SINGULAR PERTURBATION PROBLEM FOR THE INCOMPRESSIBLE REYNOLDS EQUATION

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1 Electronic Journal of Differential Equations, Vol ), No. 83, pp ISSN: URL: or ftp ejde.math.txstate.edu login: ftp) SINGULAR PERTURBATION PROBLEM FOR THE INCOMPRESSIBLE REYNOLDS EQUATION IONEL CIUPERCA, IMAD HAFIDI, MOHAMMED JAI Abstract. We study the asymptotic behavior of the solution of a Reynolds equation which describe the behavior of the fluid between two closes surfaces as the distance between the two surfaces locally tends to zero. 1. Introduction The field of lubricated contact deals with dynamical systems which consist of two or more) bodies in relative motion. The contact between the bodies is mediated by a lubricant fluid, which in this work is assumed incompressible. The simplest such contact is the wedge or plane slider), used in thrust bearings. It consists of two planar, rigid surfaces which are not mutually parallel. It is sketched in Fig. 1, in which the bottom surface is assumed to be moving horizontally towards the right. This movement entrains lubricant towards the right into the convergent gap between the surfaces. In turn, this generates a pressure field and consequently a thrust force, which allows to equilibrate a load applied to the top of the device. Under the thin-film hypothesis the gap thickness h much smaller than the inplane dimensions of the contact, with the variations in h also assumed small) the fluid pressure does not depend on the vertical coordinate, which is taken across the gap. Upon normalization and assuming that the system is in a time-independent state, the pressure satisfies the normalized Reynolds equation [5] [hx) 3 p] = h x = x 1,..., x n ) 1.1) p = 0 x 1.2) where R n n = 1 or 2) is the domain in which the two surfaces are in proximity, p is the normalized pressure, hx) is the normalized gap thickness and the relative motion is assumed along the x 1 -direction. Assume, as in Fig. 1, that a vertical force F is applied to the upper surface of the bearing at a point x 0 = x 0 1, x 0 2,..., x 0 n). To equilibrate this load the upper surface 2000 Mathematics Subject Classification. 76D08, 35B25, 35B40. Key words and phrases. Reynolds lubrication equation; weighted Sobolev Spaces; variational inequality; singular perturbation analysis; maximum principle. c 2006 Texas State University - San Marcos. Submitted January 5, Published July 21,

2 2 I. CIUPERCA, I. HAFIDI, M. JAI EJDE-2006/83 F h 1 0 Figure 1. Sketch of a slider bearing changes its position, intuitively getting closer to the lower surface as the applied load increases. Let us define the reference shape of the upper surface through a non negative function h 0. Into this function we incorporate the so-called attitude of the slider pitch and roll angles), so that the gap thickness becomes, simply, hx) = h 0 x) + ɛ 1.3) where ɛ represents the minimal distance between the surfaces. With h 0 fixed the pressure becomes a function of ɛ satisfying the problem If for any ɛ > 0, we denote [ h 0 x) + ɛ ) 3 ] h 0 p = p = 0 on gɛ) = on 1.4) p dx 1.5) then for equilibrium to hold in a system in which the only degree of freedom is the vertical position, the upper surface must be placed so as to satisfy gɛ) = F 1.6) It is easy to show both that lim ɛ + gɛ) = 0 and that g is a continuous function, so that an equilibrium position exists for any positive load F smaller than max ɛ gɛ). It is thus extremely important to analyze the behavior of g in the vicinity of zero, in particular the conditions under which lim ɛ 0 gɛ) = +. This guarantees the existence of an equilibrium position for any positive F. A finite limit, on the other hand, guarantees the existence of an equilibrium position for any 0 < F g0). It is also important to study the moments of the force exerted by the pressure for each minimal gap thickness ɛ defined with respect to the point x 0, because equilibrium also requires that m i ɛ) = px i x 0 i )dx i = 1,..., n 1.7) m i ɛ) = 0 i = 1,..., n 1.8)

3 EJDE-2006/83 SINGULAR PERTURBATION PROBLEM 3 and systems with the pitch and roll angles as additional degrees of freedom have their attitudes defined by these conditions. In this article we will assume n = 2, =]a 1, b 1 [ ]a 2, b 2 [, h 0 C 0 ) with h 0 x) > 0 a.e x, and min h 0 x) = 0. x The goal is to find the limits of gɛ) and m i ɛ), i = 1, 2 as ɛ 0 +. Beside its intrinsic importance, this study provides crucial tools for a forthcoming analysis on the existence of equilibria for the dynamical equations of slider bearings. As we will see later, the results strongly depend on the shape function h 0. We will consider two situations: i) when h 0 vanishes on a segment of type {x 1 = d 1 } only, with d 1 [a 1, b 1 ], which will be called line contact case. In this case we will assume h 0 x 1 d 1 α in a neighborhood of {x 1 = d 1 }, with α > 0. ii) when h 0 vanishes only at a single point d = d 1, d 2 ) of, which will be called point contact case. We will assume h 0 x d α in a neighborhood of {x = d}, with α > 0. In both the line-contact case and the point-contact case we obtain two types of results: i) convergence of load and momenta to some finite limits which will be made precise in Section 3 and ii) divergence to + of load and momenta. Problem 1.4) can be seen as a singular perturbation of the corresponding problem ɛ = 0 ) with small parameter ɛ. This kind of problem has been studied in [10, 2, 8, 6]. We can apply here singular perturbation results to obtain the convergence of p to the solution of the limit problem denoted p 0 in a weighted Sobolev space of type H 1 0, h 2δ1 0, h 2δ2 0 ) see Section 2 for the definition). This is not sufficient for the convergence of load and momenta; we also need a continuous embedding of H0 1, h 2δ1 0, h 2δ2 0 ) into L 1 ) which is obvious if δ 1 is not large. This singular perturbation approach works only for α < 1 in the line-contact case and for α < 3 2 in the point-contact case. To have a well-posed limit problem we need a Poicaré-like inequality for weighted Sobolev spaces. This subject is well studied in the literature see [3, 9, 4]). We prefer to give here a new elementary result Lemma 2.2) well-adapted to our problem. In cases α 1 for line-contact and α 3 2 for point-contact divergent cases) the singular perturbations results cited above are no longer applicable. This part is more difficult and we use extensively the maximum principle in order to find an appropriate lower bound for p whose integral tends to infinity. This proves the divergence to infinity of the load and we also prove that this divergence, in the line-contact case, is of order greater than ɛ 2/α 2 for α > 1 and greater than log1/ɛ) for α = 1. In the point-contact case the same result holds with α replaced by 2 3 α. In order to prove the divergence of momenta we also need to prove that p is bounded far from the annulation points of h 0. This result is again proved using the maximum principle, assuming h 0 to be a tensor product. We remark that in the point-contact case for α [ 3 2, 2[ we have divergence of load and momenta while the limit problem of 1.4) exists in a weighted Sobolev space. This is because the continuous embedding of this space in L 1 ) does not hold. In some cases the solution of 1.4) is negative, which does not correspond to the actual fluid behavior since cavitation takes place for p < 0. To account for

