Strong and linear convergence of projection-type method with an inertial term for finding minimum-norm solutions of pseudomonotone variational inequalities in Hilbert spaces

Abstract

The purpose of this paper is to investigate pseudomonotone variational inequalities in real Hilbert spaces. For solving this problem, we introduce a new method. The proposed algorithm combines the advantages of the subgradient extragradient method and the projection and contraction method. We establish the strong convergence of the proposed algorithm under conditions pseudomonotonicity and Lipschitz continuity assumptions. Moreover, under additional strong pseudomonotonicity and Lipschitz continuity assumptions, the linear convergence of the sequence generated by the proposed algorithm is obtained. Numerical examples provide to illustrate the potential of our algorithms as well as compare their performances to several related results.

1 Introduction

Let H be a real Hilbert space with inner product $\langle \cdot ,\cdot \rangle$ and norm $\Vert \cdot \Vert$. Let C be a nonempty, closed and convex subset of H. Let $F: H\rightarrow H$ be a single-valued continuous mapping. We consider classical variational inequality (VI) in the sense of Fichera [14] and Stampacchia [30] (see also Kinderlehrer and Stampacchia [21]) which is formulated as follows: find a point $x^{*}\in C$ such that

$$\begin{aligned} \langle Fx^{*}, x-x^{*}\rangle \ge 0 \ \ \forall x\in C. \end{aligned}$$

(1)

We denote by Sol(C, F) the solution set of the VI (1), which is assumed to be nonempty.

In this work, we assume that the following conditions hold:

Condition 1

The solution set $\text {Sol}(C,F)$ is nonempty.

Condition 2

The mapping $F:H\rightarrow H$ is pseudomonotone on H, that is,

$$\begin{aligned} \langle Fx,y-x \rangle \ge 0 \Longrightarrow \langle Fy,y-x \rangle \ge 0 \; \,\,\, \forall x,y \in H. \end{aligned}$$

Condition 3

The mapping $F:H\rightarrow H$ is Lipschitz continuous with constant $L>0$, that is, there exists a number $L>0$ such that

$$\begin{aligned} \Vert Fx-Fy\Vert \le L \Vert x-y\Vert \; \,\,\, \forall x,y \in H. \end{aligned}$$

Variational inequality (VI) is a very general mathematical model with numerous applications in economics, engineering mechanics, transportation, and many more, see, for example, [2, 13, 21, 23]. During the last decades, many algorithms for solving VIs have been proposed in the literature, see, e.g., [3, 10, 13, 15, 16, 21, 34, 37].

The most well-known one is extragradient method proposed by Korpelevich [22] (also by Antipin [1] independently). However, the extragradient method requires the evaluation of two orthogonal projections onto C per iteration. The first method which overcomes this obstacle is the projection and contraction method (PC) of He [18] and Sun [33]. Their algorithm is of the form:

$$\begin{aligned} y_{n}=P_{C}(x_{n}-\tau _{n} Fx_{n}), \end{aligned}$$

and then the next iterate $x_{n+1}$ is generated via the following

$$\begin{aligned} x_{n+1}=x_{n}-\gamma \eta _{n} d(x_{n},y_{n}), \end{aligned}$$

where $\gamma \in (0,2)$,

$$\begin{aligned} \eta _{n}:= \dfrac{\langle x_{n}-y_{n}, d(x_{n},y_{n})\rangle }{\Vert d(x_{n},y_{n})\Vert ^{2}}, \end{aligned}$$

and

$$\begin{aligned} d(x_{n},y_{n}):= x_{n}-y_{n}-\tau _{n} (Fx_{n}-Fy_{n}), \end{aligned}$$

where $F: C\rightarrow \mathbb {H}$ be monotone and L-Lipschitz continuous operator and $\tau _{n}\in (0,1/L)$ or $\tau _{n}$ is updated by some adaptive rule as follows:

$$\begin{aligned} \tau _{n}\Vert Fx_{n}-Fy_{n}\Vert \le \mu \Vert x_{n}-y_{n}\Vert , \ \ \mu \in (0,1). \end{aligned}$$

(2)

Recently, projection and contraction type methods for solving VI have received great attention by many authors, see, e.g., [4, 11, 12, 23, 27, 36].

The second extension of the extragradient method is known as the subgradient extragradient method proposed by Censor et al. [6,7,8]. In this algorithm, the second projection onto the feasible set C is replaced by a projection onto an easy and constructible set that contains C. For each $n\in \mathbb {N}$ generate the following sequences,

$$\begin{aligned} \left\{ \begin{array}{lll} y_{n}=P_{C}(x_{n}-\tau Fx_{n}),\\ T_{n}=\{x\in H \mid \langle x_{n} -\tau Fx_{n}-y_{n},x-y_{n}\rangle \le 0\},\\ x_{n+1}=P_{T_{n}}(x_{n}-\tau Fy_{n}),\\ \end{array}\right. \end{aligned}$$

where $\tau \in (0,1/L)$.

Since the projection and contraction and the subgradient extragradient methods require calculating only one projection onto C per iteration, their computational efforts and performance have an advantage over other existing results in the literature. Recently, [12] introduced a modification of the subgradient extragradient method by using the direction of the projection and contraction method and stepsize rule $\tau _{n}$ satisfying (2).

This paper is motivated and inspired by the work of Censor et al. [6], He [18] and Sun [33], first, we investigate the strong convergence for solving the problem (VI) by our new algorithm which is a combination of the subgradient extragradient method and the projection and contraction method in Hilbert spaces. In the proposed method, we show that an advantage of the proposed algorithm is the computation of only two values of the variational inequality mapping and one projection onto the feasible set per one iteration, which distinguishes our method from most other projection-type methods for variational inequality problems with pseudomonotone mappings. Second, the convergence rate of the algorithm is presented under strong pseudomonotonicity and Lipschitz continuity of the cost operator. Specifically, the proposed algorithm improves the results in the literature in the following ways:

• at each iteration step, a single projection is required to perform;

• an inertial term for speeding up convergence;

• step-sizes are not decreasing;

• without knowledge of the Lipschitz constant of the underline operator;

• without the assumption on the sequential weak continuity of the underline operator;

• strong convergence and moreover, a convergence estimate is established.

This paper is organized as follows: Section 2 consists of the notations and basic definitions which are useful throughout the paper. In Section 3, we propose our algorithm and prove the strong convergence of the iterative sequence to a solution of the variational inequality (1). The convergence rate of the proposed algorithm is presented in Section 4. In Section 5, some numerical results in optimal control problems are reported to demonstrate the performance of the proposed method. Final conclusions are given in Section 6.

2 Preliminaries

Let H be a real Hilbert space with inner product $\left\langle \cdot , \cdot \right\rangle$ and norm $\Vert \cdot \Vert$. The weak convergence of $\{x_{n}\}$ to x is denoted by $x_{n}\rightharpoonup x$ as $n\rightarrow \infty$, while the strong convergence of $\{x_{n}\}$ to x is written as $x_{n}\rightarrow x$ as $n\rightarrow \infty .$ For all $x,y\in H$ we have

$$\begin{aligned} \Vert x+y\Vert ^{2}\le \Vert x\Vert ^{2}+2\langle y,x+y\rangle . \end{aligned}$$

Definition 2.1

Let $T:H\rightarrow H$ be an operator. Then

• 1. T is called L-Lipschitz continuous with constant $L>0$ if

  $$\begin{aligned} \Vert Tx-Ty\Vert \le L \Vert x-y\Vert \ \ \ \forall x,y \in H, \end{aligned}$$

  if $L=1$ then the operator T is called nonexpansive and if $L\in (0,1)$, T is called a contraction.

• 2. T is called monotone if

  $$\begin{aligned} \langle Tx-Ty,x-y \rangle \ge 0 \ \ \ \forall x,y \in H; \end{aligned}$$

• 3. T is called pseudomonotone in the sense of Karamardian [19] if

  $$\begin{aligned} \langle Tx,y-x \rangle \ge 0 \Longrightarrow \langle Ty,y-x \rangle \ge 0 \ \ \ \forall x,y \in H; \end{aligned}$$

  (3)

• 4. T is called $\alpha$-strongly monotone if there exists a constant $\alpha >0$ such that

  $$\begin{aligned} \langle Tx-Ty,x-y\rangle \ge \alpha \Vert x-y\Vert ^{2} \ \ \forall x,y\in H; \end{aligned}$$

• 5. T is called $\alpha$-strongly pseudomonotone if there exists a constant $\alpha >0$ such that

  $$\begin{aligned} \langle Tx,y-x \rangle \ge 0 \Longrightarrow \langle Ty,y-x \rangle \ge \alpha \Vert x-y\Vert ^{2} \ \ \ \forall x,y \in H; \end{aligned}$$

• 6. The operator T is called sequentially weakly continuous if for each sequence $\{x_{n}\}$ we have: $x_{n}$ converges weakly to x implies ${Tx_{n}}$ converges weakly to Tx.

We note that (3) is only one of the definitions of pseudomonotonicity which can be found in the literature. For every point $x\in H$, there exists a unique nearest point in C, denoted by $P_{C}x$ such that $\Vert x-P_{C}x\Vert \le \Vert x-y\Vert \ \forall y\in C$. $P_{C}$ is called the metric projection of H onto C. It is known that $P_{C}$ is nonexpansive. For properties of the metric projection, the interested reader could be referred to Section 3 in [16].

Lemma 2.1

([16]) Let C be a nonempty closed convex subset of a real Hilbert space H. Given $x\in H$ and $z\in C$. Then $z=P_{C}x\Longleftrightarrow \langle x-z,z-y\rangle \ge 0 \ \ \forall y\in C.$ Moreover,

$$\begin{aligned} \Vert P_{C}x-P_{C}y\Vert ^{2}\le \langle P_{C} x-P_{C} y,x-y\rangle \ \forall x,y\in C. \end{aligned}$$

Lemma 2.2

Let H be a real Hilbert space. Then the following results hold:

• i) $\Vert x+y\Vert ^{2}=\Vert x\Vert ^{2}+2\langle x,y\rangle +\Vert y\Vert ^{2} \ \forall x,y\in H;$

• ii) $\Vert x+y\Vert ^{2}\le \Vert x\Vert ^{2}+2\langle y,x+y\rangle \ \forall x,y\in H.$

Lemma 2.3

([9]) Consider the problem Sol(C, F) with C being a nonempty, closed, convex subset of a real Hilbert space H and $F: C \rightarrow H$ being pseudomonotone and continuous. Then, $x^{*}$ is a solution of Sol(C, F) if and only if

$$\begin{aligned} \langle Fx, x - x^{*}\rangle \ge 0 \ \ \forall x \in C. \end{aligned}$$

Lemma 2.4

([28]) Let $\{a_{n}\}$ be sequence of nonnegative real numbers, $\{\alpha _{n}\}$ be a sequence of real numbers in (0, 1) with $\sum _{n=1}^{\infty }\alpha _{n}=\infty$ and $\{b_{n}\}$ be a sequence of real numbers. Assume that

$$\begin{aligned} a_{n+1}\le (1-\alpha _{n})a_{n}+\alpha _{n} b_{n}\ \ \forall n\ge 1. \end{aligned}$$

If $\limsup _{k\rightarrow \infty } b_{n_{k}} \le 0$ for every subsequence $\{a_{n_{k}}\}$ of $\{a_{n}\}$ satisfying $\liminf _{k\rightarrow \infty }(a_{n_{k}+1}-a_{n_{k}})\ge 0$ then $\lim _{n\rightarrow \infty }{a_{n}} = 0$.

