# Two Nonmonotonic Self-Adaptive Strongly Convergent Projection-Type Methods for Solving Pseudomonotone Variational Inequalities

Authors: Chainarong Khunpanuk; Bancha Panyanak; Nuttapol Pakkaranang
Journal: Journal of Function Spaces
 (2021)

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## Abstract

The primary objective of this study is to introduce two novel extragradient-type iterative schemes for solving variational inequality problems in a real Hilbert space. The proposed iterative schemes extend the well-known subgradient extragradient method and are used to solve variational inequalities involving the pseudomonotone operator in real Hilbert spaces. The proposed iterative methods have the primary advantage of using a simple mathematical formula for step size rule based on operator information rather than the Lipschitz constant or another line search method. Strong convergence results for the suggested iterative algorithms are well-established for mild conditions, such as Lipschitz continuity and mapping monotonicity. Finally, we present many numerical experiments that show the effectiveness and superiority of iterative methods.


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## Body

## 1. Introduction

The primary objective of this research is to investigate the iterative methodologies used to estimate the solution of variational inequalities in a real Hilbert space. To establish the convergence analysis theorems, the following conditions need to be satisfied:Condition 1.
The solution set of the problem (VIP) denoted byΩ and it is nonempty.Condition 2.
A mappingL:Z⟶Z is said to be pseudomonotone if
(1)Lp1,p2−p1≥0⇒Lp2,p1−p2≤0,∀p1,p2∈APM.Condition 3.
A mappingL:Z⟶Z is said to be Lipschitz continuous with constant L>0 if
(2)Lp1−Lp2≤Lp1−p2,∀p1,p2∈ALC.Condition 4.
A mappingL:Z⟶Z is said to be weakly sequentially continuous if Lun converges weakly to Lu for each sequence un converges weakly to an element u.
LetZ be any real Hilbert space and A be any nonempty convex closed subset of a Hilbert space Z. Assume that L:Z⟶Z be an arbitrary mapping. The variational inequality problem for an operator L on A is defined in the following manner [1, 2]:
(3)Findq∗∈AsuchthatLq∗,y−q∗≥0,∀y∈AVIP.LetΩ stand for the solution set for the problem (VIP). The mathematical model of variational inequalities covers many mathematical problems, such as partial differential equations, optimization, optimal control, mechanics, finance, and mathematical programming (see for details [3–9]) and others in [10–20]. Since it is a fundamental problem in the applied sciences and nonlinear functional analysis, many researchers are investigating not only the stability and existence of solutions to such problems but also iterative methods for solving them numerically. In order to solve variational inequalities numerically, projection iterative methods are a very important tool. Many researchers have provided various projection method extensions and modifications to solve the problem (VIP) (see [21–33]). The extragradient method described below was developed by Korpelevich [25] and Antipin [34]. Their method takes the form of
(4)u1∈A,pn=PAun−δLun,un+1=PAun−δLpn,where 0<δ<1/L. For each iteration of the above iterative scheme, two projections on the feasible set A are required to be figured out. Of course, if the feasible set A has a complicated framework, this can affect the method’s computational effectiveness. The first one is to follow the subgradient extragradient method designed by Censor et al. [22] to overcome this deficiency. This method is in the form of
(5)u1∈A,pn=PAun−δLun,un+1=PZnun−δLpn,where 0<δ<1/L and
(6)Zn=z∈Z:un−δLun−pn,z−pn≤0.It is a key point to note that the above-mentioned well-established methods have two major drawbacks. The first is the fixed constant step size, which needs knowledge or approximation of the appropriate operator Lipschitz constant, and also is only weakly convergent in Hilbert spaces. Using a fixed step size can be difficult in terms of computation, affecting the method convergence rate and efficiency.Hence, a natural question arises:“Is it possible to propose two new strongly convergent subgradient extragradient algorithms with a nonmonotone self-adaptive step size rule to solve the problem (VIP)?”The primary objective of this study is to introduce two new strongly convergent subgradient extragradient methods for enhancing the convergence rate of an iterative sequence. The answer to the above question is given in this study, which would be the subgradient extragradient algorithms, which set up a strong convergent iterative sequence by letting a variable nonmonotone step size rule. The suggested methods are employed to solve variational inequality problems involving pseudomonotone and Lipschitz regular operators in real Hilbert space. The proposed methods are based on the projection method [22] as well as the methods proposed in [26, 35]. The established method only needs to compute one projection onto the feasible set and one projection onto the half-space for each iteration. The iterative sequences established by the proposed method strongly converge to some solution of the underlined problem in the framework of some appropriate conditions on control parameters. A number of numerical examples are also added to elaborate on the computational effectiveness of the new methods over some existing methods presented in [36, 37].The paper is arranged in the following manner: In Section2, we provide some basic identities and preliminary results that were used in this paper. Section 3 includes the proposed methods and proves their convergence analysis. Finally, Section 4 presents some numerical results to illustrate the convergence and the effectiveness of the proposed methods.


## 2. Preliminaries

This section contains a number of important identities, as well as useful lemmas and definitions. For allu,y∈Z, we have
(7)u+y2=u2+2u,y+y2.A metric projectionPAp1 of an element p1∈Z is evaluated by
(8)PAp1=argminp1−p2:p2∈A.Next, we list some of the important identities that are used to prove the convergence analysis.Lemma 1 (see [38]).
LetPA:Z⟶A be a metric projection on set A. For each p1,p2∈Z and ℓ∈ℝ, then the following inequalities are satisfied:
(i)
p3=PAp1 is true if and only if(9)p1−p3,p2−p3≤0,∀p2∈A,(ii)(10)p1−PAp22+PAp2−p22≤p1−p22,p1∈A,p2∈Z,(iii)(11)p1−PAp1≤p1−p2,p2∈A,p1∈Z,(iv)(12)ℓp1+1−ℓp22=ℓp12+1−ℓp22−ℓ1−ℓp1−p22,(v)(13)p1+p22≤p12+2p2,p1+p2.Lemma 2 (see [39]).
Assuming thatcn⊂0,+∞, a sequence meets the following criteria:
(14)cn+1≤1−dncn+dnen,∀n∈ℕ.
Moreover,dn⊂0,1 and en⊂ℝ are two sequences such that
(15)limn⟶+∞dn=0,∑n=1+∞dn=+∞andlimsupn⟶+∞en≤0.
Then,limn⟶+∞cn=0.Lemma 3 (see [40]).
Assume that a sequencecn of real numbers and there is ni subsequence of n such that
cni<cni+1,∀i∈ℕ.(102)
Thus, there exists a natural nondecreasing sequencemj with mj⟶+∞ as j⟶+∞ and satisfies the following criteria for j∈ℕ:
(16)cmj≤cmj+1,cj≤cmj+1.
Indeed,mj=maxj≤j:cj≤cj+1.Lemma 4 (see [41]).
Assume thatL:A⟶Z is a pseudomonotone and continuous mapping. Then, q∗ is a solution of the problem (VIP) if and only if q∗ is a solution of the following problem:
(17)Findu∈AsuchthatLy,y−u≥0,∀y∈A.


