Asymptotic Analysis of a Viscous Fluid in a Curved Pipe with Elastic Walls
AAsymptotic Analysis of a Viscous Fluid in a Curved Pipe withElastic Walls
G. Casti˜neira a , J. M. Rodr´ıguez b a Departamento de Matem´atica Aplicada, Univ. de Santiago de Compostela, Spain b Departamento de M´etodos Matem´aticos e Representaci´on, Univ. da Coru˜na, Spain
Abstract
This communication is devoted to the presentation of our recent results regarding the asymp-totic analysis of a viscous flow in a tube with elastic walls. This study can be applied, forexample, to the blood flow in an artery. With this aim, we consider the dynamic problemof the incompressible flow of a viscous fluid through a curved pipe with a smooth centralcurve. Our analysis leads to obtain an one dimensional model via singular perturbation ofthe Navier-Stokes system as ε , a non dimensional parameter related to the radius of cross-section of the tube, tends to zero. We allow the radius depend on tangential direction andtime, so a coupling with an elastic or viscoelastic law on the wall of the pipe is possible.To perform the asymptotic analysis, we do a change of variable to a reference domainwhere we assume the existence of asymptotic expansions on ε for both velocity and pressurewhich, upon substitution on Navier-Stokes equations, leads to the characterization of variousterms of the expansion. This allows us to obtain an approximation of the solution of theNavier-Stokes equations. Keywords:
Asymptotic Analysis, Blood flow, Navier-Stokes equations.
1. Introduction
Last decades, applied mathematics have been involved in some new fields where theyhad not been applied before. One of these fields is biomedicine, from which new methodsto improve the diagnosis and treatment of different diseases are demanded. In particular,in the case of cardiovascular problems, modeling the blood flow in veins and arteries is adifficult problem.A large number of articles have studied the flow of a viscous fluid through a pipe. Forexample, in [2, 5, 11] the flow behavior inside the pipe is related with the curvature andtorsion of its middle line. In [2] the main term of the asymptotic expansion of the solutionis compared with a Poiseuille flow inside a pipe with rigid walls. In [8], the same problembut with visco-elastic walls is considered, leading to a fluid-structure problem. In [3] the
Email addresses: [email protected] (G. Casti˜neira), [email protected] (J.M. Rodr´ıguez)
Preprint submitted to arXiv.org September 8, 2018 a r X i v : . [ m a t h . A P ] F e b econdary flow is studied, the boundary layer in [10], both depending on values of Deannumber. More recently, the non-steady case in tube structures, has been considered in [6, 7],where estimates of the error between exact solution and the asymptotic approximation areproved.There are also articles where the flow in blood vessels is modeled. An one dimensionalmodel is presented in [1], where clinical procedures where this model can be useful arehighlighted. Another model for blood flow in arteries is developed in [9], relating bloodpulse and flow patterns, and remarking how this kind of models can help with the design oftreatments for particular diseases.In this article, we shall follow the spirits of [4], where asymptotic analysis is used to finda model for a steady flow through a curved pipe with rigid walls. We shall consider, instead,an unsteady flow and elastic walls. The structure of this article is the following: in section2 we shall describe the problem in a reference domain, in section 3 we shall suppose theexistence of an asymptotic expansion of the solution and we shall identify the first terms ofthis expansion, in section 5 we shall show some examples of the tangential and transversalvelocity, and finally, we shall present some conclusions in section 6.
2. Setting the problem in a reference domain
Let us suppose that central curve of the pipe is parametrized by c ( s ), where s ∈ [0 , L ] isthe arc-length parameter, and the interior points of the pipe are given by( x, y, z ) = c ( s ) + ε r R ( t, s ) [(cos θ ) N ( s ) + (sin θ ) B ( s )] , where r ∈ [0 , θ ∈ [0 , π ], { T = c (cid:48) , N , B } is the Frenet-Serret frame of c , and εR ( t, s ) isthe radius of the cross-section of the pipe at point c ( s ) and time t . The non dimensionalparameter ε represents the different scale of magnitude between the pipe diameter and itslength, so we shall assume that ε << s := s, s := θ, s := r for the variables, and { v := T , v := N , v := B } , for the Frenet-Serret frame of c . This new notation will allowus to use Einstein summation convention in what follows.Let be the subsets of R defined by Ω ε = [0 , L ] × [0 , π ] × [0 , ε ] and Ω = [0 , L ] × [0 , π ] × [0 , φ ε : Ω → Ω ε , φ ε : Ω ε → ˆΩ εt , where φ ε and φ ε are given by theexpressions, φ ε ( s , s , s ) = ( s , s , εs ) =: ( s ε , s ε , s ε ) ,φ ε ( s ε , s ε , s ε ) = c ( s ε ) + s ε R ( t, s ε )[(cos s ε ) v ( s ε ) + (sin s ε ) v ( s ε )] , (2.1)and ˆΩ εt = φ ε ( φ ε (Ω)) represents the interior points of the pipe.We can then introduce the change of variable from the reference domain Ω, φ ε = ( φ ε ◦ φ ε ) : Ω → ˆΩ εt ,φ ε ( s , s , s ) = c ( s ) + εs R ( t, s )[(cos s ) v ( s ) + (sin s ) v ( s )] =: ( x ε , x ε , x ε ) . (2.2)2et us consider the incompressible Navier-Stokes equations in the domain ˆΩ εt given by, ∂ u ε ∂t + ( ∇ u ε ) u ε = 1 ρ div T ε + b ε , (2.3)div u ε = 0 , (2.4)where u ε stands for the velocity field, b ε is the density of body forces and T ε is the stresstensor given by T ε = − p ε I + 2 µ Σ ε , where p ε is the pressure field, µ the dynamic viscosity and Σ ε = (cid:0) ∇ u ε + ( ∇ u ε ) T (cid:1) . Let ν = µ/ρ be the kinematic viscosity, so we can write these equations, ∂ u ε ∂t + ( ∇ u ε ) u ε + 1 ρ ∇ p ε − ν ∆ u ε = b ε , (2.5)div u ε = 0 . (2.6)We shall consider continuity between the fluid and the wall of the pipe displacements.Let us suppose that only radial displacements of the wall are allowed. Then the boundarycondition at the interface of the fluid and the wall of the pipe can be expressed as u ε = (cid:18) ε ∂R∂t (cid:19) n ε at s ε = ε, (2.7)where n ε is the outward unitary normal at s ε = ε .The next step is to write the equations of the problem in the reference domain Ω. Takinginto account the change of variable (2.2), we can associate to each vector field w ε in ˆΩ εt , anew vector field w ( ε ) defined in Ω, as follows w εi = w ε · e i = ( w εk e k ) · e i = ( w k ( ε ) v k ) · e i =: w k ( ε ) v ki , (2.8)where { e , e , e } is an orthonormal basis, we are using the Einstein summation convention(where latin indices indicate sum from 1 to 3), and we denote v ki := v k · e i . As first step, we shall need to study the inverse mapping of the change of variable (2.2),in particular, of its Jacobian, which terms will be needed to write Navier-Stokes equationsin the reference domain. Let us consider the mapping:˜ φ ε : [0 , T ] × Ω −→ [0 , T ] × ˆΩ εt (2.9)˜ φ ε ( t, s , s , s ) := ( t ε , x ε ) = ( t ε , φ ε ( s , s , s )) , (2.10)3ence, the associated Jacobian denoted by J φ is J φ = ∂t ε ∂t ∂t ε ∂s ∂t ε ∂s ∂t ε ∂s ∂x ε ∂t ∂x ε ∂s ∂x ε ∂s ∂x ε ∂s ∂x ε ∂t ∂x ε ∂s ∂x ε ∂s ∂x ε ∂s ∂x ε ∂t ∂x ε ∂s ∂x ε ∂s ∂x ε ∂s = ∂t ε ∂t ∂t ε ∂s ∂t ε ∂s ∂t ε ∂s ∂ x ε ∂t ∇ s x ε , where s = ( s , s , s ). Since t ε = t, x ε = c ( s ) + εs R ( t, s )((cos s ) v ( s ) + (sin s ) v ( s )) , then, ∂t ε ∂t = 1 ,∂t ε ∂s i = 0 ,∂ x ε ∂t = εs ∂R∂t ((cos s ) v ( s ) + (sin s ) v ( s )) ,∂ x ε ∂s = c (cid:48) ( s ) + εs ∂R∂s ((cos s ) v ( s ) + (sin s ) v ( s )) + εs R ((cos s ) v (cid:48) ( s ) + (sin s ) v (cid:48) ( s )) ,∂ x ε ∂s = εs R ( − (sin s ) v ( s ) + (cos s ) v ( s )) ,∂ x ε ∂s = εR ((cos s ) v ( s ) + (sin s ) v ( s )) . The inverse mapping ( ˜ φ ε ) − : [0 , T ] × ˆΩ εt −→ [0 , T ] × Ω is such that its Jacobian, denotedby J − φ , is J − φ = ∂t∂t ε ∂t∂x ε ∂t∂x ε ∂t∂x ε ∂s ∂t ε ∂s ∂x ε ∂s ∂x ε ∂s ∂x ε ∂s ∂t ε ∂s ∂x ε ∂s ∂x ε ∂s ∂x ε ∂s ∂t ε ∂s ∂x ε ∂s ∂x ε ∂s ∂x ε = ∂ s ∂t ε ∇ x ε s . J φ J − φ = I , we find the following relations, ∇ x ε s = ( ∇ s x ε ) − , (2.11) ∂ s ∂t ε = − ( ∇ x ε s ) ∂ x ε ∂t . (2.12)In order to compute ∂s i ∂ x ε , let us write ∂s i ∂ x ε = α i v + β i v + γ i v .We know, by Frenet-Serret formulas, that the following equalities hold v (cid:48) ( s ) = κ ( s ) v ( s ) , v (cid:48) ( s ) = − κ ( s ) v ( s ) + τ ( s ) v ( s ) , v (cid:48) ( s ) = − τ ( s ) v ( s ) , (2.13)where the functions κ and τ denote the curvature and torsion of the middle line of the curvedpipe. Now, by (2.11), we have that ∂s i ∂ x ε · ∂ x ε ∂s j = δ ij , (2.14)where δ ij is the Kronecker’s delta. For i = 1 we find that1 = ( α v + β v + γ v ) · (cid:18) v + εs ∂R∂s (cos s v + sin s v )+ εs R (cos s v (cid:48) + sin s v (cid:48) ) (cid:19) = α (1 + εs R (cos s ( v (cid:48) · v ) + sin s ( v (cid:48) · v )))+ β εs (cid:18) ∂R∂s cos s + R sin s ( v (cid:48) · v ) (cid:19) + γ εs (cid:18) ∂R∂s sin s + R cos s ( v (cid:48) · v ) (cid:19) , (2.15)since v = c (cid:48) , and0 = ( α v + β v + γ v ) · ( εs R ( − sin s v + cos s v ))= − β εs R sin s + γ εs R cos s , (2.16)0 = ( α v + β v + γ v ) · ( εR (cos s v + sin s v ))= β εR cos s + γ εR sin s . (2.17)From (2.16)–(2.17) is easy to check that β = γ = 0. Hence, from (2.15) and (2.13), weobtain that α = 11 − εκ ( s ) s R ( t, s ) cos s (2.18)5ow, for i = 2 in (2.14) we find, on one hand, that0 = α (1 + εs R (cos s ( v (cid:48) · v ) + sin s ( v (cid:48) · v )))+ β εs (cid:18) ∂R∂s cos s + R sin s ( v (cid:48) · v ) (cid:19) + γ εs (cid:18) ∂R∂s sin s + R cos s ( v (cid:48) · v ) (cid:19) , (2.19)and, on the other hand, that1 = − β εs R sin s + γ εs R cos s β εR cos s + γ εR sin s . where we deduce that, β = − sin s εs R ( t, s ) , γ = cos s εs R ( t, s ) . (2.20)Therefore, from (2.13) and (2.19), we obtain that α = − τ ( s )1 − εκ ( s ) s R ( t, s ) cos s . (2.21)Finally, for i = 3 in (2.14) we find, on one hand, that0 = α (1 + εs R (cos s ( v (cid:48) · v ) + sin s ( v (cid:48) · v )))+ β εs (cid:18) ∂R∂s cos s + R sin s ( v (cid:48) · v ) (cid:19) + γ εs (cid:18) ∂R∂s sin s + R cos s ( v (cid:48) · v ) (cid:19) , (2.22)and, on the other hand, that0 = − β εs R sin s + γ εs R cos s β εR cos s + γ εR sin s . where we deduce that, β = cos s εR ( t, s ) , γ = sin s εR ( t, s ) . (2.23)Therefore, from (2.13) and (2.22), we obtain that α = − s R ( t, s ) (1 − εκ ( s ) s R ( t, s ) cos s ) ∂R∂s ( t, s ) . (2.24)6o sum up, we have obtained that ∂s ∂ x ε = 11 − εκ ( s ) s R ( t, s ) cos s v ( s ) , (2.25) ∂s ∂ x ε = − τ ( s )1 − εκ ( s ) s R ( t, s ) cos s v ( s ) − sin s εs R ( t, s ) v ( s )+ cos s εs R ( t, s ) v ( s ) , (2.26) ∂s ∂ x ε = − s R ( t, s ) (1 − εκ ( s ) s R ( t, s ) cos s ) ∂R∂s ( t, s ) v ( s )+ cos s εR ( t, s ) v ( s ) + sin s εR ( t, s ) v ( s ) , (2.27)and from the relation found in (2.12), we deduce that ∂ s ∂t ε = − s R ( t, s ) ∂R∂t ( t, s ) v ( s ) . (2.28) We are now in conditions to find the expressions of the fields in (2.5)–(2.6) in the referencedomain. Firstly, from (2.8), the chain rule and (2.28), we find that ∂u εi ∂t ε = ∂ u ε ∂t ε · e i = ∂ ( u εk e k ) ∂t ε · e i = ∂ ( u k ( ε ) v k ) ∂t ε · e i = (cid:18) ∂ ( u k ( ε ) v k ) ∂t + ∂ ( u k ( ε ) v k ) ∂s ∂s ∂t ε (cid:19) · e i = (cid:18) ∂u k ( ε ) ∂t − s R ∂R∂t ∂u k ( ε ) ∂s (cid:19) ( v k · e i ) = ( D t u k ) v ki , where D t is the operator defined by D t := (cid:18) ∂∂t − s R ∂R∂t ∂∂s (cid:19) . (2.29)The components of the non-linear term in (2.5), from (2.8) and the chain rule, can bewritten as follows, ∂u εi ∂x εj u εj = ∂ ( u ε · e i ) ∂x εj ( u ε · e j ) = ∂ (( u εk e k ) · e i ) ∂x εj (( u εm e m ) · e j )= ∂ (( u k ( ε ) v k ) · e i ) ∂x εj (( u m ( ε ) v m ) · e j ) = (cid:18) ∂ ( u k ( ε ) v ki ) ∂s q ∂s q ∂x εj (cid:19) ( u m ( ε ) v mj ) . (2.30)The Laplacian term in (2.5), from (2.8) and the chain rule, leads to∆ u εi = ∆( u ε · e i ) = ∆( u εk e k ) · e i = ∆( u k ( ε ) v k ) · e i = ∂ ∂ ( x εj ) ( u k ( ε ) v ki )= ∂∂x εj (cid:18) ∂ ( u k ( ε ) v ki ) ∂s q ∂s q ∂x εj (cid:19) = ∂∂s m (cid:18) ∂ ( u k ( ε ) v ki ) ∂s q ∂s q ∂x εj (cid:19) ∂s m ∂x εj . (2.31)7n the same way we obtain the components of the pressure gradient and of the volumeforces as follows, ∂p∂x εi = ∂p∂s q ∂s q ∂x εi , ( b ε ) i = b ε · e i = (( b ε ) k e k ) · e i = (( b ( ε )) k v k ) · e i = b k ( ε ) v ki , where b k ( ε ) := ( b ( ε )) k .Finally, the incompressibility equation (2.6) in the reference domain, using (2.8) and thechain rule, has the following expression,div u ε = ∂u εj ∂x εj = ∂ ( u ε · e j ) ∂x εj = ∂ (( u εk e k ) · e j ) ∂x εj = ∂ (( u k ( ε ) v k ) · e j ) ∂x εj = ∂ ( u k ( ε ) v kj ) ∂s q ∂s q ∂x εj . With these considerations, the incompressible Navier-Stokes equations in the referencedomain can be written as D t ( u k ( ε ) v ki ) + (cid:18) ∂ ( u k ( ε ) v ki ) ∂s q ∂s q ∂x εj (cid:19) ( u m ( ε ) v mj ) − ν ∂∂s m (cid:18) ∂ ( u k ( ε ) v ki ) ∂s q ∂s q ∂x εj (cid:19) ∂s m ∂x εj = − ρ ∂p ( ε ) ∂s q ∂s q ∂x εi + b k ( ε ) v ki , (2.32) ∂∂s q ( u k ( ε ) v kj ) ∂s q ∂x εj = 0 . (2.33)Let n ε = (cos s ) v ( s ) + (sin s ) v ( s ) the outward unit normal vector at s = 1. Then,from the boundary condition (2.7) at s ε = ε , we have that u ε = u εi e i = u i ( ε ) v i = ε ∂R∂t ((cos s ) v ( s ) + (sin s ) v ( s )) , (2.34)hence, we obtain the following boundary conditions for the scaled components of velocity: u ( ε ) = 0 at s = 1 ,u ( ε ) = ε ∂R∂t cos s at s = 1 ,u ( ε ) = ε ∂R∂t sin s at s = 1 . (2.35)
3. Asymptotic expansion of the solution ε Following [4], we assume that the solution of (2.32)-(2.33) admits a formal expansion onpowers of ε , so the components of velocity and pressure fields can be written, u k ( ε ) = u k + εu k + ε u k + ... (3.1) p ( ε ) = 1 ε p + 1 ε p + p + ... (3.2)8e must remark that this assumption implies, as we shall see later, that the pressuregradient determines the velocity field. Other assumptions can be considered by choosingdifferent order of ε in the pressure and velocity fields in (3.1)-(3.2), leading to differentconclusions, but we consider that this is the most interesting case.Substituting (3.1)-(3.2) in the boundary conditions in the reference domain (see (2.35)),we obtain the following boundary conditions for the terms of the asymptotic expansion, u k = 0 , k ≥ , at s = 1 ,u = u = 0 at s = 1 ,u = ∂R∂t cos s , u = ∂R∂t sin s at s = 1 ,u kα = 0 , k ≥ , α = 2 , s = 1 . (3.3)We need to write (2.25)–(2.27) as expansions of ε . If we remark that1 a + εb = c + c ε + c ε + ... where a, b ∈ R , such that a (cid:54) = 0, then is easy to check that c k = ( − k b k a k +1 , k ≥ . Therefore, with a = 1 and b = − κs R cos s in (2.25)–(2.27), we find that ∂s q ∂ x ε · e j = ∂s q ∂x εj = 1 ε d q − j + d q j + εd q j + ε d q j + ... (3.4)with q = 1 , , d − j = 0 , d kj = ( − k b k a k +1 v j , (3.5) d − j = − sin s Rs v j + cos s Rs v j , d kj = ( − k ( − τ ) b k a k +1 v j , (3.6) d − j = cos s R v j + sin s R v j , d kj = ( − k +1 s R ∂R∂s b k a k +1 v j , (3.7)with k ≥ . Also, we assume that the applied forces admit an asymptotic expansion of the form b k ( ε ) = b k + εb k + ε b k + ...
