Convergence of a Finite Volume scheme for an elliptic-hyperbolic system

Similar documents
g i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f

Best Approximation in the 2-norm

Convergence of Finite Volumes schemes for an elliptic-hyperbolic system with boundary conditions

ODE: Existence and Uniqueness of a Solution

Problem Set 4: Solutions Math 201A: Fall 2016

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1

Notes on length and conformal metrics

Partial Derivatives. Limits. For a single variable function f (x), the limit lim

Math Advanced Calculus II

The Regulated and Riemann Integrals

New Expansion and Infinite Series

Best Approximation. Chapter The General Case

Theoretical foundations of Gaussian quadrature

Math 426: Probability Final Exam Practice

Czechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction

Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions

ACM 105: Applied Real and Functional Analysis. Solutions to Homework # 2.

MA Handout 2: Notation and Background Concepts from Analysis

Math& 152 Section Integration by Parts

Math 554 Integration

Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors

INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Analysis Autumn 2012

Convex Sets and Functions

Math 5440 Problem Set 3 Solutions

Phil Wertheimer UMD Math Qualifying Exam Solutions Analysis - January, 2015

Math 100 Review Sheet

Math 5440 Problem Set 3 Solutions

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1

Abstract inner product spaces

REGULARITY OF NONLOCAL MINIMAL CONES IN DIMENSION 2

Review of Calculus, cont d

Math 6455 Oct 10, Differential Geometry I Fall 2006, Georgia Tech

Travelling Profile Solutions For Nonlinear Degenerate Parabolic Equation And Contour Enhancement In Image Processing

Lecture 1. Functional series. Pointwise and uniform convergence.

A PROOF OF THE FUNDAMENTAL THEOREM OF CALCULUS USING HAUSDORFF MEASURES

Integration Techniques

ON THE C-INTEGRAL BENEDETTO BONGIORNO

SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set

Calculus of Variations: The Direct Approach

7.3 Problem 7.3. ~B(~x) = ~ k ~ E(~x)=! but we also have a reected wave. ~E(~x) = ~ E 2 e i~ k 2 ~x i!t. ~B R (~x) = ~ k R ~ E R (~x)=!

Partial Differential Equations

KRASNOSEL SKII TYPE FIXED POINT THEOREM FOR NONLINEAR EXPANSION

SOME INTEGRAL INEQUALITIES OF GRÜSS TYPE

1.1. Linear Constant Coefficient Equations. Remark: A differential equation is an equation

ON BERNOULLI BOUNDARY VALUE PROBLEM

Math 1B, lecture 4: Error bounds for numerical methods

Set Integral Equations in Metric Spaces

Math 360: A primitive integral and elementary functions

(4.1) D r v(t) ω(t, v(t))

Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004

S. S. Dragomir. 2, we have the inequality. b a

1.3 The Lemma of DuBois-Reymond

Positive Solutions of Operator Equations on Half-Line

1.9 C 2 inner variations

Advanced Computational Fluid Dynamics AA215A Lecture 3 Polynomial Interpolation: Numerical Differentiation and Integration.

W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying

Analytical Methods Exam: Preparatory Exercises

We partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.

Presentation Problems 5

Variational Techniques for Sturm-Liouville Eigenvalue Problems

440-2 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam

Convergence of Fourier Series and Fejer s Theorem. Lee Ricketson

AN INTEGRAL INEQUALITY FOR CONVEX FUNCTIONS AND APPLICATIONS IN NUMERICAL INTEGRATION

New Integral Inequalities for n-time Differentiable Functions with Applications for pdfs

Approximation of functions belonging to the class L p (ω) β by linear operators

arxiv: v1 [math.ca] 11 Jul 2011

Some estimates on the Hermite-Hadamard inequality through quasi-convex functions

Bernoulli Numbers Jeff Morton

1 1D heat and wave equations on a finite interval

Math Solutions to homework 1

1. On some properties of definite integrals. We prove

Lecture 19: Continuous Least Squares Approximation

Advanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015

Heat flux and total heat

Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1

Fourier series. Preliminary material on inner products. Suppose V is vector space over C and (, )

The usual algebraic operations +,, (or ), on real numbers can then be extended to operations on complex numbers in a natural way: ( 2) i = 1

Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.

STURM-LIOUVILLE BOUNDARY VALUE PROBLEMS

Math 61CM - Solutions to homework 9

Three solutions to a p(x)-laplacian problem in weighted-variable-exponent Sobolev space

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

Functional Analysis I Solutions to Exercises. James C. Robinson

POSITIVE SOLUTIONS FOR SINGULAR THREE-POINT BOUNDARY-VALUE PROBLEMS

MatFys. Week 2, Nov , 2005, revised Nov. 23

The Wave Equation I. MA 436 Kurt Bryan

The Riemann Integral

Research Article Harmonic Deformation of Planar Curves

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014

Riemann is the Mann! (But Lebesgue may besgue to differ.)

Chapter 4. Additional Variational Concepts

II. Integration and Cauchy s Theorem

JEE(MAIN) 2015 TEST PAPER WITH SOLUTION (HELD ON SATURDAY 04 th APRIL, 2015) PART B MATHEMATICS

Section 17.2 Line Integrals

Polynomials and Division Theory

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.

HOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016

Section 3.2 Maximum Principle and Uniqueness

Math 6395: Hyperbolic Conservation Laws and Numerical Methods. Hyperbolic Conservation Laws are a type of PDEs arise from physics: fluid/gas dynamics.

