Vol.2, No.4, 338-356 (2010) Natural Science
http://dx.doi.org/10.4236/ns.2010.24042
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: ShrππΜdinger; Cauchy Problem; Navier-Stokesβ; 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 Laxβ pairs. 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 Schrodingerβs equation. Therefore we consider Schrodingerβs equation as an in-terrelation between real-valued functions and analytic functions, where real-valued functions are potentials in Schrodingerβs equation and analytic functions are the corresponding eigenfunctions of the continuous spec-trum of Schrodingerβs 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-Stokesβ equations. 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-Stokesβ equations of viscous incompressible fluid.
2. BASIC NOTIONS AND SUBSIDIARY
STATEMENT
Let us consider ShroΜdingerse equation
βΞπ₯π₯ππ + ππππ = |ππ|2ππ (1) where ππ is a bounded fast-decreasing function,
ππ β π π 3, |ππ|2= οΏ½β 3
ππ =1 ππππ2.
Definition 1.Rolnikβs Class πΉπΉ is a set of measurable functions ππ,
||ππ||ππ= οΏ½β π π 6
ππ(π₯π₯)ππ(π¦π¦)
|π₯π₯ β π¦π¦|2 πππ₯π₯πππ¦π¦ <β. It is considered to be a general definition ([7]).
ππΒ±(ππ, π₯π₯) = ππππ(ππ,π₯π₯)+ +ππΒ±ππ|ππ||π₯π₯||π₯π₯| π΄π΄Β±(ππ, ππβ²) + 0 οΏ½1
|π₯π₯|οΏ½ (2) where
π₯π₯ β π π 3, ππβ² = |ππ| π₯π₯
|π₯π₯| , (ππ, π₯π₯) = οΏ½β 3
ππ =1 πππππ₯π₯ππ,
π΄π΄Β±(ππ, ππ) =(2ππ)1 3 οΏ½β π π 3
ππ(π₯π₯)ππΒ±(ππ, π₯π₯)ππβππ(ππ,π₯π₯)πππ₯π₯.
The proof of this theorem is in [7].
Consider the operators π»π» = β Ξπ₯π₯+ ππ(π₯π₯), π»π»0= β Ξπ₯π₯ defined in the dense set ππ22(π π 3) in the space πΏπΏ2(π π 3). The operator π»π» is called Schrodingerβs operator. Povz-ner [8] proved that the functions ππ Β± (ππ, π₯π₯) form a complete orthonormal system of eigenfunctions of the continuous spectrum of the operator π»π», and the operator fills up the whole positive semi-axis. Besides the conti-nuous spectrum the operator π»π» can have a finite num-ber ππ of negative eigenvalues Denote these eigenvalues by βπΈπΈππ2 and conforming normalized egenfunctions by
ππππ(π₯π₯, βπΈπΈππ2)(ππ = 1, ππ), where ππππ(π₯π₯, βπΈπΈππ2) β πΏπΏ2(π π 3).
Theorem 2 (About Completeness). For any vec-tor-function ππ β πΏπΏ2(π π 3) and eigenfunctions of the op-erator π»π», we have Parsevalβs identity
|ππ|πΏπΏ22 = βππ β
ππ =1|ππππ|2+ β«π π 3β|ππΜ (π π )|2πππ π ,
where ππππ and ππΜ are Fourier coefficients in case of dis-crete of and continuous spectrum respectively.
The proof of this theorem is in [8].
Theorem 3 (Birman - Schwingerβs Estimate). Sup-pose ππ β π π . Then the number of discrete eigenvalues of ShrππΜdinger operator satisfies the estimate
ππ(ππ) β€(4ππ)1 2 οΏ½β π π 3
οΏ½β π π 3
ππ(π₯π₯)ππ(π¦π¦) |π₯π₯ β π¦π¦|2 πππ₯π₯πππ¦π¦. The proof of this theorem is in [9].
Definition 2. [7]
ππΒ±(ππ, ππβ²) = 1 (2ππ)3 οΏ½β
π π 3
ππΒ±(π₯π₯, ππβ²)ππβππ(ππ,π₯π₯)ππ(π₯π₯)πππ₯π₯.
ππ Β±(. , . ) is called T-matrix. Let us take into consider-ation a series for ππΒ±:
ππΒ±(ππ, ππβ²) = οΏ½β β
ππ=0
ππππΒ±(ππ, ππβ²),
where
ππ0Β±(ππ, ππβ²) =
1 (2ππ)3 οΏ½β
π π 3
ππππ(ππβ²βππ,π₯π₯)
ππ(π₯π₯)πππ₯π₯,
ππππΒ±(ππ, ππβ²) =
1 (2ππ)3
(β1)ππ (4ππ)ππ οΏ½ β
π π 3(ππ +1)
ππβππ(ππ,π₯π₯0)
Γ ππ(π₯π₯0)ππΒ±ππ|ππ
β²||π₯π₯0βπ₯π₯1|
|π₯π₯0β π₯π₯1| ππ(π₯π₯1). . . ππ(π₯π₯ππβ1)
ΓππΒ±ππ|ππ
β²||π₯π₯ππ β1βπ₯π₯ππ|
|π₯π₯ππβ1β π₯π₯ππ| ππ(π₯π₯ππ)ππππ(ππ
β²,π₯π₯ππ)
πππ₯π₯0. . . πππ₯π₯ππ. As well as in [7] we formulate.
Definition 3.Series (4) is called Bornβs series.
Theorem 4. Let ππ β πΏπΏ1(π π 3) β© πΉπΉ . If πππππππΉπΉ2β€ 4ππ, then Bornβs series for ππ(ππ, ππβ²) converges as ππ, ππβ² β π π 3.
The proof of the theorem is in [7].
Definition 4. Suppose ππ β π π ; then the function π΄π΄(ππ, ππ), denoted by the following equality
π΄π΄(ππ, ππ) =(2ππ)1 3 οΏ½β π π 3
ππ(π₯π₯)
ππ+(ππ, π₯π₯)ππβππ(ππ,π₯π₯)πππ₯π₯, is called scattering amplitude
Corollary 1.Scattering amplitude π΄π΄(ππ, ππ) is equal to ππ-matrix
π΄π΄(ππ, ππ) = ππ+(ππ, ππ) =(2ππ)1 3 οΏ½β
π π 3
ππ(π₯π₯)ππ+(ππ, π₯π₯)ππβππ(ππ,π₯π₯)πππ₯π₯.
The proof follows from definition 4.
It is a well-known fact [5] that the solutions ππ + (ππ, π₯π₯) and ππ β (ππ, π₯π₯) of Eq.1 are linearly dependent
ππ += ππππ β (3) where ππ is a scattering operator with the nucleus ππ(ππ, ππ) of the form
ππ(ππ, ππ) = οΏ½β π π 3
ππ+(ππ, π₯π₯)ππ+β(ππ, π₯π₯)πππ₯π₯.
Theorem 5. (Conservation Law of Impulse and Energy). Assume that ππ β πΉπΉ, then
ππππβ= πΌπΌ, ππβππ = πΌπΌ, where πΌπΌ is anunit operator.
The proof is in [5].
