International Journal of Analysis and Applications ISSN 2291-8639 Volume 4, Number 1 (2014), 78-86 http://www.etamaths.com SOLVABILITY OF EXTENDED GENERAL STRONGLY MIXED VARIATIONAL INEQUALITIES BALWANT SINGH THAKUR Abstract. In this paper, a new class of extended general strongly mixed vari- ational inequalities is introduced and studied in Hilbert spaces. An existence theorem of solution is established and using resolvent operator technique, a new iterative algorithm for solving the extended general strongly mixed varia- tional inequality is suggested. A convergence result for the iterative sequence generated by the new algorithm is also established. 1. Introduction and Preliminaries Variational inequality theory, which was introduced by Stampacchia [24] in 1964, has had a great impact and influence in the development of several branches on pure and applied sciences. A useful and important generalization of variational inequality is the general mixed variational inequality containing a nonlinear term ϕ. Finding fixed points of a nonlinear mapping is an equally important problem in the functional analysis. Equivalent fixed point formulation of a variational in- equality problem, has given a new dimension to the study of solution of variational inequality problems. In many problems of analysis, one encounters operators who may be split in the form S = A ± T, where A and T satisfies some conditions, and S itself has nei- ther of these properties. An early theorem of this type was given by Krasnoselskii [12], where a complicated operator is split into the sum of two simpler operators. There is another setting arises from perturbation theory. Here the operator equa- tion Tx ± Ax = x is considered as a perturbation of Tx = x (or Ax = x), and one would like to assert that the original unperturbed equation has a solution. In such a situation, there is, in general, no continuous dependence of solutions on the perturbations. For various results in this direction, please see [4, 7, 8, 11, 22, 26]. Another argument is concerned with the approximate solution of the problem: For f ∈ H, find x ∈ H such that Tx±Ax = f. Here T,A : H → H are given operators. Many boundary value problems for quasi linear partial differential equations aris- ing in physics, fluid mechanics and other areas of applications can be formulated as the equation Tx±Ax = f, see, e.g. Zeidler [28]. Combettes and Hirstoaga [5] showed that the finding of zeros of sum of two operators can be solved via the variational inequality involving sum of two operators. Several authors study this 2010 Mathematics Subject Classification. 