Well-posedness of IBVP for 1D scalar non-local conservation laws

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Well-posedness of IBVP for 1D scalar non-local

conservation laws

Paola Goatin, Elena Rossi

To cite this version:

Paola Goatin, Elena Rossi. Well-posedness of IBVP for 1D scalar non-local conservation laws. Journal of Applied Mathematics and Mechanics / Zeitschrift für Angewandte Mathematik und Mechanik, Wiley-VCH Verlag, 2019, 99 (11), �10.1002/zamm.201800318�. �hal-01929196�

Well-posedness of IBVP for 1D

scalar non-local conservation laws

Paola Goatin1 Elena Rossi1

Abstract

We consider the initial boundary value problem (IBVP) for a non-local scalar conservation laws in one space dimension. The non-local operator in the flux function is not a mere convolution product, but it is assumed to be aware of boundaries. Introducing an adapted Lax-Friedrichs algorithm, we provide various estimates on the approximate solutions that allow to prove the existence of solutions to the original IBVP. The uniqueness follows from the Lipschitz continuous dependence on initial and boundary data, which is proved exploiting results available for the local IBVP.

2010 Mathematics Subject Classification: 35L04, 35L65, 65M12, 65N08, 90B20

Keywords: Scalar conservation laws, Non-local flux, Initial-boundary value problem, Lax-Friedrichs scheme

1

Introduction

We consider the following Initial Boundary Value Problem (IBVP) on the open bounded interval ]a, b[ ⊂ R          ∂tρ + dxf (t, x, ρ, J ρ) = 0, (t, x) ∈ R+×]a, b[, ρ(0, x) = ρo(x), x ∈ ]a, b[, ρ(t, a) = ρa(t), t ∈ R+, ρ(t, b) = ρb(t), t ∈ R+, (1.1)

where J denotes a non-local operator and we use the notation dxf  t, x, ρ(t, x), J ρ(t)
(x)= ∂xf  t, x, ρ(t, x), J ρ(t)
(x) + ∂ρf  t, x, ρ(t, x), J ρ(t)
(x) ∂xρ(t, x) (1.2) + ∂Rf  t, x, ρ(t, x), J ρ(t)
(x) ∂x J ρ(t)
(x).

The same problem was studied in [10]. In that case, the choice for J is the classical convolution product J ρ = ρ ∗ η, η being a smooth convolution kernel. However, in such formulation, the non-local term may exceed the boundaries of the spatial domain. The authors address this

1

Inria Sophia Antipolis - M´editerran´ee, Universit´e Cˆote d’Azur, Inria, CNRS, LJAD, 2004 route des Lucioles - BP 93, 06902 Sophia Antipolis Cedex, France. E-mail: {paola.goatin, elena.rossi}@inria.fr

issue by extending the solution outside the spatial domain, setting it constantly equal to the corresponding boundary condition value.

Here, we propose a different approach. We follow the treatment of the boundary conditions proposed in [9], where a particular multi-dimensional system of conservation laws in bounded domains with zero boundary conditions is considered. More precisely, a non-local operator aware of the presence of boundaries is introduced. In the present one-dimensional setting, this reads J ρ(t)
(x) = 1 W (x) Z b a ρ(t, y) ω(y − x) dy , with W (x) = Z b a ω(y − x) dy , (1.3) for a suitable convolution kernel ω.

In recent years, the literature on non-local conservation laws has widely increased. These equations are indeed used to model various physical phenomena: from sedimentation mod-els [4] to granular flow [1], from vehicular traffic [5] to crowd dynamics [6, 7, 8], from conveyor belts [11] to supply chains [2].

Although physically those models might be defined in a bounded domain and numerical in-tegrations require it as well, they have been mostly studied in the whole space R or Rn_{. The}

main difficulty lies indeed in the fact that the non-local operator may need to evaluate the unknown outside the boundaries of the spatial domain, where it is not defined.

The analysis of the non-local problem (1.1) is carried out exploiting the same strategy used in both [10] and [14]. As already mentioned, [10] studies the non local IBVP (1.1) where the non-local operator is the standard convolution product, while [14] considers the local problem for a balance law, i.e. a one dimensional IBVP where the flux function has the form f (t, x, ρ) and there is also a source term. We remark that it could be possible to use the results of [14] to study the non-local problem (1.1): indeed, the link between the two problems is obtained by defining the local flux by ˜f (t, x, ρ) = f (t, x, ρ, J ρ), where J ρ = J ρ(t)
(x). However, in this way the a priori estimates on the solution would be less precise than those presented in this work. Namely, a positivity result and an L1

-bound on the solution are missing in [14]. Moreover, L∞_{-estimate recovered here depends on the first derivatives of the flux function,}

see Theorem 2.3, while using the results of [14] yields an estimate depending on the mixed second derivatives of f .

Nevertheless, the result concerning the stability with respect to the flux function proved in [14], recalled below in Theorem A.4, is of crucial importance in this work, since it con-tributes significantly in the proof of the Lipschitz continuous dependence of solutions to (1.1) on initial and boundary data, see Proposition 4.1, and thus in the proof of the uniqueness of solution to (1.1). At this regard, we remark that the stability proof provided in [10] is wrong, but could be fixed following the same strategy proposed here.

The paper is organised as follows. Section 2 presents the assumptions needed on prob-lem (1.1) and the main result of this paper, whose proof is postponed to Section 5. Section 3 is devoted to the introduction of the finite volume approximation of problem (1.1) and its analysis. The Lipschitz continuous dependence of solutions to (1.1) on initial and boundary data is proved in Section 4. The final appendix A recalls some results from [14] on the local IBVP, necessary throughout the paper.

2

Main results

We introduce the following notation: sgn+(s) = ( 1 if s > 0, 0 if s ≤ 0, sgn −_{(s) =} ( 0 if s ≥ 0, −1 if s < 0, s+ = max{s, 0}, s− _{= max{−s, 0}.}

In the rest of the paper, we will denote I(r, s) = [min {r, s} , max {r, s}], for any r, s ∈ R. We make the following assumptions on the flux function f and on the convolution kernel ω:

(f ) f ∈ C2

(R+_{× [a, b] × R × R; R) and there exist L, C > 0 such that}

f (t, x, 0, R) = 0 for t ∈ R+, x ∈ [a, b], R ∈ R, sup t,x,ρ,R  ∂ρf (t, x, ρ, R)  < L, sup t,x,R  ∂_xf (t, x, ρ, R)  < C|ρ|, sup t,x,R  ∂_Rf (t, x, ρ, R)  < C|ρ|, sup t,x,R   ∂ 2 xxf (t, x, ρ, R)    < C |ρ|, sup t,x,R   ∂ 2 xRf (t, x, ρ, R)    < C |ρ|, sup t,x,R   ∂ 2 RRf (t, x, ρ, R)    < C |ρ|. (ω) ω ∈ (C2 ∩ W2,1_{∩ W}2,∞_{)(R; R) is such that} Z R ω(y) dy = 1 and there exists Kω > 0 such that for all x ∈ [a, b]

W (x) = Z b

a

ω(y − x) dy ≥ Kω. (2.1)

The requirement (2.1) guarantees that J in (1.3) is well defined for all x ∈ ]a, b[.

We recall below two different definitions of solution to problem (1.1). Recall that the two definitions are equivalent for functions in (L∞∩ BV)(R+×]a, b[; R). We refer to [13] for further details on the link between this two definitions.

The first definition follows from [3].

Definition 2.1. A function ρ ∈ (L∞∩ BV)(R+×]a, b[; R) is an entropy weak solution to problem (1.1) if, for all ϕ ∈ C1

c(R2; R+) and k ∈ R, Z +∞ 0 Z b a  |ρ − k|∂tϕ(t, x) + sgn (ρ − k) h f t, x, ρ, R(t, x)
− f t, x, k, R(t, x)
i∂xϕ(t, x) − sgn (ρ − k)∂xf t, x, k, R(t, x)
+ ∂Rf t, x, k, R(t, x)
∂xR(t, x)  ϕ(t, x)  dx dt + Z b a  ρ_o(x) − k  ϕ(0, x) dx (2.2) + Z +∞ 0 sgn ρa(t) − k
 ft, a, ρ(t, a+), R(t, a)− f t, a, k, R(t, a)
 ϕ(t, a) dt

− Z +∞ 0 sgn ρb(t) − k
 ft, b, ρ(t, b−), R(t, b)− f t, b, k, R(t, b)
 ϕ(t, b) dt ≥ 0, where, for x ∈ [a, b],

R(t, x) = J ρ(t)
(x) = 1 W (x) Z b a ρ(t, y) ω(y − x) dy , (2.3) and W is as in (1.3).

The second definition was introduced in [12, 15].

Definition 2.2. A function ρ ∈ L∞_(R+_{×]a, b[; R) is an entropy weak solution to}

prob-lem (1.1) if, for all ϕ ∈ C1

c(R2; R+) and k ∈ R, Z +∞ 0 Z b a  (ρ − k)±∂tϕ(t, x) + sgn±(ρ − k) h f t, x, ρ, R(t, x)
− f t, x, k, R(t, x)
i∂xϕ(t, x) − sgn±(ρ − k)∂xf t, x, k, R(t, x)
+ ∂Rf t, x, k, R(t, x)
∂xR(t, x)  ϕ(t, x)  dx dt + Z b a ρo(x) − k
± ϕ(0, x) dx (2.4) + ∂ρf _L∞ Z +∞ 0 ρa(t) − k
± ϕ(t, a) dt + Z +∞ 0 ρb(t) − k
± ϕ(t, b) dt ! ≥ 0, where R is as in (2.3) and ∂ρf _L∞ = sup (t,x)∈R+_×[a,b]   ∂ρf t, x, ρ(t, x), R(t, x)
 .

We can now state our main result.

Theorem 2.3. Let (f ) and (ω) hold. Let ρo ∈ BV( ]a, b[; R+) and ρa, ρb ∈ BV(R+; R+).

Then, for all T > 0, problem (1.1) has a unique entropy weak solution ρ ∈ (L1

∩ L∞_∩

BV)([0, T ]×]a, b[; R+_{). Moreover, the following estimates hold: for any t ∈ [0, T ],}

ρ(t) L1_([a,b])≤ R1(t), ρ(t) _L∞_(]a,b[)≤ R∞(t), TV (ρ(t)) ≤ et T1(t) _{TV (ρ} o) + TV (ρa; [0, t]) + TV (ρb; [0, t])
+ T2(t) T1(t) (et T1(t)_{− 1),} and, for τ > 0, ρ(t) − ρ(t − τ ) L1_([a,b])≤ τ  Ct(t) + 3 L TV (ρa; [t − τ, t]) + TV (ρb; [t − τ, t])
 , where R1(t) = kρokL1_([a,b])+ ∂ρf _L∞  kρakL1_([0,t])+ kρbkL1_([0,t])  , (2.5) R∞(t) = et C(1+L R1(t)) max n kρokL∞_(]a,b[), kρak_L∞_([0,t]), kρbk_L∞_([0,t]) o , (2.6) T1(t) = ∂ 2 ρxf _L∞_{([0,t]×[a,b]×R}2₎+ L R1(t) ∂ 2 ρRf _L∞_{([0,t]×[a,b]×R}2₎, (2.7) T2(t) = K2(t) + 3 2C(1 + L R1(t)) R∞(t) +  K3(t) + C 2 (1 + L R1(t))  kρakL∞ ([0,t]), (2.8)

with L as in (3.12), K2(t), K3(t) as in (3.27), with C1(t) substituted by R1(t), and Ct(t) is as

3

Existence of weak entropy solutions

Fix T > 0. Fix a space step ∆x such that b − a = N ∆x, with N ∈ N, and a time step ∆t subject to a CFL condition, specified later. Introduce the following notation

yk:= (k − 1/2)∆x, yk+1/2:= k∆x for k ∈ Z,

x_j+1/2:= a + j∆x = a + y_j+1/2, for j = 0, . . . , N, xj := a + (j − 1/2)∆x = a + yj, for j = 1, . . . , N,

where x_j+1/2, j = 0, . . . , N , are the cells interfaces and xj, j = 1, . . . , N , the cells centres.

