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An implicit finite volume scheme for a scalar hyperbolic problem with measure data related to piecewise

deterministic Markov processes

Robert Eymard, Sophie Mercier, Alain Prignet

To cite this version:

Robert Eymard, Sophie Mercier, Alain Prignet. An implicit finite volume scheme for a scalar hyper- bolic problem with measure data related to piecewise deterministic Markov processes. Journal of Com- putational and Applied Mathematics, Elsevier, 2008, 222 (2), pp.293-323. �10.1016/j.cam.2007.10.053�.

�hal-00693134�

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An implicit finite volume scheme for a scalar hyperbolic problem with measure data related to piecewise deterministic Markov

processes

Robert Eymard, Sophie Mercier, Alain Prignet

Université Paris-Est, Laboratoire d’Analyse et de Mathématiques Appliquées, UMR CNRS 8050, 5 bd Descartes, 77454 Marne la Vallée Cedex 2, France

Keywords:Linear hyperbolic problems with measure solutions; Weak bounded variation inequalities; Chapman–Kolmogorov equations; Piecewise- deterministic Markov process; Growth–collapse Markov process

1. Introduction

Within a more and more competitive context, industrialists often have to assess as accurately as possible different quantities linked for example to economical and/or safety constraints. For instance, an industrial gas provider must be aware of the production availability of his gas plant, because of possible penalties to be paid in case of drop in its production rate. Lots of other quantities may of course be of interest to him, such as the mean functioning cost per unit time, the mean number of component failures up to some fixed horizon, and so on. All those quantities may be We are interested here in the numerical approximation of a family of probability measures, solution of the Chapman–

Kolmogorov equation associated to some non-diffusion Markov process with uncountable state space. Such an equation contains a transport term and another term, which implies redistribution of the probability mass on the whole space. An implicit finite volume scheme is proposed, which is intermediate between an upstream weighting scheme and a modified Lax–Friedrichs one.

Due to the seemingly unusual probability framework, a new weak bounded variation inequality had to be developed, in order to prove the convergence of the discretised transport term. Such an inequality may be used in other contexts, such as for the study of finite volume approximations of scalar linear or nonlinear hyperbolic equations with initial data in L1. Also, due to the redistribution term, the tightness of the family of approximate probability measures had to be proven. Numerical examples are provided, showing the efficiency of the implicit finite volume scheme and its potentiality to be helpful in an industrial reliability context.

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written as expectations of some functional of the underlying stochastic process which describes the time evolution of the studied system, or equivalently, as integrals with respect to the marginal distribution of the process. From an industrial point of view, such distributions then are essential to evaluate. Unfortunately, they often are unreachable in closed form, especially in the modern context of dynamic reliability, which is concerned with the study of so-called hybrid systems (details below). Such distributions (or their derivatives) are hence numerically evaluated, most of the time by Monte Carlo simulations which often entail very long computation times. New methods for their numerical assessment then are to be developed, from where most of the quantities with industrial interest may be derived, in the context of dynamic reliability.

The point of this paper hence is to develop some numerical scheme, to assess the marginal distribution of some process describing the time evolution of a hybrid system. Such a hybrid system is governed by two different types of dynamics: a discrete dynamic, which is related to the occurrence of events such as failures of components, some button switching, and so on, and a continuous dynamic, linked to the evolution of continuous characteristics, such as pressure, temperature, liquid level in a tank, and so on. The time evolution of such hybrid systems is modelled with so-called Piecewise-Deterministic Markov Processes (PDMPs), which are non-diffusion Markov processes with uncountable state space; see [10] or [17] for details. A PDMP hence is a Markov hybrid process(It,Xt)t≥0, where the discrete part It takes range in a finite set E, and where the continuous partXt takes range inRN, withN ∈ N. Due to its Markovian characteristic, one may write its associated Chapman–Kolmogorov equation. A PDMP jumps at countable isolated times and such an equation comprises a transport term, which corresponds to a deterministic evolution of the process between jumps, and a redistribution term, which corresponds to jumps and entails redistribution of the probability mass on the whole space. In order to make the paper clearer, we now make the choice to specialise our exposure to so-called Markov Growth–Collapse processes (GCPs); see [5]. Such processes are PDMPs whereE is reduced to a singleton, so that the discrete part(It)t≥0 is constant and hence unnecessary. They typically describe the time evolution of a quantity, for instance the size of a queue, with successive phases of deterministic growth and random instantaneous collapse (or jump in the vocabulary of PDMPs). Their behaviour hence is typical of a PDMP and such a specialisation allows us to be much clearer in our exposure. Extension of our results for GCPs to PDMPs is straightforward and is exposed at the end of the paper, in Section6.

