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Abstract and Applied Analysis
Volume 2015 (2015), Article ID 890289, 8 pages
Research Article

An Obstacle Problem for Noncoercive Operators

Dipartimento di Matematica e Applicazioni “R. Caccioppoli”, Università di Napoli Federico II, Via Cintia, 80126 Napoli, Italy

Received 15 May 2014; Accepted 31 August 2014

Academic Editor: Adrian Petrusel

Copyright © 2015 Luigi Greco et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


We study the obstacle problem for second order nonlinear equations whose model appears in the stationary diffusion-convection problem. We assume that the growth coefficient of the convection term lies in the Marcinkiewicz space -.

1. Introduction

Let be a bounded domain in with -boundary, , and let be a Carathèodory function; that is, We assume that there exist such that for almost every we have for any vectors and in . Moreover, we assume that is a Carathèodory function verifying the following properties. (i)There exists a nonnegative function in the Lorentz space such that for almost every   and for any .(ii)Consider

The space is also known as the Marcinkiewicz space weak-.

Let and let . We define

Definition 1. Given , one says that is a solution of the obstacle problem OP if for every .

For a classical treatment of obstacle problem we refer to [1, 2]. See also [3, 4] and references therein.

Under assumptions (2) and (4) the left hand side of (7) is finite by the Sobolev embedding theorem.

We point out that assumptions (1)–(5) do not guarantee that the operator for any , , and almost every is coercive and monotone.

The aim of this paper is to establish existence and uniqueness of solutions of in the sense of Definition 1. Our first result is the following.

Theorem 2. Assume that assumptions (3) and (4) are verified, and let Then, there exists at most one solution of problem (7).

We also prove the following.

Theorem 3. Let assumptions (1)–(5) be verified and let . Assume that Then, for every , problem (7) admits a solution . Here is the Sobolev constant.

We remark that is not dense in . Moreover, condition (10) does not give any smallness control on the norm of in (see Section 2.1). This fact is very relevant when we have to prove a priori estimates for the solutions of . Indeed, in order to prove our results we follow a classical approach. First, we construct a coercive and monotone operator. Then we reduce the existence to applying a fixed point theorem.

Theorem 3 is new also in case of equations. In [58], the authors considered operators with a lower order term having the growth coefficient in spaces in which the bounded functions are dense.

A condition similar to (10) has been used in [9] for proving the existence of solutions to linear equations. In that paper, an example shows that, in general, condition (10) cannot be dropped in order to achieve existence of solutions. Regularity results for solutions have been obtained in [10].

2. Preliminary Results

2.1. Some Functional Spaces

Let be a bounded domain in . For a measurable , we denote by its Lebesgue measure. For a measurable function we denote by its distribution function and by its decreasing rearrangement; see [11]. Clearly, if . For and , we consider the quantity and for the obvious modification The Lorentz space consists of all measurable functions satisfying . The space is also known as Marcinkiewicz space or weak-. The quantity is equivalent to a norm which makes a Banach space; see [11, 12]. For , the space coincides with the usual Lebesgue space. Moreover, with continuous injections. In particular, if , The following Hölder-type inequality holds. For and , , if then See [9]. An elementary but often useful property is expressed by the equality which holds for .

We note the equality for every measurable . Here, for we assume .

We remark that, for any , is not dense in . We consider the distance of a given to :

To find a formula for the distance, we consider the truncation operator. For , we set Then Indeed, and , we have, for almost every , For other comments on the distance to and some applications, we refer to [13].

Example 4. Let be the unit ball of and . The function belongs to . Setting , for and , we compute Hence does not depend on .

On the contrary, for all , starting with the definition of , a simple application of Lebesgue dominated convergence theorem shows that is dense in . Hence, for , , and in particular the Lebesgue space , is contained in the closure of in . The closure of coincides with the closure of . The elements of the closure can be characterized by the condition of having absolutely continuous norm; see [11, Section 1.3].

Fundamental to us will be the Sobolev embedding theorem in Lorentz spaces (see [12]; see also [14, 15]).

Theorem 5. Let one assume that , ; then every function verifying actually belongs to , where , and where .

2.2. Monotone Operators

Let be a reflexive Banach space with dual . Let denote the pairing between and . Let be a closed convex set.