4 4 I. CIUPERCA, I. HAFIDI, M. JAI EJDE-2006/83 cavitation, problem P1) is replaced by the variational inequality [1] Find p K = {v H0 1 ) : v 0} h 0 x) + ɛ) 3 p ϕ p)dx h 0 ϕ p) ϕ K 1.9) The goal is also to find limits of g and m i, i = 1, 2 defined as in 1.5) and 1.7) with p now the solution of 1.9)) when ɛ goes to 0. We obtain the same kind of results as in the equation case. The paper is organized as follows. In Section 2 we present some preliminary results concerning weighted Sobolev spaces, in particular the Poincaré-like inequality. In Section 3 we study the limits of load and momenta in the equation case problem 1.4)) for the different cases cited above. Finally, Section 4 presents the same study in the inequality case problem 1.9)). 2. Preliminaries Let us consider f 0, f 1 C 0 ) with f k > 0 a.e. x, k = 0, 1. We introduce the weighted Sobolev space H 1, f 0, f 1 ) as the set of all measurable functions ϕ = ϕx) defined on with generalized) derivatives D α ϕ for α = α 1, α 2 ) N 2 with α 1 + α 2 1 such that f 0 x)ϕ 2 x)dx + f 1 x) ϕ 2 x)dx < 2.1) H 1, f 0, f 1 ) is a pre-hilbert space equipped with the scalar product ) ϕ1, ϕ 2 = f H 1,f 0 x)ϕ 1 x)ϕ 2 x)dx + f 1 x) ϕ 1 x) ϕ 2 x)dx. 0,f 1) Let H 1 0, f 0, f 1 ) be the closure of D) with respect to the norm of H 1, f 0, f 1 ), which is a Hilbert space endowed with the same scalar product as H 1, f 0, f 1 ). Remark 2.1. If 1/f k L 1 loc ), k = 0, 1 then H1, f 0, f 1 ) is a Hilbert space [7]. We have the following general Poincaré-like inequality. Lemma 2.2. Let f C 0 ) with f > 0 a.e. x. Assume that a real d 1 [a 1, b 1 ] exists such that f is non-increasing in x 1 on [a 1, d 1 ] [a 2, b 2 ] and non-decreasing in x 1 on [d 1, b 1 ] [a 2, b 2 ], with the obvious convention that for d 1 = a 1 resp. d 1 = b 1 ) the function f is only non-decreasing resp. non-increasing). Then for any δ 1, δ 2 R + such that we have K = sup x 2 [a 2,b 2] + sup x 2 [a 2,b 2] d1 x1 a 1 b1 b1 d 1 a 1 f 2δ1 δ2) s, x 2 )dsdx 1 x 1 f 2δ1 δ2) s, x 2 )dsdx 1 < f 2δ1 u 2 K f 2δ2 u 2, u H 1 0, f 2δ1, f 2δ2 )

5 EJDE-2006/83 SINGULAR PERTURBATION PROBLEM 5 Proof. For any u D) we have for x 1 < d 1 : f δ1 x 1, x 2 ) ux 1, x 2 ) f δ1 x 1, x 2 ) x1 since f is x 1 -non-increasing. We then have a 1 x1 a 1 u s, x 2 ) ds f δ1 s, x 2 ) u s, x 2 ) ds, f 2δ1 x 1, x 2 )u 2 x 1, x 2 ) x 1 ) b 1 u ) 2dx1 ) f 2δ1 δ2) s, x 2 )ds f 2δ2 x 1, x 2 ) x 1, x 2 ) a 1 a 1 By integrating in x 1 on [a 1, d 1 ] first and then in x 2 we obtain x 1 d 1 d1 b2 a 1 a 2 sup x 2 [a 2,b 2] f 2δ1 u 2 dx d1 a 1 x1 a 1 f 2δ1 δ2 )s, x 2 ) ds dx 1 ) f 2δ2 u ) 2dx ) 2.2) In the same manner, using the fact that f is x 1 -non-decreasing on [d 1, b 1 ] [a 2, b 2 ], we obtain b1 b2 f 2δ1 u 2 dx d 1 a 2 sup x 2 [a 2,b 2] b1 b1 d 1 x 1 f 2δ1 δ2 )s, x 2 ) ds dx 1 ) f 2δ2 u ) 2dx ) 2.3) Adding 2.2) and 2.3) we obtain the desired inequality and by a density argument we obtain the result. Corollary 2.3. For f, δ 1, δ 2 satisfying assumptions of Lemma 2.2 the semi-norm f δ2 L 2 ) is a norm on H0 1, f 2δ1, f 2δ2 ) and is equivalent to the norm of H 1, f 2δ1, f 2δ2 ). 3. Asymptotic behavior of the equation case problem 1.4)) In this section we set, for the sake of simplicity, =] 1, 0[ 2 and we suppose that h 0 0, h 0 C 0 ) and h 0 is non increasing in x ) We study the asymptotic behavior, as ɛ tends to 0, of the solution p of 1.4) which is non negative by the maximum principle. We introduce the following limit problem for any δ 1 R + : Proposition 3.1. Suppose that Find p 0 H0 1, h 2δ1 0, h 3 0) such that h 3 ϕ 0 p 0 ϕ = h 0 ϕ H0 1, h 2δ1 0, h 3 x 0) 1 0 sup x 2 [ 1,0] 1 x1 1 dx h 0 < + and δ 1 R + is such that h 2δ1 3 0 s, x 2 ) ds dx 1 < + 3.2)

