
Citation: Ziqi Zhu, Kaiye Zheng, Shenghua Wang. A new double inertial subgradient extragradient method for solving a non-monotone variational inequality problem in Hilbert space[J]. AIMS Mathematics, 2024, 9(8): 20956-20975. doi: 10.3934/math.20241020
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Let H be a real Hilbert space, C be a nonempty, closed and convex subset of H, and F:H→H be a mapping. The variational inequality problem (VIP in short) is to find a point x∗∈C such that
⟨F(x∗),y−x∗⟩≥0,∀y∈C. | (1.1) |
Denote the set of solutions of the VIP (1.1) by S. VIP is an important branch of nonlinear analysis, closely related to many topics and has a large application in mechanics, optimization, traffic network problems, equilibrium problems, and so on; see [1,2,3].
The projection method is one of the effective methods for solving the VIP (1.1). The simplest projection method is the following single projection method:
x1∈C,xn+1=PC(xn−λnF(xn)),n∈N, | (1.2) |
where N denotes the set of positive integers, PC:H→C is the metric projection, F:H→H is an η-strongly monotone and L-Lipschitz continuous mapping, and λn∈(0,2ηL2). The sequence {xn} generated by (1.2) converges strongly to the unique solution of VIP (1.1).
However, the strong monotonicity imposed on F in (1.2) is a relatively strict condition, which is generally difficult to satisfy. To weaken the restriction of strong monotonicity on F, Korpelevich[4] introduced the following famous extragradient method (EGM in short): x1∈C, and
{yn=PC(xn−λnF(xn)),xn+1=PC(xn−λnF(yn)), n∈N, | (1.3) |
where F:H→H is a monotone and L-Lipschitz continuous, and λn∈(0,1L). The author proved that the sequence {xn} generated by (1.3) converges weakly to a solution of the problem (1.1). In recent two decades, many modified EGM have been proposed; see, e.g., [6,7,8,9,10,11].
Recently, Noinakorn et al. [12] introduced a new EGM with inertial technique [9,10,13] for solving VIP (1.1) as follows: x0,x1∈H, and
{tn=(1−ψn)(xn+αn(xn−xn−1)),yn=PC(tn−τF(tn)),xn+1=PC(tn−τF(yn)), n∈N, | (1.4) |
where F:H→H is a pseudomonotone and L-Lipschitz continuous mapping, {ψn}⊂(0,1), 0<τ<1L, α>0, and {αn}⊂[0,ˆαn] with
ˆαn={min{ϵn‖xn−xn−1‖,α2},if xn≠xn−1,α2,otherwise. |
The authors proved that, under some mixed conditions, the sequence {xn} generated by (1.4) converges strongly to the solution of the VIP (1.1).
Tan et al.[14] proposed the following inertial subgradient EGM solving VIP (1.1): x0,x1∈H, and
{tn=(1−ψn)(xn+αn(xn−xn−1)),yn=PC(tn−λnF(tn)),Tn={x∈H|⟨tn−λnF(tn)−yn,x−yn⟩≤0},xn+1=PTn(tn−γλnρnF(yn)), n∈N, | (1.5) |
where F:H→H is a pseudomonotone and L-Lipschitz continuous mapping, λ1>1, μ∈(0,1), σ>1, α>0, γ∈(0,2σ), {ψn}⊂(0,1),
αn={min{ξn‖xn−xn−1‖,α},if xn≠xn−1,α,otherwise, |
and
ρn=(1−μ)‖tn−yn‖2‖gn‖2, with gn=tn−yn−λn(F(tn)−F(yn)). |
At each step, the new stepsize λn+1 is generated by
λn+1=min{λn+pn,qn‖tn−yn‖‖F(tn)−F(yn)‖}, |
where {pn}⊂[0,∞) with ∑∞n=1pn<∞ and {qn}⊂[1,∞) with limn→∞qn=1. The authors proved that {xn} generated by (1.5) converges strongly to a solution of VIP (1.1).
In the inertial projection methods mentioned above, only one inertial step is included. Some authors presented the new projection methods with double inertial steps for solving VIP (1.1). For example, Yao et al. [15] proposed a subgradient exgragradient method with double inertial steps for solving a pseudomonotone VIP. The authors proved the strong and weak convergence and obtained the linear convergence rate of their method under some suitable conditions. Thong et al. [16] investigated a single projection with double inertial extrapolation steps for solving a pseudomonotone VIP. The weak convergence and linear convergence rate of their method are obtained under some suitable conditions. Li and Wang [17] introduced a subgradient extragradient method with double inertial steps for solving a quasi-monotone VIP and proved the weak convergence of the proposed method under some suitable conditions. On the recent double inertial methods for solving VIP, the reader may refer to [18,19].
In this paper, inspired by the results [12,14,15,16,17,18,19], we introduce a new double inertial subgradient extragradient method to solve the VIP (1.1). In our method, we use a new manner to compute tn which includes tn in (1.4) and (1.5) as the special case. In [15,16,17,18,19], the mapping F is required to be pseudomontone or quasi-monotone, while in our method the mapping F needs not to satisfy any assumption of monotonicity. Two different self-adaptive step sizes are used to deal with the Lipschitz constant of the mapping F. The strong convergence of the proposed method is proved under some new conditions. Finally, some numerical examples are given to illustrate the convergence of our method and compare with some related methods in literature. The numerical results show that our method has certain advantages over the related methods.
In this section, we give some definitions and lemmas which will be used in the next section. In the next definitions and lemmas, we assume H be a real Hilbert space and C be a nonempty closed convex subset of H.
