#### Affine Discontinuous Galerkin Method Approximation of Second-Order Linear Elliptic Equations in Divergence Form with Right-Hand Side in

Affine Discontinuous Galerkin Method Approximation of Second-Order Linear Elliptic Equations in Divergence Form with Right-Hand Side in
Abdeluaab Lidouh and Rachid Messaoudi
Department of Mathematics and Computer Science, Laboratory LACSA, Faculty of Sciences, Mohammed 1st University, BV Mohammed VI, P.O. Box 717, 60000 Oujda, Morocco
Correspondence should be addressed to Rachid Messaoudi; rf.oohay@sne_dihcar_m
Received 4 February 2018; Accepted 21 May 2018; Published 2 July 2018
Academic Editor: Patricia J. Y. Wong
Copyright © 2018 Abdeluaab Lidouh and Rachid Messaoudi. 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 the standard affine discontinuous Galerkin method approximation of the second-order linear elliptic equation in divergence form with coefficients in and the right-hand side belongs to ; we extend the results where the case of linear finite elements approximation is considered. We prove that the unique solution of the discrete problem converges in for every with ( or ) to the unique renormalized solution of the problem. Statements and proofs remain valid in our case, which permits obtaining a weaker result when the right-hand side is a bounded Radon measure and, when the coefficients are smooth, an error estimate in when the right-hand side belongs to verifying for every , for some
1. Introduction
In this work we consider, in dimension or , the discontinuous Galerkin (dG) method approximation of the Dirichlet problemwhere is an open bounded set of , is a coercive matrix with coefficients in , and belongs to .
The solution of (1) does not belong to for a general right-hand side in . Actually, in order to correctly define the solution of (1), one has to consider a specific framework, the concept of renormalized (or equivalently entropy) solution (see for example [1, 2]). These definitions allow one to prove that in this new sense problem (1) is well-posed in the terminology of Hadamard.
For this problem the standard -nonconforming finite elements approximation, related to a triangulation of , namely,
where
with the discrete bilinear form yet to be designed, has a unique solution, since the right-hand side (2) is correctly defined for and the bilinear form is consistent.
Using the ideas which are at the root of the SWIP (Symmetric Weighted Interior Penalty) method, in the case , D. A. Di Pietro and A. Ern have proved, in [3], that the unique solution of (2) converges to the unique solution of (1) in the following sense:
with the broken gradient and the jump seminorm yet to be designed.
To do that, the authors in [3] assume that the family of triangulations belong to an admissible mesh sequence in the sense of 17 and is compatible with the partition (see Assumption 3).
The framework in this paper is the same as in [3]. The unique difference here is that is considered instead of ; and we ourselves focus on the two cases and The same convergence results are proved.
Notations. In the present work, denotes an open bounded subset of with or . A particular case is the case where is an open bounded polyhedron. We use the notation for the scalar product of the vector by the vector (which is often denoted by ). For a measurable set , we denote by the measure of and by the complement of .
For and , we have
We define also the following function spaces:
For , we define the broken polynomial space with polynomial degree k 1.
In that case, , which leads us to define the broken gradient such that, ,
and the broken divergence operator such that, ,
Moreover, for any mesh element , we denote
And for a scalar-valued function v defined on (which can admit two possible traces) the average of v is defined as
and the jump of v as
For any face F and for any integer , we define the (local) lifting operator as follows. For all ,
and for any function , we define the (global) lifting of its interface and boundary jumps as
We also introduce the normal diffusion coefficient to one face F as
the diffusion-dependent penalty parameter (harmonic average of normal diffusion) as
the weighted average operator for all such that as
the weighted average operator for all and for a.e. as
on boundary faces F such that F , we setand the skew-weighted average operator for all and for a.e. , as
The SWIP bilinear form is defined by (see Lemma 4.47 in [3])
where the quantity denotes a user-dependent penalty parameter which is independent of the diffusion coefficient.
