An Accelerated Extragradient Method for Solving Pseudomonotone Equilibrium Problems with Applications

Abstract

Several methods have been put forward to solve equilibrium problems, in which the two-step extragradient method is very useful and significant. In this article, we propose a new extragradient-like method to evaluate the numerical solution of the pseudomonotone equilibrium in real Hilbert space. This method uses a non-monotonically stepsize technique based on local bifunction values and Lipschitz-type constants. Furthermore, we establish the weak convergence theorem for the suggested method and provide the applications of our results. Finally, several experimental results are reported to see the performance of the proposed method.

1. Introduction

Assume $\mathcal{K}$ to be a subset of a Hilbert space $\mathcal{E}$ with $f:\mathcal{E}\times \mathcal{E}\to \mathcal{R}$ a bifunction with $f(z_1,z_1)=0,$ for every $z_1\in \mathcal{K}.$ An equilibrium problem [1,2] for f on $\mathcal{K}$ is formulated in the following way:

$$\mathrm{Find}{q}^{*}\in \mathcal{K}\mathrm{such}\mathrm{that}f(q^{*},z_1)\ge 0,\forall z_1\in \mathcal{K}.$$

(1)

The equilibrium problem (1) has many mathematical problems as particular instances such as the variational inequality problems (VIP), the minimization problems, the fixed point problems, the complementarity problems, the Nash equilibrium of non-cooperative games, the saddle point problems, and the vector optimization problem (see [1,3,4] for more details). The term “equilibrium problem” in a particular format was established in 1992 by Muu and Oettli [2], and it was further promoted by Blum and Oettli in the article [1]. Some of the most popular ones, interesting and worthwhile research fields in equilibrium problem theory, are to develop new iterative schemes, improve the convergence rate and efficiency of the already existing methods, and study their converging analysis with optimal conditions. Several methods have been established in the past few years to solve the equilibrium problems in real Hilbert spaces, i.e., the extragradient methods [5,6,7,8,9,10,11,12,13,14], the inertial methods [15,16,17,18,19,20], for particular classes of equilibrium problems [21,22,23,24,25,26,27,28], and others in [29,30,31,32,33,34,35,36,37].

The proximal-like method [38] is one of the famous and efficient techniques to solve equilibrium problems. This technique is equivalent to solving minimization problems on each iteration. This approach was also considered as the two-step extragradient method in [5] due to the previous contribution of the Korpelevich method [39] to numerically solve the saddle point problems. Tran et al. in [5] generated the sequence $\left\{u_n\right\}$ in the following way:

$$\left\{\begin{array}{c}{u}_0\in \mathcal{K},\hfill \\ v_n=\underset{v\in \mathcal{K}}{argmin}\{\lambda f(u_n,v)+\frac{1}{2}\parallel u_n-v{\parallel }^2\},\hfill \\ u_{n+1}=\underset{v\in \mathcal{K}}{argmin}\{\lambda f(v_n,v)+\frac{1}{2}\parallel u_n-v{\parallel }^2\},\hfill \end{array}\right.$$

where $0<\lambda <min\left\{\frac{1}{2c_1},\frac{1}{2c_2}\right\}$. The iterative sequence generated by the above-written method provides the weak convergence of the iterative sequence, and in order to operate it, previous knowledge of Lipschitz-type constants is required that help to choose the value of the stepsize.

Recently, the authors introduced an inertial iterative scheme in [19] to determine a numerical solution of pseudomonotone equilibrium problems. The key contribution is an inertial factor that has helped to enhance the rate of convergence of the iterative sequence $\left\{u_n\right\}$. The detailed method is provided as follows:

Step 1: Choose $u_{-1},u_0\in \mathcal{K},$ $\theta \in [0,1),$ $0<\lambda <\{\frac{1}{2c_1},\frac{1}{2c_2}\}$ and a sequence $\left\{{ϵ}_n\right\}\subset [0,+\infty )$ such that:

$$\sum _{n=0}^{+\infty }{ϵ}_n<+\infty ,$$

(2)

holds. Let ${\theta }_n$ satisfy $0\le {\theta }_n\le \overline{{\theta }_n}$ such that:

$$\overline{{\theta }_n}=\left\{\begin{array}{cc}min\left\{\theta ,\frac{{ϵ}_n}{\parallel u_n-u_{n-1}\parallel }\right\}\hfill & \mathrm{if}{u}_n\ne u_{n-1},\hfill \\ \theta \hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

(3)

Step 2: Compute:

$$\left\{\begin{array}{c}{\rho }_n=u_n+{\theta }_n(u_n-u_{n-1}),\hfill \\ v_n=\underset{v\in \mathcal{K}}{argmin}\{\lambda f({\rho }_n,v)+\frac{1}{2}\parallel {\rho }_n-v{\parallel }^2\},\hfill \\ u_{n+1}=\underset{v\in \mathcal{K}}{argmin}\{\lambda f(v_n,v)+\frac{1}{2}\parallel {\rho }_n-v{\parallel }^2\}.\hfill \end{array}\right.$$

In this study, we concentrate on projection methods that are well known and practically straightforward to operate due to their simple numerical computation. Motivated by the works of [19,40], we propose an inertial explicit extragradient method to solve pseudomonotone equilibrium problems and other particular classes of equilibrium problems such as the fixed point problems and the variational inequality problems. The proposed method can be considered as the modification of the methods that appeared in [5,19,39,40]. Under certain mild conditions, the weak convergence results are established corresponding to the proposed method. Numerical studies have been demonstrated that show that the suggested method is more efficient than the existing method in [19].

The remainder of the paper is arranged in the following way: Section 2 includes some preliminary and necessary results that will be used throughout the paper. Section 3 contains our main method, as well as the weak convergence theorem. Section 4 covers the applications of the proposed method. Section 5 demonstrates the numerical results that provide the computational performance of our proposed method.

2. Preliminaries

    Assume $\mathcal{K}\subset \mathcal{E}$ to be a convex and closed subset of a real Hilbert space $\mathcal{E}$, and $\mathcal{R}$ and $\mathcal{N}$ denote the set of real numbers and the set of a natural numbers, respectively. Let $f:\mathcal{E}\times \mathcal{E}\to \mathcal{R}$ be a bifunction and $EP(f,\mathcal{K})$ be the solution set of an equilibrium problem on the set $\mathcal{K}$, $q^{*}$ being an arbitrary element of $EP(f,\mathcal{K}).$ Next, we consider the definitions of a bifunction monotonicity (see [1,41] for more details). A bifunction $f:\mathcal{E}\times \mathcal{E}\to \mathcal{R}$ on $\mathcal{K}$ for $\gamma >0$ is said to be:

(1)

strongly monotone if:

$$f(z_1,z_2)+f(z_2,z_1)\le -\gamma {\parallel z_1-z_2\parallel }^2,\forall z_1,z_2\in \mathcal{K};$$

(2)

monotone if:

$$f(z_1,z_2)+f(z_2,z_1)\le 0,\forall z_1,z_2\in \mathcal{K};$$

(3)

strongly pseudomonotone if:

$$f(z_1,z_2)\ge 0⟹f(z_2,z_1)\le -\gamma {\parallel z_1-z_2\parallel }^2,\forall z_1,z_2\in \mathcal{K};$$

