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1 INSTITUTE OF MATHEMATICS THE CZECH ACADEMY OF SCIENCES A rigorous derivation of the stationary compressible Reynolds equation via the Navier Stokes equations Ionel S. Ciuperca Eduard Feireisl Mohammed Jai Adrien Petrov Preprint No PRAHA 2017

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3 A rigorous derivation of the stationary compressible Reynolds equation via the Navier Stokes equations I. S. Ciuperca E. Feireisl M. Jai A. Petrov May 17, 2017 Abstract We provide a rigorous derivation of the compressible Reynolds system as a singular limit of the compressible (barotropic) Navier Stokes system on a thin domain. In particular, the existence of solutions to the Navier Stokes system with non homogeneous boundary conditions is shown that may be of independent interest. Our approach is based on new a priori bounds available for the pressure law of hard sphere type. Finally, uniqueness for the limit problem is established in the 1D case. Key words. compressible fluids, stationary Navier Stokes equations, thin films, Bogovskii s operator, lubrication, Reynolds equation AMS Subject Classification. 35Q30, 76A20, 76D05, 76D08, 76N10 1 Introduction The asymptotic behavior of fluids in thin domains is formally described by a system of equations proposed by O. Reynolds [20]. If the fluid is compressible, the resulting system is nonlinear involving the density, the pressure and the vertical derivative of the horizontal component of the velocity as unknowns. A rigorous derivation of the Reynolds system in the compressible case is substantially hampered by a lack of analytical results concerning solvability of the compressible (stationary) Navier Stokes system, where the tangential component of the velocity is prescribed on the boundary of the physical domain. Let R N, N = 2, 3, be a bounded domain of class C 2+ν. We denote by u, p and ϱ the velocity field, pressure and density, respectively. We introduce also the symmetric part of the velocity gradient e(u) def = 1 2 ( u + ( u)t ), the density of the fluid ϱ 0 and the viscosity coefficients λ > 0 and µ > 0. Here ( ) T denotes the transpose of a tensor. Then the compressible (barotropic) Navier Stokes system describing the distribution of the density and the velocity field reads: div(ϱu) = 0 in, div(ϱu u) + p(ϱ) = div(s( u)) in, (1.1a) (1.1b) Université de Lyon, CNRS, Institut Camille Jordan UMR 5208, 43 boulevard du 11 novembre 1918, F Villeurbanne Cedex, France (ciuperca@math.univ-lyon1.fr) Institute of Mathematics of the Academy of Sciences of the Czech Republic, Žitná 25, CZ Praha 1, Czech Republic (feireisl@math.cas.cz) Université de Lyon, CNRS, INSA de Lyon Institut Camille Jordan UMR 5208, 20 Avenue A. Einstein, F Villeurbanne, France (mohammed.jai@insa-lyon.fr, apetrov@math.univ-lyon1.fr) 1

4 2 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov where S( u) def = 2µe(u) + λ div(u)i is the viscous part of the stress tensor with I denoting the identity matrix. The system (1.1) is endowed with the following boundary conditions u = ū on, (1.2) where denotes the boundary of and ū : R N such that div(ū) = 0 and ū n = 0 on. (1.3) Furthermore, in accordance with the latter condition in (1.3), we assume that the total mass of the fluid is given, namely ϱdx = M > 0. (1.4) Solvability of problem (1.1) (1.4) is largely open, in particular when ū is large compared to the inverse value of the Reynolds and Mach numbers. Strong (classical) solutions have been constructed by several authors on condition of smallness of the data, see e.g. Plotnikov, Ruban, and Sokolowski [17, 18] or Piasecki and Pokorný [16] for slightly different boundary conditions. If the boundary data are large, the theory of weak solutions must be used. Although the issue of compactness of solutions to problem (1.1) is nowadays relatively well understood, see the seminal monograph by Lions [12] as well as other numerous recent extensions of the theory listed in Plotnikov and Weigant [19], the problem of suitable a priori bounds in the case on the non homogeneous boundary conditions and a proper construction of solutions seems largely open. Chupin and Sard [5] applied the framework proposed by Bresch and Desjardins [6], where the viscosity coefficients depend on the density in a specific way. This approach requires additional friction term in the momentum equation, as well an extra boundary condition for the density, div(ϱu u) + p(ϱ) = divs( u) rϱ u u, (1.5) ϱ = ϱ b (1.6) where ϱ b is constant on each connected component of. Although this ansatz provides (formally) very strong estimates, notably on ϱ, the resulting problem is obviously overdetermined and definitely not solvable for any boundary data (1.2), (1.3) and (1.6). Indeed we may consider the fluid domain to be the space between two concentric balls, = { x R N : 0 < R 1 < x < R 2 }, with the boundary conditions { ϱ 1 if x = R 1, ū = 0 and ϱ b = ϱ 2 if x = R 2 with two different constants ϱ 1 ϱ 2. (1.7) Here R 1 > 0 and R 2 > 0 are two positive constants. Taking the scalar product of (1.5) with u, integrating the resulting equation by parts, and using (1.1a), we check easily that u 0 in. This in turn yields p(ϱ) = 0 in, meaning ϱ constant in in contrast with (1.7). The lack of rigorous results in the framework of weak solutions of (1.1) (1.4) is mainly caused by very poor a priori estimates available for the density. On the other hand, the density of real fluids admits natural bounds, 0 ϱ ϱ, (1.8)

5 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations 3 where the lower bound is obvious while the upper bound is imposed by the molecular theory - the volume of a real fluid cannot be made arbitrarily small, see e.g. [10]. This restriction is reflected by a general equation of state of a real fluid in the form p(ϱ) = ϱθz(ϱ), (1.9) where Θ > 0 is the absolute temperature and Z the compressibility factor, Z(ϱ) + for ϱ ϱ, (1.10) see e.g. [11]. The analysis in the present paper is based on the hypotheses (1.9) and (1.10). The paper is organized as follows. In Section 2, the mathematical formulation of the problem in terms of the velocity field and the density is given. Then a family of approximate problems is introduced and a fixed point procedure applied to show an existence result for these problems. Next, under appropriate assumptions on the data, we show that the problem possesses a weak solution which is obtained as a limit of a sequence of solutions of the approximate problems. In Section 3, further properties of the solution are exhibited by using more specific techniques like the Bogovskii s and extension operators, and an anisotropic version of the Sobolev interpolation inequality (see Section A). Then the justification of the Reynods system is derived. Finally, the uniqueness result for one dimensional problem follows from the strong monotonicity of the pressure. 2 Existence of weak solutions The weak formulation associated to (1.1) (1.2) is obtained by multiplying (1.1a) by ϕ C 1 ( ; R) and (1.1b) by ψ W 1,2 0 (; RN ), respectively. Integrating formally these identities over, we end up with the following problem: Find u W 1,2 (; R N ) and ϱ L (; R + ), with ϱ [0, ϱ), satisfying u = ū on, ϱ dx = M, and p(ϱ) L 2 (; R) such that ϕ C 1 ( ; R) : ϱu ϕ dx = 0, ψ W 1,2 0 (; ( ) RN ): (ϱu u): ψ + p(ϱ) div ψ dx= S( u): ψ dx. (2.1a) (2.1b) When there is no confusion, we will use simply the notation X() instead of X(; Y) where X is a functional space and Y a vectorial space. Furthermore, we denote def ϱ M = M. As ϱ < ϱ a compatibility condition ϱ M < ϱ must be imposed. The functional spaces used throughout the present paper are commonly used in the literature; the reader may consult e.g. [1, 4]. We assume that the pressure p(ζ) : [0, ϱ) R is continuously on [0, ϱ), differentiable on (0, ϱ) and satisfies assumptions: p(0) 0 and p (ζ) 0 for all ζ (0, ϱ), lim(p(ζ)) = +. ζ ϱ (2.2a) (2.2b)

6 4 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov As a matter of fact, the construction we propose below requires the estimates available for elliptic operators, notably Laplacian, on bounded domain. Accordingly, we impose rather strong regularity properties on. Our goal is to show the following existence result: Theorem 2.1. Let R N, N = 2, 3 of class C 2+ν be a bounded domain. Let the pressure be given as p = p(ϱ) satisfying (2.2). Let M > 0 be given such that ϱ M < ϱ. Let the boundary datum ū be given by a function ū W 1,q () such that q > N and ū n = 0 on. Then problem (1.1) (1.4) admits a weak solution in the class: ϱ L () with ϱ [0, ϱ) a.e. in, u W 1,2 (), p(ϱ) L 2 (). 2.1 Approximate problems We consider for any d N large enough the space X d def = span{w 1,..., w d } W 1,2 0 () with {w i } i=1 a complete orthogonal system in L2 () such that w i belongs to C 2,η () for all i N. For any R > 1 ϱ large enough, let T : R R and p R : [0, + [ R be the cut off function and a truncation of the pressure function, respectively, defined as follows: and p R (ϱ) def = T (ϱ) def = 0 if ϱ 0, ϱ if ϱ [0, ϱ], ϱ if ϱ ϱ, { p(ϱ) if ϱ [0, ϱ 1 R ], p ( ϱ 1 R )(ϱ ( ϱ 1 R )) + p( ϱ 1 R ) if ϱ > ϱ 1 R. Let us introduce a small parameter δ > 0. Then we consider the following family of approximate problems Find u = v + ū, v X d and ϱ W 2,2 () such that S( u) w i dx = T (ϱ)u u): w i dx + (p R (ϱ)+ δϱ) div w i dx δ (ϱu) w i dx δϱuw i dx for all i = 1,..., d, δϱ δ ϱ + div(t (ϱ)u) = δϱ M in, ϱ n = 0 on, (2.3a) (2.3b) (2.3c) Obviously, u = u on, and ϱ dx = M. (2.4) Notice that solutions to problem (2.3) depend on d, R and δ but, for simplicity, we denote these solutions by u and ϱ. In the sequel, we denote by the symbol the weak convergence and by the strong convergence and by and the continuous and compact embeddings, respectively.

