General Comparison Principle for Variational-Hemivariational Inequalities

Abstract

We study quasilinear elliptic variational-hemivariational inequalities involving general Leray-Lions operators. The novelty of this paper is to provide existence and comparison results whereby only a local growth condition on Clarke's generalized gradient is required. Based on these results, in the second part the theory is extended to discontinuous variational-hemivariational inequalities.

1. Introduction

Let , , be a bounded domain with Lipschitz boundary . By and , we denote the usual Sobolev spaces with their dual spaces and , respectively, where is the Hölder conjugate satisfying . We consider the following elliptic variational-hemivariational inequality. Find such that

(1.1)

where denotes the generalized directional derivative of the locally Lipschitz functions at in the direction given by

(1.2)

(cf. [1, Chapter 2]). We denote by a closed convex subset of , and is a second-order quasilinear differential operator in divergence form of Leray-Lions type given by

(1.3)

The operator stands for the Nemytskij operator associated with some Carathéodory function defined by

(1.4)

Furthermore, we denote the trace operator by which is known to be linear, bounded, and even compact.

The aim of this paper is to establish the method of sub- and supersolutions for problem (1.1). We prove the existence of solutions between a given pair of sub-supersolution assuming only a local growth condition of Clarke's generalized gradient, which extends results recently obtained by Carl in [2]. To complete our findings, we also give the proof for the existence of extremal solutions of problem (1.1) for a fixed ordered pair of sub- and supersolutions in case has the form

(1.5)

In the second part we consider (1.1) with a discontinuous Nemytskij operator involved, which extends results in [3] and partly of [4]. Let us consider next some special cases of problem (1.1), where we suppose .

(1)If and are smooth, problem (1.1) reduces to

(1.6)

which is equivalent to the weak formulation of the nonlinear boundary value problem

(1.7)

where denotes the conormal derivative of . The method of sub- and supersolution for this kind of problems is a special case of [5].

(2)For , and , (1.1) corresponds to the variational-hemivariational inequality given by

(1.8)

which has been discussed in detail in [6].

(3)If and , then (1.1) is a classical variational inequality of the form

(1.9)

whose method of sub- and supersolution has been developed in [7, Chapter 5].

(4)Let or and not necessarily smooth. Then problem (1.1) is a hemivariational inequality, which contains for as a special case the following Dirichlet problem for the elliptic inclusion:

(1.10)

and for the elliptic inclusion

(1.11)

where the multivalued functions stand for Clarke's generalized gradient of the locally Lipschitz function given by

(1.12)

Problems of the form (1.10) and (1.11) have been studied in [5, 8], respectively.

Existence results for variational-hemivariational inequalities with or without the method of sub- and supersolutions have been obtained under different structure and regularity conditions on the nonlinear functions by various authors. For example, we refer to [9–16]. In case that is the whole space or , respectively, problem (1.1) reduces to a hemivariational inequality which has been treated in [17–25].

Comparison principles for general elliptic operators , including the negative -Laplacian , Clarke's generalized gradient , satisfying a one-sided growth condition in the form

(1.13)

for all , for a.a. , and for all with can be found in [7]. Inspired by results recently obtained in [8, 26], we prove the existence of (extremal) solutions for the variational-hemivariational inequality (1.1) within a sector of an ordered pair of sub- and supersolutions without assuming a one-sided growth condition on Clarke's gradient of the form (1.13).

2. Notation of Sub- and Supersolution

For functions we use the notation , and and introduce the following definitions.

Definition 2.1.

A function is said to be a subsolution of (1.1) if the following holds:

(1);

(2)

Definition 2.2.

A function is said to be a supersolution of (1.1) if the following holds:

(1);

(2).

In order to prove our main results, we additionally suppose the following assumptions:

(2.1)

3. Preliminaries and Hypotheses

Let , and assume for the coefficients the following conditions.

(A1) Each satisfies Carathéodory conditions, that is, is measurable in for all and continuous in for a.e. . Furthermore, a constant and a function exist so that

(3.1)

for a.e. and for all , where denotes the Euclidian norm of the vector .

(A2) The coefficients satisfy a monotonicity condition with respect to in the form

(3.2)

for a.e. , for all , and for all with .

(A3) A constant and a function exist such that

(3.3)

for a.e. , for all and for all .

Condition (A1) implies that is bounded continuous and along with (A2); it holds that is pseudomonotone. Due to (A1) the operator generates a mapping from into its dual space defined by

(3.4)

where stands for the duality pairing between and , and assumption (A3) is a coercivity type condition.

