On the Existence of Solutions of Equilibria in Lubricated Journal Bearings

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SIAM J. MATH. ANAL. c© 2009 Society for Industrial and Applied MathematicsVol. 40, No. 6, pp. 2316–2327

ON THE EXISTENCE OF SOLUTIONS OF EQUILIBRIA INLUBRICATED JOURNAL BEARINGS∗

I. CIUPERCA† , M. JAI‡ , AND J. I. TELLO§

Abstract. In this paper we study a system of equations concerning equilibrium positions ofjournal bearings. The problem consists of two surfaces in relative motion separated by a smalldistance filled with a lubricant. The shape of the inlet surface is circular, while the other surfacehas a more general shape. Our result shows the existence of at least one equilibrium by using degreetheory.

Key words. Reynolds variational inequality, inverse problem, existence of solutions, degreetheory

AMS subject classifications. 35J20, 47H11, 49J10

DOI. 10.1137/080724228

1. Introduction. We consider in this paper a lubricated system called a journalbearing consisting of two cylinders in relative motion. An incompressible fluid, thelubricant, is introduced in the narrow space between the cylinders. An exterior forceF = (F1, F2) ∈ R

2 is applied on the inner cylinder (shaft) which turns with a givenvelocity ω.

The wedge between the two cylinders is assumed to satisfy the thin-film hypothe-sis, so that the pressure (assumed time-independent) does not depend on the normalcoordinate to the bodies and obeys the Reynolds equation.

In order to introduce the Reynolds equation we need to describe the geometry andthe dynamic of the system. We suppose that the interior cylinder has a circular formof constant radius (assumed 1) which rotates with known velocity. The transversalaxis of the shaft is assumed to have only two degrees of freedom in the transversalplane, i.e., parallel to the exterior cylinder (bush) which is fixed and not necessarilyof constant radius.

Let us consider (O, y1, y2) a reference system in the transversal plane to thecylinders, and suppose that the distance between O and the surface of the bushis larger than the radius of the shaft (see Figure 1.1). We also assume that therepresentation in polar coordinates (r, θ) of the bush is given by

(1.1) r = 1 + δρ(θ),

where ρ : [0, 2π] → [1,+∞[ is a known function and δ > 0 (the clearance) is a smallparameter, representing the distance between the two cylinders when O and the centerof the shaft coincide.

∗Received by the editors May 14, 2008; accepted for publication (in revised form) November 12,2008; published electronically February 20, 2009.

http://www.siam.org/journals/sima/40-6/72422.html†Universite de Lyon, Universite Lyon 1, CNRS, UMR 5208, Institut Camille Jordan, Bat. Bracon-

nier, 43, blvd du 11 novembre 1918, F-69622 Villeurbanne Cedex, France ([email protected]).

‡ICJ CNRS-UMR 5208, INSA de Lyon, Bat. Leonard de Vinci, 20 A. A. Einstein, 69621 Villeur-banne Cedex, France ([email protected]).

§Matematica Aplicada, E.U.I. Informatica, Universidad Politecnica de Madrid, 28031 Madrid,Spain ([email protected]). This author was partially supported by project MTM2005-03463 of DG-ISGPI of Spain and projects of CAM: CCG07-UPM/000-3199 at UPM and CCG07-UCM/ESP-2787at UCM.

2316

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ON REYNOLDS EQUATIONS 2317

Fig. 1.1. Scheme of the journal bearing.

Remark 1.1. The particular case ρ ≡ 1 corresponds to a circular bush with radius1 + δ and O the center of the bush.

Let us now denote by Os the center of the shaft. The position of Os is given incartesian coordinates by (δη1, δη2) and in polar coordinates by (δη, α), that is,

(1.2)η1 = η cosα,

η2 = η sinα.

It is well known (see, for instance, [5]) that the distance between the two cylinders isgiven by δh(θ, η, α) +O(δ2) with

(1.3) h(θ, η, α) = ρ(θ) − η cos(θ − α) = ρ(θ) − η1 cos θ − η2 sin θ.

The formulation of the problem is complete when the distance between the sur-faces is of order δ (i.e., h ∈ O(1)) and second order terms are neglected. Admissibleforces in that case are of order 1

δ2 or smaller. If the distance becomes smaller (i.e.,h� 1) or forces are large (larger than O( 1

δ2 )), second order terms cannot be neglectedand the formulation loses its physical meaning.

