Existence of equilibria in articulated bearings in presence of cavity

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1 J. Math. Anal. Appl ) Existence of equilibria in articulate bearings in presence of cavity G. Buscaglia a, I. Ciuperca b, I. Hafii c,m.jai c, a Centro Atómico Bariloche an Instituto Balseiro, 8400 Bariloche, Argentina b CNRS UMR 5208 Université Lyon 1, MAPLY, Bât. 101, Villeurbanne Ceex, France c CNRS UMR 5208, INSA e Lyon, Bât. Léonar e Vinci, 69621, France Receive 20 July 2006 Available online 1 February 2007 Submitte by P.G.L. Leach Abstract The existence of equilibrium solutions for a lubricate system consisting of an articulate boy sliing over a flat plate is consiere. Here we consier the case when cavity can occur Elsevier Inc. All rights reserve. Keywors: Lubrication; Tilting-pa bearing; Equilibrium; Reynols variational inequality 1. Introuction This work is a continuation of a recent paper [1] treating the existence of equilibrium in a tilting-pa thrust bearings with cavitation effects isregare. Articulate sliers have two egrees of freeom: The first one is the vertical isplacement a uner the effect of a force F, applie at a position x 0, an of the pressure loa px where p = px) is the pressure of the flui between the surfaces of the slier. The secon egree of freeom is the tilt or pitch) angle θ see Fig. 1) see [1] for more explanations on the physical moel). * Corresponing author. aresses: gustavo@cab.cnea.gov.ar G. Buscaglia), ciuperca@maply.univ-lyon1.fr I. Ciuperca), ima.hafii@insa-lyon.fr I. Hafii), mohamme.jai@insa-lyon.fr M. Jai) X/$ see front matter 2007 Elsevier Inc. All rights reserve. oi: /j.jmaa

2 842 G. Buscaglia et al. / J. Math. Anal. Appl ) Fig. 1. Scheme of an articulate slier, containing efinitions of the ifferent components of the evice an some geometrical variables an non-imensional coorinates use in the text. In [1] we treate the case in which h, the non-imensional istance between the surfaces, is non-increasing with x. This guarantees positivity of the pressure, p, on all the omain. Thus p satisfies the Reynols equation [h 3 x) p ] = h, x, x 1 p = 0, x. 1.1) In the present work we consier that the system can have a situation with h non-increasing firstly an then non-ecreasing convergent ivergent case). In such situation the solution of 1.1) is not always non-negative an we must replace Eq. 1.1) by the corresponing variational inequation. In all this paper we restrict our stuy to the one-imensional case. Thus the equilibrium problem becomes: fin p,a,θ) K R R satisfying h 3 p ϕ p) p = F, xp = Fx 0, hx) = h 0 x) + a + θx, h ϕ p) ϕ p), ϕ K, 1.2) with =], 1[, K ={ϕ H0 1), ϕ 0}, x 0, F>0an h 0 : R a given function. We consier the following hypothesis on the function h 0 : h 0 C 2 ), h 0 > 0, x, h 0 is a non-ecreasing function, 1.6) h 0 0, x, h 0 0) = h 0 0) = ) 1.4) 1.5)

3 G. Buscaglia et al. / J. Math. Anal. Appl ) A typical example of such a function is h 0 x) = 1 x) 2. In [1] we prove the existence of at least a solution of 1.2) 1.5) with θ<0inwhich case the variational inequality 1.2) becomes the Reynols equation. The result prove in [1] says that for any F>0there exists a solution of 1.2) 1.5) with θ<0provie that the articulation point x 0 is situate not far from the right en sie x = 1) of. We then can hope to improve the result by enlarging the omain where we search for the unknown θ; it is clear that for θ h 0 ) the unique solution p of 1.2) is p 0 then 1.3) an 1.4) are impossible. In the present paper we prove that for any F>0an any x 0 there exists at least a solution of the problem 1.2) 1.5) with θ< h 0 ); this improves the result in [1] in the one-imensional case. Nevertheless, this stronger result is obtaine uner the stronger hypothesis 1.6) than that of [1]. In Section 2 we give some preliminary results on the behaviour of the solution of a variational inequality with respect to a parameter. In Section 3 we give the esire existence result by reucing our problem to an algebraic equation with an unknown θ. 2. Preliminaries In this section we shall give some preliminary results concerning the behaviour of the mapping: G : A ]0, + [ Px)x ]0, + [ with P = PA) is the unique solution of the Reynols variational inequality P K, H 0 + A) 3 p ϕ P) H 0 x ϕ P), ϕ K, 2.1) where H 0 : [0, + [ is a given boune non-negative function. We shall consier two principal situations: The function H 0 is assume ecreasing in. In this case the variational inequality becomes an equality Reynols equation 1.1) with h replace by H 0 + A). The function H 0 is assume ecreasing in [,] an increasing in [,1] for some number. In this case we must consier the general Reynols variational inequality. We give now the first general result on the behaviour of GA) when G goes to infinity. Proposition 2.1. Suppose that H 0 is boune an non-negative. Then lim GA) = 0. A + More precisely, we have 2 GA) 2 H 0 L 2 ) A 3.

