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Inspiring and engaging high school students with science and technology education in regional Australia

  • In the last two decades, there was a continuing declining participation rate in STEM education, especially in secondary schools in regional Australia. To reverse this trend and inspire rural school students with science and technology education, both federal and state governments identify the new strategies to promote STEM engagement of school students. In this study, with Queensland government's Engaging Science funding support, Central Queensland University researchers collaborated with rural school to deliver a demonstration with hand-on experiences of drone technologies to students. The activity led students to understand the application of drone technologies in daily life, especially agriculture sector. These activities impressed local communities including both teachers and students by demonstrating real-world problem-solving skills, with increasing over 25% participating students' interest in STEM education. This also leads more future collaboration opportunities to deliver other projects to supporting rural schools' STEM education. Future challenge for conducting these activities would be preparing the activity materials that fit the learning style and time schedule for different knowledge levels of students.

    Citation: Yujuan Li, Robert N. Hibbard, Peter L. A. Sercombe, Amanda L. Kelk, Cheng-Yuan Xu. Inspiring and engaging high school students with science and technology education in regional Australia[J]. STEM Education, 2021, 1(2): 114-126. doi: 10.3934/steme.2021009

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  • In the last two decades, there was a continuing declining participation rate in STEM education, especially in secondary schools in regional Australia. To reverse this trend and inspire rural school students with science and technology education, both federal and state governments identify the new strategies to promote STEM engagement of school students. In this study, with Queensland government's Engaging Science funding support, Central Queensland University researchers collaborated with rural school to deliver a demonstration with hand-on experiences of drone technologies to students. The activity led students to understand the application of drone technologies in daily life, especially agriculture sector. These activities impressed local communities including both teachers and students by demonstrating real-world problem-solving skills, with increasing over 25% participating students' interest in STEM education. This also leads more future collaboration opportunities to deliver other projects to supporting rural schools' STEM education. Future challenge for conducting these activities would be preparing the activity materials that fit the learning style and time schedule for different knowledge levels of students.



    Assume that D is a nonempty, closed and convex subset of a real Hilbert space E. Let f:E×ER be a bifunction with f(v,v)=0 for all vD. An equilibrium problem (shortly, EP) for f on D is defined in the following manner: Find ζD such that

    f(ζ,v)0,vD. (EP)

    Moreover, the solution set of an equilibrium is denoted by SEP. In this study, the problem (EP) is studied based on the following conditions. A bifunction f:E×ER is said to be (see for more details [3,4]):

    (C1) pseudomonotone on D if

    f(v1,v2)0f(v2,v1)0,v1,v2D; (1.1)

    (C2) Lipschitz-type continuous [15] on D if there exists two constants k1,k2>0 such that

    f(v1,v3)f(v1,v2)+f(v2,v3)+k1v1v22+k2v2v32,v1,v2,v3D; (1.2)

    (C3) For any weakly convergent sequence {vn}D (vnv) the following inequality holds

    lim supnf(vn,v)f(v,v),vD; (1.3)

    (C4) f(v,) is convex and subdifferentiable on E for each fixed vE.

    The general format of the problem (EP) has become attractive and has received a lot of attention from several authors in recent years. Mathematically, the problem (EP) can be considered as a generalization of many mathematical models, such as the fixed-point problems, scalar and vector minimization problems, the complementarity problems, the variational inequalities problems, the Nash equilibrium problems in non-cooperative games, the saddle point problems and the inverse minimization problems [4,12,17]. The equilibrium problem (EP) has applications in economics [8] or the dynamics of offer and demand [1], continuing to exploit the theoretical structure of non-cooperative games and the Nash equilibrium idea [18,19]. To the best of our knowledge, the term "equilibrium problem" was first used in the literature in 1992 by Muu and Oettli [17] and was later studied further by Blum and Oettli [4].