7 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations Preliminary results This section is devoted to some preliminary results that will play a crucial role in the present work. First, we shall need the following result: Lemma 2.2. Assume that u W 1,2 () such that u n = 0 on holds. Then there exists a unique solution ϱ W 2,2 () to problem (2.3b) (2.4) such that ϱ 0. If, in addition, div u L (), then there exists ϱ > 0 such that ϱ ϱ. Furthermore, the following stability result holds; if u n u in W 1,2 (), and ϱ n is the solution to (2.3b) corresponding to u n, then we have where ϱ is the solution corresponding to u. ϱ n ϱ in W 2,2 () weak, Proof. The proof of existence and uniqueness results can be obtained by using an approach analogous to [15]. However for reader s convenience, a detailed proof is given in a more general setting in the Appendix. Now taking ϱ > 0 such that ϱ(δ div u) δϱ M, we get by the comparison principle that ϱ ϱ. On the other hand, assume that u n converges strongly to u in W 1,2 (). It follows from (2.4) that ϱ n L 1 () = M. Furthermore, by means of the standard elliptic estimates, there exists a constant C 1 > 0, C 1 depending on δ, such that Using the following identity ϱ n W 2,2 () C 1 ( 1 + div(t (ϱn )u n ) L 2 ()). (2.5) as well as (2.5), we get div(t (ϱ n )u n ) = T (ϱ n ) div u n + T (ϱ n ) ϱ n u n, ϱ n W 2,2 () C 1 ( 1 + ϱ un W 1,2 () + ϱ n W 1,4 () u n W 1,2 ()). (2.6) According to the interpolation theory, we infer that for α ]0, 1[, there exists a constant C α > 0 depending on α such that ϱ n W 1,4 () C α ϱ n α L 1 () ϱ n 1 α W 2,2 (). (2.7) Clearly, introducing (2.7) into (2.6) we get that ϱ n is bounded in W 2,2 (), which proves the Lemma. Let us recall the following version of Schæffer s fixed point: Lemma 2.3. Let X be a Banach space and A : X X be a continuous and compact mapping. Assume that for any θ [0, 1] any fixed point u θ X of θa(u θ ) = u θ is bounded in X uniformly with respect to θ. Then A possesses at least one fixed point in X. For a detailed proof of the above lemma, the reader is referred to [8, Theorem 9.2.4]. Notice that Poincaré and Korn inequalities, enable us to deduce that there exists a constant C P > 0 such that v W 1,2 0 () : v 2 W 1,2 () C P S( v) : v dx. (2.8)

8 6 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov 2.3 Existence result associated to approximate problem and uniform bounds In this section, the existence result to Problem (2.3) is proved by using fixed point precedure and some uniform bounds are highlighted that will play a crucial role in the next sections. To this aim, we define the mapping A by A : X d X d ṽ v = A(ṽ) v = u u, where u is a solution of the problem: S( u) w i dx = T (ϱ)ũ ũ): w i dx + (p R (ϱ)+ δϱ) div w i dx δ (ϱũ) w i dx δϱũw i dx for all i = 1,..., d, u = ū on, (2.9a) (2.9b) where we have set ũ def = ṽ + ū. Here ϱ W 2,2 (), ϱ 0 is the unique solution of the problem δϱ δ ϱ + div(t (ϱ)ũ) = δϱ M in, ϱ n = 0 on, ϱdx = M. (2.10a) (2.10b) (2.10c) Here above the finite dimensional space X d may be endowed with the Hilbert topology of (). Thus the system (2.9a) has the following form: W 1,2 0 S( u) w i dx = F i, w i W 1,2 (),W 1,2 0 () for all i = 1,..., d, where F i is an element of the dual W 1,2 (). Consequently, we may deduce from Lemma 2.2, (2.8) and Lax Milgram theorem, that A is well defined. Note that the continuity of A follows from Lemma 2.2 while the compactness of A comes from the fact that X d is a finite dimensional space. Finally, to prove existence of a fixed point of A, it suffices to verify the last assumption of Lemma 2.3. Suppose that for an arbitrary θ ]0, 1], there exists v θ a fixed point of θa. For simplicity, the subscript θ will be omitted in the sequel. Then we consider the following problem: δϱ δ ϱ + div(t (ϱ)u) = δϱ M in, S( v + θ ū) : w i dx = θ (T (ϱ)u u) : w i dx + θ (p R (ϱ) + δϱ) div w i dx θδ (ϱu) w i dx θ δϱuw i dx for all i = 1,..., d, ϱ n = 0 and u = ū on, ϱ dx = M. (2.11a) (2.11b) (2.11c) (2.11d)

9 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations 7 Since div ū = 0, we may deduce from (2.11b) that (S( u) S( ū)) : (u ū) dx + θ div(t (ϱ)u u) (u ū) dx = θ (p R (ϱ) + δϱ) div u dx θδ (ϱu) : (u ū) dx θ δϱu (u ū) dx θ S( ū) : (u ū) dx. Notice that the renormalized form of (2.11a) is given by div(h(ϱ)u) + (G (ϱ)t (ϱ) H(ϱ))divu δg (ϱ) ϱ + δg (ϱ)ϱ = δϱ M G (ϱ), (2.12) (2.13) where H and G are Lipschitz functions satisfying H (ϱ) = G (ϱ)t (ϱ). Now we choose for any ϱ > 0, G and H as follows: where Then we infer that We also set G(ϱ) = G R,δ (ϱ) and H(ϱ) def = p R (ϱ M ) δϱ M + ϱ G def p R R,δ (ϱ) = (y) + δ dy. ϱ M T (y) ϱ ϱ M G R,δ (y)t (y) dy, G R,δ (ϱ)t (ϱ) H(ϱ) = p R(ϱ) + δϱ. (2.14) ϱ G def p (y) + δ δ (ϱ) = ϱ M T (y) Consequently, integrating (2.13), we find (p R (ϱ)+ ( δϱ)divu dx = δg R,δ (ϱ) ϱ δg R,δ (ϱ)ϱ+δϱ MG R,δ (ϱ)) dx. (2.15) Since v is regular enough, we may deduce from Lemma 2.2 that ϱ ϱ with ϱ > 0 and it follows that G R,δ (ϱ) L (). We multiply (2.15) by θ and we use the resulting expression in (2.12) to get (S( u) S( ū)) : (u ū) dx + δθ ϱg R,δ (ϱ)dx + δθ ϱ 2 G R,δ (ϱ) dx = θδϱ M G R,δ (ϱ) dx (2.16) θ div(t (ϱ)u u) (u ū) dx δθ (ϱu) : (u ū) dx θδ ϱu (u ū) dx θ S( ū) : (u ū) dx. Notice that u n = 0 on implies that div(t (ϱ)u u) u dx = 1 div(t (ϱ)u) u 2 dx. (2.17) 2 dy.