Let us use the following definitions πποΏ½(ππ) = οΏ½β
π π 3
ππ(π₯π₯)ππππ(ππ,π₯π₯)πππ₯π₯,
πποΏ½(ππ β ππ) = οΏ½β π π 3
ππ(π₯π₯)ππππ(ππβππ,π₯π₯)πππ₯π₯,
πποΏ½mv(ππ) = οΏ½β π π 3
π΄π΄mv(ππ) = οΏ½β π π 3
π΄π΄(ππ, ππ))πΏπΏ(|ππ|2β |ππ|2)ππππ,
οΏ½βππ(ππ, ππ)ππππππ= οΏ½β π π 3
ππ(ππ, ππ)πΏπΏ(|ππ|2β |ππ|2)ππππ,
οΏ½βππ(ππ, ππ)ππππππ = οΏ½β π π 3
ππ(ππ, ππ)πΏπΏ(ππ2β |ππ|2)ππππ,
where ππ, ππ β π π 3 and ππ
ππ=|ππ|ππ, ππππ =|ππ|ππ.
3. ESTIMATE OF AMPLITUDE MAXIMUM
Let us consider the problem of estimating the maximum of amplitude, i.e., maxππβπ π 3|π΄π΄(ππ, ππ)|. Let us estimate the ππ term of Bornβs series |ππππ(ππ, ππ)|.Lemma 1. |ππππ(ππ, ππ)| satisfies the inequality
|ππππ+1(ππ, ππ)| β€(2ππ)1 3 1 (4ππ)ππ+1 Γ(2ππ)πΎπΎ2(ππ+1)ππ οΏ½β
π π 3
|πποΏ½(ππ)|2 |ππ|2 ππππ,
πΎπΎ = πΆπΆπΏπΏ||ππ|| + 4πππππποΏ½πΏπΏ, πΆπΆπΏπΏ = 2βππ βπΏπΏ,
where πΏπΏ-is a small value, πΆπΆ is a positive number, πππποΏ½ = πππππ₯π₯
ππβπ π 3|πποΏ½|.
Theorem 6. Suppose that πΎπΎ < 16ππ3 , then πππππ₯π₯
ππβπ π 3|π΄π΄(ππ, ππ)| satisfies the following estimate
maxππβπ π 3|π΄π΄(ππ, ππ)| β€(2ππ)1 316ππ13β πΎπΎ οΏ½β π π 3
|πποΏ½(ππ)|2 |ππ|2 ππππ, where πΎπΎ = πΆπΆπΏπΏ||ππ|| + 4πππππποΏ½πΏπΏ, πΏπΏ is a small value,
πΆπΆπΏπΏ = 2οΏ½πππΏπΏ, πππΏπΏ = πππππ₯π₯ ππβπ π 3|πποΏ½|.
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 οΏ½β
π π 3
πποΏ½(π‘π‘)πΏπΏ(|π‘π‘ β ππ|2β |ππ|2)πππ‘π‘ = ππ(2ππ),
where ππ, π‘π‘ β π π 3, πΏπΏ is Diracβs delta function,
ππ β ππ22(π π 3), |ππ|2= βππ=13 βππππ2, (ππ, π‘π‘) = β3ππ=1βπππππ‘π‘ππ. Let us formulate the basic result.
Theorem 7.Suppose that ππ β ππ22(π π 3), then (2ππ)2πποΏ½(ππ, ππ, ππ)
= β1ππβππβ22οΏ½β ππ
0 οΏ½ β
2ππ
0
οΏ½ππ((ππ2ππ ππ, πππ π ) , ππππ)
+ ππ((ππππ2ππ, πππ π ) , βππππ)οΏ½(ππππππ, πππ π )2 2 sinππ ππππ ππππ, where
ππ((ππππ, πππ π ) , ππ2ππ ππ) = πποΏ½((ππππ, πππ π ) , ππ2ππ ππ), sinππ ππππ ππππ = ππππππ, sinππ ππππππππ = πππππ π , ππ = |π‘π‘|.
Theorem 8.Fourier transformation of the function q satisfies the following estimate
|πποΏ½|πΏπΏ1β€14 οΏ½π§π§βπποΏ½ππππβπ§π§2 οΏ½ πΏπΏ1
+ 2 οΏ½βπποΏ½ππππβπ§π§2 οΏ½ πΏπΏ1
+ οΏ½πποΏ½πππππ§π§ οΏ½πΏπΏ
1
,
5. CORRELATION OF AMPLITUDE AND
WAVE FUNCTIONS
We take the relationship for ππ+, ππβ from (3) ππ+(ππ, π₯π₯) = ππβ(ππ, π₯π₯)
β2ππππ β«π π 3βπΏπΏ(|ππ|2β |ππ|2) (4) Γ π΄π΄(ππ, ππ)ππβ(ππ, π₯π₯)ππππ.
Let us denote new functions and operators we will use further
ππ0(βπ§π§ππππ, π₯π₯) = ππππ(βπ§π§ππππ,π₯π₯),
Ξ¦0(βπ§π§ππππ, π₯π₯) = ππ0(βπ§π§ππππ, π₯π₯) + ππ0(ββπ§π§ππππ, π₯π₯), Ξ¦+(βπ§π§ππππ, π₯π₯) = ππ+(βπ§π§ππππ, π₯π₯) β ππππ(βπ§π§ππππ,π₯π₯)
+ππ+(ββπ§π§ππππ, π₯π₯) β ππβππ(βπ§π§ππππ,π₯π₯),
Ξ¦β(βπ§π§ππππ, π₯π₯) = ππβ(βπ§π§ππππ, π₯π₯) β ππππ(βπ§π§ππππ,π₯π₯)
+ππβ(ββπ§π§ππππ, π₯π₯) β ππβππ(βπ§π§ππππ,π₯π₯),
π·π·1ππ = β2ππππ οΏ½β π π 3
π΄π΄(ππ, ππ)πΏπΏ(π§π§ β ππ)ππ(ππ, π₯π₯)ππππ,
π·π·2ππ = β2ππππ οΏ½β π π 3
π΄π΄(βππ, ππ)πΏπΏ(π§π§ β ππ)ππ(ππ, π₯π₯)ππππ,
π·π·3ππ = π·π·1ππ + π·π·2ππ,
where π§π§ = |ππ|2, ππ = |ππ|2, Β±ππ = Β±βπ§π§ππππ. Let us intro-duce the operators ππΒ±, ππ for the function ππ β ππ21(π π ) by the formulas
ππ+ππ =ππππ lim1 πΌπΌπππ§π§ β0 οΏ½β β
ββ ππ(βπ π )
π π β π§π§ πππ π , where πΌπΌπππ§π§ > 0,
ππβππ =ππππ lim1 πΌπΌπππ§π§ β0 οΏ½β β
ββ ππ(βπ π )
ππππ =12 (ππ++ ππβ)ππ
Use (4) and the symbols ππππ=|ππ|ππ to come to Rie-mannβ problem of finding a function π·π·+, which is ana-lytic by the variable z in the top half plane, and the func-tion π·π·β, which is analytical on the variable z in the bot-tom half plane by the specified jump of discontinuity ππ onto the positive semi axis.
For the jump the discontinuity of an analytical func-tion, we have the following equations
ππ = π·π·+β π·π·β (5) ππ = π·π·3[Ξ¦β] β π·π·3[ππβ] (6) where ππβ= ππβ(βππ, π₯π₯).
Theorem 9. Suppose that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0;
then the functions
πΉπΉ1= π·π·Β±(βπ§π§ππππ, π₯π₯)|π₯π₯=0β π·π·0(βπ§π§ππππ, π₯π₯)|π₯π₯=0, πΉπΉ2= ππΒ±ππ|π₯π₯=0
are coincided according to the class of analytical func-tions, coincide with bounded derivatives all over the complex plane with a slit along the positive semi axis.
Lemma 2.There exists 0 < |ππ| < β such that it sa-tisfies the following condition ππ+|π₯π₯=0,π§π§=0= 0 holds for the potential of the form ππ = ππππ, where ππ β πΉπΉ.
