Kruzkov's estimates for scalar conservation laws. revisited. (*) Universite d'orleans et CNRS, UMR 6628, Dept. de Mathematiques.

Transcription

1 Kruzkov's estimates for scalar conservation laws revisited F. Bouchut (), B. Perthame () (*) Universite d'orleans et CNRS, UMR 668, Dept. de Mathematiques BP 6759, Orleans cedex, France (**) Universite Pierre et Marie Curie, Laboratoire d'analyse Numerique et URA 189 BC 187, 4 place Jussieu, 755 Paris cedex 05, France perthame@ann.jussieu.fr Abstract We give a synthetic statement of Kruzkov-type estimates for multi-dimensional scalar conservation laws. We apply it to obtain various estimates for dierent approximation problems. In particular we recover for a model equation the rate of convergence in h 1=4 known for nite volume methods on unstructured grids. Key-words. Multi-dimensional scalar conservation laws, error estimates, entropy inequalities, nite volumes. Les estimations de Kruzkov pour les lois de conservation scalaires revisitees Resume Nous donnons un enonce synthetique des estimations de type Kruzkov pour les lois de conservation scalaires multidimensionnelles. Nous l'appliquons pour obtenir de nombreuses estimations pour dierents problemes d'approximation. En particulier, nous retrouvons pour une equation modele la vitesse de convergence en h 1=4 connue pour les methodes de volumes nis sur des maillages non structures. Mots-cles. Lois de conservation scalaires multidimensionnelles, estimations d'erreur, inegalites d'entropie, volumes nis Mathematics Subject Classication. Primary 65M15, 35L65, 35B30. 1

2 1 Introduction and main results This paper deals with error estimates for a multi-dimensional scalar conservation law v + div f(v) = 0 in ]0; 1[R N ; v(0; x) = v 0 (x): (1:1) Here the ux function f : R! R N is assumed to be Lipschitz continuous for simplicity. The equation has to be supplemented with the entropy inequalities, for any S : R! R convex and t S(v) + div (v) 0 in ]0; 1[R N ; (1:) with 0 = S 0 f 0. Throughout the paper, all partial dierential equations and inequations are understood in the usual sense of distributions. After the works of P.D. Lax [1], O.A. Oleinik [15], A.I. Vol'pert [0], S.N. Kruzkov [10], it is well-known that for any v 0 L 1 loc(r N ), there exists a unique solution v C([0; 1[; L 1 loc(r N )) to (1.1)-(1.). We are given an approximate solution u and we wish to estimate the dierence u v. The famous method of S.N. Kruzkov [10] has been used in many situations to obtain such estimates. It relies on a doubling of the variables and on a penalization procedure. When applied to numerical approximations (such as nite volumes or nite elements), following the ideas of N.N. Kuznetsov [11], it becomes extremely intricated because the technicalities due to the large number of variables are added to notational diculties for the numerical approximation itself. This is especially true for unstructured grids. In this context, the derivation of the rate of convergence for rst-order nite volumes methods, in several dimensions and for unstructured grids, is still complicated. On the other hand the nal result, convergence rate in h 1=4 rather than h 1= in one space dimension (see R. Sanders [17]) is very easy to explain. This loss just comes from the BV estimate which blows like h 1= (while it is bounded in one dimension), see S. Champier, T. Gallouet, R. Herbin [3], B. Cockburn, F. Coquel, P. Le Floch [4][5], A. Szepessy [18], J.-P. Vila [19], B. Cockburn, P.-A. Gremaud [7], R. Eymard, T. Gallouet, R. Herbin [9]. For uctuation splitting schemes this also applies, see B. Perthame [16]. In order to simplify these proofs, we propose to formalize Kruzkov's method as follows. We assume that u solves the entropy inequalities with error terms which are partial derivatives. For k t ju kj+div sgn(u k)[f(u) f(k)] L t G k +div H k +K k + in ]0; 1[R N i 1i;jN (1:3) where G k ; H (j) k ; K k ; L k are local Radon measures, and satisfy in the sense of measures jg k (t; x)j G (t; x); jk k (t; x)j K (t; x); jh (j) k jl k (t; x)j (j) H (t; x)j L (t; x); (t; x); with G, (j) H, K, L some non-negative k-independent Radon measures. (1:4)

3 in terms of Using Kruzkov's method, we claim that one can estimate ku vk L 1 k G k Mt;x, k H k Mt;x... And it is the very striking and fundamental idea behind all the error estimates that the right estimates are always of the type ku vk L 1 Cr k sup k jg k jk Mt;x + Cr k sup k jh k jk Mt;x + ::: (see the precise statement in Theorem.1). The derivatives in the error terms are paid, for the nal estimate, only by a square root of the appropriate norm, which is taken in M t;x (L 1 k ). This is the main contribution of this paper to indicate how to reduce the estimates to the form above. More precisely, we claim that in many situations, it is not necessary to perform the doubling of variables explicitly, it is enough to use the abstract estimate of Theorem.1, and this greatly improves the understanding. This strategy was initiated in a paper by the authors and C. Bourdarias []. This approach is shown to be successful in the most classical situations which we develop independently in x 3, x 4, x 5. Moreover, we are able to improve the corresponding results. For multidimensional monotone nite volume methods, it is also possible to write an equation of the form (1.3), see the recent paper of R. Eymard, T. Gallouet, R. Herbin [9]. In x 3, we study the following problem. We assume that u is an entropy solution t u + div g(u) = 0 in ]0; 1[R N ; (1:5) with g another Lipschitz ux function. Then using Theorem.1, we are able to estimate ku vk L 1 in terms of g f. We indeed recover the estimate in Lip(g f) that was used by C.M. Dafermos [8] and by B.J. Lucier [14] in the context of front tracking methods. But our formalism also enables to get a new estimate, in kg fk 1= L1. We refer to x 3 for a precise statement. In x 4, we consider the nonlinear diusion t u + div f(u) (u) = 0 in ]0; 1[R N ; (1:6) with : R! R a Lipschitz and nondecreasing function. We refer to Ph. Benilan, R. Gariepy [1] for recent results on that equation. We here consider an entropy solution in the sense that for any S : R! R convex and t S(u) + div (u) (u) 0 in ]0; 1[R N ; (1:7) with 0 = S 0 f 0 and 0 = S 0 0. It is easy to see that it is equivalent to require that (1.7) holds for the entropies S k () = j kj; k t ju kj + div sgn(u k)[f(u) f(k)] j(u) (k)j 0 in ]0; 1[R N : (1:8) Notice also that the equation (1.6) is recovered by letting k! 1. 3