And the SWIP norms are defined by
with the diffusion-dependent jump seminorms
The discrete Galerkin norm is defined by
with the jump seminorm
For every with , we denote by the Marcinkiewicz space whose norm is defined by
For every real number we define the truncation by
For every in , we adopt the following notations:(i), , denote the vertices of .(ii), , denote the centers of the faces .(iii), , designate the barycentric coordinates with respect to the ’s.(iv)for every we put where are the shape functions related to ; it is known that with .(v)If N designate the number of all interior centers of faces F in we define the interpolation operator and the truncated interpolation operator by with .(vi)Finally, we define the stiffness matrix ; namely, As in [4], the main assumption of the present paper is that is a diagonally dominant matrix; namely,
2. Statement of the Main Result
We consider a matrix such thatfor some , and a right-hand side such that
A function is the renormalized solution of the problem (1) if satisfies
It is known (see [1, 5]) that when belongs to , the usual weak solution of (1), namely,
is a renormalized solution of (1) and conversely the main interest of definition of renormalized solution is the following existence, uniqueness, and continuity theorem (see [1, 4]).
Theorem 1. Assume that and satisfy (33), (34), and (35). Then there exists a renormalized solution of (1). This solution is unique. Moreover this unique solution belongs to for every with . It depends continuously on the right-hand side in the following sense: if is a sequence which satisfieswhen tends to zero, then the sequence of the renormalized solutions of (1) for the right-hand sides satisfies for every and for every with when tends to zero, where is the renormalized solution of (1) for the right-hand side . Finally, if and belong to , and if and are the renormalized solutions of (1) for the right-hand sides and , then, for every , the function belongs to and for every with one hasWhere the constant only depends on , , and .
Remark 2. Throughout all this paper, we denote by any real constant which only depends on the parameters , , and …. We can use the same notation for different constants.
Now we consider a family of triangulations satisfying for each the following assumption:
Note that because of (iv) the triangulations are conforming. A particular case is the case where is a polyhedron of , and where coincides with for every .
In practice, the diffusion coefficient (i.e., matrix A) has more regularity than just belonging to . Henceforth, we make the following assumption (assumption 4.43 [3]):
An important assumption on the mesh sequence is its compatibility with the partition in the following sense (assumption 4.45 [3]).
Assumption 3 (mesh compatibility). We suppose that the admissible mesh sequence is such that, for each , each is a subset of only one set of the partition . In this situation, the meshes are said to be compatible with the partition .
For every , we denote by the diameter of and by the diameter of the ball inscribed in We setand we assume that tends to zero.
We also assume that the family of triangulations is regular in the sense of P. G. Ciarlet [6]; namely, there exists a constant such that
For every triangulation , we consider the discrete problem:
Note that the right-hand side of (48) makes sense since and . The discrete bilinear forme is consistent and coercive (see (128)) on , so a straightforward consequence of the Lax-Milgram Lemma is that the discrete problem (48) is well-posed. The solution of (48) exists and is unique.
As in [4], the main result of this paper is the following.
Theorem 4. Assume that , , and satisfy (33), (34), (35), (44), (46), (47), and (32). Then the unique solution of (48) satisfies for every and for every with when tends to zero, where is the unique renormalized solution of (1).
This theorem will be proved in Section 4, using the tools that we will prepare in Section 3. In Section 5, we will explain why the results of [4] when is a bounded Radon measure remain valid in our case. In Section 6 we also show that if we assume in addition that for every , we obtain for smooth solutions an error estimate in -norm (Section 6.1), and for Low-Regularity solutions an error estimate in -norm (Section 6.2). Finally, in Section 7 we show that in the case where A is the identity matrix, condition (32) remains satisfied when every inner angle of every d-simplex of is acute.
3. Tools
We are going to prove Theorem 4 in several steps. We begin by proving the following result which is a piecewise variant of a result of L. Boccardo & T. Gallouët [2, 5].
Theorem 5. Assume that satisfiesfor some . Then, for every with ,where the constant only depends on , , and
As in [4], to prove Theorem 5, we use the following lemmas.
Lemma 6. Under assumption (47), for all T and all , one has
Proof. Indeed, let and , soand by (47) one haswhich combined with the fact thatwhere in 2D, and in 3D, implies (52).
Lemma 7. Under assumption (47), for every q such that 1 q, the following bound holds for any :
Proof. For every , we denote by the (constant) gradient of the restriction of to . With this notation, using the continuity of across any in at the mass center of any internal , the fact that vanishes at the mass center of any external , and the known inequalityand using (52) we getwhich is (56) with
Lemma 8. Let and let If for some there exists with , then there exists a -simplex with such thatand the strictly positive constant only depends on .
Proof. Let be a simplex from the triangulation , , and , such that Consider (i)If , thus so , and such that .(ii)If , thus , such that so , and .In both cases, there exists an element in such that .
