DERIVATION OF THE REYNOLDS TYPE EQUATION WITH MEMORY EFFECTS, GOVERNING TRANSIENT FLOW OF LUBRICANT. Antonija Duvnjak University of Zagreb, Croatia

Transcript

1 GLASNIK MATEMATIČKI Vol. 39(59)(24), 77 1 DERIVATION OF THE REYNOLDS TYPE EQUATION WITH MEMORY EFFECTS, GOVERNING TRANSIENT FLOW OF LUBRICANT Antonija Duvnjak University of Zagreb, Croatia Abstract. In this paper we derive the Reynolds type lubrication equation with memory effects for lubrication of a rotating shaft. Starting from the nonstationary Stokes equations and using the asymptotic expansion we obtain a law that is nonlocal in time and gives a linear connection between velocity and pressure gradient. Our law tends to the classical Reynolds equation when time tends to infinity. We prove a convergence theorem for velocity and pressure in appropriate functional space. 1. Introduction We consider the lubrication process of journal bearings. A circular shaft of radius R and length L rotates on a lubricated support with angular velocity ω. Between the shaft and the support there is a thin domain completely filled with a viscous incompressible lubricant injected by some prescribed velocity. The flow regime is assumed to be laminar, and heating effects are neglected. Let the length scale in the plane of the film be denoted by L ϕz (as L ϕz we can choose L or 2Rπ), and let l be the film thickness; for conventional bearing geometries l/l ϕz = O(1 3 ). We use these length scales to normalize the equations of motion. Nondimensional coordinates, denoted by overbar, we define as follows {r, ϕ, z} = 1 { L ϕz r, ϕ, z}. L ϕz l Let U represent the characteristic velocity in the plane of the film. l The equation of continuity then requires U to be the velocity scale across 2 Mathematics Subject Classification. 35B25, 35Q3. Key words and phrases. Nonstationary Stokes system, nonlocal Reynolds equation. L ϕz 77

2 78 A. DUVNJAK the film, and we arrive the following definition for normalized velocity: u = {u r, u ϕ, u z } = 1 { L ϕz u r, u ϕ, u z }, U l where the overbar again denotes normalized, i.e. O(1), nondimensional quantity. The nondimensional pressure and time are chosen to be p = p { l }Re, t = Ωt. L ϕz ρu 2 Here Ω is the characteristic frequency of the flow, and the Reynolds number has the definition Re = lu /ν. Substituting into the Navier-Stokes equations and neglecting terms of order ( l L ϕz ) 2, we obtain Ω u ϕ t + Re u u ϕ = 1 R p ϕ + 2 u ϕ r 2, Ω u z t + Re u u z = p z + 2 u z r 2, Here Ω = l2 Ω ν and Re = Re l L are the reduced frequency and the reduced Reynolds number, respectively, and p = p(ϕ, z). Contrary to 4] where, starting from the full Navier-Stokes equations, we derived rigorously the classical Reynolds equation for lubrication of a rotating shaft, in this paper we consider temporal inertia limit of the Navier-Stokes equations, which is characterized by Re, Ω > 1, and derive a law which is nonlocal in time and properly describes the transient effects. According to Szeri 1], when one of the bearing surfaces undergoes rapid small-amplitude oscillation, the condition Ω >> Re is approximately satisfied. In this case, we retain the temporal inertia terms but drop the terms representing convective inertia. Expressed in primitive variables, the equation of motion have now the reduced form u ϕ = 1 p t R ϕ + u ϕ µ 2 r 2, u z = p t z + u z µ 2 r 2. From the asymptotic analysis point of view, these are equations for the lowest approximation of the velocity u. For the above-described situation the fluid flow in a thin film is described by the nonstationary Stokes system where the term with u is equal µε 2. Starting from nonstationary Stokes system and performing asymptotic analysis of this perturbed problem we study the behavior of the flow as ε. At the limit, we find the Reynolds equation with memory effects governing the 2-dimensional macroscopic flow, as an approximation of the Stokes system in thin 3-dimensional domain. The law is nonlocal in time and gives a linear

3 DERIVATION OF THE REYNOLDS TYPE EQUATION 79 connection between velocity and the pressure gradient. According to the described physical situation, memory effects appear only for short period of time. We prove that our law tends to the ordinary Reynolds equation when time tends to infinity. To achieve these results we use some properties of the Laplace transform in a similar way it has been done in 8], where the Darcy type law with memory effects was derived governing transient flow through porous media. The mathematical justification of the Reynolds equation from the fundamental hydrodynamic equations has been subject of a lot papers. The derivation of the classical Reynolds equations for a flow between two plain surfaces was given in 1] und 9]. A precise study of asymptotic behavior of the Newtonian flow in a thin domain between a rotating shaft and lubricated support was given in 4] and the difference between the solution of the Navier- Stokes system in thin domain and the solution of the Reynolds equation in terms of the domain thickness was estimated. In 2] we considered lubrication of a rotating shaft with non-newtonian fluid and starting from Navier-Stokes (Stokes) equations with nonlinear viscosity, being a power of the shear rate (power law), we derived non-linear Reynolds lubrication equation. 2. Description of the ε-problem We consider the nonstationary Newtonian flow through a thin cylinder C ε = {Ξ 1 (r, ϕ, z) R 3 ; ϕ ], 2π, z ], L, R < r < R + εh(ϕ, z)}. Function h :], 2π ], L R + is of class C 1, 2π-periodic in variable ϕ, and < β 1 h(ϕ, z) β 2, ϕ ], 2π, z ], L. The flow in cylinder C ε is described by the nonstationary Stokes system u ε t µε2 u ε + p ε = in C ε (, T ), div u ε = in C ε (, T ), u ε (x, ) = w (ϕ, z) in C ε, (2.1) u ε =, for r = R + εh(ϕ, z), u ε = ω(t) e ϕ, for r = R, u ε = g ( r R, ϕ, t), for z =, t (, T ), ε u ε = g L ( r R, ϕ, t), for z = L, ε where p ε and u ε are the pressure and the velocity, respectively. In order to have a well-posed problem we suppose that the functions g α (, ; H1 (S α )), α =, L, S α = {(ρ, ϕ); ρ ], h(ϕ, α), ϕ ], 2π}, are W 1, loc