Let be an ordered pair of sub- and supersolutions of problem (1.1). We impose the following hypotheses on and the nonlinearity in problem (1.1).

(j1) and are measurable in and , respectively, for all .

(j2) and are locally Lipschitz continuous in for a.a. and for a.a. , respectively.

(j3) There are functions and such that for all the following local growth conditions hold:

(3.5)

(F1)

(i) is measurable in for all .

(ii) is continuous in for a.a. .

(iii)There exist a constant and a function such that

(3.6)

for a.e. , for all , and for all .

Note that the associated Nemytskij operator defined by is continuous and bounded from to (cf. [27]). We recall that the normed space is equipped with the natural partial ordering of functions defined by if and only if , where is the set of all nonnegative functions of .

Based on an approach in [8], the main idea in our considerations is to modify the functions . First we set for

(3.7)

By means of (3.7) we introduce the mappings and defined by

(3.8)

The following lemma provides some properties of the functions and .

Lemma 3.1.

Let the assumptions in (j1)–(j3) be satisfied. Then the modified functions and have the following qualities.

() and are measurable in and , respectively, for all , and and are locally Lipschitz continuous in for a.a. and for a.a. , respectively.

() Let be Clarke's generalized gradient of . Then for all the following estimates hold true:

(3.9)

() Clarke's generalized gradients of and are given by

(3.10)

and the inclusions and are valid for .

Proof.

With a view to the assumptions (j1)–(j3) and the definition of in (3.8), one verifies the lemma in few steps.

With the aid of Lemma 3.1, we introduce the integral functionals and defined on and , respectively, given by

(3.11)

Due to the properties and Lebourg's mean value theorem (see [1, Chapter 2]), the functionals and are well defined and Lipschitz continuous on bounded sets of and , respectively. This implies among others that Clarke's generalized gradients and are well defined, too. Furthermore, by means of Aubin-Clarke's theorem (see [1]), for and we get

(3.12)

An important tool in our considerations is the following surjectivity result for multivalued pseudomonotone mappings perturbed by maximal monotone operators in reflexive Banach spaces.

Theorem 3.2.

Let be a real reflexive Banach space with the dual space , a maximal monotone operator, and . Let be a pseudomonotone operator, and assume that either is quasibounded or is strongly quasibounded. Assume further that is -coercive, that is, there exists a real-valued function with as such that for all one has . Then is surjective, that is, .

The proof of the theorem can be found, for example, in [28, Theorem ]. The notation and stand for and , respectively. Note that any bounded operator is, in particular, also quasibounded and strongly quasibounded. For more details we refer to [28]. The next proposition provides a sufficient condition to prove the pseudomonotonicity of multivalued operators and plays an important part in our argumentations. The proof is presented, for example, in [28, Chapter 2].

Proposition 3.3.

Let be a reflexive Banach space, and assume that satisfies the following conditions:

(i)for each one has that is a nonempty, closed, and convex subset of ;

(ii) is bounded;

(iii)if in and in with and if , then and .

Then the operator is pseudomonotone.

We denote by and the adjoint operators of the imbedding and the trace operator , respectively, given by

(3.13)

Next, we introduce the following multivalued operators:

(3.14)

where are defined as mentioned above. The operators have the following properties (see, e.g., [5, Lemmas and ]).

Lemma 3.4.

The multivalued operators and are bounded and pseudomonotone.

Let be the cutoff function related to the given ordered pair of sub- and supersolutions defined by

(3.15)

Clearly, the mapping is a Carathéodory function satisfying the growth condition

(3.16)

for a.e. , for all , where and . Furthermore, elementary calculations show the following estimate:

(3.17)

where and are some positive constants. Due to (3.16) the associated Nemytskij operator defined by

(3.18)

is bounded and continuous. Since the embedding is compact, the composed operator is completely continuous.

For , we define the truncation operator with respect to the functions and given by

(3.19)

The mapping is continuous and bounded from into which follows from the fact that the functions and are continuous from to itself and that can be represented as (cf. [29]). Let be the composition of the Nemytskij operator and given by

(3.20)

Due to hypothesis (F1)(iii), the mapping is bounded and continuous. We set , and consider the multivalued operator

(3.21)

where is a constant specified later, and the operator is given by

(3.22)

We are going to prove the following properties for the operator .

Lemma 3.5.

The operator is bounded, pseudomonotone, and coercive for sufficiently large.

Proof.