6 6 I. CIUPERCA, I. HAFIDI, M. JAI EJDE-2006/83 Then problem 3.2) admits a unique solution p 0 H 1 0 ; h 2δ1 0, h 3 0) which is independent on δ 1. Proof. From Lemma 2.2 with d 1 = 0 and δ 2 = 3 2 we deduce that the seminorm h 3 0 L 2 ) is a norm in H0 1, h 2δ1 0, h 3 0) equivalent to the norm in H 1, h 2δ1 0, h 3 0). On the other hand the application ϕ H 1 0, h 2δ1 0, h 3 0) h 0 ϕ dx R is in the dual of H0 1, h 2δ1 0, h 3 0) since we have ϕ dx ) 1/2[ h 0 h h 0 3 ϕ ) ] 2 1/2 0. By Lax-Milgram theorem we have classically the existence and the uniqueness for any fixed δ 1 R +. To prove the independence of p 0 with respect to δ 1 we consider p 1 0 and p 2 0 two solutions corresponding respectively to δ1 1 and δ1 2 with δ1 2 < δ1. 1 We remark that H0 1, h 2δ2 1 0, h 3 0) H0 1, h 2δ1 1 0, h 3 0) with dense and continuous embedding. Then p 2 0 satisfies by density the same problem as p 1 0 which by uniqueness gives the result. Problem 1.4) can be seen as a singular perturbation of problem 3.2) with the small parameter ɛ. This kind of problem has been studied in [10, 2, 8, 6]. Let us recall a simplified version of a result given in [6] for a more general problem, which allows us to obtain the convergence of p to p 0. Lemma 3.2. Let V, W, H be three Hilbert spaces, with continuous embeddings V W H and V dense in W and in H. Let bɛ; u, v), 0 < ɛ ɛ 0, be a sequence of continuous bilinear forms on V, bu, v) a continuous bilinear form on W and f H. Under the hypotheses 1) ɛ bɛ; u, v) is continuous and lim ɛ 0 bɛ; u, v) = bu, v) u, v) V V 2) αɛ) > 0 with αɛ) 0 and β > 0 such that bɛ; u, u) αɛ) u 2 V + β u 2 W, u V 3) γ > 0 such that bu, u) γ u 2 W, u W 4) For any sequence w ɛ V for which bɛ; w ɛ, w ɛ ) is bounded, bɛ; w ɛ, v) bw ɛ, v) 0, v V 5) δɛ) with δɛ) 0 such that bɛ; v, v) bv, v) + δɛ)bv, v) 0 the solution u ɛ of the problem bɛ; u ɛ, v) = f, v) v V, ɛ ɛ 0 converges strongly in W to the solution u of the problem bu, v) = f, v) v W. Proposition 3.3. Under the hypotheses of Proposition 3.1, the solution p of 1.4) converges, as ɛ tends to 0, to the solution p 0 of 3.2), strongly in H 1 0, h 2δ1 0, h 3 0). Proof. It suffices to apply Lemma 3.2 with V = H0 1 ), W = H = H0 1, h 2δ1 0, h 3 0) endowed with the norm h 3 0 L2 ), u ɛ = p, and bɛ; u, v) = h 0 + ɛ) 3 u vdx, bu, v) = h 3 0 u v dx.

7 EJDE-2006/83 SINGULAR PERTURBATION PROBLEM 7 We choose f in the following manner: Since the application v W h 0 v an element of W, from Riesz s theorem there exists f in W such that v h 0 = f, v) W, v W. Assumptions 1)-3) and 5) are immediately verified with αɛ) = ɛ 3, β = 1, γ = 1 and δɛ) = 0. Let us now verify assumption 4). Let w ɛ be a sequence in H0 1 ) for which h 0 + ɛ) 3 w ɛ ) 2 is bounded, so w ɛ W is bounded. For all v H0 1 ), we have h 0 + ɛ) 3 w ɛ v h 3 0 w ɛ v = [ h 0 + ɛ) 3/2 h 3/2 0 ] h0 + ɛ ) 3/2 wɛ v + [ h 0 + ɛ) 3/2 h 3/2 0 is ] h 3/2 0 w ɛ v Since h 0 + ɛ) 3/2 h 3/2 0 0 in L ) we easily obtain the result. For simplicity we shall distinguish here two different situations: i) the function h 0 vanishes on the entire segment x 1 = 0 namely the line-contact case, ii) the function h 0 vanishes on a unique interior point, supposed to be 0,0), namely pointcontact case Line-contact case. We assume in this section that, in addition to hypothesis 3.1) h 0 x) = 0 for x 1 = 0 h 0 x) > 0 for x 1 0 We shall prove, under some supplementary hypotheses, that: For dx h 0x) < + the load and momenta have finite limits paragraph 3.1.1). For dx h 0x) = + the load and momenta have infinite limits paragraph 3.1.2). Remark 3.4. If h 0 behaves as x 1 ) α with α > 0 in a neighbourhood of x 1 = 0, then dx h 0x) is finite for α < 1 and infinite for α Finite limit case. Theorem 3.5. If there exists α ]0, 1[ and m 0 > 0 such that h 0 x) m 0 x 1 ) α x then, for any x 0 1, x 0 2) and ɛ 0, we have p dx p 0 dx, x k x 0 k)p x k x 0 k)p 0, k = 1, 2 where p is the solution of 1.4) and p 0 the solution of problem 3.2) Proof. Since α ]0, 1[, the hypotheses in Propositions 3.1 and 3.3 are satisfied with δ 1 = 1 2. Then there exists p 0 H0 1, h 0, h 3 0) unique solution of 3.2) with δ 1 = 1 2 such that p p 0 strongly in H0 1, h 0, h 3 0). On the other hand, all u H0 1, h 0, h 3 0), we have dx ) 1/2 ) 1/2 u dx h 0 u h 0 2 dx that is, the continuous embedding of H0 1, h 0, h 3 0) in L 1 ) holds. This implies that p converges to p 0 strongly in L 1 ) which ends the proof.

