Discontinuous finite volume methods for

INTERNATIONAL JOURNAL FOR NUMERICAL METHODS IN ENGINEERING Int J Numer Meth Engng (2015) Published online in Wiley Online Library (wileyonlinelibrary.com) DOI: 10.1002/nme.5171 Discontinuous finite volume methods for the stationary Stokes–Darcy problem Gang Wang, Yinnian He*,† and Rui Li School of Mathematics and Statistics, Xi’an Jiaotong University, Xi’an, 710049, China SUMMARY In this paper, we analyze discontinuous finite volume methods for the stationary Stokes–Darcy problem that models coupled fluid flow and porous media flow The discontinuous finite volume methods are combinations of finite volume method and discontinuous Galerkin method with three interior penalty types (incomplete symmetric, nonsymmetric, and symmetric), briefly, using discontinuous functions as trial functions in the finite volume method Optimal error estimates in broken H norm are obtained for the three discontinuous finite volume methods Optimal error estimates in the standard L2 norm are derived for the symmetric interior penalty discontinuous finite volume method Numerical experiments are presented to confirm the theoretical results with non-matching meshes across the common interface of Stokes region and Darcy region Copyright © 2015 John Wiley & Sons, Ltd Received 16 July 2015; Revised 22 October 2015; Accepted November 2015 KEY WORDS: discontinuous Galerkin; finite volume method; Stokes equation; Darcy law; non-matching meshes; optimal error estimates INTRODUCTION The discontinuous Galerkin (DG) finite element method was originally introduced for a linear hyperbolic problem by Reed and Hill in 1973 [1] Because of the use of discontinuous functions, DG method has the advantages of a high order of accuracy, high parallelizability, local mass conservation, allowed hanging nodes, and easy handling of complicated geometries Since then, the study of DG methods has been an active research field Recognizing that the interelement continuity of approximating functions could be weakly imposed, in the 1970s, Arnold [2] and Wheeler [3] introduced interior penalty DG method for elliptic and parabolic problems Recently, Arnold et al [4] provided a framework for the analysis of a large class of DG methods The framework has given profound understanding and comparison of most of the DG methods that have been proposed over the past three decades for the numerical treatment of elliptic problems Well-known shortcoming of DG methods is a large number of degrees of freedom For the coercivity of bilinear term, we need to select large enough penalty parameters according to the specific problems More research works concerning DG methods can be found in [5–11] As another classical discretization technique for solving the partial differential equations, finite volume method (FVM) deals with integral equations obtained over a control volume on a dual mesh The attractive property of FVM is the local conservation of a quantity of interest, such as mass, momentum, or energy Because of its physical conservation property and simplicity, FVM is widely used in computational fluid mechanics and other applications But we cannot achieve a high order of accuracy when using FVM Recently, Cui [12] established a general framework for analyzing the class of FVMs for the Stokes equations Under the framework, optimal convergence order for velocity and pressure in various Sobolev norms *Correspondence to: Yinnian He, School of Mathematics and Statistics, Xi’an Jiaotong University, Xi’an, 710049, China † E-mail: heyn@mail.xjtu.edu.cn Copyright © 2015 John Wiley & Sons, Ltd G WANG, Y HE AND R LI is obtained in a natural and systematic way Many researchers have focused on this area and gained some significant results; some works can be found in [13–16] for elliptic problem, in [17–20] for Stokes and Navier–Stokes problems, and in [21, 22] for some overviews Combining DG and FVM, Ye [23] proposed a discontinuous finite volume method (DFVM) for the second-order elliptic problem and obtained a priori error estimates in a mesh-dependent norm Like we expect, combining the benefits of DG and FVM, such method using discontinuous piecewise polynomials for the trial functions has the flexibility, a high order of accuracy of DG, and the simplicity and conservative property of FVM The fascinating nature of DFVM reflects on the smaller conservation control volume that is less than half the size of control volume applied in the existing FVMs The localizability of the discontinuous element and its dual partition in DFVM should provide an advantage for parallel computing Then Ye [24] spread DFVM to the Stokes problem, using discontinuous P1 P approximations of velocity and pressure DFVM has been used to solve elliptic problem with adaptive technique or other applications [25–29] Recently, there are a few papers applying DFVM to Stokes problem [30–32] Kumar [33] has applied DFVM to solve nonlinear problem