    By using the idea of the Korpelevich extragradient method [13], Flam et al. [10] and Quoc et al. [21] introduced the following method for solving equilibrium problems involving pseudomonotone and Lipschitz-type bifunction. Choose a random starting point of u0D; looking the given iteration un and choose the next iteration using the iterative scheme:

    {vn=argminvD{ρf(un,v)+12unv2},un+1=argminvD{ρf(vn,v)+12unv2}, (1.4)

    where 0<ρ<min{12k1,12k2} and k1,k2 are two Lipschitz-type constants of a bifunction (1.2). The method (1.4) has been extended and modified in various ways [16,24,25,26,27,29] and others in [2,6,22,30,32,34,35,36].

    It deserves mention that the above well-established method carries two significant drawbacks. The first is the constant stepsize that requires the knowledge or approximation of the Lipschitz constant of the relevant bifunction and it only converges weakly in Hilbert spaces. From the computational point of view, it might be hard to use a fixed stepsize, and hence, the convergence rate and usefulness of the method could be influenced. The inertial-type algorithms are of particular interest here. These algorithms are derived from an oscillator equation with damping and a conservative restoring force. This second-order dynamical system is known as Heavy Ball with Friction, and it was first studied by Polyak [20]. In general, the main feature of the inertial-type algorithms is that we can use the two previous iterations to construct the next one. Recently, inertial-type algorithms have been widely studied for the special cases of the problem (EP).

    A natural question therefore arises:

    Is it possible to introduce a new inertial strongly convergent extragradient method with a non-monotone stepsize rule to determine the numerical solution of the problem (EP) involves a pseudomonotone bifunction?

    In this study, we provide a positive answer to this question, i.e., the gradient method still operate in the case of a non-monotonic stepsize rule for solving equilibrium problems accompanied by a pseudomonotone bifunction and obtain a strong convergence of the iterative sequence. We introduce a new extragradient-type method to solve the problem (EP) in the context of an infinite-dimensional real Hilbert space, inspired by the works of [7,20,21]. The key contributions to this research are given below:

    () We introduce a new self-adaptive subgradient extragradient method by using an inertial scheme and a non-monotone stepsize rule to solve equilibrium problems. Also, we confirm that the generated sequence is strongly convergent. This result can be regarded as a modification of the method (1.4).

    () The applications of our main results are considered in order to solve particular classes of equilibrium problems in a real Hilbert space.

    () The numerical experiments regarding Algorithm 1 with Algorithm 3.1 in [11], Algorithm 1 in [28] and Algorithm 3 in [31]. The numerical results have indicated that the suggested method is appropriate and performed better compared to the existing ones.

    The rest of the study has been arranged as follows: Section 2 includes basic definitions and key lemmas that are used throughout this manuscript. Section 3 consists of the proposed iterative scheme with a variable stepsize rule and a theorem of convergence analysis. Section 4 sets out the application of the proposed results to solve the problems of variational inequalities and fixed point problems. Section 5 gives numerical results to illustrate the performance of the new algorithms and equate them with the two existing algorithms.

    Let D be a nonempty, closed and convex subset of a real Hilbert space E. The metric projection PD(u) of uE onto a closed and convex subset D of E is defined by

    PD(u)=argminvDvu.

    Definition 2.1. Let D be a subset of a real Hilbert space E and ϰ:DR a given convex function.

    (1). The subdifferential of set ϰ at uD is defined by

    ϰ(u)={zE:ϰ(v)ϰ(u)z,vu,vD}.

    (2). The normal cone at uD is defined by

    ND(u)={zE:z,vu0,vD}.

    Lemma 2.2. [23] Assume that ϰ:DR is a convex, subdifferentiable and lower semicontinuous function on D. An element uD is a minimizer of a function ϰ if and only if

    0ϰ(u)+ND(u),

    where ϰ(u) stands for the subdifferential of ϰ at uD and ND(u) the normal cone of D at u.

    Lemma 2.3. [33] Assume that {an}(0,+) is a sequence satisfying the following inequality

    an+1(1bn)an+bncn,nN.

    Moreover, {bn}(0,1) and {cn}R are sequences such that

    limn+bn=0,+n=1bn=+andlim supn+cn0.

    Then, limnan=0.