Moreover, set NT = ⌊T /∆t⌋ and, for n = 0, . . . , NT let tn = n∆t be the time mesh. Set

λ = ∆t/∆x.

Approximate the initial datum ρo and the boundary data as follows:

ρ0_j := 1 ∆x Z x_j+1/2 xj−1/2 ρo(x) dx , j = 1, . . . , N, ρn_a := 1 ∆t Z tn+1 tn ρa(t) dt , ρnb := 1 ∆t Z tn+1 tn ρb(t) dt , n = 0, . . . , NT − 1.

Introduce moreover the notation ρn₀ = ρn_a and ρn_{N +1}= ρn_b. For n = 0, . . . , NT − 1, set

ωk := ω(yk) = ω (k − 1/2)∆x
for k ∈ Z, Wj+1/2:= ∆x N X k=1 ωk−j for j = 0, . . . , N, (3.1) Rn_j+1/2:= ∆x Wj+1/2 N X k=1 ωk−jρn_k for j = 0, . . . , N.

Introduce the following modified Lax-Friedrichs flux adapted to the present setting: for n = 0, . . . , NT − 1 and j = 0, . . . , N , F_j+1/2n (ρn_j, ρn_j+1) = 1 2  f (tn, xj+1/2, ρnj, Rnj+1/2) + f (tn, xj+1/2, ρnj+1, Rj+1/2n ) − α  ρn_j+1− ρn_j  , (3.2) where α ≥ 1 is the viscosity coefficient.

We define a piecewise constant approximate solution ρ∆ to (1.1) as

ρ∆(t, x) = ρnj for ( t ∈ [tn, tn+1[ , x ∈ [x_j−1/2, x_j+1/2[ , where n = 0, . . . , NT − 1, j = 1, . . . , N, (3.3) through the finite volume scheme

ρn+1_j = ρn_j − λF_j+1/2n (ρn_j, ρn_j+1) − F_j−1/2n (ρn_j−1, ρn_j). (3.4) Remark 3.1. Concerning the first formula in (3.1), observe that a different (more accurate) choice for the approximation of the kernel function ω is possible: indeed, one may define

ωk= 1 ∆x

Z k∆x

(k−1)∆x

ω(y) dy , which ensures that W_j+1/2= ∆xPN

k=1ωk−j = W (xj+1/2). This choice wouldn’t result in any

3.1 Positivity

In the case of positive initial and boundary data, we prove that under a suitable CFL condition the scheme (3.4) preserves the positivity.

Lemma 3.2. Let ρo ∈ L∞( ]a, b[; R+) and ρa, ρb ∈ L∞(R+; R+). Let (f ) and (ω) hold.

Assume that α ≥ L, λ ≤ 1 3 min  1 α, 1 2 L + C ∆x  . (3.5)

Then, for all t > 0 and x ∈ ]a, b[, the piecewise constant approximate solution ρ∆(3.3) is such

that ρ∆(t, x) ≥ 0.

Proof. We closely follow [10, Lemma 1]. Fix j between 1 and N , n between 0 and NT − 1.

Suppose that ρn_j ≥ 0 for all j = 1, . . . , N . Rewrite (3.4) as follows:

ρn+1_j = ρn_j − λhF_j+1/2n (ρn_j, ρn_j+1) ± F_j+1/2n (ρn_j, ρn_j) ± F_j−1/2n (ρn_j, ρn_j) − F_j−1/2n (ρn_j−1, ρn_j)i = (1 − βn_j − γ_jn) ρn_j + β_jnρn_j−1+ γ_jnρn_j+1− λF_j+1/2n (ρn_j, ρn_j) − F_j−1/2n (ρn_j, ρn_j), with β_jn=        λF n j−1/2(ρnj, ρnj) − Fj−1/2n (ρnj−1, ρnj) ρn_j − ρn_j−1 if ρ n j 6= ρnj−1, 0 if ρn_j = ρn_j−1, (3.6) γ_jn=        −λF n j+1/2(ρnj, ρnj+1) − Fj+1/2n (ρnj, ρnj) ρn j+1− ρnj if ρn_j 6= ρn_j+1, 0 if ρn j = ρnj+1. (3.7)

Using the explicit expression of the numerical flux (3.2) and (f ), we obtain   F n j+1/2(ρnj, ρnj) − Fj−1/2n (ρnj, ρnj)    = f (t n_{, x} j+1/2, ρnj, Rnj+1/2) − f (tn, xj−1/2, ρnj, Rnj−1/2) ± f (tn, xj−1/2, ρnj, Rnj+1/2)    ≤ f (t n_{, x} j+1/2, ρnj, Rnj+1/2) − f (tn, xj−1/2, ρnj, Rnj+1/2)    + f (t n_{, x} j−1/2, ρnj, Rnj+1/2) − f (tn, xj−1/2, ρnj, Rnj−1/2)    ≤ C ρn_j ∆x +   f (t n_{, x} j−1/2, ρnj, Rj+1/2n ) − f (tn, xj−1/2, 0, Rnj+1/2)    + f (t n_{, x} j−1/2, ρnj, Rnj−1/2) − f (tn, xj−1/2, 0, Rnj−1/2)    ≤ C ρn_j ∆x + 2 L ρn_j.

Observe that, whenever ρn_j 6= ρn_j−1,

β_jn= λ 2 (ρn_j − ρn_j−1)

h

= λ 2  ∂ρf (tn, xj−1/2, rj−1/2n , Rj−1/2n ) + α  ,

with r_j−1/2n ∈ Iρn_j−1, ρn_j. Similarly, whenever ρn_j 6= ρn_j+1,

γ_jn= − λ 2 (ρn j+1− ρnj) h f (tn, xj+1/2, ρnj+1, Rnj+1/2) − f (tn, xj+1/2, ρnj, Rnj+1/2) − α (ρnj+1− ρnj) i = λ 2  α − ∂ρf (tn, xj+1/2, rj+1/2n , Rnj+1/2)  , with rn j+1/2∈ I  ρn j, ρnj+1 

. By the conditions (3.5), we get

β_jn, γ_jn∈  0,1 3  , (1 − β_jn− γ_jn) ∈ 1 3, 1  , λ   F n j+1/2(ρnj, ρnj) − Fj−1/2n (ρnj, ρnj)    ≤ 1 3ρ n j,

which, using the inductive hypothesis, leads to

ρn+1_j ≥ (1 − β_jn− γ_jn)ρn_j + βn_j ρn_j−1+ γ_jnρn_j+1−1 3ρ n j ≥ 1 3ρ n j − 1 3ρ n j ≥ 0.  3.2 L1 bound

Lemma 3.3. Let ρo ∈ L∞( ]a, b[; R+) and ρa, ρb ∈ L∞(R+; R+). Let (f ), (ω) and (3.5)

hold. Then, for all t > 0, ρ∆ in (3.3) satisfies

ρ_∆(t, ·) _L1 (]a,b[)≤ C1(t), (3.8) where C1(t) = kρokL1_(]a,b[)+ α  kρakL1_([0,t])+ kρbkL1_([0,t])  . (3.9)

Proof. By Lemma 3.3, we know that the scheme (3.4) preserves the positivity. Therefore, for n = 0, . . . , NT − 1, compute kρn+1kL1_(]a,b[)= ∆x N X j=1 ρn+1_j = ∆x N X j=1  ρn_j − λF_j+1/2n (ρn_j, ρn_j+1) − F_j−1/2n (ρn_j−1, ρn_j)  = ∆x N X j=1 ρn_j − λ ∆xF_{N +1/2}n (ρn_N, ρn_{N +1}) − F_1/2n (ρn₀, ρn₁) = kρnkL1_([a.b]) −∆t 2 h f (tn, x_{N +1/2}, ρn_N, Rn_{N +1/2})+f (tn, x_{N +1/2}, ρn_b, Rn_{N +1/2}) − α ρn_b − ρn_N
i +∆t 2  f (tn, x1/2, ρna, Rn1/2) + f (tn, x1/2, ρn1, Rn1/2) − α ρn1 − ρna


= kρnkL1_([a.b])+ ∆t 2  −α − ∂ρf (tn, xN +1/2, rN,0n , RN +1/2n )  ρn_N +∆t 2  α − ∂ρf (tn, xN +1/2, rb,0n , RnN +1/2)  ρn_b +∆t 2  α + ∂ρf (tn, x1/2, ra,0n , Rn1/2)  ρn_a +∆t 2  −α + ∂ρf (tn, x1/2, rn1,0, R1/2n )  ρn₁, where rn_N,0∈ I 0, ρn_N
, rn

b,0 ∈ I 0, ρnb
, rna,0∈ I 0, ρna
and rn1,0 ∈ I 0, ρn1
. By (f ) and the

assumption (3.5) on α, the coefficients of ρn

N and ρn1 are negative. Thus

kρn+1kL1_(]a,b[)≤ kρnkL1_(]a.b[)+ α ∆t ρn_a+ ρn_b
.

An iterative argument yields the thesis. 

3.3 L∞ bound

Lemma 3.4. Let ρo ∈ L∞( ]a, b[; R+) and ρa, ρb ∈ L∞(R+; R+). Let (f ), (ω) and (3.5)

hold. Then, for all t > 0, ρ∆ in (3.3) satisfies

ρ_∆(t, ·) L∞ (]a,b[)≤ max n kρokL∞_(]a,b[), kρak_L∞_([0,t]), kρbk_L∞_([0,t]) o eC2(t)t_{, (3.10)} where C2(t) is given by (3.13).

