Existence, Uniqueness and Smoothness of a Solution of the Problem
Navier-Stokes for the Incompressible Fluid with Viscosity
Taalaibek D.Omurov
Doctor of Physics and Mathematics, professor of Z. Balasagyn Kyrgyz National University,
Bishkek, Kyrgyzstan, E-mail: [email protected]
Abstract. Existence, uniqueness and smoothness (or conditional-smoothness) of а solution of the
Navier-Stokes equation is one of the most important problems in mathematics. The Problem is well-known
as "Navier-Stokes Millennium Problem". In this paper we offer the analytical methods of solutions of the
Navier-Stokes equations for viscous incompressible fluid with Cauchy condition. Therefore in this work
are offered methods of analytical solutions of equations Navier-Stokes for fluid with viscosity, which
meet the requirements of the specified problem of a millenium [1].
Key words: 6th millennium problem, Navier-Stokes equations, viscous fluid, incompressible fluid,
uniqueness and smoothness of solution, Beale-Kato-Majda, Reynolds number
Preface
The research is devoted to the development of a method for solving 3D and nD Navier-Stokes
equations that describe the flow of a viscous incompressible fluid. The study includes a requirements
"Navier-Stokes Millennium Problem", as developed method of solution contains a proof of the
existence and smoothness (conditional smoothness) of solutions of the Navier-Stokes equations, where
laminar flow is separated from the turbulent flow when the critical Reynolds number: Re = 2300. The
decision is obtained for the velocity and pressure in an analytical form, as required by the
"Navier-Stokes problem Millennium". The method of solution is supported by examples for different
viscosity ranges corresponding applications.
In sections 4.3, 4.4, 7.3 and paragraphs 5, 6 new law of the pressure distribution has been found.
This law is derived from the equation of Poisson type and differs from the known laws of Bernoulli,
Darcy at all. Most importantly, the author has opened a special space for the study of the existence and
smoothness (including conditional smoothness) equations Navier-Stokes for viscous incompressible
fluid. For brevity, these spaces can be called Omurov's spaces with different metrics:
a) in the case of smoothness a space with the norms of Chebyshev type has been obtained,
b) the weighted space arises in the case of conditional-smoothness, for example the weighted
space of Sobolev type: Wn2 3; ( D0 ) .
K. Jumaliev, Academician, Director of the Institute of Physics NAS Kyrgyz Republic
August 1, 2014
1. Introduction
The difficulty of study of the 3D and nD Navier-Stokes equations is caused by nonlinear nature of the
equation and by the necessity to find the speed and pressure for any values of parameter viscosity [1].
If to designate components of vectors of speed and external force, as
v( x ,t )  [1( x ,t ),2 ( x ,t ),3 ( x ,t )], f ( x ,t )  [f1 ( x ,t ), f 2 ( x ,t ), f 3 ( x ,t )],
that corresponding problem Navier-Stokes is represented in a kind
3
it   jix  fi 
j 1
j
1

Pxi  i ,( i  1,3 ),
(1.1)
div  0, ( x, t )  T  R 3  [0, T0 ],
(1.2)
i
(1.3)
t 0
 i0 ( x1 ,x2 ,x3 ), ( x1 ,x2 ,x3 )  R 3 ,
  0 is kinematic viscosity,  is density,  is Laplace operator. Here, the condition incompressibility
(1.2) fluid it's a the additional equation. Unknown are speed  and pressure P.
To answer this question, in this article the following way proposed to solve for the Navier-Stokes
equations. For this purpose (1.1) we will transform to a kind [7, 10]:
1
1
Pxi  Qxi  i ,( i  1,3 ),

2
it  i  fi 
3
1
2
(1.4)
i  (  jix  Qx ),
j 1
j
(1.5)
i
where
i t 0  i0 ( x1 ,x2 ,x3 ),( x1 ,x2 ,x3 )  R 3 ,

3
3
3

2
0
Q(
x
,x
,x
,t
)


(
x
,x
,x
,t
);
Q

2


;
Q

2
 j0 j0 xi ,( i  1,3 ),



1
2
3
i
1
2
3
x
j
jx
x

i
i
i
i 1
j 1
j 1

without breaking equivalence of system (1.1) and (1.4), (1.5). The received systems (1.4), (1.5) contain
unknown functions
i , i and pressure P. Here  i0 – known functions because are known  j0 , j0 x .
i
The developed method of the decision of systems (1.4) and (1.5) connected with functions
i ,( i  1,3 ) , i.e.
A1) rot   0, [   (  1, 2, 3 );  i  x i, i 1,3 ] ;
rot  0, or
А2) div   0, rot   0, or
А3) i ,(i  1,3) is any functions if, accordingly, as necessary conditions, take place:
а01) rot  0  0, 0  (  10 , 20 , 30 ), а02) div 0  0, а03)  0 is any functions.
To study of many problems of theoretical and mathematical physics lead us to find solutions in various
special weight spaces. In the papers [7, 9] have been offered the methods which give solution of 3D
problem Navier-Stokes in Gn2 3; ( D0 ),
T0
1
3
3

2
k
2
  Gn23; ( D0 )   i G2 ( D0 ) {  D i C( T )  [sup  ( t ) it ( x1 ,x2 ,x3 ,t ) dt ] },
R3 0
i 1
i 1 0  k  3


T0
1

3
0
(1.61)
 D0  R  ( 0,T0 ); 0  ( t ) :  ( t ) dt  q0  const ; k  0 : D i  i ,
t
0

k

3
 i
k  0 : D ki 
;
k

 i , (  i  0,3; i0  C 3 ( R 3 ); i  1,3 ).

3
1
2

x1 x2 x3
i 1

Note, by lemma K. Friedrichs [15], the suitable solutions of Navier-Stokes problem in space
Wn2 3; ( D0 ) were constructed in [8]. Here the norm of this space is defined by the following

3
Wn23; (
D0 )
  i
i 1
3
2
W
 {

T0
T0
1
sup  [D i ( x1 ,x2 ,x3 ,t )] dt  sup  ( t ) it ( x1 ,x2 ,x3 ,t ) dt}2 .
k
3
i 1 0  k  3 R
2
2
R3
0
(1.62)
0
The work purpose. The main aim of this work is to prove the existence, uniqueness and smoothness
solution of the 3D Navier-Stokes problem in the space Cn3,13 (T ) for a viscous incompressible fluid, where
the norm of this space is defined by the following

3
Cn3,13
  i
i 1
3
C
3 ,1
{

i 1 0  k  3
D ki
C
3,3,3,1
 it C }, [Cn3,1 3 (T )  Cn3,3,3,1
 3 (T )  Cn  3 (T )].
(1.63)
But in the case of a conditional smoothness the solution we will consider in the space Gn1 3 ( D0 ) :

3
Gn13 ( D0 )
  i
i 1
3
G1 ( D0 )
 {

i 1 0  k  3
D i
T0
k
C( T )
 sup  it ( x1 ,x2 ,x3 ,t )dt}.
R3
(1.64)
0
Since i0  C 3 ( R3 ) , then the solution of 3D Navier-Stokes problem can be found also in the
weight space of Sobolev’s type Wn2 3; ( D0 ).
It is known that from uniform convergence of sequence continuous functions on [a, b] is followed
by its convergence on the average on [a, b]. Therefore, so as norm: 

Cn3,13
Wn23;
it is subordinated to norm
, that is natural describe an analytical solution in Cn3,13 (T ) . Hence, gives also feasibility to
construct the decision in Wn23; ( D0 ) , the converse is not true.
The scientific value. The offered analytical methods of solution 3D Navier-Stokes problem transforms
this problem to inhomogeneous linear equations of heat conductivity type under Cauchy condition with
enough smooth initial conditions, without attraction of any additional conditions. The analytical solutions
of transformed equations are regular with respect to viscosity factor   0 and simplify analysis of the
initial problem in mathematical and physical sense [6, 11 and 12].
The important addition to study of Navier-Stokes problem is the research of fluid with very small
viscosity (Reynolds number is great: Re ≥ 2300). In this case the obtained results of paper give the
analytical solution of Navier-Stokes problem and allow us to get full understanding of turbulence [4,12].
In the paper, fluid with average viscosity are also studied, in which all inertia members are preserved.
Note, that in the above fluids, if convective acceleration is not equal to zero then there are problems to
integrate the equations of Navier-Stokes in general [12].
Further we will show that at certain mathematical transformations nonlinear problem Navier-Stokes
will be transformed to linear a problem of conductivity of high temperature [13], when
v  R n ,( x  R n ;t  [0,T0 ] ) .
The obtained results allow us to solve nD Navier-Stokes problem for incompressible fluid with any
viscosity factor and our problem does not include a derivation of an equation in a physical meaning, since
there is a big amount of works reflecting these questions [3-6, 11 and 12].
2. Fluid with Very Small Viscosity by Condition (A1)
In this paragraph and in the subsequent points with the specified restrictions at the entrance data,
the strict substantiation of compatibility of systems (1.4), (1.5) will be given with very small viscosity
0    1 (for large Reynolds numbers). Note that the solution of Navier-Stokes equation has the
property that the flow can be divided in two parties. In the inner part there is a thin layer in which
friction forces are appeared. In the outer part the flow does not depend on friction forces, that is, it is
free from rotation of the particles, and hence, it is described by Euler equation [12]. In the paper [9],
the behaviour of the solution on the 3D Navier-Stokes equation in CN2  3, ( D0 ), where the viscosity
aspires to a zero has been studied.
Here we Let study the behaviour of the solution on the 3D Navier-Stokes equation in where the
viscosity aspires to a zero.
2.1. Fluid with the Condition (A1)
Let functions i0 ,( i  1,3 ) satisfy to a condition (a01). Then relatively i ,( i  1,3 ) we suppose a
condition (A1) and
d i fv  0 ,
(2.1)
where from system (1.4) and (1.5), accordingly we will receive following systems
1
2
it   x  Qx  fi 
i
i
1

Pxi  i ,( i  1,3 ),
(2.2)
3
1
2
i   x :  x  (  jix  Qx ),( i  1,3 ).
i
i
j 1
j
(2.3)
i
Theorem 1. Let conditions (1.2), (1.3), (A1) and (2.1) are satisfied. Then systems (2.2) and (2.3)
equivalent will be transformed to a kind
3
1
1


J


F
,
[
F
(
x
,x
,x
,t
)


f ixi ; J( x1 ,x2 ,x3 ,t )  P  Q   ],

0
0
1 2
3


2
i 1

it  fi  i  J x ,( i  1,3 ),
i

3

0
0




,
[

(
x
,x
,x
,t
)


 ixi ],

1
2
3

i 1

3
1
1
1
ds1ds2 ds3
F
(
s
,s
,s
,t
)
,(
r

( xi  si )2 ).
 P  Q   

0
1 2 3


2
4

r
i 1

R3
(2.4)
Thereby the problem (1.1) - (1.3) has the only solution which satisfies to a condition (1.2).
Proof of the theorem 1 consists of four stages.
1) For this we take the partial derivations of the system (2.2) at xi ,( i  1,2,3 ), that is, the
1-equation (2.2, i=1) is differentiated on x1 , 2-equation on x2 (2.2, i=2), 3-equation on x3 (2.2,
i=3) and summarizing the obtained equations, and by based on the formula:
3
 3 
 3
2
(2.2)
;
div


0
:
(

)

0;

( 1x1  2 x2  3 x3 )  0,
 ix


2
t i1 i
 i1 xi
i 1 xi
 3
3
 1
1
1
1

[
P



Q
]


[
P



Q
]
;
F


f ixi ,

xi
xi
x
0

2 i

2
i 1 xi 
i 1
from here we will receive the equation of Poisson [13]: [
(2.5)
1
1
P    Q]   F0 , i.e.

2
 J   F0 ,

 J  1 F0 ( s1 ,s2 ,s3 ;t ) ds1ds2 ds3 ,

4 R3
r

1
i
Jx 
F0 ( x1   1 ,x2   2 ,x3   3 ;t )
d 1d 2 d 3 ,

i
4 R3

(  12   2 2   32 )3

 si  xi   i ,( i  1,3 ),
(2.6)
so as
1
1
P   Q  J.

2
(2.7)
The algorithm for getting the Poisson equation (2.6) we call "algorithm poissonization system",
shortly APS [8]. Therefore, if J – the decision of the equation (2.6), then substituting
1
1
Pxi  Qxi   xi  J xi ,( i  1,3 ),

2
(2.8)
in (2.2), we have
it   i  i ,( i  1,3 ),

3
(2.9)


(
x
,x
,x
,t
)

f

J
,(
i

1,3
);


F


J

0,

(
x
,x
,x
,t
)

T
,

 i 1 2 3
i
xi
ixi
0
1
2
3
i 1

i.e. system (2.2) it is equivalent by (2.9). This means that the system (2.2) is converted in inhomogeneous
linear equations of heat conduction. Here the equations (2.6), (2.9) is there are first and second equations
of system (2.4).
2) From the received results follows that the system (1.1) is transformed in the linear equations of
heat conductivity with a condition of Cauchy. Consequently, Cauchy problem with sufficiently smooth
initial data t  0 in the class of bounded functions is solvable [13, 14]. Accordingly, the problem of
the Navier-Stokes equations has a unique, conditional smooth solution in the space Gn1 3 ( D0 ).
Really from system (2.9), follows
r2
1
i 
exp( 
)i0 ( s1 ,s2 ,s3 )ds1ds2 ds3 
3 
4 t
8( t ) R3
8 3
1
 i ( s1 ,s2 ,s3 ,s )ds1ds2 ds3ds 
2 3 t )d 1d 2 d 3 

3
 exp( ( 
3
  exp( ( 
0 R
2
1
r2
1
0 3 exp(  4 ( t  s ) ) ( ( t  s ))3 
R
  22   32 ))i0 ( x1  2 1 t ,x2  2 2 t ,x3 
R3
t
1

1
t
2
1
  22   32 )) i ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),
(2.10)
3
x3  2 3  ( t  s );s )d 1d 2 d 3ds  H i ( x1 ,x2 ,x3 ,t ),
si  xi  2 i t ;si  xi  2 i ( t  s ),( i  1,3 ),
where H i is known functions. The found decision (2.10) satisfies system (2.9). Really, considering
partial derivative systems (2.10):
( 0,1 )   ; h j  x j  2 j t ; l j  x j  2 j ( t  s ), ( j  1,3 ),( x1 ,x1 ,x1 ,t )  T :


1
exp( (  12   22   32 ))i0h j ( x1  2 1  t ,x2  2 2  t ,x3  2 3 t )d 1d 2 d 3 
ix j 
3 
 R3

t
 1
exp( (  12   22   32 )) il j ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 

3  
  0 R3

 ( t  s );s )d 1d 2 d 3ds,

1
ix2 
exp( (  12   22   32 ))i0h2 ( x1  2 1 t ,x2  2 2 t ,x3  2 3 t )d 1d 2 d 3 +
3 
j
j

 R3
 1 t
exp( (  12   22   32 )) il 2 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3  2 3 

3  
j
  0 R3

  ( t  s );s )d 1d 2 d 3 ds,
( x ,x ,x )  R 3 ; t  ( 0,T ] :
1
2
3
0


3

1

2
2
2


exp
(

(





))

(
 j i0 h j ( x1  2 1 t ,x2  2 2 t ,x3 

1
2
3
 it

t
j 1
 3 R3

t
3

j
1
2
2
2
2 3 t )d 1d 2 d 3   i 
exp
(

(





))

 il ( x1 

1
2
3
3  
ts j

j 1
 0 R3


2 1  ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s )d 1d 2 d 3ds, ( i  1,3 )
(2.11)
and substituting (2.11) in (2.10), we have
i t  0  i0 ( x1 ,x2 ,x3 ), ( x1 ,x2 ,x3 )  R 3 ; ( 0,1 )   ; ( x1 ,x2 ,x3 )  R 3 ; t  ( 0,T0 ] :

3

1

2
2
2
0







exp
(

(





))

(
 j i0 h j ( x1  2 1 t ,x2  2 2 t ,x3 

it
i
i
1
2
3


t
j 1
 3 R3

t
3

j
1
2
2
2
2 3 t ))d 1d 2 d 3   i 
exp
(

(





)
)(

 il ( x1  2 1 

1
2
3

ts j

j 1
 3 0 R3

  ( t  s ),x2  2 2  ( t  s ),x3  2 3 ( t  s );s ))d 1d 2 d 3ds   i 

  { 1
exp( (  12   22   32 ))i0 ( x1  2 1 t ,x2  2 2 t ,x3  2 3 t )d 1d 2 d 3 

3

 R3

t
 1

exp( (  12   22   32 )) i ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 

3  
  0 R3
3


1
2
2
2
  ( t  s );s )d 1d 2 d 3 ds} 
exp
(

(





))(
 j i0 h j ( x1  2 1 t ,x2 

1
2
3

t

j 1
 3 R3

t
3
j
2
2
2
2 t ,x  2 t ))d d d  1
exp
(

(





))(

 il ( x1 

2
3
3
1
2
3
1
2
3



ts j
j 1
 3 0 R3

2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 ( t  s );s ))d 1d 2 d 3 ds 

1
1
 1
2
2
2
 2  { 3  exp( (  1   2   3 )) t i0 h12 ( x1  2 1 t ,x2  2 2 t ,x3  2 3 t ) 
 R3


1
1
exp( (  12   22   32 )) i0 h2 ( x1  2 1 t ,x2  2 2 t ,x3 
d( x1  t )d 2 d 3 
3 
2
t

 R3

1
2 t )d d( x  2 t )d d  1
exp( (  12   22   32 )) i0 h2 ( x1  2 1 t ,
3
1
2
2
2
3
3 
3

t
 R3

t

1
1
x

2


t
,x

2


t
)d

d

d(
x

2


t
)

[  exp( (  12   22   32 )) 
 2
2
3
3
1
2
3
3

3
 0 t  s R3

 ( x  2  ( t  s ),x  2  ( t  s ),x  2 ( t  s );s )d( x  2 ( t  s ) ) 
1
2
2
3
3
1
1
 il12 1
d 2 d 3  exp( (  12   22   32 )) 2 ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 

il2

R3

2
2
2
  ( t  s );s )d 1d( x2  2 2 ( t  s ) )d 3   exp( (  1   2   3 )) il32 ( x1  2 1 ( t  s ),

R3

 x2  2 2  ( t  s ),x3  2 3 ( t  s );s )d 1d 2 d( x3  2 3 ( t  s ) )]ds} 

3

2
2
2
 1
exp
(

(





))(
 j i0 h j ( x1  2 1 t ,x2  2 2 t ,x3  2 3 t )) 

1
2
3

3

t
j 1
 R3

t
3

j
2
2
2
d 1d 2 d 3  1
exp
(

(





))(

 il ( x1  2 1 ( t  s ),x2  2 2 

1
2
3


ts j
j 1
 3 0 R3

3
j
1

2
2
2


(
t

s
),x

2


(
t

s
)
;
s
))d

d

d

ds

exp
(

(





))(



3
3
1
2
3
1
2
3


3
t
3
j

1

R

t

1
exp( (  12   22   32 )) 
i0 h j ( x1  2 1 t ,x2  2 2 t ,x3  2 3 t ))d 1d 2 d 3 
3  
 0 R3

 3
(    j  ( x  2  ( t  s ),x  2  ( t  s ),x  2 ( t  s );s ))d d d ds}  0,
il
1
1
2
2
3
3
1
2
3
(*)
 j  1
ts j
on the right side integrals of the formula (*) the integration method in parts is used. That it was
required to prove.
Further we will show that (2.10) satisfies (1.2). For this purpose considering partial derivatives
of 1st order and summarizing, with taking into account (1.2), we have
3
3
1


2
2
2
0


(
x
,x
,x
,t
)

exp
(

(





))
i0 ( x1  2 1 t ,x2  2 2 t ,x3  2 3 


xi
1
2
3
1
2
3


3

x
3
i 1
i

1

i
R

t
3

1

2
2
2
 t )d 1d 2 d 3 
exp[

(





)
]
 i ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3 

1
2
3
3  
i  1 xi
 0 R3


2 3  ( t  s );s )d 1d 2 d 3 ds  0,
3
 3 

 i0  0;   i  F0   J  0.
 i 1 xi
i  1 xi
The system (2.10) satisfies to the equation (1.2).
The limiting case in Gn1 3 ( D0 ) , when the decision of system (1.1) is representing in the form of
(2.10) with conditions (1.2), (1.3), (A 1), (2.1) and

( x1 ,x2 ,x3 ,t )  T ; f i ;i0 : sup D ki0  1 ; sup D k i   1 ,( i  1,3;k  0,3 ),
T

R3

t
1

exp( (  12   22   32 )) D k  i ( l1 ,l2 ,l3 ;s ) d 1d 2 d 3ds   1T0   2 ,
sup


3
 T  0 R3

t
3
1
2
2
2
sup 1
exp
(

(





))
  j   il j ( l1 ,l2 ,l3 ;s ) d 1d 2 d 3ds  3 1 2T0  3 ,
1
2
3
 T  3 0 R3
t  s j 1
(2.12)
 T0
1
exp( (  12   22   32 ) )d 1d 2 d 3  1,
sup   i ( x1 ,x2 ,x3 ,s )ds   1T0   2 , ( i  1,3 );
3 
3
 R3
R 0

3
1
1
2
2
2
sup
exp
(

(





))(
 j  i0 l j ( l1 ,l2 ,l3 ) )d 1d 2 d 3  1


1
2
3
 R3  3 3
i 1
3
R

1
1
 3
1
{ (   i2 exp( (  12   22   32 ) )d 1d 2 d 3 ) 2 (  exp( (  12   22   32 ) )d 1d 2 d 3 ) 2 }  3 1
,
2
3
 i  1 R3
R

i ; 0   ( 3 2 T0  1  T0  ).
li  xi  2 i ( t  s ); li  xi  2 i t , ( i  1,3 );   max
1 i  3
Really, estimating (2.10) in Gn1 3 ( D0 ) , we have
3

v

 Gn13 ( D0 )  [ i C 3 ,0 ( T )  it L1 ]  3[N1  0 ]  M*,
i 1

 i 3 ,0   D ki
 N1  40  , ( i C( T )  2  ; i  1,3 ),
C( T )
 C ( T ) 0 k 3

T0



sup
 it L1
 it ( x1 ,x2 ,x3 ,t )dt   ( 3 2 T0  1  T0  )  0 ,( i  1,3 ),
R3 0


1
sup exp( (  12   22   32 )) i0 ( x1  2 1 t ,x2  2 2 t ,x3  2 3 t ) 
 i 
3 T 

R3

t

1
d 1d 2 d 3 
sup exp( (  12   22   32 ))  i ( x1  2 1  ( t  s ),x2  2 2 
3 T  


0 R3

  ( t  s ),x3  2 3 ( t  s );s ) d 1d 2 d 3 ds  1   2  2  ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).
The uniqueness of the solution i  C 3,0 ( T ) the system (2.10) is obvious on the basis of proof by
contradiction [13]. Results (2.10) with a condition ((A1), (2.1)) are received where smoothness of
functions is required only on xi as the derivative of 1st order is in time has t>0.
Remark 1. Alternatively, we can consider, e.g., a class of suitable solutions constructed in
Wn2 3; ( D0 ). Let the solution of system (1.1) is representing in the form of (2.10) with conditions (1.2),
(1.3), (A1), (2.12) and
( x1 ,x2 ,x3 ,t )  T : sup D k i   1 , ( i  1,3;k  0,3 ),
T

T0
T0

1
2
1

2
( s )  i ( x1 ,x2 ,x3 ,s ) ds )   1 q1   4 ; 0  ( t ) :  ( t ) dt  q0 ;
( sup

3
t
0
 R 0
   max(  ,  );    ( 3  q  1   q ),
4
0
*
0
1
 *
T0
 ( t )dt  q ,
1
0
that solution (2.10) of problem Navier-Stokes (1.1) - (1.3) belongs in Wn2 3; ( D0 ) .
(2.13)
Really, estimating (2.10) in Wn2 3; ( D0 ) , we have [8]:

3



 3[N1 T0  0 ]  3M*, ( D0  R 3  ( 0,T0 ),  ( 1 ,2 ,3 )),
2
 W  i W 2
( i , )
i 1


 N1 T0  0  M*, ( N1  40  ; i  1,3 ),
 i W 2
( i , )

1

T0
  2  ( sup ( t )  ( x ,x ,x ,t ) 2 dt ) 2   ( 3  q  1   q )   ,( i  1,3 ).
it 1 2 3
*
0
1
0

 it L
R3 0

Let's notice that in work [8] similar results in case of (2.10) also are received in the weight space
Gn23; ( D0 ) .
3) The essence of this subparagraph to define the solution (2.9) in Cn3,1 3 (T ). For this purpose of
problem (2.9), (1.3) it is possible to solve differently if conditions are satisfied:
i0  0; i0  C 3 ( R 3 ); i : sup D ki0  1 ; sup D k i ( x1 ,x2 ,x3 ,t )   1 ,
T

R3

t
sup 1
exp( (  12   22   32 )) D k  i ( l1 ,l2 ,l3 ;s ) d 1d 2 d 3 ds   1T0   2 ,( i  1,3 ),


3
 T  0 R3

t
3
1
1

2
2
2
exp( (  1   2   3 ) )
 j   il( kk ) ( l1 ,l2 ,l3 ;s ) d 1d 2 d 3 ds 



sup
3
j
t  s j 1
 T  0 R3

t
1
3
1
1

2
2
2
2
2


sup
{
(

exp
(

(





)
)d

d

d

)


j
1
2
3
1
2
3

 1
3 T 
t

s
3
j

1

0
R

t
1

1
1
sup 
ds  3 1 2T0   3 ,
(  exp( (  12   22   32 ) )d 1d 2 d 3 ) 2 }ds  3 1
2 [0 ,T0 ] 0 t  s
 R3

(  i ; 1 ),  0   ( 1   ).
l j  x j  2 j ( t  s ),( j  1,3; k  0,3 ),   max
1 i  3
(2.12)*
Then speed components  are defined by a rule

i  i0 ( x1 ,x2 ,x3 )  Vi ( x1 ,x2 ,x3 ,t ), ( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),

3

Vi t 0  0, ( x1 ,x2 ,x3 )  R .
(2.14)
Then the system (2.9) will be transformed referring to
Vit  i  Vi ,( i  1,3 ),
(2.15)
where Vi ,( i  1,3 ) new unknown functions which defines the solution of problem Navier-Stokes. Hence
Vi 
t
1
8 
3
  exp( 
0 R3
ds ds ds ds
r2
1
) i ( s1 ,s2 ,s3 ,s ) 1 2 3

4 ( t  s )
( ( t  s ))3
3
t
  exp( ( 
0 R3
2
1
  22   32 )) 
 i ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s )d 1d 2 d 3ds 
(2.16)
 H i ( x1 ,x2 ,x3 ,t ), ( si  xi  2 i ( t  s );i  1,3 ).
The found solution (2.16) satisfies system (2.15). Really, having calculated partial derivative of
system (2.16):
( 0,1 )   ; ( x ,x ,x ,t )  T :
1
2
3

t
3

j
1
2
2
2
V


(
x
,x
,x
,t
)

exp
(

(





))

 il j ( x1  2 1 ( t  s ),x2 
 it

i
1
2
3
1
2
3


3
t

s
3
j

1

0 R


2 2  ( t  s ),x3  2 3  ( t  s );s )d 1d 2 d 3 ds,
t

1

V

exp( (  12   22   32 )) il j ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3 
 ix j
3  
 0 R3


2 3  ( t  s );s )d 1d 2 d 3 ds,
t

1
Vix2 
exp( (  12   22   32 )) il 2 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3 
3  
j
j

 0 R3


2 3  ( t  s );s )d 1d 2 d 3 ds, (l j  x j  2 j  ( t  s ); i  1,3; j  1,3 )
(2.17)
and substituting (2.17) in (2.15), we have (see (*)):
Vi t 0  0, ( x1 ,x2 ,x3 )  R 3 ; ( 0,1 )   ; ( x1 ,x2 ,x3 ,t )  T :

t
3
j
1

2
2
2
0

V




V



exp
(

(





))

(

 il j ( x1  2 1  ( t  s ),

it
i
i
i
1
2
3



3
t

s
3
j 1

0
R

t

1
 x2  2 2  ( t  s ),x3  2 3  ( t  s );s ))d 1d 2 d 3ds   i  {
exp( (  12   22   32 )) 
3  

 0 R3
 3
(  2 ( x  2  ( t  s ),x  2  ( t  s ),x  2 ( t  s );s ))d d d ds}=
1
2
2
3
3
1
2
3
 j 1 il j 1

t
3
j
1

2
2
2

exp
(

(





))

(

 il ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3 

1
2
3


ts j
j 1
 3 0 R3

t

1
1
1

2


(
t

s
)
;
s
))d

d

d

ds


{
[  exp( (  12   22   32 )) il 2 ( x1  2 1 

3
1
2
3

3
1
2
 0 t  s R3

  ( t  s ),x  2  ( t  s ),x  2 ( t  s );s )d( x  2 ( t  s ) )d d 
2
2
3
3
1
1
2
3

2
2
2
 exp( (      )) 2 ( x  2  ( t  s ),x  2  ( t  s ),x  2 ( t  s );s ) 
1
2
3
1
1
2
2
3
3
il2
 3
R

d 1d( x2  2 2  ( t  s ) )d 3  exp( (  12   22   32 )) 2 ( x1  2 1  ( t  s ),x2  2 2 

il3

R3

  ( t  s ),x3  2 3  ( t  s );s )d 1d 2 d( x3  2 3 ( t  s ) )]ds} 
t
3
j

1
2
2
2

exp
(

(





))(

 il j ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3 


1
2
3


3
t

s
3
j

1

0
R

t

3
1
1

2
2
2
exp
(

(





))

(
 j il j ( x1 
2 3  ( t  s );s ))d 1d 2 d 3 ds   { 3 

1
2
3

j 1
 0 t  s R3


2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 ( t  s );s ))d 1d 2d 3 ds  0.

