) V ȧ a V ȧ ( y + ) ȧ ( a a a y 1+ 1 ), t t 0. C 1 (3.15) Lemma 24 is proved. Lemma 25. Let n 0 > 0 satisfy the inequality: > 1 a n0+1 C 1 y, a n0 C > 1. (3.16) Then, F (V n0+1) f C, (3.17) ( V n y 1+ 2 ), 0 n n 0 +1, (3.18) C 1

19 The DSM for Solving NOE 75 and V n V n+1 a ( n a n+1 y 1+ 2 ), 0 n n (3.19) a n+1 C 1 Proof. The number n 0, satisfying (3.16), exists and is unique since a n > 0 monotonically decays to 0 as n. One has 2, n 0. This and inequality (3.16) imply a n a n y > C 1 a n0 a n0+1 > 1 C 1 y a n0, C > 1. (3.20) Thus, 2 C 1 y >, n n (3.21) a n It follows from Lemma 22 that there exists n 1 > 0 such that F (V n1+1) f C < F (V n1 ) f, (3.22) where V n solves the equation F (V n )+a n V n f =0.We claim that n 1 [0,n 0 ]. Indeed, one has F (V n1 ) f = a n1 V n1,and V n1 y + a n1 (cf. (3.30)), so ( C < a n1 V n1 a n1 y + ) = a n1 y +, C > 1. (3.23) Therefore, < a n 1 y C 1. (3.24) From (3.24) and (3.16) one gets < y a n1 C 1 <. (3.25) a n0+1 Since a n decreases monotonically, inequality (3.25) implies n 1 n 0.This,thefirst inequality in (3.22) and Lemma 20 imply a n1 F (V n0+1) f F (V n1+1) f C. (3.26) 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, n 0. By (3.30), V n y + a n, and, by (3.21), implies (3.18). From (3.18) and (3.27) one obtains V n V n+1 a n a n+1 V n a n a n+1 a n+1 a n+1 a n 2 y C 1 (3.27) for all n n This y ( 1+ 2 C 1 ), n n (3.28)

20 76 N. S. Hoang & A. G. Ramm Lemma 25 is proved. Lemma 26. Let V a := V,a =0, so F (V a )+av a f = 0. Let y be the minimal-norm solution to equation (1.1). Then and V,a V a a, V a y, a > 0, (3.29) V,a V a + a y +, a > 0. (3.30) a Proof. From (2.1) one gets F (V,a ) F (V a ) + a(v,a V a ) = f f. Multiply this equality by (V,a V a ) and use (1.2) to obtain V,a V a f f, V,a V a = F (V,a ) F (V a ) + a(v,a V a ), V,a V a a V,a V a 2. This implies the first inequality in (3.29). Let us derive a uniform, with respect to a, bound on V a. From the equation and the monotonicity of F one gets This implies the desired bound: F (V a ) + av a F (y) = 0, 0 = F (V a ) + av a F (y), V a y a V a, V a y. V a y, a > 0. (3.31) Similar arguments can be found in [15], p Inequalities (3.30) follow from (3.29) and (3.31) and the triangle inequality. Lemma 26 is proved. Lemma 27 ([8] Lemma 2.11). Let a(t) = has e t 2 t 0 Lemma 28 ([9] Lemma 9). Let a(t) = Define ϕ(t) = t a 2 (s) 0 2 ds. Then, one has d (c+t) b where d, c, b > 0, c 6b. One e s 2 ȧ(s) V (s) ds 1 2 a(t) V (t), t 0. (3.32) d (c+t) b where b (0, 1 4 ], d2 c 1 2b 6b. t e ϕ(t) e ϕ(t) ȧ(s) V (s) ds 1 2 a(t) V (t). (3.33) 0

