analytical solutions to turbulent flow models

Analytical Solutions to Turbulent Flow Models

A Project

Presented to

the Graduate School of

Clemson University

In Partial Fulfillment

of the Requirements for the Degree

Master’s of Mathematical Science

Mathematical Sciences

by

Stacey Watro

May 2013

Accepted by:

Dr. Leo Rebholz, Committee Chair

Dr. Eleanor Jenkins

Dr. Hyesuk Lee

Table of Contents

Title Page . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Preliminary Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 The Chorin Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.1 Leray . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 NS-α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 The Ethier-Steinman Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.1 Leray . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.2 NS- α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 Kovasznay Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.1 Leray . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.2 NS-α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

6 Conclusion and Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24A The Chorin Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25B The Ethier-Steinman Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31C Kovasznay Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

ii

Chapter 1

Introduction

The Navier Stokes equations describe the motion of viscous Newtonian fluid flow and are

given by,

ut + u · ∇u+∇p− ν∆u = 0 (1.1)

∇ · u = 0 (1.2)

Here, u represents the velocity, p pressure, and ν viscosity term. This last term can also be written

as 1Re where Re is the Reynold number, a non-dimensional ratio of inertial forces to viscous forces.

As with most real world models, solutions to the Navier Stokes equations (NSE) can rarely

be found analytically. In fact, there are only a handful of known analytical solutions to the NSE.

Solutions can be computed numerically, however, this poses its own set of complications and chal-

lenges. For turbulent flows, numerical solutions can be computationally expensive and inaccurate

[7]. In cases of high Reynolds numbers, computational solutions of the NSE are infeasible. Due

to these limitations, new models were developed, that could be more easily resolved numerically,

One class of such models is the so called α-models. These models can often times fully resolve the

physical solution on a practical grid where NSE cannot [8, 10, 12, 11, 1]. In many setting these

models can even give an accurate approximate to NSE.

In this paper we will focus on two α-models: Leray-α (1.3) and NS-α (1.4). Notice that

these models are more complex partial differential equations than the NSE. Each α-model employs

a filter equation that, in this setting, is used to describes the filtered velocity. Many filters have

1

been developed, however, in this paper the filter equation is defined as,

−α2∆u+ u = u (1.3)

where u is the filtered term and u is the unfiltered term. In many instance we will re-write this

equation as an operator in the form of F (u) = (−α2∆ + I)−1u = u.

ut + u · ∇u+∇p− ν∆u = 0, ∇ · u = 0 (1.4)

ut + (∇× u)× u+∇p− ν∆u = 0, ∇ · u = 0 (1.5)

Additionally, this paper will consider replacing the filter equation with the deconvolution

filter in each of the models. The deconvolution filter is defined as for a fixed nonnegative integer N ,

DN =∑Nn=0(I − F )n. However, in this paper we will use an alternative definition. The following

identity is proven in [4] and shows that the deconvolution operator can be considered as,

DN (u) = u+ (−1)Nα2N+2∆N+1uN+1. (1.6)

Notice that if N = 0, the deconvolution filter is the original filter equation, since D0 = I. Therefore,

when considering analytical solutions for these models, we will give a solution in terms of a general

N .

This thesis will focus on deriving analytical solutions to these specific α-models namely for

the Chorin, Ethier-Steinman, and Kovasznay flow problems. By understanding the structure of these

solutions for the NSE, we were able to discover solutions for Leray and NS-α. From our analytical

solutions, insight can be gained into how these α-models work and how their solutions different from

NSE solutions [9, 2].

In the sections to follow, we will explore analytical solutions to these flow problems and the

strategies used to develop solutions for α-models. Additionally, located in the Appendices, there are

full derivations of each solution and verification in Maple.

2

Chapter 2

Preliminary Lemmas

We start by introducing preliminary lemmas that will be used throughout this thesis. These

lemmas all pertain to decomposing the deconvolution operator. A key component to the construction

of solutions for these problems is identifying that the filtered velocity can be related to the unfiltered

velocity by a constant multiple. Hence to verify the exact solution of the models, we give some

preliminary lemmas that assume this relationship.

This first lemma in a series of four defines a relationship between a velocity that has been

filtered an arbitrary number of times and the original velocity vector.

Lemma 2.0.1. Let u ∈ Rd such that d ∈ N and u = uη where η is a fixed constant and u is the

solution to u = −α2∆u + u. Define uN := FN (u). Then uN = uηN

is the solution to the filter

equation, uN = −α2∆uN+1 + uN+1.

Proof. The N = 0 case is trivial. By assumption the N = 1 case holds. Now assume for a fixed

N > 1 that uN = uηN,

3

uN+1 = F (uN )

= (−α2∆ + I)−1uN

= (−α2∆ + I)−1 u

ηN

=u

ηN

=u

ηN+1

From here, it is clear that uN = −α2∆uN+1 + uN+1.

The next lemma begins the process of decomposing the deconvolution operator. This lemma

will be used throughout this thesis as it is a more practical way of writing the deconvolution operator,

especially when using induction proof techniques.

Lemma 2.0.2. For all N ∈ Z, N ≥ 0, α2N+2∆N+1uN+1 = α2N∆NuN+1 − α2N∆NuN .

Proof. Recall uN = uN+1 − α2∆uN+1. Therefore by substitution we can say,

α2N∆NuN = α2N∆N (uN+1 − α2∆uN+1)

= α2N∆NuN+1 − α2N+2∆N+1uN+1

By rearrange the terms above we have,

α2N+2∆N+1uN+1 = α2N∆NuN+1 − α2N∆NuN

Using the two previous lemmas we can now build two more important lemmas. These will

show that the deconvolution operator is a constant multiple of the unfiltered velocity, provided the

unfiltered velocity is also.


solution to u = −α2∆u+ u. Then for all N ∈ Z, N ≥ 0, α2N+2∆N+1uN+1 = u(

1−ηη

)N+1

.

4

Proof. For N = 0, we have

α2∆u = u− u = u

(1

η− 1

)= u

(1− ηη

).

Assume for a fixed N that α2N∆NuN = u(

1−ηη

)N. Then using the definition of the filter

and reducing gives,

α2N+2∆N+1uN+1 = α2N∆NuN+1 − α2N∆NuN

=1

ηα2N∆NuN − α2N∆NuN

= u

(1− ηη

)N (1− ηη

)= u

(1− ηη

)N+1

.


solution to u = −α2∆u+ u. Then for all N ∈ Z, N ≥ 0, DN (u) = u

(1−

(η−1η

)N+1).

Proof. Using the definition of the deconvolution operator and then applying Lemma 2.0.3 we have,

DN (u) = u+ (−1)Nα2N+2∆N+1uN+1

= u+ (−1)Nu

(1− ηη

)N+1

= u

(1 + (−1)N

(1− ηη

)N+1)

= u

(1− (−1)N+1

(1− ηη

)N+1)

= u

(1−

(η − 1

η

)N+1).

In the sections to follow, these lemmas will be used to help derive solutions for Leray and

NS-α, particularly in the Chorin problem and the Ethier-Steinman problem. Even in Kovasnzay

flow, where these lemmas will not be used, a similar structure to these conclusions can be observed.

