Electronic Journal of Differential Equations, Vol. 2022 (2022), No. 03, pp. 1–10. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu DUALITY ARGUMENTS FOR WELL-POSEDNESS OF HISTORY-DEPENDENT VARIATIONAL INEQUALITIES RONG HU, MIRCEA SOFONEA Abstract. In this article we introduce a concept of dual problems in metric spaces. Then we state and prove an equivalence result concerning their well- posedness with respect to appropriate Tykhonov triples. We exemplify this result in the study of a history-dependent variational inequality with time- dependent constraints, for which the dual problem is in a form of a history- dependent inclusion. This allows us to deduce a convergence result which pro- vides the continuous dependence of the solution with respect to the data. We end this paper with an example which represents an evidence of our abstract results. 1. Introduction The concept of well-posedness with respect to a Tykhonov triple was introduced in [14]. It extends the concept of well-posedness for a minimization problem, in- troduced in the pioneering work [12] as well as the concepts of well-posedness used in [1, 8, 15] for various optimization problems and [2, 3, 4, 5, 6, 7, 13] for vari- ous classes of inequalities. This abstract concept was applied in [10] in the study of variational inequalities governed by a history-dependent operator, the so-called history-dependent variational inequalities. There, the well-posedness with respect to several Tykhonov triples was studied and a strategy which allows to deduce con- vergence results was discussed. An existence and uniqueness result for a class of history-dependent inclusions was obtained in the recent paper [9]. This article represents a continuation of [9, 10, 14]. Here, we complete the theoretical study initiated in [14] by studying the well-posedness of a couple of problems in duality. The idea is to deduce the well-posedness of a problem P by using the well-posedness of a different problem Q, called the dual of P. The interest in this method arises in the fact that in several cases we have a number of results in the study of Problem Q which can be useful in the analysis of Problem P. This represents the first trait of novelty of the current paper. The second novelty is that we use these arguments in the study of a history-dependent variational inequality which is more general than the inequality in [10]. Indeed, in contrast with [10], the inequality we consider in this paper involves a time-dependent set of constraints. As a consequence, the dual problem of this inequality is given by a history-dependent inclusion, already studied in [9]. 2010 Mathematics Subject Classification. 35M86, 49J40, 47J20. Key words and phrases. History-dependent variational inequality; dual problem; history-dependent inclusion; Tykhonov well-posedness; convergence results. ©2022. This work is licensed under a CC BY 4.0 license. Submitted August 19, 2021. Published January 6, 2022. 