21 The DSM for Solving NOE 77 Lemma 29 ([10] Lemma 9). Let a(t) = Define ϕ(t) = t 0 a(s) 2 ds. Then, one has 4. Proofs of the Basic Results d (c+t) b where b (0, 1 2 ], dc1 b 6b. t e ϕ(t) e ϕ(t) ȧ(s) V (s) ds 1 2 a(t) V (t). (3.34) 4.1. Proofs of the discrepancy principles Proof of Theorem 1 0 Proof. The existence and uniqueness of a() follow from Lemma 22. Let us show that lim a() = 0. (4.1) 0 The triangle inequality, the first inequality in (3.29), equality (2.2) and equality (2.1) imply a() V a() a() ( V,a() V a() + V,a() ) where V a solves (2.1) with = 0. From inequality (4.2), one gets + a() V,a() = + C γ, (4.2) lim a() V a() = 0. (4.3) 0 It follows from Lemma 20 with f = f, i.e., = 0, that the function φ 0 (a) := a V a is non-negative and strictly increasing on (0, ). This and relation (4.3) imply: From (2.2) and (3.30), one gets Thus, one gets: lim a() = 0. (4.4) 0 C γ = a V,a a() y +. (4.5) C γ a() y. (4.6) If γ < 1 then C 1 γ > 0 for sufficiently small. This implies: 0 lim 0 a() lim 0 1 γ y = 0. (4.7) C 1 γ By the triangle inequality and the first inequality (3.29), one has V,a() y V a() y + V a() V,a() V a() y + Relation (2.3) follows from (4.4), (4.7), (4.8) and Lemma 19. a(). (4.8)

22 78 N. S. Hoang & A. G. Ramm Proof of Theorem 3 Proof. By Lemma 23 a u v F (u) F (v)+au av, v, u H, a >0. (4.9) Using inequality (4.9) with v = v and u = V,α(), equation (1.4) with a = α(), and inequality (2.9), one gets α() v V,α() F(v ) F (V,α() )+α()v α()v,α() = F (v )+α()v f θ. Therefore, (4.10) v V,α() θ α(). (4.11) Using the triangle inequality, (3.30) and (4.11), one gets: α() v α() V,α() + α() v V,α() α() y + + θ. (4.12) From the triangle inequality and inequalities (2.9) and (2.10) one obtains: α() v F (v ) f F (v )+α()v f C 1 γ θ. (4.13) Inequalities (4.12) and (4.13) imply C 1 γ θ θ + α() y +. (4.14) This inequality and the fact that C 1 1 γ 2θ 1 γ > 0 for sufficiently small and 0 <γ<1imply α() 1 γ y C 1 1 γ, 0 < 1. (4.15) 2θ1 γ Thus, one obtains lim 0 =0. (4.16) α() From the triangle inequality and inequalities (2.9), (2.10) and (4.11), one gets α() V,α() F(v ) f + F (v )+α()v f + α() v V,α() C 2 γ + θ + θ. This inequality implies lim α() V,α() =0. (4.17) 0 The triangle inequality and inequality (3.29) imply α V α α ( V,α V α + V,α ) (4.18) + α V,α.

23 The DSM for Solving NOE 79 From (4.18) and (4.17), one gets lim α() V α() =0. (4.19) 0 It follows from Lemma 20 with f = f, i.e., = 0, that the function φ 0 (a) :=a V a is non-negative and strictly increasing on (0, ). This and relation (4.19) imply lim α() =0. (4.20) 0 From the triangle inequality and inequalities (4.11) and (3.29) one obtains v y v V,α() + V,α() V α() + V α() y θ α() + α() + V (4.21) α() y, where V α() solves equation (3) with a = α() andf = f. The conclusion (2.11) follows from (4.16), (4.20), (4.21) and Lemma 19. Theorem 3 is proved Proofs of convergence of the dynamical systems method ProofofTheorem5 Proof. Denote Let C := C (4.22) 2 w := u V, g(t) := w. (4.23) From (4.23) and (2.15) one gets ẇ = V A 1 a(t)[ F (u ) F (V )+a(t)w ]. (4.24) We use Taylor s formula and get: F (u ) F (V )+aw = A a w + K, K M 2 2 w 2, (4.25) where K := F (u ) F (V ) Aw, andm 2 is the constant from the estimate (2.13) and A a := A + ai. Multiplying (4.24) by w and using (4.25) one gets gġ g 2 + M 2 2 A 1 a(t) g3 + V g. (4.26) Since 0 <a(t) 0, there exists t 0 > 0 such that a(t 0 ) = 1 y, C > 1. (4.27) C 1 This and Lemma 24 imply that inequalities (3.11) and (3.12) hold. Since g 0, inequalities (4.26) and (3.12) imply, for all t [0,t 0 ], that ġ g(t)+ c 0 a(t) g2 + ȧ a(t) c 1, c 0 = M ( 2 2, c 1 = y 1+ 1 ). (4.28) C 1

