Journal of Applied Mathematics

Volume 2013 (2013), Article ID 893182, 21 pages

http://dx.doi.org/10.1155/2013/893182

Research Article

Maximum Norm Analysis of an Arbitrary Number of Nonmatching Grids Method for Nonlinears Elliptic PDES

Abida Harbi

Department of Mathematics, Laboratory of Applied Mathematics, Badji Mokhtar-Annaba University, P.O. Box 12, El Hadjar, 23000 Annaba, Algeria

Received 11 August 2013; Revised 7 October 2013; Accepted 9 October 2013

Academic Editor: Srinivasan Natesan

Copyright © 2013 Abida Harbi. 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 provide a maximum norm analysis of a finite element Schwarz alternating method for a nonlinear elliptic PDE on an arbitrary number of overlapping subdomains with nonmatching grids. We consider a domain which is the union of an arbitrary number of overlapping subdomains where each subdomain has its own independently generated grid. The meshes being mutually independent on the overlap regions, a triangle belonging to one triangulation does not necessarily belong to the other ones. Under the a Lipschitz assumption on the nonlinearity, we establish, on each subdomain, an optimal error estimate between the discrete Schwarz sequence and the exact solution of the PDE.

1. Introduction

The Schwarz alternating method can be used to solve elliptic boundary value problems on domains which consist of two or more overlapping subdomains. The solution is approximated by an infinite sequence of functions which result from solving a sequence of elliptic boundary value problems in each of the subdomains. The effectiveness of the Schwarz methods for various classes of nonlinear elliptic PDE problems has been demonstrated in many papers; see [1–4] and the references therein. Also the effectiveness of the Schwarz methods for these problems, especially those in fluid mechanics, has been demonstrated in many papers. See proceedings of the annual domain decomposition conference beginning with [5].

In [6, 7], an optimal convergence order is obtained for nonlinear elliptic PDE, in the context of two overlapping nonmatching grids, in the sense that each subdomain has its own independent discretization by finite element method. This kind of discretization is very interesting as it can be applied to solve many practical problems which cannot be handled by global discretizations. Discretizations are earning particular attention of computational experts and engineers as they allow the choice of different mesh sizes and different orders of approximate polynomials in different subdomains according to the different properties of the solution and different requirements of the practical problems. Quite a few works on maximum norm error analysis of overlapping nonmatching grids methods for elliptic problems are known in the literature; compare and confer, for example, [8–10].

This paper is a continuation of previous work [6], attempting to generalize the obtained result related to convergence order also in the context of nonmatching grids in domain which consists of a union of an arbitrary number of subdomains. It is proved that the error estimate remains true also for more than two subdomains.

The proof of the main result which consists of estimating the error in the maximum norm between the continuous solution of the problem and the discrete Schwarz iterates stands on a Lipschitz continuous dependency with respect to both the boundary condition and the source term for linear elliptic equations. The optimal convergence order is then derived making use of standard finite elementerror estimate for linear elliptic equations.

Now, we give an outline of the paper. In Section 2 we state the continuous alternating Schwarz sequences and define their respective finite element counterparts in the context of nonmatching overlapping grids. Section 3 is devoted to theerror analysis of the method.

2. Preliminaries

We begin by laying down some definitions and classical results related to linear elliptic equations.

2.1. Linear Elliptic Equations

Letbe a bounded polyhedral domain oforwith sufficiently smooth boundary. We consider the following: the bilinear form

the linear form

the right hand side

the space

whereis a regular function defined on.

We consider the linear elliptic equation: Findsuch that

where, such that

Letbe the space of finite elements consisting of continuous piecewise linear functionsvanishing on, and letbe the basis function of. The discrete counterpart of (5) consists of findingsuch that

where

andis an interpolation operator on.

Theorem 1 (cf. [11]). Under the suitable regularity of the solution of problem (5), there exists a constantindependent ofsuch that

Lemma 2 (cf. [4]). Letsatisfynonnegative, andon. Thenon.

The proposition below establishes a Lipschitz continuous dependency of the solution with respect to the data.

Notation. Let;be a pair of data and;the corresponding solutions to (5).

Proposition 3 (cf. [6]). Under conditions of the preceding lemma, one has:

Remark 4. Lemma 2 stays true in the discrete case.

Indeed, assume that the discrete maximum principle (d.m.p) holds; that is, the matrix resulting from the finite element discretization is an M-Matrix. Then one has the following.

Lemma 5. Letsatisfyandon. Then,on.