4 844 G. Buscaglia et al. / J. Math. Anal. Appl ) Proof. Taking ϕ = 0 in 2.1) an since H 0 is non-negative we obtain ) P 2 A 3 H 0 x L 2) P x L 2 ) which gives easily the result by Poincaré inequality. In this section we are mainly intereste on the behaviour of GA) when A tens to 0 an on the monotony of G. These kin of problems have been consiere in [1,2] in the context of Reynols equation only. We recall these results in the next subsection an then we exten them to the case of Reynols variational inequality Equation case In this subsection the following hypothesis on H 0 will be consiere: H 0 W 1, ), min H 0 x) = 0, x H 0 x) 0 a.e.x, mes{ x : H 0 x) < 0} > 0 an we remark that in this case the problem 2.1) becomes [ H 0 + A) 3 P ] = H 0, x =], 1[, x P) = P1) = 0. We now recall a result prove in [1,2]. This result claims that GA) goes to infinity when A goes to 0 if H 0 x) is equivalent to 1 x)α with α>0 in a neighbourhoo of x = 1. Proposition 2.2. Assume, in aition to hypothesis 2.2), that there exist 0 <M 0 <M 1, α 0 an δ 0 ]0, 1[ such that 2.2) 2.3) M 0 1 x) α H 0 x) M 11 x) α a.e. x ]1 2δ 0, 1[. 2.4) Then there exists c = cm 0,M 1,δ 0,α)such that the solution P of 2.3) satisfies an 1 1 B 1 1 B ) ) B Px)x c log 1 + M 1 2, B ]0,δ 0 ], A>0, if α = 0, A Px)x ca 2 α 2, B ]0,δ 0 ], A<B α, if α>0. Corollary 2.3. Uner hypotheses of Proposition 2.2, we have lim GA) =+. A 0 We now give the following technical result.

5 G. Buscaglia et al. / J. Math. Anal. Appl ) Lemma 2.4. Uner the hypothesis 2.2) an in aition Θ + 2M 0 1 x) H 0 x) Θ + 2M 11 x), x [1 2δ 0, 1], 2.5) with 0 <Θ 1, 0 <M 0 <M 1 an δ 0 ]0, 1[, then there exists C = CM 0,M 1,δ 0 )>0 such that Px C A + Θ 2 for all A>0, such that A + Θ 2 <δ0 2. Proof. We follow the lines of the proof of Proposition 2.2. Step 1: We take φ C 2 ) with φ 1on[1 δ 0, 1] an φ1 2δ 0 ) = 0. We first show that there exists C 1 > 0 epening only on M 0,M 1 an δ 0 such that Px) C 1 q 1 x)φx), x [1 2δ 0, 1], with q 1 x) = Θ1 x) 2 + M 0 1 x) 3 [Θ1 x) + M 1 1 x) 2 + A] 3. Proof of Step 1. We apply the maximum principle on [1 2δ 0, 1]. Since q 1 1) = 0 an φ1 2δ 0 ) = 0, it suffices to prove that [ C 1 A + H 0 ) 3 q ] 1φ) H 0 x x x), x [1 2δ 0, 1]. Accoring to 2.5), it is enough to show the existence of C 2 > 0 epenent on M 0,M 1 an δ 0 such that [ A + H 0 ) 3 q ] 1φ) C 2 Θ + 2M0 1 x) ). x x By eveloping, it is enough to show with I k C 2 Θ + 2M0 1 x) ), k= 1, 2,...,5, 2.6) 5 I 1 = 3 A + H 0 ) 2 H 0 q 1 φ, I 2 = 3 A + H 0 ) 2 H 0 φ q 1, I 3 = A + H0 ) 3 q 1, I 4 = 2 A + H0 ) 3 q 1 φ, I 5 = A + H0 ) 3 q 1 φ. Some elementary calculations yiels q 1 x) Θ1 x) + M 0 1 x) 2 C3 [Θ1 x) + M 1 1 x) 2 + A] 3, x [1 δ 0, 1[, 2.7) q 1 x) Θ + 1 x C4 [Θ1 x) + M 1 1 x) 2 + A] 3, x [1 δ 0, 1[, 2.8) with C 3,C 4 > 0 epening only on M 0 an M 1.