Calculus of Variations

Transcription:

Convergence of Finite Volume scheme for n elliptic-hyperbolic system M.H. Vignl UMPA. CNRS-UMR No 18 E.N.S. Lyon 46, Allée d Itlie 69364 Lyon Cedex 07 Abstrct We study here the convergence of finite volume scheme for coupled system of n hyperbolic nd n elliptic equtions defined on n open bounded set of IR. On the elliptic eqution, four points finite volume scheme is used then n error estimte on discrete H 1 norm of order h is proved, where h defines the size of the tringultion. On the hyperbolic eqution, one uses n upstrem scheme with respect to the flow, then using n estimte on the vrition of the pproximte solution, the convergence of the pproximte solution towrd solution of the coupled system is shown, under stbility condition. Résumé On étudie ici l convergence d un schém de type volumes finis pour un système formé d une éqution elliptique et d une éqution hyperbolique linéires, définies sur un ouvert borné de IR. Pour l éqution elliptique un schém volumes finis à qutre points est utilisé, une inéglité discrète de Poincré pour les fonctions à moyenne nulle, est étblie fin de montrer une estimtion d erreurs en norme H 1 discrète, de l ordre de l tille des milles. Sur l éqution hyperbolique, on utilise un schém décentré vers l mont de l écoulement, à l ide d une estimtion fible sur l vrition de l solution pprochée, on montre, sous une condition de stbilité, l convergence de cette solution vers l solution fible du problème. 1 Introduction One considers problem coming from the modeliztion of diphsic flow in porous medium. In simplified cse it leds to the determintion of the velocity u of one of the two phses nd of the pressure P. Let Ω be n open bounded polygonl connected set of IR, one notes Γ = Ω. Let g L Γ, u 0 L Ω nd u L Γ IR, be given, with Γ = {γ Γ ; gγ 0}, one ssumes gγ dγ = 0 ; then one considers the problem defined by : Γ 1 Px = 0, x Ω u t x,t divux,t Px = 0, x Ω,t IR with following boundry conditions nd initil condition : 3 4 5 Pγ.nγ = gγ, γ Γ uγ,t = uγ,t, γ Γ,t IR ux,0 = u 0 x, x Ω where n is the outwrd unit norml to Γ. 1

More precisely, one serches u in L Ω IR nd P in H 1 Ω solutions of 1-5 in the following wek sense : Px. Φxdx gγφγdγ = 0 for ll Φ H 1 Ω nd Ω IR ux,t ϕ t Ω x,tdx dt Ω for ll ϕ C c Ω IR with Ω = Ω Γ Γ ux,t Px. ϕx,tdx dt IR u 0 xϕx,0dx uγ,tϕγ,tg γdγ dt = 0 Ω Γ IR Note tht the test functions ϕ re equl to zero on Γ = {γ Γ ; gγ < 0} but not necessrily on Γ. To discretize these equtions, finite volume scheme is used, then the results presented by R. Eymrd nd T. Gllouët in [5] nd R. Herbin in [8] re generlized. Indeed the system considered in [5] is the sme s the one presented here but the uthors use coupled finite element-finite volume scheme, then discrete unknowns re loclized t the vertices of the meshes wheres in this note they re loclized t the cells centers. For this scheme they prove convergence property towrd wek solution of system 1-5. The scheme used on the pressure is finite element scheme, hence the convergence of the pproximte solution of the elliptic eqution follows from the finite element frmework. In this note one uses four points finite volume scheme for the pressure, then convergence proof is given by generlizing the results of R. Herbin in [8]. One of the two essentil differences comes from boundry condition, since in [8] the boundry condition is Dirichlet condition wheres here Neumnn condition is considered, then estimtes re chnged by boundry terms. In prticulr, for our estimte it is necessry to count the number of tringles which re fter given tringle in ll directions, while in [8] only one direction is sufficient. In sme wy, to prove the error estimte in discrete H 1 norm, the discrete L norm of the error is mjorized by its discrete H0 1 norm, ssuming the solution s men vlue equl to zero since problem s solutions differ from constnt, wheres in [8] the discrete L norm of the error is mjorized by the sum of its discrete H0 1 norm nd of its discrete L norm on the boundry. The second difference comes from the ssumptions on the meshes. In [8] ll the meshes must hve mesure of sme order, wheres here deformtions of the meshes re uthorized see section.1. On the velocity n upstrem finite volume scheme with respect to the flow is used, then, under stbility condition, the convergence of the pproximte solution towrd solution of the hyperbolic eqution is shown. To prove this result, one needs n estimte on the vrition of the pproximte solution, it uses n estimte on discrete H 1 norm of the pproximte solution of the elliptic eqution, this result in [5] is given by the finite element frmework. Other results on the existence nd the uniqueness of solutions of hyperbolic equtions re given in [7], [3], [1], [10]. Numericl experiments on the comprison between finite element scheme nd finite volume scheme, done by J.M. Fird nd R. Herbin in [6] for conduction problem nd by R. Herbin nd O. Lbergerie in [9] for diffusion-convection problem, hve shown tht the pproximtion of fluxes is better for the finite volume scheme. Furthermore, comprison between the scheme presented here nd the weighted finite volume scheme of R. Eymrd nd T. Gllouët hve lso been done in [6] on system more generl thn 1-5, where 1 is chnged by div fux,t Px = 0, x Ω. This numericl test shows tht the scheme presented here gives better results thn those given by the weighted finite volume scheme. Other uthors hve been interested by finite volume scheme on tringulr meshes, see for instnce [11]. In [11], results re restricted to prticulr meshes, wheres, here, s it hs been lredy remrk, the ssumptions re most generl.

