Asymptotic behaviour of a class of incompressible,quasi-Newtonian fluids in thin pipes

1. Introduction of the problem

We deal with incompressible complex fluids flowing through thin 3D pipes, with prescribed pressure at the ends. Since, in anisotropic domains, the equations are not exactly solvable, it is very natural to look for an approximate solution. The very nature of the thin domain makes it computationally undesirable to employ standard numerical techniques. We show that, by formally expressing the rescaled solution as a power series of a small parameter ε, the problem of finding the its coefficients is considerable less complex. Finally, we prove error estimates of all orders to justify the validity of the above mentioned expansion. There exists a considerable amount of literature published on subjects related to fluid flow in thin domains – we shall mention only a few connected to our problem. The asymptotic behaviour of the Navier–Stokes flow in 2D thin domains has been studied in [2]. In [6] the curvature effects on the quasi-Newtonian flow in thin pipes are considered, while the inertial term is neglected. A more general quasi-Newtonian flow in a thin 3D slab has been studied in [7]. Finally, it is worth mentioning that the first mathematical investigation of the quasi-Newtonian fluids was done in [3] and [4]. While we restrict ourselves to the stationary case and we disregard the curvature effects, let us emphasise the strengths of this paper:

Taking into account the inertial term – which makes it more difficult to obtain an existence result.

Consider now a small parameter ε > 0 (throughout this material we shall always have ε ⩽ 1 ). Suppose that the fluid flows through a simply connected, thin domain Ω ε of the form Ω ε = { ( x 1 , x 2 , x 3 ) ∈ R 3 ∣ x 1 ∈ ( 0 , 1 ) , ( x 2 , x 3 ) ∈ ε S ( x 1 ) } , where S ( x 1 ) are sufficiently regular domains – we also suppose the application x 1 ↦ S ( x 1 ) is also smooth enough ( C 2 regularity is sufficient). For the sake of simplicity we are going to assume Ω ε ⊂ D ε = [ 0 , 1 ] × [ 0 , ε ] × [ 0 , ε ]  – and the three-part boundary of Ω ε is given by the following: Γ ε = { ( x 1 , x 2 , x 3 ) ∈ R 3 ∣ x 1 ∈ ( 0 , 1 ) , ( x 2 , x 3 ) ∈ ε ∂ S ( x 1 ) } , Σ 0 ε = ε S ( 0 ) , Σ 1 ε = ε S ( 1 ) (see Fig. 1). Moreover, we suppose that Σ 0 ε and Σ 1 ε are perpendicular on e 1 = ( 1 , 0 , 0 ) .

OPEN IN VIEWER

Fig. 1.

Thin, anisotropic domain.

We make the conventions Ω 1 = Ω , Γ 1 = Γ , Σ i 1 = Σ i . With respect to the boundary conditions, we impose u = 0 on Γ ε , u × n = 0 on Σ 0 ε ∪ Σ 1 ε , p = p i on Σ i ε(2) with p 0 , p 1 > 0 given constants. Note that the second condition is equivalent to u 2 = u 3 = 0 on Σ 0 ε ∪ Σ 1 ε .

In the next section we prove the existence of the solution using a classical approximation technique, while in the final section we derive a complete asymptotic expansion with respect to the small parameter ε.

The proof of the existence of the solution is based on classical compactness and monotony arguments, and is very similar to the one in [5]. There are two main differences; the first one is that we use different boundary conditions, the importance of which shall become apparent when discussing the asymptotic expansion. On the other hand, there is the problem of controlling the inertial term. While the author of [5] uses a specific property of the viscosity function, we shall use the smallness of ε together with some sharp Poincaré and Sobolev type of inequalities for thin domains.

We call the system given by (1) and (2) problem (P). We say that ( u , p ) is a solution of (P) if it satisfies (1) in the sense of distributions and (2) in the sense of traces.