Now, we can formulate Riemannβs problem. Find the analytic function π·π·Β± that satisfies (5), (6) and its solu-tion is set by the following theorem.
Theorem 10. Assume that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0,
then
π·π·Β±= ππΒ±ππ + π·π·0, ππ = π·π·3[ππ[ππβππ + π·π·0]] β π·π·3ππβ, where ππβ= ππβ(βππ, π₯π₯).
Lemma 3. Suppose that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0; then
Ξπ₯π₯ππΒ±[ππ]|π₯π₯=0= ππΒ±Ξπ₯π₯[ππ]|π₯π₯=0. Theorem 11.Suppose that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0, ππ(0) β 0, then
ππ(0)ππ|π₯π₯=0= π·π·3ππβ[ππππ|π₯π₯=0
βπ·π·3[ππππβ]|π₯π₯=0+ π·π·3οΏ½β β
0
πππππ π |π₯π₯=0.
6. AUXILIARY PROPOSITIONS
For wave functions let us use integral representations following from Lippman-Schwingerβs theorem
ππΒ±(ππ, π₯π₯) = ππππ(ππ,π₯π₯)
+4ππ οΏ½1 β π π 3
ππΒ±ππβπ§π§|π₯π₯βπ¦π¦|
|π₯π₯ β π¦π¦| ππ(π¦π¦)ππΒ±(ππ, π¦π¦)πππ¦π¦, ππΒ±(βππ, π₯π₯) = ππβππ(ππ,π₯π₯)
+4ππ οΏ½1 β π π 3
ππβππβπ§π§|π₯π₯βπ¦π¦|
|π₯π₯ β π¦π¦| ππ(π¦π¦)ππΒ±(βππ, π¦π¦)πππ¦π¦.
Lemma 4. Suppose that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0;
then
π΄π΄(ππ, ππβ²) = ππ0πποΏ½(ππ β ππβ²)
+4ππ οΏ½ππ0 β π π 3
οΏ½β π π 3
ππβππ(ππβ²,π₯π₯)ππ(π₯π₯)ππππβπ§π§|π₯π₯βπ¦π¦| |π₯π₯ β π¦π¦| Γ ππ(π¦π¦)ππππ(ππ,π¦π¦)πππ¦π¦πππ₯π₯ + π΄π΄
3(ππ, ππβ²), π΄π΄(βππ, ππβ²) = ππ0πποΏ½(βππ β ππβ²)
+4ππ οΏ½ππ0 β π π 3
οΏ½β π π 3
ππβππ(ππβ²,π₯π₯)
ππ(π₯π₯)ππβππβπ§π§|π₯π₯βπ¦π¦||π₯π₯ β π¦π¦|
Γ ππ(π¦π¦)ππβππ(ππ,π¦π¦)πππ¦π¦πππ₯π₯ + π΄π΄
3(βππ, ππβ²),
where ππ0=(2ππ)12, and π΄π΄3(ππ, ππβ²), π΄π΄3(βππ, ππβ²) are terms of order higher than 2 with regards to ππ.
Theorem 12(Parseval). The functions ππ, ππ β πΏπΏ2(π π 3) satisfy the equation
(ππ, ππ) = ππ0(ππΜ, πποΏ½β),
where (β ,β ) is a scalar product and ππ0=(2ππ)1 3.
Lemma 5.Suppose that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0, then π΄π΄(ππ, ππβ²) = ππ
0πποΏ½(ππ β ππβ²)
βππ02 οΏ½β π π 3
πποΏ½(ππ + ππ)πποΏ½(ππ β ππβ²) |ππ|2β π§π§ β ππ0 ππππ +π΄π΄3(ππ, ππβ²),
π΄π΄(βππ, ππβ²) = ππ0πποΏ½(βππ β ππβ²)
βππ02 οΏ½β π π 3
πποΏ½(βππ + ππ)πποΏ½(ππ β ππβ²) |ππ|2β π§π§ β ππ0 ππππ +π΄π΄3(βππ, ππβ²).
Corollary 2.Suppose that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0,
then
π΄π΄mv(ππ) = ππ0πποΏ½mv(ππ)
βππ02βπ§π§2 οΏ½β ππ
0 οΏ½ β 2βππ
0 οΏ½β π π 3
ππΜ(ππ + ππ)ππΜ(ππ β ππβ²)
where
π΄π΄3mv(ππ) = οΏ½β π π 3
π΄π΄3(ππ, ππβ²)πΏπΏ(π§π§ β |ππβ²|2)ππππβ²
and
π΄π΄mv(βππ) = ππ0πποΏ½mv(βππ)
βππ02βπ§π§2 οΏ½β ππ
0 οΏ½ β 2βππ
0 οΏ½β π π 3
πποΏ½(βππ + ππ)πποΏ½(ππ β ππβ²) |ππ|2β π§π§ β ππ0 ππππππππππβ²
+π΄π΄3mv(βππ), where
π΄π΄3mv(βππ) = οΏ½β π π 3
π΄π΄3(βππ, ππβ²)πΏπΏ(π§π§ β |ππβ²|2)ππππβ².
Lemma 6.Suppose that ππ β π π and π₯π₯ = 0, then
ππΒ±(ππ, 0) = 1 +4ππ οΏ½1 β π π 3
ππΒ±ππβπ§π§|π¦π¦|
|π¦π¦| ππ(π¦π¦)ππππ(ππ,π¦π¦)πππ¦π¦
+(4ππ)1 2 οΏ½β π π 3
οΏ½β π π 3
ππΒ±ππβπ§π§|π¦π¦| |π¦π¦| ππ(π¦π¦)
ππΒ±ππβπ§π§|π¦π¦βπ‘π‘| |π¦π¦ β π‘π‘|
Γ ππ(π‘π‘)ππππ(ππ,π‘π‘)πππ‘π‘πππ¦π¦ + ππ
Β±(3)(ππ, 0),
where ππΒ±(3)(ππ, 0) are terms of order higher than 2 with regards to ππ., i.e.,
ππΒ±(3)(ππ, π₯π₯) =(4ππ)1 3 οΏ½β π π 3
οΏ½β π π 3
οΏ½β π π 3
ππΒ±ππβπ§π§|π₯π₯βπ¦π¦| |π₯π₯ β π¦π¦| ππ(π¦π¦)
ΓππΒ±ππβπ§π§|π¦π¦βπ‘π‘||π¦π¦ β π‘π‘| ππ(π‘π‘)ππΒ±ππβπ§π§|π‘π‘βπ π ||π‘π‘ β π π | ππ(π π )ππΒ±(ππ, π π )πππ π πππ‘π‘πππ¦π¦. and
ππΒ±(βππ, 0) = 1 +4ππ οΏ½1 β π π 3
ππβππβπ§π§|π¦π¦|
|π¦π¦| ππ(π¦π¦)ππβππ(ππ,π¦π¦)πππ¦π¦
+ 1
(4ππ)2 οΏ½β π π 3
οΏ½β π π 3
ππβππβπ§π§|π¦π¦| |π¦π¦| ππ(π¦π¦)
ππβππβπ§π§|π¦π¦βπ‘π‘| |π¦π¦ β π‘π‘| ππ(π‘π‘)
Γ ππβππ(ππ,π‘π‘)πππ‘π‘πππ¦π¦ + ππ
Β±(3)(βππ, 0),
where ππΒ±(3)(βππ, 0) are terms of order higher than 2
with regards to ππ., i.e.,
ππΒ±(3)(βππ, π₯π₯) = 1 (4ππ)3 οΏ½β
π π 3
οΏ½β π π 3
οΏ½β π π 3
ππβππβπ§π§|π₯π₯βπ¦π¦| |π₯π₯ β π¦π¦| ππ(π¦π¦)
Γππβππβπ§π§|π¦π¦βπ‘π‘||π¦π¦ β π‘π‘| ππ(π‘π‘)ππβππβπ§π§|π‘π‘βπ π ||π‘π‘ β π π | ππ(π π )ππΒ±(βππ, π π )πππ π πππ‘π‘πππ¦π¦.
