Abstract:
The nonstationary Navier–Stokes equations are studied in the infinite cylinder II=.
under the additional condition of the prescribed time-periodic flow rate (flux) F(t). It is assumed that the flow rate F belongs to the space 
, only. The time-periodic Poiseuille solution has the form 
, where
is a solution of an inverse problem for the time-periodic heat equation with a specific over-determination condition. The existence and uniqueness of a solution to this problem is proved.
Keywords: Navier–Stokes equations, cylindrical domain, time-periodic Poiseuille-type solution, inverse problem, minimal regularity.
Articles
Time-periodic Poiseuille-type solution with minimally regular flow rate.
Vilniaus Universitetas
Esta obra está bajo una Licencia Creative Commons Atribución 4.0 Internacional.
Recepción: 23 Diciembre 2020
Revisado: 06 Julio 2021
Publicación: 01 Septiembre 2021
Mathematical modelling is very useful in many practical applications. For example, in medicine, it can be helpful by choosing the optimal strategy of medical treatment. In such modelling, very important issues are multiscale mathematical models of the blood circulation in a network of vessels. The full 3D computations are nowadays very time consuming and may be applied only for small parts of the blood circulation system. Therefore, a new trend is related to the creation of hybrid dimension models that combine the 1D reduction in the regular zones (mostly in straight vessels) with 3D zooms in small zones of singular behaviour. This method of asymptotic partial decomposition of a do- main was proposed by Panasenko (see [12]) and developed in [13–16]. It mathematically justifies a size of zoomed areas and prescribes asymptotically exact junction conditions. These hybrid-dimension models require significantly smaller computational resources.
The 1D Poiseuille-type flows in straight vessels play a very important role in hybrid- dimension models.
The steady-state Poiseuille flow in an infinite straight pipe 
of constant cross-section σ was invented by Jean Louis Poiseuille in 1841 (see [2,11,24]). The Poiseuille flow is described by the fact that the associated velocity field has only one nonzero component 
directed along the xn-axis of Π, which depends only on 
, and the pressure function 
is linear. The Poiseuille-type solutions can be also defined in the nonstationary case (see [7, 17–21, 23, 26]). Moreover, in [14] the behaviour of the nonstationary Poiseuille flow was studied in a thin cylinder (with the cross-section of radius ε), and the asymptotics of it as
was found.
In the time-periodic case, such flow is usually called Womersley’s flow (see [28]). The time-periodic Poiseuille-type solutions were studied in [3] and [8]. The time periodic solutions for the full Navier–Stokes problem were considered in many papers (see, e.g., [4–6]). Notice that the time-periodic case is very important because of applications to hemodynamics.
In mentioned above papers the Poiseuille-type solutions were studied in the case when data is sufficiently regular. However, in real applications, one usually does not have data defined by smooth functions, and it is important to study the case of minimal regularity of data. The nonstationary Poiseuille-type solution with a prescribed initial condition and given flow rate F(t) belonging to 
was studied in [22], where a new class of weak solutions was introduced, and the unique existence of the solution in such class was proved. The goal of the present paper is to extend the result obtained in [22] to the case of time-periodic Poiseuille-type solutions.
Let us consider the time-periodic Navier–Stokes problem describing the motion of a viscous incompressible fluid in the infinite cylinder II:
where u is the fluid velocity, p is the pressure function, and 
is the constant kinematic viscosity of the fluid.
We look for the solution u of (1) in Π satisfying the additional condition of prescribed flow rate (flux) F(t):
where 
The solution 
of problem (1) has the following expression:
with an arbitrary function p0(t). Putting (2) into (1), we obtain the following problem on the cross-section σ
where 
and q(t) are unknown functions, 
is the Laplace operator with respect to 
.
The Poiseuille flow can be uniquely determined either prescribing the pressure drop q(t) or the flow rate F(t). In the first case the problem is reduced to the standard time- periodic problem for the heat equation for unknown velocity 
with time- periodic forcing q(t). Problems of such type are well studied (see, e.g., [9]). However, in the real word applications the pressure is unknown, and only the flow rate (flux) of the fluid is given. Therefore, it is necessary to prescribe the additional condition
In this case the solution of problem (3), (4) is a pair of functions 
, and one has to solve for 
and q(t) more complicated inverse parabolic problem: for given F(t), to find a pair of functions 
solving problem (3) with 
satisfying the flux condition (4).
Thus, in the second case the relation between q(t) and F(t) depends on the solution of the inverse problem (in the stationary case the flux F and the pressure gradient q are proportional, and the problem remains very simple). The solvability of the time-periodic problem with the assumption that the flux F(t) is from the Sobolev space 
was proved by Beirão da Veiga [3], and in [8] an elementary relationship between the pressure drop q(t) and the flux F(t) was found. However, in applications and numerical computations, the data usually is not regular. Therefore, in this paper, we study problem (3), (4) assuming only that 
.