(4)

pseudomonotone if:

$$f(z_1,z_2)\ge 0⟹f(z_2,z_1)\le 0,\forall z_1,z_2\in \mathcal{K}.$$

The following implications can be seen from the definitions mentioned above:

$$\left(1\right)⟹\left(2\right)⟹\left(4\right)\mathrm{and}\left(1\right)⟹\left(3\right)⟹\left(4\right).$$

Generally speaking, the converse is not true. We say that a bifunction $f:\mathcal{E}\times \mathcal{E}\to \mathcal{R}$ satisfies the Lipschitz-type condition [42] on set $\mathcal{K}$ if there exist two constants $c_1,c_2>0,$ such that:

$$f(z_1,z_3)\le f(z_1,z_2)+f(z_2,z_3)+c_1\parallel z_1-z_2{\parallel }^2+c_2{\parallel z_2-z_3\parallel }^2,\forall z_1,z_2,z_3\in \mathcal{K}.$$

Let $h:\mathcal{K}\to \mathcal{R}$ be a convex function, and the subdifferential of h at $z_1\in \mathcal{K}$ is defined by:

$$\partial h\left(z_1\right)=\{z_3\in \mathcal{E}:h\left(z_2\right)-h\left(z_1\right)\ge \langle z_3,z_2-z_1\rangle ,\forall z_2\in \mathcal{K}\}.$$

A normal cone of $\mathcal{K}$ at $z_1\in \mathcal{K}$ is defined by:

$$N_{\mathcal{K}}\left(z_1\right)=\{z_3\in \mathcal{E}:\langle z_3,z_2-z_1\rangle \le 0,\forall z_2\in \mathcal{K}\}.$$

The metric projection $P_C\left(z_1\right)$ for $z_1\in \mathcal{E}$ on $\mathcal{K}$ of $\mathcal{E}$ is defined by:

$$P_C\left(z_1\right)=\underset{}{argmin}\{\parallel z_2-z_1\parallel :z_2\in \mathcal{K}\}.$$

Lemma 1.

 [43] Let $h:\mathcal{K}\to \mathcal{R}$ be a subdifferentiable, convex, and lower semi-continuous function on $\mathcal{K},$ where $\mathcal{K}$ is a nonempty, convex, and closed subset of a real Hilbert space $\mathcal{E}.$ An element $z_1\in \mathcal{K}$ is a minimizer of a function h if and only if $0\in \partial h\left(z_1\right)+N_{\mathcal{K}}\left(z_1\right),$ where $\partial h\left(z_1\right)$ and $N_{\mathcal{K}}\left(z_1\right)$ denote the subdifferential of h at $z_1\in \mathcal{K}$ and the normal cone of $\mathcal{K}$ at $z_1$, respectively.

Lemma 2.

 [44] Suppose that a sequence $\left\{{\eta }_n\right\}$ in $\mathcal{E}$ and $\mathcal{K}\subset \mathcal{E}$ such that the following conditions are true.

(i)

For every $\eta \in \mathcal{K},$${lim}_{n\to \infty }\parallel {\eta }_n-\eta \parallel$ exists;

(ii)

each sequentially weak cluster limit point of $\left\{{\eta }_n\right\}$ belongs to set $\mathcal{K}.$

Then, $\left\{{\eta }_n\right\}$ weakly converges to a point in $\mathcal{K}.$

Lemma 3.

 [45] For $u,v\in \mathcal{E}$ and $\tau \in \mathcal{R},$ then the following equality holds.

$${\parallel \tau u+(1-\tau )v\parallel }^2={\tau \parallel u\parallel }^2+{(1-\tau )\parallel v\parallel }^2-\tau (1-\tau ){\parallel u-v\parallel }^2.$$

Lemma 4.

 [46] Assume that the sequence $\left\{q_n\right\}$ and $\left\{r_n\right\}$ of nonnegative real numbers satisfies $q_{n+1}\le q_n+r_n,$ $\forall n\in \mathcal{N}.$ If $\sum r_n<\infty ,$ then ${lim}_{n\to \infty }{q}_n$ exists.

Lemma 5.

 [40] Assume that $\left\{a_n\right\},\left\{b_n\right\}$ are real numbers sequences such that $a_n\le b_n,$ $\forall n\in \mathcal{N}.$ Let $\varrho ,\sigma \in (0,1)$ and $\mu \in (0,\sigma ).$ Then, there exists a sequence ${\lambda }_n$ such that ${\lambda }_n{a}_n\le \mu b_n$ and ${\lambda }_n\in (\varrho \mu ,\sigma ).$

Corollary 1.

Assume that f satisfies a Lipschitz-type condition on $\mathcal{K}$ with constants $c_1>0$ and $c_2>0.$ Let $\varrho \in (0,1),$$0<\sigma <min\left\{1,\frac{1}{2c_1},\frac{1}{2c_2}\right\}$, and $\mu \in (0,\sigma ).$ Then, there exists $\lambda >0$ such that:

$$\lambda \left(f(z_1,z_3)-f(z_1,z_2)-c_1\parallel z_1-z_2{\parallel }^2-c_2{\parallel z_2-z_3\parallel }^2\right)\le \mu f(z_2,z_3)$$

where $\varrho \mu <\lambda <\sigma$ with $z_1,z_2,z_3\in \mathcal{K}.$

Assume that a bifunction f satisfies the following conditions:

($\mathsf{\Psi }$1)

$f(z_2,z_2)=0,$ for all $z_2\in \mathcal{K}$, and f is pseudomonotone on $\mathcal{K};$

($\mathsf{\Psi }$2)

f satisfies the Lipschitz-type condition on $\mathcal{E}$ through $c_1>0$ and $c_2>0;$

($\mathsf{\Psi }$3)

$\underset{n\to \infty }{lim\; sup}f(z_n,v)\le f(z^{*},v)$ for each $v\in \mathcal{K}$ and $\left\{z_n\right\}\subset \mathcal{K}$ satisfy $z_n\rightharpoonup z^{*}$;

($\mathsf{\Psi }$4)

$f(z_1,.)$ is subdifferentiable and convex on $\mathcal{K}$ for each $z_1\in \mathcal{K}.$

3. An Accelerated Method for Pseudomonotone Equilibrium Problems and Its Convergence Analysis

  Now, we present a method that consists of two strongly convex optimization problems with an inertial term and an explicit formula for stepsize evaluation. The detailed method is provided below:

Algorithm 1 (Accelerated method for pseudomonotone equilibrium problems)