9e substitute (3.1)-(3.2) into the equations (2.32)-(2.33) and use (3.4). Hence, we obtainthe incompressible Navier-Stokes equations in the reference domain in powers of ε : D t (( u k + εu k + ε u k + ... ) v ki )+ (cid:18) ∂ (( u k + εu k + ε u k + ... ) v ki ) ∂s q (cid:18) ε d q − j + d q j + εd q j + ε d q j + ... (cid:19)(cid:19) (cid:0) ( u m + εu m + ε u m + ... ) v mj (cid:1) − ν ∂∂s m (cid:18) ∂ (( u k + εu k + ε u k + ... ) v ki ) ∂s q (cid:18) ε d q − j + d q j + εd q j + ε d q j + ... (cid:1)(cid:1) (cid:18) ε d m − j + d m j + εd m j + ε d m j + ... (cid:19) = − ρ ∂ ( ε p + ε p + p + ... ) ∂s q (cid:18) ε d q − i + d q i + εd q i + ε d q i + ... (cid:19) + (cid:0) b k + εb k + ε b k + ... (cid:1) v ki , (3.8) ∂∂s q (( u k + εu k + ε u k + ... ) v kj ) (cid:18) ε d q − j + d q j + εd q j + ε d q j + ... (cid:19) = 0 . (3.9) In the next proposition we shall show some conditions satisfied by the asymptotic expan-sion of the flow, that will be used later to characterize the terms of the asymptotic expansionof the solution. Firstly, let Q ε ( t, s ) denotes the flow at position s = s and at time t , definedin the original domain by Q ε ( t, s ) := (cid:90) s = s u ε · v dA. (3.10)where (cid:90) s = s φ dA represents the surface integral of φ on the transversal section of ˆΩ εt at s = s .Using the mapping (2.2), we define the flow in the reference domain by Q ( ε ), that is Q ε ( t, s ) = ε Q ( ε )( t, s ) , (3.11)where Q ( ε ) = R (cid:90) π (cid:90) s u ( ε ) ds ds . (3.12)We also define the cross-sectional area in the original domain by A ε := ε A = ε πR , (3.13)where A denotes the cross-sectional area in the reference domain, A = πR .10 roposition 3.1. Let us consider a fluid inside the curved pipe ˆΩ εt , which movement isdescribed by the incompressible Navier-Stokes equations (2.5)–(2.6) with the boundary con-dition (2.7). Let us assume that there exists an asymptotic expansion of the form (3.1)–(3.2)of the problem in the reference domain. Then there exists an asymptotic expansion for thescaled flow in the reference domain of the form Q ( ε ) = Q + εQ + ε Q + ... where the term Q k is defined by Q k = R (cid:90) π (cid:90) s u k ds ds . (3.14) Moreover, the following relations hold: ∂Q ∂s + ∂A ∂t = 0 , ∂Q k ∂s = 0 ( k ≥ . (3.15) Proof.
Let ˜Ω (cid:15)t be a portion of the original domain ˆΩ εt between s = a and s = b ( a < b ).From (2 .
6) and the Gauss Theorem, we deduce that0 = (cid:90) ˜Ω (cid:15)t div u ε dV = (cid:90) ∂ ˜Ω (cid:15)t u ε · n ε dA = (cid:90) s = a u ε · n ε dA + (cid:90) s = b u ε · n ε dA + (cid:90) s ε = εR ( t,s ) u ε · n ε dA. (3.16)At the beginning and end of ˜Ω (cid:15)t we have that u ε · n ε = ( u k ( ε ) v k ) · ( − v ) = − u ( ε ) at s = a, u ε · n ε = ( u k ( ε ) v k ) · ( v ) = u ( ε ) at s = b. Therefore, from (3.10) and (3.16) we obtain0 = − Q ε ( t, a ) + Q ε ( t, b ) + (cid:90) s ε = εR ( t,s ) u ε · n ε dA. (3.17)At s ε = εR , we must consider continuity between the fluid and the wall of the pipedisplacements (see (2.7)), hence u ε · n ε = ∂∂t [ c ( s ) + εR ( t, s )(cos s ε v ( s ) + sin s ε v ( s ))] · n ε = ε ∂R∂t , and then, (cid:90) s ε = εR ( t,s ) u ε · n ε dA = (cid:90) s = bs = a (cid:90) s =2 πs =0 ε R ∂R∂t ds ds = (cid:90) ba πε R ∂R∂t ds . b − a , we obtain that0 = Q ε ( t, b ) − Q ε ( t, a ) b − a + 1 b − a (cid:90) ba πε R ∂R∂t ds . Taking the limit when b tends to a and using (3.13), we obtain the following relation,0 = ∂Q ε ∂s + 2 πε R ∂R∂t = ∂Q ε ∂s + ∂A ε ∂t . Now, since A = A ε /ε = πR and Q ( ε ) = Q ε /ε , we deduce that ∂Q ( ε ) ∂s + ∂A ∂t = 0 . (3.18)On the other hand, taking into account the expansion for u ( ε ) in (3.1) and (3.12), wecan deduce that there exists an asymptotic expansion of the form Q ( ε ) = Q + εQ + ε Q + ... (3.19)where Q k = R (cid:90) π (cid:90) s u k ds ds . Finally, upon substitution of (3.19) in (3.18), we conclude that ∂Q ∂s + ∂A ∂t = 0 , ∂Q k ∂s = 0 ( k ≥ . Remark 3.2.
The previous result is a direct consequence of the law of conservation of mass.Similar results can be found in previous works, for instance, see [1].3.3. Identifying the terms of the asymptotic expansion of the solution
In order to identify some of the terms of the asymptotic expansion proposed in (3.1)-(3.2), we shall group the terms multiplied by the same power of ε in (3.8)–(3.9), obtainingin this way new equations, easier than the original one, that can be solved to identify thementioned terms of the asymptotic expansion. With this aim, we shall recall a result from[12] (Theorem 2.4), that will be used in the following. Theorem 3.3.
Let Ω be an open bounded set of class C in R n and Γ = ∂ Ω . Let there begiven f ∈ H − (Ω) , g ∈ L (Ω) , ϕ ∈ H / (Γ) , such that (cid:90) Ω g d x = (cid:90) Γ ϕ · n d Γ , hen there exists u ∈ H (Ω) , p ∈ L (Ω) , which are solutions of the Stokes problem − ν ∆ u + ∇ p = f in Ω , div u = g in Ω , u = ϕ on Γ . u is unique and p is unique up to the addition of a constant. Let us introduce the local cartesian coordinates of the cross section of the pipe at s , asthe points z ∈ R defined by z = ( z , z ) = ( s cos s , s sin s ) , (3.20)and let ω = { ( z , z ) ∈ R /z + z < } . We can prove now the following theorem, wherethe first terms of the asymptotic expansion are identified. Theorem 3.4.
