Abstract: The paper deals with a stationary non-Newtonian flow of a viscous fluid in unbounded domains with cylindrical outlets to infinity. The viscosity is assumed to be smoothly dependent on the gradient of the velocity. Applying the generalized Banach fixed point theorem, we prove the existence, uniqueness and high order regularity of solutions stabilizing in the outlets to the prescribed quasi-Poiseuille flows. Varying the limit quasi-Poiseuille flows, we prove the stability of the solution.
Keywords: non-Newtonian flow, strain rate dependent viscosity, quasi-Poiseuille flows, domains with outlets to infinity.
Articles
Steady state non-Newtonian flow with strain rate dependent viscosity in domains with cylindrical outlets to infinity*
Vilniaus Universitetas
Esta obra está bajo una Licencia Creative Commons Atribución 4.0 Internacional.
Recepción: 15 Diciembre 2020
Aprobación: 13 Julio 2021
Fuente: This project has received funding from European Social Fund
Nº de contrato: 09.3.3-LMT-K-712-01-0012
Beneficiario: Steady state non-Newtonian flow with strain rate dependent viscosity in domains with cylindrical outlets to infinity*
Asymptotic behaviour of solutions of elliptic and parabolic equations in domains with noncompact boundaries was considered in [12], where the first theorems on stabilization of solutions were proved. They were called Phrägmen–Lindelöf theorems. The stationary elasticity equations in unbounded domains are studied in [15], and the stabilization the- orems were associated there with the Saint-Venant principle. For the stationary and non- stationary Stokes and Navier–Stokes equations with no-slip condition at the boundary of the outlets, these questions were studied in [1, 9–11, 21–24, 28], and for the viscoelastic flows, in [25]. For the non-Newtonian flows with viscosity depending on the gradient of the velocity, the existence, uniqueness and asymptotic behaviour in the outlets were studied in [13, 20]. Note that this non-Newtonian rheology governs the blood circulation in vessels (see [2, pp. 84–89, 196–200]).
A part of theoretical interest for partial differential equations, this set of questions is important for construction of asymptotic expansions of solutions in thin domains. Namely, matching of the asymptotic solutions via the boundary layer method leads ex- actly to the scaled partial differential equations in unbounded domains with cylindrical outlets (see, e.g., [16–19] for Newtonian flows and [14] for the power law fluids). In particular, results of the present paper are used for the construction of an asymptotic ex- pansion of a non-Newtonian flow in a network of thin cylinders, modeling blood vessels. In the present paper the results obtained in [20] will be extended and generalized.
First, we reconstruct the pressure, while in [20], only the weak formulation of the problem without pressure was studied. Second, in order to reconstruct the pressure, we need to have more regularity for the solution, so we will prove the third-order regularity of the velocity and second-order regularity of the pressure in weighted spaces with exponential decay at infinity. Of course, we need more regularity (..) for the viscosity ., depending on the shear rate .. However, we will rid of a restrictive condition of boundedness of (.(.).), which was assumed in [20]. Finally, we will focus on the questions of stability of solutions with respect to the quasi-Poiseuille flows to which they stabilize in the outlets. These new theorems are important for the construction of boundary layers of non-Newtonian flows.
The paper has the following structure. In Section 2, we give the definition of the domain with outlets. In Section 3, we cite and prove some auxiliary results: embed- ding inequalities in domains with cylindrical outlets and a lemma on the stabilization to a constant for functions with exponentially decaying gradient. In the same Section 3, we recall some results for the stationary Stokes equation and prove the weak Banach contraction principle. This theorem generalizes the classical Banach fixed point theorem. This result is well known in the mathematical community and is widely used. However, we could not find the proof in literature. Therefore, for the reader’s convenience, we present a proof. This generalization is used in the proofs of the regularity of the solutions. The main problem for the stationary non-Newtonian flow in unbounded domains with outlets is formulated in Section 4. In Section 5 the quasi-Poiseuille flow for the stationary non-Newtonian equations in an infinite tube is studied. A Poiseuille flow is an exact solution to the equations of the fluid motion (Stokes, Navier–Stokes) in an infinite cylinder with the no-slip condition at the boundary, with a linear pressure with respect to the longitudinal variable, and with the velocity vector having only longitudinal component (called normal velocity) different from zero; this normal velocity depends only on the transversal variables. A quasi-Poiseille (or Hägen–Poiseuille) flow is an exact solution having the same structure and corresponding to some non-Newtonian rheology. Such flows for various rheologies were studied in [2, 6, 7, 26, 27]. Contrary to [20], where also the quasi-Poiseuille flow was studied, we focus on the regularity issues. Finally, Section 6 contains the main results of the paper: existence and uniqueness of a regular classical solution (velocity and pressure) and continuity of the solution with respect to the data of problem (stability). The proof of continuity of the solution in the norm W2,2for the velocity and L2 norm for the gradient of the pressure needs the regularity “plus one” of the solution. It explains the difference of norms in Theorems 5 and 6
Consider the domain
, with J cylindrical outlets to infinity: 
where 
is a bounded domain,
for
and the outlets to infinity 
in some coordinate systems 
, having the origins within the boundary of domain 
, are given by the relations
where 
are some bounded domains in
, cross-sections of the cylinders (see Fig. 1). Assume that for any 
there exists a 
such that the cylinder 
Denote dσ the maximal diameter of the cross-sections 
. We assume that the boundary 
regular and that 
has a positive measure. Evidently, there exists a positive real number R > dσ such that the ball 
contains
We introduce the following notation:
where
is an integer
Figure 1
Domain Ω.
