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Research article

Viscosity-type method for solving pseudomonotone equilibrium problems in a real Hilbert space with applications

  • Received: 14 September 2020 Accepted: 13 November 2020 Published: 23 November 2020
  • MSC : 47H05, 47H10

  • The aim of this article is to introduce a new algorithm by integrating a viscosity-type method with the subgradient extragradient algorithm to solve the equilibrium problems involving pseudomonotone and Lipschitz-type continuous bifunction in a real Hilbert space. A strong convergence theorem is proved by the use of certain mild conditions on the bifunction as well as some restrictions on the iterative control parameters. Applications of the main results are also presented to address variational inequalities and fixed-point problems. The computational behaviour of the proposed algorithm on various test problems is described in comparison to other existing algorithms.

    Citation: Habib ur Rehman, Poom Kumam, Kanokwan Sitthithakerngkiet. Viscosity-type method for solving pseudomonotone equilibrium problems in a real Hilbert space with applications[J]. AIMS Mathematics, 2021, 6(2): 1538-1560. doi: 10.3934/math.2021093

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  • The aim of this article is to introduce a new algorithm by integrating a viscosity-type method with the subgradient extragradient algorithm to solve the equilibrium problems involving pseudomonotone and Lipschitz-type continuous bifunction in a real Hilbert space. A strong convergence theorem is proved by the use of certain mild conditions on the bifunction as well as some restrictions on the iterative control parameters. Applications of the main results are also presented to address variational inequalities and fixed-point problems. The computational behaviour of the proposed algorithm on various test problems is described in comparison to other existing algorithms.


    Assume that C is a closed and convex subset of a real Hilbert space H with the inner product and the induced norm are denoted by , and , respectively. Moreover, R be a set of real numbers during whole article. Let f:H×HR be a bifunction and satisfy f(v,v)=0 for all vC, the equilibrium problem (EP) [6,11] for a bifunction f on C is defined in the following way:

    FinduCsuchasf(u,v)0,vC.(EP)

    Moreover, SEP(f,C) denotes the solution set of an equilibrium problem over the set C and u is an arbitrary element of SEP(f,C). A metric projection PC(u) of uH onto a closed and convex subset C of H is defined by PC(u)=argminvCvu.

    In this article, the equilibrium problem is studied based on the following hypothesis:

    (a1) A bifunction f:H×HR is said to be (see [3,6]) pseudomonotone on C if

    f(u1,u2)0f(u2,u1)0,u1,u2C.

    (a2) A bifunction f:H×HR is said to be Lipschitz-type continuous [21] on C if there exist two constants c1,c2>0 such that

    f(u1,u3)f(u1,u2)+f(u2,u3)+c1u1u22+c2u2u32,u1,u2,u3C.

    (a3)lim supnf(un,v)f(p,v) for all vC and {un}C satisfies unp;

    (a4)f(u,) is subdifferentiable and convex on H for every each uH.

    The above-defined problem (EP) is a general mathematical problem in the sense that it unifies a number of mathematical problems, i.e., the fixed point problems, the vector and scalar minimization problems, the problems of variational inequalities (VIP), the complementarity problems, the saddle point problems, the Nash equilibrium problems in non-cooperative games and the inverse optimization problems [4,6,24,41]. The problem (EP) is also known as the well-known Ky Fan inequality due to his initial contribution [11]. Many authors have developed and generalized many results on the existence and nature of the solution of an equilibrium problem (see for more detail [2,4,11]). Due to the significance of the problem (EP) and its applications in both pure and applied sciences, many researchers have studied it extensively in recent years [5,10,14].

    A proximal point method is used to solve the problem (EP) based on mathematical programming [13]. This method was also known as the two-step extragradient method in [34] due to initial contribution of Korpelevich [18] to solve saddle point problems. Tran et al. in [34] established an iterative sequence {un} in the following way:

    {u0C,vn=argminvC{ρf(un,v)+12unv2},un+1=argminvC{ρf(vn,v)+12unv2},

    where 0<ρ<min{12c1,12c2}. Recently, many existing methods have been extended in the case of problem (EP) in finite and infinite-dimensional spaces, such as the proximal point-like methods [13,22], the extragradient-like methods [17,19,26,27,32], the subetaadient extragradient methods [1,29,36,37], the inertial methods [35,39] and others in [9,12,16,25,30,33,38].