We will suppose that η : R + → R is a function satisfying the following properties: η ∈ C ( R + ) , (3) 0 < η 0 ⩽ η ( x ) ⩽ η 1 ∀ x ∈ R + , (4) ( η ( x 2 ) x − η ( y 2 ) y ) ( x − y ) ⩾ 0 ∀ x , y ∈ R + . (5) Let us note that (5) is equivalent to the fact that the mapping x ↦ η ( x 2 ) x is increasing. In particular, this is true if η is differentiable and d d x ( η ( x 2 ) x ) ⩾ 0 ∀ x ∈ R + .(6) In order to introduce the variational problem, let us consider the space V = { v ∈ H 1 ( Ω ε ) 3 | div v = 0 , v = 0 on Γ ε , v × n = 0 on Σ 0 ε ∪ Σ 1 ε } . We will say that u ∈ V is the solution of the variational problem henceforth denoted ( PV ) if it satisfies ∫ Ω ε ( u · ∇ ) u · v + 2 ∫ Ω ε η ( I ( u ) ) D i j ( u ) D i j ( v ) + p 0 ∫ Σ 0 ε v · n + p 1 ∫ Σ 1 ε v · n = 0 ∀ v ∈ V(7) (we are systematically going to use Einstein’s summation convention). Consider now the space X = { v ∈ ( H 1 ( Ω ε ) ) 3 | v = 0 on Γ ε , v × n = 0 on Σ 0 ε ∪ Σ 1 ε } . Define the application L : X → X ′ by ⟨ L ( u ) , h ⟩ = ∫ Ω ε ( u · ∇ ) u · h + 2 ∫ Ω ε η ( I ( u ) ) D i j ( u ) D i j ( h ) and F : X → R by ⟨ F , h ⟩ = − p 0 ∫ Σ 0 ε h · n − p 1 ∫ Σ 1 ε h · n , where ⟨ · , · ⟩ represents the duality between X ′ and X. It is clear that L is well defined and F ∈ X ′ . The problem ( PV ) can be rewritten as: find u ∈ V such that ⟨ L ( u ) − F , v ⟩ = 0 ∀ v ∈ V .

It is a simple matter to verify that if ( u , p ) is a smooth solution of (P) then u is a solution of ( PV ). Conversely, the following is true.

Returning to the proof of Theorem 1, consider T be the restriction of L ( u ) − F to W. Clearly T ∈ W ′ and so by Lemma 1 there is p ∈ L 2 ( Ω ε ) with div ∗ p = T . Let h ∈ X with the decomposition h = v + w with v ∈ V , w ∈ W . ⟨ L ( u ) − F , h ⟩ = ⟨ T , w ⟩ = ⟨ div ∗ p , w ⟩ = ∫ Ω ε p div w = ∫ Ω ε p div h . By taking in the above equation h ∈ ( C 0 ∞ ( Ω ε ) ) 3 we get readily that u satisfies the first equation of (1) in the sense of distributions. □

Henceforth, ∥ · ∥ will always mean the L 2 norm on Ω ε unless otherwise specified. In virtue of a Poincaré inequality – see further Lemma 4 – it follows that ∥ ∇ u ∥ is a norm on V equivalent to the H 1 norm – and from now on we will consider V endowed with this norm.

A simple calculation – integrating twice by parts – shows that if u ∈ X = { v ∈ ( C 2 ( Ω ε ‾ ) ) 3 ∣ v = 0 on Γ ε , v × n = 0 on Σ 0 ε ∪ Σ 1 ε } then ∫ Ω ε ∂ x i u j ∂ x j u i = ∫ Ω ε ( div u ) 2 . Since X is dense in X it follows that for all u ∈ V ∫ Ω ε ∂ x i u j ∂ x j u i = 0 , so we can write ∫ Ω ε ( D i j ( u ) ) 2 = 1 2 ∫ Ω ε ( ∂ x i u j ) 2 = ∥ ∇ u ∥ 2 2 .

Let us consider the mapping A : V → V ′ defined by ⟨ A ( u ) , v ⟩ = 2 ∫ Ω ε η ( I ( u ) ) D i j ( u ) D i j ( v ) ∀ u , v ∈ V .

Define now L 0 : V → V ′ by ⟨ L 0 ( u ) , v ⟩ = ∫ Ω ε ( u · ∇ ) u · v ∀ v ∈ V .

We continue with a technical result, that will prove very useful lemma.

Henceforth, we shall denote by C various constants independent of ε. Theorem 1.