In this paper, we study DFVM for solving Stokes–Darcy model describing the coupling fluid flow with porous media flow This model is composed of Stokes equations for the fluid flow and Darcy’s law for the porous media flow, coupling through three interface conditions A lot of methods have been developed to numerically solve the Stokes–Darcy model, such as finite element methods [34], two-grid methods [35], DG methods [36], decoupled methods [37, 38], domain decomposition methods [39, 40], and so on Compared with these existing methods, DFVM has the property of local mass conservation at a so-called diamond control volume level Because of using low-order polynomial as the trial function, DFVM is easy to code, but also, a high order of accuracy can be achieved Because discontinuous functions are used in the approximation, the number of degrees of freedom is larger, but still in acceptable range So, parallel computing is a good choice in the future In this article, we obtained the optimal order error estimates corresponding the variables in L2 norm and broken H norm Some experiments are presented to illustrate the correctness of the theory Our numerical examples show that non-matching meshes are allowed across the interface of Stokes domain and Darcy domain The rest of the paper is organized as follows In Section 2, we describe the Stokes–Darcy model and give the abstract function spaces, the preparations of DG In Section 3, we discuss the DFVM for the Stokes–Darcy model and some lemmas that are used in the error analysis In Section 4, we present optimal order error estimation In Section 5, some experiments are made to validate the correctness and applicability of the algorithms Throughout the paper, the letter C denotes a positive constant independent of the mesh size and may indicate different values at its different appearances PRELIMINARY AND NOTATION We consider a coupled Stokes–Darcy model in a bounded domain Rd d DT2 or 3/, consisting of a fluid region f and a porous medium region p , with interface € D @ f @ p , as depicted in Figure Both f and p have Lipschitz continuous boundaries nf and np denote the unit outward normal vectors on @ f and @ p , respectively, and i i D 1; 2; : : : ; d 1/ the unit tangential vectors on the interface € Note that np D nf on € In f , the fluid flow is assumed to be governed by the Stokes equations: ² uf C rp D gf r uf D in in ; f; f (2.1) where > is the kinetic viscosity, uf denotes the fluid velocity, p denotes the kinematic pressure, and gf denotes a general body force term that includes gravitational acceleration In p , the flow is governed by the Darcy law: Copyright © 2015 John Wiley & Sons, Ltd Int J Numer Meth Engng (2015) DOI: 10.1002/nme DISCONTINUOUS FINITE VOLUME METHODS FOR THE STATIONARY STOKES–DARCY PROBLEM Figure A sketch of the porous medium domain p, < r q D gp q D Kr' : u D q p n in in in fluid domain f , and interface € p; p; (2.2) p; where q is the specific discharge, gp is a sink/source term, K is the hydraulic conductivity tensor, p n is the volumetric porosity, up is the fluid velocity in p , and the hydraulic head ' D ´ C gp , the sum of elevation head plus pressure head, where pp is the pressure of the fluid in p , is the density of the fluid, g is the gravitational acceleration, and ´ is the elevation from a reference level Without loss of generality, we assume ´ D Furthermore, we assume that K D diag.K; : : : ; K/ with K L1 p /, K > 0, which implies that the porous media is homogeneous Simplifying (2.2) results in r Kr'/ D gp p: in (2.3) The key part for this coupled model is the interface conditions that describe how different types of flow interact at the fluid/porous medium interface €: p u f n f D u p nf @uf D g' nf @nf @uf ˛ Dp i @nf i K on €; on €; uf i (2.4) on €; i where ˛ is a positive parameter depending on the properties of the porous medium Equation (2.4) is the Beavers–Joseph–Saffman–Jones condition For simplicity, we assume that the hydraulic head ' and the fluid velocity uf satisfy the homogeneous Dirichlet boundary condition except on €, that is, uf D on @ ' D on @ n €; n p €: f (2.5) We will use the standard notations for the Sobolev spaces H s D/ equipped with their usual inner products ; /s;D , norms k ks;D , and seminorms j js;D , s > The space H D/ coincides with L2 D/, in which case the norm and inner product are denoted by k kD and ; /D , respectively Let L20 D/ denote the subspace of L2 D/ consisting of functions with mean integral value zero In this paper, D denotes f or p Without loss of generality, if no specified, we will drop subscript D throughout the paper And jj jje is L2 integration on edge e Copyright © 2015 John Wiley & Sons, Ltd Int J Numer Meth Engng (2015) DOI: 10.1002/nme G WANG, Y HE AND R LI Figure An example of primal mesh and its dual mesh Let Rf;h be a regular triangulation of f with diam f / hf;K and hf;K the diameter of element Every triangle K Rf;h is