Lemma 7.Suppose that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0, then
ππΒ±(ππ, 0) = 1 β ππ0 οΏ½β π π 3
πποΏ½(ππ + ππ) |ππ|2β π§π§ β ππ0 ππππ
+ππ02β«π π 3β(|ππ|πποΏ½(ππ+ππ)2βπ§π§βππ0) (7)
Γ οΏ½β π π 3
πποΏ½(ππ + ππ1)
(|ππ1|2β π§π§ β ππ0) ππππ1ππππ + ππΒ± (3)(ππ, 0)
ππΒ±(βππ, 0) = 1 β ππ0 οΏ½β π π 3
πποΏ½(βππ + ππ) |ππ|2β π§π§ β ππ0 ππππ
+ππ02β«π π 3β(|ππ|πποΏ½(βππ+ππ)2βπ§π§βππ0) (8)
Γ οΏ½β π π 3
πποΏ½(ππ + ππ1)
(|ππ1|2β π§π§ β ππ0) ππππ1ππππ + ππΒ±(3)(βππ, 0)
Lemma 8.Suppose that ππ β π π , π₯π₯ = 0; then
πΉπΉ(ππ, 0) = βππππππ0βπ§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
πποΏ½(ππ β βπ§π§ππππ)ππππππ
+ππππππ02βπ§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
ππ. ππ. οΏ½β π π 3
πποΏ½(ππ β βπ§π§ππππ) |ππ1|2β π§π§ Γ πποΏ½(ββπ§π§ππππβ ππ1)ππππ1ππππππ
+ππππππ02βπ§π§ππ. ππ. οΏ½β π π 3
οΏ½β ππ
0 οΏ½ β
2ππ
0
πποΏ½(ππ β ππ) |ππ|2β π§π§
Γ πποΏ½(βππ β βπ§π§ππππ1)ππππππ1ππππ
+ππ+(3)(ππ, 0) β ππβ(3)(ππ, 0). and
πΉπΉ(βππ, 0) = βππππππ0βπ§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
πποΏ½(βππ β βπ§π§ππππ)ππππππ
+ππππππ02βπ§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
ππ. ππ. οΏ½β π π 3
πποΏ½(βππ β βπ§π§ππππ) |ππ1|2β π§π§ Γ πποΏ½(ββπ§π§ππππβ ππ1)ππππ1ππππππ
+ππππππ02βπ§π§ ππ. ππ. οΏ½β π π 3
οΏ½β ππ
0 οΏ½ β
2ππ
0
πποΏ½(βππ β ππ) |ππ|2β π§π§ Γ πποΏ½(βππ β βπ§π§ππππ1)ππππππ1ππππ
+ππ+(3)(βππ, 0) β ππβ(3)(βππ, 0).
7. TWO REPRESENTATIONS OF
SCATTERING AMPLITUDE
Lemma 9.Suppose that ππ β ππ21(π π ), thenππΒ±ππ = βππ + ππππ.
Lemma 10. Suppose that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0, then
Lemma 11. Suppose that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0, then
π΄π΄mv(ππ) + π΄π΄mv(βππ) = ππ0(πποΏ½mv(ππ) + πποΏ½mv(βππ))
+ππππππ02βπ§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ πποΏ½mv(βπ§π§ππππ)ππππππ
+ππππππ02βπ§π§ 2 οΏ½β
ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ πποΏ½mv(ββπ§π§ππππ)ππππππ
βππππππ02βπ§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ (ππ[πποΏ½mv](βπ§π§ππππ) + ππ[πποΏ½mv](ββπ§π§ππππ))ππππππ
βππ02βπ§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ ππ. ππ. οΏ½β π π 3
πποΏ½(ββπ§π§ππππβ ππ) |ππ|2β π§π§ ππππππππππ
+ππ02βπ§π§ 2 ππ. ππ. οΏ½β
π π 3
οΏ½β ππ
0 οΏ½ β
2ππ
0
πποΏ½(ππ β ππ) + πποΏ½(βππ β ππ) ππ β π§π§
Γ πποΏ½(βππ β βπ§π§ππππ)ππππππππππ β 2ππππ(πΉπΉ(3)(ππ, 0) +πΉπΉ(3)(βππ, 0) + ππ3(ππ, 0) + ππ(3)(ππ, 0)), where ππ3(ππ, 0), ππ(3)(ππ, 0) are defined by formulas
ππ3(ππ, 0) = β4ππ2ππ02 οΏ½β π π 3
(π΄π΄2(ππ, ππ) + π΄π΄2(βππ, ππ))
Γ πΏπΏ(π§π§ β ππ)(πποΏ½mv(ππ) + πποΏ½mv(βππ))ππππ +2ππππππ0 οΏ½β
π π 3
(π΄π΄2(ππ, ππ) + π΄π΄2(βππ, ππ))πΏπΏ(π§π§ β ππ)
Γ ππ2(ππ, 0)ππππ + 4ππ2ππ 02 οΏ½β
π π 3
(π΄π΄2(ππ, ππ) + π΄π΄2(βππ, ππ))
Γ πΏπΏ(π§π§ β ππ)(ππ[πποΏ½mv](ππ) + ππ[πποΏ½mv](βππ))ππππ β2ππππππ0 οΏ½β
π π 3
(π΄π΄2(ππ, ππ) + π΄π΄2(βππ, ππ))
Γ πΏπΏ(π§π§ β ππ)ππ[ππ2](ππ, 0)ππππ. (9) ππ(3)(ππ, 0) = 2ππππππ
02 οΏ½β π π 3
(πποΏ½(ππ β ππ) + πποΏ½(βππ β ππ))
Γ πΏπΏ(π§π§ β ππ)ππβ(2)(βππ, 0)ππππ +2ππππππ02 οΏ½β
π π 3
(π΄π΄2(ππ, ππ) + π΄π΄2(βππ, ππ))
Γ πΏπΏ(π§π§ β ππ)( οΏ½β π π 3
πποΏ½(βππ β ππ) |ππ|2β ππ + ππ0 ππππ
+ππβ(2)(βππ, 0))ππππ. (10) correspondingly,
πΉπΉ(3)(ππ, 0) = ππ
+(3)(ππ, 0) β ππβ(3)(ππ, 0), πΉπΉ(3)(βππ, 0) = ππ
+(3)(βππ, 0) β ππβ(3)(βππ, 0), and ππΒ±(3)(Β±ππ, 0) are terms of order 3 and higher w.r.t. πποΏ½ in the representations (7), (8).
Lemma 12. Suppose that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0, then
π΄π΄mv(ππ) + π΄π΄mv(βππ)
= β4ππππ(0) οΏ½ππβπ§π§ β ππ
0 οΏ½ β
2ππ
0
(π΄π΄(ππ, βπ§π§ππππ) + π΄π΄(βππ, βπ§π§ππππ))
Γ οΏ½β β
0
ππ(π π ππππ, 0)πππ π ππππππ.
8. NONLINEAR REPRESENTATION OF
POTENTIAL
Let us proceed to the construction of potential nonlinear representation.
