International Journal of Differential Equations
Volume 2015 (2015), Article ID 392479, 11 pages
http://dx.doi.org/10.1155/2015/392479

Research Article

On the Convergence of a Nonlinear Boundary-Value Problem in a Perforated Domain

Yulia Koroleva

Department of Higher Mathematics, Gubkin Russian State University of Oil and Gas, Leninsky Prospect 65-1, Moscow 119991, Russia

Received 16 July 2015; Accepted 7 September 2015

Academic Editor: Elena I. Kaikina

Copyright © 2015 Yulia Koroleva. 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.

Abstract.

We consider a family with respect to a small parameter of nonlinear boundary-value problems as well as the corresponding spectral problems in a domain perforated periodically along a part of the boundary. We prove the convergence of solution of the original problems to the solution of the respective homogenized problem in this domain.

1. Introduction

The paper is devoted to study of convergence of nonlinear boundary-value problems in a domain perforated along the boundary. There exist a lot of literatures, where boundary-value problems in perforated domains were studied. We refer to works [1–33]. In these papers and monographs the authors studied different kinds of perforation for linear as well as for nonlinear differential operators. Usually it is considered a family of problems depending on small parameter that characterizes the size of perforation. The main goal of the research is to find a homogenized (limit) model which is close to originally considered problems posed in the perforated domain. The general technique of homogenization method can be found in [23, 24, 27, 28].

The present paper will deal with convergence of boundary-value problems in perforated domains for nonlinear -Laplace operator. Some problems for nonlinear operators were homogenized, for example, in [1, 20, 21, 30–33]. We consider a family of boundary-value problems in -dimensional domain, , which is periodically perforated along the boundary by small sets. It is assumed that the diameter of each set and the distance between them have the same order. In our problem we suppose that the Dirichlet condition holds on the boundary of cavities, while the Naumann boundary condition is fulfilled on the boundary of the domain. We derive the limit (homogenized) problem for the original problems when the small parameter characterizing the size of perforation tends to zero. Moreover, we establish the strong convergence in of the solutions for the considered problems to the corresponding solution of the limit problem. In addition we have obtained an estimate of the solution in a neighborhood of the eigenvalue of a corresponding spectral problem.

One of our goals is to prove the asymptotic behavior for the eigenvalue problem for -Laplace operator in our perforated domain. Many authors considered spectral problems for -Laplace operator; see, for example, [33–38]. These papers contain the results on qualitative properties of the -Laplace spectral problems, convergence of eigenvalue problems, and some estimates for the difference between considered eigenvalues. The applications of our problem do not require the knowledge about the full spectrum of eigenvalues. Therefore we have proved the homogenization theorems only for the first eigenelement of the spectral problem in perforated domain. More precisely, we have proved that the first eigenelement of the spectral problems converges to the corresponding eigenelement of the spectral limit problem. An analogous problem for linear elliptic operators for the two-dimensional domain was considered in [15] and for dimension three in [10].

The crucial point in our analysis is the validity of the Friedrichs inequality for functions in perforated domains. We prove this nontrivial result which is of an independent interest. Some papers devoted to this inequality in domains with microinhomogeneous structure are [8–10, 18, 19].

2. Preliminaries and the Main Results

Let , , be a domain with boundary . We assume that is piece-wise smooth and consists of the parts , , where , , are orthogonal to and belong to the planes and correspondingly, and is a smooth surface. In the sequel is a small parameter, , .

Consider the set belonging to the ball

and having a smooth boundary. If one multiplies each coordinate of with parameter and does integer translations of this set along , we obtain the set denoted by Let . Define the perforated domain as See the illustration for cut of on Figure 1.

Figure 1: Structure of .

Denote by a set of functions from with zero trace on . Analogously, by we define a set of functions from with zero trace on . We also consider the space , which is the set of functions from , vanishing in the neighborhood of . Analogously, denotes the set of functions from , vanishing in a neighborhood of .