Discretiztion Before discretizing the elliptic nd the hyperbolic equtions, one first gives ssumptions on the tringultion..1 Assumptions on the tringultion Let T = K j 1 j L be tringultion of Ω which stisfies : there exists η such tht for ny θ ngle of n element of T, one hs : 6 η < θ < π η One defines h j = SK j, where SK j denotes the D Lebesgue mesure of K j, then the ssumption 6 gives the following result : there exists α 1 > 0 nd α > 0, depending only on η, such tht for ll side of the tringultion, the length l of verifies : 7 α 1.h j l α.h j if is n edge of the cell K j. Then one defines h IR by h = L mx h j. Some nottions will be useful to describe the numericl scheme : NOTATIONS { } T ext = j {1,...,L} ; K j T, K j Γ Ø A ext the set of the edges of the tringultion which re on the boundry Γ of Ω A int the set of the edges of the tringultion which re in Ω g γ = mxgγ,0 nd g γ = g For ll K j T, 1 j L one notes : c i j the edges of K j, i = 1, or 3 x j the intersection of the orthogonl bisectors of the edges of K j g ij = g γ dγ nd g ij = g γ dγ, i = 1, or 3, if c i j A ext c ij c ij g ij = g ij g ij, i = 1, or 3, if c ij A ext τ ext j the set of the suffix i = 1, or 3 such tht c i j A ext τ j the set of the suffix of the neighbours of K j c jk = K j K k for ll k τ j = dx j,c jk dx k,c jk, for ll k τ j, where d is the euclidin distnce of IR x jk the center of the side c jk 3

. Discretiztion of the elliptic eqution To discretize 1, four points finite volume scheme is used ; the principle of the finite volume schemes, see [4], is to integrte equtions on ech control volume here K j T, so one hs : K j Pγ.n Kj γ dγ = 0 where n Kj denotes the outwrd unit norml to K j. One pproximtes P by P T with P T x = P j if x K j, so the discretized eqution cn be given by pproximting the flux of P through one edge c i j of K j by : lc jk P k P j if k {1,...,L} such tht c i j = c jk gγ dγ if c i j A ext Then one hs : 8 c ij lc jk P k P j i τ extj g ij = 0 for ll j {1,...,L} with the convention = 0. One cn remrk tht on the domin s boundry the pproximtion of the flux is exct..3 Discretiztion of the hyperbolic eqution Before discretizing, one defines the time step δ, so let T be tringultion of Ω which stisfies the ssumption 6 nd α ]0,1[, then one chooses δ IR which stisfies the following conditions : 5 1 δ SK k lc jk P j P k > α j,k S 9 1 δ 5 SK j g ij > α j T ext where S = { j,k {1,...,L} ; K j,k k T T, k τ j nd P j > P k } One notes t n = n δ for ll n IN. To discretize, first n Euler scheme explicit in time is used, nd s for the elliptic eqution, one integrtes on ech control volume : ux,t n1 ux,t n K j δ dx uγ,t n Pγ.n Kj dγ = 0 K j One pproximtes u by u T,δ with u T,δx,t = u n j if x K j nd t [t n,t n1 [. Then n upstrem discrete vlue with respect to the flow is chosen for u t the interfces of meshes, nd t the boundry. One defines u 0 j for ll j {1,...,L} by u0 j = 1 u 0 x dx SK j K j nd u n ji for ll j T ext nd for i τ ext j by u n ji = 1 t n1 uγ,tdγ dt δ lc i j t n c ij 4

So, the discretized eqution is given by : SK j u n1 j u n j δ u n jk lc jk P k P j 10 where u n jk = u n j u n k if P j > P k else δ i τ extj u n ji g ij un j g ij = 0 for ll j {1,...,L} nd ll n IN 3 Convergence of the four points finite volume scheme for the elliptic eqution Let T be tringultion of Ω which stisfies the property 6. One proves in this section the existence of solutions P j 1 j L of 8 nd tht these solutions differ only from constnt, proving the following result : Proposition 1 Let g L Γ, one defines for ll j {1,...,L} : g ij = gγdγ if c i j A ext. Then : 1. if i τ extj g ij = 0 j {1,...,L} nd P j 1 j L stisfy 8 then c ij P j = P k j,k {1,...,L} L. if P j 1 j L stisfy 8 then g ij = 0 3. if Γ gγdγ = 0 then there exists P j 1 j L solutions of 8 nd solutions differ only from constnt i τ extj Furthermore one proves the numericl scheme s convergence proving n error estimte on discrete H 1 norm of order h : Theorem 1 Let g L Γ, one denotes by P the wek solution of 1, 3 such tht L SK j Px j = 0, where x j is the intersection of the orthogonl bisectors of the edges of K j, one supposes g such tht P is in C Ω. L Let P j 1 j L stisfy 8 nd SK j P j = 0, one defines the error by e j = P j Px j for ll j {1,...,L}. Then there exists C 1 nd C positive, independent of T such tht : L 1/ e k e j lc jk C 1.h nd d jk L 1/ SK j e j C.h 5