24 80 N. S. Hoang & A. G. Ramm Inequality (4.28) is of the type (2.87) with γ(t) =1, α(t) = c 0 a(t), β(t) =c ȧ 1 a(t). (4.29) Let us check assumptions (2.92) (2.94). Take μ(t) = a(t), (4.30) where = const > 0 and satisfies conditions (2.16) (2.19) in Lemma 4. It follows that inequalities (2.92) (2.94) hold. Since u 0 satisfies the first inequality in (2.20), one gets g(0) a(0), by Remark 9. This, inequalities (2.92) (2.94), and Corollary 17 yield g(t) < a(t), t t 0, g(t) := u (t) V (t). (4.31) Therefore, F (u (t)) f F(u (t)) F (V (t)) + F (V (t)) f M 1 g(t)+ F (V (t)) f (4.32) M 1a(t) + F (V (t)) f, t t 0. From (3.11) one gets F (V (t 0 )) f C. (4.33) This, inequality (4.32), the inequality M1 y (see (2.16)), the relation (4.27), and the definition C 1 =2C 1 (see (4.22)), imply F (u (t 0 )) f M 1a(t 0 ) + C M (4.34) 1(C 1) + C (C 1) + C = C 1. y Thus, if F (u (0)) f >C 1 ζ, 0 <ζ 1, (4.35) then, by the continuity of the function t F (u (t)) f on [0, ), there exists t (0,t 0 ) such that F (u (t )) f = C 1 ζ (4.36) for any given ζ (0, 1], and any fixed C 1 > 1. Let us prove (2.22). From (4.32) with t = t, and from (3.30), one gets C 1 ζ a(t ) M 1 + a(t ) V (t ) (4.37) a(t ) M 1 + y a(t )+.

25 The DSM for Solving NOE 81 Thus, for sufficiently small, onegets C ζ a(t )( M1 where C <C 1 is a constant. Therefore, lim 0 We claim that a(t ) lim 0 1 ζ C ( M1 ) + y, C >0, (4.38) ) + y =0, 0 <ζ<1. (4.39) lim t =. (4.40) 0 Let us prove (4.40). Using (2.15), one obtains: d ( ) F (u )+au f = Aa u +ȧu = ( ) F (u )+au f +ȧu. (4.41) dt This and (2.12) imply: d [ F (u ) F (V )+a(u V ) ] = [ F (u ) F (V )+a(u V ) ] +ȧu. (4.42) dt Denote v := v(t) :=F (u (t)) F (V (t)) + a(t)(u (t) V (t)), h := h(t) := v. (4.43) Multiplying (4.42) by v, one obtains Thus, hḣ = h2 + v, ȧ(u V ) +ȧ v, V h 2 + h ȧ u V + ȧ h V, h 0. (4.44) Note that from inequality (3.3) one has Inequalities (4.45) and (4.46) imply ( ḣ h ḣ h + ȧ u V + ȧ V. (4.45) a u V h, F (u ) F (V ) h. (4.46) 1 ȧ a ) + ȧ V. (4.47) Since 1 ȧ a 1 2 because c 2b, it follows from inequality (4.47) that ḣ 1 2 h + ȧ V. (4.48) Inequality (4.48) implies: h(t) h(0)e t 2 + e t 2 t 0 e s 2 ȧ V ds. (4.49)