6 846 G. Buscaglia et al. / J. Math. Anal. Appl ) Now, integrating 2.5) between x an 1 with x ], 1[ an aing A we obtain A + Θ1 x) + M 0 1 x) 2 A + H 0 x) A + Θ1 x) + M 1 1 x) 2, x [1 2δ 0, 1[. Using now the expression of q 1, the inequalities 2.7) 2.9) an the hypothesis 2.5), we easily euce 2.6) which ens the proof of Step 1. Step 2: We prove the esire inequality. Using the elementary inequality Θ1 x) 1 x) 2 + Θ 2, we euce Θ1 x) + M 1 1 x) 2 + A M 1 + 1)1 x) 2 + A + Θ 2. From the non-negativity of P an Step 1 we euce 1 1 P P C 1 q 1 x) x, 1 δ 0 1 δ 0 then P C 1 M δ 0 1 x) 3 x M 1 + 1)1 x) 2 + A + Θ 2 ) 3. Let x ]1 δ 0, 1[ such that 1 x = A + Θ 2. Then P C 1 M 0 M 1 + 2) 3 A + Θ 2 ) 3 1 x 1 x) 3 x C 1 M 0 = A + Θ 2 ) 2, 4M 1 + 2) 3 A + Θ 2 ) 3 which ens the proof. We finally give the regularity an monotonicity result which is prove in [1]. Proposition 2.5. The function G is of class C an G A) < 0, A> Inequation case In all this subsection the following hypothesis on the given function H 0 is consiere H 0 C 2 ) an H 0 x) > 0, x, an there exists ], 1[ such that H 0 ) = H ) ) = 0, H 0 x) < 0 for x< an H 0 x) > 0 for x>. The hypothesis 2.10) implies H 0 x) > 0 for any x {}. We recall that we consier P epening on A) the solution of 2.1). The analoguous of Proposition 2.2 is the following 2.9)

7 G. Buscaglia et al. / J. Math. Anal. Appl ) Proposition 2.6. Let 0 ], 1[. Then there exists C>0epening only on 0, min x H 0 x) an max x H 0 x) such that for any > 0, we have Px)x C ) A for any A with 0 <A< Proof. From results of [3], we have that P>0on],[, an P satisfies H 0 + A) 3 P ) = H 0 x, x ],[. Since P) = 0 an P) 0, we have by maximum principle P q, x [,], with q the solution of the problem H 0 + A) 3 q ) = H 0 x, q) = q) = 0. x ],[, Using now the Taylor evelopment of H 0 x) aroun we easily obtain 2.11) 1 2 min H 0 y) x) H 0 x) 1 y ],[ 2 min H 0 y) x) for any x ],[. y ],[ 2.12) By an affine a change of variables: x ], 1[ x ],[, x x = x + 1 2, 2.13) the problem 2.11) can be written in ], 1[ an we can easily see that the hypothesis 2.4) is satisfie with α = 1. Applying Proposition 2.2 we euce the result since P 0on. Corollary 2.7. lim GA) =+. A 0 We now stuy the monotony of GA). From the results of [3] an assumptions 2.10), we have two possibilities: Px)>0, x, an P satisfies H 0 + A) 3 P ) = H 0, x, x P) = P1) = ) There exists β ],1[ such that Px)>0, x ],β[, Pβ)= P β) = 0 an Px)= 0, x [β,1].