Assume that η satisfies (3)–(5). Provided ε > 0 is small enough, there exists a solution u of the variational problem ( PV ) satisfying ∥ ∇ u ∥ ⩽ C ε 2 .(13)

By Lemma 4 we get | ⟨ L 0 ( w ) , w ⟩ | = ∫ Ω ε ( w · ∇ ) w · w ⩽ C ε 1 / 2 ∥ ∇ w ∥ 3 ⩽ C ∥ ∇ w ∥ 3 , | ⟨ F , w ⟩ | ⩽ C ε 2 ∥ ∇ w ∥ . Also we have ⟨ A ( w ) , w ⟩ = ∫ Ω ε η ( I ( w ) ) I ( w ) ⩾ η 0 2 ∥ ∇ w ∥ 2 , so we obtain ( P n ( w ) , w ) ⩾ η 0 2 ∥ ∇ w ∥ 2 − C ∥ ∇ w ∥ 3 − C ε 2 ∥ ∇ w ∥ . Hence if ε ⩽ η 0 4 C , then for ∥ ∇ w ∥ = ρ ε = η 0 − η 0 2 − 16 C 2 ε 2 4 C ( P n ( w ) , w ) ⩾ 0 , and so we can apply Lemma 5 to find u n ∈ V n , ∥ ∇ u n ∥ ⩽ ρ ε with P n ( u n ) = 0 . Clearly ρ ε is independent of n so we can choose a subsequence still denoted by u n such that u n → u weakly in V , u n → u strongly in L 4 . Moreover, we have ∥ ∇ u ∥ ⩽ lim inf ∥ ∇ u n ∥ ⩽ ρ ε = 16 C ε 2 η 0 2 + 4 C 2 ε 2 ⩽ 16 C η 0 ε 2 .(14) Clearly A ( u n ) is bounded in V ′ , hence we can choose another subsequence such that A ( u n ) → χ weakly in V ′ .(15) Let us now fix n ⩾ 1 . Then for all m ⩾ n we have ⟨ A ( u m ) , z n ⟩ + ⟨ L 0 ( u m ) , z n ⟩ = ⟨ F , z n ⟩ . By Lemma 3 and (15) we can pass to the limit as m → ∞ to get ⟨ χ , z n ⟩ + ⟨ L 0 ( u ) , z n ⟩ = ⟨ F , z n ⟩ . Since this is true for all n and { z 1 , z 2 , … } is dense in V, it easily follows ⟨ χ , v ⟩ + ⟨ L 0 ( u ) , v ⟩ = ⟨ F , v ⟩ ∀ v ∈ V .(16) To conclude, it rests to prove A ( u ) = χ . Since u n is the solution of the approximate problem, we can write ⟨ A ( u n ) , u n ⟩ + ⟨ L 0 ( u n ) , u n ⟩ = ⟨ F , u n ⟩ . Using similar arguments as those in the proof of Lemma 3 we derive ⟨ L 0 ( u n ) , u n ⟩ → ⟨ L 0 ( u ) , u ⟩ , and so lim n → ∞ ⟨ A ( u n ) , u n ⟩ = ⟨ F , u ⟩ − ⟨ L 0 ( u ) , u ⟩ .(17) By Lemma 2, we have ⟨ A ( u n ) , u n − v ⟩ ⩾ ⟨ A ( v ) , u n − v ⟩ ∀ v ∈ V . Using (17) we can pass to the limit in the above equation to get ⟨ F , u ⟩ − ⟨ χ , v ⟩ ⩾ ⟨ A ( v ) , u − v ⟩ ∀ v ∈ V . By taking v = u in (16) we get ⟨ χ , u ⟩ = ⟨ F , u ⟩ , so that ⟨ χ − A ( v ) , u − v ⟩ ⩾ 0 ∀ v ∈ V , which can be further written as t ⟨ χ − A ( u + t v ) , v ⟩ ⩽ 0 ∀ t ∈ R , v ∈ V . Using the continuity of A, we derive the desired result. □