But , so In other words and since one obtainsso and as soon as For this purpose we define the -simplex such that soand to estimate the measure of , it is clear to verify first that . Let be the reference unit -simplex with vertices , , where is the canonical basis of . Let be the invertible affine mapping that maps onto and set .
Since is affine, it is easy to check that for are the barycentric coordinates with respect to the ’s and thatwhere is a constant that depends only on . This proves the result.
Lemma 9. Assume that satisfies (50), thenfor every , where if and is any real number with if ; is defined byand is a constant depending only on , , , and
Proof. Discrete Sobolev’s theorem (see theorem 5.3 in [3]) asserts thatand we will also need (56) with Here or so if and can be any real number with .
Fix If , from Lemma 8, we know that there exists , with andThereforeHenceCombining with (73) and (74) one hasFinally using (50), we obtainwhich is (71) with
Proof (proof of Theorem 5; see [4]). Fix with Take in the case , and verify , in the case
The embedding inequality () writesand by (56) one can haveSo to prove Theorem 5, it suffices to estimateSo let For every , we can writeBut coincides, up to a set of measure zero, withOn the other hand, if , then , soBy the use of (50), one hasWe fix to have ; using (71) we obtainso, for to have , in both cases or , one hasand thenfor every , which finishes the proof.
Lemma 10. Let For every , with , the set , defined bysatisfies
Proof. If , , and , soTherefore HenceThe estimate (92) follows.
Lemma 11. Let and , then the set , defined by satisfies
Proof. Indeed, if then there are two possibilities: (i)If so .(ii)If so and obviously In the two cases The estimate (97) follows.
Remark 12. It is then clear that, under hypothesis (50),
In addition, one has the following.
Proposition 13. Let and If satisfies (50), then
Proof. Fix and , For such that , we can writeOn the one hand, with in (71), one hasOn the other hand,Indeed, if and such that , then , , and, for every in , , which means , and
Therefore, with Lemma 10, and (50), one hasThe convergence (99) is then a consequence of (101) and (103).
Lemma 14. Let . For every and every with , one has
Proof. Let such that , and with It is easily checked that , , and So there are three possibilities.(i) and , and then ,(ii) and , and then ,(iii) and , and then In all cases , and (104) follows.
Proposition 15. Assume that satisfies (50). Then, for every , one haswhen tends to zero.
Proof. Fix and such that and considerLet and with It is easily checked thatwhich implies that . So there are four possibilities.(i) changes sign in ; then, by continuity, (ii) changes sign in ; then(iii) and (or ) so (a)If , then and (b)If , then and (iv) and (or ). So(a)If . There are three possibilities:case 1:, so andcase 2:, so ;case 3:, so and(b)If , thensoWe can then conclude thatConvergence (106) is then consequence of (98) and (99).
The result and the proof of the Proposition (2.7) in [4] can be conserved without changes.
Proposition 16. Under assumption (32), one has for every and every
Proof (proof of Proposition 2.7 in [4]). Sinceone haswhereFix ; (i)if , then et and ,(ii)if , then and thereforein the light of the assumption (32). This proves that for every in ; and so (117).
4. Proof of the Main Theorem
We first show an a priori estimate (compared with (50)) on the solution of (48).
Lemma 17. Let , , and such that Then
Proof. Let , , and such that For every ,the inequality (123) is then proved.
Proposition 18. Under the assumption of Theorem 4, the solution of (48) satisfies for every and every and, in particular, satisfieswhere the constant only depends on , and d.
The proof makes use of results appearing in [3] that we reproduce as follows.
Lemma 19 (Lemma 1.46 in [3], discrete trace inequality). Under assumption (47), one has for all , all T , and all F where only depends on and d.
Lemma 20 (Lemma 4.51 in [3], discrete coercivity). For all , the SWIP bilinear form is coercive on with respect to the ; i.e.,where
Proof (of Proposition 18). Using as a test function in (48) one hasand from (117) we obtain (125).
On the other hand, by combining (125) with coercivities (34) and (128) one obtains successivelyand by (123) the estimation (126) is then proved.
Theorem 21. Under the assumptions of Theorem 4, the solution of (48) satisfies for every with when tends to zero, where is the unique renormalized solution of (1).