    Lemma 2.4. [14] Assume that {an}R is a sequence and there exists a subsequence {ni} of {n} such that ani<ani+1 for all iN. Then, there exists a nondecreasing sequence mkN such that mk as k, and the subsequent conditions are fulfilled by all (sufficiently large) numbers kN:

    amkamk+1andakamk+1.

    Indeed, mk=max{jk:ajaj+1}.

    Now, we introduce a new variant of Algorithm (1.4) in which the constant stepsize ρ is chosen adaptively.

    Algorithm 1 (Inertial Non-monotone Strongly Convergent Iterative Scheme)
    Step 0: Choose u0,u1D, ϕ>0, 0<σ<min{1,12k1,12k2},μ(0,1), ρ1>0. Moreover, select {ψn}(0,1) satisfies the conditions, i.e.,
            n+ψn=0and+n=1ψn=+.
    Step 1: Compute χn=un+ϕn(unun1)ψn[un+ϕn(unun1)] where
          0ϕn^ϕnand^ϕn={min{ϕ2,ϵnunun1}ifunun1,ϕ2otherwise,(3.1)
    where ϵn=(ψn) is a positive sequence such that limn+ϵnψn=0.
    Step 2: Compute vn=argminvD{ρnf(χn,v)+12χnv2}.
    If χn=vn, then STOP. Otherwise, go to Step 3.
    Step 3: Firstly choose ωn2f(χn,vn) satisfying χnρnωnvnND(vn) and generate a half-space
            En={zE:χnρnωnvn,zvn0}
    and compute
            un+1=argminvEn{ρnf(vn,v)+12χnv2}.
    Step 4: Next, the stepsize rule ρn+1 is updated as follows:
          ρn+1={min{σ,μf(vn,un+1)f(χn,un+1)f(χn,vn)k1χnvn2k2un+1vn2+1},ifμf(vn,un+1)f(χn,un+1)f(χn,vn)k1χnvn2k2un+1vn2+1>0,σotherwise.(3.2)
    Set n=n+1 and go back to Step 1.

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    Remark 3.1. By the use of ρn+1 in expression (3.2), we obtain

    ρn+1[f(χn,un+1)f(χn,vn)k1χnvn2k2vnun+12]μf(vn,un+1). (3.3)

    Lemma 3.1. Suppose that the conditions (C1)(C4) are satisfied and {un} be a sequence generated by Algorithm 1. Then, we have

    un+1ζ2χnζ2(1ρn+1)un+1χn2ρn+1(12k1ρn)χnvn2ρn+1(12k2ρn)un+1vn2. (3.4)

    Proof. By the use of definition of un+1, we obtain

    02{ρnf(vn,)+12χn2}(un+1)+ND(un+1).

    Thus, there exists ωn2f(vn,un+1) and ¯ωnND(un+1) such that

    ρnωn+un+1χn+¯ωn=0.

    The above expression implies that

    χnun+1,vun+1=ρnωn,vun+1+¯ωn,vun+1,vD.

    Due to ¯ωnND(un+1) imply that ¯ωn,vun+10, for every vD. Thus, we obtain

    ρnωn,vun+1χnun+1,vun+1,vD. (3.5)

    By given ωn2f(vn,un+1), we have

    f(vn,v)f(vn,un+1)ωn,vun+1,vD. (3.6)

    From expressions (3.5) and (3.6), we obtain

    ρnf(vn,v)ρnf(vn,un+1)χnun+1,vun+1,vD. (3.7)

    In the similar way, vn gives that

    ρn{f(χn,v)f(χn,vn)}χnvn,vvn,vD. (3.8)

    By the use of v=un+1 into expression (3.8), we get

    ρn{f(χn,un+1)f(χn,vn)}χnvn,un+1vn. (3.9)

    By the use of v=ζ into expression (3.7), we obtain

    ρnf(vn,ζ)ρnf(vn,un+1)χnun+1,ζun+1. (3.10)

    Since ζSEP implies that f(ζ,vn)0 and pseudomonotonicity of a bifunction f gives that f(vn,ζ)0. Thus, expression (3.10) implies that

    χnun+1,un+1ζρnf(vn,un+1). (3.11)