5

Chapter 3

The Chorin Problem

The Chorin NSE solution is defined below in terms of n, any positive integer, and ν = 1Re ,

the viscosity term [3].

u1(x, y, t) = − cos(nπx) sin(nπy)e−2n2π2νt (3.1)

u2(x, y, t) = sin(nπx) cos(nπy)e−2n2π2νt (3.2)

p(x, y, t) = −1

4(cos(2nπx) + cos(2nπy)) e−4n2π2νt (3.3)

This particular solution has a special property which makes it an intuitive beginning point.

It is a straight forward calculation to show that the Chorin solution satisfies the incompressible NSE

and that

u · ∇u = −∇p (3.4)

ut = ν∆u (3.5)

This property details a natural path for extending the Chorin solution to the solutions for

α-models. In each of the cases to follow, we will show there exists a pressure term q such that q

satisfies a modified version of (3.4). These modifications will naturally be dependent on which model

is employed.

6

3.1 Leray

Recall the Leray model is of the form,

ut +DN (u) · ∇u+∇p− ν∆u = 0, ∇ · u = 0.

Lemma 3.1.1. Define the velocity, u, and pressure, q, as the following.



q(x, y, t) =−Γ

4(cos(2nπx) + cos(2nπy))e−4n2π2νt (3.8)

Γ =

(1−

(2n2π2α2

1 + 2n2π2α2

)N+1)

(3.9)

Then for all N ∈ Z, N ≥ 0, u and q satisfy the Leray-α with Deconvolution and incompressible flow

model.

When constructing this solution, it is instructional to look at the N = 0 case first, i.e. Leray

without deconvolution. We look for a pressure term, q such that q solves u · ∇u = −∇q. To begin,

we introduce a lemma that solves the filter equation.

Lemma 3.1.2. Let u be the velocity vector defined in the Chorin solution then u = u1+2n2π2α2 is the

solution to the filter equation u = −α2∆u+ u.

7

Proof.

−α2∆u+ u = −α2∆u

1 + 2n2π2α2+

u

1 + 2n2π2α2

=1

1 + 2n2π2α2(−α2∆u+ u)

=e−2n2π2νt

1 + 2n2π2α2

(−2n2π2α2)

cos(nπx) sin(nπy)

− sin(nπx) cos(nπy)

+

− cos(nπx) sin(nπy)

sin(nπx) cos(nπy)

=e−2n2π2νt

1 + 2n2π2α2


sin(nπx) cos(nπy)

(2n2π2α2 + 1)

=e−2n2π2νt

1 + 2n2π2α2


sin(nπx) cos(nπy)

(2n2π2α2 + 1)

= e−2n2π2νt


sin(nπx) cos(nπy)

= u

Now we are able to define the solution for the Leray (without deconvolution) model.




q(x, y, t) =−γ4

(cos(2nπx) + cos(2nπy))e−4n2π2νt (3.12)

γ =1

1 + 2n2π2α2(3.13)

Then u and q satisfy the Leray-α with incompressible flow model.

Proof. As u is the Chorin NSE velocity, it is clear that ut = ν∆u. Thus we need only to show that

u · ∇u = −∇q. This will be done using that u · ∇u = −∇p where p is the Chorin NSE pressure.

8

Thus we have,

u · ∇u =u

1 + 2n2π2α2· ∇u (By Lemma 3.1.2)

=1

1 + 2n2π2α2(u · ∇u)

=−1

1 + 2n2π2α2∇p

= −∇(

p

1 + 2n2π2α2

)= −∇q.

This solution can be thought of as the Leray with deconvolution solution when N = 0. Now

we will extend this solution for all N ∈ N. Recall that u = u1+2n2π2α2 , thus by Lemma 2.0.4, the

deconvolution operator can be written as,

DN (u) = u

(1−

(2n2π2α2

1 + 2n2π2α2

)N+1). (3.14)

Now the proof for Lemma 3.1.1 is straight forward. If we let u be defined as the NSE

velocity, it is clear that ut = ν∆u. We must show that DN (u) · ∇u = −∇q. Again, this will be done

using that u · ∇u = −∇p where p is the Chorin NSE pressure.

DN (u) · ∇u = u

(1−

(2n2π2α2

1 + 2n2π2α2

)N+1)· ∇u

= Γ(u · ∇u)

= −Γ (∇p)

= −∇Γp

= −∇q.

There are two important features to notice with these solutions. First, consider the effect

of α on the solution. If α is equal to 0, then the solution is exactly the NSE solution. Intuitively

this is expected, since there is no filtering, then the model becomes the NSE model. In addition,

notice that as N approaches infinity the constant Γ approaches 1 and the solution becomes the NSE

9

solution again. These behaviors are what is theoretically expected when working with Leray.

3.2 NS-α

Lemma 3.2.1. Define the velocity, u, and pressure, Q, as the following.



Q(x, y, t) = −Γ

4(cos(2nπx)(1 + cos(2nπy)) + cos(2nπy)) e−4n2π2νt (3.17)

Γ =

(1−

(2n2π2α2

1 + 2n2π2α2

)N+1)

(3.18)

For all N ∈ Z, N ≥ 0, u and Q satisfy the NS-α with Deconvolution and incompressible flow model.

Similar to the Leray case, for NS-α we start by seeking a pressure Q such that (∇ × u) ×

DN (u) = −∇Q. As the velocity vector will not change, we know the filtered velocity and the

deconvolution operator will be the same. This means that all the basic tools necessary to derive the

NS-α solution are already determined.

However, before working with NS-α it is important to highlight a feature of the model. NS-α

is based on the rotational form of NSE. This means the strategy of relating the NS-α pressure back

to the NSE pressure will change. Instead, we solve NS-α for N = 0 and use this as a building block

for the solution given a general N.

In order to prove the conjecture in Lemma 3.2.1, we must first derive the solution when

N = 0. This turns out to be a simple algebra problem since the filtered velocity is already defined.

The next lemma defines the NS-α solution without deconvolution.

Lemma 3.2.2. Define the velocity, u, and pressure, P , as the following.



P (x, y, t) = −γ4

(cos(2nπx)(1 + cos(2nπy)) + cos(2nπy)) e−4n2π2νt (3.21)

γ =1

1 + 2n2π2α2(3.22)

10

Then u and P satisfy the NS-α with incompressible flow model.

Proof. See Appendix A

Now we are able to discuss the proof for Lemma 3.2.1. If we let u be the NSE velocity, it is

clear that ut = ν∆u. Thus we show that (∇× u)×DN (u) = −∇Q. This will be done using the fact

that (∇× u)× u = −∇P where P is the NS-α pressure. In addition recall that

u = (1 + 2n2π2α2)u

⇒ Γu = Γ(1 + 2n2π2α2)u.

Thus we can conclude,

(∇× u)×DN (u) = (∇× u)× Γ(1 + 2n2π2α2)u

= (1 + 2n2π2α2)Γ ((∇× u)× u)

= (1 + 2n2π2α2)Γ(−∇P )

= −∇(1 + 2n2π2α2)ΓP

= −∇Q

Observe that the NS-α solution is a nontrivial change in pressure. Indeed this is the result

we would theoretically expect for NS-α. Unlike Leray, the NS-α model makes the pressure term

far more complicated. Also, notice that NS-α constant behaves much like in Leray. That is, as N

approaches infinity, Γ approaches 1, and the pressure becomes the same solution as when α = 0.