Problem (3), (4) can be reduced to the case when all involved functions have zero mean values. Let us denote by 
the mean value of a function H. Let 
be a solution of the following problem on σ (the stationary Poiseuille solution corresponding to the flux 
):
The solution 
of (5) can be represented in the form 
, where
and
Let us represent the solution 
in the form
Then, obviously, 
, and 
is the solution of the problem
where 
. So, without loss of generality, we assume 
, that is 
.
Below, we deal with a weak solution of problem (3), (4). The reasoning about the reduction to the case of functions with zero mean values remains valid for weak solutions as well, and we will study only this case.
The rest of the paper is organized in the following way. In Section 2, function spaces are defined and the main result is formulated. In Section 3 the Galerkin approximations of the solution are constructed, and in Section 4 a priori estimates for these approximations are proved. In Section 5 the main result of the paper, that is the existence and the uniqueness of the solution, is proved.
Below, we will use the following notation. If G is the domain in 
means, as usual, the set of all infinitely differentiable functions in G, and 
is the subset of functions from 
with compact supports in G. We use the usual (see [1,9]) notation for Lebesgue and Sobolev spaces: 
, and 
. The norm of an element u in the function space V is denoted by 
is the Bochner space of functions u such that 
for almost all 
, and the norm
is finite
Let us consider the set of smooth periodic functions 
defined on the interval 
. Let 
be a Lebesgue space on the interval 
. We extend the functions from 
to the whole line R by putting 
for any t. To emphasize that functions are periodically extended to R, we used notation 
. Let 
. Clearle 
is a closure of 
in
- norm, and it is a proper subspace of 
. Let 
be the closure of the set 
in 
-norm. Since function f from 
coincides with a continuous function on a set whose complement is of measure zero, we may assume that 
. Let 
be dual of 
.
For any function 
, denote by 
its primitive:
Clearly, 
.
If 
, then 
. Moreover,
and 
is a period function:
Thus, 
. Note that functions 
defined by (9) with various t0 differ from each other by a constant.
Note that the L2-limit of a sequence 
) is not necessary a primitive of some function from 
. We will prove that an element 
possesses a primitive in the distributional sense.
Lemma 1.Any functional 
can be represented in the form
with the uniquely determined 
.
Proof. Obviously, the functional given by formula (10) obeys the estimate
and hence, 
.
Let us take an arbitrary functional 
and show that it can be represented in the form (10). Consider the operator 
due to periodicity, 
. Since for any 
, the equality 
holds with
we have 
, and the operator ∂ is an isomorphism from 
to 
, where the bounded operator 
is given by (11).
For 
, define the functional 
. Clearly
Hence, there exists a uniquely defined 
such that
Thus
and 
is represented in the form (10).
Remark 1. Note that if the functional h can be represented in the form 
=
with 
and arbitrary 
, then H(t)= 
. Therefore, also in the case of a general functional h, for the distributional primitive H, we use the notation 
.
Definition of a weak solution
Let 
. By a weak solution of problem (8) we understand a pair 
such that 
satisfies the flux condition
and the pair 
satisfies the integral identity
for any test function 
such that 
.
For a regular solution 
, taking into account that 
, identity (13) can be easily obtained multiplying equation (8)1 by η, integrating over σ and over the interval 
and integrating by parts with respect to 
and t. On the other hand, by uniqueness of such weak solution
(see Theorem 1 below) it follows that for 
, the solution 
coincides with the regular one, that is 
. Thus, the proposed definition is an extension of the concept of weak solutions.
Theorem 1.Let 
. Then problem (8) admits a unique weak solution (V, s). Then there holds the estimate
where the constant c depends only on σ.
Remark 2. Since 
and all primitives of the function 
differ from each other by a function independent of t, the integral identity (13) remains valid for any primitive function SV, and we can assume, for example, that SV is taken to be zero at the point 
Theorem 1 will be proved applying some version of Galerkin approximations (see Sections 3 and 4). Notice that in order to get estimates of approximate solutions 
, we have used primitive functions defined over the integrals 
with specially chosen points 
and 
. Thus, estimate (14) is valid only for the primitive function SV obtained as a limit of the sequence 
Let 
and 
be eigenfunctions and eigenvalues of the Laplace operator:
Note that 
and 
. The eigenfunctions 
are orthogonal in 
, and we assume that 
are normalized in 
. Then
Moreover, 
is a basis in 
and 
.
We look for an approximate solution of problem (8) in the form
The coefficients 
and the function 
are obtained by solving the following linear problems
which, in virtue of the orthonormality of functions 
, are equivalent to ordinary differential equations for the functions 
:
where 
. Note that 
.
The solution of problem (17) has the form
where
It is easy to see that 
.
Substituting expression (18) into (15), we obtain
Now the flux condition yields
Thus, for the function 
, we derived Fredholm integral equation of the first kind:
It is well known (see, e.g., [10, 25]) that such equations, in general, are ill-posed in L2 setting. In order to regularize equation (19), we consider the following Fredholm integral equation of the second kind:
where α later will tend to 0, i.e., instead of problem (16), (19), we study the regularized problem
where
Lemma 2.Let 
. Then equation (20) admits a unique solution 
.