10 8 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov Carrying (2.11a) into (2.17), we get div(t (ϱ)u u) u dx = 1 2 ( δϱ δ ϱ δϱm ) u 2 dx. (2.18) Furthermore, the boundary condition (2.11c) leads to 1 u 2 ϱ dx = u ( u ϱ) dx. (2.19) 2 Inserting (2.19) into (2.17), we find div(t (ϱ)u u) u dx = δ ϱ u 2 dx 2 + δ u ( u ϱ) dx δϱ M u 2 dx. 2 On the other hand, we have div(t (ϱ)u u) ū dx = ( δϱ + δ ϱ + δϱ M )u ū dx + T (ϱ)(u u) ū dx = δ ϱu ū dx δ ( ϱ u) ū dx δ ( ϱ ū) u dx + δϱ M u ū dx + T (ϱ)(u u) ū dx. We may also observe that (ϱu) : (u ū) dx = u (u ū) ϱ dx + ϱ u : (u ū) dx. According to (2.20), (2.21) and (2.22) in (2.16), we get (S( u) S( ū)) : (u ū) dx + δθ ϱg R,δ (ϱ) dx + δθ + θδϱ M u 2 dx + δθ ϱ u 2 dx + δθ ϱ u 2 dx = θδϱ M θδϱ M u ū dx δθ ( ϱ u) ū dx + θ T (ϱ)(u u) ū dx + δθ ϱ u : ū dx θ S( ū) : (u ū) dx. ϱ 2 G R,δ (ϱ) dx G R,δ (ϱ) dx By using (2.8), we obtain 1 u ū 2 W C 1,2 () + δθ ϱ u 2 dx + δθ ϱg R,δ (ϱ) dx P ( p (ϱ) + δ ) + δθ ϱ 2 dx + θδϱ M u 2 dx θδϱ M G R,δ (ϱ) dx T (ϱ) 2 δθ ( ϱ u) ū dx + θ T (ϱ)(u u) ū dx + θδϱ M u ū dx + δθ ϱ u : ū dx θ S( ū) : (u ū) dx. (2.20) (2.21) (2.22) (2.23)

11 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations 9 The next observation is that ϱ M G R,δ (ϱ) dx ϱ M G R,δ (ϱ) dx {ϱ ϱ M } ϱ M G R,δ (ϱ) dx + ϱ M {ϱ M ϱ λϱ M } MG R,δ (λϱ M) + 1 ϱg R,δ (ϱ) dx λ {ϱ>λϱ M } MG R,δ (λϱ M) + 1 ϱg R,δ (ϱ) dx, λ {ϱ ϱ M } {ϱ>λϱ M } G R,δ (ϱ) dx where λ > 1 has been chosen in such a way that λϱ M < ϱ. Note that G R,δ is independent of R in the interval [0, λϱ M ] as soon as ϱ 1 R > λϱ M. Consequently, going back to (2.23), we obtain 1 u ū 2 W C 1,2 () + δθ ϱ u 2 dx + δωθ ϱg R,δ (ϱ) dx P {ϱ ϱ M } ( p (ϱ) + δ ) + δθ ϱ 2 dx + θδϱ M u 2 dx δθϱ M u ū dx T (ϱ) 2 δθ ( ϱ u) ū dx + θ T (ϱ)(u u) ū dx + δθ ϱ u: ū dx δθ ϱg R,δ (ϱ) dx θ S( ū) : (u ū) dx + δθc(m, p), {0 ϱ ϱ M } (2.24) for certain ω > 0, where C(M, p) depends only on M and the structural properties of the function p. In particular, the estimate is independent of R and δ. We focus now on the difficult terms that should be controlled in (2.24). First, we observe that there exists a constant C 1 > 0 independent of R and δ such that T (ϱ)u u ū dx ϱ u L 2 () u L 4 () ū L 4 () C 1 ū L 4 () u 2 W 1,2 () 2C 1 ū L 4 ()( u ū 2 W 1,2 () + ū W 1,2 ()). as long as N = 2, 3. In accordance with Lemma A.1 established below, the extension ū can be chosen by such a way that ū L 4 () is arbitrarily small provided ū W 1,p () with p > N. We will therefore consider such an extension supposing Secondly, we have ū W 1,p () with p > N and ū n = 0 on. (2.25) δθ ( ϱ u) ū dx θδ 3 2 +r ū 2 L () ϱ 2 L 2 () (2.26) + θδ 1 2 r ū L ()( u ū 2 W 1,2 () + ū 2 L 2 () ) with r (0, 1 2 ). The first term on the right hand side is controlled by θδ 3 ϱ 2 2 T (ϱ) dx (take δ small enough) while the second one is controlled by 1 C P u ū 2 W 1,2 (). On the other hand, we have δθ ϱ u : ūdx δθ ϱ u 2 dx + δθ 2 4 ū 4 L 4 () + δθ 4 ϱ 2 L 2 ().

12 10 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov Now from the definition of G R,δ, we have G R,δ (ϱ) 1 ϱ( pr (ϱ) p R (ϱ M ) + δ(ϱ ϱ M ) ) for ϱ ϱ M. (2.27) The expression of p R leads to G R,δ (ϱ) ηϱ ϱ for ϱ ϱ, with η > 0 and R taken large enough. We have ϱ 2 L 2 () = ϱ 2 dx + ϱ 2 dx ϱ 2 + ϱ ϱ ϱ ϱ> ϱ η ϱ> ϱ ϱg R(ϱ) dx. We conclude applying Lemma 2.3 that a solution (u, ϱ) to problem (2.3) satisfies as well the following inequality: ( p u ū 2 W 1,2 () + δ ϱ u 2 dx + δ R (ϱ) + δ T (ϱ) + δ G R,δ (ϱ)ϱ dx C 2, {ϱ ϱ M } ) ϱ 2 dx (2.28) where C 2 > 0 is a constant independent on d, R and δ. This proves the uniform bound of v θ = v, solution of (2.11), and thus the existence result to problem (2.3) by using Lemma 2.3. Besides, the standard elliptic theory implies that there exists a constant C δ > 0 depending only on δ such that ϱ W 2,2 () C δ. (2.29) 2.4 Limit d + The passage to the limit as d tends to + in (2.3) follows from estimates (2.28) and (2.29) in (2.3), the verification is let to the reader. Then, we may conclude that the limit solution, denoted once again by (u, ϱ), satisfies the following system: Find u = v + ū, v W 1,2 0 () and ϱ W2,2 () such that for all w W 1,2 0 (), S( u) w dx = (T (ϱ)u u) : w dx + (p R (ϱ) + δϱ) div w dx δ (ϱu) w dx δϱuw dx, δϱ δ ϱ + div(t (ϱ)u) = δϱ M in, u = ū and ϱ n = 0 on, ϱ dx = M. (2.30a) (2.30b) (2.30c) (2.30d) 2.5 Limit R + In the present section, our goal consists to perform the asymptotic limit for R tending to + in (2.30a) (2.30c). Here the family of solutions of (2.30a) (2.30c) is denoted by (u R, ϱ R ) R> 1. ϱ Now, keeping δ > 0 fixed, the compactness provided by the artificial viscosity approximation

13 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations 11 plays a crucial role as well as the inequalities (2.28) and (2.29) which remain valid when (u, ϱ) is replaced by (u R, ϱ R ). Since (2.29) holds and N = 2, 3, we may assume Next we observe that the limit density satisfies ϱ R ϱ in C 0 ( ) and ϱ 0. (2.31) Indeed, let us take for any Z such that ϱ M < Z < ϱ: ( 2 F Z (ϱ) = min 1, ϱ Z (ϱ Z)+). 0 ϱ < ϱ a.e. in. (2.32) We proceed by contradiction; let us suppose that meas{ϱ ϱ} > 0. Then for any Z ]ϱ M, ϱ[, we get F Z (ϱ) dx 1 dx > 0. (2.33) Now we have F Z (ϱ R )dx 1 G R,δ (Z) 1 ZG R,δ (Z) {ϱ ϱ} {ϱ R Z} {ϱ R Z} G R,δ (Z)dx ϱ R G R,δ (ϱ R)dx. In view of the uniform bounds established in (2.28), we may let R tends to + obtaining that there exists a constant C δ > 0 independent on R such that C δ F Z (ϱ)dx ZG δ (Z). By using (2.27) (remains still valid if G R,δ (Z) is replaced by G δ (Z) and p R by p), we deduce that G δ (Z) tends to + as z tends to ϱ. According to the above inequality, we may conclude that this fact contradicts the estimate (2.33). Furthermore, we have 0 ϱ ϱ δ where ϱ δ is a constant such that 0 < ϱ δ < ϱ. Finally, we need to exhibit bounds on the pressure. To this aim, we remark that (2.30a) and the estimate (2.28) lead to p R (ϱ R ) W 1,2 () c(δ). (2.34) In addition, as ϱ RG R,δ (ϱ R)dx is uniformly bounded and using (2.27), we deduce that p R (ϱ R ) L 1 () c(δ), (2.35) where c(δ) > 0 is a constant depending on δ. Using (2.34) and (2.35), there exists a limit denoted by p(ϱ) such that In view of (2.31) and (2.32), we have p R (ϱ R ) p(ϱ) in L 2 () weak. p(ϱ) = p(ϱ). Letting R tends to + in (2.30a) (2.30c), we obtain the following system of equations: div(ϱu) δ ϱ + δϱ = δϱ M in, div(ϱu u) + p(ϱ) + δ ϱ = div(s( u)) + δ (ϱu) δϱu in, ϱ n = 0 and u = ū on, ϱ dx = M, (2.36a) (2.36b) (2.36c) (2.36d)