Initialization: Choose $u_{-1},u_0\in \mathcal{E},$ $\varrho \in (0,1),$ $\sigma <min\left\{1,\frac{1}{2c_1},\frac{1}{2c_2}\right\},$ $\mu \in (0,\sigma ),$ ${\lambda }_0>0,$ $\theta \in [0,1)$, and a sequence $\left\{{ϵ}_n\right\}\subset [0,+\infty )$ such that: ${\sum }_{n=0}^{+\infty }{ϵ}_n<+\infty .$ Iterative steps: Let ${\theta }_n$ satisfy $0\le {\theta }_n\le \overline{{\theta }_n}$ such that: $$\overline{{\theta }_n}=\left\{\begin{array}{cc}min\left\{\theta ,\frac{{ϵ}_n}{\parallel u_n-u_{n-1}\parallel }\right\}\hfill & \mathrm{if}{u}_n\ne u_{n-1},\hfill \\ \theta & \mathrm{otherwise}.\end{array}\right.$$ (4) Step 1: Compute: $$v_n=\underset{y\in \mathcal{K}}{argmin}\{{\lambda }_{n}f({\rho }_n,y)+\frac{1}{2}\parallel {\rho }_n-y{\parallel }^2\},$$ where ${\rho }_n=u_n+{\theta }_n(u_n-u_{n-1}).$ If ${\rho }_n=v_n$; STOP. Otherwise, go to the next step. Step 2: Set $0<\beta \le {\beta }_n\le 1$, and compute $$u_{n+1}=(1-{\beta }_n){\rho }_n+{\beta }_n{z}_n,$$ where $z_n=\underset{}{argmin}\{\mu {\lambda }_{n}f(v_n,y)+\frac{1}{2}\parallel {\rho }_n-y{\parallel }^2:y\in \mathcal{K}\}.$ Step 3: Revised the stepsize in the following way: $$\begin{array}{cc}\hfill {\lambda }_{n+1}=& min\left\{\sigma ,\frac{\mu f(v_n,z_n)}{f({\rho }_n,z_n)-f({\rho }_n,v_n)-c_1\parallel {\rho }_n-v_n{\parallel }^2-c_2{\parallel z_n-v_n\parallel }^2+1}\right\}.\hfill \end{array}$$ (5) Set $n:=n+1$, and return back to Iterative steps.

Remark 1.

From Corollary 1, the definition of ${\lambda }_{n+1}$ in (5) is well-defined such that:

$${\lambda }_{n+1}\left[f({\rho }_n,z_n)-f({\rho }_n,v_n)-c_1\parallel {\rho }_n-v_n{\parallel }^2-c_2{\parallel v_n-z_n\parallel }^2\right]\le \mu f(v_n,z_n).$$

(6)

Remark 2.

Due to the summability of ${\displaystyle \sum _{n=0}^{+\infty }{ϵ}_n}$, the expression (3) provides that:

$$\sum _{n=1}^{\infty }{\theta }_n\parallel u_n-u_{n-1}\parallel \le \sum _{n=1}^{\infty }\overline{{\theta }_n}\parallel u_n-u_{n-1}\parallel \le \sum _{n=1}^{\infty }\theta \parallel u_n-u_{n-1}\parallel <\infty ,$$

(7)

which implies that:

$$\underset{n\to \infty }{lim}\theta \parallel u_n-u_{n-1}\parallel =0.$$

(8)

Lemma 6.

If $v_n={\rho }_n$ in Algorithm 1, then ${\rho }_n\in EP(f,\mathcal{K}).$

Proof. 

From the value of $v_n$ and Lemma 1, we have:

$$0\in {\partial }_2\left\{{\lambda }_{n}f({\rho }_n,y)+\frac{1}{2}{\parallel {\rho }_n-y\parallel }^2\right\}\left(v_n\right)+N_{\mathcal{K}}\left(v_n\right).$$

Thus, there exists ${\upsilon }_n\in \partial f({\rho }_n,v_n)$ and $\overline{\omega }\in N_{\mathcal{K}}\left(v_n\right)$ such that:

$${\lambda }_n{\upsilon }_n+v_n-{\rho }_n+\overline{\omega }=0.$$

Thus, we have:

$$\langle {\rho }_n-v_n,y-v_n\rangle ={\lambda }_n\langle {\upsilon }_n,y-v_n\rangle +\langle \overline{\omega },y-v_n\rangle ,\forall y\in \mathcal{K}.$$

Given that ${\rho }_n=v_n$, $\overline{\omega }\in N_{\mathcal{K}}\left(v_n\right)$ implies that:

$${\lambda }_n\langle {\upsilon }_n,y-v_n\rangle \ge 0.$$

(9)

Due to ${\upsilon }_n\in f({\rho }_n,v_n)$ and using the subdifferential definition, we obtain:

$$f({\rho }_n,y)-f({\rho }_n,v_n)\ge \langle {\upsilon }_n,y-v_n,\rangle ,\forall y\in \mathcal{K}.$$

(10)

Combining Expressions (9) and (10) and due to ${\lambda }_n>0,$ we get:

$$f({\rho }_n,y)-f({\rho }_n,v_n)\ge 0.$$

(11)

By $v_n={\rho }_n$, the condition ($\mathsf{\Psi }$1) implies that $f({\rho }_n,y)\ge 0,$ for all $y\in \mathcal{K}.$ □

Lemma 7.

Suppose that $f:\mathcal{E}\times \mathcal{E}\to \mathcal{R}$ satisfies the conditions($\mathsf{\Psi }$1)–($\mathsf{\Psi }$4). For each $q^{*}\in EP(f,\mathcal{K})\ne \varnothing ,$ we have:

$$\begin{array}{cc}\hfill \parallel z_n-q^{*}{\parallel }^2& \le \parallel {\rho }_n-q^{*}{\parallel }^2-(1-{\lambda }_{n+1}){\parallel z_n-{\rho }_n\parallel }^2\hfill \\ \hfill & -{\lambda }_{n+1}(1-2c_1{\lambda }_n)\parallel {\rho }_n-v_n{\parallel }^2-{\lambda }_{n+1}(1-2c_2{\lambda }_n){\parallel z_n-v_n\parallel }^2.\hfill \end{array}$$

Proof. 

From Lemma 1 and the value of $z_n$, we have:

$$0\in {\partial }_2\left\{\mu {\lambda }_{n}f(v_n,y)+\frac{1}{2}{\parallel {\rho }_n-y\parallel }^2\right\}\left(z_n\right)+N_{\mathcal{K}}\left(z_n\right).$$

Thus, we have:

$$\mu {\lambda }_n\omega +z_n-{\rho }_n+\overline{\omega }=0,$$

where $\omega \in {\partial }_{2}f(v_n,z_n)$ and $\overline{\omega }\in N_{\mathcal{K}}\left(z_n\right).$ Thus, we have:

$$\langle {\rho }_n-z_n,y-z_n\rangle =\mu {\lambda }_n\langle \omega ,y-z_n\rangle +\langle \overline{\omega },y-z_n\rangle ,\forall y\in \mathcal{K}.$$

Since $\overline{\omega }\in N_{\mathcal{K}}\left(z_n\right)$, then $\langle \overline{\omega },y-z_n\rangle \le 0,$ $\forall y\in \mathcal{K}.$ This implies that:

$$\mu {\lambda }_n\langle \omega ,y-z_n\rangle \ge \langle {\rho }_n-z_n,y-z_n\rangle ,\forall y\in \mathcal{K}.$$

(12)

Given that $\omega \in {\partial }_{2}f(v_n,z_n)$ and due to the definition of the subdifferential, we have:

$$f(v_n,y)-f(v_n,z_n)\ge \langle \omega ,y-z_n\rangle ,\forall y\in \mathcal{K}.$$

(13)

Combining Expressions (12) and (13):

$$\mu {\lambda }_{n}f(v_n,y)-\mu {\lambda }_{n}f(v_n,z_n)\ge \langle {\rho }_n-z_n,y-z_n\rangle ,\forall y\in \mathcal{K}.$$

(14)

By letting $y=q^{*}$ in Expression (14), we obtain:

$$\mu {\lambda }_{n}f(v_n,q^{*})-\mu {\lambda }_{n}f(v_n,z_n)\ge \langle {\rho }_n-z_n,q^{*}-z_n\rangle ,\forall y\in \mathcal{K}.$$

(15)

