# sghiar Navier Stokes .pdf

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International Journal of Engineering and Advanced Technology (IJEAT)
ISSN: 2249 – 8958, Volume-6, Issue-1, October 2016

Turbulent Functions and Solving the Navier-Stokes
Equation by Fourier series
M. Sghiar
And:

Abstract: I give a resolution of the Navier-Stokes  equation
by using the series of Fourier. Résumé: Je donne une résolution
de l'équation de Navier-Stokes  par les séries de Fourier.

∇=

1

Keywords: Navier-Stokes, Fourier, Séries de Fourier.

I.

(u . ∇u )= ∑ ui
1
n

∇2 u= ∑

first step to understanding the elusive phenomenon of
turbulence, the Clay Mathematics Institute in May 2000
made this problem  one of its seven Millennium Prize
Navier-Stokes equation has a solutions and I will give
techniques to resolve this beautiful equation. The NavierStokes equation, established in the nineteenth century by the
French Navier and the British Stokes. It is an equation that
describes the velocity field of a fluid. More specifically, it is
a differential equation whose velocity field is unknown.
The Navier-Stokes equation is also used to predict the
weather, the oceans simulate, optimize aircraft wings ...
Knowing that a link between the Boltzmann equation and
the Navier-Stokes equation was established, by studying
the latter problem, I found that for to solve it we can reduce
the problem of the heat-equation which is known can be
solved by several methods : one of the first methods of
solving the heat-equation was proposed by Joseph Fourier in
his treatise analytical Theory of heat  in 1822. After
giving a specific solution to the Navier-Stokes equation, I
will demonstrate how to find all solutions of this equation if
they exist, and I give the necessary and sufficient conditions
for their existence. It will be seen in a remark that if the
turbulence function is negligible, then the fluid will tend to
behave like an ideal gas.

1
n

div u= ∑
1

∂u
∂ xi

∂2 u
∂ x 2i
∂u
∂ xi

In the following, by dividing by ρ , the Navier-Stokes
equation is of the form:

∂u
2
+ (u . ∇u )= α ∇p + β ∇ u
∂t
div u= 0
III.

EXISTENCE OF THE SOLUTIONS FOR THE
NAVIER-STOKES EQUATION:

On each axis i, try to find the solutions of the form:

∂u i
∂u
∂ pi
∂2 u
+ ui i = α
+ β 2i
∂t
∂ xi
∂ xi
∂ xi

( )

This is equivalent to:

∂u i
+
∂t

(12 u −α p )= β ∂ u
2
i

2

i

∂ xi

∂x

i
2
i

If ui is a solution of the equation:

∂u i
∂2 ui

∂t
∂ x 2i

RECALL, NOTATIONS AND DEFINITIONS

Such solutions ui exist because the equation is analogous to
the heat-equation which is resolvable by the Fourier series
.

Here are the Navier-Stokes equation:

ρ

n

n

INTRODUCTION

The Navier–Stokes equations is considered to be the

II.

(∂∂x , … , ∂∂x )

(∂u∂ t +(u . ∇u ))= − ∇p+µ ∇ u
2

1

2

If pi is such 2 ui − α pi= f i (t ) , then

div u= 0
Where u is the velocity field, p is the pressure, the density
of the fluid, and µ its viscosity.

(12 u − α p )= 0
2
i

i

, and the equation is solved.

∂ xi

We do the same for all axes i until i= n −1 .
For the axe i= n :
n −1

Let be un= − ∑ ui .
1

Revised Version Manuscript Received on October 19, 2016.
M. Sghiar, 9, Allée Capitaine J. B. Bossu, 21240, Talant, France.

79

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Turbulent Functions and Solving the Navier-Stokes Equation by Fourier series
n

∂2 u n
∂u n
pn is such
=
β
We have: ∂ t
∂ x2i , and if
1 2
∂ un −α pn
1 2
un − α pn = f n (t ) , then
2
2
= 0 , and the
∂ xn

(

2
And consequently ∑ ui − α p i= g (t ) because div u= 0 .
1

We deduce therefore that:

)

2

ui − α pi= hi (t ), ∀i ∈{1, … , n } because:

∀i ∈{1, … , n }, u2i − α pi= l i (x i , t ) , and we must have
∂ (u2i − α pi )
= 0 ∀j ∈{1, … , n } .
∂ xj

equation is solved for the axe n.
It is clear that if e i is the vector for

the axes i, then

n

u= ∑ u i e i
1

is one solution of
div
u= 0 .
equation if

the

Naviers-Stokes

V.

Theorem:
The Navier-stokes equation have a solution, Moreover, any
2
solution (u, p) must check: ui − α pi= f i (t ) and

Else, to have div u= 0 , we take ui of the form:

n
∂u i
∂2 ui

i
{
1,

,
n
}
u=
ui ei and

. where
∂t
∂ x 2i ,
1

n

( ), ∀i ∈{1, … , n− 1 }

βt+

ui = e

∑1

xi

So we have solutions of the Navier-Stokes equation.
IV.

n

p= ∑ pi e i .

NECESSARY CONDITIONS

1

Conversely any pair (n, p) satisfying these conditions with
div u= 0 is solution of the Navier-Stokes equation.

Any solution (u , p ) of the Navier-Stokes equation verifies
that:
2
ui − α pi= f i (t ), ∀i ∈{1, … , n }
Indeed:

REMARKS
1- We note in the above equations the dependence between
pressure, density, speed vector fields and viscosity
2
2- By dividing by α in the equation ui − α pi= f i (t ) we

If (u , p ) is a solution of the Navier-Stokes equation , we
must have :

∂u i
+
∂t

(12 u −α p )= β ∂ u
2
i

2

2

i

∂ xi

∂x

deduce

i
2
i

∂ ui

=
∂t

∂u i
1 2
ui −α pi − β
2
∂ xi

(

²

And the equation m u i − Vpi= mf i (t ) is linking energy,
mass, pressure, temperature, volume and time ... This may
not be surprising since a link between the Boltzmann
equation and the Navier-Stokes has been established.
²
When f i (t ) tends to 0, we will have m u i ≃ Vp i , there is

∂ xi

∂ ui
∂ ui
1 2
= ∂ ui − α pi − β
∂t
2
∂ xi

)

therefore a tendency towards the law of an ideal gas, and the
function f i (t ) can be regarded as a turbulent function.

When the fluid flows in one direction, then the space-time
flows in the opposite direction with the same speed value:
We deduce that:

−ui ∂ ui= ∂

∂ ui
1 2
ui − α pi − β
2
∂ xi

(

REFERENCES

)

1. Joseph Fourier, Théorie analytique de la chaleur, Firmin Didot Père et
Fils (Paris-1822). Réédition Jacques Gabay, 1988 (ISBN 2-87647-0462)
2. http://www.claymath.org/sites/default/files/navierstokes.pdf

Therefore:

(

∂ ui
∂ xi

2

0= ∂ ui −α p i − β

1
= ρ is the
α

m

)

(

ρ ui − p i= ρ f i (t ) where

²
Let's ρ= V , we will have : m u i − Vpi= mf i (t ) .

And:

− ∂ xi

that:

density of the fluid.

Therefore:

CONCLUSION

)

And:
2

ui − α pi − β

∂ui
= f (t )
∂ xi i

So:
n

∂u

∑ u2i − α p i − β ∂ x i = f (t )
1

i

80

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