14 12 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov 2.6 Limit δ 0 Our ultimate goal is to perform the limit δ tending to 0 in system (2.36). Let (u δ, ϱ δ ) be the approximate solutions of problem (2.36). In addition to the estimates (2.28) that hold uniformly for δ tending to 0, we have at hand a rather strong bound on the density, namely, 0 ϱ δ < ϱ a.e. in. (2.37) Consequently, passing to suitable subsequences as the case may be, we may suppose ϱ δ ϱ in L () weak-* and u δ u in W 1,2 () weak, (2.38) where 0 ϱ ϱ, ϱdx = M and u = ū on A preliminary result Lemma 2.4. Assume that p : [0, ϱ) R belongs to C[0, ϱ), is non decreasing and lim ϱ ϱ p(ϱ) =. Let ϱ δ L () be a sequence such that 0 ϱ δ < ϱ and p(ϱ δ ) L β () C, with C a constant independent on δ and β (1, + ). Suppose that ϱ δ ϱ in L () weak-*, p(ϱ δ ) p(ϱ) and ϱ δ p(ϱ δ ) ϱp(ϱ) in L β () weak. Then we have 0 ϱ < ϱ a.e. in, If, in addition, ϱp(ϱ) = ϱp(ϱ) then p(ϱ) = p(ϱ). (2.39a) (2.39b) Proof. Clearly, 0 ϱ ϱ. Let us show the strict inequality in (2.39a) by contradiction. To this aim, we denote by A def = {x : ϱ(x) = ϱ} and we suppose that A > 0. Taking ϕ(x) = 1 A, the indicator function on A, as a test function in the weak-* convergence of ϱ δ to ϱ. Since 0 ϱ δ < ϱ, A (ϱ δ ϱ) dx tends to 0 that is ϱ δ converges strongly to ϱ in L 1 (A). We deduce (up to a subsequence of δ ) that which implies ϱ δ (x) ϱ a.e. in A, A (p(ϱ δ (x))) β dx +. This contradicts the hypothesis and proves (2.39a). Let us now establish (2.39b). For any η > 0 small enough, we consider the continuous function p η C 0 ([ η 2, ϱ)) given by { p(z) for z (0, ϱ), p η (z) def = p(0) for z [ η 2, 0).

15 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations 13 Furthermore, it is convenient to define def η = {x : 0 ϱ(x) ϱ η}. Now for an arbitrary ζ belonging to L () with η 2 ζ ϱ η 2, we have immediately that (p η (ϱ δ ) p η (ζ))(ϱ δ ζ) 0. Therefore passing to the limit as δ tends to 0 and using the assumption ϱp(ϱ) = ϱp(ϱ), we conclude that (p(ϱ) p η (ζ))(ϱ ζ) 0 in. Integrating over η and taking ζ = ϱ + ɛψ with ψ L () arbitrary and ɛ > 0 a small parameter and passing to the limit as ɛ tends to 0, we find (p(ϱ) p(ϱ))ψ dx 0. η Therefore we obtain p(ϱ) = p(ϱ) on η for any η > 0 small enough. Since (2.39a) holds, it follows that = η>0 η A which concludes the proof Limit in the field equations Revisiting (2.28), we see that δ 3/2 ϱ δ 2 L 2 () c. Consequently, in view of (2.38) and compactness of the embedding W 1,2 () L q () with q [1, 6) if N = 3 and q 1 arbitrary finite if N = 2, it is easy to perform the limit δ 0 in (2.36a) obtaining ϕ C 1 ( ) : ϱu ϕdx = 0 which implies that div(ϱu) = 0 in D (). (2.40) In addition, as ϱ L () and u W 1,2 (), we may apply the regularizing technique of DiPerna and Lions [7] to deduce the renormalized version of (2.40), namely, we have ϕ C 1 ( ( ) : b(ϱ)u ϕ + (b(ϱ) b (ϱ)ϱ)(divu)ϕ ) dx = 0, (2.41) and any continuously differentiable b. In order to perform the limit in (2.36b), we have to control the pressure. To this end, we multiply (2.36b) by B[ϱ δ ], div(b[ϱ δ ]) = ϱ δ ϱ M, B = 0, where B is the Bogovskii operator introduced in A. As B vanishes on the boundary, we may integrate by parts obtaining ( p(ϱδ ) + ) ( δϱ δ ϱδ dx = p(ϱδ ) + ) δϱ δ ϱm dx ϱ δ u δ u δ : B[ϱ δ ] dx + S( u δ ) : B[ϱ δ ] dx + δ (ϱ δ u δ ) B[ϱ δ ] dx + δ ϱ δ u δ B[ϱ δ ] dx. (2.42) The Bogovskii operator B : L q () W 1,q 0 (), 1 < q <, is bounded; whence we may use the uniform bounds (2.28) and (2.37) to find that p(ϱ δ )ϱ δ dx c + p(ϱ δ )ϱ M dx.

16 14 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov Now, similarly to the previous part, we choose λ > 1 such that λϱ M < ϱ. Accordingly, p(ϱ δ )ϱ δ c + c + p(ϱ δ )ϱ M dx p(ϱ δ )ϱ M dx + {ϱ δ λϱ M } c + Mp(λϱ M ) + 1 λ ϱ δ p(ϱ δ ) dx; {ϱ δ >λϱ M } p(ϱ δ )ϱ M dx whence According to (2.42) and (2.43), we get ϱ δ p(ϱ δ ) dx c. (2.43) p(ϱ δ ) dx c. (2.44) Now, we repeat the same procedure with the multiplier B[p α (ϱ δ )], where α > 0 will be fixed below. Similarly to (2.42), we have By virtue of (2.44), we get ( p(ϱδ ) + ) δϱ δ p α (ϱ δ ) dx = p(ϱδ ) + ( ) δϱ δ p α (ϱ δ ) dx 1 ϱ δ u δ u δ : B[p α (ϱ δ )] dx + S( u δ ) : B[p α (ϱ δ )] dx + δ (ϱu δ ) B[p α (ϱ δ )] dx + δ ϱu δ B[p α (ϱ δ )] dx. α (0, 1] : p α (ϱ δ ) L 1 α () c(α), (2.45) therefore all integrals on the right hand side of (2.45) remain bounded uniformly for δ tending to 0 for a suitably small α > 0. Accordingly p(ϱ δ ) L α+1 () c for a certain α > 0, (2.46) and p(ϱ δ ) p(ϱ) in L α+1 () weak. (2.47) Letting δ tends to 0 in (2.36b), we obtain ( ) ϱu u : ϕ + p(ϱ) divϕ dx = for any ϕ C 1 c(), or, equivalently, div(ϱu u) + p(ϱ) = div(s( u)) in D (). S( u) : ϕ dx (2.48)

17 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations Compactness of the density (pressure) The existence proof will be complete as soon as we show that p(ϱ) = p(ϱ) a.e. in. (2.49) To this aim, we employ the nowadays standard method based on the weak continuity of the effective viscous flux developed by Lions [12]. Our goal consists in showing that p(ϱ δ )ϱ δ dx = p(ϱ)ϱ dx. (2.50) lim δ 0 and applying Lemma 2.4 to deduce (2.49). To prove (2.50), we first renormalize equation (2.36a) to get div(b(ϱ δ )u δ ) + (b (ϱ δ )ϱ δ b(ϱ δ ))divu δ = δ ϱ δ b (ϱ δ ) δϱ δ b (ϱ δ ) + δϱ M b (ϱ δ ). Thus, integrating by parts the above expression, we obtain (b (ϱ δ )ϱ δ b(ϱ δ ))divu δ dx = δ b (ϱ δ ) ϱ δ 2 dx + δ If b is convex, we have (b (ϱ δ )ϱ δ b(ϱ δ ))divu δ dx δm max ϱ [0, ϱ] (b (ϱ)) δ In particular, for b(ϱ) = ϱ log(ϱ) we get ϱ δ divu δ dx δm(1 + log( ϱ) ) δ (ϱ M ϱ δ )b (ϱ δ ) dx. ϱ δ b (ϱ δ ) dx. ϱ δ (1 + log(ϱ δ )) dx. Finally, letting δ 0, we may conclude that lim sup δ 0 ϱ δ divu δ dx 0. (2.51) The next step consists in multiplying (2.36b) by ϕ 1 [ϕϱ δ ] where 1 denotes the inverse of the Laplacian on R 3, specifically a pseudodifferential operator with Fourier s symbol 1 ξ 2 and ϕ C c (). After a bit tedious but straightforward computation, we obtain ϕ 2 p(ϱ δ )ϱ δ dx ϕs( u δ ) : 1 [ϕϱ δ ] dx R N R N + ϕ(ϱ δ u δ u δ ) : 1 [ϕϱ δ ] dx R N = (ϱ δ u δ u δ ) ϕ 1 [ϕϱ δ ] dx R N + S( u δ ) ϕ 1 [ϕϱ δ ] dx p(ϱ δ ) ϕ 1 [ϕϱ δ ] dx (2.52) R N R N δ ϱ δ ϕ 1 [ϕϱ δ ] dx δ ϕ 2 ϱ 2 δ dx R N R N + δ (ϱu) ϕ 1 [ϕϱ δ ] dx + δ ϱ δ u δ ϕ 1 [ϕϱ δ ] dx R N R N + δ ϕ (ϱu) : 1 [ϕϱ δ ] dx. R N