26 82 N. S. Hoang & A. G. Ramm From (4.49) and the second inequality in (4.46), one gets F (u (t)) F (V (t)) h(0)e t 2 + e t 2 This and the triangle inequality imply t F (u (t)) f F (V (t)) f F (V (t)) F (u (t)) By Lemma 27 one gets a(t) V (t) h(0)e t 2 e t 2 t 0 t 0 e s 2 ȧ V ds. (4.50) e s 2 ȧ V ds. (4.51) 1 2 a(t) V (t) e t 2 0 e s 2 ȧ V (s) ds. (4.52) From the second inequality in (2.20), one gets h(0)e t a(0) V (0) e t 2, t 0. (4.53) Since a(t) = Therefore, d (c+t) b, b (0, 1], c 1, 2b <c,onegets e t 2 a(0) a(t). (4.54) e t 1 2 h(0) 4 a(t) V (0) 1 4 a(t) V (t), t 0, (4.55) where we have used the inequality V (t) V (t ) for t<t, established in Lemma 20. From (4.36) and (4.51), (4.52), (4.55), one gets C 1 ζ = F (u (t )) f 1 4 a(t ) V (t ). (4.56) It follows from the triangle inequality and the first inequality in (3.29) that a(t) V (t) a(t) V (t) +. This and (4.56) imply ( ) 0 lim a(t ) V (t ) lim 4C 1 ζ + =0. (4.57) 0 0 Since V (t) increases (see Lemma 20), the above formula implies lim 0 a(t )=0. Since 0 <a(t) 0, it follows that lim 0 t =, i.e., (4.40) holds. It is now easy to finish the proof of Theorem 5. From the triangle inequality and inequalities (4.31) and (3.29) one obtains u (t ) y u (t ) V (t ) + V (t ) V (t ) + V (t ) y a(t ) + a(t ) + V (t ) y. (4.58)

27 The DSM for Solving NOE 83 Note that V (t) :=V (t) =0 and V (t) solves (2.12). Note that V (t )=V 0,a(t ) (see equation (2.12)). From (4.39), (4.40), inequality (4.58) and Lemma 19, one obtains (2.22). Theorem 5 is proved. Remark 10. The trajectory u (t) remains in the ball B(u 0,R):={u : u u 0 < R} for all t t,wherer does not depend on as 0. Indeed, estimates (4.31), (3.30) and (3.13) imply: u (t) u 0 u (t) V (t) + V (t) + u 0 a(0) + C y C 1 + u 0 := R, t t. (4.59) Here we have used the fact that t <t 0 (see Lemma 24). Since one can choose a(t) and so that a(0) is uniformly bounded as 0 regardless of the growth of M 1 (see Remark 7) one concludes that R can be chosen independent of and M ProofofTheorem7 Proof. Denote Let C := C (4.60) 2 From (4.61) and (2.24) one gets We use Taylor s formula and get: w := u V, g(t) := w. (4.61) ẇ = V A a(t)[ F (u ) F (V )+a(t)w ]. (4.62) F (u ) F (V )+aw = A a w + K, K M 2 2 w 2, (4.63) where K := F (u ) F (V ) Aw, andm 2 is the constant from the estimate (2.13) and A a := A + ai. Multiplying (4.62) by w and using (4.63) one gets gġ a 2 g 2 + M 2(M 1 + a) g V g, g := g(t) := w(t), (4.64) where the estimates: A aa a w, w a 2 g 2 and A a M 1 + a were used. Note that the inequality A aa a w, w a 2 g 2 is true if A 0. Since F is monotone and differentiable (see (1.2)), one has A := F (u ) 0. Let t 0 > 0 be such that a(t 0 ) = 1 y, C > 1, (4.65) C 1 as in (3.10). It follows from Lemma 24 that inequalities (3.11) and (3.12) hold.

