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**finite-volume methods**

### Pierre Bousquet, Franck Boyer, Flore Nabet

**To cite this version:**

### Pierre Bousquet, Franck Boyer, Flore Nabet. On a functional inequality arising in the analysis of

### finite-volume methods. Calcolo, Springer Verlag, 2016, 53 (3), pp.363-397. �10.1007/s10092-015-0153-

### 0�. �hal-01134988v3�

FINITE-VOLUME METHODS

P BOUSQUET^{∗}, F. BOYER^{†},ANDF. NABET^{†}

Abstract.We establish a Poincaré–Wirtinger type inequality on some particular domains with a precise estimate of the constant depending only on the geometry of the domain. This type of inequality arises, for instance, in the analysis of finite volume (FV) numerical methods.

As an application of our result, we prove uniform a priori bounds for the FV approximate solutions of the
heat equation with Ventcell boundary conditions in the natural energy space defined as the set of those functions in
H^{1}(Ω)whose traces belong toH^{1}(∂Ω). The main difficulty here comes from the fact that the approximation is
performed on non-polygonal control volumes since the domain itself is non-polygonal.

Key words.Poincaré–Wirtinger inequality - Finite Volume methods - Ventcell boundary condition AMS subject classifications.35A23,35K05,65M08

1. Introduction. The main goal of this paper is to study functional inequalities of the following form

(1.1)

1 mσ

Z

σ

u− 1 mK

Z

K

u

≤Cdiam(K) 1 mK

Z

K

|Du|, ∀u∈W^{1,1}(R^{n}),

whereKis a bounded connected Lipschitz domain inR^{n}andσa non empty open subset of

∂K. We have denoted bymKthe volume ofKandm_{σ} the surface measure ofσ, namely its
(n−1)-dimensional Hausdorff measure. Those notations will be used all along this paper.

The fact that such an inequality holds is straightforward, for instance by applying the Bramble-Hilbert lemma (see for instance [2]). Our main purpose is to estimate the depen- dence of the constantCwith respect to the geometry of the domainK. In the particular case of a convex domainKthe mean-value inequality immediately implies that

(1.2)

1
m_{σ}

Z

σ

u− 1 mK

Z

K

u

≤diam(K) Å

sup

K

|Du|

ã

, ∀u∈W^{1,∞}(R^{n}),

with no geometric constant in the right-hand side. The inequality (1.1) has to be seen as a
generalisation of (1.2) to less regular functions u. This loss of regularity induces that the
constant in the inequality may depend on the shape ofK. Observe that, ifKis not convex,
one has to replacesup_{K}|Du|bysup_{Conv(K)}|Du|in (1.2), whereConv(K)is the convex hull
ofK.

Inequalities of the form (1.1) play an important role in the analysis of finite volume
numerical methods for elliptic or parabolic equations on general meshes, which is our main
motivation. They are meant to be applied to each cell (control volume)Kin a mesh of a given
computational domain. They allow to prove stability estimates in (discrete) Sobolev spaces
for the naturalL^{2}projections of the functions defined onΩand the projections of their traces.

To our knowledge, such inequalities have only been established up to now in the framework of polygonal setsK. However, for more complex situations, like for the discretisation of the heat equation with dynamic Ventcell boundary conditions, we are interested in proving such inequalities for non polygonal domains. We detail such an application in Section 6.

∗pierre.bousquet@math.univ-toulouse.fr, Institut de Mathématiques de Toulouse, UMR 5219, Université de Toulouse, F–31062 Toulouse Cedex 9, France

†{franck.boyer,flore.nabet}@univ-amu.fr, Aix Marseille Université, CNRS, Centrale Marseille, I2M, UMR 7373, 13453 Marseille, France

1

Let us mention some references where such inequalities are proved and/or used in the finite volume framework.

• In [7, Lemma 3.4] (see also [4, Lemma 7.2] and [5, Lemmas 6.2 and 6.3]), (1.1) is
proved (in 2D for simplicity) whenK is polygonal and convex, with a constantC
depending only on the number of edges/faces ofK, and on the shape-regularity ratio
(diam(K))^{2}/mK.

• In [6, Lemma 6.6], the inequality is slightly generalized to a polygonalKwhich is simply supposed to be star-shaped with respect to a suitable ball.

• In [1], such inequalities are used for the convergence and error analysis of some approximation of non-linear Leray-Lions type operators (a model of which is the p-Laplace problem).

Finally, we refer to [9] for an example of analysis of a more complex model of a non-linear evolution equation associated with a non-linear dynamical boundary condition. This refer- ence was in fact our main motivation for the present work. Indeed, compared to the other references above, the numerical method in [9] is derived on non-polygonal control volumes, so that an inequality like (1.1) is needed on non-polygonal open setsK, see also Section 6.

The outline of the paper is the following. In Section 2, we state our main result (Theorem 2.1 in 2D) and the geometric assumptions that we shall work with in the sequel. Section 3 is devoted to the proof of the main result whereas in Section 4, we state and prove a sort of Poincaré-Wirtinger inequality related to the functional inequality proved in Theorem 2.1.

Section 5 is dedicated to the extension of our main inequality in the higher dimensional case
(i.e. inR^{n}, forn ≥3). Finally, in order to illustrate this work, we provide an application,
as simple as possible, of Theorem 2.1 to the proof of uniform discrete energy estimates for a
finite volume approximation of a toy system on a non-polygonal domainΩ.

2. Main result. Given aC^{1}curveσ ⊂R^{2}and a pointz_{∗} ∈R^{2}\σ, we consider the
following domainT:

T ={z_{∗}+t(γ(θ)−z_{∗}) :t∈]0,1[, θ∈]0,1[},

whereγ: [0,1]→R^{2}is aC^{1}parametrization ofσ: γ([0,1]) =σ,γis one-to-one and|γ^{0}|
does not vanish. Without loss of generality, we choose the parametrizationγin such a way
that|γ^{0}(θ)|=mσfor everyθ∈[0,1].

We say thatT is a pseudo-triangle if for everyx∈σ,
(2.1) {z∗+t(x−z_{∗}) :t≥0} ∩σ={x}.

σ

T ˜σ

T˜σ

FIG. 2.1.The pseudo-triangleTwith its curved edgeσand one of its sub-triangle

Without loss of generality, we shall assume that the vertexz_{∗}ofT opposite toσis the
origin(0,0)ofR^{2}.

THEOREM2.1.LetTbe a pseudo-triangle as above. We assume that there existµ, ν >0
such that for any sub-arc˜σ⊂σ, the corresponding sub-triangleT_{σ}_{˜}(see Figure 2.1) satisfies

(2.2) µ≤ m_{T}_{σ}_{˜}
m_{σ}_{˜} ≤ν.

Then for everyp∈[1,+∞[and everyu∈W^{1,p}(T),
(2.3)

1
m_{σ}

Z

σ

u− 1
m_{T}

Z

T

u

p

≤C^{p}(m_{σ}+ diam(T))^{p} 1
m_{T}

Z

T

|Du|^{p},
whereConly depends on the ratio_{µ}^{ν}.