18 16 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov Similarly, we consider ϕ 1 [ϕϱ] as a test function in the limit equation (2.48) and using an analogous approach as we did for (2.52), we get ϕ 2 p(ϱ)ϱ dx ϕs( u) : 1 [ϕϱ] dx R N R N + ϕ(ϱu u) : 1 [ϕϱ] dx = (ϱu u) ϕ 1 [ϕϱ] dx (2.53) R N R N S( u) ϕ 1 [ϕϱ] dx p(ϱ) ϕ 1 [ϕϱ] dx. R N R N Next observe that (i) all the terms in (2.52) containing δ or δ vanish in the asymptotic limit when δ tends to 0, (ii) in view of the compactification effect of 1 : L q (R N ) W 1,q loc (RN ), all integrals on the right hand side of (2.52) converge to their counterparts in (2.53). We have shown that ϕ 2 p(ϱ δ )ϱ δ dx + ϕ(ϱ δ u δ u δ S( u δ )) : 1 [ϕϱ δ ] dx R N R N (2.54) δ 0 R N ϕ 2 p(ϱ)ϱ dx + Now we have ϕ(ϱ δ u δ u δ ) : 1 [ϕϱ δ ] dx = R N R N ϕ(ϱu u S( u)) : 1 [ϕϱ] dx. N ϕu j δ R N j=1 i=1 N ( ϱδ u i δ x i 1 [ xj [ϕϱ δ ] ) dx, where div(ϱ δ u δ ) = δ ϱ δ δϱ δ + δϱ M is a precompact subset of W 1,2 (). On the other hand, as curl[ 1 [ xj (ϕϱ δ )] vanishes, we may apply the celebrated Div Curl lemma of Murat and Tartar [14, 21] to conclude that ϕ(ϱ δ u δ u δ ) : 1 [ϕϱ δ ] dx ϕ(ϱu u) : 1 [ϕϱ] dx. R N δ 0 R N Accordingly, relation (2.54) reduces to ϕ 2 p(ϱ δ )ϱ δ dx ϕs( u δ ) : 1 [ϕϱ δ ] dx R N R N ϕ 2 p(ϱ)ϱ dx ϕs( u) : 1 [ϕϱ] dx. δ 0 R N R N Finally, we check that lim ϕs( u δ ) : 1 [ϕϱ δ ] dx ϕs( u) : 1 [ϕϱ] dx δ 0 R N R N = lim 1 [ϕs( u δ )]ϕϱ δ dx 1 [ϕs( u)]ϕϱ dx δ 0 R N R N = (λ + 2µ) ϕ 2 div(u δ )ϱ δ dx (λ + 2µ) ϕ 2 div(u)ϱ dx; R N R N (2.55) whence, in combination with (2.55), we obtain ϕ 2( p(ϱ)ϱ p(ϱ)ϱ ) dx = (λ + 2µ) ϕ 2( divuϱ divuϱ ) dx. (2.56) R N R N

19 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations 17 As the limit functions (ϱ, u) satisfy the renormalized equation of continuity, we have ϱdivu dx = 0, which, together with (2.51), yields ( divuϱ divuϱ ) dx 0. Moreover, since relation (2.56) is satisfies for any ϕ C c (), we easily deduce that ( ) p(ϱ)ϱ p(ϱ)ϱ dx 0. However, as p is non decreasing, we have p(ϱ)ϱ p(ϱ)ϱ 0; whence p(ϱ)ϱ = p(ϱ)ϱ a.e. in. In particular, we have shown (2.50) and therefore (2.49) holds. 3 Justification of the Reynolds system We consider now the following situation: def = Q ε = { x = (x h, z) : x h T D, 0 < z < εh(x h ), h C 2+ν (T D ) }, (3.1) where T D is the D dimensional torus, D = 1, 2 and ε > 0. In other words, the domain is periodic in the horizontal variable x h and bounded above in the vertical variable z by a graph of a smooth function. We consider a rescaled version of the compressible Navier Stokes system, div(ϱ ε u ε ) = 0 in Q ε, (3.2a) div(ϱ ε u ε u ε ) + 1 ε 2 p(ϱ ε) = div(s( u ε )) in Q ε, supplemented with the boundary conditions u ε = ū def = { (s, 0) if z = 0, 0 if z = εh(x h ), (3.2b) (3.3) where s is a constant vector field. The pressure p(ϱ ε ) is of the same type as in Theorem 2.1, and we fix the total mass of the fluid, we get ϱ ε dx = M ε > 0 where 0 < inf Q ε ε>0 ϱε M sup ϱ ε M < ϱ, (3.4) ε>0 def where ϱ ε M = Mε Q ε. In accordance with Theorem 2.1, problem (3.2) (3.4) admits a weak solution (ϱ ε, u ε ) for any ε tending to 0. Our goal is to study the asymptotic limit as ε tends to 0. Remark 3.1. As a matter of fact, the geometry of the underlying spatial domain is slightly different from that considered in Theorem 2.1. The existence proof, however, may be performed exactly as in Theorem 2.1.

20 18 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov 3.1 Uniform bounds We choose an extension ū ε of the boundary velocity ū such that ū ε = ū on Q ε, divū ε = 0, ū ε L (Q ε) c, h ū ε L (Q ε) c, z ū ε L (Q ε) cε 1. (3.5) Here and hereafter, the symbols h, div h, h denote the differential operators acting on the horizontal variable x h T D Energy estimates Similarly to the previous section, we use u ε ū ε as a test function in (3.2b) obtaining (ϱ ε u ε u ε : u ε + 1 ) Q ε ε 2 p(ϱ ε)divu ε dx ϱ ε u ε u ε : ū ε dx Q ε ( = S( uε ) S( ū ε ) ) : (u ε ū ε ) dx + S( ū ε ) : (u ε ū ε ) dx. Q ε Q ε In addition, as (ϱ ε, u ε ) satisfy the renormalized equation of continuity (see (2.41)), we get Q ε p(ϱ ε )divu ε dx = 0, and we have also Consequently, we find Q ε ϱ ε u ε u ε : u ε dx = 0, ( S( uε ) S( ū ε ) ) : (u ε ū ε ) dx Q ε = ϱ ε u ε u ε : ū ε dx S( ū ε ) : (u ε ū ε ) dx. Q ε Q ε Now, we recall the Poincaré inequality on Q ε, (3.6) v L 2 (Q ε) ε v L 2 (Q ε) whenever v W 1,2 (Q ε ) and v z=εh(xh ) = 0, (3.7) and Korn s inequality v 2 L 2 (Q ε) c Combining (3.5) (3.8), we get Q ε S( v) : v dx if v W 1,2 0 (Q ε). (3.8) 1 ε 2 u ε ū ε 2 L 2 (Q + u ε) ε ū ε 2 L 2 (Q ε) c ( ϱ ū ε L (Q ε) u ε 2 L 2 (Q + ū ) ε) ε L 2 (Q ε) u ε ū ε L 2 (Q ε) c ( ε ϱ u ε 2 L 2 (Q + ū ) ε) ε L 2 (Q ε) u ε ū ε L 2 (Q ε) c ( ε ϱ u ε ū ε 2 L 2 (Q ε) + ε ϱ ū ε 2 L 2 (Q ε) + ū ε L 2 (Q ε) u ε ū ε L 2 (Q ε)). Thus, by Hölder s inequality, it follows that u ε ū ε 2 L 2 (Q ε) + ε2 u ε ū ε 2 L 2 (Q ε) cε2 ū ε 2 L 2 (Q ε). (3.9)