Remark 1. One can extend the functions into by zero. For the extended function we keep the same notation. It is true that belongs to ; see [24].

Definition 2. For define the operator

We consider the following spectral problem:

where is the unit outward normal vector to the boundary of

Definition 3. One says that is an eigenfunction to problem (3) if there exists , satisfying the integral identity

for every The couple is called the solution to (3).

We will show that the problem

is homogenized (the limit one) for (3).

As usual, we understand the solution to this boundary-value problem in the weak sense, that is, iff satisfies

for every

Moreover, we prove the following results.

Theorem 4. Assume that , , , , and is an arbitrary compact set belonging to the complex plane ; does not contain the eigenvalues of problem (5). Then the following statements hold:

(1)There exists a number , such that the unique solution to the problem

 does exist for all and for all Moreover, the uniform (in and ) estimate

 is valid, where does not depend on and .(2)It yields that

 where is the unique solution of the problem

Here the solutions to problems (7) and (10) are understood in the weak sense, that is, , satisfies the integral identity

for every

Theorem 5. The spectrum of problems (3) and (5) is nonempty closed set. Let be the first eigenvalues of problems (3) and (5), respectively. Then

Moreover, if are corresponding eigenfunctions, normalized in , then up to a subsequence,

The proofs of analogous theorems for linear boundary-value problems were given in [6, 12–15] for different types of singular perturbations. The following lemma, which is proved in Section 3.1, is necessary for our analysis.

Lemma 6. Let be a sequence of functions from and assume that weakly in when . Then

For the questions on existence of solutions to the discussed problems we will refer to the following general result (see [39]).

Theorem 7. Let be reflexive separable Banach space. Assume that the operator has the following properties:

(i)is bounded and semicontinuous: , ; the function is continuous as a function mapping into .(ii) is monotone: (iii)Consider

Then for any there exists such that

3. The Friedrichs Inequality

In our analysis we will need the Friedrichs inequality for functions We prove this result.

Theorem 8. The inequality

holds for any functions , where the constant does not depend on

Proof. To demonstrate the technique of proving and avoiding the heavy -dimensional notations we assume that The case of arbitrary can be done by repeating all lines of the present proof.
Let the length of projection of on axis equal . Denote

Represent the domain as follows:

where is the domain (a part of the vertical strip) of the length , bounded from below by the upper part of the boundary and bounded from above by a part of the boundary ; the domain is the strip of the length , bounded from above by the lower part of and bounded from below by the line ; , , is the strip of the length , bounded from below by the line and bounded from above by a part of and having the common vertical bounds with (see Figure 2); the domains and are vertical strips bounded from below by and bounded from above by a part of and having the common vertical bounds with and , correspondingly. The sum of widths for and is Finally, is the remaining part of having the bound belonging to .
Let . Define by a domain (see Figure 2). Moreover, let be the segment of the line , which is upper bound for Analogously we define the sets and . Without loss of generality we can assume that is real-valued function. Denote by functions from , vanishing in the neighborhood of . Assume first that . Let , and point belongs to the bottom of . Since , then the Newton-Leibnitz formula gives

Taking the power on both sides of that equality and using the Hölder inequality, we have

Integrate both sides of that inequality over with respect to and assume that the function is extended by zero into . One gets

Consider now the rectangle , , which touches from the right. Draw the tangential lines to from both ends of segment (see Figure 3).
It is easy to see that the angle between the tangential lines and belongs to , where does not depend on and We remember that our assumption was . This follows from the fact that the diameter of the set and the distance between them are of the same order. Connect all points of with boundary of such that the intersection of these lines coincides with the intersection point of tangents. Thus, we have a beam of lines with directors The angle between each line and belongs to . Let , and . Since , then

Analogously to (19), taking the power on both sides of that formula and using the Hölder inequality, one obtains

Integrating both sides of that inequality over with respect to and replacing the right-hand side by the greater integral, we get

It remains to estimate the integral over . We use the same technique as for , In this case one needs to consider the domain which is bordered with the strip from the right. Analogously, we get

Let be the segment connecting the points and ; it means that . Summing up inequalities (20), (23), and (24), one gets

Finally, integrating (25) with respect to , we obtain that

Approximating the functions from by smooth functions, we conclude that inequality (26) is valid for The proof is complete.