## 3. Main Results

In this part of the research article, we propose two new methods and the corresponding strong convergence theorems. Both methods are presented in the following manner. The first method is of the following form.Algorithm 1: Nonmonotonic explicit Mann-type subgradient extragradient method.
Step 0: takeu1∈A,δ1>0 and select a nonnegative sequence of real numbers φn such that ∑n=1+∞φn<+∞. Moreover, ϕn⊂a,b⊂0,1−ψn and ψn⊂0,1 meet the following criteria:limn⟶+∞ψn=0and∑n=1+∞ψn=+∞.Step 1: evaluatepn=PAun−δnLun.Ifun=pn, then STOP. Otherwise, go to Step 2.Step 2: firstly, construct a half-spaceZn=z∈Z:un−δnLun−pn,z−pn≤0,and computeqn=PZnun−δnLpn.Step 3: evaluateun+1=1−ϕn−ψnun+ϕnqn.Step 4: evaluateδn+1=minδn+φn,μ∥un−pn∥2+μ∥qn−pn∥2/2Lun−Lpn,qn−pnifLun−Lpn,qn−pn>0,φn+δn,otherwise.3.1Setn≔n+1 and go back to Step 1.Assume thatg:Z⟶Z is a contraction having constant ξ∈0,1. The second major contribution of this study work is as follows. The second main algorithm has the following form.Algorithm 2: Nonmonotonic explicit viscosity-type subgradient extragradient method.
Step 0: takeu1∈A,δ1>0 and select a nonnegative sequence of real number φn such that ∑n=1+∞φn<+∞. Moreover, ψn⊂0,1 meet the following criteria:limn⟶+∞ψn=0and∑n=1+∞ψn=+∞.Step 1: evaluatepn=PAun−δnLun.Ifun=pn, then STOP. Otherwise, go to Step 2.Step 2: evaluateZn=z∈Z:un−δnLun−pn,z−pn≤0,and computeqn=PZnun−δnLpn.Step 3: evaluateun+1=ψngun+1−ψnqn.Step 4: evaluateδn+1=minδn+φn,μun−pn2+μqn−pn2/2Lun−Lpn,qn−pnifLun−Lpn,qn−pn>0,φn+δn,otherwise.3.2Setn≔n+1 and go back to Step 1.Lemma 5.
A sequenceδn generated by (3.1) is convergent to δ and satisfies the following inequality:
(18)minμL,δ1≤δ≤δ1+PwhereP=∑n=1+∞φn.Proof.
LetLun−Lpn,qn−pn>0 such that
(19)μun−pn2+qn−pn22Lun−Lpn,qn−pn≥2μun−pnqn−pn2Lun−Lpnqn−pn≥2μun−pnqn−pn2Lun−pnqn−pn≥μL.
By using mathematical induction on the definition ofδn+1, we have
(20)minμL,δ1≤δn≤δ1+P.
Let(21)δn+1−δn+=max0,δn+1−δn,δn+1−δn−=max0,−δn+1−δn.
Due to expression ofδn, we can write
(22)∑n=1+∞δn+1−δn+=∑n=1+∞max0,δn+1−δn≤P<+∞.
Thus,∑n=1+∞δn+1−δn+ is convergent. Next, we have to prove the convergence of the following series:
(23)∑n=1+∞δn+1−δn−.
Let∑n=1+∞δn+1−δn−=+∞. Thus, we have δn+1−δn=δn+1−δn+−δn+1−δn−. Thus, we have
(24)δk+1−δ1=∑n=0kδn+1−δn=∑n=0kδn+1−δn+−∑n=0kδn+1−δn−.
Lettingk⟶+∞ in (24), we obtain δk⟶−∞ as k⟶∞. This is a contradiction. Due to the convergence of the series ∑n=0kδn+1−δn+ and ∑n=0kδn+1−δn− taking k⟶+∞ in (24), we obtain limn⟶∞δn=δ. This completes the proof.☐☐Lemma 6.
LetL:Z⟶Z be an operator satisfies the criteria Condition 1–Condition 4. For a given q∗∈Ω≠∅, we have
(25)qn−q∗2≤un−q∗2−1−μδnδn+1un−pn2−1−μδnδn+1qn−pn2.Proof.
We have to evaluate(26)qn−q∗2=PZnun−δnLpn−q∗2=PZnun−δnLpn+un−δnLpn−un−δnLpn−q∗2=un−δnLpn−q∗2+PZnun−δnLpn−un−δnLpn2+2PZnun−δnLpn−un−δnLpn,un−δnLpn−q∗.
It is given thatq∗∈Ω⊂A⊂Zn such that
(27)PZnun−δnLpn−un−δnLpn2+PZnun−δnLpn−un−δnLpn,un−δnLpn−q∗=un−δnLpn−PZnun−δnLpn,q∗−PZnun−δnLpn≤0.
Furthermore, it implies that(28)PZnun−δnLpn−un−δnLpn,un−δnLpn−q∗≤−PZnun−δnLpn−un−δnLpn2.
By using expressions (26) and (28), we obtain
(29)qn−q∗2≤un−δnLpn−q∗2−PZnun−δnLpn−un−δnLpn2≤un−q∗2−un−qn2+2δnLpn,q∗−qn.
Sinceq∗ is the solution of problem (VIP), we have
(30)Lq∗,y−q∗≥0,∀y∈A.
We have condition on mappingL with feasible set A, we get
(31)Ly,y−q∗≥0,∀y∈A.
By substitutingy=pn∈A, we get
(32)Lpn,pn−q∗≥0.
Therefore, we have(33)Lpn,q∗−qn=Lpn,q∗−pn+Lpn,pn−qn≤Lpn,pn−qn.
From (29) and (33), we get
(34)∥qn−q∗∥2≤∥un−q∗∥2−∥un−qn∥2+2δnLpn,pn−qn≤∥un−q∗∥2−∥un−pn+pn−qn∥2+2δnLpn,pn−qn≤∥un−q∗∥2−∥un−pn∥2−∥pn−qn∥2+2un−δnLpn−pn,qn−pn.