Let us now be more specific and let (Xt)t≥0 be a GCP: as already said, the process(Xt)t≥0 jumps at isolated countable times. Between jumps, the deterministic evolution of(Xt)t≥0follows an ordinary differential equation:

dXt

dt =v(Xt) , ∀t ∈ [t1,t2) (1)

wherevis an application fromRN toRN, andt1,t2are two successive jump times. At jump times, transitions from Xt=x∈RNtoXt =y∈RNare governed by a transition rateλ (x)and by a probability measureµ(x)(dy)which stands for the conditional distribution ofXt given thatXt =x.

We make the following assumptions on the data, which will be referred to as assumptionsH0in the following:

• the transition rateλ:RN →R+is continuous and bounded byΛ= kλk,

• the velocityv:RN →RN is Lipschitz continuous and bounded byV = kvk>0,

• let P(RN) be the set of probability measures on RN; the function µ : RN → P(RN) is such that for all ψ ∈ Cb(RN) (continuous and bounded from RN toR), the function x 7→ R

RNψ(y)µ(x)(dy)is continuous (and bounded) fromRNtoR,

• the measureρini(dx)is a probability measure onRN, and stands for the initial distribution of(Xt)t≥0.

We may now write the Chapman–Kolmogorov (CK) equation associated to the Markov process(Xt)t≥0and we set ρ(t)(dx)to be the marginal distribution at timetof the process(Xt)t≥0. UnderH0, it is then known that(ρ(t)(dx))t≥0

is the single family of probability measures solution of the CK equation, which here is written as Z

RNϕ(x)ρ (t) (dx)− Z

RNϕ(x)ρini(dx)− Z t

0

Z

RN(v(x)· ∇ϕ(x))ρ(s)(dx)ds

= Z t

0

Z

RNλ(x) Z

RNϕ(y)µ(x)(dy)−ϕ(x)

ρ (t) (dx)ds ∀t ∈R+,∀ϕ∈Cc1(RN) (2)

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whereCc1(RN)stands for the set of continuously differentiable functions fromRN toRwith a compact support;

see [10] or [17] for the CK equation associated to a PDMP, and [7] for the uniqueness result. The third term in the left side of(2)is the transport term whereas the right side is the redistribution term.

This paper is dedicated to the study of the convergence of an implicit finite volume scheme towards the solution of the CK equation(2)and we now discuss the mathematical nature of some of the arising problems. To make it clearer and only for the purpose of this introduction, we consider the case where the data measures involved in the CK equation admit density with respect to Lebesgue measure onRN: we hence assume thatµ(x)(dy)=dµ(x,y)dy andρini(dx)=uini(x) dx, whereuini ∈ L1(RN)becauseρiniis a probability measure. It may then be proved that the solutionρ(t)(dx)of the CK equation admits a density and we setρ(t)(dx)=u(x,t) dx. Writing again the CK equation, the functionuthen happens to be, at least formally, a weak solution of the equation

du

dt(x,t)+div(u(x,t)v(x))= Z

RN

λ(y)dµ(y,x)u(y,t)dy−λ(x)u(x,t) (3) for(x,t)∈R+×RN with the initial condition

u(x,0)=uini(x), forx∈RN. (4)

Problem(3)–(4) can be seen as a linear hyperbolic problem with an integral form right-hand side, and an initial condition inL1(RN)instead of the standard frameworkuini ∈ L(RN). Only for the purpose of this introduction again, let us setuh,k to be an approximate solution, whereh is a space step andk is a time step. Looking at the convergence study (see(62)–(63)in the proof ofLemma 9), one may see that the following condition is used for proving the convergence ofuh,k:

∀R,T >0, lim

h→0h Z T

0

|uh,k(·,t)|

B V(B(0,R))dt =0, (5)

where, for a functionv:RN→Rand for any setΩ ⊂RN, we denote by

|v|B V()=sup

Z

RNv(x)divϕ(x)dx

, ϕ∈C1(RN)N,kϕk≤1, ϕ(x)=0 for allx∈RN\Ω

(6) the BV-semi-norm ofv.

