EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 4, Article Number 6680 ISSN 1307-5543 – ejpam.com Published by New York Business Global Some Convergence Results for the Solution of New Type of Variational Inequalities Associated with Generalized Pseudo-Monotone Mappings in Complete CAT (0) Spaces Maliha Rashid1, Amna Kalsoom1, Nida Masood1, Ahmad Aloqaily2, Nabil Mlaiki2,∗ 1 Department of Mathematics and Statistics, International Islamic University, Islamabad, Pakistan 2 Department of Mathematics and Sciences, Prince Sultan University, Saudi Arabia Abstract. The basic purpose of this article is to introduce a generalized version of pseudo- monotone variational inequality in the setting of complete CAT (0) spaces and to present some strong and ∆-convergence results for the existence of solutions for the respective variational in- equality problem. Algorithm 1 and 2 are proposed in accordance with pseudo-monotone and α-strongly pseudo-monotone mappings to prove our results under some conditions. A numerical implication of our proposed algorithm is also presented. 2020 Mathematics Subject Classifications: 47H10, 47H09, 47J25 Key Words and Phrases: Projection type method, variational inequality, pseudo-monotone mapping 1. Introduction Variational inequalities originated in the beginning of 1960s through the revolutionary work of the Italian mathematician Guido Stampacchia [1], who analyse free boundary problems occuring in elasticity theory and mechanics by using the variational inequality as an analytic tool. From 1960-1975 many foundational articles presented in the literature emphasizing the association between the complementarity problems and the variational inequalities. For the early advancement on variational inequalities readers are referred to [2–6]. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6680 Email addresses: maliha.rashid@iiu.edu.pk (M. Rashid), amna.kalsoom@iiu.edu.pk (A. Kalsoom), nida.msma673@iiu.edu.pk (N. Masood), maloqaily@psu.edu.sa (A. Aloqaily), nmlaiki@psu.edu.sa (N. Mlaiki) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 2 of 24 A large number of articles, proposed in second half of 1990s, was dedicated to the reformulation of the nonlinear complementarity problem in terms of the algorithms pro- duced through a globally convergent Newton method. After that, many iterative schemes have been formulated for finding the solutions of variational inequalities and their relevant optimization problems (see [7, 8] and literature cited in). One of the numerical methods for solving variational inequality problems (VIP’s) is known as projection method [9–11] which is further expanded to gradient, extragradient and subgradient methods (see, e.g., [12–25] and the references therein). Extragradient method is not practically useful for the solution of variational inequali- ties having non-Lipschitz mapping. In case of extragradient method, the reforms regarding [26, 27] mentioned in [28, 29] ensured the convergence without Lipschitz continuity. In [30], the authors discussed strong and weak convergence results for a VIP in the context of a pseudo-monotone, classical non-Lipschitzian, continuous mapping in Hilbert spaces over R. Korpelevich [11], introduced an extragradient method in finite dimensional Euclidean space to obtain solution of variational inequality problem under the mapping of mono- tone and Lipschitz continuous. The extragradient method has been further extended in infinite dimensional spaces by many researchers (see [12–15, 22–25] and the refferences therein). The modification in [28, 29], enables convergence in finite dimensional Euclidean space without Lipschitz continuity of the mappings associated variational inequality. in [30], the extragradient method has been expanded in infinite dimensional space to get weak and strong convergence results for VIP under the condition of classical non-lipschitz, pseudo-monotone and continuous mapping. The following article is dedicated to the analysis of a pseudo-monotone VIP in the setting of CAT (0) space, which gives a clear modification of extragradient algorithm for strong and ∆-convergence. CAT (0) spaces, established by Alexandrov in the 1950’s, were given recognition by M. Gromov, who displayed that a great deal of the theory of manifolds of non-positive sec- tional curvature could be designed without using much more than the CAT (0) condition. Gromov described the key aspects of the global geometry of manifolds of non-positive curvature, primarily relying on the CAT (0) inequality(see [31]). Let (Y, ϱ) be a metric