(2.18)
That it was required to prove.
Therefore on the basis of (2.14), (2.16) we will receive
t

1




H



exp( (  12   22   32 )) i ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3 
 i
i0
i
i0
3  
 0 R3


(2.19)
2 3  ( t  s );s )d 1d 2 d 3 ds  H i ,( i  1,3 ),

3
3
i0  0; i0  C 3 ( R 3 );  x   H ix  0,( i  1,3 ).
i
i

i 1
i 1

Limitation of functions ( 1 ,2 ,3 ) in Cn3,1 3 (T ). The limiting case which we will consider
concern results of the theorem 1. Then the solution of system (1.1) is representing in the form of
(2.19) with conditions (1.2), (1.3), (A1), (2.1) and (2.12)*.
Really, estimating (2.19) in Cn3,13 (T ) , we have
3

v

 Cn3,13 ( T )  { i C 3 ,0 ( T )  it C( T ) }  3[N1   0 ]  M*,
i 1


 it C( T )   0   ( 1   ),( i  1,3 ),

k
 i C 3 ,0 ( T )   D i C( T )  N1  40 ; i C( T )  2 ,( i  1,3 ),
0  k 3

(2.20)
so as
 i  i0  H i  1   2  2 ,( max( 1 ; 2 )   ),

t

1
H

sup
exp( (  12   22   32 ))  i ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 
 i
3 T  

0 R3


 ( t  s );s ) d 1d 2 d 3ds   2 ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).
Uniqueness is obvious, as a method by contradiction. Therefore from (2.19) uniqueness of the
solution follows in Cn3,1 3 (T ).
4) Coming back to the proof the theorem 1, thus considering (2.3), (2.19), and their partial
derivatives on xi ,( i  1,2,3 ) , we find
3
 x  ( H j  Hix  H j  H jx )   i ( x1 ,x2 ,x3 ,t ),i  1,3.
i
j 1
j
(2.21)
i
As  i – is known functions, hence from system (2.21) differentiating 1-equation on x1 [(2.21): i=1],
2- equations on x2 [(2.21): i=2], 3-equations on x3 [(2.21): i=3], summing up, we will receive
3
   0 ,( 0   ixi ( x1 , x2 , x3 , t )),
(2.22)
i 1
at that
  C 2 (T ) :

ds ds ds
1
 0 ( s1 , s2 , s3 , t ) 1 2 3 .

4 R3
r
The equation (2.22) it have the third equation of system (2.4). Therefore, from the received results, taking
into account (2.6), follows
1
1
1
P  Q   

2
4
 F ( s ,s ,s ,t )
0
R3
1
2
3
ds1ds2 ds3
,
r
(2.23)
i.e. (2.23) – is the fourth equation of system (2.4).
The formula (2.23) can be transformed in equivalent form
1
1 3 2
1
P

i    ( s1 ,s2 ,s3 ,t )ds1ds2 ds3 ,


2 i 1
r

R3

3
  1  0 ( s ,s ,s ,t ) ds1ds2 ds3 , ( Q   2 ;   1 ( F   0 )),

1 2 3
i
0

4 R3
r
4
i 1

where (2.23)* – the equation of Bernoulli’s type [12]. Then function (2.23)*:
satisfies the equation:
(2.23)*
1

P
1 3 2
i  I
2 i 1
I   4  , and function I is called Newton’s potential [13], at that on
infinity aspires to a zero.   is called density of this potential.
Hence functions
i , , are defined from systems (2.19), (2.22), (2.23) and these functions are smooth
on set to the variables, that the system (2.4) has the unique smooth solution. The theorem 1 – is proved. ■
As a consequence the theorem 1 we will receive following statements:
Theorem 2. In conditions of the theorem 1 and (2.10), (2.12) the Navier-Stokes problem
(1.1)-(1.3), (A1) has a unique and conditional-smooth solution in Gn1 3 ( D0 ) .
Proposition 1. In the conditions of (1.2), (1.3), (А1) and (2.12), (2.13) the Navier-Stokes problem
has a unique solution in the space Wn2 3; ( D0 ) .
Theorem 2*. In conditions of the theorem 1 and (2.12)*, (2.20) the problem (1.1) - (1.3), (A1) is
solvable at Cn3,1 3 (T ).
The essential factor of researches of this paragraph are results of the theorem 2*. In this case the
solution of system (1.1) is considered as the strict decision of a problem (1.1)-(1.3), (А1).
It is obvious that small changes i0 ,( i  1,3 ) or fi ,( i  1,3 ) influence the solution (2.19) a little,
i.e. continuous depends on this data. Therefore, a question on a statement correctness problems
(1.1)-(1.3), (A1) are considered at once with results of the theorem 2*.
2.2. Inequality Beale-Kato-Majda
The Beale-Kato-Majda regularity criterion originally derived for solutions to the 3D Euler equations
[2] and holds for solutions to the 3D equations Navier-Stokes [5] and the criterion can be viewed as a
continuation principle for strong solutions. A further generalization was presented in [8] where the
regularity condition is expressed in terms of the time integrability.
Note that there are some inequalities for a priori estimates depending on the spaces. To prove this
criterion, for example, enough fulfill the inequality [5]:
T0
sup  rot ( x1 ,x2 ,x3 ,t ) dt  M  const  .
R3
0
(2.24)
On the basis of results of the theorem 1 the solution of systems (1.1) it is presented in a kind (2.19),
where global existence of solutions is received in a class Cn3,13 (T ) from the point of view of the initial
data satisfying (2.19). It is pleasant that results of this theorem leads to such global classical solution
Navier-Stokes, besides it is known that in [5] classical solution is received, if the criterion of
Beale-Kato-Majda is executed.
Really, at performance of conditions of the theorem 2* takes place
t

1
exp( (  12   22   32 ))
sup


3
 T  0 R3


  ( t  s );s ) 
 2 ( x1  2 1
x3

d d d ds  h ,
1
2
3
1

t

1
exp( (  12   22   32 ))
sup
3  
T
 0 R3



  ( t  s );s ) 
 3 ( x1  2 1
x1

d d d ds  h ,
1
2
3
2

t

1
exp( (  12   22   32 ))
sup


3
 T  0 R3

 3 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3  2 3 
x2
( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s ) 

 1( x1  2 1  ( t  s ), x2  2 2  ( t  s ),x3  2 3 
x3
( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s ) 

 2 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3  2 3 
x1


 ( t  s );s )  x  1( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s ) 
2

d 1d 2 d 3 ds  h3 ,



   ( x ,x ,x )    ( x ,x ,x )  h 0 ;
10 ( x1 ,x2 ,x3 ) 
30 ( x1 ,x2 ,x3 )  h20 ,
20
1 2
3
1
 x 30 1 2 3

x

x

x
2
3
3
1

3

 
0
0
 x 20 ( x1 ,x2 ,x3 )  x 10 ( x1 ,x2 ,x3 )  h3 , ( ( hi  hi )  M 0  const   ).
i 1

1
2
Then we will receive estimation
rot  0 : sup rot ( x1 ,x2 ,x3 ,t )  M 0  .
T
In a consequence and (see (2.24)):
T0
T0
sup  rot ( x1 ,x2 ,x3 ,t ) dt  sup  {
R3

R3
0

3 ( x1 ,x2 ,x3 ,s ) 
x1
0


3 ( x1 ,x2 ,x3 ,s ) 
2 ( x1 ,x2 ,x3 ,s ) 
x2
x3

1( x1 ,x2 ,x3 ,s ) 
x3


2 ( x1 ,x2 ,x3 ,s ) 
1( x1 ,x2 ,x3 ,s ) }ds  M 0T0  M  .
x1
x2
2.3. Estimation of affinity of solutions of the equations Navier-Stokes and Euler
I. Incompressible Streams Without the Friction. For incompressible currents without a friction [2, 6
and 12]:   0 the equations of Navier-Stokes become simpler, as there are no members: i . Therefore
the problem (1.1) - (1.3) is led to a kind
1
2
3
it  (   j2 )x  fi 
j 1
i
1

Pxi ,i  1,3,
(2.25)
i ( x1 ,x2 ,x3 ,t )|t 0  0i ( x1 ,x2 ,x3 ),i  1,3,
(2.26)
div  0,  ( 1 ,2 ,3 ); rot  0.
(2.27)
The system (2.25) with conditions (2.26), (2.27) has the strict solution [12] with preservation of all
convective members at performance of a condition of Stokes. Really, for incompressible currents without
a friction, the vector of speed is represented as a gradient of potential  and this potential satisfies the
Laplace equation.
Therefore for potential currents the member in the equation (1.1), depending on viscosity, identically
disappears. At that the system (2.25) has the smooth unique solution in Cn3,13 (T ) [9, see pp.147-151]:
3

k


3 ,1
 Cn3 ( T )  i C 3 ,1 ( T )  N0 ; i C 3 ,1 ( T )   D i C( T )  it C( T ) ,
i 1
0 k 3


k
3
 i
k  0 : D0   ;k  0 : D k 
;
k

 i ,(  i  0,1,2,3;i  1,3 ),

i
i
i

x11 x2 2 x33
i 1

(2.28)
and is harmonious functions. The specified method of theorem 2* can be used in particular and for the
decision of a problem (2.25) - (2.27) as a test example. Really, on a basis APS from system (2.25), follows

3
 J   F0 , (J  1 P  1 Q, F0   f ix ),
i


2
i 1

1
1

F0 ( s1 ,s2 ,s3 ;t ) ds1ds2 ds3 ,
J 

4 R3
r


1
i
 J xi 
F0 ( x1   1 ,x2   2 ,x3   3 ;t )
d 1d 2 d 3 ,( i  1,3 ).

2
2
2 3
4 R3

(





)
1
2
3

(2.29)
So as J - the solution of the equation (2.29), with the account:
J xi 
1
1
Pxi  Qxi

2
(2.30)
from (2.25) we will receive
it  fi  J x ,( i  1,3 ).
(2.31)
i
From system (2.31) on the basis of (2.26), follows
t




(
x
,x
,x
)

 i
i0
1
2
3
0  i ( x1 ,x2 ,x3 , )d ,( i  1,3 ),


3
  f  J ,( i  1,3 );
 ixi  F0   J  0, ( x1 ,x2 ,x3 ,t )  T .

i
xi
 i
i 1
Hence from (2.29) and (2.30) we have
1
1
1
P Q

2
4
1
 F ( s ,s ,s ,t ) r ds ds ds ,
0
1
2
3
1
2
3
(2.32)
(2.33)
R3
i.e. (2.33) there is an equation of type Bernoulli that is similar with (2.23)*.
From the received results follows that the system (2.32) satisfies to a condition (2.27), and it means
that the found solutions i ,i  1,3 satisfies to the of Laplace equation, i.e. are harmonious functions.
As it has been proved.
II. It is known that limit transition to very small viscosity should be executed not in the equations of
Navier-Stokes, but in the solutions of these equations by approach of factor of viscosity to zero [12].
Then the solution of system (1.1) is representing in the form of (2.19) with conditions of theorem 2*.
To estimate affinity of solutions (2.19), (2.32) in sense Cn3,13 (T ) , when [9]:   0 , conditions are
required is:
( x1 ,x2 ,x3 ,t )  T ; i , i ,i0  C 3 ( R 3 ) : D k ( i0  i0 )  1 1 ,( x1 ,x2 ,x3 )  R 3 ,

 k

 D [ i ( x1 ,x2 ,x3 ,t )   i ( x1 ,x2 ,x3 ,t )]  2 2  ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),
t
 1
sup
exp( (  12 + 22 + 32 )){ D k [ i ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3 
 3


T
0 R3
 

2 3  ( t  s );s )   i ( x1 ,x2 ,x3 ;s ) ] }d 1d 2d 3ds  3  ,( i  1,3 ),

t
3
j
1

sup   exp( (  12 + 22 + 32 ))[
 il j ( x1  2 1  ( t  s ),x2  2 2 
 
ts
j 1
 3 T 0 R3


  ( t  s ),x3  2 3  ( t  s );s ) ]d 1d 2 d 3 ds  4  ,
l  x  2  ( t  s ); 0      const   ,( k  1,4; i  1,3 ),        .
i
i
k
0
1
2
i

(2.34)
Lemma 1. If conditions of the theorem 2* and (2.34) are satisfied, an admissible error between
solutions of system (2.19), (2.32) in Cn3,13 (T ) , when    , will be an order  (  ).
To prove to affinity of decisions (2.19) and (2.32) in Cn3,13 (T ) , at first we will prove to
affinity of solutions C(T ); C 3,0 (T ) . At that obviously that estimations relatively i  i
C 3 ,1 ( T )
will
be an order  (  ) . Really estimating (2.19), (2.32), we have
t
1
i  i  i0  i0 

3
  exp( ( 
2
1
  22   32 ))  i ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),
0 R3
x3  2 3 ( t  s );s )   i ( x1 ,x2 ,x3 ;s ) d 1d 2d 3ds 
t
1

3
  exp( ( 
2
1
  22   32 )) 
0 R3
  i ( x1 ,x2 ,x3 ;s )  i ( x1 ,x2 ,x3 ;s ) ]d 1d 2d 3ds  1 1   3   2 2 T0  C0  ,( i  1,3 ),
or
i  i
C( T )
 C0  ,( i  1,3;C0  1  3  2T0 ),
here (see.(2.32)):
t
i  i0 ( x1 ,x2 ,x3 )   i ( x1 ,x2 ,x3 , )d  i0 
0
t
1

3
  exp( ( 
2
1
  22   32 )) 
0 R3
1
(2.32)*
 exp( (      ))d 1d 2d 3  1 ).
 3 R3
Similarly, we will receive also estimations concerning expressions where partial derivative functions
 i ( x1 ,x2 ,x3 , )d 1d 2 d 3d  H i ,( i  1,3;
2
1
2
2
2
3
i and i to the third order, inclusive, contain i , i , i.e.
i  i
C 3 ,0 ( T )


0  k 3
D k ( i  i )
C( T )
 20C0  ,( i  1,3 ).
Hence, as C 3,1(T )  i ,i , estimating (2.19) and (2.32)* in sense of norm C 3,1 ( T ) we will receive:
k
  
i C 3 ,1 ( T )   D ( i  i ) C( T )  it  it C( T )  ( 20C0  20 )   N 0  ,
 i
0 k 3

 it  it
 20  ,( i  1,3 ),
C( T )



it   i , ( i  1,3 ), it  [i0 ( x1 ,x2 ,x3 )  H i ( x1 ,x2 ,x3 ,t )]  H it , (Vit  H it ; see.(2.19)),
t

t
3


1
   exp( (  12 + 22 + 32 ))[ j  il j ( x1  2 1  ( t  s ),x2 
 H it   i ( x1 ,x2 ,x3 ,t ) 
j 1 t  s
3
0 R3


2 2  ( t  s ),x3  2 3  ( t  s );s )]d 1d 2 d 3ds,( i  1,3 ).
(2.35)
Then taking into account (2.35) and   ( 1 ,2 ,3 ),  ( 1 ,2 ,3 ) , we have
 
3
Cn3,13 ( T
)
  i  i
i 1
3
C
3 ,1
(T )
 {

i 1 0  k  3
D k ( i  i )
C( T )
 it  it
C( T )
}  3N0  .
(2.36)
And it means that if    the admissible error of an estimation will be order  (  ) in
Cn3,13 (T ) . The lemma 1 – is proved. ■
Remark *. So as i0  C 3 ( R3 ),( i  1,2,3 ), that to estimate affinity of solutions (2.10), (2.32) in
sense Gn1 3 ( D0 ) , when   0 , it is required conditions of theorem 2 and (2.34). Then we have
i  i 
1

3
 exp( ( 
2
1
  22   32 )) i0 ( x1  2 1 t ,x2  2 2 t ,x3  2 3 t )  i0 ( x1 ,x2 ,x3 ) 
R3
d 1d 2 d 3  i0  i0 
t
1

3
  exp( ( 
2
1
  22   32 ))  i ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3 
0 R3
2 3  ( t  s );s )   i ( x1 ,x2 ,x3 ;s ) d 1d 2 d 3ds 
t
1

3
  exp( ( 
2
1
  22   32 ))  i ( x1 ,x2 ,x3 ;s ) 
0 R3
 i ( x1 ,x2 ,x3 ;s ) ]d 1d 2 d 3ds  ( 3 1 2T0  1  3  2T0 )   C0  ,( i  1,3 ),
or
 i  i


  i

 i
C( T )
 C0  ,( i  1,3 ),
C 3 ,0 ( T )
 20C0  ,( i  1,3 ).
Similarly, we will receive estimating it  it
(2.37)
L1 ( 0,T0 )
, i.e.
 it  it 1
 ( 31 2T0  20T0 )   C1  ,( i  1,3 ),
L ( 0,T0 )


3

j
1
2
2
2
exp
(

(





))

(
 i0h j ( x1  2 1 
it   i , ( i  1,3 ); it 

1
2
3

t

 3 R3
j 1
t

1
exp( (  12   22   32 )) 
 t ,x2  2 2 t ,x3  2 3 t )d 1d 2d 3   i 


3

 0 R3

(2.38)
j
 3
 il j ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s )d 1d 2d 3ds.
  
ts
 j 1
Then taking into account (2.37), (2.38) and estimating   in the space Gn1 3 ( D0 ) we obtain

3
Gn13 ( D0 )
 {

i 1 0  k  3
D i
T0
k
C( T )
 sup  it ( x1 ,x2 ,x3 ,t )dt}  3N0  ,( N0 =20C0  C1 ).
R3
(2.39)
0
Proposition 2. In conditions of the theorem 2 and (2.34), (2.39) an admissible error between
solutions of system (2.10), (2.32) in Gn1 3 ( D0 ) will be an order  (  ).
3. Fluid Average Viscosity with a Condition (A2)
Let's consider a fluid with viscosity with Reynolds small number where all inertial participants
contain in equations Navier-Stokes. Theoretically, it is not investigated till now [12]. Hence, here we
will consider, methods of integration of the equations Navier-Stokes, therefore the decision of the
method, from where follows of equations integration of Navier-Stokes in a case (А2), is a major factor
of this point. The developed method of the decision of system (1.1) is connected with  i , where these
functions will transform (1.1) to systems (1.4), (1.5) with conditions (а02) and
(1.3) : i t 0  0,( x1 ,x2 ,x3 )  R3 ,( i0 ( x1 ,x2 ,x3 )  0;i  1,3 ),
i t 0  0,( x1 ,x2 ,x3 )  R3 ,( i  1,3 ),
(1.3)*
(3.1)
where the current is considered with average size of viscosity.
Theorem 3. Systems (1.4), (1.5) it is equivalent will be transformed to a kind
3
1
1


J


F
,
(
J

P

Q;
F


f ixi ; divf  0; 1    0  const   ),

0
0
0
 0

2
i 1

it  fi  i  J 0 xi  i ,

i  Di 1 , 2 , 3  ,( i  1,3 ),
1
ds ds ds
1
1
 P Q
F0 ( s1 ,s2 ,s3 ,t ) 1 2 3 ,( r  ( x1  s1 )2  ( x2  s2 )2  ( x3  s3 )2 ),

2
4 R3
r
 
(3.2)
when conditions (1.2), (1.3)*, (3.1), (А2) are satisfied. Hence, the nonstationary problem of
Navier-Stokes (1.1)-(1.3)* has the smooth unique solution.
Proof. Really, from system (1.4), considering conditions (1.2), (1.3)*, (3.1) and having entered APS,
i.e. differentiating the equations of system (1.4) accordingly on хi and, then summing up, we have the
equation


 J 0   F0 ,

1 1

(3.3)
F0 ( s1 ,s2 ,s3 ,t )ds1ds2 ds3 ,
 J0 
4 R3 r


1
 i F0 ( x1   1 ,x2   2 ,x3   3 ;t )
1
1
 J 0 xi 
d 1d 2 d 3 ,( si  xi   i ; Pxi  Qxi  J 0 xi ;i  1,3 ).

4 R3

2

(  12   2 2   32 )3
If J 0 – the solution of the equation (3.3), then substituting: J 0 xi ,( i  1,3 ), in system (1.4), we have
it   i  i  i , ( i  1,3 ),

3


(
x
,x
,x
,t
)

f

J
,(
i

1,3
),
 ixi  F0  J 0  0, ( x1 ,x2 ,x3 ,t )  T .

i
0 xi
 i 1 2 3
i 1

The solution of a problem (1.1) - (1.3)* is represented in a kind
(3.4)
t

1
r2
ds ds ds d
0
exp( 
)i ( s1 ,s2 ,s3 , ) 1 2 3 3   ii ,( i  1,3 ),
i  H i 
3  
4 ( t   )
( ( t   ))
8  0 R3


t
r2
ds ds ds d
 H 0 ( x ,x ,x ,t )  1
exp( 
) i ( s1 ,s2 ,s3 , ) 1 2 3 3 ,( i  1,3 ),
 i 1 2 3
3  
4 ( t   )
( ( t   ))
8  0 R3

(3.5)
therefore concerning functions i ,(i  1,2,3) , we will receive
3
i  {[H 
0
j
j 1
t
1

3
  exp( ( 
2
1
  22   32 )) j ( x1  2 1  ( t   ),x2  2 2  ( t   ),x3 
0 R3
t
1
2 3  ( t   ); )d 1d 2 d 3 d ]  [H ix0 j 

3
  exp( ( 
0 R
2
1
  22   32 ))
3
j
( t   )
 i ( x1 
2 1  ( t   ),x2  2 2  ( t   ),x3  2 3  ( t   ); )d 1d 2 d 3d ]  [H 0j 

t
1

3
  exp( ( 
2
1
  22   32 )) j ( x1  2 1  ( t   ),x2  2 2  ( t   ),x3  2 3 
0 R3
  ( t   ); )d 1d 2 d 3 d ]  [H 0jxi 
t
1

3

0 R3
i
( t   )
exp( (  12   22   32 )) j ( x1  2 1 
  ( t   ),x2  2 2  ( t   ),x3  2 3  ( t   ); )d 1d 2 d 3d ]}  Di [1 , 2 , 3 ],
si  xi  2 i  ( t   ),( i  1,3 ),
where for example, partial derivatives of functions i are defined:
t
( x j  s j )

1
1
r2
0


H
(
x
,x
,x
,t
)

exp
(

)i ( s1 ,s2 ,s3 , ) 
 ix j
ix j
1 2
3
3
3   2 ( t   )
4

(
t


)
(

(
t


))
3
8

0 R


t
j
ds ds ds d  H 0  1
exp( (  12   22   32 ))
i ( x1  2 1 ( t   ),x2 
1
2
3
ix
3


j


(
t


)
3

0

R
(3.6)
2 2 ( t   ),x3  2 3 ( t   ); )d 1d 2 d 3d ,

t
j
 0
1
H

exp( (  12   22   32 ))
 i ( x1  2 1 ( t   ),x2  2 2 ( t   ),x3 
 ix j
3  

(
t


)
3

0

R

2 3 ( t   ); )d 1d 2 d 3d , ( si  xi  2 i ( t   );i  1,3; j  1,3 ).
(3.7)
The system (3.6) consists of three the nonlinear integral equations of Volterra and Volterra-Abel [13] on
a variable t  [0,T0 ] and contains in itself of three i ,( i  1,2,3 ) unknown functions. The theory of
the specified system is well developed in section of mathematics [13]. Therefore there is no necessity
to think out various algorithms for the decision of this system. To solve this system, it is enough to
find conditions that allow us to use the Banach Principle. Really, if the input functions satisfy to


0
D k i ( x1 ,x2 ,x3 ,t )   0 , ( i  1,3; k  0,2 ),
( x1 ,x2 ,x3 ,t )  T ; i ;H i : sup
T

t
 1
exp( (  12   22   32 ))d 1d 2 d 3 d  T0 ,


 3 [sup
0 ,T0 ] 0 3
R
 
t
 1
 3 sup   exp( (  12   22   32 ))  i ( x1  2 1  ( t   ),x2  2 2  ( t   ),x3  2 3 
  T 0 R3

  ( t   ); ) d 1d 2 d 3 d   0T0 ,

t
t
i
1
1
 1
2
2
2
1
sup
exp
(

(





))
d

d

d

d


(

)
sup


1
2
3
1
2
3



3 T
3 T
t


t


3



0
0

R

1
1
{(  2 exp( (  2   2   2 ) )d d d ) 2 ( exp( (  2   2   2 ) )d d d ) 2 }d 
1
2
3
1
2
3
1
2
3
1
2
3

 R3 i
R3

t

1
1 1
sup 
d  (  )1 2T0 ,
 (  )
[ 0 ,T0 ] 0
2
t




t
j
1
 H ix0 
sup
exp( (  12   22   32 ))
 i ( x1  2 1 ( t   ),x2  2 2 


j
(3.8)

t 
  3 T 0 R3

  ( t   ),x3  2 3  ( t   ); ) d 1d 2 d 3d  (  )1  0 2T0
and if the operators Di ,( i  1,2,3 ) satisfy to conditions of a principle of compression, that is,
3
d

D
:
d

,(
d

1
),
(
i

1,3
;
d i  d  12(  )1  *  1 ),

 i i 3
i 1

d  4(  )1[2 2 T 3  2r T 3 ]  4(  )1   1, ( i  1,3 ),
0
0
1
0
*
 i
  2 2 T 3  2r T 3 , ( k       , [k  max( 1; 144 2 )] )
0
0
1
0
0
0
0
4
 *
and
(3.9)
 Di [10 , 20 , 30 ]  i0  r1( 1  d ) : Di [1 , 2 ,3 ]  i0 
C
C


0
0
0
0
0
0
0
 Di [1 , 2 , 3 ]  Di [1 , 2 , 3 ] C  Di [1 , 2 , 3 ]  i C  di 3r1  r1( 1  d )  r1 ,

0
0
0
0
 Di : Sr1 ( i )  Sr1 ( i ), ( i  1,3 ),Sr1 ( i )  {i : i  i  r1 , ( x1 ,x2 ,x3 ,t )  T }.
(3.10)
Then, by Banach Principle, the system (3.6) is solvable at C 2,0 ( T ) and the solution of this system can
be found by Picard’s method as
i,n 1  Di [1,n ,2,n ,3,n ],( n  0,1,...;i  1,3 ),
(3.11)
where 1,0 ,2,0 ,3,0  initial estimates. Received the sequence of functions {i,n }0 ,( i  1,3 ) is
converging and fundamental in Sr1 ( i0 ) , so as
3
3

E




;
E

i ,n  i ,n 1 C , ( i  1,3 ) :

n
 n  1  i ,n  1 i ,n C
i 1
i 1

3

d 1
n




d
 0,
i   i ,n  i ,n 1 C  di En ; En  1  dEn  ...  d E1 
 i ,n  1 i ,n C
n 
i 1


k 1
k 1
3
k 1
 i ,n  k  i ,n   i ,n  j 1  i ,n  j   di  i ,n  j  i ,n  j 1   di En  j ,
C
C
C

j 0
j 0
i 1
j 0

k 1
k 1
k 1
d 1
 E  d E  ...  d d n  j 1 E  E d n d j  E d n 1 
 0.



nk
n j
1
1
1
n 

1 d
j 0
j 0
j 0
Thus elements of the constructed sequences belong Sr1 ( i0 ) . Really, for n , rather i , n 1 we have
n 1

j
0
0
0
0








...









d
(
 i , n1 i ,0
i ,n 1
i ,n
i ,2
i ,1
i ,1
i ,0
i  d )E1  Di [1 , 2 , 3 ]   i
j 0

n

1
n

j
j
 ( 1  d )r1d  d  ( 1  d )r1  ( 1  d )r1  d  r1 ,
j 0
j 0



1
 d j  ( 1  d )1 ; d i  d ; E1  3( 1  d )r1 ; i ,n1  S r1 ( i0 ), ( i  1,3 ).
3
 j 0
C

It means that sequence {i,n }0 ,( i  1,3 ) is converged to i ,( i  1,3 ), that is,
3
3

d 1
U




;
U

i  i ,0 C : U n 1  dU n  ...  d n  1U 0 
 0,

0
 n  1  i ,n  1 i C
n 
i 1
i 1

d 1

i  i ,( x1 ,x2 ,x3 ,t )  T , ( i  1,3 ).
n 
 i ,n  1 
Then according to results of the theorem 3, functions i ,i  1,3 are defined from system (3.5)
(3.12)
t

1


exp( (  12   22   32 )) i0 ( x1  2 1 ( t   ),x2  2 2 ( t   ),x3  2 3 
 i
3  
 0 R3


 ( t   ); )d 1d 2 d 3 d  H i ( x1 ,x2 ,x3 ,t ), ( i  1,3 ),

0
 si  xi  2 i ( t   );  i ( x1 ,x2 ,x3 ,t )   i ( x1 ,x2 ,x3 ,t )  i ( x1 ,x2 ,x3 ,t ), ( i  1,3 ),


(3.5)*
here i ,i ,H i - known functions and
( x1 ,x2 ,x3 ,t )  T ; i0 : sup D k i0 ( x1 ,x2 ,x3 ,t )  1 , ( i  1,3;k  0,2 ),
T

t

1
sup
exp( (  12   22   32 ))  i0 ( l1 ,l2 ,l3 ; ) d 1d 2 d 3 d   2  1T0 ,


3
 T  0 R3

t
3
1
1

2
2
2
sup
exp
(

(





)
)
 j   il0(k k ) ( l1 ,l2 ,l3 ; ) d 1d 2 d 3 d  1 2T0   3 ,

1
2
3



3
j
(3.13)
t  s j 1
 T  0 R3

i ,
l j  x j  2 j ( t  s ),( j  1,3; k  0,2 ), 0   ( 1   );   max
1 i  3

t
   1 sup exp( (  2   2   2 ))  0 ( x  2  ( t   ),x  2  ( t   ),
1
2
3
i
1
1
2
2
 i
3 T  

0 R3

 x3  2 3  ( t   ); ) d 1d 2 d 3 d  1T0   , ( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).