8 848 G. Buscaglia et al. / J. Math. Anal. Appl ) In this case P satisfies the problem H 0 + A) 3 P ) = H 0 x, x ],β[, P) = Pβ)= 0, P β) = 0, Px)= 0, x [β,1]. 2.15) We remark that β is here an unknown. By integrating 2.15), we obtain H 0 + A) 3 P x = H 0 C an P β) = 0givesC = H 0 β). We then euce P 1 x) = H 0 x) + A) 2 H 0β) + A H 0 x) + A) 3. Due to P) = 0 we obtain x Px)= s H 0 s) + A) 2 H 0 β) + A ) x s H 0 s) + A) ) Finally the conition Pβ)= 0, allows us to obtain a scalar equation satisfie by β. It is then natural to introuce the following application: ψ : [, 1] ]0, + [ R, efine by β ψβ,a)= s H 0 s) + A) 2 H 0 β) + A ) β s H 0 s) + A) ) Then for any A, the problem 2.15) reuces to the existence of β ],1[ such that ψβ,a)= 0. It is clear that ψ C 2 ], 1[ ]0, + [) an we have ψ β β β, A) = H 0 β) s H 0 s) + A) ) Hypotheses 2.10) imply that for any A,max β [,1] ψβ,a)= ψ,a), ψ β > 0forβ ],[ an ψ β < 0forβ ],1[. Since ψ,a) = 0 an from the above consierations, for any A>0 the existence an uniqueness of a solution of 2.15) is equivalent to the conition ψ1,a)<0. Then we have prove: Proposition 2.8. If ψ1,a)<0, then there exists β = βa) ],1[ such that P satisfies 2.15). If ψ1,a) 0, then P satisfies 2.14).

9 G. Buscaglia et al. / J. Math. Anal. Appl ) The next lemma is useful in the stuy of the monotony of G. Lemma 2.9. Assume that H 0 is non-ecreasing. Let ν ],1] such that H 0)>H 0 ν) an φ : [0, + [ ]0, + [ an arbitrary continuous function. Let q be the solution of the problem φh 0 ) q q) = qν) = 0. Then ν qx)x > 0. ) = H 0 x, x ],ν[, 2.19) Remark This result can be seen as a kin of maximum principle in the sense: if 1 H 0 x < 0, then 1 qx>0. Proof. By integrating 2.19) there exists c R such that φh 0 )q = H 0 c. 2.20) Due to the assumptions 2.10), three ifferent cases are possible: Case 1: q x) 0, x ],ν[ or q x) 0, x ],ν[ with mes{x: q x) = 0}=0; this is impossible because q) = qν) = 0. Case 2: There exists ν 1 ],ν[ such that q x) > 0 for x ],ν 1 [, q x) < 0 for x ]ν 1,ν[, q ν 1 ) = 0. It is clear that qx) > 0, x ],ν[ which proves the result. Case 3: There exist ν 1 ],[ an ν 2 ],ν[ such that q x) > 0 if x ],ν 1 [ ]ν 2,ν[, q x) < 0 if x ]ν 1,ν 2 [, q ν 1 ) = q ν 2 ) = 0. Let ν ],ν[ such that H 0 ν ) = H 0 ν). It is clear that ν <ν 1. We have: ν ν 0 x)qx) x = H 0 q x = H ν ν H 0 q x From 2.20) an using also H 0 ν ) = H 0 ν) we euce ν ν H 0 q x = 0. ν H 0 q x. 2.21)