8 8 I. CIUPERCA, I. HAFIDI, M. JAI EJDE-2006/ Infinite-limit case. For any δ > 0, we shall define δ =] δ, 0[ ] 1, 0[ In this paragraph we need the following supplementary hypothesis: There exist δ 0 ]0, 1 2 [, α 1 and 0 < m 1 M 1 such that h 0 W 1, ) and αm 1 x 1 ) α 1 h 0 αm 1 x 1 ) α 1, x 2δ0 3.3) Lemma 3.6. For any φ C 2 [ 2δ 0, 0]) with φ 2δ 0 ) = 0 and q 2 C 2 [ 1, 0]) with q 2 1) = q 2 0) = 0, there is c > 0 small enough such that: with px) cq 1 x 1 )φx 1 )q 2 x 2 ) q 1 x 1 ) = x = x 1, x 2 ) 2δ0 x 1 ) α+1 M1 x 1 ) α + ɛ) 3 Proof. We apply the maximum principle. Since the function q 1 x 1 )φx 1 )q 2 x 2 ), vanishes on 2δ0, it suffices to prove [ ] c a + h 0 ) 3 q 1 x 1 )φx 1 )q 2 x 2 )) h 0 x 2δ0 Dividing by x 1 ) α 1 and using 3.3) it suffices to prove the existence of a positive constant K 1 independent of ɛ such that [a + h 0 ) 3 q 1 x 1 )φx 1 )q 2 x 2 )) ] x 1 ) α 1 K 1 which is equivalent to 3ɛ + h 0 ) 2 [ h0 q 2 φq ] 1 ɛ + h0 x 1 ) α 1 ) 3 q 2 φ q 1 x 1 ) α 1 2ɛ + h 0) 3 q 2 φ h0 + 3ɛ + h 0 ) 2 [ x q 1 q 2 φ 1 ] ɛ + h0 x 1 ) α 1 ) 3 q 2 φ q 1 x 1 ) α 1 ɛ + h 0 ) 3 φq 2 q 1 x 1 ) α 1 3ɛ + h 0) 2 [ q 1 φq 2 q 1 x 1 ) α 1 h 0 x 2 x 1 ) α 1 ] K1 3.4) On the other hand, it is easy to see that a constant K 2 independent of ɛ exists such that: q 1x 1 ) K 2 x 1 ) α M1 x 1 ) α + ɛ ) 3 x 1 [ 2δ 0, 0], q 1 x 1 ) K 2 x 1 ) α 1 M1 x 1 ) α + ɛ ) 3 x 1 [ 2δ 0, 0]. Integrating 3.3) on [x 1, 0] we obtain m 1 x 1 ) α h 0 x) M 1 x 1 ) α x 2δ0 3.5) Using the above inequalities we obtain 3.4) which concludes the proof. Now we are able to give a first result in the case α 1.

9 EJDE-2006/83 SINGULAR PERTURBATION PROBLEM 9 Theorem 3.7. For α 1 we have p dx + as ɛ 0. Moreover there exists K > 0 such that for ɛ small enough we have p dx Kɛ 2 α 2 for α > 1, p dx K log 1 ) ɛ for α = 1. Proof. We apply Lemma 3.6 with φ = 1 on [ δ 0, 0] and q 2 0 with q 2 C 2 [ 1, 0]) H 1 0 ] 1, 0[) and 0 1 q 2x 2 ) > 0. We deduce that there exists a constant c > 0 independent of ɛ such that px 1, x 2 ) cq 2 x 2 )φx 1 )q 1 x 1 ), x 2δ0 with q 1 given in Lemma 3.6. Taking into account the fact that p is non negative on all of we obtain 0 0 p dx c q 2 x 2 )q 1 x 1 )dx = c q 2 x 2 )dx 2 q 1 x 1 )dx ) δ0 1 δ 0 On the other hand, for α > 1 and ɛ small enough we have 0 δ 0 q 1 x 1 )dx 1 0 ɛ 1/α q 1 x 1 )dx 1 which, with 3.6), gives the result for α > 1. For α = 1 an elementary calculation gives 0 δ 0 q 1 x 1 )dx 1 = 1 3M 3 1 2ɛ M δ 0 0 d dx M1 1 x 1 + ɛ) 3) M 1 x 1 + ɛ) 3 dx 1 M 1 x 1 δ 0 M 1 x 1 + ɛ) 3 dx 1 ɛ2 M1 2 ɛ 1+ 2 α M 1 + 1) 3 ɛ 3 α + 2) 0 dx 1 δ 0 M 1 x 1 + ɛ) 3 We easily prove that the last two terms of the right-hand side are bounded by a constant independent of ɛ. With the help of 3.6) we obtain the result. In the particular case when h 0 is symmetric in the x 2 direction we have the following asymptotic behavior of the x 2 -momentum. Theorem 3.8. Suppose that h 0 is symmetric in x 2 with respect to x 2 = 1/2 and α 1. Then for ɛ 0, we have x 2 x 0 2)p dx + if x 0 2 ] 1, 1 2 [, x 2 x 0 2)p dx if x 0 2 ] 1 2, 0[, x 2 x 0 2)p dx = 0 if x 0 2 = 1 2. Proof. By symmetry of h 0 the function px 1, x 2 ) = px 1, 1 x 2 ) is also a solution of problem 1.4), so that by uniqueness p = p. We then have x 2 x 0 2)p dx = x )p dx x ) p dx.

