We deal with a constrained quasivariational inequality under a general form. We study existence of solutions in two situations depending on whether the set of constraints is bounded or possibly unbounded.

1. Introduction and Statement of Main Results

Let X be a real reflexive and separable Banach space assumed to be compactly embedded in a Banach space Y. We denote by X^* the dual space of X, by Y^* the dual space of Y, by 〈·, ·〉 _X the duality brackets between X^* and X, by 〈·, ·〉 _Y the duality brackets between Y^* and Y, by ∥·∥_X the norm of X, and by ∥·∥_Y the norm of Y. Given a function ψ : X → ℝ ∪ {+∞}, we denote by D(ψ): = {x ∈ X : ψ(x)<+∞} the effective domain of ψ.

In this paper we deal with the following problem

()

We describe the data entering problem (1):

(i)
K ⊂ X is a nonempty, convex, closed subset;
(ii)
A : X → X^* is a (possibly nonlinear) operator;
(iii)
Φ : X × X → ℝ ∪ {+∞} is such that, for all η ∈ K, the function Φ(η, ·) : X → ℝ ∪ {+∞} is convex with K∩D(Φ(η, ·)) ≠ ∅; moreover, we will denote by ∂Φ(η, ·) the convex subdifferential of Φ(η, ·); that is,
()
(iv)
J : Y → ℝ is a locally Lipschitz function, and the notation J⁰ stands for its generalized directional derivative in the sense of Clarke [1]; that is,
()
In addition, we will denote by ∂J the generalized gradient of J; that is,
()
(v)
f ∈ X^*.

Problem (1) is called a constrained quasivariational problem. Typically, we can choose X to be the Sobolev space defined as the closure of in H¹(Ω) for a bounded domain Ω ⊂ ℝ^N (N ≥ 1), Y to be the Lebesgue space L^p(Ω) for 1 ≤ p < 2^* (where 2^* = +∞ if N ∈ {1,2} and 2^* = 2N/(N − 2) if N ≥ 3), , A = −Δ (the negative Laplacian operator), Φ(u, v) = ∫_Ω g(u, v)dx where g : ℝ² → ℝ₊ is convex in the second variable (then ), and J(u) = ∫_Ω j(x, u(x))dx where j : Ω × ℝ → ℝ is locally Lipschitz in the second variable. Constrained quasivariational problems were extensively studied; we refer, for example, to [2–5] and to the references therein. We point out three aspects which make our approach natural and general. First, we deal with the general setting of a pair of Banach spaces (X, Y) instead of focusing on spaces of functions; in particular, our results can be applied to problems with different boundary conditions. Second, the set of constraints K may be unbounded. Third, the form of the studied problem allows both variational and hemivariational constraints as it involves both a convex term Φ(u, ·) and a generalized directional derivative J⁰; this type of problems models important processes in mechanics and engineering (see [6, 7]).

In this paper, we consider the following hypotheses on the data described above:

(H₁) for every sequence {u_n} _n≥1 ⊂ K with u_n⇀u in X, for some u ∈ K, one has
()
(H₂) whenever {(η_n, u_n)} _n≥1 ⊂ (K × K)∩D(Φ), η_n⇀η in X, u_n⇀u in X, one has (η, u)∈(K × K)∩D(Φ) and
()
(H₃) given η ∈ K, if u₁, u₂ ∈ K satisfy (η, u₁) ∈ D(Φ), (η, u₂) ∈ D(Φ) and
()
then u₁ = u₂.

Remark 1. We emphasize certain situations when hypotheses (H₁)– (H₃) are satisfied.

(a) Hypothesis (H₁) is satisfied, for instance, if A is weakly strongly continuous, that is, A is continuous from X endowed with the weak topology to X^* endowed with the norm topology.

(b) Note that (H₁) is satisfied, for instance, for , any closed, convex subset K ⊂ X, and defined by A = −Δ, where is the Laplacian operator, with Ω ⊂ ℝ^N (N ≥ 1) a bounded domain. Indeed, let a sequence {u_n} _n≥1 ⊂ K with u_n⇀u in , for some u ∈ K. Using the weak lower semicontinuity of the norm, we can write

()

for all

. Here, 〈·, ·〉 are the duality brackets for the pair

and

denotes the scalar product on

. Whence (H₁) holds in this case.