    Inspired by the results in [8,16,23], in this paper, we introduce a viscosity-type subetaadient extragradient algorithm to solve the equilibrium problems involving pseudomonotone bifunction. A strong convergence theorem for the proposed algorithm is well-established by considering certain mild conditions on bifunction and control parameters. Some applications for our main results are studied to solve two particular classes of an equilibrium problem. In the end, the computational studies show that the new method is more efficient than the existing ones [16,34].

    The remainder of this article has been organized as follows: Section 2 includes some preliminary and basic results. Section 3 contains proposed algorithm and corresponding strong convergence result. Section 4 contains applications of our main results. Section 5 involves the numerical discussion of the proposed method compared to existing ones.

    A normal cone of C at uC is defined by

    NC(u)={wH:w,vu0,vC}.

    Let φ:CR is convex function. The subdifferential of φ at uC is defined by

    φ(u)={wH:φ(v)φ(u)w,vu,vC}.

    Lemma 2.1. [15] Assume that PC:HC is a metric projection such that

    (ⅰ)

    u1PC(u2)2+PC(u2)u22u1u22,u1C,u2H.

    (ⅱ) u3=PC(u1) if and only if

    u1u3,u2u30,u2C.

    (ⅲ)

    u1PC(u1)u1u2,u2C,u1H.

    Lemma 2.2. [40] Assume that {an}(0,+) is a sequence satisfying

    an+1(1γn)an+γnδn,nN,

    where {γn}(0,1) and {δn}R satisfies the following conditions:

    limnγn=0,+n=1γn=+andlim supnδn0.

    Then, limnan=0.

    Lemma 2.3. [20] Assume that a sequence {an}R and there exists a subsequence {ni} of {n} such that ani<ani+1, for all iN. Then, there is a non decreasing sequence {mk}N such that mk+ as k, and the following conditions are fulfilled by all (sufficiently large) numbers kN,

    amkamk+1andakamk+1.

    In fact mk is the largest number n in the set {1,2,,k} such that anan+1.

    Lemma 2.4. [15] For each u1,u2H and δR, then the following relationships are true.

    (ⅰ)

    δu1+(1δ)u22=δu12+(1δ)u22δ(1δ)u1u22.

    (ⅱ)

    u1+u22u12+2u2,u1+u2.

    Lemma 2.5. (Theorem 27.4 [28]) Let φ:CR be a proper convex, subdifferentiable and lower semi-continuous function on C. An element uC is a minimizer of a function φ iff

    0φ(u)+NC(u),

    where φ(u) stands for the sub-differential of φ at uC and NC(u) the normal cone of C at u.

    In this section, we present an iterative scheme for solving pseudomonotone equilibrium problems that is based on Tran et al. in [34] and viscosity scheme [23]. It is important to note that the proposed method has a straightforward structure for achieving strong convergence. Suppose that g:HH be a strict contraction function with constant ξ[0,1). The main algorithm has been presented as follows:

    Remark 3.1. It can be easily prove that CHn. By vn and Lemma 2.5, we have

    02{ρf(un,v)+12unv2}(vn)+NC(vn).

    Indeed, for ωnf(un,vn) there exists ¯ωnNC(vn) such that

    ρωn+vnun+¯ωn=0.

    Thus, we have

    unvn,vvn=ρωn,vvn+¯ωn,vvn,vC.

    Due to ¯ωnNC(vn) means that ¯ωn,vvn0, for all vC. It implies that

    unvn,vvnρωn,vvn,vC,

    which imply that unρωnvn,vvn0,vC. It proves that CHn for each nN.

    Theorem 3.1. Assume that {un} is a sequence generated by Algorithm 1 and for some uSEP(f,C). Then, {un} converges strongly to u=PSEP(f,C)g(u).

    Algorithm 1 (A Viscosity Method for Pseudomonotone Equilibrium Problems)
    Step 0: Let u0C, 0<ρ<min{12c1,12c2} and a sequence χn(0,1) satisfies the conditions, i.e.,
    limnχn=0and+nχn=+.
    Step 1: Compute
    vn=argminvC{ρf(un,v)+12unv2}.
    If un=vn, then stop the sequence. Otherwise, go to Step 2.
    Step 2: Compute
    Hn={wH:unρωnvn,wvn0},
    where ωn2f(un,vn) and evaluate
    tn=argminvHn{ρf(vn,v)+12unv2}.
    Step 3: Compute
    un+1=χng(un)+(1χn)tn,
    where g is a contraction. Set n:=n+1 and go back to Step 1.