10 10 I. CIUPERCA, I. HAFIDI, M. JAI EJDE-2006/83 The first integral of the right-hand side is equal to 0 by symmetry. Then we have the result by Theorem 3.7. Now we shall give the behavior of the x 1 -moment in the particular case when h 0 is a tensor product. This is often the case in practice for the contact line. We begin by the following lemma which means that p is bounded uniformly in ɛ far from the line contact. Lemma 3.9. Suppose that h 0 x 1, x 2 ) = a 1 x 1 )a 2 x 2 ) with a 2 C 0 [ 1, 0]), a 2 > 0, a 1 H 1 ] 1, 0[), a 1 0, a1 0. Then there exists C > 0 such that px) C x1 1 ds h 0 s) + ɛ) 2, x where h 0 x 1 ) = a 2m a 1 x 1 ) with a 2m = min x2 [ 1,0] a 2 x 2 ) Proof. We apply again the maximum principle. Let qx 1 ) = C x 1 1 1/ h 0 s)+ɛ) 2 ds. Since q 0 and p = 0 on it suffices to show that for C > 0 large enough we have [ h0 + ɛ) 3 q ] h 0 x, that is with Now we have Ex) = We easily obtain C h 0 h 0 + ɛ) 3 h 0 + ɛ) 3 Ex) h 0 3.7) Ex) = 3 h 0 + ɛ h 0 + ɛ 2 a 2m a 2 x 1 ) a 2m [ a 1 + ɛ 3 a 2m a 2 x 2 ) a 1 + ɛ 2 ] = a 2m a 1 + ɛ 3 a 2m 2 a 2 ) a 2 a 2 x 2 ) a 1 + ɛ. a 2 Ex) a 2m a 2M with a 2M = max x 2 [ 1,0] a 2x 2 ) which proves 3.7) by taking C a 2M /a 2m since h 0 + ɛ h 0 + ɛ and h0 0. Theorem Assuming 3.3) to hold, and assuming further that the function h 0 is of the form h 0 x) = g 1 x 1 )g 2 x 2 ) with g 1, g 2 W 1, ] 1, 0[), then x 1 x 0 1)p dx + as ɛ 0 for any x 0 1 ] 1, 0[. Proof. We choose δ > 0 such that δ > maxx 0 1, δ 0 ). Then we have δ 0 x 1 x 0 1)p dx δ x 0 1) p dx + x 1 x 0 1)p dx δ 1 1 We prove as in Theorem 3.7 that the first integral of the right-hand side tends to + since δ x 0 1 > 0. Applying Lemma 3.9 with a 1 = x 1 ) α g 1 and a 2 = g 2 we easily prove that the second integral is bounded by a constant independent of ɛ. We then have the result.

11 EJDE-2006/83 SINGULAR PERTURBATION PROBLEM 11 Remark An interesting open question is to obtain an upper bound for p which allows to say that p is bounded uniformly in ɛ far from x 1 = 0, without the global hypotheses that h 0 is a tensor product Point-contact case. We now assume that h 0 is, in a neighbourhood of x = 0, equivalent to x α with α > 0, where denotes the Euclidean norm. For simplicity we make here the following non-essential) hypothesis h 0 x) = x α h 1 x) with h 1 W 1, ) and h 1 > 0. We denote m = inf h 1 x), x M = sup h 1 x) x We shall prove that for 0 < α < 3 2 paragraph 3.2.1) we have finite limits of load and momenta while for α 3 2 paragraph 3.2.2) they tend to +. We begin by the following existence, uniqueness and convergence results. Proposition If 0 < α < 4 3 then for any 0 < δ 1 < 3 4 there is a unique solution p 0 H0 1, h 2δ1 0, h 3 0) of 3.2) and p p 0 in H0 1, h 2δ1 0, h 3 0). If 4 3 α < 2 then for any δ 1 > α there is a unique solution p 0 H0 1, h 2δ1 0, h 3 0) of 3.2) and p p 0 in H0 1, h 2δ1 0, h 3 0). Proof. The first hypothesis of Proposition 3.1 is obvious. The second one is evident for 4 3 α < 2 and δ α. For 0 < α < 4 3 and 0 < δ 1 < 3 4 we use the inequality s2 + x 2 2 s and the result is immediate Finite-limit case α < 3 2 ). In the following we prove that the limits of load and momenta are finite for α < 4 3 without supplementary hypotheses. For 4 3 α < 3 2 we prove the same result but adding a restrictive supplementary assumption on h 0. Theorem For 0 < α < 4 3 we have for any x0 1, x 0 2) : p p 0, x k x 0 k)p x k x 0 k)p 0 ; k = 1, 2 with p 0 solution of the limit problem 3.2). Proof. From Proposition 3.12 we have p p 0 in H 1 0, h 2δ1 0, h 3 0)-strongly for any δ 1 such that On the other hand we remark that if 0 < δ 1 < < δ 1 < 1 α 3.8) 3.9) then h 2δ1 0 is finite, which by Cauchy-Schwartz inequality gives the continuous embedding of H0 1, h 2δ1 0, h 3 0) in L 1 ). This will prove the three desired convergence. Now the existence of at least a δ 1 satisfying 3.8) and 3.9) is assured if 3/4 < 1/α which is equivalent to α < 4/3. Theorem For 4/3 α < 3/2, under the supplementary hypothesis h 0 0 on we have the same convergence as in Theorem 3.13.