Definition 2.1. Let x,y∈H. A mapping F:H→H is said to be
(ⅰ) monotone if ⟨F(x)−F(y),x−y⟩≥0;
(ⅱ) pseudomonotone if
⟨F(x),y−x⟩≥0⇒⟨F(y),y−x⟩≥0; |
(ⅲ) quasimonotone if
⟨F(x),y−x⟩>0⇒⟨F(y),y−x⟩≥0; |
(ⅳ) paramonotone on C with respect to the subset B⊂C, if F is pseudomonotone on C and
y∈C,x∈B,⟨F(y),y−x⟩=0⇒y∈B. |
It is obvious that (ⅰ)⇒(ⅱ)⇒(ⅲ) and (ⅱ) ⇒ (ⅳ).
Definition 2.2. For any x∈H, there exists a unique z∈C such that z=argminy∈C‖x−y‖. The element z is denoted by PC(x), called the metric projection of x∈H onto C. That is,
PC(x)=argminy∈C‖y−x‖. |
For any given ˉx,v∈H with v≠0, let T={x∈H:⟨v,x−ˉx⟩≤0}. Then for all y∈H, the projection PT(y) is defined by
PT(y)=y−max{0,⟨v,y−ˉx⟩‖v‖2}v. | (2.1) |
By (2.1), we see that the projection of any point onto a half-space can be computed explicitly; see [20,Lemma 1.2] for the details.
Two important properties of PC are given in the following lemma. The other properties of PC can be found in [21,Chapter 4].
Lemma 2.1. [21,Chapter 4] Let PC:H→C be the metric projection. Then, for all x∈H, the following hold:
(ⅰ) z=PC(x) if and only if ⟨x−z,z−y⟩≥0 for all y∈C.
(ⅱ) ‖PC(x)−y‖2≤‖x−y‖2−‖PC(x)−x‖2 for all y∈C.
Lemma 2.2. For any x,y∈H, it holds that
‖x+y‖2≤‖x‖2+2⟨y,x+y⟩. |
Lemma 2.3. [22] Assume that the sequences {an}⊂[0,∞), {bn}⊂(0,1) and {cn}⊂(0,∞) satisfy
an+1≤(1−bn)an+cn,∀n≥1. |
If ∑∞n=1bn=∞ and lim supn→∞cnbn≤0 hold, then limn→∞an=0.
Lemma 2.4. [23] Let {an} be a sequence of non-negative real numbers such that there exists a subsequence {anj} of {an} such that anj<anj+1 for all j∈N. Then there exists a non-decreasing sequence {mk} of N such that limk→∞mk=∞ and the following properties are satisfied by all (sufficiently large) number k∈N:
amk≤amk+1 and ak≤amk+1. |
In this section, let C be a nonempty, closed and convex subset of a real Hilbert space H and F:H→H be a mapping. Consider the following Minty VIP [24]:
find x∗∈C such that ⟨F(y),y−x∗⟩≥0, ∀y∈C. |
Denote the set of solution of Minty VIP above by Ω. Clearly, Ω is closed and convex. Since C is convex, then Ω⊂S when F is continuous on C. Moreover, it is easy to show that if F is pseudomonotone on C, then S=Ω; see [25].
In this section, we always assume that the following conditions hold:
(A1) Ω≠∅.
(A2) If y∈C and z∈Ω satisfy ⟨F(y),y−z⟩=0, then y∈Ω.
(A3) F is L-Lipschitz continuous and sequentially weakly continuous on H.
Remark 3.1. (A1) and (A2) do not imply that F is paramonotone on C with respect to Ω because F is not assumed to be pseudomonotone on C.
In the following example, F satisfies (A1)–(A3) but does not satisfy any monotone property on H.
Example 3.1. Let H=R2, C=[0,1]×[0,1] and F(x)=−(u+x) for all x∈H, where u=(1,1)T. It is easy to see that S=Ω={u} and F is Lipschitz continuous with the constant L=1. For every y∈C, ⟨F(y),y−u⟩=⟨u+y,u−y⟩=0 implies that y=u. Hence (A1)–(A3) hold. For x=(0.8177,0.5999)T and y=(0.5629,0.8247)T, we have ⟨F(x),y−x⟩=0.10349 and ⟨F(y),y−x⟩=−0.01197, and hence F is not quasimonotone on H.
Now we introduce the following double inertial subgradient extragradient method (DISEM) for solving the VIP (1.1).
Algorithm 3.1 (DISEM)
Initialization. Choose the initial points x−1,x0,x1∈H, the constants γ∈(0,1), θ1>0,θ2>0, and λ1>0, the sequences {αn}∞n=1⊂[α,1] with α∈(0,1), {ϵ1,n}∞n=1⊂(0,∞),{ϵ2,n}∞n=1⊂(0,∞) and {ψn}∞n=1⊂(0,1) satisfying
limn→∞ψn=0, ∞∑n=1ψn=∞, limn→∞ϵi,nψn=0,i=1,2. |
Set n=1.
Step 1. Given xn−2,xn−1,xn, compute tn=ψn(1−αn)xn+(1−ψn)wn, where wn=xn+θ1,n(xn−xn−1)+θ2,n(xn−1−xn−2) with
θ1,n={min{θ1,ϵ1,n‖xn−xn−1‖},if xn≠xn−1,θ1,otherwise | (3.1) |
and
θ2,n={min{θ2,ϵ2,n‖xn−1−xn−2‖},if xn−1≠xn−2,θ2,otherwise. | (3.2) |
Step 2. Compute yn=PC(tn−λnF(tn)). If yn=tn, stop and tn∈Ω; otherwise, go to Step 3.