Lemma 13. Assume that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0;
then
πποΏ½mv(ππ) + πποΏ½mv(βππ)
= βππππππ0βπ§π§ οΏ½β ππ
0
οΏ½ (πποΏ½(ππ β βπ§π§ππππ) 2ππ
0
+πποΏ½(βππ β βπ§π§ππππ))πποΏ½mv(βπ§π§ππππ)ππππππ
βππππππ0βπ§π§2 οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ πποΏ½mv(ββπ§π§ππππ)ππππππ
+ππππππ0βπ§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ (ππ[πποΏ½mv](βπ§π§ππππ) + ππ[πποΏ½mv](ββπ§π§ππππ))ππππππ
βππ0βπ§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ ππ. ππ. οΏ½β π π 3
πποΏ½(ββπ§π§ππππ β ππ) |ππ|2β π§π§ ππππππππππ
βππ0βπ§π§2 ππ. ππ. οΏ½β π π 3
οΏ½β ππ
0 οΏ½ β
2ππ
0
Γ πποΏ½(βππ β βπ§π§ππππ)ππππππππππ
β4ππππ0ππ(0) οΏ½ππβπ§π§ β ππ
0 οΏ½β
2ππ
0
(π΄π΄(ππ, βπ§π§ππππ) + π΄π΄(βππ, βπ§π§ππππ))
Γ οΏ½β β
0
ππ(π π ππππ, 0)πππ π ππππππ +2ππππππ0 (πΉπΉ(3)(ππ, 0)
+πΉπΉ(3)(βππ, 0) + ππ3(ππ, 0) + ππ(3)(ππ, 0)),
where ππ3(ππ, 0), ππ(3)(ππ, 0) are defined by Eqs.9 and 10 accordingly,
πΉπΉ(3)(ππ, 0) = ππ
+(3)(ππ, 0) β ππβ(3)(ππ, 0), πΉπΉ(3)(βππ, 0) = ππ
+(3)(βππ, 0) β ππβ(3)(βππ, 0), and ππΒ±(3)(Β±ππ, 0) are term of order 3 and higher w.r.t. πποΏ½ in representations (7), (8).
Lemma 14. Suppose that ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0, then
ππ. ππ. οΏ½β π π 3
οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β ππ) + πποΏ½(βππ β ππ)) ππ β π§π§
Γ πποΏ½(βππ β βπ§π§ππππ)ππππππππππ
= ππππ οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ πποΏ½mv(ββπ§π§ππππ)ππππππ.
Lemma 15. Let πποΏ½ β ππ21(π π ) and ππ β π π , then
οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ (ππ[πποΏ½mv](βπ§π§ππππ) + ππ[πποΏ½mv](ββπ§π§ππππ))ππππππ
= οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ (πποΏ½mv(βπ§π§ππππ) + πποΏ½mv(ββπ§π§ππππ))ππππππ,
οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ ππ. ππ. οΏ½β π π 3
πποΏ½(ββπ§π§ππππβ ππ) |ππ|2β π§π§ ππππππππππ
= ππππ οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ πποΏ½mv(ββπ§π§ππππ)ππππππ.
Theorem 14. Let ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0, then πποΏ½mv(ππ) + πποΏ½mv(βππ)
= βππππππ0βπ§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
(πποΏ½(ππ β βπ§π§ππππ) + πποΏ½(βππ β βπ§π§ππππ))
Γ πποΏ½mv(ββπ§π§ππππ)ππππππ+ ππ(ππ), ππ(ππ) =2ππππ
ππ0 (πΉπΉ(3)(ππ, 0) + πΉπΉ(3)(βππ, 0) +ππ3(ππ, 0) + ππ(3)(ππ, 0)),
where ππ0= 4ππ.
Theorem 15. Suppose ππ β πΉπΉ, ππΒ±|π₯π₯=0,π§π§=0= 0; then
ππ(ππ) = βπ§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0 οΏ½β
ππ
0 οΏ½β
2ππ
0
(πποΏ½(βππ β βπ§π§ππππ)
+πποΏ½(ππ β βπ§π§ππππ))πποΏ½(βπ§π§ππππβ βπ§π§πππ π ) Γ ππ0(βπ§π§πππ π )πππππππππππ π , where |ππ0| < πΆπΆ|ππmv|
9. THE CAUCHY PROBLEM FOR
NAVIER-STOKESβ EQUATIONS
Let us apply the obtained results to estimate the solu-tions of Cauchy problem for Navier-Stokesβ set of equa-tions
πππ‘π‘β ππΞππ + οΏ½β 3
ππ=1 πππππππ₯π₯ππ
= ββππ + πΉπΉ0(π₯π₯, π‘π‘), ππππππππ = 0, (11) ππ|π‘π‘=0 = ππ0(π₯π₯) (12) in the domain of ππππ = π π 3Γ (0, ππ). With respect to ππ0, assume
ππππππ ππ0= 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
ππ0β ππ21(π π 3), ππ β πΏπΏ2(ππππ)
then there exists a unique weak solution of problem (11), (12), (13), in ππππ1, ππ1β [0, ππ], that satisfies
πππ‘π‘, πππ₯π₯π₯π₯, βππ β πΏπΏ2(ππππ) Note that ππ1 depends on ππ0, ππ.
Lemma 16.If ππ0β ππ21(π π 3), ππ β πΏπΏ2(ππππ), then
sup
0β€π‘π‘β€ππ||ππ||πΏπΏ2(π π 3) 2 + οΏ½β
π‘π‘
0
||πππ₯π₯||πΏπΏ22(π π 3)ππππ
β€ ||ππ0||πΏπΏ22(π π 3)+ ||πΉπΉ0||πΏπΏ2(ππππ)
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
πποΏ½(π§π§(ππππβ ππππ), π‘π‘) = πποΏ½0(π§π§(ππππβ ππππ))
+ οΏ½β π‘π‘
0
ππβπππ§π§2|ππππβππππ|(π‘π‘βππ)
([(ππ, β)ππ]οΏ½ + πΉπΉοΏ½)
Γ (π§π§(ππππβ ππππ), ππ)ππππ, (14) where πΉπΉ = ββππ + πΉπΉ0.
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
πποΏ½ = οΏ½β ππ,ππ
ππππππππ
|ππ|2πποΏ½ + ππ οΏ½ππππππ β ππ
ππππ |ππ|2πΉπΉοΏ½ππ and the following estimates
||ππ||πΏπΏ2(π π 3)β€ 3||πππ₯π₯||πΏπΏ2(π π 3)
3
2 ||ππ|| πΏπΏ2(π π 3)
1
2 ,
οΏ½βπποΏ½ βπποΏ½ β€
|πποΏ½2| |ππ| +
|πΉπΉοΏ½| |ππ|2+
1 |ππ| οΏ½
βπΉπΉοΏ½ βπποΏ½ + 3 οΏ½
βπποΏ½2 β|ππ|οΏ½ ; Proof. We obtain the equation for ππ using ππππππ 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 0β€π‘π‘β€ππ[ οΏ½β
π π 3
|π₯π₯|2|ππ(π₯π₯, π‘π‘)|2πππ₯π₯
+ οΏ½β π‘π‘
0 οΏ½β π π 3
|π₯π₯|2|πππ₯π₯(π₯π₯, ππ)|2πππ₯π₯ππππ] β€ πππππππ π π‘π‘,
sup 0β€π‘π‘β€ππ[ οΏ½β
π π 3
|π₯π₯|4|ππ(π₯π₯, π‘π‘)|2πππ₯π₯
+ οΏ½β π‘π‘
0 οΏ½β π π 3
|π₯π₯|4|πππ₯π₯(π₯π₯, ππ)|2πππ₯π₯ππππ] β€ πππππππ π π‘π‘,
or
sup 0β€π‘π‘β€ππ[οΏ½οΏ½
βπποΏ½ βπ§π§οΏ½οΏ½
πΏπΏ2(π π 3)
+ οΏ½β π‘π‘
0 οΏ½β π π 3
π§π§2|πππποΏ½(ππ, ππ)|2ππππππππ] β€ πππππππ π π‘π‘,
sup 0β€π‘π‘β€ππ[οΏ½οΏ½
β2πποΏ½ βπ§π§2οΏ½οΏ½
πΏπΏ2(π π 3)
+ οΏ½β π‘π‘
0 οΏ½β π π 3
π§π§2|πππππποΏ½ (ππ, ππ)|2ππππππππ] β€ πππππππ π π‘π‘.