divided into three subtriangles by connecting the barycenter of the triangle K to its corner nodes, as shown in Figure Then we define the dual partition Tf;h of the primal partition Rf;h to be the union of the triangles T as shown in Figure for triangular mesh Respectively, we define primal mesh Rp;h and dual mesh Tp;h for Darcy domain p with mesh size hp;K We define the mesh parameter hf D max hf;K and hp D max hp;K In the K2Tf;h K2Tp;h theoretical analysis, we will drop subscript f for hf ; hf;K , respectively, p for hp and hp;K And he denotes the length of edge e The same as the usual practice, Pk T / denotes the space of polynomials with degree less than or equal to k defined on T We first define two function spaces: ® V D v H ® XD H f ¯ n€ ; ¯ D on @ p n€ : //2 j v D on p /j @ f We define the following finite dimensional trial function space for velocity on triangular partition by ® Vh D v L2 f ¯ //2 W vjK P1 K//2 ; 8K Rf;h : Define the finite dimensional test function space Wh for velocity associated with the dual partition Tf;h as ¯ ® Wh D v L2 f //2 W vjT P0 T //2 ; 8T Tf;h : Let Qh be the finite dimensional space for pressure ® Qh D q L2 f ¯ / W qjK P0 K/; 8K Rf;h : Respectively, we define trial function space Xh and test function space Yh for piezometric head '.x/: ® ¯ W jK P1 K/; 8K Rp;h ; ® ¯ Yh D L2 p / W jT P0 T /; 8T Tp;h : Xh D L2 p/ Let Efo denote the set of all interior edges e in Rf;h , EfD denote the set of all boundary edges that belong to Dirichlet boundary in Rf;h , and Ef€ denotes the set of all boundary edges that belong to interface € in Rf;h , so we have Ef D Efo [ EfD [ Ef€ For Rp;h , we define Ep D Epo [ EpD [ Ep€ Copyright © 2015 John Wiley & Sons, Ltd Int J Numer Meth Engng (2015) DOI: 10.1002/nme DISCONTINUOUS FINITE VOLUME METHODS FOR THE STATIONARY STOKES–DARCY PROBLEM For vectors v and n, let v ˝ n denote the matrix whose ij th component is vi nj For two matrixP valued variables and , we define W D 2i;j D1 ij ij Let e be an interior edge shared by two elements K1 and K2 in Th here; Tf;h or Tp;h /, and let n1 and n2 be unit normal vectors on e pointing exterior to K1 and K2 , respectively We define the average ¹ º and jump Œ  on e for scalar q, vector v; and matrix , respectively qj@K1 C qj@K2 ; Œq D qj@K1 n1 C qj@K2 n2 ; vj@K1 C vj@K2 ; Œv D vj@K1 n1 C vj@K2 n2; ¹vº D ¹qº D and ¹ ºD j@K1 C j@K2 ; Œ  D j@K1 n1 C j@K2 n2 : We also define a matrix-valued jump an edge on the boundary of here; f for a vector as v D vj@K1 ˝ n1 C vj@K2 ˝ n2 on e If e is or p /, define ¹qº D q; Œv D v n; ¹ º D ; v D v ˝ n: Let E denotes the union of the boundaries of the triangles K of Rh and E0 WD E n@ A straightforward computation gives X Z qv nds D @K K2Rh X Z K2Rh XZ nds D v @K XZ e2E0 Œq ¹vº ds C e e2E0 e Œ  ¹vº ds C XZ XZ e2E ¹qº Œvds; (2.6) ¹ º W v ds: (2.7) e e2E e Let V.h/ D Vh C H f //2 \ V, X.h/ D Xh C H p / \ X For connecting trial and test function spaces of fluid velocity and the piezometric head, we define mapping operators f W V.h/ ! Wh , p W X.h/ ! Yh , Z vjT ds T Tf;h ; he Ze jT D jT ds T Tp;h ; he e f vjT D p as shown in Figure Throughout the paper, we will drop the subscript f and p of the mapping operators f and p In fact, the mapping operator satisfies the following assumptions: Assumption For vf V.h/, K Rf;h , Z vf vf /dx D 8vf Vh ; (2.8) vf /ds D 8e @K; 8vf Vh ; (2.9) K Z vf e Copyright © 2015 John Wiley & Sons, Ltd Int J Numer Meth Engng (2015) DOI: 10.1002/nme G WANG, Y HE AND R LI if vf D 0; then vf D 0; (2.10) vf kK C hK jvf j1;K ; k vf (2.11) 8e Ef : k vf ke k vf ke (2.12) And we have similar results for variable These results can be found in [12] DISCONTINUOUS FINITE VOLUME FORMULATIONS Multiplying (2.1) by vf Vh and q Qh , respectively, we have Z X n T 2Tf;h ruf nvf ds C @T Z X n T 2Tf;h pn vf ds D @T X Z n gf vf dx (3.1) T T 2Tf;h and X Z n r uf qdx D 0: (3.2) K K2Rf;h where n is the unit outward normal vector on @T Let Tj Tf;h j D 1; 2; 3/ be the triangles in K Rf;h , as shown in Figure Then we have X T 2Tf;h Z n ruf nvf ds D @T D Z X X n ruf nvf ds K2Rf;h j D1 @Tj K2Rf;h j D1 Aj C1 CAj Z X X n X K2Rf;h Z n ruf nvf ds (3.3) ruf nvf ds: @K Figure A triangular element K and its dual elements Tj j D 1; 2; 3/ Copyright © 2015 John Wiley & Sons, Ltd Int J Numer Meth Engng (2015) DOI: 10.1002/nme DISCONTINUOUS FINITE VOLUME METHODS FOR THE STATIONARY STOKES–DARCY PROBLEM where A4 D A1 Making use of the second and third terms of interface conditions (2.4), we have Z X n ruf nvf ds @K K2Rf;h Z X D n ruf nvf ds e o D e2Ef [Ef  @uf C i @nf X Z n p à g'/nf vf (3.4) n  ruf nvf ds e à ˛ p i nf nf Z o [E D e2Ef f e € e2Ef e X vf ds D à @uf @nf n € e2Ef à i Z  X K uf i i vf /ds: i Then, using (2.7) and (3.4) and the fact that ruf D 0, for uf V \ H we rewrite (3.3) as follows: n K2Rf;h X o e2Ef //2 on Efo [ EfD , Z X D f ruf nvf ds @K Z n ® ¯ ruf W vf ds e D [Ef X Z n p € e2Ef  g'/nf vf e à ˛ p i K uf i i vf /ds: i (3.5) By same computation, we