10 850 G. Buscaglia et al. / J. Math. Anal. Appl ) We then easily obtain from 2.21) ν H 0 qx)x > 0. Let ν 3 ]ν 1,ν 2 [ such that qν 3 ) = 0, qx) > 0ifx ],ν 3 [ an qx) < 0ifx ]ν 3, 1[. Since H 0 is non-ecreasing, we have ν qx)x > 1 H 0 ν 3) ν 3 H 0 x)qx) x + 1 H Together with 2.22) this proves the claime result. 0 ν 3) We have a first result concerning the monotony of G. ν ν 3 H 0 x)qx) x. 2.22) Proposition Suppose that H 0 is non-ecreasing an A>0 such that ψ1,a) 0. Then G is erivable in A an we have G A) < 0. A Proof. Due to Proposition 2.8, we istinguish two cases following the sign of ψ1,a). Case 1: ψ1,a)>0. Uner this hypothesis P satisfies 2.14) equation case). The continuity of ψ implies ψ1,a )>0 for any A in a neighborhoo of A. From the Implicit Function Theorem, the solution P is erivable in A. Denoting q 1 = P A,wehaveG A) = q 1x) x where q 1 is the solution of the problem H 0 + A) 3 q ) 1 = 3 H 0 + A) 2 P ), x, x x q 1 ) = q 1 1) = ) By integrating 2.23) there exists a constant C 1 A) R such that H 0 + A) 3 q 1 x = 3H 0 + A) 2 P x + C 1A). 2.24) Multiplying 2.24) by H 0 + A an ifferentiating in x we obtain H 0 + A) 4 q ) 1 = 3 H 0 + A) 3 P ) + C 1 A) H 0 x x x. Using 2.14) we euce H 0 + A) 4 q ) 1 = C 1 A) 3 ) H 0 x x x. By linearity we can write q 1 = C 1 A) 3 ) ˆq ) with ˆq 1 solution of

11 H 0 + A) 4 ˆq ) 1 x x ˆq 1 ) =ˆq 1 1) = 0. G. Buscaglia et al. / J. Math. Anal. Appl ) = H 0, x, x We now nee to know the sign of C 1 A) 3 an of 1 ˆq 1x) x. Due to 2.24) we obtain q 1 x = 3 P H 0 + A x + C 1A) H 0 + A) ) an ue to q 1 ) = q 1 1) = 0, we have 1 P x C 1 A) = 3 H 0 + A) 1 H. 2.27) 0 + A) 3 Integrating 2.14) there exists C 2 A) R such that H 0 + A) 3 P x = H 0 + A C 2 A). 2.28) Diviing by H 0 + A) 4 an integrating we have 1 P x H 0 + A) x = 1 H 0 + A) 3 C 2 A) Substituting in 2.27) we obtain 1 C 1 A) 3 = 3C 2 A) H 0 + A) 4 1 H 0 + A). 3 1 H 0 + A) 4. Diviing 2.28) by H 0 + A) 3, integrating in an using P) = P1) = 0 we euce C 2 A) > 0 which gives C 1 A) < 3. Let us now prove by absur that 2.29) H 0 1)<H 0 ). Assuming that max s ],1[ H 0 s) = H 0 1) we euce H0 1) + A ) 1 s 1 H 0 s) + A) 3 s H 0 s) + A) ) which will contraict hypothesis ψ1,a)>0. This implies 2.30). We now apply Lemma 2.9 with ν = 1 an φz)= z + A) 4 an we euce that the solution ˆq 1 of 2.26) satisfies ˆq 1 x > 0. With 2.25) an 2.29) we obtain the result in this case.

12 852 G. Buscaglia et al. / J. Math. Anal. Appl ) Case 2: ψ1,a)<0. We have ψ1,a )<0 for any A in a neighborhoo of A. We euce that P = PA ) satisfies 2.15) with A replace by A an β = βa ) ],1[. From 2.18) we obtain ψ β βa), A) < 0 because βa) >. Then the Implicit Function Theorem implies that the application β is of class C 1 in A. An elementary calculus gives an ψ β β, A) = 3 A s H 0 s) + A) H 0 β) + A ) β β A) = 3 H 0β) + A) β H 0s) + A) 4 s β On the other han we have GA) = β Px)x H 0 β) β H 0s) + A) 3 s s H 0 s) + A) ) H 0s) + A) 3 s. 2.32) with P = PA)given by 2.16) an β = βa). It is clear that G is erivable in A an since Pβ)= 0, an elementary calculus gives G A) = β q 2 x) x with q 2 : [,β] R given by q 2 x) = 3 + H 0 β)β A) ) x H0 s) + A ) 3 s + 3 H 0 β) + A ) x H0 s) + A ) 4 s. Differentiating in x an multiplying by H 0 s) + A) 4 we obtain that q 2 satisfies [ H 0 + A) 4 q ] 2 = [ 3 + H 0 x x β)β A) ] H 0 x, q 2 ) = q 2 β) = 0, where q 2 β) = 0 is a irect consequence of 2.32). By linearity we have as in Case ) 2.34) q 2 = [ 3 + H 0 β)β A) ] ˆq 2, 2.35) where ˆq 2 satisfies [ H 0 + A) 4 ˆq ] 2 = H 0 x, ˆq 2 ) =ˆq 2 β) = 0. x ],β[, 2.36)

