SOLUTION TO NAVIER-STOKES EQUATIONS FOR TURBULENT CHANNEL FLOWS

. In this article, we continue the work done in [12] for turbulent channel ﬂows described by the Navier-Stokes and the Navier-Stokes-α equations. We study non-stationary solutions in special function spaces. In particular, we show the term representing the sum of pressure and potential is harmonic in the space variable. We ﬁnd an optimal choice for the function class.


Introduction
Turbulence is a fluid regime with the characteristics of being unsteady, irregular, seemingly random and chaotic [14].It can be used for modeling the weather, ocean currents, water flow in a pipe and air flow around aircraft wings.Studying turbulent fluid flows involves some of the most difficult and fundamental problems in classical physics, and is also of tremendous practical importance.The Navier-Stokes equations (NSE) have been widely used to describe the motion of turbulent fluid flows [9,19].However, solving NSE using the direct numerical simulation method for turbulent flows is difficult, since accurate simulation of turbulent flows should account for the interactions of a wide range of scales which leads to high computational costs.
Turbulence modeling could provide qualitative and in some cases quantitative measures for many applications [18].There are several types of turbulence modeling methods, for example Reynolds Average Navier-Stokes (RANS) and Large Eddy Simulation (LES).As done in [6,7], we accept that the Navier-Stokes-α (NS-α) (also called viscous Camassa-Holm equations or Lagrangian averaged Navier-Stokes equations) is a well-suited mathematical model for the dynamics of appropriately averaged turbulent fluid flows.The possibility that the NS-α is an averaged version of the NSE, first considered in [3,4], was entailed by several auspicious facts.Namely, the NS-α analogue of the Poiseuille, resp, Hagen, solution in a channel, resp, a pipe, displays both the classical Von Kármán and the recent Barenblatt-Chorin laws [16].In addition, the NS-α analogue of the Hagen solution, when suitably calibrated, yields good approximations to many experimental data [4].Moreover, the NS-α have been proved to have regular solutions [1].Therefore, continuing the study of NS-α is extremely useful and important in the aspects of both mathematical theories and down-to-earth applications.
To understand the connection for fluid flows described by NSE and by NS-α, in [12], we used a simple Reynolds type averaging.We restricted our consideration to channel flows having special function forms prescribed as a function class called P. This function class P was inspired by the concept of regular part of the weak attractor of the 3D NSE ( [10,11]) as well as by that of the sigma weak attractor introduced in [2].This led us to consider the solution for the channel flows whose averaged form has both the second and third velocity component to be zero.This will be our assumption for the discussion done in current work.Starting from there, a physical model for the wall roughness of the channel was subsequently provided to show that the NS-α model occurs naturally as the fluid flows.Moreover, by restricting to consider functions from P, a rational explanation was given to facilitate the understanding of why, as the Reynolds number increases, the fluid becomes in favor of the NS-α model instead of the NSE.The class P was composed by five assumptions, each assumption plays an unique and important role.
In this article, we study the properties of solutions in class P. We first try to find the explicit formula of the non-stationary solutions for NSE and NS-α.This particular solution has the form which only the first velocity component is nonzero.From there, we can recover the classic Poiseuille flow.Moreover, we prove the symmetric property of the integration form of this Poiseuille flow.Explicit and detailed energy estimate of the velocity field in class P is presented and is of use to show the connection between P and the weak global attractor of the equations.Moreover, for the sum of the pressure and potential, we also prove that it is actually harmonic in the space variable.Studying the properties of the sum of the pressure and potential, we find an alternative weaker condition of the last assumption in class P. Therefore, we have found an optimal choice of the class P.
The article is organized as follows.Section 2 gives elementary results on the Navier-Stokes equations and the Navier-Stokes-α model as well as the definition of the class P. In section 3, we solve the channel flows whose velocity fields have a special form.Section 4 contains the energy estimate for solutions in class P. In section 5, we discuss the harmonicity of the sum of the pressure and potential in the space variable and its consequences.The last section contains some basic inequalities together with their proofs.

Preliminaries
satisfies the NSE equation where P := p + Φ, t denotes the time, ν > 0 the kinematic viscosity, and p = p(x, t) the pressure.
The NS-α equation is where and Q in (2.3) (like P in (2.2)) may depend on the time t.
As mentioned in the introduction, we additionally assume that u 3 (x, t) = 0.The reality condition on u becomes, when viewed in the Fourier space, û * (t; and because u 3 = 0, the divergence free condition in(2.2) reduces to Using u 3 = 0, we see that (2.2) becomes We define the Reynolds type average of a scalar function φ = φ(x) as (2.15)

Proposition 2.2 ([12]
).For all u(x, t) ∈ P, we have ) Then the averaged velocity field takes the form (2.17) The following kernel representation of the averaged velocity component u 1 (t) (x 3 ) is also given in [12].