As already noticed in the introduction, the main issue is to understand how the constant in the inequality depends on the geometry of this pseudo-triangleT.

REMARK2.1.For a real flat triangleT, the quantity ^{m}_{m}^{T}^{σ}^{˜}

˜

σ does not depend onσ˜and is
equal to ^{m}_{m}^{T}

σ. In this particular case, we haveµ=νandmσ≤diam(T). Hence, we recover exactly the inequality proved in [7].

PROPOSITION 2.2. Under assumption(2.1), the mapθ 7→ det(γ(θ), γ^{0}(θ))is either
nonnegative everywhere or nonpositive everywhere.

In the sequel, we assume that the orientation is chosen such that det (γ, γ^{0})≥0. Then,
assumption(2.2)is equivalent to the following inequality

(2.4) 2µmσ≤ det (γ(θ), γ^{0}(θ))≤2νmσ, ∀θ∈[0,1].

Proof . One can assume without loss of generality thatγ([0,1])is contained in the half plane
{(x, y)∈R^{2}:x >0}. There exists aC^{1}mapϕ: [0,1]→]−π/2, π/2[such that for every
θ∈[0,1],

γ(θ)

|γ(θ)| = (cosϕ(θ),sinϕ(θ)).

By (2.1), the mapϕis one-to-one. Hence, it is either strictly increasing or strictly decreasing.

Assume for instance thatϕis strictly increasing.

This implies that for everyθ∈[0,1), for everyh >0such thatθ+h∈[0,1], det

Å

γ(θ),γ(θ+h)−γ(θ) h

ã

= 1

hdet(γ(θ), γ(θ+h))

= 1

h|γ(θ)||γ(θ+h)|sin(ϕ(θ+h)−ϕ(θ))≥0.

Passing to the limith→0yields the desired result.

Assume now that (2.2) holds true, then let0≤a < b≤1, and considerσ˜ =γ([a, b]).

Then

m_{σ}_{˜}= (b−a)m_{σ}andm_{T}_{σ}_{˜} = 1
2

Z b a

det (γ(θ), γ^{0}(θ))dθ,
thus thanks to (2.2),

µ≤ Z b

a

det (γ(θ), γ^{0}(θ))dθ
2(b−a)mσ

≤ν,

and we obtain (2.4) when b tends toa. Conversely, assume that (2.4) holds. Then (2.2)

follows by integration of (2.4) on the segment[a, b].

REMARK2.2.For some particular cases, we can estimate the constant in the inequality (2.3)even if the pseudo-triangleT does not satisfy assumption(2.2). In order to illustrate such a situation, we consider the pseudo-triangleTdefined as follows:

T ={tγ(θ) : 0< t <1,−θ0< θ < π+θ0}, withγ(θ) = Å

Rcosθ, Rsinθ+ R sin(θ0)

ã
,
where0< θ_{0}< ^{π}_{2},R >0. Observe thatγandγ^{0}are colinear forθ=−θ0orθ =π+θ_{0}
so that assumption(2.2)is not satisfied here. We decompose the pseudo-triangleT into the
piece of diskP of radiusRand centerA= (0, R/sin(θ_{0}))and the quadrilateralQdefined
by:Q=T \P.

R

θ0

O θ0

P

Q A

FIG. 2.2.A particular case which does not satisfy assumption(2.2)

First, we remark that assumption (2.2)is satisfied for the pseudo-triangleP with the
ratio ^{ν}_{µ} equal to1(see Remark 2.1). Then we can apply Theorem 2.1 to the pseudo-triangle
P, so that there exists a constantC0>0which does not depend onRandθ0such that

1
m_{σ}

Z

σ

u− 1
m_{P}

Z

P

u

≤C0R 1
m_{P}

Z

P

|Du|.

Moreover, sinceP ⊂T andTis convex, we can apply [4, Lemma 7.1],

1 mT

Z

T

u− 1 mP

Z

P

u

≤Cf0

(diam(T))^{3}
mPmT

Z

T

|Du|.

Now, we want to control the volume of the pseudo-triangleT by the volume of P. We note that

mQ= R^{2}

tanθ_{0} , mP =R^{2}

2 (π+ 2θ0) , diam(T) =R Å

1 + 1
sinθ_{0}

ã . Hence

m_{Q}≤diam(T)
R

2
πm_{P}
and then using thatR≤diam(T),

mP ≥ πR

πR+ 2diam(T)mT ≥C1

R diam(T)mT. This implies

1
m_{σ}

Z

σ

u− 1
m_{P}

Z

P

u

≤C0

C_{1}diam(T) 1
m_{T}

Z

T

|Du|, and

1 mT

Z

T

u− 1 mP

Z

P

u

≤Cf_{0}
C1

(diam(T))^{4}
Rm^{2}_{T}

Z

T

|Du|.

SinceR≤diam(T),mT ≤(diam(T))^{2}, the above two inequalities yield

1 mT

Z

T

u− 1 mσ

Z

σ

u

≤C(diam(T))^{4}
Rm^{2}_{T}

Z

T

|Du|

≤ C^{0}

sin(θ0)^{2}diam(T) 1
mT

Z

T

|Du|,

whereC^{0}is a universal constant. As expected we observe that the above inequality does not
depend onRand blows up whenθ_{0}goes to0.

3. Proof of Theorem 2.1. By Jensen’s inequality, we only need to establish the case p = 1. We begin by proving a change of variables formula (Proposition 3.1) that let us express the differences between the mean values of a function onT and onσas a weighted integral of its gradient onT. Then, we prove in Proposition 3.2 that this change of variables can be realized with a diffeomorphism satisfying suitable estimates. This proposition relies on two technical Lemmas 3.3 and 3.4. Theorem 2.1 readily follows from those two propositions.

We notice that the existence of such a diffeomorphism is ensured by a general result of [3] but we have to be able to estimate the derivatives of this diffeomorphism in function of the geometry ofT. That is why we resume explicitly the steps of [3] that allow us to control all the constants involved in the estimates.

In the sequel, we denote byQ^{2}=]0,1[^{2}the unit cube inR^{2}. By a standard approximation
argument, one can assume thatγ∈C^{2}(0,1).

PROPOSITION 3.1. Assume that there exists a Lipschitz continuous mapΦ :Q^{2} → T
such that

1. Φis aC^{1}diffeomorphism fromQ^{2}ontoT,
2. Φ(0, θ) = (0,0),

3. Φ(1, θ) =γ(θ),

4. JacΦ(s, θ) = 2m_{T}s.

Then for everyu∈W^{1,1}(T), we have
1

mσ

Z

σ

u− 1 mT

Z

T

u= 1 2mT

Z

T

Du(x, y) [s∂sΦ(s, θ)]_{(s,θ)=Φ}−1(x,y) dx dy.

Proof . It follows from (2.4) that the pseudo-triangleT is biLipschitz homeomorphic to a (true) triangle, see the proof of Lemma 4.2 in section 4. In particular,Tis a Lipschitz domain.

By a standard density argument, we can thus assume thatu∈C^{1}(T). Let
v(s, θ) =u◦Φ(s, θ)JacΦ(s, θ).