    Proof. Claim 1: The {un} sequence is bounded.

    By Lemma 2.5, we have

    02(ρf(vn,v)+12unv2)(tn)+NHn(tn).

    Thus, there exists ωnf(vn,tn) and ¯ωnNHn(tn) such that ρωn+tnun+¯ωn=0. Thus,

    untn,vtn=ρωn,vtn+¯ωn,vtn,vHn.

    Since ¯ωnNHn(tn) follows that ¯ωn,vtn0, for all vHn. Thus, we have

    ρωn,vtnuntn,vtn,vHn. (3.1)

    Since ωnf(vn,tn) and using the subdifferential definition, we get

    f(vn,v)f(vn,tn)ωn,vtn,vH. (3.2)

    Combining expressions (3.1) and (3.2), we obtain

    ρf(vn,v)ρf(vn,tn)untn,vtn,vHn. (3.3)

    Substituting v=u in expression (3.3), we obtain

    ρf(vn,u)ρf(vn,tn)untn,utn. (3.4)

    Since uSEP(f,C), it follows that f(u,vn)0, implies that f(vn,u)0 due to the pseudomonotonicity of the bifunction f. From expression (3.4), we have

    untn,tnuρf(vn,tn). (3.5)

    Due to the Lipschitz-type condition on a bifunction f, we obtain

    f(un,tn)f(un,vn)+f(vn,tn)+c1unvn2+c2vntn2. (3.6)

    Combining expressions (3.5) and (3.6), we have

    untn,tnuρ{f(un,tn)f(un,vn)}c1ρunvn2c2ρvntn2. (3.7)

    Since tnHn and it gives that unρωnvn,tnvn0, which implies that

    unvn,tnvnρωn,tnvn. (3.8)

    Since ωn2f(un,vn) and using the subdifferential definition, we obtain

    f(un,v)f(un,vn)ωn,vvn,vH.

    Substituting v=tn in the above expression, we get

    f(un,tn)f(un,vn)ωn,tnvn. (3.9)

    It follows from inequalities (3.8) and (3.9) that

    ρ{f(un,tn)f(un,vn)}unvn,tnvn. (3.10)

    From (3.7) and (3.10), we have

    untn,tnuunvn,tnvnc1ρunvn2c2ρvntn2. (3.11)

    We have the following equalities:

    2untn,tnu=unu2tnun2tnu2

    and

    2vnun,vntn=unvn2+tnvn2untn2.

    The above facts and (3.11) implies that

    tnu2unu2(12c1ρ)unvn2(12c2ρ)tnvn2. (3.12)

    It is given that uSEP(f,C). From the definition of sequence {un+1} and due to the fact that g is a contraction with ξ[0,1), we have

    un+1u=χng(un)+(1χn)tnu=χn[g(un)u]+(1χn)[tnu]=χn[g(un)+g(u)g(u)u]+(1χn)[tnu]χng(un)g(u)+χng(u)u+(1χn)tnuχnξunu+χng(u)u+(1χn)tnu. (3.13)

    Combining expressions (3.12) and (3.13) and χn(0,1), we deduce that

    un+1uχnξunu+χng(u)u+(1χn)unu=[1χn+ξχn]unu+χn(1ξ)g(u)u(1ξ)max{unu,g(u)u(1ξ)}max{u0u,g(u)u(1ξ)}. (3.14)

    Thus, we conclude that the {un} is bounded sequence. Due to the reflexivity of H and the boundedness of {un} guarantees that there exists a subsequence {unk} of {un} such that {unk}uH as k.

    Claim 2: The sequence {un} is strongly convergent.

    Next, we prove the strong convergence of the iterative sequence {un} generated by Algorithm 1. The Lipschitz-continuity and pseudomonotone property of the bifunction f implies that the solution set SEP(f,C) is a closed and convex set (for more details see [34]). Since the mapping is a contraction and so does PSEP(f,C)g. Now, we are in position to use the Banach contraction theorem for the existence of a unique fixed point uSEP(f,C) such that

    u=PSEP(f,C)(g(u)).