space. A geodesic segment connecting u1 ∈ Y to u2 ∈ Y is a mapping Υ : [0, ϱ(u1, u2)] → Y such that Υ(0) = u1,Υ(ϱ(u1, u2)) = u2 and ϱ(Υ(g1),Υ(g2) = |g1 − g2|), ∀g1, g2 ∈ [0, ϱ(u1, u2)]. A geodesic segment linking any two different points u1, u2 ∈ Y is an isometry with Υ(0) = u1,Υ(u1, u2) = u2. A unique geodesic segment is expressed by [u1, u2]. The metric space (Y, ϱ) is known as a geodesic metric space if any two points are joined by a geodesic segment and the metric (Y, ϱ) is a uniquely geodesic if there is exactly one geodesic segment to link them. A subset L ⊆ Y is called convex if any two points in Y can be joined by a geodesic and the image of every such geodesic is lying in L. Suppose (Y, ϱ) be the geodesic metric space. In a geodesic metric space, a geodesic triangle has three corners u1, u2, u3 ∈ Y and three geodesic segments ([u1, u2], [u2, u3], [u3, u1]) join- M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 3 of 24 ing them. For this triangle there exist a comparison (Alexandrov) triangle ∆(u1, u2, u3) ⊂ R2 such that ∗ ϱ(u1, u2) = ϱ(u1, u2), ∗ ϱ(u2, u3) = ϱ(u2, u3), ∗ ϱ(u3, u1) = ϱ(u3, u1). When all geodesic triangles in a geodesic metric space satisfy the following CAT (0) com- parison axiom then geodesic metric space is known as CAT (0) space (this term is due to M.Gromov [32]) if . Let ∆ and ∆ be a geodesic and comparison triangle in Z, respec- tively. If the following inequality is satisfied for all u1, u2 ∈ ∆ and all comparison points u1, u2 ∈ ∆, ϱ(u1, u2) ≤ ϱ(u1, u2), then ∆ is said to satisify CAT (0) inequality. The motivation for the conversion thus stems from the need to improve robustness and applicability of results. By transiting to CAT (0) spaces, researchers can better handle nonlinearities, ensuring more meaningful and reliable insights across a broad range of applications. We introduce two algorithms in Hadamard spaces that does not require to have previous knowledge of Lipschitz- like constants. Our proposed algorithms converges to a solution of VIP. Moreover, we present a numerical example in a Hadamard space to demonstrate the performance of our method. Question 1. Can we obtain convergence results for VIP using an Extragradient algorithm in Hadamard space under the condition of pseudo-monotone. our contributions in this paper are briefly highlighted as: 1. Our work extends algorithm of extragradient for pseudomonotone from linear spaces to nonlinear spaces. 2. Our algorithm not depends on the Lipschitz constant. 2. Preliminaries In this section, we display some notations, familiar definitions, and relevent results that will be required in the proof of our main results. Definition 1. A geodesically connected metric space Y is known as CAT (0) space and every geodesic triangle in Y is at least as ’thin’ as its comparison triangle in the Euclidean plane. For a systematic study of geodesic spaces and CAT (0) spaces the readers are referred to [31]. According to Bruhat and Tits [33], the (CN) inequality is defined as follows: Definition 2. If u, u1, u2 ∈ CAT (0) space and if u0 ∈ [u1, u2] be the middle point of the segment, then the CAT (0) inequality yields ϱ(u, u0) 2 ≤ 1 2 ϱ(u, u1) 2 + 1 2 ϱ(u, u2) 2 − 1 4 ϱ(u1, u2) 2. M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 4 of 24 In recent past, CAT (0) spaces have appealed many mathematicians, due to their geo- metrical relevance in multiple directions. Hadamard spaces [34] are originally the complete CAT (0) spaces. In 2008, the notion of quasilinearization was initiated by Berg and Nikolaev [35], given as follows: Definition 3. Denoting a vector as a pair (ξ, η) ∈ Y × Y by −→ ξη, the quasilinearization is defined as a mapping ⟨., .⟩ : (Y × Y)× (Y × Y) → R satisfying ⟨ −→ ξη, −→ γδ⟩ = 1 2 [ ϱ2(ξ, δ) + ϱ2(η, γ)− ϱ2(ξ, γ)− ϱ2(η, δ) ] , where ξ, η, γ, δ ∈ Y. Remark 1. It can easily verify that for all ξ, η, γ, δ, ζ ∈ Y, (i) ⟨ −→ ξη, −→ γδ⟩ = ⟨ −→ γδ, −→ ξη⟩, (ii) ⟨ −→ ξη, −→ γδ⟩ = −⟨ −→ ηξ, −→ γδ⟩, (iii) ⟨ −→ ξζ, −→ γδ⟩+ ⟨ −→ ζη, −→ γδ⟩ = ⟨ −→ ξη, −→ γδ⟩, (iv) Y satisfies the Cauchy-Schwarz inequality if ⟨ −→ ξη, −→ γδ⟩ ≤ ϱ(ξ, η)ϱ(γ, δ). Remark 2. A geodesically connected metric space is a CAT (0) space if and only if it satisfies the Cauchy-Schwarz inequality ([35], Corollary 3). In 2010 Kakavandi and Amini[36] develop dual space of Hadamard space Z by using the concept of quasilinearization and by initiating the concept of pseudometric space. Definition 4. To explain the conjugate space of Hadamard space Z, consider the map Φ : R×Z ×Z → H(Z,R) defined by Φ(t, ξ, η) = t⟨ −→ ξη, −→ ξα⟩ (t ∈ R, ξ, η, γ ∈ Z) (1) where H(Z,R) is