The boundedness of follows directly from the boundedness of the specific operators , , , , and . As seen above, the operator is completely continuous and thus pseudomonotone. The elliptic operator is pseudomonotone because of hypotheses (A1), (A2), and (F1), and in view of Lemma 3.4 the operators and are bounded and pseudomonotone as well. Since pseudomonotonicity is invariant under addition, we conclude that is bounded and pseudomonotone. To prove the coercivity of , we have to find the existence of a real-valued function satisfying

(3.23)

such that for all and the following holds

(3.24)

for some . Let ; that is, is of the form

(3.25)

where with for a.a. and with for a.a. . Applying (A1), (A3), (F1)(iii), (3.17), and (), the trace operator and Young's inequality yield

(3.26)

where are some positive constants. Choosing and such that yields the estimate

(3.27)

Setting for and provides the estimate in (3.24) satisfying (3.23). This proves the coercivity of and completes the proof of the lemma.

4. Main Results

Theorem 4.1.

Let hypotheses (A1)–(A3), (j1)–(j3), and (F1) be satisfied, and assume the existence of sub- and supersolutions and , respectively, satisfying and (2.1). Then, there exists a solution of (1.1) in the order interval .

Proof.

Let be the indicator function corresponding to the closed convex set given by

(4.1)

which is known to be proper, convex, and lower semicontinuous. The variational-hemivariational inequality (1.1) can be rewritten as follows. Find such that

(4.2)

for all . By using the operators and the functions introduced in Section 3, we consider the following auxiliary problem. Find such that

(4.3)

for all . Consider now the multivalued operator

(4.4)

where is as in (3.21), and is the subdifferential of the indicator function which is known to be a maximal monotone operator (cf. [28, page 20]). Lemma 3.5 provides that is bounded, pseudomonotone, and coercive. Applying Theorem 3.2 proves the surjectivity of meaning that Since , there exists a solution of the inclusion

(4.5)

This implies the existence of , and such that

(4.6)

where it holds in view of (3.12) and (3.14) that

(4.7)

with

(4.8)

Due to the Definition of Clarke's generalized gradient , one gets

(4.9)

Moreover, we have the following estimate:

(4.10)

From (4.6) we conclude

(4.11)

Using the estimates in (4.9) and (4.10) to the equation above where is replaced by , yields for all

(4.12)

Hence, we obtain a solution of the auxiliary problem (4.3) which is equivalent to the problem. Find such that

(4.13)

In the next step we have to show that any solution of (4.13) belongs to . By Definition 2.2 and by choosing , we obtain

(4.14)

and selecting in (4.13) provides

(4.15)

Adding these inequalities yields

(4.16)

Let us analyze the specific integrals in (4.16). By using (A2) and the definition of the truncation operator, we obtain

(4.17)

Furthermore, we consider the third integral of (4.16) in case ; otherwise it would be zero. Applying (1.12) and (3.8) proves

(4.18)

Proposition in [1] along with (3.7) shows

(4.19)

In view of (4.18) and (4.19) we obtain

(4.20)

and analog to this calculation

(4.21)

Due to (4.17), (4.20), and (4.21), we immediately realize that the left-hand side in (4.16) is nonpositive. Thus, we have

(4.22)

which implies and hence, . The proof for is done in a similar way. So far we have shown that any solution of the inclusion (4.5) (which is a solution of (4.3) as well) belongs to the interval . The latter implies , and , and thus from (4.5) it follows

(4.23)

where and , which proves that is also a solution of our original problem (1.1). This completes the proof of the theorem.

Let denote the set of all solutions of (1.1) within the order interval . In addition, we will assume that has lattice structure, that is, fulfills

(4.24)

We are going to show that possesses the smallest and the greatest element with respect to the given partial ordering.

Theorem 4.2.

Let the hypothesis of Theorem 4.1 be satisfied. Then the solution set is compact.

Proof.

First, we are going to show that is bounded in . Let be a solution of (4.2), and notice that is -bounded because of . This implies , and thus, is also bounded in . Choosing a fixed in (4.2) delivers

(4.25)

Using (A1), (j3), (F1)(iii), Proposition in [1], and Young's inequality yields

(4.26)

where the left-hand side fulfills the estimate

(4.27)

Thus, one has

(4.28)

where the choice proves that is bounded. Hence, we obtain the boundedness of in . Let . Since is reflexive, there exists a weak convergent subsequence, not relabelled, which yields along with the compact imbedding and the compactness of the trace operator

(4.29)

As solves (4.2), in particular, for , we obtain

(4.30)

Since is upper semicontinuous and due to Fatou's Lemma, we get from (4.30)

(4.31)

The elliptic operator satisfies the ()-property, which due to (4.31) and (4.29) implies

(4.32)

Replacing by in (1.1) yields the following inequality:

(4.33)

Passing to the limes superior in (4.33) and using Fatou's Lemma, the strong convergence of in , and the upper semicontinuity of , we have

(4.34)

Hence, . This shows the compactness of the solution set .