    By using Lemma 2.1 (ⅱ), we have

    g(u)u,vu0,vSEP(f,C). (3.15)

    By Lemma 2.4 (ⅰ) and (3.12), we have

    un+1u2=χng(un)+(1χn)tnu2=χn[g(un)u]+(1χn)[tnu]2=χng(un)u2+(1χn)tnu2χn(1χn)g(un)tn2χng(un)u2+(1χn)[unu2(12c1ρ)unvn2(12c2ρ)tnvn2]χn(1χn)g(un)tn2χng(un)u2+unu2(1χn)(12c1ρ)unvn2(1χn)(12c2ρ)tnvn2. (3.16)

    The remainder of the proof shall be split into the following two parts:

    Case 1: Assume that there is a fixed number N1N such that

    un+1uunu,nN1. (3.17)

    Thus, above implies that limnunu exists and let limnunu=l. From (3.16), we get

    (1χn)(12c1ρ)unvn2+(1χn)(12c2ρ)tnvn2χng(un)u2+unu2un+1u2. (3.18)

    Due to the existence of limnunu=l and χn0, we infer that

    limnunvn=limntnvn=0. (3.19)

    It follows that

    limnuntnlimnunvn+limnvntn=0. (3.20)

    We can also obtain

    un+1un=χng(un)+(1χn)tnun=χn[g(un)un]+(1χn)[tnun]χng(un)un+(1χn)tnun0. (3.21)

    The above expression implies that

    limnun+1un=0. (3.22)

    Thus, implies that the sequences {vn} and {tn} are bounded. Let {unk} be subsequence of {un} such that {unk} converges weakly to ˆuH. Next, we need to prove that ˆuSEP(f,C). Due to the inequality (3.3), the Lipschitz-like condition of f and (3.10), we obtain

    ρf(vnk,v)ρf(vnk,tnk)+unktnk,vtnkρf(unk,tnk)ρf(unk,vnk)c1ρunkvnk2c2ρvnktnk2+unktnk,vtnkunkvnk,tnkvnkc1ρunkvnk2c2ρvnktnk2+unktnk,vtnk, (3.23)

    where v is an arbitrary member in Hn. From (3.19), (3.20) and the boundedness of {un} imply that right-hand side goes to zero. From ρ>0, the condition (a3) and vnkˆu, we obtain

    0lim supkf(vnk,v)f(ˆu,v),vHn. (3.24)

    It follows that f(ˆu,v)0, for all vC, and hence ˆuSEP(f,C). Next, we consider

    lim supng(u)u,unu=lim supkg(u)u,unku=g(u)u,ˆuu0. (3.25)

    We have limnun+1un=0. We can deduce that

    lim supng(u)u,un+1ulim supng(u)u,un+1un+lim supng(u)u,unu0. (3.26)

    From Lemma 2.4(ⅱ) and (3.12), we have

    un+1u2=χng(un)+(1χn)tnu2=χn[g(un)u]+(1χn)[tnu]2(1χn)2tnu2+2χng(un)u,(1χn)[tnu]+χn[g(un)u]=(1χn)2tnu2+2χng(un)g(u)+g(u)u,un+1u=(1χn)2tnu2+2χng(un)g(u),un+1u+2χng(u)u,un+1u(1χn)2tnu2+2χnξunuun+1u+2χng(u)u,un+1u(1+χ2n2χn)unu2+2χnξunu2+2χng(u)u,un+1u=(12χn)unu2+χ2nunu2+2χnξunu2+2χng(u)u,un+1u=[12χn(1ξ)]unu2+2χn(1ξ)[χnunu22(1ξ)+g(u)u,un+1u1ξ]. (3.27)

    It follows from expressions (3.26) and (3.27), we obtain

    lim supn[χnunu22(1ξ)+g(u)u,un+1u1ξ]0. (3.28)

    Choose nN2N (N2N1) large enough such that 2χn(1ξ)<1. By using (3.27), (3.28) and applying Lemma 2.2, we conclude that unu0 as n.

    Case 2: Assume there is a subsequence {ni} of {n} such that

    uniuuni+1u,iN.