Then considering norm of space Cn3,13 (T ) we will receive
3

v

{ i C 3 ,0 ( T )  it C( T ) }  3[N1  0 ]  M*,
3 ,1

 Cn3 ( T )
i 1

 N1  20  , ( i C( T )   ); it
 i C 3 ,0 ( T )   D ki
C( T )
0 k 3

C( T )
 0   ( 1   ),( i  1,3 ).
(3.14)
Thus (3.5)* satisfies the equation (3.4):

   0   , ( i  1,3 ),
i
i
 it
3
3
3
 0





;


H

0
:
 ix0i  0, ( x1 ,x2 ,x3 ,t )  T ,
 i



i
i
ixi
ixi
i 1
i 1
i 1

3

div  0, (   ( 1 ,2 ,3 ); i  i );  i  f i  J 0 xi ,( i  1,3 ),  ixi  F0   J 0  0.
i 1

Really, having calculated partial derivative of system (3.5)*:
k      const   , ( k  max( 1; 144 2 )); ( x ,x ,x ,t )  T :
0
0
*
1 2
3
 0
t
t
2

1
r
ds1ds2 ds3 d
1
0


exp
(

)

(
s
,s
,s
,

)

exp( (  12   22   32 )) 
 i
i
1 2 3
3  
3
3  
4 ( t   )
( ( t   ))
8  0 R3
 0 R3

 0
 i ( x1  2 1 ( t   ),x2  2 2 ( t   ),x3  2 3 ( t   ); )d 1d 2d 3 d ,( i  1,3 ),
(3.4)*
t
3
j

1
0
2
2
2



(
x
,x
,x
,t
)

exp
(

(





))

 il0j ( x1  2 1 ( t   ),x2 
 it

i
1 2
3
1
2
3
3  
t 
j 1
 0 R3


2 2  ( t   ),x3  2 3  ( t   ); )d 1d 2 d 3d , ( l j  x j  2 j  ( t   ) ; i  1,3; j  1,3 ),
t

( x j  s j )
1
1
r2
ix 
exp( 
) i0 ( s1 ,s2 ,s3 , )ds1ds2 ds3d 
j
3
3   2 ( t   )
4 ( t   )

(  ( t   ))
8  0 R3

t
j
 1
2
2
2
exp
(

(





))
 i0 ( x1  2 1  ( t   ),x2  2 2  ( t   ),x3 
1
2
3
3



( t   )
 0 R3


2 3  ( t   ); )d 1d 2 d 3d ,

t
j
2
2
2
 2  1
exp
(

(





))
 il0j ( x1  2 1  ( t   ),x2  2 2  ( t   ),x3 
1
2
3


ix
3
j

( t   )
 0 R3

2 3  ( t   ); )d 1d 2 d 3 d ,

t
3
j
1

2
2
2


exp
(

(





))

 il0j ( x1  2 1  ( t   ),x2  2 2  ( t   ),

i
1
2
3

3  
t 
j 1
 0 R3

 x  2  ( t   ); )d d d d
3
1
2
3
(3.15)
 3

and substituting (3.15) in (3.4)*, we have
i t 0  0,( x1 ,x2 ,x3 )  R 3 ; 1    0  ; ( x1 ,x2 ,x3 ,t )  T :

t
3

 j 0
1
0
0
2
2
2
exp
(

(





))
 il j ( x1  2 1  ( t   ),x2 
0  it   i  i   i 

1
2
3
3  
j 1 t  
 0 R3


t
(3.16)

1
0
2
2
2
exp
(

(





))

2 2  ( t   ),x3  2 3  ( t   ); )d 1d 2 d 3 d   i 
1
2
3

 3 0 R3

 3 
j

 il0j ( x1  2 1 ( t   ),x2  2 2 ( t   ),x3  2 3 ( t   ); )d 1d 2 d 3d  0.

 j  1 t  
That it was required to show.
From the received results, on the basis of (3.3) follows
1
1
1
P Q

2
4

R3
F0 ( s1 ,s2 ,s3 ,t )ds1ds2 ds3
.
r
(3.17)
Then according to results of the theorem 3, functions i ,i  1,3 are defined from system (3.5)* and satisfies
the equation (1.2). For a problem Navier-Stokes (1.1)-(1.3)*, (A2), are proved: existence of the smooth
unique solution in area Cn3,13 (T ) , and we will notice that the received decision (3.5)* continuously
depends on the initial data fi ,( i  1,3 ) . The theorem is proved. ■
4. Fluid with Very Small Viscosity with a Condition (A3)
In the theory of the differential equations in partial derivatives there are various mathematical
transformations which simplify investigated problems and does possible to find the solution in certain
spaces [6, 12, 13 and 15]. Here in a case 0    1 (Reynolds number [12]: Re ≥ 2300) we will show
that at certain mathematical transformations of the equation Navier-Stokes is led to a linear kind. At that the
new system has an analytical solution, which is based on the Picard's method.
So, inverse Fourier transform plays great part while deciding boundary problem for the solution of some
integral equations, integration; Laplace transformation – while solution of simple differential equation of
multiple of N with the constant rate and their systems, some differential equation in the partial derivative,
Volterra equation of the second and first type with the difference kernel, and integral equation with the
logarithmic kernel and etc. But at the decision of equations Navier-Stokes in the general form these
transformations yet have not brought desirable results, if we do not consider special cases.
Therefore for the last decades in the mathematics the methods, connected with using of integral
of the transform, became widely spread. Different formulas of integral transform arise while a
concrete problem solving, but in the sequel they can be applied to the solution of other problems when
researching the differential and integral equation, integration.
From the received results follows that system Navier-Stokes (1.1) in the conditions of (1.2), (1.3) and
(A3) can have the analytical smooth unique solution. At least, such solution answers a mathematical
question, and possibility to construct the solution on a problem Navier-Stokes (1.1) - (1.3) for an
incompressible liquid with viscosity.
4.1. Fluid with Viscosity 0    1 , when divf  0
Let components of a vector of speed  for t  0 be defined by (1.3) as
i
t 0
 i0 ( x1 ,x2 ,x3 )  
i 0 ( x1 ,x2 ,x3 ),( i  1,3 ),
where 0  i –
the known constants. Then speed components 
(4.1)
are defined by a rules
i  iV ( x1 ,x2 ,x3 ,t ),( i  1,3 ),

3
V t 0  0 ( x1 ,x2 ,x3 ), ( x1 ,x2 ,x3 )  R ,

divf  0; div  0 :
 3
3
3
  V  0;




V

j xj
j ix j
i   jVx j  0.

j 1
j 1
j 1
(4.2)
Hence, the system (1.1) is transformed to the system of inhomogeneous linear equations
iVt  fi 
1

Pxi  i V ,( i  1,3 ),
(4.3)
where V new unknown function which defines the solution of the Navier-Stokes problem. At that
(4.1)-(4.3) are investigated in work [8] in Gn23; ( D0 ) or Wn23; ( D0 ). Here the problem (4.1)-(4.3) it is
investigated in Gn1 3 ( D0 ) . For this purpose that to solve the system (2.3) at first we define a pressure P .
Really, considering APS from system (4.3) we will receive:
 3
3
  (4.3) : 1  P   F ,( F   f ( x ,x ,x ,t )),
0
0
ixi
1
2
3
 i1 xi

i 1

ds ds ds
1
1
F0 ( s1 ,s2 ,s3 ,t ) 1 2 3 ,
 P

4 R3
r

1
 i d 1d 2 d 3
 Px  1
F0 ( x1   1 ,x2   2 ,x3   3 ;t )
,( si  xi   i ;i  1,3 ).

i
2
2
2 3
4 R3

(





)
1
2
3

Therefore
Vt   0 ( x1 ,x2 ,x3 ,t )  V , ( x1 ,x2 ,x3 ,t )  T ,
 3

 0 xi  0, ( x1 ,x2 ,x3 ,t )  T ,
 i 1
(  )1( f   1 P )  (  )1( f   1 P )  (  )1( f   1P )   ,
1
x1
2
2
x2
3
3
x3
0
 1
(4.4)
(4.5)
i.e. is the system (4.3) is transformed to the linear equations of heat conductivity with a condition of
Cauchy in a kind (4.5), and in a class of functions with smooth enough initial data is correctly put
[13, 14]. Accordingly there is an the conditional-smooth unique solution of a problem Navier-Stokes
in G 1( D0 ). Really from system (4.13), follows:
V
1
8( t )3
 exp( 
R3
r2
1
)0 ( s1 ,s2 ,s3 )ds1ds2 ds3 
4 t
8 3
 0 ( s1 ,s2 ,s3 ,s )ds1ds2 ds3 ds 
2 3 t )d 1d 2 d 3 

3
 exp( ( 
3
2
1
  exp( 
0 R3
r2
1
)

4 ( t  s ) ( ( t  s ))3
  22   32 ))0 ( x1  2 1 t ,x2  2 2 t ,x3 
R3
t
1

1
t
  exp( ( 
2
1
  22   32 )) 0 ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3 
0 R3
2 3 ( t  s );s )d 1d 2 d 3 ds  H 0 ( x1 ,x2 ,x3 ,t ),( si  xi  2 i t ;si  xi  2 i ( t  s );i  1,3 ),
where H0 – known function and the found solution (4.6) satisfies system (4.5).
For this purpose, considering partial derivative systems (4.6):
( 0,1 )   ; h j  x j  2 j t ; l j  x j  2 j ( t  s ),( j  1,3 ),( x1 ,x1 ,x1 ,t )  T :


1
exp( (  12   22   32 ))0 h j ( x1  2 1 t ,x2  2 2 t ,x3  2 3 t )d 1d 2 d 3 
Vx j 
3 
 R3


t
 1
exp( (  12   22   32 )) 0l j ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 

3  
3
  0R
  ( t  s );s )d d d ds,
1
2
3

(4.6)

1
exp( (  12   22   32 ))0 h2 ( x1  2 1 t ,x2  2 2 t ,x3  2 3 t ) 
Vx2j 
3 
j
 R3

t

d 1d 2 d 3  1
exp( (  12   22   32 )) 0l 2 ( x1  2 1 ( t  s ),x2  2 2 


3
j

 0 R3

  ( t  s ),x3  2 3  ( t  s );s )d 1d 2 d 3ds,

3
( x1 ,x2 ,x3 )  R ; t  ( 0,T0 ] :

3
j
2
2
2
V  1
e
xp
(

(





))

(

0 h ( x1  2 1 t ,x2  2 2 t ,x3 

t
1
2
3


t j
j 1
 3 R3

t
3
j
1

2
2
2

2


t
)d

d

d




exp
(

(





))

 0l j ( x1 

3
1
2
3
0
1
2
3

3  
t

s
3
j

1

0 R

2  ( t  s ),x  2  ( t  s ),x  2 ( t  s ) ;s )d d d ds
1
2
2
3
3
1
2
3

and substituting (4.7) in (4.5), we have
V t  0  0 ( x1 ,x2 ,x3 ), ( x1 ,x2 ,x3 )  R 3 ;( 0,1 )   ; ( x1 ,x2 ,x3 )  R 3 ; t  ( 0,T0 ] :

3

1

2
2
2
0

V




V

exp
(

(





))

(
 j 0 h j ( x1  2 1  t ,x2 

t
0
0
1
2
3

3 
t
j 1
 R3

t
3

j
1
2
2
2
2 2  t ,x3  2 3  t ))d 1d 2 d 3   0 
exp
(

(





))

(



1
2
3


ts

j 1
 3 0 R3

 0l j ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 ( t  s ) ;s ))d 1d 2d 3ds   0 

  { 1
exp( (  12   22   32 ))0 ( x1  2 1 t ,x2  2 2 t ,x3  2 3 t )d 1d 2 d 3 

3 
 R3

 1 t
exp( (  12   22   32 )) 0 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ) ,x3  2 3 

3  
  0 R3

3


1
2
2
2


(
t

s
)
;
s
)d

d

d

ds
}

exp
(

(





))(
 j 0 h j ( x1  2 1 t ,


1
2
3
1
2
3

t
j 1
 3 R3


t
3
j
2
2
2
 x2  2 2 t ,x3  2 3 t ))d 1d 2 d 3  1
exp
(

(





))(

 0l j 

1
2
3


3

ts
j 1
 0 R3

( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s ) ;s ))d 1d 2 d 3 ds 

1
 1  { 1
exp( (  12   22   32 )) 0 h2 ( x1  2 1  t ,x2  2 2 t ,x3  2 3 t ) 
 2
3 
t 1
 R3


1
1
exp( (  12   22   32 )) 0 h2 ( x1  2 1 t ,x2  2 2 t ,
d( x1  2 1 t )d 2 d 3 
3 
t 2
 R3


1
 x3  2 3  t )d 1d( x2  2 2 t )d 2 d 3  1
exp( (  12   22   32 )) 0 h2 ( x1 


t 3
 3 R3
(4.7)
t

1
1

2 1 t ,x2  2 2 t ,x3  2 3 t )d 1d 2 d( x3  2 3 t ) 
3 
 0 ts


2
2
2
[  exp( (  1   2   3 )) 0l12 ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 ( t  s );s ) 
 R3
d( x  2  ( t  s ) )d d  exp( (  2   2   2 )) ( x  2 ( t  s ),x  2 
1
1
2
3
1
2
3
1
1
2
2

0l22

R3

  ( t  s ),x3  2 3  ( t  s );s )d 1d( x2  2 2 ( t  s ) )d 3  exp( (  12   22   32 )) 


R3

 0l 2 ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 ( t  s );s )d 1d 2 d( x3 
3

3


1
2
2
2

2


(
t

s
)
)
]
ds
}

exp
(

(





))(
 j 0 h j ( x1  2 1 t ,x2  2 2 


3
1
2
3

t
j 1
 3 R3


t
3
j
2
2
2
 t , x3  2 3 t ))d 1d 2 d 3  1
exp
(

(





))(

 0l j ( x1 

1
2
3



ts
j 1
 3 0 R3

2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 ( t  s );s ))d 1d 2 d 3ds 

3

2
2
2
 1
exp
(

(





))(
 j 0 h j ( x1  2 1 t ,x2  2 2 t ,x3  2 3 t )) 

1
2
3

 3 3
t
j 1
R

t
3

j
1
2
2
2
exp
(

(





))(

 0l j ( x1  2 1  ( t  s ),x2 
d 1d 2 d 3 

1
2
3

ts
j 1
 3 0 R3


2 2  ( t  s ),x3  2 3  ( t  s );s ))d 1d 2 d 3ds}  0,
i.e. the system (4.6) satisfies a problem (4.5). That it was required to prove.
The limiting case in G 1 ( D0 ) , when the decision of (4.5) is representing in the form of (4.6), when
( x1 ,x2 ,x3 ,t )  T ; li  xi  2 i ( t  s ),( i  1,3 );V0 ; 0 : sup D kV0  1 , ( k  0,3 ),

R3

T0
sup D k ( x ,x ,x ,t )   ; sup  ( x ,x ,x ,s )ds   T   ,
0
1
2
3
1
1 0
2
 0 1 2 3
 R3
R3 0

t
1

sup
exp( (  12   22   32 ) D k  0 ( l1 ,l2 ,l3 ;s ) d 1d 2 d 3 ds   1T0   2 ,
 T
3  
 0 R3

t
3

1
1
2
2
2
exp
(

(





)
sup
  j  0l j ( l1 ,l2 ,l3 ;s ) d 1d 2 d 3ds  3 1 2T0  3 ,
1
2
3
3  
t  s j 1
 T  0 R3

3
1
1
2
2
2
sup
exp
(

(





))(
 j  V0 l j ( l1 ,l2 ,l3 ) )d 1d 2 d 3  1


1
2
3

 R3  3 R3
i 1
3
 3
1
1
{ (  2 exp( (  2   2   2 ) )d d d ) 2 ( exp( (  2   2   2 ) )d d d ) 2 }  3  1 ,
1
2
3
1
2
3
1
2
3
1
2
3
1
 i

 
2
i  1 R3
R3

 li  xi  2 i t , ( i  1,3 ),   max  i ; 0   ( 3 2 T0  1  T0  ).
1 i  3

(4.8)
Really, estimating (4.6) in G 1 ( D0 ) , we have

V


 Vt

V


G1 ( D0 )
 V
C 3 ,0 ( T )
 Vt
L1
 N 3  0 ,
T0
L1
 sup  Vt ( x1 ,x2 ,x3 ,t )dt   ( 3 2 T0  1  T0  )  0 ,
R3
C 3 ,0 ( T )

0

0  k 3
D kV
C( T )
 N 3  40  , ( V
C( T )
 2  ).
The uniqueness of the solution the system (4.6) in G 1 ( D0 ) is obvious on the basis of proof by
contradiction [13]. Results (4.6) with a condition (4.2), (4.8) are received where smoothness of
functions is required only on xi as the derivative of 1st order is in time has t>0.
Therefore, on a basis transformation (4.2) we will receive decisions of system (1.1), which satisfies
a condition (1.2), i.e.
i  i H 0 ( x1 ,x2 ,x3 ,t ), ( i  1,3 ),

3
3
3
1
 3
2
2
2



H

0;

H

exp
(

(





))
iV0 hi ( x1  2 1 t ,x2 




ix
i
0
x
i
0
x
1
2
3

i
i
i
3 
3
i 1
i 1
i 1
i 1

R

t
3

1
2
2
2

2


t
,x

2


t
)d

d

d


exp
(

(





))



2
3
3
1
2
3
1
2
3
i 0li ( x1  2 1 
3  
3
i

1

0 R


  ( t  s ),x2  2 2  ( t  s ),x3  2 3  ( t  s );s )d 1d 2 d 3ds  0,
h  x  2 t ; l  x  2 ( t  s ),( i  1,3 ).
i
i
i
i
i
 i

(4.9)
In the conclusion estimating (4.9) it is had
3

v

(

,

,

)
:
v

[ iV C 3 ,0 ( T )  iVt L1 ]  d0 [N 3  0 ]  d0 M 0 ,

1
2
3

Gn13 ( D0 )

i 1

3
V 1

V

V

N


,(
d

i ;M 0  N 3  0 ).
3 ,0
1

t L
3
0
0
C (T )
 G ( D0 )
i 1
(4.10)
Theorem 4. In the conditions of (1.2), (4.1), (4.8) and (4.10) the problem (1.1), (1.2), (4.1) has a
unique solution in Gn1 3 ( D0 ) , which is defined by a rule (4.9).
Remark 2. If i0 ,( i  1,2,3 ) initial components of a vector of speed  at the moment of time t  0 it
is set in a kind:
i
t 0
 0,( x1 ,x2 ,x3 )  R 3 ,( i  1,3 ).
(4.1)*
Then with that end in view we will assume that there are functions 0  iV0 ,( i  1,2,3 ) , which
satisfy conditions
C 3 ( R 3 )  0 ( x1 ,x2 ,x3 );0  i  const, (i  1,3 ):

3

k
sup
D



(

),(
0



1;


0;

0 ).

0
0
0
i 0  xi  0;  0 (  ) 
 3
 0
i 1
R
Hence we will use transformation of a kind
i  i [V ( x1 ,x2 ,x3 ,t )  0 ( x1 ,x2 ,x3 )], ( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),

3
V t 0  0 ( x1 ,x2 ,x3 ), ( x1 ,x2 ,x3 )  R ,

3
3
div  0 :   jVx j  0;   j0 x j  0,
j 1
j 1

 3
3
3
3
3
 jix  iV   jVx  i0    jVx  iV   j0  x  i0    j0 x  0,
j
j
j
j
j
 j 1
j 1
j 1
j 1
j 1
(4.2)*
that, the system (1.1) will be transformed to a kind
iVt  fi 
1

Pxi  i V ,( i  1,3 ).
(4.3)*
Further, from system (4.3)*, considering condition (4.2)*, and having entered [8] APS we have the
equation (4.4). Therefore the system (4.13) will be transformed to (4.5). Hence from system (4.5), we
have the of equation (4.6). In the conclusion, with the account (4.2)*, follows
i  i [ 0 ( x1 ,x2 ,x3 )  H 0 ( x1 ,x2 ,x3 ,t )]  Yi ( x1 ,x2 ,x3 ,t ),( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),

1

exp( (  12   22   32 ))0 ( x1  2 1 t ,x2  2 2 t ,x3 
 0 ( x1 ,x2 ,x3 ) 
3 
 R3


2 3 t )d 1d 2 d 3  0 ( x1 ,x2 ,x3 ),

t
H  1
exp( (  12   22   32 )) 0 ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 
3  
 0
 0 R3

  ( t  s );s )d 1d 2 d 3 ds,( si  xi  2 i t ;si  xi  2 i ( t  s );i  1,3 )

(4.9)*
and summarising, takes place an estimation of a kind (4.10), where Yi ,( i  1,2,3 ) – known functions.
From here we will receive similar results in the conditions of the theorem 4 in Gn1 3 ( D0 ) .
4.2. Fluid with Small Viscosity, when 0  0; divf  0
I. The overall objective of this point: to change a method (4.2) so that the received anal ytical
solution of the Navier-Stokes problem with viscosity, belonged in Cn3,1 3 (T ).
If takes place
i t 0  i0 ( x1 ,x2 ,x3 )  
i 0 ( x1 ,x2 ,x3 ),i  1,3,
3

3
3
k
  j0 x j  0; 0  0; 0  C ( R ); divf  0; sup D f i  N0  const , ( i  1,3;k  0,4 ),
T
 j 1
(4.11)
that we will use transformation of a kind
i  i [0 ( x1 ,x2 ,x3 )  Z( x1 ,x2 ,x3 ,t )],( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),

3
 Z t 0  0, ( x1 ,x2 ,x3 )  R ,

3
3
div  0 :   j Z x j  0;   j0 x j  0,
j 1
j 1

 3
3
3
3
3
 jix  
i 0   j0 x j  
i 0   j Z x j  i Z   j0 x j  i Z   j Z x j  0,
j
 j 1
j 1
j 1
j 1
j 1
(4.12)
where 0  i  the known constants. Hence, the system (1.1) will be transformed to a kind
i Zt  fi 
1

Pxi  i Z ,i  1,3.
(4.13)
From system (4.13), considering conditions (4.11), (4.12), and having entered [8] APS we have the
equation (4.4). Hence the system (4.13) will be transformed to a kind
 Z    Z , ( x ,x ,x ,t )  T ,
0
1 2
3
 t
 Z t  o  0,

1
1
1
1
1
1
( 1 ) ( f1   Px1 )  ( 2 ) ( f 2   Px2 )  ( 3 ) ( f 3   Px3 )   0 ( x1 ,x2 ,x3 ,t ).
(4.14)
Then the decision of a problem (4.14) is presented in a kind
Z

1
8 3
  exp( 
0 R3
t
1

t
3
  exp( ( 
2
1
r2
1
)
0 ( s1 ,s2 ,s3 ,s )ds1ds2 ds3 ds 
4 ( t  s ) ( ( t  s ))3
  22   32 )) 0 ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s ) 
0 R3
(4.15)
d 1d 2 d 3 ds  H( x1 ,x2 ,x3 ,t ),( x1 ,x2 ,x3 ,t )  T , ( si  xi  2 i ( t  s );i  1,3 ),
here H – known function. The solution (4.15) satisfies system (4.14).
Really, having calculated partial derivative of system (4.15):
3
j

1 t
2
2
2
Z


(
x
,x
,x
,t
)

exp
(

(





))

 0l j ( x1  2 1 
 t

0
1
2
3
1
2
3


3
3
t

s
j

1

0
R

  ( t  s ),x  2 ( t  s ),x  2 ( t  s );s )d d d ds,
2
2
3
3
1
2
3

t

1
exp( (  12   22   32 )) 0l j ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3 

Z x j 
3  
 0 R3


2 3  ( t  s );s )d 1d 2 d 3 ds,

1 t
 Z x2 
exp( (  12   22   32 )) 0l 2 ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3 
j
j
3  

 0 R3


2 3  ( t  s );s )d 1d 2 d 3 ds, ( l j  x j  2 j ( t  s ); j  1,3 )
and substituting (4.16) in (4.14), we
(4.16)
Z
 0, ( x1 ,x2 ,x3 )  R 3 ; ( 0,1 )   ; l j  x j  2 j ( t  s ); ( x1 ,x2 ,x3 ,t )  T :
t 0

t
3
j

1
2
2
2
 0l j ( x1  2 1 

))





(

(
exp



Z




Z

0


3
2
1
0
0
t


3
s

t
3
1

j

0
R

  ( t  s ),x  2 ( t  s ),x  2 ( t  s );s )d d d ds   
0
3
2
1
3
3
2
2

t
3

1
exp( (  12   22   32 )) 0l 2 ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3 
  {
3  
j
j 1

 0 R3

t
3
j
2
2
2
2  ( t  s );s )d d d ds}  1
 0l j ( x1 

))





(

(
exp

3
2
1
3
2
1
3


3

ts
j 1
 0 R3

2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 ( t  s );s )d 1d 2 d 3ds 

t
1
1
 1
[  exp( (  12   22   32 )) 0l 2 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),
{



 2
3
1
 0 t  s R3


2
2
2
 x3  2 3  ( t  s );s )d( x1  2 1 ( t  s ) )d 2 d 3  3 exp( (  1   2   3 )) 0l22 ( x1  2 1 
R

  ( t  s ),x  2 ( t  s ),x  2 ( t  s );s )d d( x  2 ( t  s ) )d 
3
2
2
1
3
3
2
2

2
2
2
 exp( (  1   2   3 )) 2 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s ) 
0l3
 R3

t
3
j
1

2
2
2
 0l j ( x1 

))





(

(
exp

}
ds
]
)
)
s

t
(


2

x
d(

d

d


3
2
1
3
3
2
1



3
ts
j 1

0 R3

2  ( t  s ),x  2  ( t  s ),x  2 ( t  s );s )d d d ds 
3
2
1
3
3
2
2
1

t
3

1
1
exp( (  12   22   32 )) j 0l j ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),
  { 3 

j 1
 0 t  s R3


 x3  2 3  ( t  s );s )d 1d 2 d 3 ds  0.
That it was required to prove.
From the received results follows that functions i are defined on the basis of (4.12), i.e.
i  i [0 ( x1 ,x2 ,x3 )  H( x1 ,x2 ,x3 ,t )],( x1 ,x2 ,x3 ,t ) T ,( i  1,3 ).
(4.17)
Further, considering partial derivatives of 1st order systems (4.17) and summing up with acceptance in
attention (1.2), (4.12) we have, that the system (4.17) satisfies to a condition (1.2)
3
3
3
3
 3
1 t
2
2
2





H



exp
(

(





))
{

 ixi  i 0 xi  i xi  i 0 xi

1
2
3
i 0li ( x1 

i 1
i 1
i 1
i 1
 3 0 R3
 i1

2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s )}d 1d 2 d 3 ds  0,

3
3
li  xi  2 i ( t  s ),( i  1,3 ),  

0;


i 0 xi
i 0 xi  0.

i 1
i 1

II. So as i0  i0 , ( 0  C 3 ( R 3 )) , that solution of problem (1.1)-(1.3) belongs in Cn3,13 (T ) :
3

k


 Cn3,13 {  D i C  it C },
i 1 0  k  3

 3,1
3,3,3,1
3,3,3,1
Cn  3 ( T )  Cn  3 ( T )  Cn  3 ( T );v  ( 1 ,2 ,3 ); i  i [0  Z ],( i  1,3 ); 0  0,

k
3
 Z
k
0
k
 Z 3 ,1 
D
Z

Z
;
k

0
:
D
Z

Z
;
k

0
:
D
Z

;
k

 i , (  i  0,3 ).
k  3

t C
3
1
2
C
 C

x

x

x
0

i

1
1
2
3

Really, if
i0  C 3 ( R 3 ); 0 : sup D ki0   1 ; sup D k 0 ( x1 ,x2 ,x3 ,t )  1 , ( i  1,3; k  0,3 ),
T