Then for every(t, θ)∈Q^{2},

v(1, θ)−v(t, θ) = Z 1

t

∂sv(s, θ)ds.

Hence, thanks to the assumptions onΦ, we obtain
2m_{T}u(γ(θ))−u◦Φ(t, θ)JacΦ(t, θ) =

Z 1 t

(∂_{s}(u◦Φ)JacΦ + 2m_{T}(u◦Φ))ds.

By integrating this equality onQ^{2}and using an obvious change of variables, we get
(3.1) 2m_{T}

mσ

Z

σ

u− Z

T

u= Z 1

0

dθ Z 1

0

dt Z 1

t

(∂_{s}(u◦Φ)JacΦ + 2m_{T}(u◦Φ))ds.

By Fubini theorem, Z 1

0

dθ Z 1

0

dt Z 1

t

(∂s(u◦Φ)JacΦ + 2m_{T}(u◦Φ)) ds

= Z

Q^{2}

s(∂_{s}(u◦Φ)JacΦ + 2m_{T}(u◦Φ))ds dθ.

Since JacΦ(s, θ) = 2mTs, the same change of variables gives Z

Q^{2}

s(∂s(u◦Φ)JacΦ + 2mT(u◦Φ))ds dθ

= Z

T

Du(x, y) [s∂sΦ(s, θ)]_{(s,θ)=Φ}−1(x,y)dx dy+
Z

T

u.

The claim now follows from this identity together with (3.1).

The proof of Theorem 2.1, that is the one of the inequality (2.3), is now a straightforward consequence of the following proposition that claims that we can build a suitableΦand apply Proposition 3.1.

PROPOSITION3.2.Under assumption(2.2), there exists a universal constantC >0and
a mapΦ :Q^{2} →T satisfying the properties required in Proposition 3.1 and the additional
estimate

(3.2) |∂sΦ(s, θ)| ≤Cν^{3}

µ^{3}(diam(T) +m_{σ}), ∀(s, θ)∈Q^{2}.

The sequel of this section is thus devoted to the proof of Proposition 3.2. We first observe that, in the case whenTis a real triangle, namely whenσis a segment, the mapΦ =φ2with

φ2: (s, θ)∈Q^{2}7→sγ(θ)∈T ,
satisfies the assumptions of Proposition 3.1 as well as estimate (3.2).

In the general case whereσis curved, we have

Jacφ2(s, θ) =sdet (γ(θ), γ^{0}(θ)),

and this quantity depends onθ: we cannot chooseΦ = φ2anymore. Thus, we are going
to right compose φ2 with a diffeomorphism of the unit cube to construct aΦ : Q^{2} → T
satisfying Proposition 3.2, see Figure 3.1.

To simplify the notation, we definegby

g: (s, θ)∈Q^{2}7→Jacφ_{2}(s, θ) =sdet (γ, γ^{0})(θ).

Then, we construct a first diffeomorphismφ1ofQ^{2}such thatφ2◦φ1satisfies the first three
assumptions of Proposition 3.1 as a well as a weaker version (integrated with respect tos) of
the third assumption. Obtaining the strong version of this property (that is a point-by-point
equality) will be the purpose of Lemma 3.4.

LEMMA3.3.There exists aC^{1}diffeomorphismφ1:Q^{2}→Q^{2}such that
1. for everyx∈∂Q^{2},φ1(x) =x,

2. for everyθ∈[0,1], (3.3)

Z 1 0

Jac(φ2◦φ1)(s, θ)ds= Z 1

0

g◦φ1(s, θ)Jacφ1(s, θ)ds=mT.
3. for every(s, θ)∈Q^{2},

(3.4) µ

4ν ≤Jacφ1(s, θ)≤ 5ν µ. Proof . Let

(3.5) ε= µ

10ν

andζ∈C_{c}^{∞}(0,1)be a cut-off function such that0≤ζ≤1 +ε,|ζ^{0}|L^{∞} ≤^{10}_{ε} and
(3.6)

Z 1 0

ζ(s)ds= 1 and Z 1

0

|ζ(s)−1|ds < ε.

We introduce the map

G: (a, b)7→

Z 1 0

s ds

Z a+ζ(s)b 0

det (γ, γ^{0})(θ)dθ.

ThenGis well-defined andC^{2}on the set{(a, b) : 0 ≤a≤1,_{1+ε}^{−a} ≤b ≤ ^{1−a}_{1+ε}}(here, we
use the fact thatγ∈C^{2}([0,1])so that det (γ, γ^{0})isC^{1}([0,1])). Moreover, we have

∂aG(a, b) = Z 1

0

det (γ, γ^{0})(a+ζ(s)b)s ds

∂bG(a, b) = Z 1

0

det (γ, γ^{0})(a+ζ(s)b)ζ(s)s ds.

By (2.4), we have 2mσµ

Z 1 0

ζ(s)s ds≤∂bG(a, b)≤2mσν Z 1

0

ζ(s)s ds.

We can bound the right hand side by2mσνwhile 2mσµ

Z 1 0

ζ(s)s ds≥2mσµ ÇZ 1

0

s ds− Z 1

0

|ζ(s)−1|ds å

≥2mσµ ÇZ 1

0

s ds−ε å

≥4mσµ 5 . Hence we conclude that

(3.7) 0<4mσµ

5 ≤∂bG(a, b)≤2mσν.

We claim that

(3.8) G

Å a, −a

1 +ε ã

≤mTa.

Indeed, by (2.4), G

Å a, −a

1 +ε ã

≤2m_{σ}ν
Z 1

0

s(a− a

1 +εζ(s))ds≤ 2m_{σ}νa
1 +ε

Z 1 0

(1 +ε−ζ(s))ds.

By (3.6), this implies G

Å a, −a

1 +ε ã

≤2mσνaε

1 +ε ≤2mσνaε.

Since

m_{T} = 1
2

Z 1 0

det (γ, γ^{0})(θ)dθ≥m_{σ}µ,
it follows from (3.5) that

G Å

a, −a 1 +ε

ã

≤m_{T}a,
which proves our claim.

Similarly, one can prove that Z 1

0

s ds Z 1

a+ζ(s)^{1−a}_{1+ε}

det (γ, γ^{0})(θ)dθ≤mT(1−a).

This can be written as

G(1,0)−G Å

a,1−a 1 +ε ã

≤m_{T}(1−a).

Since

G(1,0) = Z 1

0

s ds Z 1

0

det (γ, γ^{0})(θ)dθ=m_{T},
this implies

(3.9) G

Å a,1−a

1 +ε ã

≥mTa.

We deduce from (3.7), (3.8) and (3.9) that for everya∈[0,1], there exists a uniquew(a)∈ [−a/(1 +ε),(1−a)/(1 +ε)]such that

G(a, w(a)) =mTa.

By the implicit function theorem, the functionwisC^{2}on[0,1]and satisfies
(3.10) ∂aG(a, w(a)) +∂bG(a, w(a))w^{0}(a) =m_{T}.
SinceG(0,0) = 0andG(1,0) =mT, we havew(0) = 0 =w(1).