obtain that X T 2Tf;h D Z n pn vf ds @T Z X X n K2Rf;h j D1 pn vf ds C Aj C1 CAj X o e2Ef Z ¹pº Œvf ds C n e D [Ef X € e2Ef Z n pnf vf ds: e (3.6) Multiplying Darcy’s Equation (2.3) by Yh , we have Z X g @T T 2Tp;h Z X Kr' n ds D g T 2Tp;h gp dx: (3.7) T Let Tj Tp;h j D 1; 2; 3/ be the triangles in K Rp;h , as shown in Figure Then we have X T 2Tp;h D Z g Kr' n ds @T X X K2Rp;h j D1 Z g Kr' n ds Aj C1 CAj X K2Rp;h (3.8) Z g Kr' n ds: @K Using (2.6) and the third interface condition in (2.4) and the fact that Œr' D for ' X \H on Epo [ EpD , (3.7) and (3.8) become Copyright © 2015 John Wiley & Sons, Ltd p/ Int J Numer Meth Engng (2015) DOI: 10.1002/nme G WANG, Y HE AND R LI Z g X X Aj C1 CAj K2Rp;h j D1 D g e o D e2Ep [Ep Z X Z Z X g ¹Kr'º Œ ds gn uf nf ds X Kr' n ds e € e2Ep gp dx: T T 2Tp;h (3.9) So, 8uf ; vf V.h/; '; X.h/; p; q L20 following bilinear and linear terms: A1 uf ; '/; vf ; // Z X X n D Z C X n g'nf vf C p e o e2Ef K i X D [Ef he ˇ f ¹Kr'º Œ uf i X vf /ds i X uf W vf ds C gK e ds e i Z ˛f Z g à ˛ vf ds e D [Ef o [E D e2Ep p  ® ¯ ruf W n X ds Aj C1 CAj Z € e2Ef Cn Kr' n Z X o e2Ef g K2Rp;h j D1 /, by the mapping operators We define the ruf n vf ds Aj C1 CAj K2Rf;h j D1 X X f ˛p o [E D e2Ep p Z gn Z u f nf ds e € e2Ep Œ'Œ ds; he ˇ p e (3.10) Z X X n B vf ; /; p/ D pn n A uf ; '/; vf ; // Z X X n D K2Rf;h j D1 r uf qds K K2Rf;h X Aj C1 CAj X X (3.11) e Z ¹qº Œ uf ds; n (3.12) e o D e2Ef [Ef ruf n vf ds ¹pº Œ vf ds; n o [E D e2Ef f Z X C uf ; '/; q/ D vf ds C Aj C1 CAj K2Rf;h j D1 Z X Z g Kr' n ds; Aj C1 CAj K2Rp;h j D1 (3.13) Z X X n B vf ; p/ D K2Rf;h j D1 F vf ; /D X T 2Tf;h Copyright © 2015 John Wiley & Sons, Ltd pn Z n gf vf dx C T vf ds; (3.14) Aj C1 CAj X T 2Tp;h Z g gp dx; (3.15) T Int J Numer Meth Engng (2015) DOI: 10.1002/nme DISCONTINUOUS FINITE VOLUME METHODS FOR THE STATIONARY STOKES–DARCY PROBLEM where ˛f > 0, ˛p > are penalty parameters to be determined later In this article, we take ˇf D ˇp D It is clear that the exact solution uf ; '; p/ of the Stokes–Darcy problem satisfies the following: A1 uf ; '/; vf ; // C B vf ; /; p/ D F vf ; C uf ; '/; q/ D / 8.vf ; / Vh Xh ; 8q Qh : (3.16) (3.17) Algorithm 1: The discontinuous finite volume scheme for Stokes–Darcy problem finds uf;h ; 'h /; ph / Vh Xh / Qh such that A1 uf;h ; 'h /; vf ; // C B vf ; /; ph / D F vf ; C uf;h ; 'h /; q/ D / 8.vf ; / Vh 8q Qh : Xh ; (3.18) (3.19) We define two norms for V.h/ and X.h/ as follows: X e2Ef ˇ he f jkvkj21 D jvj21;h C jkvkj D jkvkj21 C X jj v jj2e ; h2K jvj22;K ; K2Rf;h jk kj2 D j j21;h C X e2Ep ˇ he p jjŒ jj2e C X h2K j j22;K ; K2Rp;h P where j j21;h D K j j21;K In fact, it is easy to see that jk kj and jk kj1 are equivalent by the standard inverse inequality For ease of analysis, we define the mixed L2 norm and broken H norm as follows: k.vf ; /k2 D nkvf k2 C gk k2 ; jk.vf ; /kj2 D n jkvf kj2 C gKjk kj2 : We will take advantage of the following trace inequalities in the later theoretical analysis Let T be an element and e be an edge of T ; there exists a constant C that depends only on the minimum angle of T such that for any function u H T /, jjujj2e C hT jjujj20;T C hT juj21;T ; (3.20) ˇˇ ˇˇ2 ˇˇ @u ˇˇ ˇˇ ˇˇ C h juj2 C hT juj2 : T 1;T 2;T ˇˇ @n ˇˇ e (3.21) Firstly, we give the following two equivalence relations Lemma ([24]) 8uf ; vf V.h/; '; X.h/, we have Copyright © 2015 John Wiley & Sons, Ltd Int J Numer Meth Engng (2015) DOI: 10.1002/nme G WANG, Y HE AND R LI A uf ; '/; vf ; // D n ruf ; rvf / C n n uf ; vf K2Rf;h ruf n vf vf /ds @Kn€ K2Rf;h X vf /K C gK.r'; r / Z X C Z X g Kr' n /ds @Kn€ K2Rp;h X g.r Kr'; /K (3.22) K2Rp;h C Z X n p g'/nf vf vf / e € e2Ef  à ˛ uf i KZ i X C gn uf nf p i vf vf //ds i /ds: e € e2Ep Furthermore, if uf ; '/; vf ; / Vh Xh , and p Qh , A uf ; '/; vf ; // D n ruf ; rvf / C gK.r'; r /  X Z ˛ n g'nf vf vf /C p e € e2Ef C i à K uf i i vf vf //ds i Z X gn uf n f /ds: e € e2Ep (3.23) Lemma ([24]) 8vf V.h/; q L20 B vf ; q/ D /, we have Z X n.r vf ; q/ C n f K2Rf;h Z Cn vf vf @Kn€ R € ds represents P X n.rq; vf vf /K K2Rf;h vf /nqds; € where vf / nqds C € e2Ef R e ds or P € e2Ep R e (3.24) ds Furthermore, if q Qh ; vf Vh , then B vf ; q/ D r vf ; q/ (3.25) C.vf ; q/ D B.vf ; q/: (3.26) and Next, we set up the continuity and coercivity of the bilinear form A1 ; / as well as the boundedness of the bilinear C ; / Lemma 8uf ; vf V.h/; '; X.h/, there exists constant C D C ; g; n; ; K; ˛f ; ˛p / such that A1 uf ; '/; vf ; // C jk.uf ; '/kj jk.vf ; /kj: (3.27) Proof Firstly, using Hölder inequality, Poincaré inequality, (3.20), and (2.12), we estimate A uf ; '/; vf ; // and the interface integration in the bilinear form A1 uf ; '/; vf ; //, Copyright © 2015 John Wiley & Sons, Ltd Int J Numer Meth Engng (2015) DOI: 10.1002/nme DISCONTINUOUS FINITE VOLUME METHODS FOR THE STATIONARY STOKES–DARCY PROBLEM  vf C p Z A uf ; '/; vf ; //Cn € g'nf ˛ K i à uf Z i vf /ds i gn ds uf nf € i C @n juf j1;h jvf j1;h X Cn hK1 jj vf vf jj2 K C hK j