In what follows, we make the following convention: if v : Ω ε → R 3 ( R ) then we will denote v ˜ : Ω → R 3 ( R ) the application defined by v ˜ ( x 1 , x 2 , x 3 ) = v ( x 1 , ε x 2 , ε x 3 ) . We will denote by ∥ · ∥ 0 the L 2 norm on Ω. Moreover, we will also systematically use the dummy index k to represent either summation for k ∈ { 2 , 3 } , either the repeated index. We can write u ˜ 1 ∂ x 1 u ˜ 1 + u ˜ k ∂ x k u ˜ 1 ε + ∂ x 1 p ˜ = 2 ∂ x 1 ( η ˜ ∂ x 1 u ˜ 1 ) + ∂ x k ( η ˜ ( ∂ x 1 u ˜ k + ( ∂ x k u ˜ 1 ) / ε ) ) ε , u ˜ 1 ∂ x 1 u ˜ 2 + u ˜ k ∂ x k u ˜ 2 ε + ∂ x 2 p ˜ ε = ∂ x 1 ( η ˜ ( ∂ x 1 u ˜ 2 + ∂ x 2 u ˜ 1 ε ) ) + ∂ x k ( η ˜ ( ∂ x 2 u ˜ k + ∂ x k u ˜ 2 ) ) ε 2 , u ˜ 1 ∂ x 1 u ˜ 3 + u ˜ k ∂ x k u ˜ 3 ε + ∂ x 3 p ˜ ε = ∂ x 1 ( η ˜ ( ∂ x 1 u ˜ 3 + ∂ x 3 u ˜ 1 ε ) ) + ∂ x k ( η ˜ ( ∂ x k u ˜ 3 + ∂ x 3 u ˜ k ) ) ε 2 , ∂ x 1 u ˜ 1 + ∂ x k u ˜ k ε = 0 , with η ˜ = η ( I ( u ˜ ) ) . An elementary calculation shows that ∥ u ˜ ∥ 0 = 1 ε ∥ u ∥ ⩽ ∥ ∇ u ∥ ⩽ C ε 2 . The above estimate, as well as fair intuitions leads us to formally write u ˜ 1 = ε 2 u ˜ 1 0 + ε 4 u ˜ 1 1 + ⋯ + ε 2 n + 2 u ˜ 1 n + ⋯ , u ˜ k = ε 3 u ˜ k 0 + ε 5 u ˜ k 1 + ⋯ + ε 2 n + 3 u ˜ k n + ⋯ ∀ k ∈ { 2 , 3 } , p ˜ = p ˜ 0 + ε 2 p ˜ 1 + ⋯ + ε 2 n p ˜ n + ⋯ , where u ˜ α m , p ˜ m are independent of ε for all m ⩾ 0 , α ∈ { 1 , 2 , 3 } . The boundary conditions are expected to become u ˜ α m = 0 on Γ , ∀ m ⩾ 0 , α ∈ { 1 , 2 , 3 } , u ˜ k m = 0 on Σ 0 ∪ Σ 1 , ∀ m ⩾ 0 , k ∈ { 2 , 3 } , p ˜ 0 ( i ) = p i on Σ i , p ˜ m ( i ) = 0 on Σ i , ∀ m ⩾ 1 .(18) Writing (formally) the Taylor expansion of η in 0 and plugging in the above expansion, we arrive at η ˜ = A ˜ 0 + ε 2 A ˜ 1 + ⋯ + ε 2 n A ˜ n + ⋯ , where A ˜ n depends only on u ˜ i for i < n . Note A ˜ 0 = η ( 0 ) : = η 0 .

