DISCONTINUOUS VARIATIONAL-HEMIVARIATIONAL INEQUALITIES INVOLVING THE P-LAPLACIAN

. We deal with discontinuous quasilinear elliptic variational-hemivariational inequalities. By using the method of sub-and supersolution and based on the results of S. Carl, we extend the theory for discontinuous problems. The proof of the existence of extremal solutions within a given order interval of sub-and supersolutions is the main goal of this paper. In the last part, we give an example of the construction of sub-and supersolutions.


Introduction
Let Ω ⊂ R N , N ≥ 1, be a bounded domain with Lipschitz boundary ∂Ω. As V = W 1,p (Ω) and V 0 = W 1,p 0 (Ω), 1 < p < ∞, we denote the usual Sobolev spaces with their dual spaces V * = (W 1,p (Ω)) * and V * 0 = W −1,q (Ω), respectively (q is the Hölder conjugate of p). In this paper, we consider the following elliptic variationalhemivariational inequality where j o (s; r) denotes the generalized directional derivative of the locally Lipschitz function j : R → R at s in the direction r given by j o (s; r) = lim sup y→s,t↓0 j(y + tr) − j(y) t , (cf. [7,Chapter 2]), and K ⊂ V 0 is some closed and convex subset. The operator ∆ p u = div(|∇u| p−2 ∇u) is the p-Laplacian, 1 < p < ∞, and F denotes the Nemytskij operator related to the function f : Ω × R × R → R given by F (u)(x) = f (x, u(x), u(x)).
In [3] the method of sub-and supersolution was developed for variational-hemivariational inequalities of the form (1.1) with F (u) ≡ f ∈ V * 0 . The aim of this paper is the generalization for discontinuous Nemytskij operators F : L p (Ω) → L q (Ω). Let us consider some special cases of problem (1.1) as follows.
(ii) If f : Ω × R → R is a Carathéodory function satisfying some growth condition and j = 0, then (1.1) is a classical variational inequality of the form for which the method of sub-and supersolution has been developed in [5,Chapter 5].
for which the sub-supersolution method is well known.

Notations and hypotheses
For functions u, v : Ω → R, we use the notation Definition 2.1. A function u ∈ V is called a subsolution of (1.1) if the following holds: (1) u ≤ 0 on ∂Ω and F (u) ∈ L q (Ω); Definition 2.2. A function u ∈ V is called a supersolution of (1.1) if the following holds: We impose the following hypotheses for j and the nonlinearity f in problem (1.1).
(C) (i) x → f (x, r, u(x)) is measurable for all r ∈ R and for all measurable functions u : Ω → R. (ii) r → f (x, r, s) is continuous for all s ∈ R and for almost all x ∈ Ω. (iii) s → f (x, r, s) is decreasing for all r ∈ R and for almost all x ∈ Ω. (iv) For a given ordered pair of sub-and supersolutions u, u of problem (1.1), there exists a function k 1 ∈ L q + (Ω) such that |f (x, r, s)| ≤ k 1 (x) for all r, s ∈ [u(x), u(x)] and for almost all x ∈ Ω. By [2] the mapping x → f (x, u(x), u(x)) is measurable for x → u(x) measurable, but the associated Nemytskij operator F : L p (Ω) → L q (Ω) needs not necessarily be continuous. In this paper we assume K has lattice structure, that is, K fulfills We recall that the normed space L p (Ω) is equipped with the natural partial ordering of functions defined by u ≤ v if and only if v − u ∈ L p + (Ω), where L p + (Ω) is the set of all nonnegative functions of L p (Ω).