vf vf j2 1;K Á1 2 hK1 juf j2 1;K C hK juf j2;K Á1 K2Rf;h X Cn hjuf j2;K jvf j1;K C gKj'j1;h j j1;h K2Rf;h X C gK hK1 jj jj2 K C hK j j2 1;K Á1 2 hK1 j'j2 1;K C hK j'j2;K Á1 K2Rp;h X C gK hj'j2;K j j1;K A C C g n @ K2Rp;h 2 A jj'jj2 0;K C hK j'j1;K / ÂZ € K2Rp;h 11 Á ÂZ 2 A jjuf jj2 C h ju j 0;K K f 1;K X C C gn@ 11 X Œ 2 ds he € K2Rf;h 11 Á ÂZ 2 A jjuf jj2 C h ju j 0;K K f 1;K X n˛ CCp @ K K2R f;h € vf he vf he Ã1 ds Ã1 Ã1 ds n˛ C jk.uf ; '/kj jk.vf ; /kj C C g n.jk'kj jkvf kj C jkuf kj jk kj/ C C p jkuf kj jkvf kj: K (3.28) By means of (2.12), (3.20), Hölder inequality, and Cauchy–Schwarz inequality, we have Z X ® n e o D e2Ef [E f X Cn ˛f he o D e2Ef [E f B 6n @ B C gK @ Z e C B he jruf j ds A @ B C gK˛p @ e o D e2Ep [E p 6n @ X uf he juf j2 1;K C B ds A @ e C B Œ '2 ds A @ he C h2 K juf j2;K Á A X C gK @ € à ÂZ Ã1 C ds A C  ds A 11 vf he e Z e vf he Á ÂZ 2 A C h j'j j'j2 1;K K 2;K K2Rp;h ÂZ e Z 11 2 o D e2Ep [E p € vf Œ he X ÂZ Œ'Œ ds e 11 Z o D e2Ef [E f K2Rf;h e X 11 Z X he o D e2Ep [E p 11 Z o D e2Ef [E f C B he jr'j ds A @ Z X e X B C n ˛f @ 11 Z ˛p he 11 Z o D e2Ef [E f o D e2Ep [E p o D e2Ep [E p X Copyright © 2015 John Wiley & Sons, Ltd C ds A 11 Œ he C  ds A Ã1 ds Œ 2 ds he Ã1 2 1 uf ds vf ds C gK˛p € he € he n C ˛f /jkuf kj jkvf kj C gK.1 C ˛p /jk'kj jk kj: C n ˛f ds e X uf W vf ds C gK X ¹Kr'º Œ g o D e2Ep [E p e o D e2Ef [E f Z X vf ds 11 Z X ¯ ruf W ÂZ € Œ'2 ds he à ÂZ € Œ 2 ds he Ã1 (3.29) Int J Numer Meth Engng (2015) DOI: 10.1002/nme G WANG, Y HE AND R LI The sum of (3.28) and (3.29) yields the continuity Lemma ([24]) For vf V.h/; X.h/; q L20 f /, 12 X C B h2K jqj21;K A A : C vf ; /; q/ jk.vf ; /kj @kqk C @ 0 (3.30) K2Rf;h Furthermore, if vf Vh ; Xh ; q Qh , then C vf ; /; q/ jk.vf ; /kj kqk: Lemma For vf ; / Vh h small enough (3.31) Xh , there is a constant C such that for ˛f ; ˛p large enough and for mesh size A1 vf ; /; vf ; // > C jk.vf ; /kj2 : (3.32) Proof From Lemma 3, we know that Z X o e2Ef ¯ rvf W 12 jvf j21;K C h2K jvf j22;K A ÂZ C gK @ j j21;K C h2K j j22;K A ÂZ € K2Rp;h ds e vf he € 12 X ¹Kr º Œ g o D e2Ep [Ep K2Rf;h Z X vf ds e D [Ef X 6n @ ® n à 12 ds Œ 2 ds he (3.33) à 12 : So, using Young inequality, inverse inequality, and (3.33), we have A1 vf ; /; vf ; // à  Z ˛ g nf vf C p vf i > A vf ; /; vf ; //Cn € i K i X ˛f Z X ˛p Z Cn vf ds C gK Œ 2 ds h h e e e e o D o D e2Ef [Ef e2Ep [Ep 12 n @ X jvf j21;K C h2K jvf j22;K A ÂZ K2Rf;h gK @ X 12 j j21;K C h2K j j22;K A K2Rp;h Copyright © 2015 John Wiley & Sons, Ltd € € Œ 2 ds he ÂZ i gn vf nf ds € à 12 vf he Z vf /ds ds à 12 Int J Numer Meth Engng (2015) DOI: 10.1002/nme DISCONTINUOUS FINITE VOLUME METHODS FOR THE STATIONARY STOKES–DARCY PROBLEM X > n jjrvf jj C gKjjr jj C n o [E D e2Ef f n 2"1 X ˛f he "1 jvf j21;K C h2K jvf j22;K K2Rf;h gK 2"2 X j j21;K C h2K j "2 j22;K K2Rp;h Z vf X ds C gK e vf he n e o D e2Ef [Ef X o [E D e2Ep p Z X Z gK e o D e2Ep [Ep Z ˛p he Œ 2 ds e ds Œ 2 ds he > C jk.vf ; /kj2 : (3.34) For the aforementioned inequality, in order to maintain n 2"1 n jjrvf jj2 C gKjjr jj2 X jvf j21;K C h2K jvf j22;K K2Rf;h gK 2"2 X j j21;K C h2K j j22;K > 0: K2Rp;h We must take "1 > 0, "2 > large enough and hK small enough And then ˛f ; ˛p must hold larger enough in the contest with "1 ; "2 This completes the proof of the lemma In fact, in the definition of A1 ; /, we make full use of the idea of incomplete interior penalty discontinuous Galerkin finite element method As we all know, the drawback of incomplete interior penalty discontinuous Galerkin method is the dependence of penalty parameter To absorb the classical theory of interior penalty DG method, we set up the following two discontinuous finite volume schemes The second scheme is parameter insensitive with reference to nonsymmetric interior penalty discontinuous Galerkin method The third scheme (symmetric interior penalty discontinuous Galerkin (SIPG) method) depends on the penalty parameters Then we define A2 uf ; '/; vf ; // D A1 uf ; '/; vf ; // C Z X o e2Ef n D [Ef ¯ ® rvf W uf ds C e Z X ¹Kr º Œ 'ds: g o [E D e2Ep p e (3.35) Algorithm 2: The second discontinuous finite volume scheme for Stokes–Darcy problem seeks uf;h ; 'h /; ph / Vh Xh / Qh such that A2 uf;h ; 'h /; vf ; // C B vf ; /; ph / D F vf ; C uf;h ; 'h /; q/ D / 8.vf ; / Vh Xh ; 8q Qh : (3.36) (3.37) We can see that A2 vf ; /; vf ; // > A vf ; /; vf ; // C n X o D e2Ef [Ef ˛f he Z vf e ds C gK X o D e2Ep [Ep ˛p he Z Œ 2 ds e > C jk.vf ; /kj2 : So, Algorithm is unconditionally stable with respect to the choice of the penalty parameters Algorithm 3: The third discontinuous finite volume scheme for Stokes–Darcy problem seeks uf;h ; 'h /; ph / Vh Xh / Qh such that Copyright © 2015 John Wiley & Sons, Ltd Int J Numer Meth Engng (2015) DOI: 10.1002/nme G WANG, Y HE AND R LI A3 uf;h ; 'h /; vf ; // C B vf ; /; ph / D F vf ; C uf;h ; 'h /; q/ D / 8.vf ; / Vh Xh ; 8q Qh : (3.38) (3.39) Here, we define A3 uf ; '/; vf ; // Z X D A1 uf ; '/; vf ; // ® n ¯ rvf W uf ds e