Proof. Fix n between 0 and NT − 1. For j = 1, . . . N , rearrange (3.4) as in Lemma 3.2, with

the notation (3.6)–(3.7): ρn+1_j = (1 − β_jn− γ_jn) ρ_jn+ β_jnρn_j−1+ γ_jnρ_j+1n − λF_j+1/2n (ρn_j, ρn_j) − F_j−1/2n (ρn_j, ρn_j). (3.11) Compute   F n j+1/2(ρnj, ρnj) − Fj−1/2n (ρnj, ρnj)    = f (t n_{, x} j+1/2, ρnj, Rnj+1/2) − f (tn, xj−1/2, ρnj, Rj−1/2n ) ± f (tn, xj−1/2, ρnj, Rnj+1/2)    ≤   ∂xf (t n_{, ˜x} j, ρnj, Rnj+1/2)   |xj+1/2− xj−1/2| +   ∂Rf (t n_{, x} j−1/2, ρnj, ˜Rnj)      R n j+1/2− Rnj−1/2   . By (3.1), we have   R n j+1/2− Rnj−1/2    =        ∆x W_j+1/2   N X k=1 ωk−jρn_k  − ∆x W_j−1/2   N X k=1 ωk−j+1ρn_k          ≤ ∆x |Wj+1/2| N X k=1 |ωk−j− ωk−j+1| ρn_k + ∆x   N X k=1 |ωk−j+1| ρn_k        1 Wj+1/2 − 1 Wj−1/2     

≤ ∆x Kω N X k=1 ρn_k      Z y_k−j+1 yk−j ω′(y) dy      +∆x 2 K2 ω kωkL∞_(R)   N X k=1 ρn_k         N X k=1 (ωk−j+1− ωk−j)       ≤ ∆x Kω ω′ _L∞_(R)kρ n_k L1_(]a,b[)+ ∆x K2 ω kωkL∞ (R)kρnkL1_(]a,b[) N X k=1      Z y_k−j+1 yk−j ω′(y) dy      ≤ ∆x Kω ω′ _L∞_(R)kρ n_k L1_(]a,b[)+ ∆x K2 ω kωkL∞_(R) ω′ _L1 (R)kρ n_k L1_(]a,b[) ≤ ∆x   ω′ _L∞ (R) Kω +kωkL ∞_(R) ω′ _L1 (R) K2 ω  C₁(tn),

where C1(t) is defined in (3.9). Setting

L =   ω′ _L∞_(R) Kω +kωkL ∞ (R) ω′ _L1 (R) K2 ω  , (3.12) we obtain   F n j+1/2(ρnj, ρnj) − Fj−1/2n (ρnj, ρnj)    ≤ C ∆x  ρn_n  1 + L C₁(tn)
. Inserting the above estimate into (3.11) and exploiting the bounds on βn

j and γjnobtained in

the proof of Lemma 3.2, we get

ρn+1_j ≤ (1 − βn_j − γn_j) ρn_j + β_jnρn_j−1+ γ_jnρn_j+1+ λ   F n j+1/2(ρnj, ρnj) − Fj−1/2n (ρnj, ρnj)    ≤ (1 − βn_j − γn_j)kρnkL∞_(]a,b[)+ β n j max n kρnkL∞_(]a,b[), ρ n a o + γ_jn maxnkρnkL∞_(]a,b[), ρ n b o + λ ∆x C(1 + L C1(tn))kρnkL∞_(]a,b[) ≤ maxnkρnkL∞ (]a,b[), ρna, ρnb o  1 + ∆t C 1 + L C1(tn)
 ≤ eC2(tn)∆t_max n kρnkL∞_(]a,b[), ρ n a, ρnb o , where C2(t) = C(1 + L C1(t)), (3.13)

L being as in (3.12). An iterative argument, together with the fact that C2(tn−1) ≤ C2(tn) for

all n = 1, . . . , NT, yields the thesis. 

3.4 BV estimates

Proposition 3.5. (BV estimate in space) Let ρo∈ BV( ]a, b[; R+), ρa, ρb ∈ BV(R+; R+).

Let (f ), (ω) and (3.5) hold. Then, for all n = 1, . . . , NT. the following estimate holds N X j=0   ρ n j+1− ρnj    ≤ Cx(t n_{), (3.14)}

where Cx(tn) = eK1(t n_{) t}n   N X j=0   ρ 0 j+1− ρ0j    + n X m=1   ρ m a − ρm−1a    + n X m=1   ρ m b − ρm−1b      + K4(t n₎ K1(tn)  eK1(tn) tn_{− 1}  , (3.15)

with K1(tn) and K4(tn) are defined in (3.27) and (3.30).

Remark 3.6. Estimate (3.14) is defined also for n = 0, setting P0

m=1am = 0, with some

abuse of notation.

Proof. Consider the inner terms and the boundary ones separately.

For j = 1, . . . , N − 1 and n = 0, . . . , NT − 1, focus on the difference ρn+1_j+1 − ρn+1_j ,

exploit-ing (3.4): ρn+1_j+1 − ρn+1_j = ρn_j+1− ρn_j − λhF_j+3/2n (ρn_j+1, ρ_j+2n ) − F_j+1/2n (ρn_j, ρn_j+1) − F_j+1/2n (ρn_j, ρn_j+1) + F_j−1/2n (ρn_j−1, ρn_j)i ± λ F_j+3/2n (ρn_j, ρn_j+1) ± λF_j+1/2n (ρn_j−1, ρn_j) = ρn_j+1− ρn_j − λhF_j+3/2n (ρn_j+1, ρ_j+2n ) − F_j+1/2n (ρn_j, ρn_j+1) + F_j+1/2n (ρn_j−1, ρn_j) − F_j+3/2n (ρn_j, ρn_j+1)i − λhF_j+3/2n (ρn_j, ρ_j+1n ) − F_j+1/2n (ρn_j−1, ρn_j) + F_j−1/2n (ρn_j−1, ρn_j) − F_j+1/2n (ρn_j, ρn_j+1)i = An_j − λ Bn_j, where we set An_j = ρn_j+1− ρn_j − λhF_j+3/2n (ρn_j+1, ρ_j+2n ) − F_j+1/2n (ρn_j, ρn_j+1) + F_j+1/2n (ρn_j−1, ρn_j) − F_j+3/2n (ρn_j, ρn_j+1)i, Bn_j = F_j+3/2n (ρn_j, ρn_j+1) − F_j+1/2n (ρn_j−1, ρ_jn) + F_j−1/2n (ρn_j−1, ρn_j) − F_j+1/2n (ρn_j, ρn_j+1). Rearrange An_j as follows: An_j = ρn_j+1− ρn_j − λF n j+3/2(ρnj+1, ρnj+2) − Fj+3/2n (ρnj+1, ρnj+1) ρn j+2− ρnj+1  ρn_j+2− ρn_j+1 − λF n j+3/2(ρnj+1, ρnj+1) − Fj+3/2n (ρnj, ρnj+1) ρn_j+1− ρn_j  ρn_j+1− ρn_j + λF n j+1/2(ρnj, ρnj+1) − Fj+1/2n (ρnj, ρnj) ρn j+1− ρnj  ρn_j+1− ρn_j + λF n j+1/2(ρnj, ρnj) − Fj+1/2n (ρnj−1, ρnj) ρn j − ρnj−1  ρn_j − ρn_j−1 = δ_jn(ρn_j − ρn_j−1) + γn_j+1(ρn_j+2− ρn_j+1) + (1 − γ_jn− δn_j+1)(ρn_j+1− ρn_j),

where δn_j =        λF n j+1/2(ρnj, ρnj) − Fj+1/2n (ρnj−1, ρnj) ρn j − ρnj−1 if ρn_j 6= ρn_j−1, 0 if ρn_j = ρn_j−1, (3.16) while γn

j is as in (3.7). It can be proven that δnj ∈0, 1/3. Thus, N −1 X j=1   A n j    ≤ N −1 X j=1   ρ n j+1− ρnj    + N −2 X j=0 δ_j+1n   ρ n j+1− ρnj    − N −1 X j=1 δ_j+1n   ρ n j+1− ρnj    + N X j=2 γ_jn ρ n j+1− ρnj    − N −1 X j=1 γ_jn ρ n j+1− ρnj    = N −1 X j=1   ρ n j+1− ρnj    + δ n 1  ρn₁ − ρn_a − δn_N  ρn_N − ρn_{N −1} + γ_Nn  ρn_b − ρn_N − γ₁n  ρn₂ − ρn₁ . (3.17) Focus now on Bn j: B_jn= 1 2 h f (tn, x_j+3/2, ρn_j, R_j+3/2n ) + f (tn, x_j+3/2, ρn_j+1, Rn_j+3/2) − f (tn, x_j+1/2, ρn_j, Rn_j+1/2) − f (tn, x_j+1/2, ρn_j+1, Rn_j+1/2) + f (tn, x_j−1/2, ρn_j−1, Rn_j−1/2) + f (tn, x_j−1/2, ρn_j, Rn_j−1/2) −f (tn, x_j+1/2, ρn_j−1, Rn_j+1/2) − f (tn, x_j+1/2, ρn_j, Rn_j+1/2)i = 1 2 h f (tn, x_j+3/2, ρn_j+1, Rn_j+3/2) − f (tn, x_j+1/2, ρn_j+1, Rn_j+1/2)i + 1 2 h f (tn, x_j−1/2, ρn_j−1, Rn_j−1/2) − f (tn, x_j+1/2, ρn_j−1, Rn_j+1/2)i + 1 2 h f (tn, xj+3/2, ρnj, Rj+3/2n ) − 2 f (tn, xj+1/2, ρnj, Rnj+1/2) + f (tn, xj−1/2, ρnj, Rnj−1/2) i = 1 2 h ∂Rf (tn, xj+3/2, ρnj+1, ˜Rnj+1) (Rnj+3/2− Rnj+1/2) + ∆x ∂xf (tn, ˜xj+1, ρnj+1, Rnj+1/2) i − 1 2 h ∂Rf (tn, xj−1/2, ρnj−1, ˜Rnj) (Rnj+1/2− Rj−1/2n ) + ∆x ∂xf (tn, ˜xj, ρnj−1, Rj+1/2n ) i + 1 2 h f (tn, x_j+3/2, ρn_j, R_j+1/2n ) − 2 f (tn, x_j+1/2, ρ_jn, Rn_j+1/2) + f (tn, x_j−1/2, ρn_j, Rn_j+1/2)i + 1 2 h f (tn, x_j+3/2, ρn_j, R_j+3/2n ) − f (tn, x_j+3/2, ρn_j, Rn_j+1/2)i + 1 2 h f (tn, x_j−1/2, ρn_j, R_j−1/2n ) − f (tn, x_j−1/2, ρn_j, Rn_j+1/2)i = ∆x 2 h (˜xj+1− ˜xj) ∂xx2 f (tn, ˆxj+1/2, ρnj+1, Rnj+1/2) + ∂ρx2 f (tn, ˜xj, ˜ρnj, Rj+1/2n ) (ρnj+1− ρnj−1) i + 1 2 h 2 ∆x ∂_xR2 f (tn, ˇx_j+1/2, ρn_j+1, ˜Rn_j+1) (Rn_j+3/2− Rn_j+1/2) + ∂2_ρRf (tn, x_j−1/2, ¯ρn_j, ˜R_j+1n ) (Rn_j+3/2− Rn_j+1/2) (ρn_j+1− ρn_j−1)