R3

t
sup 1
exp( (  12   22   32 )) D k  0 ( l1 ,l2 ,l3 ; ) d 1d 2 d 3 d  1T0   2 ,


3
 T  0 R3

t
3
1
1

2
2
2
sup
exp
(

(





)
)
 j   0l( kk ) ( l1 ,l2 ,l3 ; ) d 1d 2 d 3 d 


1
2
3
3  
j
T
t   j 1
 0 R3


t
3
 1 1 sup exp( (  12   22   32 )) 1   j d 1d 2 d 3 d  31 2T0   3 ,


t   j 1
 3 T 0 R3

l j  x j  2 j ( t   ),( j  1,3 ),   max(  i ; 1 ), 0   ( 1   ),
1 i  3

(4.18)
that on a basis (4.15) we will receive:
Z
C3
 N1   0 , ( 
0
,1
N
1
2 0
).
In the conclusion estimating (4.17) it is had

v  (  , , );    [  Z ],( i  1,3 ) :
1
2
3
i
i
0

3

 v Cn3,13 ( T )   [ i Z C 3 ,0 ( T )  i0 C 3 ( T )  i Zt C( T ) ]  d0 [N1  N 2  0 ]  d0 [N0  0 ],
i 1

3

Z

Z

Z

N


,
(
d

i ; N0  N1  N 2  40  ).
 C 3 ,1 ( T )

t C( T )
1
0
0
C 3 ,0 ( T )
i 1

Lemma 2. In the conditions of (1.2), (4.11), (4.12) and (4.18) the equation (4.15) has a unique
solution in C 3,1 ( T ) .
Theorem 4*. At performance of conditions of the lemma 2 the problem (1.1), (1.2), (4.11)
is solvable at Cn3,13 (T ) of the defined by a rule (4.17).
Remark 3. Results 4.2 taking into account transformation (4.12) can be used concerning
equations Navier-Stokes (1.1):
i 3

1
  j i  fi  Pxi  i ,( i  1,3 )
t j 1 x j

and
(1)
div  0, ( x, t )  T  U 3  [0, T0 ],
(2)
in the limited area U 3 spaces of variables ( x1 ,x2 ,x3 ) the limited surface S , and for t , satisfying
to an inequality 0  t  T0 , under conditions [13]:
i S  0, ( i  1,3 ),
(3)
i t 0  
i 0 ( x1 ,x2 ,x3 ),( 0 S  0 ).
(4)
Really, applying transformation (4.12), i.e.:
i  i [0 ( x1 ,x2 ,x3 )  Z( x1 ,x2 ,x3 ,t )],( x1 ,x2 ,x3 ,t ) T  U 3  [0,T0 ],( i  1,3 ),

 Z S  0, t  [0,T0 ],

3
 Z t 0  0, ( x1 ,x2 ,x3 ) U ,
we will receive the equation (4.13), i.e.:
i Zt  fi 
1

(5)
Pxi  i Z ,( i  1,3 ).
(6)
Hence (4.14):
1

 P  4 F0 , ( F0  
1
4
3
f
i 1
ixi
(7)
( x1 ,x2 ,x3 ,t )).
The equation (7) is the equation of Poisson and for descriptive reasons decisions we put a boundary
problem in a kind:
 P S   1  0, t  [0,T0 ],

P
 n   2  0.
S

(8)
Then equation (7) we obtain explicit representations for the unknown pressure
P :
1
r dS  1
1
n
4
3
(9)

 Z t   0  Z , ( x1 ,x2 ,x3 ,t )  T ,

1
1
1
1
1
1

( 1 ) ( f1   Px1 )  ( 2 ) ( f 2   Px2 )  ( 3 ) ( f 3   Px3 )   0 ( x1 ,x2 ,x3 ,t ),
(10)
ds ds ds
1
P   F0 ( s1 ,s2 ,s3 ,t ) 1 2 3 

r
4
U3
1

S

1
 r 
S
2
dS 
1
 r F ( s ,s ,s ,t ) 
0
1
2
U3
ds1ds2 ds3 , [see. (0.8) :  1 , 2  0; r  ( x1  s1 )2  ( x2  s2 )2  ( x3  s3 )2 ] .
Thus, considering (6), (9) we will receive [13]:

Z


Z
S
 0, t  [0,T0 ],
t 0
 0, ( x1 ,x2 ,x3 ) U 3 .
(11)
Full research of a problem (10), (11) is well shown in work [13, pp.349-351]. And it means that the
generalized solution will be the decision in usual sense, at sufficient smoothness 0 [13, pp.311-318,
XXII]. Therefore similar conclusions we can make concerning a problem (1)-(4). That it was required
to prove.
Remark 4. In point 2.2 Beale-Kato-Majda inequality is resulted. Therefore, here we will specify
communication criterion of a regularity of Beale-Kato-Majda with results of the theorem 4*. On the
basis of results of the theorem 4* the solution of system (1.1) it is presented in a kind (4.17), where
global existence of solution is received in a class Cn3,13 (T ) from the point of view of the initial data
satisfying (4.17).
Really, at performance of conditions of theorem 4* and
rot  0, ( v  ( 1 ,2 ,3 )), i  i [0  Z ] : 30 x ( x1 ,x2 ,x3 )  20 x ( x1 ,x2 ,x3 )  h10 ,
2
3

0
 10 x ( x1 ,x2 ,x3 )  10 x ( x1 ,x2 ,x3 )  h2 ; 20 x ( x1 ,x2 ,x3 )  10 x ( x1 ,x2 ,x3 )  h30 ,
3
1
1
2

t

1
exp( (  12   22   32 )) 3 0l2 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3  2 3 
sup


3
 T  0 R3

  ( t  s );s )  2 0l3 ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s ) 

d 1d 2 d 3 ds  h1 ,( li  xi  2 i ( t  s ),i  1,3 ),
t

1
exp( (  12   22   32 )) 1 0l3 ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 
sup
3  
T

 0 R3

  ( t  s );s )  3 0l1 ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s ) 
d d d ds  h ,
1
2
3
2

t

1
exp( (  12   22   32 )) 2 0l1 ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 
sup
3  
T
 0 R3


  ( t  s );s )  1 0l2 ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s ) d 1d 2 d 3 ds  h3 ,
takes place
3
rot  0 : sup rot ( x1 ,x2 ,x3 ,t )  M 0   , (  ( hi  hi0 )  M 0 ).
(4.19)
i 1
T
In a consequence and (see (2.24)):
T0
T0
sup  rot ( x1 ,x2 ,x3 ,t ) dt  sup  { 3 x2 ( x1 ,x2 ,x3 ,s )  2 x3 ( x1 ,x2 ,x3 ,s ) 
R3
0
R3
0
(4.20)
 1x3 ( x1 ,x2 ,x3 ,s )  3 x1 ( x1 ,x2 ,x3 ,s )  2 x1 ( x1 ,x2 ,x3 ,s )  1x2 ( x1 ,x2 ,x3 ,s ) }ds  M 0T0  M  .
Well then, we obtain an estimation of type Beale-Kato-Majda [5].
4.3. Modified Variant of a Method (4.12) with the Conditions divf  0,rot f  0
Let’s consider updating of the basic method 4.1. Here we will see that components of speed grow
than in rules (4.2) or (4.12), when 0    1 , on the basis of regulatory functions i . Therefore in
this case the problem (1.1)-(1.3) does not contain in itself restriction in kinds
(А1), (А2) and in it urgency of research in a case (А3) and divf  0, rot f  0 .
Offered methods of integrated transformations are entered so that to transform nonlinear of the
Navier-Stokes problems to linear problems of heat conductivity. Questions solvability of the problem
are proved on the basis of the developed methods. These methods are initial problems lead to the
integral equations where possibility to find the analytical solution taking into account the theory of the
integral equations of Volterra-Abel [13].
For incompressible currents with a friction, when
i t 0  i0 ( x1 ,x2 ,x3 )  
i 0 ( x1 ,x2 ,x3 ),( i  1,3 ),
 3

3
3
i 0 xi  0; 0  0; 0  C ( R ),
 
 i 1
divf  0, rot f  0,(  f  0; i  1,3 ), sup D k f  N  const , ( T  R 3  [0,T ] ),
i
i
0
0

T

t
 ( x ,x ,x ,t )   f ( x ,x ,x ,s )ds, ( i  1,3 ),
0 i 1 2 3
 i 1 2 3

(4.21)
functions i ,i  1,2,3 is represented in a kind

i  i [0 ( x1 ,x2 ,x3 )  Z( x1 ,x2 ,x3 ,t )]  i ( x1 ,x2 ,x3 ,t ),( x1 ,x2 ,x3 ,t ) T ,( i  1,3 ),

3

 Z t 0  0, ( x1 ,x2 ,x3 )  R ,
(4.22)
where 0  i  the known constants. From the entered functions i , is required
it   fi ( x1 ,x2 ,x3 ,t ),( i  1,3 ),

t

k
k
 fi  0 : i  0, ( D i    D f i ( x1 ,x2 ,x3 ,t,s )ds  N 0 T0 ; i  1,3 ),
0

at that all functions i are regular concerning viscosity parameter ( 0,1 )   , (here  in a role of
small parameter), i.e.:
lim i ( x1 ,x2 ,x3 ,t )  0,( x1 ,x2 ,x3 ,t ) T.
 0
Further, supposing (4.21), (4.22) and
3
3
3

div


0
:

Z

0;


0;
ixi  0,



i xi
i 0 xi

i 1
i 1
i 1

3
3
1 3 2
1

2
U    j ;   j  i x j  2 (   j )xi  2 U xi ,
j 1
j 1
j 1

3
3
 3
1
 ji x j    j ( 0  Z ) i x j    j i ( 0 x j  Z x j )  U xi ,
2
j 1
j 1
 j 1
3
 3
  j ( 0  Z )i ( 0 x j  Z x j )  i ( 0  Z )  j ( 0 x j  Z x j )  0,
j 1
 j 1

it  i Zt  it ; i  [i ( 0   Z )   i ]  i  Z ,( i  1,3; 0  0 ),

(4.23)
for incompressible currents with a friction the equations of Navier-Stokes (1.1) become simpler as take
place (4.21)-(4.23). Therefore the problem (1.1) - (1.3), is led to a kind
3
3
1
2
i Zt    j ( 0  Z )i x    j i ( 0 x  Z x )  U x  ( 1   ) fi 
j 1
j
j 1
j
j
i
1

Pxi  i Z ,( i  1,3 ).
(4.24)
The system (1.1) is equivalent transformed to a kind (4.24). Then a new system of equations is called
of inhomogeneous linear system a transfer of vortices [12]. Here the inertial terms in the equations
(1.1) are linearized using regulatory functions i ,( i  1,2,3 ) , which were first introduced in [8] and
in the method (4.22).
From system (4.24), considering conditions (4.21)-(4.23) and having entered APS we have the equation:
1
1
 3 
 x (4.24) : (  P  2 U )  {F0  B[Z x1 ,Z x2 ,Z x3 ]},
 i 1 i
1
1
1 1
{F0 ( s1 ,s2 ,s3 ,t )  ( B[Z s1 ,Z s2 ,Z s3 ] )( s1 ,s2 ,s3 ,t )}ds1ds2ds3 ,
 P U 
2
4 R3 r

1
 Px  1 U x  1
{F0 ( x1   1 ,x2   2 ,x3   3 ;t )  ( B[Z h1 ,Z h2 ,Z h3 ] )( x1   1 ,x2   2 ,x3 
  i 2 i 4 R3

i
 ; t )}
d 1d 2 d 3 , ( si  xi   i ;hi  xi   i ;i  1,3 ),
3
2
2
2 3

(





)
1
2
3

(4.25)
so as takes place
3
3
3
3

(
B
[
Z
,Z
,Z
]
)(
x
,x
,x
,t
)

(


)Z

(

 j ix j xi 
  jxi Z x j )i ,
x1
x2
x3
1 2
3

i 1 j 1
i 1 j 1

3
3
3
3

div
f

0;
F
(
x
,x
,x
,t
)


(


)


(



0
1 2
3
i 
j xi 0 x j
0 xi   j  i x j ),
i

1
j

1
i

1
j 1

3
3
3
3
3
3
3


 {  j ( 0  Z ) i x j    j i ( 0 x j  Z x j )}   i (   j xi 0 x j )  0 xi (   j  i x j ) 
j 1
i 1
j 1
i 1
j 1
 i 1 xi j 1
 3 3
3
3
  (   j  ix j )Z xi   (   jxi Z x j )i ,
i 1 j 1
 i 1 j 1
3
3
3

(   j x j )xi  0; (   j0 x j )xi  0, (   j Z x j )xi  0,( i  1,3 ),
j 1
j 1
 j 1
 3
3
3
3

 1
1

1
1
 [ i Z x ]  0;   ( i Z )  0;  ( U x )  U ;  (  Px )    P,
i
i
i
2


 t i 1
i  1 xi
i  1 xi 2
i  1 xi

B[Z x1 ,Z x2 ,Z x3 ]  0,( x1 ,x2 ,x3 ,t )  T .
lim
 0

There are various methods [12] which give communication of speed and pressure. For example, if
the law of the pressure distribution obtained from the Bernoulli equation. Here is received the new law
of distribution of pressure in the form of (4.25), for the first time similar results are received in work
[8]. Then on a basis (4.25) system (4.24) it is equivalent, will be transformed to a kind:

Zt  0  ( Q[Z ,Z x1 ,Z x2 ,Z x3 ] )( x1 ,x2 ,x3 ,t )  Z ,( i  1,3 ),

3

Z t 0  0,( x1 ,x2 ,x3 )  R ,
where
(4.26)
3
3
3
3

1

(
x
,x
,x
,t
)





d
[
(
1


)
f


(


j 0xj
0 
i
0   j i x j ) 
 0 1 2 3
j 1
i 1
i 1
j 1

3
3
 1
i

{
(
F
(
x


,x


,x


;t
))
}
d

d

d

]
;
d

i  0,


0
1
1
2
2
3
3
1
2
3
0
3 
2
2
2 3
4

i

1
i

1
(





)
R

1
2
3

3
3

1
(
Q
[
Z
,Z
,Z
,Z
]
)(
x
,x
,x
,t
)


{
d
Z(
x
,x
,x
,t
)
[
(
 j  i x j ( x1 ,x2 ,x3 ,t ))] 



x1
x2
x3
1
2
3
0
1
2
3
i

1
j

1

3
 3
i
1
(
( B[Z h1 ,Z h2 ,Z h3 ] )( x1 
  Z x j ( x1 ,x2 ,x3 ,t ) j ( x1 ,x2 ,x3 ,t )  d 01[


4 R3 i 1 (  12   22   32 )3
 j 1

 1 ,x2   2 ,x3   3 ;t ))d 1d 2 d 3 ]}, (hi  xi   i ,i  1,3 ).

The problem (4.26) is led to system of the integrated equations quite regular rather   ( 0,1 ) , in a kind
 Z x  Wi ( x1 ,x2 ,x3 ,t ),( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),
 i
t

1
r2
1
Z

M

exp
(

)
( Q[Z ,W1 ,W2 ,W3 ] )( s1 ,s2 ,s3 ,s )ds1ds2 ds3ds 

1


3
3
4

(
t

s
)
(

(
t

s
))
3
8

0 R

t

1
 M 1 
exp( (  12   22   32 ))( Q[Z ,W1 ,W2 ,W3 ] )( x1  2 1  ( t  s ),x2  2 2 
3  

 0 R3

  ( t  s ),x3  2 3  ( t  s );s )d 1d 2 d 3ds  ( 0 [Z ,W1 ,W2 ,W3 ] )( x1 ,x2 ,x3 ,t ),

t
r2
( xi  si )
1
W  M  1
(
exp
(

))
( Q[Z ,W1 ,W2 ,W3 ] ) 
1xi
 i


4  ( t  s ) 2 ( t  s ) (  ( t  s ))3
8  3 0 R3


t
i
( s ,s ,s ,s )ds ds ds ds  M  1
exp( (  12   22   32 ))

1 2 3
1
2
3
1xi


3

( t  s )
 0 R3

( Q[Z ,W1 ,W2 ,W3 ] )( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s ) 

d 1d 2 d 3 ds  ( i [Z ,W1 ,W2 ,W3 ] )( x1 ,x2 ,x3 ,t ), ( si  xi  2 i ( t  s );i  1,3 ),

t
 M ( x ,x ,x ,t )  1
exp( (  12   22   32 )) 0 ( x1  2 1 ( t  s ),x2  2 2 
3  
 1 1 2 3
 0 R3
(4.27)

  ( t  s ) ,x3  2 3  ( t  s );s )d 1d 2 d 3 ds.
The received system (4.27) consists of four integral equations, i.e. Volterra and Volterra-Abel on a
variable t  [0,T0 ] and thus contains in itself of four unknown functions. Therefore there is no
necessity to think out various algorithms for the decision of this system and it is enough to show conditions
which provide contraction mapping principle and for the decision of this system to use a of Picard’s method.
Therefore, if takes place:
( x1 ,x2 ,x3 ,t )  T ;M 1 , ,i : sup D k M 1 ( x1 ,x2 ,x3 ,t )  1 , ( k  0,3; t  s   ),
T

3
3

sup  ( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, )  sup{d 01 (   j  i s ( x1  2 1  ,x2  2 2  ,
j
T T
i 1 j 1
 T T

3
 x3  2 3  ;t   ) )    j ( x1  2 1  ,x2  2 2  ,x3  2 3  ;t   ) 

j 1

3
3
3
i
 d 1[ 1 (
{ (   j  il ( x1  2 1    1 ,x2  2 2    2 ,
 0 4 3 
j
i 1
(  12   22   32 )3 i 1 j 1
R

3
3

x

2




;t


)
)

(
 jl ( x1  2 1    1 ,x2  2 2    2 ,x3 
 3


3
3
i
i

1
j

1

2    ;t   ) ) } )d d d ]}    ,
3
3
i
1
2
3
2

t

1
k0 
sup exp( (  12   22   32 )) ( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, )d 1d 2 d 3d   2 T0 ,
3 T  


0 R3

t
k  1 sup exp( (  2   2   2 )) ( x ,x ,x , , , ;t, )  i d d d d 
1
2
3
1 2
3 1 2 3
1
2
3

 i

 3 T 0 R3

 2T0  2  ,( i  1,3 ),   max(  2T0 ; 3 2T0  2 )


(4.28)
and if operators:  i ,( i  0,3 ) compressing with a compression factor k i ,
h

 i ,( i  0,3 ) : ki  4 ,( h  1 ),
 3
 ki   (  2T0   3 2T0  2 )   (   1 )  h  1
 i 0
(4.29)
and:
 Sr ( 0 )  {Z ,W j : V ; W j  r1 ,( x1 ,x2 ,x3 ,t )  T },( j  1,3 );
 1
  i [0,0,0,0] C  r1 ( 1  h ) :  i [Z ,W1 ,W2 ,W3 ] C   i [Z ,W1 ,W2 ,W3 ]  i [0,0,0,0 ] C 
(4.30)



[
0,0,0,0
]

k
4r

r
(
1

h
)

hr

r
(
1

h
)

r
,
i
i
1
1
1
1
1

C

 i : Sr1 ( 0 )  Sr1 ( 0 ), ( i  0,3 ).
Then on the basis of a contraction mapping principle system (4.27) is solvable and solution of this
system we can find on the basis of Picard’s method

( x1 ,x2 ,x3 ,t )  T : Z n1  0 [Z n ,W1,n ,W2,n ,W3,n ],
(4.31)

W


[
Z
,W
,W
,W
]
,(
n

0,1,...;Z

0;
W

0;
i

1,3
).

i
,n

1
i
n
1,n
2,n
3,n
0
i
,0

Received the sequence of functions Z n1 ,Wi , n1 ,( i  1,2,3 ) is converging and fundamental in S r1 ( 0 ) ,
at that elements of the constructed sequences belong S r1 ( 0 ) , so as
3
3

E

Z

Z

W

W
;
E

Z

Z

Wi ,n  Wi ,n 1 C ,


n 1
n C
i ,n  1
i ,n C
n
n
n 1 C
 n 1
i 1
i 1

 Z n  1  Z n  k0 En ; Wi ,n  1  Wi ,n  ki En ,( i  1,3 ) :
C
C

h 1
n
 En  1  hEn  ...  h E1 
 0,
n 

k 1
k 1
 Z Z  k E ; W

W

ki En  j ,( i  1,3 ),


n C
0 n j
i ,n  k
i ,n C
 nk
j 0
j 0

k 1
k 1
k 1

1
h 1
 En  k  h En  j  ...  h h n  j 1 E1  E1h n  h j  E1h n

 0,
n 
1 h

j 0
j 0
j 0

n 1
n 1
 Z n  1  Z n  1  Z n  ...  Z 2  Z1  Z1  k0 (  h j )E1   0 [0,0,0,0 ]  ( 1  h )r1h  h j 
C

j 0
j 0

n

( 1  h )r  ( 1  h )r h j  r , [ h j  ( 1  h )1 ; k  1 h; E  4( 1  h )r ; Z  S ( 0 )],

1
1
1
0
1
1
n 1
r1

4
j 0
j 0

n 1
n
1
W
j
j

k
(
h
)E


[
0,0,0,0
]

(
1

h
)r
h; i  1,3;Wi , n  1  S r1 ( 0 )].
i 
1
i
1  h  r1 , [k i 
C
 i , n 1
4
j

0
j

0

Therefore
3
3

h1
U

Z

W
;
U

Z

Z

Wi ,n1  Wi C : U n 1  h n1U 0 
 0,


0
i
n

1
n

1
n

C
C
C
i 1
i 1

h1
h1
3,1
 Z 
Wi ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).
n
 n1 n  Z  H  C (T ); Wi ,n1 
Hence on the basis of (4.22)
i ,n  1  i [0 ( x1 ,x2 ,x3 )  Z n  1( x1 ,x2 ,x3 ,t )]  i ( x1 ,x2 ,x3 ,t ),( n  0,1,2,...;i  1,3 ),

h 1
n 1
 0,( i  1,3 ).
 i ,n  1  i C  i Z n  1  Z C  i hU n  i h U 0 
n 
(4.32)
(4.33)
And it means that sequence {i ,n }0 converging to a limit i ,( i  1,3 ) :
h 1
i ,n 1 
i  C 3,1(T ),( i  1,3 ).
n 
(4.34)
Theorem 5*. Under conditions (1.2), (1.3), (4.21), (4.22), (4.28)-(4.30) and (4.34) the Navier-Stokes
problem is solvable at Cn3,13 (T ) .
Remark 5. If takes place 0    21 , that 0    1. In a case   2 1 , then
0    ( 2 1[ 1  4  1  1] )2  1.
4.4. Updating of a Method (4.2), When divf  0, rot f  0, 0  0
Let consider updating of the basic method of paragraph 4.1: (4.2), when:
(4.35)
i t 0  i0 ( x1 ,x2 ,x3 )  i0 ( x1 ,x2 ,x3 ),( i  1,3 ),

3

3
3


C
(
R
);


0;
 i0 xi  0,
 0
0
i 1


3
 fi  i  it ; divf  0 : div  0;   it  0,


i 1 xi

it ( x1 ,x2 ,x3 ,t )   fi ( x1 ,x2 ,x3 ,t ), ( i  1,3 ),   ( 1 ,2 ,3 ); f  ( f 1 , f 2 , f 3 ),
divf  0; rot f  0 :  f  0,(   0 ),
i
i

t
3
3
3

1
1
2
2
U    j ;   j ix j  (   j )xi  U xi ; i    fi ( x1 ,x2 ,x3 ,s )ds,( i  1,3 ).
2 j 1
2

j 1
j 1
0
Then functions i ,i  1,3 is represented in a kind
(4.36)
i  iV ( x1 ,x2 ,x3 ,t )  i ( x1 ,x2 ,x3 ,t ),( i  1,3 ),

V t 0  0 ( x1 ,x2 ,x3 ),( x1 ,x2 ,x3 )  R 3 ,

(4.37)
div  0 :
3
t 3
3
  V  0;



 ixi  (  fixi ( x1 ,x2 ,x3 ,s ))ds  0.
 i xi
i 1
0 i 1
i  1
Further, supposing (4.36), (4.37) and
3
3
3
3

1
   ji x j  V   j ix j  i  Vx j  j  U xi ,( iV   jVx j  0 ),
2
(4.38)
 j 1
j 1
j 1
j 1

it  iVt  it , i  [i V  i ]  i V ,( i  1,3; i  0 ),
for incompressible currents with a friction the equations of Navier-Stokes (1.1) become simpler as take
place (4.36)-(4.38). Therefore the problem (1.1)-(1.3), is led to a kind
3
3
1
1
(4.39)
iVt  V   j ix j  i  Vx j  j  U xi  i  Pxi  i V ,( i  1,3 ).
2

j 1
j 1
From system (4.39), considering conditions (4.36)-(4.38) and having entered APS we have the
equation:
3 
1
1
 (4.39) : ( P  U )   B[Vx1 ,Vx2 ,Vx3 ],

2
i 1 xi

3
3
3
3
( B[V ,V ,V ] )( x ,x ,x ,t )  V (


)

(
 xi  j ix j    jxiVx j )i ,
x1 x2 x3
1 2 3

i 1
j 1
i 1 j 1

1
1
1
1
{( B[Vs1 ,Vs2 ,Vs3 ] )( s1 ,s2 ,s3 ,t )}ds1ds2 ds3 ,
(4.40)
 P U 

2
4 3 r

R

i
 1 P  1U  1
{( B[Vh1 ,Vh2 ,Vh3 ] )( x1   1 ,x2   2 ,x3 
xi
xi


2
4 3 (  2   2   2 )3
1
2
3
R

 ;t )}d d d , ( s  x   ;h  x   ;i  1,3 ),
1 2 3
i
i
i i
i
i
 3

so as takes place
3
3
3
 3

 1
1

1
1
( i V )  0;  ( U xi )  U ;  (  Pxi )   P,
 [ iVxi ]  0;  
2


 t i 1
i 1 xi
i 1 xi 2
i 1 xi
3
3
3
3
3
3
3

 3

(V


)


(
V

)


(

V
)


(
 j ix j  i  x j jxi  j x  i xi x j  i  Vx j  jxi ),
i
xi j 1
j i 1

1
i 1
j 1
j 1
i 1
j 1

3
3
3
3
3
3
3
3


(

V

)

V
(


)

V

(

)

V
(
 xi  j ix j
 j  ixi x j  xi   j ix j ),
 x i  x j j
j 1
i 1
j 1
j 1
i 1
i 1
j 1
i 1 i
3
3
 3
 3
  j
(  iVxi )x j  0; V   j (  ixi )x j  0; div  0; lim B[Vx1 ,Vx2 ,Vx3 ]  0.
 0
x j i 1
 j 1
j 1
i 1
Then on a basis (4.40) system (4.39) it is equivalent, will be transformed to a kind:
3
3
3
3

i
1
1 1
(

Vt  d V [ (   j ix j )]   Vx j  j   0  d [


4 3 i 1 (  2   2   2 )3

i 1 j 1
j 1
1
2
3
R

{( B[Vh1 ,Vh2 ,Vh3 ] )( x1   1 ,x2   2 ,x3   3 ;t )} )d 1d 2 d 3 ]  V ,

3
3

1
d



0;

(
x
,x
,x
,t
)

d
 i
 i ( x1 ,x2 ,x3 ,t ),
0 1 2 3

i 1
i 1

or from (4.41), follows:
Vx  Wi ( x1 ,x2 ,x3 ,t ),( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),
 i
t

1
r2
V  M 1( x1 ,x2 ,x3 ,t ) 
exp
(

)( Q[V ,W1 ,W2 ,W3 ] )( s1 ,s2 ,s3 ,s ) 

4 ( t  s )

8  3 0 R3

t

1
1
ds
ds
ds
ds

M

exp( (  12   22   32 ))( Q[V ,W1 ,W2 ,W3 ] )( x1 

1


3 1 2 3
3
 (  ( t  s ))
 0 R3

2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3  ( t  s );s )d 1d 2 d 3ds   0 [V ,W1 ,W2 ,W3 ],

t

( xi  si )
1
r2
( exp( 
))( Q[V ,W1 ,W2 ,W3 ] )( s1 ,s2 ,s3 ,s ) 
Wi  M 1xi 

4 ( t  s )

8  3 0 R3 2  ( t  s )

t

i
1
1
ds ds ds ds  M 1xi 
exp( (  12   22   32 ))




3 1 2 3
( t  s )
 3 0 R3
 ( ( t  s ) )

( Q[V ,W1 ,W2 ,W3 ] )( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s ) 
d d d ds   [V ,W ,W ,W ],
i
1 2 3
 1 2 3
where

1
exp( (  12   22   32 ))0 ( x1  2 1 t ,x2  2 2 t ,x3  2 3 t )d 1d 2d 3 
M 1 

3

 R3

t
 1

exp( (  12   22   32 )) 0 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3  2 3 