We claim that for everys∈[0,1], for everya∈[0,1],
1 +ζ(s)w^{0}(a)>0.

To this end, it would be sufficient to prove that∂_{b}G(a, w(a))(1 +ζ(s)w^{0}(a))>0but for a
further use we shall derive more precise bounds on this quantity.

• By (3.10), we can write

(3.11) ∂bG(a, w(a))(1+ζ(s)w^{0}(a)) =∂bG(a, w(a))+ζ(s)(mT−∂aG(a, w(a))),
and thus

∂bG(a, w(a))(1 +ζ(s)w^{0}(a)) =∂bG(a, w(a))−∂aG(a, w(a))

+ζ(s)m_{T} + (1−ζ(s))∂_{a}G(a, w(a)).

But

∂bG(a, w(a))−∂aG(a, w(a)) = Z 1

0

det (γ, γ^{0})(a+ζ(s)w(a))(ζ(s)−1)s ds

≥ −2νm_{σ}
Z 1

0

|ζ−1| ≥ −2νm_{σ}ε=−µmσ

5 . In the last inequality, we have used (3.6). Moreover, since0≤ζ≤1 +ε,

ζ(s)mT + (1−ζ(s))∂aG(a, w(a))

≥min (∂_{a}G(a, w(a)),(1 +ε)m_{T}−ε∂_{a}G(a, w(a))).
Sincem_{σ}µ≤∂_{a}G(a, w(a))≤m_{σ}νandm_{T} ≥m_{σ}µ, we get by (3.5)

ζ(s)m_{T} + (1−ζ(s))∂aG(a, w(a))≥ 9mσµ
10 .
This implies

(3.12) ∂bG(a, w(a))(1 +ζ(s)w^{0}(a))≥mσµ
2 .

• Using thatm_{T} ≤m_{σ}νand that∂aG >0, we obtain by (3.7) and (3.11)
(3.13) ∂_{b}G(a, w(a))(1 +ζ(s)w^{0}(a))≤∂_{b}G(a, w(a)) +ζ(s)m_{T}

≤2νmσ+ (1 +ε)mσν ≤4mσν.

Gathering (3.7), (3.12) and (3.13), we deduce that

(3.14) µ

4ν ≤1 +ζ(s)w^{0}(a)≤5ν
µ.
We now define

φ_{1}(s, θ) = (s, θ+ζ(s)w(θ)) , (s, θ)∈Q^{2}.

It appears thatφ_{1}isC^{1} onQ^{2} and satisfies Jacφ_{1}(s, θ) = 1 +ζ(s)w^{0}(θ) > 0. Since for
everys∈[0,1], the functionθ7→θ+ζ(s)w(θ)is continuous and increasing, it maps[0,1]

onto[0,1]. Henceφ1is aC^{1}diffeomorphism fromQ^{2}ontoQ^{2}. Moreover,φ1agrees with
the identity map on∂Q^{2}. We now turn to the proof of (3.3).

Leta∈[0,1]. By definition ofwandg, Z 1

0

ds

Z a+ζ(s)w(a) 0

g(s, θ)dθ=mTa.

By definition ofφ1, this can be written Z

φ_{1}((0,1)×(0,a))

g=mTa.

By the change of variables formula, this gives Z a

0

dθ Z 1

0

g◦φ1(s, θ)Jacφ1(s, θ)ds=mTa.

Since this holds true for anya∈[0,1], by derivation we deduce Z 1

0

g◦φ1(s, θ)Jacφ1(s, θ)ds=mT, ∀θ∈(0,1);

which completes the proof of (3.3).

Since Jacφ1(s, θ) = 1 +ζ(s)w^{0}(θ), we can use (3.14), to obtain the estimate (3.4). The

lemma is proven.

We proceed with the construction of the diffeomorphismΦthat we search in the form
φ_{2}◦φ_{1}◦φ_{0}. To simplify the notation, we consider the mapg_{1}:Q^{2}→Rdefined as follows:

g_{1}(s, θ) =Jac(φ_{2}◦φ_{1})(s, θ)

=g◦φ1(s, θ)Jacφ1(s, θ)

=sdet (γ, γ^{0})(θ+ζ(s)w(θ))(1 +ζ(s)w^{0}(θ)).

Observe that for everys∈(0,1],θ∈[0,1],g_{1}(s, θ)>0.

LEMMA 3.4. There exists a Lipschitz homeomorphismφ_{0} :Q^{2} →Q^{2}which isC^{1} on
Q^{2}and such that

1. for everyθ∈[0,1],φ_{0}(0, θ) = (0, θ)andφ_{0}(1, θ) = (1, θ),

2. for every(s, θ)∈Q^{2},

Jac(φ2◦φ1◦φ0)(s, θ) =g1◦φ0(s, θ)Jacφ0(s, θ) = 2mTs.

3. for every(s, θ)∈Q^{2},

|∂sφ_{0}(s, θ)| ≤C,
whereConly depends onν/µ.

Proof . For every(s, θ)∈Q^{2}, we denote byv(s, θ)the unique element of[0,1]such that
(3.15)

Z v(s,θ) 0

g1(s^{0}, θ)ds^{0}=mTs^{2}.

The mapvis well-defined sinceg1(s, θ)>0for everys∈(0,1],θ∈[0,1]and also because Z 1

0

g1(s^{0}, θ)ds^{0}=mT.

This is exactly the reason why we constructedφ1 in Lemma 3.3. Moreover, v(0, θ) = 0
andv(1, θ) = 1. By the implicit function theorem,v isC^{1} on(0,1]×[0,1]and satisfies
g1(v(s, θ), θ)∂sv(s, θ) = 2mTs; that is,

(3.16) v(s, θ) det (γ, γ^{0})(θ+ζ◦v(s, θ)w(θ))(1 +ζ◦v(s, θ)w^{0}(θ))∂sv(s, θ) = 2mTs.

In particular, for every(s, θ)∈Q^{2},∂_{s}v(s, θ)>0.

We deduce from Lemma 3.3 that for every(s, θ)∈Q^{2},
mσ

µ^{2}

2νs≤g1(s, θ)≤ 10mσν^{2}
µ s.

Integrating with respect tosthose inequalities between0andv(s, θ), using (3.15), and the
fact thatm_{σ}µ≤m_{T} ≤m_{σ}ν, we get

(3.17) µ

√5νs≤v(s, θ)≤2ν µs.

From (3.16) and (3.14) ,

(3.18) 0< v(s, θ)∂sv(s, θ)≤4ν^{2}
µ^{2}s.

In view of (3.17), this also implies that

(3.19) 0< ∂_{s}v(s, θ)≤4√

5ν^{3}
µ^{3}.
A similar argument proves that∂θv∈L^{∞}(Q^{2}). We now define

φ0(s, θ) = (v(s, θ), θ).