Now, the problem will be posed in a fixed domain Ω = ]0, 2π[× ]0, 1[ whichparametrizes the space between shaft and bush. In fact the gap is approximatedby the following domain given in cylindrical coordinates by (r, θ, x) with respect toOs:

1 ≤ r ≤ 1 + δh(θ, η, α), θ ∈ [0, 2π[, x ∈ [0, 1],

where Os = (δη cos (α), δη sin (α)).Since the wedge between the two cylinders satisfies the thin-film hypothesis, the

pressure of the lubricant fluid (assumed time-independent) does not depend on the

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2318 I. CIUPERCA, M. JAI, AND J. I. TELLO

normal coordinate to the bodies and obeys the Reynolds equation (see [9]). We alsoconsider that there is an alimentation region along the circles {x = 0} and {x = 1},respectively, where the pressure in the fluid equals the atmospheric pressure supposedto be 0 by translation. Then the pressure p : (θ, x) ∈ Ω → R satisfies the followingproblem written in nondimensional form:

(1.4)

⎧⎪⎨⎪⎩

∇ · (h3∇p) = ∂h∂θ on Ω,

p is 2π-periodic in θ,p = 0 on ]0, 2π[× {0} ∪ ]0, 2π[ × {1}.

In general the solution of (1.4) is not always nonnegative and we must replace (1.4)by the corresponding variational inequality.

In this work we are interested in an equilibrium problem which entails findingthe position (η1, η2) of the shaft such that the hydrodynamic force (load) created bythe pressure film equilibrates the exterior force F . Thus the problem is formulated asfollows:

Find p ∈ K, (η1, η2) ∈ A such that

∫Ω

h3∇p · ∇(ϕ− p) ≥∫

Ω

h∂

∂θ(ϕ− p) ∀ϕ ∈ K,(1.5)

∫Ω

p cos θdθdx = F1,(1.6)

∫Ω

p sin θdθdx = F2,(1.7)

where h is given in (1.3),

K ={ϕ ∈ H1

0 (Ω) : ϕ ≥ 0},

and A ⊂ R2. The set of admissible positions of the shaft (to be defined later) is such

that h(θ, η, α) > 0 ∀θ ∈ [0, 2π].As far as we know, very few works can be found in the literature concerning exis-

tence of equilibrium in the lubricated devices in spite of a larger number of referencesto numerical simulations; see, for instance, [3] or [4].

Some results exist in the case of sliders, that is, mechanisms consisting of analmost plane surface sliding above a horizontal plane surface.

Exact solutions for the case in which the upper surface is an inclined plane of angleθ and infinite width have been known for a long time, since the problem becomes one-dimensional and is easily integrated [7].

For more general shapes of the upper surfaces an existence result is obtained in [1]for the equation case and in [2] for the variational inequality in the one-dimensionalcase.

The content of the paper is as follows: In section 2 we give the main result. Insection 3 we recall some elements of the degree theory which will be used. In section 4some preliminary results are given, and section 5 is devoted to the proof of the mainresult.

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ON REYNOLDS EQUATIONS 2319

2. Main results and assumptions. For the rest of the paper we assume that

ρ ∈ C3(R), ρ, ρ′, and ρ′′ are 2π-periodic,(2.1)

ρ′′(θ) + ρ(θ) > 0 ∀ θ ∈ R,(2.2)

min0≤θ≤2π

ρ(θ) = 1.(2.3)

We denote the following:

(2.4)

ρM = max0≤θ≤2π

ρ(θ),

m = min0≤θ≤2π

(ρ′′(θ) + ρ(θ)) > 0,

M = max0≤θ≤2π

(ρ′′(θ) + ρ(θ)) > 0.

Remark 2.1. All these assumptions are clearly satisfied in the particular case ofa circular bush, corresponding to ρ ≡ 1.

Remark 2.2. Assumption (2.2) is the most restrictive of the assumptions. It isintroduced for technical reasons and guarantees that in the limit case the set whereh = 0 is a single line.

Let us introduce the function a(α) : R → R+ given by

(2.5) a(α) := minα−π

2 <θ<α+π2

{ρ(θ)

cos(θ − α)

}.

The fact that

limθ→α±π

2

ρ(θ)cos(θ − α)

= ∞

and the continuity and boundedness of ρ guarantee the existence of at least oneminimum.