3  
  0 R3
(4.41)
(4.42)
  ( t  s );s )d 1d 2 d 3ds, ( s j  x j  2 j t ; s j  x j  2 j ( t  s ); j  1,3 ),

3
3

1
(
Q
[
V
,V
,V
,V
]
)(
s
,s
,s
,s
)


{
d
[
V
(
s
,s
,s
,s
)
(

   j is j ( s1 ,s2 ,s3 ,s ))] 
s1 s2 s3
1 2 3
1 2 3

i 1 j 1
 3
3
1

1 1
{( B[Vh ,Vh ,Vh ] )( s1 
  Vs j ( s1 ,s2 ,s3 ,s ) j ( s1 ,s2 ,s3 ,s )  d [ 4  (  i
1
2
3
2
2
2 3
3
(





)
j

1
i

1

1
2
3
R

 1 ,s2   2 ,s3   3 ;s )} )d 1d 2 d 3 ]}, ( hi  si   i ; i  1,3 ).
Here (4.42) – system of the nonlinear integrated equations of Volterra-Abel of the second sort
concerning V ,Wi on a variable t  [0,T0 ] . Therefore, if takes place:
 M ;  ;  ; t  s   : sup D k M ( x ,x ,x ,t )   , ( k  0,3 ),
i
1 1 2 3
1
 1
T

3
3

1
sup

(
x
,x
,x
,

,

,

;t,

)

sup{
d
(
   j il j ( x1  2 1  ,x2  2 2  ,x3 
1 2 3 1 2 3

T

T
T

T
i
1 j 1


3
2 3  ;t   ) )    j ( x1  2 1  ,x2  2 2  ,x3  2 3  ;t   ) 

j 1

3
3
3
 1 1
1
(

{
(
 j il j ( x1  2 1    1 ,x2  2 2  
d [



i
2
2
2 3
4 3 i 1

(





)
i

1
j

1
1
2
3
R

3
3

 2 ,x3  2 3    3 ;t   ) )   (   jli ( x1  2 1    1 ,x2  2 2    2 ,x3 

i 1 j 1

2 3    3 ;t   ) )i } )d 1d 2 d 3 ]}   2  ,

t
k  1 sup
exp( (  12   22   32 )) ( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, )d 1d 2 d 3d   2 T0 ,
 0
3 T  

0 R3


t

k  1 sup
exp( (  12   22   32 )) ( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, ) i d 1d 2 d 3d 
i




 3 T 0 R3
(4.43)

 2T   ,( i  1,3 ),   max(  T ;3 2T  )
0 2
2 0
0 2



and
3

h
 i ,( i  0,3 ) : ki  , [  ki   (  2T0   3 2T0  2 )   (   1 )  h  1],
(4.44)
4 i 0


 i : Sr1 ( 0 )  Sr1 ( 0 ), ( i  0,3 ), [S r1 ( 0 )  {V ,W j : V ; W j  r1 ,( x1 ,x2 ,x3 ,t )  T }; j  1,3],

  i [0,0,0,0 ] C  r1( 1  h ) :  i [V ,W1 ,W2 ,W3 ] C   i [V ,W1 ,W2 ,W3 ]  i [0,0,0,0 ] C 
  [0,0,0,0 ]  k 4r  r ( 1  h )  r ,
i
i 1
1
1
C

Then on the basis of (4.31) - (4.34) there is a unique solution of system (4.42), and this solution is
found on the basis of a Picard’s method. It means that sequence {i ,m }0 is converged to i , that is,
h 1
i ,n 1 
i ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).
n 
(4.45)
From the received results follows that the Navier-Stokes problem (1.1)-(1.3) in a case (4.36) and (4.45), with
smooth enough initial data has the conditional-smooth unique solution in Gn13 ( D0 ) , so as V  G1( D0 ).
Therefore results are formulated by the following theorem.
Theorem 5. Under conditions (1.2), (1.3), (4.36), (4.37) and (4.45) the Navier-Stokes problem has a
conditional-smooth unique solution in Gn13 ( D0 ) .
Remark 6. Updating of a Method (4.2), when divf  0 . The algorithm (4.37) also is applicable in
a case, if
3


f




,
(
i

1,3
),
div
f

0
:
div


0;
 i
 x it  0,
i
it

i 1 i

(4.46)
divf  0; rot f  0 :  fi  0,( i  0;i  1,3 ),

3



(
x
,x
,x
)



(
x
,x
,x
),
(
 i0 xi  0 ),
i0 1 2 3
i 0 1 2 3
 i t 0
i 1

that on a basis (4.39)-(4.41):
3 
1
1
 (4.39) : ( P  U )  {B[Vx1 ,Vx2 ,Vx3 ]  div},

2
i 1 xi

3
3
3
3
( B[V ,V ,V ] )( x ,x ,x ,t )  (


)V

(
 jxi Vx j )i ,




x1 x2 x3
1 2 3
j ix j
xi

i 1 j 1
i 1 j 1

3
1
1
1
1
{
(
B
[
V
,V
,V
]
)(
s
,s
,s
,t
)

 P U 
isi ( s1 ,s2 ,s3 ,t )}ds1ds2ds3 ,
s1 s2 s3
1 2 3
2
4 3 r

i 1
R

i
 1 P  1U  1
{( B[Vh1 ,Vh2 ,Vh3 ] )( x1   1 ,x2   2 ,x3   3 ;t ) 
x
x

  i 2 i 4
2
2
2 3
3 (1   2   3 )
R

 3
 ih ( x1   1 ,x2   2 ,x3   3 ;t )}d 1d 2 d 3 , ( si  xi   i ;hi  xi   i ;i  1,3 ),

i
 i 1

3
3
3
Vt  d 1V [ (   j ix )]   Vx  j   0  ( Q[V ,Vx ,Vx ,Vx ] )( x1 ,x2 ,x3 ,t )  V ,
j
j
1
2
3

i 1 j 1
j 1

3

i
1 1
Q
[
V
,V
,V
,V
]


d
[
(
( B[Vh1 ,Vh2 ,Vh3 ] )( x1   1 ,x2 


x1 x2 x3

2
2
2 3
4


R 3 i 1 (  1   2   3 )

 2 ,x3   3 ;t ))d 1d 2 d 3 ],

3
3
3
3
i
d     0; div  0;   d 1[   1 {
(
ih ( x1   1 ,
i
0
i

4 3 i 1 (  2   2   2 )3 i 1 i
i 1
i 1
1
2
3
R

3
3

1
 x2   2 ,x3   3 ;t ))}d 1d 2 d 3 ]; U    2j ;   j ix  U x ,( i  1,3 ).
j
2 i

j 1
j 1
we will receive (4.42):
V   0 [V ,W1 ,W2 ,W3 ],


(4.42)*
Wi   i [V ,W1 ,W2 ,W3 ],( i  1,3 ),


Vxi  Wi ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).
Hence, as takes place (4.47), that the solution of system (4.42)* we can find on the basis of Picard’s
method. Then on a basis (4.31)-(4.34) we have (4.45).
Further, we will receive similar results in the conditions of the theorem 5, i.e. under conditions (1.2),
(1.3), (4.45), (4.46) problem Navier-Stokes has the unique solution in Gn13 ( D0 ) in a kind (4.37).
5. Fluid with Viscosity, When 1  0    
In the Navier-Stokes problems with great values of viscosity components the speed can be so small
that in many researches do not consider the forces of inertia.
Let's consider a fluid with viscosity with Reynolds small number where all inertial participants
contain in equations Navier-Stokes. As has been said, theoretically, it is not investigated [12]. Hence,
here we will consider, methods of integration of the equations Navier-Stokes. However, it is necessary
to notice that technical applications of the theory of a liquid with very great values of viscosity, short
of the theory of greasing [12], it is rather limited. But theoretically researches in this direction from the
scientific point of view are very important. Therefore in this paragraph methods are developed for the
decision of these problems in space Cn3,13 (T ) . Alternatively, we can consider, e.g., a class of suitable
solutions constructed in Gn1 3 ( D0 ) on the basis of lemma K. Friedrichs [15].
The basic methods (4.2) and (4.12) of paragraph 4 are for this purpose modified on the basis of
regulating functions Ki ,( i  1,2,3 ) , and the problem (1.1) - (1.3) does not suppose restrictions (А1),
(А2) in what the urgency of researches of this paragraph consists.
5.1. Fluid with Average Viscosity by Conditions (A3), divf  0
Therefore, in this case, if the initial data ( i0 ; fi ) is set in a kind:

 i0 ( x1 ,x2 ,x3 )  i0 ( x1 ,x2 ,x3 ),( x1 ,x2 ,x3 )  R 3 ,( i  1,3 ),
 i t 0
3
 i0 xi  0, ( 0  i  const ;1  0     ),
i 1

3
1


f



K
,
[
K

f
;
i

1,3
]
,
div
f

0
:
div


0;
K it  0,
 i

i
it
it
i

x

i

i 1

divf  0; rot f  0 :  fi  0,(  K i  0;i  1,3 ),   ( 1 ,2 ,3 ); f  ( f1 , f 2 , f 3 ),
t

3
3
3
U   K 2j ;  K j Ki x  1 (  K 2j )x  1 U x ,( i  1,3 ); K i  1 fi ( x1 ,x2 ,x3 ,s )ds,
j
i

2 j 1
2 i
 0
j 1
j 1

we enter for definition a component of speeds:
(5.1)

i  iV( x1 ,x2 ,x3 ,t )  Ki ( x1 ,x2 ,x3 ,t ),( i  1,3 ),

3

V t 0  0 ( x1 ,x2 ,x3 ), ( x1 ,x2 ,x3 )  R ,
at that
3
3

div  0 :  iVxi  0;  Kixi  0,

i 1
i 1
 3
3
3
3
1

   ji x j  V   j Ki x j  i  Vx j K j  U xi ,( iV   jVx j  0 ),
2
 j 1
j 1
j 1
j 1

it  iVt  Kit ; i  {i V   Ki }  i V ,( i  1,3 ).

Then on the basis (5.2), (5.3) we will receive
3
3
1
1
iVt  V   j Ki x j  i  Vx j K j  U xi  i  Pxi  i V ,( i  1,3 ).
2

j 1
j 1
Hence on a basis (5.4) and having entered APS, we have
1
 1
(  P  2 U )   B* [Vx1 ,Vx2 ,Vx3 ],

3
3
3
3

( B* [Vx1 ,Vx2 ,Vx3 ] )( x1 ,x2 ,x3 ,t )   (   j K ix j )Vxi   (  K jxi Vx j )i ,

i 1 j 1
i 1 j 1

1
1
1
1
{( B* [Vs1 ,Vs2 ,Vs3 ] )( s1 ,s2 ,s3 ,t )}ds1ds2 ds3 ,
 P U 


2
4

r
3

R

i
 1 P  1U  1
{( B* [Vh1 ,Vh2 ,Vh3 ] )( x1   1 ,x2   2 ,x3 
xi
xi


2
4 3 (  2   2   2 )3
1
2
3
R

 ;t )}d d d , ( s  x   ;h  x   ;i  1,3 ),
1 2 3
i
i
i i
i
i
 3
so as
3

 3

(5.4) :
[ iVxi ( x1 ,x2 ,x3 ,t )]  0,

 xi
t i 1
i 1

3
3  1

1
1
U xi  U ;  (  Pxi )    P,



i 1 xi 2
i 1 xi
 3
3
3
3

 3

( i V )  0;  K j
(  iVxi )x j  0; V   j (  K i xi )x j  0; div  0,
 
x j i 1
 i 1 xi
j 1
j 1
i 1
3
3
3
3
3
3
3

 3

(V

K
)


(
V
K
)

K
(

V
)


(
 j i x j  i  x j j xi  j x  i xi x j  i  Vx j K j xi ),
 x
i
j i 1
j 1
i 1
j 1
j 1
i 1
j 1
 i 1
3
3
3
3
3
3
3
3 

( i  Vx j K j )  Vxi (   j Ki x j )  V   j (  Ki xi )x j  Vxi (   j Ki x j ).
i 1 xi
j 1
i 1
j 1
j 1
i 1
i 1
j 1
Then system (5.4) it is equivalent will be transformed to a kind
(5.2)
(5.3)
(5.4)
(5.5)
3
3
3
3

i
1
1 1
V

d
V
[
(

K
)
]

V
K



d
[
(

 t
  j ixj  xj j 0


2
2
2
3
4


i 1 j 1
j 1
R 3 i 1 (  1   2   3 )

{( B* [Vh1 ,Vh2 ,Vh3 ] )( x1   1 ,x2   2 ,x3   3 ;t )} )d 1d 2 d 3 ]  V ,

3
3

1
hi  xi   i ,( i  1,3 ),  0  d  i ; d   i  0.
i 1
i 1

For consideration of unknown function V we have
t

1
r2
exp
(

)( Q[V ,Vs1 ,Vs2 ,Vs3 ] )( s1 ,s2 ,s3 ,s ) 
V  M 1( x1 ,x2 ,x3 ,t ) 
3  
4

(
t

s
)

8  0 R3

1

ds ds ds ds,
 ( ( t  s ) )3 1 2 3

t

1
r2
1
r2
M 1 
exp
(

)

(
s
,s
,s
)ds
ds
ds

exp
(

)
0 1 2 3
1 2 3


4 t
4 ( t  s )

8 (  t )3 R3
8  3 0 R3


1
1
ds ds ds ds 
exp( (  12   22   32 ))0 ( x1  2 1 
 0 ( s1 ,s2 ,s3 ,s )

3 1 2 3
3
( ( t  s ) )

 R3


t
 t ,x  2 t ,x  2 t )d d d  1
exp( (  12   22   32 )) 0 ( x1 
2
2
3
3
1 2 3

3  
 0 R3

2  ( t  s ),x  2 ( t  s ),x  2 ( t  s ) ;s )d d d ds,
2
2
3
3
1 2 3
 1
 s  x  2 t ; s  x  2  ( t  s ); j  1,3,
j
j
j
j
j
 j
3
3

( Q[V ,Vs ,Vs ,Vs ] )( s1 ,s2 ,s3 ,s )  {d 1[V ( s1 ,s2 ,s3 ,s )(   j K is ( s1 ,s2 ,s3 ,s ))] 
1
2
3
j

i 1 j 1

3
 3
1
1 1

V
(
s
,s
,s
,s
)

K
(
s
,s
,s
,s
)

d
[
(
i


j 1 2 3
  sj 1 2 3

2
2
2 3
4

(1   2   3 )
 j 1
R 3 i 1

{( B* [Vh1 ,Vh2 ,Vh3 ] )( s1   1 ,s2   2 ,s3   3 ;s )} )d 1d 2 d 3 ]},( h j  s j   j , j  1,3 ).
Hence, differentiating (5.7) on xi and having entered designation:
Vxi  Wi ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),
from (5.7) we will receive
t

1
exp( (  12   22   32 ))( Q[V ,W1 ,W2 ,W3 ] )( x1  2 1 ( t  s ),
V  M 1 



 3 0 R3

 x2  2 2 ( t  s ),x3  2 3 ( t  s );s )d 1d 2d 3ds   0 [V ,W1 ,W2 ,W3 ],

t

i
1
W

M

exp( (  12   22   32 ))
( Q[V ,W1 ,W2 ,W3 ] )( x1  2 1 ( t  s ),
1xi
 i


3

(
t

s
)
3
 0R


 x2  2 2 ( t  s ),x3  2 3 ( t  s );s )d 1d 2 d 3ds   i [V ,W1 ,W2 ,W3 ], ( i  1,3 ).
(5.6)
(5.7)
(5.8)
(5.9)
If takes place:

k
( x1 ,x2 ,x3 ,t )  T ;M 1 ;Y ;K i ,( t  s   ) : sup D M 1 ( x1 ,x2 ,x3 ,t )  1 , ( k  0,3 ),
T

3
3

1
sup Y( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, )  sup{d  (   j K is ( x1  2 1  ,x2  2 2 
j
T T
T T
i 1 j 1

3



,x

2


;t


)
)

 K j ( x1  2 1  ,x2  2 2  ,x3  2 3  ;t   ) 
3
3

j 1


3
3
3
1
 d 1[ 1
(  i
{ (   j K il ( x1  2 1    1 ,x2  2 2  
j
2
2
2 3
(5.10)
4 3 i 1

(





)
i 1 j 1
1
2
3
R

3
3

 2 ,x3  2 3    3 ;t   ) )   (  K jli ( x1  2 1    1 ,x2  2 2    2 ,x3 

i 1 j 1

2 3    3 ;t   ) )i } )d 1d 2 d 3 ]}  1  2 ,



t

1
1
sup   exp( (  12   22   32 ))Y( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, )d 1d 2 d 3d 
 2T0 ,
k0 

 3 T 0 R3


t

1

sup   exp( (  12   22   32 ))Y( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, ) i d 1d 2 d 3d 
ki 

 3 T 0 R3


1
 2T0  2 ,( i  1,3 ),   max(  2T0 ; 3 2T0  2 ); 1    0  


and
3

h
1
1
1
1

,(
i

0,3
)
:
k

,
[
ki 
(  2T0  3 2T0  2
)
(1
)  h  1],
 i

i
4





i 0

 i : Sr1 ( 0 )  S r1 ( 0 ),( i  0,3 ); S r1 ( 0 )  {V ,W j : V ; W j  r1 ,( x1 ,x2 ,x3 ,t )  T },( j  1,3 ), (5.11)

  i [0,0,0,0 ] C  r1( 1  h ) :  i [V ,W1 ,W2 ,W3 ] C   i [V ,W1 ,W2 ,W3 ]  i [0,0,0,0 ] C 
  [0,0,0,0]  k 4r  r ( 1  h )  r .
i
i 1
1
1
C

Then the solution (5.9) exists and is unique, and we can find this solution on the basis of a Picard’s method
( x1 ,x2 ,x3 ,t )  T : Vn 1   0 [Vn ,W1,n ,W2,n ,W3,n ],



Wi ,n  1   i [Vn ,W1,n ,W2,n ,W3,n ],( n  0,1,...; V0  0; Wi ,0  0; i  1,3 ).
(5.12)
Therefore using known mathematical conclusions concerning of Picard’s method we can tell the
following. On the basis of conclusions of a method of Picard’s the sequence of functions
{Vn }0 ,{Wi ,n}0 ,( i  1,3 ) is converging and fundamental in S r1 ( 0 ) and at that elements of the
constructed sequences belong S r1 ( 0 ) (see. §4.3). Then we have
3
3

h 1


V

W
;


V

V

Wi ,n  1  Wi C :  n  1  h n  1 0 
 0,


i C
n 1
n 1
 0
n 
C
C
i

1
i

1

h 1
h 1
V 
Wi ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).
n 
 n  1 n  V  H ; Wi ,n  1 
Hence on the basis of (5.2) we will receive
i ,n  1  i [0 ( x1 ,x2 ,x3 )  Vn 1( x1 ,x2 ,x3 ,t )]  K i ( x1 ,x2 ,x3 ,t ),( n  0,1,2,...;i  1,3 ),

h 1
n 1
 0,( i  1,3 ),
 i ,n  1  i C  i Vn  1  V C  i h n  i h  0 
n 
(5.13)
(5.14)
i.e. it means that sequence {i ,n }0 converging to a limit i ,( i  1,3 ) :
h 1
i ,n 1 
i ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).
n 
(5.15)
Theorem 6. Under conditions (1.2), (1.3), (A3), (5.1), (5.15) the Navier-Stokes problem has a
continuous unique solution in Gn1 3 ( D0 ) and can be defined by the (5.2).
Remarks:
I. Uniqueness is obvious, as a method by contradiction. Results (5.13) with a condition ((A3), (5.1),
(5.11), (5.12)) are received where smoothness of functions is required only on xi ,( i  1,2,3 ) as the
derivative of 1st order is in time has t  0 . Then taking into account (5.13) the system (5.9) has the
continuous unique solution: V  C 3,0 (T ). Therefore: V  G1( D0 ). Hence, on the basics (5.2), (5.14),
(5.15):   Gn1 3 ( D0 ).
II. The algorithm (5.2) also is applicable in a case, if
3


f



K
,
div
f

0
:
div


0;
 i
 x Kit  0,
i
it

i 1 i

3
3
1

2
div
f

0;
rot
f

0
:

f

0,(

K

0;i

1,3
);
U

K
;

 j  K j Ki x j  2 U xi ,
i
i

j 1
j 1

i t 0  i0 ( x1 ,x2 ,x3 )  i0 ( x1 ,x2 ,x3 ),( i  1,3 ).

That on a basis (5.16), (5.2) we will receive (5.4). Hence, we have
3


1
1
div  0;  (5.4) : ( P  U )  {B* [Vx ,Vx ,Vx ]  div},
1
2
3

2

i 1 xi

3
3
3
3
( B [V ,V ,V ] )( x ,x ,x ,t )  (  K )V  ( K V ) ,
  j ix j xi   jxi x j i
1 2 3
 * x1 x2 x3
i
1 j 1
i 1 j 1

1
3
1
1
1

P

U

{
(
B
[
V
,V
,V
]
)(
s
,s
,s
,t
)


isi ( s1 ,s2 ,s3 ,t )}ds1ds2ds3 ,
* s1 s2 s3
1 2 3
2
4 3 r

i
1
R

i
 1 P  1U  1
{( B* [Vh1 ,Vh2 ,Vh3 ] )( x1   1 ,x2   2 ,x3   3 ;t ) 
xi
xi


2
4 3 (  2   2   2 )3
1
2
3
R

 3
 ih ( x1   1 ,x2   2 ,x3   3 ;t )}d 1d 2 d 3 , ( si  xi   i ;hi  xi   i )
 i 1 i
and
(5.16)
(5.17)
3
3
3
3

i
1
1 1
V

d
V
[
(

K
)
]

V
K



d
[
(

 t
  j ixj  xj j 0


2
2
2 3
4


i 1 j 1
j 1
R 3 i 1 (  1   2   3 )

{( B* [Vh1 ,Vh2 ,Vh3 ] )( x1   1 ,x2   2 ,x3   3 ;t )} )d 1d 2d 3 ]  V ,

3
3
(5.18)

1
1
1
{

(
F
(
x


,x


,x


 0 ( x1 ,x2 ,x3 ,t )  d [ i 
i
0
1
1
2
2
3
4 3 i 1
(  12   22   32 )3
i 1

R

3
3

 3 ;t ))}d 1d 2 d 3 ], ( d   i  0; F0 ( x1 ,x2 ,x3 ,t )  ixi ).
i 1
i 1

Here (5.17) differs from (5.5), as div  0. Then considering (5.18), (5.7), (5.8) we will receive system (5.9):
V   0 [V ,W1 ,W2 ,W3 ],
(5.19)

Wi  i [V ,W1 ,W2 ,W3 ],( i  1,3 ).
Therefore, as takes place (5.10), (5.11), (5.13) in a consequence (5.14), (5.15), i.e.
h 1
 i ,n  1  i  i Vn 1  V  i h n  i h n  1 0 
 0,( i  1,3 ),
n 
C
C
(5.20)

h 1


,

(
x
,x
,x
,t
)

T
,(
i

1,3
).
i ,n  1 
i
1
2
3
n 
Further, we will receive similar results in the conditions of the theorem 6.
5.2. Updating of a Method (5.2) in a Case 0  0; divf  0; rot f  0
The decision method from where follows of equations integration of Navier-Stokes in Cn3,13 (T ) in a case
i t 0  i0 ( x1 ,x2 ,x3 )  
i 0 ( x1 ,x2 ,x3 ),( 0  i  const ;i  1,3 ),
 3

3
3
i 0 xi  0, ( i  1,3 ), 0  0; 0  C ( R ),
 
 i 1
divf  0; rot f  0, (  f  0; f  ( f , f , f )); sup D k f  N , ( x ,x ,x ,t )  T ,
i
1
2
3
i
0
1 2
3

T

t
 K ( x ,x ,x ,t )  1 f ( x ,x ,x ,s )ds, (  K  0; i  1,3 ),
i
1 2
3
i
 i 1 2 3
 0

is a major factor of this point.
Therefore we enter for definition a component of speeds:

i  i [0 ( x1 ,x2 ,x3 )  Z( x1 ,x2 ,x3 ,t )]  K i ( x1 ,x2 ,x3 ,t ),( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),

3

 Z t 0  0,( x1 ,x2 ,x3 )  R ,
at that
3
3
3


0;
K ixi  0,
div  0 :  i Z xi  0;  

i 0 xi
i 1
i 1
i 1

3
3

1
1
fi ( x1 ,x2 ,x3 ,t ); U   K 2j ;  K j K i x j  U xi , ( i  1,3 ),
 K it 
2

j 1
j 1

3
3
 3
1
 ji x j    j ( 0  Z )K i x j   i ( 0 x j  Z x j )K j  U xi ,
2
 j 1
j 1
j 1
(5.21)
(5.22)
3
 3

(


Z
)

(


Z
)


(


Z
)
 j ( 0 x j  Z x j )  0,

i
0xj
xj
i
0
 j 0
j

1
j

1

   Z  K ;   [ (   Z )   K ]   Z , ( i  1,3 ).
i t
it
i
i
0
i
i
 it
(5.23)
Hence on the basis of (5.21)-(5.23) we will receive
3
3
1
2
i Zt    j ( 0  Z )Ki x   K j i ( 0 x  Z x )  U x  ( 1 
j 1
j
j 1
j
j
i
1

) fi 
1

Pxi  i Z ,( i  1,3 ).
(5.24)
Then for incompressible currents with a friction the equations of Navier-Stokes (1.1) become simpler as take
place (5.21), (5.23). Therefore a problem (1.1)-(1.3) with account [(5.5)-(5.9) here instead of i we
will consider K i ] we will receive
t

1
r2
1
exp( 
)( Q[Z ,W1 ,W2 ,W3 ] )( s1 ,s2 ,s3 ,s )
ds1ds2 ds3ds 
Z  M 1 
3  
4 ( t  s )
(  ( t  s ))3
8  0 R3

t

 M 1  1
exp( (  12   22   32 ))( Q[Z ,W1 ,W2 ,W3 ] )( x1  2 1 ( t  s ),x2  2 2 


3

 0 R3

  ( t  s ),x3  2 3 ( t  s );s )d 1d 2 d 3ds  (  0 [Z ,W1 ,W2 ,W3 ] )( x1 ,x2 ,x3 ,t ),

t
1
r2
( xi  si )

(5.25)
W

M

(
exp
(

))
( Q[Z ,W1 ,W2 ,W3 ] )( s1 ,s2 ,s3 ,s ) 
 i
1xi


3
4

(
t

s
)
2

(
t

s
)
3
8

0 R


t
i
 ds1ds2 ds3 ds  M 1x  1
exp( (  12   22   32 ))
( Q[Z ,W1 ,W2 ,W3 ] )( x1 


i
3
 ( ( t  s ) )
( t  s )
 3 0 R3

2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3  ( t  s );s )d 1d 2 d 3 ds 

 (  i [Z ,W1 ,W2 ,W3 ] )( x1 ,x2 ,x3 ,t ),( i  1,3 ),


where
1
 1
(  P  2 U )  {F0  B* [Z x1 ,Z x2 ,Z x3 ]},

1
1 1
1
  P  2 U  4  r {F0 ( s1 ,s2 ,s3 ,t )  ( B* [Z s1 ,Z s2 ,Z s3 ] )( s1 ,s2 ,s3 ,t )}ds1ds2 ds3 ,
R3

1
1
1
i
{F0 ( x1   1 ,x2   2 ,x3   3 ;t )  ( B* [Z h1 ,Z h2 ,Z h3 ] )( x1 
 Pxi  U xi 

2
4 R3 (  12   22   32 )3


 1 ,x2   2 ,x3   3 ;t )}d 1d 2 d 3 ,( si  xi   i ;hi  xi   i ; i  1,3 ),

 Z xi  Wi ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),

 Z t   0  ( Q[Z ,Z x1 ,Z x2 ,Z x3 ] )( x1 ,x2 ,x3 ,t )  Z ,( i  1,3 ),
(5.26)
t

1
M
(
x
,x
,x
,t
)

exp( (  12   22   32 )) 0 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),
 1 1 2 3
3  
 0 R3


 x3  2 3  ( t  s );s )d 1d 2 d 3 ds, ( si  xi  2 i ( t  s );t  s   ;i  1,3 ),
3
3

1
( Q[Z ,Z s1 ,Z s2 ,Z s3 ] )( s1 ,s2 ,s3 ,s )  {d 0 [Z( s1 ,s2 ,s3 ,s )(   j K i s j ( s1 ,s2 ,s3 ,s ))] 
i 1 j 1

3
 3
1
i
  Z s j ( s1 ,s2 ,s3 ,s )K j ( s1 ,s2 ,s3 ,s )  d 01[
(


4 R3 i 1 (  12   22   32 )3
 j 1

{( B* [Z h1 ,Z h2 ,Z h3 ] )( s1   1 ,s2   2 ,s3   3 ;s )} )d 1d 2 d 3 ]}, hI  si   i ,( i  1,3 ),