Let us assume that there exists an asymptotic expansion of the form (3.1)-(3.2). Then:(i) The term of order zero of velocity, u , verifies u = R ρ ν ∂p ∂s ( s − , (3.21) u = u = 0 , (3.22) while zeroth order term of pressure, p , is the solution of the problem, ∂∂s (cid:18) R ∂p ∂s (cid:19) = 16 νρ R ∂R∂t , (3.23) with suitable boundary conditions.(ii) The components of the first order term of velocity, u , are u = (cid:20) R κs cos s νρ ∂p ∂s + R νρ ∂p ∂s (cid:21) ( s − , (3.24) u = s R ρ ν (cid:20) ∂∂s (cid:18) R ∂p ∂s (cid:19) − R s ∂ p ∂s (cid:21) cos s , (3.25) u = s R ρ ν (cid:20) ∂∂s (cid:18) R ∂p ∂s (cid:19) − R s ∂ p ∂s (cid:21) sin s . (3.26) The first order term of pressure, p , is the solution of the problem, ∂∂s (cid:18) R ∂p ∂s (cid:19) = 0 , (3.27) with the appropriate boundary conditions. iii) The first component of the second order term of velocity, u , is given by u = R (cid:20) R ρ ν ∂ p ∂t∂s − R ρ ν ∂p ∂s ∂ p ∂s − R ρ ν ∂ p ∂s + 11 κ R ρ ν ∂p ∂s (cid:21) ( s − R (cid:20) − ρ ν ∂∂t (cid:18) R ∂p ∂s (cid:19) + R ρ ν ∂p ∂s ∂∂s (cid:18) R ∂p ∂s (cid:19) + 14 ρ ν ∂ ∂s (cid:18) R ∂p ∂s (cid:19) − κ R ρ ν ∂p ∂s + 1 ρ ν ∂p ∂s − b ν (cid:21) ( s − R ρ ν ∂p ∂s ∂ p ∂s ( s −
1) + 3 κR ρ ν ∂p ∂s ( s − s ) cos s + 5 κ R ρ ν ∂p ∂s ( s − s ) cos(2 s ) , (3.28) and the second order term of pressure, p , is p = − R ∂ p ∂s s + p ( t, s ) , (3.29) where p ( t, s ) is the solution, with the adequate boundary conditions, of the problem ∂∂s (cid:18) R ∂p ∂s (cid:19) = ∂∂s (cid:20) − R ρ ν ∂p ∂s ∂ p ∂s − R ∂ p ∂s − κ R ∂p ∂s + R ν ∂R∂t ∂p ∂s − R ρ ν ∂R∂s (cid:18) ∂p ∂s (cid:19) − R (cid:18) ∂R∂s (cid:19) ∂p ∂s − R ∂ R∂s ∂p ∂s − R ∂R∂s ∂ p ∂s + R ν ∂ p ∂t∂s + R ρ b (cid:21) . (3.30) Let us consider the local cartesian coordinates at cross section of the pipe at s , defined by z = ( z , z ) = ( s cos s , s sin s ) (and then, ( s , s ) are the local polar coordinates at thesame cross section). Let be U = ( u , u ) . Then ( U , p ) is the solution of the followingproblem ∆ z U = Rρ ν ∇ z p + F in ω, div z U = g in ω, U = on ∂ω, (3.31) where ω = { ( z , z ) /z + z < } and the fields g and F are defined, respectively, by g = (cid:18) − κR ρ ν ∂ p ∂s − κ (cid:48) R ρ ν ∂p ∂s (cid:19) z (cid:0) z + z (cid:1) − κτ R ρ ν ∂p ∂s z (cid:0) z + z (cid:1) + (cid:18) κR ρ ν ∂R∂s ∂p ∂s + 9 κR ρ ν ∂ p ∂s + 3 κ (cid:48) R ρ ν ∂p ∂s (cid:19) z + 3 κτ R ρ ν ∂p ∂s z − R ρ ν ∂ p ∂s (cid:0) z + z (cid:1) + R ρ ν ∂∂s (cid:18) R ∂p ∂s (cid:19) , (3.32)14 = (cid:32) κR ρ ν (cid:18) ∂p ∂s (cid:19) (cid:0) ( z + z ) + 1 (cid:1) + κ (cid:48) R ρ ν ∂p ∂s + 5 R κ ρ ν ∂∂s (cid:18) R ∂p ∂s (cid:19) + (cid:32) − κR ρ ν (cid:18) ∂p ∂s (cid:19) − R κ ρ ν ∂ p ∂s − κ (cid:48) R ρ ν ∂p ∂s (cid:33) ( z + z ) − κR ρ ν ∂ p ∂s z − R ν b , − κτ R ρ ν ∂p ∂s ( z + z − − R κ ρ ν ∂ p ∂s z z − R ν b (cid:19) . (3.33) Problem (3.31) has a unique solution ( U , p ) , with U unique and p unique up to a functiondepending on t and s , that can be computed explicitly (see appendix B).Proof. After substitution of (3.1)–(3.2) in (2.32)–(2.33), we have obtained equations (3.8)–(3.9), and we proceed now to identify the terms of the asymptotic expansion.(i) Grouping the terms multiplied by ε − in the equation (3.8), we find these two equa-tions related with the zeroth-order term of pressure, − sin s s ∂p ∂s + (cos s ) ∂p ∂s = 0 , cos s s ∂p ∂s + (sin s ) ∂p ∂s = 0 . Therefore, it is clear that ∂p ∂s = ∂p ∂s = 0, so p = p ( t, s ) . (3.34)This is, the zeroth-order term of pressure does not depend on the cross-sectional variablesand only depends on time and on the point s of the midle line of the curved pipe.If we group now the terms multiplied by ε − in the equation (3.8), we obtain the equations1( Rs ) ∂ u ∂s + 1 R s ∂u ∂s + 1 R ∂ u ∂s = 1 νρ ∂p ∂s , (3.35)1( Rs ) ∂ u ∂s + 1 R s ∂u ∂s + 1 R ∂ u ∂s = 1 νρ (cid:18) − sin s Rs ∂p ∂s + cos s R ∂p ∂s (cid:19) , (3.36)1( Rs ) ∂ u ∂s + 1 R s ∂u ∂s + 1 R ∂ u ∂s = 1 νρ (cid:18) cos s Rs ∂p ∂s + sin s R ∂p ∂s (cid:19) . (3.37)Using the change of variable (3.20) in (3.35), and taking into account the boundary conditionin (3.3), we obtain the following problem for the axial component of the zeroth-order termof velocity: ∆ z u = R νρ ∂p ∂s in ω,u = 0 on ∂ω. (3.38)15he problem (3.38) has a unique solution, which expression is u = R ρ ν ∂p ∂s ( s − . Now, from the first relation in (3.15), we have that ∂∂s (cid:18) R (cid:90) π (cid:90) s u ds ds (cid:19) = − πR ∂R∂t . Hence, using the expression for u that we have obtained, we deduce that the left-hand sideof last equality verifies ∂∂s (cid:18) R (cid:90) π (cid:90) R ρ ν ∂p ∂s s ( s − ds ds (cid:19) = ∂∂s (cid:18) − πR ρ ν ∂p ∂s (cid:19) Since p does not depend on the cross-sectional variables (see (3.34)), we obtain that p satisfies the following equation, ∂∂s (cid:18) R ∂p ∂s (cid:19) = 16 νρ R ∂R∂t , that has a unique solution with the appropriate initial and boundary conditions.If we now group the terms multiplied by ε − in the equation (3.9), we find that − sin s s ∂u ∂s + (cos s ) ∂u ∂s + cos s s ∂u ∂s + (sin s ) ∂u ∂s = 0 . Using the change of variable (3.20) in this last equation and in (3.36)–(3.37), and consideringthe boundary conditions in (3.3), the cross-sectional components of the zeroth-order termof velocity, denoted by U = ( u , u ), and the first order term of pressure, p , are solutionof the problem, ∆ z U = Rνρ ∇ z p in ω, div z U = 0 in ω, U = on ∂ω, Applying Theorem 3.3, this problem has a unique solution (up to an arbitrary