21 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations Pressure estimates First of all, we estimate the integral mean of the pressure. To this end, we use the Bogovskii operator B ε [ϱ ε ], specifically, div(b ε [ϱ ε ]) = ϱ ε 1 ϱ ε dx with B ε [ϱ ε ] = 0 on Q ε, Q ε Q ε as a test function in (3.2b), we obtain p(ϱ ε )ϱ ε dx = ϱ ε Mp(ϱ ε ) dx Q ε Q ε ε 2 ϱ ε u ε u ε : B ε [ϱ ε ] dx + ε 2 S( u ε ) : B ε [ϱ ε ] dx. Q ε Q ε Consequently, the Hölder s inequality leads to p(ϱ ε )ϱ ε dx ϱ ε Mp(ϱ ε ) dx Q ε Q ε + ε 2( ϱ u ε 2 L 4 (Q + u ε) ε L 2 (Q ε)) Bε [ϱ ε ] L 2 (Q ε). (3.10) On the other hand, we may observe that (see [13, Lemma 9]) ε B ε [ϱ ε ] L 2 (Q ε) c ϱ ε L 2 (Q ε) Then (3.9) and (3.10) give rise to p(ϱ ε )ϱ ε dx ϱ ε Mp(ϱ ε ) dx Q ε Q ε + εc ( u ε 2 L 4 (Q ε) + ū ε L 2 (Q ε)) ϱε L 2 (Q ε). (3.11) Similarly to Section 2.1, we take λ > 1 and λϱ ε M < ϱ, and estimate ϱ ε Mp(ϱ ε ) dx ϱ ε Mp(ϱ ε ) dx + 1 ϱ ε p(ϱ ε ) dx Q ε {ϱ ε<λϱ ε M } λ Q ε M ε p ( λϱ ε ) 1 M + ϱ ε p(ϱ ε ) dx cεp(λϱ ε λ M) + 1 ϱ ε p(ϱ ε ) dx. Q ε λ Q ε Finally, we may deduce that (3.11) reduces to 1 p(ϱ ε )ϱ ε dx c ( 1 + ( u ε 2 L Q ε 4 (Q + ū ) ) ε) ε L 2 (Q ε) ϱε L 2 (Q ε). (3.12) Q ε On the other hand, again by interpolation, we find if D = 2 and v L 4 (Q ε) v 1/4 L 2 (Q ε) v 3/4 L 6 (Q ε) cε1/4 v 1/4 L 2 (Q ε) v 3/4 L 2 (Q ε) = cε1/4 v L 2 (Q ε), v L 4 (Q ε) v 1 β L 2 (Q ε) v β L q (Q ε) cε1 β v 1 β L 2 (Q ε) v β L q (Q ε) = cε1 β v L 2 (Q ε)

22 20 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov for any β > 1 2 if D = 1, whenever v W1,2 (Q ε ) and v z=εh(xh ) = 0. Indeed v can be extended to be zero for εh(x h ) z < 1 and, consequently, we may use the standard Sobolev embedding on the strip T N (0, 1), namely W 1,2 L 6 if N = 3, W 1,2 L β for any β finite if N = 2. In both cases, relation (3.12) together with the energy bound (3.9) give rise to 1 p(ϱ ε )ϱ ε dx c ( 1 + ( u ε 2 L Q ε 4 (Q + ū ) ε) ε L 2 (Q ε) ϱε L 2 (Q ε)) Q ε c ( 1+ ( ε 1/2 u ε 2 L 2 (Q + ū ) ε) ε L 2 (Q ε)) ϱε L 2 (Q ε) c ( 1 + ( (3.13) ε 1/2 ū ε 2 L 2 (Q + ū ) ε) ε L 2 (Q ε)) ϱε L 2 (Q ε) c ( 1 + ū ε L 2 (Q ε) ϱ ε L 2 (Q ε)) ϱc. With (3.13) at hand, we may estimate the pressure p(ϱ ε ) in the L 2 norm. Indeed, repeating the above procedure with B[p(ϱ ε )], divb ε [p(ϱ ε )] = p(ϱ ε ) 1 p(ϱ ε ) dx, Q ε Q ε we get p(ϱ ε ) 2 dx = 1 ( ) 2 p(ϱ ε ) dx ε 2 ϱ ε u ε u ε : B ε [p(ϱ ε )] dx Q ε Q ε Q ε Q ε + ε 2 Q ε S( u ε ) : B ε [p(ϱ ε )] dx. Thus, exactly as in (3.10) (3.11), we have p(ϱ ε ) 2 dx 1 ( ) 2 p(ϱ ε ) dx + cε ūε Q ε Q ε L 2 (Q ε) p(ϱ ε ) L 2 (Q ε); Q ε which implies by using (3.13) that p(ϱ ε ) 2 L 2 (Q 1 ( ) 2 ε) p(ϱ ε ) dx + ε 2 ū ε 2 L Q ε 2 (Qε) Q ε cε ( (3.14) 1 + ε ū ε 2 ) L 2 (Q ε). 3.2 Limit ε 0 In order to perform the limit as ε tends to 0, it is convenient to work on a fixed spatial domain. Introducing a new vertical variable Z = 1 εz, we get Q def = { x = (x h, Z) : x h T D, 0 < z < h(x h ), h C 2+ν (T 1 ) }. (3.15) For u ε = (u h,ε, V ε ), the system (3.2) written in the new coordinates reads: div h (ϱu h,ε ) + 1 ε Z(ϱV ε ) = 0, div h (ϱu h,ε u h,ε ) + 1 ε Z(ϱV ε u h,ε ) + 1 ε 2 hp(ϱ) = µ ( h u h,ε + 1 ) ε 2 2 Zu h,ε + (λ + µ) h div h u h,ε + (λ + µ) 1 ε h Z V ε, div h (ϱu h,ε V ε ) + 1 ε Z(ϱV 2 ε ) + 1 ε 3 Zp(ϱ) (3.16a) (3.16b) (3.16c) = µ h V ε + 1 ε 2 µ 2 ZV ε + 1 ε Zdiv h u h,ε + (λ + µ) 1 ε 2 2 ZV ε.

23 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations 21 Thus the uniform bounds (3.9) yield u h,ε u h in L 2 (Q) weak, V ε V in L 2 (Q) weak, Z u h,ε Z u h in L 2 (Q) weak, Z V ε Z V in L 2 (Q) weak, ε ( h u ε L 2 (Q) + h V ε L (Q)) 2 c. (3.17a) (3.17b) (3.17c) (3.17d) On the other hand, it follows from (3.4) that V = 0 for z = 0 or z = h, u h = s for z = 0 and u h = 0 for z = h. In addition, we have (see (A.5)), ( ) ε uh,ε L 4 (Q) + V h,ε L 4 (Q) c, (3.18) and by interpolation, we may deduce that Moreover, by virtue of (3.14), ε 1/6( u h,ε L 12/5 (Q) + V h,ε L 12/5 (Q)) c. (3.19) p(ϱ ε ) p(ϱ) in L 2 (Q) weak, (3.20) and, finally, ϱ ε ϱ in L (Q) weak-* and 0 ϱ ϱ. (3.21) Observe that all limits exhibited in (3.17), (3.20) and (3.21) hold up to a suitable subsequence Limit in the field equations As (ϱ ε, u h,ε, V ε ) satisfy (3.16b) and (3.16c) in D (Q), we easily deduce from (3.17) (3.21)) that ϕ C c (Q) : Q ( µ Z u h Z ϕ + p(ϱ) y ϕ ) dx = 0. (3.22) On the other hand, we have ϕ C c (Q) : Q p(ϱ) Z ϕ dx = 0, (3.23) meaning that p(ϱ) = p(ϱ)(x h ) is independent of the vertical coordinate Z. Finally, we take ϕ = ϕ(x h ) as a test function in (3.16a), we find ϕ C c (T D ) : T D ( h(x h ) 0 ) ϱu h dz h ϕ dx h = 0.