+ ∂2_RRf (tn, xj−1/2, ρnj−1, ˆRj+1/2n ) (Rnj+3/2− Rnj+1/2) ( ˜Rnj+1− ˜Rjn) +∂Rf (tn, xj−1/2, ρnj−1, ˜Rnj) (Rj+3/2n − Rj+1/2n − Rj+1/2n + Rj−1/2n ) i + ∆x 2 h ∂xf (tn, ¯xj+1, ρnj, Rnj+1/2) − ∂xf (tn, ¯xj, ρnj, Rnj+1/2) i + 1 2 h ∂Rf (tn, xj+3/2, ρnj, ¯Rnj+1) (Rnj+3/2− Rj+1/2n ) −∂Rf (tn, xj−1/2, ρnj, ¯Rnj) (Rnj+1/2− Rnj−1/2) i = ∆x 2 h (˜xj+1− ˜xj) ∂xx2 f (tn, ˆxj+1/2, ρnj+1, Rnj+1/2) + ∂ρx2 f (tn, ˜xj, ˜ρnj, Rj+1/2n ) (ρnj+1− ρnj−1) i + 1 2 h 2 ∆x ∂_xR2 f (tn, ˇxj+1/2, ρnj+1, ˜Rnj+1) (Rnj+3/2− Rnj+1/2) + ∂2_ρRf (tn, x_j−1/2, ¯ρn_j, ˜R_j+1n ) (Rn_j+3/2− Rn_j+1/2) (ρn_j+1− ρn_j−1) + ∂2_RRf (tn, x_j−1/2, ρn_j−1, ˆR_j+1/2n ) (Rn_j+3/2− Rn_j+1/2) ( ˜Rn_j+1− ˜R_jn) +∂Rf (tn, xj−1/2, ρnj−1, ˜Rnj) (Rj+3/2n − Rj+1/2n − Rj+1/2n + Rj−1/2n ) i + ∆x 2 (¯xj+1− ¯xj) ∂ 2 xxf (tn, ¯xj+1/2, ρnj, Rnj+1/2) + 1 2 h 2 ∆x ∂_xR2 f (tn, xj+1/2, ρnj, ¯Rnj+1) (Rnj+3/2− Rnj+1/2) + ∂2_RRf (tn, x_j−1/2, ρn_j, ¯R¯n_j+1/2) (R_j+3/2n − R_j+1/2n ) ( ¯Rn_j+1− ¯Rn_j) +∂Rf (tn, xj−1/2, ρnj, ¯Rnj) (Rnj+3/2− Rnj+1/2− Rnj+1/2+ Rnj−1/2) i , where ˜ Rn_j+1, ¯Rn_j+1∈IRn_j+1/2, R_j+3/2n , x˜j+1, ¯xj+1∈ ]xj+1/2, xj+3/2[, ˜ Rn_j, ¯Rn_j ∈IRn_j−1/2, R_j+1/2n , x˜j, ¯xj ∈ ]xj−1/2, xj+1/2[, ˆ Rn_j+1/2∈I ˜Rn_j, ˜Rn_j+1, xˆ_j+1/2∈ ]˜xj, ˜xj+1[, ¯ ¯ Rn_j+1/2∈I ¯Rn_j, ¯Rn_j+1, xˇ_j+1/2, x_j+1/2∈ ]x_j−1/2, x_j+3/2[, ˜ ρn_j, ¯ρn_j ∈Iρn_j−1, ρn_j+1, x¯_j+1/2∈ ]¯xj, ¯xj+1[. Notice that  x˜_j+1− ˜x_j  ≤ 2 ∆x,  x¯_j+1− ¯x_j  ≤ 2 ∆x,   R n j+3/2− Rnj+1/2    ≤ L C1(t n_{) ∆x,}

where L is as in (3.12). Moreover, by their very definition, for ϑn_j+1, εn_j ∈ [0, 1],   R˜ n j+1− ˜Rnj    =   ϑ n j+1Rnj+3/2+ (1 − ϑnj+1) Rnj+1/2− εnj Rnj+1/2− (1 − εnj) Rj−1/2n    ≤ R n j+3/2− Rnj+1/2    +   R n j+1/2− Rnj−1/2    ≤ 2 L C1(tn) ∆x,

and similarly R¯ n j+1− ¯Rjn    ≤ 2 L C1(t n_{) ∆x. Compute now} Rn_j+3/2− 2 Rn_j+1/2+ Rn_j−1/2= ∆x   N X k=1 ρn_k ω k−j−1 W_j+3/2 − ωk−j W_j+1/2 − ωk−j W_j+1/2 + ωk−j+1 W_j−1/2 ! . (3.18)

Observe that, for k fixed: ωk−j−1 W_j+3/2 − ωk−j W_j+1/2 = ∆x Wj+3/2Wj+1/2   N X ℓ=1  ωk−j−1ωℓ−j− ωk−jωℓ−j−1   = ∆x W_j+3/2W_j+1/2   N X ℓ=1  ωk−j−1ωℓ−j− ωℓ−j−1+ωk−j−1− ωk−jωℓ−j−1    = (∆x) 2 W_j+3/2W_j+1/2   N X ℓ=1  ωk−j−1ω′(ξ_{ℓ−j−1/2}) − ωℓ−j−1ω′(ξ_k−j−1/2)  , (3.19)

where ξℓ−j−1/2∈ ]yℓ−j−1, yℓ−j[, ξk−j−1/2∈ ]yk−j−1, yk−j[, and similarly

ωk−j Wj+1/2 −ω k−j+1 Wj−1/2 = (∆x) 2 Wj+1/2Wj−1/2   N X ℓ=1  ωk−jω′(ξ_ℓ−j+1/2) − ωℓ−jω′(ξ_k−j+1/2)  . (3.20)

In order to compute the difference between (3.19) and (3.20), which appears in (3.18), we add and subtract the term

(∆x)2 W_j+3/2W_j+1/2   N X ℓ=1  ωk−jω′(ξℓ−j+1/2) − ωℓ−jω′(ξk−j+1/2)   .

The terms with common denominator yield (∆x)2 W_j+3/2W_j+1/2   N X ℓ=1  ωk−j−1ω′(ξ_{ℓ−j−1/2}) − ωℓ−j−1ω′(ξ_k−j−1/2) −ωk−jω′(ξ_ℓ−j+1/2) + ωℓ−jω′(ξ_k−j+1/2)   = (∆x) 2 W_j+3/2W_j+1/2   N X ℓ=1  (ωk−j−1− ωk−j) ω′(ξ_{ℓ−j−1/2}) + ωk−j(ω′(ξ_{ℓ−j−1/2}) − ω′(ξ_ℓ−j+1/2)) −(ωℓ−j−1− ωℓ−j) ω′(ξk−j−1/2) − ωℓ−j(ω′(ξk−j−1/2) − ω′(ξk−j+1/2))   

= (∆x) 2 W_j+3/2W_j+1/2   N X ℓ=1 −∆x ω′(ξk−j−1/2) ω′(ξℓ−j−1/2) − ωk−j Z ξ_ℓ−j+1/2 ξℓ−j−1/2 ω′′(y) dy +∆x ω′(ξℓ−j−1/2) ω′(ξk−j−1/2) + ωℓ−j Z ξ_k−j+1/2 ξk−j−1/2 ω′′(y) dy !  = (∆x) 2 Wj+3/2Wj+1/2   N X ℓ=1 ωℓ−j Z ξ_k−j+1/2 ξk−j−1/2 ω′′(y) dy − ωk−j Z ξ_ℓ−j+1/2 ξℓ−j−1/2 ω′′(y) dy ! . (3.21)

We are left with   N X ℓ=1  ωk−jω′(ξℓ−j+1/2) − ωℓ−jω′(ξk−j+1/2)    (∆x)2 Wj+3/2Wj+1/2 − (∆x) 2 Wj+1/2Wj−1/2 ! . (3.22)

In particular, observe that 1 W_j+3/2W_j+1/2 − 1 W_j+1/2W_j−1/2 = Wj−1/2− Wj+3/2 W_j+3/2W_j+1/2W_j−1/2 = ∆x Wj+3/2Wj+1/2Wj−1/2   N X β=1  ωβ−j+1− ωβ−j−1   = ∆x Wj+3/2Wj+1/2Wj−1/2   N X β=1 Z y_β−j+1 yβ−j−1 ω′(y) dy   ≤ 2 ∆x ω′ _L1 Wj+3/2Wj+1/2Wj−1/2 . (3.23)

Coming back to (3.18), exploiting (3.21), (3.22) and (3.23), we get   R n j+3/2− 2 Rj+1/2n + Rj−1/2n    ≤ (∆x) 3   Wj+3/2Wj+1/2             N X k=1 ρn_k Z ξ_k−j+1/2 ξk−j−1/2 ω′′(y) dy     N X ℓ=1 ωℓ−j   −   N X k=1 ρn_kωk−j     N X ℓ=1 Z ζ_ℓ−j+1/2 ξℓ−j−1/2 ω′′(y) dy          + 2 (∆x) 4 ω′ _L1   Wj+3/2Wj+1/2Wj−1/2             N X k=1 ρn_kωk−j     N X ℓ=1 ω′(ξ_ℓ−j+1/2)   −   N X k=1 ρn_kω′(ξ_k−j+1/2)     N X ℓ=1 ωℓ−j         

≤ 2 (∆x) 2 Kω ω′′ _L∞kρ n_k L1+ (∆x)2 K2 ω kωkL∞ ω′′ _L1kρnk_L1 +2 (∆x) 2 K3 ω kωkL∞ ω′ 2 L1kρ n_k L1+ 2 (∆x)2 K2 ω ω′ _L∞ ω′ _L1kρnk_L1 = ∆x2W kρnkL1, (3.24) where we set W = 2 Kω ω′′ _L∞+ 1 K2 ω kωkL∞ ω′′ _L1+ 2 K3 ω kωkL∞ ω′ 2 L1+ 2 K2 ω ω′ _L∞ ω′ _L1. (3.25) Hence,   B n j    ≤ (∆x) 2_C ρ n j+1    + ∆x 2 ∂ 2 ρxf _L∞   ρ n j+1− ρnj−1    + (∆x) 2_{C L C} 1(tn)   ρ n j+1    +∆x 2 L C1(t n₎ ∂ 2 ρRf L∞   ρ n j+1− ρnj−1    + (∆x) 2_{C L}2_(C 1(tn))2   ρ n j−1    + (∆x) 2_C ρ n j    + (∆x)2C L C1(tn)   ρ n j    + (∆x) 2_{C L}2_(C 1(tn))2   ρ n j    +1 2(∆x) 2_{C W C} 1(tn)   ρ n j−1    +   ρ n j     . Therefore, N −1 X j=1 λ B n j    ≤ ∆t 2  ∂ 2 ρxf _L∞+ L C1(t n₎ ∂ 2 ρRf _L∞ N −1 X j=1   ρ n j+1− ρnj−1    + ∆t C1 + 2 L C1(tn) + 2 L2(C1(tn))2+ W C1(tn)  ∆x N X j=1   ρ n j    + ∆t ∆x C  L2(C1(tn))2+ 1 2W C1(t n₎   ρn_a   ≤ K1(tn) ∆t   N −1 X j=1   ρ n j+1− ρnj    + 1 2  ρn₁ − ρn_a   + K₂(tn) ∆t + K₃(tn) ∆t  ρn_a , (3.26) with K1(t) = ∂ 2 ρxf _L∞ + L C1(t) ∂ 2 ρRf _L∞ , K2(t) = C C1(t) 1 + 2 L C1(t) + 2 K3(t)
, (3.27) K3(t) = C C1(t)  L2C1(t) + 1 2W  ,

C1(t) is as in (3.9) and L is as in (3.12). Observe that the two norms of f appearing in K1(t)

are bounded due to (f ), Lemma 3.3 and Lemma 3.4, since they are evaluated on the compact set [0, t] × [a, b] × [− ρ(t) _L∞, ρ(t) _L∞] × [−J(t), J(t)], with J(t) = kωkL∞ Kω C1(t).