    From expression (3.2), we have

    f(vn,un+1)ρn+1[f(χn,un+1)f(χn,vn)k1χnvn2k2vnun+12]. (3.12)

    Combining expressions (3.11) and (3.12) gives that

    χnun+1,un+1ζρn+1[ρn{f(χn,un+1)f(χn,vn)}k1ρnχnvn2k2ρnun+1vn2]. (3.13)

    From expressions (3.9) and (3.13), we obtain

    2χnun+1,un+1ζρn+1[2χnvn,un+1vn2k1ρnχnvn22k2ρnun+1vn2]. (3.14)

    By the use of following formulas:

    2χnun+1,un+1ζ=χnζ2un+1χn2un+1ζ2.
    2χnvn,un+1vn=χnvn2+un+1vn2χnun+12.

    Finally, we have

    un+1ζ2χnζ2(1ρn+1)un+1χn2ρn+1(12k1ρn)χnvn2ρn+1(12k2ρn)un+1vn2. (3.15)

    Theorem 3.2. Assume that conditions (C1)(C4) are satisfied. Then, the sequence {un} generated by Algorithm 1 converges strongly to an element ζ=PSEP(0).

    Proof. Thus, expression (3.1) implies that

    limn+ϕnψnunun1limn+ϵnψn=0. (3.16)

    By the use of definition of {χn} and inequality (3.16), we obtain

    χnζ=un+ϕn(unun1)ψnunϕnψn(unun1)ζ=(1ψn)(unζ)+(1ψn)ϕn(unun1)ψnζ (3.17)
    (1ψn)unζ+(1ψn)ϕnunun1+ψnζ(1ψn)unζ+ψnK1, (3.18)

    where

    (1ψn)ϕnψnunun1+ζK1.

    By the use of Lemma 3.1, we obtain

    un+1ζ2χnζ2,n>1. (3.19)

    Combining (3.18) with (3.19), we obtain

    un+1ζ(1ψn)unζ+ψnK1max{unζ,K1}max{u2ζ,K1}. (3.20)

    Thus, we infer that the sequence {un} is bounded. Indeed, by (3.18) we have

    χnζ2(1ψn)2unζ2+ψ2nK21+2K1ψn(1ψn)unζunζ2+ψn[ψnK21+2K1(1ψn)unζ]unζ2+ψnK2, (3.21)

    for some K2>0. Combining the expressions (3.4) with (3.21), we have

    un+1ζ2unζ2+ψnK2(1ρn+1)un+1χn2ρn+1(12k1ρn)χnvn2ρn+1(12k2ρn)un+1vn2. (3.22)

    Due to the Lipschitz-continuity and pseudomonotonicity of f implies that the solution set SEP is a closed and convex set (for further details see [21]). It is given that ζ=PSEP(0) such that

    0ζ,vζ0,vSEP. (3.23)

    The remainder of the proof is divided into the following two cases:

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

    un+1ζunζ,nN1. (3.24)

    Thus, above expression implies that limn+unζ exists and let limn+unζ=l, for some l0. From the expression (3.22), we have

    (1ρn+1)un+1χn2+ρn+1(12k1ρn)χnvn2+ρn+1(12k2ρn)un+1vn2unζ2+ψnK2un+1ζ2. (3.25)

    Due to existence of limit of the sequence unζ and ψn0, we conclude that

    χnvn0andun+1vn0asn+. (3.26)

    It continues from (3.25) that

    limn+un+1χn=0. (3.27)

    Next, we have to compute

    χnun=un+ϕn(unun1)ψn[un+ϕn(unun1)]unϕnunun1+ψnun+ϕnψnunun1=ψnϕnψnunun1+ψnun+ψ2nϕnψnunun10. (3.28)

    The above expression implies that

    limn+unun+1limn+unχn+limn+χnun+1=0. (3.29)

    he above explanation guarantees that the sequences {χn} and {vn} are also bounded. Due to the reflexivity of E and the boundedness of {un} guarantees that there exists a subsequence {unk} such that {unk}ˆuE as k+. Next, we have to prove that ˆuSEP. Due to the inequality (3.7) we have