In order to prove (5), ifuini were in L2loc(RN), one might proceed as is done in [6] or in [14] for explicit or implicit finite volumes schemes on general meshes anduini ∈ L(RN), and prove some weak bounded variation (BV) inequality of the following shape:

Z T 0

|uh,k(·,t)|

B V(B(0,R))dt ≤ Ckuinik

L2(B(0,R0))

h ,

whereCandR0depend onRandT. Unfortunately, such an inequality is no longer valid here for the present scheme anduini ∈ L1(RN), which is imposed by our probabilistic context. We have hence been led to develop a new weak BV inequality:

Z T 0

|uh,k(·,t)|

B V(B(0,R))dt ≤ C h1/q,

for some 1 < q < 2 (seeLemma 4). This inequality, which is sufficient to get (5), is shown using the analogy between an upstream weighting scheme or a modified Lax–Friedrichs scheme, and a continuous problem including a vanishing viscosity term. We can hence use the tools developed in [11] or [16] for the convergence of finite volume approximations to the solution of elliptic problems with measure data, which themselves mimic the continuous framework provided in [4,3] for continuous parabolic problems. Nevertheless, the proofs given in such papers do not hold in the case of the classical upstream weighting scheme, since a non-vanishing viscosity term was required on the whole mesh, which has led us to introduce an intermediate scheme between an upstream weighting one and a modified Lax–Friedrichs scheme.

Also, in order to complete the convergence proof, a last technical problem is due to the redistribution term in the CK equation. Due to that term, we indeed have to control the probability mass which escapes outside compact sets for

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the family of approximate distributions. In other words, we have to prove tightness of this family. With that aim, we explicitly construct a Liapounov function, which classically allows us to conclude (see [12] e.g.).

Finally, note that we had already studied the convergence of an explicit finite volume scheme towards the solution of the CK equation(2)in a previous paper [8]. The convergence proof was established there only under an inverse CFL conditionh/k→0, this condition being used in order to replace a weak bounded variation inequality which we did not have. The tools developed in the present paper would however not hold in the framework of [8], where the scheme is a purely upstream weighting one.

This paper is organised as follows. The numerical scheme is provided in Section 2 (in the case when E is a singleton). Some properties of the scheme are developed in Section 3, which are required to get some weak BV inequalities given in Section4. Such weak BV inequalities are used for proving the convergence of the scheme, which is done in Section5. We then present in Section6extensions of our results to the case of a generalE, with possible jumps between discrete states ofE. We finally conclude this paper in Section7with numerical experiments showing the efficiency and precision of the method, and its relevance in an industrial reliability context.

2. The finite volume scheme

Let us first give the definition of admissible meshes ofRN.

Definition 1. An admissible mesh ofRNfor problem(2)is a partitionMofRN such that

(1) for allK ∈M,K is bounded, the interior ofKis an open convex subset ofRNand theN dimensional measure ofK, denoted by m(K), is strictly positive,

(2) for all K ∈ M, denoting by∂K the boundary of K, and, for allL ∈ M, denoting by K|L =∂K ∩∂L, there existsNK ⊂Msuch thatK 6∈NK and∂K =S

L∈NK K|L, and, for allL ∈NK,K|L, called an edge ofK, is included in a hyperplane ofRN, with a strictly positive N−1 dimensional measure equal to m(K|L); we then denote bynK Lthe unit normal vector toK|Loriented fromK toL,

(3) the size of the mesh, defined byhM=supK∈Mdiam(K), is finite, (4) there existsC1≥1 with

1 C1

hM X

L∈NK

m(K|L)≤m(K)≤C1hM X

L∈NK

m(K|L), ∀K ∈M, (7)

1

C1hM≤diam(K)≤hM, ∀K ∈M. (8)

and 1

C1hNM≤m(K)≤C1hMN , ∀K ∈M. (9)

We then denote byCMthe infimum of allC1such that(7)–(9)hold.

Note that all classical regular meshes ofRNare admissible in the sense of the previous definition.

Now, letMbe a fixed admissible mesh ofRN. For such a mesh, we set vK,L = 1

m(K|L) Z

K|L

v(x)·nK Lds(x), ∀K ∈M,∀L∈NK, (10) where ds(x)stands for theN−1 dimensional measure onK|L and

wK,L =max(|vK,L|, ε), ∀K ∈M,∀L ∈NK (11)

for a givenε∈ [0,V]. We also set λK = 1

m(K) Z

K

λ (x)dx aK,L = 1

m(K) Z

K

λ(x) Z

L

µ(x) (dy)

dx

(6)

forK,L∈Mwith X

L∈M

aK,LK. (12)

For a given time stepk>0, the scheme then is written as u(K0)= 1

m(K) Z

K

ini(x), ∀K ∈M (13)

and

m(K)(u(Kn+1)−u(Kn))+k X

L∈NK

m(K|L) vK,L

u(Kn+1)+u(Ln+1)

2 +wK,L

2 (u(Kn+1)−u(Ln+1))

!