From hypothesis $q^{*}\in EP(f,\mathcal{K})$ such that $f(q^{*},v_n)\ge 0$ and due to the condition ($\mathsf{\Psi }$1) implying that $f(v_n,q^{*})\le 0,$ we have:

$$\langle {\rho }_n-z_n,z_n-q^{*}\rangle \ge \mu {\lambda }_{n}f(v_n,z_n).$$

(16)

Combining Expressions (6) and (16), we obtain:

$$\begin{array}{cc}\hfill \langle {\rho }_n-z_n,z_n-q^{*}\rangle \ge & {\lambda }_{n+1}[{\lambda }_n\left\{f({\rho }_n,z_n)-f({\rho }_n,v_n)\right\}\hfill \\ \hfill & -c_1{\lambda }_n\parallel {\rho }_n-v_n{\parallel }^2-c_2{\lambda }_n{\parallel z_n-v_n\parallel }^2].\hfill \end{array}$$

(17)

In a similar way as Expression (14), we obtain:

$${\lambda }_{n}f({\rho }_n,y)-{\lambda }_{n}f({\rho }_n,v_n)\ge \langle {\rho }_n-v_n,y-v_n\rangle ,\forall y\in \mathcal{K}.$$

(18)

By substituting $y=z_n,$ we obtain:

$${\lambda }_n\left\{f({\rho }_n,z_n)-f({\rho }_n,v_n)\right\}\ge \langle {\rho }_n-v_n,z_n-v_n\rangle .$$

(19)

The expressions (17) and (19) imply that:

$$\begin{array}{cc}\hfill 2\langle {\rho }_n-z_n,z_n-q^{*}\rangle & \ge {\lambda }_{n+1}[2\langle {\rho }_n-v_n,z_n-v_n\rangle \hfill \\ \hfill & -2c_1{\lambda }_n\parallel {\rho }_n-v_n{\parallel }^2-2c_2{\lambda }_n{\parallel z_n-v_n\parallel }^2].\hfill \end{array}$$

(20)

We have the following formulas:

$$2\langle {\rho }_n-z_n,z_n-q^{*}\rangle =\parallel {\rho }_n-q^{*}{\parallel }^2-\parallel z_n-{\rho }_n{\parallel }^2-{\parallel z_n-q^{*}\parallel }^2.$$

$$2\langle {\rho }_n-v_n,z_n-v_n\rangle =\parallel {\rho }_n-v_n{\parallel }^2+\parallel z_n-v_n{\parallel }^2-{\parallel {\rho }_n-z_n\parallel }^2.$$

Combining the above equalities with Expression (20), we have:

$$\begin{array}{cc}\hfill \parallel z_n-q^{*}{\parallel }^2& \le \parallel {\rho }_n-q^{*}{\parallel }^2-(1-{\lambda }_{n+1}){\parallel z_n-{\rho }_n\parallel }^2\hfill \\ \hfill & -{\lambda }_{n+1}(1-2c_1{\lambda }_n)\parallel {\rho }_n-v_n{\parallel }^2-{\lambda }_{n+1}(1-2c_2{\lambda }_n){\parallel z_n-v_n\parallel }^2.\hfill \end{array}$$

□

Theorem 1.

Let $\left\{{\rho }_n\right\},$ $\left\{u_n\right\}$, and $\left\{v_n\right\}$ be the sequences generated by Algorithm 1 converging weakly to $q^{*}\in EP(f,\mathcal{K}).$

Proof. 

From the value of $u_{n+1}$ with Lemma 3, we have:

$$\begin{array}{cc}\hfill \parallel u_{n+1}-q^{*}{\parallel }^2& =\parallel (1-{\beta }_n){\rho }_n+{\beta }_n{z}_n-q^{*}{\parallel }^2\hfill \\ \hfill & =\parallel (1-{\beta }_n)({\rho }_n-q^{*})+{\beta }_n(z_n-q^{*}){\parallel }^2\hfill \\ \hfill & =(1-{\beta }_n)\parallel {\rho }_n-q^{*}{\parallel }^2+{\beta }_n\parallel z_n-q^{*}{\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel {\rho }_n-z_n\parallel }^2\hfill \\ \hfill & \le (1-{\beta }_n)\parallel {\rho }_n-q^{*}{\parallel }^2+{\beta }_n{\parallel z_n-q^{*}\parallel }^2.\hfill \end{array}$$

(21)

By Lemma 7 and Expression (21), we obtain:

$$\begin{array}{cc}\hfill \parallel u_{n+1}-q^{*}{\parallel }^2& \le \parallel {\rho }_n-q^{*}{\parallel }^2-{\beta }_n(1-{\lambda }_{n+1}){\parallel z_n-{\rho }_n\parallel }^2\hfill \\ \hfill & -{\beta }_n{\lambda }_{n+1}(1-2c_1{\lambda }_n)\parallel {\rho }_n-v_n{\parallel }^2-{\beta }_n{\lambda }_{n+1}(1-2c_2{\lambda }_n){\parallel z_n-v_n\parallel }^2.\hfill \end{array}$$

(22)

From Corollary 1, we have $\varrho \mu <{\lambda }_n<\sigma ,$ for all $n\ge 1.$ From Lemma 7, we have:

$$\parallel z_n-q^{*}{\parallel }^2\le {\parallel {\rho }_n-q^{*}\parallel }^2,\forall n\ge 1.$$

(23)

Combining Expressions (21) and (23), we have:

$$\parallel u_{n+1}-q^{*}{\parallel }^2\le {\parallel {\rho }_n-q^{*}\parallel }^2,\forall n\ge 1.$$

(24)

Due to definition of ${\rho }_n$, we have:

$$\begin{array}{cc}\hfill \parallel {\rho }_n-q^{*}{\parallel }^2& =\parallel u_n+{\theta }_n(u_n-u_{n-1})-q^{*}{\parallel }^2\hfill \\ \hfill & =\parallel (1+{\theta }_n)(u_n-q^{*})-{\theta }_n(u_{n-1}-q^{*}){\parallel }^2\hfill \\ \hfill & =(1+{\theta }_n)\parallel u_n-q^{*}{\parallel }^2-{\theta }_n\parallel u_{n-1}-q^{*}{\parallel }^2+{\theta }_n(1+{\theta }_n){\parallel u_n-u_{n-1}\parallel }^2\hfill \end{array}$$

(25)

$$\le (1+{\theta }_n)\parallel u_n-q^{*}{\parallel }^2-{\theta }_n\parallel u_{n-1}-q^{*}{\parallel }^2+2\theta {\parallel u_n-u_{n-1}\parallel }^2.$$

(26)

From the definition of ${\rho }_n$, we can obtain:

$$\begin{array}{c}\hfill \parallel {\rho }_n-q^{*}\parallel =\parallel u_n+{\theta }_n(u_n-u_{n-1})-q^{*}\parallel \le \parallel u_n-q^{*}\parallel +{\theta }_n\parallel u_n-u_{n-1}\parallel \end{array}$$

(27)

Combining Expressions (24) and (27), we have:

$$\begin{array}{cc}\hfill \parallel u_{n+1}-q^{*}\parallel & \le \parallel u_n-q^{*}\parallel +\theta \parallel u_n-u_{n-1}\parallel ,\forall n\ge 1.\hfill \end{array}$$

(28)