11

Chapter 4

The Ethier-Steinman Problem

The Ethier-Steinman problem is the natural next step after working with the Chorin prob-

lem. Like the Chorin solution, Ethier-Steinman (E-S) also satisfies (3.4) and (3.5), only now in three

dimensions. The solution is defined using two fixed constants a, d and the viscosity term ν [5].

u1 = −a (eax sin(ay + dz) + eaz cos(ax+ dy)) e−νd2t (4.1)

u2 = −a (eay sin(az + dx) + eax cos(ay + dz)) e−νd2t (4.2)

u3 = −a (eaz sin(ax+ dy) + eay cos(az + dx)) e−νd2t (4.3)

p = −a2

2(e2ax + e2ay + e2az + 2 sin(ax+ dy) cos(az + dx)ea(y+z) (4.4)

+ 2 sin(ay + dz) cos(ax+ dy)ea(z+x) + 2 sin(az + dx) cos(ay + dz)ea(x+y)) (4.5)

As the E-S solution has this same special property as Chorin, the strategy for extending the NSE

solution to a solution for the α-models will be the same as in the previous section.

4.1 Leray

We begin by seeking a q which solves Dn(u) · ∇u = −∇q where u is the NSE velocity. We

define this solution in the next lemma.

12





q = −Γa2

2(e2ax + e2ay + e2az + 2 sin(ax+ dy) cos(az + dx)ea(y+z) (4.9)

+ 2 sin(ay + dz) cos(ax+ dy)ea(z+x) + 2 sin(az + dx) cos(ay + dz)ea(x+y)) (4.10)

Γ =

(1−

(α2d2

1 + α2d2

)N+1)

(4.11)

Then for all N ∈ Z, N ≥ 0, u and q satisfy the Leray-α with Deconvolution and incompressible flow

model.

We find that proving this solution is the same process as proving the Chorin Leray solution,

so we omit the N = 0 case. However, we still must derive the filtered velocity. This then allows us

to use the preliminary lemmas and very easily derive a proof for the above lemma.

Lemma 4.1.2. Let u be defined as in the Ethier-Steinman solution, then u = u1+α2d2 solves the

filter equation u = −α2∆u+ u.

13

Proof. By substituting and expanding we find that,

−α2∆u+ u = −α2∆

(u

1 + α2d2

)+

u

1 + α2d2

=−α2

1 + α2d2∆u+

u

1 + α2d2

=−α2

1 + α2d2∆

−a (eax sin(ay + dz) + eaz cos(ax+ dy)) e−νd

2t

−a (eay sin(az + dx) + eax cos(ay + dz)) e−νd2t

−a (eaz sin(ax+ dy) + eay cos(az + dx)) e−νd2t

+u

1 + α2d2

=−α2

1 + α2d2∆

d2a (eax sin(ay + dz) + eaz cos(ax+ dy)) e−νd

2t

d2a (eay sin(az + dx) + eax cos(ay + dz)) e−νd2t

d2a (eaz sin(ax+ dy) + eay cos(az + dx)) e−νd2t

+u

1 + α2d2

=−α2

1 + α2d2(−d2u) +

u

1 + α2d2

= uα2d2 + 1

1 + α2d2

= u.

Now by lemma 2.0.4 we are able to define the deconvolution operator as

DN (u) = u

(1−

(α2d2

1 + α2d2

)N+1). (4.12)

Thus the proof of Lemma 4.1.1 becomes analogous to that of Lemma 3.1.1.

The distinct similarities between the Chorin solution and E-S solution cannot go unnoticed.

The solutions are almost identical in appearance, derivation and convergence. Like the Chorin Leray

solution, when α = 0 the solution reverts back to the NSE solution. Also, as N approaches infinity

the constant Γ approaches 1 and the NSE solution is recovered again. These similarities beg the

question, will the E-S solution behave the same as Chorin for all α models? Unfortunately, we will

see in the next section that this answer is no.

14

4.2 NS- α

For NS-α we begin in much the same way we began with Chorin NS-α. We start by seeking

a pressure term Q such that Q satisfies (∇ × u) × DN (u) = −∇Q. As the deconvolution filter is

already defined, we see that this problem will turn into an algebra problem. In fact, the algebra

returns an interesting result.





Q = 0 (4.16)

Then for all N ∈ Z, N ≥ 0, u and Q satisfy the NS-α with Deconvolution and incompressible flow

model.

Proof. See Appendix B.

This is a curious result as the Chorin solution is similar to E-S in many ways. But with E-S

the curl of the velocity is a multiple of velocity, this is not the case with the Chorin solution. This

fact is what makes the NS-α solution so much different for these two seemly similar models.

This interesting solution also leads to the discussion of validity of the solutions in this paper.

That is, how well do these solutions match physical results. In the previous section, the α-model

solutions were similar to NSE, thus with the correct choice of N or α the solution could be physically

accurate. However, do to the nature of the NS-α pressure term it is acceptable that the pressure is

0.

15

Chapter 5

Kovasznay Flow

The final solution we will be examining is the Kovasznay Flow exact NSE solution. Unlike

the previous two sections, the derivation of the α-model solutions will not be dependent on the

preliminary lemmas. In addition, this solution does not maintain the same special property that

provided our road map in the other two sections. In fact, the Kovasznay Flow α-model derivation

will be far less elegant than the previous two solutions.

The NSE Kovasznay Flow exact solution (5.1)-(5.4) is defined in terms of a fixed constant λ

which depends on the Reynolds number [6]. In the first lemma of this section, we will demonstrate

how this constant can be manipulated. This will be a key feature in developing the α-model solutions

for Kovasznay Flow.

u1 = 1− eλx cos(2πy) (5.1)

u2 =λ

2πeλx sin(2πy) (5.2)

p = p0 −1

2e2λx where p0 is an arbitrary constant (5.3)

λ =Re

2−√Re2

4+ 4π2 (5.4)

Lemma 5.0.2. Let λ = Re2 −

√Re2

4 + 4π2 then, −λ − ν(−λ2 + 4π2) = 0. Therefore we also have,

λ2

2π − ν(λ3

2π + 2πλ) = 0.

16

Proof. By expanding λ we see that,

−λ− ν(−λ2 + 4π2) = −

(Re

2−√Re2

4+ 4π2

)− ν

−(Re2−√Re2

4+ 4π2

)2

+ 4π2

= −

(Re

2−√Re2

4+ 4π2

)− 1

Re

(−Re2

2+Re

√Re2

4+ 4π2

)

= −Re2

+

√Re2

4+ 4π2 +

Re

2−√Re2

4+ 4π2

= 0.

Unfortunately, it it less clear how to extend the Kovasznay Flow exact solution to solutions

for α models. The only tool we have is to manipulate terms such that Lemma 5.0.2 can be used to

eliminate extra terms.

5.1 Leray

The exact solution for Leray with deconvolution is as follows.

Lemma 5.1.1. Define the velocity, u, and pressure, p, as the following.

u1 = 1− eλx cos(2πy) (5.5)

u2 =λ


p = p0 −1−

(−α2λRe1−α2λRe

)N+1

2e2λx where p0 is an arbitrary constant (5.7)

λ =Re

2−√Re2

4+ 4π2 (5.8)

Then for all N ∈ Z, N ≥ 0, u and p satisfy the Leray-α with incompressible flow model and

Deconvolution.