Definition 2.2

([26]) Let $\{x_{n}\}$ be a sequence in H.

• i) $\{x_{n}\}$ is said to converge R-linearly to $x^{*}$ with rate $\rho \in [0, 1)$ if there is a constant $c>0$ such that

  $$\begin{aligned} \Vert x_{n}-x^{*}\Vert \le c\rho ^{n} \ \ \forall n\in \mathbb {N}. \end{aligned}$$

• ii) $\{x_{n}\}$ is said to converge Q-linearly to $x^{*}$ with rate $\rho \in [0, 1)$ if

  $$\begin{aligned} \Vert x_{n+1}-x^{*}\Vert \le \rho \Vert x_{n}-x^{*}\Vert \ \ \forall n\in \mathbb {N}. \end{aligned}$$

3 Strong convergence analysis

First, we introduce the proposed algorithm:

Observe that the projection onto half-space $T_{n}$ in Step 2 is explicit [5, Section 4.1.3, p. 133], therefore, Algorithm 1 requires only one projection in Step 1. Moreover, the stepsize $\tau _{n}$ is updated adaptively in Step 3 without requiring the knowledge of the Lipschitz constant L. We start the convergence analysis by proving the following Lemmas.

Lemma 3.5

([24]) Assume that F is L-Lipschitz continuous on H. Let $\left\{ \tau _{n}\right\}$ be the sequence generated by (5). Then

$$\begin{aligned} \lim _{n\rightarrow \infty } \tau _{n}=\tau \text { with } \tau \in \bigg [ \min \left\{ \tau _{1},\dfrac{\mu }{L}\right\} ,\tau _{1}+\alpha \bigg ], \end{aligned}$$

where $\alpha =\sum _{n=1}^{\infty } \alpha _{n}$. Moreover

$$\begin{aligned} \Vert Fw_{n}-Fv_{n}\Vert \le \dfrac{\mu }{\tau _{n+1}}\Vert w_{n}-v_{n}\Vert . \end{aligned}$$

(6)

Lemma 3.6

Assume that F is Lipschitz continuous on H and pseudomonotone on C. Then for every $x^{*} \in Sol(C,F)$, there exists $n_{0} > 0$ such that

$$\begin{aligned} \Vert u_{n+1}-x^{*}\Vert ^{2}\le \Vert w_{n}-x^{*}\Vert ^{2}-\Vert w_{n}- u_{n+1}-\gamma \eta _{n}d_{n}\Vert ^{2}-(2-\gamma )\gamma {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2}\ \ \forall n\ge n_{0}. \end{aligned}$$

Proof

Using (6), we have

$$\begin{aligned} \Vert d_{n}\Vert= & {} \Vert w_{n}-v_{n}-\tau _{n}(Fw_{n}-Fv_{n})\Vert \nonumber \\\ge & {} \Vert w_{n}-v_{n}\Vert -\tau _{n}\Vert Fw_{n}-Fv_{n}\Vert \nonumber \\\ge & {} \Vert w_{n}-v_{n}\Vert -\dfrac{\mu \tau _{n}}{\tau _{n+1}}\Vert w_{n}-v_{n}\Vert \nonumber \\= & {} \left( 1- \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) \Vert w_{n}-v_{n}\Vert . \end{aligned}$$

(7)

Since $\lim _{n\rightarrow \infty }\left( 1- \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) =1-\mu >0$, there exists $n_{0}\in \mathbb {N}$ such that

$$\begin{aligned} 1- \dfrac{\mu \tau _{n}}{\tau _{n+1}}>\dfrac{1-\mu }{2} \ \ \forall n\ge n_{0}. \end{aligned}$$

Therefore, it follows from (7) that for all $n\ge n_{0}$ we get

$$\begin{aligned} \Vert d_{n}\Vert \ge \dfrac{1-\mu }{2}\Vert w_{n}-v_{n}\Vert >0. \end{aligned}$$

(8)

Since $x^{*}\in Sol(C,F) \subset C\subset T_{n}$, using Lemma 2.1 we have

$$\begin{aligned} \Vert u_{n+1}-x^{*}\Vert ^{2}= & {} \Vert P_{T_{n}}(w_{n}-\gamma \eta _{n} \tau _{n} Fv_{n})-P_{T_{n}}x^{*}\Vert ^{2}\\\le & {} \langle u_{n+1}-x^{*},w_{n}-\gamma \eta _{n} \tau _{n} Fv_{n}-x^{*}\rangle \\= & {} \dfrac{1}{2}\Vert u_{n+1}-x^{*}\Vert ^{2}+\dfrac{1}{2}\Vert w_{n}-\gamma \eta _{n} \tau _{n} Fv_{n}-x^{*}\Vert ^{2}-\dfrac{1}{2}\Vert u_{n+1}-w_{n}+\gamma \eta _{n} \tau _{n} Fv_{n}\Vert ^{2}\\= & {} \dfrac{1}{2}\Vert u_{n+1}-x^{*}\Vert ^{2}+\dfrac{1}{2}\Vert w_{n}-x^{*}\Vert ^{2}+\dfrac{1}{2}\gamma ^{2} {\eta ^{2}_{n}} {\tau ^{2}_{n}} \Vert Fv_{n}\Vert ^{2}- \langle w_{n}-x^{*},\gamma \eta _{n}\tau _{n} Fv_{n} \rangle \\&- \dfrac{1}{2}\Vert u_{n+1}-w_{n}\Vert ^{2}-\dfrac{1}{2}\gamma ^{2} {\eta ^{2}_{n}} {\tau ^{2}_{n}} \Vert Fv_{n}\Vert ^{2}-\langle u_{n+1}-w_{n},\gamma \eta _{n} \tau _{n} Fv_{n}\rangle \\= & {} \dfrac{1}{2}\Vert u_{n+1}-x^{*}\Vert ^{2}+\dfrac{1}{2}\Vert w_{n}-x^{*}\Vert ^{2} -\dfrac{1}{2}\Vert u_{n+1}-w_{n}\Vert ^{2}-\langle u_{n+1}-x^{*}, \gamma \eta _{n}\tau _{n} Fv_{n}\rangle . \end{aligned}$$

This implies that

$$\begin{aligned} \Vert u_{n+1}-x^{*}\Vert ^{2}\le \Vert w_{n}-x^{*}\Vert ^{2} -\Vert u_{n+1}-w_{n}\Vert ^{2}-2\gamma \eta _{n} \tau _{n}\langle u_{n+1}-x^{*}, Fv_{n}\rangle . \end{aligned}$$

(9)

Since $v_{n}\in C$ and $x^{*}\in Sol(C,F)$,we get $\langle Fx^{*},v_{n}-x^{*}\rangle \ge 0$. By the pseudomonotonicity of F, we have $\langle Fv_{n},v_{n}-x^{*}\rangle \ge 0$, which implies

$$\begin{aligned} \langle Fv_{n},u_{n+1}-x^{*}\rangle =\langle Fv_{n},u_{n+1}-v_{n}\rangle +\langle Fv_{n},v_{n}-x^{*}\rangle \ge \langle Fv_{n},u_{n+1}-v_{n}\rangle . \end{aligned}$$

Thus, we obtain

$$\begin{aligned} -2\gamma \eta _{n} \tau _{n} \langle Fv_{n},u_{n+1}-x^{*}\rangle \le -2\gamma \eta _{n} \tau _{n} \langle Fv_{n},u_{n+1}-v_{n}\rangle . \end{aligned}$$

(10)

On the other hand, from $u_{n+1}\in T_{n}$ we have

$$\begin{aligned} \langle w_{n}-\tau _{n} Fw_{n}-v_{n},u_{n+1}-v_{n}\rangle \le 0. \end{aligned}$$

This implies that

$$\begin{aligned} \langle w_{n}-v_{n}-\tau _{n} (Fw_{n}-Fv_{n}),u_{n+1}-v_{n}\rangle \le \tau _{n} \langle Fv_{n},u_{n+1}-v_{n}\rangle , \end{aligned}$$

thus

$$\begin{aligned} \langle d_{n},u_{n+1}-v_{n}\rangle \le \tau _{n} \langle Fv_{n},u_{n+1}-v_{n}\rangle . \end{aligned}$$

Hence

$$\begin{aligned} -2\gamma \eta _{n} \tau _{n} \langle Fv_{n},u_{n+1}-v_{n}\rangle \le -2\gamma \eta _{n} \langle d_{n},u_{n+1}-v_{n}\rangle . \end{aligned}$$

(11)

On the other hand, we have

$$\begin{aligned} -2\gamma \eta _{n} \langle d_{n},u_{n+1}-v_{n}\rangle = -2\gamma \eta _{n} \langle d_{n},w_{n}-v_{n}\rangle +2\gamma \eta _{n} \langle d_{n},w_{n}- u_{n+1}\rangle . \end{aligned}$$

(12)

From (8), we have $d_{n}\ne 0 \ \ \forall n\ge n_{0}$, thus $\eta _{n}=\dfrac{\langle w_{n}-v_{n},d_{n}\rangle }{\Vert d_{n}\Vert ^{2}}$, which means

$$\begin{aligned} \langle w_{n}-v_{n},d_{n}\rangle =\eta _{n} \Vert d_{n}\Vert ^{2}\ \ \forall n\ge n_{0}. \end{aligned}$$

(13)

Moreover

$$\begin{aligned} 2\gamma \eta _{n} \langle d_{n},w_{n}- u_{n+1}\rangle= & {} 2 \langle \gamma \eta _{n}d_{n},w_{n}- u_{n+1}\rangle \nonumber \\= & {} \Vert w_{n}- u_{n+1}\Vert ^{2}+\gamma ^{2} {\eta ^{2}_{n}}\Vert d_{n}\Vert ^{2}-\Vert w_{n}- u_{n+1}-\gamma \eta _{n}d_{n}\Vert ^{2}. \end{aligned}$$

(14)

Substituting (13) and (14) into (12) we get for all $n \ge n_{0}$ that

$$\begin{aligned} -2\gamma \eta _{n} \langle d_{n},u_{n+1}-v_{n}\rangle\le & {} -2\gamma {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2}+\Vert w_{n}- u_{n+1}\Vert ^{2}+\gamma ^{2} {\eta ^{2}_{n}}\Vert d_{n}\Vert ^{2}-\Vert w_{n}- u_{n+1}-\gamma \eta _{n}d_{n}\Vert ^{2}\nonumber \\= & {} \Vert w_{n}- u_{n+1}\Vert ^{2}-\Vert w_{n}- u_{n+1}-\gamma \eta _{n}d_{n}\Vert ^{2}-(2-\gamma )\gamma {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2}. \end{aligned}$$