Proof. The proof is a direct consequence of the discrete maximum principle.

Let;be a pair of data and;the corresponding solutions to (7).

Proposition 6 (cf. [6]). Let the d.m.p hold. Then, under conditions of Lemma 5, one has

2.2. The Schwarz Alternating Methods for Nonlinear PDEs

Consider the nonlinear PDE:

or in its weak form

whereis a nondecreasing nonlinearity. We assume thatis a Lipschitz continuous on, that is

such that

whereis the constant defined in (6).

We suppose that the problem (12) has a subsolutionand aersolutionwhich satisfyDefine the sector of smooth functions as follows:

Finally we assume that

where

With these assumptions cf. [12], the problem (12) has a unique solution in the sectorWe can easily see that (12) is equivalent to

whereand we note that the functionalsatisfies (17) and (18). We introduce the following comparison lemma.

Lemma 7 (cf. [3]). Suppose thatandonwith bothpositive inandonThenon

We decomposeintooverlapping smooth subdomainssuch that

We denote bythe boundary ofand the interior boundaries by

We assume that the intersection ofand,is empty. Let

We associate problem (13) with the following system. Findsolution of

where

and (cf. [13])

In the sequel, we define a subsolution and a supersolution of (23)

Definition 8. A smooth functionis a subsolution of (23) if

where

Similarly, a supersolutionof (23) satisfies

2.3. The Continuous Schwarz Sequences

Letbe an initialization data:

We define the alternating Schwarz sequencesonsuch that each term of each sequence,solves

where (cf. [13])

Theorem 9. The Schwarz sequences,converge to the solution of (23).

Proof. We first show by induction, that each term of the Schwarz sequences is well defined inIndeed, for, we begin by subdomain one, since,andsatisfy

respectively with

Lemma 7 implies that

In subdomain 2, the condition of the interior boundaryis given by

so (37) implies that

On the other hand,, andsatisfy, respectively,

Then Lemma 7 implies that

The same idea is used sequentially in the remainder of the proof related to the rest of subdomains,. Indeed we obtain forand for

Now, let us assume that

and prove that

Indeed, in subdomain 1 (43) implies that

Since

then Lemma 7 implies that

Using (47) and (43), we can write

We can easily obtain

using (48) and by adopting the same approach used in subdomain one. The result related to rest of subdomains,, is obtained sequentially by similar way; that is

We demonstrate, by induction, that the Schwarz sequences are nondecreasing. Indeed forwe have demonstrated in (50) that, for,

Now, let us assume that

and prove that

Begining by subdomain one, (52) implies that

Making use of (54),

We get by using Lemma 7

On the other hand, (52) and (56) imply

then making use of (57),

We get by using Lemma 7 the following result related to the second subdomain:

The same idea is sequentially applied to the rest of subdomains. Finally, we obtain

The sequencesare bounded and nondecreasing, so they are monotone converging pointwise to,. By using an elliptic regularity argument, we can see that functionssatisfy the same PDE oncf. [3, 12].

2.4. The Discretization

Forletbe a standard regular and quasiuniform finite element triangulation in;being the meshsize. Themeshes is mutually independent onwhere; that is a triangle belonging to one triangulation does not necessarily belong to the other ones.

We consider the following discrete spaces:

and for every, we set

wheredenotes the interpolation operator on.

The Discrete Maximum Principle (cf. [14, 15]). We assume that the respective matrices resulting from the discretiztions of problems (30) are M-matrices.

Note that as the meshesare independent over the overlapping subdomains, it is impossible to formulate a global approximate problem which would be the direct discrete counterpart of problem (13).

2.5. The Discrete Schwarz Sequences

Now, we define the discrete counterparts of the continuous Schwarz sequences defined in (30). Indeed, letbe the discrete analog of, defined in (29) and we define forthe discrete sequencesonsuch thatsolves

where

Notation. From now on, we will adopt the following notations:

3.   Error Analysis

3.1. The Auxiliary Schwarz Problems

This section is devoted to the proof of the main result of the present paper. To that end, we begin by introducingdiscrete auxiliary problems.

We define the following problems; forsolves

It is then clear thatis the finite element approximation ofsolutions of (23). Therefore, making use of standard maximum norm estimates for linear elliptic problems, we have

whereis a constant independent of.