Notice that this solution looks very similar to previous solutions we have seen before; the

pressure term is only affected by a constant and the constant is of the same form as previous models.

However deriving this solution will be more computationally challenging than in previous sections.

17

This stems from the solution to the filter equation. In this model, the filtered velocity is no longer

a constant multiple of original velocity. Thus we must first solve the filtered velocity and then walk

through the steps of breaking down the deconvolution operator given this filtered velocity.

Lemma 5.1.2. Let u be the velocity vector of the Kovasznay Flow solution. Then

u =

1− 11−α2λRee

λx cos(2πy)

λ2π(1−α2λRe)e

λx sin(2πy)

solves the filter equation, −α2∆u+ u = u.

Proof. We begin by substituting u into the filter equation and expanding the terms. Then we

recombine terms with the intension of using Lemma 5.0.2. Thus we have,

−α2∆u+ u = −α2

−λ2

(1−α2λRe)eλx cos(2πy) + 4π2

(1−α2λRe)eλx cos(2πy)

λ3

2π(1−α2λRe)eλx sin(2πy)− 2πλ

(1−α2λRe)eλx sin(2πy)

+

1− 11−α2λRee

λx cos(2πy)

λ2π(1−α2λRe)e

λx sin(2πy)

=

α2λ2

(1−α2λRe)eλx cos(2πy)− 4π2α2

(1−α2λRe)eλx cos(2πy) + 1− 1

1−α2λReeλx cos(2πy)

−α2λ3

2π(1−α2λRe)eλx sin(2πy) + 2πα2λ

(1−α2λRe)eλx sin(2πy) + λ

2π(1−α2λRe)eλx sin(2πy)

=

−eλx cos(2πy)(

−α2λ2

(1−α2λRe)) + 4π2α2

(1−α2λRe) + 11−α2λRe

)+ 1

λ2π e

λx sin(2πy)(− α2λ2

(1−α2λRe) + 4π2α2

(1−α2λRe) + 1(1−α2λRe)

)

=

−eλx cos(2πy)(

−α2λ2+4π2α2+1(1−α2λRe))

)+ 1

λ2π e

λx sin(2πy)(

−α2λ2+4π2α2+1(1−α2λ2+4π2α2)

)

=

−eλx cos(2πy)(

1+α2(−λ2+4π2)(1−α2λRe))

)+ 1

λ2π e

λx sin(2πy)(

1+α2(−λ2+4π2)(1−α2λRe)

)

=

−eλx cos(2πy)(

1+α2 −λν

(1−α2λRe))

)+ 1

λ2π e

λx sin(2πy)(

1+α2 −λν

(1−α2λRe)

)

=

−eλx cos(2πy) + 1

λ2π e

λx sin(2πy)

By Lemma 5.0.2

= u.

Now we extend this solution by considering a velocity that has been filtered an arbitrary

number of times.

18

Lemma 5.1.3. Let u be the velocity for the Kovasznay Flow exact NSE solution. Then for all

N ∈ Z, N ≥ 0, uN = βNu+

1− βN

0

where β = 11−α2λRe and uN = −α2∆uN+1 + uN+1.

Proof. See Appendix C

Now that we have defined uN it is useful to also define a corollary that relates sequential

filtered terms.

Corollary 5.1.1. Let u be the velocity vector in the Kovasznay Flow solution. Then for all N ∈ Z,

N ≥ 0, uN = βuN−1 +

1− β

0

where β = 11−α2λRe and uN = −α2∆uN+1 + uN+1.

Proof.

uN−1 = βN−2u+

1− βN−2

0

⇒ βN−2u = uN−1 −

1− βN−2

0

uN = βN−1u+

1− βN−1

0

= β

(βN−2u

)+

1− βN−1

0

= β

uN−1 −

1− βN−2

0

+

1− βN−1

0

= βuN−1 −

β − βN−1

0

+

1− βN−1

0

= βuN−1 +

1− β

0

Based on the previous lemma and corollary we can now decompose the deconvolution op-

erator in the same manner as in the preliminary lemmas. Note that in this proof use will use the

19

result from Lemma 2.0.2.

Lemma 5.1.4. Let u be the velocity vector in the Kovasznay Flow solution. Then for all N ∈

Z, N ≥ 0, α2N+2∆N+1uN+1 = (β − 1)N+1u −

(β − 1)N+1

0

where β = 11−α2λRe and uN =

−α2∆uN+1 + uN+1.

Proof. Let N = 0, then

α2∆u = u− u

= βu+

1− β

0

− u= u(β − 1)−

β − 1

0

.

Assume for a fixed N , α2N+2∆N+1uN+1 = (β − 1)N+1u−

(β − 1)N+1

0

.

α2N+4∆N+2uN+2 = α2N+2∆N+1uN+2 − α2N+2∆N+1uN+1

= α2N+2∆N+1

βuN+1 +

1− β

0

− α2N+2∆N+1uN+1

= βα2N+2∆N+1uN+1 − α2N+2∆N+1uN+1

= α2N+2∆N+1uN+1 (β − 1)

=

(β − 1)N+1u−

(β − 1)N+1

0

(β − 1)

= (β − 1)N+2u−

(β − 1)N+2

0

.

Now we have all the tools to define the deconvolution operator for Kovasznay Flow.

20

Lemma 5.1.5. Let u be the velocity vector in the Kovasznay Flow solution. Then for all N ∈ Z,

N ≥ 0, DN (u) = u(1− (1− β)N+1

)+

(1− β)N+1

0

where β = 11−α2λRe .

Proof.

DN (u) = u+ (−1)Nα2N+2∆N+1uN+1

= u+ (−1)N

(β − 1)N+1u−

(β − 1)N+1

0

= u(1 + (−1)N (β − 1)N+1

)+ (−1)N+1

(β − 1)N+1

0

= u

(1− (1− β)N+1

)+

(1− β)N+1

0

At this point we now have everything necessary to prove Lemma 5.1.2. The proof, in

Appendix C, is not particularly elegant but surprisingly the solution is not that far removed from

the Chorin and E-S Leray solutions. The pressure is not a constant multiple of the NSE pressure,

but its not a completely new term. Also, the constant that is developed is of the same form as

the Chorin and E-S Leray constant. In addition, if α = 0 then the NSE pressure is restored and

if N approaches infinity, then the NSE pressure is also resolved. Once again these are the types of

behaviors we expect from a Leray model.

5.2 NS-α

As with all the other models finding the NS-α solution is much more of an algebra problem.

As we already know the deconvolution operator, we simply seek a Q such that (∇× u)×DN (u) =

−∇Q. There is seemingly no clever way to derive this solution, so use brute forced approach.

21


u1 = 1− eλx cos(2πy) (5.9)

u2 =λ


Q =λ

8π2ν

(1−

(−α2λRe

1− α2λRe

)N+1)

sin2(2πy)− eλx cos(2πy) (5.11)

λ =Re

2−√Re2

4+ 4π2 (5.12)

Then for all N ∈ Z, N ≥ 0, u and Q satisfy the N-α with incompressible flow model and Deconvo-

lution.