Step 3. Compute xn+1=PTn(tn−λnF(yn)), where
Tn={z∈H:⟨tn−λnF(tn)−yn,z−yn⟩≤0}. |
Step 4. Update the new step size λn+1 by
λn+1={λn,if ‖F(tn)−F(yn)‖=0,min{λn,γ‖tn−yn‖‖F(tn)−F(yn)‖},otherwise | (3.3) |
or
λn+1={λn,if λn‖F(tn)−F(yn)‖≤γ‖tn−yn‖,γλn,otherwise. | (3.4) |
Set n=n+1 and go to Step 1.
Remark 3.2. If αn≡1 in Algorithm 3.1, then tn=(1−ψn)wn is defined as a similar manner with (1.4) and (1.5). In addition, for each n∈N, by Lemma 2.1 (ⅰ) it is easy to show that C⊂Tn for each n∈N. Moreover, since Tn is a half-space, xn+1 can be explicitly computed by (2.1) and so only one projection for yn is computed at each iteration.
Remark 3.3. By (3.1), (3.2) and the hypothesis on {ϵi,n}(i=1,2) and {ψn}, we have
limn→∞θi,n‖xn−i+1−xn−i‖ψn≤limn→∞ϵi,nψn=0, i=1,2. | (3.5) |
Since limn→∞ψn=0, αn>α, by (3.5) we see that for each z∈Ω, there exists Mz>0 such that
supn∈N{‖z‖+1−ψnαn(θ1,nψn‖xn−xn−1‖+θ2,nψn‖xn−1−xn−2‖)}≤Mz. | (3.6) |
Note that (3.3) and (3.4) are the different manners of computing λn+1. The following lemma shows that {λn} generated by (3.3) or (3.4) has the limit which is a crucial result for proving the convergence of Algorithm 3.1.
Lemma 3.1. The sequence {λn} has the limit λ>0.
Proof. Obviously, the sequence {λn} defined by (3.3) or (3.4) is nonnegative and nonincreasing, which implies that the limit of {λn} exists. Set limn→∞λn=λ. Now we show the desired result by the following cases:
(ⅰ) {λn} is defined as in (3.3). If ‖F(tn)−F(yn)‖≠0, by the L-Lipschitz continuity of F, we have
γ‖tn−yn‖‖F(tn)−F(yn)‖≥γ‖tn−yn‖L‖tn−yn‖=γL, |
which together with the definition of λn+1 implies that λn+1≥min{λn,γL}. If ‖F(tn)−F(yn)‖=0, then λn+1=λn≥min{λn,γL}. So by a mathematical induction we get λn≥min{λ1,γL} for all n∈N. Therefore, we have λ≥min{λ1,γL}.
(ⅱ) {λn} is defined as in (3.4). Assume that λ=0. Then there must exist a subsequence λnk of {λn} such that
λnk‖F(tnk)−F(ynk)‖>γ‖tnk−ynk‖, ∀k∈N. |
Since F is L-Lipschitz continuous, we have
λnkL‖tnk−ynk‖>γ‖tnk−ynk‖ |
and hence λnk>γL for all k∈N. Letting k→∞, from limk→∞λnk=0 we get γL≤0. It is a contradiction. Therefore, by the two cases above we can conclude λ>0. The proof is complete.
Remark 3.4. By the proof process of Lemma 3.1, we see that there exists n0∈N such that λn+1 defined by (3.3) or (3.4) satisfies that
λn+1≤γ‖tn−yn‖‖F(tn)−F(yn)‖, ∀n≥n0. |
Lemma 3.2. There exists n0∈N such that for each z∈Ω and n≥n0,
‖xn+1−z‖2≤‖tn−z‖2−(1−γλnλn+1)(‖xn+1−yn‖2+‖yn−tn‖2). | (3.7) |
Proof. For each z∈Ω and n∈N, by Remark 3.2 we have z∈C⊂Tn for each n∈N. From Lemma 2.1 (ⅰ) and the definition of xn+1 it follows that
⟨z−xn+1,tn−λnF(yn)−xn+1⟩≤0. |
That is
⟨xn+1−tn,xn+1−z⟩≤λn⟨F(yn),z−xn+1⟩. | (3.8) |
Since xn+1∈Tn, we have
⟨yn−tn,yn−xn+1⟩≤λn⟨F(tn),xn+1−yn⟩. | (3.9) |
On the other hand, we have
‖xn+1−z‖2=‖yn−z‖2−‖xn+1−yn‖2+2⟨xn+1−yn,xn+1−z⟩=‖tn−z‖2−‖xn+1−yn‖2−‖yn−tn‖2+2⟨yn−tn,yn−z⟩+2⟨xn+1−yn,xn+1−z⟩=‖tn−z‖2−‖xn+1−yn‖2−‖yn−tn‖2+2⟨yn−tn,yn−xn+1⟩+2⟨xn+1−tn,xn+1−z⟩. | (3.10) |
Substituting (3.8) and (3.9) into (3.10) we get
‖xn+1−z‖2≤‖tn−z‖2−‖xn+1−yn‖2−‖yn−tn‖2+2λn⟨F(yn),z−xn+1⟩+2λn⟨F(tn),xn+1−yn⟩=‖tn−z‖2−‖xn+1−yn‖2−‖yn−tn‖2+2λn⟨F(yn),z−yn⟩+2λn⟨F(tn)−F(yn),xn+1−yn⟩. | (3.11) |
By Remark 3.4, there exists n0∈N such that
λn+1≤γ‖tn−yn‖‖F(tn)−F(yn)‖, ∀n≥n0. |
Hence by (3.11) we have
‖xn+1−z‖2≤‖tn−z‖2−‖xn+1−yn‖2−‖yn−tn‖2+2λn⟨F(yn),z−yn⟩+2λn‖F(tn)−F(yn)‖‖xn+1−yn‖≤‖tn−z‖2−‖xn+1−yn‖2−‖yn−tn‖2+2λn⟨F(yn),z−yn⟩+2γλnλn+1‖tn−yn‖‖xn+1−yn‖≤‖tn−z‖2−(1−γλnλn+1)(‖xn+1−yn‖2+‖yn−tn‖2)+2λn⟨F(yn),z−yn⟩, ∀n≥n0. | (3.12) |
Since yn∈C and z∈Ω, we have ⟨F(yn),yn−z⟩≥0. Substituting this result into (3.12) we can get (3.7). The proof is complete.