(15)

Combining (11) and (15), we obtain

$$\begin{aligned} -2\gamma \eta _{n} \tau _{n} \langle Fv_{n},u_{n+1}-v_{n}\rangle\le & {} -2\gamma {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2}+\Vert w_{n}- u_{n+1}\Vert ^{2}+\gamma ^{2} {\eta ^{2}_{n}}\Vert d_{n}\Vert ^{2}-\Vert w_{n}- u_{n+1}-\gamma \eta _{n}d_{n}\Vert ^{2}\nonumber \\= & {} \Vert w_{n}- u_{n+1}\Vert ^{2}-\Vert w_{n}- u_{n+1}-\gamma \eta _{n}d_{n}\Vert ^{2}-(2-\gamma )\gamma {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2}. \end{aligned}$$

(16)

Again, combining (10) and (16), we get

$$\begin{aligned} -2\gamma \eta _{n} \tau _{n} \langle Fv_{n},u_{n+1}-x^{*}\rangle\le & {} -2\gamma {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2}+\Vert w_{n}- u_{n+1}\Vert ^{2}+\gamma ^{2} {\eta ^{2}_{n}}\Vert d_{n}\Vert ^{2}-\Vert w_{n}- u_{n+1}-\gamma \eta _{n}d_{n}\Vert ^{2}\nonumber \\= & {} \Vert w_{n}- u_{n+1}\Vert ^{2}-\Vert w_{n}- u_{n+1}-\gamma \eta _{n}d_{n}\Vert ^{2}-(2-\gamma )\gamma {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2}. \end{aligned}$$

(17)

Substituting (17) into (9) we get

$$\begin{aligned} \Vert u_{n+1}-x^{*}\Vert ^{2}\le \Vert w_{n}-x^{*}\Vert ^{2}-\Vert w_{n}- u_{n+1}-\gamma \eta _{n}d_{n}\Vert ^{2}-(2-\gamma )\gamma {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2}\ \ n \ge n_{0}. \end{aligned}$$

(18)

Theorem 3.1

Assume that Conditions 1–3 hold. In addition, we assume that the mapping $F:H\rightarrow H$ satisfies the following condition

$$\begin{aligned} \text { whenever } \{u_{n}\} \subset C, u_{n} \rightharpoonup z, \text { one has } \Vert Fz\Vert \le \liminf _{n\rightarrow \infty }\Vert Fu_{n}\Vert . \end{aligned}$$

(19)

Then the sequence $\{u_{n}\}$ is generated by Algorithm 1 converges strongly to an element $z\in Sol(C,F)$, where $z=P_{Sol(C,F)}(0)$.

Proof

Claim 1. The sequence $\{u_{n}\}$ is bounded. Indeed, we have

$$\begin{aligned} \Vert w_{n}-z\Vert= & {} \Vert (1-\gamma _{n})(u_{n}+\theta _{n}(u_{n}-u_{n-1}))-z\Vert \nonumber \\= & {} \Vert (1-\gamma _{n})(u_{n}-z)+(1-\gamma _{n})\theta _{n}(u_{n}-u_{n-1})-\gamma _{n} z\Vert \nonumber \\\le & {} (1-\gamma _{n})\Vert u_{n}-z\Vert +(1-\gamma _{n})\theta _{n}\Vert u_{n}-u_{n-1}\Vert +\gamma _{n} \Vert z\Vert \nonumber \\= & {} (1-\gamma _{n})\Vert u_{n}-z\Vert +\gamma _{n} [ (1-\gamma _{n})\dfrac{\theta _{n}}{\gamma _{n}}\Vert u_{n}-u_{n-1}\Vert +\Vert z\Vert ]. \end{aligned}$$

(20)

On the other hand, since (4) we have

$$\begin{aligned} \dfrac{\theta _{n}}{\gamma _{n}}\Vert u_{n}-u_{n-1}\Vert \le \dfrac{\epsilon _{n}}{\gamma _{n}}\rightarrow 0 \end{aligned}$$

this implies that $\lim _{n\rightarrow \infty }\bigg [ (1-\gamma _{n})\dfrac{\theta _{n}}{\gamma _{n}}\Vert u_{n}-u_{n-1}\Vert +\Vert z\Vert \bigg ]=\Vert z\Vert$, thus there exists $M>0$ such that

$$\begin{aligned} (1-\gamma _{n})\dfrac{\theta _{n}}{\gamma _{n}}\Vert u_{n}-u_{n-1}\Vert +\Vert z\Vert \le M. \end{aligned}$$

(21)

Combining (20) and (21) we obtain

$$\begin{aligned} \Vert w_{n}-z\Vert \le (1-\gamma _{n})\Vert u_{n}-z\Vert +\gamma _{n} M. \end{aligned}$$

Moreover, we have $\lim _{n\rightarrow \infty }(1-\mu \dfrac{\tau _{n}}{\tau _{n+1}})=1-\mu >\dfrac{1-\mu }{2}$, thus there exists $n_{0}\in \mathbb {N}$ such that $1-\mu \dfrac{\tau _{n}}{\tau _{n+1}}>0 \ \ \forall n\ge n_{0},$ by Claim 1 we obtain

$$\begin{aligned} \Vert u_{n+1}-z\Vert \le \Vert w_{n}-z\Vert \ \ \forall n\ge n_{0}. \end{aligned}$$

(22)

Thus

$$\begin{aligned} \Vert u_{n+1}-z\Vert\le & {} (1-\gamma _{n})\Vert u_{n}-z\Vert +\gamma _{n} M\\= & {} \max \{\Vert u_{n}-z\Vert , M \}\le ...\le \max \{\Vert u_{n_{0}}-z\Vert ,M\}. \end{aligned}$$

Therefore, the sequence $\{u_{n}\}$ is bounded.

Claim 2.

$$\begin{aligned} \Vert w_{n}- u_{n+1}-\gamma \eta _{n}d_{n}\Vert ^{2}+(2-\gamma )\gamma {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2} \le \Vert u_{n}-z\Vert ^{2}-\Vert u_{n+1}-z\Vert ^{2}+\gamma _{n} M_{1}. \end{aligned}$$

Indeed, we have $\Vert w_{n}-z\Vert \le (1-\gamma _{n})\Vert u_{n}-z\Vert +\gamma _{n} M$, this implies that

$$\begin{aligned} \Vert w_{n}-z\Vert ^{2}\le & {} (1-\gamma _{n})^{2}\Vert u_{n}-z\Vert ^{2}+2\gamma _{n}(1-\gamma _{n})M \Vert u_{n}-z\Vert +{\gamma _{n}^{2}}M^{2}\nonumber \\\le & {} \Vert u_{n}-z\Vert ^{2}+\gamma _{n} [2(1-\gamma _{n})M \Vert u_{n}-z\Vert +\gamma _{n} M^{2}]\nonumber \\\le & {} \Vert u_{n}-z\Vert ^{2}+\gamma _{n} M_{1}, \end{aligned}$$

(23)

where $M_{1}:=\max \{2(1-\gamma _{n})M \Vert u_{n}-z\Vert +\gamma _{n} M^{2}:\ \ n\in \mathbb {N}\}$. Substituting (23) into (18) we get

$$\begin{aligned} \Vert u_{n+1}-x^{*}\Vert ^{2}\le \Vert u_{n}-z\Vert ^{2}+\gamma _{n} M_{1}-\Vert w_{n}- u_{n+1}-\gamma \eta _{n}d_{n}\Vert ^{2}-(2-\gamma )\gamma {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2}. \end{aligned}$$

Or equivalently

$$\begin{aligned} \Vert w_{n}- u_{n+1}-\gamma \eta _{n}d_{n}\Vert ^{2}+(2-\gamma )\gamma {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2} \le \Vert u_{n}-z\Vert ^{2}-\Vert u_{n+1}-z\Vert ^{2}+\gamma _{n} M_{1}. \end{aligned}$$

Claim 3.

$$\begin{aligned} \dfrac{\left( 1-\dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}{\left( 1+ \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}\Vert w_{n}-v_{n}\Vert ^{2}\le {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2}\ \ \forall n\ge n_{0}. \end{aligned}$$

We have

$$\begin{aligned} \Vert d_{n}\Vert \le \Vert w_{n}-v_{n}\Vert +\tau _{n}\Vert Fw_{n}-Fv_{n}\Vert \le \left( 1+ \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) \Vert w_{n}-v_{n}\Vert . \end{aligned}$$

Hence

$$\begin{aligned} \Vert d_{n}\Vert ^{2}\le \left( 1+ \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}\Vert w_{n}-v_{n}\Vert ^{2}, \end{aligned}$$

or equivalently

$$\begin{aligned} \dfrac{1}{\Vert d_{n}\Vert ^{2}} \ge \dfrac{1}{\left( 1+\dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}\Vert w_{n}-v_{n}\Vert ^{2} }. \end{aligned}$$

Again, we find

$$\begin{aligned} \langle w_{n}-v_{n},d_{n}\rangle= & {} \Vert w_{n}-v_{n}\Vert ^{2}-\tau _{n} \langle w_{n}-v_{n},Fw_{n}-Fv_{n}\rangle \nonumber \\\ge & {} \Vert w_{n}-v_{n}\Vert ^{2}-\tau _{n} \Vert w_{n}-v_{n}\Vert \Vert Fw_{n}-Fv_{n}\Vert \nonumber \\\ge & {} \Vert w_{n}-v_{n}\Vert ^{2}- \dfrac{\mu \tau _{n}}{\tau _{n+1}}\Vert w_{n}-v_{n}\Vert ^{2}\nonumber \\= & {} \left( 1- \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) \Vert w_{n}-v_{n}\Vert ^{2}. \end{aligned}$$

Hence for all $n\ge n_{0}$

$$\begin{aligned} \eta _{n} \Vert d_{n}\Vert ^{2} = \langle w_{n}-v_{n}, d_{n}\rangle \ge \left( 1- \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) \Vert w_{n}-v_{n}\Vert ^{2} \end{aligned}$$

(24)

and

$$\begin{aligned} \eta _{n}=\dfrac{\langle w_{n}-v_{n}, d_{n}\rangle }{\Vert d_{n}\Vert ^{2}} \ge \dfrac{\left( 1- \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) }{\left( 1+ \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}. \end{aligned}$$

(25)

Combining (24) and (25), we get

$$\begin{aligned} {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2} \ge \dfrac{\left( 1-\dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}{\left( 1+ \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}\Vert w_{n}-v_{n}\Vert ^{2}\ \ \forall n\ge n_{0}. \end{aligned}$$

(26)

Claim 4.

$$\begin{aligned} \Vert u_{n+1}-z\Vert ^{2}\le & {} (1-\gamma _{n})\Vert u_{n}-z\Vert ^{2}+\gamma _{n}\bigg [2(1-\gamma _{n})\Vert u_{n}-z\Vert \dfrac{\theta _{n}}{\gamma _{n}} \Vert u_{n}-u_{n-1}\Vert \\&+\theta _{n} \Vert u_{n}-u_{n-1}\Vert \dfrac{\theta _{n}}{\gamma _{n}}\Vert u_{n}-u_{n-1}\Vert +2 \Vert z\Vert \Vert u_{n}-u_{n+1}\Vert +2 \langle -z,u_{n+1}-z \rangle \bigg ] \end{aligned}$$

