Vol.2, No.4, 338-356 (2010) Natural Science
http://dx.doi.org/10.4236/ns.2010.24042
Copyright © 2010 SciRes. OPEN ACCESS
Application of analytic functions to the global solvabilty
of the Cauchy problem for equations of Navier-Stokes
Asset Durmagambetov
Ministry of Education and Science of the Republic of Kazakhstan, Buketov Karaganda State University, Institute of Applied Mathemat-
ics, Buketov Karaganda, Kazakhstan; aset.durmagambetov@gmail.com
Received 14 December 2009; revised 29 January 2010; accepted 3 February 2010.
ABSTRACT
The interrelation between analytic functions and
real-valued functions is formulated in the work.
It is shown such an interrelation realizes nonli-
near representations for real-valued functions
that allow to develop new methods of estimation
for them. These methods of estimation are ap-
proved by solving the Cauchy problem for equ-
ations of viscous incompressible liquid.
Keywords: Shrdinger; Cauchy Problem;
Navier-Stokes’; Inverse; Analytic Functions;
Scattering Theory
1. INTRODUCTION
The work of L. Fadeyev dedicated to the many- dimen-
sional inverse problem of scattering theory inspired the
author of this article to conduct this research. The first
results obtained by the author are described in the works
[1-3]. This problem includes a number of subproblems
which appear to be very interesting and complicated.
These subproblems are thoroughly considered in the
works of the following scientists: R. Newton [4], R.
Faddeyev [5], R. Novikov and G. Khenkin [6], A. Ramm
[3] and others. The latest advances in the theory of SIPM
(Scattering Inverse Problem Method) were a great sti-
mulus for the author as well as other researchers. Anoth-
er important stimulus was the work of M. Lavrentyev on
the application of analytic functions to Hydrodynamics.
Only one-dimensional equations were integrated by
SIPM. The application of analytic functions to Hydro-
dynamics is restricted only by bidimensional problems.
The further progress in applying SIPM to the solution of
nonlinear equations in R3 was hampered by the poor
development of the three-dimensional inverse problem
of scattering in comparison with the progress achieved in
the work on the one-dimensional inverse problem of
scattering and also by the difficulties the researchers
encountered building up the corresponding Laxpairs. It
is easy to come to a conclusion that all the success in
developing the theory of SIPM is connected with ana-
lytic functions, i.e., solutions to Schrodingers equation.
Therefore we consider Schrodingers equation as an in-
terrelation between real-valued functions and analytic
functions, where real-valued functions are potentials in
Schrodinge rs equation and analytic functions are the
corresponding eigenfunctions of the continuous spec-
trum of Schrodingers operator. The basic aim of the
paper is to study this interrelation and its application for
obtaining new estimates to the solutions of the problem
for Navier-Stokesequations. We concentrated on for-
mulating the conditions of momentum and energy con-
servation laws in terms of potential instead of formulat-
ing them in terms of wave functions. As a result of our
study, we obtained non-trivial nonlinear relationships of
potential. The effectiveness and novelty of the obtained
results are displayed when solving the notoriously diffi-
cult Chauchy problem for Navier-Stokesequations of
viscous incompressible fluid.
2. BASIC NOTIONS AND SUBSIDIARY
STATEMENT
Let us consider Shrodingerse equation
Δ+= ||2 (1)
where is a bounded fast-decreasing function,
3,||2=
3
=1 2.
Definition 1. Rolnik’s Class is a set of measurable
functions ,
||||=
6()()
||2<.
It is considered to be a general definition ([7]).
Theorem 1. Suppose that ; then a exists a
unique solution of Eq.1, with asymptotic form (2) as
|| .
A. Durmagambetov / Natural Science 2 (2010) 338-356 339
Copyright © 2010 SciRes. OPEN ACCESS
±(,)= (,)+
+±||||
||±(,)+0 1
|| (2)
where 3,= ||
||, (,) =
3
=1 ,
±(,) =1
(2)3
3()±(,)(,).
The proof of this theorem is in [7].
Consider the operators = +(), 0=
defined in the dense set 2
2(3) in the space 2(3).
The operator is called Schrodingers operator. Povz-
ner [8] proved that the functions ± (,) form a
complete orthonormal system of eigenfunctions of the
continuous spectrum of the operator , and the operator
fills up the whole positive semi-axis. Besides the conti-
nuous spectrum the operator can have a finite num-
ber of negative eigenvalues Denote these eigenvalues
by 2 and conforming normalized egenfunctions by
(,2)(=1, ),
where (,2)2(3).
Theorem 2 (About Completeness). For any vec-
tor-function 2(3) and eigenfunctions of the op-
erator , we have Parsevals identity
||2
2=
=1 ||2+
3|()|2,
where and are Fourier coefficients in case of dis-
crete of and continuous spectrum respectively.
The proof of this theorem is in [8].
Theorem 3 (Birman - Schwinger’s Estimate). Sup-
pose . Then the number of discrete eigenvalues of
Shrdinger operator satisfies the estimate
()1
(4)2
3
3()()
||2.
The proof of this theorem is in [9].
Definition 2. [7]
±(,) =1
(2)3
3±(,)(,)().
±(.,.) is called T-matrix. Let us take into consider-
ation a series for ±:
±(,)=
=0 ±(,),
where 0±(,)=1
(2)3
3(,)(),
±(,)=1
(2)3
(1)
(4)
3(+1) (,0)
×(0)±|||01|
|01|(1)...(1)
×±|||1|
|1|()(,)0....
As well as in [7] we formulate.
Definition 3. Series (4) is called Borns series.
Theorem 4. Let 1(3) . If 
24,
then Borns series for (,) converges as ,3.
The proof of the theorem is in [7].
Definition 4. Suppose ; then the function
(,), denoted by the following equality
(,)=1
(2)3
3()
+(,)(,),
is called scattering amplitude
Corollary 1. Scattering amplitude (,) is equal to
-matrix (,) =+(,)
=1
(2)3
3()+(,)(,).
The proof follows from definition 4.
It is a well-known fact [5] that the solutions + (,)
and (,) of Eq.1 are linearly dependent
+= (3)
where is a scattering operator with the nucleus
(,) of the form
(,)=
3+(,)+
(,).
Theorem 5. (Conservation Law of Impulse and
Energy). Assume that , then
=,=,
where is anunit operator.
The proof is in [5].
Let us use the following definitions
()=
3()(,),
()=
3()(,),
mv () =
3())(||2||2),
340 A. Durmagambetov / Natural Science 2 (2010) 338-356
Copyright © 2010 SciRes. OPEN ACCESS
mv () =
3(,))(||2||2),
(,)=
3(,)(||2||2),
(,) =
3(,)(2||2),
where ,3 and =
||, =
||.
3. ESTIMATE OF AMPLITUDE MAXIMUM
Let us consider the problem of estimating the maximum
of amplitude, i.e., max
3|(,)|. Let us estimate the
term of Borns series |(,)|.
Lemma 1. |(,)| satisfies the inequality
|+1 (,)| 1
(2)3
1
(4)+1
×
(2)2(+1)
3
|()|2
||2,
=|||| + 4,=2
,
where -is a small value, is a positive number,
=
3||.
Theorem 6. Suppose that <163, then