Proof. First, we show that equation (20)is well defined in the space 
.Obviously, if F is periodic and 
is the solution of (20), then 
also is a periodic function. Assume that mean value 
. Then
Since
the second term in (22) is equal to 
and from (22) it follows that
and thus, 
.
From this it follows that the mean value 
also vanishes: 
.
It is well known that Fredholm integral equations of the second kind satisfy the Fredholm alternative (e.g., [27]). So, it is enough to prove the uniqueness of the solution to (20). Let 
. By construction
and the homogeneous equation (20) gives
Multiplying (21) by 
, summing by k from 1 to N and integrating over the interval 
yield
Integrating by parts with respect to t and taking into account the time-periodicity of 
, we obtain
Thus, 
for a.a. 
, and the lemma is proved.
Let the pair 
be the solution of problem (21) and 
be the solution of problem (6). Consider the integral 
. Since the mean value 
, we have
Therefore, there exists 
such that 
. By periodicity we also have 
Let 
define the notation 
, define the notation
Since the mean value of F nabishes, we hace 
. Moreover, 
Lemma 3. The following estimate
holds with a constant c independent of α and N.
Proof. Define
. Multiplying (21) by 
summing the obtained relation from k = 1 to k = N and integrating them over the interval 
, we obtain
Using the relation
and (21), we derive
Therefore, (24) can be written as
Then
Calculating similarly two integrals J1 and J2 on the left-hand side of (25), we derive that
Hence, relation (25) takes the form
Now multiply (21) by 
, sum from k = 1 till k = N and integrate over the interval 
:
Evaluating each of the three integrals in (27) and having in mind that mean values of participating functions are zero, we obtain
From (26) and (28) it follows that
and
Let us estimate the integral 
. Let 
be a solution of problem (6). Remind that the flux of U0 is nonzero, 
(see (7)). Since 
is a basis in 
, U0 can be expressed as a Fourier series in 
:
Let us multiply relations (21) by ak and sum them over k. This gives
On the other hand, multiplying (6) by 
and integrating by parts in σ, we obtain
Substituting (32) into (31) yields
i.e.,
Integrating (33) with respect to t from τ to 
, we obtain
Here we have used the choice of the point t∗, that is
and hence,
From (34) it follows that
Substituting (35) into (29) yields
and choosing ε sufficiently small, we obtain
Estimates (36) and (35) give
Finally, from (30) and (37) it follows that
The constants in (36)–(38) are independent of α and N.
The constructed approximate solutions 
satisfy equalities (21). Multiplying these relations by arbitrary functions 
such that 
, summing over k from k = 1 to 
, integrating with respect to t and then integrating by parts, we obtain the integral identity
for test functions η having the form 
.
Recall that 
obey a priori estimate (23) with a constant c independent of α and N.
Since all functions in (23) are 2π-periodic, inequality (23) is equivalent to
Let us fix N and choose a subsequences 
and 
such that l
converges weakly in 
to some 
converges weakly in 
to SV(N)5, while 
converges weakly in 
to s(N). The last convergence means that
where , and 
, and 
is a primitive of s(N) in the distributional sense.
Obviously, for the limit functions V(N) and Ss(N), estimate (40) remains valid with a constant c independent of N. In (39), taking 
and passing to the limit as 
, we get
Let us show that 
satisfy the flux condition
Integrating equation (21) for 
from t to 2π yields
Obviously, the sequence 
is bounded in 
. So we may assume, without loss of generality, that 
is weakly convergent to 
in 
. Then the sequence of primitives 
for all 
, and hence, 
. From (43) we have
Therefore,
and differentiating this equality with respect to t, we get (42).
Now we choose a subsequence 
such that 
converges weakly in 
to some 
converges weakly in 
; to SV and 
converges weakly in 
to s. In (41) ,passing to the limit over the subsequence as 
yields
Exactly as above, we can prove that 
satisfies the flux condition (12). However, the integral identity (44) is proved, up to now, only for test functions η, which can be represented as the sums: 
with 
such that 
. After we have passed to the limit in (41) as 
, the subscript M in these sums can be arbitrary large natural number. Such sums are dense in the space
. This can be proved exactly in the same way as it is done in the book [9] for the case of an initial boundary value problem. Thus, they also are dense in the subspace 
, and therefore, (41) remains valid for all 
This proves that 
satisfies the integral identity (13), and thus, it is a weak solution of problem (8). Estimate (14) for 
follows from estimate (40) for the approximate solutions.
Assume that 
. Since 
, we take
Obviously, 
. Putting this η into identity (13), we obtain
Since
by Fubini’s theorem and the time-periodicity of the function 
we have
Therefore, from (45) it follows that
Then
, and hence, 
. Integrating over σ, we get
Thus, 
, and since
, we conclude that
, that is 
for a.a. 
and t.
This implies 
. From identity (13) it follows that
In (46), we take 
, where 
is arbitrary, and 
is the solution of problem (6). Recall that 
. Then (46) takes the form
Thus, Ss(t)=const. Since 
,i.e. the mean value of Ss(t) is equal to zero, we get that Ss(t) = 0. Therefore, the functional s=0.