$\forall n\ge n_{0}$. Indeed, using Lemma 2.2 ii) and (22) we get

$$\begin{aligned} \Vert x_{n+1}-z\Vert ^{2}\le & {} \Vert w_{n}-z\Vert ^{2}\ \ \forall n\ge n_{0}\\= & {} \Vert (1-\gamma _{n}) (u_{n}-z)+(1-\gamma _{n}) \theta _{n} (u_{n}-u_{n-1})-\gamma _{n} z\Vert ^{2}\ \ \forall n\ge n_{0} \\\le & {} \Vert (1-\gamma _{n})(u_{n}-z)+(1-\gamma _{n})\theta _{n} (u_{n}-u_{n-1})\Vert ^{2}+2\gamma _{n} \langle -z,w_{n}-z\rangle \ \ \forall n\ge n_{0}\\= & {} (1-\gamma _{n})^{2}\Vert u_{n}-z\Vert ^{2}+2(1-\gamma _{n})\theta _{n}\Vert u_{n}-z\Vert \Vert u_{n}-u_{n-1}\Vert +{\theta _{n}^{2}}\Vert u_{n}-u_{n-1}\Vert ^{2}\\&+2 \langle -z,u_{n}-u_{n+1} \rangle +2 \langle -z,u_{n+1}-z \rangle \ \ \forall n\ge n_{0} \\\le & {} (1-\gamma _{n})\Vert u_{n}-z\Vert ^{2}+\gamma _{n}\bigg [2(1-\gamma _{n})\Vert u_{n}-z\Vert \dfrac{\theta _{n}}{\gamma _{n}} \Vert u_{n}-u_{n-1}\Vert \\&+\theta _{n} \Vert u_{n}-u_{n-1}\Vert \dfrac{\theta _{n}}{\gamma _{n}}\Vert u_{n}-u_{n-1}\Vert +2 \Vert z\Vert \Vert u_{n}-u_{n+1}\Vert +2 \langle -z,u_{n+1}-z \rangle \bigg ]\ \ \forall n\ge n_{0}. \end{aligned}$$

Claim 5. $\{\Vert u_{n}-z\Vert ^{2}\}$ converges to zero. Indeed, by Lemma 2.4 it suffices to show that $\limsup _{k\rightarrow \infty }\langle -z, u_{n_{k}+1}-z\rangle \le 0$ and $\limsup _{k\rightarrow \infty }\Vert u_{n_{k}}-u_{n_{k}+1}\Vert \le 0$ for every subsequence $\{\Vert u_{n_{k}}-z\Vert \}$ of $\{\Vert u_{n}-z\Vert \}$ satisfying

$$\begin{aligned} \liminf _{k\rightarrow \infty }(\Vert u_{n_{k}+1}-z\Vert -\Vert u_{n_{k}}-z\Vert )\ge 0. \end{aligned}$$

For this, suppose that $\{\Vert u_{n_{k}}-z\Vert \}$ is a subsequence of $\{\Vert u_{n}-z\Vert \}$ such that $\liminf _{k\rightarrow \infty }(\Vert u_{n_{k}+1}-z\Vert -\Vert u_{n_{k}}-z\Vert )\ge 0.$ Then

$$\begin{aligned} \liminf _{k\rightarrow \infty }(\Vert u_{n_{k}+1}-z\Vert ^{2}-\Vert u_{n_{k}}-z\Vert ^{2})=\liminf _{k\rightarrow \infty }[(\Vert u_{n_{k}+1}-z\Vert -\Vert u_{n_{k}}-z\Vert )(\Vert u_{n_{k}+1}-z\Vert +\Vert u_{n_{k}}-z\Vert )]\ge 0. \end{aligned}$$

By Claim 2 we obtain

$$\begin{aligned} \limsup _{k\rightarrow \infty }&\bigg [\Vert w_{n_{k}}- u_{n_{k}+1}-\gamma \eta _{n_{k}}d_{n_{k}}\Vert ^{2}+(2-\gamma )\gamma \eta ^{2}_{n_{k}} \Vert d_{n_{k}}\Vert ^{2}\bigg ]\\\le & {} \limsup _{k\rightarrow \infty }\bigg [ \Vert u_{n_{k}}-z\Vert ^{2}-\Vert u_{{n_{k}}+1}-z\Vert ^{2}+\gamma _{n_{k}} M_{1}\bigg ]\\\le & {} \limsup _{k\rightarrow \infty }\bigg [ \Vert u_{n_{k}}-z\Vert ^{2}-\Vert u_{n_{k}+1}-z\Vert ^{2}\bigg ]+\limsup _{k\rightarrow \infty }\gamma _{n_{k}} M_{1}\\= & {} - \liminf _{k\rightarrow \infty }\bigg [ \Vert u_{n_{k}+1}-z\Vert ^{2}-\Vert u_{_{n_{k}}}-z\Vert ^{2}\bigg ]\\\le & {} 0. \end{aligned}$$

This implies that

$$\begin{aligned} \lim _{k\rightarrow \infty } \Vert w_{n_{k}}- u_{n_{k}+1}-\gamma \eta _{n_{k}}d_{n_{k}}\Vert =0 \text { and } \lim _{k\rightarrow \infty }\gamma \eta ^{2}_{n_{k}} \Vert d_{n_{k}}\Vert =0. \end{aligned}$$

We have

$$\begin{aligned} \Vert w_{n_{k}}- u_{n_{k}+1}\Vert\le & {} \Vert w_{n_{k}}- u_{n_{k}+1}-\gamma \eta _{n_{k}}d_{n_{k}}\Vert +\gamma \Vert \eta _{n_{k}}d_{n_{k}}\Vert \nonumber \\= & {} \Vert w_{n_{k}}- u_{n_{k}+1}-\gamma \eta _{n_{k}}d_{n_{k}}\Vert + \gamma \dfrac{1}{\eta _{n_{k}}} \eta _{n_{k}}^{2}\Vert d_{n_{k}}\Vert . \end{aligned}$$

(27)

Combining (25) and (28) we get

$$\begin{aligned} \Vert w_{n_{k}}- u_{n_{k}+1}\Vert\le & {} \Vert w_{n_{k}}- u_{n_{k}+1}-\gamma \eta _{n_{k}}d_{n_{k}}\Vert +\gamma \eta _{n_{k}}\Vert d_{n_{k}}\Vert \nonumber \\= & {} \Vert w_{n_{k}}- u_{n_{k}+1}-\gamma \eta _{n_{k}}d_{n_{k}}\Vert +\gamma \dfrac{1}{\eta _{n_{k}}} \eta ^{2}_{n_{k}}\Vert d_{n_{k}}\Vert \nonumber \\= & {} \Vert w_{n_{k}}- u_{n_{k}+1}-\gamma \eta _{n_{k}}d_{n_{k}}\Vert + \gamma \dfrac{\left( 1+ \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}{\left( 1-\mu \dfrac{\tau _{n}}{\tau _{n+1}}\right) } \eta _{n_{k}}^{2}\Vert d_{n_{k}}\Vert . \end{aligned}$$

(28)

This implies that

$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert w_{n_{k}}- u_{n_{k}+1}\Vert =0. \end{aligned}$$

(29)

From Claim 3 we also have

$$\begin{aligned} \lim _{k\rightarrow \infty } \Vert v_{n_{k}}-w_{n_{k}}\Vert =0. \end{aligned}$$

(30)

Moreover, we have

$$\begin{aligned} \Vert u_{n_{k}}-w_{n_{k}}\Vert =\theta _{n_{k}} \Vert u_{n_{k}}-u_{n_{k}-1}\Vert =\gamma _{n_{k}}.\dfrac{\theta _{n_{k}}}{\gamma _{n_{k}}} \Vert u_{n_{k}}-u_{n_{k}-1}\Vert \rightarrow 0. \end{aligned}$$

(31)

From (29), (30) and (31), we get

$$\begin{aligned} \Vert u_{n_{k}+1}-u_{n_{k}}\Vert \le \Vert u_{n_{k}+1}-w_{n_{k}}\Vert +\Vert w_{n_{k}}-v_{n_{k}}\Vert +\Vert v_{n_{k}}-u_{n_{k}}\Vert \rightarrow 0. \end{aligned}$$

(32)

Since the sequence $\{u_{n_{k}}\}$ is bounded, it follows that there exists a subsequence $\{u_{n_{k_{j}}}\}$ of $\{u_{n_{k}}\}$, which converges weakly to some $z^{*}\in H$, such that

$$\begin{aligned} \limsup _{k\rightarrow \infty }\langle -z,u_{n_{k}}-z\rangle =\lim _{j\rightarrow \infty }\langle -z,u_{n_{k_{j}}}-z\rangle =\langle -z,z^{*}-z\rangle . \end{aligned}$$

(33)

Using (31), we get

$$\begin{aligned} w_{n_{k}} \rightharpoonup z^{*} \text { as } k\rightarrow \infty , \end{aligned}$$

Using (30), we obtain

$$\begin{aligned} v_{n_{k}} \rightharpoonup z^{*} \text { as } k\rightarrow \infty . \end{aligned}$$

Now, we show that $z^{*}\in Sol(C,F)$. Indeed, since $v_{n_{k}}=P_{C}(w_{n_{k}}-\tau _{n_{k}}Fw_{n_{k}})$, we have

$$\begin{aligned} \langle w_{n_{k}}-\tau _{n_{k}} Fw_{n_{k}}-v_{n_{k}},x-v_{n_{k}}\rangle \le 0 \ \ \forall x\in C, \end{aligned}$$

or equivalently

$$\begin{aligned} \dfrac{1}{\tau _{n_{k}}}\langle w_{n_{k}}-v_{n_{k}},x-v_{n_{k}}\rangle \le \langle Fw_{n_{k}},x-v_{n_{k}}\rangle \ \ \forall x\in C. \end{aligned}$$

Consequently

$$\begin{aligned} \dfrac{1}{\tau _{n_{k}}}\langle w_{n_{k}}-v_{n_{k}},x-v_{n_{k}}\rangle +\langle Fw_{n_{k}},v_{n_{k}}-w_{n_{k}}\rangle \le \langle Fw_{n_{k}},x-w_{n_{k}}\rangle \ \ \forall x\in C. \end{aligned}$$

(34)

Being weakly convergent, $\{w_{n_{k}}\}$ is bounded. Then, by the Lipschitz continuity of F, $\{Fw_{n_{k}}\}$ is bounded. As $\Vert w_{n_{k}}-v_{n_{k}}\Vert \rightarrow 0$, $\{v_{n_{k}}\}$ is also bounded and $\tau _{n_{k}} \ge \min \{\tau _{1},\dfrac{\mu }{L}\}$. Passing (34) to limit as $k\rightarrow \infty$, we get

$$\begin{aligned} \liminf _{k\rightarrow \infty }\langle Fw_{n_{k}},x-w_{n_{k}}\rangle \ge 0\ \ \forall x\in C. \end{aligned}$$