(c) Hypothesis (H₂) is fulfilled in the case where Φ is sequentially weakly lower semicontinuous, D(Φ) is weakly closed, and Φ(·, u) is weakly strongly continuous on its effective domain for all u ∈ X.

(d) If A is strongly monotone, that is, there exists a constant m > 0 such that

()

and ∂J is bounded on K in the sense that

()

with a positive constant

, where

is the best constant satisfying

, for all u ∈ X (which exists by the continuity of the embedding of X in Y), then condition (H₃) is satisfied.

(e) If A is strictly monotone and J is Gâteaux differentiable and regular (see [1, Definition 2.3.4]), then condition (H₃) is satisfied. In particular, if A is strictly monotone and J is continuously differentiable, then (H₃) is satisfied.

In this paper, we distinguish two cases depending on whether the set K is bounded or not necessarily bounded. The following result concerns the former situation.

Theorem 2. Assume that conditions (H₁)– (H₃) are satisfied and that the closed, convex set K is bounded in X. Then problem (1) has at least one solution.

Remark 3. Note that the existence of a solution of problem (1), which is the conclusion of Theorem 2, forces the intersection diag (K)∩D(Φ) to be nonempty, where the notation diag (K) stands for the diagonal of the set K; that is, diag (K) = {(v, v) : v ∈ K}. The nonemptiness of this intersection is not directly implied by the hypotheses (H₁)– (H₃), nor by the assumption made that K∩D(Φ(η, ·)) ≠ ∅ for all η ∈ K. However, Theorem 4 below incorporates hypothesis (H₄) which assumes in particular that diag (K)∩D(Φ) ≠ ∅.

Now, we deal with the case where K is not assumed to be bounded. In this case, we additionally suppose the following:

(H₄) there exist an element v₀ ∈ K with (η, v₀) ∈ D(Φ) for all η ∈ K and a real p ≥ 1 such that
()
(H₅) there exists a constant c₀ > 0 such that we have
()
for all u ∈ K with (u, u) ∈ D(Φ), where v₀ and p ≥ 1 are as in (H₄).

We state now our main result for problem (1) dealing with the case where the set K is possibly unbounded.

Theorem 4. Assume that conditions (H₁)– (H₅) are satisfied. Then problem (1) has at least a solution.

The rest of the paper is organized as follows. In Section 2, we present the proof of Theorem 2, where we apply a version of the Schauder fixed point theorem. In Section 3, we give the proof of Theorem 4, which is actually based on Theorem 2.

2. Proof of Theorem 2

For each η ∈ K, we consider the auxiliary problem

()

Our first purpose, accomplished in Lemma 6 below, is to show that problem (13) has a unique solution. To do this, we need Fan’s lemma (see [8, page 208]) which we recall in the following statement.

Theorem 5. Let W be a Hausdorff topological vector space, let Z be a nonempty subset of W, and let F : Z → 2^W be such that

(i)
F(x) is a nonempty, closed subset of W, for all x ∈ Z;
(ii)
for all {x₁, …, x_n} ⊂ Z;
(iii)
there is for which is compact.

Then

Lemma 6. Assume that hypotheses (H₁)– (H₃) are fulfilled and that the closed, convex set K is bounded in X. Then, for every η ∈ K, problem (13) has a unique solution.

Proof. Fix η ∈ K. Consider the set-valued mapping G : K∩D(Φ(η, ·)) → 2^X defined by

()

for all v ∈ K∩D(Φ(η, ·)). We show that the assumptions of Theorem 5 are satisfied for W = X endowed with the weak topology, Z = K∩D(Φ(η, ·)), and F = G.

For every v ∈ K∩D(Φ(η, ·)), we clearly have v ∈ G(v); hence G(v) is nonempty.

We check that G(v) is weakly compact for every v ∈ K∩D(Φ(η, ·)). To this end, we first prove that G(v) is sequentially weakly closed in X. Let a sequence {u_n} _n≥1 ⊂ G(v) with u_n⇀u in X, for some u ∈ X. Taking into account that X is compactly embedded in Y it follows that u_n → u in Y. Using the first part of assumption (H₂), we have that u ∈ K∩D(Φ(η, ·)). As u_n ∈ G(v), we know that

()

Passing to the limsup as n → ∞, we find

()

Here we made use of the weak convergence u_n⇀u in X, the continuity of J⁰(η; ·) on Y, and the second part of (H₂). Combining with (H₁), we obtain that u ∈ G(v), thereby G(v) is sequentially weakly closed in X.