Now we define the set A by

A ={

(η1, η2) ∈ R2 : 0 ≤ η < a(α)

},

where (η, α) are given by (1.2).It is clear that h(θ) > 0 ∀θ ∈ [0, 2π] if and only if (η1, η2) ∈ A.For any fixed (η1, η2) ∈ A, problem (1.5) has been studied by several authors. The

existence and uniqueness of solutions can be obtained by using the direct methodsin the calculus of variations along with strict convexity of the associated functional.Then (1.5) admits a unique classical solution p ∈ K; see, for instance, Kinderlehrerand Stampacchia [6].

Thus problem (1.5)–(1.7) is equivalent to the following problem:

(2.6)

{Find (η1, η2) ∈ A such that

G(η1, η2) = (0, 0),

where G = (G1, G2) : A→ R2 is given by

G1(η1, η2) =∫

Ω

p cos θdθdx − F1,(2.7)

G2(η1, η2) =∫

Ω

p sin θdθdx − F2,(2.8)

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2320 I. CIUPERCA, M. JAI, AND J. I. TELLO

where p (depending on η1, η2) is the unique solution of (1.5).The main result of the paper is presented in the following theorem.Theorem 2.1. Under assumptions (2.1)–(2.3) and for any F ∈ R

2 there existsat least one solution (η1, η2) ∈ A of (2.6).

3. Known results on degree theory. In order to prove Theorem 2.1 we usethe topological degree theory, which is rapidly recalled in the following.

The topological degree for continuous mappings between n-dimensional Euclideanspaces was first introduced by L. E. J. Brouwer in 1912. We first introduce thedefinition of degree for C1 maps.

Let S be a bounded open subset of Rn and a C1(S) function f : S → R

n. Lety0 ∈ R

n such that y0 ∈ f(∂S), and suppose that f ∈ C1(S) and that Df(x) isinvertible for all x ∈ f−1(y0). Then f(x) = y0 has either no solutions in S or a finitenumber r of solutions, say x1, x2, . . . , xr and det(Df(xi)) = 0 for i = 1, 2, . . . , r.

Definition 3.1. We define the degree of f in S at y0 as follows:If r = 0, then d(f, S, y0) := 0; else

d(f, S, y0) :=r∑

i=1

sign(det(Df(xi))).

Remark 3.1. In the particular case where f is a linear function, i.e., f(x) = Ax,where A is an invertible matrix, the general formula for calculating the degree of linearfunctions is the following:

deg(Ax, S, 0) = sgn(detA) if 0 is contained in S.

The definition of degree can be extended to continuous maps; see [8, ExtensionLemma, p. 60]. For the reader’s convenience we state the following result (see, forinstance, [8, Corollary 4 to Theorem 1.12, p. 81]) adapted to the finite-dimensionalcase.

Theorem 3.1. Let S be a bounded open set in Rn. Let f0 and f1 be two con-

tinuous functions from S to Rn. We assume moreover that S is a star domain with

respect to the point y0 ∈ S and that

[f0(x) − y0] · [f1(x) − y0] > 0 for any x ∈ ∂S.

Then

d(f0, S, y0) = d(f1, S, y0).

Remark 3.2. It is clear that if d(f, S, y0) = 0, then there exists at least a solutionx ∈ S of the equation f(x) = y0.

4. Preliminary results.Lemma 4.1. Let a(α) be the function defined in (2.5). Then

1 = min0≤θ≤2π

ρ ≤ a(α) ≤ max0≤θ≤2π

ρ = ρM .

Proof. Since

min0≤θ≤2π

ρ ≤ ρ ≤ max0≤θ≤2π

ρ

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ON REYNOLDS EQUATIONS 2321

we have that

min0≤θ≤2π

{ρ} minα−π

2 <θ<α+π2

{1

cos(θ − α)

}≤ a(α)

and

a(α) ≤ max0≤θ≤2π

{ρ} minα−π

2 <θ<α+π2

{1

cos(θ − α)

}.

The fact that

minα−π

2 <θ<α+π2

{1

cos(θ − α)

}= 1

implies

min{ρ} ≤ a(α) ≤ max{ρ},and the proof is complete.

Lemma 4.2. There exists s with 0 < s < π2 such that

|θα − α| < s

for any α ∈ R and for any θα ∈ ]α− π2 , α+ π

2 [ satisfying

ρ(θα)cos(θα − α)

= a(α).

Proof. From Lemma 4.1 we deduce

cos(θα − α) ≥ 1ρM

,

which implies

(4.1) |θα − α| ≤ arccos1ρM

<π

2.