(35)

Moreover, we have

$$\begin{aligned} \langle Fv_{n_{k}},x-v_{n_{k}}\rangle =\langle Fv_{n_{k}}- Fw_{n_{k}},x-w_{n_{k}}\rangle +\langle Fw_{n_{k}},x-w_{n_{k}}\rangle +\langle Fv_{n_{k}},w_{n_{k}}-v_{n_{k}}\rangle . \end{aligned}$$

(36)

Since $\lim _{k\rightarrow \infty }\Vert w_{n_{k}}-v_{n_{k}}\Vert =0$ and F is L-Lipschitz continuous on H, we get

$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert Fw_{n_{k}}-Fv_{n_{k}}\Vert =0 \end{aligned}$$

which, together with (35) and (36) implies that

$$\begin{aligned} \liminf _{k\rightarrow \infty }\langle Fv_{n_{k}},x-v_{n_{k}}\rangle \ge 0. \end{aligned}$$

Next, we choose a sequence $\{\epsilon _{k}\}$ of positive numbers decreasing and tending to 0. For each k, we denote by $N_{k}$ the smallest positive integer such that

$$\begin{aligned} \langle Fv_{n_{j}},x-v_{n_{j}}\rangle +\epsilon _{k} \ge 0 \ \ \forall j\ge N_{k}. \end{aligned}$$

(37)

Since $\{ \epsilon _{k}\}$ is decreasing, it is easy to see that the sequence $\{N_{k}\}$ is increasing. Furthermore, for each k, since $\{v_{N_{k}}\}\subset C$ we can suppose $Fv_{N_{k}}\ne 0$ (otherwise, $v_{N_{k}}$ is a solution) and, setting

$$\begin{aligned} t_{N_{k}} = \dfrac{Fv_{N_{k}}}{\Vert Fv_{N_{k}}\Vert ^{2} }, \end{aligned}$$

we have $\langle Fv_{N_{k}}, t_{N_{k}}\rangle = 1$ for each k. Now, we can deduce from (37) that for each k

$$\begin{aligned} \langle Fv_{N_{k}}, x+\epsilon _{k} t_{N_{k}}-v_{N_{k}}\rangle \ge 0. \end{aligned}$$

From F is pseudomonotone on H, we get

$$\begin{aligned} \langle F(x+\epsilon _{k} t_{N_{k}}), x+\epsilon _{k} t_{N_{k}}-v_{N_{k}}\rangle \ge 0. \end{aligned}$$

This implies that

$$\begin{aligned} \langle Fx, x-v_{N_{k}}\rangle \ge \langle Fx-F(x+\epsilon _{k} t_{N_{k}}), x+\epsilon _{k} t_{N_{k}}-v_{N_{k}} \rangle -\epsilon _{k} \langle Fx, t_{N_{k}}\rangle . \end{aligned}$$

(38)

Now, we show that $\lim _{k\rightarrow \infty }\epsilon _{k} t_{N_{k}}=0$. Indeed, since $w_{n_{k}}\rightharpoonup z$ and $\lim _{k\rightarrow \infty }\Vert w_{n_{k}}-v_{n_{k}}\Vert =0,$ we obtain $v_{N_{k}}\rightharpoonup z \text { as } k \rightarrow \infty$. By $\{v_{n}\}\subset C$, we obtain $z^{*}\in C$. Since F satisfies Condition (19), we have

$$\begin{aligned} 0 < \Vert Fz^{*}\Vert \le \liminf _{k\rightarrow \infty }\Vert Fv_{n_{k}}\Vert . \end{aligned}$$

Since $\{v_{N_{k}}\}\subset \{v_{n_{k}}\}$ and $\epsilon _{k} \rightarrow 0$ as $k \rightarrow \infty$, we obtain

$$\begin{aligned} 0 \le \limsup _{k\rightarrow \infty } \Vert \epsilon _{k} t_{N_{k}} \Vert = \limsup _{k\rightarrow \infty } \left( \dfrac{\epsilon _{k}}{\Vert Fv_{n_{k}}\Vert }\right) \le \dfrac{\limsup _{k\rightarrow \infty }\epsilon _{k} }{\liminf _{k\rightarrow \infty }\Vert Fv_{n_{k}}\Vert }=0, \end{aligned}$$

which implies that $\lim _{k\rightarrow \infty } \epsilon _{k} t_{N_{k}} = 0.$

Now, letting $k\rightarrow \infty$, then the right-hand side of (38) tends to zero by F is uniformly continuous, $\{w_{N_{k}}\}, \{t_{N_{k}}\}$ are bounded and $\lim _{k\rightarrow \infty }\epsilon _{k} t_{N_{k}}=0$. Thus, we get

$$\begin{aligned} \liminf _{k\rightarrow \infty }\langle Fx,x-v_{N_{k}}\rangle \ge 0. \end{aligned}$$

Hence, for all $x\in C$ we have

$$\begin{aligned} \langle Fx, x-z^{*}\rangle =\lim _{k\rightarrow \infty } \langle Fx, x-v_{N_{k}}\rangle =\liminf _{k\rightarrow \infty } \langle Fx, x-v_{N_{k}}\rangle \ge 0. \end{aligned}$$

By Lemma 2.3, we get

$$\begin{aligned} z^{*}\in \text {Sol(C,F)}. \end{aligned}$$

Since (33) and the definition of $z=P_{Sol(C,F)}(0)$, we have

$$\begin{aligned} \limsup _{k\rightarrow \infty }\langle -z,u_{n_{k}}-z\rangle =\langle -z,z^{*}-z\rangle \le 0. \end{aligned}$$

(39)

Combining (32) and (39), we have

$$\begin{aligned} \limsup _{k\rightarrow \infty }\langle -z,u_{n_{k}+1}-z\rangle\le & {} \limsup _{k\rightarrow \infty }\langle -z,u_{n_{k}}-z\rangle \nonumber \\= & {} \langle -z,z^{*}-z\rangle \nonumber \\\le & {} 0. \end{aligned}$$

(40)

Hence, by (40), $\lim _{n\rightarrow \infty }\dfrac{\theta _{n}}{\gamma _{n}}\Vert u_{n}-u_{n-1}\Vert =0$, $\lim _{k\rightarrow \infty }\Vert u_{n_{k}+1}-u_{n_{k}}\Vert =0$, Claim 5 and Lemma 2.4, we have $\lim _{n\rightarrow \infty }\Vert u_{n}-z\Vert =0$. That is the desired result.

4 Convergence rate

In this section, we provide a result on the convergence rate of the iterative sequence generated by Algorithm 1 with $\gamma _{n}=0$ and the mapping F is L-Lipschitz continuous on H and $\kappa$-strongly pseudomonotone on C. The algorithm is of the form:

Using the technique in [35] we obtain the following result.

Theorem 4.2

Assume that F is L-Lipschitz continuous on H and $\kappa$-strongly pseudomonotone on C. Let $\delta \in (0,1)$ be arbitrary and $\theta$ be such that

$$\begin{aligned} 0 \le \theta \le \min \left\{ \frac{\xi }{2+\xi }, \frac{\sqrt{(1+\delta \xi )^{2}+4\delta \xi }-(1+\delta \xi )}{2}, (1-\delta ) \left( 1-\dfrac{(1-\gamma )\gamma (1-\mu )^{2}}{2(1+\mu )^{2}} \right) \right\} , \end{aligned}$$

(41)

where $\xi :=\dfrac{1-\beta }{\beta }.$ Then the sequence $\{u_{n}\}$ generated by Algorithm 2 converges strongly to the unique solution $x^{*}$ of (1) with an R-linear rate.

Proof

First, we will shall that, there exists $\rho ,\xi \in (0,1)$ such that

$$\begin{aligned} \Vert u_{n+1}-x^{*}\Vert ^{2}\le&\rho \Vert w_{n}-x^{*}\Vert ^{2}-\xi \Vert u_{n+1}-w_{n}\Vert ^{2}. \end{aligned}$$

Indeed, under assumptions made, it was proved that (1) has a unique solution [20]. From the $\kappa$-strong pseudomonotonicity of F, we have $\langle Fv_{n},v_{n}-x^{*}\rangle \ge \kappa \Vert v_{n}-x^{*}\Vert ^{2}$. This implies that

$$\begin{aligned} \langle Fv_{n},u_{n+1}-x^{*}\rangle =\langle Fv_{n},u_{n+1}-v_{n}\rangle +\langle Fv_{n},v_{n}-x^{*}\rangle \ge \langle Fv_{n},u_{n+1}-v_{n}\rangle \\ \ge \kappa \Vert v_{n}-x^{*}\Vert ^{2}+\langle Fv_{n},u_{n+1}-v_{n}\rangle . \end{aligned}$$

Therefore

$$\begin{aligned} -2\gamma \eta _{n} \tau _{n} \langle Fv_{n},u_{n+1}-x^{*}\rangle \le -2\gamma \eta _{n} \tau _{n} \kappa \Vert v_{n}-x^{*}\Vert ^{2}-2\gamma \eta _{n} \tau _{n} \langle Fv_{n},u_{n+1}-v_{n}\rangle . \end{aligned}$$

(42)

Substituting (42) into (9), we get

$$\begin{aligned} \Vert z_{n}-x^{*}\Vert ^{2}\le \Vert w_{n}-x^{*}\Vert ^{2} -\Vert z_{n}-w_{n}\Vert ^{2}-2\gamma \eta _{n} \tau _{n} \kappa \Vert v_{n}-x^{*}\Vert ^{2}-2\gamma \eta _{n} \tau _{n} \langle Fv_{n},z_{n}-v_{n}\rangle . \end{aligned}$$

(43)

Again, substituting (16) into (43), we obtain

$$\begin{aligned} \Vert z_{n}-x^{*}\Vert ^{2}\le \Vert w_{n}-x^{*}\Vert ^{2}-\Vert w_{n}- z_{n}-\gamma \eta _{n}d_{n}\Vert ^{2}-(2-\gamma )\gamma {\eta ^{2}_{n}} \Vert d_{n}\Vert ^{2}-2\gamma \eta _{n} \tau _{n} \kappa \Vert v_{n}-x^{*}\Vert ^{2}. \end{aligned}$$

(44)

Combining (26) and (44), we get

$$\begin{aligned} \Vert z_{n}-x^{*}\Vert ^{2}&\le \Vert w_{n}-x^{*}\Vert ^{2}-\Vert w_{n}- z_{n}-\gamma \eta _{n}d_{n}\Vert ^{2}-(2-\gamma )\gamma \dfrac{\left( 1-\dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}{\left( 1+ \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}\Vert w_{n}-v_{n}\Vert ^{2}\nonumber \\&\quad -2\gamma \eta _{n} \tau _{n} \kappa \Vert v_{n}-x^{*}\Vert ^{2} \ \ \forall n\ge n_{0}. \end{aligned}$$

(45)