Preliminaries
Here we consider (1.1) for a Carathéodory function h : Ω × R → R (i.e., x → h(x, s) is measurable in Ω for all s ∈ R and s → h(x, s) is continuous on R for almost all x ∈ Ω), which fulfills the following growth condition: and for a.e. x ∈ Ω, (3.1) where k 2 ∈ L q + (Ω) and [u, u] is some ordered pair in L p (Ω), specified later. Note that the associated Nemytskij operator H defined by H(u)(x) = h(x, u(x)) is continuous and bounded from [u, u] ⊂ L p (Ω) to L q (Ω) (cf. [9]). Next we introduce the indicator function I K : V 0 → R ∪ {+∞} related to the closed convex set K = ∅ given by which is known to be proper, convex and lower semicontinuous. The variationalhemivariational inequality (1.1) can be rewritten as follows: Find u ∈ V 0 such that is a special case of the elliptic variationalhemivariational inequality in [5,Corollary 7.15] for which the method of sub-and supersolutions was developed. In the next result, we show the existence of extremal solutions of (3.2) for a Carathéodory function h = h(x, s).
Lemma 3.1. Let hypotheses (A),(B) and (2.1) be satisfied and assume the existence of sub-and supersolutions u and u satisfying u ≤ u, u ∨ K ⊂ K and u ∧ K ⊂ K. Furthermore we suppose that the Carathédory function h : Ω × R → R satisfies (3.1). Then, (3.2) has a greatest solution u * and a smallest solution u * such that that is, u * and u * are solutions of (3.2) that satisfy (3.3), and if u is any solution of (3.2) such that u ≤ u ≤ u, then u * ≤ u ≤ u * .
Proof. The proof follows the same ideas as in the proof for H(u) ≡ h ∈ V * 0 with an additional modification. We only introduce a truncation operator related to the functions u and u defined by The mapping T is continuous and bounded from V into V which follows from the fact that the functions min(·, ·) and max(·, ·) are continuous from V to itself and that T can be represented as T u = max(u, u) + min(u, u) − u (cf. [8]). In the auxiliary problems of the proof of [5,Corollary 7.15], we replace h ∈ V * 0 by (H • T )(u) and argue in an analogous way.
An important tool in extending the previous result to discontinuous Nemytskij operators is the next fixed point result. The proof of this Lemma can be found in [4, Theorem 1.1.1].
Lemma 3.2. Let P be a subset of an ordered normed space, G : P → P an increasing mapping and G[P ] = {Gx | x ∈ P }.
(1) If G[P ] has a lower bound in P and the increasing sequences of G[P ] converge weakly in P , then G has the least fixed point x * , and has an upper bound in P and the decreasing sequences of G [P ] converge weakly in P , then G has the greatest fixed point x * , and x * = max{x | x ≤ Gx}.