Now taking s = 12 (π

2 + arccos 1ρM

) we obtain the result.Lemma 4.3. Under assumptions (2.1)–(2.3), for any α ∈ R, the set of θ ∈

]α− π2 , α+ π

2 [, which minimizes the function θ �→ ρ(θ)cos(θ−α) , is a single point which we

denote by θα satisfying

ρ(θα)cos(θα − α)

= a(α)

and

ρ′(θα) cos(θα − α) + ρ(θα) sin(θα − α) = 0.

Moreover the function α �→ θα belongs to C1(R).Proof. Let α be a fixed value of R and consider the critical points of the function

θ ∈ ]α− π2 , α+ π

2 [ �→ ρ(θ)cos(θ−α) , which satisfies

(4.2)(

ρ(θ)cos(θ − α)

)′=

ρ′(θ)cos(θ − α)

+ρ(θ) sin(θ − α)

cos(θ − α)2= 0.

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2322 I. CIUPERCA, M. JAI, AND J. I. TELLO

Consider

f(θ, α) := ρ′ cos(θ − α) + ρ sin(θ − α).

Notice that if θ is a minimum of

ρ(θ)cos(θ − α)

,

then f(θ, α) = 0. We now have that

∂f

∂θ= (ρ′′ + ρ) cos(θ − α),

which is positive in |θ − α| < π2 (by hypothesis (2.2)). The assertion follows using

also the implicit function theorem on the open set (θ, α) ∈ ]−s, s[×R with s given inLemma 4.2.

We now introduce the function h0 : R2 → R defined by

h0(θ, α) = h(θ, a(α), α) = ρ(θ) − a(α) cos(θ − α).

Remark 4.1. We have from Lemma 4.3

h0(θα, α) =∂h0

∂θ(θα, α) = 0,(4.3)

∂2h0

∂θ2(θα, α) = ρ′′(θα) + ρ(θα).(4.4)

Lemma 4.4. Let us denote for any α ∈ [0, 2π] and ε > 0 small enough

Iα,ε := [θα − 2√ε, θα −√

ε].

Then for any α ∈ [0, 2π] and θ ∈ Iα,ε we have(i) h(θ, a(α) − ε, α) ≤ (2M + 2)ε,(ii) −∂h

∂θ (θ, a(α) − ε, α) ≥ m2

√ε,

with m and M defined in (2.4).Proof.(i) We have

h(θ, a(α) − ε, α) = h0(θ, α) + ε cos(θ − α).

From Taylor development of h0 at θ = θα, using (4.3) and (4.4) we have (i).(ii) We have

−∂h∂θ

(θ, a(α) − ε, α) = −∂h0

∂θ(θ, α) + ε sin(θ − α).

Using the Taylor development of ∂h0∂θ at θ = θα we obtain

−∂h∂θ

(θ, a(α) − ε, α) =∂2h0

∂θ2(θ, α)(θα − θ) + ε sin(θ − α)

with θ ∈ Iα,ε.

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ON REYNOLDS EQUATIONS 2323

From (4.4), the uniform continuity of ∂2h0∂θ2 , and (2.4) we deduce

∂2h0

∂θ2(θ, α) ≥ 2

3m.

Since θα − θ ≥ √ε, we obtain for ε small enough

−∂h∂θ

(θ, a(α) − ε, α) ≥ 23m√ε− ε,

which ends the proof.Lemma 4.5. Let ε be small enough and η := a(α) − ε. There exists a constant

c > 0 independent of ε and α such that

inf0≤α≤2π

∫Ω

h3(θ, η, α)|∇p|2dθdx > cε−12 as ε→ 0.

Proof. We take in (1.5) φ := p+ ϕ with ϕ ∈ K arbitrary. Then∫Ω

h∂ϕ

∂θdθdx ≤

∫Ω

h3∇p∇ϕdθdx.

By the Cauchy–Schwarz inequality we have

∫Ω

h∂ϕ

∂θdθdx ≤

∣∣∣∣∫

Ω

h3|∇p|2dθdx∣∣∣∣12

∣∣∣∣∫

Ω

h3|∇ϕ|2dθdx∣∣∣∣12

and we deduce

(4.5)∣∣∣∣∫

Ω

h3|∇p|2dθdx∣∣∣∣12

≥ supϕ∈K,ϕ �=0

− ∫Ω ϕ

∂h∂θ dθdx∣∣∫

Ω h3|∇ϕ|2dθdx∣∣ 1

2.