Lemma 3.3. The sequences {xn} and {tn} are bounded.
Proof. By Lemma 3.1 we have
limn→∞(1−γλnλn+1)=1−γ<1. |
Hence there exist γ′∈(γ,1) and n1∈N such that
1−γλnλn+1>1−γ′, ∀n≥n1. | (3.13) |
Set n2=max{n0,n1}. For each z∈Ω, by (3.7) and (3.13) we have
‖xn+1−z‖2≤‖tn−z‖2−(1−γ′)(‖xn+1−yn‖2+‖yn−tn‖2)≤‖tn−z‖2, n≥n2. | (3.14) |
From (3.7) and the definition of tn it holds
‖tn−z‖=‖ψn(1−αn)xn+(1−ψn)wn−z‖=‖ψn((1−αn)xn−z)+(1−ψn)(wn−z)‖≤‖ψn((1−αn)xn−z)‖+(1−ψn)‖wn−z‖=‖ψn((1−αn)xn−z)‖+(1−ψn)‖xn+θ1,n(xn−xn−1)+θ2,n(xn−1−xn−2)−z‖≤ψn[(1−αn)‖xn−z‖+αn‖z‖]+(1−ψn)[‖xn−z‖+θ1,n‖xn−xn−1‖+θ2,n‖xn−1−xn−2‖]=(1−ψnαn)‖xn−z‖+ψnαn‖z‖+(1−ψn)[θ1,n‖xn−xn−1‖+θ2,n‖xn−1−xn−2‖]=(1−ψnαn)‖xn−z‖+ψnαn[‖z‖+1−ψnαn(θ1,nψn‖xn−xn−1‖+θ2,nψn‖xn−1−xn−2‖)]≤(1−ψnαn)‖xn−z‖+ψnαnMz, ∀n≥n0. | (3.15) |
Substituting (3.15) into (3.14) we have
‖xn+1−z‖≤‖tn−z‖≤(1−ψnαn)‖xn−z‖+ψnαnMz≤max{‖xn−z‖,Mz}≤…≤max{‖xn0−z‖,Mz},∀n≥n2. | (3.16) |
It follows that {xn} is bounded and so is {tn} by (3.16). The proof is complete.
Let x∗=PΩ(0). We have x∗∈S because of Ω⊂S. In the position we give the main result on Algorithm 3.1 as follows.
Theorem 3.1. The sequence {xn} generated by Algorithm 3.1 converges strongly to the element x∗.
Proof. For each n∈N, by the definition of tn and Lemma 2.2 we have
‖tn−x∗‖2=‖ψn(1−αn)xn+(1−ψn)wn−x∗‖2=‖ψn((1−αn)xn−x∗)+(1−ψn)(wn−x∗)‖2≤(1−ψn)2‖wn−x∗‖2+2ψn⟨(1−αn)xn−x∗,tn−x∗⟩=(1−ψn)2‖xn+θ1,n(xn−xn−1)+θ2,n(xn−1−xn−2)−x∗‖2+2ψn[(1−αn)⟨xn−x∗,tn−x∗⟩+αn⟨−x∗,tn−x∗⟩]. | (3.17) |
Note that
‖xn+θ1,n(xn−xn−1)+θ2,n(xn−1−xn−2)−x∗‖2=‖xn−x∗‖2+θ21,n‖xn−xn−1‖2+θ22,n‖xn−1−xn−2‖2+2θ1,n⟨xn−xn−1,xn−x∗⟩+2θ2,n⟨xn−1−xn−2,xn−x∗⟩+2θ1,nθ2,n⟨xn−xn−1,xn−1−xn−2⟩≤‖xn−x∗‖2+θ21,n‖xn−xn−1‖2+θ22,n‖xn−1−xn−2‖2+2θ1,n‖xn−xn−1‖‖xn−x∗‖+2θ2,n‖xn−1−xn−2‖‖xn−x∗‖+2θ1,nθ2,n‖xn−xn−1‖‖xn−1−xn−2‖. | (3.18) |
Substituting (3.18) into (3.17) we get
‖tn−x∗‖2≤(1−ψn)2[‖xn−x∗‖2+θ21,n‖xn−xn−1‖2+θ22,n‖xn−1−xn−2‖2+2θ1,n‖xn−xn−1‖‖xn−x∗‖+2θ2,n‖xn−1−xn−2‖‖xn−x∗‖+2θ1,nθ2,n‖xn−xn−1‖‖xn−1−xn−2‖]+2ψn(1−αn)‖xn−x∗‖‖tn−x∗‖+2ψnαn⟨−x∗,tn−x∗⟩≤(1−ψn)2[‖xn−x∗‖2+θ21,n‖xn−xn−1‖2+θ22,n‖xn−1−xn−2‖2+2θ1,n‖xn−xn−1‖‖xn−x∗‖+2θ2,n‖xn−1−xn−2‖‖xn−x∗‖+2θ1,nθ2,n‖xn−xn−1‖‖xn−1−xn−2‖]+ψn(1−αn)(‖xn−x∗‖2+‖tn−x∗‖2)+2ψnαn⟨−x∗,tn−x∗⟩. |