From Expressions (7) and (8), we deduce that:

$$\begin{array}{c}\hfill \sum _{n=1}^{\infty }\parallel u_n-u_{n-1}\parallel <+\infty \mathrm{and}\underset{n\to \infty }{lim}\parallel u_n-u_{n-1}\parallel =0.\end{array}$$

(29)

Using Lemma 4 with Expressions (28) and (29), we have:

$$\underset{n\to \infty }{lim}\parallel u_n-q^{*}\parallel =l,\mathrm{for}\mathrm{some}\mathrm{finite}l\ge 0.$$

(30)

By using (29) and (30) and letting $n\to \infty$ in (25), it is implied that:

$$\underset{n\to \infty }{lim}\parallel {\rho }_n-q^{*}\parallel =l.$$

(31)

Combining Expressions (22) and (26), we have:

$$\begin{array}{cc}\hfill \parallel u_{n+1}-q^{*}{\parallel }^2& \le (1+{\theta }_n)\parallel u_n-q^{*}{\parallel }^2-{\theta }_n\parallel u_{n-1}-q^{*}{\parallel }^2+2\theta {\parallel u_n-u_{n-1}\parallel }^2\hfill \\ \hfill & -{\beta }_n(1-{\lambda }_{n+1}){\parallel z_n-{\rho }_n\parallel }^2\hfill \\ \hfill & -{\beta }_n{\lambda }_{n+1}(1-2c_1{\lambda }_n)\parallel {\rho }_n-v_n{\parallel }^2-{\beta }_n{\lambda }_{n+1}(1-2c_2{\lambda }_n){\parallel z_n-v_n\parallel }^2,\hfill \end{array}$$

(32)

which further implies that (for $n\ge 1$):

$$\begin{array}{cc}\hfill 0& \le \beta (1-\sigma )\parallel z_n-{\rho }_n{\parallel }^2+\beta \sigma (1-2c_1\sigma ){\parallel {\rho }_n-v_n\parallel }^2\hfill \\ \hfill & +\beta \sigma (1-2c_2\sigma ){\parallel z_n-v_n\parallel }^2\hfill \\ \hfill & \le \parallel u_n-q^{*}{\parallel }^2-{\parallel u_{n+1}-q^{*}\parallel }^2+{\theta }_n\left(\parallel u_n-q^{*}{\parallel }^2-{\parallel u_{n-1}-q^{*}\parallel }^2\right)\hfill \\ \hfill & +2\theta \parallel u_n-u_{n-1}{\parallel }^2.\hfill \end{array}$$

(33)

By letting $n\to \infty$ in Expression (33), we obtain:

$$\underset{n\to \infty }{lim}\parallel z_n-{\rho }_n\parallel =\underset{n\to \infty }{lim}\parallel {\rho }_n-v_n\parallel =\underset{n\to \infty }{lim}\parallel v_n-z_n\parallel =0.$$

(34)

From Expressions (31) and (34), we obtain:

$$\underset{n\to \infty }{lim}\parallel v_n-q^{*}\parallel =l.$$

(35)

It follows from Expressions (30), (31), and (35) that the sequences $\left\{{\rho }_n\right\},$$\left\{u_n\right\}$ and $\left\{v_n\right\}$ are bounded, and for each $q^{*}\in EP(f,\mathcal{K}),$ the limit of $\parallel {\rho }_n-q^{*}\parallel ,$ $\parallel u_n-q^{*}\parallel$ and $\parallel v_n-q^{*}\parallel$ exists. Next, for using Lemma 2, we need to show that any sequential weak limit point of the sequence $\left\{u_n\right\}$ belongs to the set $EP(f,\mathcal{K}).$ Suppose z to be an arbitrary weak cluster point of $\left\{u_n\right\}$, i.e., a subsequence $\left\{u_{n_k}\right\}$ of $\left\{u_n\right\}$ weakly converges to $z.$ Due to $\parallel u_n-v_n\parallel \to 0$, then $\left\{v_{n_k}\right\}$ also weakly converges to z and $z\in \mathcal{K}.$ Next, we need to show that $z\in EP(f,\mathcal{K}).$ From (14) and the definition of ${\lambda }_{n+1}$ and (19), we have:

$$\begin{array}{cc}\hfill \mu {\lambda }_{n_k}f(v_{n_k},y)& \ge \mu {\lambda }_{n_k}f(v_{n_k},z_{n_k})+\langle {\rho }_{n_k}-z_{n_k},y-z_{n_k}\rangle \hfill \\ \hfill & \ge {\lambda }_{n_k}{\lambda }_{n_k+1}f({\rho }_{n_k},z_{n_k})-{\lambda }_{n_k}{\lambda }_{n_k+1}f({\rho }_{n_k},v_{n_k})-c_1{\lambda }_{n_k}{\lambda }_{n_k+1}{\parallel {\rho }_{n_k}-v_{n_k}\parallel }^2\hfill \\ \hfill & -c_2{\lambda }_{n_k}{\lambda }_{n_k+1}{\parallel v_{n_k}-z_{n_k}\parallel }^2+\langle {\rho }_{n_k}-z_{n_k},y-z_{n_k}\rangle \hfill \\ \hfill & \ge {\lambda }_{n_k+1}\langle {\rho }_{n_k}-v_{n_k},z_{n_k}-v_{n_k}\rangle -c_1{\lambda }_{n_k}{\lambda }_{n_k+1}{\parallel {\rho }_{n_k}-v_{n_k}\parallel }^2\hfill \\ \hfill & -c_2{\lambda }_{n_k}{\lambda }_{n_k+1}{\parallel v_{n_k}-z_{n_k}\parallel }^2+\langle {\rho }_{n_k}-z_{n_k},y-z_{n_k}\rangle .\hfill \end{array}$$

(36)

By letting $n\to \infty ,$ in the above expression, we obtain:

$$0\le \underset{k\to \infty }{lim\; sup}f(v_{n_k},y)\le f(z,y),\forall y\in \mathcal{K}.$$

It is concluded that $z\in EP(f,\mathcal{K}).$ Finally, by Lemma 2, $\left\{{\rho }_n\right\},$ $\left\{u_n\right\}$, and $\left\{v_n\right\}$ weakly converge to $q^{*}$ as $n\to \infty .$ □

4. Applications of the Main Results

We consider the implementation of our results to solve the variational inequality problems involving the pseudomonotone and Lipschitz-type continuous operator. A variational inequality problem is formulated in the following way:

$$\mathrm{Find}{q}^{*}\in \mathcal{K}\mathrm{such}\mathrm{that}〈G\left(q^{*}\right),y-q^{*}〉\ge 0,\forall y\in \mathcal{K}.$$

An operator $G:\mathcal{E}\to \mathcal{E}$ is said to be:

(i)

L-Lipschitz continuous on $\mathcal{K}$ if:

$$\parallel G\left(z_1\right)-G\left(z_2\right)\parallel \le L\parallel z_1-z_2\parallel ,\forall z_1,z_2\in \mathcal{K};$$

(ii)

pseudomonotone on $\mathcal{K}$ if:

$$〈G\left(z_1\right),z_2-z_1〉\ge 0⟹〈G\left(z_2\right),z_1-z_2〉\le 0,\forall z_1,z_2\in \mathcal{K}.$$

Assume that G satisfies the following conditions:

(G1)

G is pseudomonotone on $\mathcal{K}$ with solution set $VI(G,\mathcal{K})\ne \varnothing$;

(G2)

G is L-Lipschitz continuous on $\mathcal{K}$ with $L>0$;

(G3)