the space of all continuous real-valued functions on Z. Then the Cauchy- Schwartz inequality implies that Φ(t, ξ, η) is a Lipschitz function with Lipschitz semi-norm H(Φ(t, ξ, η)) = tϱ(ξ, η), for all t ∈ R and ξ, η ∈ Z, where H(Ψ) = sup{Ψ(ξ)−Ψ(η) ϱ(ξ,η) ; ξ, η ∈ Z, ξ ̸= η} is the Lipschitz semi-norm, for any function Ψ : Z → R. Now, we present the pseudometric D on R×Z ×Z by D((t, ξ, η), (s, γ, δ)) = H(Φ(t, ξ, η)− Φ(s, γ, δ)) (t, s ∈ R, ξ, η, γ, δ ∈ Z). (2) M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 5 of 24 Lemma 1. [36] D((t, ξ, η), (s, γ, δ)) = 0 if and only if t⟨ξη,−→er⟩ = s⟨γδ,−→er⟩, for all e, r ∈ Z. Proof. By (1) and (2) and formulation of Lipschitz semi-norm, D((t, ξ, η), (s, γ, δ)) = 0 if and only if there exist a constant κ ∈ R such that t⟨ξη,−→ξe⟩ = s⟨γδ,−→γe⟩+κ, for all e ∈ Z. Therefore, for all e, r ∈ Z t⟨ξη,−→er⟩ = t⟨ξη, −→ ξr⟩ − t⟨ξη, −→ ξe⟩ = s⟨γδ,−→γe⟩ − s⟨γδ,−→γe⟩ = s⟨γδ,−→er⟩. Conversely if t⟨ξη,−→er⟩ = s⟨γδ,−→er⟩, for all e, r ∈ Z, then Φ(t, ξ, η)(e) = t⟨ξη, ξe⟩ = s⟨γδ, ξe⟩ = Φ(s, γ, δ)(e)− s⟨γδ, γξ⟩, for all e ∈ Z, which yields D((t, ξ, η), (s, γ, δ)) = 0. Definition 5. For a Hadamard space (Z, ϱ), the pseudometric space (R×Z ×Z ,D) can be considered as a subspace of the pseudometric space (Lip(Z,R),H) of all real-valued Lipschitz functions. Also, D explain an equivalence relation on R × Z × Z, where the equivalence class of (t, ξ, η) is [t −→ ξη] = {s −→ γδ; t⟨ −→ ξη, −→ γδ⟩ = s⟨ −→ γδ, −→ γδ⟩ (γ, δ ∈ Z)}. The set Y ∗ = {t −→ ξη; (t, γ, δ) ∈ R×Z ×Z} is a metric space with metric D, which is called the dual metric space of (Z, ϱ). In [37], Theorem 2.3, the projection operator is utilized for the existence of solution of the respective variational inequality in a Hilbert space over R. By using the concept of quasilinearization, authors in [38] extended the above mentioned result in CAT (0) space that is as follows: Theorem 1. [38] Let (Z, ϱ) be a complete CAT (0) space and ∅ ̸= L ⊆ Z is convex. Then w = PLw2 ⇔ ⟨−−→w1w, −−→ww2⟩ ≥ 0, ∀ w1 ∈ L, w2 ∈ Z, and w ∈ L. In the following, (Y, ϱ) and (Z, ϱ) will represent CAT (0) space and complete CAT (0) space respectively. These are some lemma’s taken from literature which are helpful in our main results. Lemma 2. [39] Let u1, u2, u ∈ Y and τ ∈ [0, 1]. Then (i) ϱ(τu1 ⊕ (1− τ)u2, u) ≤ τϱ(u1, u) + (1− τ)ϱ(u2, u), (ii) ϱ2(τu1 ⊕ (1− τ)u2, u) ≤ τϱ2(u1, u) + (1− τ)ϱ2(u2, u)− τ(1− τ)ϱ2(u1, u2). Lemma 3. [39] Let u1, u2, u ∈ Y and τ ∈ [0, 1]. Then (i) ϱ(τu1 ⊕ (1− τ)u2, γu1 ⊕ (1− γ)u2) = |τ − γ|ϱ(u1, u2), M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 6 of 24 (ii) ϱ(τu1 ⊕ (1− τ)u2, τu1 ⊕ (1− τ)q) ≤ (1− τ)ϱ(u2, u). Lemma 4. [40] In (Z, ϱ) space, every bounded sequence always has a ∆-convergent sub- sequence. Lemma 5. [41] Assume {ℵn}, {ℑn}, {cn} and {σn} be nonnegative sequences such that ℵn+1 ≤ (1− σn)ℵn + σnℑn + cn, n ≥ 0 with {σn} ⊂ [0, 1],Σ∞ n=oσn = ∞, limn→∞ℑn = 0 and σ∞n=0cn <∞. Then limn→∞ ℵn = 0. Lemma 6. [42] For (Z, ϱ), the inequality stated below holds ϱ2(u,w) ≤ ϱ2(r, w) + 2⟨−→ur,−→uw⟩, ∀ u, r, w ∈ Z. Lemma 7. [42] For any ℓ ∈ (0, 1) and s, t ∈ Y, assume sℓ = ℓs⊕ (1− ℓ)t. Then, for all u, v ∈ Y, (i) ⟨−→sℓu,−→sℓv⟩ ≤ ℓ⟨−→su,−→sℓv⟩+ (1− ℓ)⟨−→tu,−→sℓv⟩ (ii) ⟨−→sℓu,−→sv⟩ ≤ ℓ⟨−→su,−→sv⟩+ (1− ℓ)⟨−→tu,−→sv⟩ and ⟨−→sℓu, −→ tv⟩ ≤ ℓ⟨−→su,−→tv⟩+ (1− ℓ)⟨−→tu,−→tv⟩. Lemma 8. [43] Assume a non-negative sequence {bn} of real numbers, such that there exist a subsequence {bnl } of the sequence {bn} satisfying bnl < bnl+1 , for all l ∈ N. So there is a non-decreasing sequence {ak} of natural numbers in such a way that ak → ∞ as k → ∞, and for all k ∈ N satisfy the conditions stated below: bak ≤ bak+1 and bk ≤ bak+1 . Indeed, ak = max{l ≤ k : bl ≤ bl+1}. Lemma 9. [44] Consider a sequence {bn} ∈ Z and if a nonempty subset L ⊆ Z satisfying the following conditions: (i) for every ω ∈ L, limn→∞ ϱ(bn, ω) exists; (ii) if {bnj} is a subsequence of {bn} which is ∆-convergent to v, then v ∈ L. Then {bn} ∆-converges to an element of L. M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 7 of 24 3. Variational Inequality and Some Crucial Lemmas In this section we introduce variational inequality and several lemmas which are es- sential for our main results. Consider a closed, convex subset L ⊆ Z, and define a map A1 : L → Z∗ , A2 : Z∗ → L and A : L → L. Finding a point w∗ ∈ L such that ⟨ −−−−→ wAw∗, −−→ ww∗⟩ ≥ 0, for all w ∈ L. (3) Problem (3) is referred as variational inequality and denoted by V I(L, A). Definition 6. The map A : L → L is known as (i) monotone if ⟨ −−−−−−→ Aw1Aw2, −−−→w1w2⟩ ≥ 0, ∀ w1, w2 ∈ L. (ii) pseudo-monotone if ⟨ −−−−→ w1Aw ∗ 1, −−−→ w1w ∗ 1⟩ ≥ 0 ⇒ ⟨ −−−−→ w∗ 1Aw1, −−−→ w∗ 1w1⟩ ≥ 0, ∀ w1, w ∗ 1 ∈ L. Definition 7. Consider the space (Z, ϱ). For α > 0, map A is known as α−strongly pseudo-monotone if ⟨ −−−−−−→ Aw1Aw2, −−−→w1w2⟩ ≥ αϱ2(w1, w2), ∀ w1, w2 ∈ L. The convergence of the approaches is assumed to meet the following conditions. Condition 1. The subset L of a Hadamard space (Z, ϱ) is nonempty, closed and convex. Condition 2. The mapping A : L → L is a pseudo-monotone, uniformly continuous on L. Condition 3. The solution set of VI(3) is non-empty, that is V I(L, A) ̸= ϕ. Condition 4. Let ϖ : L → Z be a contraction map. Let’s say there’s a sequence {ξn} of real numbers in an open interval (0, 1) in such a way that lim n→∞ ξn = 0, Σ∞ n=1ξn = ∞. Now we will discuss some lemmas which are crucial for our main results. These lemmas has been established by authors in the framework of Hilbert space. Here, we explain these lemmas in a complete CAT (0) space setting and provide the proof. M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 8 of 24 Lemma 10. Let u1 ∈ Z. Then ϱ2(PLu1, u2) ≤ ϱ2(u1, u2)− ϱ2(u1, PLu2), for all u2 ∈ L. Proof. Consider ⟨−−→u1u2, −−→u1u2⟩ = ⟨ −−−−−→ u1PLu1, −−→u1u2⟩+ ⟨ −−−−−→ PLu1u2, −−→u1u2⟩, = ⟨ −−−−−→ u1PLu1, −−−−−→ u1PLu1⟩+ ⟨ −−−−−→ u1PLu1, −−−−−→ PLu1u2⟩+ ⟨ −−−−−→ PLu1u2, −−−−−→ u1PLu1⟩ +⟨ −−−−−→ PLu1u2, −−−−−→ PLu1u2⟩, = ⟨ −−−−−→ u1PLu2, −−−−−→ u1PLu2⟩+ ⟨ −−−−−→ PLu1u2, −−−−−→ PLu1u2⟩+ 2⟨ −−−−−→ u1PLu1, −−−−−→ PLu1u2⟩, = ⟨ −−−−−→ u1PLu1, −−−−−→ u1PLu1⟩+ ⟨ −−−−−→ PLu1u2, −−−−−→ PLu1u2⟩+ 2⟨ −−−−−→ u2PLu1, −−−−−→ PLu1u1⟩. By Theorem 1, we have ⟨ −−−−−→ u2PLu1, −−−−−→ PLu1u1⟩ ≥ 0. We have ⟨−−→u1u2, −−→u1u2⟩ ≥ ⟨ −−−−−→ u1PLu1, −−−−−→ u1PLu1⟩+ ⟨ −−−−−→ PLu1u2, −−−−−→ PLu1u2⟩, ϱ2(u1, u2) ≥ ϱ2(u1, PLu1) + ϱ2(u2, PLu1), ϱ2(u2, PLu1) ≤ ϱ2(u1, u2)− ϱ2(u1, PLu1). Lemma 11. Consider a closed, convex subset L ⊂ Z and defined C := {u ∈ Z : ψ(u) ≤ 0}. If L is nonempty and a real valued function ψ is Lipschitz continuous on Z with modulus Θ > 0, then ϱ(u,C) ≥ Θ−1max{ψ(u), 0}, for all u ∈ L, (4) the distance from u to C is denoted by d(u,C). Proof. Clearly (4) holds for all u ∈ C and we are left to proof that (4) holds for every u ∈ L/C. Assume u ̸∈ C but u ∈ L. Since C is closed, there exist ω(u) ∈ C such that ϱ(u, ω) = ϱ(u,C). Since ψ is Lipschitz continuous, we have ϱ(ψ(u), ψ(ω(u))) ≤ Θϱ(u, ω), = Θϱ(u,C). Since u ̸∈ C and ω(u) ∈ L, we have ψ(u) > 0 and ψ(y(u)) ≤ 0. Then ψ(u) ≤ ψ(u)− ψ(ω(u)) ≤ |ψ(u)− ψ(ω(u))|, = ϱ(ψ(u), ψ(ω(u))) ≤ Θϱ(u,C). Lemma 12. Let L be a nonempty, closed and convex subset of a complete CAT (0) space Z and A be a pseudo-monotone map. If ⟨ −−→ uAu, −−→ uu∗⟩ ≥ 0, ∀ u ∈ L. (5) Then u∗ is the solution of V I(L, A). M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 9 of 24 Proof. Suppose ⟨ −−→ uAu, −−→ uu∗⟩ ≥ 0 holds for all u ∈ L. Thus ⟨ −−−−−→ uλ ∗Auλ ∗, −−−→ uλ ∗u∗⟩ ≥ 0, uλ ∗ ∈ L ⟨ −−−→ uλ ∗u∗, −−−−−→ uλ ∗Auλ ∗⟩ ≥ 0. By using Lemma 7 and applying limit, we obtain ⟨ −−−→ uλ ∗u∗, −−−−−→ uλ ∗Auλ ∗⟩ ≤ λ⟨ −−→ uu∗, −−−−−→ uλ ∗Auλ ∗⟩+ (1− λ)⟨ −−→ u∗u∗, −−−−−→ uλ ∗Auλ ∗⟩. = λ⟨ −−→ uu∗, −−−−−→ uλ ∗Auλ ∗⟩ ≤ ⟨ −−−−−→ uλ ∗Auλ ∗, −−→ uu∗⟩, ≤ λ⟨ −−−−→ uAuλ ∗, −−→ uu∗⟩+ (1− λ)⟨ −−−−−→ u∗Auλ ∗, −−→ uu∗⟩, = λ⟨ −−−→ uAu∗, −−→ uu∗⟩+ (1− λ)⟨ −−−−→ u∗Au∗, −−→ uu∗⟩, = λ⟨ −−−→ uAu∗, −−→ uu∗⟩+ (1− λ)⟨ −−→ u∗u, −−→ uu∗⟩+ (1− λ)⟨ −−−→ uAu∗, −−→ uu∗⟩, = ⟨ −−−→ uAu∗, −−→ uu∗⟩+ (1− λ)⟨ −−→ u∗u, −−→ uu∗⟩, ≤ ⟨ −−−→ bAu∗, −−→ uu∗⟩+ (1− λ)ϱ2(u, u∗). This implies ⟨ −−−→ uAu∗, −−→ uu∗⟩ ≥ 0. Thus u∗ is a solution of (5). Now, we introduce our algorithm as follows: Algorithm 1. Initialization: Given µ, ν, ς ∈ (0, 1). Let b1 ∈ L be arbitrary Iterative Steps: For the given iteration bn, we first calculate bn+1 as stated below: Step 1. Compute sn = PL(ξnbn ⊕ (1− ξn)Abn), where ξn := ςνmn, with mn is the minimal nonnegative integer satisfying ⟨ −−−−−→ AbnAsn, −−→ bnsn⟩ ≤ µϱ2(bn, sn). If Asn = 0 or bn = sn holds then algorithm stops and sn is a solution of VI. Else Step 2. Calculate bn+1 = PLn(bn), where Ln := {L ∈ Z : hn(b) ≤ 0} and hn(b) = ⟨ −→ bsn, −−−−−−−−−−−−−−−−−−→ (ξnbn ⊕ (1− ξn)Abn)Abn⟩. (6) Place n := n+ 1 and repeat the Step 1. M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 10 of 24 Lemma 13. Suppose that Conditions 1-3 hold. Let b∗ be a solution of V I(L, A) and the function hn be defined by (6). Then hn(b ∗) ≤ 0 and hn(bn) ≥ (1− µ)ϱ2(bn, sn). Proof. Since b∗ ∈ V I(L, A), we have ⟨ −−−→ snAb ∗, −−→ b∗sn⟩ ≤ 0. (7) It is implied from Lemma 7 and (7) that hn(b ∗) = ⟨ −−→ b∗sn, −−−−−−−−−−−−−−−−−−→ (ξnbn ⊕ (1− ξn)Abn)Abn⟩, ≤ ξn⟨ −−→ b∗sn, −−−→ bnAbn⟩+ (1− ξn)⟨ −−→ b∗sn, −−−−−→ AbnAbn⟩, ≤ ξn⟨ −−→ b∗sn, −−→ bnsn⟩+ ⟨ −−→ b∗sn, −−−→ snAb ∗⟩+ ξn⟨ −−→ b∗sn, −−−−−→ Ab∗Abn⟩, by taking limit, we get hn(b ∗) ≤ 0. Thus Claim 1 of Lemma 13 holds. Now, To prove Claim 2, from (6), we have hn(bn) = ⟨ −−→ bnsn, −−−−→ snAbn⟩ = ⟨ −−→ bnsn, −−→ snbn⟩+ ⟨ −−→ bnsn, −−−−→ bnAsn⟩+ ⟨ −−→ bnsn, −−−−−→ AsnAbn⟩. As ⟨ −−→ bnsn, −−−−→ bnAsn⟩ ≥ 0, we have hn(bn) ≥ ⟨ −−→ bnsn, −−→ snbn⟩+ ⟨ −−→ bnsn, −−−−−→ AsnAbn⟩, ≥ −ϱ2(bn, sn)− µϱ2(bn, sn), = (−1− µ)ϱ2(bn, sn). Lemma 14. Consider a nonempty, closed and convex subset of a Hadamard space Z be L. The map A is uniformly continuous and pseudo-monotone on L. The solution set of the V I(L, A) is nonempty and suppose a sequence produced by Algorithm 1 is {bn}. If there is a subsequence {bnk } of {bn} such that {bnk } ∆-converges to z ∈ Z and limk→∞ ϱ(bnk , snk ) = 0, then z ∈ V I(L, A). Proof. From ∆− limn→∞ bnk = z, limk→∞ ϱ(bnk , snk ) = 0, and sn ⊂ L, we have z ∈ L and snk = PL(ξnk bnk ⊕ (1− ξnk )Abnk ) thus, 0 ≤ ⟨ −−→ bsnk , −−−−−−−−−−−−−−−−−−−−−→ snk (ξnk bnk ⊕ (1− ξnk )Abnk )⟩, ∀b ∈ L. By Lemma 7 and Remark 1, we get ⟨ −−−−−−−−−−−−−−−−−−−−−→ (ξnk bnk ⊕ (1− ξnk )Abnk )snk , −−→ snk b⟩ ≤ ξnk ⟨ −−−−→ bnk snk , −−→ snk b⟩+ (1− ξnk )⟨ −−−−−→ Abnk snk , −−→ snk b⟩, = ξnk ⟨ −−−−→ bnk snk , −−→ snk b⟩+ (1− ξnk )⟨ −−→ snk b, −−→ bsnk ⟩+ (1− ξnk )⟨ −−−→ bAbnk , −−→ bsnk ⟩, M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 11 of 24 = ξnk ⟨ −−−−→ bnk snk , −−→ snk b⟩+ (1− ξnk )⟨ −−→ snk b, −−→ bsnk ⟩+ (1− ξnk )⟨ −−−→ bAbnk , −−→ bbnk ⟩+ (1− ξnk )⟨ −−−→ bAbnk , −−−−→ bnk snk ⟩, ≤ ⟨ −−−−→ bnk snk , −−→ snk b⟩ − ϱ2(snk , b) + ⟨ −−−→ bAbnk , −−→ bbnk ⟩+ ⟨ −−−→ bAbnk , −−−−→ bnk snk ⟩. This implies that ⟨ −−−−→ bnk snk , −−→ snk b⟩+ ⟨ −−−→ bAbnk , −−→ bbnk ⟩+ ⟨ −−−→ bAbnk , −−−−→ bnk snk ⟩ ≥ 0, ∀b ∈ L. (8) Now, we will prove lim inf⟨ −−−→ bAbnk , −−→ bbnk ⟩ ≥ 0. (9) Taking k → ∞ in (8), we get lim inf⟨ −−−→ bAbnk , −−→ bbnk ⟩ ≥ 0. Since A is uniformly continuous map, thus we have ϱ(Abnk , Asnk ) → 0 as k → ∞. (10) On the other, hand we have ⟨ −−−→ bAsnk , −−→ bsnk ⟩ = ⟨ −−−→ bAbnk , −−→ bsnk ⟩+ ⟨ −−−−−−→ Abnk Asnk , −−→ bsnk ⟩ = ⟨ −−−→ bAbnk , −−→ bbnk ⟩+ ⟨ −−−→ bAbnk , −−−−→ bnk snk ⟩+ ⟨ −−−−−−→ Abnk Asnk , −−→ bsnk ⟩, which together with (9) and (10) gives lim inf⟨ −−−→ bAsnk , −−→ bsnk ⟩ ≥ 0. Now, we have to prove that v ∈ V I(L, A). Since A is pseudo-monotone, so we get ⟨ −−→ vAb, −→ vb⟩ = lim k→∞ ⟨ −−−→ snk Ab, −−→ snk b⟩ = lim k→∞ inf⟨ −−−→ snk Ab, −−→ snk b⟩ ≥ 0. Using Lemma 12, we get v ∈ V I(L, A) and the proof is finished. 4. convergence results In this section we prove the results of strong and ∆-convergence. Theorem 2. Consider a nonempty, closed and convex subset of a Hadamard space Z be L. The map A : L → L is a pseudo-monotone, uniformly continuous on L. The solution set of the V I(L, A) is nonempty, that is V I(L, A) ̸= ϕ. Then any sequence {bn} ∆-convergent in V I(L, A). M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 12 of 24 Proof. Claim 1. The sequence {bn} is a bounded. To proof the claim, assume v ∈ V I(L, A), we have ϱ2(bn+1, v) = ϱ2(PLnbn, v) ≤ ϱ2(bn, v)− ϱ2(PLnbn, bn), (11) ϱ2(bn+1, v) ≤ ϱ2(bn, v)− ϱ2(bn,Ln). This implies that ϱ(bn+1, v) ≤ ϱ(bn, v). Thus limn→∞ ϱ(bn, v) exists. Therefore, the sequence {bn} is bounded and implies that the sequence {sn} is also bounded. Claim 2. [ 1 M (1− µ)ϱ2(bn, sn) ]2 ≤ ϱ2(bn, v)− ϱ2(bn+1, v), for some M > 0. Indeed, the bounded sequences {bn}, {sn} implies that {Abn}, {Asn} are also bounded, so for all n there exists M > 0 such that ξnd(bn, Abn) ≤ M. Therefore, for all u, q ∈ Z, we have ϱ(hn(u), hn(q)) = |hn(u)− hn(q)| , = ∣∣∣⟨−−→usn,−−−−−−−−−−−−−−−−−−→(ξnbn ⊕ (1− ξn)Abn)Abn⟩ −⟨−→qsn, −−−−−−−−−−−−−−−−−−→ (ξnbn ⊕ (1− ξn)Abn)Abn⟩ ∣∣∣ . By using Remark 1, we have ϱ(hn(u), hn(q)) = ∣∣∣⟨−→uq,−−−−−−−−−−−−−−−−−−→(ξnbn ⊕ (1− ξn)Abn)Abn⟩ ∣∣∣ . By Lemma 7 and Cauchy schwartz inequality, we get ⟨−→uq, −−−−−−−−−−−−−−−−−−→ (ξnbn ⊕ (1− ξn)Abn)Abn⟩ ≤ ξn⟨−→uq, −−−→ bnAbn⟩+ [ (1− ξn) ×⟨−→uq, −−−−−→ AbnAbn⟩ ] , = ξn⟨−→uq, −−−→ bnAbn⟩ ≤ ξnd(u, q)d(bn, Abn), ≤ Mϱ(u, q). Thus we have hn(.) is M-Lipschitz continuous on Z and by Lemma 11, we get ϱ(bn,Ln) ≥ 1 M hn(bn), (12) which, together with Lemma 13, we get ϱ(bn,Ln) ≥ 1 M (−1− µ)ϱ2(bn, sn). M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 13 of 24 Combining (11) and (12), ϱ2(bn+1, v) ≤ ϱ2(bn, v)− [ 1 M (−1− µ)ϱ2(bn, sn) ]2 . Thus, Claim 2 is proven. Claim 3 The sequence {bn} ∆-converges in V I(L, A). Indeed, by Lemma 4, there exists the subsequence {bnk } of bounded sequence {bn} such that the subsequence {bnk } ∆- converges to v ∈ Z. Using Claim 2, we can find lim n→∞ ϱ(bn, sn) = 0. It is implied from Lemma 