Focus now on the boundary terms. From the definition of the scheme (3.4), with the notation (3.6)–(3.7) and (3.16), we have

ρn+1₁ − ρn+1_a = (1 − β₁n− γ₁n)ρn₁ + βn₁ ρn_a+ γ₁nρn₂ − λhF_3/2n (ρn₁, ρn₁) − F_1/2n (ρ₁n, ρn₁)i− ρn+1_a ± ρn_a = γ₁n(ρn₂ − ρn₁) + (1 − β₁n)(ρn₁ − ρn_a) + (ρn+1_a − ρ_an) − λhF_3/2n (ρn₁, ρ₁n) − F_1/2n (ρn₁, ρn₁)i = γ₁n(ρn₂ − ρn₁) + (1 − δ₁n)(ρn₁ − ρn_a) + (ρn+1_a − ρ_an) − λhF_3/2n (ρn_a, ρn₁) − F_1/2n (ρn_a, ρn₁)i, since β₁n(ρn₁ − ρn_a) + λhF_3/2n (ρn₁, ρn₁) − F_1/2n (ρn₁, ρn₁)i = λhF_1/2n (ρn₁, ρn₁) − F_1/2n (ρn_a, ρn₁) + F_3/2n (ρn₁, ρn₁) − F_1/2n (ρn₁, ρ₁n) ± F_3/2n (ρn_a, ρn₁)i = δ₁n(ρn₁ − ρn_a) + λhF_3/2n (ρ_an, ρn₁) − F_1/2n (ρn_a, ρn₁)i. Observing that λhF_3/2n (ρn_a, ρn₁) − F_1/2n (ρn_a, ρn₁)i = λ 2 h f (tn, x_3/2, ρn_a, Rn_3/2) + f (tn, x_3/2, ρn₁, Rn_3/2) − f (tn, x_1/2, ρn_a, Rn_1/2) − f (tn, x_1/2, ρn₁, Rn_1/2)i = λ 2 h ∂xf (tn, ˜x1, ρna, Rn3/2) ∆x + ∂Rf (tn, x1/2, ρna, ˜Rn1) (Rn3/2− R1/2n ) +∂xf (tn, x1, ρn1, R3/2n ) ∆x + ∂Rf (tn, x1/2, ρn1, ¯Rn1) (Rn3/2− Rn1/2) i , where ˜x1, x1∈ ]x1/2, x3/2[ and ˜R1n, ¯Rn1 ∈ I(R1/2n , Rn3/2), we conclude

λ   F n 3/2(ρna, ρn1) − F1/2n (ρna, ρn1)    ≤ ∆t C 2 (1 + L C1(t n_{)) (|ρ}n a| + |ρn1|).

By the positivity of the coefficients involved, we obtain   ρ n+1 1 − ρn+1a    ≤ γ n 1 |ρn2 − ρn1| + (1 − δn1) |ρn1 − ρna| + |ρn+1a − ρna| + ∆tC 2 (1 + L C1(t n_{)) (|ρ}n a| + |ρn1|). (3.28)

Concerning the other boundary term, we have

ρn+1_b − ρn+1_N = ρn+1_b ± ρn_b − (1 − βn_N− γ_Nn)ρn_N + β_Nn ρn_{N −1}+ γn_Nρn_b + λhF_{N +1/2}n (ρn_N, ρn_N) − F_{N −1/2}n (ρn_N, ρn_N)i = (ρn+1_b − ρn_b) + (1 − γ_Nn)(ρn_b − ρn_N) + βn_N(ρn_N − ρn_{N −1}) + λhF_{N +1/2}n (ρn_N, ρn_N) − F_{N −1/2}n (ρn_N, ρn_N)i = (ρn+1_b − ρn_b) + (1 − γ_Nn)(ρn_b − ρn_N) + δn_N(ρn_N − ρn_{N −1}) + λhF_{N +1/2}n (ρn_{N −1}, ρn_N) − F_{N −1/2}n (ρn_{N −1}, ρn_N)i,

since β_Nn(ρn_N − ρn_{N −1}) + λhF_{N +1/2}n (ρn_N, ρn_N) − F_{N −1/2}n (ρn_N, ρn_N)i = λhF_{N −1/2}n (ρn_N, ρn_N) − F_{N −1/2}n (ρn_{N −1}, ρn_N) + F_{N +1/2}n (ρn_N, ρn_N) −F_{N −1/2}n (ρn_N, ρn_N) ± F_{N +1/2}n (ρn_{N −1}, ρn_N)i = δ_Nn(ρn_N − ρn_{N −1}) + λhF_{N +1/2}n (ρ_{N −1}n , ρn_N) − F_{N −1/2}n (ρn_{N −1}, ρn_N)i. Observing that λhF_{N +1/2}n (ρn_{N −1}, ρn_N) − F_{N −1/2}n (ρn_{N −1}, ρn_N)i = λ 2 h f (tn, x_{N +1/2}, ρn_{N −1}, Rn_{N +1/2}) + f (tn, x_{N +1/2}, ρn_N, Rn_{N +1/2}) −f (tn, x_{N −1/2}, ρn_{N −1}, Rn_{N −1/2}) − f (tn, x_{N −1/2}, ρn_N, R_{N −1/2}n )i = λ 2 h ∂xf (tn, ˜xN, ρnN −1, RnN +1/2) ∆x + ∂Rf (tn, xN −1/2, ρnN −1, ˜RnN) (RnN +1/2− RnN −1/2) +∂xf (tn, xN, ρnN, RnN +1/2) ∆x + ∂Rf (tn, xN −1/2, ρnN, ¯RnN) (RnN +1/2− RN −1/2n ) i , where ˜xN, xN ∈ ]xN −1/2, xN +1/2[ and ˜RNn, ¯RnN ∈ I(RnN −1/2, RN +1/2n ), we conclude

λ   F n N +1/2(ρnN −1, ρnN) − FN −1/2n (ρnN −1, ρnN)    ≤ ∆t C 2 (1 + L C1(t n_{)) (|ρ}n N −1| + |ρnN|).

By the positivity of the coefficients involved, we obtain   ρ n+1 b − ρ n+1 N    ≤ |ρ n+1 b − ρnb| + (1 − γNn) |ρnb − ρnN| + δNn |ρnN − ρnN −1| + ∆tC 2 (1 + L C1(t n_{)) (|ρ}n N −1| + |ρnN|). (3.29)

Collect now the estimates (3.17), (3.26), (3.28) and (3.29):

N X j=0   ρ n+1 j+1 − ρn+1j    =   ρ n+1 1 − ρn+1a    + N −1 X j=1   ρ n+1 j+1 − ρn+1j    +   ρ n+1 b − ρn+1n    ≤ γ₁n|ρn₂ − ρn₁| + (1 − δ₁n)|ρn₁ − ρn_a| + |ρn+1_a − ρn_a| + ∆tC 2 (1 + L C1(t n_{)) (|ρ}n a| + |ρn1|) + N −1 X j=1   ρ n j+1− ρnj    + δ n 1  ρn₁ − ρn_a  − δ_Nn  ρn_N − ρn_{N −1}  + γ_Nn  ρn_b − ρn_N  − γn₁  ρn₂ − ρn₁   + K1(tn) ∆t   N −1 X j=1   ρ n j+1− ρnj    + 1 2  ρn₁ − ρn_a    + K₂(tn) ∆t + K₃(tn) ∆t  ρn_a   + |ρn+1_b − ρn_b| + (1 − γ_Nn)|ρn_b − ρn_N| + δ_Nn|ρn_N − ρn_{N −1}| + ∆tC 2 (1 + L C1(t n_{)) (|ρ}n N −1| + |ρnN|)

≤ |ρn+1_a − ρn_a| + |ρn+1_b − ρn_b| + 1 + K1(tn) ∆t
N X j=0   ρ n j+1− ρnj    + K2(t n_{) ∆t} +3 2C(1 + L C1(t n_{)) kρ}n_k L∞ (]a,b[)∆t +  K3(tn) + C 2 (1 + L C1(t n₎₎  kρakL∞ ([0,tn_])∆t.

Exploiting (3.10) and setting K4(tn) = K2(tn) + 3 2C(1 + L C1(t n_{)) e}C2(tn) tn_max n kρokL∞ (]a,b[), kρakL∞ ([0,t]), kρbkL∞ ([0,t]) o +  K3(tn) + C 2 (1 + L C1(t n₎₎  kρakL∞ ([0,tn_]), (3.30)

we deduce from the previous estimate by a standard iterative procedure

N X j=0   ρ n j+1− ρnj    ≤ e K1(tn) tn   N X j=0   ρ 0 j+1− ρ0j    + n X m=1   ρ m a − ρm−1a    + n X m=1   ρ m b − ρm−1b      +K4(t n₎ K1(tn)  eK1(tn) tn_{− 1}  ,

concluding the proof. 

Corollary 3.7. (BV estimate in space and time) Let ρo ∈ BV( ]a, b[; R+) and ρa, ρb ∈

BV(R+_{; R}+_{). Let (f ), (ω) and (3.5) hold. Then for all n = 1, . . . , N}

T, the following estimate

holds n−1 X m=0 N X j=0 ∆t ρ m j+1− ρmj    + n−1 X m=0 N +1 X j=0 ∆x ρ m+1 j − ρmj    ≤ Cxt(t n_{), (3.31)} where Cxt(n ∆t) is given by (3.36).

Proof. By Proposition 3.5, we have

n−1 X m=0 N X j=0 ∆t ρ m j+1− ρmj    ≤ n ∆t Cx(n ∆t). (3.32)

By the definition of the scheme (3.4), for m ∈ {0, . . . , n − 1} and j ∈ {1, . . . , N }, we have   ρ m+1 j − ρmj    ≤ λ α 2   ρ m j+1− ρmj    +   ρ m j − ρmj−1     +λ 2   f (t m_{, x} j+1/2, ρmj , Rmj+1/2) + f (tm, xj+1/2, ρmj+1, Rmj+1/2) −f (tm, x_j−1/2, ρm_j−1, Rm_j−1/2) − f (tm, x_j−1/2, ρm_j , Rm_j−1/2)  ≤ λ α 2   ρ m j+1− ρmj    +   ρ m j − ρmj−1     +λ 2   ∂xf (t m_{, ˜x} j, ρmj , Rmj+1/2)   ∆x + ∂ρf (t m_{, x} j−1/2, ˜ρmj−1/2, Rmj+1/2)      ρ m j − ρmj−1   

+ ∂Rf (t m_{, x} j−1/2, ρmj−1, ˜Rmj )      R m j+1/2− Rj−1/2m    + ∂xf (t m_{, ˜x} j, ρmj+1, Rmj+1/2)   ∆x + ∂ρf (t m_{, x} j−1/2, ˜ρmj+1/2, Rmj+1/2)      ρ m j+1− ρmj    + ∂Rf (t m_{, x} j−1/2, ρmj , ˜Rmj )      R m j+1/2− Rmj−1/2     ≤ λ 2(α + L)   ρ m j+1− ρmj    +   ρ m j − ρmj−1     +λ 2C∆x "   ρ m j+1    +   ρ m j    + L C1(t m₎   ρ m j    +   ρ m j−1    # , where ˜ xj ∈ ]xj−1/2, xj+1/2[, R˜mj ∈ I(Rmj−1/2, Rmj+1/2), ˜ ρ_j−1/2m ∈ I(ρm_j−1, ρm_j ), ρ˜m_j+1/2∈ I(ρm_j , ρm_j+1). Therefore, by Lemma 3.3 and Proposition 3.5

N X j=1 ∆x   ρ m+1 j − ρmj    ≤ ∆t (α + L) N X j=0   ρ m j+1− ρmj    + ∆t C (1 + L C1(m ∆t)) ∆x N X j=1   ρ m j    + ∆t ∆xC 2  ρm_b  + L C₁(m ∆t)  ρm_a   ≤ ∆t (α + L) Cx(m ∆t) + ∆t C (1 + L C1(m ∆t)) C1(m ∆t) + ∆tC 2  kρbkL∞_([0,tm_])+ L C1(m ∆t) kρak_L∞_([0,tm_])  = ∆t Ct(m ∆t), where we set Ct(τ ) = (α + L) Cx(τ ) + C C1(τ ) 1 + L C1(τ )
+ C 2  kρbkL∞_{([0,τ ])}+ L C1(τ ) kρak_L∞_{([0,τ ])}  . (3.33) In particular, N +1 X j=0 ∆x ρ m+1 j − ρmj    = ∆x   ρ m+1 a − ρma    + ∆x   ρ m+1 b − ρmb    + N X j=1 ∆x ρ m+1 j − ρmj    ≤ ∆x ρ m+1 a − ρma    + ∆x   ρ m+1 b − ρmb    + ∆t Ct(m ∆t), (3.34) which, summed over m = 0, . . . , n − 1, yields

n−1 X m=0 N +1 X j=0 ∆x ρ m+1 j − ρmj    ≤ ∆x n−1 X m=0   ρ m+1 a − ρma    +   ρ m+1 b − ρmb     + n ∆t Ct(n ∆t). (3.35)

Summing (3.32) and (3.35) we obtain the desired estimate (3.31), with

+ n ∆tC 2  kρbkL∞ ([0,tn_])+ L C1(n ∆t) kρakL∞ ([0,tn_])  (3.36) + ∆x n−1 X m=0   ρ m+1 a − ρma    +   ρ m+1 b − ρmb     , concluding the proof. Notice that the last sum in (3.36) is bounded by

∆x TV (ρa; [0, T ]) + TV (ρb; [0, T ])
.