1 2 R. HU, M. SOFONEA EJDE-2022/03 The rest of this article is structured as follows. In Section 2 we introduce the concept of dual problems in metric spaces. Then we state and prove an equivalence result, Theorem 2.2. In Section 3 we introduce a history-dependent variational inequality P, then we use Theorem 2.2 to prove its well-posedness. We complete our study in Section 4 where we prove a convergence result which states the continuous dependence of the solution of P with respect to the data. We end this section by recalling some basic definitions which will be crucial in the rest of the paper. Consider an abstract mathematical objectM, called generic “problem”, defined in a metric space (Z, d). Problem M could be an equation, a minimization problem, a fixed point problem, an inclusion or an inequality problem, for instance. We associate to Problem M the concept of “solution” which follows from the context. The concept of well-posedness for ProblemM is provided by the following definition. Definition 1.1. (a) A Tykhonov triple is a mathematical object of the form T = (I,Ω, C) where I is a given nonempty set, Ω : I → 2Z is a nonempty set-valued operator and C is a nonempty subset of sequences with elements in I. (b) Given a Tykhonov triple T = (I,Ω, C), a sequence {zn} ⊂ Z is called a T -approximating sequence if there exists a sequence {θn} ∈ C such that zn ∈ Ω(θn) for each n ∈ N. (c) Given a Tykhonov triple T = (I,Ω, C), Problem M is said to be T -well- posed if it has a unique solution and every T -approximating sequence con- verges in Z to its solution. For a Tykhonov triple T = (I,Ω, C) we refer to I as the set of parameters; the family of sets {Ω(θ)}θ∈I represents the family of approximating sets; besides, we say that C defines the criterion of convergence. We remark that approximating sequences always exist since, by assumption, C 6= ∅ and, moreover, for any sequence {θn} ∈ C and any n ∈ N, the set Ω(θn) is not empty. In addition, we recall that the concept of approximating sequence depends on the Tykhonov triple T and, for this reason, we use the terminology “T -approximating sequence”. As a consequence, the concept of well-posedness depends on T and, therefore, we refer to it as “well- posedness with respect to T ” or “T -well-posedness”, for short. 2. An abstract equivalence result Consider two abstract problems P and Q formulated in the metric spaces (U, dU ) and (Σ, dΣ), respectively. We start with the following definition. Definition 2.1. Problems P and Q are said to be dual of each other if there exists a mapping D : U → Σ such that: (a) D is bijective; (b) both D : U → Σ and its inverse D−1 : Σ→ U are continuous; (c) u ∈ U is solution of Problem P if and only if σ = Du is solution of Problem Q. If (a)–(c) hold we say that Problem Q is a dual problem of Problem P (with D) and, conversely, Problem P is a dual problem of Problem Q (with D−1). In this framework we consider two Tykhonov triples TP = (I,ΩP , C) and TQ = (I,ΩQ, C) EJDE-2022/03 DUALITY ARGUMENTS FOR WELL-POSEDNESS 3 in which the approximating sets ΩP : I → 2U and ΩQ : I → 2Σ are such that ΩP (θ) = {u ∈ U : Du ∈ ΩQ(θ)}, ΩQ(θ) = {σ ∈ Σ : D−1σ ∈ ΩP (θ)}, (2.1) for all θ ∈ I. The interest in considering these Tykhonov triples follows from the following equivalence result. Theorem 2.2. Let P and Q be dual problems (with D and D−1, respectively). Then Problem P is TP -well-posed if and only if Problem Q is TQ-well-posed. Proof. Assume that Problem P is TP -well-posed. This implies that Problem P has a unique solution u ∈ U . Moreover, it follows from properties (a) and (c) in Definition 2.1 that σ = Du is the unique solution of Problem Q. Let {σn} ⊂ Σ be a TQ-approximating sequence for Problem Q. Then there exists a sequence {θn} ∈ C such that, for each n ∈ N, σn ∈ ΩQ(θn), i.e., D−1σn ∈ ΩP (θn), which means that {D−1σn} ⊂ U is a TP -approximating sequence for Problem P. Therefore, using the TP -well-posedness of Problem P