4 8 A. DUVNJAK continuous in variables ρ and ϕ. They are 2π-periodic in variable ϕ and for all t satisfy the hypothesis (H1) (H2) 2π h(ϕ,) 2π h(ϕ,) ρ e z g (ρ, ϕ, t)dρdϕ = e z g (ρ, ϕ, t)dρdϕ = 2π h(ϕ,l) 2π h(ϕ,l) (H3) g α (h(ϕ), ϕ, t) =, g α (, ϕ, t) = ω(t) e ϕ, α =, L. ρ e z g L (ρ, ϕ, t)dρdϕ, e z g L (ρ, ϕ, t)dρdϕ, We suppose that ω W 1, loc (, ), w L 2 (Ω) is 2π-periodic in variable ϕ and div ϕ,z w = w ϕ + R w z Under the assumptions =, where Ω =], 2π ], L. g α (t) g α H 1 (S α), for α =, L, ω(t) ω, when t, we want to investigate the behavior of the limit problem when time tends to infinity. Lemma 6.1 gives the existence of a solenoidal vector valued function φ ε W 1, loc (, ; H1 (C ε )) such that φ ε / C ε = u ε / C ε. Classical theory (see Temam 11]) gives the existence of a unique weak solution u ε L 2 (, T ; H 1 (C ε )), p ε H 1 (, T ; L 2 (C ε)) such that u ε φ ε L 2 (, T ; V ε ), u ε L (, T ; L 2 (C ε )) and t (uε φ ε ) L 4/3 (, T ; V ε ), where V ε is the functional space given by and T >. V ε = {u H 1 (C ε ), div u = u C ε } 3. Statement of the main results Let us now state the main results of this paper whose proofs will be found in the following sections. The first theorem shows that the Reynolds type equation with memory effects is an approximation of the Stokes system. We recall that Ω =], 2π ], L. Theorem 3.1. Let (u ε, p ε ) be a solution of the nonstationary Stokes problem (2.1). Let p be a solution of the nonstationary Reynolds problem 1 p ( ξ R ϕ ϕ + ξw ϕ) + R p ( ξ z z + ξw z) = η ω in Ω ], T, ξ p ] (3.1) z + ξw z = 1 h(ϕ,) e z g (ρ, ϕ, t)dρ, z= R ξ p ] z + ξw z = 1 h(ϕ,l) e z g L (ρ, ϕ, t)dρ, z=l R p is 2π-periodic in variable ϕ,

5 DERIVATION OF THE REYNOLDS TYPE EQUATION 81 and let u be given by (3.2) Then 1 ε 1 ε u ϕ = G 1 p R ϕ ] + Gw ϕ + g ω, u z = G p z + Gw z. R+εh R R+εh R u ε h u weakly in L 2 (Ω (, T )), p ε p weakly in H 1 (, T ; L 2 (Ω)). This theorem is a direct consequence of the Propositions 7.1 and 7.2 whose proof will be found in Section 7. Remark 3.2. The existence and uniqueness of the problem (3.1) is given in Section 5, while the functions ξ, η, g and G are defined in Section 4. The following theorems establish the connection between the nonlocal Reynolds-type problem (3.1) and the ordinary Reynolds problem from 4]. Theorem 3.3. Let (u, p) be a solution of the problem (3.1) (3.2). Let p H 1 (Ω) L 2 (Ω) be a solution of the problem 1 p (h3 R ϕ ϕ ) + R p (h3 z z ) = 6 h ϕ µω in Ω, p (3.3) z = 12µ h(ϕ,) h 3 e z g (ρ, ϕ)dρ, for z =, (ϕ, ) p z = 12µ h(ϕ,l) h 3 e z gl (ρ, ϕ)dρ, for z = L, (ϕ, L) p is 2π-periodic in ϕ, and let u be defined by (3.4) Then u ϕ = 1 (ρ h)ρ p 2µR ϕ + ω(1 ρ h ) u z = 1 (ρ h)ρ p 2µ z. { p(t) p in H 1 (Ω) L 2 (Ω), when t. u(t) u in L 2 (C),

6 82 A. DUVNJAK Theorem 3.4. Let p, u be a solution of the problem (3.3)-(3.4). Under assumptions (3.5) (3.6) for some λ 1 >, we have g α (t) g α H 1 (S α) C exp( λ 1 t), for α =, L, ω(t) ω C exp( λ 1 t), p(t) p H1 (Ω) L 2 (Ω) C exp( λt), h u(t)dρ for t t and λ < min{λ 1, π2 β 2 2 }. h u dρ L2 (Ω) C exp( λt), Therefore, when time tends to infinity the nonlocal Reynolds equation (3.1) tends to the ordinary Reynolds equation that is valid for stabilized flow. That stabilization occurs in a very short time when data satisfy (3.5) (3.6). 4. Asymptotic expansion Due to the geometry of the domain, we work in the cylindrical coordinate system. We read the Stokes equations in cylindrical coordinate system and try to find an ansatz that fits the system and the boundary conditions on r = R, R + εh. We seek expansions for u ε and p ε in the form (4.1) (4.2) u ε u (ρ, ϕ, z, t) + εu 1 (ρ, ϕ, z, t) +..., p ε p (ρ, ϕ, z, t) + εp 1 (ρ, ϕ, z, t) +..., where ρ = r R ε. Substituting these expansions into the Stokes equations and collecting equal powers of ε leads to the lowest terms in the form 1 ε : p ρ =, t u r u µ 2 r ρ 2 ε : t u z u µ 2 z ρ 2 + p1 ρ =, t u ϕ u µ 2 ϕ ρ p R ϕ =, From the incompressibility equation we get 1 ε : u r ρ =, + p z =. ε : u ϕ ϕ + R u z z + u r =.