    ρnkf(vnk,v)ρnkf(vnk,unk+1)+χnkunk+1,vunk+1ρnkρnk+1f(χnk,unk+1)ρnkρnk+1f(χnk,vnk)k1ρnkρnk+1χnkvnk2k2ρnkρnk+1vnkunk+12+χnkunk+1,vunk+1ρnk+1χnkvnk,unk+1vnkk1ρnkρnk+1χnkvnk2k2ρnkρnk+1vnkunk+12+χnkunk+1,vunk+1, (3.30)

    where v is an arbitrary element in En. It continues from that (3.26)–(3.29) and the boundedness of {un} that the right-hand side goes to zero. From ρn>0, the condition (1.3) and vnkˆu, we have

    0lim supk+f(vnk,v)f(ˆu,v),vEn. (3.31)

    It implies that f(ˆu,v)0,vD, and hence ˆuSEP. Next, we have

    lim supn+ζ,ζun=limk+ζ,ζunk=ζ,ζˆu0. (3.32)

    By the use of limn+un+1un=0. Thus, expression (3.32) implies that

    lim supn+ζ,ζun+1lim supn+ζ,ζun+lim supn+ζ,unun+10. (3.33)

    By the use of expression (3.17), we have

    χnζ2=un+ϕn(unun1)ψnunϕnψn(unun1)ζ2=(1ψn)(unζ)+(1ψn)ϕn(unun1)ψnζ2(1ψn)(unζ)+(1ψn)ϕn(unun1)2+2ψnζ,χnζ=(1ψn)2unζ2+(1ψn)2ϕ2nunun12+2ϕn(1ψn)2unζunun1+2ψnζ,χnun+1+2ψnζ,un+1ζ(1ψn)unζ2+ϕ2nunun12+2ϕn(1ψn)unζunun1+2ψnζχnun+1+2ψnζ,un+1ζ=(1ψn)unζ2+ψn[ϕnunun1ϕnψnunun1+2(1ψn)unζϕnψnunun1+2ζχnun+1+2ζ,ζun+1]. (3.34)

    From expressions (3.19) and (3.34) we obtain

    un+1ζ2(1ψn)unζ2+ψn[ϕnunun1ϕnψnunun1+2(1ψn)unζϕnψnunun1+2ζχnun+1+2ζ,ζun+1]. (3.35)

    By the use of (3.27), (3.33), (3.35) and applying Lemma 2.3, conclude that limn+unζ=0.

    Case 2: Suppose that there exists a subsequence {ni} of {n} such that

    uniζuni+1ζ,iN.

    By using Lemma 2.4 there exists a sequence {mk}N as {mk}+ such that

    umkζumk+1ζandukζumk+1ζ,for allkN. (3.36)

    As similar to Case 1, the expression (3.25) implies that

    (1ρmk+1)umk+1χmk2+ρmk+1(12k1ρmk)χmkvmk2+ρmk+1(12k2ρmk)umk+1vmk2umkζ2+ψmkK2umk+1ζ2. (3.37)

    Due to ψmk0, we deduce the following:

    limk+χmkvmk=limk+umk+1vmk=0. (3.38)

    It follows that

    limk+umk+1χmk=0. (3.39)

    Next, we have to evaluate

    χmkumk=umk+ϕmk(umkumk1)ψmk[umk+ϕmk(umkumk1)]umkϕmkumkumk1+ψmkumk+ϕmkψmkumkumk1=ψmkϕmkψmkumkumk1+ψmkumk+ψ2mkϕmkψmkumkumk10. (3.40)

    It follows that

    limk+umkumk+1limk+umkχmk+limk+χmkumk+1=0. (3.41)

    By using the same explanation as in the Case 1, such that

    lim supk+ζ,ζumk+10. (3.42)

    By using the expressions (3.35) and (3.36), we obtain

    umk+1ζ2(1ψmk)umkζ2+ψmk[ϕmkumkumk1ϕmkψmkumkumk1+2(1ψmk)umkζϕmkψmkumkumk1+2ζχmkumk+1+2ζ,ζumk+1](1ψmk)umk+1ζ2+ψmk[ϕmkumkumk1ϕmkψmkumkumk1+2(1ψmk)umkζϕmkψmkumkumk1+2ζχmkumk+1+2ζ,ζumk+1]. (3.43)