Proposition 2.3 ([12]
).For u(x, t) ∈ P we have where, the kernel K(x, t) is defined by the series,

Channel flows with velocity field of a particular form
As shown in Proposition 2.2, the averaged velocity field in the solution of (2.2) has a special form, namely, both the second and the third components vanish (see (2.17)).Thus, it is worth to consider the solutions with this form of the NSE and NS-α in a more general setting.See also (2.7) and (2.8) for time independent solutions of this form.
Using the first and fourth equations in (3.2), we obtain that U is independent of x 1 and ∂ 2 P ∂x 2 1 = 0, which combined with the second and third equations in (3.2) for P , imply that P must be of the form Solving (3.2).Based on the periodicity (2.9) in (A2), we can expand U in Fourier series where So equation (3.4) can be written as where (2.5) implies the boundary condition We first consider the case when n = 0: Equation (3.6) implies Taking the dot product with Û (n, x 3 , t) in (3.8), and then integrating with respect to x 3 from 0 to h, we obtain 1 2 Integrating by parts and applying the boundary condition (3.7), we have 1 2 (3.9) Using Poincaré inequality (6.1), Equation (3.9) becomes 1 2 (3.10) By assumption (A3), W (n, t 0 ) is bounded for all t 0 ∈ R. Hence, by taking t 0 → −∞ in (3.10), we obtain W (n, t) = 0, for all t ∈ R. Therefore, Û (n, x 3 , t) = 0, for all n = 0.
So, we only need to consider the case when n = 0.In this case (3.6) implies and, the expansion of U (x 2 , x 3 , t) in (3.5) becomes The equation satisfied by U (x 3 , t) follows from equation (3.4) (or (3.11)): Since U (0, t) = U (h, t) = 0, we can take the Fourier sine expansion for U (x 3 , t): Equation (3.13) gives the following equation for the coefficients Û (k, t): Thus, we obtain the explicit form U (x 3 , t) in the following theorem.
The above theorem shows that non-stationary solutions for (2.2) exist, and are given by (2.6) and (2.7).In particular, if p1 (t) is a constant, the Poiseuille flow (time independent) is recovered.This matches the form given in (2.7).
Corollary 3.2.If we assume that p1 (t) = p10 , where p10 ∈ R is a constant, then (3.16) 3.1.1.A symmetry property.Consider the averaged quantity which, from (2.5), satisfies We can prove that U 2 (x 3 , t) satisfies the following symmetry property, which is a priori assumption in [3]- [5] for the study of the steady solutions.
Proof.By taking average in x 2 as defined in(3.17) of the equation (3.4), and invoking the periodicity condition (2.9), we see that U 2 (x 3 , t) must satisfy with boundary conditions from (3.18): Taking the dot product of (3.22) with U and then integrating with respect to x 3 from 0 to h, together with the boundary conditions (3.23), we have Invoking the Poincaré inequality (6.1), we obtain 1 2 Therefore, from (3.24) we obtain for −∞ < t 0 < t < ∞, where ψ(t) := h 0 ( U (x 3 , t)) 2 dx 3 is nonnegative for all t ∈ R.Under the assumption (A3), we could let t 0 → −∞ in (3.25) to obtain ψ(t) ≡ 0 for all t ∈ R. Hence, U (x 3 , t) ≡ 0 for all x 3 ∈ [0, h] and all t ∈ R. Consequently, the symmetry property (3.19) on U 2 (x 3 , t) can be obtained.

3.2.
The NS-α case.We assume the following form of solution for (2.3), Simple form of the NS-α.Using (3.26), equation (2.3) becomes Using the first and fourth equations in (3.27), we obtain By taking partial derivative in the second and third equations in (3.27) with respect to x 1 , we have Thus, Q must be of the form Hence, the first equation in (3.27) can be further simplified to be 3 x2 . Hence, Similarly, we have the equations for Û (n, x 3 , t), which follow from (3.29): when Using arguments similar to those in the previous section, we obtain Û (n, x 3 , t) = 0, when n = 0, so U = Û (0, x 3 , t) = U (x 3 , t) is also independent of x 2 , and satisfies, From (3.32) and (3.3), we obtain the following theorem.
Theorem 3.4.Let u = (U (x, t), 0, 0) be a solution of the NSE (2.2) with P = P (x, t) given, satisfying (A1)-(A3).Then, u is also a solution of the NS-α (2.3) with We apply From (3.32), it follows that the Fourier coefficient Û (k, t) satisfies: Hence, using the same procedure as in previous section, we obtain the following form of the solution U = U (x 3 , t).
Theorem 3.5.The solution of the NSE (2.2) has the form The above theorem shows that non-stationary solutions for NS-α (2.3) exist, besides those stationary solutions given by (2.6) and (2.8).In the particular case when q1 (t) is a constant, we obtain the steady state solution mentioned in [3]- [5], which is basically of the form (2.8).

