Research Article | Open Access
On the Convergence of a Nonlinear Boundary-Value Problem in a Perforated Domain
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.
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  and for dimension three in .
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.
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 .
Definition 2. For define the operator
We consider the following spectral problem: where is the unit outward normal vector to the boundary of
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
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 ).
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)ConsiderThen 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.
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 identitywhere 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 ).
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, thatThen, 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 thatDue to convergences (63) the right-hand side tends to zero in the limit; thereforeBy 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 subsequenceMoreover, 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).
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 ) 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.
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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