12 12 I. CIUPERCA, I. HAFIDI, M. JAI EJDE-2006/83 Proof. We need here an estimation of p in a stronger norm than h 3/2 0 L2 ) in order to obtain h δ1 0 p L 2 ) bounded with a better parameter δ 1 than in Theorem We prove in the following that h 3 δ)/2 0 p L2 ) is bounded for an appropriate δ > 0. Taking ϕ = h 0 + ɛ ) δ p with 0 < δ < 2 α as a test function in the variational formulation of 1.4) we obtain h0 + ɛ ) 3 δ p 2 = δ h0 + ɛ ) 2 δ h 0 h0 p p h0 + ɛ ) δ p =: I1 + I 2 Using Green s formula we deduce I 1 = δ 2 = δ 2 [ h0 + ɛ ) 2 δ h0 ] p 2 [2 δ) h 0 + ɛ ) 1 δ h0 2 + h0 + ɛ ) 2 δ h0 ] p ) which is negative, thanks to the additional hypothesis h 0 0. On the other hand I δ p h 0 + ɛ) 1 δ = 1 h 0 + ɛ) 3 δ)/2 p 1 δ h 0 + ɛ) 1+δ)/2 1 h 0 + ɛ) 3 δ p ) 2 1/2 h 0 + ɛ) 1 δ 1 δ) 1/2 1 h 0 + ɛ) 3 δ p dx dx δ) 2 dx h 0 + ɛ) 1+δ The last integral of the above inequality is bounded uniformly in ɛ due to the hypotheses on δ. We deduce from 3.10) that h 3 δ 0 p dx) 2 1/2 C Applying Lemma 2.2 with f = h 0, δ 2 = 3 2 δ 2 and δ 1 > 3 2 δ 2 1 α we deduce that p is bounded in H0 1, h 2δ1 0, h 3 δ 0 ). We then infer the existence of ξ H0 1, h 2δ1 0, h 3 δ 0 ) and of a subsequence of ɛ such that p ξ weakly in H0 1, h 2δ1 0, h 3 δ 0 ). From the continuous embedding of H0 1, h 2δ1 0, h 3 δ 0 ) in H0 1, h 2δ1 0, h 3 0) and by identification and uniqueness of p 0 we deduce that p p 0 weakly in H0 1, h 2δ1 0, h 3 δ 0 ) for the entire sequence. Choosing now δ = 2 α 1 η, δ 1 = 3 2 δ 2 1 α + η 2 = 2 2 α + η with 0 < η < 3 α 2 we obtain h 2δ1 0 < + which implies the continuous embedding of H0 1, h 2δ1 0, h 3 δ 0 ) in L 1 ). We then obtain p p 0 weakly in L 1 ) which gives the desired convergence Infinite-limit caseα 3/2). In this paragraph we use the polar coordinates r, θ: x 1 = r cos θ, x 2 = r sin θ and we denote, the image of by this change of variables. For simplicity notations we use the same notation as in cartesian coordinates for example h 0 r, θ) means

13 EJDE-2006/83 SINGULAR PERTURBATION PROBLEM 13 h 0 r cos θ, r sin θ)). We set: The problem 1.4) becomes r =]0, r[ ] π, π 2 [, r ]0, 1[ [ a + h0 r, θ)) 3 r p ] [a + h 0 r, θ)) 3 + dr r = r h 0 cos θ sin θ h 0 in p = 0 in. p] 3.11) Let us remark that from the relation h 0 = r α h 1 r, θ) there exists a positive constant K > 0 and r 1 ]0, 1[ such that h 0 Krα 1, r, θ) r ) We recall that h 0 is non-increasing in x 1 which is equivalent in polar coordinates to r h 0 cos θ sin θ h 0 on 3.13) We need here the following supplementary local condition: There exists r 2 > 0 and β ]0, 1[ such that βr h 0 cos θ sin θ h 0 on r ) Remark Hypothesis 3.14) is a little stronger locally than 3.13) and is true if for example h0 0 locally in particular if h 0 is radial) since h0 > 0 locally and cos θ 0 We now give an analog of Lemma 3.6 in the line-contact case. Lemma Suppose that 3.14) is fulfilled and set r 0 = min{r 1, r 2 }. Then for any φ C 2 [0, r 0 ]) with φr 0 ) = 0, a constant c > 0 exists such that pr, θ) cq 1 r)φr)q 2 θ) r, θ) r0 with q 1 r) = r α+1 Mr α + ɛ) 3, Proof. It suffices to prove the inequality [ ɛ + h 0 ) 3 r cq1 q 2 φ )] + [ ɛ + h0 ) 3 r From 3.14), we have and q 2θ) = cos 3 θ sin θ cq1 q 2 φ )] r h 0 r h 0 cos θ sin θ h 0 1 β)r h 0 cos θ. cos θ sin θ h )

14 14 I. CIUPERCA, I. HAFIDI, M. JAI EJDE-2006/83 Carrying out the differentiations in the left-hand side of 3.15) and dividing by r h0 cos θ, we obtain the following inequality in r0, q 2 θ) [ cos θ φ h0 ) 1 3ɛ + h 0 ) 2 q 1r) + a + h 0 ) 2 q 1r) ɛ + h 0) r + ɛ + h 0 ) 3 q 1 h0 ) 1 ] r) + q [ 2θ) r) 3 ) 2q1 ɛ + h cos θ φ 0 + ) 3 q 1 ɛ + h 0 r + q 2θ) cosθ ɛ + h 0) 3 q 1 φ h ) 1 0 q + 3 2θ) cos θ φɛ + h 0) 2 h 0 q 1 + q 2 cos θ φɛ + h 0) 3 q 1 r 2 1 β. c h0 ) 1 Remark also that there is a constant K 1 > 0 such that h0 ) 1 ) 3q + 2 ɛ + h0 1 h0 ) 1 r 2 h0 ) 1 ] 1r) 3.16) q 1r) K 1 r α Mrα + ɛ ) 3, q 1 r) K 1 r α 1 Mrα + ɛ ) ) Using now 3.12), 3.17) and the expression of q 2 we obtain that the absolute value of the left-hand side of 3.16) is bounded by a constant. Taking c small enough we obtain the result. Theorem Under hypothesis 3.14) we have p dx + for ɛ 0 Moreover there exists K > 0 such that for ɛ small enough we have p dx Kɛ 3 α 2 for α > 3 2, p dx K log 1 ɛ ) for α = 3 2. Proof. Using polar coordinates and the non-negativity of p we have pdx rpr, θ) dr dθ, ρ ]0, 1] ρ Applying Lemma 3.16 with φ = 1 on [0, r0 2 ] we show that there exists a c > 0 such that π/2 r0/2 pdx c q 2 θ)dθ rq 1 r)dr. π As in the proof of Theorem 3.7 with some elementary computations we obtain the result. Theorem Under hypothesis 3.14) if moreover h 1 r, θ) = g 1 r)g 2 θ) with g 1 C 1 [0, 2], g 2 C 1 [ π, π 2 ], g 1r) > 0, g 2 θ) > 0 and d dr r α g 1 r) ) 0 we have x k x 0 k)p dx +, as ɛ 0, k = 1, 2. 0