3|(,)| satisfies the following estimate
max
3|(,)| 1
(2)3
1
163
3
|()|2
||2,
where =|||| + 4, is a small value,
= 2
, =
3||.
4. REPRESENTATION OF FUNCTIONS
BY ITS SPHERICAL AVERAGES
Let us consider the problem of defining a function by its
spherical average. This problem emerged in the course
of our calculation and we shall consider it hereinafter.
Let us consider the following integral equation
3()(||2||2)=(2),
where ,3, is Diracs delta function,
2
2(3), ||2=
3
=1 2, (,) =
3
=1 .
Let us formulate the basic result.
Theorem 7. Suppose that 2
2(3), then
(2)2(,,)
=1
2
2
0
2
0(2
(,),)
+(2
(,),)2
(,)2 sin  ,
where (2
(,),)= (2
(,),),
sin  =,
sin =,= ||.
Theorem 8. Fourier transformation of the function q
satisfies the following estimate
||11
4
21
+2 
21
+
1
,
5. CORRELATION OF AMPLITUDE AND
WAVE FUNCTIONS
We take the relationship for +, from (3)
+(,)=(,)
2
3(||2||2) (4)
×(,)(,).
Let us denote new functions and operators we will use
further 0(,)=(,),
0(,) =0(,) + 0(,),
+(,) =+(,)(,)
++(,)(,),
(,) =(,)(,)
+(,)(,),
1=2
2
3
(2(,) + 2(,))
×()([mv ]() + [mv ]())
20
3
(2(,) + 2(,))
×()[2](, 0). (9)
(3)(, 0)=20
2
3
(()+ ())
×()
(2)(, 0)
+20
2
3
(2(,) + 2(,))
×()(
3()
||2+0
+
(2)(, 0)). (10)
correspondingly,
(3) (, 0)=+
(3)(, 0)
(3)(, 0),
(3) (,0)=+
(3)(, 0)
(3)(, 0),
and ±
(3), 0) are terms of order 3 and higher w.r.t.
 in the representations (7), (8).
Lemma 12. Suppose that , ±|=0,=0= 0,
then mv () + mv ()
=
4(0)
0
2
0
((,) + (,))
×
0(, 0).
8. NONLINEAR REPRESENTATION OF
POTENTIAL
Let us proceed to the construction of potential nonlinear
representation.
Lemma 13. Assume that , ±|=0,=0=0 ;
then mv () + mv ()
=0
0(()
2
0
+())mv ()
0
2
0
2
0
(() + ())
×mv ()
+0
0
2
0
(() + ())
× ([mv ]() + [mv ]())
0
0
2
0
(() + ())
×..
3()
||2 
0
2..
3
0
2
0
(()+ ())