= −km(K)λKu(Kn+1)+k X

L∈M

m(L)aL,Ku(Ln+1), ∀K ∈M,∀n ∈N.

(14)

We first prove the existence and uniqueness of a solution to this numerical scheme in the following lemma.

Lemma 2. Let us assume hypothesesH0and letMbe an admissible mesh of RN in the sense of Definition1. Let k>0andε∈ [0,V]be given. Then there exists one and only one family of real numbers(u(Kn))K∈M,n∈Nsuch that (13)–(14)hold andP

K∈Mm(K)|u(Kn)|<∞for all n∈N. Moreover, the following properties hold:

u(Kn)≥0, ∀K ∈M,∀n ∈N, (15)

and X

K∈M

m(K)u(Kn)=1, ∀n∈N. (16)

Proof. LetkkL

1 be the following norm onL1= {u :=(uK)K∈M s.t. P

K∈Mm(K)|uK|<+∞}:

kukL

1 = X

K∈M

m(K)|uK|. (17)

Foru∈L1fixed, let us considerψudefined byψu(p)=rfor p∈L1with Cm(K)rK =Cm(K)pK−m(K)

k (pK−uK)

− X

L∈NK

m(K|L) vK,L

pK+pL

2 +wK,L

2 (pK−pL)

−m(K)λKpK+ X

L∈M

m(L)aL,KpL

(18)

= m(K)

k uK+m(K) C−1

k− 1

2m(K) X

L∈NK

m(K|L) vK,L +wK,L

−λK

! pK

+1 2

X

L∈NK

m(K|L) wK,L−vK,L

pL + X

L∈M

m(L)aL,KpL (19)

whereC>0 is a constant to be chosen such that the coefficient of pK in(19)is non-negative.

Due to(7),vK,L+wK,L ≤2V andλK ≤Λ, we know 1

k + 1

2m(K) X

L∈NK

m(K|L) vK,L+wK,L

K ≤ 1

k+VCM

hM

+Λ. We then takeCsuch that

C≥ 1

k +VCM hM +Λ.

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As the coefficients of pL andpKin(19)now are non-negative (rememberwK,L−vK,L >0), we derive that kψu(p)kL

1

≤ 1 C

X

K∈M

"

m(K)

k |uK| +m(K) C−1 k − 1

2m(K) X

L∈NK

m(K|L) vK,L+wK,L

−λK

!

|pK|

+1 2

X

L∈NK

m(K|L) wK,L−vK,L

|pL| + X

L∈M

m(L)aL,K|pL|

#

(20)

= 1 C

"

1 kkukL

1+

C−1

k

kpkL

1−1 2

X

K∈M

X

L∈NK

m(K|L) vK,L+wK,L

|pK| − X

K∈M

m(K) λK|pK|

+1 2

X

K∈M

X

L∈NK

m(K|L) wK,L−vK,L

|pL| + X

L∈M

m(L)λL|pL|

#

(21) using(12).

Noting thatP

K∈M

P

L∈NK =P

L∈M

P

K∈NL,vL,K = −vK,L, m(K|L)=m(L|K)and thatwK,L =wL,K, we easily get

X

K∈M

X

L∈NK

m(K|L) wK,L−vK,L

|pL| = X

K∈M

X

L∈NK

m(K|L) wK,L+vK,L

|pK| (22) and hence

u(p)kL

1 ≤ 1

C 1

kkukL

1+

C−1

k

kpkL

1

<+∞. (23)

As a consequence, the functionψumapsL1intoL1for anyu ∈L1. Besides, similar computations also give

ψu(p)−ψu p0 L

1

≤ C−1

k

C p−p0

L

1 (24)

for all p,p0∈L1. The functionψuthen is a contraction on the Banach spaceL1and, for allu ∈L1, the functionψu

admits a unique fixed pointp∈L1. Noting thatψu(n) u(n+1)

=u(n+1)is equivalent to(14), the existence and uniqueness of u(n)

n∈NinL1such that (13)–(14)are true is then clear, recursively.

Let us now set C=

u:=(uK)K∈M∈L1s.t.uK ≥0 for allK ∈Mand kukL

1 =1 .

Takingu ∈ Cand p ∈ C, it is easy to check thatr := ψu(p)∈C, using the non-negativeness of the coefficients in (19), and noting that(20)–(21)and consequently the left inequality from(23)now are equalities.