Let 
, be domain with . outlets to infinity. We define in
weighted function spaces. Denote
define a smooth function 
and set 
Denote by 
, the space of functions obtained as the closure of 
and set 
Notice that for 
, the elements of the space 
exponentially vanish as 
Lemma 1 [Poincaré’s inequality]. There exists a constant 
independent of 
such that for any function 
the following inequality holds:
Proof. For the proof, see, e.g., [20].
Lemma 2.
(i) For any function 
the inequality holds:
(1)(ii) For any function 
, the inequality holds:
(2)
Proof. (i) Let us represent the domain 
as a union of bounded domains:
In every 
we have the inequalities
(3)with the constant c independent of k. Multiplying inequalities (3) by 
and having in mind that 
in 
we obtain
Here the constants depend on 
only. Summing these inequalities over k and j and adding the inequality 
we obtain (1).
(ii) By same token, using the inequalities
we get (2).
Lemma 3. Let us define the half-cylinder 
, where σ is a bounded domain in 
with Lipschitz boundary. Suppose that p 
and
Then there exists a constant p. such that the following estimate holds:
Proof. First, we prove that the mean value 
is a bounded function. Let z > 0. Since 
and since
we have
It is easy to prove that there exists a constant p0 such that 
. Indeed,since 
is bounded, there is a sequence {zk} such that
for some constant 
Consider
Passing to the limit as 
we get
Now, by Poincaré’s inequality,
Let
Integrating the last inequality from ξ to r yields
Multiplying both sides of the last inequality by
and integrating with r from z to 2z yields
From this it follows that
Here we used that 
and so, ξ → +∞ implies 
By the triangle inequality we get 
. In order to finish the proof of the lemma, we need an auxiliary inequality
(4)(for the proof see [18, Cor. 7.1]), which holds for any function f such that
Applying (4) to 
, we obtain
Remark 1. From the last lemma it follows that if 
, then there exist constants 
= 1, 2, . . . , J, such tha
Consider now the Stokes problem in the domain 
with J outlets to infinity:
(5)Let H(
) be the space of divergence-free functions of
It is well known (see [3, 4, 8]) that there exists a unique weak solution v ∈ H(
) to (5), which satisfies the integral identity 
for all η ∈ H(Ω) and the estimate 
The following Agmon–Duglis–Nirenberg (ADN)-type theorem is proved in [24] (see Theorem III.3.2).
Theorem 1. Let l be an integer, l > 0. Let ∂Ω ∈ C+2. There exists a positive β∗ such that for all
, the weak solution v belongs to the space 
, and there exists a pressure function p with 
) such that the pair (v(x), p(x)) satisfies equations (5) almost everywhere in Ω. The following estimate holds:
Moreover, the local estimate
(6)holds with the constant c independent of K.
Theorem 2. Let X and Y be reflexive Banach spaces,
for all x ∈ X. Suppose that M ⊂ X is a closed, bounded set, 
, and the mapping 
satisfies the inequalit
(7)Then T admits exactly one fixed point 
Proof. Let us define a sequence
by the recurrent formulas
(8)Since T maps the bounded set M to itself, there exists a positive constant c0 such that
and 
. Since the space X is reflexive, there exists a subse- quence 
such that
(9)For simplicity, we will not distinguish in notation the subsequence 
and the se- quence
. From (7) it follows that
Therefore,
is strongly convergent in Y and 
. From (8) we obtai
(10)Thus,
as n → +∞, and hence,
(11)Relations (10) and (11) yield 
. The uniqueness of the fixed point is obvious.