7 DERIVATION OF THE REYNOLDS TYPE EQUATION 83 First we conclude that u r =, p = p (ϕ, z) and p 1 = p 1 (ϕ, z). We compute only the first term in the asymptotic expansion and we further denote that term by (u, p) instead of (u, p ). So, the first term (u, p) in the asymptotic expansion is a solution of the following linear problem u ϕ µ 2 u ϕ t ρ p R ϕ =, u z t u z µ 2 ρ 2 + p z =, (4.3) u ϕ ϕ + R u z z =, u ϕ =, for ρ =, u ϕ = ω(t), for ρ = h, u z =, for ρ =, h, u(, ) = w To solve this problem we use the Laplace transform. Here û denotes the Laplace transform of function u in variable t. Extending functions u and p by zero for t < and using the Laplace transform in variable t we get the system û ϕ wϕ µ 2 û ϕ ρ ˆp R ϕ =, û z wz µ 2 û z ρ 2 + ˆp z =, (4.4) û ϕ ϕ + R û z z =, û ϕ =, for ρ =, û ϕ = ˆω(), for ρ = h, û z =, for ρ =, h. We can compute û ϕ and û z as û ϕ = Ĝ(ρ, ϕ, z, ) û z = Ĝ(ρ, ϕ, z, ) ˆp z + w z 1 ] ˆp R ϕ + w ϕ + ĝ(ρ, ϕ, z, )ˆω, ], where Ĝ and ĝ are complex functions of three real variables (ρ, ϕ, z) and a complex variable : (4.5) Ĝ(ρ, ϕ, z, ) = 1 + eh /µ e (h ρ) /µ e ρ /µ (1 + e h, /µ )/µ ĝ(ρ, ϕ, z, ) = e(h ρ) /µ ( 1 + e 2ρ /µ ) 1 + e 2h. /µ

8 84 A. DUVNJAK We define the functions ˆξ and ˆη as follows: and ˆξ(ϕ, z, ) = h(ϕ,z) 1 e h /µ Ĝ(ρ, ϕ, z, )dρ = 2 (1 + e h + h /µ )(/µ) 3/2 /µ ˆη(ϕ, z, ) = 2e h /µ (1 + e h /µ ) 2 h ϕ. Let us note that ˆξ and ˆη are analytic functions in the half plane Re > π2 β 2 2 µ. Integrating the equation û ϕ ϕ + R û z z = with respect to ρ over ], h(ϕ, z), we get the equation 1 R ϕ ( ˆξ(ϕ, z, ) ˆp ϕ + ˆξw ϕ) + R z ( ˆξ(ϕ, z, ) ˆp z + ˆξw z) = ˆη(ϕ, z, )ˆω() in Ω =], 2π ], L. Using the inverse Laplace transform we come back to the domain t and find function u in the form u ϕ = G 1 p R ϕ ] + Gw ϕ + g ω, u z = G p z + Gw z, where G = G(ρ, ϕ, z, t) is the inverse Laplace transform of Ĝ and g = g(ρ, ϕ, z, t) is the inverse Laplace transform of ĝ. The pressure p satisfies Reynolds equation 1 p ( ξ R ϕ ϕ + ξw ϕ ) + R p ( ξ z z + ξw z ) = η ω (a.e.) in Ω ], T where ξ = ξ(ϕ, z, t) is the inverse Laplace transform of ˆξ and η = η(ϕ, z, t) is the inverse Laplace transform of ˆη. One boundary condition for p is 2πperiodicity with respect to ϕ. The other boundary condition follows from e z h(ϕ,α) u(ρ, ϕ, α, t)dρ = and should be of the form R ξ p ] z + ξw z z=α = h(ϕ,α) h(ϕ,α) e z g α (ρ, ϕ, t)dρ, for α =, L, e z g α (ρ, ϕ, t)dρ, for α =, L.

9 DERIVATION OF THE REYNOLDS TYPE EQUATION Solvability of the Reynolds equation Now, we study the solvability of the Reynolds-type equation (3.1) which determines the effective flow. Function Ĝ given by (4.5) satisfies the equations Ĝ Ĝ µ 2 ρ 2 = 1, Ĝ(, ϕ, z, ) =, Ĝ(h, ϕ, z, ) =. From these equations we get that h Ĝdρ = µ for Re >, as well as the estimate h h 1 C( + 1) Ĝ 2 dρ + µ Ĝ ρ 2 h h Ĝ ρ 2. Ĝ ρ 2 As Re ˆξ = Re h Ĝdρ, for Re > we have the estimate Re ˆξ C ( + 1) 2, where C > is a constant. Therefore, for every C, Re >, the bilinear form a(u, v) = ˆξ ϕ,z u ϕ,z vdϕdz, Ω is continuous and elliptic on H 1 (Ω)/R] H 1 (Ω)/R]. Note that ˆη =. Ω For every C, Re >, the compatibility condition 2π h(ϕ,) e z ĝ (ρ, ϕ, )dρdϕ = 2π h(ϕ,l) e z ĝ L (ρ, ϕ, )dρdϕ is satisfied. Functions ĝ (., ) and ĝ L (., ) are in H 1/2 (S 1 ). Thus the problem 1 ˆp ( ˆξ R ϕ ϕ + ˆξw ϕ) + R ˆp ( ˆξ z z + ˆξw z) = ˆηˆω in Ω, ˆξ ˆp ] (5.1) z + ˆξw z = 1 h(ϕ,) e z ĝ dρ, z= R ˆξ ˆp ] z + ˆξw z = 1 h(ϕ,l) e z ĝ L dρ, z=l R ˆp is 2π-periodic in variable ϕ,