344 A. Durmagambetov / Natural Science 2 (2010) 338-356
Copyright © 2010 SciRes. OPEN ACCESS
×()
40(0)
0
2
0
((,) + (,))
×
0(, 0)+2
0
((3) (, 0)
+(3) (,0)+3(, 0)+(3 )(, 0)),
where 3(, 0), (3 )(, 0) are defined by Eqs.9 and
10 accordingly,
(3) (, 0)=+
(3)(, 0)
(3)(, 0),
(3) (,0)=+
(3)(, 0)
(3)(, 0),
and ±
(3), 0) are term of order 3 and higher w.r.t. 
in representations (7), (8).
Lemma 14. Suppose that , ±|=0,=0= 0,
then
..
3
0
2
0
(()+ ())

×()
=
0
2
0
(() + ())
×mv ().
Lemma 15. Let 2
1() and , then
0
2
0
(() + ())
× ([mv ]() + [mv ]())
=
0
2
0
(() + ())
× (mv () + mv ()),
0
2
0
(() + ())
×..
3()
||2 
=
0
2
0
(() + ())
×mv ().
Theorem 14. Let , ±|=0,=0 =0, then
mv () + mv ()
=0
0
2
0
(() + ())
×mv ()+(),
() =2
0
((3) (, 0)+(3)(, 0)
+3(, 0)+( 3)(, 0)),
where 0=4.
Theorem 15. Suppose , ±|=0,=0= 0; then
() =
0
2
0
0
2
0
(()
+())()
×0(),
where |0|< |mv |
9. THE CAUCHY PROBLEM FOR
NAVIER-STOKESEQUATIONS
Let us apply the obtained results to estimate the solu-
tions of Cauchy problem for Navier-Stokesset of equa-
tions +
3
=1
=+0(,), = 0, ( 11)
|=0=0() (12)
in the domain of =3×(0, ). With respect to 0,
assume  0= 0. (13)
Problem (11), (12), (13) has at least one weak solution
(q,p) in the so-called Leray-Hopf class, see [3].
Let us mention the known statements proved in [10].
Theorem 16. Suppose that
02
1(3), 2()
then there exists a unique weak solution of problem (11),
(12), (13), in 1, 1[0, ], that satisfies
, ,2()
Note that 1 depends on 0, .
Lemma 16. If 02
1(3), 2(), then
sup
0||||2(3)
2+
0
||||2(3)
2
||0||2(3)
2+ ||0||2()
Our goal is to prove the global unicity weak solution
of (11), (12), (13) irrespective of initial velocity and
A. Durmagambetov / Natural Science 2 (2010) 338-356 345
Copyright © 2010 SciRes. OPEN ACCESS
power smallness conditions.
Therefore let us obtain uniform estimates.
Statement 1. Weak solution of problem (11), (12),
(13), from Theorem 16 satisfies the following equation
((), ) =0(())
+
02||()([(,)]
+)
× ((), ), (14 )
where =+0.
Proof. The proof follows from the definition of
Fourier transformation and the formulas for linear diffe-
rential equations.
Lemma 17. The solution of the problem (11), (12),
(13) from Theorem 16, satisfies the following equation
=
,
||2
+
||2
and the following estimates
||||2(3)3||||2(3)
3
2||||2(3)
1
2,

|2|
||+||
||2+1
||
+3 2
||;
Proof. We obtain the equation for using  and
Fourier transformation. The estimates follow from the
obtained equation.
This completes the proof of Lemma 17.
Lemma 18. Weak solution of problem (11), (12), (13),
from Theorem 16 satisfies the following inequalities
sup
0[
3
||2|(,)|2
+
0
3
||2|(,)|2],
sup
0[
3
||4|(,)|2
+
0
3
||4|(,)|2],
or
sup
0[
2(3)
+
0
32|
(,)|2],
sup
0[2
22(3)
+
0
32|
(,)|2].
Proof. The proof follows from Navier-Stokesequa-
tion, the first priori estimate formulated in Lemma 16
and obtained from Lemma 17.
This completes the proof of Lemma 18.
Lemma 19. Weak solution of problem (11), (12), (13),
from Theorem 16, satisfies the following inequalities
max
||max
|0|
+
2sup
0||||2(3)
2+
0
||||2(3)
2,
max