14 that v ∈ V I(L, A). Therefore, we proved that: (i) limn→∞ ϱ(bn, v) exists, for every v ∈ V (L, A); (ii) Each ∆-limit of the sequence {bn} ∈ V I(L, A). Thus, by Lemma 9, the sequence {bn} is ∆-convergent in V I(L, A). Now we introduce an algorithm for strong convergence: Algorithm 2. Initialization: Given µ, ν, ς ∈ (0, 1). Let b1 be the arbitrary element of L. Iterative Steps: For the given iteration bn, first calculate bn+1 as stated below: Step 1. Compute sn = PL(ξnbn ⊕ (1− ξn)Abn), where ξn := ςνmn, with mn is the minimal nonnegative integer satisfying ⟨ −−−−−→ AbnAsn, −−→ bnsn⟩ ≤ µϱ2(bn, sn). If bn = sn or Asn = 0 then algorithm stops and sn is a solution of VI. Otherwise Step 2. Calculate bn+1 = ξnϖ(bn)⊕ (1− ξn)PLn(bn), where Ln := {b ∈ Z : hn(b) ≤ 0} and hn(b) = ⟨ −→ bsn, −−−−−−−−−−−−−−−−−−→ (ξnbn ⊕ (1− ξn)Abn)Abn⟩. Place n := n+ 1 and move to Step 1. Theorem 3. Consider a nonempty, closed and convex subset of a Hadamard space Z be L. The mapping A : L → L is a pseudo-monotone, uniformly continuous on Z. The solution set of VI is nonempty, that is V I(L, A) ̸= ∅. Let {ξn} be the sequences of real numbers in (0, 1) such that lim n→∞ ξn = 0,Σ∞ n=1ξn = ∞. Then any sequence {bn} converges strongly to v ∈ V I(L, A), where v = PV I(L,A)ϖ(v). M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 14 of 24 Proof. Claim 1. The sequence {bn} is bounded. To prove the claim, assume zn = PLn(bn), using Lemma 10, we have ϱ2(zn, v) = ϱ2(PLnbn, v) ≤ ϱ2(bn, v)− ϱ2(PLnbn, bn), according to Claim 1 in Theorem 1, we get ϱ2(zn, v) = ϱ2(PLnbn, v) ≤ ϱ2(bn, v)− [ 1 M (−1− µ)ϱ2(bn, sn) ]2 . This implies that ϱ(zn, v) ≤ ϱ(bn, v). (13) Consider ϱ(bn+1, v) = ϱ(ξnϖ(bn)⊕ (1− ξn)PLn(bn), v) and by using Lemma 2-(i),we get ϱ(ξnϖ(bn)⊕(1− ξn)PLn(bn), v) ≤ ξnϱ(ϖ(bn), v) + (1− ξn)ϱ(PLnbn, v), ≤ ξn(ϱ(ϖ(bn), ϖ(v) + ϱ(ϖ(v), v)) + [ (1− ξn)ϱ(PLnbn, v) ] , ≤ ξnϱ(ϖ(bn), ϖ(v)) + ξnϱ(ϖ(v), v) + [ (1− ξn)ϱ(PLnbn, v) ] , ≤ ξnρϱ(bn, v) + ξnϱ(ϖ(v), v) + (1− ξn)ϱ(zn, v), ≤ ξnρϱ(bn, v) + ξnϱ(ϖ(v), v) + (1− ξn)ϱ(bn, v), = (ξnρ+ (1− ξn))ϱ(bn, v) + ξnϱ(ϖ(v), v), = (1− ξn(1− ρ))ϱ(bn, v) + ξn(1− ρ) ϱ(ϖ(v), v) (1− ρ) , ≤ max { ϱ(bn, v), ϱ(ϖ(v), v) (1− ρ) } , ... ≤ max { ϱ(b1, v), ϱ(ϖ(v), v) (1− ρ) } , for 0 ≤ ξn(1− ρ) ≤ 1. Thus we proved the Claim 1. Claim 2. To prove ϱ2(zn, bn) ≤ ϱ2(bn, v)− ϱ2(bn+1, v) + 2ξn⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩. Let sn = ξnv ⊕ (1− ξn)zn. It follows from Lemmas 6 and 7 that ϱ2(bn+1, v) = ϱ2(ξnϖ(bn)⊕ (1− ξn)zn, v) ≤ ϱ2(sn, v) + 2⟨ −−−−→ bn+1sn, −−−→ bn+1v⟩, = [ϱ(ξnv + (1− ξn)zn, v)] 2 M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 15 of 24 +2⟨( −−−−−−−−−−−−−−−−−−→ ξnϖ(bn)⊕ (1− ξn)zn))sn, −−−→ bn+1v⟩, ϱ2(bn+1, v) ≤ [ξnd(v, v) + (1− ξn)ϱ(zn, v)] 2 + 2[ξn⟨ −−−−−→ ϖ(bn)sn, −−−→ bn+1v⟩ +(1− ξn)⟨−−→znsn, −−−→ bn+1v⟩], = (1− ξn) 2ϱ2(zn, v) + 2[ξn⟨ −−−−−−−−−−−−−−−−−−→ ϖ(bn)(ξnv ⊕ (1− ξn)zn), −−−→ bn+1v⟩ +(1− ξn)⟨ −−−−−−−−−−−−−−−→ zn(ξnv ⊕ (1− ξn)zn), −−−→ bn+1v⟩], ≤ (1− ξn) 2ϱ2(zn, v) + 2 [ ξ2n⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩+ { ξn(1− ξn) ×⟨ −−−−−→ ϖ(bn)zn, −−−→ bn+1v⟩ } + ξn(1− ξn)⟨−→znv, −−→ bn+1⟩+ { (1− ξn) 2 ×⟨−−→znzn, −−−→ bn+1v⟩ }] , = (1− ξn) 2ϱ2(zn, v) + 2ξ2n⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩ +2ξn(1− ξn)⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩, = (1− ξn) 2ϱ2(zn, v) + 2ξ2n⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩ +2ξn⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩ − 2ξ2n⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩, = (1− ξn) 2ϱ2(zn, v) + 2ξn⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩. ϱ2(bn+1, v) ≤ (1− ξn) d 2(zn, v) + 2ξn⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩ ϱ2(bn+1, v) ≤ ϱ2(zn, v) + 2ξn⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩. (14) On the other hand, we have ϱ2(zn, v) ≤ ϱ2(bn, v)− ϱ2(zn, bn), by putting above in (4.3), we have ϱ2(bn+1, v) ≤ ϱ2(bn, v)− ϱ2(bn, zn) + 2ξn⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩, ϱ2(bn, zn) ≤ ϱ2(bn, v)− ϱ2(bn+1, v) + 2ξn⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩. Claim 3. To prove (1− ξn) [ 1 M ξn(−1− µ)ϱ2(bn, sn) ]2 ≤ ϱ2(bn, v)− ϱ2(bn+1, v) + ξnϱ 2(ϖ(bn), v). Consider ϱ2(bn+1, v) = ϱ2(ξnϖ(bn)⊕ (1− ξn)zn, v) and by using Lemma 2-(ii), we have ϱ2(bn+1, v) ≤ ξnϱ 2(ϖ(bn), v) + (1− ξn)ϱ 2(zn, v) −ξn(1− ξn)ϱ 2(ϖ(bn), zn), ≤ ξnϱ 2(ϖ(bn), v) + (1− ξn)ϱ 2(zn, v). By using (13), we obtain ϱ2(bn+1, v) ≤ ξnϱ 2(ϖ(bn), v) + (1− ξn)ϱ 2(bn, v) M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 16 of 24 −(1− ξn) [ 1 M (−1− µ)ϱ2(bn, sn) ]2 , ≤ ξnϱ 2(ϖ(bn), v) + ϱ2(bn, v)− (1− ξn) × [ 1 M (−1− µ)ϱ2(bn, sn) ]2 . This