Thenφ_{0}is a homeomorphism fromQ^{2}ontoQ^{2}which isC^{1}onQ^{2}. Moreover,φ_{0}(0, θ) =
(0, θ),φ_{0}(1, θ) = (1, θ)and Jacφ_{0}(s, θ) =∂_{s}v(s, θ). By differentiation of (3.15), we get

2m_{T}s=g_{1}(v(s, θ), θ)∂_{s}v(s, θ) =g_{1}◦φ_{0}(s, θ)Jacφ_{0}(s, θ).

This completes the proof of the lemma.

Gathering the previous results we can finally conclude the proof.

Proof of Proposition 3.2.

Lemmas 3.3, 3.4 clearly show that, as announced, the map
Φ =φ_{2}◦φ_{1}◦φ_{0},

satisfies all the assumptions of Proposition 3.1. The structure is summarized in Figure 3.1:

• The side{s= 0}ofQ^{2}(in red solid on the figure) is pointwise preserved byφ0and
φ1and mapped to the vertex ofT(also in red) byφ2.

• The side{s= 1}ofQ^{2}(in blue dashdotted on the figure) is pointwise preserved by
φ0andφ1and mapped toσbyφ2.

• The horizontal segments{θ = cte} ofQ^{2} (in magenta dashed) are preserved as a
whole byφ_{0}then deformed byφ_{1}andφ_{2}.

• The vertical segments{s = cte}(in green dotted) are preserved as a whole byφ_{1}
and deformed byφ^{−1}_{0} andφ_{2}.

Q^{2} Q^{2} Q^{2} T

Φ

φ0 φ1 φ_{2}

FIG. 3.1.Construction of the diffeomorphismΦ

Moreover, by definition, we have (using the same notation as in the proofs of the above two lemmas)

Φ(s, θ) =v(s, θ)γ(θ+ζ◦v(s, θ)w(θ)).

Hence,

∂_{s}Φ(s, θ) = (∂_{s}v(s, θ))γ(θ+ζ◦v(s, θ)w(θ))

+v(s, θ)∂sv(s, θ)ζ^{0}(v(s, θ))w(θ)γ^{0}(θ+ζ◦v(s, θ)w(θ)).

By construction,|ζ^{0}|L^{∞} ≤10/ε= 100ν/µand|w|L^{∞} ≤1. It then follows from (3.18)
and (3.19) that

|∂_{s}Φ(s, θ)| ≤Cν^{3}

µ^{3}(diam(T) +m_{σ}).

The proof is complete.

4. A Poincaré inequality. In this section, we derive a Poincaré inequality related to Theorem 2.1.

THEOREM 4.1. We consider the same assumption(2.2)as in Theorem 2.1. Then for
everyp∈[1,+∞[and everyu∈W^{1,p}(T),

1 mT

Z

T

u− 1 mσ

Z

σ

u

p

≤C(diam(T) +mσ)^{p} 1
mT

Z

T

|Du|^{p},

whereConly depends onpand on the ratio ^{ν}_{µ}.

This result is a consequence of the inequality proved in the previous section and of the following lemma whose proof is postponed at the end of the section.

LEMMA4.2.Under the assumption(2.2), for everyu∈W^{1,p}(T), we have
1

m^{2}_{T}
Z

T

Z

T

|u(x)−u(x^{0})|^{p}dx dx^{0}≤C(diam(T) +m_{σ})^{p} 1
mT

Z

T

|Du|^{p},
whereConly depends onpand on the ratio ^{ν}_{µ}.

Proof of Theorem 4.1. By the triangle inequality, 1

mT

Z

T

u− 1 mσ

Z

σ

u

p

≤C_{p} 1
mT

Z

T

u− 1 mT

Z

T

u

p

+C_{p}

1 mT

Z

T

u− 1 mσ

Z

σ

u

p

. One can estimate the second term with Theorem 2.1:

1 mT

Z

T

u− 1 mσ

Z

σ

u

p

≤C^{p}(mσ+ diam(T))^{p} 1
mT

Z

T

|Du|^{p},

whereConly depends on the ratio^{ν}_{µ}. By Jensen’s inequality, the first term is not larger than
the quantity

1
m^{2}_{T}

Z

T

Z

T

|u(x)−u(x^{0})|^{p}dx dx^{0},

which is, in turn, estimated by using Lemma 4.2.

It remains to prove the lemma.

Proof of Lemma 4.2. Let us introduce the reference unit triangleT_{0}defined by
T0={(a, b)∈]0,1[^{2}, b < a}.

On this domain, the following inequality classicaly holds (4.1)

Z

T0

Z

T0

|v(y)−v(y^{0})|^{p}dy dy^{0} ≤Cp

Z

T0

|Dv(y)|^{p}dy, ∀v∈W^{1,p}(T0),
with a value ofC_{p}>0depending only onp.

We introduce (see Figure 4.1) the following diffeomorphism fromT_{0}ontoT
Ψ : (a, b)∈T07→aγ(b/a)∈T.

T_{0} T

Ψ

FIG. 4.1.The diffeomorphismΨ

We proceed now with the estimate of the derivatives ofΨ. An immediate computation shows that

DΨ(a, b) = (γ(b/a)−(b/a)γ^{0}(b/a), γ^{0}(b/a)),

so that we get

JacΨ(a, b) = det (γ(b/a), γ^{0}(b/a)),
and thus, by (2.4) (which is a consequence of (2.2)), we deduce that
(4.2) 0<2µm_{σ} ≤JacΨ(a, b)≤2νm_{σ}, ∀(a, b)∈T_{0},
and

(4.3) kDΨk∞≤(kγk∞+kγ^{0}k∞)≤(diam(T) +mσ).

For anyu∈C^{1}( ¯T)we setv=u◦Ψ∈W^{1,p}(T0)and we useΨas a change of variables
Z

T

Z

T

|u(x)−u(x^{0})|^{p}dx dx^{0}=
Z

T0

Z

T0

|v(y)−v(y^{0})|^{p}JacΨ(y)JacΨ(y^{0})dy dy^{0}

≤ kJacΨk^{2}_{∞}
Z

T_{0}

Z

T_{0}

|v(y)−v(y^{0})|^{p}dy dy^{0}.
Then by (4.1) and the change of variablesx= Ψ(y)again, we get

Z

T

Z

T

|u(x)−u(x^{0})|^{p}dx dx^{0}≤CpkJacΨk^{2}_{∞}
Z

T0

|Dv(y)|^{p}dy

≤C_{p} kJacΨk^{2}_{∞}
inf_{T}_{0}JacΨ

Z

T_{0}

|Dv(y)|^{p}JacΨ(y)dy

=Cp

kJacΨk^{2}_{∞}
infT_{0}JacΨ

Z

T

|(Dv)(Ψ^{−1}(x))|^{p}dx

=Cp

kJacΨk^{2}_{∞}
inf_{T}_{0}JacΨ

Z

T

|Du(x)|^{p}kDΨ(Ψ^{−1}(x))k^{p}dx

≤C_{p}kJacΨk^{2}_{∞}kDΨk^{p}_{∞}
infT0JacΨ

Z

T

|Du(x)|^{p}dx.