Let 
be positive constants. Let 
be a bounded C3 -smooth function 
such that for all 
(12)where A is a positive constant independent of y.
Consider the steady state boundary value problem for the non-Newtonian fluid motion equations in the domain
(13)where 
is the strain rate matrix with the elements
, 
and
, 
.
We look for the solution v having prescribed fluxes Fj over the cross sections 
of outlets to infinity
(14)where
(15)Here and below an integral over 
. is understood as an integral over any orthogonal cross- section of 
Note that this integral for a divergence-free vector function is independent of the position of this cross-section.
Since div 
, equations (13) can be written in the form
The non-Newtonian Poiseuille flow with the strain rate dependent viscosity was studied in the book [2] and recently in [20]. We will need below some extended versions of theorems proved there.
Let us recall the definition of a quasi-Poiseuille flow for equations (13). Let σ be a bounded domain with Lipschitz boundary in 
. Consider in the infinite cylinder Π = R × σ the Dirichlet boundary value problem
where
(below we will see that for the quasi-Poiseuille flow, dii = 0).
Define a quasi-Poiseuille flow as a couple
such that 
, 0, . . . , 0)T, and 
, where
is the solution of the following problem:
(16)Here 
, and α is the given pressure slope
Theorem 3. Let 
. For any
, there exists
such that for all 
and any 
, problem (16) admits a unique1 solution 
The solution vPα satisfies the estimate
(17)where the constant c depends only on σ.
Proof. Let 
be an operator 
such that for any 
is a solution of the Poisson problem
(18)Where
Using the embedding 
and conditions (12), we obtain
Analogously,
and using, in addition, the embedding. 
, we derive
(20)Define in 
a closed bounded set
. Assume that
Then (19) and (20) yield the estima
(21)Then for the solution of the Poisson equation (18), the following estimate holds:
(22)Set 
and suppose that
Then from (22) it follows that 
. The last inequality implies that the operator
maps the closed bounded set 
onto itself.
Let us show that 
is a contraction in 
. Multiplying equations (18) by an arbitrary η ∈ 
and integrating by parts, we get
Thus, for any 
, the following equality holds:
(23)Using Young’s inequality, we obtain
Using (12),
we have
Therefore, taking in (23)
we derive the inequality
and it follows that
Let
Then for any
operato 
is a contraction in 
with the contraction factor
and by Theorem 2 there exists a unique fixed point 
of the operator 
, which is a solution of problem (16).
From estimates (20), (21) applied to the fixed point vPα it follows that
and thus ,by (22)
If 
, the last estimate implies (17)
Define
the flux corresponding to the pressure slope 
. Note that in the case of the steady Newtonian flow (the steady form of Navier–Stokes or Stokes equations), F(α) is proportional to α. This case corresponds to the value 
, and so, 
, where 
and 
is a solution of the Poisson equation
We consider as well the operator (function) corrector of the non-Newtonian flux with respect to the Newtonian one: 
, and prove that for sufficiently small 
is a contraction.
The next lemma is an extension of Lemma 2.3 and Corollary 2.4 [20].
Lemma 4. For any 
, there exists a number
such that for any 
and every 
, the solution 
of problem (16) is a Lipschitz-continuous function with respect to α in the norm
. Moreover F(α) is a Lipschitzcontinuous function with respect to α.
Proof. Let
be two solutions of problem (16) corresponding to α = α1 and α = α2, respectively. By Theorem 3 these solutions exist if 
Moreover, the following estimates hold:
Using the integral identities
(24)subtracting (24) with 
from (24) with 
, taking 
and using arguments similar to those at the end of the proof of Theorem 3, we ge
where 
, then from the last inequality it follows that
(25)Further,
Importar imagen and this estimate completes the proof.
Lemma 5. For any 
, there exists a number 
such that for all
the operator 
is a contraction on the interval
Proof. Denote
where 
. Then 
satisfy the following problems for
Subtracting one problem from another, we get for 
the following relations:
Applying a standard a priori estimate for the solution of the Poisson equation with Dirich- let conditions, we obtain
and by using similar arguments as before we obtain from inequalities (17), (25) for 
So, finally,
Since
we have
So, if 
is a contraction with the factor q = c8κ −1λ <
Remark 2. The same proof shows that if the constant κ−1 is replaced by another constant
then for any α0 > 0, there exists a number 
such that for all
the operator 
) is a contraction on the interval 
Lemma 6. sgn(F(α)) = sgn(α).