implies that (1− ξn) [ 1 M (−1− µ)ϱ2(bn, sn) ]2 ≤ ξnϱ 2(ϖ(bn), v) + ϱ2(bn, v)− ϱ2(bn+1, v). Claim 4. To prove ϱ2(bn+1, v) ≤ (1− (1− ρ)ξn) 1− ξnρ ϱ2(bn, v) + 2ξn 1− ξnρ ⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩. Consider ϱ2(bn+1, v) ≤ (1− ξn)ϱ 2(zn, v) + 2ξn⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩. (15) By Cauchy schwartz inequality, we have ⟨ −−−−→ ϖ(bn)v, −−−→ bn+1v⟩ = ⟨ −−−−−−−→ ϖ(bn)ϖ(v), −−−→ bn+1v⟩+ ⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩, ≤ d(ϖ(bn), ϖ(v))d(bn+1, v) + ⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩, ≤ ρd(bn, v)d(bn+1, v) + ⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩, ≤ ρ 2 [ϱ2(bn, v) + ϱ2(bn+1, v)] + ⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩, by putting above in (15), we have ϱ2(bn+1, v) ≤ (1− ξn)ϱ 2(zn, v) + ξnρϱ 2(bn, v) + ξnρϱ 2(bn+1, v) + 2ξn⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩, ϱ2(bn+1, v)− ξnρϱ 2(bn+1, v) ≤ (1− ξn + ξnρ)ϱ 2(bn, v) + 2ξn⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩ (1− ξnρ)ϱ 2(bn+1, v) ≤ (1− (1− ρ)ξn)ϱ 2(bn, v) + 2ξn⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩, ϱ2(bn+1, v) ≤ (1− (1− ρ)ξn) 1− ξnρ ϱ2(bn, v) + 2ξn 1− ξnρ ⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩. Claim 5. The sequence ϱ2(bn, v) converges to zero. There are two scenarios for the proof of convergence of sequence {ϱ2(bn, v)}. Case 1. There exist a natural number N such that ϱ2(bn+1, v) ≤ ϱ2(bn, v) for all n ≥ N. This implies that limn→∞ ϱ2(bn, v) exist. From Claim 2, we have lim n→∞ ϱ2(bn, zn) = 0. M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 17 of 24 Now, according to Claim 3, lim n→∞ ϱ2(bn, sn) = 0. Since the sequence {bn} is bounded. It is implied by Lemma 4, that every bounded sequence in (Z, ϱ) always has a ∆-convergent subsequence, say bnk ∆-converges to z such that lim n→∞ sup⟨ −−−−→ ϖ(v)v, −→ bnv⟩ = lim n→∞ ⟨ −−−−→ ϖ(v)v, −−→ bnk v⟩ = ⟨ −−−−→ ϖ(v)v,−→zv⟩. (16) Since bnk ∆-converges to z and ϱ(bn, sn) = 0, it implies from Lemma14 that z ∈ V I(L, A). On the other hand, ϱ(bn+1, zn) = ϱ(ξnϖ(bn)⊕ (1− ξn)zn, zn) ≤ ξnϱ(ϖ(bn), zn) + (1− ξn)ϱ(zn, zn) = ξnϱ(ϖ(bn), zn) → 0 as n→ ∞. Thus, ϱ(bn+1, bn) ≤ ϱ(bn+1, zn) + ϱ(zn, bn) → 0 as n→ ∞. Since v = PV I(L,A)ϖ(v) and bnk ∆-converges to z ∈ V I(L, A), using (16), we get lim n→∞ sup⟨ −−−−→ ϖ(v)v, −→ bnv⟩ = ⟨ −−−−→ ϖ(v)v,−→zv⟩ ≤ 0. This implies that lim n→∞ sup⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩ ≤ lim n→∞ sup⟨ −−−−→ ϖ(v)v, −−−−→ bn+1bn⟩+ lim n→∞ sup⟨ −−−−→ ϖ(v)v, −→ bnv⟩ ≤ 0. From Claim 4 ϱ2(bn+1, v) ≤ 1− (ξn − 2ξnρ) 1− ξnρ ϱ2(bn, v) + 2ξn 1− ξnρ ⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩ = 1− (1− 2ρ)ξn 1− ξnρ ϱ2(bn, v) + 2 (1− 2ρ)ξn (1− ξnρ)(1− 2ρ) ⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩. Now, taking λn = (1− 2ρ)ξn 1− ξnρ ,ℑn = 2⟨ −−−−→ ϖ(v)v, −−−→ bn+1v⟩ 1− 2ρ , by Lemma 5, we can conclude that ϱ2(bn, v) = 0. ⇒ bn → v as n→ ∞. Case 2. There exist a subsequence {ϱ2(bnj , v)} of {ϱ2(bn, v)} such that ϱ2(bnj , v) < ϱ2(bnj+1 , v), for all j ∈ N. In this case, from Lemma 8 that there exist a nondecreasing sequence of natural numbers {ak} such that lim k→∞ ak = ∞, and the inequalities stated below holds for all values of k ∈ N: M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 18 of 24 (i) ϱ2(bak , v) ≤ ϱ2(bak+1 , v), (ii) ϱ2(bk, v) ≤ ϱ2(bak+1 , v). According to Claim 2, we have ϱ2(zak , bak) ≤ ϱ2(bak , v)− ϱ2(bak+1 , v) + 2ξak⟨ −−−−−→ ϖ(bak)v, −−−−→ bak+1 v⟩, ≤ ϱ2(bak+1 , v)− ϱ2(bak+1 , v) + 2ξak⟨ −−−−−→ ϖ(bak)v, −−−−→ bak+1 v⟩, ≤ ξak⟨ −−−−−→ ϖ(bak)v, −−−−→ bak+1 v⟩, ≤ ξakϱ(ϖ(bak), v)ϱ(bak+1 , v) → 0 as k → ∞. According to Claim 3, we have (1− ξak) [ 1 M ξn(−1− µ)ϱ2(bak , sak) ]2 ≤ ϱ2(bak , v)− ϱ2(bak+1 , v) + ξakϱ 2(ϖ(bak), v), ≤ ϱ2(bak+1 , v)− ϱ2(bak+1 , v) + ξakd 2(ϖ(bak), v), ≤ ξakϱ 2(ϖ(bak), v) → 0 as k → ∞. Using the same argument as in the proof of Case 1, we obtain ϱ(bak+1 , bak) → 0 and lim sup k→∞ ⟨ −−−−→ ϖ(v)v, −−−→ ba+1v⟩ ≤ 0. From Claim 4 ϱ2(bmk+1 , v) ≤ (1− (1− ρ)ξmk ) (1− ξmk ρ) ϱ2(bmk , v) + 2ξmk 1− ξmk ρ ⟨ −−−−→ ϖ(v)v, −−−−→ bmk+1 v⟩, since ϱ2(bk, v) ≤ ϱ2(bmk+1 , v), we have ϱ2(bmk+1 , v) ≤ (1− (1− ρ)ξmk ) 1− ξmk ρ ϱ2(bmk+1 , v) + 2ξmk 1− ξmk ρ ⟨ −−−−→ ϖ(v)v, −−−−→ bmk+1 v⟩, ξmk (1− 2ρ)ϱ2(bmk+1 , v) ≤ 2ξmk ⟨ −−−−→ ϖ(v)v, −−−−→ bmk+1 v⟩, ϱ2(bk, v) ≤ 2 1− 2ρ ⟨ −−−−→ ϖ(v)v, −−−−→ bmk+1 v⟩. Therefore, lim k→∞ sup ϱ2(bk, v) ≤ 0, that is, bk → v. M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 19 of 24 5. Some consequences and numerical illustrations In this section, we first derive 1 and 2 corollaries of Theorem 2 and 3 respectively. We then illustrate