4 Our approach allows to recover the usual estimate of ku vk L 1 in \" 1= " i.e. Lip() 1= for v; u L 1 t (BV (R N )), and also the recent result of B. Cockburn, P.-A. Gremaud [6] which states that v L 1 t (BV (R N )) and only u L 1 t (L 1 (R N )) is enough to get the same estimate. Moreover our result (see (ii) below) explains how the regularity of is involved in this matter and we also give a new result for \irregular" diusions. We have the following results. Theorem 1.1 Assume that u L 1 loc([0; 1[; L 1 (R N )) is right-continuous with values in L 1 loc(r N ) and is an entropy solution in the sense of (1.8). Let v be an entropy solution of (1.1) with initial data v 0 L 1 \ BV (R N ). Then, for any T 0, we have (i) If T V (u(t; :)) V for any t 0 then k u(t; :) v(t; :) k L 1k u 0 v 0 k L 1 +C q T V (v 0 )V q T Lip(); (1:9) (ii) if k u(t; :) k L 1 U for any t 0, then denoting Q = sup j() (0)j=jj, 6=0 k u(t; :) v(t; :) k L 1k u 0 v 0 k L 1 +C N T V (v 0 ) =3 (T QU) 1=3 ; (1:10) q (iii) k u(t; :) v(t; :) k L 1k u 0 v 0 k L 1 +C T V (v 0 ) T Lip(): (1:11) Here and throughout this paper, the total variation T V (v 0 ) of a function v 0 BV (R N ) is dened by T V (v 0 ) = NX j=1 R : The letter C denotes various absolute constants, while C N denotes a possible dependence in the dimension N. Notice that if v is bounded, say a v b; then the Lipschitz constant of in (1.11) can be replaced by Lip [a;b] () by using Appendix (ii) instead of (i) in the proof (see x 4). For the sake of completeness, let us also recall that another classical analysis leading to local estimates in Lip() 1=4 can be carried out when () = ; > 0; by using the entropy dissipation equation (for u 0 L (R N t u + div (u) u = jruj ; with 0 () = f 0 () (see [3], [18], [4], [5], [19], [7], [9] and x 5). To deal with a general, the inequalities (1.7) are not sucient since the entropy dissipation is neglected. In x 5, we present a relaxation model towards piecewise constant functions. It does not involve notations as complicated as those of nite volumes but leads to basically the same kind of theoretical diculties. We show that error estimates can again be reduced to studying (1.3)-(1.4). We are given a general grid (C i ) ii of R N : I is a countable 4

5 set; for any i I, C i is a Borel set; for i 6= j jc i T Cj j = 0; R N = N [ ([ ii C i ) with jnj = 0; and h = sup ii diam(c i ) < 1: (1:1) For technical reasons we assume that the following L 1 BV condition is satised: there exists a constant K h 0 such that 8i I 1I Ci BV (R N ); T V (1I Ci ) K h jc i j: (1:13) Notice that it is not a \regularity" assumption on the grid since K h is allowed to blow up very fast when h! 0. It only means that there is no degeneracy at innity. We dene the piecewise constant projector in space P 0 : L 1 loc(r N )! L 1 loc(r N ) by (P 0 u) jci = 1 jc i j C i u : (1:14) We consider for a given " > 0 the unique solution u C([0; 1[; L 1 (R N )) of 8 t u + div f(u) = P 0 u u in ]0; 1[R N ; " >: u(0; :) = u 0 L 1 (R N ); (1:15) with the entropy t S(u) + div (u) S 0 (u) P 0 u u " in ]0; 1[R N (1:16) for any S convex, Lipschitz and C 1 (or equivalently for any entropy S k () = j kj). Again we may estimate the dierence with the exact solution. Theorem 1. With the above notations and assumptions, if u 0 L T 1 L (R N ) then s P 0 u u " Tj!j " k k u0 L (1:17) ]0;T [! for any T > 0 and any bounded subset! of R N. Denoting by v an entropy solution of (1.1) with initial data v 0 BV (R N ), we have for any T > 0, x 0 R N and R > 0 ju(t; x) v(t; x)j dx ju jx x 0 j<r jx x 0 j<r+mt +p 0 (x) v 0 (x)j dx br` + C N p` T V (v )q 0 br+ k u0 k L b R N=4 1= ( R b (1:18) N=4 + `N=4 ) with M = Lip(f), R b q = R + MT + h and ` = h T=". 5

6 In other words, the estimate is of order h 1= =" 1=4. When " h, which is similar to the limitation involved by a CFL condition for nite volumes, we recover the usual rate in h 1=4. However, the estimate (1.18) is not really interesting because we have in mind to approximate the solution v by a piecewise constant function. Thus, it is more appropriate to estimate P 0 u v. This can be performed combining (1.17) and (1.18), and rises an estimate in L 1 (]0; T [B(x 0 ; R)). Notice also that it is very natural to choose u 0 = P 0 v 0. Then, assuming that v 0 L 1 T L T BV (R N ), the L norm of u 0 in (1.18) can be majorized by k v 0 k L. Thus, if the grid is regular in the sense that C i is open and convex, and then the initial error can be estimated by diam(c i ) N +1 jc i jh; (1:19) k u 0 v 0 k L 1 (R N ) C NT V (v 0 )h: (1:0) This is just a straightforward application of Poincare-Wirtinger's inequality for a convex domain. The paper is organized as follows. In x we state and prove our abstract version of Kruzkov's estimates. In x 3, x 4 and x 5 we treat in detail the above independent applications. The appendix is devoted to two independent lemmas. The rst deals with a special \inverse" of the \div" operator which makes the link between nite volumes and the formulation (1.3). It is needed for the proof of Theorem 1.. The second deals with a general estimate on the gradient r[g(u)] where g is Lipschitz and u BV: It is used frequently in this paper. An error estimate for partial derivatives in the right-hand side This section is devoted to the main estimate of this paper, the reformulation of Kruzkov type estimates. We give two versions of the result. Theorem.1 Let u; v L 1 loc([0; 1[; L 1 loc(r N )) be right-continuous with values in L 1 loc(r N ). Assume that u solves the entropy inequalities with right-hand side (1.3)- (1.4), and that v is an exact solution: for k R Moreover we assume t jv kj + div sgn(v k)[f(v) f(k)] 0 in ]0; 1[R N : (:1) G L 1 loc([0; 1[; L 1 loc(r N )): (:) 6