In order to prove the existence of extremal elements of the solution set , we drop the -dependence of the operator . Then, our assumptions read as follows.

() Each satisfies Carathéodory conditions, that is, is measurable in for all and continuous in for a.e. . Furthermore, a constant and a function exist so that

(4.35)

for a.e. and for all , where denotes the Euclidian norm of the vector .

() The coefficients satisfy a monotonicity condition with respect to in the form

(4.36)

for a.e. , and for all with .

() A constant and a function exist such that

(4.37)

for a.e. , and for all .

Then the operator acts in the following way:

(4.38)

Let us recall the definition of a directed set.

Definition 4.3.

Let be a partially ordered set. A subset of is said to be upward directed if for each pair there is a such that and . Similarly, is downward directed if for each pair there is a such that and . If is both upward and downward directed, it is called directed.

Theorem 4.4.

Let hypotheses ()–() and (j1)–(j3) be fulfilled, and assume that (F1) and (4.24) are valid. Then the solution set of problem (1.1) is a directed set.

Proof.

By Theorem 4.1, we have . Let be given solutions of (1.1), and let . We have to show that there is a such that . Our proof is mainly based on an approach developed recently in [26] which relies on a properly constructed auxiliary problem. Let the operator be given basically as in (3.15)–(3.18) with the following slight change:

(4.39)

We introduce truncation operators related to and modify the truncation operator as follows. For , we define

(4.40)

and we set

(4.41)

as well as

(4.42)

Moreover, we define

(4.43)

for and introduce the functions and defined by

(4.44)

Furthermore, we define the functions and for as follows:

(4.45)

and for

(4.46)

where . (Note that for we understand the functions above being defined on .) Apparently, the mappings are Carathéodory functions which are piecewise linear with respect to . Let us introduce the Nemytskij operators and defined by

(4.47)

Due to the compact imbedding and the compactness of the trace operator , the operators and are bounded and completely continuous and thus pseudomonotone. Now, we consider the following auxiliary variational-hemivariational inequality. Find such that

(4.48)

for all . The construction of the auxiliary problem (4.48) including the functions and is inspired by a very recent approach introduced by Carl and Motreanu in [26]. The first part of the proof of Theorem 4.1 delivers the existence of a solution of (4.48), since all calculations in Section 3 are still valid. In order to show that the solution set of (1.1) is upward directed, we have to verify that a solution of (4.48) satisfies . By assumption , that is, solves

(4.49)

for all . Selecting in the inequality above yields

(4.50)

Taking the special test function in (4.48), we get

(4.51)

Adding (4.50) and (4.51) yields

(4.52)

The condition () implies directly

(4.53)

and the second integral can be estimated to obtain

(4.54)

In order to investigate the third integral, we make use of some auxiliary calculation. In view of (4.44) we have for

(4.55)

Applying Proposition in [1] and (3.7) results in

(4.56)

Furthermore, we have in case

(4.57)

Thus, we get

(4.58)

The same result can be proven for the boundary integral meaning

(4.59)

Applying (4.53)–(4.59) to (4.52) yields

(4.60)

and hence, meaning that for . This proves . The proof for can be shown in a similar way. More precisely, we obtain a solution of (4.48) satisfying which implies and . The same arguments as at the end of the proof of Theorem 4.1 apply, which shows that is in fact a solution of problem (1.1) belonging to the interval . Thus, the solution set is upward directed. Analogously, one proves that is downward directed.

Theorems 4.2 and 4.4 allow us to formulate the next theorem about the existence of extremal solutions.

Theorem 4.5.

Let the hypotheses of Theorem 4.4 be satisfied. Then the solution set possesses extremal elements.

Proof.

Since and are separable, is also separable; that is, there exists a countable, dense subset of . We construct an increasing sequence as follows. Let and select such that

(4.61)

By Theorem 4.4, the element exists because is upward directed. Moreover, we can choose by Theorem 4.2 a convergent subsequence (denoted again by ) with in and in . Since is increasing, the entire sequence converges in and further, . One sees at once that which follows from

(4.62)

and the fact that is closed in implies

(4.63)

Therefore, as , we conclude that is the greatest element in . The existence of the smallest solution of (1.1) in can be proven in a similar way.

Remark 4.6.

If depends on , we have to require additional assumptions. For example, if satisfies in a monotonicity condition, the existence of extremal solutions can be shown, too. In case , a Lipschitz condition with respect to is sufficient for proving extremal solutions. For more details we refer to [7].