28 84 N. S. Hoang & A. G. Ramm Since g 0, inequalities (4.64) and (3.12) imply, for all t [0,t 0 ], that ġ(t) a 2 (t)g(t)+c 0 (M 1 +a(t))g 2 (t)+ ȧ(t) a(t) c 1, c 0 = M ( 2 2,c 1 = y 1+ 1 ). C 1 (4.66) Inequality (4.66) is of the type (2.87) with γ(t) =a 2 ȧ(t) (t), α(t) =c 0 (M 1 + a(t)), β(t) =c 1 a(t). (4.67) Let us check assumptions (2.92) (2.94). Take μ(t) = a 2, =const. (4.68) (t) ByLemma6thereexist and a(t) such that conditions (2.25) (2.29) hold. This implies that inequalities (2.92) (2.94) hold. Thus, Corollary 17 yields g(t) < a2 (t), t t 0. (4.69) Note that inequality (4.69) holds for t = 0 since (2.29) holds. Therefore, F (u (t)) f F(u (t)) F (V (t)) + F (V (t)) f M 1 g(t)+ F (V (t)) f (4.70) M 1a 2 (t) + F (V (t)) f, t t 0. It follows from Lemma 20 that F (V (t)) f is decreasing. Since t 1 t 0,onegets F (V (t 0 )) f F(V (t 1 )) f = C. (4.71) This, inequality (4.70), the inequality M1 y (see (2.26)), the relation (4.65), and the definition C 1 =2C 1 (see (4.60)) imply F (u (t 0 )) f M 1a 2 (t 0 ) + C M (4.72) 1(C 1) + C (2C 1) = C 1. y We have used the inequality a 2 (C 1) (t 0 ) a(t 0 )= (4.73) y which is true if is sufficiently small, or, equivalently, if t 0 is sufficiently large. Thus, if then there exists t (0,t 0 ) such that F (u (0)) f >C 1 ζ, 0 <ζ 1, (4.74) for any given ζ (0, 1], and any fixed C 1 > 1. F (u (t )) f = C 1 ζ (4.75)

29 The DSM for Solving NOE 85 Let us prove (2.32). If this is done, then Theorem 7 is proved. First, we prove that lim 0 a(t ) =0. From (4.70) with t = t, (2.12) and (3.30), one gets C 1 ζ a 2 (t ) M 1 + a(t ) V (t ) a 2 (t ) M 1 + y a(t )+. Thus, for sufficiently small, onegets ( C ζ C 1 ζ M1 a(0) a(t ) where C <C 1 is a constant. Therefore, ( lim 0 a(t ) lim M1 a(0) 0 Secondly, we prove that 1 ζ C (4.76) ) + y, C >0, (4.77) ) + y =0, 0 <ζ<1. (4.78) lim t =. (4.79) 0 Using (2.24), one obtains: d ( ) F (u )+au f = Aa u +ȧu = A a A ( ) a F (u )+au f +ȧu. (4.80) dt This and (2.12) imply: d [ F (u ) F (V )+a(u V ) ] = A a A [ a F (u ) F (V )+a(u V ) ] +ȧu. (4.81) dt Denote v := F (u ) F (V )+a(u V ), h := h(t) := v(t). (4.82) Multiplying (4.81) by v and using monotonicity of F,oneobtains hḣ = A aa a v, v + v, ȧ(u V ) +ȧ v, V h 2 a 2 + h ȧ u V + ȧ h V, h 0. (4.83) Again, we have used the inequality A a A a a 2, which holds for A 0, i.e., monotone operators F.Thus, From inequality (3.3) we have Inequalities (4.84) and (4.85) imply ( ḣ h ḣ ha 2 + ȧ u V + ȧ V. (4.84) a u V h, F (u ) F (V ) h. (4.85) a 2 ȧ a ) + ȧ V. (4.86)

30 86 N. S. Hoang & A. G. Ramm Since a 2 ȧ a 3a2 4 > a2 2 Inequality (4.87) implies: Denote h(t) h(0)e t a 2 (s) 0 2 ds + e t a 2 (s) 0 2 ds From (4.88) and (4.85), one gets by inequality (2.25), it follows from inequality (4.86) that ḣ a2 2 h + ȧ V. (4.87) ϕ(t) := t 0 t 0 e s 0 a 2 (ξ) 2 dξ ȧ(s) V (s) ds. (4.88) a 2 (s) ds. (4.89) 2 t F (u (t)) F (V (t)) h(0)e ϕ(t) + e ϕ(t) e ϕ(s) ȧ(s) V (s) ds. (4.90) This and the triangle inequality imply F (u (t)) f F(V (t)) f F(V (t)) F (u (t)) t a(t) V (t) h(0)e ϕ(t) e ϕ(t) e ϕ(s) ȧ V ds. From Lemma 28 it follows that 1 2 a(t) V (t) e ϕ(t) From (2.30) one gets t (4.91) e ϕ(s) ȧ V (s) ds. (4.92) h(0)e ϕ(t) 1 4 a(0) V (0) e ϕ(t), t 0. (4.93) If c 1and2b c 2 1, then it follows that e ϕ(t) a(0) a(t). (4.94) Indeed, inequality a(0) a(t)e ϕ(t) is obviously true for t =0,and ( a(t)e ϕ(t)) t 0, provided that c 1and2b c 2 1. Inequalities (4.93) and (4.94) imply e ϕ(t) h(0) 1 4 a(t) V (0) 1 4 a(t) V (t), t 0, (4.95) where we have used the inequality V (t) V (t ) for t t, established in Lemma 20. From (4.75) and (4.91), (4.92), (4.95), one gets C ζ = F (u (t )) f 1 4 a(t ) V (t ). (4.96) It follows from the triangle inequality and the first inequality in (3.29) that a(t) V (t) a(t) V (t) +. (4.97)