3.1 Proof of Proposition 1 Proof of the first prt of the Proposition 1 One supposes tht j {1,...,L} g ij = 0 nd P j 1 j L stisfy 8. i τ extj { } Let j 0 {1,...,L} such tht P j0 = min P k ; k {1,...,L} Thnks to 8, one hs : But ccording to ssumptions k τ j0, so one hs : lc P k P j0 j0k g ij0 = 0 d j0k 0 i τ extj 0 g ij0 = 0 ; nd lc j0k > 0, d j0k > 0 nd P k P j0 0 i τ extj 0 P k = P j0 k τ j0 Using the fct tht Ω is connected, one obtins by induction : P j = P k Proof of the second prt of the Proposition 1 Supposing tht P j 1 j L stisfy 8 nd summing these equtions one gets : L i τ extj Proof of the third prt of the Proposition 1 One supposes tht gγdγ = 0. Γ g ij = 0 j,k {1,...,L} Then thnks to the first prt of this Proposition, 8 is liner system which hs kernel of dimension 1. So, thnks to the second prt of this Proposition, the imge spce of this liner system is the set of the L B IR L, B = t b 1,b,...,b L such tht b j = 0. Then, s L b j = from constnt. 3. Proof of Theorem 1 Γ gγdγ = 0, there exists P j 1 j L solutions of 8 nd these solutions differ only Definition of the consistency error on the fluxes As it hs been remrk in section., fluxes re exct on the domin s boundry, then one defines the consistency error only t the interfces of meshes. The exct flux on the side c jk in the direction of K j to K k is : F cjk K j = 1 Pγ.n Kj γ dγ lc jk c jk nd the pproximte flux on the side c jk in the sme direction is : F cjk K j = P k P j 6

One defines the consistency error, denoted by R cjk K j, by : R cjk K j = F cjk K j Px k Px j One cn remrk tht the conservtivity of exct nd pproximte fluxes implies : Now the following result is proved : R cjk K j = R cjk K k Lemm 1 Under the ssumptions of Theorem 1, there exists constnt C R 0 independent of T such tht : R cjk K j C R.h j,k {1,...,L} Proof of Lemm 1 Using first order Tylor expnsion one proves tht there exists C 1 0 nd C 0 depending only on α 1, α nd on the second order derivtive of P such tht : 1 R cjk K j Pγ.n Kj dγ Px jk.n Kj lc jk c jk Px jk.n Kj Px k Px j C 1.h C.h So the proof of Lemm 1 is completed. Let e k = P k Px k be the error on the cell K k for ll k {1,...,L}, then one proves the first result of Theorem 1, i.e. the following property : There exists C 0 independent of T such tht : 11 L 1/ e k e j lc jk C.h d jk Proof of the inequlity 11 Thnks to 1 one hs : 1 lc jk F cjk K j i τ extj g ij = 0 One subtrcts 8 from 1, one multiplies by e j nd one sums over j, then using the conservtivity of the exct nd pproximte fluxes, the properties 6 nd 7, the Lemm 1 nd the Young inequlity for more detils see [8], one gets : L e k e j lc jk 6 α α 1 C R SΩh where SΩ is the D Lebesgue mesure of the domin Ω. Then the proof of the inequlity 11 is completed. Let s complete the proof of Theorem 1. Proof of Theorem 1 The L discrete error estimte is shown by using discrete Poincré-Wirtinger inequlity, i.e. the following result : 7

There exists C > 0, independent of T, such tht : 13 L SK j e j m C A int e K e K with m = 1 L SK k e k, where e SΩ K nd e K re the errors on the both elements of T for which k=1 is n edge nd d is defined s follows : for ll A int there exists j nd k in {1,...,L} such tht = c jk then d =. This result will be proved in four steps : Step 1 : Let P be squre included in Ω nd the both directions D 1 nd D defined by P, see figure 1. d l 4,18190,18 18.000,16.000190.000,18.00018.000,0.000 4.00071,4971,18 4.000 Figure 1: One notes d = b c nd one chooses for coordinte system the coordinte system defined by ny point of IR nd the both directions D 1 nd D. Let A int,p be the set of the sides of A int such tht P. Let K j nd K k in T such tht K j P nd K k P, x = x 1,x K j P nd y = y 1,y K k P, one denotes by [x,y] the line segment delimited by x nd y, nd one defines : A x1,x,y respectively A x1,y 1,y the set of the sides of A int such tht the intersection with the line segment [x 1,x,x 1,y ] respectively [x 1,y,y 1,y ] is point. A x 1 respectively A 1 y the set of the sides of A int,p such tht the intersection with the line defined by x 1,0 respectively 0,y nd prllel to D respectively D 1 is point. Then one hs : e j e k e K e K e K e K A x1,x,y A x1,y 1,y.e. y P Ω, x P Ω 8

Integrting over K k P nd summing over k, one obtins for ll x P Ω : d e j m P with m P = 1 SP c c b b A x 1 L SK k Pe k. k=1 A x 1 e K e K dy1 dy c c b b A 1 y e K e K dy1 dy For ll A int, let θ be the ngle between nd D 1, then swpping the summtion nd the integrl in the second terms, one hs : d e j m P d e K e K e K e K l cos θ A int,p Using the Cuchy Schwrz inequlity, one gets : d e j m P d e K e K sin θ d sinθ d A x 1 A x 1 e K e K l A int,p nd then : e j m P d α h l K j R 1 x A x 1 e K e K sin θ d 3α K j D x 1 e j m P C 1 d A int,p 1 4α h e d K e K A int,p Integrting over [b,c] with respect to x 1 nd swpping the summtion nd the integrl, one hs for ll x [b,c] : e K e K 14 A int,p where D x 1 is the line defined by the point 0,x nd the direction D 1, R 1 x is the set of the tringles which re cut by D x 1 nd C 1 = d α h 6 α d 4α h. α 1 With the sme rguments, one cn prove the following result for ll x 1 [b,c] : 15 l K j R x 1 K j D x 1 e j m P C 1 d One concludes by integrting 14 with respect to x so : 16 Step : L SK j P e j m P C 1 d A int,p A int,p d e K e K d e K e K d l l l 9