    Thus, above expression implies that

    umk+1ζ2[ϕmkumkumk1ϕmkψmkumkumk1+2(1ψmk)umkζϕmkψmkumkumk1+2ζχmkumk+1+2ζ,ζumk+1]. (3.44)

    Since ψmk0, and umkζ is a bounded sequence. Therefore, expressions (3.42) and (3.44) implies that

    umk+1ζ20,ask+. (3.45)

    It implies that

    limn+ukζ2limn+umk+1ζ20. (3.46)

    As a consequence unζ. This completes the proof of the theorem.

    In this section, we have written about the new results from our main proposed methods to solve variational inequalities. In the last few years, variational inequalities have drawn a considerable amount of attention from both researchers and readers. It is well established that variational inequalities deal with a large variety of topics in partial differential equations, optimal control, optimization techniques, applied mathematics, engineering, finance, and operational science. The variational inequality problem for an operator A:EE is defined as follows:

    FindζDsuch thatA(ζ),vζ0,vD. (VIP)

    We consider the following conditions to study the variational inequalities.

    (A1) The solution set of the problem (VIP) is denoted by VI(A,D) and it is nonempty;

    (A2) An operator A:EE is said to be a pseudomonotone if

    A(u),vu0A(v),uv0,u,vD;

    (A3) An operator A:EE is said to be a Lipschitz continuous if there exists a constants L>0 such that

    A(u)A(v)Luv,u,vD;

    (A4) An operator A:EE is said to be sequentially weakly continuous, i.e., {A(un)} weakly converges to A(u) for every sequence {un} converges weakly to u.

    On the other hand, we have also developed some results to solve fixed point problems. The existence of a solution to a theoretical or real-world problem should be analogous to the existence of a fixed point for an appropriate map or operator. Fixed-point theorems are thus extremely important in many fields of mathematics, engineering, and science. In many cases, it is not difficult to find an exact solution. Therefore, it is crucial to create effective techniques to approximate the desired result. The fixed point problem for an operator B:EE is defined as follows:

    FindζDsuch thatB(ζ)=ζ. (FPP)

    The following conditions are required to study fixed point theorems.

    (B1) The solution set of the problem (FPP) is denoted by Fix(B,D) is nonempty;

    (B2) B:DD is said to be a κ-strict pseudocontraction [5] on D if

    BuBv2uv2+κ(uBu)(vBv)2,u,vD;

    (B3) B:EE is said to be weakly sequentially continuous, i.e., {B(un)} weakly converges to B(u) for every sequence {un} converges weakly to u.

    Corollary 4.1. Assume that an operator A:DE satisfies the conditions (A)–(A) and the solution set VI(A,D). Choose u0,u1D, ϕ>0, 0<σ<min{1,1L},μ(0,1), ρ1>0. Moreover, select {ψn}(0,1) meet the conditions, i.e.,

    limn+ψn=0and+n=1ψn=+.

    (i) Compute

    χn=un+ϕn(unun1)ψn[un+ϕn(unun1)],

    where ϕn modified on each iteration as follows:

    0ϕn^ϕnand^ϕn={min{ϕ2,ϵnunun1}ifunun1,ϕ2otherwise,

    where ϵn=(ψn) is a positive sequence such that limn+ϵnψn=0.

    (ii) Compute

    {vn=PD(χnρnA(χn)),un+1=PEn(χnρnA(vn)),

    where

    En={zE:χnρnA(χn)vn,zvn0}.

    (iii) Compute

    ρn+1={min{σ,μAvn,un+1vnAχn,un+1vnL2χnvn2L2un+1vn2+1},ifμAvn,un+1vnAχn,un+1vnL2χnvn2L2un+1vn2+1>0,σotherwise.

    Then, the sequences {un} converge strongly to ζVI(A,D).