3.1.1. The Main Results

Theorem 10. Let. Then, there exists a constantindependent of bothandsuch that

Proof. The proof of (68) will be carried by induction and is decomposed in three principal parts where each part is devoted to one of the following situations: situation (A), situation (B), and situation (C), respectively. The three situations are defined by

and in the last situation we assume that a nonempty subsetof the setexists such that

To this end, we apply Proposition 6 and Theorem 1 for each subdomainas follows:

The last inequality implies that

The intervalwhichbelongs to is divided into two subintervals as follows:

and in each situation (A), (B), or (C) we deal with each subinterval separately.

Part 1. We consider situation (A). We begin the proof by the first subinterval that is, so

Forapplying (73) to subdomain one, we get

We have to distinguish between two cases:

or

Case  1 implies that

Then,

Case  2 implies that

By multiplying (81) bywe get

Sois bounded by bothandthen

or

That is,

or

which implies that

or

It is clear that both cases (a) and (b) are true because both coincide with (69). So for either of them there is a contradiction, and thus case 2 is impossible or case 2 is possible only if

Then case 2 implies that

Hence in both cases 1 and 2, we obtain

Similarly applying (73) to the second subdomain, we get

We have to distinguish the three following cases

or

or

Case  1  implies that

Then,

We note that the two last inequalities coincide with (91) and (69), respectively. Case  2 implies that

By multiplying (98) by, we get

We remark thatis bounded by bothandthen

or

Which implies that

or

That is

or

It is clear that both cases (a) and (b) are true because they coincide with (91). So there is either contradiction and case 2 is impossible or cases 2 is possible and we must have

Then case 2 implies

With

which coincides with (69). Case  3 implies

By multiplying (109) bywe get

Sois bounded by bothandthen w have

or

Which implies

or

That is,

or

It is clear that both cases (c) and (d) are true because they coincide with (69). So there is contradiction and case  3 is impossible or case 3 is possible only if

Then case 3 implies that

with

Hence, in the three cases 1, 2, and 3 we get

Equations (91) and (120) imply

Similarly applying the same idea for the rest of subdomain, orderly, we get

Now, let us assume that

and prove that

Equation (124) is obtained sequentially in the order of the numbering of subdomains. We begin by subdomain 1; indeed applying (73) to subdomain one, for iteration, we get

We have to distinguish the two following cases

or

Case  1 implies

Then

Case  2 implies

By multiplying (130) bywe get

Sois bounded by bothandthen

or

That is

or

Which implies

or

It is clear that both cases (a) and (b) are true because both coincide with (123). So either there is a contradiction and thus case 2 is impossible or case 2 is possible only if

Then case 2 implies

Hence in both cases 1 and 2 we obtain

Similarly applying (73) to the second subdomain, for iterationwe get

We have to distinguish the three following cases

or

or

Case  1 implies

Then

We note that the two last inequalities coincide with (140) and (123) respectively. Case  2 implies that

By multiplying (147) bywe get

We remark thatis bounded by bothandthen

or

Which implies

or

That is

or

It is clear that both cases are true because they both coincide with (140). So there is either contradiction and case 2 is impossible or cases 2 is possible and we must have

Case  2 implies that

With

which coincides with (123). Case  3 implies

By multiplying (158) bywe get

Sois bounded by bothandthen

or

Which implies

or

That is

or

It is clear that both cases (c) and (d) are true because they both coincide with (123). So there is contradiction and case 3 is impossible or case 3 is possible only if

Then case 3 implies

With

Hence in three cases 1, 2, and 3 we get

Equations (140) and (169) imply

Similarly applying the same idea for the rest of subdomains,orderly we get

which is the desired result. In the sequel we deal with the second subintervalindeedso

Forapplying (73) to subdomain one we get

We have to distinguish the two following cases

or

Case  1 implies

Then

Case  2 implies

By multiplying (178) bywe get

Sois bounded by bothandthen

or

That is

or

Which implies

or

It is clear that only the case (a) is true because it coincides with (69) whereas (b) contradicts (172). Then case (a) with case 2 imply

Hence in both cases 1 and 2 we get

By the same way we can obtain orderly

The remainder of the proof is also by induction by addopting the same idea applied in iteration one.