Proof. Inde
Lemma 7. For any F0 > 0, there exists 
such that for all 
and every
, there is a unique pair (
, α) satisfying (16) and such that 
. Moreover, the following estimate hold
(26)Proof. F is a Lipschitz continuous function. So, for any fixed F0, we can find a number α0 = α0(F0) such that for all λ ∈ (0, min{λ1(α0), λ2(α0)}), Lemma 4 holds, and for α ∈ [−α0, α0], we have 
Since by Lemma 4, F(α) is Lipschitz continuous and, by Lemma 5, κ−1G(α) is a contraction, we conclude that there exist constants 0 < a1 < a2 suc
Therefore, for every
there exists at least one 
such that F(a) = F
In order to prove the uniqueness of α, we argue by contradiction: suppose that there are two such number α1 and α2, i.e., F(α1) = F(α2) = F. Then, by definition, |κ−1G(α1) − κ−1G(α2)| = |α1 − α2|. But this contradicts the fact that κ−1G(α) is a contraction, and so α1 = a2.
Theorem 4
(i)For any α0, there exists 
such that for all 
and every 
, there holds the estimate
(27)(ii) For any F0, there exists 
such that for all 
] and every 
, there holds the estimate
(28)where
Proof. (i) Let 
such that 
with c from (26), and let 
be the number defined in Theorem 
be the number defined in Lemma 7. Then due to these theorem and lemma, for
and every 
, there exist solutions (vPα1 , α1) and (vPα2 , α2) of problem (16) such that 
, and the following estimates
hold. Moreover,
The difference 
satisfies the equations
(29)Where
It is easy to calculate that
By using Sobolev embedding theorems we get the inequality
Therefore, the classical estimate for the Poisson equation (29) yields
(30)If 
(30)
Thus, inequality (27) is proved.
(ii) Let 
be the number defined in Lemma 7. By the definition of the function 
we have
Thus,
Since for sufficiently small 
, the operator 
is a contraction (see Lemma 5), the last estimate yields
with 
and thus,
From (31) and (32) follows (28).
Consider the domain
with J cylindrical outlets to infinity. We assume that the boundary 
regular. Consider in 
problem (13)–(15). Denote 
. Let F0 be a nonnegative number. By Lemma 7 there exists a number λ00 depending on F0 such that for every
and for any set of fluxes (F1, . . . , FJ ) such that 
, there exist J pressure slopes αj and corresponding J quasi-Poiseuille flows VPαj (x) = (vPαj (x´), 0, . . . , 0)T ∈ W3,2 (σj ), defined in cylinders Πj = {x (j) ∈ R n, x(j)0 ∈ σj , x(j) 1 ∈ R}, j = 1, . . . , J, such that F(αj ) = Fj
We define cut-off functions
associated to each outlet Ωj as C (3)-smooth functions vanishing everywhere in Ω except for the outlet Ωj , where they depend on the local longitudinal variable 
only, are equal to zero if 
< 1, and equal to one if 
> 2, and put
It is easy to see that for 
Moreover, from the condition 
it follows that
. Finally, estimates (17) and (26) yield
Since h ∈ 
by results in [5], there exits a vector field
such that div
and
(33)Moreover, since supp
, W can be constructed such that supp
. Extend the functions W and Vχ by zero into the whole Ω and set
(34)Then
and for 
, the vector-field 
coincides with the velocity part 
of the corresponding Poiseuille flow. Note that the vector field W has zero flux.
By denoting in (13)
(35)where
, we obtain the following problem:
(36)
Theorem 5. Assume that
Then for any 
there exist numbers 
such that for all 
and for any 
satisfying
and any set (F1, . . . , FJ ) with 
, problem (13), (14), (15) possesses a unique solution (v, p)2 admitting representation (35) with 
The following estimate holds:
(37)Moreover, there exist constants q1, q2, . . . , qJ such tha
(38)Proof. Define K as the operator 
such that for any 
is a solution of the problem
(39)Where
After subtracting and adding the expression
, we write . in the form
where 
The function g has compact support, supp 
. and
(40)Note that
where 
˙ is the Jacobian matrix of
and
Since
by using (12) we obtain the estimate
Using the embedding inequalities (1), (2) and estimates (26), (33), we obtain
Let us estimate the integrals containing the terms 
and 
. Inequalities (1), (2) and (3) yield
Similar considerations give us also the estimates
Collecting the above inequalities and adding (40), we derive
(41)Similarly,
The 
norm of this expression is evaluated according to the following scheme: in each product of gradients, the first-order terms
are evaluated by 
the second-order terms
are evaluated in the 
-norm, finally, the third-order terms 
are evaluated in the 
norm. Then we apply the embedding inequalities of Lemma 2. So, for the gradient of H, we obtain the estimat
(42)From (41) and (42) it follows that the right-hand side
of system (39) satisfies the estimate
Then, by Theorem 1, for sufficiently small
, the solution (KU, q) of the Stokes problem (39) is subject to
(43)Assume that
Then from (43) it follows
and if
satisfies
we obtain the estimate
Thus, by (44), the operator K maps the ball 
into itsel
Let us show that K is a contraction in the space H(
). Multiplying equations (39) by arbitrary
and integrating by parts, we get
(45)From (45) it follows that for any
the following equality holds:
(46)Applying Young’s inequality, we have
Since, by (12),
we get
taking in (46)
we derive the inequality
Therefore,
Let
Then for any 
, the operator K is a contraction with the contraction factor
and, by Theorem 2, there exists a unique fixed point U of the operator K , which is a solu- tion (together with the corresponding pressure function q) of problem (36). Estimate (37) for the fixed point . and the pressure . follows from the fact that
(see inequality (44)).