This solution, derived in Appendix C, is very interesting because of its change from the

NSE solution. The pressure now includes both exponential and trigonometric terms. In addition,

it is now dependent on both the x and y directions. This shows how much the pressure changed by

using the NS-α model. Observe that when N approaches infinity, the solution approaches the α = 0

solution. Overall, we can see that this NS-α definitely complicates this pressure term.

22

Chapter 6

Conclusion and Future Work

In conclusion, we can see that α-models have the potential to drastically change a pressure

term or simply modify a pressure by a constant. It is clear that the Leray model maintains the

structure of the NSE pressure term, in many cases only changing the term by a constant. In

addition, the NSE solution could always be resolved by letting α = 0 or letting N approach infinity.

This model could be a very appropriate model to use if the user simply wished to scale the pressure

term, as in these three cases that is really what Leray achieved.

Whereas the Leray solution maintained the integrity of the NSE pressure, NS-α gave signif-

icant changes in the pressure term. From a range of far more complicated to oversimplifying, NS-α

always drastically changed the pressure term in these three models.

From these conclusions we can see the inherent differences between Leray and NS-α. Though

it is important to notice that each change was highly relative to the solution in question. For instance,

for the Chorin problem both the Leray and NS-α pressure seemed highly logical results whereas with

Ethier-Steinman the NS-α solution was clearly not as physically accurate. These conclusions lead

to an understanding that both models could be useful in many areas, but in many cases Leray may

be advantageous over NS-α.

This paper has simply scratched the surface of what is to come in the way of research on

analytical solution of turbulence flow models. Not only can more α-models be looked at, but also

more analytical solutions. In addition, an important next step for the work done in this paper is to

verify the results done with computational simulations.

23

Appendices

24

Appendix A The Chorin Problem

A.1 Proof of Lemma 3.2.1

Proof. Recall that the velocity vector is the same as in NSE. Thus it is enought to show (∇×u)×u =

−∇P.

∇× u =

∣∣∣∣∣∣∣∣∣∣i j k

∂∂x

∂∂y

∂∂z

u1 u2 0

∣∣∣∣∣∣∣∣∣∣= (u2x − u1y )k

(u2x − u1y ) = nπ cos(nπx) cos(nπy)e−2n2π2νt + nπ cos(nπx) cos(nπy)e−2n2π2νt

= 2nπ cos(nπx) cos(nπy)e−2n2π2νt

(∇× u)× u = −(u2(u2x − u1y )i− u1(u2x − u1y )j

)= −

2nπu2 cos(nπx) cos(nπy)e−2n2π2νt

−2nπu1 cos(nπx) cos(nπy)e−2n2π2νt

From Lemma 3.1.1, u = u

1+2n2π2α2 thus we have

=−2nπ

1 + 2n2π2α2

u2 cos(nπx) cos(nπy)e−2n2π2νt

−u1 cos(nπx) cos(nπy)e−2n2π2νt

=

−2nπ

1 + 2n2π2α2

sin(nπx) cos(nπx) cos2(nπy)e−4n2π2νt

cos2(nπx) cos(nπy) sin(nπy)e−4n2π2νt

.

Using the trigonometric identities 2 sin(nπx) cos(nπx) = sin(2nπx) and 1+cos(2nπx)2 = cos2(nπx)

25

in both the x and y coordinates we obtain,

=−nπ

1 + 2n2π2α2e−4n2π2νt

sin(2nπx) cos2(nπy)

cos2(nπx) sin(2nπy)

=

−nπ1 + 2n2π2α2

e−4n2π2νt

sin(2nπx)(

1+cos(2nπy)2

)sin(2nπy)

(1+cos(2nπx)

2

)

=−nπ

2(1 + 2n2π2α2)e−4n2π2νt

sin(2nπx) (1 + cos(2nπy))

sin(2nπy) (1 + cos(2nπx))

⇒ (∇× u)× u =

−nπ2(1 + 2n2π2α2)

e−4n2π2νt



Let P =−1

4(1 + 2n2π2α2)[cos(2nπx)(1 + cos(2nπy)) + cos(2nπy)]e−4n2π2νt

⇒ ∇P =−1

4(1 + 2n2π2α2)e−4n2π2νt

−2nπ sin(2nπx) (1 + cos(2nπy))

(−2nπ) cos(2nπx) sin(2nπy)− 2nπ sin(2nπy)

=

2nπ

4(1 + 2n2π2α2)e−4n2π2νt


cos(2nπx) sin(2nπy) + sin(2nπy)

=

nπ

2(1 + 2n2π2α2)e−4n2π2νt



⇒ −∇P = (∇× u)× u

A.2 Maple

This u (velocity vector) will be the same for all four models. (Note that in

these equations v– ν, a– α)

u1 := − cos (nπ x) sin (nπ y) e−2n2π2vt

u2 := sin (nπ x) cos (nπ y) e−2n2π2vt

u := VectorField ( (u1 , u2 , 0) , cartesianx,y,z)

The terms below are for pressure: (c*p) is the pressure for Leray and (c*q) is

the pressure for NS-alpha. The constant C replaces c when we are using deconvolution.

c :=(1 + 2n2π2a2

)−1

p := −1/4 (cos (2nπ x) + cos (2nπ y)) e−4n2π2 vt

q := −1/4 (cos (2nπ x) (1 + cos (2nπ y)) + cos (2nπ y)) e−4n2π2 vt

C := 1−(

2n2π2a2

1 + 2n2π2a2

)N+1

Now we check the filter equation. As the velocity vector is not changing we will

have the same filter equation over all four models.

ubar := cu

simplify(−a2Laplacian (ubar) + ubar − u

)0

Now we will check the deconvolution operator. Similarly to the filter equation,

the deconvolution operator won’t change between models. (Note that we only check

this operator for N=2, N=3. In the code below N is replaced with r so we can keep N

a general variable.

r := 2

cr :=((

1 + 2n2π2a2)r+1

)−1

ubar := cr u

Cr := 1−(

2n2π2a2

1 + 2n2π2a2

)r+1

usolve := Cr u

simplify(u+ (−1)

ra2 r+2Laplacian (Laplacian (Laplacian (ubar))) − usolve

)0

r := 3

cr :=((

1 + 2n2π2a2)r+1

)−1

ubar := cr u

Cr := 1−(

2n2π2a2

1 + 2n2π2a2

)r+1

usolve := Cr u

simplify(u+ (−1)

ra2 r+2Laplacian (Laplacian (Laplacian (Laplacian (ubar)))) − usolve

)0

Now we begin to check that this velocity vector and pressure term satisfy our

models. In order to do this recall that the equation we are looking to solve will be of

the form: ut + X + Y − ν∆u = 0. In these four models the only terms that will change

will be the X and Y terms. Thus we start by showing that ut = ν∆u.

upartial :=

(d

dtu1 ,

d

dtu2 , 0

)nudeltau := −v (Laplacian (u1 , [x, y]) ,Laplacian (u2 , [x, y]) , 0)

upartial + nudeltau

0

Now we show that the pressure term is equal to the nonlinear term, for each

individual model. Note for each of these models our maple output will never be zero,

as with the problems above. Below each final maple output will be a series of steps for

how to show th

We start with regular Leray.

ubar := cu

nonlinear := simplify (DotProduct (Jacobian (u, [x, y, z]) , ubar))

pressure :=

(∂

∂x(cp) ,

∂

∂y(cp) , 0

)simplify (nonlinear + pressure, trig)

0

Now consider Leray with Deconvolution.

nonlinear := DotProduct (Jacobian (u, [x, y, z]) , Cu)

pressure :=

(∂

∂x(Cp) ,

∂

∂y(Cp) , 0


12nπ

(−1 +

(2n2π2a2

1+2n2π2a2

)N+1)(

2 sin(nπx)(e−2n2π2vt

)2cos(nπx)− sin(2nπx)e−4n2π2vt

)12nπ

(−1 +

(2n2π2a2

1+2n2π2a2

)N+1)(

2 sin(nπy)(e−2n2π2vt

)2cos(nπy)− sin(2nπy)e−4n2π2vt

)

Steps to Simplify the above term to zero: Consider the last term in the x-component.