It follows that
‖tn−x∗‖2≤1−ψn+ψ2n−ψnαn1−ψn+ψnαn‖xn−x∗‖2+(1−ψn)21−ψn+ψnαn[θ21,n‖xn−xn−1‖2+θ22,n‖xn−1−xn−2‖2+2θ1,n‖xn−xn−1‖‖xn−x∗‖+2θ2,n‖xn−1−xn−2‖‖xn−x∗‖+2θ1,nθ2,n‖xn−xn−1‖‖xn−1−xn−2‖]+2ψnαn1−ψn+ψnαn⟨−x∗,tn−x∗⟩=(1−2ψnαn1−ψn+ψnαn)‖xn−x∗‖2+ψ2n1−ψn+ψnαn‖xn−x∗‖2+(1−ψn)21−ψn+ψnαn[θ21,n‖xn−xn−1‖2+θ22,n‖xn−1−xn−2‖2+2θ1,n‖xn−xn−1‖‖xn−x∗‖+2θ2,n‖xn−1−xn−2‖‖xn−x∗‖+2θ1,nθ2,n‖xn−xn−1‖‖xn−1−xn−2‖]+2ψnαn1−ψn+ψnαn⟨−x∗,tn−x∗⟩=(1−γn)‖xn−x∗‖2+φn, | (3.19) |
where γn=2ψnαn1−ψn+ψnαn and
φn=γnψn2αn‖xn−x∗‖2+γn(1−ψn)22αnψn[θ21,n‖xn−xn−1‖2+θ22,n‖xn−1−xn−2‖2+2θ1,n‖xn−xn−1‖‖xn−x∗‖+2θ2,n‖xn−1−xn−2‖‖xn−x∗‖+2θ1,nθ2,n‖xn−xn−1‖‖xn−1−xn−2‖]+γn⟨−x∗,tn−xn⟩+γn⟨−x∗,xn−x∗⟩. |
Substituting (3.19) into (3.7) we have
‖xn+1−x∗‖2≤(1−γn)‖xn−x∗‖2+φn−(1−γ′)(‖xn+1−yn‖2+‖yn−tn‖2)≤(1−γn)‖xn−x∗‖2+φn, ∀n≥n2. | (3.20) |
Next we prove the strong convergence of {‖xn−x∗‖2} to zero by considering the following two cases.
Case 1. Suppose there exists N∈N with N>n2 such that {‖xn−x∗‖2} is monotonically nonincreasing for n≥N. Then {‖xn−x∗‖2} is convergent and hence
‖xn−x∗‖2−‖xn+1−x∗‖2→0. | (3.21) |
Since limn→∞ψn=0, by the boundedness of {xn} and (3.5) we get
‖tn−xn‖=‖ψn(1−αn)xn+(1−ψn)wn−xn‖≤ψn(1−αn)‖xn‖+‖(1−ψn)wn−xn‖=ψn(1−αn)‖xn‖+‖(1−ψn)[xn+θ1,n(xn−xn−1)+θ2,n(xn−1−xn−2)]−xn‖=ψn(1−αn)‖xn‖+‖(1−ψn)[θ1,n(xn−xn−1)+θ2,n(xn−1−xn−2)]−ψnxn‖≤ψn(1−αn)‖xn‖+(1−ψn)(θ1,n‖xn−xn−1‖+θ2,n‖xn−1−xn−2‖)+ψn‖xn‖=ψn(2−αn)‖xn‖+(1−ψn)ψn(θ1,n‖xn−xn−1‖ψn+θ2,n‖xn−1−xn−2‖ψn)→0, as n→∞. | (3.22) |
On the other hand, by (3.7) and (3.16) with z=x∗ we have
‖xn+1−x∗‖2≤[(1−ψnαn)‖xn−x∗‖+ψnαnMx∗]2−(1−γ′)(‖xn+1−yn‖2+‖yn−tn‖2)≤(1−ψnαn)‖xn−x∗‖2+ψnαnM2x∗−(1−γ′)(‖xn+1−yn‖2+‖yn−tn‖2)≤‖xn−x∗‖2+ψnαnM2x∗−(1−γ′)(‖xn+1−yn‖2+‖yn−tn‖2) |
and hence
(1−γ′)(‖xn+1−yn‖2+‖yn−tn‖2)≤‖xn−x∗‖2−‖xn+1−x∗‖2+ψnαnM2x∗ | (3.23) |
for all n≥N. Since limn→∞ψn=0, letting n→∞ in (3.23), by (3.21) we get
limn→∞‖yn−xn+1‖2=0andlimn→∞‖tn−yn‖2=0. | (3.24) |
From Lemma 3.2 it follows that {yn} is bounded.