o [E D e2Ef f X o [E D e2Ep p Z ¹Kr º Œ 'ds: g e (3.40) The continuity and coercivity of bilinear A2 ; /, A3 ; / can be obtained by means of the proof of bilinear A1 ; / Remark In the variational formulation (3.38)–(3.39), we take ˇf D ˇp D 1, corresponding to the SIPG method in the context of DG methods applied to the second-order linear elliptic problems Optimal L2 error estimates can be derived for the SIPG method by the Aubin–Nitsche lift technique used in the analysis of the classical DG finite element method So, in Section 4, we prove the optimal L2 error estimates for variational formulation (3.38)–(3.39) On the other two cases, they have been observed numerically on uniform meshes that L2 convergence rate is optimal while using linear approximation functions Note that if using super-penalty techniques and taking ˇf D ˇp D 3, the optimal L2 error estimates can be achieved in the existing theoretical analysis of DG method ERROR ESTIMATES It has been proved in [24] that bilinear form C ; / satisfies the following discrete inf condition That is, there exists a positive constant ˇ , such that 8q Qh , sup C vf ; /; q/ /2.Vh ;Xh / jk.vf ; /kj sup 0Ô.vf ; D C vf ; /; q/ 0Ô.vf ;0/2.Vh ;Xh / jk.vf ; /kj D C vf ; 0/; q/ > kqk; jkvf kj 0Ô.vf ;0/2.Vh ;Xh / sup (4.1) sup where ˇ is a positive constant depending only on f Define operators …1 W V ! V h , …2 W X ! Xh , …3 W L20 f / ! Qh ; the following results can be seen in [24], C vf jvf j …1 vf ; /; q/ D …1 vf js;k C h2 s jvf j2;k …2 js;k C h2 s j j2;k Theorem 4.1 Let uf;h ; 'h ; ph / Vh ; Xh / H f //2 \ V H p / \ X such that 8q Qh ; 8vf V 8K Rf;h ; s D 0; 1; 8vf V 8K Rp;h ; s D 0; 1; X: (4.2) (4.3) (4.4) Qh be the unique solution of (3.18)–(3.19) and uf ; '; p/ L20 f / be the solution of (3.16)–(3.17); then there exists C > Copyright © 2015 John Wiley & Sons, Ltd Int J Numer Meth Engng (2015) DOI: 10.1002/nme DISCONTINUOUS FINITE VOLUME METHODS FOR THE STATIONARY STOKES–DARCY PROBLEM jk.uf uf;h ; ' 'h /kj C kp ph k C h.kuf k2 C k'k2 C kpk1 /: (4.5) Proof Let uf;h D uf uf …1 uf uf;h …1 uf / DW h; (4.6) ' 'h D ' …2 ' 'h …2 '/ DW Á Áh ; (4.7) p ph D p …3 p ph …3 p/ DW  Âh : (4.8) Subtracting (3.18) and (3.19) from (3.16) and (3.17) gives A1 ; Á/; vf ; // C B vf ; /; Â/ D A1 h ; Áh /; vf ; // C B vf ; /; Âh / C ; Á/; q/ D C h ; Áh /; q/ D 0: (4.9) (4.10) for any vf Vh ; Xh ; q Qh Taking vf D h ; D Áh , and using (4.9), (4.10), and Lemma 2, one can find that A1 h ; Áh /; h ; Áh // D A1 ; Á/; h ; Áh // C h ; Áh /; Áh /: (4.11) Using Lemmas 3–5, we can show that jk h ; Áh /kj2 0 B B C @jk h ; Áh /kj jk ; Á/kj C jk h ; Áh /kj @kÂk C @ X K2Rf;h 12 11 CC h2K jÂj21;K A AA ; (4.12) which implies jk.uf;h …1 uf ; 'h B C @jk uf …2 '/kj …1 uf ; ' …2 ' kj C kp …3 pk C @ X K2Rf;h h2K jp 12 C (4.13) …3 pj21;K A A : Using triangular inequality and the properties of interplant operators …1 ; …2 ; …3 , that is, (4.2)–(4.4), we obtain jk.uf uf;h ; ' 'h /kj jk.uf …1 uf ; ' …2 '/kj C jk.uf;h …1 uf ; 'h …2 '/kj C @jk.uf …3 pk C@ X …1 uf ; ' h2K jp K2Rf;h Copyright © 2015 John Wiley & Sons, Ltd …2 '/kj C kp (4.14) 12 C …3 pj21;K A A C h.kuf k2 C k'k2 C kpk1 /: Int J Numer Meth Engng (2015) DOI: 10.1002/nme G WANG, Y HE AND R LI By Lemmas and and (4.1), we obtain that kph ˇ …3 pk ˇ D ˇ D ˇ C vf ; /; ph …3 p/ jk.vf ; /kj /2.Vh ;Xh / sup 0Ô.vf ; B vf ; /; …3 p ph / jk.vf ; /kj /2.Vh ;Xh / sup 0Ô.vf ; B vf ; /; p sup ph / C B vf ; /; …3 p jk.vf ; /kj 0Ô.vf ; /2.Vh ;Xh / A uf sup uf;h ; ' 0Ô.vf ; /2.Vh ;Xh / 0 B B C @jk @uf uf;h ; ' 'h /kj C kp p/ 'h /; vf ; // C B vf ; /; …3 p p/ jk.vf ; /kj 12 11 X CC …3 pk @ h2K jp …3 pj21;K A AA (4.15) K2Rf;h C h.kuf k2 C k'k2 C kpk1 /: Finally, (4.5) can be derived by the triangle inequality and aforementioned inequalities (4.14) and (4.15) In fact, (4.5) can be also obtained for Algorithms and through similar proofs with Theorem 4.1 Consider the dual problem of Stokes–Darcy model: Find w; ˆ; / Uh Xh Qh , such that 8.vf ; ; q/ Vh Xh Qh , a vf ; /; w; ˆ// C b vf ; /; / D n.uf ® where Uh D vf Vj divvf D 0; in f a vf ; /; w; ˆ// D n rvf ; rw/ n o D e2Ef [Ef ® g e ˛f he Z vf w ds e ¹Kr º Œˆds e ¹Krˆº Œ ds C gK o [E D e2Ep p e X o [E D e2Ep p Z C gn X o [E D e2Ef f Z X o [E D e2Ep p Z X (4.16) ¯ rvf W w ds ¹rwº W vf ds C n C g.Kr ; rˆ/ g Z e o [E D e2Ef f 'h ; /; and X n Z X ¯ uf;h ; vf / C g.' ˛p he Z Œ Œˆds e ˆnf vf ds nf w € Z Cn ˛ p € i K vf i / i w/ds; (4.17) i b vf ; /; / D n ; r vf / C n X o D e2Ef [Ef Copyright © 2015 John Wiley & Sons, Ltd Z ¹ º Œvf ds: e (4.18) Int J Numer Meth Engng (2015) DOI: 10.1002/nme DISCONTINUOUS FINITE VOLUME METHODS FOR THE STATIONARY STOKES–DARCY PROBLEM Assume that the dual problem of Stokes–Darcy model has the H f //2 H p // H regularity property in the sense that the solution w; ˆ; / H f //2 H p // H and the following a priori estimate hold true: kwk2 C kˆk2 C k k1 C.kuf uf;h k C k' 'h k/: / f/ f (4.19) Let wI Vh , ˆI Xh be the usual continuous piecewise linear interplant; it