    Corollary 4.2. Assume that B:DD is a κ-strict pseudocontraction and weakly continuous with solution set Fix(B,D). Choose u0,u1D, ϕ>0, 0<σ<min{1,1κ32κ},μ(0,1), ρ1>0. Moreover, select {ψn}(0,1) meet the conditions, i.e.,

    limn+ψn=0and+n=1ψn=+.

    (i) Compute

    χn=un+ϕn(unun1)ψn[un+ϕn(unun1)],

    where ϕn modified one each iteration as follows:

    0ϕn^ϕnand^ϕn={min{ϕ2,ϵnunun1}ifunun1,ϕ2otherwise.

    where ϵn=(ψn) is a positive sequence such that limn+ϵnψn=0.

    (ii) Compute

    {vn=PD[χnρn(χnB(χn))],un+1=PEn[χnρn(vnB(vn))],

    where

    En={zE:(1ρn)χn+ρnB(χn)vn,zvn0}.

    (iii) Evaluate stepsize rule for next iteration is evaluated as follows:

    ρn+1={min{σ,μvnBvn,un+1vnχnB(χn),un+1vn(32κ22κ)χnvn2(32κ22κ)un+1vn2+1},ifμvnBvn,un+1vnχnB(χn),un+1vn(32κ22κ)χnvn2(32κ22κ)un+1vn2+1>0,σotherwise.

    Then, the sequence {un} converges strongly to ζFix(B,D).

    In this section, the numerical performance of the proposed method is described in contrast with some similar works in the literature.

    Example 5.1. The test problem here taken from the Nash-Cournot Oligo-polistic equilibrium model in [9,21]. Suppose that the set D is defined by

    D:={uRN:10ui10}

    and f:D×DR is defined as follows

    f(u,v)=Mu+Nv+r,vu,u,vD,

    where rRN and M, N matrices of order N. The matrix M is symmetric positive semi-definite and the matrix NM is symmetric negative semi-definite with Lipschitz-type criteria k1=k2=12MN (see [21] for details). Two matrices M,N are taken randomly [Two diagonal matrices randomly A1 and A2 with elements from [0,2] and [2,0], respectively. Two random orthogonal matrices O1=RandOrthMat(N) and O2=RandOrthMat(N) are generated. Thus, a positive semi-definite matrix B1=O1A1OT1 and a negative semi-definite matrix B2=O2A2OT2 is obtained. Finally, set N=B1+BT1,S=B2+BT2 and M=NS].

    Experiment 1: In the first experiment, we take into account the numerical efficiency of the Algorithm 1 using different starting point choices. This experiment helps the reader see how much the starting points influenced the efficiency of the Algorithm 1. For these numerical results, we have use N=20, u0=u1, ρ1=0.50,σ=12.2k1,μ=0.22,ϕ=1.00,ϵn=1(n+1)2,ψn=120(n+2),Dn=Error=un+1un for Algorithm 1 (Alg-4). Figures 1 and 2 demonstrate the numerical efficacy of the proposed mehod.

    Figure 1.  Numerical illustration of Algorithm 1 for different starting points while N=20 and the number of iterations are 104,133,151,178,164, respectively.
    Figure 2.  Numerical illustration of Algorithm 1 for different starting points while N=20 and the execution time are 1.0100, 1.2801, 1.8741, 1.8852, 1.8766, respectively.

    Experiment 2: In the second experiment, we consider the numerical efficiency of the Algorithm 1 using different inertial parameter ϕ choices. This experiment helps the reader see how much the inertial parameter ϕ influenced the efficiency of the Algorithm 1. For these numerical results, we have use N=20, u0=u1=(1,1,,1,1), ρ1=0.30,σ=12.5k1,μ=0.33,ϵn=1(n+1)2,ψn=110(n+2),Dn=Error=un+1un for Algorithm 1 (Alg-4). Figures 3 and 4 demonstrate the numerical efficacy of the proposed method.

    Figure 3.  Numerical illustration of Algorithm 1 for ϕ=1,0.8,0.6,0.4,0.2 and the number of iterations are 114, 93, 78, 63, 54, respectively.
    Figure 4.  Numerical illustration of Algorithm 1 for ϕ=1,0.8,0.6,0.4,0.2 and the execution time are 1.0763, 0.8257, 1.0927, 0.5489, 0.5748, respectively.