    Thus, by Lemma 2.3 there is a sequence {mk}N as {mk}, such that

    umkuumk+1uandukuumk+1u,for allkN. (3.29)

    Similar to Case 1, the expression (3.16) provides that

    (1χmk)(12c1ρ)umkvmk2+(1χmk)(12c2ρ)tmkvmk2χmkg(umk)u2+umku2umk+1u2. (3.30)

    Due to χmk0, we deduce the following:

    limkumkvmk=limktmkvmk=0. (3.31)

    Also, we can obtain

    umk+1umk=χmkg(umk)+(1χmk)tmkumk=χmk[g(umk)umk]+(1χmk)[tmkumk]χmkg(umk)umk+(1χmk)tmkumk0. (3.32)

    We have to use the same justification as in the Case 1, such that

    lim supkg(u)u,umk+1u0. (3.33)

    By using expressions (3.27) and (3.29), we have

    umk+1u2[12χmk(1ξ)]umku2+2χmk(1ξ)[χmkumku22(1ξ)+g(u)u,umk+1u1ξ][12χmk(1ξ)]umk+1u2+2χmk(1ξ)[χmkumku22(1ξ)+g(u)u,umk+1u1ξ]. (3.34)

    It follows that

    umk+1u2χmkumku22(1ξ)+g(u)u,umk+1u1ξ. (3.35)

    Since χmk0, and umku is a bounded. Thus, expressions (3.33) and (3.35) implies that

    umk+1u20,ask. (3.36)

    The above implies that

    limkuku2limkumk+1u20. (3.37)

    Consequently, unu. This completes the proof of the theorem.

    Remark 3.2. If we define g(u)=u0 in Algorithm 1, we obtain the Halpern subetaadient extragradient method in [16].

    Now, we consider the application of our main results to solve the problem of classic variational inequalities [31] for an operator G:HH is defined in the following way:

    FinsuCsuch thatG(u),vu0,vC.(VIP)

    We consider the following conditions to study variational inequalities.

    (b1) The solution set of the problem (VIP) denoted by VI(G,C) is nonempty.

    (b2)G:HH is said to be a pseudomonotone, i.e.,

    G(u),vu0G(v),uv0,u,vC.

    (b3) G:HH is said to be a Lipschitz continuous if there exits a constants L>0 such that

    G(u)G(v)Luv,u,vC;

    (b4)G:HH is called to be sequentially weakly continuous, i.e., {G(un)} weakly converges to G(u) for every sequence {un} converges weakly to u.

    If we define f(u,v):=G(u),vu, for all u,vC. Then, the equilibrium problem becomes the problem of variational inequalities described above where L=2c1=2c2. From the above value of the bifunction f, we have

    {vn=argminvC{ρf(un,v)+12unv2}=PC(unρG(un)),tn=argminvHn{ρf(vn,v)+12unv2}=PHn(unρG(vn)). (4.1)

    As a consequence of the results in Section 3, we have the following results:

    Corollary 4.1. Let G:CH be a mapping satisfying the conditions (b1)(b4). Choose u0C, 0<ρ<1L and a sequence χn(0,1) satisfying the conditions, i.e.,

    limnχn=0and+nχn=+.

    Assume that {un} is the sequence generated in the following way:

    {vn=PC(unρG(un)),tn=PHn(unρG(vn)),un+1=χng(un)+(1χn)tn,

    where

    Hn={zH:unρG(un)vn,zvn0}.

    Then, the sequence {un} converges strongly to uVI(G,C).

    Note that condition (b4) can be deleted in the case when G is monotone. Indeed, this condition is a particular case of condition (a3) is used to prove (3.24). Without condition (b4), the inequality (3.23) is obtained by imposing the monotonocity on G. In that case, we obtain

    G(v),vvnG(vn),vvn,vC. (4.2)

    By the use f(u,v)=G(u),vu in (3.23), we get

    lim supkG(vnk),vvnk0,vHn. (4.3)

    Combining (4.2) with (4.3), we obtain

    lim supkG(v),vvnk0,vC. (4.4)

    Let vs=(1s)ˆu+sy, for all s[0,1]. Due to convexity of C implies that vsC for any s(0,1). Since vnkˆuC and G(v),vˆu0 for every vC. Thus, we have

    0G(vs),vsˆu=sG(vs),vˆu. (4.5)

    Therefore, G(vs),vˆu0, for all s(0,1). Since vsˆu as s0 and due to continuity of G, we have G(ˆu),vˆu0, for all vC, which implies that ˆuVI(G,C).

    Remark 4.1. From the above discussion, it can be concluded that the Corollary 4.1 still holds, even if we remove the condition (b4) in case of monotone variational inequaltiy.