It follows from (45) that

$$\begin{aligned} \Vert z_{n}-x^{*}\Vert ^{2}\le \Vert w_{n}-x^{*}\Vert ^{2}-(2-\gamma )\gamma \dfrac{\left( 1-\dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}{\left( 1+ \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}\Vert w_{n}-v_{n}\Vert ^{2}-2\gamma \eta _{n} \tau _{n} \kappa \Vert v_{n}-x^{*}\Vert ^{2} \ \ \forall n\ge n_{0}. \end{aligned}$$

(46)

On the other hand, we have

$$\begin{aligned} \Vert u_{n+1}-x^{*}\Vert ^{2}= & {} \Vert (1-\beta )w_{n}-\beta z_{n}-x^{*}\Vert ^{2}\\= & {} (1-\beta )\Vert w_{n}-x^{*}\Vert ^{2}+\beta \Vert z_{n}-x^{*}\Vert ^{2}-\beta (1-\beta )\Vert w_{n}-z_{n}\Vert ^{2}. \end{aligned}$$

Since the definition of $\{u_{n}\}$ we deduce $w_{n}-z_{n}=\dfrac{1}{\beta }(u_{n+1}-w_{n})$. Thus,

$$\begin{aligned} \Vert u_{n+1}-x^{*}\Vert ^{2}=(1-\beta )\Vert w_{n}-x^{*}\Vert ^{2}+\beta \Vert z_{n}-x^{*}\Vert ^{2}-\dfrac{1}{\beta }(1-\beta )\Vert u_{n+1}-w_{n}\Vert ^{2}. \end{aligned}$$

(47)

Substituting (46) into (47), we get

$$\begin{aligned} \Vert u_{n+1}-x^{*}\Vert ^{2}=&\Vert w_{n}-x^{*}\Vert ^{2}-\beta (2-\gamma )\gamma \dfrac{\left( 1-\dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}{\left( 1+ \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}\Vert w_{n}-v_{n}\Vert ^{2}-2\beta \gamma \eta _{n} \tau _{n} \kappa \Vert v_{n}-x^{*}\Vert ^{2}\nonumber \\ \quad&-\dfrac{1}{\beta }(1-\beta )\Vert u_{n+1}-w_{n}\Vert ^{2}\ \ \forall n\ge n_{0}. \end{aligned}$$

(48)

Substituting (25) into (48), we deduce

$$\begin{aligned} \Vert u_{n+1}-x^{*}\Vert ^{2}=&\Vert w_{n}-x^{*}\Vert ^{2}-\beta (2-\gamma )\gamma \dfrac{\left( 1-\dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}{\left( 1+ \dfrac{\mu \tau _{n}}{\tau _{n+1}}\right) ^{2}}\Vert w_{n}-v_{n}\Vert ^{2}-2\beta \gamma \dfrac{1-\mu \dfrac{\tau _{n}}{\tau _{n+1}}}{(1+\mu \dfrac{\tau _{n}}{\tau _{n+1}})^{2}}\tau _{n} \kappa \Vert v_{n}-x^{*}\Vert ^{2}\nonumber \\ \quad&-\dfrac{1}{\beta }(1-\beta )\Vert u_{n+1}-w_{n}\Vert ^{2}\ \ \forall n\ge n_{0}. \end{aligned}$$

(49)

Setting

$$\begin{aligned} \beta ^{*}:=\min \left\{ \dfrac{(2-\gamma )\gamma (1-\mu )^{2}}{2(1+\mu )^{2}}, \beta \gamma \dfrac{1-\mu }{(1+\mu )^{2}}\kappa \tau \right\} \text { where } \tau =\lim _{n\rightarrow \infty }\tau _{n}, \end{aligned}$$

we have

$$\begin{aligned} 1 > \lim _{n\rightarrow \infty }\dfrac{(2-\gamma )\gamma \left( 1-\mu \dfrac{\lambda _{n}}{\lambda _{n+1}}\right) ^{2}}{ \left( 1+\mu \dfrac{\lambda _{n}}{\lambda _{n+1}}\right) ^{2}}=\dfrac{(2-\gamma )\gamma (1-\mu )^{2}}{(1+\mu )^{2}}\ge 2\beta ^{*}. \end{aligned}$$

and

$$\begin{aligned} \lim _{n\rightarrow \infty }\beta \gamma \dfrac{1-\mu \dfrac{\tau _{n}}{\tau _{n+1}}}{(1+\mu \dfrac{\tau _{n}}{\tau _{n+1}})^{2}}\tau _{n} \kappa =\beta \gamma \dfrac{1-\mu }{(1+\mu )^{2}}\kappa \tau \ge \beta ^{*}. \end{aligned}$$

Using (49), hence there exists $n_{1}>n_{0}$ such that for all $n \ge n_{1}$

$$\begin{aligned} \Vert u_{n+1}-x^{*}\Vert ^{2}\le & {} \Vert w_{n}-x^{*}\Vert ^{2}-\xi \Vert u_{n+1}-w_{n}\Vert ^{2}- 2\beta ^{*}\Vert w_{n}-v_{n}\Vert ^{2}-2\beta ^{*}\Vert v_{n}-x^{*}\Vert ^{2}\nonumber \\\le & {} (1-\beta ^{*})\Vert w_{n}-x^{*}\Vert ^{2}-\xi \Vert u_{n+1}-w_{n}\Vert ^{2}\nonumber \\= & {} \rho \Vert w_{n}-x^{*}\Vert ^{2}-\xi \Vert u_{n+1}-w_{n}\Vert ^{2}, \end{aligned}$$

(50)

where $\rho := 1 - \beta ^{*} \in (0,1)$ and $\xi :=\dfrac{1-\beta }{\beta }$.

Next, we prove that the sequence $\{u_{n}\}$ converges strongly to $x^{*}$ with an R linear rate. Indeed, we have

$$\begin{aligned} \Vert w_{n}-x^{*}\Vert ^{2}= & {} \Vert (1+\theta )(u_{n}-x^{*})-\theta (u_{n-1}-x^{*})\Vert ^{2}\\= & {} (1+\theta )\Vert u_{n}-x^{*}\Vert ^{2}-\theta \Vert u_{n-1}-x^{*}\Vert ^{2}+\theta (1+\theta )\Vert u_{n}-u_{n-1}\Vert ^{2} \end{aligned}$$

and

$$\begin{aligned} \Vert u_{n+1}-w_{n}\Vert ^{2}= & {} \Vert u_{n+1}-u_{n}-\theta (u_{n} -u_{n-1})\Vert ^{2}\\= & {} \Vert u_{n+1}-u_{n}\Vert ^{2} + \theta ^{2} \Vert u_{n} - u_{n-1}\Vert ^{2}- 2 \theta \left\langle u_{n+1} -u_{n}, u_{n} -u_{n-1}\right\rangle \\\ge & {} \Vert u_{n+1}-u_{n}\Vert ^{2} + \theta ^{2} \Vert u_{n} - u_{n-1}\Vert ^{2}- 2 \theta \Vert u_{n+1} -u_{n}\Vert \Vert u_{n} -u_{n-1}\Vert \\\ge & {} \Vert u_{n+1}-u_{n}\Vert ^{2} + \theta ^{2} \Vert u_{n} - u_{n-1}\Vert ^{2}- \theta \Vert u_{n+1} -u_{n}\Vert ^{2} -\theta \Vert u_{n} -u_{n-1}\Vert ^{2}\\\ge & {} (1-\theta )\Vert u_{n+1}-u_{n}\Vert ^{2} - \theta (1-\theta ) \Vert u_{n} - u_{n-1}\Vert ^{2}. \end{aligned}$$

Combining these inequalities with (50) we obtain for all $n \ge n_{1}$ that

$$\begin{aligned} \Vert u_{n+1}-x^{*}\Vert ^{2}\le & {} \rho (1+\theta )\Vert u_{n}-x^{*}\Vert ^{2}-\rho \theta \Vert u_{n-1}-x^{*}\Vert ^{2}+\rho \theta (1+\theta )\Vert u_{n}-u_{n-1}\Vert ^{2}\\&- \xi (1-\theta )\Vert u_{n+1}-u_{n}\Vert ^{2} + \xi \theta (1-\theta ) \Vert u_{n} - u_{n-1}\Vert ^{2}, \end{aligned}$$

or equivalently

$$\begin{aligned}&\Vert u_{n+1}-x^{*}\Vert ^{2} - \rho \theta \Vert u_{n}-x^{*}\Vert ^{2} + \xi (1-\theta )\Vert u_{n+1}-u_{n}\Vert ^{2}\\\le & {} \rho \left[ \Vert u_{n}-x^{*}\Vert ^{2}-\theta \Vert u_{n-1}-x^{*}\Vert ^{2}+ \xi (1-\theta ) \Vert u_{n} - u_{n-1}\Vert ^{2}\right] \\&- \left( \rho \xi (1-\theta )-\rho \theta (1+\theta )- \xi \theta (1-\theta )\right) \Vert u_{n} - u_{n-1}\Vert ^{2}. \end{aligned}$$

Setting

$$\begin{aligned} a_{n}:= \Vert u_{n}-x^{*}\Vert ^{2}-\theta \Vert u_{n-1}-x^{*}\Vert ^{2}+ \xi (1-\theta )\Vert u_{n} - u_{n-1}\Vert ^{2}, \end{aligned}$$

since $\rho \in (0,1)$ we can write

$$\begin{aligned} a_{n+1}\le & {} \Vert u_{n+1}-x^{*}\Vert ^{2} - \rho \theta \Vert u_{n}-x^{*}\Vert ^{2} + \xi (1-\theta )\Vert u_{n+1}-u_{n}\Vert ^{2}\\\le & {} \rho a_{n} - \left( \rho \xi (1-\theta )-\rho \theta (1+\theta )- \xi \theta (1-\theta )\right) \Vert u_{n} - u_{n-1}\Vert ^{2}. \end{aligned}$$

Note that from (41) we have

$$\begin{aligned} \theta\le & {} (1-\delta ) \left( 1-\dfrac{(1-\gamma )\gamma (1-\mu )}{2(1+\mu )} \right) \\\le & {} (1-\delta ) \bigg (1-\min \left\{ \dfrac{(2-\gamma )\gamma (1-\mu )^{2}}{2(1+\mu )^{2}}, \beta \gamma \dfrac{1-\mu }{(1+\mu )^{2}}\kappa \tau \right\} \bigg ) = (1-\delta ) \rho , \end{aligned}$$

which implies

$$\begin{aligned} \xi \theta (1-\theta ) \le (1-\delta ) \rho \xi (1-\theta ). \end{aligned}$$

(51)

Since

$$\begin{aligned} \theta \le \frac{\sqrt{(1+\delta \xi )^{2}+4\delta \xi }-(1+\delta \xi )}{2} \end{aligned}$$

it holds

$$\begin{aligned} \theta ^{2} +(1+\delta \xi ) \theta -\delta \xi \le 0, \end{aligned}$$

or equivalently

$$\begin{aligned} \theta (1+\theta ) \le \delta \xi (1-\theta ). \end{aligned}$$

Hence

$$\begin{aligned} \rho \theta (1+\theta ) \le \delta \rho \xi (1-\theta ). \end{aligned}$$