functiondepending only on t and s , for the pressure term), and this solution is u = u = 0 , p = p ( t, s ) . ε − in the equation (3.8), we obtain1( Rs ) ∂ u ∂s + 1 R s ∂u ∂s + 1 R ∂ u ∂s = 1 νρ (cid:18) ∂p ∂s + Rκs cos s ∂p ∂s (cid:19) + κ cos s R ∂u ∂s , (3.39)1( Rs ) ∂ u ∂s + 1 R s ∂u ∂s + 1 R ∂ u ∂s = 1 νρ (cid:18) − sin s Rs ∂p ∂s + cos s R ∂p ∂s (cid:19) , (3.40)1( Rs ) ∂ u ∂s + 1 R s ∂u ∂s + 1 R ∂ u ∂s = 1 νρ (cid:18) cos s Rs ∂p ∂s + sin s R ∂p ∂s (cid:19) . (3.41)Using (3.21) in (3.39), and then the change of variable (3.20) and the boundary condition(3.3), we obtain that u is the unique solution of the problem ∆ z u = R νρ (cid:18) ∂p ∂s + 3 Rκ z ∂p ∂s (cid:19) in ω,u = 0 on ∂ω. (3.42)Now, it is easy to check (for example, by substitution in (3.39)) that the unique solution is u = (cid:20) R κs cos s νρ ∂p ∂s + R νρ ∂p ∂s (cid:21) ( s − . From the second relation in (3.15), for k = 1, we have that ∂∂s (cid:18) R (cid:90) π (cid:90) s u ds ds (cid:19) = 0 , and, substituting the expression of u , we obtain that0 = ∂∂s (cid:18) R (cid:90) π (cid:90) (cid:20) R κs cos s νρ ∂p ∂s + R νρ ∂p ∂s (cid:21) s ( s − ds ds (cid:19) = ∂∂s (cid:18) R (cid:90) π (cid:90) R ρ ν ∂p ∂s s ( s − ds ds (cid:19) = ∂∂s (cid:18) − πR ρ ν ∂p ∂s (cid:19) , so we conclude that p (remember that it is only function of t and s ) is solution of ∂∂s (cid:18) R ∂p ∂s (cid:19) = 0 , with suitable initial and boundary conditions.17f we now group the terms multiplied by ε in (3.9), we obtain − sin s s ∂u ∂s + (cos s ) ∂u ∂s + cos s s ∂u ∂s + (sin s ) ∂u ∂s + R (cid:18) ∂u ∂s − τ ∂u ∂s − s R ∂R∂s ∂u ∂s (cid:19) = 0 . Let be U = ( u , u ). Applying the change of variable (3.20) to the previous equation, to(3.40)–(3.41) and taking into account the boundary conditions (3.3), we find that ( U , p )is solution of the problem ∆ z U = Rνρ ∇ z p in ω, div z U = R ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − ( z + z ) R ∂ p ∂s (cid:19) =: g in ω, U = ∂R∂t (cos s , sin s ) =: ϕ on ∂ω. (3.43)Theorem 3.3 ensures the existence and uniqueness of solution of this problem (up to anarbitrary function depending only on t and s , for the pressure term) if the compatibilitycondition given by (cid:90) ω g = (cid:90) ∂ω ϕ · n , (3.44)where n = (cos s , sin s ) is the unit outward normal vector on ∂ω , is fulfilled. On one hand,we have (cid:90) ω g = (cid:90) ω R ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − ( z + z ) R ∂ p ∂s (cid:19) dz dz = (cid:90) (cid:90) π s R ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − s R ∂ p ∂s (cid:19) ds ds = 2 πR ρ ν (cid:90) (cid:18) s ∂∂s (cid:18) R ∂p ∂s (cid:19) − s R ∂ p ∂s (cid:19) ds = 2 πR ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − R ∂ p ∂s (cid:19) . On the other hand, (cid:90) ∂ω ϕ · n = (cid:90) π ∂R∂t ds = 2 π ∂R∂t . Therefore, the compatibility condition (3.44) is equivalent to R ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − R ∂ p ∂s (cid:19) = ∂R∂t , (3.45)18nd it is easy to deduce from (3.23) that (3.45) is verified. Then, we can conclude that thereexists a solution ( U , p ) of (3.43) such that U is unique and p is unique up to a functiondepending on t and s . This solution can be computed explicitly (see appendix A for thedetails), and it is U = R ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − ( z + z ) R ∂ p ∂s (cid:19) ( z , z ) ,p = − R ∂ p ∂s (cid:0) z + z (cid:1) + p ( t, s ) , where the unknown function p will be determined later (see (3.51)).(iii) Grouping the terms multiplied by ε = 1 in (3.8), we find ∂ ( u k v ki ) ∂t − s R ∂R∂t ∂ ( u k v ki ) ∂s + (cid:18) − sin s Rs u + cos s Rs u (cid:19) ∂ ( u k v ki ) ∂s + (cid:18) cos s R u + sin s R u (cid:19) ∂ ( u k v ki ) ∂s + ∂ ( u k v ki ) ∂s u − τ ∂ ( u k v ki ) ∂s u + (cid:18) − sin s Rs u + cos s Rs u (cid:19) ∂ ( u k v ki ) ∂s − s R ∂R∂s ∂ ( u k v ki ) ∂s u + (cid:18) cos s R u + sin s R u (cid:19) ∂ ( u k v ki ) ∂s − ν (cid:18) κ cos s sin s ∂ ( u k v ki ) ∂s − κ s cos s ∂ ( u k v ki ) ∂s + ∂ ( u k v ki ) ∂s − τ (cid:48) ∂ ( u k v ki ) ∂s − s R (cid:32) ∂ R∂s − R (cid:18) ∂R∂s (cid:19) (cid:33) ∂ ( u k v ki ) ∂s − τ ∂ ( u k v ki ) ∂s ∂s − s R ∂R∂s ∂ ( u k v ki ) ∂s ∂s + 2 τ s R ∂R∂s ∂ ( u k v ki ) ∂s ∂s + τ ∂ ( u k v ki ) ∂s + s R (cid:18) ∂R∂s (cid:19) ∂ ( u k v ki ) ∂s + κ sin s Rs ∂ ( u k v ki ) ∂s − κ cos s R ∂ ( u k v ki ) ∂s + 1( Rs ) ∂ ( u k v ki ) ∂s + 1 R s ∂ ( u k v ki ) ∂s + 1 R ∂ ( u k v ki ) ∂s (cid:19) = DP k v ki + b k v ki , (3.46)19here, DP = − ρ (cid:18) κ s R cos s ∂p ∂s + κs R cos s ∂p ∂s − τ κs R cos s ∂p ∂s − κs cos s ∂R∂s ∂p ∂s + ∂p ∂s − τ ∂p ∂s − s R ∂R∂s ∂p ∂s (cid:19) ,DP = − ρ (cid:18) − sin s Rs ∂p ∂s + cos s R ∂p ∂s (cid:19) ,DP = − ρ (cid:18) cos s Rs ∂p ∂s + sin s R ∂p ∂s (cid:19) . Now, we use the expressions obtained in steps ( i ) and ( ii ) (see (3.21)–(3.27) and (3.29)), andwe replace them into equation (3.46). Since { v , v , v } is an orthonormal basis, we obtainthree equations by grouping the terms multiplied by each vector in (3.46). Therefore, fromthe terms multiplied by v in (3.46), we obtain1( Rs ) ∂ u ∂s + 1 R s ∂u ∂s + 1 R ∂ u ∂s = (cid:18) R ρ ν ∂ p ∂t∂s − R ρ ν ∂p ∂s ∂ p ∂s − R ρ ν ∂ p ∂s + 7 κ R ρ ν ∂p ∂s (cid:19) s + R ρ ν ∂p ∂s ∂ p ∂s s − ρ ν ∂∂t (cid:18) R ∂p ∂s (cid:19) + R ρ ν ∂p ∂s ∂∂s (cid:18) R ∂p ∂s (cid:19) + 14 ρ ν ∂ ∂s (cid:18) R ∂p ∂s (cid:19) − κ R ρ ν ∂p ∂s + 1 ρ ν ∂p ∂s + 3 κR ρ ν ∂p ∂s s cos s + 15 κ R ρ ν ∂p ∂s s cos s − b ν . This equation, together with the boundary condition at s = 1 (see (3.3)), and