    Next, we consider the application of our results to solve fixed-point problems involving κ-strict pseudocontraction mapping and the fixed point problem for an operator T:HH is defined in the following way:

    FinduCsuch thatT(u)=u.(FPP)

    We assume that the following requirements have been met to study fixed point problems.

    (c1)T:CC is said to be a κ-strict pseudo-contraction [7] on C if

    TuTv2uv2+κ(uTu)(vTv)2,u,vC;

    (c2) weakly sequentially continuous on C if

    T(un)T(u)for any sequence inCsatisfyunu.

    If we consider that the mapping T is a κ-strict pseudocontraction and weakly continuous then f(u,v)=uTu,vu satisfies the conditions (a1)(a4) and 2c1=2c2=32κ1κ. The values of vn and tn in Algorithm 1, we have

    {vn=argminvC{ρf(un,v)+12unv2}=PC[unρ(unT(un))],tn=argminvHn{ρf(vn,v)+12unv2}=PHn[unρ(vnT(vn))]. (4.6)

    Corollary 4.2. Let C be a nonempty, convex and closed subset of a Hilbert space H and T:CC is a κ-strict pseudocontraction and weakly continuous with solution set Fix(T). Choose u0C, 0<ρ<1κ32κ and sequence χn(0,1) satisfies the conditions, i.e.,

    limnχn=0and+nχn=+.

    Let {un} be the sequence generated in the following way:

    {vn=PC[unρ(unT(un))],tn=PHn[unρ(vnT(vn))],un+1=χng(un)+(1χn)tn.

    where

    Hn={wH:(1ρ)un+ρT(un)vn,wvn0}.

    Then, sequence {un} strongly converges to uFix(T).

    Numerical results are presented in this section to show the efficiency of the proposed method. The MATLAB codes was run in MATLAB version 9.5 (R2018b) on the Intel(R) Core(TM)i5-6200 CPU PC @ 2.30GHz 2.40GHz, RAM 8.00 GB.

    Example 5.1. Assume that f:C×CR is defined by

    f(u,v)=Pu+Qv+c,vu,u,vC,

    where cRn and P, Q are matrices of order n. The matrix P is symmetric positive semi-definite and the matrix QP is symmetric negative semi-definite with Lipschitz-type constants c1=c2=12PQ (see [34] for details). The matrices P,Q and vector c are defined by

    P=(3.1200023.6000003.5200023.3000003)Q=(1.6100011.6000001.5100011.5000002)c=(12121).

    The constraint set CRn is considered as C:={uRn:5ui5}. Furthermore, control parameters conditions are taken as follows: ρ=14c1 and Dn=unvn for Algorithm 1 (EgM) in [34]; ρ=14c1, χn=1100(n+2) and Dn=unvn for Algorithm 2 (H-EgM) in [16]; ρ=14c1,χn=1100(n+2), g(u)=u2 and Dn=unvn for Algorithm 1 (V-EgM). The numerical and graphical results of three methods are shown in Figures 14 and Table 1.

    Figure 1.  Numerical behaviour of Algorithm 1 with Algorithm 1 in [34] and Algorithm 3.2 in [16] while u0=(0,0,0,0,0)T.
    Figure 2.  Numerical behaviour of Algorithm 1 with Algorithm 1 in [34] and Algorithm 3.2 in [16] while u0=(2,2,2,2,2)T.
    Figure 3.  Numerical behaviour of Algorithm 1 with Algorithm 1 in [34] and Algorithm 3.2 in [16] while u0=(1,0,1,2,1)T.
    Figure 4.  Numerical behaviour of Algorithm 1 with Algorithm 1 in [34] and Algorithm 3.2 in [16] while u0=(2,1,3,4,5)T.
    Table 1.  Numerical results values for Figures 14.
    Number of Iterations Execution Time in Seconds
    u0 EgM H-EgM V-EgM EgM H-EgM V-EgM
    (0,0,0,0,0)T 27 32 21 0.217359 0.242564 0.226170
    (2,2,2,2,2)T 27 60 21 0.264663 0.553107 0.175542
    (1,0,1,2,1)T 31 49 24 0.292574 0.576171 0.219645
    (2,1,3,4,5)T 32 80 25 0.303238 0.635728 0.204971

     | Show Table
    DownLoad: CSV

    Example 5.2. Let f:C×CR defined in the following way:

    f(u,v)=5i=2(viui)u,u,vR5,

    where C={(u1,,u5):u11,ui1,i=2,,5}. Thus, the bifunction f is Lipschitz-type continuous with c1=c2=2, and satisfies the conditions (a1)(a4). The solution set of an equilibrium problem is EP(f,C)={(u1,1,1,1,1):u1>1}. Furthermore, the control conditions ρ=16c1 for Algorithm 1 (EgM) in [34]; ρ=16c1 and χn=1200(n+2) for Algorithm 2 (H-EgM) in [16]; ρ=16c1,χn=1100(n+2) and g(u)=u3 for Algorithm 1 (V-EgM). The numerical results of three methods are shown in Tables 24.

    Table 2.  Example 5.2: Numerical study of Algorithm 1 in [34] with TOL = 103 and u0=(2,10,13,5,3)T.
    Iter (n) u1 u2 u3 u4 u5
    1 1.9999999902 8.9034597770 11.903459749 3.9034598398 1.9034598209
    2 2.0000001112 7.9193730581 10.919372983 2.9193733082 1.0000084500
    3 2.0000001050 7.0300765337 10.030076426 2.0300768272 1.0000000224
    4 2.0000002139 6.2245554051 9.2245552403 1.2245575794 1.0000004995
    5 2.0000002018 5.4892537351 8.4892535376 1.0000000391 1.0000000272
    6 2.0000001984 4.8171736152 7.8171733928 1.0000000297 1.0000000297
    26 2.0000015507 1.0000019717 1.0384612829 1.0000019717 1.0000019717
    27 2.0000016692 1.0000021759 1.0000028248 1.0000021759 1.0000021759
    28 2.0000017877 1.0000021759 1.0000021759 1.0000021759 1.0000021759
    CPU time is seconds 0.850380

     | Show Table
    DownLoad: CSV
    Table 3.  Example 5.2: Numerical study of Algorithm 3.2 in [16] with TOL = 103 and u0=(2,10,13,5,3)T.
    Iter (n) u1 u2 u3 u4 u5
    1 1.999999942 8.7515611443 11.751561124 3.7515611399 1.7515612532
    2 1.999999938 7.6461082189 10.646108180 2.6461082675 1.0016667229
    3 1.999999939 6.6610509620 9.6610509019 1.6610510862 1.0012500202
    4 2.000000061 5.7776478420 8.7776477159 1.0020019229 1.0010004543
    5 2.000000055 4.9810038039 7.9810036568 1.0016666917 1.0008333584
    6 2.000000048 4.2622453450 7.2622451730 1.0014285992 1.0007143134
    38 2.0000028210 1.0005787475 1.0007710569 1.0002582317 1.0001300255
    39 2.0000029393 1.0005643243 1.0007518260 1.0002518214 1.0001268203
    40 2.0000030576 1.0005506046 1.0007335331 1.0002457238 1.0001237715
    CPU time is seconds 1.191645

     | Show Table
    DownLoad: CSV
    Table 4.  Example 5.2: Numerical study of Algorithm 1 with TOL = 103 and u0=(2,10,13,5,3)T.
    Iter (n) u1 u2 u3 u4 u5
    1 1.9950000119 8.5308876762 11.523387660 3.5433877312 1.54838779261
    2 1.9916750267 7.2594419489 10.246954421 2.2802545702 0.99924734095
    3 1.9891855417 6.1501278861 9.1339059323 1.1771670518 0.99874942183
    4 1.9871963569 5.1700590512 8.1508532923 0.9991771920 0.99899876985
    5 1.9855403611 4.3036279625 7.2819381754 0.9991660040 0.99916585540
    6 1.9841221147 3.5358595487 6.5120423683 0.9992851446 0.99928514452
    19 1.9741765217 0.99975135569 1.0030932482 0.99975135569 0.99975135569
    20 1.9737065994 0.99976333843 0.9997641538 0.99976333843 0.99976333843
    21 1.9732581496 0.99977416665 0.99977416684 0.99977416665 0.99977416665
    CPU time is seconds 0.711785

     | Show Table
    DownLoad: CSV

    Example 5.3. Suppose that H=L2([0,1]) is a Hilbert space with the inner product u,v=10u(t)v(t)dt, for all u,vH and the induced norm is

    u=10|u(t)|2dt.