Since {xn} is bounded, there exists a subsequence {xnk} of {xn} such that xnk⇀ˆx, where ⇀ denotes the weak convergence. From (3.22) and (3.24) we see that tnk⇀ˆx, ynk⇀ˆx and ˆx∈C. Without loss generality we may assume that
lim supn→∞⟨−x∗,xn−x∗⟩=limk→∞⟨−x∗,xnk−x∗⟩=⟨−x∗,ˆx−x∗⟩. | (3.25) |
From (3.24) we have
‖tn−xn+1‖≤‖tn−yn‖+‖yn−xn+1‖→0, as n→∞. | (3.26) |
Let ˉx∈Ω be a fixed point. By the definition of ynk and Lemma 2.1 (ⅰ) we have
⟨ynk−(tnk−λnkF(tnk))⟩≥⟨ˉx−ynk⟩. | (3.27) |
Letting k→∞ in (3.27), by Lemma 3.1, (A3) and (3.24) we get
⟨F(ˆx),ˉx−ˆx⟩≥0. |
In addition, since ˆx∈C and ˉx∈Ω, we have
⟨F(ˆx),ˉx−ˆx⟩≤0. |
Hence ⟨F(ˆx),ˉx−ˆx⟩=0. From (A2) it follows that ˆx∈Ω. This result with (3.25) and Lemma 2.1 (ⅰ) leads to that
lim supn→∞⟨−x∗,xn−x∗⟩=⟨−x∗,ˆx−x∗⟩≤0. | (3.28) |
Since limn→∞ψn=0, there exists n1∈N such that ψn<12 for all n≥n1, which implies that
γn=2ψnαn1−ψn+ψnαn∈(0,1), ∀n≥n1. |
In addition, since αψn≤γn≤4ψn for all n∈N, by the hypothesis on {ψn} we get
limn→∞γn=0and∞∑n=1γn=∞. | (3.29) |
By the definitions of γn and {φn} we have
φnγn=(1−ψn)22αn[ψnθ21,nψ2n‖xn−xn−1‖2+ψnθ22,nψ2n‖xn−1−xn−2‖2+2θ1,nψn‖xn−xn−1‖‖xn−x∗‖+2θ2,nψn‖xn−1−xn−2‖‖xn−x∗‖+ψn2θ1,nθ2,nψ2n‖xn−xn−1‖‖xn−1−xn−2‖]+⟨−x∗,tn−xn⟩+⟨−x∗,xn−x∗⟩+ψn2αn‖xn−x∗‖2. | (3.30) |
Letting n→∞ in (3.30), since {xn} is bounded, αn>α>0, ψn→0, by (3.5), (3.22) and (3.28) we get
lim supn→∞φnγn≤0. | (3.31) |
Applying Lemma 2.3 to (3.21) and using (3.29) and (3.31), we obtain limn→∞‖xn−x∗‖=0.
Case 2. Suppose there is a subsequence {ni} of {n} such that
‖xni−x∗‖≤‖xni+1−x∗‖,∀i∈N. |
From Lemma 2.4 it follows that there exists a sequence {mk}⊂N with mk→∞ such that
‖xmk−x∗‖≤‖xmk+1−x∗‖ and ‖xk−x∗‖≤‖xmk+1−x∗‖,∀k∈N. | (3.32) |
Replacing n in (3.20) with mk and using (3.32) we obtain
‖xmk+1−x∗‖2≤(1−γmk)‖xmk−x∗‖2+φmk≤(1−γmk)‖xmk+1−x∗‖2+φmk |
and hence
‖xmk+1−x∗‖2≤φmkγmk, ∀k∈N with mk≥n0. | (3.33) |
Note that by a similar process of showing (3.32) as in Case 1 we can get
lim supk→∞φnkγnk≤0, |
which together with (3.33) leads to that
lim supk→∞‖xmk+1−x∗‖2=0. | (3.34) |
Combining (3.32) with (3.34) we have
lim supk→∞‖xk−x∗‖2≤lim supk→∞‖xmk+1−x∗‖2=0. |
Hence lim supk→∞‖xk−x∗‖2=0. The proof is complete.
Remark 3.5. If replacing the conditions (A1) and (A2) with the following
(A1') S≠∅ and (A2') F is pseudomonotone on C with respect to S, and replacing Ω with S in the proof lines of Lemma 3.2, Lemma 3.3 and Theorem 3.1, we still can obtain the same results. In fact, the conditions (A1') and (A2') are often used in the related literature.
In this section, we use some numerical examples to illustrate the convergence of Algorithm 3.1 and compare the numerical results with Algorithm 3.1 of Noinakorn et al.[12] (denoted by Algorithm EXN), Algorithm 3.1 of Tan et al.[14] (denoted by Algorithm EXT). The codes of the three algorithms for the following numerical examples are written by Matlab R2016b running on a MacBook air Desktop with Core(TM) i5-4260U CPU @1.40GHz 2.00GHz, RAM 4GB.
In the following Examples 4.1–4.3, we use the following parameters and control sequences for Algorithm 3.1, Algorithm EXN and Algorithm EXT:
∙ Algorithm 3.1: ϵ1,n=ϵ2,n=100(1+n)2, {αn}, γ, λ1, θ1,θ2 are as follows:
Case 1. ψn=90100+n,αn=0.7+0.09n,γ=0.2,λ1=0.3,θ1=0.3,θ2=0.5;Case 2. ψn=80100+n,αn=0.5+0.09n,γ=0.8,λ1=0.8,θ1=0.6,θ2=0.1;Case 3. ψn=70100+n,αn=0.7+1n,γ=0.5,λ1=0.5,θ1=0.5,θ2=0.5,Case 4. ψn=60100+n,αn=0.5+1n,γ=0.2,λ1=0.3,θ1=0.3,θ2=0.5;Case 5. ψn=50100+n,αn=0.5,γ=0.5,λ1=0.5,θ1=0.5,θ2=0.5. |
∙ Algorithm EXT: λ1=0.5,μ=0.4,γ=1.5,α=0.4,ψn=1n+1,ξn=100(n+1)2,pn=1(n+1)1.1, and qn=1+nn.