In this step the inequlities 15, 14 nd 16 re proved over hlf of the squre P denoted T, i.e the tringle A,B,D see figure 1. Using the orthogonl symmetry in reltion to [BD], the problem is the sme s the one of the step 1, then with the sme nottions, one hs the following results : nd : Step 3 : K j R 1 b K j R b l K j D 1 b e j m T C 1 d l K j D b e j m T C 1 d L SK j T e j m T C 1 d A int,t A int,t A int,t e K e K d e K e K d e K e K One proves the result over n ordinry tringle of Ω. So let n ordinry tringle T of Ω nd right-ngle tringle T, see figure. d l l l 44.000,117.0004.000,15.00040.000,117.000 4,154,18187,18 179.000,16.000187.000,18.000179.000,0.00 Figure : Then there exists liner ppliction F which trnsforms T in T defined s follow : x y x x l 1 l 1 tn θ F : = y y y l sin θ As T = L K j T then T = L F K j T Let the function e defined from T to IR by ex,y = e j if x,y F K j T. One hs the following result : there exists η such tht for ll tringultion T = K j 1 j L of Ω one hs, for ll ngle of n element of T : η < θ < π η 10

Then there exists ηη,f such tht for ll ngle of n element of F K j : 1 j L but one cn hve : θ π. θ > ηη,f So one defines T k 1 k L tringultion of FΩ such tht : 1. k {1,...,L }, there exists j {1,...,L} such tht : T k F K j. for ll θ ngle of n element of T k 1 k L, one hs : ηη,f < θ < π ηη,f According to the step, one hs: L S F K j T ej m L T = S T k T ej m T C k=1 A int,t e K e K d l where m T is the men vlue of e over T. Remrking tht : nd tht A int,t m T = m T = 1 ST Thus : So, one obtins : e K e K d l C α1,α,η,f L SK j Te j, one gets : L A int,t S F K j T e j m T C α1,α,η,f e K e K = Cα1,α,η,F S F K j T 1 = dx dy = F K j T l 1 l sin θ K j T L SK j T e j m T l 1 l sin θ C α1,α,η,f With the sme rguments, one proves the following inequlities : 17 K j R I l K j I e j m T l 1 C α1,α,η,f A int,t dx dy = A int,t A int,t A int,t e K e K e K e K 1 l 1 l sin θ SK j T e K e K e K e K 11

18 K j R J l K j J e j m T l sinθ C α1,α,η,f A int,t e K e K where R I respectively R J is the set of the tringles K of T such tht K I respectively K J is not emptyset. Step 4 : Let subset T of Ω which is the union of two tringles T 1 nd T, one denotes T 1 T by I. As m = ST 1m 1 ST m, one cn write : SΩ L L SK j T 1 e j m T SK j T 1 e j m 1 ST 1 ST SΩ m m 1 According to the step 3 one hs : L SK j T 1 e j m T C ST 1 A int,t1 Then it just remins to estimte the difference between m nd m 1. e K e K ST1 ST SΩ m m 1 Let x = x 1,x I, so : m m 1 ex 1,x m ex 1,x m 1 Integrting over I nd thnks to the inequlities 17 nd 18, one hs : li m m 1 C e K e K e K e K Then : A int,t1 L SK j T 1 e j m T C A int,t1 A int,t e K e K A int,t e K e K With the sme rguments, one cn prove the following result : L SK j T e j m T C e K e K nd then : A int,t1 L SK j T e j m T C A int As we hve supposed tht Ω is finite union of tringles, one hs : L SK j e j m C A int where C does not depend of the tringultion T. α 1 One concludes remrking tht l for ll A int, so : α d A int,t e K e K e K e K e K e K 1

L SK j e j m α C α 1 A int e K e K Remrk 1 If the boundry condition is Dirichlet condition, then this proof cn be generlized, it is closed to those given by R. Herbin in [8]. In this cse one considers direction D which is prllel to none edges of the mesh. Let j {1,...,L} nd x K j, then one denotes by A jx, the set of the edges such tht the intersection between these sides nd the line which contins x nd prllel to D is not empty set. One cn write : e e j K C d e K Ω d cos θ where θ is the ngle between nd D. A jx Integrting over K j, summing over j nd swpping summtions nd integrls, one gets : L e SK j e j C d K e K Ω l l A int d d l e K d A ext nd then one concludes with the error estimte in discrete H 1 0 norm see [8]. 4 Convergence of the numericl scheme for the hyperbolic eqution In this section one will prove the convergence of the solution of 10 towrd wek solution of problem, 4, 5, proving the following Theorem : Theorem Let T q,δ q q IN be sequence of tringultions of Ω nd time steps which stisfy the properties 6 nd the stbility condition 9. One notes T q = K j q 1 j L q nd t n q = nδ q. Let the sequence u Tq,δ q q IN with u Tq,δ q x,t = u qn j if x K q j nd t [t n q,tn1 q {u [, where qn j, } j {1,...,L q }, n IN, is solution of the discretized eqution 10 ssocited to the tringultion T q nd the time step δ q. Then : 1 there exists subsequence still denoted u Tq,δ q q IN, which converges towrd u when q, i.e. when h goes to 0, in L Ω IR for the wek topology, i.e. one hs : lim q u Tq,δ q x,tϕx,tdx dt = Ω IR ux,tϕx,tdx dt Ω IR for ll ϕ L 1 Ω IR. u is wek solution of problem, 4, 5, i.e. one hs : ux,t ϕ x,tdx dt ux,t Px. ϕx,tdx dt Ω IR t Ω IR u 0 xϕx,0dx uγ,tϕγ,tg γdγ dt = 0 Ω Γ IR for ll ϕ C c Ω IR with Ω = Ω Γ. 13