31 The DSM for Solving NOE 87 From (4.97) and (4.96) one gets ( 0 lim a(t ) V (t ) lim 4C ζ + ) =0. (4.98) 0 0 Since V (t) is increasing, this implies lim 0 a(t ) = 0. Since 0 <a(t) 0, it follows that (4.79) holds. From the triangle inequality and inequalities (4.69) and (3.29) one obtains u (t ) y u (t ) V + V (t ) V (t ) + V (t ) y a2 (t ) + a(t ) + V (t ) y, (4.99) where V (t) := V (t) =0 and V (t) solves (2.12). From (4.78), (4.79), inequality (4.99) and Lemma 19, one obtains (2.32). Theorem 7 is proved. By the arguments, similar to the ones in the proof of Theorems or in Remark 10, one can show that the trajectory u (t) remains in the ball B(u 0,R):= {u : u u 0 <R} for all t t,wherer does not depend on as ProofofTheorem9 Proof. Denote Let C := C (4.100) 2 From (4.101) and (2.35) one gets w := u V, g := g(t) := w(t). (4.101) ẇ = V [ F (u ) F (V )+a(t)w ]. (4.102) Multiplying (4.102) by w and using (1.2) one gets gġ ag 2 + V g. (4.103) Let t 0 > 0 be such that a(t 0 ) = 1 y, C 1 C > 1. (4.104) This t 0 exists and is unique since a(t) > 0 monotonically decays to 0 as t. It follows from inequality (4.104) and Lemma 24 that inequalities (3.11) and (3.12) hold. Since g 0, inequalities (4.103) and (3.12) imply ġ a(t)g(t)+ ȧ(t) a(t) c 1, ( c 1 = y 1+ 1 ). C 1 (4.105)

32 88 N. S. Hoang & A. G. Ramm Inequality (4.105) is of the type (2.87) with ȧ(t) γ(t) =a(t), α(t) =0, β(t) =c 1 a(t). (4.106) Let us check assumptions (2.92) (2.94). Take μ(t) =, =const. (4.107) a(t) By Lemma 8 there exist and a(t) such that conditions (2.37) (2.40) hold. It follows that inequalities (2.92) (2.94) hold. Thus, Corollary 17 yields g(t) < a(t), t t 0. (4.108) The triangle inequality and inequality (4.108) imply F (u (t)) f F(u (t)) F (V (t)) + F (V (t)) f M 1 g(t)+ F (V (t)) f (4.109) M 1a(t) + F (V (t)) f, t t 0. Inequality (3.11), inequality (4.109), the inequality M1 y (see (2.37)), the relation (4.104), and the definition C 1 =2C 1 (see (4.100)) imply F (u (t 0 )) f M 1a(t 0 ) + C M (4.110) 1(C 1) + C (C 1) + C = C 1. y Thus, if then there exists t (0,t 0 ) such that F (u (0)) f >C 1 ζ, 0 <ζ 1, (4.111) F (u (t )) f = C 1 ζ (4.112) for any given ζ (0, 1], and any fixed C 1 > 1. Let us prove (2.43). If this is done, then Theorem 9 is proved. First, we prove that lim 0 a(t ) =0. From (4.109) with t = t, and from (3.30), one gets C 1 ζ a(t ) M 1 + a(t ) V (t ) a(t ) M 1 + y a(t )+. Thus, for sufficiently small, onegets C ζ a(t )( M1 (4.113) ) + y, C >0, (4.114)