Replace 2 sin(nπx) cos(nπx) with sin(2nπx) and replace(e−2n2π2νt

)2with e−4n2π2νt. Thus is whole

term goes to zero. The same can be done in the y-component.

Now we consider NS-alpha. Notice that we now use q for our pressure function.

nonlinear := CrossProduct (Curl (u) , cu)

pressure := simplify

(VectorField

((∂

∂x(cq) ,

∂

∂y(cq) , 0

), cartesianx,y,z

))simplify (nonlinear + pressure, exp )

− 2 cos(2πx) cos2(nπy)nπe−4n2π2vt sin(nπx)1+2n2π2a2 + sin(2nπx)nπ(1+cos(2nπy))e−4n2π2vt

2(1+2n2π2a2)

− 2 cos(2πy) cos2(nπx)nπe−4n2π2vt sin(nπy)1+2n2π2a2 + sin(2nπy)nπ(1+cos(2nπx))e−4n2π2vt

2(1+2n2π2a2)

Steps to simplify the above term to zero: Consider the first term in the x-component.

First replace 2 cos(nπx) sin(xπx) with sin(2nπx). Then replace cos2(nπy) with (1/2)(cos(2nπy)+1).

Then the x-component with be 0. The same can be down in the y-component.

Now consider NS-alpha with Deconvolution.

nonlinear := simplify (CrossProduct (Curl (u) , Cu))


(VectorField

((∂

∂x(Cq) ,

∂

∂y(Cq) , 0

), cartesianx,y,z

))simplify (nonlinear + pressure, exp )

1

2(1+2n2π2a2)

((1 + 2n2π2a2 − 2N+1π2N+2

(n2a2

1+2n2π2a2

)Nn2a2

)sin(2nπx)nπ(1 + cos(2nπy))e−4n2π2vt

)1

2(1+2n2π2a2)

((1 + 2n2π2a2 − 2N+1π2N+2

(n2a2

1+2n2π2a2

)Nn2a2

)sin(2nπy)nπ(1 + cos(2nπx))e−4n2π2vt

)

+

−1

1+2n2π2a2

(2 cos(nπx) cos2(nπy)nπe−4n2π2vt

(1 + 2n2π2a2 − 2N+1π2N+2

(n2a2

1+2n2π2a2

)Nn2a2

)sin(nπx)

)−1

1+2n2π2a2

(2 cos(nπy) cos2(nπx)nπe−4n2π2vt

(1 + 2n2π2a2 − 2N+1π2N+2

(n2a2

1+2n2π2a2

)Nn2a2

)sin(nπy)

)

Steps to simplify the above term to zero: Using the same trigonometric substitutions

as in NS-alpha, we can show that both the x and y components go to zero.

Appendix B The Ethier-Steinman Problem

B.1 Proof of Lemma 4.2.1

Show that the following solves the NS-α with Deconvolution and incompressible flow model.

u1 = −a (eax sin(ay + dz) + eaz cos(ax+ dy)) e−νd2t

u2 = −a (eay sin(az + dx) + eax cos(ay + dz)) e−νd2t

u3 = −a (eaz sin(ax+ dy) + eay cos(az + dx)) e−νd2t

Q = 0

Proof. As u is the NSE velocity we no that ∇ · u = 0 and ut = ν∆u therefore it is enough to show

(∇× u)×DN (u) = 0.

(∇× u) =

−ad (eax sin(ay + dz) + eaz cos(ax+ dy)) e−νd

2t

−ad (eay sin(az + dx) + eax cos(ay + dz)) e−νd2t

−ad (eaz sin(ax+ dy) + eay cos(az + dx)) e−νd2t

= du

Recall that DN (u) = Γu where Γ =

(1−

(α2d2

1+α2d2

)N+1)

for all N ∈ Z, N ≥ 0. Therefore,

(∇× u)×DN (u) = du× uΓ

=

∣∣∣∣∣∣∣∣∣∣i j k

du1 du2 du3

Γu1 Γu2 Γu3

∣∣∣∣∣∣∣∣∣∣= dΓ

(i (u3u2 − u2u3)− j (u3u1 − u1u3) + k (u2u1 − u1u2)

)= 0

B.2 Maple Verification

This u (velocity vector) will be the same for all four models.

u1 := ‘-‘(a (ea·x · sin (a · y + d · z) + ea·z · cos (a · x+ d · y)) · e‘-‘(v·d2·t)

)u2 := ‘-‘

(a (ea·y · sin (a · z + d · x) + ea·x · cos (a · y + d · z)) · e‘-‘(v·d2·t)

)u3 := ‘-‘

(a (ea·z · sin (a · x+ d · y) + ea·y · cos (a · z + d · x)) · e‘-‘(v·d2·t)

)u := VectorField ((u1 , u2 , u3 ) , cartesianx,y,z)

p := −a2

2 (e2ax + e2ay + e2az + 2 sin(ax+ dy) cos(az+ dx)ea(y+z) + 2 sin(ay+ dz) cos(ax+ dy)ea(z+x)

+ 2 sin(az + dx) cos(ay + dz)ea(x+y))

Now we begin to check that this velocity vector and pressure term satisfy our

models. In order to do this recall that the equation we are looking to solve will be of

the form: ut + X + Y − ν∆u = 0. In these four models the only terms that will change

will be the X and Y terms. Thus we start by showing that ut = ν∆u.

upartial :=

(d

dtu1 ,

d

dtu2 ,

d

dtu3

)nudeltau := simplify (‘-‘ (v · (Laplacian (u1 , [x, y, z]) ,Laplacian (u2 , [x, y, z]) ,Laplacian (u3 , [x, y, z]))))

simplify (upartial + nudeltau)

0 · ex

Now we check the filter equation. As the velocity vector is not changing we

will have the same filter equation over all four models. (Here A–α)

c := 1 ·(1 +A2 · d2

)−1

ubar := c · u

simplify(‘-‘

(A2 · Laplacian (ubar)

)+ ubar + ‘-‘ (u)

)0 · x

Now we will check the deconvolution operator. Similarly to the filter equation,

the deconvolution operator won’t change between models. (Note that we only check

this operator for N=2, N=3. In the code below N is replaced with r so we can keep N

a general variable.

r := 2

cr := 1 ·((

1 +A2 · d2)r+1

)−1

ubar := cr · u

Cr := 1 + ‘-‘

((A2 · d2 · 1 ·

(1 +A2 · d2

)−1)r+1

)usolve := Cr · u

simplify(u+ (−1)

r ·A2·r+2 · Laplacian (Laplacian (Laplacian (ubar))) + ‘-‘ (usolve))

0 · x

r := 3

cr := 1 ·((

1 +A2 · d2)r+1

)−1

ubar := cr · u

Cr := 1 + ‘-‘

((A2 · d2 · 1 ·

(1 +A2 · d2

)−1)r+1

)

usolve := Cr · u

simplify(u+ (−1)

r ·A2·r+2 · Laplacian (Laplacian (Laplacian (Laplacian (ubar)))) + ‘-‘ (usolve))

0 · x

Now we show that the pressure term is equal to the nonlinear term, for each

individual model.