In order to prove the first prt of this Theorem, one will prove in the subsection 4.1 n L Ω IR estimte on the pproximte solution. Then to prove the second prt, one will first show, in subsection 4., wek estimte on the vrition of the pproximte solution, nd then, using this estimte, one will prove tht u is wek solution of problem, 4, 5. 4.1 L Ω IR estimte on the pproximte solution Here one proves tht the fmily u Tq,δ q q IN is bounded in L Ω IR, then, thnks to the sequentil wek reltive compctness of the bounded sets of L Ω IR, it shows the existence of subsequence, still denoted u Tq,δ q q IN, which converges in L Ω IR for the wek topology when h goes to 0. Let T be tringultion of Ω nd δ IR which stisfy the property 6 nd the stbility condition 9. Another expression of the eqution 10 is : u n1 j = u n j 1 δ lc jk P k P j g ij SK j, P j>p k i τ extj δ u n k lc jk P k P j u n ji g ij SK j, P k >P j i τ extj for ll j {1,...,L}. So u n1 j is liner combintion of the u n k, 1 k L nd un ji i = 1, or 3. Then like in [5], thnks to 8 nd to the stbility condition 9, one hs the following properties : 1. the sum of the coefficients of the combintion is equl to 1. the coefficients of the combintion re ll positive nd one cn write : u n1 j mx sup K j T u n j, sup u n ji... mx sup j T ext, i=1,,3 K j T mx u 0 L Ω, u L Γ IR Then u Tq,δ q q IN is bounded in L Ω IR. 4. Convergence of the numericl scheme u 0 j, u L Γ IR The L Ω IR stbility gives the existence of subsequence which converges to u in L Ω IR for the wek topology. One will show tht u is solution of, 4, 5 in wek sense. Let T be tringultion of Ω nd δ IR which stisfy the property 6 nd the stbility condition 9. One considers ϕ Cc Ω IR with Ω = Ω Γ, nd one defines T such tht, x Ω, supp ϕx,. [0,T 1] nd such tht there exists N IN such tht N δ T < N 1δ. 1 Multiplying 10 by SK j ϕx,tn nd integrting over K j, then summing over j nd n, one gets : E 1h E h = 0 14

with : E 1h = E h = N L n=0 N n=0 K j SK j u n1 j u n j L δ 1 SK j ϕx,tn dx u n jk P k P j lc jk K j 1 SK j ϕx,tn dx i τ extj g ij un ji g ij un j In clssicl wy one could show tht : lim E 1h = h 0 Ω IR ux,t ϕ t The proof of the following result will be given : lim h = h 0 Ω ux,t Px. ϕx,tdx dt IR One defines E 3h nd E 4h by : N E 3h = δ u n j u n k P k P j ϕγ,t n dγ c jk E 4h = n=0 j,k S N δ u T,δx,t n Px. ϕx,t n dx Ω n=0 x,tdx dt u 0 xϕx,0dx Ω Γ L Γ i τ extj IR uγ,tϕγ,tg γdγ dt u n j u n ji u T,δγ,t n ϕγ,t n gγdγ c ij ϕγ,t n gγdγ The result will be proved in three steps. The first step is the proof of the existence of constnt C 0, independent of T nd δ, such tht : E 3h E h C h 1/ According to 8 one hs : N L E h = δ u n jk u n j P j P k lc jk d n=0 jk i τ extj u n j u n jig ij 1 K j SK j ϕx,tn dx E h = N δ n=0 j,k S lc jk SK k un j u n k P j P k ϕx,t n dx K k L 1 SK j un j u n jig ij ϕx,t n dx K j i τ extj 15

thus : E h E 3h N δ u n j u n k P j P k ϕγ,t n dγ lc jk ϕx,t n dx d n=0 jk j,k S c jk SK k K k L u n j u n g ij ji ϕx,t n dx ϕγ,t n gγdγ SK j K j i τ extj c ij One defines : D 1 = ϕγ,t n dγ lc jk ϕx,t n dx c jk SK k K k g ij D = ϕx,t n dx ϕγ,t n gγdγ SK j K j c ij Remrking tht if ϕ C is constnt then D 1 = 0, one gets : D 1 = ϕγ,t n ϕx k,t n dγ lc jk ϕx,t n ϕx k,t n dx c jk SK k K k C h j lc jk Similrly one could show tht : nd then : N E h E 3h C δ n=0 j,k S D C h j g ij h j lc jk P j P k u n j u n k N δ h j g ij un j u n ji n=0 j T ext i τ extj To conclude, the following Lemm is proved : Lemm Let T be tringultion of Ω nd δ IR which stisfy the property 6 nd the stbility condition 9. Let T > 0, one defines N by N δ T < N 1δ nd one supposes N 1, then one notes EF h T = δ N n=0 j,k S h j lc jk P j P k u n j u n k Then there exists constnt C 0 independent of T nd δ such tht : Proof of Lemm EF h T C h 1/ j T ext i τ extj h j g ij un j u n ji Let n IN, K j T, one multiplies 10 by u n j nd one sums over n nd j, then using the following property : u n1 j u n j u n j = 1 un1 j u n j 1 un j 1 un1 j 16

one gets : 19 1 N L n=0 SK j u n1 j Let B 1 nd B be defined by : B 1 = B = δ N L n=0 N Then one hs : N B = δ n=0 Remrking tht : n=0 j,k S j,k S nd thnks to 8 : u n j δ N n=0 SK j u n1 j u n j L L u n j u n jk lc jk P k P j i τ extj u n j u n jk lc jk P k P j u n j u n j u n k lc jk P k P j u n j u n ji g ij un j g ij 1 i τ extj L i τ extj L SK j u 0 j u n j u n ji g ij un j g ij u n j u n j u n k = 1 un j u n k 1 un j 1 un k j,k S u n j u n k lc jk P k P j = u n j lc jk P k P j i τ extj u n j u n ji g ij un j g ij L u n j lc jk P k P j g ij ij g = 0 j {1,...,L} one cn write : B = δ N u n j u n k lc jk P k P j n=0 L i τ extj L i τ extj g ij u n j u n ji u n ji g ij un j g ij Furthermore ccording to 10 one hs : N L n=0 SK j u n1 j u n j = N L n=0 δ u n jk u n j lc jk P k P j SK j i τ extj g ij u n ji u n j 17