5. Generalization to Discontinuous Nemytskij Operators

In this section, we will extend our problem in (1.1) to include discontinuous nonlinearities of the form . We consider again the elliptic variational-hemivariational inequality

(5.1)

where all denotations of Section 1 are valid. Here, denotes the Nemytskij operator given by

(5.2)

where we will allow to depend discontinuously on its third argument. The aim of this section is to deal with discontinuous Nemytskij operators by combining the results of Section 4 with an abstract fixed point result for not necessarily continuous operators, cf. [30, Theorem ]. This will extend recent results obtained in [3]. Let us recall the Definitions of sub- and supersolutions.

Definition 5.1.

A function is called a subsolution of (5.1) if the following holds:

(1);

(2)

Definition 5.2.

A function is called a supersolution of (5.1) if the following holds:

(1)

(2).

The conditions for Clarke's generalized gradient and the functions are the same as in (j1)–(j3). We only change the property (F1) to the following.

(F2)

(i) is measurable for all , for all , and for all measurable functions .

(ii) is continuous in for all and for a.a. .

(iii) is decreasing for all and for a.a. .

(iv)There exist a constant and a function such that

(5.3)

for a.e. , for all , and for all .

By [31] the mapping is measurable for ; however, the associated Nemytskij operator is not necessarily continuous. 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 [30, Theorem ].

Lemma 5.3.

Let be a subset of an ordered normed space, an increasing mapping, and .

(1)If has a lower bound in and the increasing sequences of converge weakly in , then has the least fixed point , and .

(2)If has an upper bound in and the decreasing sequences of converge weakly in , then has the greatest fixed point , and .

Our main result of this section is the following theorem.

Theorem 5.4.

Assume that hypotheses ()–(), (j1)–(j3), (F2), and (4.24) are valid, and let and be sub- and supersolutions of (5.1) satisfying and (2.1). Then there exist extremal solutions and of (5.1) with .

Proof.

We consider the following auxiliary problem:

(5.4)

where , and we define the set , and is a supersolution of (5.1) satisfying . On we introduce the fixed point operator by , that is, for a given supersolution , the element is the greatest solution of (5.4) in , and thus, it holds for all . This implies . Because of (4.24), is also a supersolution of (5.4) satisfying

(5.5)

for all . By the monotonicity of with respect to its third argument, , and using the representation for any we obtain

(5.6)

for all . Consequently, is a supersolution of (5.1). This shows .

Let , and assume that . Then we have the following.

(5.7)

(5.8)

Since , it follows that and due to (4.24), is also a subsolution of (5.7), that is, (5.7) holds, in particular, for , that is,

(5.9)

for all . Using the monotonicity of with respect to its third argument yields

(5.10)

for all . Hence, is a subsolution of (5.8). By Theorem 4.5, we know that there exists the greatest solution of (5.8) in . But is the greatest solution of (5.8) in and therefore, . This shows that is increasing.

In the last step we have to prove that any decreasing sequence of converges weakly in . Let be a decreasing sequence. Then a.e. for some . The boundedness of in can be shown similarly as in Section 4. Thus the compact imbedding along with the monotony of as well as the compactness of the trace operator implies

(5.11)

Since , it follows . From (5.4) with replaced by and by , and using the fact that is upper semicontinuous, we obtain by applying Fatou's Lemma

(5.12)

The -property of provides the strong convergence of in . As is also a supersolution of (5.4) Definition 5.2 yields

(5.13)

for all . Due to and the monotonicity of we get

(5.14)

for all , and since the mapping is continuous from to itself (cf. [29]), we can pass to the upper limit on the right-hand side for . This yields

(5.15)

which shows that is a supersolution of (5.1), that is, . As is an upper bound of , we can apply Lemma 5.3, which yields the existence of the greatest fixed point of in . This implies that must be the the greatest solution of (5.1) in . By analogous reasoning, one shows the existence of the smallest solution of (5.1). This completes the proof of the theorem.

Remark 5.5.

Sub- and supersolutions of problem (5.1) have been constructed in [32] under the conditions ()–(), (j1)–(j2) and (F2)(i)–(F2)(iii), where the gradient dependence of has been dropped, meaning that . Further, it is assumed that which is the negative -Laplacian defined by

(5.16)

The coefficients are given by

(5.17)

Thus, hypothesis () is satisfied with and . Hypothesis () is a consequence of the inequalities from the vector-valued function (see [7, page 37]), and () is satisfied with and . The construction is done by using solutions of simple auxiliary elliptic boundary value problems and the eigenfunction of the -Laplacian which belongs to its first eigenvalue.