    Experiment 3: In third experiment, we provide the numerical comparison of Algorithm 1 with Algorithm 1 in [28] and Algorithm 3.1 in [11] and Algorithm 3 in [31]. For these numerical studies we have assumed that starting points are u0=u1=v0=(1,1,,1), N=5,10,40,100 and error term Dn=un+1un. Figures 512 have shown a number of results for Tolerance = 104. Information regarding the control parameters shall be considered as described in the following:

    Figure 5.  Numerical illustration of Algorithm 1 with Algorithm 1 in [28] and Algorithm 3.1 in [11] and Algorithm 3 in [31] for N = 5.
    Figure 6.  Numerical illustration of Algorithm 1 with Algorithm 1 in [28] and Algorithm 3.1 in [11] and Algorithm 3 in [31] for N = 5.
    Figure 7.  Numerical illustration of Algorithm 1 with Algorithm 1 in [28] and Algorithm 3.1 in [11] and Algorithm 3 in [31] for N = 10.
    Figure 8.  Numerical illustration of Algorithm 1 with Algorithm 1 in [28] and Algorithm 3.1 in [11] and Algorithm 3 in [31] for N = 10.
    Figure 9.  Numerical illustration of Algorithm 1 with Algorithm 1 in [28] and Algorithm 3.1 in [11] and Algorithm 3 in [31] for N = 40.
    Figure 10.  Numerical illustration of Algorithm 1 with Algorithm 1 in [28] and Algorithm 3.1 in [11] and Algorithm 3 in [31] for N = 40.
    Figure 11.  Numerical illustration of Algorithm 1 with Algorithm 1 in [28] and Algorithm 3.1 in [11] and Algorithm 3 in [31] for N = 100.
    Figure 12.  Numerical illustration of Algorithm 1 with Algorithm 1 in [28] and Algorithm 3.1 in [11] and Algorithm 3 in [31] for N = 100.

    (i)ρn=14k1,ϕn=1(n+1)0.5, and Error=un+1un for Algorithm 3.1 in [11] (Alg-1).

    (ii)ρ=14k1,θ=0.50, ϵn=1(n+1)2,γn=120(n+2),βn=710(1γn),Error=un+1un for Algorithm 3 in [31] (Alg-2).

    (iii)ρ=15k1, ϕn=1100(n+2), f(u)=u2 and Error=un+1un for Algorithm 1 (Alg-3) in [28].

    (iv)ρ1=0.50,σ=12.2k1,μ=0.22,ϕ=1.00,ϵn=1(n+1)2,ψn=120(n+2),Error=un+1un for Algorithm 1 (Alg-4).

    Then numerical results are reported in Table 1.

    Table 1.  Numerical finding and their values for Figures 512.
    Number of iterations Elapsed time in seconds
    N Alg-1 Alg-2 Alg-3 Alg-4 Alg-1 Alg-2 Alg-3 Alg-4
    5 57 44 38 27 0.4910193 0.4805127 0.5035203 0.3568740
    10 63 58 47 35 0.5564982 0.6466944 0.4253435 0.3249474
    40 75 82 66 52 0.6889176 1.0781367 1.1124397 0.6130995
    100 118 143 81 66 1.4267609 1.8261292 1.767970 1.448753

     | Show Table
    DownLoad: CSV

    We constructed an explicit, inertial extragradient-type method to find a numerical solution to the pseudomonotone equilibrium problems in a real Hilbert space. This method is seen as a modification of the two-step gradient method. A strongly convergent result is well-proven, corresponding to the proposed algorithm. Numerical findings were presented to demonstrate our algorithm's numerical superiority over existing methods. These computational findings have indicated that the variable stepsize rule continues to increase the performance of the iterative sequence in this context.

    Pongsakorn Yotkaew was financial support by Khon Kaen University. Nuttapol Pakkaranang would like to thank Phetchabun Rajabhat University.

    No potential conflict of interest was reported by the authors.



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