Now, starting fromu ∈C, the sequence(pn)n∈Nrecursively defined byp0=uand pn+1u(pn)is such that pn∈Cfor alln∈Nand converges inL1towards the single fixed point ofψu. We derive that for allu∈C, the single fixed point ofψuinL1actually is an element ofC.

Consequently, starting withu(0)∈C, the single sequence u(n)

n∈NinL1such that(13)–(14)are true has elements inC, which achieves the proof.

3. Properties of the scheme

For alls∈R, we denote bybscthe greatest integer lower than or equal tosand for allR ≥0, letB(0,R)⊂RN be the open ball with centre 0 and radius R, withB(0,0)= ∅.

Lemma 3. Let us assumeH0and letMbe an admissible mesh ofRNin the sense ofDefinition1. Let m∈(0,+∞), k >0andε ∈(0,V]be given and let(u(Kn))K∈M,n∈Nbe the family of real numbers defined by(10),(11),(13)and

(8)

(14)such that(15)and(16)hold. Let R>0and T >0be given and letθ(R):R+→ [0,1]be defined byθ(R)(s)=1 for all s∈ [0,R],1+R−s for all s∈ [R,R+1]and 0 for all s≥ R+1. Let us denote

θK(R)= 1 m(K)

Z

K

θ(R)(|x|)dx, and let us define(bu(Kn))K∈M,n∈Nby

bu(Kn)K(R)u(Kn), ∀K ∈M,∀n∈N. (25) Let C1be such that CM≤C1, let C2be such that hM<C2 and k<C2 and let T1be the term defined by

T1=

bT/kc

X

n=0

k X

K∈M

X

L∈NK

m(K|L) (bu(Kn+1)

bu(Ln+1))2

(1+max(bu(Kn+1),bu(Ln+1)))m+1. (26) Then there exists C3, only depending on N , R, T , m,v, C1,Λ,εand C2, such that

T1≤C3. (27)

Proof. Let us introduce, as in [4], the functionφm : R→Rsuch thatφm(s)=(1−1/(1+ |s|)m)×sign(s)with φm(0)=0 andφm0 (s)=m/(1+ |s|)m+1, for alls∈R. We define the real functionΦmbyΦm(s)=Rs

0 φm(u)du, for alls∈R. We have 0≤φm(s)≤1 and 0≤Φm(s)≤sfors≥0. From(14), we get

m(K)(u(Kn+1)−u(Kn))+km(K)u(Kn+1)DK+k 2

X

L∈NK

m(K|L) wK,L−vK,L

(u(Kn+1)−u(Ln+1))

= −km(K)λKu(Kn+1)+k X

L∈M

m(L)aL,Ku(Ln+1), ∀K ∈M,∀n ∈N (28) denoting by

DK = 1 m(K)

Z

K

div(v(x))dx = 1 m(K)

X

L∈NK

m(K|L)vK,L, ∀K ∈M. (29)

Thanks toH0, we get the existence ofC4, only depending onv, such that

|DK| ≤C4. (30)

Let us multiply (28) by θK(R)φmK(R)u(Kn+1)), and sum the result on K ∈ M and n = 0, . . . ,bT/kc. We get T2+T3+T4+T5=0, with

T2=

bT/kc

X

n=0

X

K∈M

θK(R)φmK(R)u(Kn+1))m(K)(u(Kn+1)−u(Kn)),

T3=

bT/kc

X

n=0

k X

K∈M

m(K)θK(R)φm(KR)u(Kn+1))u(Kn+1)DK,

T4= 1 2

bT/kc

X

n=0

k X

K∈M

θK(R)φmK(R)u(Kn+1)) X

L∈NK

m(K|L) wK,L−vK,L

(u(Kn+1)−u(Ln+1)), and

T5=k

bT/kc

X

n=0

X

K∈M

θK(R)φmK(R)u(Kn+1)) m(K)λKu(Kn+1)− X

L∈M

m(L)aL,Ku(Ln+1)

! . We havebu(Kn+1) ≥ 0 and P

K∈Mm(K)bu(Kn+1) ≤ 1 from Lemma 2, and bu(Kn+1) = 0 for all K ∈ M such that K 6⊂ B(0,R+1+C2). The Taylor–Lagrange expansion formula yields that, for alla,b ∈ R, there exists

(9)

Ψa,b∈ [min(a,b),max(a,b)]such that φm(a)(a−b)=Φm(a)−Φm(b)+1

m0a,b)(a−b)2≥Φm(a)−Φm(b). (31) Applying(31)fora =

bu(Kn+1)andb=

bu(Kn), we get that T2≥ X

K∈M

m(K)Φm(bu(KbT/kc+1))− X

K∈M

m(K)Φm(bu(K0)), (32)

with

0≤ X

K∈M

m(K)Φm(bu(K0))≤1 due toΦm(s)≤sfors≥0.