Sinceqn=PZnun−δnLpn and from δn+1, we have
(35)2un−δnLpn−pn,qn−pn=2un−δnLun−pn,qn−pn+2δnLun−Lpn,qn−pn≤δnδn+12δn+1Lun−Lpn,qn−pn≤μδnδn+1un−pn2+μδnδn+1qn−pn2.
Combining expressions (34) and (35), we obtain
(36)qn−q∗2≤un−q∗2−un−pn2−pn−qn2+δnδn+1μun−pn2+μqn−pn2≤un−q∗2−1−μδnδn+1un−pn2−1−μδnδn+1qn−pn2.
☐☐Lemma 7.
LetL:Z⟶Z be a mapping meet the items Condition 1–Condition 4. If there exists a subsequence unk convergent weakly to u^ and limk⟶∞unk−pnk=0, then u^ is the solution of (VIP).Proof.
We need to prove thatu^∈Ω. Indeed, we have
(37)pnk=PAunk−δnkLunk,that is equivalent to
(38)unk−δnkLunk−pnk,y−pnk≤0,∀y∈A.
The above inequality implies that(39)unk−pnk,y−pnk≤δnkLunk,y−pnk,∀y∈A.
Thus, we obtain(40)1δnkunk−pnk,y−pnk+Lunk,pnk−unk≤Lunk,y−unk,∀y∈A.
Sinceminμ/L,δ1≤δ≤δ1+P and unk is a bounded sequence. By taking limk⟶∞unk−pnk=0 and k⟶∞ in (18), we obtain
(41)liminfk⟶∞Lunk,y−unk≥0,∀y∈A.
Moreover, we have(42)Lpnk,y−pnk=Lpnk−Lunk,y−unk+Lunk,y−unk+Lpnk,unk−pnk.
Sincelimk⟶∞unk−pnk=0 and L is L-Lipschitz continuity on Z, we have
(43)limk⟶∞Lunk−Lpnk=0,which together with (42) and (43), we obtain
(44)liminfk⟶∞Lpnk,y−pnk≥0,∀y∈A.
Next, we have(45)Luni,y−uni+εk≥0,∀i≥mk.
Due toεk decreasing, this implies that mk is increasing.☐☐Case I.
Suppose that a subsequenceunmkj of unmk such as Lunmkj=0 (∀j). Let j⟶∞, we get
(46)Lu^,y−u^=limj⟶∞Lunmkj,y−u^=0.
Then,u^∈A which further implies that u^∈Ω.Case II.
Suppose that there existsN0∈ℕ such that for all nmk≥N0, Lunmk≠0. Consider that
(47)Ξnmk=Lunmk∥Lunmk∥2,∀nmk≥N0.
Due to the above definition, we obtain(48)Lunmk,Ξnmk=1,∀nmk≥N0.
Moreover, expressions (45) and (48) for all nmk≥N0, we have
(49)Lunmk,y+εkΞnmk−unmk≥0.
Due to the pseudomonotonicity ofL for nmk≥N0, we have
(50)Ly+εkΞnmk,y+εkΞnmk−unmk≥0.
For allnmk≥N0, we have
(51)Ly,y−unmk≥Ly−Ly+εkΞnmk,y+εkΞnmk−unmk−εkLy,Ξnmk.
Let us consider thatLu^≠0; we obtain
(52)Lu^≤liminfk⟶∞Lunk.
Thus, we obtain(53)0≤limk⟶∞εkΞnmk=limk⟶∞εkLunmk≤0Lu^=0.
By lettingk⟶∞ in (51), we obtain
(54)Ly,y−u^≥0,∀y∈A.
Due to the Minty Lemma in [41], we infer that u^∈Ω.Theorem 8.
LetL:Z⟶Z be a mapping that satisfies the conditions Condition 1–Condition 4. Then, un sequence generated by Algorithm 1 strongly converges to an element q∗∈Ω.Proof.
Sinceδn⟶δ, there exists a fixed number ε∈0,1−μ such that
(55)limn⟶∞1−μδnδn+1=1−μ>ε>0,∀n≥n0.
Thus, expression (36) gives that
(56)qn−q∗2≤un−q∗2,∀n≥n0.
It is given thatq∗∈Ω; we obtain
(57)un+1−q∗=1−ϕn−ψnun+ϕnqn−q∗=1−ϕn−ψnun−q∗+ϕnqn−q∗−ψnq∗≤1−ϕn−ψnun−q∗+ϕnqn−q∗+ψnq∗.
Next, we have to evaluate the following:(58)1−ϕn−ψnun−q∗+ϕnqn−q∗2=1−ϕn−ψn2un−q∗2+ϕn2qn−q∗2+21−ϕn−ψnun−q∗,ϕnqn−q∗≤1−ϕn−ψn2un−q∗2+ϕn2qn−q∗2+2ϕn1−ϕn−ψnun−q∗qn−q∗,(59)≤1−ϕn−ψn2qn−q∗2+ϕn2qn−q∗2+ϕn1−ϕn−ψnun−q∗2+ϕn1−ϕn−ψnqn−q∗2≤1−ϕn−ψn1−ψnun−q∗2+ϕn1−ψnqn−q∗2.
Substituting (56) into (59), we obtain
(60)1−ϕn−ψnun−q∗+ϕnqn−q∗2≤1−ϕn−ψn1−ψnun−q∗2+ϕn1−ψnun−q∗2=1−ψn2un−q∗2.
Next, we have(61)1−ϕn−ψnun−q∗+ϕnqn−q∗≤1−ψnun−q∗.
From expressions (57) and (61), we obtain
(62)un+1−q∗≤1−ψnun−q∗+ψnq∗≤maxun−q∗,q∗≤maxun0−q∗,q∗.
Thus, from the above relation, we obtain thatun is bounded sequence. From sequence un+1, we can write
(63)un+1−q∗2=1−ϕn−ψnun+ϕnqn−q∗2=1−ϕn−ψnun−q∗+ϕnqn−q∗−ψnq∗2=1−ϕn−ψnun−q∗+ϕnqn−q∗2+ψn2q∗2−21−ϕn−ψnun−q∗+ϕnqn−q∗,ψnq∗.
By the use of expression (59), we have
(64)1−ϕn−ψnun−q∗+ϕnqn−q∗2≤1−ϕn−ψn1−ψnun−q∗2+ϕn1−ψnqn−q∗2.
Combining expressions (63) and (64) (for some K2>0), we obtain