3
3
3
3
( B [Z ,Z ,Z ] )( x ,x ,x ,t )  (  K )Z  ( K Z ) ,




1 2
3
j ix j
xi
jxi x j
i
 * x1 x2 x3
i 1 j 1
i 1 j 1

3
3
3
3
3

F
(
x
,x
,x
,t
)


(
K

)


(

K
);
div
f

0;
d

i  0,



i 
j xi 0 x j
0 xi  j i x j
0
 0 1 2 3
i 1
j 1
i 1
j 1
i 1

3
3
3
3
3

1
1
i
1



K


d
[
(
1

)
f


(

K
)

{
( F0 ( x1 



j 0xj
0 
i
0  j ixj
 0

2
2
2 3
4


3
j 1
i 1
i 1
j 1
i 1
(





)
R
1
2
3

 ,x   ,x   ;t ))}d d d ] .
2
3
3
1
2
3
 1 2
If the known functions entering into system (5.25) satisfy conditions
( x1 ,x2 ,x3 ,t )  T ;M 1 ;Y ;K i , ( t  s   ) :

k
sup D M 1 ( x1 ,x2 ,x3 ,t )  1 , ( k  0,3 ),
 T
3
3

1
sup
Y(
x
,x
,x
,

,

,

;t,

)

sup{
d
(
1 2
3 1 2 3
0    j K i s j ( x1  2 1  ,x2  2 2  ,x3 
 T T
T T
i 1 j 1

3

2 3  ;t   ) )   K j ( x1  2 1  ,x2  2 2  ,x3  2 3  ;t   ) 
j 1

3
3
3

i
1
 d01[
(
{
(

   j K il j ( x1  2 1    1 ,x2  2 2  
4 R3 i 1 (  12   22   32 )3 i 1 j 1


3
3



,x

2




;t


)
)

(
K jl ( x1  2 1    1 ,x2  2 2    2 ,x3 
 2 3


3
3
i
i

1
j

1

2    ;t   ) ) } )d d d ]}  (  )1  ,
3
3
i
1
2
3
2

t

1
k0 
sup exp( (  12   22   32 ))Y( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, )d 1d 2 d 3d 
3 T  


0 R3

t
 1  T ; k  1 sup exp( (  2   2   2 ))Y( x ,x ,x , , , ;t, )  i 
2 0
i
1
2
3
1 2
3 1 2 3




 3 T 0 R3

1

d 1d 2 d 3 d  2T0  2  ,( i  1,3 ), *  max(  2T0 ;3 2T0  2 )

and
(5.27)
h


,(
i

0,3
)
:
k

,( h  1 ),
i
i

4
 3
 k  1 (  T  3 2T  1 )  1 ( 1  1 )  h  1,( 1       ),
i
2 0
0 2
*
0





i 0

 S r1 ( 0 )  {Z ,W j : Z ; W j  r1 ,( x1 ,x2 ,x3 ,t )  T },( j  1,3 ),

  i [0,0,0,0 ] C  r1 ( 1  h ) :  i [Z ,W1 ,W2 ,W3 ] C   i [Z ,W1 ,W2 ,W3 ]   i [0,0,0,0 ] C 

  i [0,0,0,0 ] C  ki 4r1  r1 ( 1  h )  hr1  r1 ( 1  h )  r1 ,

 i : S r1 ( 0 )  S r1 ( 0 ), ( i  0,3 ).
(5.28)
That the solution of this system we can find on the basis of Picard’s method
( x1 ,x2 ,x3 ,t )  T : Z n 1   0 [Z n ,W1,n ,W2,n ,W3,n ],



Wi ,n  1   i [Z n ,W1,n ,W2,n ,W3,n ],( Z0  0; Wi ,0  0; i  1,3; n  0,1,...).
(5.29)
Then considering results (5.13)-(5.15) we will receive, i.e. that sequence {i ,n }0 converging to a limit i :
h 1
i ,n 1 
i ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).
n 
(5.30)
Theorem 6*. If the conditions (1.2), (1.3), (5.21)-(5.23) and (5.30) are fulfilled, then the
Navier-Stokes problem is solvable at Cn3,13 (T ) .
Remark 7. If takes place 0  *  2 1 , that 1    0  . In a case *  2 1 , then
m0    0  const  , ( m0  max[1;4( 1  4 *1  1)2 ] ).
(1)*
In a case   1 , that
3
k
i 0
i
 ( T0  3 2 ) 2 T0  2 *  h  1, (see. (5.28)).
(2)*
Let's note, as the equations of system (   1 , (5.25)) are the equations of Volterra-Abel [13] on a
variable t [0,T0 ] , discussing in language of the Volterra’s equations, we can find the solution in
C 3,1 ( T ) , when the condition (5.28) is not satisfied. For simplicity, let's assume
   1; C0  ( T0  3 2 ) 2 :
 3

 ki  h  C0 T0  1.
 i 1
(3)*
And in this case all results of the theorem 6* are carried out, i.e. that is we will prove that under
condition of (3)* system (5.25) correct in C 3,1( T ).
1
1
Really, as in case of (3)*, the interval [0,T0 ] we will divide on two parts: [0, T0 ], [ T0 ,T0 ].
2
2
1
1
1
Thus a step: h  T0 . Then in area T1  R 3  [0, T0 ] and in T2  R 3  [ T0 ,T0 ], we will receive
2
2
2
systems

 Z1   0 ,1[Z1 ,W1,( 1 ) ,W2,( 1 ) ,W3,( 1 ) ],( x1 ,x2 ,x3 ,t )  T1 ,


Wi ,( 1 )   i ,1[Z1 ,W1,( 1 ) ,W2,( 1 ) ,W3,( 1 ) ],( x1 ,x2 ,x3 ,t )  T1 ,( i  1,3 ).
(4)*
Thus, operators  0 ,1 ;  i ,1 are compressing
3
d0   ki  h*  (
i 1
T0
T
 3 2 ) 2 0  1
2
2
(5)*
and display ranges of definition in it self, i.e. conditions of contraction mapping principle [13] are
satisfied. Then is under condition of (5)* system (4)* correct in C 3,1(T1 ).
1
Further, we will consider T2  R 3  [ T0 ,T0 ], accordingly  0,2 ;  i ,2 , at that
2
1
1

3
 Z1( x1 ,x2 ,x3 , 2 T0 )  Z 2 ( x1 ,x2 ,x3 , 2 T0 ), ( x1 ,x2 ,x3 )  R ,

W ( x ,x ,x , 1 T )  W ( x ,x ,x , 1 T ), ( x ,x ,x )  R 3 ,( i  1,3 ),
i ,( 2 )
1 2
3
0
1 2
3
 i ,( 1 ) 1 2 3 2 0
2
(6)*
and hence, operators  0 ,2 ;  i ,2 satisfy to conditions of a principle of compression. Then system of
the integral equations corresponding to the operator  0 ,2 ;  i ,2 it is correct in C 3,1(T2 ).
Therefore the system (5.25) correct in C 3,1(T  R3  [0,T0 ] ) . The offered method the decision of
system (5.25) in the theory of the Volterra’s equations is called [13] «a method pasting» or «a method
of subareas».
5.3. Fluid with Viscosity, When Re  2300; fi  0, ( i  1,3 )
Area of the fluid with average viscosity, when fi  0,( i  1,3 ) it is studied in 5.1. Here we will
consider a the general methods of the equations integration of Navier-Stokes with conditions, i.e. with
average and with Reynolds small number ([12]: Re  2300 ), where all inertial participants contain in
equations Navier-Stokes, when fi  0,( i  1,3 ).
The overall objective of this paragraph: to change a method (5.2) so that the received
analytical decision the Navier-Stokes problem with viscosity, belonged in Cn3,1 3 (T ).
In particular in many scientific literatures, investigates a problem [12]:
3
it   jix  i 
j
j 1
1

Pxi ,( i  1,3 ),
(5.31)
div  0, ( x, t )  T  R 3  [0, T0 ],
(5.32)
i
(5.33)
t 0
3
 
i 0 ( x1 ,x2 ,x3 ),( x1 ,x2 ,x3 )  R ,( i  1,3 ),
where 0  i  the known constants and in the equations (1.1) are assumed
3
3
0  n0    0   , ( i0  
i 0 ; 0  C ( R ); n0 , 0 
(5.34)
c ).
onst
Hence, here we will consider a methods of the equations integration of Navier-Stokes (5.31) with a
conditions (5.32) - (5.34).
With that end in view we will assume that there are functions 0  fi ,( i  1,3 ) , which satisfy conditions
C 3,0 ( T )  f i : sup D k f i   i (  )   (  )  1, ( 0    1; i  1,3 ),
T

divf  0; rot f  0 :



 fi  0, ( f  ( f1 , f 2 , f 3 )),

t

 I i ( x1 ,x2 ,x3 ,t )   f i ( x1 ,x2 ,x3 ,s )ds , ( 0  0; I i  0; I i t  f i ; i  1,3 ).
0

Hence is offered the method:

   [  Z( x ,x ,x ,t )]  I ( x ,x ,x ,t ),( i  1,3 ),
i
0
1
2
3
i
1
2
3
 i

3
 Z t 0  0,( x1 ,x2 ,x3 )  R ,

3
3
3
3
div  0 :  i0 x  0;  i Z x  0, ( I i x  0;U   I 2j ;
i
i
i

i 1
i 1
i 1
j 1
(5.35)
(5.36)
3
I I
j 1
j
i xj
1
 U  xi ).
2
Thus takes place conditions
3
3
3
1
 3 
I

0;




(


Z
)I

I j i ( 0 x j  Z x j )  U  xi ,




i

t
j
i
x
j
0
i

x
 x
j
j
2
j 1
j 1
j 1
 i1 i
3
 3
  j ( 0  Z )i ( 0 x j  Z x j )  i ( 0  Z )  j ( 0 x j  Z x j )  0,
j 1
 j 1
   Z  I ;   [ (   Z )   I ]    Z ,( i  1,3 ).
i t
i t
i
i
0
i
i
 it

(5.37)
Then for incompressible currents with a friction the equations of Navier-Stokes (5.31) become simpler as
take place (5.32)-(5.34). Therefore the problem (5.31) - (5.33), is led to a kind
3
3
1
2
i Zt    j ( 0  Z )Ii x   I j i ( 0 x  Z x )  U x   fi 
j 1
j
j 1
j
j
i
1

Pxi  i Z ,( i  1,3 ).
(5.38)
From system (5.38), considering conditions (5.31)-(5.33) and having entered APS, we will
receive the equation:
1
1
 3 
 x (5.38) : (  P  2 U  )  {F0  B [Z x1 ,Z x2 ,Z x3 ]},
 i 1 i
3
3
3
3

(
B
[
Z
,Z
,Z
]
)(
x
,x
,x
,t
)

(

I
)Z

(
I j xi Z x j )i ,
  x1 x2 x3




1 2
3
j i x j
xi
i

1
j

1
i

1
j

1

3
3
3
3

 F0 ( x1 ,x2 ,x3 ,t )   i (  I j xi 0 x j )  0 xi (   j I i x j ); divf  0,
i 1
j 1
i 1
j 1

1
1
1 1
 P  U 
{F0 ( s1 ,s2 ,s3 ,t )  ( B [Z s1 ,Z s2 ,Z s3 ] )( s1 ,s2 ,s3 ,t )}ds1ds2 ds3 ,
2
4 R3 r


 1 P  1U  1
{F0 ( x1   1 ,x2   2 ,x3   3 ;t )  ( B [Z h1 ,Z h2 ,Z h3 ] )( x1   1 ,x2   2 ,x3 
  xi 2  xi 4 3
R

i

d 1d 2 d 3 ,si  xi   i ;hi  xi   i ,( i  1,3 ),
 3 ;t )}
2
2
2 3
(





)
1
2
3

(5.39)
so as takes place
3
3
3
3
3
 3  3
{

(


Z
)I

I

(


Z
)
}


(
I

)


(
 j I i x j ) 







j
0
i

x
j

i
0
x
x
i
j

x
0
x
0
x
 x
j
j
j
i
j
i
i 1
j 1
j 1
i 1
j 1
i 1
j 1
i

3
3
 3 3
  (   j I i x j )Z xi   (  I j xi Z x j )i ,
i 1
j 1
 i1 j 1
 3
3
3
( I )  0; (   )  0, (  Z )  0, ( i  1,3 ),



j 0 x j xi
j x j xi
 j 1 j x j xi
j 1
j 1

3
3
3
3

 1
1

1
1
  [  Z ]  0; 
(


Z
)

0;
(
U
)


U
;
(  Pxi )    P.




i
 xi

 t i 1 i xi
2


i 1 xi
i 1 xi 2
i 1 xi

Then on a basis (5.39) of system (5.38) it is equivalent, will be transformed to a kind:
 Z t   0 ( x1 ,x2 ,x3 ,t )  ( Q[Z ,Z x ,Z x ,Z x ] )( x1 ,x2 ,x3 ,t )  Z ,( i  1,3 ),
1
2
3

3
 Z t  0  0,( x1 ,x2 ,x3 )  R ,

3
3
3
3
3
d    0;    I   d 1[ (  f )   (  I ) 


i
0
j 0 x j
0 
i
0  j i x j
 0 
i 1
j 1
i 1
i 1
j 1

3
i
 1
( F0 ( x1   1 ,x2   2 ,x3   3 ;t ))}d 1d 2 d 3 ],
 4  {
(  12   22   32 )3

R3 i  1

3
3
( Q[Z ,Z ,Z ,Z ] )( x ,x ,x ,t )  {d 1Z( x ,x ,x ,t )[ (  I ( x ,x ,x ,t ))] 


x1
x2
x3
1 2
3
0
1 2
3
j i x j
1 2
3

i 1 j 1

3
1
i
 3
1

Z
(
x
,x
,x
,t
)I
(
x
,x
,x
,t
)

d
[
(


j
1 2
3
0

  xj 1 2 3
2
4 R3 i 1 (  1   22   32 )3
j 1

{( B [Z ,Z ,Z ] )( x   ,x   ,x   ;t )} )d d d ]},

1
1 2
2
3
3
1
2
3
h1
h2
h3

h  x   , ( i  1,3 ).
i
i
 1
(5.40)
The problem (5.40) is led to system of the integrated equations in a kind
t

r2
1
)( Q[Z ,W1 ,W2 ,W3 ] )( s1 ,s2 ,s3 ,s ) 

(
exp

M

Z

1
3  
)
s

t
(

4
3

8
0
R

t

1
1

exp( (  12   22   32 ))( Q[Z ,W1 ,W2 ,W3 ] )( x1 
ds1ds2 ds3 ds  M 1 


3
3
 ( ( t  s ) )
 0 R3

2 1  ( t  s ),x2  2 2 ( t  s ),x3  2 3 ( t  s );s )d 1d 2d 3ds 

 (  0 [Z ,W1 ,W2 ,W3 ] )( x1 ,x2 ,x3 ,t ),
t

1
( xi  si )
r2
1

))

(
exp
(
Wi  M 1 xi 
3  
4  ( t  s ) 2 ( t  s ) (  ( t  s ))3

8  0 R3

t
1

exp( (  12   22   32 )) 

M

ds
ds
ds
)ds
,s
,s
,s
s
)(
]
,W
,W
,W
Z
[
Q
(


1 xi
3
2
1
1 2 3
3
2
1
3  
 0 R3


i

( Q[Z ,W1 ,W2 ,W3 ] )( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 
 ( t  s )

  ( t  s );s )d 1d 2 d 3 ds  (  i [Z ,W1 ,W2 ,W3 ] )( x1 ,x2 ,x3 ,t ),

 Z xi  Wi ( x1 ,x2 ,x3 ,t ),( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),

t
 M 1 ( x1 ,x2 ,x3 ,t )  1
exp( (  12   22   32 )) 0 ( x1  2 1 ( t  s ),x2  2 2 
3  
(5.41)

 0 R3

  ( t  s ),x3  2 3 ( t  s );s )d 1d 2 d 3 ds,( si  xi  2 i ( t  s );i  1,3 ).


To solve a problem (5.40) concerning this problem we will receive system (5.41) of four integral equations.
Let concerning known functions M 1 ,  ,I i takes place:
 M 1 ,  ,I i : sup D k M 1 ( x1 ,x2 ,x3 ,t )  1 (  ), ( k  0,3; t  s   ),

T
3
3

sup   ( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, )  sup{d 01  (   j I i s j ( x1  2 1  ,x2  2 2  ,x3  2 3 
T T
 T T
i 1 j 1

3
  ;t   ) )   I j ( x1  2 1  ,x2  2 2  ,x3  2 3  ;t   ) 

j 1

3
3
3
i
 1 1

d
[
(
{
(
 0

   j Ii l j ( x1  2 1    1 ,x2  2 2  
4 R3 i 1 (  12   22   32 )3 i 1 j 1


3
3
 2 ,x3  2 3    3 ;t   ) )   (  I j l ( x1  2 1    1 ,x2  2 2    2 ,x3 
i

i 1 j 1

2 3    3 ;t   ) )i } )d 1d 2 d 3 ]}   2 (  ),

t
k  1 sup exp( (  2   2   2 )) ( x ,x ,x , , , ;t, )d d d d 
1
2
3

1
2
3 1 2 3
1
2
3

 0
 3 T 0 R3

t

i
1
2
2
2



(

)T
;
k

sup
exp
(

(





))

(
x
,x
,x
,

,

,

;t,

)

 2
0
i
1
2
3

1
2
3
1
2
3


 3 T 0 R3


d d d d  2T   (  ) 1 ,( i  1,3 ),   max(  T ; 3 2T  )
1
2
3
0 2
2 0
0 2



(5.42)
and if operators:  i ,( i  0,3 ) compressing with a compression factor k i ,
h

 i ,( i  0,3 ) : ki  4 ,( h  1 ),
 3
 k   (  )(  T  3 1 2T  )   (  )( 1  1 )  h  1,
i
2 0
0 2


n0
i 0


1
 1 ) ]1 , ( 0    1; 0  n0    0   ),
 (  )   [(
n0


 Sr1 ( 0 )  {Z ,W j : Z ; W j  r1 ,( x1 ,x2 ,x3 ,t )  T },( j  1,3 )
(5.43)
and:
  i [0,0,0,0] C  r1( 1  h ) :

  i [Z ,W1 ,W2 ,W3 ] C   i [Z ,W1 ,W2 ,W3 ]   i [0,0,0,0 ] C   i [0,0,0,0 ] C 

 ki 4r1  r1( 1  h )  hr1  r1( 1  h )  r1 ,
 : S ( 0 )  S ( 0 ), ( i  0,3 ).
r1
 i r1
(5.44)
Hence on the basis of contraction mapping principle system (5.41) is solvable and for which makes
Picard’s method

( x1 ,x2 ,x3 ,t )  T : Z n1   0 [Z n ,W1,n ,W2,n ,W3,n ],


Wi ,n1   i [Z n ,W1,n ,W2,n ,W3,n ],( n  0,1,...;Z0  0; Wi ,0  0; i  1,3 ).
(5.45)
The resulting sequence of functions {Z n }0 ,{Wi ,n }0 is converging and fundamental in S r1 ( 0 ) . Thus
we have
3
3

h1


Z

W
;


Z

Z

Wi ,n1  Wi C :  n1  h n1  0 
 0,


i C
n 1
n 1
n
 0
C
C
i

1
i

1

h1
h1
 Z 
Wi ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).
n
 n1 n  Z  H ; Wi ,n1 
(5.46)
Then on the basis of (5.36) and
i,n 1  i [0 ( x1 ,x2 ,x3 )  Zn 1( x1 ,x2 ,x3 ,t )]  I i ( x1 ,x2 ,x3 ,t ),( n  0,1,2,...;i  1,3 ),
we will receive
i ,n  1  i
C
 i Z n  1  Z
C
h 1
 i h n  i h n  1 0 
 0,( i  1,3 ).
n 
And it means that sequence {i ,n }0 converging to a limit i ,( i  1,3 ) :
(5.47)
h 1
i ,n 1 
i  C 3,1(T ),( i  1,3 ).
n 
(5.48)
It is obvious that small changes i0  
i 0 or f i ,( i  1,3 ) influence the solution (5.36) a little,
i.e. continuous depends on this data. Therefore, a question on a statement correctness problems
(5.31)-(5.33) are considered at once with results of the theorem 6* in Cn3,13 (T ) .
Let's notice that if
3
k
i 0
i
  1 , at that (see. (5.44)):
  (  )(  2T0  3 2T0  2 )  2 (  )  h  1, ( (  )   ( 2  )1 ; 0    1 ),
that, obviously, the Navier-Stokes problem (5.31) - (5.33) is solvable at Cn3,13 (T ) .
6. Updating of a Method (4.12) on the Basis of Integrals of Poisson’s Type
In the previous paragraphs we researched various variants of a method (4.12) for the Navier-Stokes
problem in spaces Cn3,13 (T ) .
There are the various partial experimental methods connecting speed and pressure. For example,
method Betz, A., Jones B.M. and others [12] establish these connections on the basis of the equation of
Bernoulli. We offer the method using regulatory functions for getting connections between pressure
and speed for   ( 1 ,2 ,3 ) . This distribution law allow us to express speed in the integral form.
In this connection in this paragraph the method of integrated transformation on the basis of
Poisson’s integrals is considered, when divf  0; 0    1 or divf  0; 1    0   . From the
received results it follows that system Navier-Stokes (1.1) in the conditions of (1.2) and (1.3) can have the
analytical smooth unique solution in Cn3,13 (T ) .
6.1. Fluid with Very Small Viscosity
0    1,
when divf  0
I. For incompressible currents with a friction, when

i 0 ( x1 ,x2 ,x3 ),( i  1,3 ),
i t 0  i0 ( x1 ,x2 ,x3 )  
 3
3
3
 
i 0 xi  0; 0  0; 0  C ( R ),
 i 1
divf  0; sup D k f  N  const , ( T  R 3  [0,T ] ), ( i  1,3 ),
i
0
0

T
functions i ,i  1,2,3 is represented in a kind
(6.1)

i  i [0 ( x1 ,x2 ,x3 )  Z( x1 ,x2 ,x3 ,t )]  i ( x1 ,x2 ,x3 ,t ),( x1 ,x2 ,x3 ,t ) T ,( i  1,3 ),

3

 Z t 0  0, ( x1 ,x2 ,x3 )  R ,
where 0  i  the known constants. From the entered functions i , is required:
lim  ( x ,x ,x ,t )  0,( x ,x ,x ,t )  T ; r  ( x  s )2  ( x  s )2  ( x  s )2 ,
1 2
3
1
1
2
2
3
3
  0 i 1 2 3
t

r2
1
 i ( x1 ,x2 ,x3 ,t )   1
exp
(

)
f i ( s1 ,s2 ,s3 ,s )ds1ds2 ds3ds 



4  ( t  s ) (  ( t  s ))3
8  3 0 R3

t

1


exp( (  12   22   32 )) fi ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 

3  
 0 R3

  ( t  s );s )d d d ds, ( s  x  2 ( t  s ); i  1,3 ),
1
2
3
i
i
i

t

3 
1
j 
2
2
2
 it   fi ( x1 ,x2 ,x3 ,t )  
exp
(

(





))
[
f il ( x1  2 1 ( t  s ),x2 

1
2
3


ts j

j 1
 3 0 R3

2 2  ( t  s ),x3  2 3  ( t  s );s )]d 1d 2 d 3ds,( i  1,3 ),

t
( x j  s j )
1
r2
   1
exp
(

) fi ( s1 ,s2 ,s3 ,s )ds1ds2 ds3ds 

3
 ix j
4 ( t  s )
8  3 0 R3 2  ( t  s ) (  ( t  s ))

t


1
exp( (  12   22   32 )) j f i ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3 
 
3  
ts

 0 R3

2 3  ( t  s );s )d 1d 2 d 3ds,( i  1,3; j  1,3 ),

t

 2   1
exp( (  12   22   32 )) j fil j ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),
3  
ix j

ts
 0 R3

 x3  2 3  ( t  s );s )d 1d 2 d 3ds,

t
j
   1
exp( (  12   22   32 ))
f 2 ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),
3
3


ix
 j
t  s il j

0 R3

 x3  2 3  ( t  s );s )d 1d 2 d 3ds,( i  1,3; j  1,3 ),

t

1



exp( (  12   22   32 )) f i ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3  2 3 
 i
3  
 0 R3

  ( t  s );s ) d d d ds  N ,
1
2
3
0

 N (  )   f T ; f  f  const , ( i  1,3 ),
0 0
i
0
 0
here ( 0,1 )   in a role of small parameter. At that
3
3
3

div


0
:

Z

0;


0;
 ixi  0,



i xi
i 0 xi

i 1
i 1
i 1


t
3
j
2
2
2
     1
exp
(

(





))
[
fil ( x1  2 1  ( t  s ),x2 

i
1
2
3


ts j
j 1
 3 0 R3
(6.2)
2 2  ( t  s ),x3  2 3  ( t  s );s )]d 1d 2 d 3ds,( l j  x j  2 j ( t  s );i  1,3; j  1,3 ),

3
3
3
 3




(


Z
)




(


Z
)

 j i x j ,


ixj
j i
0xj
xj
 j i x j  j 0
j 1
j 1
j 1
j 1

3
 3

(


Z
)

(


Z
)


(


Z
)
 j ( 0 x j  Z x j )  0,

 j 0
i
0xj
xj
i
0
j

1
j

1

   Z   ;   [ Z   ], (   0; i  1,3 ),
i t
it
i
i
i
0
 it

t
3
j
2
2
2
     Z   f    1
exp
(

(





))
[
fil ( x1  2 1 

it
i
i t
i
1
2
3


3

ts j
j 1
 0 R3

  ( t  s ),x2  2 2  ( t  s ),x3  2 3 ( t  s );s )]d 1d 2d 3ds  {i Z 

t
3
j
  1
2
2
2
exp
(

(





))
[
f il ( x1  2 1  ( t  s ),x2  2 2 

1
2
3

3  
ts j
j 1

0 R3

  ( t  s ),x  2  ( t  s );s )]d d d ds}   Z   f   Z , ( i  1,3 ).
3
3
1
2
3
i t
i
i

(6.3)
Then for incompressible currents with a friction the equations of Navier-Stokes (1.1) become simpler as take
place (6.1)-(6.3). Therefore the system (1.1), is led to a kind
3
3
3
i Zt    j ( 0  Z )i x    j i ( 0 x  Z x )    j i x  ( 1   ) fi 
j 1
j
j 1
j
j
j 1
j
1

Pxi 
(6.4)
 i Z ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).
From system (6.4), considering conditions (6.1)-(6.3) and having entered APS we have the equation:
1
 3 
 x (6.4) :   P  {F0  B[Z x1 ,Z x2 ,Z x3 ]},
 i 1 i
3
3
3
3

(
B
[
Z
,Z
,Z
]
)(
x
,x
,x
,t
)

(


)Z

(
 jxi Z x j )i ,





x1
x2
x3
1 2
3
j ix j
xi
i

1
j

1
i

1
j

1

3
3
3
3
3

 3
divf  0; F0 ( x1 ,x2 ,x3 ,t )   (   j  i x j )   i (   j xi 0 x j )  0 xi (   j  i x j ),
i  1 xi
j 1
i 1
j 1
i 1
j 1

1
1 1
 P
3 r {F0 ( s1 ,s2 ,s3 ,t )  ( B[Z s1 ,Z s2 ,Z s3 ] )( s1 ,s2 ,s3 ,t )}ds1ds2ds3 ,

4


R
1
1
i
 Pxi 
{F0 ( x1   1 ,x2   2 ,x3   3 ;t )  ( B[Z h1 ,Z h2 ,Z h3 ] )( x1   1 ,

2
4 R3 (  1   22   32 )3


 x2   2 ,x3   3 ;t )}d 1d 2 d 3 , ( si  xi   i ;hi  xi   i ;i  1,3 ),
so as takes place
3
3
3
3
 3  3
{

(


Z
)




(


Z
)



}


(
 j xi 0 x j ) 






j
0
ixj
j i
0xj
xj
j ixj
i
 x
j 1
j 1
i 1
j 1
 i 1 i j 1
 3
3
3
3
3
3
  (   )  (   )Z  (  Z ) ,

 j ix j xi 
 jxi x j i
0 xi  j i x j
 
i 1
j 1
i 1 j 1
i 1 j 1
(6.5)
3
3
 3
(

)

0;
(


)

0,
(
 j Z x j )xi  0, ( i  1,3 ),


j 0 x j xi
  j x j xi
j 1
j 1
 j 1

3
3
3

1
1

  [  Z ]  0;
(

P
)



P;

( i Z )  0.


i xi
xi
 t 


i 1
i  1 xi
i  1 xi
Then on a basis (6.5) system (6.4) it is equivalent, will be transformed to a kind:


 Z t   0  ( Q[Z ,Z x1 ,Z x2 ,Z x3 ] )( x1 ,x2 ,x3 ,t )  Z ,( i  1,3 ),

3
 Z t 0  0,( x1 ,x2 ,x3 )  R ,

3
3
3
3
3
d     0;  ( x ,x ,x ,t )      d 1[ ( 1   ) f   (    ) 
0
1 2
3
j 0xj
0
i
0
j ixj
 0 i 1 i
j 1
i 1
i 1
j 1
 3 3
3
1
i