Energy estimates
In this Section, we will use inequality (6.6) in Lemma 6.2 to obtain an inequality estimating the energy of the velocity field u(x, t) ∈ P.
Taking the dot product of (2.2) with u and integrating over Ω : Note that here, the nonlinear term Ω (u • ∇)u • udx vanishes.Indeed, using integration by parts and the periodicity conditions (A2), one gets, for j = 1, 2, similarly, using the boundary condition (2.5) and integration by parts, we obtain Therefore, Hence Remark 4.1.Observe that the above proof can be applied to show for u, v ∈ P.
We have the following energy estimates.EJDE-2020/05 Proposition 4.2.For u(x, t) ∈ P, Proof.For the term on the right hand side of (4.1), by (2.11), we have Similarly, From the no-slip boundary condition (2.5), Therefore, where in the last line, the incompressibility condition (i.e., the second equation in (2.2)) is used.Therefore, using (6.6) in Lemma 6.2, relations (2.11), and denoting j = 3 − j for j = 1, 2, (4.1) becomes 1 2 where • j denotes the average in the x j direction, i.e., We then use Young's inequality and (A4) to obtain Hence, Remark 4.3.Following the general procedure in [8], one can start from (4.4) and show the existence of the weak global attractor of the NSE (2.2).

5.
Harmonicity of P = P (x, t) in the space variables 5.1.Harmonicity of P .The following property of P , namely, harmonicty in the space variable, could be deduced.Recall that P = p + Φ, where p is the pressure and Φ is the potential.Notice that from (2.14), P is independent of x 3 .
Lemma 5.1.Let u(x, t) ∈ P, then P = P (x 1 , x 2 , t) is harmonic in the space variables x 1 and x 2 .
Proof.From (2.14), by taking ∂/∂x 1 in the first equation and ∂/∂x 2 in the second equation and then summing the two resulting equations, we can obtain where P = P (x 1 , x 2 , t) is independent of x 3 (see the third equation in (2.14)).In (5.1), the left-hand side (LHS) takes values zero at x 3 = 0 and x 3 = h, while the right hand side (RHS) is independent of x 3 , hence for all x 1 , x 2 ∈ R.
Now we have an intriguing corollary for the harmonicity of P .
Proof.Equation (5.3) is obtained by using the fourth relation in (2.14), and the fact that the LHS in (5.1) equals zero.
Remark 5.3.For u = (u 1 , u 2 , u 3 ) being the velocity field for channel flows, after we assume u 3 = 0. Corollary 5.2 tells us that the two nonzero components, u 1 and u 2 , are not totally independent, one of them is, at least locally, a function of the other component.
Lemma 5.4.For the term P = P (x 1 , x 2 , t), we have for all t ∈ R.

5.3.
Simple form of P .From (2.11), we observe that Similar arguments apply to the case when y ≤ 0, and we obtain the next result.
Lemma 5.5.Let u(x, t) ∈ P be a solution of (2.2).Then for all y, x 2 , t ∈ R, where p is as given in (A4).
Next, we explore the harmonicity of P = P (x, t) = P (x 1 , x 2 , t) in x 1 and x 2 , and get the following explicit formula for P (x, t), namely, P (x, t) is linear in the variable x 1 .
Lemma 5.6.The term P = P (x 1 , x 2 , t) in (2.14) is of the form (5.8) Proof.Using (5.4) in Lemma 5.4 given in Appendix C, and Liouville's theorem for harmonic function ∂ for some function p1 (t) of time t, so that and, by the harmonicity of P (x 1 , x 2 , t) in x 1 and x 2 , we have Therefore, p0 (x 2 , t) is a linear function in x 2 , but then periodicity of P (x 1 , x 2 , t) in x 2 would imply that p0 (x 2 , t) is only a function of time t, i.e., p0 (x 2 , t) = p0 (t).
It follows from the harmonicity of P and Lemma 5.6 that we can replace (A5) by a weaker condition.for any given x 1 , x 3 and t ∈ R.
We have found an optimal choice of the function class.

2. 1 .
Mathematical background.Throughout, we consider an incompressible viscous fluid in an immobile region O ⊂ R 3 subjected to a potential body force F = −∇Φ, with a time independent potential Φ = Φ(x) ∈ C ∞ (O).The velocity field of such flows,