∙ Algorithm EXN: τ=0.7L,α=0.6,ϵn=1(n+1)2,ψn=1n+2, where L is the Lipschitz constant of the mapping F.
Remark 4.1. The parameters and control sequences above for Algorithm EXN and Algorithm EXT were used in [12] and [14]. For the better comparison we continue to use the parameters and control sequences for Algorithm EXN and Algorithm EXT in the following numerical examples.
In the first example, since F is not pseudomonotone on C, Algorithm EXN, Algorithm EXT and many other related algorithms in the literature can not be applied to the example.
Example 4.1. Let H=Rm, C=[0,π]×[0,π]×[0,1]×⋯×[0,1], F:H→H be a mapping defined by
F(x)=(x2+cos(x2),x1+sin(x1),x3,⋯,xm), ∀x=(x1,⋯,xm)∈H. |
It follows that S=Ω={0} and so (A1) holds. We show that (A2) and (A3) hold. For x=(x1,⋯,xm)T∈C, since ∑mi=3x2i≥0, x2+cos(x2)>0, and x1+sin(x1)≥0, it is easy to see that ⟨F(x),x⟩=∑mi=3x2i+x1(x2+cos(x2))+x2(x1+sin(x1))=0 is if and only if x=0. So (A2) holds. For x,y∈H, we have
‖F(x)−F(y)‖2=(x2−y2+cos(x2)−cos(y2))2+(x1−y1+sin(x1)−sin(x1))2+m∑i=3(xi−yi)2≤4m∑i=1(xi−yi)2=4‖x−y‖2. |
It follows that F is 2-Lipschitz continuous on H and so (A3) holds.
For x=(2.8677,0.9806,0.9005,0,⋯,0)T and y=(1.0688,1.9091,0.8044,0,⋯,0)T, we have
⟨F(x),y−x⟩>0and⟨F(y),y−x⟩<0. |
Hence, F is not quasimonotone on C.
We take the initial point x−1=x0=x1=(1,1,⋯,1)T∈C and use ‖xn‖≤10−4 as the stopping criterion of Algorithm 3.1. In the process of performing Algorithm 3.1, the step size λn is computed by (3.3). The computed results for Algorithm 3.1 with the different dimension m are shown in Figure 1. From the curves in Figure 1 we see that xn→0.
For comparing the computation results with other algorithms, the mappings F in the following examples are pseudomonotone on C and Ω≠∅. By Remark 3.5 Algorithm 3.1 still can be applied to the examples.
Example 4.2. [9,12,Example 4.1] Let F:Rm→Rm be a mapping defined by
F=NNT+B+D, |
where N=rand(m) is a random matrix, B=0.5K−0.5KT with K=rand(m) is a skew-symmetric matrix, and D=diag(rand(m,1)) is a diagonal matrix. The feasible set
C={x∈Rm:−2≤xi≤5,i=1,2,⋯,m}. |
It follows that F is monotone (hence it is pseudomonotone) and L-Lipschitz continuous with L=‖F‖. The unique solution of the VIP (1.1) with the mapping F in this example is x∗=0. Since F is continuous and psuedomonotone on C, and C is convex, it has Ω=S and so (A1) holds. In addition, since the matrix NNT+B+D is positive definite [26], ⟨F(y),y⟩=yT(NNT+B+D)y=0 implies that y=0. So (A2) holds.
We choose the initial points x0,x−1,x1=(1,⋯,1)T for Algorithm 3.1 and x0=x1=(1,⋯,1)T for Algorithm EXN and Algorithm EXT, and use ‖xn‖<10−4 as the common stop criterion for all the algorithms. In the process of performing Algorithm 3.1, the step size λn is computed by (3.4). Figure 2 illustrates the curves and Table 1 gives the CPU time in seconds of the numerical results for these algorithms with the different dimension.
Algorithm 3.1 | Algorithm EXN | Algorithm EXT | |||||
Case 1 | Case 2 | Case 3 | Case 4 | Case 5 | |||
m=50 | 0.02153 | 0.02071 | 0.01973 | 0.01627 | 0.01771 | 0.01526 | 0.02115 |
m=100 | 0.02013 | 0.02114 | 0.02257 | 0.02331 | 0.03007 | 0.04877 | 0.01776 |
m=200 | 0.03685 | 0.03234 | 0.03367 | 0.03213 | 0.03281 | 0.08586 | 0.02367 |
m=500 | 0.21339 | 0.21312 | 0.23498 | 0.15074 | 0.20881 | 1.77414 | 0.27757 |
Example 4.3. [12, Example 4.3] Let F:Rm→Rm be defined by
F(x)=(5−‖x‖)x, ∀x∈Rm. |
It follows that F is L-Lipschitz continuous with L=11 and pseudomonotone but not monotone on Rm (see for more details [27]). The feasible set is C={x∈Rm:‖x‖≤3}. The unique solution of the VIP (1.1) with the mapping F in this example is x∗=0. It is obvious that the conditions (A1) and (A3) hold. For x∈C, since ‖x‖≤3, ⟨F(x),x⟩=(5−‖x‖)‖x‖2=0 is if and only if x=0 and so (A2) holds.
We take the initial points x−1=x0=x1=(1,1,⋯,1)T for our Algorithm 3.1 and x0=x1=(1,1,⋯,1)T for Algorithm EXN and Algorithm EXT, and use ‖xn‖<10−5 as the common stop criterion for all the algorithms. In the process of performing Algorithm 3.1, the step size λn is computed by (3.4). Figure 3 illustrates the curves and Table 2 gives the CPU time in seconds of the numerical results for these algorithms with the different dimension m.