(65)un+1−q∗2≤1−ϕn−ψn1−ψnun−q∗2+ϕn1−ψnqn−q∗2+ψnK2≤1−ϕn−ψn1−ψnun−q∗2+ψnK2+ϕn1−ψnun−q∗2−1−μδnδn+1un−pn2−1−μδnδn+1qn−pn2=1−ψn2un−q∗2+ψnK2−ϕn1−ψn1−μδnδn+1un−pn2+1−μδnδn+1qn−pn2≤un−q∗2+ψnK2−ϕn1−ψn1−μδnδn+1un−pn2+1−μδnδn+1qn−pn2.The remainder of the proof is now split into two parts:Case 1.
Suppose that there exists a fixed numbern1∈ℕ (n1≥n0) such that
(66)un+1−q∗≤un−q∗,∀n≥n1.
Then,limn⟶∞un−q∗ exists. From (65), we have
(67)ϕn1−ψn1−μδnδn+1un−pn2+1−μδnδn+1qn−pn2≤un−q∗2+ψnK2−un+1−q∗2.
Due to the existence oflimn⟶+∞un−q∗ and ψn⟶0, we infer that
(68)limn⟶∞un−pn=limn⟶∞qn−pn=0.
It follows that(69)limn⟶∞un−qn≤limn⟶∞un−pn+limn⟶∞pn−qn=0.
It follows from expression (69) and ψn⟶0 that
(70)un+1−un=1−ϕn−ψnun+ϕnqn−un=un−ψnun+ϕnqn−ϕnun−un≤ϕnqn−un+ψnun,which implies that
(71)un+1−un⟶0asn⟶+∞.
We haveq∗=PΩ0 and by using Lemma 1 (i), we can write
(72)0−q∗,y−q∗≤0,∀y∈Ω.
By using expression (72) and Lemma 7, we obtain
(73)limsupn⟶∞q∗,q∗−un=limsupk⟶∞q∗,q∗−unk=q∗,q∗−u^≤0.
Due tolimn⟶∞un+1−un=0. It gives that
(74)limsupn⟶∞q∗,q∗−un+1≤limsupn⟶∞q∗,q∗−un+limsupn⟶∞q∗,un−un+1≤0.
Next, we assume that(75)tn=1−ϕnun+ϕnqn.
Thus, we have(76)un+1=tn−ψnun=1−ψntn−ψnun−tn=1−ψntn−ψnϕnun−qn,where
(77)un−tn=un−1−ϕnun−ϕnqn=ϕnun−qn.
Thus, we obtain(78)un+1−q∗2=1−ψntn+ϕnψnqn−un−q∗2=1−ψntn−q∗+ϕnψnqn−un−ψnq∗2≤1−ψn2tn−q∗2+2ϕnψnqn−un−ψnq∗,1−ψntn−q∗+ϕnψnqn−un−ψnq∗=1−ψn2tn−q∗2+2ϕnψnqn−un−ψnq∗,tn−ψntn−ψnun−tn−q∗=1−ψntn−q∗2+2ϕnψnqn−un,un+1−q∗+2ψnq∗,q∗−un+1≤1−ψntn−q∗2+2ϕnψnqn−unun+1−q∗+2ψnq∗,q∗−un+1.
Next, we have to compute(79)tn−q∗2=1−ϕnun+ϕnqn−q∗2=1−ϕnun−q∗+ϕnqn−q∗2=1−ϕn2un−q∗2+ϕn2qn−q∗2+21−ϕnun−q∗,ϕnqn−q∗≤1−ϕn2un−q∗2+ϕn2qn−q∗2+2ϕn1−ϕnun−q∗qn−q∗≤1−ϕn2un−q∗2+ϕn2qn−q∗2+ϕn1−ϕnun−q∗2+ϕn1−ϕnqn−q∗2=1−ϕnun−q∗2+ϕnqn−q∗2≤1−ϕnun−q∗2+ϕnun−q∗2=un−q∗2.
From expressions (78) and (79), we have
(80)un+1−q∗2≤1−ψnun−q∗2+ψn2ϕnqn−unun+1−q∗+2q∗,q∗−un+1.
By the use of expressions (74) and (80) and Lemma 2, we can derive that un−q∗⟶0 as n⟶+∞.Case 2.
Suppose that there exists a subsequenceni of n in order that
(81)uni−q∗≤uni+1−q∗,∀i∈ℕ.
By Lemma3, we have
(82)umk−q∗≤umk+1−q∗anduk−q∗≤umk+1−q∗,∀k∈ℕ.
By the use of expression (67), we have
(83)ϕmk1−ψmk1−μδmkδmk+1umk−pmk2+1−μδmkδmk+1qmk−pmk2≤umk−q∗2+ψmkK2−umk+1−q∗2.
Due toψmk⟶0, we can deduce as follows:
(84)limk⟶∞umk−pmk=limk⟶∞qmk−pmk=0.
It follows that(85)limk⟶∞umk−qmk≤limk⟶∞umk−pmk+limk⟶∞pmk−qmk=0.
Further, it implies that(86)umk+1−umk=1−ϕmk−ψmkumk+ϕmkqmk−umk=umk−ψmkumk+ϕmkqmk−ϕmkumk−umk≤ϕmkqmk−umk+ψmkumk⟶0.
Similar to Case1, we obtain
(87)limsupk⟶∞q∗,umk+1−q∗≤0.
By the use of expressions (80) and (82), we have
(88)umk+1−q∗2≤1−ψmkumk−q∗2+ψmk2ϕmkqmk−umkumk+1−q∗+2q∗,q∗−umk+1≤1−ψmkumk+1−q∗2+ψmk2ϕmkqmk−umkumk+1−q∗+2q∗,q∗−umk+1.
It follows that(89)umk+1−q∗2≤2ϕmkqmk−umkumk+1−q∗+2q∗,q∗−umk+1.
Sinceψmk⟶0 and umk−q∗ is a bounded sequence with (87) and (89), we have
(90)umk+1−q∗2⟶0,ask⟶∞.
The above expression implies that(91)limk⟶∞uk−q∗2≤limk⟶∞umk+1−q∗2≤0.
From the above discussions, we haveun⟶q∗ as n⟶∞.Theorem 9.
Assume thatL:Z⟶Z is a mapping that meets the conditions Condition 1–Condition 4. Then, the un sequence generated by Algorithm 2 strongly converges to an element q∗∈Ω.Proof.
Sinceδn⟶δ, there exists a positive number ε∈0,1−μ such that
(92)limn⟶∞1−μδnδn+1=1−μ>ε>0.
Thus, there exists a numberN1∈ℕ such that
(93)1−μδnδn+1>ε>0,∀n≥N1.
From expression (36), we obtain
(94)qn−q∗2≤un−q∗2,∀n≥N1.
Next, we have(95)un+1−q∗=ψngun+1−ψnqn−q∗=ψngun−q∗+1−ψnqn−q∗=ψngun+gq∗−gq∗−q∗+1−ψnqn−q∗≤ψngun−gq∗+ψngq∗−q∗+1−ψnqn−q∗≤ψnξun−q∗+ψngq∗−q∗+1−ψnqn−q∗.
From expressions (94) and (95) and ψn⊂0,1, we obtain
(96)un+1−q∗≤ψnξun−q∗+ψngq∗−q∗+1−ψnun−q∗=1−ψn+ξψnun−q∗+ψn1−ξgq∗−q∗1−ξ≤maxun−q∗,gq∗−q∗1−ξ≤maxuN1−q∗,gq∗−q∗1−ξ.