Let m be given by (2.4) and ψ ∈ C2(R) such that(i) supp(ψ) ⊂ [−2,−1];(ii) ψ ≥ 0;(iii)

∫Rψ > 0.

As a consequence of (i)–(iii) we have

(4.6)

∫Rψ∣∣∫

R|ψ′|2∣∣ 1

2= c1 > 0.

Let

ϕε(θ, x) := ψ

(θ − θα√

ε

)x(1 − x)

with θα defined in Lemma 4.3.It is clear that supp(ϕε) = Iα,ε × [0, 1] with Iα,ε as in Lemma 4.4. Using

Lemma 4.4 (ii) we deduce

(4.7)

−∫

Ω

∂h

∂θϕεdθdx ≥ min

θ∈Iα,ε

(−∂h∂θ

) ∫Ω

ϕεdθdx

≥ m

2ε

∫R

ψdy

∫ 1

0

x(1 − x)dx =m

12ε

∫R

ψdy.

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2324 I. CIUPERCA, M. JAI, AND J. I. TELLO

On the other hand, using Lemma 4.4 (i) we have

∣∣∣∣∫

Ω

h3|∇ϕε|2dθdx∣∣∣∣12

≤ maxIα,ε

{h3} 12

∣∣∣∣∫

Ω

|∇ϕε|2dθdx∣∣∣∣12

≤ ((2M + 2)ε)3/2

∣∣∣∣∫

Ω

|∇ϕε|2dθdx∣∣∣∣12

.

A simple calculation gives ∫Ω

|∇ϕε|2dθdx ≤ c3ε−1/2.

We then have

(4.8)∣∣∣∣∫

Ω

h3|∇ϕε|2dθdx∣∣∣∣12

≤ c4ε5/4,

which, also using (4.7), gives us

(4.9)− ∫

Ω ϕε∂h∂θ dθdx∣∣∫

Ωh3|∇ϕε|2dθdx

∣∣1/2≥ c5ε

− 14 .

Finally from (4.5) and (4.9) we have the result.We now introduce for any ε small the function h : R

2 → R defined by

(4.10) h(θ, α) := h(θα, a(α) − ε, α) + [ρ′′(θα) + ρ(θα)](1 − cos(θ − θα)).

Notice that

(4.11)

⎧⎪⎪⎨⎪⎪⎩h(θα, a(α) − ε, α) = h(θα, α) = ε cos(θα − α),∂h∂θ (θα, a(α) − ε, α) − ∂h

∂θ (θα, α) = ∂h∂θ (θα, a(α) − ε, α) = −ε sin(θα − α),

∂2h∂θ2 (θα, a(α) − ε, α) − ∂2h

∂θ2 (θα, α) = −ε cos(θα − α).

Lemma 4.6. There exists a constant c independent of ε and α such that

∫ 2π

0

(h(θ, a(α) − ε, α) − h(θ, α)

)2

h3(θ, a(α) − ε, α)dθ ≤ c.

Proof. By the Taylor polynomial of h− h and h at θα and (4.11) we have

(4.12) |h(θ, a(α) − ε, α) − h(θ, α)| ≤ ε|θ − θα| + ε|θ − θα|2 + c|θ − θα|3

and

(4.13)

h(θ, a(α) − ε, α) = ε cos(θα − α) − ε sin(θα − α)(θ − θα)

+12

(ρ′′(θα) + ρ(θα) − ε cos(θα − α)

)(θ − θα)2

+16∂3h

∂θ3(θ, a(α) − ε, α)(θ − θα)3

with θ ∈ R.

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ON REYNOLDS EQUATIONS 2325

From Lemma 4.2, (4.13), and (2.2) there exists m1,m2, and m3 positive andindependent of ε and α such that

(4.14) h(θ, a(α) − ε, α) ≥ m1ε+m2(θ − θα)2

∀ θ such that |θ − θα| ≤ m3.We have from (4.12)

∫ 2π

0

(h− h)2

h3dθ ≤ 2(ε2I1 + ε2I2 + c2I3)

with Ik =∫ 2π

0(θ−θα)2k

h3 dθ, k = 1, 2, 3.Now from (4.14) it is clear that I3 is bounded uniformly in ε and α.We will prove the uniform estimate for ε2I1. For ε2I2 the proof is similar.We have

I1 = I11 + I2

1 + I31 ,

where I11 , I

21 , I

31 are the subintegrals, respectively, in the intervals |θ−θα| ≥ m3, ε

1/3 ≤|θ − θα| ≤ m3, and 0 ≤ |θ − θα| ≤ ε1/3.