and the continuity of operator D to deduce that the sequence {σn} converges to the unique solution σ ∈ Σ of Problem Q. We conclude from above that Problem Q is TQ-well-posed. Similar arguments show that if Problem Q is TQ-well-posed then Problem P is TP -well-posed, which completes the proof. � 3. Problem statement and its well-posedness Everywhere in the rest of this article V will be a real Hilbert space. We use (·, ·)V and ‖ · ‖V for the inner product and the associated norm of space V . For any nonempty closed convex set K ⊂ X we denote by PK : V → K and NK : V → 2V the projection operator on K and the outward normal cone of K, respectively. Moreover, we use notation X = C([0, T ];V ) for the space of continuous functions on [0, T ] with values in V , equipped with the norm of the uniform convergence. Finally, we mention that, unless stated otherwise, all the limits below are considered as n→∞, even if we do not mention it explicitly. Let K : [0, T ]→ 2V , A : V → V , S : C([0, T ];V )→ C([0, T ];V ) and f : [0, T ]→ V . With these data we consider the following time-dependent inequality. Problem P. Find a function u ∈ C([0, T ];V ) such that the following inequality holds: u(t) ∈ K(t) and (Au(t), v − u(t))V + (Su(t), v − u(t))V ≥ (f(t), v − u(t))V (3.1) for all v ∈ K(t) and t ∈ [0, T ]. Note that here and below when no confusion arises, we use the shorthand notation Su(t) to represent the value of the function Su at the point t, i.e., Su(t) = (Su)(t). Moreover, for an element v ∈ V we shall still write v for the constant function t 7→ v for all t ∈ [0, T ] and, therefore, notation Sv used below in this section defines an element of X. Our aim in what follows is to associate Problem P with a dual problem Q and to deduce its well-posedness with respect to a specific Tykhonov triple. To this end, we consider the following hypotheses. (H1) (a) K : [0, T ]→ 2V has nonempty closed and convex values. (b) For each u ∈ V , t ∈ [0, T ] and each sequence {tn} ⊂ [0, T ], tn → t implies PK(tn)u→ PK(t)u in V . 4 R. HU, M. SOFONEA EJDE-2022/03 (H2) A : V → V is a linear continuous and coercive operator, i.e. there exist LA, mA > 0 such that ‖Au‖V ≤ LA‖v‖V , (Au, u)V ≥ mA‖u‖2V ∀u ∈ V. (H3) S : C([0, T ];V ) → C([0, T ];V ) is a history-dependent operator, i.e., there exists LS > 0 such that ‖Su(t)− Sv(t)‖V ≤ LS ∫ t 0 ‖u(s)− v(s)‖V ds ∀u, v ∈ C([0, T ];V ), t ∈ [0, T ]. (H4) f ∈ C([0, T ];V ). Note that assumption (H2) implies that the operator A is invertible and its inverse A−1 : V → V satisfies the inequalities ‖A−1u‖V ≤ 1 mA ‖u‖V , (A−1u, u)V ≥ mA L2 A ‖u‖2V ∀u ∈ V. (3.2) Next, we consider the Tykhonov triple TP = (I,ΩP , C) defined as follows: I = R+, C = {{θn}n : θn ∈ I ∀n ∈ N, θn → 0 as n→∞} (3.3) and, for each θ ≥ 0, the set ΩP (θ) is defined as follows: ΩP (θ) = { u ∈ C([0, T ];V ) : u(t) ∈ K(t), (Au(t), v − u(t))V + (Su(t), v − u(t))V + θ ≥ (f(t), v − u(t))V ∀ v ∈ K(t), t ∈ [0, T ] } . (3.4) Note that ΩP (θ) 6= ∅ for each θ ∈ R+, as it will result from the proof of Theorem 3.2 below. To proceed, we need the following preliminary result. Proposition 3.1. Under assumptions (H2)–(H4), the operator D : C([0, T ];V )→ C([0, T ];V ) defined by Du(t) = Au(t) + Su(t)− f(t) ∀u ∈ C([0, T ];V ), t ∈ [0, T ] (3.5) is bijective and has inverse of the form A−1+R, where R : C([0, T ];V )→ C([0, T ];V ) is a history-dependent operator with constant LR > 0. A proof of Proposition 3.1 can be found in [11, p.55]. Based on