3.5 Discrete entropy inequality

We introduce the following notation: for j = 1, . . . , N , n = 0, . . . , NT − 1, k ∈ R,

H_jn(u, v, z) = v − λ F_j+1/2n (v, z) − F_j−1/2n (u, v), Gn,k_j+1/2(u, v) = F_j+1/2n (u ∧ k, v ∧ k) − F_j+1/2n (k, k),

Ln,k_j+1/2(u, v) = F_j+1/2n (k, k) − F_j+1/2n (u ∨ k, v ∨ k),

where F_j+1/2n (u, v) is defined as in (3.2). Observe that, due to the definition of the scheme, ρn+1_j = Hn

j(ρnj−1, ρnj, ρnj+1). Notice moreover that the following equivalences hold true: (s −

k)+_{= s ∧ k − k and (s − k)}−_{= k − s ∨ k.}

Lemma 3.8. Let (f ), (ω) and (3.5) hold. Then the approximate solution ρ∆in (3.3) satisfies

the following discrete entropy inequalities: for j = 1, . . . , N , n = 0, . . . , NT − 1 and k ∈ R,

(ρn+1_j − k)+− (ρn_j − k)++ λGn,k_j+1/2(ρn_j, ρn_j+1) − Gn,k_j−1/2(ρn_j−1, ρn_j) +λ sgn+(ρn+1_j − k)f (tn, xj+1/2, k, Rnj+1/2) − f (tn, xj−1/2, k, Rnj−1/2)  ≤ 0, (3.37) and (ρn+1_j − k)−− (ρn_j − k)−+ λLn,k_j+1/2(ρn_j, ρn_j+1) − Ln,k_j−1/2(ρn_j−1, ρn_j) +λ sgn−(ρn+1_j − k)f (tn, xj+1/2, k, Rnj+1/2) − f (tn, xj−1/2, k, Rnj−1/2)  ≤ 0. (3.38)

Proof. Consider the map (u, v, z) 7→ H_jn(u, v, z). By the CFL condition (3.5), it holds ∂H_jn ∂u (u, v, z) = λ 2  ∂ρf (tn, xj−1/2, u, Rnj−1/2) + α  ≥ 0, ∂H_jn ∂v (u, v, z) = 1 − λ α − λ 2  ∂ρf (tn, xj+1/2, v, Rnj+1/2) − ∂ρf (tn, xj−1/2, v, Rnj−1/2)  ≥ 0 ∂H_jn ∂z (u, v, z) = λ 2  α − ∂ρf (tn, xj+1/2, z, Rj+1/2n )  ≥ 0. Notice that H_jn(k, k, k) = k − λf (tn, x_j+1/2, k, R_j+1/2n ) − f (tn, x_j−1/2, k, R_j−1/2n ).

The monotonicity properties obtained above imply that H_jn(ρn_j−1∧ k, ρn_j ∧ k, ρn_j+1∧ k) − H_jn(k, k, k) ≥ H_jn(ρn_j−1, ρn_j, ρn_j+1) ∧ H_jn(k, k, k) − H_jn(k, k, k) = H_jn(ρn_j−1, ρ_jn, ρn_j+1) − H_jn(k, k, k)+ =  ρn+1_j − k + λf (tn, x_j+1/2, k, Rn_j+1/2) − f (tn, x_j−1/2, k, R_j−1/2n ) + . Moreover, we also have

H_jn(ρn_j−1∧ k, ρn_j ∧ k, ρn_j+1∧ k) − H_jn(k, k, k) = (ρn_j ∧ k) − k − λhF_j+1/2n (ρn_j ∧ k, ρn_j+1∧ k) − F_j−1/2n (ρn_j−1∧ k, ρn_j ∧ k) − F_j+1/2n (k, k) + F_j−1/2n (k, k)i = (ρn_j − k)+− λGn,k_j+1/2(ρn_j, ρn_j+1) − Gn,k_j−1/2(ρn_j−1, ρn_j). Hence, (ρn_j − k)+− λGn,k_j+1/2(ρn_j, ρn_j+1) − Gn,k_j−1/2(ρn_j−1, ρn_j) ≥  ρn+1_j − k + λf (tn, x_j+1/2, k, Rn_j+1/2) − f (tn, x_j−1/2, k, Rn_j−1/2) + = sgn+  ρn+1_j − k + λf (tn, x_j+1/2, k, Rn_j+1/2) − f (tn, x_j−1/2, k, Rn_j−1/2)  ×  ρn+1_j − k + λf (tn, xj+1/2, k, Rnj+1/2) − f (tn, xj−1/2, k, Rnj−1/2)  ≥ ρn+1_j − k++ λ sgn+ρn+1_j − k f (tn, x_j+1/2, k, Rn_j+1/2) − f (tn, x_j−1/2, k, Rn_j−1/2), proving (3.37), while (3.38) is proven in an entirely similar way. 

3.6 Convergence towards an entropy weak solution

The uniform L∞-bound provided by Lemma 3.4 and the total variation estimate of Corol-lary 3.7 allow to apply Helly’s compactness theorem, ensuring the existence of a subsequence of ρ∆, still denoted by ρ∆, converging in L1 to a function ρ ∈ L∞([0, T ]×]a, b[), for all T > 0.

We need to prove that this limit function is indeed an entropy weak solution to (1.1), in the sense of Definition 2.2.

Lemma 3.9. Let ρo ∈ BV( ]a, b[; R+) and ρa, ρb ∈ BV(R+; R+). Let (f ), (ω) and (3.5)

hold. Then the piecewise constant approximate solutions ρ∆in (3.3) resulting from the adapted

Lax–Friedrichs scheme (3.4) converge, as ∆x → 0, towards an entropy weak solution of the initial boundary value problem (1.1).

Proof. We consider the discrete entropy inequality (3.37), for the positive semi-entropy, and we follow [10], see also [15]. Add and subtract Gn,k_j+1/2(ρn_j, ρn_j) in (3.37) and rearrange it as follows

+ λGn,k_j+1/2(ρn_j, ρn_j) − Gn,k_j−1/2(ρn_j−1, ρn_j)

+ λ sgn+(ρn+1_j − k)f (tn, x_j+1/2, k, R_j+1/2n ) − f (tn, x_j−1/2, k, Rn_j−1/2)

Let ϕ ∈ C1

c([0, T [×[a, b]; R+) for some T > 0, multiply the inequality above by ∆x ϕ(tn, xj)

and sum over j = 1, . . . , N and n ∈ N, so to get 0 ≥ ∆x +∞ X n=0 N X j=1 h (ρn+1_j − k)+− (ρn_j − k)+iϕ(tn, xj) (3.39) + ∆t +∞ X n=0 N X j=1   Gn,k_j+1/2(ρn_j, ρn_j+1) − Gn,k_j+1/2(ρn_j, ρn_j) (3.40) −Gn,k_j−1/2(ρn_j−1, ρn_j) − Gn,k_j+1/2(ρn_j, ρn_j)  ϕ(tn, xj) (3.41) + ∆t +∞ X n=0 N X j=1 sgn+(ρn+1_j − k)f (tn, x_j+1/2, k, Rn_j+1/2)−f (tn, x_j−1/2, k, R_j−1/2n )ϕ(tn, xj). (3.42) Summing by parts, we obtain

[(3.39)] = − ∆x N X j=1 (ρ0_j − k)+ϕ(0, xj) − ∆x ∆t +∞ X n=1 N X j=1 (ρn_j − k)+ϕ(t n_{, x} j) − ϕ(tn−1, xj) ∆t −→ ∆x→0+− Z b a (ρo(x) − k)+ϕ(0, x) dx − Z +∞ 0 Z b a (ρ(t, x) − k)+∂tϕ(t, x) dx dt , and [(3.42)] = ∆x ∆t +∞ X n=0 N X j=1 sgn+(ρn+1_j − k)f (t n_{, x} j+1/2, k, Rnj+1/2)−f (tn, xj−1/2, k, Rnj−1/2) ∆x ϕ(t n_{, x} j) −→ ∆x→0+ Z +∞ 0 Z b a sgn+(ρ(t, x) − k) d dxf (t, x, k, R(t, x)) ϕ(t, x) dx dt ,

by the Dominated Convergence Theorem. Concerning (3.40)–(3.41), we get [(3.40) − (3.41)] = ∆t +∞ X n=0 N X j=1  Gn,k_j+1/2(ρn_j, ρn_j+1) − Gn,k_j+1/2(ρn_j, ρn_j) ϕ(tn, xj) − ∆t +∞ X n=0 N −1 X j=0  Gn,k_j+1/2(ρn_j, ρn_j+1) − Gn,k_j+3/2(ρn_j+1, ρn_j+1)ϕ(tn, xj+1) = ∆t +∞ X n=0 N −1 X j=1   Gn,k_j+1/2(ρn_j, ρn_j+1) − Gn,k_j+1/2(ρn_j, ρn_j)ϕ(tn, xj) −Gn,k_j+1/2(ρn_j, ρn_j+1) − G_j+3/2n,k (ρn_j+1, ρn_j+1) ϕ(tn, xj+1) 

+ ∆t +∞ X n=0   Gn,k_{N +1/2}(ρn_N, ρn_b) − Gn,k_{N +1/2}(ρn_N, ρn_N)ϕ(tn, xN) −Gn,k_1/2(ρn_a, ρn₁) − G_3/2n,k(ρn₁, ρn₁)ϕ(tn, x1)  = Tint+ Tb= T, where we set Tint= ∆t +∞ X n=0 N −1 X j=1   Gn,k_j+1/2(ρn_j, ρn_j+1) − Gn,k_j+1/2(ρn_j, ρn_j) ϕ(tn, xj) −Gn,k_j+1/2(ρn_j, ρn_j+1) − G_j+3/2n,k (ρn_j+1, ρn_j+1)ϕ(tn, xj+1)  , Tb = ∆t +∞ X n=0   Gn,k_{N +1/2}(ρn_N, ρn_b) − Gn,k_{N +1/2}(ρn_N, ρn_N)ϕ(tn, xN) −Gn,k_1/2(ρn_a, ρn₁) − G_3/2n,k(ρn₁, ρn₁) ϕ(tn, x1)  . Define S = − ∆x ∆t +∞ X n=0 N X j=1 Gn,k_j+1/2(ρn_j, ρn_j)ϕ(t n_{, x} j+1) − ϕ(tn, xj) ∆x − α ∆t +∞ X n=0  (ρn_a− k)+ϕ(tn, a) + (ρn_b − k)+ϕ(tn, b). (3.43) Observe that Gn,k_j+1/2(ρn_j, ρn_j) = F_j+1/2n (ρn_j ∧ k, ρn_j ∧ k) − F_j+1/2n (k, k) = f (tn, xj+1/2, ρnj ∧ k, Rj+1/2n ) − f (tn, xj+1/2, k, Rnj+1/2) = sgn+(ρn_j − k)f (tn, x_j+1/2, ρ_jn, Rn_j+1/2) − f (tn, x_j+1/2, k, Rn_j+1/2).