47J20, 65K10, 65K15, 90C33. Key words and phrases. Extended general strongly mixed variational inequality; fixed point problem; resolvent operator technique; relaxed cocoercive mapping; maximal monotone operator. c©2014 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 78 STRONGLY MIXED VARIATIONAL INEQUALITIES 79 type of situations, see, e.g. [6, 21] and references therein. Motivated by these facts, in this paper we study a variational inequality problem involving operator of the form T −A. Let H be a real Hilbert space whose inner product and norm are denoted by 〈·, ·〉 and ‖·‖, respectively. Let ϕ : H → R∪{+∞} be a proper convex lower semi- continuous function. Let T : H → H be a nonlinear operator and g,h : H → H are any mappings. We consider the problem of finding x∗ ∈ H such that (1) 〈T(x∗) −A(x∗),h(y∗) −g(x∗)〉 + ϕ(h(y∗)) −ϕ(g(x∗)) ≥ 0 , ∀y∗ ∈ H , where A is a nonlinear continuous mapping on H and ∂ϕ denotes the subdifferential of ϕ. We call inequality (1) as extended general strongly mixed variational inequality. We now consider some special cases of the problem (1) : (1) If A ≡ 0, then the problem (1) reduces to the extended general mixed variational inequality problem considered in [20] (2) If h is an identity mapping on H, then the problem (1) reduces to the problem studied by [10]. (3) If A ≡ 0 and h ≡ g, then the problem (1) reduces to the general mixed variational inequality problem considered in [2, 17, 18, 19]. (4) If h,g be identity mappings on H, then the problem (1) reduces to a class of variational inequality studied by [25]. (5) If A ≡ 0 and h,g be identity mappings on H, then the problem (1) reduces to the mixed variational inequality or variational inequality of second kind see [1, 9, 15, 16]. For a multivalued operator T : H → H, we denote by D(T) = {u ∈ H : T(u) 6= ∅} , the domain of T , R(T) = ⋃ u∈H T(u) , the range of T , Graph(T) = {(u,u∗) ∈ H ×H : u ∈ D(T) and u∗ ∈ T(u)} , the graph of T. Definition 1.1. T is called monotone if and only if for each u ∈ D(T), v ∈ D(T) and u∗ ∈ T(u), v∗ ∈ T(v), we have 〈v∗ −u∗,v −u〉≥ 0 . T is maximal monotone if it is monotone and its graph is not properly contained in the graph of any other monotone operator. T−1 is the operator defined by v ∈ T−1(u) ⇔ u ∈ T(v) . Definition 1.2 (See [3]). For a maximal monotone operator T , the resolvent op- erator associated with T , for any σ > 0, is defined as JT (u) = (I + σT) −1(u) , ∀u ∈ H . 80 THAKUR It is known that a monotone operator is maximal if and only if its resolvent operator is defined everywhere. Furthermore, the resolvent operator is single-valued and nonexpansive i.e. ‖JT (x) −JT (y)‖ ≤ ‖x−y‖ , ∀x,y ∈ H. In particular, it is well known that the subdifferential ∂ϕ of ϕ is a maximal monotone operator; see [13]. Lemma 1.3. [3] For a given z ∈ H , u ∈ H satisfies the inequality 〈u−z,x−u〉 + λϕ(x) −λϕ(u) ≥ 0 , ∀x ∈ H if and only if u = Jϕ(z), where Jϕ = (I + λ∂ϕ) −1 is the resolvent operator and λ > 0 is a constant. Inequality (1), can be written in an equivalent form as follows: Find x∗ ∈ H such that (2) 〈ρ(T(x∗) −A(x∗)) + g(x∗) −h(x∗),h(y∗) −g(x∗)〉 + ρϕ(h(y∗)) −ρϕ(g(x∗)) ≥ 0 , for all y∗ ∈ H . This equivalent formulation plays an important role in the development of iter- ative methods for solving the mixed variational inequality problem (1). Using Lemma 1.3, we will establish following important relation: Lemma 1.4. x∗ ∈ H is a solution of (2) if and only if x∗ satisfies the following relation (3) g(x∗) = Jϕ (h(x ∗) −ρ(T(x∗) −A(x∗))) , where ρ > 0 is a constant and Jϕ = (I + ρ∂ϕ) −1 is the proximal mapping, I stands for the indentity operator on H. Proof. Let x∗ ∈ H be a solution of problem (2), then (4) 〈g(x∗) − (h(x∗) −ρ (T(x∗) −A(x∗))) ,h(y∗) −g(x∗)〉+ ρϕ(h(y∗))−ρϕ(g(x∗)) ≥ 0 , for all y∗ ∈ H. Applying Lemma 1.3 for λ = ρ, inequality (4) is equivalent to g(x∗) = Jϕ (h(x ∗) −ρ (T(x∗) −A(x∗))) , the required result. � Lemma 1.4 implies that the problem (2) is equivalent to the fixed point problem (3). This alternative equivalent formulation provides a natural connection between variatonal inequality problem (2) and the fixed point theory which will be used to prove existence result. The following lemma is in this sense : Lemma 1.5. x∗ ∈ H is a solution of (2) if and only if x∗ is a fixed point of the mapping F given by (5) F(u) = u−g(u) + Jϕ (h(u) −ρ(T(u) −A(u))) , u ∈ H . Proof. Let x∗ ∈ H be a fixed point of the mapping F. Then g(x∗) = Jϕ (h(x ∗) −ρ(T(x∗) −A(x∗))) . From Lemma 1.4, x∗ is a solution of (2). � STRONGLY MIXED VARIATIONAL INEQUALITIES 81 We now recall some some definitions: Definition 1.6. An operator T : H → H is said to be : (i) strongly monotone, if for each x ∈ H, there exists a constant ν > 0 such that 〈T(x) −T(y),x−y〉≥ ν‖x−y‖2 holds, for all y ∈ H; (ii) φ−cocoercive, if for each x ∈ H, there exists a constant φ > 0 such that 〈T(x) −T(y),x−y〉≥−φ‖T(x) −T(y)‖2 holds, for all y ∈ H; (iii) relaxed (φ,γ)−cocoercive or relaxed cocoercive with respect to constant (φ,γ), if for each x ∈ H, there exists constants γ > 0 and φ > 0 such that 〈T(x) −T(y),x−y〉≥−φ‖T(x) −T(y)‖2 + γ‖x−y‖2 holds, for all y ∈ H; (iv) µ−Lipschitz continuous or Lipschitz with respect to constant µ, if for each x,y ∈ H, there exists a constant µ > 0 such that ‖T(x) −T(y)‖≤ µ‖x−y‖ . 2. Main results Lemma 1.5, is the main motivation for our next result: Theorem 2.1. Let H be a real Hilbert space and T,A,g,h : H → H are operators. Suppose that the following assumptions are satisfied : (i) T,g,h are relaxed cocoercive with constants (φT ,γT ), (φg,γg), (φh,γh) re- spectively, (ii) T,A,g,h are Lipschitz mappings with constants µT ,µA,µg,µh respectively. If 1 + µ2g(1 + 2φg) > 2γg , 1 + µ 2 h(1 + 2φh) > 2γh , and (6) ρ ∈ (( γT −φT µ2T ) − √ d µ2T + µ 2 A , ( γT −φT µ2T ) + √ d µ2T + µ 2 A ) , where d := (φT µ 2 T −γT ) 2 − 1 2 (µ2T + µ 2 A)(1 + κ(2 −κ)) > 0 κ = √ 1 − 2γg + µ2g(1 + 2φg) + √ 1 − 2γh + µ2h(1 + 2φh) , then the problem (2) has a unique solution. Proof. It is enough to show that the mapping F defined by (5) has a fixed point. For u ∈ H, set p(u) = T(u) −A(u). 82 THAKUR For all x 6= y ∈ H, we have ‖F(x) −F(y)‖≤‖x−y − (g(x) −g(y))‖ + ‖Jϕ (h(x) −ρ(p(x))) −Jϕ (h(y) −ρ(p(y)))‖ ≤‖x−y − (g(x) −g(y))‖ + ‖h(x) −h(y) −ρ (p(x) −p(y))‖ ≤‖x−y − (g(x) −g(y))‖ + ‖x−y − (h(x) −h(y))‖ + ‖x−y −ρ (p(x) −p(y))‖ .(7) Since g is relaxed (φg,γg)−cocoercive and µg-Lipschitz mapping, we can compute the following: ‖x−y − (g(x) −g(y))‖2 = ‖x−y‖2 − 2〈g(x) −g(y),x−y〉 + ‖g(x) −g(y)‖2 ≤ (1 + µ2g)‖x−y‖ 2 + 2φg ‖g(x) −g(y)‖ 2 − 2γg ‖x−y‖ 2 ≤ ( 1 − 2γg + µ2g(1 + 2φg) ) ‖x−y‖2 .