Proposition 3.1 we consider the following problem. Problem Q. Find a function σ ∈ C([0, T ];V ) such that the following inclusion holds: − σ(t) ∈ NK(t)(A −1σ(t) +Rσ(t)) ∀ t ∈ [0, T ]. (3.6) Our main result in this section is the following. Theorem 3.2. Assume (H1)–(H4) hold. Then Problem P is TP -well-posed. Proof. The proof will be carried out in several steps, as follows. Step 1. We prove that Problems P and Q are dual problems. Indeed, since ‖v‖X = max t∈[0,T ] ‖v(t)‖V ∀ v ∈ C([0, T ];V ), it is easy to see that any history-dependent operator H : C([0, T ];V )→ C([0, T ];V ) is continuous. Therefore, Proposition 3.1 implies that the operator D defined by (3.5) satisfies conditions (a) and (b) in Definition 2.1 with U = Σ = C([0, T ];V ). EJDE-2022/03 DUALITY ARGUMENTS FOR WELL-POSEDNESS 5 Assume now that u ∈ C([0, T ];V ) is a solution of Problem P, i.e., u(t) ∈ K(t) and (f(t)−Au(t)− Su(t), v − u(t))V ≤ 0 ∀ v ∈ K(t), t ∈ [0, T ]. (3.7) This inequality is equivalent to the inclusion f(t)−Au(t)− Su(t) ∈ NK(t)(u(t)) ∀ t ∈ [0, T ]. (3.8) We now take σ(t) = Du(t) for all t ∈ [0, T ], then we use definition (3.5) and Proposition 3.1 to see that σ(t) = Au(t) + Su(t)− f(t), u(t) = A−1σ(t) +Rσ(t) ∀ t ∈ [0, T ]. (3.9) Therefore, from (3.8) and (3.9) we deduce that − σ(t) ∈ NK(t)(A −1σ(t) +Rσ(t)) ∀ t ∈ [0, T ], (3.10) which shows that σ is a solution of Problem Q. Conversely, if σ ∈ C([0, T ];V ) is a solution of (3.10) then it is easy to see that the function u(t) = D−1σ(t) for all t ∈ [0, T ] is a solution of inequality (3.7). We deduce from here that condition (c) in Definition 2.1 is also satisfied. There- fore, P and Q are dual problems, which concludes step 1. Step 2. We now construct a Tykhonov triple TQ and prove the TQ-well-posedness of Problem Q. First, we recall that the existence of a unique solution to Problem Q follows from a recent result proved in [9]. Next, we use notation (3.3) and define the Tykhonov triple TQ = (I,ΩQ, C) as follows: ΩQ(θ) = { σ ∈ C([0, T ];V ) : A−1σ(t) +Rσ(t) ∈ K(t), (A−1σ(t) +Rσ(t)− v, σ(t))V ≤ θ ∀ v ∈ K(t), t ∈ [0, T ] } (3.11) for each θ ≥ 0. Let σ ∈ C([0, T ];V ) be the solution of Problem Q. Then A−1σ(t) +Rσ(t) ∈ K(t), (A−1σ(t) +Rσ(t)− v, σ(t))V ≤ 0 ∀ v ∈ K(t), t ∈ [0, T ]. (3.12) This implies that σ ∈ ΩQ(θ) for each θ ∈ R+ and, therefore, TQ is a Tykhonov triple. Let t ∈ [0, T ] and let {σn} ⊂ C([0, T ];V ) be a TQ-approximating sequence for Problem Q. Then there exists a sequence {θn} ∈ C such that, for each n ∈ N, σn ∈ ΩQ(θn). We now use (3.11) and (3.12) to deduce that (A−1σ(t)−A−1σn(t), σ(t)− σn(t))V ≤ θn + (Rσ(t)−Rσn(t), σn(t)− σ(t))V . Using (3.2) and the history-dependence of the operator R we find that mA−1‖σ(t)− σn(t)‖2V ≤ θn + LR (∫ t 0 ‖σ(s)− σn(s)‖V ds ) ‖σ(t)− σn(t)‖V where, here and below, mA−1 = mA L2 A . Therefore, the elementary inequality x2 ≤ ax+ b =⇒ x ≤ a+ √ b ∀x, a, b > 0 implies that ‖σ(t)− σn(t)‖V ≤ ( θn mA−1 )1/2 + LR mA−1 ∫ t 0 ‖σ(s)− σn(s)‖V ds 6 R. HU, M. SOFONEA EJDE-2022/03 and, using Gronwall’s argument, we find that ‖σ(t)− σn(t)‖V ≤ ( θn mA−1 )1/2 e LR m A−1 t . Next, since θn → 0 we deduce that σn → σ in C([0, T ];V ). We conclude from here that Problem Q is TQ-well-posed. Step 3. Completion of the proof. Let θ ∈ R+, u ∈ C([0, T ], V ) and let σ = Du. Then, using arguments similar to those used to prove that σ satisfies inequality (3.10) if and only if u satisfies (3.7), it is easy to see that u ∈ ΩP (θ) ⇐⇒ σ ∈ ΩQ(θ). We conclude from here