Thanks to(30)and 0≤φm(s)≤1 for alls≥0, we have

T3≥ −(T +C2)C4. (33)

Let us writeT4=T6+T7, with T6=1

2

bT/kc

X

n=0

k X

K∈M

X

L∈NK

m(K|L) wK,L −vK,L

φm(bu(Kn+1))(bu(Kn+1)

bu(Ln+1)), and

T7=1 2

bT/kc

X

n=0

k X

K∈M

φm(bu(Kn+1)) X

L∈NK

m(K|L) wK,L −vK,L

u(Ln+1)L(R)−θK(R)).

Changing the order of summation, we get T7=1

2

bT/kc

X

n=0

k X

L∈M

u(Ln+1) X

K∈NL

φm(bu(Kn+1))m(K|L) wK,L−vK,L

L(R)−θK(R)).

Using(8), we get|θL(R)−θK(R)| ≤2hM. Hence, using(7), we get

0≤ X

K∈NL

φm(bu(n+1K ))m(K|L) wK,L −vK,L

L(R)−θK(R)| ≤m(L)4V C1. This leads to

T7≥ −2V C1 bT/kc

X

n=0

k X

L∈M

m(L)u(n+1)L ≥ −(T +C2)2V C1. (34)

Let us apply(31)toT6. We getT6=T8+T9, with T8=1

2

bT/kc

X

n=0

k X

K∈M

X

L∈NK

m(K|L) wK,L −vK,L

m(bu(Kn+1))−Φm(bu(Ln+1))).

Thanks to the expression ofφ0m, we have T9=m

4

bT/kc

X

n=0

k X

K∈M

X

L∈NK

m(K|L) wK,L−vK,L (bu(Kn+1)

bu(Ln+1))2 (1+Ψ

bu(n+1)K ,bu(n+1)L )m+1. We remark thatwK,L−vK,L ≥0 and

(bu(Kn+1)

bu(Ln+1))2 (1+Ψ

bu(Kn+1),bu(Ln+1))m+1 ≥ (bu(Kn+1)

bu(Ln+1))2 (1+max(bu(Kn+1),bu(Ln+1)))m+1,

(10)

which implies that T9≥ m

4

bT/kc

X

n=0

k X

K∈M

X

L∈NK

m(K|L) wK,L−vK,L (bu(Kn+1)

bu(Ln+1))2 (1+max(bu(Kn+1),bu(Ln+1)))m+1. Gathering by edges the right-hand side of the above equation leads to

T9≥ m 4

bT/kc

X

n=0

k X

K∈M

X

L∈NK

m(K|L)wK,L

(bu(Kn+1)

bu(Ln+1))2 (1+max(bu(Kn+1),bu(Ln+1)))m+1.

Thanks to(11), we also knowwK,L ≥ε, and the termT1defined by(26)now satisfies T9≥ m

4εT1. (35)

We next writeT8=T10+T11, with T10=

bT/kc

X

n=0

k X

K∈M

X

L∈NK

m(K|L) vK,L

Φm(bu(n+1K ))+Φm(bu(n+1L ))

2 +wK,L

2 (Φm(bu(n+1K ))−Φm(bu(n+1L )))

! ,

and

T11= −

bT/kc

X

n=0

k X

K∈M

m(K)Φm(bu(Kn+1))DK. Gathering by edges, we remark that

T10=0. (36)

We also get that

T11≥ −(T +C2)C4. (37)

T5 = k

bT/kc

X

n=0

X

K∈M

θK(R)φm(KR)u(Kn+1)) m(K)λKu(Kn+1)− X

L∈M

m(L)aL,Ku(Ln+1)

!

≥ −k

bT/kc

X

n=0

X

K∈M

θ(KR)φmK(R)u(Kn+1)) X

L∈M

m(L)aL,Ku(Ln+1)

≥ −Λ(T+C2).

Thanks to the relationT2+T3+T7+T9+T10+T11+T5=0, and to(32),(33),(34),(36),(37), and(35), we deduce the existence ofC3, only depending onN,R,T,m,v,C1,ε,ΛandC2, such that(27)holds.

4. A weak BV inequality

In this section, we apply many times H¨older’s inequality X

i∈I

aibi ≤ X

i∈I

aiα

!α1 X

i∈I

bβi

!β1

, forai,bi ≥0, α, β >1 with 1 α +1

β =1, (38)

with various choices forI,ai,bi,αandβ that we define each time.