15 EJDE-2006/83 SINGULAR PERTURBATION PROBLEM 15 Proof. First we prove that there exists K > 0 large enough such that with 2 ds pr, θ) K r h 0 s) + ɛ), r, θ) 3.18) 2 h 0 r) = g 2m r α g 1 r), and g 2m = min θ [ π, π 2 ] g 2θ). We use the maximum principle as in the proof of Lemma 3.9. It suffices to prove the following inequality with K ɛ + h 0) 3 ɛ + h 0 ) + Kr h 0 ɛ + h 0 ) 3 2 ɛ + h 0 ) Er, θ) r h 0 3 cos θ + sin θ h 0 Er, θ) = 3 ɛ + h 0 ɛ + h 0 2 g 2m g 2 θ). 3.19) We consider two situations Case 1: r r 1 with r 1 given in 3.12). As in the proof of Lemma 3.9 we have Er, θ) g2m g 2M with g 2M = max θ [ π, π 2 ] g 2 θ). Then it suffices to prove for r r 1 Kr h 0 g 2m r h 0 g 2M cos θ + sin θ h ) with K > 0 large enough. From 3.12) the function h 0 / ) r h0 is bounded for r r 1. Now dividing by r h0 the inequality 3.20) is obvious, which proves 3.19) for r r 1. Case 2: r > r 1. We shall prove Kɛ + h 0 ) r h 0 cos θ + sin θ h 0 for r > r ) for K > 0 large enough, which implies 3.19) for r > r 1. We have, from hypothesis on h 0, h 0 r) h 0 r 1 ) so Kɛ + h 0 r)) K h 0 r 1 ) for r r 1. Since the right-hand side of 3.21) is bounded, the result is obvious. From the tow cases above, the proof of 3.18) is complete. Now we have x k x 0 k)p dx = rr cos θ x 0 1)p dr dθ + rr cos θ x 0 1)p dr dθ. 3.22) δ δ We choose 0 < δ < minr 0, x0 1 2 ) with r 0 given in Lemma We have r cos θ x 0 1 r + x 0 1 x0 1 2 for r < δ, so rr cos θ x 0 1)p dr dθ x0 1 rp dr dθ δ 2 δ and this last integral goes to + as in the proof of Theorem Now using 3.18) we easily prove that the second integral of the right-hand side of 3.22) is bounded which ends the proof for k = 1. The case k = 2 is similar.

16 16 I. CIUPERCA, I. HAFIDI, M. JAI EJDE-2006/83 4. Asymptotic behavior in the inequality case Problem 1.9)) In this section we suppose for simplicity that =] 1, 1[ 2 and that h 0 is nonincreasing in x 1 on 1 and non-decreasing in x 1 on 2 where we denote 1 =] 1, 0[ ] 1, 1[ and 2 =]0, 1[ ] 1, 1[. We study the asymptotic behaviour of the solution p of 1.9) when ɛ 0. In order to introduce the limit problem we define K δ1 as the closure of K = {ϕ H0 1 ) : ϕ 0} with respect to the norm of H0 1, h 2δ1 0, h 3 0). We remark that K δ1 is a closed convex set in H0 1, h 2δ1 0, h 3 0). We now define the limit problem Find p 0 K δ1 such that h 3 4.1) 0 p 0 ϕ p 0 )dx h 0 ϕ p) ϕ K δ1 We now give the following existence, uniqueness and convergence results. dx h 0 0 x1 sup x 2 [ 1,1] x1 Proposition 4.1. Suppose that K = + sup x 2 [ 1,1] 0 0 < + and δ 1 R + is such that h 2δ1 3 0 s, x 2 ) ds dx 1 h 2δ1 3 0 s, x 2 ) ds dx 1 < Then Problem 4.1) admits an unique solution p 0 K δ1 which is independent of δ 1. Also the solution p of problem 1.9) converges, when ɛ 0, to p 0 strongly in H0 1, h 2δ1 0, h 3 0). Proof. We apply Lemma 2.2 with d 1 = 0 and δ 2 = 3 2 and we obtain classically the first result. Taking ϕ = 0 in 1.9) we obtain h 0 + ɛ) 3 p 2 p dx h 0 = h 0 + ɛ) p which leads to h0 + ɛ) p 3/2 dx ) 1/2 L 2 ) h 0 x) 4.2) which implies h 3/2 0 p dx ) 1/2. L2 ) h 0 x) 4.3) From Lemma 2.2 with d 1 = 0 and δ 2 = 3/2 we deduce that p is bounded in H0 1, h 2δ1 0, h 3 0). Then an element ξ K δ1 exists such that, up to a subsequence, p ξ weakly in H0 1, h 2δ1 0, h 3 0). We now pass to the limit in all terms in the inequality h 0 + ɛ) 3 p ϕ Writing for any ϕ K, h 0 + ɛ) 3 p ϕ = + h 0 + ɛ) 3 p 2 + h 0 ϕ p) ϕ K. 4.4) h0 + ɛ) 3/2 h 3/2 ) 0 h0 + ɛ) 3/2 p ϕ h0 + ɛ) 3/2 h 3/2 ) 3/2 0 h 0 p ϕ + h 3 0 p ϕ

17 EJDE-2006/83 SINGULAR PERTURBATION PROBLEM 17 and using 4.2) and 4.3) we deduce h 0 + ɛ) 3 p ϕ Writing also we obtain Finally we have which gives h 0 h 0 From 4.4)-4.7) we deduce h 3 0 ξ ϕ ϕ p) = ϕ p) h 3 0 ξ ϕ ϕ K. 4.5) h 1/2 0 h 3/2 0 h 0 h 0 + ɛ) 3 p 2 ϕ p), ϕ ξ) ϕ K. 4.6) h 3 0 p 2 lim inf h 0 + ɛ) 3 p 2 h 3 0 ξ ) h 3 0 ξ 2 + h 0 ϕ ξ) ϕ K. By denseness and uniqueness we deduce that ξ = p 0 and that the entire sequence p converges to p 0. It remains to prove the strong convergence. We have h 3 0 p p 0 ) 2 h 0 + ɛ) 3 p 2 + h 3 0 p h 3 0 p p 0. Taking ϕ = 0 in 1.9) and 4.1) and passing to the limit we deduce lim h 3 0 p p 0 ) 2 p 0 2 h 0 2 h 3 ɛ 0 x 0 p The right hand-side of the above inequality is 0 take ϕ = 0 and ϕ = 2p 0 in 4.1)) which proves the result. In the following we shall use the classical notation 0 ɛ = {x : px) = 0} cavitation zone) + ɛ = {x : px) > 0} active zone) It is well known that if x 0 ɛ then h0 so that in 1, p satisfies 0 which implies the inclusion 1 + ɛ [h 0 + ɛ) 3 p ] = h ) The next lemma will be useful for the proofs in the infinite-limit cases. Lemma 4.2. Let be an open subset of 1 with Lipschitz boundary and p the solution of 4.8) with p = 0 on. Then p p on. Proof. Since p and p satisfy 4.8) on, we have the result by the maximum principle since p 0 on and p = 0 on.