$\underset{n\to \infty }{lim\; sup}\langle G\left(u_n\right),y-u_n\rangle \le \langle G\left(p\right),y-p\rangle ,$$\forall y\in \mathcal{K}$ and $\left\{u_n\right\}\subset \mathcal{K}$ satisfy $u_n\rightharpoonup p.$

Let $f(x,y):=〈G\left(x\right),y-x〉,$ $\forall x,y\in \mathcal{K}.$ Then, the problem (1) translates into the variational inequality problem with $L=2c_1=2c_2.$ From the value of $v_n,$ we have:

$$\begin{array}{cc}\hfill v_n& =\underset{y\in \mathcal{K}}{argmin}\left\{{\lambda }_{n}f({\rho }_n,y)+\frac{1}{2}{\parallel {\rho }_n-y\parallel }^2\right\}\hfill \\ \hfill & =\underset{y\in \mathcal{K}}{argmin}\left\{{\lambda }_n\langle G\left({\rho }_n\right),y-{\rho }_n\rangle +\frac{1}{2}\parallel {\rho }_n{-y\parallel }^2+\frac{{\lambda }_n^2}{2}\parallel G\left({\rho }_n\right){\parallel }^2-\frac{{\lambda }_n^2}{2}{\parallel G\left({\rho }_n\right)\parallel }^2\right\}\hfill \\ \hfill & =\underset{y\in \mathcal{K}}{argmin}\left\{\frac{1}{2}{\parallel y-({\rho }_n-{\lambda }_{n}G\left({\rho }_n\right))\parallel }^2\right\}\hfill \\ \hfill & =P_{\mathcal{K}}({\rho }_n-{\lambda }_{n}G\left({\rho }_n\right)).\hfill \end{array}$$

(37)

In a similar way as Expression (37), the value of $u_{n+1}$ is written as:

$$z_n=P_{\mathcal{K}}({\rho }_n-\mu {\lambda }_{n}G\left(v_n\right)).$$

Corollary 2.

Let $G:\mathcal{K}\to \mathcal{E}$ be an operator satisfying the conditions(G1)–(G3). Assume that $\left\{{\rho }_n\right\},$$\left\{u_n\right\}$, and $\left\{v_n\right\}$ are the sequences generated in the following way:

(i)

Choose $u_{-1},u_0\in \mathcal{E},$$\varrho \in (0,1),$$\sigma <min\left\{1,\frac{1}{L}\right\},$$\mu \in (0,\sigma ),$${\lambda }_0>0,$$\theta \in [0,1)$, and a sequence $\left\{{ϵ}_n\right\}\subset [0,+\infty )$ such that:

$$\sum _{n=0}^{+\infty }{ϵ}_n<+\infty .$$

(38)

(ii)

Let ${\theta }_n$ satisfy $0\le {\theta }_n\le \overline{{\theta }_n}$ such that:

$$\overline{{\theta }_n}=\left\{\begin{array}{cc}min\left\{\theta ,\frac{{ϵ}_n}{\parallel u_n-u_{n-1}\parallel }\right\}\hfill & \mathrm{if}{u}_n\ne u_{n-1},\hfill \\ \theta \hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

(39)

(iii)

Set ${\rho }_n=u_n+{\theta }_n(u_n-u_{n-1})$, and compute:

$$\left\{\begin{array}{c}{v}_n=P_{\mathcal{K}}({\rho }_n-{\lambda }_{n}G\left({\rho }_n\right)),\hfill \\ z_n=P_{\mathcal{K}}({\rho }_n-\mu {\lambda }_{n}G\left(v_n\right)),\hfill \\ u_{n+1}=(1-{\beta }_n){\rho }_n+{\beta }_n{z}_n,\hfill \end{array}\right.$$

(40)

where $0<\beta \le {\beta }_n\le 1.$

(iv)

Next, stepsize ${\lambda }_{n+1}$ is obtained as follows:

$$\begin{array}{cc}\hfill {\lambda }_{n+1}=& min\left\{\sigma ,\frac{\mu \langle G\left(v_n\right),z_n-v_n\rangle }{\langle G\left({\rho }_n\right),z_n-v_n\rangle -\frac{L}{2}\parallel {\rho }_n-v_n{\parallel }^2-\frac{L}{2}{\parallel z_n-v_n\parallel }^2+1}\right\}.\hfill \end{array}$$

(41)

Then, $\left\{{\rho }_n\right\},$ $\left\{u_n\right\}$ and $\left\{v_n\right\}$ are weakly convergent to $q^{*}\in VI(G,\mathcal{K}).$

Next, consider the applications of our results that are discussed in Section 3 to solve fixed point problems involving $\kappa$-strict pseudocontraction. A mapping $T:\mathcal{K}\to \mathcal{K}$ is said to be:

(i)

a $\kappa$-strict pseudo-contraction [47] on $\mathcal{K}$ if:

$$\parallel T{z}_1-T{z}_2{\parallel }^2\le \parallel z_1-z_2{\parallel }^2+\kappa {\parallel (z_1-T{z}_1)-(z_2-T{z}_2)\parallel }^2,\forall z_1,z_2\in \mathcal{K};$$

(42)

that is equivalent to:

$$〈T{z}_1-T{z}_2,z_1-z_2〉\le \parallel z_1-z_2{\parallel }^2-\frac{1-\kappa }{2}{\parallel (z_1-T{z}_1)-(z_2-T{z}_2)\parallel }^2,\forall z_1,z_2\in \mathcal{K};$$

(43)

(ii)

sequentially weakly continuous on $\mathcal{K}$ if:

$$T\left(u_n\right)\rightharpoonup T\left(p\right)\mathrm{for}\mathrm{every}\mathrm{sequence}\mathrm{in}\mathcal{K}\mathrm{satisfying}{u}_n\rightharpoonup p\left(\mathrm{weakly}\mathrm{converges}\right).$$

A fixed point problem is formulated in the following way:

$$\mathrm{Find}{q}^{*}\in \mathcal{K}\mathrm{such}\mathrm{that}T\left(q^{*}\right)=q^{*}.$$

Let $f(x,y)=\langle x-Tx,y-x\rangle ,\forall x,y\in \mathcal{K}.$ Then, the problem (1) translates into the fixed point problem with $2c_1=2c_2=\frac{3-2\kappa }{1-\kappa }.$ The value of $v_n$ in Algorithm 1 converts into the followings:

$$\begin{array}{cc}\hfill v_n& =\underset{y\in \mathcal{K}}{argmin}\{{\lambda }_{n}f({\rho }_n,y)+\frac{1}{2}\parallel {\rho }_n-y{\parallel }^2\}\hfill \\ \hfill & =\underset{y\in \mathcal{K}}{argmin}\{{\lambda }_n\langle {\rho }_n-T\left({\rho }_n\right),y-{\rho }_n\rangle +\frac{1}{2}\parallel {\rho }_n-y{\parallel }^2\}\hfill \\ \hfill & =\underset{y\in \mathcal{K}}{argmin}\{{\lambda }_n\langle {\rho }_n-T\left({\rho }_n\right),y-{\rho }_n\rangle +\frac{1}{2}\parallel {\rho }_n{-y\parallel }^2+\frac{{\lambda }_n^2}{2}\parallel {\rho }_n-T\left({\rho }_n\right){\parallel }^2\}\hfill \\ \hfill & =\underset{y\in \mathcal{K}}{argmin}\{\frac{1}{2}\parallel y-{\rho }_n+{\lambda }_n({\rho }_n-T\left({\rho }_n\right)){\parallel }^2\}\hfill \\ \hfill & =P_{\mathcal{K}}[{\rho }_n-{\lambda }_n({\rho }_n-T\left({\rho }_n\right))].\hfill \end{array}$$

(44)

Corollary 3.