7 Then, for any T 0; x 0 R N ; R > 0; > 0; > 0; 0; denoting by we have jx x 0 j<r M = Lip(f); B t = B(x 0 ; R + M(T t) + + ); (:3) ju(t; x) v(t; x)j dx B0 with C an absolute constant and E t = E K = sup t=0 or T 0<s t< 0<tT xb t B t jv(s; x) v(t; x)j dx; E x = K (t; x); E H = 1 E G = ju(0; x) v(0; x)j dx+c(e t +E x +E G +E H +E K +E L ); NX j=1 0<tT xb t 1 + T + MT + (:4) sup jv(t; x + h) v(t; x)j dx; (:5) t=0 or T B t jhj< (j) H (t; x); E L = 1 sup 0<t<T Before proving Theorem.1, some comments are in order. X 1i;jN 0<tT xb t L (t; x); (:6) B t G (t; x) dx: (:7) 1. If u is an exact solution, the right-hand side of (1.3) is zero. Hence, the only error terms E t and E x remain in (.4). Therefore, we can choose = 0 and let ;! 0 so that by the assumed regularity of v we get E t ; E x! 0. We hence recover Kruzkov's estimate [10] ju(t; x) v(t; x)j dx ju(0; x) v(0; x)j dx: (:8) jx x 0 j<r jx x 0 j<r+mt When the right-hand side of (1.3) is not zero, we can use the well-known T V D property on the exact solution v: We obtain if v 0 BV (R N ) E t M T V (v 0 ); E x T V (v 0 ): (:9) Then, we choose the time and space regularization parameters and so as to minimize the error terms in (.4). The parameter is either chosen to be 0 for a local estimate or tends to +1 for a global one.. Our assumption that u and v are L 1 loc right-continuous is motivated by numerical schemes, it is well suited for such problems. We refer to T.-P. Liu, M. Pierre [13] for uniqueness results in one dimension when the initial data is recovered only in weak L For numerical schemes, the assumption (.) on G is somehow related to an estimate on the time modulus of continuity of u. In fact, G is comparable with t u, with t the timestep (see []). However, when no such estimate is available, it is possible to get an estimate of k u v k L 1 t;x with the only regularity of G stated in (1.4) and an estimate of G only for small t. In order to obtain this result, just average (.4) with 7

8 respect to T and then observe that when averaged, the last inequality of (.8) in the proof below can be replaced by a suitable one since j 0 (t)j Y 0 " (t)+y 0 "(t T ). It is also possible to choose a dierent test function, see R. Eymard, T. Gallouet, R. Herbin [9]. Proof of Theorem.1. Let us introduce two test functions C 1 c (]0; 1[R N ), 0, and C 1 c (] 1; 0[R N ), 0, to be chosen later on, and set '(t; x; s; y) = (t; x)(t s; x y): (:10) This choice is slightly dierent from that of S.N. Kruzkov in [10], and is inspired by that of N.N. Kuznetsov [11] (who takes = 1I 0<t<T ). Notice that ' C 1 c ((]0; 1[R N ) ). For each (s; y) ]0; 1[R N and k R, let us take ' as a test function in (t; x) for (1.3). We get with (1.4) ]0;1[R N ]0;1[R N ]0;1[R N h ju kj@t ' + sgn(u k)[f(u) f(k)] r x ' i (t; x) dtdx 4 G t ' H k r x ' + K k ' + X 4 G j@ t 'j + NX j=1 (j) H L k X + K ' + L 1i;jN ' 5 (t; x) i i 5 (t; x) dtdx: (:11) Similarly, for each (t; x) ]0; 1[R N and l R, we take ' as a test function in (s; y) for (.1), and obtain ]0;1[R N h jv lj@s ' + sgn(v l)[f(v) f(l)] r y ' i (s; y) dsdy 0: (:1) Now we take k = v(s; y) in (.11) and integrate with respect to (s; y), we take l = u(t; x) in (.1) and integrate with respect to (t; x). By summing up the results we obtain h ju(t; x) v(s; y)j@t (t; x) + sgn(u(t; x) v(s; y))[f(u(t; x)) f(v(s; y))] r x (t; x) i (t s; x y) dsdtdxdy " G (t; t (t; x)(t s; x y) + (t; x)@ t (t s; x y) NX + (j) H (t; (t; x)(t s; x y) + (t; x)@ j (t s; x y) R : j=1 + K (t; x)(t; x)(t s; x y) + X 1i;jN L (t; ij (t; x)(t s; x y) i (t; x)@ j (t s; x y) # j (t; x)@ i (t s; x y) + (t; x)@ ij (t s; x y) dsdtdxdy (:13) 8