using thechange of variable (3.20), shows that u is the unique solution of the problem ∆ z u = (cid:18) R ρ ν ∂ p ∂t∂s − R ρ ν ∂p ∂s ∂ p ∂s − R ρ ν ∂ p ∂s + 7 κ R ρ ν ∂p ∂s (cid:19) ( z + z )+ R ρ ν ∂p ∂s ∂ p ∂s ( z + z ) − R ρ ν ∂∂t (cid:18) R ∂p ∂s (cid:19) + R ρ ν ∂p ∂s ∂∂s (cid:18) R ∂p ∂s (cid:19) + R ρ ν ∂ ∂s (cid:18) R ∂p ∂s (cid:19) − κ R ρ ν ∂p ∂s + R ρ ν ∂p ∂s + 3 κR ρ ν ∂p ∂s z + 15 κ R ρ ν ∂p ∂s z − R b ν in ω,u = 0 on ∂ω. (3.47)20he solution of (3.47) must be polynomial on z and z and, by inspection in this kind offunctions, we can find that u is u = R (cid:18) R ρ ν ∂ p ∂t∂s − R ρ ν ∂p ∂s ∂ p ∂s − R ρ ν ∂ p ∂s + 11 κ R ρ ν ∂p ∂s (cid:19) (( z + z ) − R (cid:18) − ρ ν ∂∂t (cid:18) R ∂p ∂s (cid:19) + R ρ ν ∂p ∂s ∂∂s (cid:18) R ∂p ∂s (cid:19) + 14 ρ ν ∂ ∂s (cid:18) R ∂p ∂s (cid:19) − κ R ρ ν ∂p ∂s + 1 ρ ν ∂p ∂s − b ν (cid:19) ( z + z −
1) + R ρ ν ∂p ∂s ∂ p ∂s (( z + z ) − κR ρ ν ∂p ∂s ( z + z − z + 5 κ R ρ ν ∂p ∂s ( z + z − z − z ) , (3.48)that, using the change of variable (3.20), can be written u = R (cid:18) R ρ ν ∂ p ∂t∂s − R ρ ν ∂p ∂s ∂ p ∂s − R ρ ν ∂ p ∂s + 11 κ R ρ ν ∂p ∂s (cid:19) ( s − R (cid:18) − ρ ν ∂∂t (cid:18) R ∂p ∂s (cid:19) + R ρ ν ∂p ∂s ∂∂s (cid:18) R ∂p ∂s (cid:19) + 14 ρ ν ∂ ∂s (cid:18) R ∂p ∂s (cid:19) − κ R ρ ν ∂p ∂s + 1 ρ ν ∂p ∂s − b ν (cid:19) ( s −
1) + R ρ ν ∂p ∂s ∂ p ∂s ( s − κR ρ ν ∂p ∂s ( s − s ) cos s + 5 κ R ρ ν ∂p ∂s ( s − s ) cos(2 s ) . (3.49)Using now (3.15) for k = 2, we have that ∂∂s (cid:18) R (cid:90) π (cid:90) s u ds ds (cid:19) = 0 , (3.50)and, if we substitute the expression of u given by (3.49) in (3.50), we obtain that p is thesolution of the problem ∂∂s (cid:18) R ∂p ∂s (cid:19) = ∂∂s (cid:18) − R ρ ν ∂p ∂s ∂ p ∂s − R ∂ p ∂s − κ R ∂p ∂s + R ν ∂R∂t ∂p ∂s − R ρ ν ∂R∂s (cid:18) ∂p ∂s (cid:19) − R (cid:18) ∂R∂s (cid:19) ∂p ∂s − R ∂ R∂s ∂p ∂s − R ∂R∂s ∂ p ∂s + R ν ∂ p ∂t∂s + R ρ b (cid:19) , (3.51)with the adequate initial and boundary conditions. With this equation, we have completedthe description of p (see (3.29)–(3.30)). 21dentifying now the terms multiplied by v in (3.46), we obtain1( Rs ) ∂ u ∂s + 1 R s ∂u ∂s + 1 R ∂ u ∂s = κR ρ ν (cid:18) ∂p ∂s (cid:19) (cid:0) s + 1 (cid:1) + κ (cid:48) R ρ ν ∂p ∂s + 5 κ ρ ν ∂∂s (cid:18) R ∂p ∂s (cid:19) + (cid:32) − κR ρ ν (cid:18) ∂p ∂s (cid:19) − κ ρ ν ∂∂s (cid:18) R ∂p ∂s (cid:19) − κ (cid:48) R ρ ν ∂p ∂s + κRρ ν ∂R∂s ∂p ∂s − κR ρ ν ∂ p ∂s (cid:33) s − κR ρ ν ∂ p ∂s s cos s + 1 ρ ν (cid:18) − sin s Rs ∂p ∂s + cos s R ∂p ∂s (cid:19) − b ν , and if we identify the terms of (3.46) multiplied by v ,1( Rs ) ∂ u ∂s + 1 R s ∂u ∂s + 1 R ∂ u ∂s = − κ sin s cos s s ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) s − R ∂ p ∂s s (cid:19) + κ sin s cos s ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − R ∂ p ∂s s (cid:19) − κτ R ρ ν ∂p ∂s ( s − ρ ν (cid:18) cos s Rs ∂p ∂s + sin s R ∂p ∂s (cid:19) − b ν . Using the change of variable (3.20) and doing some simplifications, we deduce that u and u verify∆ z u = κR ρ ν (cid:18) ∂p ∂s (cid:19) (cid:0) ( z + z ) + 1 (cid:1) + κ (cid:48) R ρ ν ∂p ∂s + 5 κR ρ ν ∂∂s (cid:18) R ∂p ∂s (cid:19) + (cid:32) − κR ρ ν (cid:18) ∂p ∂s (cid:19) − κR ρ ν ∂ p ∂s − κ (cid:48) R ρ ν ∂p ∂s (cid:33) ( z + z ) − κR ρ ν ∂ p ∂s z + Rρ ν ∂p ∂z − R ν b , (3.52)∆ z u = − κτ R ρ ν ∂p ∂s ( z + z − − κR ρ ν ∂ p ∂s z z + Rρ ν ∂p ∂z − R ν b . (3.53)Now, if we group the terms multiplied by ε in (3.9), we obtain − sin s Rs ∂u ∂s + cos s Rs ∂u ∂s + cos s R ∂u ∂s + sin s R ∂u ∂s = − κs R cos s (cid:18) ∂u ∂s − s R ∂R∂s ∂u ∂s (cid:19) − ∂u ∂s + κu + τ ∂u ∂s + s R ∂R∂s ∂u ∂s , i ) and ( ii ) (see (3.21)–(3.27)), − sin s s ∂u ∂s + cos s s ∂u ∂s + (cos s ) ∂u ∂s + (sin s ) ∂u ∂s = − κR s cos s (cid:18) ρ ν ∂∂s (cid:18) R ∂p ∂s (cid:19) ( s − − R ρ ν ∂R∂s ∂p ∂s s (cid:19) − R ρ ν ∂∂s (cid:18) κR ∂p ∂s (cid:19) s ( s −
1) cos s − R ρ ν ∂∂s (cid:18) R ∂p ∂s (cid:19) ( s − κR ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − R s ∂ p ∂s (cid:19) s cos s − κτ R ρ ν ∂p ∂s s ( s −
1) sin s + 3 κR ρ ν ∂R∂s p s (3 s − s cos s + R ρ ν ∂R∂s ∂p ∂s s =: g. (3.54)Let be U = ( u , u ). Using the change of variable (3.20), we obtain from the equations(3.52), (3.53), (3.54) and the boundary conditions (3.3), that ( U , p ) solves the followingproblem, ∆ z U = Rρ ν ∇ z p + F in ω, div U = g in ω, U = on ∂ω, where, F := (cid:32) κR ρ ν (cid:18) ∂p ∂s (cid:19) (cid:0) ( z + z ) + 1 (cid:1) + κ (cid:48) R ρ ν ∂p ∂s + 5 R κ ρ ν ∂∂s (cid:18) R ∂p ∂s (cid:19) + (cid:32) − κR ρ ν (cid:18) ∂p ∂s (cid:19) − R κ ρ ν ∂ p ∂s − κ (cid:48) R ρ ν ∂p ∂s (cid:33) ( z + z ) − κR ρ ν ∂ p ∂s z − R ν b , − κτ R ρ ν ∂p ∂s ( z + z − − R κ ρ ν ∂ p ∂s z z − R ν b (cid:19) , and g is given by (3.32) (from the definition of g in (3.54), and using (3.20) after somesimplification (see (B.3)), we can obtain that g is given by (3.32)).Applying Theorem 3.3, this problem has a unique solution if the compatibility condition (cid:90) ω g dz dz = 0 , (3.55)is fulfilled, or equivalently, using the change of variable (3.20), if (cid:90) π (cid:90) s g ds ds = 0 . (cid:90) π (cid:90) s g ds ds = πR ρ ν ∂∂s (cid:18) R ∂p ∂s (cid:19) + πR ρ ν ∂R∂s ∂p ∂s = πR ρ ν ∂R∂s ∂p ∂s + πR ρ ν ∂ p ∂s . From (3.27) we deduce that 4 R ∂R∂s ∂p ∂s + R ∂ p ∂s = 0 , so the compatibility condition (3.55) is verified and the problem (3.31) has uniqueness ofsolution. Furthermore, since g and F are polynomial on s (see (3.32)–(3.33)), the solutionof (3.31) must be also polynomial on s and we can compute it explicitly (see appendix Bfor the details).