Assume that $\mathcal{K}$ is a nonempty, closed, and convex subset of a Hilbert space $\mathcal{E}$ and $T:\mathcal{K}\to \mathcal{K}$ is a κ-strict pseudocontraction and weakly continuous with $Fix\left(T\right)\ne \varnothing .$

(i)

Choose $u_{-1},u_0\in \mathcal{E},$ $\varrho \in (0,1),$ $\sigma <min\left\{1,\frac{1-\kappa }{3-2\kappa }\right\},$ $\mu \in (0,\sigma ),$ ${\lambda }_0>0,$ $\theta \in [0,1)$, and a sequence $\left\{{ϵ}_n\right\}\subset [0,+\infty )$ such that:

$$\sum _{n=0}^{+\infty }{ϵ}_n<+\infty .$$

(45)

(ii)

Let ${\theta }_n$ satisfy $0\le {\theta }_n\le \overline{{\theta }_n}$ such that:

$$\overline{{\theta }_n}=\left\{\begin{array}{cc}min\left\{\theta ,\frac{{ϵ}_n}{\parallel u_n-u_{n-1}\parallel }\right\}\hfill & \mathrm{if}{u}_n\ne u_{n-1},\hfill \\ \theta \hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

(46)

(iii)

Set ${\rho }_n=u_n+{\theta }_n(u_n-u_{n-1})$, and compute:

$$\left\{\begin{array}{c}{v}_n=P_{\mathcal{K}}[{\rho }_n-{\lambda }_n({\rho }_n-T\left({\rho }_n\right))],\hfill \\ z_n=P_{\mathcal{K}}[{\rho }_n-\mu {\lambda }_n(v_n-T\left(v_n\right))],\hfill \\ u_{n+1}=(1-{\beta }_n){\rho }_n+{\beta }_n{z}_n,\hfill \end{array}\right.$$

(47)

where $0<\beta \le {\beta }_n\le 1.$

(iv)

Next, stepsize ${\lambda }_{n+1}$ is obtained as follows:

$$\begin{array}{cc}\hfill {\lambda }_{n+1}=& min\left\{\sigma ,\frac{\mu \langle v_n-T{v}_n,z_n-v_n\rangle }{\langle {\rho }_n-T{\rho }_n,z_n-v_n\rangle -\left(\frac{3-2\kappa }{2-2\kappa }\right)\parallel {\rho }_n-v_n{\parallel }^2-\left(\frac{3-2\kappa }{2-2\kappa }\right){\parallel z_n-v_n\parallel }^2+1}\right\}.\hfill \end{array}$$

(48)

Then, the $\left\{{\rho }_n\right\},$ $\left\{u_n\right\}$, and $\left\{v_n\right\}$ sequences are weakly convergent to $q^{*}\in Fix\left(T\right).$

5. Numerical Experiments

MATLAB Version 9.5 (R2018b) was run on a PC (with Intel(R) Core(TM)i3-4010U CPU @ 1.70 GHz 1.70 GHz, RAM 4.00 GB). We used the built-in MATLAB fmincon function to solve the minimization problems. In this section, we discuss a number of test problems and explain the effectiveness of the proposed methodology.

Example 1.

Let $f:\mathcal{K}\times \mathcal{K}\to \mathcal{R}$ be defined by:

$$f(u,v)=\sum _{i=2}^5(v_i-u_i)\parallel u\parallel ,\forall u,v\in {\mathcal{R}}^5$$

where $\mathcal{K}=\left\{(u_1,\cdots ,u_5):u_1\ge -1,u_i\ge 1,i=2,\cdots ,5\right\}.$ The bifunction f is Lipschitz-type continuous with constants $c_1=c_2=2,$ satisfying the conditions($\mathsf{\Psi }$1)–($\mathsf{\Psi }$4). The solution set is $EP(f,\mathcal{K})=\{(u_1,1,1,1,1):u_1>1\}$ (see [48] for more details). In addition, to estimate the optimal values of the control parameters, two experiments are performed by assuming the variation of the control parameters $\lambda ,$ ${\lambda }_0$ and inertial factor $\theta .$ The values of the control parameters for Algorithm 1 (Alg2) are $\sigma =\frac{1}{2.3c_1},$ $\mu =\frac{1}{2.4c_1},$ $u_0=u_{-1}=(2,3,2,5,5),$ ${\beta }_n=0.80$, and ${ϵ}_n=\frac{1}{n^2}$; for Algorithm 1 (Alg1) in [19], they are $u_0=u_{-1}=(2,3,2,5,5)$ and ${ϵ}_n=\frac{1}{n^2}.$ Numerical results are shown in

Table 1. Example 1: Algorithm 1’s (Alg2) numerical comparison with Algorithm 1 in [19].

			Number of Iterations		CPU Time in Seconds
$\theta$	$\lambda$	${\lambda }_0$	Alg1	Alg2	Alg1	Alg2
0.40	0.22	0.45	14	2	0.9134	0.5424
0.40	0.17	0.35	18	2	0.8615	0.5223
0.40	0.12	0.32	20	2	1.0815	0.5112
0.40	0.07	0.25	22	2	1.4219	0.5367
0.40	0.02	0.05	26	2	1.7329	0.5181

and

Table 2. Example 1: Algorithm 1’s numerical comparison with Algorithm 1 in [19].

			Number of Iterations		CPU Time in Seconds
$\theta$	$\lambda$	${\lambda }_0$	Alg1	Alg2	Alg1	Alg2
0.80	0.22	0.16	21	2	1.0482	0.0811
0.60	0.22	0.16	15	2	0.8676	0.0626
0.40	0.22	0.16	12	2	1.0545	0.0791
0.20	0.22	0.16	11	2	0.09923	0.0892
0.05	0.22	0.16	19	2	1.09151	0.0788

by using the stopping criterion $D_n=\parallel {\rho }_n-v_n\parallel \le ϵ={10}^{-4}.$

Example 2.

Let us consider the Nash–Cournot equilibrium model that appeared in the paper [5,49]. A bifunction f can be written in the following manner:

$$f(u,v)=\langle Pu+Qv+c,v-u\rangle ,$$

where $c\in {\mathcal{R}}^5$, and the matrices P, Q are:

$$P=\left(\begin{array}{ccccc}3.1& 2& 0& 0& 0\\ 2& 3.6& 0& 0& 0\\ 0& 0& 3.5& 2& 0\\ 0& 0& 2& 3.3& 0\\ 0& 0& 0& 0& 3\end{array}\right)Q=\left(\begin{array}{ccccc}1.6& 1& 0& 0& 0\\ 1& 1.6& 0& 0& 0\\ 0& 0& 1.5& 1& 0\\ 0& 0& 1& 1.5& 0\\ 0& 0& 0& 0& 2\end{array}\right)c=\left(\begin{array}{c}1\\ -2\\ -1\\ 2\\ -1\end{array}\right)$$

where the Lipschitz constants are $c_1=c_2=\frac{1}{2}\parallel P-Q\parallel .$ The set $\mathcal{K}\subset {\mathcal{R}}^5$ is considered as follows:

$$\mathcal{K}:=\{u\in {\mathcal{R}}^5:-5\le u_i\le 5\}.$$

To see the suitable values of the control parameters, different tests are performed by assuming the variation of the control parameters’ inertial factor $\theta .$ The values of the control parameters for Algorithm 1 are $\sigma =\frac{1}{2.3c_1},$ $\mu =\frac{1}{2.4c_1},$ $u_0=u_{-1}=(1,1,1,1,1),$ $\lambda =\frac{1}{3c_1},$ ${\beta }_n=0.80$, and ${ϵ}_n=\frac{1}{n^2}$; for Algorithm 1 in [19], they are $u_0=u_{-1}=(1,1,1,1,1),$ ${ϵ}_n=\frac{1}{n^2}$, and $\lambda =0.20.$

Table 3. Example 2: Algorithm 1’s numerical comparison with Algorithm 1 in [19].