10 86 A. DUVNJAK has a unique solution ˆp() H 2 (Ω) L 2 (Ω) for all C, Re >. Applying the isomorphism theorem 2.1. in 7], pages 99-1 (for details see 3]), we get the existence of the unique solution p D + (t, H1 (Ω) L 2 (Ω)) of the problem (3.1). For every C, Re >, ˆp() H 2 (Ω) L 2 (Ω) thus we can conclude that the problem (3.1) has the unique solution p H 2 (Ω) L 2 (Ω) for (a.e.) t ], T. Let us now study the regularity in time. We introduce the function P (ϕ, z, t) = t p(ϕ, z, δ)dδ. Using the Laplace transform we get a problem for the Laplace transform ˆP (ϕ, z, ) of the function P (instead of ˆP (ϕ, z, ) we use abbreviated form ˆP ()) : (5.2) 1 ˆP ( ˆξ R ϕ ϕ + ˆξw ϕ ) + R ˆP ( ˆξ z z + ˆξw z ) = ˆηˆω in Ω, ] ˆP ˆξ z + ˆξw z = 1 h(ϕ,) e z ĝ dρ, R z= ] ˆP ˆξ z + ˆξw z = 1 h(ϕ,l) e z ĝ L dρ, R z=l ˆP is 2π-periodic in ϕ. This problem has a unique solution ˆP () H 2 (Ω) L 2 (Ω) for every C, Re >. Note that ˆξ is the analytic function in variable C for Re > π2 µ. β2 2 Let µ = a + ib, where Re µ = a2 b 2 >. As ˆξ = 2 th ( h 2 ) 1 + h 1/2 ] sh ah = 2 ch ah + cos bh + i sin bh a ib ch ah + cos bh a 2 + b 2 + h, the real part of ˆξ is equal Re ˆξ a sh ah + b sin bh = 2 (a 2 + b 2 )(ch ah + cos bh) + h. We have h (Re ˆξ) = sh 2 ah sin 2 bh (ch ah + cos bh) 2 >

11 DERIVATION OF THE REYNOLDS TYPE EQUATION 87 and Re ˆξ = for h =, therefore Re ˆξ > for Re µ = a2 b 2 >. Furthermore, (5.3) Re ˆξ Re ˆξ(β ash aβ 1 + b sin bβ 1 1 ) = 2 (a 2 + b 2 )(ch aβ 1 + cos bβ 1 ) + β 1 = c(β 1, ) >. We notice that for Re >, if is large enough, instead of a constant c(β 1, ) in (5.3) we can take a constant C 1 which does not depend on. For Re >, close to, a constant c(β 1, ) depends on and is of the form c(β 1, ) = C 2, C 2 >. We need to know the behavior of ˆP () when. We have ˆξ = O( 1 and ˆη = O( 1 ) for a large. It is easy to see that lim ˆξ = ξ() = h lim lim ˆη = η() = in ˆω = ω() =, in L (Ω), L (Ω), lim ĝ α = g α () in H 1/2 (S α ), for α =, L. Let P () be a solution of the problem (5.4) div ϕ,z ( h ϕ,z P () + w ) = ] P () h + wz = 1 z z= R ] P () h + wz = 1 z z=l R P () is 2π-periodic in ϕ. h(ϕ,) h(ϕ,l) e z g (ρ, ϕ, )dρ, e z g L (ρ, ϕ, )dρ, We have P () H 2 (Ω) L 2 (Ω). Note that P () = if and only if ) h(ϕ,α) e z g α (ρ, ϕ, )dρ = wz (ϕ, α), for α =, L. For the difference ϕ,z ( ˆP () P () ) we have the following estimate: Lemma 5.1. ϕ,z( ˆP () P () ) L2 (Ω) C 2

12 88 A. DUVNJAK Proof. From (5.2) and (5.4) we get a problem for the difference ˆP () P () : (5.5) div ϕ,z ˆξ ϕ,z ( ˆP () P () ] ) 1 = div ϕ,z ˆηˆω + 1 ( ˆξ h) w + 1 (h ˆξ) ϕ,zp () ˆξ ϕ,z ( ˆP () P () ˆP () P () 1 (h ˆξ) ϕ,zp () ) ) 1 ˆηˆω 1 ( ˆξ h) w ] is 2π-periodic in ϕ, h(ϕ,α) ν = 1 ( ˆB α B α() ), for α =, L, where ˆB α = 1 R e z ĝ α dρ for α =, L. Multiplying the problem (5.5) with ˆP () P (), after integration over Ω, for < we get ϕ,z ( ˆP () P () ) L 2 (Ω) C ( 2 ˆη ˆω L 2 (Ω) + ˆξ h L 2 (Ω) + h ˆξ L 2 (Ω) ϕ,z P () L 2 (Ω) ) + g () ĝ H 1/2 (S ) + g l () ĝ L H 1/2 (S L) C 2, and in that case we have the result. For a large we have the estimate ϕ,z ( ˆP () P () ) L2 (Ω) C ( (5.6) ˆη ˆω L2 (Ω) + ˆξ h L2 (Ω) + h ˆξ L2 (Ω) ϕ,z P () L2 (Ω) + g () ĝ H 1/2 (S ) + g l () ĝ l H 1/2 (S L)). Note that ˆξ h L 2 (Ω) C for a large. Analogous estimates are verified for other terms on the right side of the inequality (5.6). Therefore the claim of Lemma 5.1 follows for every, Re >. For every, Re >, ˆP () P () is the element of the vector space L 2 (Ω), therefore ˆP () P () H1 (Ω) C ϕ,z ( ˆP () P ()) L2 (Ω). From this inequality and Lemma 5.1 we have lim ˆP () = P () in H 1 (Ω). ]