13 G. Buscaglia et al. / J. Math. Anal. Appl ) On the other han, from 2.32) we obtain 3 + H 0 β)β A) = 3 H 0 β) + A ) β H 0 + A) 4 β H 0 + A) 3 which implies 3 + H 0 β)β A) > 0. We now use Lemma 2.9 in orer to fin the sign of 1 ˆq 2x) x. Let us first prove by absur that H 0 β) < H 0 ). Assume that max ],β[ H 0 = H 0 β). This implies H0 β) + A ) β s β H 0 s) + A) 3 > s H 0 s) + A) 2 which contraicts the equality that ψβ,a)= 0, which shows 2.38). Applying Lemma 2.9 to problem 2.36) with ν = β an φz)= z + A) 4, we obtain β ˆq 2 x > 0. Combining 2.39), 2.35) an 2.37) the claime result is prove. 2.37) 2.38) 2.39) Proposition If A>0 satisfies ψ1,a)= 0, then ψ A 1,A)>0. Proof. From the efinition 2.17) of ψ we have by elementary calculus ψ 1 A 1,A)= 3 H 0 1) H 0 x) H 0 x) + A) 4 x. Exactly as in 2.30), we prove that H 0 1)<H 0 ), which implies the existence of ν ],[ such that H 0 ν) = H 0 1). We now write ψ ν A 1,A)= 3 φx) 1 H 0 x) + A x + 3 with φx)= H 01) H 0 x) H 0 x)+a) 3. It is clear that ψ1,a)= 0 means 1 Since we have φx)x = 0. ν φx) H 0 x) + A x 2.40)

14 854 G. Buscaglia et al. / J. Math. Anal. Appl ) φx)<0 for x ],ν[ an φx)>0 forx ]ν,1[, we euce ψ A 1,A)> 3 H 0 1) + A ν which gives the result by using 2.40). φx)x + 3 H 0 1) + A 1 ν φx)x We are now in a position to state the main result of this section. Proposition Assume in aition to hypothesis 2.10) that H 0 is non-ecreasing. Then if the function G is continuous, then it is strictly ecreasing. Proof. From Proposition 2.12, we euce that it can exists at most a single point A>0 such that ψ1,a)= 0. Then Proposition 2.11 gives the claime result. 3. The existence result We point out that we want to show the existence of a solution p,a,θ) of the system 1.2) 1.5). We will search the unknown a, θ) in the biimensional set { Q = a, θ) R 2 : θ< h 0 ), a > min h0 x) + θx )}. x Let us introuce the following applications g 1,g 2 : Q R efine by g 1 a, θ) = px F, g 2 a, θ) = xp x Fx 0 with p = pa,θ) the unique solution of the variational inequality 1.2) where h is given by 1.5). It is clear that the original problem 1.2) 1.4) is reuce to fin a, θ) Q solution of the system { g1 a, θ) = 0, 3.1) g 2 a, θ) = 0. The regularity of g 1 an g 2 are not obvious since we cannot use there the Implicit Function Theorem. In the next proposition, we show the continuity of g 1 an g 2 by a irect metho. Proposition 3.1. The functions g 1 an g 2 are continuous on Q. Proof. Let a, θ) in Q an a n,θ n ) a, θ) such that there exists b 0 > 0 with h 0 x) + θ n x + a n b 0, x. 3.2) Let p n = pa n,θ n ). Taking φ = 0 in 1.2) an using 3.2) we euce p n H 1 0 ) C with C inepenent of n.