max
0

+
2sup
0
2(3)
+
0
32|
(,)|2,
max
2
2max
20
2
+
2sup
02
22(3)
+
0
32|
(,)|2.
Proof. We obtain these estimates using representation
(14), Parsevals equality, Cauchy - Bunyakovskiy in-
equality (14) by Lemma 18.
This proves Lemma 19.
Lemma 20. Weak solution of problem (11), (12), (13),
from Theorem 16 satisfies the following inequalities
|mv (,)|1,mv (,)
 2,
2mv (,)
23,
where 1,2,3 are limited.
Proof. Let us prove the first estimate. These inequali-
ties
|mv (,)|
2
0
2
0
|((), )|
2max
||1,
where 1=.
Follows from definition (2) for the average of and
from Lemmas 18, 19.
The rest of estimates are proved similarly.
346 A. Durmagambetov / Natural Science 2 (2010) 338-356
Copyright © 2010 SciRes. OPEN ACCESS
This proves Lemma 20.
Lemma 21. Weak solution of problem (11), (12), (13),
from Theorem 16 satisfies the following inequalities
, (= 0,2,4), where
0=
0
|1|2,1= (,)+,
2=
01
2,4=
021
22.
The proof follows from the apriori estimate of Lemma
16 and the statement of Lemma 18.
This completes the proof of Lemma 21.
Lemma 22. Suppose that , 
||< , then
3
3()()
||2(||2+ max
||)2
Proof. Using Plansherels theorem, we get the state-
ment of the lemma.
This proves Lemma 22.
Lemma 23. Weak solution of problem (11), (12), (13),
from Theorem 16 satisfies the following inequalities
|((), )| |0(())|
+1
21
20
1
2
||, (1 5)
where
0=
0
|1|2,1= (,)+.
Proof. From Formula (14) we get
|((), )| |0(())|
+|
02||2() (16)
×1((), )|
where 1= (,)+.
Using the denotation
=|
02||2()
×1((), )|,
Taking into account Holders inequality in we obtain

0
|2||2()|
1
×
0
|1|
1
where , satisfies the equality 1
+1
= 1.
Suppose = = 2. Then
1
21
2
0|1|21
2
||.
Taking into consideration the estimate in (16), we
obtain the statement of the lemma.
This proves Lemma 23.
Now, we have the uniform estimates of Rolnik norms
for the solution of problems (11), (12), (13). Our further
and basic aim is to get the uniform estimates |
|1(3),
a component of velocity components in the Cauchy
problem for Navier-Stokesequations. In order to
achieve the aim, we use Theorem 8 it implies to get es-
timates of spherical average.
Lemma 24. Weak solution of problem (11), (12), (13),
from Theorem 16 satisfies the following inequalities
|mv |1(3)
20
(1) +1|mv |1(3)
+||13 (17)
the function is defined in Theorem 15,
0
(1) =
3
0
2
0
|0(())|
× |mv (,)|,1=1
1
280
1
2,
and 0 is defined in Lemma 23.
Proof. From the statement of Theorem 14, we get the
estimate
|mv |1(3)
2
3
0
2
0
|((), )|
× |mv (,)|+ ||1(3).
(15) in the integral, we obtain
|mv |1(3)
2(
3
0
2
0
|0(()|
× |mv (,)|
+1
1
20
1
2
3
0
2
0
|mv (,)|
×
||)+ ||1(3).
A. Durmagambetov / Natural Science 2 (2010) 338-356 347
Copyright © 2010 SciRes. OPEN ACCESS
Let us use the notation
0
(1) =
3
0
2
0
|0(())|
× |mv (,)|,
then
|mv |1(3)
2(0
(1) +1
1
20
1
2
×
3
0
2
0
|mv (,)| 
||)+ ||1(3).
Let us use the notation
0=
0
2
0

||
and obtain 0. Since
|| = ((,))
1
2=(1 cos)
1
2
where is the angle between the unit vectors ,, it
follows that
0= 4
0
sin
(1 cos)
1
2= 2
7
2.
Using 0 in the estimate |mv |1(3), we obtain the
statement of the lemma.
This completes the proof of Lemma 24.
Theorem 17. Weak solution of problem (11), (12),
(13), from Theorem 16 satisfies the following inequali-
ties mv
1(3)
20+1mv
1(3)
+
1(3), (18)
where
0=
3
0
2
0
|0(())|
× |mv (,)|
and 1 is defined in Lemma 24.
Proof. Proof follows from (16), (17).
Corollary 3. Weak solution of problem (11), (12), (13),
from Theorem 16 satisfies the following inequalities
mv
1(3)
20+
1(3),
where
=1
2
1
240
1
2
.
Let’s consider the influence of the following large
scale transformations in Navier-Stokesequation on
=,=
,=
,0
=0
2.
Statement 2. Let
=4
1
3(0+ 1)
2
3
,
then 8
7.
Proof. By the definitions and 0, we have
=(
)
1
2((
)
1
240
2)1
=1
2(1
240
3
2
)1<8
7.
This proves Statement 2.
Lemma 25. Weak solution of problem (11), (12), (13),
from Theorem 16 satisfies the following inequalities
((), )
 0(())

+41
1
20
1
2
2|| (19)
+1
21
22
1
2
||,
where
2=
01
2.
Proof. The underwritten inequalities follows from re-
presentation (14)
((), )
 0(())