It follows then easily that S −→ ∆x→0+ − Z +∞ 0 Z b a sgn+(ρ(t, x) − k) f (t, x, ρ, R(t, x)) − f (t, x, k, R(t, x))
∂xϕ(t, x) dx dt − α Z +∞ 0 (ρa(t) − k)+ϕ(t, a) dt + Z +∞ 0 (ρb(t) − k)+ϕ(t, b) dt ! . Let us rewrite S (3.43) as follows:

S = − ∆t +∞ X n=0 N X j=1 Gn,k_j+1/2(ρn_j, ρ_jn) ϕ(tn, xj+1) − ϕ(tn, xj
− α ∆t +∞ X n=0  (ρn_a− k)+ϕ(tn, a) + (ρn_b − k)+ϕ(tn, b)

= ∆t +∞ X n=0   N X j=1 Gn,k_j+1/2(ρn_j, ρn_j) ϕ(tn, xj+1) − N −1 X j=0 Gn,k_j+3/2(ρn_j+1, ρn_j+1) ϕ(tn, xj+1)   − α ∆t +∞ X n=0  (ρn_a− k)+ϕ(tn, a) + (ρn_b − k)+ϕ(tn, b) = − ∆t +∞ X n=0 N −1 X j=1  Gn,k_j+1/2(ρn_j, ρn_j) − Gn,k_j+3/2(ρn_j+1, ρn_j+1) ϕ(tn, xj+1) − ∆t +∞ X n=0  Gn,k_{N +1/2}(ρn_N, ρn_N) ϕ(tn, xN +1) − Gn,k_3/2(ρn1, ρn1)) ϕ(tn, x1)  − α ∆t +∞ X n=0  (ρn_a− k)+ϕ(tn, a) + (ρn_b − k)+ϕ(tn, b) = Sint+ Sb, where we set Sint= − ∆t +∞ X n=0 N −1 X j=1  Gn,k_j+1/2(ρn_j, ρn_j) − Gn,k_j+3/2(ρn_j+1, ρn_j+1)ϕ(tn, xj+1), Sb = − ∆t +∞ X n=0  Gn,k_{N +1/2}(ρn_N, ρn_N) ϕ(tn, xN +1) − Gn,k_3/2(ρn1, ρn1)) ϕ(tn, x1)  − α ∆t +∞ X n=0  (ρn_a− k)+ϕ(tn, a) + (ρn_b − k)+ϕ(tn, b).

Focus on Sint_{: by adding and subtracting G}n,k

j+1/2(ρnj, ρnj) in the brackets, we can rewrite it as

Sint= − ∆t +∞ X n=0 N −1 X j=1  Gn,k_j+1/2(ρn_j, ρn_j+1) − Gn,k_j+1/2(ρn_j, ρn_j)ϕ(tn, xj+1) − ∆t +∞ X n=0 N −1 X j=1  Gn,k_j+1/2(ρn_j, ρn_j+1) − Gn,k_j+3/2(ρn_j+1, ρn_j+1) ϕ(tn, xj+1),

We evaluate now the distance between Tint and Sint:

  T int_{− S}int  ≤ ∆t +∞ X n=0 N −1 X j=1   G n,k j+1/2(ρnj, ρnj+1) − Gn,kj+1/2(ρnj, ρnj)     ϕ(tn, x_j+1) − ϕ(tn, x_j)  . Observe that   G n,k j+1/2(ρ n j, ρnj+1) − Gn,kj+1/2(ρ n j, ρnj)    =  F n j+1/2(ρnj ∧ k, ρnj+1∧ k) − Fj+1/2n (ρnj ∧ k, ρnj) ∧ k    = 1 2   f (t n_{, x} j+1/2, ρnj ∧ k, Rnj+1/2) + f (tn, xj+1/2, ρnj+1∧ k, Rnj+1/2)

−2 f (tn, x_j+1/2, ρn_j ∧ k, R_j+1/2n ) − α (ρn_j+1∧ k − ρn_j ∧ k)  = 1 2   f (t n_{, x} j+1/2, ρnj+1∧ k, Rnj+1/2) − f (tn, xj+1/2, ρnj ∧ k, Rnj+1/2) −α (ρn_j+1∧ k − ρn_j ∧ k)  ≤ 1 2(L + α)   ρ n j+1∧ k − ρnj ∧ k    ≤ α   ρ n j+1− ρnj   . Therefore,   T int_{− S}int  ≤ α ∆x ∆t k∂xϕkL∞ +∞ X n=0 N −1 X j=1   ρ n j+1− ρnj    ≤ α ∆x T k∂xϕkL∞ max 0≤n≤T /∆tTV (ρ∆(t n_{, ·)) = O(∆x), (3.44)}

thanks to the uniform BV estimate (3.14). Pass now to the terms Tb _{and S}b_:

Sb− Tb = − ∆t +∞ X n=0  Gn,k_{N +1/2}(ρn_N, ρn_N) ϕ(tn, xN +1) − Gn,k_3/2(ρn1, ρn1)) ϕ(tn, x1)  − α ∆t +∞ X n=0  (ρn_a− k)+ϕ(tn, a) + (ρn_b − k)+ϕ(tn, b) − ∆t +∞ X n=0  Gn,k_{N +1/2}(ρn_N, ρn_b) − Gn,k_{N +1/2}(ρn_N, ρn_N)ϕ(tn, xN) + ∆t +∞ X n=0  Gn,k_1/2(ρn_a, ρn₁) − Gn,k_3/2(ρn₁, ρn₁)ϕ(tn, x1) = ∆t +∞ X n=0  Gn,k_1/2(ρn_a, ρn₁) ϕ(tn, x1) − α (ρna− k)+ϕ(tn, a)  (3.45) − ∆t +∞ X n=0  α (ρn_b − k)+ϕ(tn, b) + Gn,k_{N +1/2}(ρn_N, ρn_b) ϕ(tn, xN)  (3.46) − ∆t +∞ X n=0 Gn,k_{N +1/2}(ρn_N, ρn_N) ϕ(tn, xN +1) − ϕ(tn, xN)
. (3.47) Observe that ∂F_j+1/2n ∂u (u, v) = 1 2  ∂ρf (tn, xj+1/2, u, Rnj+1/2) + α  ≥ 1 2(−L + α) ≥ 0, ∂F_j+1/2n ∂v (u, v) = 1 2  ∂ρf (tn, xj+1/2, v, Rnj+1/2) − α  ≤ 1 2(L − α) ≤ 0,

meaning that the numerical flux is increasing with respect to the first variable and decreasing with respect to the second one. Thus,

≥ F_j+1/2n (k, v ∧ k) − F_j+1/2n (k, k) = 1 2  f (tn, x_j+1/2, v ∧ k, Rn_j+1/2) − f (tn, x_j+1/2, k, Rn_j+1/2) − α(v ∧ k − k) ≥ − L + α 2 |v ∧ k − k| ≥ − α (v − k)+ and Gn,k_j+1/2(u, v) = F_j+1/2n (u ∧ k, v ∧ k) − F_j+1/2n (k, k) ≤ F_j+1/2n (u ∧ k, k) − F_j+1/2n (k, k) = 1 2  f (tn, x_j+1/2, u ∧ k, Rn_j+1/2) − f (tn, x_j+1/2, k, Rn_j+1/2) − α(k − u ∧ k) ≤ L + α 2 |u ∧ k − k| ≤ α (u − k)+. Hence, [(3.45)] = ∆t +∞ X n=0 Gn,k_1/2(ρn_a, ρ₁n) ϕ(tn, x1) − ϕ(tn, a)
+ ∆t +∞ X n=0  Gn,k_1/2(ρn_a, ρn₁) − α (ρn_a− k)+ϕ(tn, a) ≤ α T ∆x k∂xϕkL∞ sup 0≤n≤T /∆t (ρn_a− k)+ + α ∆t +∞ X n=0  (ρn_a− k)+− (ρn_a− k)+ϕ(tn, a) ≤ α T ∆x k∂xϕkL∞kρak_L∞ = O(∆x), [(3.46)] = − ∆t +∞ X n=0  α (ρn_b − k)++ Gn,k_{N +1/2}(ρn_N, ρn_b)ϕ(tn, b) − ∆t +∞ X n=0 Gn,k_{N +1/2}(ρn_N, ρn_b) ϕ(tn, xN) − ϕ(tn, b)
≤ − α ∆t +∞ X n=0  (ρn_b − k)+− (ρn_b − k)+ϕ(tn, b) + α T ∆x k∂xϕkL∞ sup 0≤n≤T /∆t (ρn_b − k)+ ≤ α T ∆x k∂xϕkL∞kρbk_L∞ = O(∆x), [(3.47)] = ∆t       +∞ X n=0 Gn,k_{N +1/2}(ρn_N, ρn_N) ϕ(tn, xN +1) − ϕ(tn, xN)
      ≤ ∆t ∆x k∂xϕkL∞ +∞ X n=0   G n,k N +1/2(ρ n N, ρnN)   

= ∆t ∆x k∂xϕkL∞ +∞ X n=0   F n N +1/2(ρnN∧ k, ρnN∧ k) − FN +1/2N (k, k)    = ∆t ∆x k∂xϕkL∞ +∞ X n=0   f (t n_{, x} N +1/2, ρnN ∧ k, RnN +1/2) − f (tn, xN +1/2, k, RnN +1/2)    ≤L ∆t ∆x k∂xϕkL∞ +∞ X n=0  ρn_N∧ k − k  = L ∆t ∆x k∂xϕkL∞ +∞ X n=0 (ρn_N− k)+ ≤ L T ∆x k∂xϕkL∞ sup 0≤n≤T /∆t kρnkL∞ = O(∆x),

thanks to the uniform L∞ estimate (3.10). Hence, Sb− Tb ≤ O(∆x), so that we finally get 0 ≥ [(3.39) . . . (3.42)]

= [(3.39)] + [(3.42)] + T ± S ≥ [(3.39)] + [(3.42)] + S − O(∆x),

concluding the proof. 

4

Lipschitz continuous dependence on initial and boundary

data

Proposition 4.1. Fix T > 0. Let (f ), (ω) and (3.5) hold. Assume moreover ∂2_xρf, ∂_ρR2 f ∈ L∞_{([0, T ] × [a, b] × R}2_{; R). Let ρ}

o, σo ∈ BV( ]a, b[; R+) and ρa, ρb, σa, σb ∈ BV( ]0, T [; R+).

Call ρ and σ the corresponding solutions to (1.1). Then the following estimate holds ρ(T ) − σ(T ) _L1 (]a,b[) ≤  kρo− σokL1_(]a,b[)+ L  kρa− σakL1_{([0,T ])}+ kρb− σbkL1_{([0,T ])}  1 + B(T ) T eB(T ) T, and B(T ) is defined in (4.25).