By using the previous estimates (4.2) and (4.3) we conclude that Z

T

Z

T

|u(x)−u(x^{0})|^{p}dx dx^{0}≤C_{p}2m_{σ}ν^{2}

µ (diam(T) +m_{σ})^{p}
Z

T

|Du(x)|^{p}dx

≤C_{p}2m_{σ}ν^{2}

µ^{2} µ(diam(T) +m_{σ})^{p}
Z

T

|Du(x)|^{p}dx.

Sinceµmσ≤mT, we finally obtain Z

T

Z

T

|u(x)−u(x^{0})|^{p}dx dx^{0}≤Cm_{T}(diam(T) +m_{σ})^{p}
Z

T

|Du(x)|^{p}dx

for aCdepending only onpandν/µ. Dividing this inequality bym^{2}_{T} gives the claim.

5. The higher dimensional case. Letn≥2. Letγ:Q^{n−1}→R^{n}be aC^{1}map on the
closure of the unit cubeQ^{n−1}= (0,1)^{n−1}such thatγis one-to-one and|∂1γ∧· · ·∧∂_{n−1}γ|>

0onQ^{n−1}. We denote byσ=γ(Q^{n−1})the corresponding hypersurface and byT the set:

T ={sγ(θ) :s∈(0,1), θ∈Q^{n−1}}.

We assume that for everyθ∈Q^{n−1},

{sγ(θ), s≥0} ∩σ={γ(θ)}.

THEOREM5.1.We assume that there existµ, ν >0such that for everyθ∈Q^{n−1},
(5.1) µ≤ det (γ, ∂1γ, . . . , ∂_{n−1}γ)(θ)

n|∂1γ∧ · · · ∧∂_{n−1}γ|(θ) ≤ν.

Then for everyp∈[1,+∞[and everyu∈W^{1,p}(T),

1 mσ

Z

σ

u− 1 mT

Z

T

u

p

≤C^{p}(kDγkL^{∞}+ diam(T))^{p} 1
mT

Z

T

|Du|^{p},
whereConly depends on the ratio_{µ}^{ν}.

REMARK5.1.Observe that the quantity in(5.1)is invariant with respect to the parametriza-
tion ofσ. More precisely, let ψ : Q^{n−1} → Q^{n−1} be a C^{1} map such that |Jacψ| > 0
everywhere andeγ=γ◦ψ. Then

∂1eγ∧ · · · ∧∂_{n−1}eγ= ((∂1γ∧ · · · ∧∂_{n−1}γ)◦ψ)Jacψ.

This implies

det (eγ, ∂_{1}γ, . . . , ∂e _{n−1}eγ) =heγ, ∂_{1}eγ∧ · · · ∧∂_{n−1}eγi

=heγ,((∂1γ∧ · · · ∧∂_{n−1}γ)◦ψ)Jacψi

= ( det (γ, ∂_{1}γ, . . . , ∂_{n−1}γ)◦ψ)Jacψ.

Hence,

det (eγ, ∂1eγ, . . . , ∂_{n−1}eγ)

|∂_{1}eγ∧ · · · ∧∂_{n−1}eγ| = det (γ, ∂1γ, . . . , ∂_{n−1}γ)◦ψ

|∂_{1}γ∧ · · · ∧∂_{n−1}γ| ◦ψ .

In particular, whenψis aC^{1}diffeomorphism,γsatisfies(5.1)if and only ifγesatisfies(5.1).

In view of Lemma A.1 given in the appendix, one can assume without loss of generality that

(5.2) |∂1γ∧ · · · ∧∂n−1γ|(θ) =mσ, ∀θ∈Q^{n−1}.

Exactly as in the2dimensional case, the proof of Theorem 5.1 is a consequence of the following two propositions.

PROPOSITION 5.2. Assume that there exists a Lipschitz continuous mapΦ : [0,1]×
Q^{n−1}→T such that

1. Φis aC^{1}diffeomorphism from(0,1)×Q^{n−1}ontoT,
2. Φ(0, θ) = (0,0),

3. Φ(1, θ) =γ(θ),

4. JacΦ(s, θ) =nm_{T}s^{n−1}.
Then for everyu∈W^{1,1}(T), we have

1 mσ

Z

σ

u− 1 mT

Z

T

u= 1 nmT

Z

T

Du(x, y) [s∂sΦ(s, θ)]_{(s,θ)=Φ}−1(x,y) dx dy.

The proof is very similar to the proof of Proposition 3.1 and we omit it. In particular, we observe that by (5.2), we have

Z

σ

u= Z

Q^{n−1}

u(γ(θ))|∂_{1}γ∧ · · · ∧∂_{n−1}γ|(θ)dθ=m_{σ}
Z

Q^{n−1}

u(γ(θ))dθ

=mσ

Z

Q^{n−1}

u(Φ(1, θ))dθ.

The construction of a suitableΦthen follows the lines of the two dimensional case. More precisely, the following proposition is the analogue of Proposition 3.2:

PROPOSITION5.3.There exists aΦ : [0,1]×Q^{n−1}→T satisfying the assumptions of
Proposition 5.2 and such that

|∂sΦ(s, θ)| ≤C(diam(T) +kDγkL^{∞})
whereConly depends onν/µ.

Proof of Theorem 5.1. By Jensen’s inequality, we only need to prove the casep= 1and
by a standard approximation argument, one can further assume thatγ ∈ C^{2}(Q^{n−1}). The
required inequality is then a consequence of the equality given by Proposition 5.2, and of the
construction and estimate of the mapΦgiven by Proposition 5.3.

It remains to prove Proposition 5.3. As in the 2D case, we will look forΦin the following form

Φ =φ_{2}◦φ_{1}◦φ_{0},
where we still denote byφ2the map

φ2: (s, θ)∈[0,1]×Q^{n−1}7→sγ(θ)∈T .¯

The mapsφ1andφ0are built in a similar way as in Section 3 so that we only proceed to indicate the major changes in the following two lemmas.

LEMMA 5.4. There exists aC^{1} diffeomorphismφ1 : [0,1]×Q^{n−1} → [0,1]×Q^{n−1}
such that

1. for everyθ∈Q^{n−1},φ1(0, θ) = (0, θ)andφ1(1, θ) = (1, θ),
2. for everyθ∈Q^{n−1},

Z 1 0

Jac(φ2◦φ1)(s, θ)ds=mT.
3. for everys∈[0,1]and everyθ∈Q^{n−1},

(5.3) |∂sφ1(s, θ)| ≤C , C^{0}≤Jacφ1(s, θ)≤C^{00}
whereC, C^{0}andC^{00}>0only depend on the ratioν/µ.

Proof .

It is divided into three steps. In the first one we detail an auxiliary construction that will be used in Step 2 in order to build, by induction, the diffeomorphismφ1. In the third and final step, we establish the required estimates (5.3).