+2||2|
0
()2||2()
×1((), )|
+|
02||2()
×1
((), )|.
Let us introduce the following denotation
1= 2||2|
0
()2||2()
×1((), )|,
348 A. Durmagambetov / Natural Science 2 (2010) 338-356
Copyright © 2010 SciRes. OPEN ACCESS
2=|
02||2()
×1
((), )|,
then
((), )
 0(())

+1+2.
Estimate 1 by means of
sup
|| <,
where > 0 we obtain
14
|
02||2
2
×1((), )|.
On applying Holders inequality, we get
14

0
|2||2
2|
1
×
0
|1|
1
,
where , satisfy the equality 1
+1
=1.
For == 2 we have
141
1
20
1
2
2||,
21
21
22
1
2
||,
2=
01
2.
Inserting 1,2 in to 
, we obtain the statement of
the lemma.
This completes the proof of Lemma 25.
Theorem 18. Weak solution of problem (11), (12),
(13), from Theorem 16 satisfies the following inequali-
ties mv
 1(3)
2(0+1+2
+3|mv |1(3)+ (1+2)mv
1(3)
+1mv
1(3)) + 
1(3), (20)
where
1=
3
0
2
00(())

× |mv (,)|,
2=
3
0
2
0
|0(())|
×mv (,)
 ,
2=1
1
22
11
20
1
2,3=1
1
282
1
2,
and 2 is defined in Lemma 25, =.
Proof. From the statement of Theorem 14 we get the
following estimate
mv
 1(3)
2(
3
0
2
0
|((), )|
× |mv (,)|
+
3
0
2
0((), )

× |mv (,)|
+
3
0
2
0
|((), )|
×mv (,)
 ) + 
1(3)
.
Let us introduce the following denotation
1=
3
0
2
0
|((), )|
× |mv (,)|,
2=
3
0
2
0((), )

× |mv (,)|,
3=
3
0
2
0
|((), )|
×mv (,)
 ,
then mv
 1(3)
2(1+2+3) +
1(3)
.
The estimate of 1 was obtained in theorem 16,
therefore from (15), (18), it follows that
A. Durmagambetov / Natural Science 2 (2010) 338-356 349
Copyright © 2010 SciRes. OPEN ACCESS
10+1mv
1(3)
.
Inserting inequality (19) into 2, we get
2 
3
0
2
00(())

× |mv (,)|
+41
1
20
1
20
3
|mv (,)|

+1
21
22
1
20
3
|mv (,)|,
Let us take into account the estimate of 0 obtained
in Lemma 25,
0=
0
2
0

||= 2
7
2.
Inserting this value in 2, we obtain
2 
3
0
2
00(())

× |mv (,)|
+1
1
22
11
20
1
2
3
|mv (,)|

+1
1
282
1
2
3
|mv (,)|.
Let us introduce the following denotation
1=
3
0
2
00(())

× |mv (,)|,
then
21+2mv
1(3)
+3|mv |1(3),
where
2=1
1
22
11
20
1
2,3=1
1
282
1
2.
Using inequality (16) in 3, we get
3 
3
0
2
0
|0(())|
×mv (,)
 
+1
21
20
1
20
3mv (,)
 .
Similarly as we estimated 2, obtain
32+1mv
 1(3)
,
where
2=
3
0
2
0
|0(())|
×mv (,)
 .
Inserting 1,2,3 mv
1(3), we obtain the state-
ment of the theorem.
This completes the proof of Theorem 18.
Lemma 26. Weak solution of problem (11), (12), (13),
from Theorem 16 satisfies the following inequalities
2((), )
220(())
2
+1
1
2160
1
2
3||+1
1
282
1
2
2||
+1
21
24
1
2
||, ( 21)
where
sup
|| <,
as > 0,
4=
021
22.
Proof. From (14) we have the following inequalities
2((), )
220(())
2
+422||4|
0
()2
×2||2()1((), )|
+4||2|
0
()
×2||2()1
((), )|
+
02||2()21
2((), ).
350 A. Durmagambetov / Natural Science 2 (2010) 338-356
Copyright © 2010 SciRes. OPEN ACCESS
Let us introduce the following denotation
1= 422||4|
0
()2
×2||2()1((), )|,
2= 4||2|
0
()
×2||2()1
((), )|,
3=|
02||2()
×21
2((), )| ,
then 2((), )
220(())
2
+1+2+3.
Using the estimate
sup
|| <,
as > 0, we estimate 1,2
116
2|
02||2
2
×1((), )|,
28
|
02||2
2
×1
((), )|.
Using Holders inequality
116
2
0
|2||2
2|
1
×
0
|1|
1
,
28