15 G. Buscaglia et al. / J. Math. Anal. Appl ) We euce the existence of p K an of a subsequence of p n enote also by p n such that p n pin H0 1)-weak. We also have for any ϕ K, Using h 0 + a n + θ n x) 3 p n x ϕ ) 2 h 0 + a n + θ n x) 3 pn + h 0 + a n + θ n x) ϕ p n) x. 3.3) x x h 0 + a n + θ n x h 0 + a + θx in L ) strongly an lim inf n + ) 2 ) h 0 + a n + θ n x) 3 pn p 2 h 0 + a + θx) 3 x x an passing to the limit in 3.3), we obtain p = pa,θ). Since p is unique, we euce that the entire sequence p n converges to p in H0 1 ) weakly which easily gives the result. Now in orer to reuce 3.1) to a single equation, we nee the following result: Proposition 3.2. For any θ< h 0 ) there exists an unique solution a of the equation g 1 a, θ) = 0 such that a, θ) Q. Proof. We istinguish the two cases. Case 1: θ 0. We apply the results of Section 2.1 which H 0 x) = h 0 x) + θx θ, which satisfies hypothesis 2.2) an we classically euce the result as a consequence of Propositions 2.1, 2.5 an Corollary 2.3 with α = 0forθ<0an α = 1forθ = 0. Case 2: θ ]0, h 0 )[. Since h 0 is increasing because h 0 > 0) there exists a unique solution θ ], 1[ of the equation h 0 θ ) = θ 3.4) an we have that the function x h 0 x) + a + θx is ecreasing on ], θ [ an increasing on ] θ, 1[. It is convenient to write hx) = h 0 x) + a + θx in the form hx) = b + rx) with b = h 0 θ ) + a + θ θ 3.5) an rx) = h 0 x) h 0 θ ) h 0 x θ )x θ ). 3.6) It is clear that

16 856 G. Buscaglia et al. / J. Math. Anal. Appl ) r C 2 ), r θ ) = r θ ) = 0, r x) < 0 forx< θ, r x) > 0 forx> θ, r x) = h 0 x) > 0, x. It is also clear that the require problem becomes: show the existence of a unique b>0 an p K, satisfying: b + r) 3 p ϕ p) ϕ p) b + r), x ϕ K, 3.7) p = F. 3.8) It is natural to introuce the application b ]0, + [ G 1 b) = px ]0, + [ with p the solution of 3.7) an the problem 3.7) 3.8) is equivalent to: fin b>0 such that G 1 b) = F. 3.9) In fact G 1 b) = g 1 b h 0 θ ) θ θ,θ).) From 3.5) an Proposition 3.1 the function G 1 is continue since θ is fixe. We apply the results of Section 2.2 with H 0 = r an A = b. Now the result is a consequence of Proposition 2.1, Corollary 2.7 an Proposition Now for any θ< h 0 ), we enote by a = aθ) the unique solution of g 1 a, θ) = 0. We introuce the application S : ], h 0 )[ R efine by ) Sθ) = g 2 aθ),θ an it is clear that the problem 1.2) 1.5) or 3.1) is reuce to search for at least a zero of S. We remark that the restriction of S on ], 0[ is exactly the function S introuce in [1] since the variational inequality becomes equation in this case). We prove in [1] an the result is of course vali here): lim Sθ) > 0. θ The next result proves that S takes a negative value in at least a point. We obtain the following result. 3.10) Proposition 3.3. There exists θ 0 ]0, h 0 )[ such that Sθ 0 )<0. Proof. Using the notations of the proof of Case 2 of Proposition 3.2, we have Sθ) = px x 0 ) with px),b > 0 the unique solution of 3.7) 3.8), with r given by 3.6). 3.11)

17 G. Buscaglia et al. / J. Math. Anal. Appl ) It is clear that if θ h 0 ), then θ an we obtain r) 0. Since there exists θ in a neighborhoo of h 0 such that r)<r1) 3.12) we euce the existence of β θ ], 1[ such that the solution p of 3.7) satisfies r + b) 3 p ) = r x on ],β θ [, 3.13) p) = pβ θ ) = 0, 3.14) p β θ ) = 0, 3.15) px) = 0 on]β θ, 1[ 3.16) if not one woul euce exactly as for 2.38) that r1)<r) which contraict 3.12)). We then have exactly as in 2.38) that rβ θ )<r) which implies that rβ θ ) 0 when θ h 0 ). On the other han, from 3.6), we easily euce 3.17) rx) 1 2 inf h 0 x)x θ) 2 x an the positivity of h 0 implies β θ θ 0. We then euce β when θ h 0 ). Then for any x 0 ], 1[, we can choose θ in a neighborhoo of h 0 ) such that β θ <x 0. Since p = 0forx β θ,wehave 3.18) β θ Sθ) = px x 0 ) an finally the positivity of p on ],β[ an 3.18) give the claime result. It now remains to show the continuity of the function S. Proposition 3.4. The function S is continuous on ], h 0 )[. Proof. The ifficulty of the proof of the continuity of S comes from the change in the nature of problem satisfie by a, p) when θ varies in the mentione interval. We fixe θ ], h 0 )[ an we consier three cases. Case 1: θ 0 < 0. The continuity of S in θ 0 was prove in [1] see Proposition 3.3). Case 2: θ 0 ]0, h 0 )[. Letθ n θ 0 an we can suppose that θ n [0,θ 1 ] with 0 <θ 1 < h 0 ) fixe. We enote by θn the solution of 3.4) with θ = θ n an r n x) = h 0 x) h 0 θn ) h 0 θ n )x θn ). We also enote b n > 0,p n x) the unique solution of 3.7) 3.8) with r = r n.