Step 1: Construction of an auxiliary function

Fix1 ≤k ≤n−1. Givenθ∈Q^{n−1}, we use the notationθ^{k−1} = (θ_{1}, . . . , θ_{k−1})and
θ^{0} = (θ_{k+1}, . . . , θ_{n−1}). Assume that there exists aC^{1}functionh: [0,1]×Q^{n−1}→Rsuch
that

1. for everyθ^{0}= (θk+1, . . . , θ_{n−1})∈Q^{n−k−1},
Z

(0,1)×Q^{k}

h(s, θ^{k}, θ^{0})dsdθ^{k} =m_{T},
2. there existC <1< C^{0}only depending onν/µsuch that

Cnm_{σ}µs^{n−1}≤h(s, θ)≤C^{0}nm_{σ}νs^{n−1}, ∀(s, θ)∈[0,1]×Q^{n−1}.
We introduce

ε= Cµ
2nC^{0}ν

and a cut-off function:ζ∈C_{c}^{∞}((0,1)×Q^{k−1})such that0≤ζ≤1 +εand
Z

(0,1)×Q^{k−1}

ζ(s, θ^{k−1})dsdθ^{k−1}= 1,
Z

(0,1)×Q^{k−1}

|ζ(s, θ^{k−1})−1|dsdθ^{k−1}< ε.

We can further require that

kDζk_{L}^{∞}≤ C_{n}
ε

for some constantCnwhich only depends onn. Consider the map
G_{θ}^{0} : (a, b)7→

Z

(0,1)×Q^{k−1}

dsdθ^{k−1}

Z a+ζ(s,θ^{k−1})b
0

h(s, θ^{k}, θ^{0})dθ_{k}.

ThenG_{θ}^{0} is well-defined andC^{2}on the set{(a, b) : 0 ≤a ≤1,_{1+ε}^{−a} ≤ b ≤ ^{1−a}_{1+ε}}. As in
the proof of Lemma 3.3, there exists aC^{2}mapw: [0,1]×Q^{n−k−1}7→[−1,1]such that for
everya∈[0,1]and everyθ^{0} ∈Q^{n−k−1},Gθ^{0}(a, w(a, θ^{0})) =mTa; that is,

Z

(0,1)×Q^{k−1}

dsdθ^{k−1}

Z a+ζ(s,θ^{k−1})w(a,θ^{0})
0

h(s, θ^{k}, θ^{0})dθk =mTa.

Moreover,w(0, θ^{0}) = 0 =w(1, θ^{0})and by differentiation of the above identity with respect
toa, one gets

(5.4)

Z

(0,1)×Q^{k−1}

h s, θ^{k−1}, a+ζ(s, θ^{k−1})w(a, θ^{0}), θ^{0}
1 +ζ(s, θ^{k−1})∂aw(a, θ^{0})

ds dθ^{k−1}=mT.
As in the proof of Lemma 3.3, this leads to the following estimate:

(5.5) D≤1 +ζ(s, θ^{k−1})∂aw(a, θ^{0})≤D^{0}

for some constantsD,D^{0}>0which only depend onν/µ. We omit the details.

Step 2 : construction ofφ_{1}

Letψn+1 = id_{[0,1]×Q}_{n−1} andhn+1 = hn = Jacφ2. We construct by induction on
k=n, . . . ,0two sequences of maps

ψk+1: [0,1]×Q^{n−1}→[0,1]×Q^{n−1} , hk : [0,1]×Q^{n−1}→R
such that fork= 0, . . . , n,

1. The mapψk+1is aC^{2}diffeomorphism from[0,1]×Q^{n−1}onto[0,1]×Q^{n−1},
2. For everyθ∈Q^{n−1},ψk+1(0, θ) = (0, θ)andψk+1(1, θ) = (1, θ),

3. We have

hk= (hk+1◦ψk+1)Jacψk+1,

4. There exist two constants0< C_{k} <1< C_{k}^{0} depending only onν/µsuch that for
everys∈[0,1]and everyθ∈Q^{n−1},

(5.6) Ck≤Jacψk+1(s, θ)≤C_{k}^{0},
Cknmσµs^{n−1}≤hk(s, θ)≤C_{k}^{0}nmσνs^{n−1}.
5. We have

Z

(0,1)×Q^{k}

h_{k}(s, θ^{k}, θ^{0})dsdθ^{k}=m_{T},
whereθ^{k}= (θ1, . . . , θk)andθ^{0} = (θk+1, . . . , θ_{n−1}).

Observe that these conditions are satisfied fork = n. We assume that for some k ≥ 1,
ψn+1, ψn. . . , ψk+1, and thushn, . . . , hkare already constructed and satisfy the above prop-
erties. Letεk = _{2nC}^{C}^{k}^{µ}0

kν andζk a function which satisfies the properties of the functionζ
introduced in Step 1, withε=εk. We apply Step 1 to the functionhkwithζk andC =Ck,
C^{0} =C_{k}^{0}. This gives a functionwk : [0,1]×Q^{n−k−1} → [−1,1]satisfying the properties
enumerated in Step 1.

We then constructψ_{k}as follows

ψk(s, θ1, . . . , θ_{n−1}) = (s, θ^{k−1}, vk(s, θ), θ^{0})

= (s, θ_{1}, . . . , θ_{k−1}, v_{k}(s, θ), θ_{k+1}, . . . , θ_{n−1})
with

vk(s, θ) =θk+ζk(s, θ^{k−1})wk(θk, θ^{0}).

Since Jacψ_{k} = 1 +ζ_{k}∂_{θ}_{k}w_{k}, (5.5) implies

C_{k−1}≤Jacψk≤C_{k−1}^{0} ,
where0< Ck−1<1< C_{k−1}^{0} only depend onν/µ.

Lethk−1= (hk◦ψk)Jacψk. By (5.4), Z

(0,1)×Q^{k−1}

h_{k−1}

= Z

(0,1)×Q^{k−1}

hk(s, θ^{k−1}, θk+ζk(s, θ^{k−1})wk(θk, θ^{0}), θ^{0})Jacψk(s, θ)dsdθ^{k−1}

=mT.

As in the proof of Lemma 3.3, one can check that hk−1 andψk satisfy all the remaining
properties, even if it means changing the actual value of the constantsCk−1andC_{k−1}^{0} . This
completes the construction by induction ofh0, . . . , hnandψ1, . . . , ψn+1. We now define

φ_{1}=ψ_{n}◦ψ_{n−1}◦ · · · ◦ψ_{1}.

Thenφ1 is aC^{1}diffeomorphism from[0,1]×Q^{n−1} onto itself andφ1 coincides with the
identity on{0} ×Q^{n−1}and{1} ×Q^{n−1}. By construction,

(5.7)

h0= (h1◦ψ1)Jacψ1= (((h2◦ψ2)Jacψ2)◦ψ1)Jacψ1

= (h_{2}◦ψ_{2}◦ψ_{1})Jac(ψ_{2}◦ψ_{1}) =. . .

= (hn◦ψn◦ · · · ◦ψ1)Jac(ψn◦ · · · ◦ψ1)

= (h_{n}◦φ_{1})Jacφ_{1}

= (Jacφ2◦φ1)Jacφ1=Jac(φ2◦φ1).

In particular, we deduce that
m_{T} =

Z

(0,1)

h0(s, θ^{0})ds=
Z

(0,1)

Jac(φ2◦φ1)(s, θ^{0})ds.