We start with regular Leray.

ubar := c · u

nonlinear := simplify (DotProduct (Jacobian (u, [x, y, z]) , ubar))

pressure :=

(∂

∂xc · p, ∂

∂yc · p, ∂

∂zc · p


0 · ex

Now consider Leray with Deconvolution.

C := 1 + ‘-‘

((A2 · d2 · 1 ·

(1 +A2 · d2

)−1)N+1

)nonlinear := simplify (DotProduct (Jacobian (u, [x, y, z]) , C · u))


((∂

∂xC · p, ∂

∂yC · p, ∂

∂zC · p

))

nonlinear + pressure

0 · ex

Now we consider NS-Alpha with Deconvolution. It is enought to show that

given any N, the nonlinear term is 0.

simplify (CrossProduct (Curl (u) , C · u))

0 · x

Appendix C Kovasznay Flow

C.1 Proof of Lemma 5.1.3

Show that for all N ∈ Z, N ≥ 0, uN = βNu+

1− βN

0


Proof. Let N = 1

u =

1− βeλx cos(2πy)

λβ2π e

λx sin(2πy)

=

β − βeλx cos(2πy) + 1− βλβ2π e

λx sin(2πy)

=

β − βeλx cos(2πy)

λβ2π e

λx sin(2πy)

+

1− β

0

= βu+

1− β

0

For a fixed N assume, uN = βNu+

1− βN

0


uN+1 = F (uN )

= (−α2∆ + I)−1uN

= (−α2∆ + I)−1

βN − βNeλx cos(2πy)− βN + 1

λβN

2π eλx sin(2πy)

By Induction Hyp.

= (−α2∆ + I)−1

βN1− eλx cos(2πy)

λ2π e

λx sin(2πy)

+

−βN + 1

0

= βN (−α2∆ + I)−1

1− eλx cos(2πy)

λ2π e

λx sin(2πy)

+ (−α2∆ + I)−1

−βN + 1

0

= βNF (u) + F

−βN + 1

0

Aside:

We will show that F

−βN + 1

0

=

−βN + 1

0

.

−α2∆

−βN + 1

0

+

−βN + 1

0

=

−βN + 1

0

⇒ (−α2∆ + I)

−βN + 1

0

=

−βN + 1

0

⇒

−βN + 1

0

= (−α2∆ + I)−1

−βN + 1

0

⇒

−βN + 1

0

= F

−βN + 1

0

Thus we can say,

uN+1 = βNF (u) + F

−βN + 1

0

= βNu+

−βN + 1

0

= βN

βu+

1− β

0

+

−βN + 1

0

= βN+1u+

βN − βN+1

0

+

−βN + 1

0

= βN+1u+

1− βN+1

0

C.2 Derivation of Leray with Deconvolution Solution

ut +∇u ·DN (u) +∇p− ν∆u

= ut +∇u ·

u (1− (1− β)N+1)

+

(1− β)N+1

0

+∇p− ν∆u (By Lemma 5.1.5)

= ut +(1− (1− β)N+1

)(∇u · u) +∇u ·

(1− β)N+1

0

+∇p− ν∆u

=(1− (1− β)N+1

)−λeλx cos(2πy) + λe2λx

λ2

2π eλx sin(2πy)

+ (1− β)N+1

−λeλx cos(2πy)

λ2

2π eλx sin(2πy)

+

−λ(1− (1− β)N+1)e2λx

0

− ν

−λ2eλx cos(2πy) + 4π2eλx cos(2πy)

λ3

2π eλx sin(2πy)− 2πλeλx sin(2πy)

=

−λeλx cos(2πy)(1− (1− β)N+1

)+ λe2λx

(1− (1− β)N+1

)+ (1− β)N+1

(−λeλx cos(2πy)

)λ2

2π eλx sin(2πy)

(1− (1− β)N+1

)+ (1− β)N+1

(λ2

2π eλx sin(2πy)

)

+

−λ(1− (1− β)N+1)e2λx

0

− ν−λ2eλx cos(2πy) + 4π2eλx cos(2πy)

λ3


=

−λeλx cos(2πy) +(1− (1− β)N+1

)λe2λx

λ2

2π eλx sin(2πy)

+

−λ(1− (1− β)N+1)e2λx

0

− ν

−λ2eλx cos(2πy) + 4π2eλx cos(2πy)

λ3


=

−λeλx cos(2πy)

λ2

2π eλx sin(2πy)

− ν−λ2eλx cos(2πy) + 4π2eλx cos(2πy)

λ3


=

eλx cos(2πy)(−λ− ν(−λ2 + 4π2)

)eλx sin(2πy)

(λ2

2π − ν(λ3

2π − 2πλ))

= 0 (By Lemma 5.0.2)

C.3 Derivation of NS-α with Deconvolution Solution

We will show that the following solves ut + (∇× u)×DN (u) +∇p− ν∆u = 0.

u1 = 1− eλx cos(2πy)

u2 =λ

2πeλx sin(2πy)

p =λ

8π2ν

(1−

(−α2λRe

1− α2λRe

)N+1)e2λx sin2(2πy)− eλx cos(2πy)

As the given u is the same as in Leray, we know that,

DN (u) = u(1− (1− β)N+1) +

(1− β)N+1

0

.

Therefore,

∇× u =

∣∣∣∣∣∣∣∣∣∣i j k

∂x ∂y ∂z

u1 u2 0

∣∣∣∣∣∣∣∣∣∣= (∂xu2 − ∂yu2) k

=

(λ2

2πeλx sin(2πy)− 2πeλx sin(2πy)

)k = eλx sin(2πy)

(λ2 − 4π2

2π

)k

(∇× u)×DN (u) =

∣∣∣∣∣∣∣∣∣∣i j k

0 0 ∂xu2 − ∂yu2

u1 u2 0

∣∣∣∣∣∣∣∣∣∣= i (−DN (u)2(∂xu2 − ∂yu2))− j (−DN (u)1(∂xu2 − ∂yu2))

=

−DN (u)2 (∂xu2 − ∂yu2)

DN (u)1 (∂xu2 − ∂yu2))

=

−u2(1− (1− β)N+1

) (eλx sin(2πy)

(λ2−4π2

2π

))(u1(1− (1− β)N+1

)+ (1− β)N+1

) (eλx sin(2πy)

(λ2−4π2

2π

))

=

− λ2π e

λx sin(2πy)(1− (1− β)N+1

) (eλx sin(2πy)