Plugging in the expansions and identifying the coefficients of ε − 1 in the second equation, of ε 0 in the first equation and taking (18) into account we get ∂ x 2 p ˜ 0 = ∂ x 3 p ˜ 0 = 0 and − η 0 ∂ x k 2 u ˜ 1 0 + ∂ x 1 p ˜ 0 = 0 on S ( x 1 ) , u ˜ 1 0 = 0 on ∂ S ( x 1 ) .(P₁⁰) Observe that we can solve for u ˜ 1 0 ( x 1 , x k ) = ∂ x 1 p ˜ 0 ( x 1 ) W ( x 1 , x k ) , where W is the unique solution of the problem Δ k v ˜ = 1 / η 0 on S ( x 1 ) , v ˜ = 0 on ∂ S ( x 1 ) , where Δ k signifies the Laplacian in two variables – Δ k = ∂ x 2 2 + ∂ x 3 2 . To find p ˜ 0 let us use (a priori) the following compatibility relation ∫ S ( x 1 ) ∂ x 1 u ˜ 1 0 = 0 .(C₀) If we define β ( x 1 ) = ∫ S ( x 1 ) W ( x 1 , x k ) then p ˜ 0 ( x 1 ) = p 0 + p 1 − p 0 ∫ 0 1 d x 1 β ( x 1 ) ∫ 0 x 1 d y 1 β ( y 1 ) , u ˜ 1 0 ( x 1 , x k ) = p 1 − p 0 ∫ 0 1 d x 1 β ( x 1 ) · W ( x 1 , x k ) β ( x 1 ) . Due to the imposed regularity requirements, we have u ˜ 1 0 ∈ C 2 ( Ω ‾ ) and p ˜ 0 ∈ C 1 ( Ω ‾ ) . It is easy to check that ( u ˜ 1 0 , p ˜ 0 ) satisfy ( P 1 0 ) in the classical sense, and that u ˜ 1 0 verifies ( C 0 ) for all x 1 . The key ingredient for that – and a property which we shall further use – is that whenever w ˜ ∈ C 1 ( Ω ‾ ) satisfies w ˜ = 0 on Γ, in virtue of Reynolds’ transport theorem ∫ S ( x 1 ) ∂ x 1 w ˜ = ∂ x 1 ∫ S ( x 1 ) w ˜ ∀ x 1 . Moreover, we have p ˜ 0 ( i ) = p i .

To determine all u ˜ α m we proceed by complete induction. Let j ⩾ 1 . Assume we have found (sufficiently) smooth u ˜ 1 i such that ∫ S ( x 1 ) ∂ x 1 u ˜ i 0 = 0 for all i < j and u ˜ k i for all i < j − 1 (nothing if j = 1 ). We show how to determine u ˜ 1 j , u ˜ k j − 1 , p ˜ j such that ∫ S ( x 1 ) ∂ x 1 u ˜ 1 j = 0 .(C _j ) Again, plugging in the expansion and identifying in the second and third equation the coefficients of ε 2 j − 1 and in the last one those of ε 2 j , together with (18), we obtain − η 0 ∂ x k 2 u ˜ 2 j − 1 + ∂ x 2 p ˜ j = f ˜ 2 j on S ( x 1 ) , − η 0 ∂ x k 2 u ˜ 3 j − 1 + ∂ x 3 p ˜ j = f ˜ 3 j on S ( x 1 ) , ∂ x k u ˜ k j − 1 = − ∂ x 1 u ˜ 1 j − 1 on S ( x 1 ) , u ˜ 2 j − 1 = u ˜ 3 j − 1 = 0 on ∂ S ( x 1 ) ,(P_k^j−1) with f ˜ l j = ∑ i + m = j − 2 ∂ x 1 ( A ˜ i ∂ x 1 u ˜ l m ) + ∑ i + m = j − 1 i ⩾ 1 ( ∂ x 1 ( A ˜ i ∂ x l u ˜ 1 m ) + 2 ∂ x l ( A ˜ i ∂ x l u ˜ l m ) ) + ∑ i + m = j − 1 i ⩾ 1 ∂ x l ( A ˜ i ∂ x 2 u ˜ 3 m + ∂ x 3 u ˜ 2 m ) − ∑ i + m = j − 3 ( u ˜ 1 i ∂ x 1 u ˜ l m + u ˜ k i ∂ x k u ˜ l m ) for l ∈ { 2 , 3 } . Observe that f ˜ l j are known, since they depend on terms u ˜ 1 i with i < j and u ˜ k i with i < j − 1  – moreover, they are smooth by the induction hypothesis. In virtue of the compatibility ∫ S ( x 1 ) ∂ x 1 u ˜ j − 1 0 = 0 , it follows (see [1]) that the Stokes system has a unique regular solution ( u ˜ k j − 1 , p ˜ j ) that satisfies ( P k j−1 ) in the classical sense. Note that p ˜ j is determined up to a constant (for every x 1 ). Hence we can write p ˜ j ( x 1 , x k ) = q ˜ j ( x 1 , x k ) + r ˜ j ( x 1 ) , where q ˜ j is fixed and r ˜ j is to be determined.