(


)

{
( F0 ( x1   1 ,x2   2 ,x3   3 ;t ))}d 1d 2 d 3 ],
   j ixj


2
2
2 3
4

3
i

1
j

1
i

1
(





)

R
1
2
3

3
3
( Q[Z ,Z x ,Z x ,Z x ] )( x1 ,x2 ,x3 ,t )  {d 01Z( x1 ,x2 ,x3 ,t )[(   j i x ( x1 ,x2 ,x3 ,t ))] 
1
2
3
j

i 1 j 1
 3
3
i
 Z ( x ,x ,x ,t ) ( x ,x ,x ,t )  d 1[ 1 (
{( B[Z l1 ,Z l2 ,Z l3 ] )( x1 


xj
1 2
3
j
1 2
3
0

 j 1
4 R3 i 1 (  12   22   32 )3

 ,x   ,x   ;t )} )d d d ]},( l  x   , i  1,3 ).
1
2
3
i
i
i
 1 2 2 3 3
(6.6)
The problem (6.6) is led to system of the integrated equations quite regular rather ( 0,1 )   , in a kind
 Z x  Wi ( x1 ,x2 ,x3 ,t ),( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),
 i
t

1
r2
ds ds ds ds
Z

M

exp
(

)( Q[Z ,W1 ,W2 ,W3 ] )( s1 ,s2 ,s3 ,s ) 1 2 3 3 

1


4 ( t  s )
(  ( t  s ))
8  3 0 R3

t

1
 M 1 
exp( (  12   22   32 ))( Q[Z ,W1 ,W2 ,W3 ] )( x1  2 1 ( t  s ),x2 
3  

 0 R3

2 2  ( t  s ),x3  2 3  ( t  s );s )d 1d 2 d 3 ds  ( 0 [Z ,W1 ,W2 ,W3 ] )( x1 ,x2 ,x3 ,t ),

t
r2
( xi  si )
1
W  M  1
( exp( 
))

i
1xi


3

4  ( t  s ) 2  ( t  s ) (  ( t  s ) )3
8  0 R3

t
1


(
Q
[
Z
,W
,W
,W
]
)(
s
,s
,s
,s
)ds
ds
ds
ds

M

exp( (  12   22   32 )) 
1
2
3
1 2 3
1
2
3
1xi

3  
 0 R3


i
( Q[Z ,W1 ,W2 ,W3 ] )( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 

 ( t  s )

  ( t  s );s )d 1d 2 d 3 ds  ( i [Z ,W1 ,W2 ,W3 ] )( x1 ,x2 ,x3 ,t ),
t

1
 M 1( x1 ,x2 ,x3 ,t ) 
exp( (  12   22   32 )) 0 ( x1  2 1 ( t  s ),x2  2 2 
3  

 0 R3

  ( t  s ),x3  2 3 ( t  s );s )d 1d 2 d 3ds,( si  xi  2 i  ( t  s );i  1,3 ).
(6.7)
Let's notice that if known functions f i satisfy conditions of submultiplications [15] and takes place:
( x1 ,x2 ,x3 ,t )  T ;M 1 ;  ;  i :

k
sup D M 1 ( x1 ,x2 ,x3 ,t )  1 , ( k  0,3; t  s   ),
 T
t

1



exp( (  12   22   32 )) f i ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3 
 i
3  
 0 R3

2  ( t  s );s )d d d ds, ( i  1,3 ),
3
1
2
3

3
3

1
sup  ( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, )  sup{d0  (   j  i s j ( x1  2 1  ,x2  2 2  ,
T T
i 1 j 1
 T T
3

 x3  2 3  ;t   ) )    j ( x1  2 1  ,x2  2 2  ,x3  2 3  ;t   ) 

j 1

3
3
3
i
 1 1

d
[
(
{
(
 j  il ( x1  2 1    1 ,x2  2 2  


 0 4 3 
2
2
2 3
j
i

1
i

1
j

1
(





)
R
1
2
3

3
3

 2 ,x3  2 3    3 ;t   ) )   (   jl ( x1  2 1    1 ,x2  2 2    2 ,x3 
i
i 1 j 1


2 3    3 ;t   ) )i } )d 1d 2 d 3 ]}   2  ,

t
k0  1 sup exp( (  12   22   32 )) ( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, )d 1d 2 d 3d   2 T0 ,


 3 T 0 R3

t
i
1

2
2
2
k

sup
exp
(

(





))

(
x
,x
,x
,

,

,

;t,

)
d 1d 2 d 3 d 
i
1
2
3
1
2
3
1
2
3

3 T  

3

0
R

 2T   ,( i  1,3 );   max(  T ; 3 2T  ),
0 2
2 0
0 2

(6.8)
at that if operators:  i ,( i  0,3 ) compressing with a compression factor k i ,
h

 i ,( i  0,3 ) : ki  4 ,( h  1 ),
 3
 ki   (  2T0   3 2T0  2 )   (   1 )  h  1,
 i 0

 S r1 ( 0 )  {Z ,W j : Z ; W j  r1 ,( x1 ,x2 ,x3 ,t )  T },( j  1,3 )
and
  i [0,0,0,0 ]  r1( 1  h ) :
C

  i [Z ,W1 ,W2 ,W3 ] C   i [Z ,W1 ,W2 ,W3 ]  i [0,0,0,0 ] C   i [0,0,0,0 ] C 

 ki 4r1  r1( 1  h )  hr1  r1( 1  h )  r1 ;  i : S r1 ( 0 )  S r1 ( 0 ), ( i  0,3 ).
(6.9)
(6.10)
Then on the basis of contraction mapping principle system (6.7) is solvable and solution of this system
we can find on the basis of Picard’s method
 Z n  1   0 [Z n ,W1,n ,W2 ,n ,W3,n ],

Wi ,n  1   i [Z n ,W1,n ,W2 ,n ,W3,n ], ( n  0,1,...; Z 0  0; Wi ,0  0; i  1,3 ),

3
3
 En  1  Z n  1  Z n   Wi ,n  1  Wi ,n ; En  Z n  Z n 1   Wi ,n  Wi ,n 1
C
C
C

i 1
i 1

 Z n  1  Z n C  k0 En ; Wi ,n  1  Wi ,n C  ki En ,( i  1,3 ),

h 1
n
 0,
n 
 En  1  hEn  ...  h E1 
k 1
k 1

 Z n  k  Z n C   k0 En  j ; Wi ,n  k  Wi ,n C   ki En  j ,( i  1,3 ),

j 0
j 0

k 1
k 1
k 1
h 1
 En  k  h En  j  ...  h h n  j  1 E1  E1h n  h j  E1h n 1 
 0,
n 

1

h
j 0
j 0
j 0

3
3
U  Z 
W
;
U

Z

Z

Wi ,n  1  Wi C ,


i C
n 1
n 1
C
C
 0
i 1
i 1

h 1
n 1
 0,
U n  1  h U 0 
n 

h 1
h 1
 Z  H ; Wi ,n  1 
Wi ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).
 Z n  1 
n 
n 
C
:
(6.11)
Hence on the basis of (6.2) and
i,n 1  i [0 ( x1 ,x2 ,x3 )  Zn 1( x1 ,x2 ,x3 ,t )]  i ( x1 ,x2 ,x3 ,t ),( n  0,1,2,...;i  1,3 ),
(6.12)
we will receive
i ,n  1  i
C
 i Z n  1  Z
C
h 1
 i hU n  i h n  1U 0 
 0,( i  1,3 ).
n 
(6.13)
And it means that sequence {i ,n }0 converging to a limit C 3,1(T )  i ,( i  1,3 ) :
h 1
i ,n 1 
i ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),
n 
(6.14)
i.e. under conditions (1.2), (6.1), (6.2), (6.8) - (6.10) and (6.14) problem Navier-Stokes has the smooth
unique solution in Cn3,13 (T ) in a kind (6.2).
6.2. Fluid with Viscosity 1  0    , when divf  0
In this paragraph we will consider a liquid with viscosity and with a Reynolds small number of
where all inertial participants contain in the equations Navier-Stokes. Thus we will consider, methods
of integrated transformations on the basis of integrals of Poisson’s type, when 1  0    .
The decision method, from where follows of equations integration of Navier-Stokes in a case
3
3
i t 0  i0 ( x1 ,x2 ,x3 )  
i 0 ( x1 ,x2 ,x3 ),( 0  C ( R ); 0  i  const ; i  1,3 ),


k
divf  0; f  ( f1 , f 2 , f 3 ); sup D f i  N 0  const ,( x1 ,x2 ,x3 ,t )  T ; 0  0;
T

3
 
i 1
i 0 xi
 0, ( i  1,3 ),
t

1
1
r2
1
K
(
x
,x
,x
,t
)

exp
(

)
f i ( s1 ,s2 ,s3 ,s )ds1ds2 ds3ds 
 i 1 2 3


4 ( t  s ) ( ( t  s ))3
 8  3 0 R3


t
 1 1

exp( (  12   22   32 )) fi ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3  2 3 

3  
  0 R3


3
 ( t  s );s )d d d ds, [r 
( xi  si )2 ; si  xi  2 i ( t  s ); i  1,3],

1
2
3

i 1
is a major factor of this point.
Therefore we enter for definition a component of speeds:
i  i [0 ( x1 ,x2 ,x3 )  Z( x1 ,x2 ,x3 ,t )]  K i ( x1 ,x2 ,x3 ,t ),( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),

3
 Z t 0  0,( x1 ,x2 ,x3 )  R ,

div  0 :
 3
3
3
  Z  0;



i xi
i 0 xi  0;  K ixi  0,
 i 1
i 1
i 1
(6.15)
(6.16)
at that
t
3 

1
1 1
j 
2
2
2
fi ( x1 ,x2 ,x3 ,t ) 
exp
(

(





))
[
f il j ( x1  2 1 ( t  s ),x2 
 K it 

1
2
3



  3 0 R3
j 1 t  s


2 2  ( t  s ),x3  2 3 ( t  s );s )]d 1d 2 d 3ds,( i  1,3 ),
t

( x j  s j )
1
1
r2
 K ix  1
exp
(

) f i ( s1 ,s2 ,s3 , s )ds1ds2 ds3 ds 
 j
4 ( t  s )
 8  3 0 R3 2  ( t  s ) (  ( t  s ))3

t
j
 1 1

exp( (  12   22   32 ))
fi ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3 
3  



(t  s )

0 R3

2  ( t  s );s )d d d ds,( i  1,3; j  1,3 ),
3
1
2
3

t

j
1 1
exp( (  12   22   32 ))
f il ( x1  2 1 ( t  s ),x2  2 2 ( t  s ),x3 
 K ix2j 
3  
  0 R3
( t  s ) j


2 3  ( t  s );s )d 1d 2 d 3ds,

t
3 
j 
2
2
2
  K  1 1
exp
(

(





))
[
fil j ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),x3 

i
1
2
3

  3 0 R3
j 1 t  s

2 3  ( t  s );s )]d 1d 2 d 3ds, ( l j  x j  2 j ( t  s ); i, j  1,3 ),

3
3
3
 3




(


Z
)K


(


Z
)K

K j Ki x j ,




j
i
x
j
0
i
x
i
0
x
x
j

j
j
j
j
j 1
j 1
j 1
j 1

3
 3
  j ( 0  Z )i ( 0 x j  Z x j )  i ( 0  Z )  j ( 0 x j  Z x j )  0,
j 1
 j 1
   Z  K ;   [ Z   K ] , ( i  1,3 ),
i t
it
i
i
i
 it
t
3 

1
1 1
j 
2
2
2
fi ( x1 ,x2 ,x3 ,t ) 
exp
(

(





))
[
f il ( x1 
it  i  i Z t 

1
2
3

  3 0 R3
ts j
j 1


2 1  ( t  s ),x2  2 2  ( t  s ),x3  2 3 ( t  s );s )]d 1d 2d 3ds 

3 
 1 1 t
j 
2
2
2
exp
(

(





))
[
f il ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),x3 


1
2
3


ts j
j 1
   3 0 R3

2 3  ( t  s );s )]d 1d 2 d 3ds  i Z  i Z t  1 f i  i Z , ( i  1,3 ).


Hence on the basis of (6.15)-(6.17) we will receive
3
3
3
1
1
i Zt    j ( 0  Z )Ki x j   K j i ( 0 x j  Z x j )   K j Ki x j  ( 1 
) f i  Pxi 
j 1
j 1
j 1


(6.17)
(6.18)
 i Z ,( i  1,3 ).
Then for incompressible currents with a friction the equations of Navier-Stokes (1.1) become simpler as take
place (6.15), (6.17). Therefore a system (1.1) with account [(6.5) - (6.7), here instead of i we will
consider Ki] we will receive
 Z x  Wi ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ),
 i
t

1
r2
1
Z

M

exp
(

)
( Q[Z ,W1 ,W2 ,W3 ] )( s1 ,s2 ,s3 ,s ) 

1
3  
4  ( t  s ) (  ( t  s ))3
8  0 R3

t

1
ds1ds2 ds3 ds  M 1 
exp( (  12   22   32 ))( Q[Z ,W1 ,W2 ,W3 ] )( x1  2 1 
3  

 0 R3

  ( t  s ),x2  2 2  ( t  s ),x3  2 3 ( t  s ); s )d 1d 2 d 3 ds 

 ( 0 [Z ,W1 ,W2 ,W3 ] )( x1 ,x2 ,x3 ,t ),

t
r2
( xi  si )
1
W  M  1
( exp( 
))
( Q[Z ,W1 ,W2 ,W3 ] ) 
i
1xi


3

4  ( t  s ) 2 ( t  s ) (  ( t  s ) )3
8  0 R3

t
1
i


(
s
,s
,s
,s
)ds
ds
ds
ds

M

exp( (  12   22   32 ))

1
2
3
1xi
 1 2 3


3

(
t

s
)
3

0 R

( Q[Z ,W ,W ,W ] )( x  2  ( t  s ),x  2 ( t  s ),x  2 ( t  s );s ) 
1
2
3
1
1
2
2
3
3

d d d ds  ( [Z ,W ,W ,W ] )( x ,x ,x ,t ), ( s  x  2 ( t  s ); i  1,3 ),
1
2
3
i
1
2
3
1 2
3
i
i
i

where

1 t
exp( (  12   22   32 )) 0 ( x1  2 1 ( t  s ),x2  2 2 
 M 1 ( x1 ,x2 ,x3 ,t ) 
3  
 0 R3


 ( t  s ),x3  2 3 ( t  s );s )d 1d 2 d 3 ds,( si  xi  2 i ( t  s ); i  1,3 ),
 1
 P  {F0  B* [Z x ,Z x ,Z x ]},
1
2
3
 

(6.19)
3
3
3
3

(
B
[
Z
,Z
,Z
]
)(
x
,x
,x
,t
)

(

K
)Z

(
K jxi Z x j )i ,




1
2
3
j ix j
xi
 * x1 x2 x3
i 1
j 1
i 1
j 1

3
3
3
3
3

 3
(  K j K i x j )   i (  K j xi0 x j )   0 xi (   j K i x j ),
divf  0; F0 ( x1 ,x2 ,x3 ,t )  
i 1 xi
j 1
i 1
j 1
i 1
j 1

1
1 1
{F0 ( s1 ,s2 ,s3 ,t )  ( B* [Z s1 ,Z s2 ,Z s3 ] )( s1 ,s2 ,s3 ,t )}ds1ds2 ds3 ,
 P


4

r
3

R
1
i
1
 Pxi 
{F0 ( x1   1 ,x2   2 ,x3   3 ;t )  ( B* [Z h1 ,Z h2 ,Z h3 ] )( x1 

2
4 R3 (  1   22   32 )3


 1 ,x2   2 ,x3   3 ;t )}d 1d 2 d 3 , ( si  xi   i ;hi  xi   i ;i  1,3 ),

3
3
3
3
3
 Z  d 1 [  (   Z )K  K K ]  K (   Z )  d 1{ [( 1  1 ) f 



t
0   j
0
ixj
j ixj
j
0xj
xj
0
i


i 1 j 1
j 1
j 1
i 1

i
 1
{F0 ( x1   1 ,x2   2 ,x3   3 ;t )  ( B* [Z h1 ,Z h2 ,Z h3 ] )( x1 


2
2
2 3
 4 R3 (  1   2   3 )

 1 ,x2   2 ,x3   3 ;t )}d 1d 2 d 3 ]}  Z ,
3
3

1
(
Q
[
Z
,Z
,Z
,Z
]
)(
s
,s
,s
,s
)


{
d
[
Z(
s
,
s
,s
,s
)
(
 j K i s j ( s1 ,s2 ,s3 ,s ))] 



s1
s2
s3
1 2 3
0
1 2 3
i

1
j

1

3
 3
i
1
  Z s j ( s1 ,s2 ,s3 ,s )K j ( s1 ,s2 ,s3 ,s )  d 01[
(
{( B* [Z h1 ,Z h2 ,Z h3 ] )( s1 

4 R3 i1 (  12   22   32 )3
 j 1

 1 ,s2   2 ,s3   3 ; s )} )d 1d 2 d 3 ]}, ( hi  si   i ; i  1,3 ),

3
3
3
3
3
3
 ( x ,x ,x ,t )   K   d 1[ ( 1  1 ) f    (   K )   (  K K ) 
0
1
2
3
j 0xj
0
i
0
j ixj
j ixj


j 1
i 1
i 1
j 1
i 1
j 1

3
3
i
 1 {
(
F
(
x


,x


,x


;t
))
}
d

d

d

]
;
d

i  0.

0
1
1
2
2
3
3
1
2
3
0
 4 3 
2
2
2 3
i 1
i 1
(





)
R
1
2
3


If takes place:

( x ,x ,x ,t )  T ;M ,Y ,K : sup D k M ( x ,x ,x ,t )   , ( k  0,3; t  s   ),
1 2
3
1
i
1
1 2
3
1

T

3
3
sup Y( x ,x ,x , , , ;t, )  sup{d 1 (  K ( x  2  ,x  2  ,x  2 
1 2
3 1 2 3
0   j
isj
1
1
2
2
3
3
 T T
T T
i 1 j 1

3

  ;t   ) )   K j ( x1  2 1  ,x2  2 2  ,x3  2 3  ;t   ) 
j 1


3
3
3
i
1
 d01[
(
{
(

   j K il j ( x1  2 1    1 ,x2  2 2    2 ,x3 
4 R3 i 1 (  12   22   32 )3 i 1 j 1


3
3
2    ;t   ) )  (

 K jli ( x1  2 1    1 ,x2  2 2    2 ,x3  2 3 
3
3

i 1 j 1

1

    3 ;t   ) )i } )d 1d 2 d 3 ]}    2 ,

t

1
1
sup   exp( (  12   22   32 ))Y( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, )d 1d 2 d 3d 
 2T0 ,
k0 

 3 T 0 R3


t

1

k

sup
exp( (  12   22   32 ))Y( x1 ,x2 ,x3 , 1 , 2 , 3 ;t, ) i d 1d 2 d 3d 


 i

 3 T 0 R3


1
 2T0  2 ,( i  1,3 ), *  max(  2T0 ;3 2T0  2 )


(6.20)
and
h

 i ,( i  0,3 ) : ki  4 ,( h  1 ),
 3
 k  1 (  T  3 2T  1 )  1 ( 1  1 )  h  1,( 1       ),
i
2 0
0 2
*
0





i 0


 S r1 ( 0 )  {Z ,W j : Z ; W j  r1 ,( x1 ,x2 ,x3 ,t )  T },( j  1,3 ),

  i [0,0,0,0 ] C  r1 ( 1  h ) :  i [Z ,W1 ,W2 ,W3 ] C   i [Z ,W1 ,W2 ,W3 ]  i [0,0,0,0 ] C 

  i [0,0,0 ,0 ] C  ki 4r1  r1 ( 1  h )  hr1  r1 ( 1  h )  r1 ,

 i : S r1 ( 0 )  S r1 ( 0 ), ( i  0,3 ).
(6.21)
The solution of this system we can find on the basis of Picard’s method (6.11). Then considering
results (6.12)-(6.14), hence we will receive, that sequence {i ,n }0 converging to a limit
C 3,1(T )  i ,( i  1,3 ) :
h 1
i ,n 1 
i  H i ,( x1 ,x2 ,x3 ,t )  T ,( i  1,3 ).
n 
(6.22)
So as consequence of paragraphs 6.1 and 6.2 we will receive following statements:
Theorem 7. The Navier-Stoke’s nonstationary problem (1.1)-(1.3) is solvable at Cn3,13 (T ) , when
are fulfilled the conditions:
1) (6.1), (6.2), (6.8) - (6.10), (6.14) and 0    1,divf  0, or
2) (6.15)-(6.17), (6.20), (6.21), (6.22) and 1    0  ,divf  0.
7. Fluid with Viscosity, When v  R n ,( x  R n ;t  [0,T0 ] )
Here we will show that at certain mathematical transformations of the equation Navier-Stokes led
to a linear kind, when v  R n . So, once again clearly confirmed [12], that is, that the Navier-Stokes
equations with viscosity have solutions in analytical form, which is based on the Picard's method.
Offered methods of integrated transformations in paragraph 4 have been based on integrals of
Poisson's type. These methods are entered so that to transform nonlinear the Navier-Stokes problems
in inhomogeneous linear problems of heat conductivity [13]. Our purpose – to apply these methods for
the equation Navier-Stokes in a case v  ( 1 ,...,n ),
n
it   jix  fi 
j
j 1
1

Pxi  i ,( i  1,n ),
div  0, ( x, t )  T*  R n  [0, T0 ],
i
t 0
 i0 ( x1 ,x2 ,...,xn ),( x1 ,x2 ,...,xn )  R n ,( i  1,n ).
(1.1)n
(1.2)n
(1.3)n
From the received results follows that system Navier-Stokes (1.1)n in the conditions of (1.2)n, (1.3)n can have
the analytical smooth unique solution in Cn3,1(T* ), [or the conditional-smooth unique solution in
Gn1 ( D0  R n  ( 0,T0 )); Wn2 ( D0 ) ].
7.1. Fluid with Very Small Viscosity 0    1 , When divf  0
Let components of a vector of speed  for t  0 be defined by (1.3)n as
i
t 0
 i0 ( x1 ,x2 ,...,xn )  
i 0 ( x1 ,x2 ,...,xn ),( i  1,n ),
(7.1)
where 0  i ,( i  1,...,n ) are known constants. Then speed components  are defined by the rules
i  iV ( x1 ,x2 ,...,xn ,t ),( i  1,n ),

n
V t 0  0 ( x1 ,x2 ,...,xn ), ( x1 ,x2 ,...,xn )  R ,

divf  0; div  0 :
 n
n
n
  V  0;




V

j xj
j ix j
i   jVx j  0.

j 1
j 1
j 1
(7.2)
Hence, the system (1.1)n will be transformed to a kind
iVt  fi 
1

Pxi  i V ,( i  1,n ),
(7.3)
here V is a new unknown function which, by (7.2), defines the solution of the Navier-Stokes problem.
Here substitution (7.2) it is equivalent will transform system (1.1)n in the inhomogeneous linear equations of a
kind (7.3).
For this purpose, at first we will define pressure P . Really, considering APS from system (7.3)
we will receive:
1
 n 
 x (7.3) :   P   n F0 ,
 i 1 i
n

n 1 n
1


2(
n

2
)
[

(
)
]

;
F


(

)
f ixi ( x1 ,x2 ,...,xn ,t ), ( n  3 ),
 n

0
n
2
i

1

1
n
ds ds ...ds
 P  F0 ( s1 ,s2 ,...,sn ,t ) 1 n22 n , ( r   ( xi  si )2 ),

r

i 1
Rn

 1 P   i ( n  2 )F0 ( x1   1 ,x2   2 ,...,xn   n ;t )d 1d 2 ...d n ,( s  x   ;i  1,n ).
i
i
i
  xi n
(  12   2 2  ...   n2 )n
R

Therefore
Vt   0 ( x1 ,x2 ,...,xn ,t )  V ,
 n

 0 xi  0, ( x1 ,x2 ,...,xn ,t )  T* ,
 i 1
(  )1( f   1 P )  (  )1( f   1 P )  ...  (  )1( f   1 P )   ,
1
x1
2
2
x2
n
n
xn
0
 1
(7.4)
(7.5)
i.e. is the system (7.3) is transformed in the linear equations of heat conductivity with a condition of
Cauchy in a kind (7.5), and in a class of functions with smooth enough initial data is correctly put
[13, 14]. Accordingly there is an the conditional smooth and unique solution of a problem
Navier-Stokes in G 1( D0 ).
Really from system (7.13), follows:
V

1
2 n ( t )n
1
( ( t  s ))
n
 exp( 
Rn
r2
1
)V0 ( s1 ,s2 ,...,sn )ds1ds2 ...dsn 
4 t
2n  n
 0 ( s1 ,s2 ,...,sn ,s )ds1ds2 ...dsn ds 
1

n
2 1 t ,x2  2 2 t ,...,xn  2 n t )d 1d 2 ...d n 
 exp( ( 
2
1
t
  exp( 
0 Rn
r2
)
4 ( t  s )
  22  ...   n2 ))V0 ( x1 
Rn
t
1

n
  exp( ( 
2
1
  22 ...   n2 )) 
(7.6)
0 Rn
 0 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),...,xn  2 n ( t  s );s )d 1d 2 ...d nds 
 H 0 ( x1 ,x2 ,...,xn ,t ), ( si  xi  2 i t ;si  xi  2 i ( t  s );i  1,n ),
H 0 – is known function. The limiting case in G 1 ( D0 ) , when the solution (7.5) is representing in the
form (7.6), when
( x1 ,x2 ,...,xn ,t )  T* ;V0 ; 0 : sup D kV0  1 , ( k  0,3 ), sup D k 0 ( x1 ,x2 ,...,xn ,t )   1 ,

T*
Rn

t
sup 1
exp( (  12   22  ...   n2 ) ) D k  0 ( l1 ,l2 ,...,ln ;s ) d 1d 2 ...d n ds   1T0   2 ,
 T*  n 0 n
R

t
n

1
1
2
2
2
sup
exp
(

(




...


)
)

  j   0l j ( l1 ,l2 ,...,ln ;s ) d 1d 2 ...d nds 
1
2
n
n  
t  s j 1
 T*  0 Rn
T0

 n 1 2T0   3 ; sup  0 ( x1 ,x2 ,...,xn ,s )ds   1T0   2 ,


Rn 0

n

1
1
2
2
2
sup
exp
(

(




...


))(
 j  V0 l j ( l1 ,l2 ,...,ln ) )d 1d 2 ...d n  1

 n

1
2
n

n
n
R
n
i

1


R

1
 n
{ (   i2 exp( (  12   22  ...   n2 ) )d 1d 2 ...d n ) 2 (  exp( (  12   22  ...   n2 ) ) 
 i 1 Rn
Rn

1
d d ...d ) 2 }  n 1 , ( l  x  2 ( t  s ); l  x  2 t ;i  1,n ),
1
2
n
1
i
i
i
i
i
i

2

   max  i ; 0   ( n 2 T0  1  T0  ).
1 i  3

(7.7)
Really, estimating (7.6) in G 1 ( D0 ) , we have
V

V


V


 Vt

 V
G1 ( D0 )
C 3 ,0 ( T* )
C( T* )
C 3 ,0 ( T* )


0 k n
 Vt
D kV
L1
C( T* )
 N*  0 ,
 N* ,
 2 ,
(7.8)
T0
L1
 sup  Vt ( x1 ,x2 ,...,xn ,t )dt   ( n 2 T0  1  T0  )  0 .
Rn
0
Uniqueness is obvious, as a method by contradiction from (7.6) in G 1 ( D0 ) . Results (7.6) with
a condition (7.2), (7.7) are received where smoothness of functions is required only on xi ,( i  1,...,n ) as
the derivative of 1st order is in time has t  0 . Hence, on a basis transformation (7.2) we will receive
solution of system (1.1)n, which satisfies a condition (1.2)n, i.e.
i  i H 0 ( x1 ,x2 ,...,xn ,t ),( i  1,n ),

n
n
n
1
 n
2
2
2



H

0;

H

exp
(

(




...


))
iV0 hi ( x1  2 1 


i 0 xi
1
2
n
 ixi  i 0 xi
n 
n
i

1
i

1
i

1
i

1


R

t
n
2
2
2
 t ,x  2 t ,...,x  2 t )d d ...d  1
exp
(

(




...