18 858 G. Buscaglia et al. / J. Math. Anal. Appl ) We prove by absur that there exists a compact ˆK ]0, + [ inepenent of n such that b n ˆK, n N. Let us suppose the contrary, then we have two cases: 3.19) Case 2.1: There exists a subsequence enote by b n such that b n +. Then by Proposition 2.1 with H 0 = r n an A = b n, we euce that p n x 0 which contraicts p n x = F. Case 2.2: There exits a subsequence enote by b n such that b n 0. We apply Proposition 2.6 with H 0 = r n, 0 = θ1 an A = b n notice that r n x) = h 0 x) is inepenent of n) an we euce that p n x + which also contraicts p n = F. Then we prove 3.19). Then there is a subsequence enote also by b n an an element b ˆK such that b n b. By a similar proof to that of Proposition 3.1 we euce the existence of a subsequence enote by p n an of an element p K such that p n p weakly in H 1 ) with p an b satisfying 3.7) 3.8) with rx) = h 0 x) h 0 θ0 ) h 0 θ 0 )x θ0 ) since r n r strongly in L )). By uniqueness of the solution of the 3.7) 3.8), we euce that all the sequence b n,p n ) converges to the limit. Then Sθ n ) Sθ 0 ). Case 3: θ 0 = 0. We also take a sequence θ n 0 an we can consier two cases: Case 3.1: θ n > 0. We show exactly as in Case 2 that Sθ n ) S0). Case 3.2: θ n < 0. We are in the hypothesis of Section 2.1, so Sθ n ) = p n x x 0 )x with b n > 0,p n x) solution of [ bn + h 0 x) θ n 1 x) ) ] 3 p n = [ h0 x) θ n 1 x) ], x p n ) = p n 1) = 0, p n x = F. Exactly as in Case 2.1 we prove that no subsequence b n can exist. Let us now suppose that a subsequence of b n satisfies b n 0. We apply Lemma 2.4 with H 0 x) = h 0 x) θ n 1 x) an A = b n.wehave H 0 x) = θ n h 0 x) an the assumptions of Lemma 2.4 are satisfie with M 0 = min x h 0, M 1 = max x h 0, δ 0 = 1 an Θ = θ n. Then we obtain p n x + which contraicts p n x = F. We then prove that there is no subsequence of b n which tens to 0. We euce the existence of a compact ˆK ]0, + [ such that b n ˆK, n N. The en of the proof is as in Case 2.

19 G. Buscaglia et al. / J. Math. Anal. Appl ) Now applying the intermeiate value theorem, we obtain from 3.10), Propositions 3.3 an 3.4 an the main result of this paper. Theorem 3.5. For any x 0 ], 1[ an F>0 there exists at least a solution p,a,θ) of 1.2) 1.5). Acknowlegments This work was partially supporte by grant A03E01 from ECOS-SECYT an by the CMIMF Action Intérée N o MA/04/94. References [1] G. Buscaglia, I. Ciuperca, I. Hafii, M. Jai, Existence of equilibria in articulate bearings, J. Math. Anal. Appl ) [2] I. Ciuperca, I. Hafii, M. Jai, Singular perturbation problem for the incompressible Reynols equation, Electron. J. Differential Equations ) 2006) [3] G. Bayaa, M. Chambat, Sur quelques moélisations e la zone e cavitation en lubrification hyroynamique Some moelling of the cavitation region in hyroynamic lubrication), J. Méc. Théor. Appl )