0
|2||2
2|
1
×
01

1
,
where , satisfy the equality 1
+1
=1.
For == 2 we get
1161
1
20
1
2
3||,
281
1
22
1
2
2||.
Taking into account Holders inequality for 3, we get
31
21
24
1
2
||,4=
021
22.
Inserting 1, 2, 3 in 2
2, we get the statement
of the lemma.
This completes the proof of Lemma 26.
Theorem 19. Weak solution of problem (11), (12),
(13), from Theorem 16 satisfies the following estimate
2mv
21(3)
2(2(1+2+3)
+4+5+ (22+4)mv
1(3)
+(23+5)|mv |1(3)+6|mv |1(3)
+2(1+2)mv
 1(3)
+ 23mv
 1(3)
+12mv
21(3)) + 2
21(3), (22)
where
3=
32
0
2
00(())

×mv (,)
 ,
4=
32
0
2
020(())
2
× |mv (,)|,
5=
32
0
2
0
|0(())|
×2mv (,)
2,
4=1
1
22
15
20
1
2,
5=1
1
22
13
22
1
2,
6=1
1
284
1
2,
A. Durmagambetov / Natural Science 2 (2010) 338-356 351
Copyright © 2010 SciRes. OPEN ACCESS
and 4 is defined in Lemma 26.
Proof. From the statement of Theorem 14 we have the
estimate
2mv
21(3)
2(2
3
0
2
0
|((), )|
×mv (,)
 
+2
3
0
2
0((), )

× |mv (,)|+
+2
32
0
2
0((), )

×mv (,)
 +
+
32
0
2
02((), )
2
× |mv (,)|+
+
32
0
2
0
|((), )|
×2mv (,)
2)
+2
21(3)
=
2
5
=1 +2
21(3)
.
Let us use the estimates for 1,2
1= 2
3
0
2
0
|((), )|
×mv (,)
 
21+2mv
1(3)
+3|mv |1(3),
2= 2
3
0
2
0((), )

× |mv (,)|
22+1mv
 1(3).
Let us use inequality (19) to estimate 3, then we get
3= 2
32
0
2
0((), )

×mv (,)
 
<2(
32
0
2
00(())

×mv (,)
 
+41
1
20
1
20
3mv (,)
 
+1
21
22
1
20
3mv (,)
 ).
Inserting the value of the integral 0, from Lemma 18,
we get
3=2(
32
0
2
00((), )

×mv (,)
 
+1
1
22
11
20
1
2
3mv (,)
 
+1
1
282
1
2
3mv (,)
 )
=2(
32
0
2
00(())

×mv (,)
 
+2
3mv (,)
 +3
3mv (,)
 ).
Let us introduce the following denotation
3=
32
0
2
00(())

×mv (,)
 ,
then 32(3+2
3mv (,)
 
+3
3mv (,)
 ).
Applying inequality (21) to estimate 4, we get
352 A. Durmagambetov / Natural Science 2 (2010) 338-356
Copyright © 2010 SciRes. OPEN ACCESS
4=
32
0
2
02((), )
2
× |mv (,)|
 
32
0
2
020(())
2
× |mv (,)|
+1
1
2160
1
20
3
1
|mv (,)|
+1
1
282
1
20
3
|mv (,)|
+1
21
24
1
20
3|mv (,)|.
Inserting the value of 0, we obtain
4 
32
0
2
020(())
2
× |mv (,)|
+1
1
22
15
20
1
2
3
1
|mv (,)|
+1
1
22
13
22
1
2
3
|mv (,)|
+1
21
284
1
2
3|mv (,)|.
Let us introduce the following denotation
4=1
1
22
15
20
1
2,5=1
1
22
13
22
1
2,
6=1
21
284
1
2,
then
4 
32
0
2
020(())
2
× |mv (,)|+4
3
1
|mv (,)|
+5
3
|mv (,)|+6
3|mv (,)|.
Introduce the denotation
4=
32
0
2
020(())
2
× |mv (,)|,
then 44+4
3
1
|mv (,)|
+5
3
|mv (,)|+6
3|mv (,)|.
Using inequality (16) to estimate 5 , we obtain
5=
32
0
2
0
|((), )|
×2mv (,)
2
 