24 22 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov Strong convergence of the pressure Our ultimate task is to show strong convergence of the pressure. First, we find from (3.16c) Z p(ϱ ε ) = ε 3 div h (ϱ ε u h,ε V ) ε 2 Z (ϱ ε V 2 ε ) + µε 3 h V ε + ε(λ + µ) 2 ZV ε + (λ + µ)ε 2 Z div h (u h,ε ). Now we claim that, thanks to the bounds established in (3.17) and (3.20), we have Indeed the most difficult term reads, z p(ϱ ε ) W 1,2 (Q) cε. (3.24) ε 2 Z (ϱ ε V 2 ε ) W 1,2 (Q) ε 2 ϱ ε V 2 ε L 2 (Q) ϱε εv ε 2 L 4 (Q) cε where we have used (3.18). On the other hand, we employ the method used in Section 2.6.3, specifically, we consider ϕ C c (Q) : ϕ 1 [ϕp(ϱ ε )] = ϕ [ h 1 [ϕp(ϱ ε )], Z 1 [ϕp(ϱ ε )] ], as a test function in (3.16b) and (3.16c). Similarly to Section 2.6.3, we obtain ϕ 2 p(ϱ ε ) 2 dx = p(ϱ ε ) ϕ 1 [ϕp(ϱ ε )] dx R N R N ε 2 ϕ(ϱ ε u h,ε u h,ε ) h 1 h [ϕp(ϱ ε )] dx ε 2 (ϱ ε u 2 h,ε) h ϕ h 1 [ϕp(ϱ ε )] dx R N R N ε 3 ϕ(ϱ ε u h,ε V ε ) h 1 Z [ϕp(ϱ ε )] dx ε 3 (ϱ ε u h,ε V ε ) h ϕ Z 1 [ϕp(ϱ ε )] dx R N R N ε ϕ(ϱ ε u h,ε V ε ) Z 1 h [ϕp(ϱ ε )] dx ε (ϱ ε u h,ε V ε ) Z ϕ h 1 [ϕp(ϱ ε )] dx R N R N ε 2 ϕ(ϱ ε Vε 2 ) Z 1 Z [ϕp(ϱ ε )] dx ε 2 (ϱ ε Vε 2 ) Z ϕ Z 1 [ϕp(ϱ ε )] dx R N R N + ε 2 β 1 ϕdiv h u h,ε div h 1 h [ϕp(ϱ ε )] dx + ε 2 β 1 div h u h,ε h ϕ 1 h [ϕp(ϱ ε )] dx R N R N + ε β 1 ϕ Z V ε div h 1 h [ϕp(ϱ ε )] dx + ε β 1 Z V ε h ϕ 1 h [ϕp(ϱ ε )] dx R N R N + ε 2 β 1 ϕ Z u h,ε h 1 Z [ϕp(ϱ ε )] dx + ε 2 β 1 h ϕ Z u h,ε 1 Z [ϕp(ϱ ε )] dx R N R N + ε β 2 ϕ Z V ε Z 1 Z [ϕp(ϱ ε )] dx + ε β 2 Z ϕ Z V ε 1 Z [ϕp(ϱ ε )] dx R N R N + ε 2 µϕ h u h,ε h 1 h [ϕp(ϱ ε )] dx + ε 2 µ h ϕ h u h,ε 1 h [ϕp(ϱ ε )] dx R N R N + ε 3 µϕ h V ε h 1 Z [ϕp(ϱ ε )] dx + ε 3 µ h ϕ h V ε 1 Z [ϕp(ϱ ε )] dx R N R N + µϕ Z u h,ε Z 1 h [ϕp(ϱ ε )] dx + µ Z ϕ Z u h,ε 1 h [ϕp(ϱ ε )] dx, R N R N def def where β 1 = λ + µ and β 2 = λ + 2µ. Our goal is to let ε tends to 0 in the above identity. According to the following Sobolev embedding: W 1,2 L 6 if D = 2 and W 1,2 L q for any finite q if D = 1, clearly all terms in the form 1 [ϕp(ϱ ε )] are uniformly bounded al least in L 6 (Q). Consequently, it is easy to check that all integrals in the above inequality containing these terms and multiplied by some power of ε vanish in the limit as ε tends to 0 (we also used

25 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations 23 her (3.19)). Next, we have to check that also the remaining integrals multiplied by some power of ε vanish in the asymptotic limit. We focus on the most difficult term, ε ϕϱ ε (u h,ε V ε ) Z 1 h [ϕp(ϱ ε )] dx R N = ε ϱ ε (u h,ε V ε ) h 1 Z [ϕp(ϱ ε )] dx R N = ε ϱ ε (u h,ε V ε ) h 1 [ Z ϕp(ϱ ε )] dx R N + ε ϱ ε (u h,ε V ε ) h 1 [ϕ Z p(ϱ ε )] dx. R N By virtue of (3.17), we have ε ϱ ε (u h,ε V ε ) h 1 [ Z ϕp(ϱ ε )] dx 0, R N while, in accordance with (3.18) and (3.24), we find ε ϱ ε (u h,ε V ε ) h 1 [ϕ Z p(ϱ ε )] dx εϱ ε u h,ε V ε L 2 R N (Q) ϕ Zp(ϱ ε ) W 1,2 (R N ) c εu L h,ε εv L 4 (Q) ε 4 (Q) ϕ Zp(ϱ ε ) W 1,2 (R N ) cε. for any α < 1. Consequently, we may let ε tends to 0, we get lim ε 0 ϕ 2 p(ϱ ε ) 2 dx = R N p(ϱ) ϕ 1 [ϕp(ϱ)] dx R N + µϕ Z u h h 1 [ Z ϕp(ϱ)] dx R N + µ Z ϕ Z u h 1 h [ϕp(ϱ)] dx. R N Finally, using ϕ 1 [ϕp(ϱ)] (3.25) as a test function in (3.22) and (3.23), then comparing the resulting expression with (3.25), we may conclude that ϕ C c (Q) : lim ϕ 2 p(ϱ ε ) 2 dx = ϕ 2 p(ϱ) 2 dx, ε 0 R N R N which immediately yields p(ϱ ε ) p(ϱ) in L 2 loc (Q). Under the supplementary hypothesis that p > 0, we get the desired conclusion Theorem 3.2. Let Q R N, N = 2, 3, defined as follows: p(ϱ) = p(ϱ). (3.26) Q def = { x = (x h, z) : x h T D, 0 < z < h(x h ), h C 2+ν (T D ) }, with D = 1, 2. Let the pressure be given as p = p(ϱ), where p C 0 [0, ϱ) C 1 (0, ϱ), p(0) 0, p (ϱ) > 0 for ϱ > 0, lim ϱ ϱ p(ϱ) = +.

26 24 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov Let M ε ε > 0 such that 0 < inf ε>0 M ε Q ε sup ε>0 M ε Q ε < ϱ. Let (ϱ ε, u h,ε, V ε ) be a family of weak solutions to the rescaled problem (3.16a) (3.16c), where the velocities satisfies the boundary conditions (3.3) and Q ϱ ε dx = Mε ε Then, up to a subsequence, we have M ε ε M, (3.27a) 0 ϱ ε ϱ, ϱ ε ϱ in L 1 (Q), (3.27b) where the limit satisfies p(ϱ ε ) p(ϱ) in L 2 loc (Q), p(ϱ ε) p(ϱ) in L 2 (Q) weak, u h,ε u h, Z u h,ε Z u h in L 2 (Q) weak, V ε V, Z V ε Z V in L 2 (Q) weak, (3.27c) (3.27d) (3.27e) in D (). ϱ = ϱ(x h ), 0 ϱ ϱ, p = p(ϱ) L 2 (T D ), ϱ dx = Q hϱdx h = M, T D u h Z=0 = s, u h Z=h(xh ) = 0, ( h(x h ) ) div h ϱu h dz = 0, µ Zu 2 h + h p(ϱ) = 0 0 (3.28a) (3.28b) (3.28c) (3.28d) 3.3 Uniqueness for the limit problem It is interesting to know if the convergence stated in Theorem 3.2 is unconditional, meaning if the limit problem (3.28d) is uniquely solvable in the class (3.28a). We show that this is indeed the case at least for D = 1, p (ϱ) > 0 with ϱ (0, ϱ). We denote v = u h and x h = y and we assume that s > 0. To this aim, we first observe that the second equation in (3.28d), and the boundary conditions satisfied v at Z = 0, h(y) lead to Integrating over Z, we find v = 1 ( 2µ yp(ϱ)(z 2 Zh) + s h 0 1 Z h v dz = h3 12µ yp(ϱ) + sh 2, ), (3.29) from which we deduce y p(ϱ) L 2 (T 1 ). In particular p(ϱ) is Hölder continuous and periodic in y. It follows that sup y T 1 ϱ(y) < ϱ. According to (3.28d), we obtain the following ODE ( h 3 ) y 12µ ϱ yp(ϱ) ( ϱhs ) = y 2 in D (T 1 ) and We show that problem (3.30) admits at most one solution. First, we have T 1 hϱ dy = M. (3.30) h 3 12µ ϱ yp(ϱ) = ϱhs + λ with λ R, (3.31) 2