Definition 6. A mapping is called monotone if The monotone mapping is called strictly monotone if

Definition 7. is called coercive on if there exists an element such that

The following existence and uniqueness result is contained in [1] (see [1], Cap. III, Theorem 1.7 and Corollary 1.8).

Theorem 8. Let and let be strictly monotone, coercive, and continuous on finite dimensional subspaces. Then, there exists Such a solution is unique.

2.3. The Leray-Schauder Theorem

We will use the well-known Leray-Schauder fixed point theorem in the following form (see [16, Theorem 11.3, page 280]).

A continuous mapping between two Banach spaces is called compact if the images of bounded sets are precompact.

Theorem 9. Let be a compact mapping of a Banach space into itself, and suppose there exists a constant such that for all and satisfying . Then, has a fixed point.

3. Uniqueness of Solutions: Proof of Theorem 2

Proof of Theorem 2. Suppose that verify (7); that is, suppose that . We will prove that a.e. in . To this aim we use as test functions in (34) and in (35) for a number . Those functions are admissible since and belong to and , a.e. on . Observing that , we obtain Now we set
We have, using (3), (36), and (4),
Then we have Now, let , so that where . Combining (39) and the last inequality we obtain
Letting we obtain , and then, by the arbitrariness of , we can conclude that for almost every .

4. Existence of Solutions: Proof of Theorem 3

As , it is not restrictive to assume a.e. in . Moreover, let us observe that if assumption (10) holds true then by (23) there exists a positive constant such that Let us fix such a value of .

Here below we denote where as above is the truncation operator at level .

Let and be Carathèodory functions satisfying (1)–(4). Let us consider the operator defined by The operator is strictly monotone and coercive on . In fact for we have

Then, by (42), we have .

The following technical lemma will be useful in the sequel. We shall follow closely the proof of Lemma 4.1 in [7]. We include some details for the sake of completeness.

Lemma 10. Let one assume (1)–(5), , , , and let . Assume that is such that and verifies for all . Then, where depends only on , , and .

Proof. Taking as a test function in (46), we obtain
By assumptions (2)–(5) we have with . Note that in the last inequality we used also Young’s inequality.
Noting that and that and combining (48) and (49), by Hölder’s inequality, we have
Then, by the elementary relation , , we obtain with . This concludes our proof, observing that by Sobolev embedding theorem with .

Proof of Theorem 3. We will obtain the existence of a solution to problem (7) by applying the Leray-Schauder fixed point theorem stated in Section 2.3 to a suitable compact operator . Hence, we will now construct such operator.
Let . Using Theorem 8 we have that the problem admits a unique solution , since the operator defined by is strictly monotone and coercive in (see (45)).
Hence we can define an operator We claim that such an operator is compact.
Let us prove the compactness. (The proof that is continuous is similar.) To this aim, suppose that is a bounded sequence in . Then, up to a subsequence, there exists such that in . Denoting and we have that
Hence, adding (59) to (58) and using (3), (50), (51), and (4) we have
Dividing the last inequality by , by (10) and (42), This implies that in so that we can conclude that is compact.
A fixed point of is a solution to problem (7). We will prove that has a fixed point. In particular, we will find a constant such that the a priori estimate holds for every and satisfying .
To this aim let and let be a solution to the equation . Then, and
Now our aim is to estimate .
We use as a test function in (62) obtaining Moreover, we observe that by (2) and (3) with . This gives, using (63),
By (65), using Hölder’s inequality and Theorem 5, we obtain Hence, we obtain, for every , with .
Now, let us denote And let us set At this point our aim is to estimate .
Let us preliminarily observe that using as a test function in (62) we obtain
Using (2) and (3) and arguing as above by (70) we obtain with . Using (42), this leads to the estimate with .
On the other hand, since is a solution to , we can apply Lemma 10 to obtain where . Moreover, by (20), we have
Then, combining (73) and (74), we can now fix , independent of and such that
By (75) and (72), we obtain with .
Now we are in a position to estimate . We obtain, combining (67) and (76),
Hence, for all and all solution to , we have , with .
Since is a compact operator, Theorem 9 implies that has a fixed point, which is a solution of (7).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.


Luigi Greco and Gioconda Moscariello are partially supported by the 2010 PRIN “Calculus of Variations.


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