Proof. Introduce the following notation: R(t, x) = 1 W (x) Z b a ρ(t, y) ω(y − x) dy , g(t, x, u) = f (t, x, u, R(t, x)), S(t, x) = 1 W (x) Z b a σ(t, y) ω(y − x) dy , h(t, x, u) = f (t, x, u, S(t, x)). (4.1)

Observe that, due to (ω) and Theorem 2.3, for any t ∈ [0, T ] and x ∈ [a, b], R(t, x) ≤ kωkL∞ Kω R1(t), S(t, x) ≤ kωkL∞ Kω S1(t),

where S1(t) is defined analogously to R1(t) in (2.5) for σ. For later use, set

J(t) = kωkL∞

Kω

maxR1(t), S1(t) ,

so that R(t, x), S(t, x) ∈ [−J(t), J(t)]. Compute for later use  ∂_xR(t, x)  ≤ ω′ _L1kωk_L∞ K2 ω + ω′ _L∞ Kω ! ρ(t) _L1 (]a,b[)≤ L R1(t), (4.2)   ∂ 2 xxR(t, x)    ≤   2 ω′ 2 L1kωkL∞ K3 ω + ω′′ _L1kωkL∞ K2 ω +2 ω′ _L1 ω′ _L∞ K2 ω + ω′′ _L∞ Kω   ρ(t) L1_(]a,b[) ≤ W R1(t), (4.3)

with L defined exactly as in (3.12) and W defined as in (3.25). Observe that L and W are finite thanks to (ω). Compute also

 R(t, x) − S(t, x)  ≤ kωkL∞ Kω Z b a  ρ(t, y) − σ(t, y)  dy , (4.4)  ∂xR(t, x) − ∂xS(t, x)  ≤ L Z b a  ρ(t, y) − σ(t, y) dy . (4.5)

We can think of ρ and σ as solutions to the following IBVPs          ∂tρ + ∂xg(t, x, ρ) = 0, ρ(0, x) = ρo(x), ρ(t, a) = ρa(t), ρ(t, b) = ρb(t),          ∂tσ + ∂xh(t, x, σ) = 0, (t, x) ∈ ]0, T [×]a, b[, σ(0, x) = σo(x), x ∈ ]a, b[, σ(t, a) = σa(t), t ∈ ]0, T [, σ(t, b) = σb(t), t ∈ ]0, T [.

Moreover, consider also the following IBVP:          ∂tπ + ∂xg(t, x, π) = 0, (t, x) ∈ ]0, T [×]a, b[, π(0, x) = σo(x), x ∈ ]a, b[, π(t, a) = σa(t), t ∈ ]0, T [, π(t, b) = σb(t), t ∈ ]0, T [. (4.6)

Thanks to (f ) and to the additional assumptions on f , the flux functions g and h defined in (4.1) satisfy the hypotheses of Theorem A.2, Proposition A.3 and Theorem A.4. Indeed, focusing on g, compute:

∂_xu2 g(t, x, u) = ∂_xρ2 f (t, x, u, R(t, x) + ∂_uR2 f (t, x, u, R(t, x)) ∂xR(t, x).

Thus, thanks also to (4.2), ∂_xu2 g is finite. Therefore, we can use the results of [14], re-called in Appendix A: by Theorem A.2, problem (4.6) admits a unique solution in (L∞_∩

BV)( ]0, T [×]a, b[; R), which satisfies, for all t ∈ [0, T [

π(t)

TV (π(t)) ≤ TV (σo) + TV (σa; [0, t]) + TV (σb; [0, t]) + K(t) t
eC4(t) t, with P(t) :=  maxnkσokL∞_(]a.b[), kσak_L∞_([0,t]), kσbk_L∞_([0,t]) o + C3(t) t  eC4(t) t_, C3(t) := ∂_xg(·, ·, 0) L∞ ([0,t]×[a,b]), C4(t) := ∂ 2 xug _L∞_{([0,t]×[a,b]×R)} , K(t) := 2 C3(t) + 2 (b − a) ∂ 2 xxg _L∞_{([0,t]×[a,b]×[−P(t), P(t)])} +1 2  3 P(t) + kσakL∞ ([0,t])  ∂ 2 xug _L∞_{([0,t]×[a,b]×[−P(t), P(t)])} . Due to the definition of g in (4.1) and (f ), we obtain

 ∂_xg(t, x, 0)  =  ∂_xf (t, z, 0, R(t, x)) + ∂_Rf (t, x, 0, R(t, x)) ∂_xR(t, x)  = 0, ∂ 2 xug L∞ ([0,t]×[a,b]×R)≤ ∂ 2 xρf L∞ ([0,t]×[a,b]×R×[−J(t),J(t)]) + ∂ 2 ρRf L∞ ([0,t]×[a,b]×R×[−J(t),J(t)])L R1(t),

so that C3(t) = 0, C4(t) ≤ C5(t), where we set

C5(t) = ∂ 2 xρf _L∞ ([0,t]×[a,b]×R×[−J(t),J(t)])+ ∂ 2 ρRf _L∞ ([0,t]×[a,b]×R×[−J(t),J(t)])L R1(t) (4.7) Hence π(t) _L∞_(]a,b[)≤ P∞(t) = e C5(t) t_max n kσokL∞ (]a.b[), kσakL∞ ([0,t]), kσbkL∞ ([0,t]) o . (4.8) Moreover, ∂ 2 xxg _L∞_{([0,t]×[a,b]×[−P(t), P(t)])} ≤ ∂ 2 xxf _L∞_{([0,t]×[a,b]×[−P(t), P(t)]×[−J(t),J(t)])} + 2 ∂ 2 xRf _L∞_{([0,t]×[a,b]×[−P(t), P(t)]×[−J(t),J(t)])}k∂xRkL ∞ ([0,t]×[a,b]) + ∂ 2 RRf _L∞ ([0,t]×[a,b]×[−P(t), P(t)]×[−J(t),J(t)])k∂xRk 2 L∞_{([0,t]×[a,b])} + k∂Rf kL∞_{([0,t]×[a,b]×[−P(t), P(t)]×[−J(t),J(t)])} ∂ 2 xxR L∞ ([0,t]×[a,b]) ≤ C P∞(t)  1 + R1(t)  2 L + L2R1(t) + W  , so that K(t) ≤ ˆK(t) where we set

ˆ K(t) = 2 (b−a) C P∞(t)  1 + R1(t)  2 L + L2R1(t) + W  +1 2  3 P∞(t) + kσakL∞_([0,t])  C3(t), (4.9)

and we then obtain TV (π(t)) ≤TV (σo) + TV (σa; [0, t]) + TV (σb; [0, t]) + ˆK(t) t  eC5(t) t_{. (4.10)} For t > 0, compute ρ(t) − σ(t) _L1 (]a,b[)≤ ρ(t) − π(t) _L1 (]a,b[)+ π(t) − σ(t) _L1 (]a,b[). (4.11)

The first term on the right hand side of (4.11) evaluates the distance between solutions to IBVPs of the type considered in the Appendix A with the same flux function, but different initial and boundary data. Therefore, we can apply Proposition A.3, to get

ρ(t) − π(t) L1_(]a,b[)≤ kρo− σokL1_(]a,b[) + k∂ugkL∞_{([0,t]×[a,b]×R)}  kρa− σakL1_([0,t])+ kρb− σbkL1_([0,t])  . Due to the definition of g in (4.1), we obtain

k∂ugkL∞ ([0,t]×[a,b]×R)= ∂_ρf _L∞_{([0,t]×[a,b]×R×R)} < L, hence ρ(t) − π(t) _L1 (]a,b[)≤ kρo− σokL1_(]a,b[)+ L  kρa− σakL1_([0,t])+ kρb− σbkL1_([0,t])  . (4.12) On the other hand, the second term on the right hand side of (4.11) evaluates the distance between solutions to IBVPs of the type considered in Appendix A with different flux functions, but same initial and boundary data. We apply Theorem A.4 to obtain

π(t) − σ(t) _L1 (]a,b[) ≤ Z t 0 Z b a ∂x(g − h)(s, x, ·) _L∞ (U (s))dx ds (4.13) + Z t 0 ∂_u(g − h)(s, ·, ·) _L∞ (]a,b[×U (s)) min n TV σ(s)
, TV π(s)
ods (4.14) + 2 Z t 0 (g − h)(s, a, ·) _L∞_{(U (s))}ds + 2 Z t 0 (g − h)(s, b, ·) _L∞_{(U (s))}ds , (4.15)

where U (s) = [−U (s), U (s)], with U (s) = maxn π(s) L∞ (]a,b[), σ(s) L∞ (]a,b[) o . Let us now estimate all the terms appearing in (4.13)–(4.15). First of all, by Theorem 2.3,

σ(t) L∞ (]a,b[)≤ S∞(t) = e t C(1+L S1(t)) max n kσokL∞_(]a.b[), kσak_L∞_([0,t]), kσbk_L∞_([0,t]) o , (4.16) so that, comparing (4.16) with (4.8) we obtain

U (t) ≤ maxnkσokL∞_(]a.b[), kσak_L∞_([0,t]), kσbk_L∞_([0,t]) o expt C 1 + L S1(t)
+ t C5(t)  . (4.17) Then, by Theorem 2.3, TV (σ(t)) ≤ et T1(t) _{TV (σ} o) + TV (σa; [0, t]) + TV (σb; [0, t])
+ T2(t) T1(t) (et T1(t)_{− 1) (4.18)}

with T1(t) = ∂ 2 ρxf _L∞_{([0,t]×[a,b]×R}2₎ + L S1(t) ∂ 2 ρRf _L∞_{([0,t]×[a,b]×R}2₎ , T2(t) = K2(t) + 3 2C(1 + L S1(t)) S∞(t) +  K3(t) + C 2 (1 + L S1(t))  kσakL∞ ([0,t]), K2(t) = C S1(t) 1 + 2 L S1(t) + 2 K3(t)
, K3(t) = C S1(t)  L2S1(t) + 1 2W  . Hence, comparing (4.10) and (4.18), we get

minnTV σ(s)
, TV π(s)
o≤ et T3(t) _{TV (σ} o) + TV (σa; [0, t]) + TV (σb; [0, t])
+ min  ˆ K(t) eC5(t) t_, T2(t) T1(t) (et T1(t)_{− 1)}  (4.19) =: T4(t), with T3(t) = ∂ 2 ρxf L∞_{([0,t]×[a,b]×R}2₎ + L minR1(t), S1(t) ∂ 2 ρRf L∞_{([0,t]×[a,b]×R}2₎ . (4.20) Focus on (4.13): by (f ), (4.4) and (4.5),   ∂x g(t, x, u) − h(t, x, u)
  =     d dx  f t, x, u, R(t, x)
− f t, x, u, S(t, x)
     ≤ ∂_xf (t, x, u, R(t, x)) − ∂_xf (t, x, u, S(t, x))   + ∂Rf (t, x, u, R(t, x)) ∂xR(t, x) − ∂Rf (t, x, u, S(t, x)) ∂xS(t, x)   ≤  ∂ 2 xRf (t, x, u, R1(t, x))     R(t, x) − S(t, x)  +   ∂ 2 RRf (t, x, u, R2(t, x))     ∂_xR(t, x)    R(t, x) − S(t, x)   +∂_Rf (t, x, u, S(t, x))    ∂_xR(t, x) − ∂_xS(t, x)   ≤ C |u|  kωkL∞ Kω 1 + L R1(t)
+ L  Z b a  ρ(t, y) − σ(t, y)  dy , with Ri(t, x) ∈ I R(t, x), S(t, x)
, i = 1, 2. Hence, ∂x(g − h)(s, x, ·) _L∞_{(U (s))} ≤ C U (s)  kωkL∞ Kω 1 + L R1(s)
+ L  Z b a  ρ(s, y) − σ(s, y) dy . (4.21) Pass to (4.14): by (f ), (4.4) and (4.5),   ∂u g(t, x, u) − h(t, x, u)
  =     ∂ρ  f t, x, u, R(t, x)
− f t, x, u, S(t, x)
     =  ∂ 2 ρRf t, x, u, R3(t, x)
    R(t, x) − S(t, x)  ,