Therefore, we infer that theun is a bounded sequence. Now, we are in a position to use the Banach contraction theorem for the existence of a unique fixed point q∗∈Ω such that
(97)q∗=PΩgq∗.
Due to the projection mapping, we can write(98)gq∗−q∗,y−q∗≤0,∀y∈Ω.
From Lemma1 and expression (36), we have
(99)un+1−q∗2=ψngun+1−ψnqn−q∗2=ψngun−q∗+1−ψnqn−q∗2=ψngun−q∗2+1−ψnqn−q∗2−ψn1−ψngun−qn2≤ψngun−q∗2+1−ψn∥un−q∗2−1−μδnδn+1un−pn2−1−μδnδn+1qn−pn2−ψn1−ψngun−qn2≤ψngun−q∗2+un−q∗2−1−ψn1−μδnδn+1un−pn2−1−ψn1−μδnδn+1qn−pn2.
The above expression implies that(100)1−ψn1−μδnδn+1un−pn2+1−ψn1−μδnδn+1qn−pn2≤ψngun−q∗2+un−q∗2−un+1−q∗2.☐Case 1.
Suppose that there exists a fixed numberN2∈ℕ (N2≥N1) such that
(101)un+1−q∗≤un−q∗,∀n≥N2.
Then,limn⟶∞un−q∗ exists and let limn⟶∞un−q∗=l. By the use of expression (99), we have
(102)1−ψn1−μδnδn+1un−pn2+1−ψn1−μδnδn+1qn−pn2≤ψngun−q∗2+∥un−q∗2−un+1−q∗2.
By the of existence of a limit of a sequencelimn⟶∞∥un−q∗∥ and ψn⟶0, we obtain
(103)limn⟶∞un−pn=limn⟶∞qn−pn=0.
Furthermore, it implies that(104)limn⟶∞un−pn≤limn⟶∞un−pn+limn⟶∞qn−pn=0.
Thus, we have(105)un+1−un=ψngun+1−ψnqn−un=ψngun−un+1−ψnqn−un≤ψngun−un+1−ψnqn−un⟶0.
Thus, we have(106)limn⟶∞un+1−un=0.
By using expression (97) and Lemma 7, we obtain
(107)limsupn⟶∞gq∗−q∗,un−q∗=limsupk⟶∞gq∗−q∗,unk−q∗=gq∗−q∗,u^−q∗≤0.
By the use oflimn⟶∞un+1−un=0, we have
(108)limsupn⟶∞gq∗−q∗,un+1−q∗≤limsupn⟶∞gq∗−q∗,un+1−un+limsupn⟶∞gq∗−q∗,un−q∗≤0.
By using Lemma1 and expression (94), we have
(109)un+1−q∗2=ψngun+1−ψnqn−q∗2=ψngun−q∗+1−ψnqn−q∗2≤1−ψn2qn−q∗2+2ψngun−q∗,1−ψnqn−q∗+ψngun−q∗=1−ψn2qn−q∗2+2ψngun−gq∗+gq∗−q∗,un+1−q∗=1−ψn2qn−q∗2+2ψngun−gq∗,un+1−q∗+2ψngq∗−q∗,un+1−q∗≤1−ψn2qn−q∗2+2ψnξun−q∗2un+1−q∗+2ψngq∗−q∗,un+1−q∗≤1+ψn2−2ψnun−q∗2+2ψnξun−q∗2+2ψngq∗−q∗,un+1−q∗=1−2ψnun−q∗2+ψn2un−q∗2+2ψnξun−q∗2+2ψngq∗−q∗,un+1−q∗=1−2ψn1−ξun−q∗2+2ψn1−ξψnun−q∗221−ξ+gq∗−q∗,un+1−q∗1−ξ.
It is clear from expressions (108) and (109) such that
(110)limsupn⟶∞ψnun−q∗221−ξ+gq∗−q∗,un+1−q∗1−ξ≤0.
By choosingn≥N3∈ℕ (N3≥N2) large enough such that 2γn1−ξ<1. By the use of (109) and (110) and through Lemma 2, we conclude that ∥un−q∗∥⟶0 as n⟶∞.Case 2.
Consider that there exists a subsequenceni of n such that
(111)uni−q∗≤uni+1−q∗,∀i∈ℕ.
Then, using Lemma3, there exists a sequence mk⊂ℕ as mk⟶∞, such that
(112)umk−q∗≤umk+1−q∗,uk−q∗≤umk+1−q∗,forallk∈ℕ.
Thus, we have(113)1−ψmk1−μδmkδmk+1umk−pmk2+1−ψmk1−μδmkδmk+1qmk−pmk2≤ψmkgumk−q∗2+umk−q∗2−umk+1−q∗2.
Sinceψmk⟶0 implies that
(114)limk⟶∞umk−pmk=limk⟶∞qmk−pmk=0.
Next, we obtain(115)umk+1−umk=ψmkgumk+1−ψmkqmk−umk=ψmkgumk−umk+1−ψmkqmk−umk≤ψmkgumk−umk+1−ψmkqmk−umk⟶0.
Similar to the Case1, we can write
(116)limsupk⟶∞gq∗−q∗,umk+1−q∗≤0.
By using (109) and (112), we have
(117)umk+1−q∗2≤1−2ψmk1−ξumk−q∗2+2ψmk1−ξψmkumk−q∗121−ξ+gq∗−q∗,umk+1−q∗1−ξ≤1−2ψmk1−ξumk+1−q∗2+2ψmk1−ξψmkumk−q∗221−ξ+gq∗−q∗,umk+1−q∗1−ξ.
Moreover, we have(118)umk+1−q∗2≤ψmk∥umk−q∗∥221−ξ+gq∗−q∗,umk+1−q∗1−ξ.
Sinceψmk⟶0 and umk−q∗ is a bounded sequence. Then, the expressions (116) and (118) imply that
(119)umk+1−q∗2⟶0,ask⟶∞.
The above expression implies that(120)limk⟶∞uk−q∗2≤limk⟶∞umk+1−q∗2≤0.
Consequently,un⟶q∗ as n⟶∞. This completes the proof of the theorem.☐Remark 10.
(i)
Two nonmonotonic explicit extragradient-type methods for finding an approximate solution of variational inequalities involving pseudomonotone mapping in a real Hilbert space have been established(ii)
Two strongly convergent results, corresponding to the proposed algorithms have been proven(iii)
It is important to note that these methods use nonmonotonic step size rules that use an operator value rather than the Lipschitz constant of an operator