32
0
2
0
|0(())|
×2mv (,)
2
+1
21
20
1
20
32mv (,)
2.
Inserting the value of the integral 0, we obtain
5 
32
0
2
0
|0(())|
×2mv (,)
2+12mv
21(3)
.
Let us introduce the following denotation
5=
32
0
2
0
|0(())|
×2mv (,)
2,
then 55+12mv
21(3)
.
Inserting ,(=1, . . . ,5) in 2mv
21(3), we ob-
tain the statement of the theorem.
This completes the proof of Theorem 19.
Lemma 27. Weak solution of problem (11), (12), (13),
from Theorem 16 satisfies the following estimate
mv
1(3)0, (23)
|mv |1(3)1, (24)
A. Durmagambetov / Natural Science 2 (2010) 338-356 353
Copyright © 2010 SciRes. OPEN ACCESS
|mv |1(3)2, (25)
where
=1
2
1
240
1
2
,0=
20+
1(3)
,
1=
20
(1) + ||1(3), (26)
2=
20
(2) + ||1(3),
0
(2) =
3
0
2
02|0(())|
× |mv (,)|.
Proof. From inequality (15) and estimate (17), we
make the sequence of estimates
|mv |1(3)
20
(+1) +1|mv |1(3)
+||1(3),
where
0
(+1) =
3
0
2
0+1|0(())|
× |mv (,)|.
1=1
1
280
1
2,
and is an exponent of . From this recurrence for-
mula, as = 0,=1, we get estimates (17) and (18)
accordingly.
For = 1 we have
|mv |1(3)
20
(2) +1|mv |1(3)
+||1(3).
Considering estimates (17), (18) and the last estimate,
we obtain the statement of the lemma.
This proves Lemma 27.
Lemma 28. Weak solution of problem (11), (12), (13),
from Theorem 16 satisfies the following estimates
mv
1(3)02+1, (27)
mv
1(3)22+3, (28)
where 0=
23
(0)1+ (1
(0) +2
(0))0,
1=
2(0+1+2)+
1(3)
,
2=
23
(0)2+ (1
(0) +2
(0))1,
3=
20
(1) +1
(1) +2
(1)+
1(3)
,
1
(1) =
32
0
2
00(())

× |mv (,)|,
2
(1) =
32
0
2
0
|0(())|
×mv (,)
 ,
1
(0) =80
1
2
1
2
,2
(0) =2
11
20
1
2
1
2
,
3
(0) =82
1
2
1
2
,
Proof. From inequality (19) and estimate (20), let us
make the sequence of estimates
mv
 1(3)
2(0
()+1
()+2
()
+3|mv |1(3)+ (1+2)mv
11(3)
+1mv
 1(3)
) + 
1(3)
,
where
0
()=
3
0
2
0|0(())|
× |mv (,)|,
1
()=
3+1
0
2
00(())

× |mv (,)|,
2
()=
3+1
0
2
0
|0(())|
×mv (,)
 ,
and is an exponent of . From this recurrence for-
mula, we get estimate (17) and (18) for = 0,= 1,
accordingly. And
mv
 1(3)
2(0
(1) +1
(1) +2
(1)
+3|mv |1(3)+ (1+2)|mv |1(3)
354 A. Durmagambetov / Natural Science 2 (2010) 338-356
Copyright © 2010 SciRes. OPEN ACCESS
+1mv
 1(3)
) + 
1(3)
,
Considering estimate (17) and the last estimate, we
obtain the statement of the lemma.
This completes the proof of Lemma 28.
Lemma 29. The solution of the problem (11), (12),
(13), from Theorem 16, satisfies the following estimate
2mv
21(3)03
+12+2, (29)
where 0=(3
(0)2+ (1
(0) +2
(0))0),
1=
2(22
(0) +4
(0))0
+(23
(0) +5
(0))1+6
(0)2
+23
(0)3+ 2(1
(0) +2
(0))1,
2=
2(2(1+2+3)
+4+5) + 2
21(3)
,
4
(0) =2
15
20
1
2
1
2
,5
(0) =2
13
22
1
2
1
2
,
6
(0) =84
1
2
1
2
.
Proof. From (22), we obtain the following estimate
2mv
21(3)
2(2(1+2+3)
+4+5+ (22
(0) +4(0)) mv
1(3)
+(23(0) + 5(0))|mv |1(3)
+6(0)|mv |1(3)+
+2(1(0) + 2(0)) mv
 1(3)
+23(0) mv
 1(3)
) + 2
21(3)
.
Using estimates (23)-(28) in the last inequality, we
obtain the statement of the lemma.
This proves Lemma 29.
Theorem 20. The solution of the problem (11), (12),
(13), from Theorem 16, satisfies the following estimate
||1(3)10+20
1
22
1
2+323
+40
1
2+52
1
2+64
1
22
+70
1
2+82
1
2+9,
where
=1
2
1
240
1
2
,0=
0
|1|2,
1= (,)+,
2=
01
2,4=
021
22,
1=2232
(1 + 2
5
2)0,
2=2242
(1 + 2
5
2)1,
3=2232
2,
4=23
1
2
((1 + 2
9
2)0+ (1+ 2
5
2)1),
5=23
1
2
((1 + 2
3
2)1+3),
6=23
1
2
,
7=22
1
21 + 2
5
20,8=22
1
21,
9=
2(1+2), 0=
20+
1(3),
1=
20
(1) + ||1(3), 2=
20
(2) + ||1(3),
1=
2(0+1+2)+
1(3)
,
3=
20
(1) +1
(1) +2
(1)+
1(3)
,
2=
2(2(1+2+3)
+4+5) + 2
21(3)
,
2=9
4(2)3,
the function is defined in Theorem 15.
Proof. From the Theorem 8
||1(3)mv
1(3)
A. Durmagambetov / Natural Science 2 (2010) 338-356 355
Copyright © 2010 SciRes. OPEN ACCESS
+2 mv
 1(3)
+1
42mv
21(3)
.
Using estimates (23), (27), (29) in the right side of this
inequality, we get
||1(3)0+ 2(02+1)
+1
4(03+12+2)
1
403+ (20+1)2+ (0+1+2),
where , are defined in Lemma 27, is defined
in Lemma 28, and is defined in Lemma 29. Taking
into account these notations and calculating the coeffi-
cients at 0,2,4, we obtain the statement of the
theorem.
This proves Theorem 20.
Lemma 30. The function , defined in Theorem 15,
satisfies the following estimates
||1(3),||1(3),