33 The DSM for Solving NOE 89 where C <C 1 is a constant. Therefore, lim 0 a(t ) lim 0 Secondly, we prove that 1 ζ C ( M1 ) + y =0, 0 <ζ<1. (4.115) lim t =. (4.116) 0 Using (2.35), one obtains: d ( ) ( ) F (u )+au f = Aa u +ȧu = A a F (u )+au f +ȧu, (4.117) dt where A a := F (u )+a. This and (2.12) imply: d [ F (u ) F (V )+a(u V ) ] [ = A a F (u ) F (V )+a(u V ) ] +ȧu. (4.118) dt Denote v := F (u ) F (V )+a(u V ), h = v. (4.119) Multiplying (4.118) by v and using monotonicity of F,oneobtains hḣ = A av, v + v, ȧ(u V ) +ȧ v, V h 2 (4.120) a + h ȧ u V + ȧ h V, h 0. Again, we have used the inequality F (u )v, v 0 which follows from the monotonicity of F.Thus, Inequalities (4.121) and (3.3) imply ( ḣ h Since a ȧ a a 2 Inequality (4.123) implies: Denote ḣ ha + ȧ u V + ȧ V. (4.121) a ȧ a ) + ȧ V. (4.122) by inequality (2.36), it follows from inequality (4.122) that h(t) h(0)e t a(s) 0 2 ds + e t a(s) 0 From (4.124) and (3.3), one gets ḣ a 2 h + ȧ V. (4.123) ϕ(t) := 2 ds t t 0 0 e s a(ξ) 0 2 dξ ȧ(s) V (s) ds. (4.124) a(s) ds. (4.125) 2 t F (u (t)) F (V (t)) h(0)e ϕ(t) + e ϕ(t) e ϕ(s) ȧ(s) V (s) ds. (4.126) 0

34 90 N. S. Hoang & A. G. Ramm Therefore, F (u (t)) f F(V (t)) f F(V (t)) F (u (t)) t a(t) V (t) h(0)e ϕ(t) e ϕ(t) e ϕ(s) ȧ V ds. From Lemma 29 it follows that 1 2 a(t) V (t) e ϕ(t) From (2.41) one gets t 0 0 (4.127) e ϕ(s) ȧ V (s) ds. (4.128) h(0)e ϕ(t) 1 4 a(0) V (0) e ϕ(t), t 0. (4.129) If c 1and2b d, then it follows that e ϕ(t) a(0) a(t). (4.130) Indeed, inequality a(0) a(t)e ϕ(t) is obviously true for t =0,and ( a(t)e ϕ(t)) 0, t provided that c 1and2b d. Inequalities (4.129) and (4.130) imply e ϕ(t) h(0) 1 4 a(t) V (0) 1 4 a(t) V (t), t 0, (4.131) where we have used the inequality V (t) V (t ) for t t, established in Lemma 20. From (4.112) and (4.127), (4.128), (4.131), one gets C ζ = F (u (t )) f 1 4 a(t ) V (t ). (4.132) It follows from the triangle inequality and the first inequality in (3.29) one obtains This and inequality (4.132) imply a(t) V (t) a(t) V (t) +. (4.133) ( 0 lim a(t ) V (t ) lim 4C ζ + ) =0. (4.134) 0 0 Since V (t) is increasing, this implies lim 0 a(t ) = 0. Since 0 <a(t) 0, it follows that (4.116) holds. From the triangle inequality and inequalities (4.108) and (3.29) one obtains: u (t ) y u (t ) V + V (t ) V (t ) + V (t ) y a(t ) + a(t ) + V (t ) y, (4.135) where V (t) :=V (t) =0 and V (t) solves (2.12). From (4.115), (4.116), inequality (4.135) and Lemma 19, one obtains (2.43). Theorem 9 is proved. By the arguments, similar to the ones in the proof of Theorems or in Remark 10, one can show that: the trajectory u (t) remains in the ball B(u 0,R):= {u : u u 0 <R} for all t t,wherer does not depend on as 0.