13 DERIVATION OF THE REYNOLDS TYPE EQUATION 89 ˆP () P () is the Laplace transform of the function P (t) P (). According to Lemma 5.1 ˆP () P () H1 (Ω) C and the representation formula 2 for the inverse Laplace transform implies that P is a continuous function in variable t. Furthermore, P C(, T ]; H 2 (Ω) L 2 (Ω)) and consequently p = P t H 1 (, T ; H 2 (Ω) L 2 (Ω)). 6. A priori estimates Lemma 6.1. Let φ ε (t) H 1 (C ε ), for (a.e.) t, be a solution of the problem div φ ε (t) =, φ ε (t) = ω(t) e ϕ, for r = R, φ ε (t) =, for r = R + εh, φ ε (t) = g ( r R, ϕ, t), for z =, ε φ ε (t) = g L ( r R, ϕ, t), for z = L. ε Then we have and φ ε (t) H 1 (C ε) C ε ( g (t) H 1/2 (S ) + g (t) H 1/2 (S L) + ω(t) ) t φε (t) H 1 (C ε) C ε ( t g (t) H 1/2 (S ) + t g (t) H 1/2 (S L) + t ω(t) ). Proof. As compatibility conditions are satisfied, conclusion follows from Lemma 4 in 2] and continuous dependence of solution on given data. Furthermore, φ ε W 1, loc (, ; H1 (C ε )). Proposition 6.2. Let u ε be a weak solution for (2.1). Then we have u ε L2 (Q εt ) C ε, (6.1) u ε L 2 (Q εt ) C ε, u ε L (,T ;L 2 (C ε) C ε, where Q εt = C ε (, T ). Proof. Let φ ε be a function from Lemma 6.1 and v ε = u ε φ ε. Then (v ε, p ε ) is a solution of the problem v ε t µε2 v ε + p ε = φε t + µε2 φ ε in Q εt, (6.2) div v ε = in Q εt, v ε (x, ) = w (x) φ ε (x, ) in C ε, v ε = on C ε (, T ).

14 9 A. DUVNJAK The energy inequality that corresponds to this problem (see Temam 11], page 291) is t v ε (t) 2 + 2µε 2 v ε 2 C ε C ε t v ε () ( C ε Cε φε t + µε2 φ ε 2 )v ε, for (a.e.) t, T ]. The inequalities t φ 2 Cε ε t t vε 2 Cε φε t t v ε 2, 2 C ε t t 2µ ε 2 φ ε v ε 2ε 2 µ φ ε t C ε C ε 2 µε2 v ε 2, C ε and the energy inequality give t v ε (t) 2 + µε 2 v ε 2 C ε C ε t v ε () C ε Cε φε t 2 + 2µε 2 t C ε φ ε 2. Backward substitution, u ε = v ε + φ ε, implies the following inequality for u ε t u ε (t) 2 + µε 2 u ε 2 u ε () 2 + φ ε () 2 + φ ε (t) 2 C ε C ε C ε C ε C ε t t + 2 Cε φε t 2 + 3µε 2 φ ε 2. C ε We use estimates for φ ε given in Lemma 6.1 and conclude that t u ε (t) 2 + µε 2 u ε 2 Cε C ε C ε for (a.e.) t, T ]. The conclusion follows. In the next step we introduce the pressure. Temam 11], page 37, we introduce the function U ε (t) = t u ε (s)ds. Following the idea from Let u ε be a weak solution of the problem (2.1), then U ε C(, T ], H 1 (C ε )), div U ε =. According to De Rham s lemma (see Temam 11]), there exists P ε C(, T ], L 2 (C ε)), P ε C(, T ], H 1 (C ε )), such that (6.3) P ε (t) + u ε (t) w µε 2 U ε (t) = in H 1 (C ε ), for every t, T ].

15 DERIVATION OF THE REYNOLDS TYPE EQUATION 91 Lemma 6.3. P ε satisfies the following estimate (6.4) P ε C(,T ],L 2 (C ε)) Cε 1/2. Proof. We suppose that C ε P ε (t) =. Let z ε be a solution of the problem { div zε = P ε (t) in C ε, From Lemma 6.1 follows z ε = on C ε. z ε H1 (C ε) C ε P ε (t) L2 (C ε). We take z ε as a test function in (6.3) and get P ε (t) 2 L 2 (C = ε) C µε2 U ε (t) z ε u ε (t)z ε + w z ε ε C ε C ε C(µε 2 u ε L 2 (Q εt ) + ε u ε (t) L 2 (C ε) + ε w L 2 (C ε)) z ε L 2 (C ε) Cε 1/2 P ε (t) L 2 (C ε). To prove a convergence it is convenient for sequences of functions u ε, U ε, P ε defined on C ε to introduce rescaling functions u(ε)(ρ, ϕ, z, t) = u ε (r, ϕ, z, t), U(ε)(ρ, ϕ, z, t) = U ε (r, ϕ, z, t), P (ε)(ρ, ϕ, z, t) = P ε (r, ϕ, z, t), where ρ = r R ε. Functions u(ε), U(ε) and P (ε) are defined on a domain Q T = C (, T ) which does not depend on ε, where C = {(ρ, ϕ, z) R 3 ; ϕ ], 2π, z ], L, < ρ < h(ϕ, z)}. The following estimates are obtained from Proposition 6.2 and preceding lemma by simply changing the variable. Lemma 6.4. For u(ε) we have the following estimates u(ε) L 2 (Q T ) C, u(ε) L (,T ;L 2 (C)) C, u(ε) ρ L 2 (Q T ) C, ϕ,z ε u(ε) L 2 (Q T ) Cε 1, u r(ε) ρ L 2 (Q T ) Cε,