## 4. Numerical Illustrations

In this section, computational results of the proposed methods are described and compared to existing related work in the literature. All computations are done in MATLAB R2018b and run on an HP i-5 Core(TM)i5-6200 8.00 GB (7.78 GB usable) RAM laptop.Example 11.
Let a mappingL:ℝm⟶ℝm be defined by
(121)Lu=Mu+q,with q∈ℝm and
(122)M=NNT+B+D.During this experiment, we have chosenN=randm to be a random matrix and B=0.5K−0.5KT to be a skew-symmetric matrix with K=randm, and D=diagrandm,1 is a diagonal matrix. The feasible set A is defined in the subsequent sense:
(123)A=u∈ℝm:Qu≤b,where Q=rand100,m and b=rand100,1. It is obvious that L is monotone and Lipschitz continuous by L=M. The starting point is u1=2,2,⋯,2 and Dn=un−pn≤10−3. The numerical results of these methods are shown in Figures 1–6 and Table 1. The control conditions are taken in the following manner: (1) Algorithm 2 in [37] (shortly, Method-1): δ1=0.12,μ=0.55,ϕn=1/50n+2; (2) Algorithm 1 in [36] (shortly, Method-2): δ1=0.12,γ=0.55,ϕn=1/2n+2,fu=u/3; (3) Algorithm 1 (shortly, Method-3): δ1=0.12,μ=0.55,ψn=1/2n+2,ϕn=5/101−ψn,φn=100/n+12; and (4) Algorithm 2 (shortly, Method-4): δ1=0.12,μ=0.55,ψn=1/2n+2,fu=u/3,φn=100/n+12.Example 12.
LetZ=L20,1 be a Hilbert space having an inner product
(124)u,y=∫01utytdt,∀u,y∈Z,and the induced norm is determined as follows:
(125)∥u∥=∫01ut2dt.Figure 1
Algorithmic descriptions of Algorithm1 and Algorithm 2 as well Algorithm 2 in [37] and Algorithm 1 in [36] when m=5.Figure 2
Algorithmic descriptions of Algorithm1 and Algorithm 2 with Algorithm 2 in [37] and Algorithm 1 in [36] when m=5.Figure 3
Algorithmic descriptions of Algorithm1 and Algorithm 2 with Algorithm 2 in [37] and Algorithm 1 in [36] when m=10.Figure 4
Algorithmic descriptions of Algorithm1 and Algorithm 2 with Algorithm 2 in [37] and Algorithm 1 in [36] when m=10.Figure 5
Algorithmic descriptions of Algorithm1 and Algorithm 2 with Algorithm 2 in [37] and Algorithm 1 in [36] when m=20.Figure 6
Algorithmic descriptions of Algorithm1 and Algorithm 2 with Algorithm 2 in [37] and Algorithm 1 in [36] when m=20.Table 1
Algorithmic study for Example11.
mMethod-1Method-2Method-3Method-4Iter.TimeIter.TimeIter.TimeIter.Time5680.798005200570.665091600420.480156900290.360117900102643.2181572002412.998657400951.121305500580.733156900203844.8842107002973.5934864001641.9301829001131.339267400306498.3564997005457.8507650003083.9866781001962.59355520040113816.9690999099715.258664403574.7847123002383.41040050050246737.429776501796224.3252350055810.326194854876.132444757LetA≔u∈L20,1:u≤1 be a unit ball. A mapping L:A⟶Z is defined by
(126)Lut=∫01ut−Ht,sfusds+gt,where
(127)Ht,s=2tset+see2−1,fu=cosu,gt=2tetee2−1.It can easily be seen thatL is Lipschitz-continuous with the constant L=2 and monotone. The starting point for this experiment is taken differently, and Dn=un−pn≤10−3. The numerical results of these methods are shown in Table 2. The control conditions are taken as follows: (1) Algorithm 2 in [37] (shortly, Method-1): δ1=0.33,μ=0.75,ϕn=1/100n+2; (2) Algorithm 1 in [36] (shortly, Method-2): δ1=0.33,γ=0.75,ϕn=1/100n+2,fu=u/2; (3) Algorithm 1 (shortly, Method-3): δ1=0.33,μ=0.75,ψn=1/100n+2,ϕn=7/101−ψn,φn=100/n+12; and (4) Algorithm 2 (shortly, Method-4): δ1=0.33,μ=0.75,ψn=1/100n+2,fu=u/2,φn=100/n+12.Example 13.
Let a mappingL:ℝ4⟶ℝ4 be defined by
(128)Lu=u1+u2+u3+u4−4u2u3u4u1+u2+u3+u4−4u1u3u4u1+u2+u3+u4−4u1u2u4u1+u2+u3+u4−4u1u2u3.
Moreover, the constraint setA is defined by
(129)A=u∈ℝ4:1≤ui≤5,i=1,2,3,4.
It is clear to see thatL is not monotone on the set A. The starting point for this experiment is u1=1,2,3,4T and Dn=un−pn≤10−3. The numerical results of these methods are shown in Tables 3–6. The control conditions are taken as follows: (1) Algorithm 2 in [37] (shortly, Method-1): δ1=0.05,μ=0.33,ϕn=1/20n+2; (2) Algorithm 1 in [36] (shortly, Method-2): δ1=0.05,γ=0.33,ϕn=1/20n+2,fu=u/4; (3) Algorithm 1 (shortly, Method-3): δ1=0.05,μ=0.33,ψn=1/20n+2,ϕn=6/101−ψn,φn=100/n+12; and (4) Algorithm 2 (shortly, Method-4): δ1=0.05,μ=0.33,ψn=1/20n+2,fu=u/4,φn=100/n+12.Example 14.
This test problem is taken from [42]. Let a mapping L:ℝ2⟶ℝ2 be defined by
(130)Lu=0.5u1u2−2u2−107−4u1−0.1u22−107,where
(131)A=u∈ℝ2:u1−22+u2−22≤1.
We can easily observe that the mappingL is not monotone on A but pseudomonotone and Lipschitz continuous with L=5. This problem has a unique solution that is u∗=2.707,2.707T. The starting point for this experiment is taken differently, and Dn=un−pn≤10−3. The numerical results of these methods are shown in Figures 7–10 and Table 7. The control conditions are taken as follows: (1) Algorithm 2 in [37] (shortly, Method-1): δ1=0.333,μ=0.90,ϕn=1/3n+2; (2) Algorithm 1 in [36] (shortly, Method-2): δ1=0.333,γ=0.90,ϕn=1/2n+2,fu=u/5; (3) Algorithm 1 (shortly, Method-3): δ1=0.333,μ=0.90,ψn=1/2n+2,ϕn=5/101−ψn,φn=100/n+12; and (4) Algorithm 2 (shortly, Method-4): δ1=0.333,μ=0.90,ψn=1/2n+2,fu=u/5,φn=100/n+12.Table 2
Algorithmic study for Example12.
mMethod-1Method-2Method-3Method-4Iter.TimeIter.TimeIter.TimeIter.Time14820.2940287004060.2118526003490.1667209003080.166098200t22680.1364829201960.1121846001550.1037283001120.094193700et3710.1452820003110.1235520002550.0997482602110.073510000sint4310.2128492003890.2016822002790.1129401002390.102749200Table 3
Example13: algorithmic description of Algorithm 2 in [37] and u1=1,2,3,4T.
Iter. (n)u1u2u3u414.5999999990233818.546499999911718.700499999915317.59449999931182-6.2676000489950616.737888658086316.967114165091415.85481014094423-5.9141818148370516.030879530064016.298643761716615.29195493877994-5.389214794663564.788016500282114.644612849747505.3871430432641951.005804065962494.732412741862974.628702401886355.2517959689412461.230130139517254.706775426477294.635929298731704.9714285714181571.449509079935504.702136559308654.659125357195774.9749999998484781.667178860071894.716600157058474.697510984048194.9777777649842691.885633184715144.748719604831094.750385811908074.97999999900850102.107078122450624.797812815123874.817594123919644.98181818166635112.333597923531434.863774679083334.899414582773164.98333333327664122.566670540079204.946671141936124.982654268795774.98294766642007132.799245401306814.960317452891424.970369493336734.98462062077961143.030334549909154.961240453793474.973562614220284.98546003308369153.260127410950864.963586869488724.975052594644894.98655189412727⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮10864.999248431191924.999442804752274.999637177453584.9998315492959810874.999249121896304.999443316849104.999637510944444.9998317041824610884.999249811332274.999443828005464.999637843822784.9998318587843410894.999250499503334.999444338223954.999638176090274.9998320131024010904.999251186412964.999444847507154.999638507748604.9998321671374310914.999251872064624.999445355857624.999638838799444.9998323208902010924.999252556461774.999445863277934.999639169244464.9998324743615010934.999253239607834.999446369770614.999639499085324.9998326275520810944.999253921506264.999446875338234.999639828323684.9998327804627310954.999254602160444.999447379983304.999640156961184.9998329330942010964.999255281573804.999447883708354.999640484999474.9998330854472710974.999255959749714.999448386515894.999640812440184.9998332375226910984.999256636691564.999448888408444.999641139284954.9998333893212210994.999257312402714.999449389388484.999641465535404.9998335408436111004.999257986886524.999449889458494.999641791193164.9998336920906311014.999258660146324.999450388620974.999642116259834.9998338430630111024.999259332185444.999450886878374.999642440737024.9998339937615211034.999260003007214.999451384233164.999642764626344.9998341441868811044.999260672614924.999451880687784.999643087929384.9998342943398411054.999261341011874.999452376244684.999643410647744.9998344442211511064.999262008201344.999452870906294.999643732782994.9998345938315411074.999262674186604.999453364675044.999644054336714.99983474317174CPU time is seconds8.339245Table 4
Example13: algorithmic description of Algorithm 1 in [36] and u1=1,2,3,4T.
Iter. (n)u1u2u3u414.5499999990233818.446499999911718.550499999915317.39449999931182-6.2422279744061613.524185113797413.621347199230412.53026190817713-0.6598761194460774.682381276469574.651943300931695.0032916246231440.9468024776258824.610846007702984.581191710177744.9236249795804451.049343900097194.557664901943214.528684930671424.8611834739502061.148542261673864.518243321234364.489868163656734.8127077953730871.245745906299484.489466981648604.461648399058124.7755068643897781.341460562341254.469330501447834.442033578361964.7472988897661991.436273118385684.456472206639724.429670807746644.72665515468439101.530652370509344.449942992249694.423618033093174.71255603858976111.624985305584094.449071783309464.423209088228004.70425974095071121.719620006244844.453382487673894.427971588785284.70123046557113131.814883647984924.462542508026354.437575647525824.70308628647431141.911092174939184.476328573518914.451800464742994.70956149682268152.008557762730814.494603890187284.470512241497274.72048089550204⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮10814.999490782315314.999490782315314.999490782315314.9994907823153110824.999491300391424.999491300391424.999491300391424.9994913003914210834.999491817414424.999491817414424.999491817414424.9994918174144210844.999492333387524.999492333387524.999492333387524.9994923333875210854.999492848313924.999492848313924.999492848313924.9994928483139210864.999493362196794.999493362196794.999493362196794.9994933621967910874.999493875039314.999493875039314.999493875039314.9994938750393110884.999494386844634.999494386844634.999494386844634.9994943868446310894.999494897615904.999494897615904.999494897615904.9994948976159010904.999495407356244.999495407356244.999495407356244.9994954073562410914.999495916068784.999495916068784.999495916068784.9994959160687810924.999496423756614.999496423756614.999496423756614.9994964237566110934.999496930422844.999496930422844.999496930422844.9994969304228410944.999497436070544.999497436070544.999497436070544.9994974360705410954.999497940702784.999497940702784.999497940702784.9994979407027810964.999498444322624.999498444322624.999498444322624.9994984443226210974.999498946933104.999498946933104.999498946933104.9994989469331010984.999499448537254.999499448537254.999499448537254.9994994485372510994.999499949138094.999499949138094.999499949138094.9994999491380910004.999500448738634.999500448738634.999500448738634.9995004487386310014.999500947341864.999500947341864.999500947341864.99950094734186CPU time is seconds7.690413Table 5
Example13: algorithmic description of Algorithm 1 and u1=1,2,3,4T.
Iter. (n)u1u2u3u414.7666666656357919.399083333240119.472749999910618.21641666594032-6.8991205933923818.161428019900118.243006386948116.83574254330783-6.7944315252030017.889626378004217.969845105601316.58603565935494-6.485779088840554.748765813466004.689574090650685.7429677075938150.9672594584264784.733860920542184.676005068238315.2860026440435861.954051938437236.828025610817806.994833399887245.1666974324046673.196835027666845.046588345502395.035347413204225.0567061070995784.881550786055834.963084028009234.965006198083674.9613388349205291.290270819908606.209429572006196.323449814341446.1058307743988810-16.897726967928439.252349306896239.541676268464438.981064196815211-16.868698816115239.228825026995539.516544616554938.959047056415012-17.59361968105975.257779486759775.050595536139375.45334598035476130.9538043945366625.243407059183215.037792216869985.2372030839448214-2.9734072784829317.292948282610217.177248044509517.289686885996715-2.9844463546054917.291179150781317.175960997006117.2879313433204⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮7184.999137219601844.999669427358914.999669412668184.999669426931457194.999138418860684.999669886209444.999669871540724.999669885782627204.999139614790234.999670343787934.999670329141144.999670343361747214.999140807404