The existence of the constants q1, . . . , qJ and estimate (38) follows from Lemma 3 and Remark 1.
Assume that we have two sets of fluxes 
satisfying condition (14) and two functions 
flux carriers corresponding to fluxes 
, respectively (see formula (34)). Denote by 
the solutions of problem (36) corresponding to flux carriers 
and right-hand sides 
Assume that
(47)Denote
Theorem 6. There exists 
such that for all 
and sufficiently small Q, for arbitrary f(i) and 
satisfying (47), the following estimate holds:
(48)Moreover, if 
are normalized by the condition 
, then the limit constants 
J at infinity of q(1)(x) and the corresponding constants 
satisfy the estimate
Proof. Due to condition (47) and inequality (37), 
where BMand bM1 are balls of the radius M and M1, respectively, M, M1, M2 are positive numbers defined by F0, f0 of condition (47). The difference
satisfies the equation
(49)Where
From (28), (33) and (40) it follows that supp
(50)Since 
, there exists an intege
(51)Obviously, for every sufficiently large K,
with the constant c7 defined by F0 and f0. In particular, condition (51) holds if 
(52)We assume without loss of generality that 
.
Let us estimate the norm 
. Consider the function
where
and
(53)Let us construct 
.
Notice tha
and
Therefore, for any 
we ha
and there exist functions 
satisfying the equation
and the estimate
with the constant c independent of k and j. Extend 
by zero to Ω. Putting Φ(x) = 
we obtain the function belonging to 
, which satisfies equation (53) and obeys the estimate
(54)with the constant c independent of kQ.
Since, by construction, 
is solenoidal and supp
, multiplying (49) by 
and integrating by parts, we obtain
Where
Let us estimate the right-hand side of (55). First of all, we notice that
where, in order to estimate the last term, we applied inequality (54). Moreover, by (50),(51) and (54),
(56)
(57)
(58)
(59)Substituting estimates (56)–(59) into (55), for sufficiently small ε and sufficiently large kQ, we obtain
(60)Consider the term K6 containing 
. We have
Therefore,
(61)Arguing as in the proof of Theorems 3 and 5 and using (17), (1), (47),
Estimating analogously the other terms in (61) and using (52), we derive
(62)Thus, similarly to (59), we get
(63)Similarly, we have
Arguing as in the proof of Theorem 5, we obtain
Estimating the integral containing 
, we have used the same argument as in the proof of (62) (see Lemma 2).
The other terms 
can be estimated similarly (for M5, we estimate |∇u (2)| in L∞-norm via W3,2 (Ω)-norm of u(2) and
norm of 
, and we derive the following estimate for the norm of M(u (1) , u (2))
(64)This gives the estimate for K5:
(65)Substituting (63), (65) into (60) and choosing ε sufficiently small, we obtain the estima
(66)Now we apply the local ADN estimate (6) to the solution (u,q) of problem (49),
(67)Applying estimates (50), (66), (64) and (62) to the right-hand side of (67), we derive
(68)Thus, if cλ < 1/2, from (68) follows estimate (48). Let us estimate the differences 
Let the pressure 
be normalized by the condition 
Then q satisfies the inequality
Denote 
. Since 
(see Lemma 3), we can assume, without loss of generality, that kQ is chosen such that 
. Then
So,
and thus
This project has received funding from European Social Fund (project No. 09.3.3-LMT-K-712-01-0012)under grant agreement with the Research Council of Lithuania (LMTLT)
Figure 1
Domain Ω.