16 92 A. DUVNJAK where ϕ,z ε u = 1 R+ερ ( ur 1 R+ερ ( uϕ 1 R+ερ ϕ u ϕ) ϕ + u r) u z ϕ u r z u ϕ z u z z. Lemma 6.5. Function U(ε) satisfies the following estimates (6.5) U(ε) C(,T ],L 2 (C)) C, U(ε) ρ C(,T ],L 2 (C)) C, ϕ,z ε U(ε) C(,T ],L 2 (C)) Cε 1, U r(ε) C(,T ],L ρ 2 (C)) Cε. Lemma 6.6. Function P (ε) satisfies estimates (6.6) P (ε) C(,T ],L 2 (C)) C ϕ,z ε P (ε) C(,T ],H 1 (C)) C, P (ε) ρ C(,T ],H 1 (C)) Cε. where ϕ,z ε P (ε) = 1 R+ερ P (ε) z P (ε) ϕ ]. We introduce functional spaces 7. Convergence W T = {φ L 2 (Q T ) 2 : φ ρ L2 (Q T ) 2 }, and Y T = {φ C(, T ], L 2 (C) 2 ) : φ ρ C(, T ], L2 (C) 2 )}. In the following, for any function u = (u r, u ϕ, u z ) we denote by ũ = (u ϕ, u z ). Proposition 7.1. There exist subsequences of u(ε), U(ε) and P (ε) (again denoted by u(ε), U(ε) and P (ε)) and functions u W T, U Y T,

17 DERIVATION OF THE REYNOLDS TYPE EQUATION 93 P L (, T ; L 2 (C)) and p = t P H 1 (, T ; L 2 (C)), such that ũ(ε) u & ũ(ε) u ρ ρ weakly in L2 (Q T ), u r (ε) & u r(ε) weakly in L 2 (Q T ), ρ h(ϕ,z) ũ(ε)(ρ, ϕ, z)dρ Ũ(ε) U & Ũ(ε) ρ U r (ε) & U r(ε) ρ h(ϕ,z) P (ε) P weak* in L (, T ; L 2 (Ω)), u(ρ, ϕ, z)dρ weakly in L 2 ((, T ) Ω), U ρ weak* in L (, T ; L 2 (C)), weak* in L (, T ; L 2 (C)), t P (ε) p weakly in H 1 (, T ; L 2 (Ω)), where Ω = {(ϕ, z) R 2 ; ϕ ], 2π, z ], L}. Further, P = P (ϕ, z, t) and p = p(ϕ, z, t). For (a.e.) t (, T ), function u satisfies the equations h(ϕ,z) div ϕ,z u(ρ, ϕ, z, t)dρ = in Ω, (7.1) ν ν h(ϕ,) h(ϕ,l) u(ρ, ϕ,, t)dρ = u(ρ, ϕ, L, t)dρ = h(ϕ,) h(ϕ,l) u(, ϕ, z, t) = ω(t) e ϕ, u(h, ϕ, z, t) =, u is 2π-periodic in variable ϕ. Function U satisfies the equations h(ϕ,z) div ϕ,z U(ρ, ϕ, z, t)dρ = in Ω, (7.2) ν ν h(ϕ,) h(ϕ,l) U(ρ, ϕ,, t)dρ = U(ρ, ϕ, L, t)dρ = t h(ϕ,) h(ϕ,l) e z g (ρ, ϕ, t)dρ, e z g L (ρ, ϕ, t)dρ, e z t t U(, ϕ, z, t) = ω(t)dt e ϕ, U(h, ϕ, z, t) =, U is 2π-periodic in variable ϕ, e z g (ρ, ϕ, s)dsdρ, g L (ρ, ϕ, s)dsdρ,

18 94 A. DUVNJAK for every t, T ] and (7.3) U(t) = t u(ρ, ϕ, z, s)ds. Proof. Weak convergence is a direct consequence, according to the weak compactness, of the estimates given by the previous lemmas. The estimate (6.6) implies that the pressure P is independent of variable ρ. From (6.5) we deduce Ur ρ =. To prove that U r =, we choose the test function q = θ(t, ϕ, z)ρ in div ε U(ε) =, where θ D((, T ) (, L); D per (, 2π)). Using the Green formula, the boundary conditions on C imply T ε (θ(t, ϕ, z)ρ) U(ε) =. We get T C T θu r (ε) = ε ( C When ε tends to zero we get T C 1 θ R + ερ ϕ U ϕ(ε) + θ z U z(ε)). C θu r =. This implies U r =. It is easy to get (7.3). As du dt = u L2 (Q T ), U is an element of the space Y T. On the other side, T T h(ϕ,z) = θ div ε U(ε) = θ div ε U(ε)dρ dϕdzdt = T C Ω θ h(ϕ,z) Ω div ϕ,z Ũ(ε) + (ρu r(ε)) ρ +ερ U z(ε) ]dρ dϕdzdt. z Due to the boundary conditions on U(ε), T h(ϕ,z) θ div ϕ,z Ũ(ε)dρ dϕdzdt = Ω = ε T Ω θ z thus we have T T ϕ,z θ Ũ(ε)dρdϕdzdt = ε C h(ϕ,z) Ω θ z + R ε U r (ε) ρ ρu z (ε)dρ dϕdzdt, h(ϕ,z) In the limit, when ε tends to zero, we get T ϕ,z θ Udρdϕdzdt =, C ρu z (ε)dρ dϕdzdt.