Proof. The proof follows from Navier-Stokesβ equa-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
maxππ |πποΏ½| β€ maxππ |πποΏ½0|
+ππ2 sup
0β€π‘π‘β€ππ||ππ||πΏπΏ2(π π 3) 2 + οΏ½β
π‘π‘
0
||πππ₯π₯||πΏπΏ2(π π 3)
2 ππππ,
maxππ οΏ½βπποΏ½βπ§π§οΏ½ β€ maxππ οΏ½βπποΏ½βπ§π§ οΏ½0
+ππ2 sup 0β€π‘π‘β€πποΏ½οΏ½
βπποΏ½ βπ§π§οΏ½οΏ½
πΏπΏ2(π π 3)
+ οΏ½β π‘π‘
0 οΏ½β π π 3
π§π§2|πππποΏ½(ππ, ππ)|2ππππππππ,
max ππ οΏ½
β2πποΏ½
βπ§π§2οΏ½ β€ maxππ οΏ½ β2πποΏ½0
βπ§π§2οΏ½
+ππ2 sup 0β€π‘π‘β€πποΏ½οΏ½
β2πποΏ½ βπ§π§2οΏ½οΏ½
πΏπΏ2(π π 3)
+ οΏ½β π‘π‘
0 οΏ½β π π 3
π§π§2|πππππποΏ½ (ππ, ππ)|2ππππππππ.
Proof. We obtain these estimates using representation (14), Parsevalβs 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
|πποΏ½mv(π§π§, π‘π‘)| β€ π§π§ππ1, οΏ½βπποΏ½mvβπ§π§(π§π§, π‘π‘)οΏ½ β€ π§π§ππ2,
οΏ½β2πποΏ½mvβπ§π§(π§π§, π‘π‘)2 οΏ½ β€ π§π§ππ3,
where ππ1, ππ2, ππ3 are limited.
Proof. Let us prove the first estimate. These inequali-ties
|πποΏ½mv(π§π§, π‘π‘)| β€2 οΏ½π§π§ β ππ
0 οΏ½ β
2ππ
0
|πποΏ½(π§π§(ππππβ ππππ), π‘π‘)|ππππππ
β€ 2πππ§π§max
ππ |πποΏ½| β€ π§π§ππ1, where ππ1= πππππππ π π‘π‘.
Follows from definition (2) for the average of ππ and from Lemmas 18, 19.
This proves Lemma 20.
Lemma 21. Weak solution of problem (11), (12), (13), from Theorem 16 satisfies the following inequalities πΆπΆππ β€ πππππππ π π‘π‘, (ππ = 0,2,4), where
πΆπΆ0= οΏ½β π‘π‘
0
|πΉπΉοΏ½1|2ππππ, πΉπΉ
1= (ππ, β)ππ + πΉπΉ,
πΆπΆ2= οΏ½β π‘π‘
0 οΏ½βπΉπΉοΏ½1
βπ§π§ οΏ½ 2
ππππ, πΆπΆ4= οΏ½β π‘π‘
0 οΏ½β2πΉπΉοΏ½1
βπ§π§2οΏ½ 2
ππππ.
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 ππ β π π , πππππ₯π₯
ππ |πποΏ½| < β, then
οΏ½β π π 3
οΏ½β π π 3
ππ(π₯π₯)ππ(π¦π¦)
|π₯π₯ β π¦π¦|2 πππ₯π₯πππ¦π¦ β€ πΆπΆ(|ππ|πΏπΏ2+ maxππ |πποΏ½|)2
Proof. Using Plansherelβs 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
|πποΏ½(π§π§(ππππβ ππππ), π‘π‘)| β€ |πποΏ½0(π§π§(ππππβ ππππ))|
+ οΏ½2ππ1οΏ½
1 2 πΆπΆ0
1 2
π§π§|ππππβππππ|, (15) where
πΆπΆ0= οΏ½β π‘π‘
0
|πΉπΉοΏ½1|2ππππ, πΉπΉ
1= (ππ, β)ππ + πΉπΉ.
Proof. From Formula (14) we get
|πποΏ½(π§π§(ππππβ ππππ), π‘π‘)| β€ |πποΏ½0(π§π§(ππππβ ππππ))|
+| β«0π‘π‘βππβπππ§π§2|ππππβππππ|2(π‘π‘βππ) (16)
Γ πΉπΉοΏ½1(π§π§(ππππβ ππππ), ππ)ππππ| where
πΉπΉ1= (ππ, β)ππ + πΉπΉ. Using the denotation
πΌπΌ = | οΏ½β π‘π‘
0
ππβπππ§π§2|ππππβππ ππ|2(π‘π‘βππ)
Γ πΉπΉοΏ½1(π§π§(ππππβ ππππ), ππ)ππππ|,
Taking into account Holderβs inequality in πΌπΌ we obtain
πΌπΌ β€ οΏ½οΏ½β π‘π‘
0
|ππβπππ§π§2|ππππβππ
ππ|2(π‘π‘βππ)|πππππποΏ½
1 ππ
Γ οΏ½οΏ½β π‘π‘
0
|πΉπΉ1|πππππποΏ½ 1 ππ
where ππ, ππ satisfies the equality 1ππ+1ππ= 1. Suppose ππ = ππ = 2. Then
πΌπΌ β€ οΏ½2πποΏ½1 1
2οΏ½β«0π‘π‘β|πΉπΉοΏ½1|2πππποΏ½ 1 2
π§π§|ππππβ ππππ| .
Taking into consideration the estimate πΌπΌ 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 |πποΏ½|ππ πΏπΏ1(π π 3),
a component of velocity components in the Cauchy problem for Navier-Stokesβ equations. In order to achieve the aim, we use Theorem8 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
|πποΏ½mv|πΏπΏ1(π π 3)β€
πΆπΆ
2 οΏ½π΄π΄0(1)+ π½π½1|πποΏ½mv|πΏπΏ1(π π 3)οΏ½
+|ππ|πΏπΏ1οΏ½π π 3οΏ½ (17) the function ππ is defined in Theorem15,
π΄π΄0(1)= οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½0(π§π§(ππππβ ππππ))|
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ, π½π½1= οΏ½1πποΏ½ 1 2
8πππΆπΆ012, and πΆπΆ0 is defined in Lemma 23.
Proof. From the statement of Theorem 14, we get the estimate
|πποΏ½mv|πΏπΏ1(π π 3) β€
πΆπΆ 2 οΏ½β
π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½(π§π§(ππππβ ππππ), π‘π‘)|
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ + |ππ|πΏπΏ1(π π 3).