A weak BV inequality is provided in the next lemma. Such an inequality is valid for general families (bu(Kn))K∈M,n∈Nand does not depend on the scheme. Its proof requires some Sobolev inequalities which are given further inLemmas 6 and7. Such Sobolev inequalities were already given in [9], with alternate assumptions and proofs however (see the introduction of this paper).

(11)

Lemma4 (AWeakBVInequality).LetN ∈ N? andletMbeanadmissiblemeshofRN inthesenseofDefinition1.

If N ≥ 2, let q ∈ (1,N+2N+1)be given and let m := [(2−q) (N+1) /N] −1 >0. If N =1, let q ∈ (1,√ 2)and m∈(0,2−qq2)be given. Let(bu(Kn))K∈M,n∈Nbe a family of non-negative real numbers such thatP

K∈Mm(K)bu(Kn)≤1 for all n ∈Nand such that there exists R >0withbu(Kn) =0, for all n∈ N, for all K ∈Msuch that K 6⊂B(0,R). Let T >0and k >0be given and let the terms T1and T12be respectively defined by(26)and

T12=

bT/kc

X

n=0

k X

K∈M

X

L∈NK

m(K|L)|

bu(Kn+1)− bu(Ln+1)|.

Let C1be such that CM ≤C1, let C3 be such that T1 ≤C3 and let C2be such that hM <C2and k <C2. Then there exists C5, only depending on N , R, T , q, m, C1, C2and C3 such that

T12≤C5h−1M/q. (39)

Remark 5. Forε∈(0,V], we know from(27)inLemma 3that the assumptions of the previous lemma are valid and hence that(39)is true for the family(bu(n)K )K∈M,n∈Nassociated to the scheme and constructed by(25).

Proof. In the course of this proof, we denote byCi, fori >5, various positive real numbers, only depending onN, R,T,q,C1,C2andC3. Let us first defineT13by

T13=h2−qM

bT/kc

X

n=0

k X

K∈M

X

L∈NK

m(K|L)|

bu(Kn+1)

bu(Ln+1)|q.

According to some ideas of [4,3], our aim is to prove thatT13 is bounded independently ofhM (inequality(44)), which then provides the conclusion(39), applying H¨older’s inequality, as we show at the end of this proof. Letθ(R) be associated toRas inLemma 3. We denote byMθ(R) the set of allK ∈Msuch thatθK(R) 6=0 or such that there exists L ∈ NK withθL(R) 6= 0. We then apply H¨older’s inequality(38)withI = {(n,K,L), n = 0, . . . ,bT/kc, K ∈Mθ(R),L ∈NK},α=2/q,β=2/(2−q),

an,K,L = km(K|L) (bu(n+1)K

bu(n+1)L )2 (1+max(bu(n+1K ),bu(n+1L )))m+1

!q/2

, and

bn,K,L =(h2Mkm(K|L))(2−q)/2

1+max(bu(Kn+1),bu(Ln+1))(m+1)q/2

. Hence we get

T13≤(T1)q/2(T14)1−q/2, (40)

definingT14by T14=h2M

bT/kc

X

n=0

k X

K∈Mθ(R)

X

L∈NK

m(K|L)

1+max(bu(n+1)K ,bu(n+1)L )(m+1)q/(2−q)

.

Note that the expression ofT14would be meaningless if we write the sum onK ∈Minstead ofK ∈Mθ(R). Thanks to q ∈(1,2)andm>0, which implyr :=(m+1)q/(2−q) >1, and thanks to the inequality(a+b)r ≤2r−1(ar+br) for allr≥1,a,b≥0, we get the existence ofC6, only depending onN,mandq, such that

T14≤hMC6(T15+T16), (41)

definingT15andT16by T15=hM

bT/kc

X

n=0

k X

K∈Mθ(R)

X

L∈NK

m(K|L)≤C7, (42)

(12)

withC7 =(T +C2)C1m(B(0,R+1C2))and T16=hM

bT/kc

X

n=0

k X

K∈Mθ(R)

X

L∈NK

m(K|L)

max(bu(Kn+1),bu(Ln+1))(m+1)q/(2−q)

.

We can then write T16≤hM

bT/kc

X

n=0

k X

K∈Mθ(R)

X

L∈NK

m(K|L) bu(Kn+1)

(m+1)q/(2−q)

+

bu(Ln+1)

(m+1)q/(2−q) ,

which leads, thanks to(7), to T16≤2C1

bT/kc

X

n=0

k X

K∈Mθ(R)

m(K) bu(n+1K )

(m+1)q/(2−q)

.