9 Let us now precise the choice of. Let > 0 and set Y (t) = R t 1 Y 0 (s) ds with Y 0 (t) = 1 Y 0 1( t ) and Y 0 1 C 1 c (]0; 1[), Y and R Y 0 1 = 1. We introduce another parameter " > 0 and a function C 1 c (]0; T + "[); 0 to be chosen later. We dene (t; x) = (t) (t; x); (t; x) = 1 Y (jx x 0 j R = M(T t)) 0: (:14) Then C 1 (R R N ) as soon as M" R + =. We t (t; x) = 0 (t) (t; x) M(t)Y 0 (); Therefore, by the Lipschitz condition on f we get r x (t; x) = (t)y 0 () x x 0 jx x 0 j : (:15) ju(t; x) v(s; y)j@ t (t; x) + sgn(u(t; x) v(s; y))[f(u(t; x)) f(v(s; y))] r x (t; x) = ju(t; x) v(s; y)j 0 (t) (t; x) (t)y 0 ()h Mju(t; x) v(s; y)j x x + sgn(u(t; x) v(s; y))[f(u(t; x)) f(v(s; y))] 0 i jx x 0 j ju(t; x) v(s; y)j 0 (t) (t; x): Together with (.13) it yields (:16) ju(t; x) v(s; y)j 0 (t) (t; x)(t s; x y) dsdtdxdy R : (:17) Now, by the triangle inequality we get with I = R t = R x = 0 I + R t + R x + R (:18) ju(t; x) v(t; x)j 0 (t) (t; x)(t s; x y) dsdtdxdy; jv(t; y) v(s; y)jj 0 (t)j (t; x)(t s; x y) dsdtdxdy; (:19) jv(t; x) v(t; y)jj 0 (t)j (t; x)(t s; x y) dsdtdxdy: We now prescribe to be a regularizing sequence (t; x) = t (t) x (x); t ; x C 1 c ; 0; t dt = x dx = 1; t (t) = 1 t 1 ( t ); supp t 1 ] 1; 0[; x (x) = 1 N x 1 We also choose x to be a product x 1 (x) = For we take x ; supp x 1 B (0; 1=4) : NY j=1 x j 1 (x j ): (:0) (t) = Y " (t) Y " (t T ) (:1) 9

10 so that 0 1. We are going to take the lim "!0 in (.18), xed. We notice that and obtain 1I jx x0 j<r+m (T t)+= (t; x) 1I jx x0 j<r+m (T t)+=+; (:) R t jv(t; y) v(s; y)j(y 0 "(t) + Y 0 " (t T )) t (t s) dsdtdy: jy x 0 j<r+m (T t)+3=4+ Then by right continuity of v at 0 and T, lim Rt X "!0 t=0;t just by choosing = =4 +. For the term R x we have R x jx x 0 j<r+m (T t)+=+ jv(t; y) v(s; y)j t (t s) dsdy E t (:3) jy x 0 j<r+m (T t)+3=4+ and by the same continuity property of v lim Rx X "!0 t=0;t For I we have using (.) I = jv(t; x) v(t; y)j(y 0 " (t) + Y 0 " (t T ))x (x y) dtdxdy; jv(t; x) v(t; y)j x (x y) dxdy E x : (:4) jx x 0 j<r+m (T t)+=+ ju(t; x) v(t; x)j[y 0 " (t) Y 0 " (t T )] (t; x) dtdx ju(t; x) v(t; x)jy 0 "(t) dtdx jx x 0 j<r+m (T t)+=+ and by right continuity of u and v at 0 and T, lim I "!0 ju(0; x) v(0; x)j dx jx x 0 j<r+mt +=+ ju(t; x) v(t; x)jy 0 "(t T ) dtdx; jx x 0 j<r+m (T t)+= ju(t; x) v(t; x)j dx: (:5) jx x 0 j<r+= It now remains to estimate R dened in (.13). Using the bounds and denoting by the set = jr x (t; x)j C ; j@ t(t; x)j j 0 (t)j + C M ; (:6) t; x; 0 < t < T + "; jx x 0 j < R + M(T t) + = + ; (:7) 10

11 we get the estimates K (t; x)(t; x)(t s; x y) (j) H (t; x)(t; x) j@ j (t s; x y)j C G (t; x)(t; x) j@ t (t s; x y)j C (j) H (t; x) j@ j (t; x)j (t s; x y) C G (t; x) j@ t (t; x)j (t s; x y) C (1 + M(T + ")=) sup K (t; x); (j) H (t; x); G (t; x); (j) H (t; x); G (t; x) dx: 0<t<T +" jx x 0 j<r+m (T t)+=+ (:8) For the second-order terms of R, we notice that in the support of r x (t; x), we have jx x 0 j R + = M" + C 0 C 0 for some C 0 > 0: It i We hence get L (t; x)(t; ij (t s; x C y) L (t; x) j@ i (t; x)j j@ j (t s; x y)j C L (t; ij (t; x) C (t s; x y) : (:9) L (t; x); L (t; x); L (t; x): Taking into account that = =4 +, we obtain by collecting (.8) and (.30) (:30) lim R C(E G + E H + E K + E L ); (:31) "!0 and the lim "!0 in (.18) gives the result together with the estimates (.3), (.4), (.5). Notice that in Theorem.1 and its proof, we never use any specic property of the exact solution v. We only use the entropy inequalities (.1) in the distributional sense for any k R. Therefore, our result actually contains Kruzkov's uniqueness theorem with the stated regularity. Moreover, it is possible to modify the proof in order to also consider error terms for v. The result is the following. Proposition. We make the same assumptions as in Theorem.1, except that we replace the inequalities (.1) on v t jv kj+div sgn(v k)[f(v) t G k +div H k +K k + X L i in ]0; 1[R N ; (:3) 11

12 where G k ; H (j) k ; K k ; L k jg k j G ; are local measures, and jh (j) k j (j) H ; jk k j K ; jl k j with G ; (j) H ; K ; L some k-independent non-negative measures. Then we have the same estimate (.4), with the additional terms where E K = 0<tT + xb 0 K (t; x); E H = 1 L ; (:33) C(E G + E H + E K + E L ); (:34) NX j=1 0<tT + xb 0 E G = 1 Proof. The inequality (.1) becomes ]0;1[R N ]0;1[R N ]0;1[R N (j) H (t; x); E L = 1 0<tT + xb 0 X 1i;jN h jv lj@s ' + sgn(v l)[f(v) f(l)] r y ' i (s; y) dsdy 4 G s ' H l r y ' + K l ' + X 4 G j@ s 'j + NX j=1 (j) H and gives a new term in (.13) R = 0<tT + xb 0 L (t; x); (:35) G (t; x): (:36) L l X + K ' j " G (s; y)(t; x)j@ t (t s; x y)j+ NX + K (s; y)(t; x)(t s; x y)+ X 1i;jN j=1 L 1i;jN ' 5 (s; y) j j 5 (s; y) dsdy; (j) H (s; y)(t; x) j@ j (t s; x y)j L (s; y)(t; ij (t s; x y) # dsdtdxdy: Since where (t; x)(t s; x y) 6= 0 we have (s; y) ; e with s; y; 0 < s < T + " + ; jy x 0 j < R + MT + 3=4 + e = (:37) (:38) ; (:39) we get R C e G (s; y)+ C NX j=1 e (j) H (s; y)+ e K (s; y)+ C X 1i;jN e L (s; y): (:40) Therefore, lim "!0 R C(E G + E H + E K + E L ); and we get the result. 1