Using that X is reflexive and separable and K is bounded, convex, and closed, we deduce that K is metrizable and weakly compact (see, e.g., [9, pages 44–50]). Since G(v) ⊂ K and using that G(v) is sequentially weakly closed, we derive that G(v) is weakly compact whenever v ∈ K∩D(Φ(η, ·)). Therefore conditions (i) and (iii) in Theorem 5 are fulfilled.

We focus now on the verification of condition (ii) in Theorem 5. Arguing by contradiction, we suppose that there exist v₁, …, v_n ∈ K∩D(Φ(η, ·)) and u₀ ∈ conv{v₁, …, v_n} such that . The convexity of the set K and of the function Φ(η, ·) ensures that u₀ ∈ K∩D(Φ(η, ·)). Then the assertion that reads as

()

Let

()

It is clear that v_i ∈ Λ for all i ∈ {1, …, n}. The convexity of the functions Φ(η, ·) and J⁰(η; ·) implies that Λ is a convex subset in X. We infer that conv{v₁, …, v_n} ⊂ Λ, so u₀ ∈ Λ, which is obviously impossible. This contradiction justifies condition (ii) in Theorem 5. Thus all the assumptions of Theorem 5 are satisfied.

Applying Theorem 5, we obtain

()

This ensures the existence of an element u ∈ K∩D(Φ(η, ·)) satisfying

()

for all v ∈ K∩D(Φ(η, ·)). The above inequality being also satisfied if v ∉ D(Φ(η, ·)), we conclude that u is a solution of problem (13).

It remains to show that the solution of problem (13) is unique. If u₁, u₂ ∈ K are solutions of (13), then we have that (η, u₁) ∈ D(Φ), (η, u₂) ∈ D(Φ), and

()

Letting v = u₂ in the first inequality and v = u₁ in the second one and then adding the obtained relations, we arrive at

()

By assumption (H₃), we conclude that u₁ = u₂. The proof is complete.

Denote by u_η ∈ K the unique solution of problem (13) corresponding to η ∈ K. Lemma 6 guarantees that u_η exists and is unique. We define π : K → K by

()

Lemma 7. Assume that hypotheses (H₁)– (H₃) are fulfilled and that the closed, convex set K is bounded in X. Then, the map π : K → K given in (23) is sequentially weakly continuous.

Proof. Let a sequence {η_n} _n≥1 ⊂ K such that η_n⇀η in X for some η ∈ K. We need to show that π(η_n )⇀π(η) as n → ∞. To do this, it suffices to check that, for any relabeled subsequence {η_n} _n≥1, there is a subsequence of {π(η_n)} _n≥1 weakly converging to π(η).

By the compactness of the embedding of X in Y, we have that η_n → η in Y. Denote, for simplicity, π(η_n) = u_n. The definition of π yields (η_n, u_n) ∈ D(Φ) and

()

Since K is bounded, {u_n} _n≥1 ⊂ K and X is reflexive, we know that along a subsequence, denoted again by {u_n} _n≥1, we have

()

for some w ∈ X. The first part of (H₂) yields (η, w)∈(K × K)∩D(Φ). Moreover, the compactness of the embedding of X in Y implies that u_n → w in Y. Letting n → ∞ in (24), by means of (H₁), (H₂), the convergences η_n → η and u_n → w in Y, and the upper semicontinuity of J⁰(·; ·) on Y × Y, we get

()

This means that w ∈ K is a solution of problem (13). Lemma 6 ensures that w is the unique solution of (13). Thus, by (23), we have π(η) = w. Taking into account (25), it follows that π(η_n)⇀π(η) as n → ∞ up to a subsequence. This completes the proof.

Remark 8. As noted in the proof of Lemma 6, the closed, bounded, convex subset K ⊂ X is metrizable for the weak topology. Therefore, Lemma 7 implies that π is weakly continuous.

We need the following version of the Schauder fixed point theorem (see [10, page 452]).

Theorem 9. Suppose that

(i)
X is a reflexive, separable Banach space;
(ii)
the map T : M ⊂ X → M is sequentially weakly continuous;
(iii)
the set M is nonempty, closed, bounded, and convex.

Then T has a fixed point.

We are now in position to prove Theorem 2.

Proof of Theorem 2. In view of Lemma 7 and the assumptions on X and K, we may apply Theorem 9 which shows that the map π : K → K admits a fixed point u ∈ K; that is, π(u) = u. Using the definition of π (see (23)), we deduce that u ∈ K is a solution of problem (1).