that ΩP (θ) 6= ∅ and, moreover, (2.1) holds. It follows now from Steps 1 and 2 that we are in a position to use Theorem 2.2 to conclude the proof of Theorem 3.2. � 4. A convergence result In this section we use the well-posedness of Problem P with respect to the Tykhonov triple TP to deduce a continuous dependence result of the solution with respect to the data. To this end we assume that (H1) holds and we consider three sequences {An}, {Sn} and {fn} such that, for each n ∈ N, the following conditions hold. (H5) An : V → V satisfies condition (H2) with mn > 0. (H6) Sn : C([0, T ];V ) → C([0, T ];V ) is a history-dependent operator, i.e., there exists Ln > 0 such that ‖Snu(t)− Snv(t)‖V ≤ Ln ∫ t 0 ‖u(s)− v(s)‖V ds for all u, v ∈ C([0, T ];V ) and t ∈ [0, T ]. (H7) fn ∈ C([0;T ];V ). Then we consider the following variational problem. Problem Pn. Find a function un ∈ C([0, T ];V ) such that the following inequality holds: un(t) ∈ K(t), and (Anun(t), v − un(t))V + (Snun(t), v − un(t))V ≥ (fn(t), v − un(t))V (4.1) for all v ∈ K(t) and t ∈ [0, T ]. Then the arguments in Section 2 imply that Problem Pn has a unique solution, for each n ∈ N. Assume now that (H8) There exists u0 ∈ V such that u0 ∈ K(t) for all t ∈ [0, T ]. (H9) (a) For each n ∈ N there exists αn > 0 such that ‖Anu−Au‖V ≤ αn(‖u‖V + 1) ∀u ∈ V. (b) αn → 0 as n→∞. (H10) (a) For each n ∈ N there exists βn > 0 such that ‖Snu(t)− Su(t)‖V ≤ βn (∫ t 0 ‖u(s)‖V ds+ 1 ) for all u ∈ C([0, T ];V ) and t ∈ [0, T ]. (b) βn → 0 as n→∞. (H11) fn → f in C([0, T ];V ). EJDE-2022/03 DUALITY ARGUMENTS FOR WELL-POSEDNESS 7 The main result of this section is the following. Theorem 4.1. Assume (H1)–(H11) hold. Then, the solution un of Problem Pn converges to the solution u of Problem P, i.e., un → u in C([0, T ];V ). (4.2) Proof. The proof is carried out in three steps, as follows. Step 1. We prove that there exists M > 0 such that ‖un(t)‖V ≤M ∀n ∈ N, t ∈ [0, T ]. (4.3) Let n ∈ N and t ∈ [0, T ]. Then, using (4.1) with v = u0 ∈ K(t) we find that (Anun(t), un(t)− u0)V ≤ (Snun(t), u0 − un(t))V + (fn(t), un(t)− u0)V . (4.4) We write (Anun(t), un(t)− u0)V = (Anun(t)−Aun(t), un(t)− u0)V + (Aun(t)−Au0, un(t)− u0)V + (Au0, un(t)− u0)V , then we use Cauchy-Schwarz inequality and assumptions (H2), (H9)(a) to see that (Anun(t), un(t)− u0)V ≥ −(αn(‖un(t)‖V + 1))‖un(t)− u0‖V +mA‖un(t)− u0‖2V − ‖Au0‖V ‖un(t)− u0‖V . (4.5) Note that (Snun(t), u0 − un(t))V = (Snun(t)− Sun(t), u0 − un(t))V + (Sun(t)− Su0(t), u0 − un(t))V + (Su0(t), u0 − un(t))V . Then, it follows from assumptions (H3) and (H10)(a) that (Snun(t), u0 − un(t))V ≤ βn (∫ t 0 ‖un(s)‖V ds+ 1 ) ‖un(t)− u0‖V + LS (∫ t 0 ‖un(s)− u0‖V ds ) ‖un(t)− u0‖V + ‖Su0(t)‖V ‖un(t)− u0‖V . (4.6) Moreover, we have (fn(t), un(t)− u0)V = (fn(t)− f(t), un(t)− u0)V + (f(t), un(t)− u0)V ≤ ‖fn(t)− f(t)‖V ‖un(t)− u0‖V + ‖f(t)‖V ‖un(t)− u0‖V . (4.7) We now combine inequalities (4.4)–(4.7) to see that mA‖un(t)− u0‖V ≤ βn (∫ t 0 ‖un(s)‖V ds+ 1 ) + LS ∫ t 0 ‖un(s)− u0‖V ds + ‖Su0(t)‖V + ‖fn(t)− f(t)‖V + ‖f(t)‖V + αn(‖un(t)‖V + 1) + ‖Au0‖V , which implies that (mA − αn)‖un(t)‖V 8 R. HU, M. SOFONEA EJDE-2022/03 ≤ βn (∫ t 0 ‖un(s)‖V ds+ 1 ) + LS ∫ t 0 ‖un(s)‖V ds+ LST‖u0‖V + ‖Su0(t)‖V + ‖fn(t)− f(t)‖V + ‖f(t)‖V + αn + ‖Au0‖V +mA‖u0‖V . Using assumptions (H9)(b), (H10)(b) and (H11), there exist N0 ∈ N and two positive constants C0 and C1 such that ‖un(t)‖V ≤ C0 + C1 ∫ t 0 ‖un(s)‖V ds for all n ≥ N0. It follows from Gronwall’s inequality that ‖un(t)‖V ≤ C0e