Figure 2: Parts of .

Figure 3: Tangential lines.

Remark 9. (1) Extending functions from by zero into we obtain the Friedrichs inequality in :

(2) The validity of Friedrichs inequality means that one can introduce in the norm which is equivalent to

3.1. Proof of Lemma 6

Proof. First we point out that with the same method of proof inequality (25) is valid also for the case of an arbitrary , where

Approximating the functions from by smooth functions, we conclude that inequality (25) is valid for Using (25) and keeping in mind the uniform boundedness of the sequence , we obtain that

Now, we can pass to the limit in (30) when and find that

Due to the fact that is an arbitrary small positive number and , it follows from (31) that on The proof is complete.

4. Proof of Theorem 4

For the proof we need the following lemma.

Lemma 10. Let be an arbitrary compact set in the complex plane and Suppose that the estimate

holds uniformly in and for any solution of the boundary-value problem (7), which is normalized in . Then estimate (32) holds also for any solution of problem (7) when

Proof. Let us remember first that we denote by the norm in space. If , then by setting , we obtain that

and the function satisfies the identity

where

Hence, due to the assumptions we see that the estimate

holds for Multiplying the last inequality by and using (33) and (35), we obtain the estimate (32) (probably with a different constant) for any The proof is complete.

4.1. Proof of Part 1

Step 1. The existence of the solution to problem (7) can be proved with help of Theorem 7. Indeed, we take Introduce the operator For we define

Moreover, for it holds that , and for we can introduce the functional on :

It is clear that is the solution to (7) iff for any Let us verify the properties of operator We take with Then

One can check the semicontinuity and monotonicity of operator either directly or by using the following general result (see [39]).

Proposition 11. If convex functional is differentiable in Gato sense, that is, there exists such continuous linear mapping of space onto such that

then the mapping is monotone and semicontinuous.

Let us define the functional

on the space It is easy to check that is convex and differentiable in Gato sense and Hence, is semicontinuous and monotone operator. Finally, taking into account Remark 9 we calculate that

Therefore by Theorem 7 for any there exists a function satisfying

Step 2. Let us derive now estimate (8). Substitute with in (11). Then it follows that

Hence,

Assume now that estimate (8) does not hold, that is, that there exists a sequence when , and such that the inequality

holds for the solutions to problem (7), where

Due to Lemma 10 we may assume without loss of generality that the sequence is normalized in , that is, that

Then, by using (44) and (45), we obtain that

when
Hence, there exists a subsequence of indexes and , such that

By (47) and (49) we have that

Moreover, since the operator is bounded, it follows that . Therefore we conclude the existence of such that weakly in Suppose that is an arbitrary fixed function from . Then for all small . Substitute in the integral identity (11) as a test function; , , and , when . We have that

Now we want to pass to the limit as Due to the definition of , we see that

From the other hand,

Let us show now that The monotonicity of operator implies

Passing to the limit, it yields that

If we take , , , we get

Let Then

This exactly means that Thus, passing to the limit in the integral identity (51) as , using (48), (49), and Lemma 6, we obtain that

From the density of the embedding into we can conclude that this inequality holds also for It follows from Lemma 6 that Due to the fact that and is an arbitrary function from , it follows that is the eigenvalue of the limit problem (5). But we assumed that did not contain the eigenvalues of the limit problem (5). This contradiction proves estimate (8).

Step 3. Let us prove now the uniqueness of the solution of (7). Let and be two different solutions. Then

This together with equation

implies that

due to the boundary conditions. The uniqueness of the solution to (10) can be proved identically.