4. Behavior of the wall of the pipe
We need to close the equations of the model presented here (see theorem 3.4) with a lawdescribing the behavior of the wall of the pipe, that is, with a equation that allows us todetermine R . There are different possibilities: a rigid wall, elastic or viscoelastic laws, etc.The simplest case is when we consider that the wall of the pipe is rigid. We have assumedthat R ( t, s ) is given in (2.1)–(2.2), so it is enough to suppose that ∂R∂t = 0to obtain a rigid wall. If we consider the steady case and a rigid wall of the pipe, our modelreduces to the model obtained in [4].Other simple case is when we consider an algebraic elastic law (see [1]): p − p e = Eh R ( R − R ) (4.1)where E is the Young modulus of the wall, h its thickness, R the radius of the cross-sectionat rest, and p e is the external pressure.More complex (elastic or viscoelastic) laws can also be considered (see [1] again).
5. Some numerical examples
In this section we shall present some numerical examples in some representative casesin order to illustrate the behavior of the approximated solution obtained in the previoussections. 24e start plotting the main tangential velocity u and its corrections u and u . Weobserve in Figure 1 that u is a Poiseuille flow (other works as [2, 8] have also shown thisbehavior). Figure 1: Plot of u field. In Figure 2 we can see that u is a correction of u that takes into account the curvatureof the middle line (the fluid is faster in the side of the cross section of the pipe pointing to N ). Figure 2: Plot of u field. The correction of order two u , has a complex dependence on various terms (see (3.28)),but it is also similar to a Poiseuille flow (see Figure 3).25 igure 3: Plot of u field. We have seen at (3.22) that, at order zero, the transversal velocity is zero, so the tan-gential velocity is dominant. The first order correction, U = ( u , u ), is related with theexpansion and contraction of the pipe wall in radial direction. We can see in Figure 4different cases depending on the value of ∂p ∂s (dp1), ∂ p ∂s (dp2) and ∂r∂s (dr).The second order correction of transversal velocity, U = ( u , u ), is related with therecirculation of the fluid in the cross section of the pipe, as we can see in Figure 5, wherewe show different cases depending on the curvature (k), its derivative (dk) and the torsion(tau) of the middle line of the pipe.
6. Conclusions
A transient model for a newtonian fluid through a curved pipe with moving walls hasbeen obtained. The asymptotic expansions have allowed us to find out the main componentsof velocity and their corrections. Furthermore, our model reduces to the obtained in [4],when steady case and rigid walls are considered. Plots presented here (see figures 1–5)compare very well with real patterns of fluid flow through a curved pipe and agree with thedata available in the literature. A simple algebraic elastic law for the pipe wall has beenconsidered in (4.1), but other more general laws can be used.
7. Acknowledgements
This research was partially supported by Ministerio de Econom´ıa y Competitividad undergrant MTM2012-36452-C02-01 with the participation of FEDER.26 igure 4: Plot of ( u , u ) field. igure 5: Plot of ( u , u ) field. . Computing U and p Let us consider ( U , p ), the solution of problem (3.43), ∆ z U = Rνρ ∇ z p in ω, div z U = R ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − ( z + z ) R ∂ p ∂s (cid:19) =: g in ω, U = ∂R∂t (cos s , sin s ) =: ϕ on ∂ω, that we have seen in the proof of theorem 3.4 that has a unique solution (in the case of p ,up to an arbitrary function of t and s ). In order to compute U and p , we are going toconsider some easier auxiliary problems. First, let us consider the problem (cid:40) ∆ z ϕ = g in ω,∂ϕ∂ n = ϕ · n = ∂R∂t on ∂ω, (A.1)which has a unique solution (up to an arbitrary function of t and s ), since the compatibilitycondition (3.44) is verified. This problem can be written, using change of variable (3.20), asfollows s ∂ ϕ∂s + 1 s ∂ϕ∂s + ∂ ϕ∂s = R ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − s R ∂ p ∂s (cid:19) in ω,∂ϕ∂s = ∂R∂t on ∂ω. (A.2)If we look for a solution of the form ϕ = a ( t, s ) s + b ( t, s ) s + c ( t, s ) , we can identify a and b substituting in (A.2). Taking into account (3.23) to verify that theboundary condition is fulfilled, we find that ϕ ( t, s , s , s ) = s R ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − s R ∂ p ∂s (cid:19) + c ( t, s ) (A.3)is the solution of (A.2). Using again (3.20), we obtain that ϕ ( t, s , z , z ) = ( z + z ) R ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − ( z + z ) R ∂ p ∂s (cid:19) + c ( t, s ) . (A.4)Let be V = U − ∇ z ϕ , and χ = ( − sin s , cos s ) the unit tangent vector on ∂ω . Then,from (3.43) and (A.1), we obtain that ( V , p ) satisfies the problem ∆ z V = Rνρ ∇ z p − ∆ z ( ∇ z ϕ ) in ω, div z V = 0 in ω, V · n = U · n − ∇ z ϕ · n = 0 on ∂ω, V · χ = U · χ − ∇ z ϕ · χ = − ∂ϕ∂ χ on ∂ω. (A.5)29ince on ∂ω we have that U · χ = 0, ∇ z ϕ · χ = ∂ϕ∂ χ = ∂ϕ∂s = 0, and ϕ verifies in ω that ∇ z ϕ = 2 R ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − ( z + z ) R ∂ p ∂s (cid:19) ( z , z ) , ∆ z ( ∇ z ϕ ) = − R ρ ν ∂ p ∂s ( z , z ) , we obtain that ( V , p ) satisfies the problem ∆ z V = Rνρ ∇ z p + R ρ ν ∂ p ∂s ( z , z ) in ω, div z V = 0 in ω, V = on ∂ω, (A.6)Then, by Theorem 3.3, problem (A.6) has a unique solution (up to an arbitrary functionof t and s , in the case of p ), which expression is V = ,p = − R ∂ p ∂s (cid:0) z + z (cid:1) + p ( t, s ) , (A.7)where p ( t, s ) = c ( t, s ) is a smooth function, which is determined in (3.51). Finally, since U = V + ∇ z ϕ , we have that U = R ρ ν (cid:18) ∂∂s (cid:18) R ∂p ∂s (cid:19) − ( z + z ) R ∂ p ∂s (cid:19) ( z , z ) . B. Computing U and p Let us consider ( U , p ), solution of the problem ∆ z U = Rρ ν ∇ z p + F in ω, div U = g in ω, U = on ∂ω, (B.1)where F and g are given, respectively, by (3.33) and (3.54).In order to compute ( U , p ), we shall consider a decomposition of this problem in someeasier ones (as done to compute ( U , p ) in appendix A).First, let us consider the problem, (cid:40) ∆ z ϕ = g in ω,∂ϕ∂ n = 0 on ∂ω, (B.2)30hich has unique solution, since the compatibility condition (3.55) is verified.Simplifying from (3.54), we have g = (cid:18) − κR ρ ν ∂ p ∂s − κ (cid:48) R ρ ν ∂p ∂s (cid:19) s cos s − κτ R ρ ν ∂p ∂s s sin s + (cid:18) κR ρ ν ∂R∂s ∂p ∂s + 9 κR ρ ν ∂ p ∂s + 3 κ (cid:48) R