a numerical experiment to demonstrate the performance of our method. Corollary 1. Consider a nonempty, closed and convex subset of a Hadamard space Z be L. Let Z∗ be a metric space and A2 : L → Z∗, A1 : Z∗ → L. The map A2 is a uniformly continuous on L and A1 is uniformly continuous on A2(L), A is a pseudo-monotone on L. The solution set of the V I(L, A) is nonempty, that is V I(L, A) ̸= ϕ. Then any sequence bn is ∆-convergent in V I(L, A). Corollary 2. Let L be a nonempty, closed and convex subset of a Hadamard space Z. Consider the mappings A1 : L → L and A2 : L → L such that A1oA2 = A. The mapping A is a pseudo-monotone, uniformly continuous on Z. The solution set of VI is nonempty, that is V I(L, A) ̸= ∅. Let {ξn} be the sequences of real numbers in (0, 1) such that lim n→∞ ξn = 0,Σ∞ n=1ξn = ∞. Then the sequence {bn} converges strongly to v ∈ V I(L, A), where v = PV I(L,A)ϖ(v). Example 1. Let Z = R2,L = {u ∈ R2 : −1 ≤ ui ≤ 1, i = 1, 2}. Define the maps A1 : L → (R2)∗, and A2 : (R2)∗ → L such that A1(b) = f, f ∈ (R2)∗, A2(b) = s′ respectively. Let b1 = (0.1, 0.2) ∈ L be arbitrary and define gi : R2 → R such that g1(b1, b2) = cos(b1), g2(b1, b2) = cos(b2). Then we have Ab1 = ( cos(b1) + cos(b2) 2 , cos(b1)− cos(b2) 2 ) = ( cos(0.1) + cos(0.2) 2 , cos(0.1)− cos(0.2) 2 ) = (0.98754, 0.00747) . Step 1 For n = 1, and ξ1 = 0.2, s1 = PL(ξ1b1 ⊕ (1− ξ1)Ab1) = PL((0.2)(0.1, 0.2)⊕ (1− 0.2)(0.98754, 0.00747)) = PL((0.02, 0.04)⊕ (0.790032, 0.005976)) = PL(0.810032, 0.045976). As (0.810032, 0.045976) ∈ L, So it’s unique nearest point in L is (0.810032, 0.045976) implies s1 = (0.810032, 0.045976). As1 = ( 0.68948 + 0.99894 2 , 0.68948− 0.99894 2 ) = (0.84421,−0.15473). M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 20 of 24 Now to prove ⟨ −−−−−→ Ab1As1, −−→ b1s1⟩ ≤ µϱ2(b1, s1). (17) Consider left hand side of (17), ⟨ −−−−−→ Ab1As1, −−→ b1s1⟩ = 1 2 [ ϱ2(Ab1, s1) + ϱ2(As1, b1)− ϱ2(Ab1, b1) −ϱ2(ΠAy1, y1) ] = −0.0932945 and ϱ2(b1, s1) = (0.1− 0.810032)2 + (0.2− 0.045976)2 = 0.527873 µϱ2(b1, s1) = 0.2(0.527873) = 0.1055746 Thus, we have ⟨ −−−−−→ Ab1As1, −−→ b1s1⟩ ≤ µϱ2(b1, s1). Step 2 Now, for b2 = PLn(b1), hn(b) = ⟨ −→ bsn, −−−−−−−−−−−−−−−−−−→ (ξnbn ⊕ (1− ξn)Abn)Abn⟩ and Ln = {b ∈ Z : hn(b) ≤ 0}. Let b = (µ1, µ2), h1(b) = 0. Then h1(b) = ⟨ −→ bs1, −−−−−−−−−−−−−−−−−→ (ξ1b1 ⊕ (1− ξ1)Ab1)Ab1⟩ = 0 1 2 [ d2(b, Ab1) + d2(s1, ξ1b1 ⊕ (1− ξ1)Ab1) −d2(b, ξ1b1 ⊕ (1− ξ1)Ab1)− d2(s1, Ab1) ] = 0 −0.355016µ1 + 0.77012µ2 + 0.2845414 = 0. (18) From (18), we have µ1 = 0.21993µ2+0.801164. If µ1 = 0, then µ2 = 3.6428 and if µ2 = 0, then µ1 = 0.801164. Since b2 = {s′ ∈ L1 : ϱ(b1, s ′) = ϱ(b1,L1)} and ϱ((0.1, 0.2),L1) = −0.355016(0.1) + 0.77012(0.2) + 0.2845414 2 √ (−0.355016)2 + (0.77012)2 = 0.727949 Let s′(µ1, µ2)t. Then s′ is point of intersection of line L1 with the perpendicular line. The equation of perpendicular line is µ1 = − 1 0.21993 µ2 + 0.801164 (19) If µ1 = 0, then µ2 = 0.176199 and if µ2 = 0, then µ1 = 0.801164. To find point of intersection, subtracting (18) and (19), we get µ2 = 0, µ1 = 0.801164 and s′ = b2 = (0.801164, 0). M. Rashid et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6680 21 of 24 Example 2. Let L = {(x1, 0, 0, . . . ) ∈ ℓ2 : 0 ≤ x1 ≤ 1}, which is convex (if x = (x1, 0, . . . ), y = (y1, 0, . . . ) ∈ L and λ ∈ [0, 1], then λx+ (1− λ)y = ( λx1 + (1− λ)y1, 0, . . . ) ∈ L. For a concrete test, we work with a simple diagonal operator A1 : L → ℓ2 given by A1(x) = ( αnxn ) , αn = 1 n+ 1 (so α1 = 1 2), and A2(y) = PL(y). Choose ξn = ξ = 0.2 and an initial vector b1 ∈ L with first coordinate x1 = 0.8. The orthogonal projection PL onto set L can be described explicitly PL(v) = (π[0,1](v1), 0, 0, . . . ), so the Algorithm 1 update sn = PL(ξbn + (1 − ξ)Abn), the entire iteration is governed solely by the first coordinate, leading to a simple one-dimensional recurrence relation. For the purpose of demonstration we will take the simplified update bn+1 = sn, where sn = PL(ξbn + (1− ξ)Abn). With ξ = 0.2 and α1 = 1/2 one gets xn+1 = 0.6xn, x1 = 0.8, hence xn = 0.6n−1x1 and bn = (xn, 0, 0, . . . ) → 0 strongly in ℓ2. This worked-out example provides a straightforward demonstration in the infinite-dimensional framework (imple- mented by truncation for computations) validating the convergence behaviour asserted in Theorem 2. Conclusion 1. The variational inequality is defined for asymptotically nonexpansive map- ping in CAT (0) spaces. We acquire strong and ∆-convergence of two projection type methods for variational inequality problem using pseudo-monotone mapping. 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