(8) Similarly, (9) ‖x−y − (h(x) −h(y))‖2 ≤ ( 1 − 2γh + µ2h(1 + 2φh) ) ‖x−y‖2 . Also, ‖x−y −ρ(p(x) −p(y))‖2 = ‖x−y −ρ(T(x) −T(y)) + ρ(A(x) −A(y))‖2 ≤ 2‖x−y −ρ(T(x) −T(y))‖2 + 2ρ2 ‖A(x) −A(y)‖2 ≤ 2‖x−y −ρ(T(x) −T(y))‖2 + 2ρ2µ2A ‖x−y‖ 2 .(10) Now, we estimate ‖x−y −ρ(T(x) −T(y))‖2 ≤‖x−y‖2 − 2ρ〈T(x) −T(y),x−y〉 + ρ2 ‖T(x) −T(y)‖2 ≤ ( 1 − 2ργT + 2ρµ2T φT + ρ 2µ2T ) ‖x−y‖2 .(11) Substituting (11) into (10), gives (12) ‖x−y −ρ(p(x) −p(y))‖≤ √ 2 (1 − 2ργT + 2ρµ2T φT + ρ2(µ 2 T + µ 2 A)) ‖x−y‖ . Substituting (8), (9), (12) into (7), we have ‖F(x) −F(y)‖≤ (κ + f(ρ))‖x−y‖ , where κ = √ 1 − 2γg + µ2g(1 + 2φg + √ 1 − 2γh + µ2h(1 + 2φh) , and f(ρ) = √ 2 (1 − 2ργT + 2ρµ2T φT + ρ2(µ 2 T + µ 2 A)) . From (6), we get that (κ + f(ρ)) < 1, thus F is a contraction mapping and therefore has a unique fixed point in H, which is a solution of variational inequality (2). � Remark 2.2. Theorem 2.1, extend and improve Theorem 3.1 of [20]. STRONGLY MIXED VARIATIONAL INEQUALITIES 83 If K is closed convex set in H and ϕ(x) = δK (x), for all x ∈ K, where δK is the indicator function of K defined by δK (x) = { 0, if x ∈ K ; +∞, otherwise , then the problem (2) reduces to the following variational inequality problem: Con- sider the problem of finding x∗ ∈ K (13) 〈ρ(T(x∗) −A(x∗)) + g(x∗) −h(x∗), h(y∗) −g(x∗)〉≥ 0 , ∀y∗ ∈ K . We immediately obtain following result from Theorem 2.1 : Corollary 2.3. Let H be a real Hilbert space, K be a nonempty closed convex subset of H and T,A : H → H and g,h : K → K are operators. Suppose that following assumptions are satisfied : (i) T,g,h are relaxed cocoercive with constants (φT ,γT ), (φg,γg), (φh,γh) re- spectively, (ii) T,A,g,h are Lipschitz mappings with constants µT ,µA,µg,µh respectively. If (6) holds, then the problem (13) has a unique solution. If we take h as identity mapping in (13), we get an inequality, equivalent to the general strongly nonlinear variational inequality studied by Siddiqi and Ansari [23]. Corollary 2.3 partially extends and improves the result of [14, 23]. 3. Iterative algorithm and convergence We rewrite the relation (3) in the following form (14) x∗ = x∗ −g(x∗) + Jϕ (h(x∗) −ρ(T(x∗) −A(x∗))) . Using the fixed point formulation (14), we now suggest and analyze the following iterative methods for solving the variational inequality problem (2). Algorithm 1. For a given x0 ∈ H, find the approximate solution xn+1 by the iterative scheme xn+1 = xn −g(xn) + Jϕ (h(xn) −ρ (T(xn) −A(xn))) , n = 0, 1, 2, . . . which is called explicit iterative method. Algorithm 2. For a given x0 ∈ H, find the approximate solution xn+1 by the iterative scheme xn+1 = xn −g(xn) + Jϕ (h(xn+1) −ρ (T(xn+1) −A(xn+1))) , n = 0, 1, 2, . . . which is an implicit iterative method. Now, we use Algorithm 1 as predictor and Algorithm 2 as a corrector to obtain the following predictor-corrector method for solving variational inequality problem (1). Algorithm 3. For a given x0 ∈ H, find the approximate solution xn+1 by the iterative scheme yn = xn −g(xn) + Jϕ (h(xn) −ρ(Txn −Axn)) xn+1 = xn −g(xn) + Jϕ (h(yn) −ρ(Tyn −Ayn)) , n = 0, 1, 2, . . . . 