ρ ν ∂p ∂s (cid:19) s cos s + 3 κτ R ρ ν ∂p ∂s s sin s − R ρ ν ∂ p ∂s s + R ρ ν ∂∂s (cid:18) R ∂p ∂s (cid:19) , (B.3)that, using (3.20), can be written as in (3.32). From (B.2), and using again (3.20), we havethat ϕ is solution of s ∂ ϕ∂s + 1 s ∂ϕ∂s + ∂ ϕ∂s = g in ω,∂ϕ∂s = 0 on ∂ω. (B.4)If we look for a solution of the form ϕ = a ( t, s ) s cos s + b ( t, s ) s sin s + c ( t, s ) s cos s + d ( t, s ) s sin s + e ( t, s ) s cos s + f ( t, s ) s sin s + j ( t, s ) s + h ( t, s ) s + i ( t, s ) , (B.5)where a, b, c, d, e, f, g, h, i are smooth unknown functions, and we substitute in (B.4), we findthat ϕ ( t, s , s , s ) = (cid:18) − R ρ ν (cid:18) κ ∂ p ∂s + 3 κ (cid:48) ∂p ∂s (cid:19) cos s − κτ R ρ ν ∂p ∂s sin s (cid:19) s − R ρ ν ∂ p ∂s s + (cid:18) R ρ ν (cid:18) κ ∂R∂s ∂p ∂s + 3 κR ∂ p ∂s + κ (cid:48) R ∂p ∂s (cid:19) cos s + 3 κτ R ρ ν ∂p ∂s sin s (cid:19) s + R ρ ν ∂∂s (cid:18) R ∂p ∂s (cid:19) s + (cid:18)(cid:18) R ρ ν (cid:18) κ ∂ p ∂s + 3 κ (cid:48) ∂p ∂s (cid:19) − R ρ ν (cid:18) κ ∂R∂s ∂p ∂s + κ (cid:48) R ∂p ∂s + 3 κR ∂ p ∂s (cid:19)(cid:19) cos s − κτ R ρ ν ∂p ∂s sin s (cid:19) s + i ( t, s ) , (B.6)where equation (3.27) has been used to guarantee that the boundary condition in (B.4) is31erified. Coming back to the local cartesian coordinates, we can write (B.6) ϕ ( t, s , z , z ) = (cid:18) − R ρ ν (cid:18) κ ∂ p ∂s + 3 κ (cid:48) ∂p ∂s (cid:19) z − κτ R ρ ν ∂p ∂s z − R ρ ν ∂ p ∂s (cid:19) ( z + z ) + (cid:18) R ρ ν (cid:18) κ ∂R∂s ∂p ∂s + 3 κR ∂ p ∂s + κ (cid:48) R ∂p ∂s (cid:19) z + 3 κτ R ρ ν ∂p ∂s z + R ρ ν ∂∂s (cid:18) R ∂p ∂s (cid:19)(cid:19) ( z + z )+ (cid:18) R ρ ν (cid:18) κ ∂ p ∂s + 3 κ (cid:48) ∂p ∂s (cid:19) − R ρ ν (cid:18) κ ∂R∂s ∂p ∂s + κ (cid:48) R ∂p ∂s + 3 κR ∂ p ∂s (cid:19)(cid:19) z − κτ R ρ ν ∂p ∂s z + i ( t, s ) . (B.7)Now, let us consider V = U − ∇ z ϕ , and χ = ( − sin s , cos s ) the unit tangent vectoron ∂ω . Then, from (B.1) and (B.2), we obtain that ( V , p ) satisfies the problem ∆ z V = Rρ ν ∇ z p + F − ∆ z ( ∇ z ϕ ) in ω, div z V = 0 in ω, V · n = U · n − ∇ z ϕ · n = 0 on ∂ω, V · χ = U · χ − ∇ z ϕ · χ = − ∂ϕ∂ χ on ∂ω (B.8)Let us consider an arbitrary smooth function ψ and let be W = V − (cid:16) ∂ψ∂z , − ∂ψ∂z (cid:17) . Then W is solution of the problem ∆ z W = Rρ ν ∇ z p + F − ∆ z ( ∇ z ϕ ) − ∆ z (cid:18) ∂ψ∂z , − ∂ψ∂z (cid:19) in ω, div z W = 0 in ω, W · n = − (cid:18) ∂ψ∂z , − ∂ψ∂z (cid:19) n = −∇ z ψ · χ = − ∂ψ∂ χ on ∂ω, W · χ = − ∂ϕ∂ χ + (cid:18) ∂ψ∂z , − ∂ψ∂z (cid:19) · χ = − ∂ϕ∂ χ + ∂ψ∂ n on ∂ω. (B.9)If we are able to find a function ψ such that ∂ψ∂ χ = 0 on ∂ω, ∂ψ∂ n = ∂ϕ∂ χ on ∂ω, (B.10)then problem (B.9) will be equivalent to ∆ z W = Rρ ν ∇ z p + F − ∆ z ( ∇ z ϕ ) − ∆ z (cid:18) ∂ψ∂z , − ∂ψ∂z (cid:19) in ω, div z W = 0 in ω, W = on ∂ω, (B.11)32nd we shall have, by Theorem 3.3, uniqueness of solution ( W , p ) ( W is unique and p isunique up to a function depending on t and s ).The next step is, following (B.10), finding a function ψ such that ∂ψ∂ χ = ∂ψ∂s = 0 on ∂ω, (B.12) ∂ψ∂ n = ∂ψ∂s = ∂ϕ∂ χ = ∂ϕ∂s on ∂ω. (B.13)From (B.6) and (3.27), we deduce that ∂ψ∂s = (cid:18) − R ρ ν (cid:18) κ ∂ p ∂s + 3 κ (cid:48) ∂p ∂s (cid:19) + 6 R ρ ν (cid:18) κ ∂R∂s ∂p ∂s + κR ∂p ∂s + 3 κR ∂ p ∂s (cid:19)(cid:19) sin s − κτ R ρ ν ∂p ∂s cos s on ∂ω, so, one possible election for ψ is ψ ( t, s , s , s ) = R ρ ν (cid:18)(cid:18) κR ∂ p ∂s + 6 κ (cid:48) R ∂p ∂s + 108 κ ∂R∂s ∂p ∂s (cid:19) sin s − κτ R ∂p ∂s cos s (cid:19) s ( s − , that, applying (3.20), can be written ψ ( t, s , z , z ) = ( ψ ( t, s ) z + ψ ( t, s ) z ) z + z −
12 (B.14)where, ψ ( t, s ) = R ρ ν (cid:18) κR ∂ p ∂s + 6 κ (cid:48) R ∂p ∂s + 108 κ ∂R∂s ∂p ∂s (cid:19) , (B.15) ψ ( t, s ) = − κτ R ρ ν ∂p ∂s . (B.16)Then, we have (cid:18) ∂ψ∂z , − ∂ψ∂z (cid:19) = (cid:18) ψ (cid:0) z + 3 z − (cid:1) + ψ z z , − ψ (cid:0) z + z − (cid:1) − ψ z z (cid:19) , (B.17)and − ∆ z (cid:18) ∂ψ∂z , − ∂ψ∂z (cid:19) = ( − ψ , ψ ) . (B.18)33ence, from (B.11), W is solution of the problem ∆ z W = ∇ z q + F in ω, div z W = 0 in ω, W = on ∂ω, (B.19)where q = Rρ ν p − g − ψ z + 4 ψ z + q , (B.20)with q = q ( t, s ) an arbitrary smooth function of t and s . From Theorem 3.3, we havethe existence and uniqueness (up to an arbitrary function of t and s , in the case of q ) of( W , q ).Once we have computed W and q (see below), we obtain U and p in the followingway, U = W + ∇ z ϕ + (cid:18) ∂ψ∂z , − ∂ψ∂z (cid:19) , (B.21) p = ρ νR (cid:0) q + g + 4 ψ z − ψ z − q (cid:1) , (B.22)where ϕ , ψ , g , ψ and ψ are given, respectively, by (B.7), (B.14), (3.32), (B.15) and (B.16).Let us now compute ( W , q ). In order to make such computation, let us remark that F = ( F , F ) is polynomial in z and z . Developing, from (3.33), F and F in powers of z and z , we obtain that F = f + f z + f z + f z z + f z + f z ,F = f + f z z + f z + f z , where f mnα denotes the coefficient that multiplies z m z n in F α . Since F is polynomial in z and z , W = ( W , W ) and q must be also polynomial in z and z , so let us suppose that W = (cid:0) w + w z + w z + w z + w z z + w z + w z + w z z + w z z + w z + w z + w z z + w z z + w z z + w z (cid:1) ( z + z − , (B.23) W = (cid:0) w + w z + w z + w z + w z z + w z + w z + w z z + w z z + w z + w z + w z z + w z z + w z z + w z (cid:1) ( z + z − , (B.24) q = q + q z + q z + q z + q z z + q z + q z + q z z + q z z + q z + q z + q z z + q z z + q z z + q z + q z + q z z + q z z + q z z + q z z + q z . (B.25)By substitution in (B.19), we obtain a linear system with a unique solution, so W and q are determined from the coefficients of the field F and our assumption about the form of W and q was correct. 34n this way we find that, w = 1192 (cid:0) f − f (cid:1) − f − f , w = 7240 f − f ,w = 596 f + 13960 f + 315760 f − f , w = − f ,w = 196 f + 7960 f − f − f , w = 1480 f + 372880 f ,w = 1360 f − f , w = − f ,w = 1480 f − f − f + 148 f , w = 196 f ,w = − f − f , w = 596 f ,w = 180 f − f , (B.26)while (cid:40) w = w = w = w = w = w = w = w = 0 ,w = w = w = w = w = w = w = w = w = 0 , (B.27)and q = 112 f − f − f − f − f , q = 112 f − f ,q = 340 f − f − f − f , q = − f − f ,q = − f − f , q = − f ,q = 112 f + 120 f − f − f − f , q = 18 f − f ,q = 11480 f − f − f , (B.28)while q = q = q = q = q = q = q = q = q = q = q = 0 , (B.29)and q is an arbitrary smooth function depending only on t and on s . References [1] L. Formaggia, D. Lamponi, and A. Quarteroni. One-dimensional models for blood flow in arteries.
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