Main results
One of our main results is the following theorem.
If f is right-continuous (resp., left-continuous) in the third argument, then there exists a greatest solution u * (resp., a smallest solution u * ) of (1.1) in the order interval [u, u].
Proof. We choose a fixed element z ∈ [u, u] which is a supersolution of (1.1) satisfying z ∧ K ⊂ K and consider the following auxiliary problem: ) is a Carathéodory function satisfying some growth condition as in (3.1). Since F z (z) = F (z), z is also a supersolution of (4.1). By Definition 2.1, we have for a given subsolution u of (1.1) Setting w = u − (u − v) + for all v ∈ K and using the monotonicity of f with respect to s, we get which shows that u is also a subsolution of (4.1). Lemma 3.1 implies the existence of a greatest solution u * ∈ [u, z] of (4.1). Now we introduce the set A given by A := {z ∈ V : z ∈ [u, u] and z is a supersolution of (1.1) satisfying z ∧ K ⊂ K} and define the operator L : A → K by z → u * =: Lz. This means that the operator L assigns to each z ∈ A the greatest solution u * of (4.1) in [u, z]. In the next step we construct a decreasing sequence as follows: . . .  As u n ∈ [u, u n−1 ], we get u n (x) u(x) for a.e. x ∈ Ω. Furthermore, the sequence u n is bounded in V 0 , that is, u n V0 ≤ C for all n and due to the monotony of u n and the compact embedding V 0 → L p (Ω), we obtain u n u in V 0 , u n → u in L p (Ω) and a.e. pointwise in Ω. (4. 3) The fact that u n is a solution of (4.1) with z = u n−1 and v = u ∈ K results in Applying Fatou's Lemma, (4.3), and the upper semicontinuity of j o (·, ·) yields lim sup The right-continuity of f and the strong convergence of the decreasing sequence (u n ) along with the upper semicontinuity of j o (·; ·) allow us to pass to the lim sup in (4.1), where u (resp., z) is replaced by u n (resp., u n−1 ). We have This shows that u is a solution of (1.1) in the order interval [u, u]. Now, we still have to prove that u is the greatest solution of (1.1) in [u, u]. Let u be any solution of (1.1) in [u, u]. Because of the fact that K has lattice structure, u is also a subsolution of (1.1), respectively, a subsolution of (4.1). By the same construction as in (4.2) we obtain.  Obviously, the sequences in (4.2) and (4.4) create the same extremal solutions u n and u n , which implies that u ≤ u n = u n for all n. Passing to the limit delivers the assertion. The existence of a smallest solution can be shown in a similar way.
In the next theorem we will prove that only the monotony of f in the third argument is sufficient for the existence of extremal solutions. The function f needs neither be right-continuous nor left-continuous. Theorem 4.2. Assume that hypotheses (A)-(C), (2.1) are valid and let u and u be sub-and supersolutions of (1.1) satisfying u ≤ u , u ∨ K ⊂ K and u ∧ K ⊂ K. Then there exist extremal solutions u * and u * of (1.1) with u ≤ u * ≤ u * ≤ u.
Proof. As in the proof of Theorem 4.1, we consider the following auxiliary problem: and z is a supersolution of (1.1) satisfying z ∧ K ⊂ K} and introduce the fixed point operator L : A → K by z → u * =: Lz. For a given supersolution z ∈ A, the element Lz is the greatest solution of (4.5) in [u, z], and thus it holds u ≤ Lz ≤ z for all z ∈ A which implies L : A → [u, u]. Because of (2.1), Lz is also a supersolution of (4.5) satisfying By the monotonicity of f with respect to Lz ≤ z and using the representation Consequently, Lz is a supersolution of (1.1). This shows L : A → A.
Let v 1 , v 2 ∈ A and assume that v 1 ≤ v 2 . Then we have Since v 1 ≤ v 2 , it follows that Lv 1 ≤ v 2 and due to (2.1), Lv 1 is also a subsolution of (4.6), that is, (4.6) holds, in particular, for v ∈ Lv 1 ∧ K, that is, Using the monotonicity of f with respect to s yields and hence Lv 1 is a subsolution of (4.7). By Lemma 3.1, we know there exists a greatest solution of (4.7) in [Lv 1 , v 2 ]. But Lv 2 is the greatest solution of (4.7) in [u, v 2 ] ⊇ [Lv 1 , v 2 ] and therefore, Lv 1 ≤ Lv 2 . This shows that L is increasing.
In the last step we have to prove that any decreasing sequence of L(A) converges weakly in A. Let (u n ) = (Lz n ) ⊂ L(A) ⊂ A be a decreasing sequence. The same argument as in the proof of Theorem 4.1 delivers u n (x) u(x) a.e. in Ω. The boundedness of u n in V 0 , and the compact imbedding V 0 → L p (Ω) along with the monotony of u n implies u n u in V 0 , u n → u in L p (Ω) and a.e. x in Ω.
Since u n ∈ K solves (4.5), it follows u ∈ K. From (4.5) with u replaced by u n and v by u and with the fact that (s, r) → j o (s; r) is upper semicontinuous, we obtain by applying Fatou's Lemma The S + -property of −∆ p provides the strong convergence of (u n ) in V 0 . As Lz n = u n is also a supersolution of (4.5), Definition 2.2 yields Due to z n ≥ u n ≥ u and the monotonicity of f , we get and, since the mapping u → u + = max(u, 0) is continuous from V 0 to itself (cf. [8]), we can pass to the upper limit on the right hand side for n → ∞. This yields which shows that u is a supersolution of (1.1), that is, u ∈ A. As u is an upper bound of L(A), we can apply Lemma 3.2, which yields the existence of a greatest fixed point u * of L in A. This implies that u * must be the greatest solution of (1.1) in [u, u]. By analogous reasoning, one shows the existence of a smallest solution u * of (1.1). This completes the proof of the theorem.
We suppose the following conditions for f and Clarke's generalized gradient of j, where λ > λ 1 is any fixed constant: f (x, s, s) |s| p−2 s = +∞, uniformly with respect to a.a. x ∈ Ω.
(iv) f is bounded on bounded sets.  1), where e ∈ int(C 1 0 (Ω) + ) is the unique solution of −∆ p u = 1 in V 0 . Moreover, −εϕ 1 is a supersolution and εϕ 1 is a subsolution of (1.1) provided that ε > 0 is sufficiently small. Proof. A sufficient condition for a subsolution u ∈ V of problem (1.1) is u ≤ 0 on ∂Ω, F (u) ∈ L q (Ω) and and thus, u is a subsolution of (1.1). Analogously, u ∈ V is a supersolution of problem (1.1) if u ≥ 0 on ∂Ω, F (u) ∈ L q (Ω), and if the following inequality is satisfied, . The main idea of this proof is to show the applicability of [6, Lemma 2.1-2.3]. We put g(x, s) = f (x, s, s) + ξ + λ|s| p−2 s for ξ ∈ ∂j(s) and notice that in our considerations the nonlinearity g needs not be a continuous function. In view of assumption (B), we see at once that |ξ| |s| p−1 ≤ c, for |s| ≥ k > 0, ∀ξ ∈ ∂j(s), where c is a positive constant. This fact and the condition (D) yield the following limit values: In order to apply Theorem 4.2, we need to satisfy the assumptions which depend on the specific K. For example, we consider an obstacle problem given by for a.e. x ∈ Ω}, ψ ∈ L ∞ (Ω), ψ ≥ C > 0, (4.10) where C is a positive constant. One can show that for the positive pair of sub-and supersolutions in Proposition 4.3, all these conditions in (4.9) with respect to the closed convex set K defined in (4.10) can be satisfied.