is not hard to see that there exists a constant C independent of mesh size h such that jkw wI kj C jkˆ ˆI kj C k …3 k C h.kuf uf;h k C k' 'h k/: (4.20) We have the following optimal order L2 error estimation Theorem 4.2 Let uf;h ; 'h ; ph / Vh ; Xh / Qh be obtained from (3.38) and (3.39) and uf ; '; p/ H f //2 \ V H p / \ X L20 f / be the unique solution of (3.16) and (3.17); then there exists a positive constant C such that k.uf uf;h ; ' 'h /k C h2 kuk2 C k'k2 C kpk1 C kgf k1 C kgp k1 /: (4.21) Proof Firstly, we need the following error equations, which can be obtained from subtracting (3.38) and (3.39) from (3.16) and (3.17), A3 uf uf;h ; ' 'h /; wI ; C uf Taking vf D uf njjuf uf;h ; uf;h jj2 C gjj' D' I // uf;h ; ' C B wI ; I /; p ph / D 0; (4.22) 'h /; …3 / D 0: (4.23) 'h in (4.16), we have 'h jj2 D a uf uf;h ; ' 'h /; w; ˆ// C b uf uf;h ; ' 'h /; /: (4.24) Subtracting (4.22) from the sum of (4.23) and (4.24), and the fact Pthat w DR 0; Œˆ D 0; 0; wI D 0; Œ ˆI  D 0; ŒˆI  D on all interior edges, and K2Rf;h n @Kn€ r.uf n wI wIR/ds D 0, P wI / n.p ph /ds D 0, we deduce that K2Rf;h n @Kn€ wI njjuf uf;h jj2 C gjj' wI D uf;h / 'h jj2 D n r.uf uf;h /; r.w wI // n uf Crp; wI wI / C n.r wI w/; p ph / C n.r uf uf;h /; …3 / C g.Kr.' 'h /; r.ˆ ˆI // C g.r Kr'; ˆI ˆI / X Z Cn ¹rwI ºW uf uf;h / ¹rwº W uf uf;h ds o D e2Ef [Ef C g X e Z ¹KrˆI º Œ ' Cn X o [E D e2Ef f 'h / ¹Krˆº Œ' 'h ds e o [E D e2Ep p Z ¹ º Œuf uf;h  ¹…3 º Œ uf uf;h /ds e Z C gn ' 'h /nf w wI / uf Z € ˛ uf uf;h / i / Cn p € i K i WD I1 C I2 C I3 C I4 ; uf;h / nf ˆ ˆI /ds i w wI //ds (4.25) Copyright © 2015 John Wiley & Sons, Ltd Int J Numer Meth Engng (2015) DOI: 10.1002/nme G WANG, Y HE AND R LI where I2 ; I3 ; I4 denote interface integration and I1 denotes the remaining integration Using the conclusions in [12], we can see that I1 C hn kuf uf;h k jkuf 'h k jk' C C h gKk' uf;h kj C C h2 nkgf k1 kuf 'h kj C C h gkgp k1 k' C C h n.jjuf jj2 C jjpjj1 /jjuf C h njjuf 'h k uf;h jj uf;h jj2 C gjj' C C h2 njjuf uf;h k 'h jj2 uf;h jj2 C gjj' C C h2 jjuf jj2 C jjpjj1 /jj.uf jk.uf 'h jj2 uf;h ; ' uf;h ; ' (4.26) 'h /kj nkgf k21 C gkgp k21 'h /jj: Then we estimate the interface integration, I2 C p ÂZ gK € 6C p Ch 'h 2 ds p 'h kj n @ gKjk' p Œ' he à 12 p n ÂZ he jw wI j2 ds à 12 € X jjw 12 wI jj20;K C h2K jw wI j21;K A (4.27) K2Rf;h gKjk' p 'h kj njjuf uf;h jj: Similarly, we have p I3 C h n jkuf p I4 C h n jkuf uf;h kj uf;h kj p gk' p nkuf 'h k; (4.28) uf;h k: (4.29) Combining the aforementioned four terms with Cauchy–Schwarz inequality, we obtain (4.21) Remark The aforementioned analysis can be spread to the case of Stokes–Darcy model with nonhomogeneous Dirichlet boundary condition NUMERICAL EXPERIMENTS In this section, we present several examples for our DFVM Consider the following three Stokes– Darcy models on ; where f D 0; 1/ 1; 2/ and p D 0; 1/ 0; 1/ with the interface € D 0; 1/ ¹1º The uniform triangular mesh Rh is constructed by (1) dividing the domain into an n n rectangular mesh and (2) connecting the diagonal line with the negative diagonal line Denote the mesh size as h D 1=n We consider two types of non-matching meshes across the common interface as illustrated in Figure with h D 1=4, on the left side of Figure 4, hf D 23 h and hp D h, and on the right side of Figure 4, hf D 12 h and hp D h We take penalty parameters ˛f D and ˛p D Example We test our numerical schemes with all the physical parameters n; ; g; ; K, and ˛ simply set to The Dirichlet boundary conditions and the force terms are determined by the given exact solution as follows: uf x/ D x y Copyright © 2015 John Wiley & Sons, Ltd 1/2 C y; x.y 1/3 C sin x//; (5.1) Int J Numer Meth Engng (2015) DOI: 10.1002/nme DISCONTINUOUS FINITE VOLUME METHODS FOR THE STATIONARY STOKES–DARCY PROBLEM 2 1.8 1.8 1.6 1.6 1.4 1.4 1.2 1.2 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0.2 0.4 0.6 0.8 0 0.2 0.4 0.6 0.8 Figure Non-matching meshes with h D 1=4 Table I Convergence rates and errors of Algorithm for Example h kuf 1=8 1=16 1=32 1=64 1=128 Rate uf;h k 6.3584E 1.5997E 4.0123E 1.0047E 2.5138E 1.9957 jkuf 3 4 uf;h kj 4.7264E 2.3657E 1.1833E 5.9177E 2.9591E 0.9994 1 2 jjp ph jj 1.4141E 5.9531E 2.7807E 1.3589E 6.7426E 1.0976 jj' 2 'h jj 1.6635E 4.4255E 1.1362E 2.8748E 7.2277E 1.9617 3 jk' 'h kj 6.7732E 3.4337E 1.7261E 8.6508E 4.3300E 0.9919 1 2 Table II Convergence rates and errors of Algorithm for Example h 1=8 1=16 1=32 1=64 1=128 Rate kuf uf;h k 6.3670E 1.6004E 4.0141E 1.0054E 2.5159E 1.9959 jkuf 3 4 uf;h kj 4.7246E 2.3650E 1.1831E 5.9172E 2.9590E 0.9992 p.x/ D Œ2 '.x/ D Œ2 1 2 jjp ph jj 1.3321E 5.8630E 2.7792E 1.3616E 6.7532E 1.0755 sin x/sin sin x/Œ1 y jj' 2 'h jj 1.2666E 3.2874E 8.3780E 2.1151E 5.3140E 1.9743 Á y ; 4 jk' 'h kj 6.7277E 3.4254E 1.7246E 8.6479E 4.3294E 0.9895 1 2 (5.2) cos y/: (5.3) As Theorems and state, on quasi-uniform mesh size h, we expect a convergence of order O.h2 / for the velocity and piezometric head in standard L2 norm and