It is clear that ε2I11 is bounded uniformly in ε and α since h is lower bounded by

a positive constant on the interval |θ − θα| ≥ m3.From (4.14) we have

I21 ≤ m−3

2

∫ε1/3≤|θ−θα|≤m3

dθ

|θ − θα|4 ≤ cε−4/3

and

I31 ≤ m−3

1 ε−3

∫0≤|θ−θα|≤ε1/3

|θ − θα|2dθ ≤ m−31 ε−2,

which proves the lemma.

5. Proof of Theorem 2.1. We apply the degree theory recalled in section 3 tothe function G.

Since G is not defined on A we introduce for any ε > 0 small enough the domain

Aε ={(η1, η2) : 0 ≤ η ≤ a(α) − ε

}.

Let us now introduce the vector field

W : (η1, η2) ∈ A→ (η sin θα,−η cos θα) ∈ R2

with (η, α) defined in (1.2) and θα as in Lemma 4.3.We observe that

W · (−η2, η1) = −η2 cos(θα − α).

From Lemmas 4.2 and 4.3 we have that

W · (−η2, η1) < 0 ∀(η1, η2) ∈ A.

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2326 I. CIUPERCA, M. JAI, AND J. I. TELLO

From Theorem 3.1 we deduce that

deg(W,Aε, 0) = deg((−η2, η1), Aε, 0) = 1.

Notice that deg(f, S, 0) = (−1)ndeg(−f, S, 0), where n is the dimension of the domain(for the application here, n = 2, so the sign does not change).

It suffices now to prove the inequality

(5.1) G(η1, η2) ·W (η1, η2) > 0 ∀(η1, η2) ∈ ∂Aε,

and the proof is finished using again Theorem 3.1.We have, ∀(η1, η2) ∈ ∂Aε, η = a(α) − ε. Then

(5.2) G(η1, η2) ·W (η1, η2) = −η∫

Ω

p sin(θ − θα)dθdx + η(−F1 sin θα + F2 cos θα).

Now taking ϕ = 0 and ϕ = 2p, respectively, in (1.5) we obtain

(5.3)∫

Ω

h3|∇p|2dθdx = −∫

Ω

∂h

∂θpdθdx.

Notice that

(5.4)∫

Ω

∂h

∂θpdθdx =

∫Ω

∂

∂θ(h− h)pdθdx +

∫Ω

∂h

∂θpdθdx

with h given by (4.10).We now have

(5.5)

∫Ω

∂

∂θ(h− h)pdθdx = −

∫Ω

(h− h)∂p

∂θdθdx

≥ −∫

Ω

|h− h|h3/2

h3/2|∇p|dθdx

≥ −12

∫Ω

|h− h|2h3

dθdx− 12

∫Ω

h3|∇p|2dθdx.

Since

∂h

∂θ= (ρ′′(θα) + ρ(θα)) sin(θ − θα)

we obtain from (5.3) and (5.5)

∫Ω

h3|∇p|2dθdx ≤ 12

∫Ω

|h− h|2h3

+12

∫Ω

h3|∇p|2dθdx

− (ρ′′(θα) + ρ(θα))∫

Ω

sin(θ − θα)pdθdx,

which implies, using also hypotheses (2.2),

−∫

Ω

sin(θ − θα)pdθdx ≥∫Ω h

3|∇p|2dθdx− ∫Ω

|h−h|2h3 dθdx

2(ρ′′(θα) + ρ(θα)).

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ON REYNOLDS EQUATIONS 2327

Since ρ is in C2 and from Lemmas 4.5 and 4.6, we obtain for ε small enough

−∫

Ω

sin(θ − θα)p ≥ c(ε−1/2 − 1)

with c > 0 a constant independent of ε and α.Since η = a(α) − ε and thanks to (5.2) we deduce

G(η1, η2) ·W (η1, η2) ≥ η(c(ε−1/2 − 1) − ‖F‖

).

Taking ε small enough, we prove the theorem.Remark 5.1. In the same manner we can prove the existence of at least one

equilibrium solution for some other similar problems in this context. For instance,the case of Dirichlet boundary conditions on every boundary of Ω, or the case of aone-dimensional domain with Dirichlet boundary conditions.

Acknowledgments. The third author thanks INSA-Lyon and Institut CamilleJordan at University of Lyon for their hospitality.

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