Definition 1. Let A : H⟶H be called the operator and also ϕ-pseudomonotone if 〈Aα, ϕ(β) − ϕ(α)〉 ≥ 0  provides 〈Aβ, ϕ(β) − ϕ(α)〉 ≥ 0,  ∀ α, β ∈ H.

It is considered [22, 24] that monotonicity implies pseudomonotonicity, but the converse does not exist.

Lemma 1. For z ∈ H, α ∈ M holds for the inequality:

if and only if where P_M is called the projection of H onto the convex set M.

Lemma 2. α is a solution of the given GVI (1) if and only if α ∈ M satisfies the relation

where ρ ≥ 0 is taken as the constant and P_M is considered the projection operator H onto M.

Lemma 3. The function α ∈ H : ϕ(α) ∈ M   satisfies inequalities (1) if and only if z ∈ H satisfies equation (9), provided

Algorithm 1.

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  Step 0. We set the parameters as follows. For α_o ∈ H,   set n = 0,   and take δ₀, δ ∈ (0,1), ϵ > 0, γ ∈ [1,2), μ ∈ (0,1), and ρ > 0.

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  Step 1. If R₁(α_n) < ϵ, then we terminate; otherwise, take $$ where m_n finds the smallest nonnegative integer that satisfies the inequality $$

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  Step 2. Compute d₁(α_n) = R₁(α_n) − ρ_nA(α_n) + ρ_nAϕ⁻¹P_M[ϕ(α_n) − ρ_nA(α_n)] and $$.

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  Step 3. Find the next iteration, ϕ(α_n+1) = ϕ(α_n) − ω_nd₁(α_n).

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  Step 4. If $$ then again take ρ = ρ_n/μ, else ρ = ρ_n. Consider n = n + 1, and start iteration from step 1.

Algorithm 2.

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  Step 0. For parameters ϵ > 0, ρ > 0, δ₀, δ ∈ (0,  1),  γ ∈ [0,1], μ ∈ (0,1), and α_o ∈ H,   we start from n = 0, η ∈ (0,1).

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  Step 1. We again take ρ_n = ρ.  If ‖R(α_n)‖ < ϵ, then computation stops; otherwise, we continue and consider $$ and find the smallest nonnegative integer m_n, which satisfies the inequality $$.

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  Step 2. Calculate the next iterate:

  $$ (20)

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  where

  $$ (21)
  $$ (22)

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  Step 3. If

  $$ (23)

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  then again we set ρ = ρ_n/μ, else by setting ρ = ρ_n. Take n = n + 1, and repeat step 1.

Here, ω_n is taken as corrector step size which contains the GWHE (9).

We consider the convergence of the main established results, which is the main target of this research.

Theorem 1. If α^∗ is a solution of inequality(1) and the operator A : H⟶H is ϕ-pseudomonotone, then

Proof. Let α^∗ ∈ H     be a solution of GVI (1). Then,

since T is ϕ-pseudomonotone. Taking ϕ(β) = ϕ(α) − ηR(α) in (25), we have

This implies that

Taking z = ϕ(α) − ρAα,     α = P_M[ϕ(y) − ρAα], and β = ϕ(α^∗) in (3), we obtain

Using(3.1),

from which, we have

Adding (27) and (30), we obtain

Using (19), (21), (23), (27), and (31), we have

which is the desired result.

Theorem 2. If α^∗ ∈ H is a solution of GVI (1) and α_n+1 is the approximate solution found from Algorithm 2, then

Proof. From (20), (21), (22), and (32), we have

which is the required result.

Theorem 3. Let α_n+1 be the approximated solution obtained from Algorithm 2 and α^∗ ∈ M   be a solution of GVI (1). If H   is a finite-dimensional space, then $$

Proof. Let α^∗ ∈ M be a solution of GVI (1). From (34), we get that the sequence {α_n} is bounded; we have

which shows both expressions are going to be zero when n⟶∞ such as and which implies (36) holds. Let α^∗ be taken as the cluster point of {α_n} , and consider the subsequence {α_ni} of the sequence {α_n} converge to point α^∗. We know continuity of R holds; we have which provides that  α^∗ is a solution of GVI (1) by Theorem 3 and

Thus, the sequence {α_n} converges exactly one cluster point and the consequences, and we obtain

Since ϕ is injective, it gives that $$ which satisfies the GVI (1).

Suppose that (37) holds and $$ If (32) does not hold, then, by taking the value of η_n, we obtain

Let α^∗ be the cluster point of {α_n} and let {α_ni} be the subsequence {α_ni} converge to α^∗. We apply the limit in (41); then,

which gives R(α^∗) = 0, that is, α^∗ ∈ H is a solution of inequality (1), and by Lemma 1, inequality (41) holds. By repeating the same process and arguments, we approach that $$, the desired result.

Problem 1. This problem is relevant to inequality (1), with ϕ(α) = Aα + q and Aα = α, where

We set the following domain for the considered problem: M = {α ∈ (Rⁿ/0) ≤ α_i ≤ 1,  for i = 1,2,3, …, n}. Results for Algorithm 1 are mentioned in Table 1. Tables 2 and 3 represent the outcomes of Algorithm 2 with the initial point α⁰ = −A⁻¹q for the order n = 100 of the generated matrix .  For all output, we consider μ, δ ∈ (0,1), γ ∈ [1,2], and ρ > 0. The iterative process will stop when we have r(α_n, ρ_n) ≤ 10⁻⁷.

From Tables 1 and 2, we observe with the change of parameters that the number of iterations also varies. Table 2 gives the results for the newly established method (Algorithm 2). From the output, we observe that the newly established method converges more quickly than Algorithm 1 for solving the main GVI.

From Tables 3 and 4, we see that, in the new iterative scheme, the number varies (iterations) by changing the parameters δ, ρ, and μ. By changing the parameters accordingly, we can reduce the number of iterations.