Proof (proof of Theorem 3.2 in [4]). Consider a sequence in , converging strongly in to (for example, ). Let to be the unique solution of (48) for the right-hand side Then satisfiesApplying estimate (126) to this problem, we obtain for every , every , and every Which implies by Theorem 5 that for every with , every , and every On the other hand, since and satisfies (44), (46), and (47), it is known (see [3]) that for every fixed when tends to zero, where is the unique solution ofFinally, the function is also the unique renormalized solution of the problemThe estimate (43) combined with the inequality allows one to havefor every with , where is the unique renormalized solution of (1).
Writing nowUsing (134), (136), (137), and (141), one has for every and every with and passing to the limit when tends to zero proves Theorem 21.
To complete the proof of Theorem 4, it remains to prove the following proposition.
Proposition 22. Under the assumptions of Theorem 4, the solution of (48) satisfiesfor every .
Proof (proof of Proposition 3.3 in [4]). First by result of Proposition 15 and the estimate (126) one can have (see theorem 5.7 in [3]) the two following convergences:for every .
On the other hand, using (123) one hasso, by Lebesgue’s dominated convergence theorem combined with discrete Rellich-Kondrachov’s compactness theorem (see theorem 5.6 in [3]) one hasTherefore passing to the limit with respect to in (125) yields ConsequentlyBy the fact thatand since is the renormalized solution of (1), it is known that (see [4])Finally, from (152) and (154), we deduce thatwhich combined with the weak convergence (148) impliesOwing to Proposition 4.36 in [3], for all and all , one can haveand, since the right-hand side tends to zero, the result (146) holds.
Finally, using the result of Proposition 4.34 in [3]with the triangle inequality that yieldsconcluding the proof of (145).
5. The Case Where f Is a Bounded Radon Measure
The materials used in [4], to handle the case where f belongs to , are not specific to the case of finite elements approximation; only the weak convergence (148) requires clarification; in our approach it is based on the result of Proposition 15 whose proof involves only properties of not . So we can also state the following convergence result.
Theorem 23 (Theorem 4.1 in [4]). Assume that f belongs to and A and satisfy (46), (33), (35), (44), and (47) and (32). Then there exist a subsequence, still denoted by h, and a function u such that for every and for every q with 1 one haswhen tends to zero along this subsequence, where satisfies
6. Convergence Rate Estimation6.1. Error Estimates for Smooth Solutions
Assumption 24 (regularity of exact solution and space ). As in [3], we assume that is compatible with the partition in the sense of Assumption 3, and the unique solution is such thatAnd we set
The convergence analysis is performed in the spirit of (Theorem 1.35 [3]) by establishing discrete coercivity, consistency, and boundedness for . The discrete bilinear form is extended to .
Without further knowledge on the exact solution v apart from the domain and the datum , Assumption 24 can be asserted for instance if the domain is convex; see Grisvard [7].
A straightforward consequence of the Lax-Milgram Lemma is that the discrete problem (48) is well-posed.
Theorem 25. Under the assumptions of Theorem 4 ( or ), if f belongs to for some r such that , a convex polyhedron , and , then with
Proof (proof of Theorem 5.1 in [4]). From [3] the unique solution of (48) with right-hand side verifiessoCombined with (142), (134), and (141) allows one to haveand by proceeding as in [4], we obtain (173).
Remark 26. If for some with andthen a small adaptation of the proof given in [4] provides an error estimate in -norm, with , since, with (168), it is known that
6.2. Error Estimates for Low-Regularity Solutions
Assumption 27 (regularity of exact solution and space ). As in [3], we assume that the mesh is compatible with the partition in the sense of Assumption 3, , and that there is such that ; the unique solution is such thatwhere designate that the mesh is compatible with the partition , and we setWe also assume since, in the case , Assumption 27 amounts to Assumption 24.
Assumption 27 requires p 1 for d = 2 and p > 6/5 for d = 3. In particular, we observe that, in two space dimensions, with p 1 holds true in polygonal domains; see, e.g., Dauge [8]. Moreover, using Sobolev embeddings (see [Evans [9], Sect. 5.6] or [Brézis [10], Sect. IX.3]), Assumption 27 implies
Theorem 28. Under the assumptions of Theorem 4 ( or ), if f belongs to for some r such that , a convex polyhedron , and A , then with
Remark 29. Moreover, under the assumption in Remark 26 one can have an error estimate in -norm, with .
7. The Case Where A Is the Identity Matrix
Returning to the definition (31) of mass matrix where , one has
for in
If , so
since , and
If , so (see Proposition 6.1 in [4]),
It is therefore concluded that, under the condition (), one can have . Thus, the matrix verifies (32).
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
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