Step 3: Proof of(5.3)

For everyk = 1, . . . , n−1,and every (s, θ) ∈ [0,1]×Q^{n−1}, let us introduce the
notation:

[(s, θ)]^{k}=θ^{k} = (θ_{1}, . . . , θ_{k}).

Then

ψ_{1}(s, θ) = (s,[ψ_{1}(s, θ)]^{1}, θ^{0}) = (s, θ_{1}+ζ_{1}(s)w_{1}(θ_{1}, θ^{0}), θ^{0})
withθ^{0}= (θ_{2}, . . . , θ_{n−1}), and for everyk= 1, . . . , n,

ψk◦ψk−1◦ · · · ◦ψ1(s, θ)

= s,[ψ_{k−1}◦ · · · ◦ψ_{1}(s, θ)]^{k−1}, θ_{k}+ζ_{k}(s,[ψ_{k−1}◦ · · · ◦ψ_{1}(s, θ)]^{k−1})w_{k}(θ_{k}, θ^{0}), θ^{0}
whereθ^{0}= (θ_{k+1}, . . . , θ_{n}).

Sincekw_{k}k_{L}^{∞} ≤ 1, it follows by induction onk = 1, . . . , nthat there existsA_{k} >0
which depends only onν/µsuch that

k∂_{s}(ψ_{k}◦ · · · ◦ψ_{1})k_{L}^{∞} ≤A_{k}.

In particular, k∂_{s}φ_{1}k_{L}^{∞} ≤ A_{n}. The fact thatC ≤ Jacφ_{1} ≤ C^{0}, for someC, C^{0} > 0
depending only onν/µ, follows from (5.6) and the identity

Jacφ_{1}=

n

Y

k=1

Jacψ_{k}.

The lemma is proved.

LEMMA5.5.There exists an homeomorphismφ_{0}: [0,1]×Q^{n−1}→[0,1]×Q^{n−1}which
isC^{1}on(0,1)×Q^{n−1}and such that

1. for everyθ∈(0,1),φ0(0, θ) = (0, θ)andφ0(1, θ) = (1, θ),
2. for every(s, θ)∈(0,1)×Q^{n−1},

Jac(φ2◦φ1◦φ0)(s, θ) =nm_{T}s^{n−1},

3. there exists a C^{1} mapv : (0,1)×Q^{n−1} → [0,1]such that for every (s, θ) ∈
(0,1)×Q^{n−1},φ0(s, θ) = (v(s, θ), θ)and

|v(s, θ)| ≤Cs , |∂sv(s, θ)| ≤C whereConly depends onν/µ.

The proof is essentially the same as the proof of Lemma 3.4. The only difference is that now by (5.6) fork= 0and (5.7)

Cnmσµs^{n−1}≤Jac(φ2◦φ1)≤C^{0}nmσνs^{n−1}.
The rest of the proof is the same and we omit it.

Proof of Proposition 5.3. By construction,

Φ(s, θ) =φ2◦φ1◦φ0(s, θ) =φ2(φ1(v(s, θ), θ)).

Hence,

∂sΦ(s, θ) =Dφ2(φ1(v(s, θ), θ)).∂sφ1(v(s, θ), θ).∂sv(s, θ).

By Lemma 5.5,|∂sv|,|v| ≤Cand by Lemma 5.4, we have|∂sφ1| ≤C. Hence

|∂sΦ(s, θ)| ≤C(diam(T) +kDγkL^{∞})

whereConly depends onν/µ.

6. Applications to the analysis of some finite volume methods.

6.1. Regular families of meshes of a smooth domain. LetΩbe a bounded domain of
R^{2}with aC^{2}boundary; we setΓ = ∂Ω. A finite volume meshMofΩis a finite family
of compact subsets ofΩwith non-empty interiors usually refered to as control volumes and
denoted by the letterK. This family is supposed to satisfy

K ∩˚ L˚=∅, ∀K,L ∈M,K 6=L,

Ω = [

K∈M

K.

Mext

Mint

Γ

FIG. 6.1.The non-polygonal meshMofΩand the two submeshesMintandMext

We assume thatMcan be split into two disjoint subsets (see Figure 6.1) as follows:

• The set of polygonal control volumesM_{int} that satisfy: for anyK ∈ M_{int},K is
polygonal andK ∩∂Ωcontains at most a finite number of points.

• The set of curved control volumesMextthat satisfy : for any K ∈ Mext,K is a pseudo-triangle whose curved edge is contained in the boundary of the domainΩ.

With any such curved control volumeK, we associate the (real) triangleKe which possesses the same vertices asK(see the dashed lines in Figure 6.1). Observe that Kemay not be included inΩ.

We may now define the approximate mesh to be the following set of control volumes fM= [

K∈Mint

{K} ∪ [

K∈Mext

{K}.e This is a finite volume mesh made of polygonal control volumes.

The size and the regularity of such a mesh are measured by the quantities size(M) = max

σ∈E m_{σ}, andreg_{1}(M) = max

L∈‹^{M}

diam(L)^{2}
m_{L} ,

whereEis the set of the edgesσof all the control volumes in the meshM. Usual convergence
results in the finite volume framework assume that size(M)goes to 0 and thatreg_{1}(M)
remains bounded. This means that control volumes are not allowed to become flat while the
mesh is refined.

The main objective of this section is to prove that, if one builds a mesh Mof Ω as described previously such thatsize(M)is small enough, then each boundary curved control volumesK ∈ Mext satisfies the assumptions of Theorem 2.1 with a ratioνK/µKwhich is independent ofK. In other words, on such curved elements, the inequality (2.3) holds with a constantCuniformly bounded as the mesh is refined.

PROPOSITION 6.1. LetΩbe a bounded domain of classC^{2} inR^{2}andξ_{0} >0. There
exists h_{0} > 0 depending only on Ω and ξ_{0}, such that for any finite volume mesh Mas
described above, if

(6.1) reg_{1}(M)≤ξ0andsize(M)≤h0,

then any exterior control volumeK ∈ Mext (which is a pseudo-triangle) satisfies the as- sumption(2.4)with two values ofµandνthat satisfyν/µ= 3.

Proof . The exterior control volumeKcan be written in the following form (see Figure 6.2) K={(1−s)γ(t) :s∈[0,1], t∈[0, h]},

where the opposite vertex which is supposed to be the origin (0,0)lies insideΩ, andγ :
[0, h] →R^{2}is a normal parametrization of the curved edgeσ⊂ Γ: kγ^{0}(t)k = 1for every
t∈[0, h].

We also introduce the associated real triangleKewith vertices(0,0), γ(0)andγ(h).

First, we claim that if we assume that

(6.2) h0< 1

2kγ^{00}k∞

, then we have

(6.3) m_{σ}=h≤2diam(K).e

Indeed, lett∈[0, h]. By the mean value inequality, there existsξ_{t}∈[0, h]such that
hγ^{0}(t), γ(h)−γ(0)i=hhγ^{0}(t), γ^{0}(ξ_{t})i

≥h−h^{2}kγ^{0}k∞kγ^{00}k∞=h−h^{2}kγ^{00}k∞

≥ h 2.