C1t. This inequality completes the proof of (4.3). Step 2. We prove that the sequence {un} ⊂ C([0, T ];V ) is a TP -approximating sequence for Problem P. Let n ∈ N, t ∈ [0, T ] and v ∈ K(t). We write (Aun(t), v − un(t))V + (Sun(t), v − un(t))V − (f(t), v − un(t))V = (Anun(t), v − un(t))V + (Snun(t), v − un(t))V − (fn(t), v − un(t))V + (Aun(t)−Anun(t), v − un(t))V + (Sun(t)− Snun(t), v − un(t))V + (fn(t)− f(t), v − un(t))V , then we use inequality (4.1) to find that (Aun(t), v − un(t))V + (Sun(t), v − un(t))V ≥ (f(t), v − un(t))V − ‖Aun(t)−Anun(t)‖V ‖v − un(t)‖V − ‖Sun(t)− Snun(t)‖V ‖v − un(t)‖V − ‖fn(t)− f(t)‖V ‖v − un(t)‖V . Therefore, (H9)(a), (H10)(a) and (4.3) imply that there exists a constant C > 0 such that (Aun(t), v − un(t))V + (Sun(t), v − un(t))V + C(αn(M + 1) + βn(MT + 1) + ‖fn(t)− f(t)‖V ) ≥ (f(t), v − un(t))V . (4.8) We define θn = C(αn(M + 1) + βn(MT + 1) + ‖fn(t)− f(t)‖V ), (4.9) and combine (4.8), (4.9), (3.4) to see that un ∈ ΩP (θn). Moreover, (H9)(b), (H10)(b) and (H11) imply that θn → 0 as n → ∞. It follows from here that {un} is a TP -approximating sequence for Problem P. Step 3. Completion of the proof. We use Theorem 3.2 and Definition 1.1(c) to deduce the convergence (4.2), which concludes the proof. � We end this section with the following example in which V = R. Example 4.2. Consider Problem P in the particular case when K(t) = [0, 2− e−t] ∀ t ∈ [0, T ], Au = u ∀u ∈ V, Su(t) = ∫ t 0 u(s) ds ∀ t ∈ [0, T ], u ∈ C([0, T ];V ), f(t) = t ∀ t ∈ [0, T ]. EJDE-2022/03 DUALITY ARGUMENTS FOR WELL-POSEDNESS 9 Note that in this case inequality (3.1) becomes: u(t) ∈ K(t) and( u(t) + ∫ t 0 u(s) ds− t )( v − u(t) ) ≥ 0 ∀ v ∈ K(t), t ∈ [0, T ]. (4.10) Assume that Anu = n+ 1 n u ∀u ∈ V, Snu(t) = n+ 1 n ∫ t 0 u(s) ds ∀ t ∈ [0, T ], u ∈ C([0, T ];V ), fn(t) = t+ 1 n ∀ t ∈ [0, T ], for each n ∈ N. Then, inequality (4.1) becomes: un(t) ∈ K(t) and(n+ 1 n un(t) + n+ 1 n ∫ t 0 un(s) ds− t− 1 n )( v − un(t) ) ≥ 0 (4.11) for all v ∈ K(t) and t ∈ [0, T ]. It is easy to see that in this particular case assumptions (H1)–(H11) are satisfied. Therefore, Theorems 3.2 and 4.1 guarantee the unique solvability of inequalities (4.10) and (4.11) as well as the convergence (4.2). This convergence can be proved directly. Indeed, consider the integral equation u(t) + ∫ t 0 u(s) ds = t ∀ t ∈ [0, T ]. The solution of this equation is u(t) = 1− e−t ∀ t ∈ [0, T ]. (4.12) and, since 0 ≤ 1 − e−t ≤ 2 − e−t for all t ∈ [0, T ], we deduce that the function (4.12) is also the solution of the history-dependent inequality (4.10). Next, using a similar argument based on the solvability of the integral equation n+ 1 n un(t) + n+ 1 n ∫ t 0 un(s) ds = t+ 1 n ∀ t ∈ [0, T ], we find that the solution of the history-dependent variational inequality (4.11) is un(t) = n n+ 1 − n− 1 n+ 1 e−t ∀ t ∈ [0, T ]. (4.13) Then, a simple calculation shows that |un(t)− u(t)| ≤ 3 n+ 1 ∀ t ∈ [0, T ], which implies the convergence (4.2). Acknowledgments. 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Rong Hu School of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu, Sichuan 611731, China Email address: hrong1130@foxmail.com Mircea Sofonea Laboratoire de Mathématiques et Physique, University of Perpignan Via Domitia, 52 Avenue Paul Alduy, 66860 Perpignan, France Email address: sofonea@univ-perp.fr 1. Introduction 2. An abstract equivalence result 3. Problem statement and its well-posedness 4. A convergence result Acknowledgments References