84 THAKUR Using Algorithm 3, we can suggest following : Algorithm 4. For a given x0 ∈ H, find the approximate solution xn+1 by the iterative scheme yn = xn −g(xn) + Jϕ (h(xn) −ρ(Txn −Axn)) xn+1 = (1 −αn)xn + αn (xn −g(xn) + Jϕ (h(yn) −ρ(Tyn −Ayn))) , where n = 0, 1, 2, . . . , {αn} is sequences in [0, 1], satisfying certain conditions. Now, we define a more general predictor-corrector iterative method for approxi- mate solvability of variational inequality problem (1). Algorithm 5. For a given x0 ∈ H, find the approximate solution xn+1 by the iterative scheme (15) yn = (1 −βn)xn + βn (xn −g(xn) + Jϕ (h(xn) −ρ(Txn −Axn))) xn+1 = (1 −αn)xn + αn (xn −g(xn) + Jϕ (h(yn) −ρ(Tyn −Ayn))) , where n = 0, 1, 2, . . . , {αn}, {βn} are sequences in [0, 1], satisfying certain condi- tions. We need following result to prove the next result : Lemma 3.1. [27] Let {an} be a non negative sequence satisfying an+1 ≤ (1 − cn)an + bn , with cn ∈ [0, 1], ∑∞ n=0 cn = ∞, bn = o(cn). Then limn→∞an = 0. Theorem 3.2. Let T,A,g,h satisfy all the assumptions of Theorem 2.1, also con- dition (6) holds and {αn}, {βn} are sequences in [0, 1] for all n ≥ 0 such that∑∞ n=0 αn = ∞. Then the approximate sequence {xn} constructed by the Algorithm 5 converges strongly to a solution x∗ of (2). Proof. For u ∈ H, set pu = Tu−Au. Since x∗ ∈ H is a solution of (1), by (14), we have x∗ = x∗ −g(x∗) + Jϕ (h(x∗) −ρ(T(x∗) −A(x∗))) . Using (15), we have ‖xn+1 −x∗‖≤ (1 −αn)‖xn −x∗‖ + αn ‖xn −x∗ − (g(xn) −g(x∗))‖ + αn ‖Jϕ (h(yn) −ρp(yn)) −Jϕ (h(x∗) −ρp(x∗))‖ ≤ (1 −αn)‖xn −x∗‖ + αn √ 1 − 2γg + µ2g(1 + 2φg)‖xn −x ∗‖ + αn ‖h(yn) −h(x∗) −ρ (p(yn) −p(x∗))‖ ≤ (1 −αn)‖xn −x∗‖ + αn √ 1 − 2γg + µ2g(1 + 2φg)‖xn −x ∗‖ + αn ‖yn −x∗ − (h(yn) −h(x∗))‖ + αn ‖yn −x∗ −ρ (p(yn) −p(x∗))‖ ≤ (1 −αn)‖xn −x∗‖ + αn √ 1 − 2γg + µ2g(1 + 2φg)‖xn −x ∗‖ + αn √ 1 − 2γh + µ2h(1 + 2φh)‖yn −x ∗‖ + αn √ 2 (1 − 2ργT + 2ρµ2T φT + ρ2(µ 2 T + µ 2 A))‖yn −x ∗‖ = (1 −αn)‖xn −x∗‖ + αnθg ‖xn −x∗‖ + αn (θh + f(ρ))‖yn −x∗‖ ,(16) STRONGLY MIXED VARIATIONAL INEQUALITIES 85 where θg = √ 1 − 2γg + µ2g(1 + 2φg) , θh = √ 1 − 2γh + µ2h(1 + 2φh) and f(ρ) = √ 2 (1 − 2ργT + 2ρµ2T φT + ρ2(µ 2 T + µ 2 A)). Similarly, we have ‖yn −x∗‖≤ (1 −βn)‖xn −x∗‖ + βn ‖xn −x∗ − (g(xn) −g(x∗))‖ + βn ‖Jϕ (h(xn) −ρp(xn)) −Jϕ (h(x∗) −ρp(x∗))‖ ≤ (1 −βn)‖xn −x∗‖ + βnθg ‖xn −x∗‖ + βn ‖h(xn) −h(x∗) −ρ (p(xn) −p(x∗))‖ ≤ (1 −βn)‖xn −x∗‖ + βnθg ‖xn −x∗‖ + βn ‖xn −x∗ − (h(xn) −h(x∗))‖ + βn ‖xn −x∗ −ρ (p(xn) −p(x∗))‖ ≤ (1 −βn)‖xn −x∗‖ + βnθg ‖xn −x∗‖ + βnθh ‖xn −x∗‖ + βnf(ρ)‖xn −x∗‖ = (1 −βn)‖xn −x∗‖ + βn(κ + f(ρ))‖xn −x∗‖ ≤ (1 −βn)‖xn −x∗‖ + βn ‖xn −x∗‖ = ‖xn −x∗‖ .(17) Substituting (17) into (16), yields that ‖xn+1 −x∗‖≤ (1 −αn)‖xn −x∗‖ + αn(θg + θh + f(ρ))‖xn −x∗‖ = (1 −αn (1 − (κ + f(ρ))))‖xn −x∗‖ .(18) By virtue of Lemma 3.1, we get from (18) that, limn→∞‖xn+1 −x∗‖ = 0, i.e. xn → x∗, as n →∞. This completes the proof. � Remark 3.3. Theorem 3.2, extend and improve Theorem 2.1 of [10] and Theorem 3.2 of [20]. 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