(15) in the integral, we obtain
|πποΏ½mv|πΏπΏ1(π π 3)β€
πΆπΆ 2 ( οΏ½β
π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½0(π§π§(ππππβ ππππ)|
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ
+ οΏ½1πποΏ½ 1 2
πΆπΆ012 οΏ½β π π 3
οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½mv(ππ, π‘π‘)|
Γ ππππππ
Let us use the notation
π΄π΄0(1)= οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½0(π§π§(ππππβ ππππ))|
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ, then
|πποΏ½mv|πΏπΏ1(π π 3)β€
πΆπΆ
2 (π΄π΄0(1)+ οΏ½1πποΏ½ 1 2
πΆπΆ012
Γ οΏ½β π π 3
οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½mv(ππ, π‘π‘)||ππππππππβ ππππππ | ππππ) + |ππ|πΏπΏ1(π π 3).
Let us use the notation
πΌπΌ0= οΏ½β ππ
0 οΏ½ β
2ππ
0
ππππππ |ππππβ ππππ| and obtain πΌπΌ0. Since
|ππππβ ππππ| = ((ππππβ ππππ, ππππβ ππππ))12= (1 β cosππ)12 where ππ is the angle between the unit vectors ππππ, ππππ, it follows that
πΌπΌ0= 4ππ οΏ½β ππ
0
sinππ
(1 β cosππ)12ππππ = 2 7 2ππ.
Using πΌπΌ0 in the estimate |πποΏ½mv|πΏπΏ1(π π 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
οΏ½πποΏ½mvπ§π§ οΏ½πΏπΏ
1(π π 3)
β€πΆπΆ2 οΏ½π΄π΄0+ π½π½1οΏ½πποΏ½mvπ§π§ οΏ½πΏπΏ
1(π π 3)
οΏ½
+ οΏ½πππ§π§οΏ½
πΏπΏ1(π π 3), (18)
where
π΄π΄0= οΏ½β π π 3
οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½0(π§π§(ππππβ ππππ))|
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ and π½π½1 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
οΏ½πποΏ½mvπ§π§ οΏ½πΏπΏ
1(π π 3)
β€ οΏ½πΆπΆ2 π΄π΄0+ οΏ½πππ§π§οΏ½πΏπΏ
1(π π 3)οΏ½ πΎπΎ,
where
πΎπΎ = ππ
1 2
ππ12β 4πππΆπΆπΆπΆ0 1 2 .
Letβs consider the influence of the following large scale transformations in Navier-Stokesβ equation on πΎπΎ
π‘π‘β² = π‘π‘π΄π΄, ππβ² =ππ
π΄π΄ , ππβ² = ππ
π΄π΄ , πΉπΉ0β² =π΄π΄πΉπΉ02. Statement 2. Let
π΄π΄ = 4
ππ13(πΆπΆπΆπΆ0+ 1)23 ,
then πΎπΎ β€87.
Proof. By the definitions πΆπΆ and πΆπΆ0, we have
πΎπΎ = (π΄π΄)ππ 12((ππ π΄π΄)
1
2β4πππΆπΆπΆπΆ0 π΄π΄2 )β1 = ππ12(ππ12β4πππΆπΆπΆπΆ0
π΄π΄32
)β1 <8 7. This proves Statement 2.
Lemma 25.Weak solution of problem (11), (12), (13), from Theorem 16 satisfies the following inequalities
οΏ½βπποΏ½(π§π§(ππππβπ§π§β ππππ), π‘π‘)οΏ½ β€ οΏ½βπποΏ½0(π§π§(ππππβπ§π§β ππππ))οΏ½
+4πΌπΌ οΏ½1πποΏ½
1 2 πΆπΆ0
1 2
π§π§2|ππππβππππ| (19)
+ οΏ½1 2πποΏ½
1 2 πΆπΆ2
1 2
π§π§|ππππβ ππππ|, where
πΆπΆ2= οΏ½β π‘π‘
0 οΏ½βπΉπΉοΏ½βπ§π§ οΏ½1
2 ππππ.
Proof. The underwritten inequalities follows from re-presentation (14)
οΏ½βπποΏ½(π§π§(ππππβπ§π§β ππππ), π‘π‘)οΏ½ β€ οΏ½βπποΏ½0(π§π§(ππππβπ§π§β ππππ))οΏ½
+2πππ§π§|ππππβ ππππ|2| οΏ½β π‘π‘
0
(π‘π‘ β ππ)ππβπππ§π§2|ππππβππππ|2(π‘π‘βππ)
Γ πΉπΉοΏ½1(π§π§(ππππβ ππππ), ππ)ππππ|
+| οΏ½β π‘π‘
0
ππβπππ§π§2|ππππβππ ππ|2(π‘π‘βππ)
ΓβπΉπΉοΏ½βπ§π§ (π§π§(ππ1 ππβ ππππ), ππ)ππππ|. Let us introduce the following denotation
πΌπΌ1= 2πππ§π§|ππππβ ππππ|2| οΏ½β π‘π‘
0
(π‘π‘ β ππ)ππβπππ§π§2|ππππβππππ|2(π‘π‘βππ)
πΌπΌ2= | οΏ½β π‘π‘
0
ππβπππ§π§2|ππππβππππ|2(π‘π‘βππ)
ΓβπΉπΉοΏ½1
βπ§π§ (π§π§(ππππβ ππππ), ππ)ππππ|, then
οΏ½βπποΏ½(π§π§(ππππβπ§π§β ππππ), π‘π‘)οΏ½ β€ οΏ½βπποΏ½0(π§π§(ππππβπ§π§β ππππ))οΏ½ +πΌπΌ1+ πΌπΌ2.
Estimate πΌπΌ1 by means of sup
π‘π‘ |π‘π‘
ππππβπ‘π‘| < πΌπΌ,
where ππ > 0 we obtain
πΌπΌ1β€4πΌπΌπ§π§ | οΏ½β π‘π‘
0
ππβπππ§π§2|ππππβππππ|2π‘π‘βππ2
Γ πΉπΉοΏ½1(π§π§(ππππβ ππππ), ππ)ππππ|. On applying Holderβs inequality, we get
πΌπΌ1β€4πΌπΌπ§π§ οΏ½οΏ½β π‘π‘
0
|ππβπππ§π§2|ππππβππππ|2π‘π‘βππ2 |πππππποΏ½ 1 ππ
Γ οΏ½οΏ½β π‘π‘
0
|πΉπΉ1|πππππποΏ½ 1 ππ ,
where ππ, ππ satisfy the equality 1ππ+1ππ= 1. For ππ = ππ = 2 we have
πΌπΌ1β€ 4πΌπΌ οΏ½1πποΏ½ 1 2 πΆπΆ0
1 2 π§π§2|ππππβ ππππ|,
πΌπΌ2β€ οΏ½2πποΏ½1 1 2 πΆπΆ2
1 2 π§π§|ππππβ ππππ|,
πΆπΆ2= οΏ½β π‘π‘
0 οΏ½βπΉπΉοΏ½βπ§π§ οΏ½1
2 ππππ.
Inserting πΌπΌ1, πΌπΌ2 in to οΏ½βπποΏ½βπ§π§οΏ½, 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
οΏ½βπποΏ½mvβπ§π§ οΏ½πΏπΏ
1(π π 3)
β€πΆπΆ2 (π΄π΄0+ π΄π΄1+ π΄π΄2
+π½π½3|πποΏ½mv|πΏπΏ1(π π 3)+ (π½π½1+ π½π½2) οΏ½πποΏ½mv
π§π§ οΏ½πΏπΏ1(π π 3)
+π½π½1οΏ½βπποΏ½mv
βπ§π§ οΏ½πΏπΏ1(π π 3)) + οΏ½
βππ
βπ§π§οΏ½πΏπΏ1(π π 3), (20)
where
π΄π΄1= οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
οΏ½βπποΏ½0(π§π§(ππβπ§π§ππβ ππππ))οΏ½
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ,
π΄π΄2= οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½0(π§π§(ππππβ ππππ))|
Γ οΏ½βπποΏ½mvβπ§π§(π§π§ππππ, π‘π‘)οΏ½ ππππππππππ,
π½π½2= οΏ½1πποΏ½ 1 2
2112πππΌπΌπΆπΆ0 1
2, π½π½3= οΏ½1 πποΏ½
1 2
8πππΆπΆ212,
and πΆπΆ2 is defined in Lemma 25, πΆπΆ = πππππππ π π‘π‘.