1(3),

1(3),2
21(3).
Proof. We can get the estimate of cubic members w.r.t.
 in if we resume all the methods for estimating
square members w.r.t. .
This completes the proof of Lemma 30.
Lemma 31. Weak solution of problem (11), (12), (13),
from Theorem 16 satisfies the following estimates
021
3
(|0()|)mv ,
0
(1) 21
3(|0()|)mv ,
0
(2) 21
32(|0()|)mv ,
121
3(0()
 )mv ,
1
(1) 21
32(0()
 )mv ,
222
3(|0()|)mv ,
2
(1) 22
32(|0()|)mv ,
322
32(0()
 )mv ,
421
32(20()
2)mv ,
523
32(|0()|)mv .
Proof. The proof follows from Lemmas 18, 19, 20.
This proves Lemma 31.
Theorem 21. Suppose that
02
1(3), 02(),
01(), 0
 1(),
20
21(), 01(3),
=
31(|0()|)mv ,
(= 1,3),
=
33(0()
 )mv ,
(= 4,5),
6=
32(20()
2)mv .
Then there exists a unique weak solution of (11), (12),
(13), satisfying the following inequalities
max
3
=1
|
|1(3),
where  depends only on the theorem conditions.
Proof. It is sufficient to get uniform estimates of the
maximum to prove that the theorem .These obvious-
ly follow from the estimate |
|1(3). Uniform esti-
mates allow to extend the rules of the local existence and
unicity local to an interval, where they are correct. To
estimate the component of velocity, we use statement 2
=
0||||2(3)
2++ 1,
=4
1
3(0+ 1)
2
3
.
Using Lemmas 21, 22 for the potential
=
0||||2(3)
2++ 1
We have ()<1, i.e., it is not necessary to take
into account normalization numbers when proving the
theorem. Now the statetement of the theorem follows
from Theorems 20, 17, Lemmas 21, 30, 31 and the con-
ditions of Theorem 21, that give uniform of velocity
maxima at a specified interval of time.
356 A. Durmagambetov / Natural Science 2 (2010) 338-356
Copyright © 2010 SciRes. OPEN ACCESS
This comletes the proof of Theorem 21.
Note. In the estimate for  the condition (0) >1
is used. This conditioncan be obviated if we use smooth
and bounded function and make all the estimates for
1=+ such that 1(0)>1 is satisfied. Using the
function , we also choose the constant A concordant
with the constant from Lemma 3.
Theorem 21 proves the global solvability and unicity
of the Cauchy problem for Navier-Stokesequation.
10. CONCLUSIONS
In Introduction we mentioned the authors whose scien-
tific researches we consider appropriate to call the pre-
history of this work. The list of these authors may be
considerably extended if we enumerate all the predeces-
sors diachronically or by the significance of their con-
tribution into this research. Actually we intended to ob-
tain evident results which were directly and indirectly
indicated by these authors in their scientific works. We
do not concentrate on the solution to the multi- dimen-
sional problem of quantum scattering theory although it
follows from some certain statements proved in this
work. In fact, the problem of over-determination in the
multi-dimensional inverse problem of quantum scatter-
ing theory is obviated since a potential can be defined by
amplitude averaging when the amplitude is a function of
three variables. In the classic case of the multi- dimen-
sional inverse problem of quantum scattering theory the
potential requires restoring with respect to the amplitude
that depends on five variables. This obviously leads to
the problem of over-determination. Further detalization
could have distracted us from the general research line
of the work consisting in application of energy and mo-
mentum conservation laws in terms of wave functions to
the theory of nonlinear equations. This very method we
use in solving the problem of the century, the problem of
solvability of the Cauchy problem for Navier-Stokes
equations of viscous incompressible fluid. Let us also
note the importance of the fact that the laws of momen-
tum and energy conservation in terms of wave functions
are conservation laws in the micro-world; but in the clas-
sic methods of studying nonlinear equations scientists
usually use the priori estimates reflecting the conserva-
tion laws of macroscopic quantities. We did not focus
attention either on obtaining exact estimates dependent
on viscosity, lest the calculations be complicated. How-
ever, the pilot analysis shows the possibility of applying
these estimates to the problem of limiting viscosity tran-
sition tending to zero.
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[2] Durmagambetov, A.A. (1996) Inverse problem of
quantum theory of scattering. International Conference
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