    Moreover, assume that C:={uL2([0,1]):u1}. Assume that G:CH is defined by

    G(u)(t)=10[u(t)H(t,s)f(u(s))]ds+g(t)

    where H(t,s)=2tse(t+s)ee21,f(u)=cos(u) and g(t)=2tetee21. We consider the bifunction f(u,v)=G(u),vu with the Lipschitz-type continuous with c1=c2=1. Furthermore, control conditions ρ=15c1 for Algorithm 1 (EgM) in [34]; ρ=15c1 and χn=1300(n+2) for Algorithm 2 (H-EgM) in [16]; ρ=15c1,χn=1100(n+2) and g(u)=u2 for Algorithm 1 (V-EgM). The numerical results of three methods are shown in Figures 5 and 6 and Table 5.

    Figure 5.  Numerical study of Algorithm 1 with Algorithm 1 in [34] and Algorithm 3.2 in [16].
    Figure 6.  Numerical study of Algorithm 1 with Algorithm 1 in [34] and Algorithm 3.2 in [16].
    Table 5.  Numerical results values for Figures 5 and 6.
    Number of Iterations Execution Time in Seconds
    u0 EgM H-EgM V-EgM EgM H-EgM V-EgM
    t 49 106 41 0.058515 0.107587 0.052661
    3t2 52 54 40 0.057655 0.058550 0.042137
    sin(t) 52 76 41 0.056489 0.083268 0.057777
    exp(t) 55 97 52 0.130014 0.106885 0.061795

     | Show Table
    DownLoad: CSV

    Example 5.4. Assume that f:H×HR is defined by

    f(u,v)=(5u)u,vu,u,vH,

    where H=l2 is a real Hilbert space having the elements are square-summable sequences and C={uH:u3}. The bifunction f is Lipschitz-type continuous and value of Lipschitz-constants are c1=c2=112. Furthermore, control conditions ρ=14c1 for Algorithm 1 (EgM) in [34]; ρ=14c1 and χn=1200(n+2) for Algorithm 2 (H-EgM) in [16]; ρ=14c1,χn=1100(n+2) and g(u)=u2 for Algorithm 1 (V-EgM). The numerical results of three methods are shown in Figures 7 and 8 and Table 6. The projection onto C is evaluated in the following way:

    PC(u)={uifu3,3uu,else.
    Figure 7.  Numerical study of Algorithm 1 with Algorithm 1 in [34] and Algorithm 3.2 in [16] when u0=(1,1,,1500,0,0,)..
    Figure 8.  Numerical study of Algorithm 1 with Algorithm 1 in [34] and Algorithm 3.2 in [16] when u0=(1,2,,1000,0,0,)..
    Table 6.  Numerical results values for Figures 7 and 8.
    Number of Iterations Execution Time in Seconds
    u0 EgM H-EgM V-EgM EgM H-EgM V-EgM
    (1,1,,1500,0,0,) 39 59 33 1.233697 1.849991 1.8499918
    (1,2,,1000,0,0,) 55 68 49 1.773837 2.065840 1.5172537

     | Show Table
    DownLoad: CSV

    We have studied a viscosity type extragradient-like method for determining the solution of pseudomonotone equilibrium problem in real Hilbert spaces and also prove that the generated sequence converges strongly to the solution. Numerical conclusions were drawn to explain the numerical efficiency of our algorithms in comparison to other methods. These numerical studies showed that viscosity influences improve the efficiency of the iterative sequence in this context.

    This research work was financially supported by King Mongkut's University of Technology Thonburi through the "KMUTT 55th Anniversary Commemorative Fund". In particular, Habib ur Rehman was financed by the Petchra Pra Jom Doctoral Scholarship Academic for Ph.D. Program at KMUTT [grant number 39/2560]. Moreover, this project was supported by Center of Excellence in Theoretical and Computational Science (TaCS-CoE), KMUTT. Also, this project was financially supported by King Mongkut’s University of Technology North Bangkok, Contract no. KMUTNB-BasicR-64-22.

    The first author would like to thank the "Petchra Pra Jom Klao Ph.D. Research Scholarship from King Mongkut’s University of Technology Thonburi". We are very grateful to the editor and the anonymous referees for their valuable and useful comments, which helps in improving the quality of this work.

    No potential conflict of interest was reported by the author(s).



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