19 DERIVATION OF THE REYNOLDS TYPE EQUATION 95 which implies T Ω h(ϕ,z) θ div ϕ,z Udρ =. From this we deduce that div ϕ,z h(ϕ,z) U(ρ, ϕ, z, t)dρ = for every t, T ]. h(ϕ,z) div ϕ,z Ũ(ε)(ρ, ϕ, z, t)dρ is a bounded sequence in C(, T ], L 2 (Ω)) and h(ϕ,z) Ũ(ε)(ρ, ϕ, z, t)dρ is a bounded sequence in C(, T ], L 2 (Ω)), therefore ν h(ϕ,z) Ũ(ε)(ρ, ϕ, z, t)dρ, for z =, L are bounded sequences in C(, T ], H 1/2 (, 2π)). Taking a limit when ε tends to zero we get ν h(ϕ,α) U(ρ, ϕ, α, t)dρ = h(ϕ,α) e z t g α(ρ, ϕ, s)dsdρ for α =, L. In the same way, working with the spaces L 2 ((, T ) Ω) and L 2 (, T ; H 1/2 (, 2π)) instead of C(, T ], L 2 (Ω)) and C(, T ], H 1/2 (, 2π)), we can see that u satisfies (7.1). Proposition 7.2. Let u and p be defined in previous lemma. Then u verifies (3.2) and p is a solution of the Reynolds equation (3.1). Proof. Multiplying the equation ε P (ε) + u(ε) w µε 2 ε U(ε) = with φ C (Q T ) we get T µε 2 ε U(ε) ε φ + C T C in H 1 (C) T T u(ε)φ + w φ + P (ε)div ε φ =. C C As U r = and u r =, we can take φ = (, φ). Taking a limit when ε tends to zero gives T U µ φ T T T C ρ ρ + u φ + w φ P ϕ,z φ =, C C C and then µ 2 U ρ 2 + u + w = ϕ,z P. Deriving this equation with respect to t gives the following equations for u: (7.4) (7.5) u t u µ 2 ρ 2 = ϕ,zp, u() = w.

20 96 A. DUVNJAK Solving these equations under conditions u(, ϕ, z, t) = ω(t) e ϕ, u(h, ϕ, z, t) =, u is 2π-periodic u ϕ we find that u as a function of ϕ,z p is given by (3.2). The other equations of (7.1) lead to the Reynolds equation (3.1) for p. Remark 7.3. Theorem 3.1 is a direct consequnce of Propositions 7.1 and 7.2. It is easy to see that these results hold true if the initial condition u ε (x, ) = w (ϕ, z) in the problem (2.1) is replaced by u ε (x, ) = w ɛ = w ( r R, ϕ, z), ε where w H 1 (C) is 2π-periodic function in variable ϕ. In that case the function u is given by (3.2) and p is the solution of the problem 1 p ( ξ R ϕ ϕ + χ(w ϕ )) + R p ( ξ z z + χ(w z )) = η ω, ξ p ] (7.6) z + χ(w z ) = 1 h(ϕ,) e z g (ρ, ϕ, t)dρ, z= R ξ p ] z + χ(w z) = 1 h(ϕ,l) e z g L (ρ, ϕ, t)dρ, z=l R p is 2π-periodic in ϕ, where (7.7) (7.8) χ(w ϕ) = χ(w z ϕ) = h(ϕ,z) h(ϕ,z) G(ρ, ϕ, z, t)w ϕ(ρ, ϕ, z)dρ, G(ρ, ϕ, z, t)w z(ρ, ϕ, z)dρ. 8. Proofs of Theorems 3.3 and 3.4 In this section we prove theorems characterizing the behaviour of problem 2.1 for large t. Proof of Theorem 3.3. If lim ˆp exists, it is equal lim t p(t). In the same way, lim û ϕ = lim t u ϕ (t) and lim û z = lim t u z (t) if these limits exist. We use the equations û ϕ = Ĝ 1 R ˆp ] ϕ + w ϕ + ĝˆω, û z = Ĝ ˆp ] z + w z.

21 DERIVATION OF THE REYNOLDS TYPE EQUATION 97 and the problem for ˆp 1 ˆp ( ˆξ R ϕ ϕ + ˆξw ϕ) + R ˆp ( ˆξ z z + ˆξw z) = ˆηˆω ˆξ ˆp ] (8.1) z + ˆξw z = 1 h(ϕ,) e z ĝ dρ, z= R ˆξ ˆp ] z + ˆξw z = 1 h(ϕ,l) e z ĝ L dρ, z=l R ˆp is 2π-periodic in ϕ. in Ω Let us first show that lim ˆp = p. The difference ˆπ = p ˆp is the solution of the problem ] 1 div ϕ,z ˆξ ϕ,z ˆπ = div ϕ,z (ˆξ 1 ] 12 h3 ) ϕ,z ˆp ξw ˆ + ˆηˆω 1 h 2 ϕ µω, ˆξ ϕ,zˆπ 1 (ˆξ 1 ] 12 h3 ) ϕ,z ˆp + ξw ˆ ν = ( ˆB α B α() ), for α =, L, ˆπ is 2π-periodic in ϕ. The estimate ϕ,z ˆπ L 2 (Ω) C( 1 (ˆξ 1 12 h3 ) ϕ,z ˆp L 2 (Ω) + ˆξw L 2 (Ω) + ˆηˆω 1 h 2 ϕ µω L 2 (Ω) + ĝ g H 1/2 (S ) + ĝ L g L H 1/2 (S L)). is valid for ˆπ. From the last result we see that for ˆπ = p ˆp holds the estimate ϕ,z ˆπ L2 (Ω) C( (ˆξ 1 12 h3 ) ϕ,z ˆp L2 (Ω) + ˆξw L2 (Ω) + ˆηˆω 1 h 2 ϕ µω L2 (Ω) + ĝ g H 1/2 (S ) + ĝ L gl H 1/2 (S L)). As lim ˆξ = 1 12 h3 and lim ˆη = 1 2 ϕ µ in L (Ω), we get that lim ϕ,z ( ˆp p ) = in L 2 (Ω). This implies that lim t p(t) = p in H 1 (Ω) L 2 (Ω). We consider now the difference û ϕ u ϕ L 2 (C). As lim Ĝ = 1 2 (ρ h)ρ and lim ĝ = 1 ρ h u L (C), we can conclude that lim t u(t) = u in L 2 (C). Proof of Theorem 3.4. We consider the difference π between p and p(t), π = p p(t). From the proof of the previous theorem for ˆπ = p ˆp h