13 3 Estimates for two dierent ux functions In this section we apply Theorem.1 to prove various estimates for the following problem. Given f; g : R! R N two globally Lipschitz ux functions and u 0, v 0 L 1 loc (RN ), we consider the entropy solutions u and v t u + div g(u) = t v + div f(v) = 0 in ]0; 1[R N (3:1) with initial data u 0 and v 0 respectively. Then, what kind of estimate can we hope for u v in terms of g f? We have the following result. Theorem 3.1 Let us assume that v 0 BV (R N ). Then, with the above notations, we have for any T 0 (i) k u(t; :) v(t; :) k L (R k 1 N ) u0 v 0 k L (R +C T V 1 N ) (v0 ) T Lip(g f); (3:) (ii)k u(t; :) v(t; :) k L 1 (R k N ) u0 v 0 k L 1 (R +C N ) N k u0 k 1(R T V 1= L N ) (v0 ) T Q(f; g) ; (3:3) with Q(f; g) = (iii) if g f L 1 (R), then for x 0 R N and R > 0 ju(t; x) v(t; x)j ju 0 (x) v 0 (x)j jx x 0 j<r sup jg() f() g(0) + f(0)j=jj; (3:4) Rnf0g jx x 0 j<r+mt 1= +C N (R + MT ) N T V jx x 0 j<r+mt (v0 ) T k g f (g f)(0) k L 1! ; with M = max(lip(f); Lip(g)). (3:5) The estimate (i) was known (see C.M. Dafermos [8], B.J. Lucier [14]), but the much stronger estimates (ii) and (iii) seem fairly new. Proof of Theorem 3.1. We have for k R Therefore, we can t jv kj + div sgn(v k)[f(v) f(k)] t ju kj + div sgn(u k)[g(u) g(k)] t ju kj + div sgn(u k)[f(u) f(k)] div k (u); (3:6) with k () = sgn( k)[(f g)() (f g)(k)] sgn( k)[(f g)(0) (f g)(k)] = sgn( k)[(f g)() (f g)(0)] +[sgn( k) sgn( k)][(f g)(0) (f g)(k)]: (3:7) 13

14 Notice that k Lip(R; R N ). (i) Assume that u 0 BV (R N ). Then k (u) C([0; 1[; L 1 loc(r N )) \ B([0; 1[; BV (R N )) (where B(X; Y ) denotes the set of bounded functions X! Y ), and by (j) i (u) Lip( (j) k Lip(g i Therefore, K k div k (u) L 1 (]0; 1[; M(R N )) with NX jk k j Lip(g K i We estimate K by R N K(t; dx) = Lip(g f)t V (u(t; :)) Lip(g f)t V (u 0 i : (3:8) Now, the inequalities (1.3)-(1.4) are satised with a single term, of order zero, K k : Hence, we can apply Theorem.1. Then, we just let! 0; then! 0 and nally R! 1 in (.4) and we obtain R N ju(t; :) v(t; :)j R N ju0 v 0 j + C T Lip(g f)t V (u 0 ): The estimate (i) follows by exchanging u and v. (ii) The right-hand side of (3.6) can also be viewed as div H k ; with H k = k (u): (3:9) We have j k ()j j(f g)() (f g)(0)j + 1I jkjjj j(f g)(0) (f g)(k)j Q(f; g)jj + 1I jkjjj Q(f; g)jkj 3 Q(f; g)jj: (3:10) Therefore, we get jh k j 3Q(f; g)juj H. Since u 0 L 1 (R N ) we nd R H(t; x) dx = 3 Q(f; g) N R ju(t; x)j dx 3 Q(f; g) k k N u0 L 1 : Then, we can apply Theorem.1 with a single divergence term in the right-hand side. Since v 0 BV (R N ), we can use (.9). We let! 0 and, for any > 0, we get jx x 0 j<r ju(t; :) v(t; :)j R N ju 0 v 0 j + C N T V (v 0 ) + 1 Q(f; g) k u0 k L 1 T : 14

15 Finally, we let R tend to 1, we choose the optimal, and we obtain (ii). (iii) This estimate is a localized version of (ii), but we have to be careful in order to get the right domain of dependence. Assume that v 0 BV loc (R N ), and let us denote br = R + MT and B b = B(x 0 ; R). b There exists an extension bv 0 of v 0 jb b such that bv 0 L 1 \ BV (R N ), supp bv 0 B(x 0 ; R), b k bv 0 k L (R C 1 N ) N k v 0 k L (b 1 B) ; T V R N (bv0 ) C N T Vb B (v 0 ) + 1 R b k v 0 k L (b 1 B) : (3:11) Let us dene bu 0 (x) = ( u0 (x) if x b B; bv 0 (x) if x = b B; and consider the entropy solutions bu and bv associated with the ux functions g and f, and with initial data bu 0 and bv 0. Since supp bu 0 B(x 0 ; b R), we have for any t 0 Therefore, for any t 0 supp bu(t; :) B(x 0 ; b R + t Lip(g)): j k (bu(t; x))j 3 k g f (g f)(0) k L 1 1I B(x0 ;b R+t Lip(g)) (x) H (t; x): Then we apply Theorem.1 to bu and bv with H k = k (bu) and obtain by letting! 0 that for any > 0 jx x 0 j<r jbu(t; x) bv(t; x)j R N +C N T V (bv 0 ) + T jbu 0 bv 0 j B(x 0 ; R b + T Lip(g)) k g f (g f)(0) kl 1 : Choosing the optimal and taking into account the denition of bu 0 and bv 0 and the nite speed of propagation for bu and bv, we get jx x 0 j<r ju(t; x) v(t; x)j jx x 0 j<b R ju 0 v 0 j+c N h br N T V (bv 0 )T k g f (g f)(0) k L 1i 1=: It remains to estimate T V R N (bv0 ). We have by Poincare-Wirtinger's inequality bb 1 v0 jbj b bb v 0 C N b R T VbB (v 0 ): Therefore, by (3.11), the result holds as soon as R b B v 0 = 0: If not, let c = R b B v 0 =j b Bj. The functions u c and v c are entropy solutions associated to the ux functions g( + c) g(0) and f( + c) f(0); and have initial data u 0 c and v 0 c. Since R bb (v 0 c) = 0, the result holds for these functions, and it exactly means that it holds for u and v. 15

16 4 The nonlinear diusion model This section is devoted to the proof of the Theorem 1.1. Let us prove (i). Since u B([0; 1[; BV (R N )) (B(X; Y ) denotes the set of bounded functions X! Y ), we have g(u) B([0; 1[; BV (R N )) for any Lipschitz function g by Appendix. Moreover, g(u) L 1 (]0; 1[; M(R N g(u) : (4:1) By choosing we get g k () = j() (k)j j(0) (k)j; g k (u) Now the error term in (1.8) can be written (j) H : (4:3) j(u) (k)j = div H k ; H k = r[g k (u)]; (4:4) and we have jh (j) k j (j) H : Therefore we can apply Theorem.1 with this single error term. We choose = 0 and we let! 0 so that E t! 0: Then, E x is estimated by (.9) and We hence obtain that for any > 0 jx x 0 j<r ju(t; x) v(t; x)j dx E H T V Lip(): ju(0; x) v(0; x)j dx + C jx x 0 j<r+mt + T V (v 0 ) + T Lip()V : We then let R! 1, and by choosing the optimal value of we get (1.9). Let us now prove (ii). We consider the error term in (1.8) as j(u) (k)j = X 1i;jN with g k dened in (4.). Since we have L L i ; ij Qjuj L k = ij g k (u); (4:5) L ; (4:6) we can apply Theorem.1. As above, we choose = 0, and let! 0 so that E t! 0. We have E L T NQU; 16

17 and E x is estimated by (.9). Therefore we get that for any > 0 ju(t; x) v(t; x)j dx T V (v 0 ) + T NQU : jx x 0 j<r ju(0; x) v(0; x)j dx + C jx x 0 j<r+mt + We then let R! 1, and by choosing the optimal value of we get (1.10). In order to prove (iii) we need a preliminary result, which is another version of Theorem.1. It formalizes the method of B. Cockburn and P.-A. Gremaud [6] to replace the BV regularity of u by the only BV regularity of v. Proposition 4.1 We make the same hypothesis as in Theorem.1, and assume moreover that H k = H(t; x; k); L k = L (t; x; k) (4:7) are Borel functions satisfying jh(t; x; k 1 ) H(t; x; k )j M H jk 1 k j; jl (t; x; k 1 ) L (t; x; k )j M L jk 1 k j: We also assume that v B([0; 1[ loc ; BV loc (R N )): Then, we can replace the terms E H and E L in the estimate (.4) by E H and E L, NX X E H = 1 + j=1 0<tT xb t (j) H (t; x); EL = 1 ( + ) provided that we add the term + ); with = M H N X j=1 0<tT + xb (t; x) ; E@L = 1 X 1i;jN 1i;jN M L 0<tT xb t 0<tT + xb 0 Moreover, if a v b, it is sucient that (4.8) holds for a k 1 ; k b. Notice that the only dierence between E H and E H (4:8) L (t; x); (t; x) : (4:10) (respectively E L and E L ) is that we replace a factor 1= by 1=( + ); which is bounded when! 0. Then for a local estimate we choose for a nite positive value, and for a global estimate we just let! 1. The result also holds when considering error terms for v as in Proposition.. Proof of Proposition 4.1. In the proof of Theorem.1, instead of using (1.4) in (.11), we keep the terms H(t; x; k) and L (t; x; k). Hence, the terms we have to estimate are R H = H(t; x; v(s; y)) rx ' dsdtdxdy; R L = L (t; x; v(s; ' (4:11) i X ij 17

18 They can be written R H = + R H ; = = X ij R L = + R L ; with H(t; x; v(s; y)) r y ' dsdtdxdy; L (t; x; v(s; j dsdtdxdy; (4:1) and R H ; R L take into account r x ' + r y ' j : Then we estimate R H and R L as in the proof of Theorem.1. The only dierence is that we do not have any longer the terms corresponding to the maximal order of derivation for. Therefore, we only retain in (.8) and (.30) terms in 1= and 1= respectively, instead of 1= and 1= as before. We hence get lim jrh j CE H ; limjr L "!0 "!0 j CE L : Now, in order to estimate and ; we notice that for xed s; t; x, the function v(s; :) belongs to BV loc : By the Lipschitz conditions (4.8) and by Appendix, we obtain that H(t; x; v(s; :)) BV loc ; L (t; x; v(s; :)) BV loc. Therefore, we integrate by parts to get X = ' div y h H(t; x; v(s; y)) i ; i Then, by Appendix (i), or (ii) if a v b; j ' M H (s; y) ; jr@l j M j h L (t; x; v(s; y)) i : (4:13) (s; y) : These terms are nally estimated as in the proof of Proposition. and we obtain which ends the proof of Proposition 4.1. lim jr@h j ; "!0 "!0 j ; Proof of Theorem 1.1 (iii). We begin as in the proof of (ii). We notice that L k = ij g k (u) can be written L k = L (t; x; k), with L (t; x; k) = ij j(u(t; x)) (k)j j(0) (k)j Therefore, the Lipschitz condition (4.8) is fullled with M L apply Proposition 4.1. We obtain jx x 0 j<r ju(t; x) v(t; x)j jx x 0 j<r+mt ++ : = ij Lip(), and we can ju 0 (x) v 0 (x)j + C(E t + E x + + E L ): 18