27 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations 25 which implies that By using the following identity: we obtain Since we have 1 12µ ϱ yp(ϱ) = ϱs 2h 2 + λ h 3 with λ R. (3.32) ϱp (ϱ) y ϱ = y f(ϱ) with f(ϱ) def = T 1 ϱ h 2 dy > 0 T 1 ϱ y p(ϱ) dy = 0. and ϱ 1 1 dy > 0, T 1 h3 sp (s) ds, we conclude from (3.32) that λ < 0. Now we prove that ϱ > 0. We proceed by contradiction; we assume that there exists y 0 T 1 such that ϱ(y 0 ) = 0. (3.33) We may deduce from (3.31) and y p(ϱ) L 2 (T 1 ) that ϱ(y) > 0 a.e. y T 1. Then it follows from (3.32) that ( 6µs ϱ = h µλ ) 1 h 3 ϱ p a.e. on T 1. (3.34) (ϱ) We observe that we can take y 0 1. Since ϱ is a continuous function, ϱ(y 0 ) = 0 and λ < 0, it comes by using (3.34) that there exists η > 0 small enough satisfying y 0 + η T 1 such that ϱ (y) < 0 a.e. y (y 0, y 0 + η). But ϱ(y 0 + η) = y0 +η y 0 ϱ (y) dy < 0 gives the contradiction. We conclude that ϱ > 0. Accordingly, equation (3.31) becomes a non degenerate first order non linear ODE with C 1 non linearities. Next, we claim that for a given λ < 0, there is at most one periodic solution ϱ of (3.31) satisfying also T 1 ϱh dy = M. (3.35) Indeed, two possible such solutions do not intersect, therefore there exists exactly one of them for which (3.35) holds. Let now ϱ 1 and ϱ 2 - two periodic solutions - of (3.31) corresponding to two different constants λ 1 λ 2 and satisfying (3.35). To fix the ideas, we suppose that λ 2 > λ 1. It follows from p > 0 on (0, ϱ) and (3.31) that ϱ 1(y) > ϱ 2(y) for any y [0, 1] such that ϱ 1 (y) = ϱ 2 (y) > 0. (3.36) Since (3.35) holds, there exists α (0, 1) such that ϱ 1 (α) = ϱ 2 (α). We conclude from (3.36) that we have y [0, α) : ϱ 1 (y) > ϱ 2 (y) and y (α, 1] : ϱ 1 (y) < ϱ 2 (y), which contradicts the periodicity of ϱ 1 and ϱ 2.

28 26 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov A Auxiliary results, Bogovskii s operator, extension lemma A.1 Bogovskii s operator The goal is to find a solution branch of the obviously undetermined problem, divb = f 1 f dx in and B = 0 on, (A.1) where is a bounded Lipschitz domain in R N. In this paper, we consider the operator B = B[f] proposed by Bogovskii [3]. The following properties of B were proved in the monograph by Galdi [9]: The operator f B[f] is linear, a priori defined from C ( ) to C c (). B can be extended to functions f L p (), B[f] W 1,p 0 () c(p) f L p () for any 1 < p <. (A.2) If, in addition, f = div(g), where g L q such that g n = 0 on holds, then B[div(g)] L q () c(q) g L q (). (A.3) A.2 Extension operator Next, we claim the following extension lemma. Lemma A.1. Let R N, N = 2, 3, be a bounded Lipschitz domain. Let ū W 1,p () be given such that p ]0, + [ and ū n = 0 on. Let q be given such that 1 < q < if p < N and 1 < q arbitrary finite otherwise. Then for any δ > 0, there exists ū δ W 1,p () with the following properties: ū δ = ū on in the sense of traces and divū δ = 0 in, ū δ L q () < δ and ū δ W 1,p () c(δ, p, q) ū W 1,p (), Proof. As W 1,p () L q (), it is easy to construct an extension ũ δ such that ũ δ = ū on, ũ δ L q () < δ and ũ δ W 1,p () c(δ, p, q) ū W 1,p (). Now, consider the Bogovkii s operator solving div(b) = div(ũ δ ) and B = 0 on. It follows from the basic properties of the operator B stated in (A.2) and (A.3) that ū δ = ũ δ B Np N p satisfies ū δ = ū on and div(ū δ ) = 0 in, ū W 1,p () ũ δ W 1,p () + B W 1,p () c 1 (δ, p, q) ( ũ δ W 1,p () + divũ δ L ()) p c2 (δ, p, q) ũ δ W 1,p (), ū δ L q () ũ δ L q () + B L q () δ + c(q) ũ δ L q () δ(1 + c(q)). (A.4a) (A.4b) (A.4c)

29 A derivation of the stationary compressible Reynolds equation via the Navier Stokes equations 27 A.3 Anisotropic interpolation Consider the domain def = { x = (x h, z) : x h T 2, 0 < z < h(x h ) }, where h is a Lipshitz function, see [2]. Then we have v L 4 () ( ) 1 h v L 2 () + v ( ) 1 L 2 2 () Z v L 2 () + v L 2 2 () (A.5) for any v W 1,2 () such that v(, z = h(x)) = 0 and v n = 0 on hold. B Existence and uniqueness results to the approximate problem We prove here the following lemma Lemma B.1. Assume that u W 1,2 () such that u n = 0 on holds and g L 2 (). Then there exists a unique solution ϱ W 2,2 () such that ϱ 0 to problem δϱ δ ϱ + div(t (ϱ)u) = g in, ϱ n = 0 on. (B.1a) (B.1b) Furthermore, we have the following comparison principle: for any g 1 L 2 () and g 2 L 2 () with g 1 g 2, we have ϱ 1 ϱ 2 where ϱ 1 and ϱ 2 are the solutions to problem (B.1) corresponding to g 1 and g 2, respectively. Proof. The existence results comes from the Schauder s fixed point theorem. To this aim, let S be the following mapping: S : W 1,2 () W 1,2 () where ϱ is assumed to be a solution to the problem ϱ ϱ = S( ϱ) δϱ δ ϱ + div(t ( ϱ)u) = g in, ϱ n = 0 on. Since u n = 0 on, the variational formulation for ϱ is given by ψ W 1,2 () : δ ϱψ dx + δ ϱ ψ dx (T ( ϱ)u) ψ dx = with ϱ W 1,2 (). Observe that T ( ϱ)u L 2 () ϱ u L 2 (), gψ dx, which implies by using Lax Milgram theorem that there exists a constant C 0 independent of ϱ such that S( ϱ) W 1,2 () C. Then S( B W 1,2 ()(0, C)) B W 1,2 ()(0, C) where B W 1,2 ()(0, C) denotes a closed ball in W 1,2 () with radius C. Since T W 1, (R) and u L 6 (), we may deduce that div(t ( ϱ)u) is bounded in L 3 2 () by a constant independently of ϱ. It follows by using a regularity argument that ϱ is bounded in W 2, 3 2 () and it comes from the Sobolev compactness that S( B W 1,2 ()(0, C)) is

30 28 I. S. Ciuperca, E. Feireisl, M. Jai, A. Petrov relatively compact in W 1,2 (). On the other hand, assume that ϱ n converges strongly to ϱ in W 1,2 () as n tends to +. Since T is Lipschitz continuous and W 1,2 () L 6 (), it is clear that T ( ϱ n ) T ( ϱ) L 6 () 0. Since u L 3 (), we may infer that ϱ n converges to ϱ in W 1,2 () as n tends to + and then S is continuous. Hence we may conclude by using Schauder s fixed point theorem (see [8]) that (B.1) possesses a solution. We establish now the comparison principle which gives also the uniqueness result associated to our problem. We introduce the following notations: ϱ def = ϱ 1 ϱ 2 and g def = g 1 g 2, then we have δϱ δ ϱ + div((t (ϱ 1 ) T (ϱ 2 ))u) = g in, (B.2a) ϱ n = 0 on, (B.2b) where g is a positive function. Let us define ϕ W 1, (R) such that ϕ(0) = 0 and ϕ (ζ) > 0 for all ζ [0, + ). Furthermore, ( ) + and ( ) denote the positive and negative part of ( ), respectively. Multiplying (B.2a) by ϕ(ϱ ) and integrating over the resulting identity, we find δ ϱ ϕ(ϱ ) dx + δ ϕ (ϱ ) ϱ 2 dx u ϱ ϕ (ϱ ) ϱ dx, which leads to Choosing ϕ(ζ) = δ ϱ ϕ(ϱ ) dx 1 u 2 (ϱ ) 2 ϕ (ϱ ) dx. 2δ ζ ζ+η, we have (ϱ ) 2 η + ϱ dx η 2δ 2 u 2 dx. We may deduce that ϱ dx = 0 which ends the proof of the comparison principle. Acknowledgments The research of E. F. leading to these results has received funding from the European Research Council under the European Union s Seventh Framework Programme (FP7/ )/ ERC Grant Agreement The Institute of Mathematics of the Academy of Sciences of the Czech Republic is supported by RVO: This research was performed during the stay of E. F. as invited professor at the INSA Lyon. References [1] R. A. Adams. Sobolev spaces. Academic Press [A subsidiary of Harcourt Brace Jovanovich, Publishers], New York-London, [2] A. Assemien, G. Bayada, M. Chambat. Inertial effects in the asymptotic behavior of a thin film flow. Asymptotic Anal., 9(3) (1994), [3] M. E. Bogovskii. Solution of some vector analysis problems connected with operator div and grad (in Russian). Trudy Sem. S. L. Sobolev, 80 (1980), [4] H. Brezis, Analyse fonctionnelle, Collection Mathématiques Appliquées pour la Maîtrise. [Collection of Applied Mathematics for the Master s Degree], Masson, Paris, [5] L. Chupin, R. Sart. Compressible flows: new existence results and justification of the Reynolds asymptotic in thin films. Asymptot. Anal., 76(3-4) (2012),