	Number of Iterations		CPU Time in Seconds
$\theta$	Alg1	Alg2	Alg1	Alg2
0.10	42	8	1.5851	0.2822
0.15	29	8	1.3148	0.2433
0.40	28	7	1.1278	0.2662
0.55	37	8	1.2211	0.2745
0.70	47	8	1.7188	0.2279
0.85	49	8	1.6188	0.2179

reports the numerical results by using the stopping criterion $D_n=\parallel {\rho }_n-v_n\parallel \le ϵ={10}^{-6}.$

To determine the suitable values of the control parameters, different tests are presented by assuming the different initial points. The control parameter values for Algorithm 1 are $\sigma =\frac{1}{2.1c_1},$ $\mu =\frac{1}{2.2c_1},$ $u_0=u_{-1},$ $\lambda =\frac{1}{3c_1},$ $\theta =0.45,$ ${\beta }_n=0.80$, and ${ϵ}_n=\frac{1}{n^2}$; for Algorithm 1 in [19], they are $u_0=u_{-1},$ $\theta =0.45,$ ${ϵ}_n=\frac{1}{n^2}$, and $\lambda =0.18.$

Table 4. Example 2: Algorithm 1’s numerical comparison with Algorithm 1 in [19].

	Number of Iterations		CPU Time in Seconds
$u_0=u_{-1}$	Alg1	Alg2	Alg1	Alg2
$(0,0,1,1,0)$	22	7	1.0321	0.1634
$(1,1,1,1,0)$	24	7	1.0945	0.1858
$(1,2,0,3,0)$	25	7	1.0328	0.2012
$(2,2,1,-4,5)$	30	7	1.0517	0.2020
$(2,-2,2,-4,5)$	35	8	1.0919	0.1428

reports the numerical results by using the stopping criterion $D_n=\parallel {\rho }_n-v_n\parallel \le ϵ={10}^{-6}.$

Example 3.

Assume that $\mathcal{E}=L^2\left([0,1]\right)$ through an inner product $\langle u,v\rangle ={\int }_0^{1}u\left(t\right)v\left(t\right)dt$ and induced norm $\parallel u\parallel =\sqrt{{\int }_0^1{u}^2\left(t\right)dt},$ $\forall u,v\in \mathcal{E}.$ The set $\mathcal{K}:=\{u\in L^2\left([0,1]\right):{\int }_0^{1}tu\left(t\right)dt=2\}.$ Assume that $f:\mathcal{E}\times \mathcal{E}\to \mathcal{R}$ is defined by:

$$f(u,v)=\langle G\left(u\right),v-u\rangle ,$$

where $G\left(u\left(t\right)\right)={\int }_0^{t}u\left(s\right)ds$ for every $u\in L^2\left([0,1]\right)$ and $t\in [0,1].$ It is observed that f is monotone, and Lipschitz-type constants $c_1=c_2=\frac{1}{\pi }$ (see [45]). The projection on set $\mathcal{K}$ is computed in the following manner:

$$P_{\mathcal{K}}\left(u\right)\left(t\right):=u\left(t\right)-\frac{{\int }_0^{1}tu\left(t\right)dt-2}{{\int }_0^1{t}^{2}dt}t,t\in [0,1].$$

The values of the control parameters for Algorithm 1 (Alg2) are $\sigma =\frac{1}{3.4c_1},$ $\mu =\frac{1}{3.6c_1},$ $\theta =0.45,$ $u_0=u_{-1},$ ${\lambda }_0=\frac{1}{3c_1},$ ${\beta }_n=0.90$, and ${ϵ}_n=\frac{1}{n^2}$; for Algorithm 1 (Alg1) in [19], they are $\theta =0.45,$ $u_0=u_{-1},$ ${ϵ}_n=\frac{1}{n^2}$, and $\lambda =0.12.$ Numerical results are reported in

Table 5. Example 3: Algorithm 1’s numerical comparison with Algorithm 1 in [19].

	Number of Iterations		CPU Time in Seconds
$u_{-1}=u_0$	Alg1	Alg2	Alg1	Alg2
$2t$	37	4	4.9566	0.5460
$2t^2$	41	2	5.2378	0.4331
$2sin\left(t\right)$	48	3	6.4556	0.3945
$2cos\left(t\right)$	51	6	6.6756	0.4945
$2exp\left(t\right)$	56	6	6.8713	0.5108

by using the stopping criterion $D_n=\parallel {\rho }_n-v_n\parallel \le ϵ={10}^{-4}.$

Example 4.

Assume that an operator $G:{\mathcal{R}}^2\to {\mathcal{R}}^2$ is defined by:

$$G\left(u\right)=\left(\begin{array}{c}0.5u_1{u}_2-2u_2-{10}^7\\ -4u_1-0.1u_2^2-{10}^7\end{array}\right)$$

on $\mathcal{K}=\{u\in {\mathcal{R}}^2:{(u_1-2)}^2+{(u_2-2)}^2\le 1\}.$ We can easily see that G is Lipschitz continuous with $L=5$ and pseudomonotone. Let the bifunction $f(u,v)=\langle G\left(u\right),v-u\rangle$ and $c_1=c_2=\frac{5}{2}.$ For these tests, we used the same initial values as seen in the table below and stepsize $\lambda =0.001,$ $u_{-1}=u_0,$ $\theta =0.45$, and ${ϵ}_n=\frac{1}{n^2}$ for Algorithm 1 in [19]. For Algorithm 1, we used $\sigma =\frac{1}{3.2c_1},$ $\mu =\frac{1}{3.4c_1},$ ${\lambda }_0=\frac{1}{3c_1},$ $u_{-1}=u_0,$ $\theta =0.45$, and ${ϵ}_n=\frac{1}{n^2}.$ Figure 1, Figure 2, Figure 3 and Figure 4 and

Table 6. Figure 1, Figure 2, Figure 3 and Figure 4: Algorithm 1’s numerical comparison with Algorithm 1 in [19].

	Number of Iterations		CPU Time in Seconds
$u_{-1}=u_0$	Alg1	Alg2	Alg1	Alg2
$(1.7,1.9)$	17	6	0.482301	0.3473423
$(2.5,3.5)$	18	8	0.789869	0.5659991
$(1.5,2.5)$	17	7	0.709038	0.3791612
$(2.7,2.0)$	15	7	0.585472	0.4401882

report the numerical results by letting the stopping criterion $D_n=\parallel {\rho }_n-v_n\parallel \le ϵ={10}^{-6}.$