18 18 I. CIUPERCA, I. HAFIDI, M. JAI EJDE-2006/ Line-contact case. We suppose for simplicity h 0 x) = x 1 ) α h 1 x), x with α > 0, h 1 W 1, ) and h 1 > 0. We have the following result. Theorem 4.3. For any α ]0, 1[ and x 0 1, x 0 2) we have p dx p 0 dx x k x 0 k)p dx x k x 0 k)p 0 dx, k = 1, 2 Proof. The hypotheses of Proposition 4.1 are satisfied with δ 1 = 1/2. Then the proof is exactly as the proof of Theorem 3.5. Now for the infinite-limit case we use Lemma 4.2 with = 1. Performing for p the same kind of estimates as for p in paragraph we easily obtain the following result. Theorem 4.4. For α 1 we have 1) p dx + 2) If h 0 is symmetric in x 2 with respect to x 2 = 0 then x 2 x 0 2)p dx + for x 0 2 < 0 x 2 x 0 2)p dx for x 0 2 > 0 x 2 x 0 2)p dx = 0 for x 0 2 = 0 3) We assume that h 1 is of the form h 1 x) = g 1 x 1 )g 2 x 2 ), g 2 C 0 [ 1, 1]), g 1 H 1 ] 1, 1[), g 2 > 0, g 1 > 0 and d dx 1 x 1 ) α g 1 ) 0. Then for any x 0 1 ] 1, 0[ we have x 1 x 0 1)p dx + as ɛ 0 Remark 4.5. In the above theorem we obtained the behaviour of the x 1 -moment for x 0 1 only. The problem is open when x 0 is such that x Point-contact case. We suppose h 0 x) = x α h 1 x) with h 1 as in Section 3.2. The analogous of Theorem 3.13 for the inequality problem is the following. Theorem 4.6. For 0 < α < 4/3 we have for any x 0 1, x 0 2) p dx p 0 dx, x k x 0 k)p dx x k x 0 k)p 0 dx, k = 1, 2. The proof the above theorem uses Proposition 4.1 and is exactly as the proof of Theorem For the infinite-limit case we pass again to polar coordinates. We have the following result which is immediate applying Lemma 4.2 with =] 1, 0[ 2 which reduces the problem to the equation case. Theorem 4.7. Suppose that r 2 > 0 and β ]0, 1[ exist such that βr h 0 cos θ sin θ h 0, r, θ) [0, r 2] ] π, π 2 [. Then for α 3/2, we have p dx +.

19 EJDE-2006/83 SINGULAR PERTURBATION PROBLEM 19 Also fir h 1 of the form h 1 r, θ) = g 1 r)g 2 θ) with g 1 C 1 [0, 2], g 2 C 1 [ π, π 2 ], g 1 > 0, g 2 > 0 and d dr rα g 1 r)) 0, we have x k x 0 k)p dx + as ɛ 0 for all x 0 k ] 1, 0[, k = 1, 2. We remark that for α [4/3, 3/2[, we are not able to obtain a result as in Theorem 3.14, since we can not take a test function ϕ in 1.9) such that ϕ p = ch 0 + ɛ) δ p with c independent of ɛ. References [1] Bayada, G.; Chambat, M.: Sur quelques modélisations de la zone de cavitation en lubrification hydrodynamique. J. Méc. Théor. Appl., 51986), [2] Caillerie, D.: Etude générale d un type de problèmes raides et de perturbation singulière. C. R. Acad. Sci. Paris Sér. I Maths., ), [3] Edmunds, D.E.; Opic, B.: Weighted Poincaré and Friedrichs inequalities. J. London Math. Soc. 2), ), [4] Franchi, B.: Weighted Sobolev-Poincaré inequalities and pointwise estimates for a class of degenerate elliptic equations. Trans. Amer. Math. Soc., ), [5] Frene, J.; Nicolas, D.; Degueurce, B.; Berthe, D.; Godet, M.: Lubrification hydrodynamique. Paliers et butées. EYROLLES, [6] Denise, H.: Phénomènes de perturbation singulière dans les problèmes aux limites. Ann. Inst. Fourier. Grenoble, ), [7] Kufner, A.; Opic, B.: The Dirichlet problem and weighted spaces. I. Časopsis Pěst. Mat., ), [8] Lions, J.L.: Perturbations singulières dans les problèmes aux limites et en contrôle optimal. Springer-Verlag, Berlin, Lecture Notes in Mathematics, Vol [9] Opic, B.; Kufner, A.: Hardy-type inequalities, volume 219 of Pitman Research Notes in Mathematics Series. Longman Scientific & Technical, Harlow, [10] Sanchez-Palencia, E.: Asymptotic and spectral properties of a class of singular-stiff problems. J. Math. Pures Appl., ), Ionel Ciuperca CNRS-UMR 5208 Université Lyon1, MAPLY, 101, Villeurbanne Cedex, France address: ciuperca@maply.univ-lyon1.fr Imad Hafidi CNRS-UMR 5208, Insa de Lyon, Bat Léonard de Vinci, 69621, France address: Imad.Hafidi@insa-lyon.fr Mohammed Jai CNRS-UMR 5208, Insa de Lyon, Bat Léonard de Vinci, 69621, France address: Mohammed.Jai@insa-lyon.fr