Let us now consider two cases:

(1) CaseN ≥2: thanks to the definition ofm, we know(m+1)q/ (2−q)=q(N+1) /N. Let us apply H¨older’s inequality(38)withI =Mθ(R), and

aK =(m(K)bu(Kn+1))q/N, bK =m(K)1−q/N(u(Kn+1))q,

α=N/q,β=N/(N−q)(this is possible sinceN ≥2) andq?=N q/(N−q). We get X

K∈Mθ(R)

m(K)

bu(Kn+1)(m+1)q/(2−q)

 X

K∈Mθ(R)

m(K)bu(Kn+1)

q N

 X

K∈Mθ(R)

m(K)

bu(Kn+1)q?

q q?

. (2) CaseN =1: let us defineq?,αandβby 1/α+1/β =1, 1/α+q?/β =(m+1)q/ (2−q)and 1β = q

q?. This leads toq? =q/(1+q−(m+1)q/(2−q))withq? ∈(q,+∞), sinceq < (m+1)q/(2−q) <1+q due to m∈(0,2−qq2)withq ∈(1,√

2). Therefore, setting

aK =(m(K)bu(Kn+1))1, bK =m(K)1(u(Kn+1))q?, we get

X

K∈Mθ(R)

m(K)

bu(Kn+1)(m+1)q/(2−q)

 X

K∈Mθ(R)

m(K)bu(Kn+1)

1

 X

K∈Mθ(R)

m(K)

bu(Kn+1)q?

q q?

. In both cases, thanks to the hypothesisP

K∈Mm(K)bu(Kn+1)≤1, we obtain T16≤2C1

bT/kc

X

n=0

k

 X

K∈Mθ(R)

m(K) bu(n+1K )

q?

q/q?

.

We now applyLemma 6or7to the values(bu(n+1K ))K∈M: there existsC8, only depending onN,q,m,RandC1 such that

 X

K∈Mθ(R)

m(K)

bu(Kn+1)q?

q/q?

≤C8h1−qM X

K∈M

X

L∈NK

m(K|L)|

bu(Kn+1)

bu(Ln+1)|q, which leads to

T16≤h1−qM 2C1C8 bT/kc

X

n=0

k X

K∈M

X

L∈NK

m(K|L)|

bu(Kn+1)

bu(Ln+1)|q,

(13)

and therefore

hMT16 ≤2C1C8T13. (43)

Hence we get from the hypothesisT1≤C3,(40)–(43) T13≤(C3)q/2(C6(hMC7 +2C1C8T13))1−q/2.

This leads, thanks tohM≤C2, to the existence ofC9, such that

T13≤C9. (44)

We now apply H¨older’s inequality(38)toT12withI = {(n,K,L), n =0, . . . ,bT/kc,K ∈M, L ∈NK},α=q, β =q/(q−1),

an,K,L =(h2−qM km(K|L))1/q|

bu(Kn+1)− bu(Ln+1)|, and

bn,K,L =h−1M/q(hMkm(K|L))(q−1)/q. Hence we get

T12≤h−1M/qT131/q hM bT/kc

X

n=0

k X

K∈M

X

L∈NK

m(K|L)

!1−1/q

.

Applying(42)and(44), we get T12≤h−1M/qC91/q(C7)1−1/q, which implies(39).

Lemma 6 (General Discrete Sobolev Inequality, Case N =1).Let N =1and letMbe an admissible mesh in the sense of Definition1. Let(buK)K∈Mbe a family of real numbers such that the number of K ∈Msuch thatbuK 6=0 is finite, let A>0such that

X

K∈M,buK6=0

m(K)≤ A,

and let q ∈(1,+∞)and q? ∈(1,+∞)be given. Let C1 >CMbe given. Then there exists C10, only depending on N , q, q?, A and C1, such that

X

K∈M

m(K)| buK|q?

!1/q?

≤C10 X

K∈M

X

L∈NK

m(K|L)hM| buK

buL|q hqM

!1q

. (45)

Proof. We have, for allK¯ ∈M,

|buK¯| ≤ X

K∈M

X

L∈NK

|buK− buL|. Hence we get

 X

K∈M¯

m(K¯)| buK¯|q?

1/q?

≤ A1/q? X

K∈M

X

L∈NK

|buK − buL|.

Recalling that m(K|L)=1 for allK ∈MandL ∈NK becauseN =1, we have X

L∈NK

1≤ C1 hM

m(K)

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