O.h/ for the pressure in L2 norm, for velocity and piezometric head in broken H norm For three DFVMs, the error results are presented in Tables I–III with mesh size hf D 23 hp And the convergence rate of error is presented in Figure with mesh size hf D 12 hp , hp D h These results are highly consistent with the results of Theorems and Copyright © 2015 John Wiley & Sons, Ltd Int J Numer Meth Engng (2015) DOI: 10.1002/nme G WANG, Y HE AND R LI Table III Convergence rates and errors of Algorithm for Example h 1=8 1=16 1=32 1=64 1=128 Rate kuf uf;h k 6.4372E 1.6144E 4.0378E 1.0093E 2.5227E 1.9988 jkuf 3 4 uf;h kj 4.7619E 2.3747E 1.1856E 5.9233E 2.9605E 1.0019 1 2 jjp ph jj 1.7376E 6.4273E 2.8335E 1.3626E 6.7396E 1.1721 2 jj' 'h jj 2.8105E 7.5206E 1.9283E 4.8699E 1.2229E 1.9611 3 4 jk' 'h kj 7.0734E 3.4988E 1.7408E 8.6854E 4.3385E 1.0068 1 2 Figure The convergence rates with three discontinuous finite volume methods for example In Examples and 3, we take mesh size hf D 23 hp and hp D h Example We take all parameters remain the same as in Example And the exact solutions are assumed as Copyright © 2015 John Wiley & Sons, Ltd Int J Numer Meth Engng (2015) DOI: 10.1002/nme DISCONTINUOUS FINITE VOLUME METHODS FOR THE STATIONARY STOKES–DARCY PROBLEM  uf x/ D cos y ÁÁ2 sin p.x/ D cos à xÁ sin y/ C y/ =4 ; xÁ ; cos  xÁ y '.x/ D à y ÁÁ2 =4; 2 cos ycos (5.4) (5.5) xÁ =4: (5.6) The domain is initially partitioned into triangles with h D 18 Successive uniform refinements are performed to numerically explain the convergence rate For three DFVMs, the error results are presented in Tables IV–VI The numerical results agree with our theory Example We consider the model with more practical significance physical parameters The test shows the convergence with varying hydraulic conductivity K Assume that other parameters n; ; g; ˛; remain the same as previous test problems We assume the exact solutions as follows: uf x/ D y 2y C 1; x p.x/ D x C y 1/ C x/; (5.7) gn ; 3K (5.8) Table IV Convergence rates and errors of Algorithm for Example h kuf 1=8 1=16 1=32 1=64 1=128 Rate uf;h k 1.2435E 3.1166E 7.7893E 1.9470E 4.8675E 1.9993 jkuf 5 uf;h kj 8.6447E 4.3086E 2.1508E 1.0746E 5.3707E 1.0022 2 2 jjp ph jj 6.4513E 3.4249E 1.7544E 8.8663E 4.4553E 0.9640 2 3 jj' 'h jj 6.3426E 1.5821E 3.9534E 9.8823E 2.4704E 2.0011 4 6 jk' 'h kj 4.3434E 2.1891E 1.0988E 5.5043E 2.7548E 0.9947 2 3 Table V Convergence rates and errors of Algorithm for Example h kuf 1=8 1=16 1=32 1=64 1=128 Rate uf;h k 1.2393E 3.1330E 7.8745E 1.9744E 4.9439E 1.9925 jkuf 5 uf;h kj 8.6064E 4.3002E 2.1489E 1.0741E 5.3696E 1.0006 2 2 jjp ph jj 6.4910E 3.4347E 1.7569E 8.8724E 4.4568E 0.9661 2 3 jj' 'h jj 6.5121E 1.6783E 4.2741E 1.0794E 2.7127E 1.9768 4 5 jk' 'h kj 4.3191E 2.1836E 1.0975E 5.5013E 2.7540E 0.9928 2 3 Table VI Convergence rates and errors of Algorithm for Example h 1=8 1=16 1=32 1=64 1=128 Rate kuf uf;h k 1.3232E 3.3302E 8.4039E 2.1169E 5.3174E 1.9898 jkuf 5 Copyright © 2015 John Wiley & Sons, Ltd uf;h kj 8.9556E 4.3925E 2.1729E 1.0803E 5.3854E 1.0139 2 2 jjp ph jj 6.4876E 3.4278E 1.7536E 8.8610E 4.4534E 0.9662 2 3 jj' 'h jj 9.7811E 2.6292E 6.8113E 1.7324E 4.3678E 1.9517 4 5 jk' 'h kj 4.5290E 2.2393E 1.1116E 5.5367E 2.7629E 1.0088 2 3 Int J Numer Meth Engng (2015) DOI: 10.1002/nme G WANG, Y HE AND R LI Table VII Convergence rates and errors of Algorithm for Example with K D 0:1 h 1=2 1=4 1=8 1=16 1=32 1=64 1=128 Rate kuf uf;h k 2.2955E 4.5363E 9.2432E 2.0817E 4.9968E 1.2322E 3.0675E 2.1449 jkuf 4 5 uf;h kj 2.5283E 1.1795E 5.7520E 2.8538E 1.4234E 7.1110E 3.5544E 1.0254 1 2 3 jjp ph jj 2.4354E 9.4442E 4.2952E 2.0421E 1.0006E 4.9655E 2.4755E 1.1034 jj' 2 2 3 'h jj 1.2611E 3.6680E 1.0656E 2.9151E 7.6485E 1.9606E 4.9645E 1.8851 jk' 2 4 'h kj 2.2025E 1.1381E 5.7780E 2.9070E 1.4569E 7.2901E 3.6461E 0.9861 0 1 2 Table VIII Convergence rates and errors of Algorithm for Example with K D 0:01 h 1=2 1=4 1=8 1=16 1=32 1=64 1=128 Rate kuf uf;h k 4.4006E 7.9100E 1.7902E 4.3017E 1.0609E 2.6387E 6.5841E 2.1177 jkuf 3 4 uf;h kj 3.4444E 1.2519E 5.8424E 2.8646E 1.4248E 7.1129E 3.5546E 1.0997 1 2 3 jjp ph jj 3.6889E 1.9952E 8.4998E 3.0118E 1.1653E 5.1997E 2.5062E 1.2003 jj' 1 2 3 'h jj 1.2928E 3.6321E 1.0511E 2.8802E 7.5669E 1.9412E 4.9174E 1.8934 jk' 1 3 'h kj 2.2056E+1 1.1374E+1 5.7771E 2.9069E 1.4568E 7.2901E 3.6461E 0.9864 Table IX Convergence rates and errors of Algorithm for Example with K D 0:001 h 1=2 1=4 1=8 1=16 1=32 1=64 1=128 Rate kuf uf;h k 3.9601E 2.3292E 3.6643E 8.8591E 2.2617E 5.7437E 1.4464E 2.4568 jkuf 4 5 n '.x/ D K uf;h kj 2.5001E 2.8966E 7.4033E 3.0842E 1.4494E 7.1405E 3.5579E 1.5761  x.1 2 3 x/.y jjp ph jj 5.9104E 1.1538E 4.7328E 1.6336E 4.7199E 1.3186E 3.9971E 1.7550 1/ C y 3 jj' 0 1 2 'h jj jk' 1.3386E+1 3.6176E 1.0282E 2.8251E 7.4420E 1.9122E 4.8485E 1.9051 à y Cy C 'h kj 2.2096E+2 1.1377E+2 5.7768E+1 2.9069E+1 1.4568E+1 7.2901E 3.6461E 0.9869 x: g (5.9) We take K D 0:1; 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