(λ2−4π2

2π

))((

1− eλx cos(2πy)) (

1− (1− β)N+1)

+ (1− β)N+1) (eλx sin(2πy)

(λ2−4π2

2π

))

=

− λ2π e

λx sin(2πy)(1− (1− β)N+1

) (eλx sin(2πy)

(λ2−4π2

2π

))(1− (1− β)N+1

)eλx sin(2πy)

(λ2−4π2

2π

)−(1− (1− β)N+1

)e2λx cos(2πy) sin(2πy)

(λ2−4π2

2π

)

+

0

(1− β)N+1eλx sin(2πy)(λ2−4π2

2π

)

=

− λ2π e

λx sin(2πy)(1− (1− β)N+1

) (eλx sin(2πy)

(λ2−4π2

2π

))eλx sin(2πy)

(λ2−4π2

2π

)−(1− (1− β)N+1


(λ2−4π2

2π

)

=

− λ2π e

2λx sin2(2πy)(1− (1− β)N+1

) (λ2−4π2

2π

)eλx sin(2πy)

(λ2−4π2

2π

)−(1− (1− β)N+1


(λ2−4π2

2π

)

=

− λ2π e

2λx sin2(2πy)(1− (1− β)N+1

) (λ

2πν

)eλx sin(2πy)

(λ

2πν

)−(1− (1− β)N+1


(λ

2πν

) (By Lemma 5.0.2)

=

− λ2

4π2ν e2λx sin2(2πy)

(1− (1− β)N+1

)(λ

2πν

)eλx sin(2πy)−

(λ

2πν

) (1− (1− β)N+1


=

− λ2


(1−


)N+1)

(λ

2πν

)eλx sin(2πy)−

(λ

2πν

)(1−


)N+1)e2λx cos(2πy) sin(2πy)

−ν∆u =

λeλx cos(2πy)

−λ2

2π eλx sin(2πy)

∇p =

λ2

4π2ν

(1−


)N+1)e2λx sin2(2πy)− λeλx cos(2πy)

λ2πν

(1−


)N+1)e2λx sin(2πy) cos(2πy) + 2πeλx sin(2πy)

ut + (∇× u)×DN (u) +∇p− ν∆u

=

− λ2


(1−


)N+1)

(λ

2πν

)eλx sin(2πy)−

(λ

2πν

)(1−


)N+1)e2λx cos(2πy) sin(2πy)

+

λ2

4π2ν

(1−


)N+1)e2λx sin2(2πy)− λeλx cos(2πy)

λ2πν

(1−


)N+1)e2λx sin(2πy) cos(2πy) + 2πeλx sin(2πy)

+

λeλx cos(2πy)

−λ2

2π eλx sin(2πy)

=

0(λ

2πν

)eλx sin(2πy) + 2πeλx sin(2πy) + −λ2

2π eλx sin(2πy)

=

0

−eλx sin(2πy)2πν

(−λ− ν(−λ2 + 4π2)

)

=

0

−eλx sin(2πy)2πν (0)

(By Lemma 5.0.2)

=

0

0

= 0

C.4 Maple Verification

This u (velocity vector) will be the same for all four models.

k := 1/2R−√

1/4R2 + 4π2

u1 := 1− cos (2π y) ekx

u2 := 1/2k sin (2π y) ekx

π

u := simplify (VectorField ( (u1 , u2 , 0) , cartesianx,y,z))

The pressure term for Leray:

p := simplify

(p0 − e2 kx

2− 2 a2kR

)Here we verify that ubar satisfies the filter equation:

ubar1 := 1− cos (2π y) ekx

1− a2kR

ubar2 := 1/2k sin (2π y) ekx

π (1− a2kR)

ubar := simplify (VectorField ( (ubar1 , ubar2 , 0) , cartesianx,y,z))

simplify(−a2Laplacian (ubar) + ubar − u

)0

Now we show that the terms satisfy Leray:

upartial :=

(d

dtu1 ,

d

dtu2 , 0

)nudeltau := simplify

(− (Laplacian (u1 , [x, y]) ,Laplacian (u2 , [x, y]) , 0)

R

)nonlinear := simplify (DotProduct (Jacobian (u, [x, y, z]) , ubar))


((d

dxp,

d

dyp, 0

))

simplify (nonlinear + pressure + nudeltau)

0

Now we test the Deconvolution operator:

r := 2

ubar := br+1u+ VectorField((

(1− b)r+1, 0, 0

), cartesianx,y,z

)b := simplify

((1− a2kR

)−1)

decon := u(

1− (1− b)r+1)

+ VectorField((

(1− b)r+1, 0, 0

), cartesianx,y,z

)simplify

(u+ (−1)

ra2 r+2Laplacian (Laplacian (Laplacian (ubar))) − decon

)0

r := 3

ubar := br+1u+ VectorField((

(1− b)r+1, 0, 0

), cartesianx,y,z

)b := simplify

((1− a2kR

)−1)

decon := u(

1− (1− b)r+1)

+ VectorField((

(1− b)r+1, 0, 0

), cartesianx,y,z

)simplify

(u+ (−1)

ra2 r+2Laplacian (Laplacian (Laplacian (Laplacian (ubar)))) − decon

)0

Now we show the Deconvolution solution holds:

q := p0 − 1/2(

1− (1− b)N+1)

e2 kx

pressure :=

(d

dxq,

d

dyq, 0

)nudeltau := simplify

(− (Laplacian (u1 , [x, y]) ,Laplacian (u2 , [x, y]) , 0)

R

)mydecon := u

(1− (1− b)N+1

)+ VectorField

(((1− b)N+1

, 0, 0), cartesianx,y,z

)nonlinear := simplify (DotProduct (Jacobian (u, [x, y, z]) ,mydecon))

simplify (nonlinear + pressure + nudeltau)

0

Now we check NS-Alpha

p := 1/8ke2 kx (sin (2π y))

2

π2 (R−1 − a2k)− ekx cos (2π y)

pressure := VectorField

((d

dxp,

d

dyp, 0

), cartesianx,y,z

)ubar := simplify (VectorField ( (ubar1 , ubar2 , 0) , cartesianx,y,z))

nonlinear := CrossProduct (Curl (u) , ubar)

nudeltau := simplify

(−VectorField ( (Laplacian (u1 , [x, y]) ,Laplacian (u2 , [x, y]) , 0) , cartesianx,y,z)

R

)simplify (nudeltau + pressure + nonlinear)

0

Now we check NS-Alpha with Deconvolution

q := 1/8 k

(1−

(− a2kR

1− a2kR

)N+1)

e2 kx (sin (2π y))2Rπ−2 − ekx cos (2π y)

pressure := VectorField

((d

dxq,

d

dyq, 0

), cartesianx,y,z

)mydecon := u

(1− (1− b)N+1

)+ VectorField

(((1− b)N+1

, 0, 0), cartesianx,y,z

)nonlinear := CrossProduct (Curl (u) ,mydecon)

nudeltau := simplify

(−VectorField ( (Laplacian (u1 , [x, y]) ,Laplacian (u2 , [x, y]) , 0) , cartesianx,y,z)

R

)simplify (nudeltau + pressure + nonlinear)

0

Bibliography

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