(52)

From (51) and (52) we deduce

$$\begin{aligned} \rho \xi (1-\theta )-\rho \theta (1+\theta )- \xi \theta (1-\theta ) \ge 0. \end{aligned}$$

Moreover, since $\theta \le \frac{\xi }{2+\xi }$, we have $\theta \le \frac{\xi (1-\theta )}{2}$, which implies

$$\begin{aligned} a_{n}= & {} (1-\xi (1-\theta )) \Vert u_{n}-x^{*}\Vert ^{2} + \xi (1-\theta ) \left( \Vert u_{n}-x^{*}\Vert ^{2}+\Vert u_{n} - u_{n-1}\Vert ^{2}\right) -\theta \Vert u_{n-1}-x^{*}\Vert ^{2}\\\ge & {} (1-\xi (1-\theta )) \Vert u_{n}-x^{*}\Vert ^{2} +\frac{\xi (1-\theta )}{2} \Vert u_{n-1}-x^{*}\Vert ^{2} -\theta \Vert u_{n-1}-x^{*}\Vert ^{2} \\\ge & {} (1-\xi (1-\theta )) \Vert u_{n}-x^{*}\Vert ^{2} \ge 0. \end{aligned}$$

Hence for all $n \ge n_{1}$ it holds

$$\begin{aligned} a_{n+1} \le \rho a_{n} \le ...\le \rho ^{n-n_{1}+1}a_{n_{1}}. \end{aligned}$$

This follows that

$$\begin{aligned} \Vert u_{n}-x^{*}\Vert ^{2}\le \dfrac{a_{n_{1}}}{\rho ^{n_{1}} (1-\xi (1-\theta ))}\rho ^{n}, \end{aligned}$$

which implies that $\left\{ u_{n} \right\}$ converges R-linearly to $x^{*}$.

5 Numerical experiments

In this section, we first present computational experiments to illustrate our newly proposed algorithm for solving variational inequality arising in optimal control problem.

Let T be a positive number. Denote by $L_{2}([0,T],\mathbb {R}^{m})$ the Hilbert space of square integrable, measurable vector functions $u:[0,T]\rightarrow \mathbb {R}^{m}$ with inner product $\langle u,v\rangle ={\int _{0}^{T}}\langle u(t),v(t)\rangle dt,$ and norm $\Vert u\Vert _{2}=\sqrt{\langle u,u\rangle }.$ We consider the following optimal control problem:

$$\begin{aligned} u^{*}(t)=\text {argmin}\{z(u):u\in U\}, \end{aligned}$$

on the interval [0, T] and assume that such a control exists. Here U is the set of admissible controls, which has the form of an m-dimensional box and consists of piecewise continuous function:

$$\begin{aligned}U=\big \{u(t)\in L_{2}([0,T],\mathbb {R}^{m}):u_{i}(t)\in [u_{i}^{-},u_{i}^{+}],i=1,2,\ldots ,m\big \}. \end{aligned}$$

Specially, the control can be bang-bang (piecewise constant function). The terminal objective has the form

$$\begin{aligned}z(u)=\phi (x(T)), \end{aligned}$$

where $\phi$ is a convex and differentiable function, defined on the attainability set.

Suppose that the trajectory $x(t) \in L_{2}([0,T])$ satisfies constraints in the form of a system of linear differential equation:

$$\begin{aligned} \dot{x}(t)=A(t)x(t)+B(t)u(t),\quad x(0)=x_{0},\quad t\in [0,T], \end{aligned}$$

where $A(t)\in \mathbb {R}^{n\times n}$, $B(t)\in \mathbb {R}^{n\times m}$ are given continuous matrices for every $t\in [0,T]$. By the Pontryagin Maximum Principle there exists a function $p^{*}\in L_{2}([0,T])$ such that the triple $(x^{*},p^{*},u^{*})$ solves for a.e. $t\in [0,T]$ the system

$$\begin{aligned} \left\{ \begin{array}{ll} \dot{x}^{*}(t)&{}=A(t)x^{*}(t)+B(t)u^{*}(t)\\ x^{*}(0)&{}=x_{0},\\ \end{array}\right. \end{aligned}$$

$$\begin{aligned} \left\{ \begin{array}{ll} \dot{p}^{*}(t)&{}=-A(t)^{\top } p^{*}(t)\\ p^{*}(T)&{}=\nabla \phi (x(T)),\\ \end{array}\right. \end{aligned}$$

$$\begin{aligned} 0\in B(t)^{\top } p^{*}(t)+N_{U}(u^{*}(t)), \end{aligned}$$

where $N_{U}(u)$ is the normal cone to U at u. Denoting $Gu(t):=B(t)^{\top } p(t)$, it is known that Gu is the gradient of the objective cost function z (see [38] and the references contained therein). We can write the inclusion problem (54) as a variational inequality problem: find $u\in U$ such that

$$\begin{aligned} \langle Gu,v-u\rangle \ge 0,~\forall v\in U. \end{aligned}$$

Discretizing the continuous functions and choosing a natural number N with the mesh size $h:=T/N$, we identify any discretized control $u^{N}:=(u_{0}, u_{1},\ldots ,u_{N-1})$ with its piecewise constant extension:

$$\begin{aligned} u^{N}(t)=u_{i}~\text {for}~t\in \left[ t_{i},t_{i+1}\right) ,~i=0,1,\ldots ,N. \end{aligned}$$

Moreover, we identity any discretized state $x^{N}:=(x_{0},x_{1},\ldots ,x_{N})$ with its piecewise linear interpolation

$$\begin{aligned} x^{N}(t)=x_{i}+\frac{t-t_{i}}{h}\left( x_{i+1}-x_{i}\right) ,~\text {for}~t \in \left[ t_{i},t_{i+1}\right) ,~i=0,1,\ldots ,N-1. \end{aligned}$$

Similarly we get the co-state variable $p^{N}:=(p_{0},p_{1},\ldots ,p_{N})$ (see [38] for more details).

Next, we consider three examples in which the terminal function is not linear.

In the numerical results listed in the following tables, ’Iter.’ denotes the number of iterations and ’CPU in s’ denotes the execution time in seconds. Besides, ’–’ means that the algorithm can’t reach the error conditions because the inner loop of the algorithm reaches infinite loop after some steps.

In the following three examples, we take $N=100$. The initial control $u_{0}(t)$ is chosen randomly in $[-1,1]$, and the termination condition is controlled by the relative solution error, defined by (Table 1)

$$\begin{aligned} \text {RSE} = \frac{\Vert u_{n}-u^{*}\Vert }{\Vert u_{n}\Vert } \end{aligned}$$

at the current $u_{n}$.

Example 1

(See [39, Rocket Car])

$$\begin{aligned} \begin{array}{ll} \text{ minimize } &{} \frac{1}{2}\left( \left( x_{1}(5)\right) ^{2}+\left( x_{2}(5)\right) ^{2}\right) \\ \text{ subject } \text{ to } &{} \dot{x_{1}}(t)=x_{2}(t), \\ &{} \dot{x_{2}}(t)=u(t),~\forall t\in [0,5], \\ &{} x_{1}(0)=6,~ x_{2}(0)=1, \\ &{} u(t)\in [-1,1]. \end{array} \end{aligned}$$

The corresponding exact optimal control is

$$\begin{aligned} u^{*}(t)=\left\{ \begin{array}{ll} &{}1\ \quad \text{ if } \, t\in (\tau , 5]\\ -&{}1 \quad \text{ if } \, t\in (0,\tau ], \end{array}\right. \end{aligned}$$

where $\tau =3.5174292$.

Fig. 1

Random initial control (green) and optimal control (red) on the left and optimal trajectories on the right computed by Alg. 1 for Example 1

Fig. 2

Comparison of Alg. 1 with Alg. 3.3 in [31] and Alg. 2 in [32] for Example 1

Table 1 Numerical results for Algorithm 1, Algorithm 3.3 in [31] and Algorithm 2 in [32] in Example 1

In Fig. 1 and Fig. 2, we take $\text {RSE} \le 0.3$. Figure 1 displays the approximate optimal control and the corresponding trajectories. Other parameters are selected as follows:

Algorithm 1: $\tau _{1}=0.5, \mu =0.7, \gamma =1.9, \theta =0.5, \gamma _{n}=\frac{1}{3000(n+1)}, \epsilon _{n}=\frac{1}{3000(n+1)^{1.1}}$ and $\alpha _{n}=\frac{1}{n^{2}}$;

Algorithm 3.3 in [31]: $\sigma =0.1, \gamma =0.6$ and $\alpha _{n}=\frac{1}{3000(n+1)}$;

Algorithm 2 in [32]: $\lambda _{0}=0.6, \mu =0.9$ and $\alpha _{n}=\frac{1}{3000(n+1)}$.

The corresponding results reported in Fig. 2 and Table 1 illustrate that Algorithm 1 behaves better than Algorithm 3.3 in [31] and Algorithm 2 in [32] in term of time and steps.

Example 2

(See [40, Example 6.3])

$$\begin{aligned} \begin{array}{ll} \text{ minimize } &{} -x_{1}(2) +\left( x_{2}(2)\right) ^{2} \\ \text{ subject } \text{ to } &{} \dot{x_{1}}(t)=x_{2}(t), \\ &{} \dot{x_{2}}(t)=u(t),~\forall t\in [0,2], \\ &{} x_{1}(0)=0,~ x_{2}(0)=0,\\ &{} u(t) \in [-1,1]. \end{array} \end{aligned}$$

The corresponding exact optimal control is

$$\begin{aligned}u^{*}= \left\{ \begin{array}{ll} &{}1\ \quad \text{ if } \, t\in [0, 6/5)\\ -&{}1 \quad \text{ if } \, t\in (6/5,2], \end{array}\right. \end{aligned}$$

Fig. 3

Random initial control (green) and optimal control (red) on the left and optimal trajectories on the right computed by Algorithm 1 for Example 2

Fig. 4

Comparison of Alg. 1 with Alg. 3.3 in [31] and Alg. 2 in [32] for Example 2

Table 2 Numerical results for Algorithm 1, Algorithm 3.3 in [31] and Algorithm 2 in [32] in Example 2

In Fig. 3 and Fig. 4, we take $\text {RSE} \le 10^{-5}$. Figure 3 displays the approximate optimal control and the corresponding trajectories. Other parameters are selected as follows:

Algorithm 1: $\tau _{1}=0.2, \mu =0.8, \gamma =1.9, \theta =0.5, \gamma _{n}=\frac{1}{10000(n+1)}, \epsilon _{n}=\frac{1}{10000(n+1)^{1.1}}$ and $\alpha _{n}=\frac{1}{n^{2}}$;

Algorithm 3.3 in [31]: $\sigma =0.1, \gamma =0.3$ and $\alpha _{n}=\frac{1}{10000(n+1)}$;

Algorithm 2 in [32]: $\lambda _{0}=0.3, \mu =0.9$ and $\alpha _{n}=\frac{1}{10000(n+1)}$.