19 We have E L NQ ( + ) ju(t; x)j dtdx; (T + ) ]0;T [R N Lip()T V (v 0 ); and E x is estimated by (.9). By letting! 0 and! 1 (so that E L! 0) we get for any > 0 jx x 0 j<r ju(t; x) v(t; x)j R N ju 0 (x) v 0 (x)j + C T V (v 0 ) + T Lip()T V (v0 ) : By choosing the optimal value of and letting R! 1 we obtain the result (iii). 5 A relaxation model for nite volume methods In this section we use Theorem.1 to deduce an estimate in h 1=4 for the nite volume type model presented in the introduction. We rst state an existence result. Proposition 5.1 With the notations of the introduction, for any " > 0 and any u 0 L 1 (R N ) there exists a unique solution u C([0; 1[; L 1 (R N )) of (1.15) satisfying the entropy inequalities (1.16) for all convex, Lipschitz and C 1 functions S, or equivalently for all Kruzkov entropies. Since this result uses very standard techniques, we only give a short sketch of the proof. First let us notice that for an arbitrary convex and Lipschitz S, S 0 (u) is not well dened because of possible jumps of S 0. That is why we require S to be C 1 : Then, if (1.16) holds for these C 1 test functions, it is easy to see that it also holds for the entropies S k, with the convention that S 0 k() = sgn( k) and sgn(0) = 0: The uniqueness is obtained by Kruzkov's method. We actually get that for two solutions u; v of t ju vj+div sgn(u v)[f(u) f(v)] jp 0 (u v)j ju vj " which yields the contraction property in ]0; 1[R N ; (5:1) k u(t; :) v(t; :) k L 1k u 0 v 0 k L 1 : (5:) For existence, we use the small diusion approximation method. Here the L 1 BV hypothesis (1.13) on the grid is involved. It is actually equivalent to assert that P 0 maps L 1 (R N ) into BV (R N ) with T V (P 0 w) K h k w k L 1; w L 1 (R N ): (5:3) 19

20 It yields the following a priori estimate for a solution u of (1.15)-(1.16) of initial data u 0 L 1 \ BV (R N ): (t; :) k M e t=" k M +(1 e t=" )K h k u 0 k L 1 : (5:4) This allows to prove existence for u 0 L 1 \ BV: Then by the contraction property (5.) we obtain the existence of a solution for any u 0 L 1 (R N ). Now, we can prove the convergence rate to the continuous solution. Proof of Theorem 1.. If u 0 L 1 \ L p (R N ) for some p, 1 p < 1; we can actually choose S() = jj p in (1.16). Integrating this inequality with respect to x we get d dt R jujp dx N R S0 (u) P 0 u u dx in ]0; 1[: N " Then, since S 0 (P 0 u) is constant in each cell, and since the integral of P 0 u u on each cell vanishes, we have Therefore, we obtain d dt R S0 (u) P 0 u u N " R N jujp + and for any T 0 R N ju(t; :)jp + = h i S 0 (u) S 0 P 0 u u (P R 0 u) N " h S 0 (P R 0 u) i S 0 P 0 u u (u) N " ]0;T [R N h S 0 (P 0 u) S 0 (u) i P 0 u u In the case where u 0 L 1 \ L (R N ) (p = ); we nd that (P 0 u u) ju 0 j R N ]0;1[R N " 0 in ]0; 1[; " 0: (5:5) R N ju0 j p : (5:6) : (5:7) The announced inequality (1.17) easily follows by Holder's inequality. Now we write the entropy inequalities, for k t ju kj + div sgn(u k)[f(u) f(k)] sgn(u k) P 0 u u " = h sgn(u k) sgn(p 0 u k) i P 0 u u " sgn(p 0 u k) P 0 u u : " + sgn(p 0 u k) P 0 u u " (5:8) 0

21 Since the integral in each cell of the right-hand side vanishes (for a xed t), we can invert the \div" operator as in Appendix 1 and therefore sgn(p 0 u k) P 0 u u " = div H k (5:9) for some H k L 1 (]0; 1[; L 1 (R N )); which can be upper bounded by (A.7) and (1.17) s TjB(x jh k (t; x)j dtdx 0 ; R + h)j h k u 0 k L : " ]0;T [B(x 0 ;R) By examining the proof of Appendix 1, it is easy to see that since the sign in (5.9) is bounded by 1, we have indeed for some H independent of k, and which also satises ]0;T [B(x 0 ;R) H (t; x) dtdx h jh k (t; x)j H (t; x) (5:10) s TjB(x 0 ; R + h)j " k u 0 k L : (5:11) Therefore, combining (5.8) and (5.9), we may apply Theorem.1 with the only error term (5.9). By choosing = 0 and letting! 0 (so that E t! 0) we obtain, using (.9) for E x and (5.11) for E H, that for any > 0 jx x 0 j<r ju(t; x) v(t; x)j dx ju 0 (x) v 0 (x)j dx jx x 0 j<r+mt + +C N T V (v 0 ) + h s T (R + MT + + h) N " k u 0 k L 1 A : (5:1) Finally, we choose and we get the inequality (1.18). = q br` (5:13) Remark. It is an open problem to obtain an estimate in h 1= (except in one dimension, see R. Sanders [17]). Notice that Proposition 4.1 cannot be applied since the above term H k is not Lipschitz continuous in k. When all the cells are star-shaped, denoting by H(t; x) the function obtained by Appendix 1 such that we actually have by the support property (A.4) Then, the bound (5.11) is obvious. div H = P 0 u u ; (5:14) " H k = sgn(p 0 u k)h: (5:15) 1

22 Appendix. Inversion of the \div" operator; composition of Lipschitz and BV functions A1 Inversion of the \div" operator We consider a question issued from the nite volume type of approximations in section 5. In this context, we need the result below in order to use the error estimates stated in x. Lemma A1.1 Let (C i ) ii be a general grid of R N as described in the introduction. Assume that m L 1 (R N ) satises C i m(x) dx = 0; i I: (A:1) Then there exists a vector eld H(x) such that div H = m in R N ; (A:) k H k L 1 (R N h k m k ) L 1 (R N : (A:3) ) Remark A1. (1). If all the cells C i are star-shaped (there is a point x i C i such that for any x C i the segment [x i ; x] is included in C i ), then the construction of H is purely local. More precisely with H = X ii H i ; H i = 0 in R N nc i ; (A:4) div H i = m1i Ci ; (A:5) k H i k L 1 (R N ) h k m1i C i k L 1 (R N ) : (A:6) (). The estimate (A.3) can be localized: for any bounded Borel set! R N ; we have k H k L (!) 1 h k m k L 1 (! ); (A:7) h! h = [ C i : dist(c i ;!)h Actually, H i = 0 in! as soon as dist(c i ;!) > h, with H i dened in the proof below. (A:8) Proof of Lemma A1.1. To solve (A.) in the distributional sense means that we look for a function H such that, for all test functions ' C 1 c (R N ), R H(x) R r'(x) dx = N = X ii = X ii = X ii N m(x)'(x) dx C i m(x) ('(x) '(x i )) dx R N 1 C i m(x) (m1i Ci ) r' (x i + (x x i )) (x x i ) ddx x i + y x i r'(y) y x i d dy : N (A:9)