3. Proof of Theorem 4

It suffices to prove Theorem 4 when the set K is unbounded because for a bounded set K the result is true according to Theorem 2. Let K_m = {x ∈ K : ∥x∥_X ≤ m}. Let m₀ ≥ 1 be an integer such that , where v₀ is the element entering (H₄). We claim that Theorem 2 can be applied with K replaced by K_m whenever m ≥ m₀.

Note that , so v₀ ∈ K_m∩D(Φ(η, ·)) for all η ∈ K, all m ≥ m₀ (using the first part of (H₄)). Thus, K_m∩D(Φ(η, ·)) ≠ ∅ for all η ∈ K_m, all m ≥ m₀. Since K is convex and closed in X, it turns out that K_m is convex, closed, and bounded in X, for all m ≥ m₀.

We check that assumptions (H₁)– (H₃) of Theorem 2 remain valid when K is replaced by K_m with m ≥ m₀. Towards this, we fix some m ≥ m₀. If {(η_n, u_n)} _n≥1 ⊂ (K_m × K_m)∩D(Φ) satisfies η_n⇀η in X and u_n⇀u in X, then assumption (H₂) (for K) implies (η, u)∈(K × K)∩D(Φ). On the other hand, the weak convergences ensure that

()

Hence, (η, u)∈(K_m × K_m)∩D(Φ). The second part of (H₂) for K_m and conditions (H₁) and (H₃) for K_m hold because (H₁), (H₂), and (H₃) have been imposed for K, which contains K_m. Thus it is permitted to apply Theorem 2 for K_m in place of K, with any m ≥ m₀.

Applying Theorem 2, we find a sequence

in X such that u_m ∈ K_m, (u_m, u_m) ∈ D(Φ), and

()

for all v ∈ K_m, all m ≥ m₀. Letting v = v₀ (see (H₄)) in (28), we obtain

()

for all m ≥ m₀. By the definition of the convex subdifferential ∂Φ(u_m, ·), we have

()

Then, invoking the growth condition for ∂Φ(u_m, ·)(v₀) in (H₅), we see that

()

Recall that

()

(see [1, Proposition 2.1.2(b)]). This fact combined with the growth condition for the generalized gradient ∂J(u_m) as stated in (H₅) enables us to write

()

for all m ≥ m₀. By the continuity of the embedding X ⊂ Y, the inequality above leads to

()

where c₁ > 0 is a constant. Combining (29), (31), and (34) yields

()

for all m ≥ m₀. Relation (35) ensures that the sequence

is bounded in X; indeed, if we suppose that we have

along a (relabeled) subsequence, then it is seen from (35) that there is a constant c > 0 such that

()

which contradicts hypothesis (H₄).

By the reflexivity of X, there exists a subsequence of

, denoted again by

, such that

()

for some u ∈ X. Using hypothesis (H₂) with η_m = u_m, we derive that (u, u)∈(K × K)∩D(Φ).

It remains to show that u verifies the inequality in problem (1). Let an arbitrary element v ∈ K and let m₁ = m₁(v) ∈ ℕ such that m₁ ≥ max {m₀, ∥v∥_X}. Then v ∈ K_m for each m ≥ m₁ and so from (28), we have that

()

The compactness of the embedding X ⊂ Y and (37) guarantee that u_m → u in Y as m → ∞. Then the upper semicontinuity of J⁰(·; ·) on Y × Y implies

()

Assumptions (H₁) and (H₂) ensure that

()

Passing to the limsup as m → ∞ in (38) and using (39) and (40), we get that u ∈ K satisfies the inequality in (1). Since v was chosen arbitrarily in K, we conclude that u solves problem (1). The proof of Theorem 4 is complete.

Acknowledgment

This work is funded by a Marie Curie Intra-European Fellowship for Career Development within the European Community’s 7th Framework Program (Grant Agreement no. PIEF-GA-2010-274519).

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Existence Results for Constrained Quasivariational Inequalities

Abstract

1. Introduction and Statement of Main Results

2. Proof of Theorem 2

3. Proof of Theorem 4

Acknowledgment

References

References

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Existence Results for Constrained Quasivariational Inequalities

Abstract

1. Introduction and Statement of Main Results

2. Proof of Theorem 2

3. Proof of Theorem 4

Acknowledgment

References

References

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