The number of the neighbours of tringle K j is lower thn 3, nd the number of its sides on the boundry is lower thn, then thnks to the Cuchy-Schwrz inequlity : N B 1 δ 5 lc jk u n j u n SK k k P j P k n=0 j,k S L 5 SK j i τ extj g ij u n ji u n j Thus using this inequlity nd the lst expression of B in 19 one gets : N δ θ jk lc jk P j P k L u n j u n d k θ j g ij jk with : n=0 j,k S L i τ extj u n j g ij θ jk = 1 δ θ j = 1 δ So one cn write the following inequlity : δ N n=0 j,k S lc jk P j P k δ N L n=0 i τ extj i τ extj 5 SK k lc jk P j P k > α 5 SK j g ij > α u n j u n k δ N u n ji u n j L u n ji g ij SK j u 0 j L n=0 i τ extj 1 α SΩ u 0 L Ω 1 α u L Γ IR T g ij u n ji u n j Γ g γ dγ = K α Using the Cuchy Schwrz inequlity nd the previous property one gets : 1/ K N EF h T h 1/ δ h j lc jk P j P k N L δ α But δ N L n=0 i τ extj j,k S h j g ij T h Γ n=0 j,k S g γ dγ = T hg with T nd G independent of T nd δ. Still using the Cuchy Schwrz inequlity, one hs : 1/ h j lc jk P j P k h lc jk j But j,k S j,k S n=0 i τ extj h j g ij 1/ lc jk P j P k α 1 α lc jk C j,k S, where C only depends on α 1, α nd on η, so : j,k S h j 1/ lc jk C 1/ h d j C SΩ 1/ jk α 1 j,k S 1/ 18

Thus to complete the proof of Lemm, it just remins to show tht there exists constnt C 0, independent of T nd δ, such tht : 0 P j P k lc jk C j,k S This result will be proved in four steps nother proof is possible, using the error estimte 11, see Remrk : Step 1 : One shows tht there exists constnt C 1 0 such tht : 1 L P j P k lc jk C 1 j T ext h j P j 1/ Multiplying 8 by P j nd summing over j, one gets : Then thnks to Cuchy-Schwrz : L i τ extj L lc jk P k P j P j d jk = L i τ extj g ij P j g ij P j α 1/ g L 1/ Γ h j P j lc i j j T ext j T ext α 1/ mesγ g L 1/ Γ h j P j j T ext where mesγ is the 1D Lebesgue mesure of Γ. Furthermore one hs : L nd then : Step : L lc jk P k P j P j = 1 L lc jk P j P k lc jk P j P k α 1/ mesγ g L Γ One shows tht there exists C nd C 3 positive such tht : j T ext h j P j 1/ 1/ h j P j C j T ext L lc jk P j P k C 3 L SK j P j Ω is bounded polygonl open set, first Ω will be supposed convex. If Ω is convex, then one defines two directions D 1 nd D nd one breks up the boundry of Ω in four prts, not necessry disjoint s on the figure 3. 19

5,3946,39 390,17390,0 4.00094,4314,3378,3385,4 34,17534,0 0,37173,37 137,16913 Figure 3: One notes T Γ1 respectively T Γ, T Γ3, T Γ4 the set of the suffix of the meshes which hve sides on Γ 1 respectively on Γ, Γ 3, Γ 4. Let j T Γ1 nd x K j, then one defines : D 1 j x = {x 1,x Ω ; x 1,x is in the line which contins x nd prllel to D 1 } A 1 j x = { A int ; D 1 j x } Let ϕ C Ω such tht x Ω, 0 ϕx 1, ϕx = 1 x Γ 1, nd ϕx = 0 x Γ 3. Then for ll j T Γ1 one hs : P j P j P j ϕ j A 1 j x P K ϕ K P K ϕ K P j0 x ϕ j0 x where j 0 x is the suffix of the element of T Γ3 such tht there exists i {1,,3} such tht c i j 0 x D 1 j x, so : P j P K ϕ K P K ϕ K P K ϕ K P K ϕ K A 1 j x A 1 j x P j P j ϕ j P j0 x ϕ j0 x P j0 x 0 But ϕ is in C Ω nd belongs to [0,1], then one hs : P j P K P K C 1 h K P K C h j0 x P j0 x h j P j 0

where C 1 = ϕ L Ω α α 1 nd C = ϕ L Ω α 1. Integrting over Γ 1, one obtins : α 1 h j P j j T Γ1 A int PK P K l C 1 h K P K l C α SK j P j SK j P j j T Γ1 j T Γ3 According to the Young inequlity : α 1 h j P j 1 PK P K P K P K l C 1 h K P K l j T Γ1 A int C α SK j P j SK j P j j T Γ1 j T Γ3 then s l d α 1 α for ll A int one obtins : h j P j α α 1 j T Γ1 L lc jk P j P k α L 3α C 1 C SK j P j α 1 Similrly one could prove the sme result for T Γ, T Γ3 nd T Γ4. Then summing these inequlities, one completes the second step with Ω convex. If Ω is not convex then one breks up its boundry Γ in p prts Γ i 1 i p disjoint nd for ll i = 1,...,p, one defines Γ i s on the figure 4. 4.00013,68130,6130,6 4.000176,63183,1183,1 15,63178,51 4.000177,146168 Figure 4: 1