23 Here x i denotes an arbitrary point of C i, but in the star-shaped case it should satisfy the property in Remark A1. (1). Therefore we choose H i (y) = 1 We may majorize the L 1 norm of H i, R N jh i(y)j dy R N 1 = diam(c i ) H = X ii H i ; (A:10) (m1i Ci ) (jmj1i Ci ) x i + y x i x i + y x i y xi R (jmj1i N C i )(x)jx x i j dxd jm(x)j dx: C i d: (A:11) N +1 jy xi j N +1 ddy (A:1) Adding these inequalities, we obtain the statement of Lemma A1.1. In the star-shaped case, we notice from formula (A.11) that if H i (y) 6= 0; then z = x i + y x i C i for some ]0; 1[. This means that y = x i + (z x i ) also belongs to C i. This proves (A.4). The properties (A.5) and (A.6) are contained in the above proof. A Composition of Lipschitz and BV functions Let be an open subset of R N and consider a function u BV loc (), and g : R! R Lipschitz continuous. The well-known theory of A.I. Vol'pert [0] allows to compute the gradient of g(u) provided that g C 1. However, for a general Lipschitz function g we have the following simple result. Lemma A.1 With the above notations, g(u) belongs to BV loc () and (i) in the sense of [g(u)] ; (A:13) (ii) if a u b [g(u)] Lip [a;b] : (A:14) Proof. Let us dene g n = n g, u n = n u, where n 0 is a standard smoothing sequence (in 1 or N dimensions). Since these functions are smooth we g n (u n ) = gn(u 0 n n k g 0 n k L n n : 3

24 From the g n (u n ) we n and hence Lip(g) : (A:15) Therefore, for any test function ' C 1 c () we have * * [g n (u n )]; ' Lip(g) and by letting n! 1 [g(u)]; ' + Lip(g) + ; j'j + ; j'j Since the right-hand side is bounded by C k ' k L 1, we get that ; : [g(u)] is locally a measure, and hence g(u) BV loc (). Then by density (A.16) still holds for any ' C c (). By approximation in L 1 (j@ j [g(u)]j + j@ j uj); (A.16) is also true for any ' measurable and bounded with compact support in. Therefore, [g(u)] E (A:17) for any Borel set E with compact closure in, and (A.13) follows from the denition of the absolute value of a measure. In order to prove (ii) we dene eg() = 8 >< g() if a b; g(a) if a; >: g(b) if b: Then Lip(eg) = Lip(g), and eg(u) = g(u). We get the result by applying (i) to eg and u. [a;b] References [1] Ph. Benilan, R. Gariepy, Strong solutions in L 1 of degenerate parabolic equations, J. Di. Eq. 119 (1995), [] F. Bouchut, Ch. Bourdarias, B. Perthame, A MUSCL method satisfying all the numerical entropy inequalities, Math. Comp. 65 (1996), [3] S. Champier, T. Gallouet, R. Herbin, Convergence of an upstream nite volume scheme for a nonlinear hyperbolic equation on a triangular mesh, Numer. Math. 66 (1993),

25 [4] B. Cockburn, F. Coquel, P. Le Floch, An error estimate for nite volume methods for multidimensional conservation laws, Math. Comp. 63 (1994), [5] B. Cockburn, F. Coquel, P. Le Floch, Convergence of the nite volume method for multidimensional conservation laws, SIAM J. Num. Anal. 3 (1995), [6] B. Cockburn, P.-A. Gremaud, A priori error estimates for numerical methods for scalar conservation laws. Part I: The general approach, Math. Comp. 65 (1996), [7] B. Cockburn, P.-A. Gremaud, Error estimates for nite element methods for scalar conservation laws, SIAM J. Numer. Anal. 33 (1996), [8] C.M. Dafermos, Polygonal approximations of solutions of the initial value problem for a conservation law, J. Math. Anal. Appl. 38 (197), [9] R. Eymard, T. Gallouet, R. Herbin, The nite volume method, book to appear, "Handbook of Numerical Analysis", Ph. Ciarlet and J.L. Lions eds. [10] S.N. Kruzkov, First order quasilinear equations in several independent variables, Math. USSR Sb. 10 (1970), [11] N.N. Kuznetsov, Accuracy of some approximate methods for computing the weak solutions of a rst-order quasi-linear equation, USSR Comp. Math. and Math. Phys. 16 (1976), [1] P.D. Lax, Hyperbolic systems of conservation laws II, Comm. Pure Appl. Math. 10 (1957), [13] T.-P. Liu, M. Pierre, Source-solutions and asymptotic behavior in conservation laws, J. Di. Eq. 51 (1984), [14] B.J. Lucier, A moving mesh numerical method for hyperbolic conservation laws, Math. Comp. 46 (1986), [15] O.A. Oleinik, On the uniqueness and stability of the generalized solution of the Cauchy problem for a quasilinear equation, A.M.S. Transl. () 33 (1963), [16] B. Perthame, Convergence of N-schemes for linear advection equations, Trends in Applications of Mathematics to Mechanics, Pitman M SPAM77, New-York (1995). [17] R. Sanders, On convergence of monotone nite dierence schemes with variable spatial dierencing, Math. Comp. 40 (1983), [18] A. Szepessy, Convergence of a shock-capturing streamline diusion nite element method for scalar conservation laws in two space dimensions, Math. Comp. (1989),

26 [19] J.-P. Vila, Convergence and error estimates in nite volume schemes for general multidimensional scalar conservation laws I. Explicite monotone schemes, Math. Modeling and Num. Anal. 8 (1994), [0] A.I. Vol'pert, The spaces BV and quasilinear equations, Math. USSR-Sbornik (1967),