324.999670800099654.999670785474744.999670799674107224.999141996716724.999671255149864.999671240546764.999671254724957234.999143182741104.999671708943804.999671694362444.999671708519517244.999144365491124.999672161486644.999672146926974.999672161062997254.999145544980344.999672612783544.999672598245504.999672612360527264.999146721222204.999673062839664.999673048323174.999673062417277274.999147894230034.999673511660124.999673497165134.999673511238357284.999149064017204.999673959249984.999673944776434.999673958828837294.999150230596994.999674405614294.999674391162124.999674405193777304.999151393982554.999674850758094.999674836327244.999674850338197314.999152554186974.999675294686384.999675280276784.999675294267097324.999153711223264.999675737404124.999675723015724.999675736985457334.999154865104394.999676178916284.999676164549014.999676178498227344.999156015843294.999676619227744.999676604881564.999676618810307354.999157163452744.999677058343424.999677044018264.999677057926607364.999158307945514.999677496268184.999677481963974.999677495851967374.999159449334264.999677933006844.999677918723534.999677932591237384.999160587631604.999678368564244.999678354301774.999678368149247394.999161722850064.999678802945164.999678788703464.999678802530767404.999162855002164.999679236154344.999679221933374.999679235740557414.999163984100284.999679668196534.999679653996224.999679667783347424.999165110156774.999680099076434.999680084896734.999680098663847434.999166233183884.999680528798744.999680514639584.999680528386747444.999167353193824.999680957368114.999680943229454.999680956956727454.999168470198764.999681384789174.999681370670954.99968138437837CPU time is seconds5.182248Table 6
Example13: algorithmic description of Algorithm 2 and u1=1,2,3,4T.
Iter. (n)u1u2u3u413.7239999992556014.572109999932714.907269999935414.28202999947552-2.1545688339750714.181951824716914.535150810126413.87434725797983-1.9188456464498014.733550567952115.069950570376814.44070293037044-3.750319652705868.600665167800678.663012306906898.547191738705205-0.6103549058636965.970305383646585.973352336840895.967738396113416-0.7601800526336886.396128020723056.399157562750616.3935757016946870.8834263574105485.463342995549895.450418155494625.4742330349203482.242751257636555.115760786949065.121031380037415.1113142907121691.889040983496736.052440569251466.055293709328796.05002755957568102.941735362038575.296415434544505.297017045936785.29590634994195113.959637527201035.106933258873125.107098891355205.10679307211866124.055391135589025.224250124788155.224388493548655.22413300798106134.486402808268405.092455712779695.092518898560435.09240223042630144.636871951659055.074278861515745.074321058702545.07424314405336154.789119038549075.035281852586415.035304573878775.03526262019362⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮6164.999138618304954.999670285843384.999670324903284.999670252780066174.999140012593414.999670818259534.999670857253204.999670785252256184.999141402375394.999671348958984.999671387886654.999671316007576194.999142787672684.999671877950024.999671916811914.999671845054306204.999144168506954.999672405240904.999672444037224.999672372400686214.999145544899764.999672930839794.999672969570764.999672898054896224.999146916872474.999673454754844.999673493420674.999673422025076234.999148284446374.999673976994104.999674015595024.999673944319296244.999149647642564.999674497565624.999674536101834.999674464945586254.999151006482014.999675016477384.999675054949094.999674983911946264.999152360985624.999675533737284.999675572144714.999675501226266274.999153711174084.999676049353224.999676087696564.999676016896436284.999155057067974.999676563333014.999676601612494.999676530930296294.999156398687774.999677075684434.999677113900254.999677043335596314.999157736053794.999677586415214.999677624567574.999677554120086324.999159069186244.999678095533024.999678133622144.999678063291436334.999160398105194.999678603045494.999678641071564.999678570857266344.999161722830574.999679108960214.999679146923444.999679076825176354.999163043382214.999679613284704.999679651185304.999679581202686364.999164359779824.999680116026454.999680153864624.999680083997286374.999165672042934.999680617192914.999680654968854.999680585216426384.999166980191044.999681116791454.999681154505364.999681084867476394.999168284243434.999681614829454.999681652481514.99968158295780CPU time is seconds4.517220Figure 7
Algorithmic descriptions of Algorithm1 and Algorithm 2 as well as Algorithm 2 in [37] and Algorithm 1 in [36] when u1=1.5,1.7T.Figure 8
Algorithmic descriptions of Algorithm1 and Algorithm 2 as well as Algorithm 2 in [37] and Algorithm 1 in [36] when u1=0,0T.Figure 9
Algorithmic descriptions of Algorithm1 and Algorithm 2 as well as Algorithm 2 in [37] and Algorithm 1 in [36] when u1=10,10T.Figure 10
Algorithmic descriptions of Algorithm1 and Algorithm 2 as well as Algorithm 2 in [37] and Algorithm 1 in [36] when u1=−5,−5T.Table 7
Algorithmic study for Example14.
u1Method-1Method-2Method-3Method-4Iter.TimeIter.TimeIter.TimeIter.Time1.5,1.7T1709.381243001478.166378500815.082575400613.4896803000,0T26714.70651731669.6950111001075.894192000643.75333920010,10T72940.43051331608.596874100894.762929600643.520466200−5,−5T48328.035652701508.5705279001186.676780500603.331014400


## 5. Conclusion

Two nonmonotonic explicit extragradient-type methods for finding an approximate solution of variational inequalities involving pseudomonotone mapping in a real Hilbert space have been established. Two strongly convergent results, corresponding to the proposed algorithms have been proven. The numerical results were interpreted to demonstrate that the proposed algorithms worked numerically better than current methods. According to these numerical findings, the nonmonotone variable step size rule improves the efficiency of the iterative sequence in this case.



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*Source: 8327694-2021-12-08.xml*