Part 2. We deal with situation (B). We begin by the first subinterval that isIndeed for, (73) implies for subdomain 1,

We have to distinguish the two following cases

or

Case  1 implies

So

We remark that (193) contradicts (70) so case 1 is impossible. Case  2 implies

By multiplying (194) bywe get

It is clear thatis bounded by bothand, so

or

Which implies

or

Then

or

We can see that only case (b) is possible and by addingin (199) we get

thus

comparing this last inequalty (203) with (194), we deduce the two following possibilities

or

which implies

or

It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if

Similarly applying (73) to subdomain 2 we get

So we have to distinguish between the three following cases

or

or

Case  1 implies

Then

Equations (214) contradicts (70) so case 1 is impossible. Case  2 implies

By multiplying (215) bywe get

It is clear thatis bounded by bothandso

or

That is

or

It is clear that only case (b) is true because it coincides with (208) and by addingin (220) we get

comparing this last inequalty with (215), we distingush the two following possibilities

or

which imply

or

It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if

Case  3 implies

By multiplying (227) bywe get

We remark thatis bounded by bothandthen

or

That is

or

Which implies

or

It is clear that only case (d) is possible. By addingin (232) we get

so

comparing the last inequality with (227), we consider the two following possibilties

or

that is

or

It is clear that the two possibilities are true so either there is a contradiction and case 3 is impossible or case 3 is possible only if

So case 1 is impossible and in both cases 2 and 3 we obtain

Equations (208) and (242) imply

We adopt the same approach for the other subdomainsin order, we get

The remainder of the proof is by induction and is by adopting the same approach used in iteration one. The last step of the proof of this Part  2 is devoted to the subintervaland is also by inductionIndeed for subdomain 1 and(73) implies

We have to distinguish the two following cases

or

Case  1 implies

So

We remark that (249) coincides with (70). Case  2 implies

By multiplying (250) bywe get

It is clear thatis bounded by bothand, so

or

Which implies

or

Then

or

It is clear that both cases (a) and (b) are true. So either there is contradiction and case 2 is impossible or case 2 is possible and we must have

Then case 2 implies

So in both cases 1 and 2 we get

Similarly applying (73) to subdomain 2 we get in this situation (B)

So we have to distinguish between the three following cases

or

or

Case  1 implies

Then

Case  2 implies

By multiplying (267) bywe get

It is clear thatis bounded by bothandso

or

That is

or

Thus

or

It is clear that both cases (a) and (b) are true because they both coincide with (208). So either there is contradiction and case 2 is impossible or case 2 is possible and we must have

With

So case  2 implies

Then

Case  3 implies

By multiplying (279) bywe get

We remark thatis bounded by bothandthen

or

That is

or

Which implies

or

It is clear that both cases (c) and (d) are true. So either there is contradiction and case 3 is impossible or case 3 is possible only if.

Then in case 3 we obtain

Thus

With

We adopt the same approach for the other subdomains,in order, we get

The remainder of the proof related to iterationsis by induction by adopting the same idea used in iteration one.

Part  3. We deal with situation (C). The proof related to this part  3 is also by induction and is so similar to the part  2 one. So we give it for the first subinterval. The one for the second subintervalwill be omitted. Indeed forandwe use (73) in conjunction with (71) we get

We have to distinguish the two following cases

or

Case  1 implies

We remark that (295) contradicts (71) so case 1 is impossible. Case  2 implies

By multiplying (296) bywe get

It is clear thatis bounded by bothand, so

or

Which implies

or

Then

or

We can see that only case (b) is possible and by addingin (301) we get

thus

comparing this last inequalty (305) with (296), we deduce the two following possibilities

or

which implies

or

It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if

Similarly applying (73) to subdomain 2 in conjunction with (71) we get

So we have to distinguish between the three following cases

or

or

Case  1 implies

Then

Equation (316) contradicts (71) so case 1 is impossible. Case  2 implies

By multiplying (317) bywe get

It is clear thatis bounded by bothandso

or

That is

or

It is clear that only case (b) is true because it coincides with (310) and by addingin (322) we get

comparing this last inequalty with (317), we distingush the two following possibilities

or

which imply

or

It is clear that the two possibilities are true so either there is a contradiction and case 2 is impossible or case 2 is possible only if

Case 3 implies

By multiplying (329) bywe get

We remark thatis bounded by bothandthen

or

That is

or

Which implies

or

It is clear that only case (d) is possible. By addingin (334) we get

so

comparing the last inequality with (329), we consider the two following possibilties

or

that is

or

It is clear that the two possibilities are true so either there is a contradiction and case 3 is impossible or case 3 is possible only if

So case 1 is impossible and in both cases 2 and 3 we obtain

Equations (310) and (344) imply

We adopt the same approach for the other subdomains,in order, we get

The remainder of the proof is by induction and is by adopting the same approach used in iteration one.

Conflict of Interests

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

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