Proof. From the statement of Theorem 14 we get the following estimate
οΏ½βπποΏ½mvβπ§π§ οΏ½πΏπΏ
1(π π 3)
β€πΆπΆ2 ( οΏ½β π π 3
οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½(π§π§(ππππβ ππππ), π‘π‘)|
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ
+ οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
οΏ½βπποΏ½(π§π§(ππππβπ§π§β ππππ), π‘π‘)οΏ½
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ
+ οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½(π§π§(ππππβ ππππ), π‘π‘)|
Γ οΏ½βπποΏ½mvβπ§π§(π§π§ππππ, π‘π‘)οΏ½ ππππππππππ) + οΏ½βππβπ§π§οΏ½πΏπΏ
1(π π 3)
.
Let us introduce the following denotation
πΌπΌ1= οΏ½β π π 3
οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½(π§π§(ππππβ ππππ), π‘π‘)|
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ,
πΌπΌ2= οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
οΏ½βπποΏ½(π§π§(ππππβπ§π§β ππππ), π‘π‘)οΏ½
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ,
πΌπΌ3= οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½(π§π§(ππππβ ππππ), π‘π‘)|
Γ οΏ½βπποΏ½mvβπ§π§(π§π§ππππ, π‘π‘)οΏ½ ππππππππππ, then
οΏ½βπποΏ½mv βπ§π§ οΏ½πΏπΏ1(π π 3)
β€πΆπΆ
2 (πΌπΌ1+ πΌπΌ2+ πΌπΌ3) + οΏ½ βππ βπ§π§οΏ½πΏπΏ1(π π 3)
.
πΌπΌ1β€ π΄π΄0+ π½π½1οΏ½πποΏ½mvπ§π§ οΏ½πΏπΏ
1(π π 3)
.
Inserting inequality (19) into πΌπΌ2, we get
πΌπΌ2β€ οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
οΏ½βπποΏ½0(π§π§(ππβπ§π§ππβ ππππ))οΏ½
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ
+4πΌπΌ οΏ½1πποΏ½ 1 2
πΆπΆ012πΌπΌ 0 οΏ½β
π π 3
|πποΏ½mv(ππ, π‘π‘)| π§π§ ππππ
+ οΏ½2πποΏ½1 1 2
πΆπΆ212πΌπΌ 0 οΏ½β
π π 3
|πποΏ½mv(ππ, π‘π‘)|ππππ,
Let us take into account the estimate of πΌπΌ0 obtained in Lemma 25,
πΌπΌ0= οΏ½β ππ
0 οΏ½ β
2ππ
0
ππππππ |ππππβ ππππ| = 2
7 2ππ.
Inserting this value in πΌπΌ2, we obtain
πΌπΌ2β€ οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
οΏ½βπποΏ½0(π§π§(ππβπ§π§ππβ ππππ))οΏ½
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ
+ οΏ½1πποΏ½ 1 2
2112πππΌπΌπΆπΆ0 1 2 οΏ½β
π π 3
|πποΏ½mv(ππ, π‘π‘)| π§π§ ππππ
+ οΏ½1πποΏ½ 1 2
8πππΆπΆ212 οΏ½β π π 3
|πποΏ½mv(ππ, π‘π‘)|ππππ.
Let us introduce the following denotation
π΄π΄1= οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
οΏ½βπποΏ½0(π§π§(ππππβπ§π§β ππππ))οΏ½
Γ |πποΏ½mv(π§π§ππππ, π‘π‘)|ππππππππππ, then
πΌπΌ2β€ π΄π΄1+ π½π½2οΏ½πποΏ½mvπ§π§ οΏ½πΏπΏ
1(π π 3)
+ π½π½3|πποΏ½mv|πΏπΏ1(π π 3),
where
π½π½2= οΏ½1πποΏ½ 1 2
2112πππΌπΌπΆπΆ0 1 2, π½π½
3= οΏ½1πποΏ½ 1 2
8πππΆπΆ212. Using inequality (16) in πΌπΌ3, we get
πΌπΌ3β€ οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½0(π§π§(ππππβ ππππ))|
Γ οΏ½βπποΏ½mvβπ§π§(π§π§ππππ, π‘π‘)οΏ½ ππππππππππ
+ οΏ½2πποΏ½1 1 2
πΆπΆ012πΌπΌ 0 οΏ½β
π π 3
οΏ½βπποΏ½mvβπ§π§(ππ, π‘π‘)οΏ½ ππππ.
Similarly as we estimated πΌπΌ2, obtain
πΌπΌ3β€ π΄π΄2+ π½π½1οΏ½βπποΏ½mvβπ§π§ οΏ½πΏπΏ
1(π π 3)
,
where
π΄π΄2= οΏ½β π π 3
π§π§ οΏ½β ππ
0 οΏ½ β
2ππ
0
|πποΏ½0(π§π§(ππππβ ππππ))|
Γ οΏ½βπποΏ½mvβπ§π§(π§π§ππππ, π‘π‘)οΏ½ ππππππππππ.
Inserting πΌπΌ1, πΌπΌ2, πΌπΌ3 οΏ½βπποΏ½βπ§π§mvοΏ½πΏπΏ
1(π π 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
οΏ½β2πποΏ½(π§π§(ππππβπ§π§β ππππ2 ), π‘π‘)οΏ½ β€ οΏ½β2πποΏ½0(π§π§(ππππβπ§π§2β ππππ))οΏ½
+ οΏ½1πποΏ½ 1
2 16πΌπΌπΆπΆ0 1 2 π§π§3|ππππβ ππππ| + οΏ½
1 πποΏ½
1
2 8πΌπΌπΆπΆ2 1 2 π§π§2|ππππβ ππππ|
+ οΏ½2ππ1οΏ½
1 2 πΆπΆ4
1 2
π§π§|ππππβππππ|, (21) where
sup π‘π‘ |π‘π‘
ππππβπ‘π‘| < πΌπΌ,
as ππ > 0,
πΆπΆ4= οΏ½β π‘π‘
0 οΏ½β2πΉπΉοΏ½1
βπ§π§2οΏ½ 2
ππππ.
Proof. From (14) we have the following inequalities
οΏ½β2πποΏ½(π§π§(ππππβπ§π§β ππππ2 ), π‘π‘)οΏ½ β€ οΏ½β2πποΏ½0(π§π§(ππππβπ§π§2β ππππ))οΏ½
+4ππ2π§π§2|ππππβ ππππ|4| οΏ½β π‘π‘
0
(π‘π‘ β ππ)2
Γ ππβπππ§π§2|ππππβππππ|2(π‘π‘βππ)
πΉπΉοΏ½1(π§π§(ππππβ ππππ), ππ)ππππ|
+4πππ§π§|ππππβ ππππ|2| οΏ½β π‘π‘
0
(π‘π‘ β ππ)
Γ ππβπππ§π§2|ππππβππππ|2(π‘π‘βππ)βπΉπΉοΏ½1
βπ§π§ (π§π§(ππππβ ππππ), ππ)ππππ|
+ οΏ½οΏ½β π‘π‘
0
ππβπππ§π§2|ππππβππππ|2(π‘π‘βππ)β2πΉπΉοΏ½1