Algorithm 3.1 | Algorithm EXN | Algorithm EXT | |||||
Case 1 | Case 2 | Case 3 | Case 4 | Case 5 | |||
m=500 | 0.00233 | 0.00147 | 0.00973 | 0.00211 | 0.00188 | 0.00249 | 0.00547 |
m=1000 | 0.01803 | 0.02136 | 0.01901 | 0.02341 | 0.02055 | 0.01568 | 0.01612 |
m=3000 | 0.01698 | 0.01766 | 0.01518 | 0.01809 | 0.01663 | 0.01044 | 0.01719 |
m=5000 | 0.01104 | 0.01113 | 0.01108 | 0.01129 | 0.01167 | 0.01276 | 0.01397 |
For the numerical results of the three algorithms for Examples 4.2 and 4.3, we summarized as follows. For Example 4.2, Algorithm 3.1 needs the less CPU time than another two algorithms and for Example 4.3, CPU time has no obvious difference for the three algorithms. From the curves in Figure 2 we see that Algorithm 3.1 has the less numbers of iterations than the another two algorithms when the algorithms stop. From the curves in Figure 3 we see that the numbers of iterations of Algorithm 3.1 are different when the algorithm stops, which implies that the numbers of iterations are sensitive to the setting of the control parameters and sequences (especially the setting of {ψn}) for Algorithm 3.1. For Example 4.3, Algorithm EXN needs the most numbers of iterations and Algorithm 3.1 needs the less or more numbers of iterations than Algorithm EXT for the different setting of control parameters and sequences. Overall, from the numerical results for Examples 4.2 and 4.3, we see that our Algorithm 3.1 has certain competitiveness than the other two algorithms.
In this paper, we constructed a new double inertial subgradient extragradient algorithm for solving a non-monotone variational inequality problem in a Hilbert space. Without prior knowledge of the Lipschitz constant of the involved mapping, we use a dynamic manner to update the step-size in our algorithm. Some new conditions are used to guarantee the strong convergence of the proposed method. Some artificially constructed numerical examples and an applied problem of image recovery are solved by our algorithm and some other related algorithms. The results show the effectiveness and competitiveness of our algorithm with other compared algorithms.
Ziqi Zhu: Methodology, Computing numerical examples, Writing-original draft; Kaiye Zheng: Formal analysis, Writing-original draft; Shenghua Wang: Design of algorithm, Proof of conclusions. All authors have read and approved the final version of the manuscript for publication.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
All authors declare no conflicts of interest in this paper.
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1. | Limei Xue, Jianmin Song, Shenghua Wang, A modified projection and contraction method for solving a variational inequality problem in Hilbert spaces, 2025, 10, 2473-6988, 6128, 10.3934/math.2025279 |
Algorithm 3.1 | Algorithm EXN | Algorithm EXT | |||||
Case 1 | Case 2 | Case 3 | Case 4 | Case 5 | |||
m=50 | 0.02153 | 0.02071 | 0.01973 | 0.01627 | 0.01771 | 0.01526 | 0.02115 |
m=100 | 0.02013 | 0.02114 | 0.02257 | 0.02331 | 0.03007 | 0.04877 | 0.01776 |
m=200 | 0.03685 | 0.03234 | 0.03367 | 0.03213 | 0.03281 | 0.08586 | 0.02367 |
m=500 | 0.21339 | 0.21312 | 0.23498 | 0.15074 | 0.20881 | 1.77414 | 0.27757 |
Algorithm 3.1 | Algorithm EXN | Algorithm EXT | |||||
Case 1 | Case 2 | Case 3 | Case 4 | Case 5 | |||
m=500 | 0.00233 | 0.00147 | 0.00973 | 0.00211 | 0.00188 | 0.00249 | 0.00547 |
m=1000 | 0.01803 | 0.02136 | 0.01901 | 0.02341 | 0.02055 | 0.01568 | 0.01612 |
m=3000 | 0.01698 | 0.01766 | 0.01518 | 0.01809 | 0.01663 | 0.01044 | 0.01719 |
m=5000 | 0.01104 | 0.01113 | 0.01108 | 0.01129 | 0.01167 | 0.01276 | 0.01397 |
Algorithm 3.1 | Algorithm EXN | Algorithm EXT | |||||
Case 1 | Case 2 | Case 3 | Case 4 | Case 5 | |||
m=50 | 0.02153 | 0.02071 | 0.01973 | 0.01627 | 0.01771 | 0.01526 | 0.02115 |
m=100 | 0.02013 | 0.02114 | 0.02257 | 0.02331 | 0.03007 | 0.04877 | 0.01776 |
m=200 | 0.03685 | 0.03234 | 0.03367 | 0.03213 | 0.03281 | 0.08586 | 0.02367 |
m=500 | 0.21339 | 0.21312 | 0.23498 | 0.15074 | 0.20881 | 1.77414 | 0.27757 |
Algorithm 3.1 | Algorithm EXN | Algorithm EXT | |||||
Case 1 | Case 2 | Case 3 | Case 4 | Case 5 | |||
m=500 | 0.00233 | 0.00147 | 0.00973 | 0.00211 | 0.00188 | 0.00249 | 0.00547 |
m=1000 | 0.01803 | 0.02136 | 0.01901 | 0.02341 | 0.02055 | 0.01568 | 0.01612 |
m=3000 | 0.01698 | 0.01766 | 0.01518 | 0.01809 | 0.01663 | 0.01044 | 0.01719 |
m=5000 | 0.01104 | 0.01113 | 0.01108 | 0.01129 | 0.01167 | 0.01276 | 0.01397 |