In Fig. 4 and Table 2, we can obtain that the performance of our Algorithm 1 in time and steps is better than Algorithm 2 in [32] and Algorithm 3.3 in [31].

Example 3

(control of a harmonic oscillator)

$$\begin{aligned} \begin{array}{ll} \text{ minimize } &{} x_{2}(3\pi )\\ \text{ subject } \text{ to } &{} \dot{x_{1}}(t)=x_{2}(t), \\ &{} \dot{x_{2}}(t)=-x_{1}(t)+u(t),~\forall t\in [0,3\pi ], \\ &{} x(0)=0,\\ &{} u(t) \in [-1,1]. \end{array} \end{aligned}$$

The exact optimal control in this problem is known:

$$\begin{aligned}u^{*}=\left\{ \begin{array}{ll} &{}1\ \quad \text{ if } \, t\in [0, \pi /2) \cup (3\pi /2,5\pi /2) \\ -&{}1 \quad \text{ if } \, t\in (\pi /2,3\pi /2) \cup (5\pi /2,3\pi ].\\ \end{array}\right. \end{aligned}$$

In Fig. 5 and Fig. 6, we take $\text {RSE} \le 0.1$. Figure 5 displays the approximate optimal control and the corresponding trajectories. Other parameters are selected as follows:

Algorithm 1: $\tau _{1}=1.9, \mu =0.8, \gamma =1.2, \theta =0.5, \gamma _{n}=\frac{1}{100(n+1)}, \epsilon _{n}=\frac{1}{100(n+1)^{1.1}}$ and $\alpha _{n}=\frac{1}{n^{2}}$;

Algorithm 3.3 in [31]: $\sigma =0.0001, \gamma =0.1$ and $\alpha _{n}=\frac{1}{100(n+1)}$;

Algorithm 2 in [32]: $\lambda _{0}=1.9, \mu =0.9$ and $\alpha _{n}=\frac{1}{100(n+1)}$.

As shown in Fig. 6 and Table 3, Algorithm 1 performs better than Algorithm 3.3 in [31] and Algorithm 2 in [32].

Fig. 5

Random initial control (green) and optimal control (red) on the left and optimal trajectories on the right computed by Algorithm 1 for Example 3

Fig. 6

Comparison of Alg. 1 with Alg. 3.3 in [31] and Alg. 2 in [32] for Example 3

Table 3 Numerical results for Algorithm 1, Algorithm 3.3 in [31] and Algorithm 2 in [32] in Example 3

Next we provide several numerical examples to demonstrate the efficiency of the proposed algorithm compared to some known algorithms.

Example 4

In the example, we study an important Nash-Cournot oligopolistic market equilibrium model, which was proposed originally by Murphy et. al. [25] as a convex optimization problem. Later, Harker reformulated it as a monotone variational inequality in [17]. We provide only a short description of the problem, for more details we refer to [13, 17, 25]. There are N firms, each of them supplies a homogeneous product in a non-cooperative fashion. Let $q_{i} \ge 0$ be the ith firm’s supply at cost $f_{i}(q_{i})$ and p(Q) be the inverse demand curve, where $Q \ge 0$ is the total supply in the market, i.e., $Q =\sum _{i=1}^{N} q_{i}$. A Nash equilibrium solution for the market defined above is a set of nonnegative output levels $(q_{1}^{*},q_{2}^{*},...,q_{N}^{*})$ such that $q_{i}^{*}$ is an optimal solution to the following problem for all $i= 1,2 ...,N$:

$$\begin{aligned} \max _{q_{i} \ge 0} {q_{i} p (q_{i} + Q_{i}^{*}) - f_{i} (q_{i})} \end{aligned}$$

(53)

where

$$\begin{aligned} Q_{i}^{*} = \sum _{j=1, j\not = i}^{N} q_{j}^{*}. \end{aligned}$$

A variational inequality equivalent to (53) is (see [17])

$$\begin{aligned} \text {find} \quad (q_{1}^{*},q_{2}^{*},...,q_{N}^{*}) \in \mathbb {R}^{N}_{+} \quad \text {such that} \quad \langle F(q^{*}), q-q^{*}\rangle \ge 0 \quad \forall q \in \mathbb {R}^{N}_{+}, \end{aligned}$$

(54)

where $F(q^{*}) = (F_{1}(q^{*}), F_{2} (q^{*}), ..., F_{N}(q^{*}))$ and

$$\begin{aligned} F_{i}(q^{*}) = f_{i}^{\prime }(q_{i}^{*}) -p \left( \sum _{j=1}^{N} q_{j}^{*}\right) - q_{i}^{*}p^{\prime } \left( \sum _{j=1}^{N} q_{j}^{*}\right) . \end{aligned}$$

As in the classical example of the Nash-Cournot equilibrium [17, 25], we consider an oligopoly with $N=5$ firms, each with the inverse demand function p and the cost function $f_{i}$ take the form

$$\begin{aligned} p(Q) = 5000^{1/1.1} Q^{-1/1.1} \quad \text {and} \quad f_{i}(q_{i}) = c_{i} q_{i} +\frac{\beta _{i}}{\beta _{i} + 1} L_{i}^{\frac{-1}{\beta _{i}}} q_{i}^{\frac{\beta _{i}+1}{\beta _{i}}} \end{aligned}$$

and the parameters $c_{i}, L_{i}, \beta _{i}$ as in [25], see Table 4.

Table 4 Parameters for experiment

The process is started with the initial $x_{0}=(10,10,10, 10)^{T}$ and $x_{1} =0.9*x_{0}$ and stopping conditions is $\text {Residual} := \Vert w_{n} - v_{n}\Vert \le 10^{-9}$ or the number of iterations $\ge$ 200 for all algorithms. Other parameters are selected as follows:

Algorithm 1: $\tau _{1}=1.8, \mu =0.7, \gamma =1.99, \theta =0.5, \gamma _{n}=\frac{1}{10000000*(n+1)}$ and $\alpha _{n}=\frac{1}{n^{2}}$

Algorithm 1 in [29]: $\nu _{0} =1, \theta = 0.9$ and $\rho _{n} = \rho = 0.4$.

Algorithm 2 in [32]: $\lambda _{0}=1, \mu =0.7$ and $\alpha _{n}=\frac{1}{10000000*(n+1)}$. The numerical results are described in Fig. 7.

Fig. 7

Comparison of Alg. 1 with Alg. 1 in [29] and Alg. 2 in [32] for Example 4. Execution times of the Algorithms respectively are 0.019, 0.048, 0.07 (seconds)

Example 5

Consider the following fractional programming problem:

$$\begin{aligned} \min f(x)=\frac{x^{T}Qx+a^{T}x+a_{0}}{b^{T}x+b_{0}} \end{aligned}$$

subject to $x \in X:=\{x \in \mathbb {R}^{m}: b^{T}x+b_{0}>0\}$,

where Q is an $m \times m$ symmetric matrix, $a, b \in \mathbb {R}^{m}$, and $a_{0}, b_{0} \in \mathbb {R}$. It is well known that f is pseudo-convex on X when Q is positive-semidefinite. We consider the following cases:

Case 1:

$$\begin{aligned} Q = \left( \begin{array}{llll} 5 &{} -1 &{} 2 &{} 0 \\ -1 &{} 5 &{} -1 &{} 3 \\ 2 &{} -1 &{} 3 &{} 0\\ 0 &{} 3 &{} 0 &{} 5 \end{array}\right) , a= \left( \begin{array}{llll} 1\\ -2\\ -2\\ 1 \end{array}\right) , b= \left( \begin{array}{llll} 2\\ 1\\ 1\\ 0 \end{array}\right) , a_{0}=-2, b_{0}=4. \end{aligned}$$

We minimize f over $C := \{x \in \mathbb {R}^{4} : 1 \le x_{i} \le 10, i = 1,..., 4\} \subset X$. It is easy to verify that Q is symmetric and positive definite in $\mathbb {R}^{4}$ and consequently f is pseudo-convex on X.

The process is started with the initial $x_{0}=(20,-20, 20,-20)^{T}$ and $x_{1} =0.9*x_{0}$ and stopping conditions is $\text {Residual} := \Vert u_{n} - q_{n}\Vert \le 10^{-9}$ or the number of iterations $\ge$ 200 for all algorithms. Other parameters are selected as follows:

Algorithm 1: $\tau _{1}=0.6, \mu =0.6, \gamma =1.5, \theta =0.5, \alpha _{n}=\frac{1}{n^{2}}$ and $\gamma _{n}=\frac{1}{C*(n+1)}$ with $C=10^{7}$.

Algorithm 1 in [29]: $\nu _{0} =1, \theta = 0.9.\mu =0.6$ and $\rho _{n} = \rho = 0.4$.

Algorithm 2 in [32]: $\lambda _{0}=0.6, \mu =0.6$ and $\alpha _{n}=\frac{1}{C*(n+1)}$. The numerical results are described in Fig. 8.

Fig. 8

Comparison of Alg. 1 with Alg. 1 in [29] and Alg. 2 in [32] for Example 5. Execution times of the Algorithms respectively are 0.18, 0.24, 0.8 (seconds)

Case 2: In the second experiment, to make the problem even more challenging. Let matrix $A:\mathbb {R}^{m \times m} \rightarrow \mathbb {R}^{m \times m}$, vectors $c, d,y_{0} \in \mathbb {R}^{m}$ and $c_{0},d_{0}$ are generated from a normal distribution with mean zero and unit variance. We put $e=(1,1,...,1)^{T} \in \mathbb {R}^{m}, Q=A^{T}A+I, a:=e+c, b:=e+d, a_{0}=1+c_{0}, b_{0}=1+d_{0}$. We minimize f over $C := \{x \in \mathbb {R}^{m} : 1 \le x_{i} \le 10, i = 1,..., m\} \subset X$. Because Matrix Q is symmetric and positive definite in $\mathbb {R}^{m}$ and consequently f is pseudo-convex on X. The process is started with the initial $x_{0}:=m*y_{0}$ and $x_{1} =0.9*x_{0}$, stopping conditions and parameters as Case 1. The numerical results are described in Fig. 9.

Fig. 9

Comparison of Alg. 1 with Alg. 1 in [29] and Alg. 2 in [32] for Example 5 with $n=100$. Execution times of the Algorithms respectively are 0.24, 0.37, 1.25 (seconds)

The corresponding results reported in Figs. 7, 8, and 9 show that Alg. 1 behaves better than Algorithm 1 in [29] and Algorithm 2 in [32].

6 Conclusions

In this paper, we introduce some results of the modified subgradient extragradient method for solving pseudomonotone variational inequalities in real Hilbert spaces. The algorithm only needs to calculate one projection onto the feasible set C per iteration and does not require the prior information of the Lipschitz constant of the cost mapping. First, we prove a sufficient condition for a strong convergence of the proposed algorithm under relaxed assumptions. Second, the proposed algorithm is proved to converge strongly with an R-linear convergence rate, under Lipschitz continuity and strong pseudomonotonicity assumptions. Finally, several numerical results are presented to illustrate the efficiency and advantages of the proposed method.