Then one defines for ll i = 1,...,p, ϕ i C Ω such tht x Ω, 0 ϕ i x 1, ϕ i γ = 1 if γ Γ i nd ϕ i γ = 0 if γ Γ i. Then similrly to the convex cse one could prove. Step 3 : One supposes tht : L SK j P j = 0 it is lwys possible since solutions of 8 differ from constnt. Then s one hs proved 11, one could show tht there exists constnt C 4 0 such tht : 3 L L SK j P j lc jk C 4 P j P k d jk

Step 4 : One concludes, remrking tht : j,k S lc jk P j P k = 1 then thnks to 1, nd 3 one hs : L L lc jk P j P k lc jk P j P k C 1 C C 3 C 4 L lc jk P j P k 1/ So the proof of the inequlity 0 nd of the Lemm re completed, nd then one hs : E 3h E h C h 1/ Remrk As P is supposed to be in C Ω, the estimte 0 cn lso be given by the error estimte in discrete H 1 0 norm 11, but our proof of 0 does not use the property P C Ω nd then it could be extended to more complex cses. The second step is the proof of the following result : E 3h E 4h C h E 3h cn lso be written : N L E 3h = δ n=0 u n j P k P j ϕγ,t n dγ c jk u n j u n ji ϕγ,t n gγdγ i τ extj c ij nd : N E 4h = δ n=0 L u n j Pγ.n Kj γϕγ,t n dγ u n j u n ji c jk i τ extj ϕγ,t n gγdγ c ij then one gets : N E 3h E 4h δ n=0 L u n j c jk P k P j But ccording to 1 nd 8 if ϕ ϕx jk,t n is constnt, E 3h E 4h T u0 L Ω C ϕ h c jk Pγ.n Kj γ ϕγ,tn dγ E 3h E 4h = 0, thus : L P k P j lc jk Px jk.n Kj x jk d jk Px jk.n Kj x jk Pγ.n Kj γ dγ 3

where C ϕ only depends on α nd the first order derivtive of ϕ. Then one hs : E 3h E 4h T u0 L Ω C ϕ h L 3C 1 h j C h j 3C 1 C SΩT u 0 L Ω C ϕ h One concludes remrking tht : lim 4h = h 0 Ω ux,t Px. ϕx,tdx dt IR which completes the proof of the numericl scheme s convergence. Γ IR uγ,tϕγ,tg γdγ dt Remrk 3 One cn prove the sme results for system little bit different thn 1-5, where 3 is chnged by Fourier condition : where λ > 0. Pγ.nγ λpγ = gγ γ Γ Then one must introduce some unknowns t the boundry. As fluxes re not exct on the boundry, the error estimte, in discrete H 1 norm, includes boundry terms. The technic used to prove the error estimte in discrete H 1 0 norm is closed to this used in this note. Acknowledgements : I wish to thnk Thierry Gllouët nd Rphele Herbin for their ttention to this work nd their precious remrks. References [1] Benhrbit S., Chlbi A., Vil J.P., Numericl Viscosity nd Convergence of Finite Volume Methods for Conservtion Lws with Boundry Conditions, to pper in SIAM journl of Num. Anl. [] Chmpier S., Gllouët T., Herbin R., 1993, Convergence of n Upstrem finite volume scheme on Tringulr Mesh for Nonliner Hyperbolic Eqution, Numer. Mth., 66, 139 157. [3] DiPern R., 1985, Mesure-Vlued Solutions to Conservtion Lws, Arch. Rt. Mech. Anl., 88, 3 70. [4] Eymrd R., Gllouët T., Herbin R., The finite volume method, in preprtion for the Hndbook of Numericl Anlysis, Ph. Cirlet nd J.L. Lions eds. [5] Eymrd R., Gllouët T., 1993, Convergence d un schém de type Eléments Finis-Volumes Finis pour un système formé d une éqution elliptique et d une éqution hyperbolique, M AN, 7, No 7, 843 861. [6] Fird J.M. nd Herbin R., 1994, Comprison between finite volume nd finite element methods for the numericl computtion of n elliptic problem rising in electrochemicl engineering, Comput. Mth. Appl. Mech. Engin., 115, 315 338. [7] Gllouët T., Herbin R., 1993, A uniqueness Result for Mesure Vlued Solutions of Non Liner Hyperbolic Equtions, Differentil Integrl Equtions, 6, No 6, 1383 1394. [8] Herbin R., 1994, An error estimte for finite volume scheme for diffusion convection problem on tringulr mesh, Num. Meth. in P.D.E. 4

[9] Herbin R., Lbergerie O., Finite volume nd finite element schemes for elliptic-hyperbolic problems, submitted. [10] Szepessy A., 1989, Mesure vlued solution of sclr conservtion lws with boundry conditions, Arch. Rt. Mech. Anl., 107, No, 18 193. [11] Vssilevski P. S., Petrov S. I., Lzrov R. D., 199, Finite Difference Schemes on Tringulr Cell- Centered Grids With Locl Refinement, SIAM J. Sci. Stt. Comput., 13, No 6, 187 1313. 5