4.2. Proof of Part 2

Let be an arbitrary fixed number and assume that the sequence when Thinking analogously as in Part 1 of the proof, we have existence of a subsequence and a function such that in (and strongly in ) when In addition, weakly in It can be shown exactly in the same way that

Now by using Lemma 6 we obtain that the limit function Passing to the limit in the integral identity (51) by means of the same reasoning as in the proof of Part 1 of the theorem, we get that

which coincides with the integral identity of problem (10). Since the solution to problem (10) is unique we conclude that In addition, from (49) we find that

From the integral identities we have that

Due to convergences (63) the right-hand side tends to zero in the limit; therefore

By using the inequality

with and we can show the strong convergence of gradients:

Thus, we have proved that, up to a subsequence,

The proof is complete.

5. Proof of Theorem 5

Define by the spectrum of problem

That is, (69)

Theorem 12. The spectrum is nonempty closed set,

Proof. To show that the spectrum is nonempty, one needs to prove the existence of weak solution to problem (69). We can make use of Theorem 7. We take and For we define

One needs to verify the properties of operator First, the boundedness is fulfilled since

To check the semicontinuity, monotonicity of the operator, and the property one can in the same way as in the proof of Theorem 4. Let us omit the details.
As soon as we conclude that the conditions of Theorem 7 are fulfilled, we conclude the existence of the solution. Thus, the spectrum is nonempty. Check now the positiveness of Let and the corresponding eigenfunction It yields that

Let correspond to eigenfunctions and Without loss of generality one may assume that Since is bounded, it follows from

that which implies the existence of such that up to a subsequence

Moreover,

and therefore there exists such that

These convergences imply that and

Thus it remains to show that

For every it holds that

Passing to the limit as , one obtains that

Take now , , Then

Passing to the limit as , we get

From this inequality we deduce the validity of (78).

Let us mention the properties of the first eigenvalue to spectral problem (69). The following theorem directly follows from the results in [34].

Theorem 13. The smallest eigenvalue

is simple and isolated. Moreover, if are two eigenfunctions corresponding to , then they are proportional:

Now we come back to our spectral problems (3) and (5). The existence of normalized solution and properties of spectrum follows by Theorems 12 and 13. The uniqueness of the normalized solution is a direct consequence of Theorem 13. Concerning the regularity, the known fact (see [40]) says that if eigenfunctions , then The statements of Theorem 5 can be proved similarly to the proof of statement (2) of Theorem 4.

Indeed, since is isolated, that is, ( is the second eigenvalue to (3)), there exists a compact containing Hence, there exists a subsequence such that The boundedness of in implies the existence of limit (strong in and weak in ). The boundedness of operator in the dual space gives the existence of weak limit

Thus, passing to the limit in the identity

we obtain

Having in mind the same arguments as in the proof of Theorem 4 one can show that , and which means that problem (5) is the limit one for (3). Analogously as in Theorem 4 one can derive that

Thus, Theorem 5 is completely proved.

The question on convergence of the full spectrum for similar -Laplace boundary-value problem is studied, for example, in [33, 35]. We do not cover this studying for our problem since in the applications of the results of the present paper we will use only the fact about converging of the first eigenelements.

6. The Estimate of the Solution in a Neighborhood of the Eigenvalue

Let us derive an estimate for the solution to problem (7) in a neighborhood of

Lemma 14. Let be close to and converges to Then the following estimate holds:

Proof. Since the function , it satisfies Friedrichs inequality Moreover, it is clear that the best constant in the inequality is , where is the first eigenvalue to (3). Thus,

By using estimates (43), (8), (89), integral identity with , and equivalent norm in , it yields that

Replacing with the equivalent norm , we obtain (88).

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The work was partially supported by RFBR (Project 15-58-45142) and by the grant of President of Russian Federation supporting young Russian scientists (Project M-4615.2015.1).

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