))
i 

2
2
n
n
1
2
n
1
2
n


i 1
 n 0 Rn

 0l ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),...,xn  2 n ( t  s );s )d 1d 2 ...d n ds  0.
i

(7.9)
In the conclusion estimating (7.9) it is had
n

v

(

,

,...,

);
d

i ; i  iV ,( i  1,n ) :

1
2
n
0

i 1

n

(7.10)
 v Gn13 ( D0 )   [ iV C 3 ,0 ( T* )  iVt L1 ]  d0 [N*  0 ]  d0 M 0 ,( M 0  N*  0 ),
i 1

V 1
 V C 3 ,0 ( T )  Vt L1  N*  0 .
*
 G ( D0 )

Theorem 8. In the conditions of (1.2)n, (7.1), (7.7) and (7.10) the problem (1.1)n, (1.2)n, (7.1) has a
unique solution in Gn1 ( D0 ) , which is defined by a rule (7.9).
Remark 8. It is known that the turbulence solutions are conditional smooth in analytical sense [4].
Therefore, we consider a class of suitable solutions of the Navier-Stokes problem in weight space of
Sobolev's type Wn2, ( D0 ) . Since i0  C 3 ( R n ),( i  1,...,n ) , that solution (7.6) of the Navier-Stokes
problem (1.1)n - (1.3)n belongs in Wn,2 ( D0 ) :
n



2
 W
 i W(2i , ) , [W(2i , )  {( x1 ,x2 ,...,xn ,t )  D0 : D ki  L2 ; it  L2 }; i  1,n],
n,

i 1

1
T0
T0
2

k
2
[D i ( x1 ,x2 ,...,xn ,t )] dt  sup  ( t ) it ( x1 ,x2 ,...,xn ,t ) dt} 2 ,

 i W(2 , )  {  sup
n
i
Rn 0
0  k 3 R 0

if takes place
( x1 ,x2 ,...,xn ,t )  T* : sup D k i   1 , ( i  1,n;k  0,3 ),
T*


T0
T0
T0
1
2
1

2
(
sup

(
s
)

(
x
,x
,...,x
,s
)
ds
)


q


;
0


(
t
)
:

(
t
)
dt

q
;
( t )dt  q1 ,
 n 
i
1 2
n
1
1
4
0


t
R
0
0
0

   max(  ,  );    ( n  q  1   q ).
4
0
*
0
1
 *

(7.11)
For this purpose it is enough to show function accessories V in W2 ( D0 ) .
i  iV ,( i  1,n ) : W2 ( D0 )  {( x1 ,x2 ,...,xn ,t )  D0 : D kV  L2 ; Vt  L2 },

1
T0
T0

2
k
2
2
 V W2  {  sup  [D V ( x1 ,x2 ,...,xn ,t )] dt  sup  ( t ) Vt ( x1 ,x2 ,...,xn ,t ) dt} .
0  k 3 Rn 0
Rn 0

Let the solution of system (7.5) to represent in the form of (7.6) with conditions (7.7), (7.13). Then
estimating (7.6) in W2 ( D0 ) , we have [8]:
V 2
 N* T0  0  M* ,
 W ( D0 )

1
T0

2
2
 Vt L2  ( sup  ( t ) Vt ( x1 ,x2 ,...,xn ,t ) dt )  * ( n  q0  1   q1 )  0 ,

Rn 0

 * ( n  q0  1   q1 )  0 .

(7.12)
Then on a basis (7.2), (7.12) it is had
n

3
d0   i ; D0  R  ( 0,T0 );   ( 1V ,...,nV ) :

i 1

n
n





i V W 2 ( D )  d0 ( N* T0  0 )  d0 M* ,
2
2


i W
 Wn, ( D0 )
( D0 )

0
( i , )
i 1
i 1

i.e. in the conditions of (1.2)n, (1.3)n, (7.1), (7.7) and (7.11) the problem (1.1)n - (1.3)n has a the limited
solution in Wn,2 ( D0 ) .
7.2. Fluid with Viscosity 0    1 , When divf  0; 0  0
I. The overall objective of this point: to change a method (7.2) so that the received analytical
solution of a problem Navier-Stokes with viscosity, belonged in Cn3,1(T* ).
If takes place

i t 0  i0 ( x1 ,...,xn )  
i 0 ( x1 ,...,xn ),i  1,n,
n

3
n
  j0 x j  0; 0  0; 0  C ( R ),
 j 1
divf  0; sup D k f  N , ( i  1,n;k  0,4 ),
i
0

T*
(7.13)
that we will use transformation of a kind
i  i [0 ( x1 ,x2 ,...,xn )  Z( x1 ,x2 ,...,xn ,t )], ( x1 ,x2 ,...,xn ,t )  T* ,( i  1,n ),

n
 Z t 0  0, ( x1 ,x2 ,...,xn )  R ,

n
n
div  0 :   j Z x j  0;   j0 x j  0,
j 1
j 1

 n
n
n
n
n
 jix  i0   j0 x  i0   j Z x  i Z   j0 x  i Z   j Z x  0,
j
j
j
j
j
 j 1
j 1
j 1
j 1
j 1
(7.14)
where 0  i – the known constants. Hence, the system (1.1)n will be transformed to a kind
i Zt  fi 
1

Pxi  i Z ,( i  1,n ).
(7.15)
From system (7.15), considering conditions (7.13), (7.14), and having entered APS we have the equation
(7.4). Therefore the system (7.15) will be transformed to a kind
 Z    Z , ( x ,x ,...,x ,t )  T ,
0
1
2
n
*
 t
 Z t o  0,

1
1
1
1
1
1
( 1 ) ( f1   Px1 )  ( 2 ) ( f 2   Px2 )  ...  ( n ) ( f n   Pxn )   0 ( x1 ,...,xn ,t ).
Then the solution of problem (7.16) is presented in a kind
(7.16)
Z

1
2n  n
  exp( 
0 Rn
t
1

t
n
  exp( ( 
2
1
r2
1
)
 0 ( s1 ,s2 ,...,sn ,s )ds1ds2 ...dsn ds 
4  ( t  s ) (  ( t  s ))n
  22  ...   n2 )) 0 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),...,xn 
(7.17)
0 Rn
2 n  ( t  s );s )d 1d 2 ...d n ds  H( x1 ,...,x, ,t ), ( si  xi  2 i ( t  s ) ;i  1,n ),
here H – known function. The found solution (7.17) satisfies system (7.16). Really, having calculated
partial derivative of system (7.17):
n
j

1 t
2
2
2
Z



exp
(

(




...


))

 0l j ( x1  2 1 ( t  s ),
 t

0
1
2
n


n
n
t

s
j

1

0 R

 x  2 ( t  s ),...,x  2 ( t  s );s )d d ...d ds,
2
n
n
1
2
n
 2
t

1
exp( (  12   22  ...   n2 )) 0l j ( x1  2 1 ( t  s ),x2  2 2 

Z x j 
n  
 0 Rn


  ( t  s ),...,xn  2 n ( t  s );s )d 1d 2 ...d n ds,

1 t
 Z x2 
exp( (  12   22  ...   n2 )) 0l 2 ( x1  2 1 ( t  s ),x2  2 2 
j
j
n  

 0 Rn


  ( t  s ),...,xn  2 n ( t  s );s )d 1d 2 ...d n ds, ( l j  x j  2 j ( t  s ); j  1,n )
(7.18)
and substituting (7.18) in (7.16), we have
 Z t 0  0, ( x1 ,x2 ,...,xn )  R n ; ( 0,1 )   ; ( x1 ,x2 ,...,xn ,t )  T* :

t
n
j
1

2
2
2
0

Z




Z



exp
(

(




...


))

 0l j ( x1 

t
0
0
1
2
n



n
t

s
n
j 1

0
R

2  ( t  s ),x  2  ( t  s ),...,x  2 ( t  s );s )d d ...d ds   
1
2
2
n
n
1
2
n
0

t
n

1
2
2
2
 { n   exp( (  1   2  ...   n )) 0l 2j ( x1  2 1  ( t  s ),x2  2 2  ( t  s ),...,xn 
j 1
 0 Rn


t
n
j
2
2
2
2  ( t  s );s )d d ...d ds}  1
exp
(

(




...


))

 0l j ( x1 

n
1
2
n
1
2
n


ts
j 1
 n 0 Rn

t
1
1
1


2


(
t

s
),x

2


(
t

s
),...,x

2


(
t

s
)
;
s
)d

d

..
.d

ds


{

1
2
2
n
n
1
2
n


n
2
t

s

0

[ exp( (  2   2  ...   2 )) 2 ( x  2  ( t  s ),x  2 ( t  s ),...,x  2 ( t  s ) ;s ) 
1
2
n
1
1
2
2
n
n
0l1
 n
R

d( x1  2 1  ( t  s ) )d 2 ...d n   exp( (  12   22  ...   n2 )) 2 ( x1  2 1 ( t  s ),x2  2 2 
0l2

Rn

  ( t  s ),...,xn  2 n ( t  s );s )d 1d( x2  2 2 ( t  s ) )...d n  ... 

2
2
2
  exp( (  1   2  ...   n )) 0ln2 ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),...,xn  2 n ( t  s ) ;s ) 
 Rn
t
n
j

1
2
2
2
exp
(

(




...


))

 0l j ( x1 
d 1d 2 ...d( xn  2 n ( t  s ) )]ds} 

1
2
n

ts
j 1
 n 0 Rn

t

1
2 1  ( t  s ),x2  2 2 ( t  s ),...,xn  2 n ( t  s );s )d 1d 2 ...d n ds   { 1

n 
 0 ts


n
  exp( (  12   22  ...   n2 )) j 0l ( x1  2 1  ( t  s ),x2  2 2 ( t  s ),...,xn  2 n 
j
 Rn
j 1

  ( t  s );s )d 1d 2 ...d n ds  0.
That it was required to prove.
From the received results follows that functions i are defined on the basis of (7.14), i.e.
i  i [0 ( x1 ,x2 ,...,xn )  H( x1 ,x2 ,...,xn ,t )],( x1 ,x2 ,...,xn ,t ) T* ,( i  1,n ).
(7.19)
Further, considering partial derivatives of 1st order systems (7.19) and summing up with acceptance in
attention (1.2)n, (7.14) we have, that the system (7.19) satisfies to a condition (1.2)n.
3
n
3,1
II. So as i0  
i 0 , ( 0  C ( R )) , that the solution of problem (1.1)n-(1.3)n belongs in Cn (T* ) :
n

v

(

,

,...,

)
:


{  D ki
 it C( T ) },

1
2
n

Cn3 ,1 ( T* )
C( T* )
*
i 1 0  k  3


3
n
3,1
3,3,...,3,1
( T* )  Cn3,3,...,3,1( T* ),
i  i [0  Z ] ; 0  C ( R ); 0  0, ( i  1,n ), Cn ( T* )  Cn

 Z 3 ,1
  Dk Z
 Z t C( T ) .
C( T* )
*
 C ( T* ) 0  k  3

Really, if
i0  C 3 ( R n ); 0 : sup D ki0   1 ; sup D k 0 ( x1 ,x2 ,...,xn ,t )  1 , ( i  1,n; k  0,3 ),
T*

Rn

t
sup 1
exp( (  12   22  ...   n2 ) D k  0 ( l1 ,l2 ,...,ln ; ) d 1d 2 ...d n d  1T0   2 ,
 T*  n 0 Rn

t
n
1
1

2
2
2
sup
exp
(

(




...


)
 j   0l( kk ) ( l1 ,l2 ,...,ln ; ) d 1d 2 ...d n d 

1
2
n
 T
n  
j
t   j 1
 *  0 Rn
 n 2T   , ( l  x  2 ( t   ); j  1,n ),   max(  ; ),    ( 1   ),
1
0
3
j
j
j
i
1
0

1 i  3
that on a basis (7.17) we will receive
Z
C 3 ,1 ( T* )
 N 1  0 .
In the conclusion estimating (7.19) it is had
n

v

3 ,1
i 0 C 3 ( T* )  i Z t C( T* ) ]  d0 [N 1  N 2   0 ]  d 0 [N 0   0 ],
 Cn ( T* ) [ i Z C 3 ,0 ( T* )  

i 1

n
 Z 3 ,1

Z

Z

N


,
(
d

i ; 0 C 3 ( T )  N 2 ; N0  N1  N 2 ).
3 ,0

t
1
0
0
C ( T* )
C( T* )
*
 C ( T* )
i 1
(7.20)
Lemma 3. In the conditions of (1.2)n, (7.13), (7.14) and (7.20) the equation (7.17) has a unique
solution in C 3,1(T* ) .
Theorem 8*. At performance of conditions of the lemma 3 the problem (1.1)n, (1.2)n, (7.13) is
solvable at Cn3,1(T* ) of the defined by a rule (7.19).
The essential factor of researches of this paragraph are results of the theorem 8*. In this case the
solution of system (1.1)n is considered as the strict solution of a problem (1.1)n-(1.3)n in Cn3,1(T* ) .
It is obvious that small changes i0 ,( i  1,n ) or fi ,( i  1,n ) influence the solution (7.17) a little,
i.e. continuous depends on this data. Therefore, a question on a statement correctness problems
(1.1)n-(1.3)n are considered at once with results of the theorem 8*.
7.3. Fluid with Viscosity, when Re  2300; fi  0,( i  1,3 )
In this paragraph we generalise results of 5.3, so as this cases has applied value in the theory of a
liquid of average and small currents [12], when:
( x,t ) T*  Rn  [0,T0 ]; fi  0,( i  1,n ), 0  n0    0  , ( n0 , 0  const )) .
(7.21)
Under a condition (7.21) the problem (1.1)n – (1.3)n is led to a kind
n
it   jix  i 
j 1
j
1

Pxi ,( i  1,n ),
(7.22)
div  0, ( x, t )  T* ,
(7.23)
i
(7.24)
t 0
n
3
n
 i0  
i 0 ( x1 ,...,xn ),( x1 ,...,xn )  R , ( 0  C ( R ); 0  0; i  1,n ),
where 0  i  the known constants. Hence, here we will consider a methods of the equations
integration of Navier-Stokes (7.22) with a conditions (7.23), (7.24).
With that end in view we will assume that there are functions: 0  fi ,( i  1,n ) , which satisfy
conditions
 3,0
k
C ( T* )  fi : sup D f i   i (  )   (  )  1, ( 0    1; i  1,n ),
T*


divf  0, rot f  0 :  f i  0; ( f  ( f 1 , f 2 ,..., f n )),

t
 I ( x ,x ,...,x ,t )  f ( x ,x ,...,x ,s )ds, ( I  0; I  f ; i  1,n ).
n
i
i t
i
0 i 1 2 n
 i 1 2
Hence is offered the method:
(7.25)

i  i [0  Z( x1 ,...,xn ,t )]  I i ( x1 ,...,xn ,t ),( i  1,n ),

n

 Z t 0  0,( x1 ,...,xn )  R .
Thus takes place conditions
(7.26)
n
n
n

div


0
:


0;

Z

0;
I i xi  0,



i 0 xi
i xi

i 1
i 1
i 1

n
n
n
n
n




0;
I

0;




(


Z
)I

I

(


Z
)

I j  I i x j ,





0
i

t
j
i
x
j
0
i

x
j

i
0
x
x

j
j
j
j
i  1 xi
j 1
j 1
j 1
j 1

(7.27)
n
 n
  j ( 0  Z )i ( 0 x j  Z x j )  i ( 0  Z )  j ( 0 x j  Z x j )  0,
j 1
 j 1

it  i Z t  I i t ; i  [i ( 0  Z )   I i ]  i Z ,( i  1,n ).
Then for incompressible currents with a friction the equations of Navier-Stokes (7.22) become simpler as
take place (7.23), (7.24). Therefore the problem (7.22)-(7.24), is led to a kind
n
n
n
i Zt    j ( 0  Z )Ii x   I j i ( 0 x  Z x )   I j Ii x   fi 
j 1
j
j 1
j
j
j 1
j
1

Pxi  i Z ,( i  1,n ).
From system (7.28), considering conditions (7.22)-(7.24) and having entered [8]:
will receive the equation:
n

1

div
f

0;
(7.28) :
 P   n  {F0  B [Z x1 ,Z x2 ,...,Z xn ]},




i  1 xi

n

n
 n  2( n  2 )[ ( )]1  n , ( n  3 ), r   ( xi  si )2 ,
2

i 1

n
n
n
n
( B [Z x ,Z x ,...,Z x ] )( x1 ,...,xn ,t )  (  n )1  [ (   j I i x )Z x   (  I j x Z x )i ],
1
2
n
j
i
i
j

i 1 j 1
i 1 j 1

n
n
n
n
n
 n

1
(  I j I i x j ))],
 F0 ( x1 ,...,xn ,t )  (  n )  [ i (  I j xi 0 x j )  0 xi (   j I i x j )   (
i 1
j 1
i 1
j 1
i  1 xi
j 1

1
1
 P   n  2 {F0 ( s1 ,s2 ,...,sn ,t )  ( B [Z s1 ,Z s2 ,...,Z sn ] )( s1 ,s2 ,...,sn ,t )}ds1ds2 ...dsn ,
r

Rn
1
i( n  2 )
 Px  
{F0 ( x1   1 ,x2   2 ,...,xn   n ;t )  ( B [Z h1 ,Z h2 ,...,Z hn ] )( x1 
i
2

(  1   22  ...   n2 )n
Rn

 1 ,x2   2 ,...,xn   n ;t )}d 1d 2 ...d n , ( si  xi   i ; hi  xi   i ; i  1,n ),

so as takes place
n
n


1
1
 n
[

Z
]

0;

(


Z
)

0;
(  Pxi )    P,


i
 t  i xi


i  1 xi
i  1 xi
 i 1
 n
n
n
n
n
n
 n

{

(


Z
)I

I

(


Z
)
}


(
I

)


(
 j 0



i x j
j i
0xj
xj
i  j xi 0 x j
0 xi   j I i x j ) 

i  1 xi j  1
j 1
i 1
j 1
i 1
j 1
(7.28)
APS, we
(7.29)
n
n
 n n

(

I
)Z

(
   j i x j xi   I j xi Z x j )i ,
i 1 j 1
 i 1 j 1
 n
n
n
( I
)

0;
(


)

0,
(
 j Z x j )xi  0, ( i  1,n ).


j x j xi
j 0 x j xi
 
j 1
j 1
j 1
Then on a basis (7.29) the system (7.28) it is equivalent, will be transformed to a kind:


 Z t   0  ( Q[Z ,Z x1 ,Z x2 ,...,Z xn ] )( x1 ,...,xn ,t )  Z ,( i  1,n ),

n
 Z t 0  0,( x1 ,...,xn )  R ,

n
n
n
n
n
n
 ( x ,...,x ,t )   I   d 1[ (  f )   (   I )   (  I I ) 
n
j 0 x j
0
i
0
j i x j
j i x j
 0 1
j 1
i 1
i 1
j 1
i 1 j 1

n
n
i( n  2 )


{
(
F
(
x


,x


,...,x


;t
))
}
d

d

...d

]
;
d

i  0,
  

0
1
1 2
2
n
n
1
2
n
0
2
2
2 n
n
i

1
i

1
(




...


)
 R
1
2
n

n
n
( Q[Z ,Z x ,Z x ,...,Z x ] )( x1 ,...,xn ,t )  {d 01Z( x1 ,...,xn ,t )[(   j I i x ( x1 ,...,xn ,t ))] 
1
2
n
j

i 1 j 1
 n
n
i( n  2 )
 Z ( x ,...,x ,t )I ( x ,...,x ,t )  d 1[ (
{( B [Z h1 ,...,Z hn ] )( x1 


xj
1
n
j
1
n
0

2
2
2 n
 j 1
n
i

1
(




...


)
R
1
2
n

 ,x   ,...,x   ;t )} )d d ...d ]}, ( h  x   ;i  1,n ).
n
n
1
2
n
1
i
i
 1 2 2
(7.30)
The problem (7.30) is led to system of the integrated equations in a kind

 Z  W ( x ,...,x ,t ),( x ,...,x ,t )  T ,( i  1,n ),
i
1
n
1
n
 xi
t

1
r2
1
exp
(

)

 Z  M 1 ( x1 ,...,xn ,t ) 


4  ( t  s ) (  ( t  s ) )n
2 n  n 0 Rn


t
( Q[Z ,W ,W ,...,W ] )( s ,s ,...,s ,s )ds ds ...ds ds  M  1
exp( (  12   22  ...   n2 )) 
1
2
n
1 2
n
1
2
n
1


n

 0 Rn

( Q[Z ,W1 ,W2 ,...,Wn ] )( x1  2 1 ( t  s ),x2  2 2 ( t  s ),...,xn  2 n ( t  s );s ) 

d 1d 2 ...d n ds  (  0 [Z ,W1 ,W2 ,...,Wn ] )( x1 ,...,xn ,t ),

t
1
r2
( xi  si )
1
W  M

(
exp
(

))

i
1 xi


n
n

4  ( t  s ) 2 ( t  s ) (  ( t  s ))n
2  0 Rn

t
1


(
Q
[
Z
,W
,W
,...,W
]
)(
s
,s
,...,
s
,s
)ds
ds
...ds
ds

M

exp( (  12   22  ...   n2 )) 
1
2
n
1 2
n
1
2
n
1 xi

n  
 0 Rn


i
( Q[Z ,W1 ,W2 ,...,Wn ] )( x1  2 1 ( t  s ),x2  2 2 ( t  s ),...,xn  2 n 


(
t

s
)

(7.31)

  ( t  s );s )d 1d 2 ...d n ds  (  i [Z ,W1 ,W2 ,...,Wn ] )( x1 ,...,xn ,t ),
where
M 1 ( x1 ,...,xn ,t ) 
t
1

n
  exp( ( 
2
1
  22  ...   n2 )) 0 ( x1  2 1 ( t  s ),x2  2 2 
0 Rn
 ( t  s ),...,xn  2 n ( t  s );s )d 1d 2 ...d nds, ( si  xi  2 i ( t  s ),i  1,n ).
To solve this system, it is enough to find conditions that allow us to use the Banach Principle.
Really, if the input functions M 1 ,  ,I i satisfy to
( x1 ,...,xn ,t )  T* : sup D k M 1 ( x1 ,x2 ,x3 ,t )  1 (  ), ( k  0,3; t  s   ),
T*


n
n
sup   ( x1 ,...,xn , 1 ,..., n ;t, )  sup{d 01 (   j I i s ( x1  2 1  ,x2  2 2 
j
T* T*
T* T*
i 1 j 1

n
  ,...,x  2  ;t   ) ) 
I j ( x1  2 1  ,x2  2 2  ,...,xn  2 n 

n
n

j 1

n
n
n
 i ( n  2 ) n1

1


;t


)

d
[
(
{
(

  j Ii l j ( x1  2 1    1 ,
0

n 
2
2
2 n
i 1
i 1 j 1
(




...


)
R
1
2
n


n
n
 x2  2 2    2 ,...,xn  2 n    n ;t   ) )   (  I j li ( x1  2 1    1 ,x2 
i 1 j 1

2    ,...,x  2    ;t   ) ) } )d d ...d ]}    (  ),
2
2
n
n
n
i
1
2
n
2

t

1
k0 
sup
exp( (  12   22  ...   n2 ))  ( x1 ,...,xn , 1 ,..., n ;t, )d 1 ...d n d   2 (  )T0 ,
n T  

 * 0 Rn

t

k  1 sup
exp( (  12   22  ...   n2 ))  ( x1 ,...,xn , 1 ,..., n ;t, ) i d 1 ...d n d 
i




 n T* 0 Rn

1

 2T0  2 (  )  ,( i  1,n ),   max(  2T0 ; n 2T0  2 ),

(7.32)
at that if the operators  i ,( i  0,1,...,n ) satisfy to conditions of a principle of compression, that is,
n
h
1
1


,(
i

0,n
)
:
k

,
[
ki   (  )(  2T0  n
2T0  2 )   (  )(
 1 )  h  1],

i
 i
n  1 i 0

n0


 (  )   [( 1  1 ) ]1 , ( 0    1; 0  n       )
0
0

n0

and
 i : S r ( 0 )  S r ( 0 ), ( i  0,n ),
1
1

 S r ( 0 )  {Z ,Wi : Z ; Wi  r1 ,( x1 ,...,xn ,t )  T* },( i  1,n );  i [0,0,0,...,0] C  r1 ( 1  h ) :
 1
  i [Z ,W1 ,W2 ,...,Wn ] C   i [Z ,W1 ,W2 ,...,Wn ]   i [0,0,0,...,0 ] C   i [0,0,0,...,0] C 

 ki ( n  1 )r1  r1( 1  h )  hr1  r1 ( 1  h )  r1 .
(7.33)
(7.34)
Then, by Banach Principle, the system (7.31) is solvable and the solution of this system can be found
by Picard’s method as

( x1 ,...,xn ,t )  T* : Z m  1   0 [Z m ,W1,m ,W2,m ,...,Wn,m ],
(7.35)

W


[
Z
,W
,W
,...,W
]
,
(
m

0,1,...;
Z

0;
W

0;
i

1,n
).

i
m
1,m
2,m
n,m
0
i ,0
 i ,m  1
Then on the basis of conclusions of a method of Picard’s the sequence of functions
{Zm }0 ,{Wi,m}0 ,( i  1,n ) is converging and fundamental in S r1 ( 0 ) , so as
n
n

E

Z

Z

W

W
;
E

Z

Z

Wi ,m  Wi ,m 1 C ,


m 1
m C
i ,m  1
i ,m C
m
m
m 1 C
 m 1
i 1
i 1

h 1
 Z m  1  Z m  k0 Em ; Wi ,m 1  Wi ,m  ki Em ,( i  1,n ) : Em 1  hEm  ...  h m E1 
 0,
m 
C
C

k 1
k 1

Z

Z

k
E
;
W

W

ki Em  j ,( i  1,n ),


m C
0 m j
i ,m  k
i ,m C
 mk
j 0
j 0

k 1
k 1
k 1

1
h 1
m  j 1
E1  E1h m  h j  E1h m

 0.
 Em  k  h Em  j  ...  h h
m 
1

h
j 0
j 0
j 0

Thus we will receive
n
n



Z

W
;


Z

Z

Wi ,m  1  Wi


i C
m 1
m 1
 0
C
C
i 1
i 1

h 1
 Z( x1 ,...,xn ,t )  H( x1 ,...,xn ,t ),
 Z m  1 
m 

h 1
Wi ( x1 ,...,xn ,t ),( x1 ,...,xn ,t )  T* ,( i  1,n ).
Wi ,m  1 
m 

Therefore on the basis of (7.26) and
C
h 1
:  m 1  h m  1 0 
 0,
m 
i,m1  i [0 ( x1 ,...,xn )  Zm1( x1 ,...,xn ,t )]  I i ( x1 ,...,xn ,t ),( m  0,1,2,...;i  1,n ),
(7.36)
(7.37)
we will receive
i ,m  1  i
C
 i Z m  1  Z
C
h 1
 i h m  i h m  1 0 
 0,( i  1,n ).
m 
And it means that sequence {i ,m }0 converging to a limit i ,( i  1,n ) :
h 1
i ,m 1 
i  C 3,1(T* ),( i  1,n ).
m 
(7.38)
Theorem 9. In the conditions of (7.23), (7.24), (7.25), (7.33), (7.34) and (7.38) the Navier-Stokes
problem is correctly put in Cn3,1(T* ).
The system (7.31) consists of n  1 the integral equations of Volterra and Volterra-Abel
on a variable t  [0,T0 ] . Thus it is proved that for this system the Banach Principle all conditions are
executed. Then the system (7.31) is unequivocally compatible, i.e. on a basis (7.36) there is the smooth
and unique function Z( x1 ,...,xn ,t )  C 3,1(T* ) . Hence considering (7.26), under conditions (7.23),
(7.24), (7.25), (7.33), (7.34) and (7.38) we will receive i  C 3,1(T* ),( i  1,n ) .
It is obvious that small changes i0  
fi ,( i  1,n ) influence the solution
i 0 ,( i  1,n ) or
(7.26) a little, i.e. continuous depends on this data. Therefore, a question on a statement correctness
problems (7.22)-(7.24) are considered at once with results of the theorem 9 in Cn3,1(T* ) . ■
8. Conclusion
The essence of the given work consists that the developed methods of an initial problem lead to
Navier-Stokes equations linearization into integral form without attraction of any conditions for solving the
task (1.1)-(1.3). That’s why the received analytical solution is regular to viscosity factor, and mathematic
meanings of variables respond to physical significance according to requirements of Clay Mathematics
Institute. So, The developed out in this article methods of solution Navier Stokes equations bring the task to
equivalent linear equations with heat conductivity type under Cauchy condition in class of functions with
smooth enough initial conditions under t  0 .
We should remind that Clay Institute suggested solving one from three variants of Navier Stokes
equation. But in this work summarising the result received in article [8] for any viscosity meanings in
unlimited area in both spaces, as exactly this variant according to Clay Institute is more preferable in
comparison with other variants received in [8]. In paragraph 4 the received results are illustrated by an
example in the limited area.
The solution of 6th millennium problem allowed to consider it that results of this work in Cn3,13 (T ) as
principle of continuation of strong solutions connected with Beale-Kato-Majda [2, 5] criterion
requirements but already for 3D Navier-Stokes equation for viscous incompressible fluid [1]. When
analytical solution of Navier-Stokes equation is found so Beale-Kato-Majda’s criterion for Euler’s
equation is automatically proved because viscosity meaning   0 in them responds to Euler’s
equation for ideal liquid.
From the received results follows the Navier-Stokes equations (1.1) in the conditions of (1.2),
(1.3) have the smooth and unique solution in Cn3,13 (T ), [Cn3,1(T* )] , or conditionally-smooth unique
solution in Gn13 ( D0 ),
millennium problem [1].
[Wn23 ( D0 ); Gn1 ( D0 ); Wn,2 ( D0 )], that reflect the requirements of 6th
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