22 98 A. DUVNJAK we have the following estimate As ϕ,z ˆπ L 2 (Ω) C( 1 (ˆξ 1 12 h3 ) ϕ,z ˆp L 2 (Ω) + ˆξw L 2 (Ω) Re ˆξ bh + bh cos bh 2 sin bh = 2b 3 cos bh 2 + ˆηˆω 1 h 2 ϕ µω L 2 (Ω) + ĝ g H 1/2 (S ) + ĝ L g L H 1/2 (S L)). bβ 1 + bβ 1 cos bβ 1 2 sin bβ 1 2b 3 cos bβ1 2 >, for µ = b2 > π2, b R, we can conclude that ˆp( λ) is defined for β2 2 close to and lim exp(λt) ϕ,z(p(t) p ) L2 (Ω) = lim t + ϕ,z(ˆp( λ) p λ ) L2 (Ω) ( C lim 1 + λ (ˆξ( λ) 1 12 h3 ) ϕ,z ˆp L2 (Ω) + ˆξ( λ)w L2 (Ω) + ˆη( λ)ˆω( λ) 1 h 2 ϕ µ ω λ L 2 (Ω) ) + ĝ ( λ) g λ H 1/2 (S ) + ĝ L ( λ) g L λ H 1/2 (S L) ( C lim + λ (ˆξ( λ) 1 12 h3 ) L2 (Ω) + ˆξ( λ) L2 (Ω) + ˆη( λ)ˆω( λ) 1 ) h 2 ϕ µ ω λ L 2 (Ω) +C lim exp(λt) ( g (t) g t H 1/2 (S ) + g L (t) gl ) H 1/2 (S L) =, for λ < λ 1, π2 β 2 2 µ. This proves the assertion of Theorem 3.4. At the end we give a regularity result for the limit problem Regularity in time. The regularity in time for the limit velocity and the limit pressure is given by the following theorem. Theorem 8.1. Let (u, p) be a solution to the problem (3.1) (3.2). Then we have tp C(, T ]; H 1 (Ω) L 2 (Ω)), (8.2) tu C(, T ]; L 2 (C)). (8.3)

23 DERIVATION OF THE REYNOLDS TYPE EQUATION 99 Moreover, lim t t p(t) H 1 (Ω) L 2 (Ω) = lim t t u(t) L 2 (Ω) =. Proof. Using the finite difference in, from the problem (5.5) we get the problem for ( ˆP P () ). Multiplying that problem by ( ˆP P () ), after integrating over po Ω, for a large we get the estimate ϕ,z ( ˆP P () ) L 2 (Ω) ( C (ˆξ + ˆξ ) ϕ,z( ˆP P () ) L 2 (Ω) + (ĝ g () ) H 1/2 (S ) + (ĝ L g ) L() ) H 1/2 (S L) ( 1 C 2 ϕ,z( ˆP P () ) L 2 (Ω) + ĝ + g () 2 H 1/2 (S ) + ĝ L + g ) L() 2 H 1/2 (S L) C( ) C 3. For <, we have the estimate ϕ,z ( ˆP P () ) L 2 (Ω) ( C (ˆξ + ˆξ ) ϕ,z( ˆP P () + ĝ L + g ) l() 2 H 1/2 (S l ) C C 3. ) L 2 (Ω) + ĝ + g () 2 H 1/2 (S ) ( ĝ + g () H 1/2 (S ) ĝ L + g L() H 1/2 (S L) As ˆp = ˆP + ˆP = ˆP P () + ( ˆP P () following estimate ˆp ϕ,z L 2 (Ω) C 2. ˆp ), for ˆp ) we can infer the is the Laplace transform of t p. Using the representation formula for the inverse Laplace transform we can conclude that tp C(, T ]; H 1 (Ω) L 2 (Ω)). As û = Ĝ ϕ,z ˆp + Ĝ ˆp ϕ,z + ĝ ˆω ],

24 1 A. DUVNJAK Ĝ = O( 1 ) and 1 2 (ĝ ˆω) = O( ) for a large, with the same argument, 2 we get the assertion (8.3). Roughly speaking this theorem says that p is singular at t =. For t >, p is regular in time, with the space regularity depending on the functions g α, α =, L. References 1] G. Bayada and M. Chambat, The Transition Between the Stokes Equations and the Reynolds Equation: A Mathematical Proof, Appl. Math. Optim. 14 (1986), ] A. Duvnjak, Derivation of non linear Reynolds type Problem for Lubrication of a Rotating Shaft, ZAMM 82 (22), ] A. Duvnjak, Neki nižedimenzionalni modeli za opisivanje toka fluida kroz tanko područje, Thesis, University of Zagreb, ] A. Duvnjak and E. Marušić-Paloka, Derivation of the Reynolds Equation for Lubrication of a Rotating Shaft, Arch. Math. 36 (2), ] A. Duvnjak and E. Marušić-Paloka, Correctors for Reynolds Equation Describing the Process of Lubrication of a Rotating Shaft, Proceedings of STAMM 98, Nica, 1998, Chapman and Hall Monographs and Surveys in Pure and Applied Mathematics 16 (2), ] H. Le Dret, Problémes variationnels dans les multidomaines, R.M.A. 19, Masson, Paris, ] J. L. Lions, Problémes aux Limites en Théorie des Distributions, Acta Mathematica 94 (1955), ] A. Mikelić, Mathematical Derivation of the Darcy type Law with Memory Effects, Governing Transient Flow Through Porous Media, Glasnik Matematički 29(49) (1994), ] S. A. Nazarov, Asimptotic Solution of the Navier Stokes Problem on the Flow of a Thin Layer of Fluid, Siberian Math. J. 31 (199), ] Szeri, A. Z., Fluid Film Lubrication, Theory and Design, Cambridge University Press, ] R. Temam, Navier Stokes Equations, Theory and Numerical Analysis, North Holland, Amsterdam, ] V. S. Vladimirov, Equations of Mathematical Physics, Mir Publishers, Moscow, Faculty of Electrical Engineering and Computing Department of Applied Mathematics University of Zagreb Unska 3 1 Zagreb CROATIA Received: Revised: