The 3D Navier–Stokes Equations: Invariants, Local and Global Solutions

In this article, I consider local solutions of the 3D Navier–Stokes equations and its properties such as an existence of global and smooth solution, uniform boundedness. The basic role is assigned to a special invariant class of solenoidal vector fields and three parameters that are invariant with respect to the scaling procedure. Since in spaces of even dimensions the scaling procedure is a conformal mapping on the Heisenberg group, then an application of invariant parameters can be considered as the application of conformal invariants. It gives the possibility to prove the sufficient and necessary conditions for existence of a global regular solution. This is the main result and one among some new statements. With some compliments, the rest improves well-known classical results.


Introduction
During the last century, the Navier-Stokes equations attracted very much attention. The first essential steps in this way were offered by C. Oseen [1], F. K. G. Oldquist [2], J. Leray [3][4][5], and E. Hopf [6]. Later, the Cauchy problem and the boundary value problem were actively studied by many authors (see, for example, [7,8], the review [9][10][11][12][13][14][15][16][17] and etc.). The main objects and tools of these works were weak solutions or fix points of integral operators. Here, a special case is connected with the existence problem of a global and regular solution in the 3D Cauchy problem. In response to the new setting of this task by Ch. Fefferman in 2000 (see [18]), O.A. Ladyzhenskaya wrote in her review [9] that she would put the main question otherwise: "Do or don't the Navier-Stokes equations give, together with initial and boundary dates, the deterministic description of fluid dynamics?" Then, this problem is more difficult and more interesting from the physical point of view. Therefore, I introduced some invariants for studying solutions properties. At least, it is natural for applications because invariants are very important and strong tools. Moreover, these invariants didn't apply earlier.
Let us describe them now. The first invariant connected with the Cauchy problem that provided initial data belongs to a special class C ∞ 6/5, 3/2 of solenoidal vector fields vanishing at infinity. Here, outer forces are trivial. Then, the class C ∞ 6/5, 3/2 is invariant (Theorem 2). This is a new result. The second invariant is a special parameter λ (see (68)) which is connected with a velocity changing of E 2 , where E is a kinetic energy of a fluid flow. If λ ≥ 1 or kinetic energy at a special moment is not less any mean depending on λ for λ < 1 ( i.e., changing of E 2 at moment t = 0 is negligible), then an ideal, global and smooth motion is determined. In other words, a global regular solution exists (Theorem 7). This is an essential and qualitative improvement of the classical result together with a new a priori estimate given by Theorems 8-10. These theorems are new results in principle.
(1) solutions are bounded with respect to a uniform norm and therefore it belongs to any class L p, q ; (2) there is a universal time interval [0, T 0 ) where bounded solutions exist; (3) more exact necessary conditions of a hypothetical turbulence phenomenon if it is; (4) a lower estimate of the kinetic energy which influences an existence of a global smooth solution.
The last two items are very important. If dissipation of kinetic energy is large (close to the unit), then blow up is probable.
To the structure of the paper. In the first part (Section 2), there are considered solutions' properties of the Cauchy problem in a local form if initial data is smooth enough. Here, there is given a modification of classical results with some supplements (see Theorem 1).The rest of this part contains technical lemmas which are proved by application of hydrodynamic potentials and multiplicative inequalities from Appendix (Appendix A). In the second part (Sections 3 and 4), there are existence conditions of global solutions studied in this problem, conditions for local solutions' extensions if the kinetic energy is small and close to the minimum. A more precise hypothetic blow up time interval is found. Here, three basic parameters λ, µ, ε are very useful.
The third part (Sections 5 and 6) contains the proof of main statements (Theorems 7-9), which are based on properties of invariant parameters λ, µ, ε.
I think, in this way, it is convenient to remove any restrictions on a smoothness in some contrast to the traditional way. The main idea is connected with an invariant form of an a priori estimate for gradient norms of a velocity. In addition, other norms are estimated in class L 6 and, after that, it is done in class L 2 . In particular, it is shown that there is a bad solution of a class L 6 with some good properties. As the corollary, this solution has many uniformly bounded norms with respect to time argument. Only after that, by routine calculations, we prove the bad solution from above belongs to a class L 2 . Precisely, this step distinguishes from classical way for the second time (see [7]).
In the considered problem, a boundedness of solutions depends on a smoothness of initial data. At least, initial data from the Sobolev class W 3 2 gives the same in principle. The offered construction doesn't permit diminishing the index of smoothness. In the final (Section 7), we explain the principal difference between the Navier-Stokes equations in space and plane.
A part of local results in modification (Section 2) and invariants as tools (Section 4) were announced by author in [20][21][22].
NOTATION. Now, let us consider the Cauchy problem (n = 3): u i u k, i = ν∆u k − P , k , k = 1, 2, . . . , n, where u is a velocity of flow, P is a pressure function, symbols D t u = ∂u ∂t , u k, i = ∂u k ∂x i , u k, ij = ∂ 2 u k ∂x i ∂x j , ..., P , k = ∂P ∂x k indicate a partial differentiation or differentiation in distributions, is the Laplace operator, and ν is a positive constant (viscosity coefficient). A mapping ϕ has all derivatives and satisfies conditions of averaged growth: ϕ ∈ L 6/5 (R 3 ), ϕ , i ∈ L 3/2 (R 3 ). The other derivatives belong to classes L r (R 3 ) for any r > 1. Furthermore, this class is denoted by symbol C ∞ 6/5, 3/2 . A class C ∞ 0 (R n ) is the class of infinitely smooth mappings with a compact support. A norm in a space L p (Ω) is defined by formula: A mixed norm is defined by equality: A symbol D α v denotes a partial differentiation or distributions with respect to a multi-index α. An order of the derivative is indicated by |α|. Jacobi matrix of a mapping v with respect to spatial variables is denoted by ∇v . Its modulus is Functions' properties from the Sobolev classes W l p (Ω) are given, for example, in [23][24][25]. A norm in this functional space is defined by Let v be a mapping that is determined on the whole space. For the Riesz potential, we apply notation: where γ(α) = π n 2 Γ( α 2 ) Γ( n−α 2 ) and Γ is the Euler gamma-function. The properties of these potentials can be found in [24].
The agreement about summation. Everywhere in this article, the repeated indices give a summation if it is not done reservation specially. For example, Furthermore, S T = [0, T] × R 3 . A number T 0 we define by formula: We apply the definition of a weak solution given in [7] everywhere.

Preliminaries. Boundedness and Smoothness Properties of Local Solutions in the Cauchy Problem
Here, with some compliments, a local result described by Theorem 1 is basic in this section. The rest contains only technical statements.
Theorem 1. Let T 0 be a number from formula (5) and a mapping ϕ ∈ C ∞ 6/5, 3/2 . Then, on the set S T 0 , there exist weak solutions u and P of problems (1) and (2) with the following properties: (1) mappings u and P uniformly continuous and bounded on the set S T for every number T, 0 < T < T 0 ; (2) the solution u belongs to Sobolev classes W 2 2 (S T ) and W 1 6 (S T ) for every number T, 0 < T < T 0 , moreover, all norms u p , ∇u p , D t u p , u , ij p , ∇D t u 2 are uniformly bounded in spaces L p (R 3 ), 2 ≤ p ≤ 6, by a constant C = C(ν, ϕ, T) depending on ν, ϕ and T only, in addition u 2 ≤ ϕ 2 ; (3) gradients ∇u i , i = 1, 2, 3, ∇P are bounded on the set S T for every number T, 0 < T < T 0 ; (4) the solution P satisfies uniform estimates: ∇P q ≤ C, 3 2 < q < ∞, ∇D t P q ≤ C, P , ij q ≤ C, for all numbers q, 3 2 < q ≤ 3, and t ∈ [0, T], T < T 0 , with constants C depending on ν, ϕ, T and q only; (5) solutions u and P are classical solutions that is for any T < T 0 they belong to the class C ∞ ((0, T 0 ) × R 3 ) C(S T ).
The proof of the theorem is given to the end of this section. We note items 1, 3, 4 compliment well-known Ladyzhenskaya's results (see [7]). Item (2) contains new uniform estimate for norms of derivatives. Hence, it follows a boundedness of weak solutions and a finiteness of its mixed norms. Moreover, we have an existence of weak solution with required properties on the interval [0, T 0 ) with the finite length. To the studying of the smoothness property for weak solutions, the mixed norms were applied by O. Ladyzhenskaya in [26] (see, also [7]). They were applied by other authors (see, for example, [8,10,14]). Item (5) is a particular case from [27]. However, from this theorem, a deeper result follows (see Theorem 7).

A Priori Estimates of Gradients' Norms
Lemma 1. Suppose that a mapping w : S T 0 → R 3 belongs to a class C 2 and w(0, x) = ϕ(x). If, for every t ∈ [0, T 0 ), Laplacian supports are subsets of some ball with a fixed radius and w ∈ L 6 (R 3 ), ∇w ∈ L 2 (R 3 ) L 6 (R 3 ), then for all mappings w satisfying condition: the following estimate holds: Proof. We take from Corollary A4 the second inequality. Then, from (6), we obtain: Let y = w 2 / ∇w 3 2 . Then, (7) can be rewritten in the form: The maximal mean on the right-hand side is 27a 4 1 256ν 3 . Therefore, integrating the inequality Furthermore, we take a number a 1 from Corollary A4 and obtain the required estimate.

Lemma 2.
Let T 0 be a constant from Lemma 1. Assume a mapping w : S T 0 → R 3 belongs to a class C 3 and w(0, x) = ϕ(x), D t w(0, x) = ψ(x). Suppose that, for every t, there are fulfilled conditions: (1) Laplacian supports w, D t w are subsets of a ball with a fixed radius; (3) with constants k 1 , l the inequalities hold: (4) the equality is true. Then, for every segment, [0, T] where T < T 0 the estimate ∇D t w 2 ≤ k 2 ∇ψ 2 holds with a constant k 2 which depends on ν, T, k 1 , l, ∇ϕ 2 only.
Proof. The integral on the right-hand side in formula (8) we rewrite with two integrals J 1 and J 2 . Applying Corollary A4, we make estimates for every integral. In integral J 1 , a triple of mappings u, v, w is the triple D t w, w, D t w. In integral J 2 , a required triple is the triple w, D t w, D t w. Therefore, condition (3) yields estimates: Hence, from (8), we get: a k 1 ∇ϕ 1/2 Let g(t) = ∇D t w 2 , h(t) = D t w 2 /g(t). Then, formula (9) can be transformed to the formula: Let us integrate over segment [0, t] this inequality. For the next step, we apply to each term the Hölder inequality for three and two factors, respectively getting quantities h 2 (t) and w 2 2 . Hence, from condition (3), we obtain: where a 1 = a k 1 ∇ϕ 2 Then, the last estimates give g(t) ≤ e 2M g(0). From the definition of function g, we have g(0) = ∇ψ 2 .

A Priori Estimates of Laplacian Norms
Lemma 3. Let w, T 0 be a mapping and a number from Lemma 1. Then, for every number T, 0 < T < T 0 , there exists a constant l = l(ν, ϕ, T) such that Proof. We transform inequality (7) applying the estimate from Lemma 1. Then, This inequality we integrate over the segment [0, t]. Then, we estimate the right-hand side applying the Hölder inequality and underlining the integral with the term The direct calculations of the integral on the right-hand side and the estimate give the inequality: Take out the first term on the left hand. Then, the required estimate for function β(t) will be obvious. If β(t) ≤ ∇ϕ 2 , then the estimate is acceptable. If β(t) ≥ ∇ϕ 2 , then we have: Hence, it follows the lemma.

Lemma 4.
Let w be a mapping from Lemma 2 and a number T 0 from Lemma 1. Then, for every number T, Proof. For the mapping w, inequality (9) is fulfilled. Its right-hand side we estimate relying on Lemma 2. Then, 1 2 Let C be a maximal coefficient of factors Therefore, from formula (10), we have inequality: This inequality we integrate over segment[0, t] and its right-hand side we estimate applying the Hölder inequality and underlining terms with norms D t w 2 . If then we have the estimate: We increase the right side using Lemma 3 and deduce the left side taking out the first positive term. Then, we obtain: Hence, we get the lemma in the same way as Lemma 3. If β 1 (t) > ∇ψ 2 , then, from we obtain the lemma inequality. If β 1 (t) ≤ ∇ψ 2 , then the estimate is acceptable.

Basic Space of Solenoidal Vector Fields and Orthogonal Systems
Let us consider solenoidal vector fields ϕ : R 3 → R 3 from class C ∞ with a compact support of ϕ. A closure of this class is defined by the norm: We denote its by J 2 0 (R 3 ). From Lemmas A1 and A2, it follows that elements u ∈ J 2 0 (R 3 ) are represented by the Riesz potentials; moreover, u, ∇u ∈ L 6 (R 3 ). Otherwise, each element is defined uniquely by its Laplacian. The class J 2 0 (R 3 ) is a separable space as a subspace of the Sobolev classes W l p (R 3 ), 1 < p < ∞. Therefore, there exists a countable system (ψ n ) n=1,... of infinite smooth vector fields satisfying conditions: (1) div ψ n = 0; (2) supports of ψ n are compact sets; (3) the closure of a linear span in norm (11) coincides with the space J 2 0 (R 3 ). Now, we apply the Sonin-Shmidt orthogonalization to the fundamental system (ψ n ) n=1,... and construct a countable system of mappings (b n ) n=1,... , which would be with the orthogonality property of Laplacians in the space L 2 (R 3 ). That is, the scalar product where δ ij is Kronecker's symbol. Then, every mapping b n is a finite linear combination of mappings (ψ k ). Therefore, a support of b n is a compact set. Let b n = a n .
The system (a n ) is complete for the space J 2 0 (R 3 ); that is, the following proposition is true.
Proof. From chosen mappings a n , the equality (u, ψ n ) = 0 for each element of the fundamental system (ψ n ) n=1,... follows. The Stokes theorem gives |x|≤r u k ψ n k dx = − |x|≤r u k, i ψ n k, i dx + |x|=r u k ψ n k, i x i r dS.
The integral over the sphere vanishes as r → ∞. Actually, from Corollary A2 of Lemma A2, we have: Furthermore, we apply Lemma A4 (α = 2, p = 6) taking into consideration a continuity of u. The passage to the limit yields the equality (u, ψ n ) = −(∇u, ∇ψ n ) or (∇u, ∇ψ n ) = 0. We take a sequence of finite and smooth mappings which converges to the vector field u in the space J 2 0 (R 3 ). Hence, (∇u, ∇u) = 0. The summability of u in the space L 6 (R 3 ) proves lemma equality. Remark 1. To the fundamental system of mappings (ψ n ) we can adjoin any solenoidal vector field ϕ ∈ C ∞ 0 (R 3 ), ϕ = 0 or any vector field from the class J 2 0 (R 3 ) as the first element of this system.

Successive Approximations of Solutions and Its Estimates: Velocity
Let (a n 1,... ) be an orthonormal system of mappings in the space L 2 (R 3 ) constructed above with the completeness property in J 2 0 (R 3 ) and conditions (12) and (13). Moreover, a n ∈ C ∞ 0 (R 3 ) for all n and a 1 = ϕ ϕ 2 where a vector field ϕ ∈ C ∞ 0 (R 3 ) is initial data in problems (1) and (2).
For successive approximations v n , we define changing Ladyzhenskaya's construction in ( [7], p. 197). Set v n (t, x) = n ∑ q=1 c qn (t)a q (x). (14) Then, an approximative solution v n is built as a hydrodynamical potential Functions c qn are solutions of a system of differential equations: with initial data: Now, we find an existence interval of a smooth solution in system (16). For every equation from (16), we multiply by functions c qn and sum them. As a result, we have: From Corollary A3, we get: From Lemmas A1 and A2 vector fields v n ∈ L 6 (R 3 ), ∇v n ∈ L 6 (R 3 ) L 2 (R 3 ). Equalities (17) and (18) are conditions of Lemma 1 for mappings v n . Therefore, in system (16), an existence of smooth solutions on some interval [0, t 0 ) is guaranteed by well-known theorems for ordinary differential equations. By Lemma 1 (see estimates), these solutions can be extended on the interval [0, T 0 ) where T 0 is the constant in Lemma 1 (see also (5)). Thus, we proved the following statement. Lemma 6. Let [0, T 0 ) be an interval from Lemma 1. Then, for every t ∈ [0, T 0 ), approximations v n constructed by formulas (14) and (16) satisfy conditions: where a constant A from Lemma A1.
Proof. Item (1) follows from Lemma 1. Item (2) is the corollary of the second representation in (A1), Lemma A1 and arguments in the proof of Corollary A4. (1) where a number k 2 = k 2 (ν, ϕ, T) depends on ν, ϕ, T only; (2) ∇D t v n 6 ≤ A D t v n 2 with the constant A from Lemma A1.
Proof. The item (2) can be proved in the same way as the estimate (2) from Lemma 6. Let us prove item (1). We differentiate equalities (16) with respect to t. Then, from each, we multiply by the derivative c qn (t) and add together in final. As a result, we have A support of D t v n is a compact set. The Stokes theorem and Corollary A3 give: From Lemmas A1 and A2, we have v n , D t v n ∈ L 6 (R 3 ), ∇v n , D t ∇v n ∈ L 6 (R 3 ) L 2 (R 3 ).
By Lemma 3, the vector field v n satisfies the inequality: Then, mappings v n satisfy Lemma 2. This implies: with some constant k 2 = k 2 (ν, ϕ, T). Let us estimate the right-hand side of (20). In (16), we take t = 0. Then, we multiply them by numbers c qn (0) respectively and add them together. As a result, formula (17) gives where We move derivatives with the factor D t v n in (21) using Corollary A3 and a finiteness of mapping ϕ. Then, we obtain ∇D t v n (0, x) 2 2 = ∇D t v n (0, x), ∇T 1 (x) .

Lemma 9.
Let T 0 be a constant of Lemma 1. Then, approximations v n from (14)- (16) are bounded by a constant C on the set S T for every T < T 0 where a constant C depends on ν, ϕ, T only.
Proof. For approximation v n , we use integral representation (A2). One should replace integration over whole space by integrations over ball |y − x| ≤ 1 and its complement. Then, v n (t, x) = 1 4π (J 1 + J 2 ). Every term is estimated by application of Hölder's inequality. We have where C 1 , C 2 are universal constants. The norm ∇v n 6 is estimated in two steps. In the first step, we apply inequality 2) from Lemma 6. After that, we use inequality (1) from lemma 8. To estimate another norm ∇v n 2 , we can apply inequality (1) from Lemma 6. Hence, we get a boundedness of all vector fields v n by a general constant.
Proof. The statement of lemma is the corollary well-known results about integral differentiation with a weak singularity (see [28]). From the second representation of Lemma A2, we obtain two equalities: where k ij are some constants, T ij are singular integral operators. Its boundedness in the space L 2 gives the required estimates. (1) v n i, j v n j, ik p ≤ C; (2) v n i, j D t v n j, ik p ≤ C D t v n 2 ; (3) D t v n i, j v n j, ik p ≤ C D t v n 3/p−2 2 ; (4) v n i, j D t v n j, i p ≤ C; where constants C depend on ν, ϕ, T, p only.
Proof. Apply Hölder's inequality. Then, Denote h = v n , g = v n . An exponent 2p/(2 − p) ∈ [2,6]. Then, the first factor in (23) is estimated by Lemma 1 with an assumption r = 2, s = 6. Uniform estimate (1) follows from Lemma 6 and Lemma 8. In the same way taking a pair h = v n , g = D t v n we get estimate (2). Now, denote h = D t v n , g = v n . To the first factor from the right-hand side of (23) we apply Lemma A5 relying on r = 2, s = 6, t = 3 − 3/p. The norm ∇D t v n 2 has a uniform estimate with respect to t and n by Lemma 7. Apply the both estimates of this lemma and obtain estimate (3). The other estimates (4) and (1) we prove in the same way.

Successive Approximations of Solutions and Its Estimates: Pressure
Let v n be an approximation from formulae (14)- (16). Fix T, T < T 0 where T 0 is the constant from Lemma 1. Consider a hydrodynamical potential . This follows from estimates of Lemma 6, Lemma 8 and Hölder's inequality. By Lemma A4 on every segment [0, T], we have: Lemma A1 implies a uniform estimate with respect to t and n: for any exponent q > 3, where 1 q = 1 p − 2 3 . Let us decompose integral in (24) by two integrals J 1 and J 2 : over ball |y − x| < 1 and over its exterior. Every integral we estimate by Hölder's inequality or a simple estimation. Then, Thus, with some constant C = C(ν, ϕ, T) on the set S T for all n, we obtain: Function P n has derivatives in distributions: The differentiation of the integral from (24), the summation and a simple estimation give: By Lemma A1 for exponents p ∈ (1, 3] and q > 3/2 where 1 q = 1 p − 1 3 , we have: The right-hand side of (29) is bounded upper by a constant C = C(ν, ϕ, T, p). Here, we apply inequalities from Lemma 6, Lemma 8 and Lemma A5. Therefore, Derivatives Thus, D t ∇P n ∈ L q (R 3 ) for any exponent q > 3/2. By Lemma A1, we obtain: Consider two cases: 1 < p ≤ 3/2 and 3/2 < p ≤ 3. Let 1 < p ≤ 3/2 . Then, the right-hand side of (31) is bounded by a constant C = C(ν, ϕ, T, p). This follows from estimate 4 of Lemma 12. Let 3/2 < p ≤ 3. Then, the exponent 6p/(6 − p) ∈ [2,6]. Applying Hölder's inequality, we get The first factor is estimated uniformly by a some constant C = C(ν, ϕ, T, p). This is proved by application Lemma A5, Lemmas 6 and 8. The second factor is estimated by inequality (2) from Lemma 7. Hence, for an exponent q, q > 3, 1 Applying the integral representation for derivative D t P n in the same way we prove another uniform estimate D t P n q ≤ C for every exponent q, q > 3.

Lemma 13.
Let T 0 be a constant from Lemma 1. Let P n be a function defined by (24). Then, on every segment [0, T], T < T 0 , with some constants C 1 = C(ν, ϕ, T), C 2 = C(ν, ϕ, T, q), there are fulfilled uniform estimates with respect to t ∈ [0, T] and n: ∇D t P n q ≤ C 2 for every q ∈ (3/2, 3]; (4) ∇D t P n q ≤ C 2 D t P n 2 , P n q ≤ C 2 , D t P n q ≤ C 2 for every q > 3.

Lemma 14.
Suppose that T 0 is the constant from Lemma 1. Let P n be a function defined by (24). Then, on every segment [0, T], T < T 0 , with some constants C 2 = C(ν, ϕ, T, q), there are fulfilled uniform estimates with respect to t ∈ [0, T] and n: for every q ∈ (3/2, 3] and every pair of numbers k, m = 1, 2, 3. Proof. These estimates follow from Lemma A1, Lemma 12 and integral representations for derivatives extracting from (24). Apply Lemma A1 and item (1) of Lemma 12. Then, we obtain the first inequality.
In the same way, we get the second inequality with an application of estimates (2) and (3) from Lemma 12.

Proof.
It is sufficient to repeat the proof of Lemma 11 with the application of formula (24).

Lemma 16.
Let T 0 be a constant from Lemma 1. Supposing that P n is the function defined by (24), then Proof. This follows from proposition A3.

Estimates of Uniform Continuity of Approximations in Spaces L 2 (R 3 ) and C(S T )
Now, we estimate the integral continuity modulus of gradients and Laplacians for approximations following [7]. Let T 0 be a constant from Lemma 1. Let T, T 1 be arbitrary numbers such that (16) we write by the following form: Every equality we multiply by difference c qn (t + h) − c qn (t) respectively and add together them.
To the scalar product on the right-hand side, we apply Cauchy-Bunyakovskii's inequality. The integral (J is its mean) we estimate by Corollary A4 for the triple v n , v n , z. Then, Every factor from the right-hand side of these inequalities is bounded by a constant C = C(ν, ϕ, T 1 ) uniformly with respect to t, n, h. This follows from estimates of Lemmas 6-8, definition of z and the choice of means h, T 1 . Since then we get inequalities: Integrating it over segments [0, h] if h > 0 and [h, 0] if h < 0 in any case we have: ∇z 2 2 ≤ 2C|h|. Thus, the following statement is proved.

Lemma 17.
Let T 0 be a constant from Lemma 1. Let T, T 1 , T < T 1 < T 0 be arbitrary but fixed numbers. Then, there exists a constant C = C(ν, ϕ, T 1 ) such that, for all approximations v n , there is a fulfilled inequality: Lemma 18. Let T 0 be a constant from 1. Let T, T 1 , T < T 1 < T 0 be arbitrary but fixed numbers. Then, there exists a constant C = C(ν, ϕ, T 1 ) such that for all approximations v n there is fulfilled inequality: Proof. Formulae (16) and (32) yield equalities: . Every equality we multiply by factor c qn (t + h) respectively and add together them. Furthermore, in the second term, we replace differentiation on variable t by differentiation on variable h. Hence, we obtain: Here, L 1 , L 2 are integrals from the first and the second products sums, respectively. Hence, The scalar products on the left-hand side of (33) are bounded uniformly. This follows from estimates of Lemmas 6-8. Therefore, we have: with some constant C = C(ν, ϕ, T 1 ). A uniform boundedness of integrals L 1 , L 2 follows from Corollary A4. For the verification, we take mappings triples z, v n (t + h, ·), D t v n (t + h, ·) and v n (t, ·), z, D t v n (t + h, ·), respectively. Finally, applying estimates from Lemma 6, Lemma 8 and Lemma 17, we obtain: Assume h > 0 without restriction of the generality. Then, from (35) after Hölder's inequality application and inequality (2) of Lemma 8, we get: where C 1 is a new constant. From (36) in the same way, we obtain another estimate: Integrating (34) and, gathering last estimates, we get lemma inequality.

Lemma 19.
Let T 0 be a number from Lemma 1. Let T, T 1 , T < T 1 < T 0 be arbitrary but fixed numbers. Then, there exists a constant C = C(ν, ϕ, T 1 ) such that for all approximations v n and P n there are fulfilled inequalities: Therefore, one should find uniform estimates for every modulus on the right-hand side considering mappings v n , P n . From representation (A2), it follows: |v To every integral, we apply again Hölder's inequality. Then, The second representation in (A2) and Lemma A1 yield estimate: Therefore, previous inequalities and estimates from Lemma 17 and Lemma 18 give formula: where C is a constant depending on ν, ϕ, T 1 only. Let us estimate the second modulus applying Poisson's formula (see (A1)). Then, From the inequality, with some constant C 1 , we obtain: Previous estimates and Lemma 8 (estimate (1)) yield: where a constant C depends on ν, ϕ, T 1 only. Thus, the first estimate follows from (37) and (38).
In the same way, we prove an inequality of the kind (38) for the function P n (formula (24)). The norm ∇v n 4 , which appears after applying Holder's inequality, we must estimate by Lemma A5. Then, ∇v n 4 ≤ ∇v n 1/2 2 ∇v n 1/6 6 .
Furthermore, Lemma 6 (estimates (1), (2)) and lemma 8 (estimate (1)) yield the inequality ∇v n 4 ≤ C, where C = C(ν, ϕ, T) is some universal constant. Then, it follows: is represented in the following form: To obtain this formula, we change summation index for a separate terms (use (24)) and apply Hölder's inequality for three factors and two factors. We make estimates separately on a ball |y − x| ≤ 1 and its In the last case, as the first step, we make a simple estimate, thereupon, we apply Hölder's inequality. The analogous arguments that are used above for the proof of the first estimate in lemma and formula (39) yield the inequality: where C = C(ν, ϕ, T 1 ) is some constant depending on ν, ϕ, T 1 only. Uniform estimates (39) and (40) prove the second inequality of lemma.

Weak Limits Properties of Approximation Sequences
Lemma 20. Let T 0 be a number from Lemma 1 and T < T 0 be a positive number. Then, the sequence of mappings (v n ) n=1,... defined by (14)- (16) is bounded in the space W 1 6 (S T ) and the sequence (P n ) n=1,... constructed by formula (24) is bounded in spaces W 1 q (S T ), q > 3.
Proof. Estimate (2) from Lemma 6 and estimate (1) from Lemma 8 yield inequality ∇v n 6 ≤ C. It is fulfilled with some constant C whenever n and t ∈ [0, T]. For all mappings v n , D t v n integral representation (A2) is true. Then, by Lemma A1, we obtain: From inequalities (1) of Lemmas 6 and 8, we conclude that there exist constants C 1 , C 2 , such that v n 6 ≤ C 1 , D t v n 6 ≤ C 2 . All norms are uniformly bounded with respect to t. Hence, the sequence (v n ) n=1,... is bounded in W 1 6 (S T ). Uniform boundedness of these norms P n q , ∇P n q , D t P n q , q > 3, with respect to t and n follows from Lemma 13. Therefore, the sequence (P n ) n=1,... is bounded in spaces W 1 q (S T ).

Remark 2.
The spaces W 1 6 (S T ), W 1 q (S T ) are reflexive. Hence, every bounded set from it is a weakly compact set (see [? ]). Then, by Lemma 20, sequences (v n ) ... , (P n ) ... are bounded in these spaces. It is possible to extract a weakly converging subsequences from its. Let be weak limits of these subsequences. Without restriction of generality, we assume that these subsequences converge to the own weak limits on every compact set of S T . This follows from Arzela's theorem and Lemma 19.

Lemma 21.
Let u and P be weak limits from (41). Then, (1) mappings u and P are uniformly continuous on a set S T , T < T 0 , moreover, u(0, x) = ϕ(x); (2) mappings u and P are bounded on a set S T ; (3) the mapping u ∈ W 1 6 (S T ) and there exists a constant C = C(ν, ϕ, T) such that following inequalities are true: (22), a constant C = C(ν, ϕ, T); (5) u has distributions of the second and third orders: u , ij , D t u , ij , in addition, for all t ∈ [0, T], there are fulfilled inequalities: the function P ∈ W 1 q (S T ) for every q > 3, in this case, there exists a constant C = C(ν, ϕ, T, q) such that, for all t ∈ [0, T] estimates P q ≤ C, D t P q ≤ C are true; (7) there exist constants C i = C i (ν, ϕ, T, q) such that ∇P q ≤ C 1 for every q > 3/2 and ∇D t P q ≤ C 2 for every q ∈ (3/2, 3]; (8) the function P has distributions of the second and third orders: P , km , P , kmj , D t P , i , in addition, there exists a number C = C(ν, ϕ, T, q) such that, for all t ∈ [0, T], the following inequalities hold: P , km q ≤ C, D t P , i q ≤ C for every q ∈ (3/2, 3] and P , kmj q ≤ C, for every q ∈ (1, 3/2]. Proof. Property (1) follows from Remark 2. A uniform continuity follows from Lemma 19 and a uniform convergence of subsequences (v n k ) k=1,... and (P n k ) k=1,... on compact subsets of S T . Property (2) follows from a uniform convergence on compact sets, Lemma 9 and Lemma 13 (item (1)).
Property (3) follows from norm semicontinuity of a weak limit in reflexive spaces. Property (4) follows from Lemma 11. A uniform boundedness of norms v n , ij 2 (see Lemma 8 and Lemma 11) and norms boundedness D t v n , ij 2 in the space W 1 2 (S T ) (see Lemma 8 and Lemma 11) guarantee an existence of distributions u , ij , D t u , ij . Estimates of its norms follow from a semicontinuity of a weak limit norm.

Lemma 22.
Weak limits from (41) satisfy equalities: Proof. The first equality is fulfilled for mappings v n and P n . The sequence (∇v n ) n=1,... is bounded in the space W 1 2 (S T ). In addition, estimates of norms ∇v n 2 , v n , ij 2 , D t v n , i 2 are uniform with respect to t and n (see . Apply Sobolev-Kondrashov's embedding theorem (see [23], pp. 83-94) to the sequence (∇v n ) n=1,... . As a bounded set, it is embedded in the space L q ([0, T] × Ω) for every ball Ω ⊂ R 3 . An exponent q satisfies condition In this case, a dimension of spatial domain [0, T] × Ω) m = 4. Thus, we can assume that a subsequence (∇v n k ) k=1,... converges strongly to a mapping ∇u in the space L q ([0, T] × Ω), q < 4, for every ball Ω ⊂ R 3 . Denote the integral from the first equality of the lemma by Q k (t, x).
we deduce: Multiply this inequality by |η| where η ∈ C 0 (S T ) an arbitrary test-function. Thereupon, integrate over the set S T and change integration order. Then, where I 2 is the Riesz potential, K 1 is the interior integral calculating over ball |y| < r, and K 2 is the interior integral calculating over exterior of this ball. Estimate every integral applying Hölder's inequality. Thus, we have The second and the third factors on the right-hand side we estimate by constants independent of t and n (see Lemmas 6,8,21 with conditions (3)-(4) and Lemma A5). A radius r is fixed so that the first factor is less an arbitrary positive number ε. Then, K 2 ≤ Cε. Integral K 1 we estimate on a subsequence. Then, The second and the third factors are uniformly bounded by a some constant C. Therefore, the inequality: is fulfilled. The middle factor is not greater ε if a number k is large enough. This follows from condition of a strong convergence on a bounded set. Combining all estimates above, we obtain the inequality This means that d n k → 0 weakly because a function η is an arbitrary. The first equality is proved. The second equality is proved in the same way. Consider the difference d n = v n , j − R j where R j is the integral of the second equality. In the integral d n (t, x), we replace the variable by y = x + z. Thereupon, we multiply the equality by a test-function η ∈ C 0 (S T ) and integrate its over set S T . Change integration order and carry over Laplace operator to function η. Then, Replace variables in the interior integral by x = y − z and change integration order. Hence, we get: Integration with respect to y we make separately over ball |y| < r and its exterior. The furthest arguments are conducted in the same way as above. A distinction in the following. In this time, we use an uniform convergence of a subsequence (v n k ) k=1,... on compact sets (see Remark 2).

Weak Solutions and Gradients Boundedness
Lemma 23. Let u and P be weak limits from (41). Then, for every solenoidal vector field ψ ∈ C ∞ 0 (R 3 ) and almost everywhere t ∈ [0, T], there is fulfilled integral identity: Proof. Equalities (16) multiply by a test-function η ∈ C ∞ 0 ([0, T]) and integrate its over segment [0, T] . If a subsequence (v n k ) k=1,... converges weakly, then, for all q = 1, . . . , n k , we have Fix a some number q . Then, the passage to the limit gives the equality This is explained by a weak convergence of a sequence (v .. to the mapping u i u , i . It is given by support compactness of a vector field a q , by uniform boundedness with respect to t and n of norms ∇v n p , 2 ≤ p ≤ 6, and a uniform convergence of subsequence (v n k ) k=1,... on compact subsets of S T . A function η is an arbitrary. Therefore, from (42), we obtain (D t u, a q ) − ν( u, a q ) + u i u j, i a q j dx = 0.
It is already fulfilled for every natural number q. The construction of vector fields a q permits this integral identity to extend on elements of the fundamental system (ψ n n=1,... ) (see (12) and (13)), i.e., We show that identity (43) is true for every solenoidal vector field ψ ∈ C ∞ 0 (R 3 ). Let (ξ m ) m=1,... be a sequence of a finite linear combinations of mappings ψ n , which converges to a vector field ψ ∈ and equality (43) for mappings ξ m is true. Mappings u, u i u , i belong to the space L 2 (R 3 ) for a.e. t. Then, as the same condition is. Consider the equality of scalar products and note that the right side tends to (D t u , j , ψ , j ) (see Lemma 21 item (4)). On the other side, −(D t u , j , ψ , j ) = (D t u , ψ). Condition (43) is true for an arbitrary ψ ∈ C ∞ 0 (R 3 ). From (∇P, ψ) = 0, we have the lemma.

Lemma 24.
(see ([7], pp. 41-44), see also [29].) Let B ⊂ R 3 be an arbitrary ball. Then, a space L 2 (B) of any vector fields has a decomposition by a direct sum L 2 (B) = G(B) ⊕ J 0 (B) of orthogonal subspaces. A subspace G(B) is the space of gradients ∇g where g : B → R is locally square-integrable function with a finite norm ∇g 2 . A space J 0 (B) is the closure with respect to the norm L 2 (B) of all solenoidal vector fields from the class C ∞ 0 (B) .

Lemma 25.
If u and P are weak limits (41), then there are fulfilled equalities: a.e. on a set S T for any T ∈ [0, T 0 ).

Proof. Let
Every vector field h p , p = 2, 3, 6, belongs to the space L p (R 3 ) (see Lemma 21). Mappings norms h p are bounded by constants independent of t ∈ [0, T]. From the first equality of Lemma 22, we gather (H, ∇g) = 0, where g ∈ C ∞ 0 (R 3 ) is an arbitrary. We assume the mapping H and its generators h p belong to the class C ∞ (R 3 ). Otherwise, we take averages with a kernel from C ∞ 0 (R 3 ) for them. For averages, the equality (H, ∇g) = 0 and the equality of Lemma 23 are kept. This follows from behind an arbitrary choice of a smooth function g and a field ψ ∈ C ∞ 0 (R 3 ). Then, div H = 0. Moreover, a smoothness H and the equality of Lemma 23 imply ( H, ψ) = 0. From Lemma 24 on every ball B ⊂ R 3 , we have H = ∇h. A function h is infinitely smooth. This is given by smoothness H . Then, div H = h. On the other hand, div H = div H = 0. Therefore, the function h is a harmonic function. Hence, and from above, there is 2 H = 0. By Lemma A7, we have H = 0. Making an average parameter tending to zero, we obtain this equality in the general case.
Proof. From Lemma 25, we conclude that Laplacian u is the linear combination of three vector fields ∇P, D t u, u i u , i . Coordinates u i are bounded on the set S T by Lemma 21 item (2). Then, from Lemma 21 (see estimates (3) and (6)), it follows the first part of the lemma. Gradients boundness ∇u i we obtain from the second integral representation of Lemma 22 and estimate u 6 ≤ C. In the next step, we repeat the proof of Lemma 9. Gradients boundedness ∇P we get from the first integral representation of Lemma 22 and gradients boundedness ∇u i with repeating of the proof from Lemma 9.

Remark 3.
The proof of this result relies on Calderon-Zygmund's theorem and a boundedness of singular integral operators of parabolic type (see [30]).

Remark 4.
Norms D α ∂ j u ∂t j are bounded by a constant that depends on exponents p, q, derivative order and the mixed norm u p, q . It follows directly from the proof of the theorem in [13].
Proof. Let T < T 0 be a positive arbitrary number. Integrate the equality of Lemma 25 over segment [0, t] where t < T. Then, continuity and absolute continuity on lines of mapping u give: Every integrable term has finite norms u 2 , ∇P 2 , u i u , i 2 .
In addition, every norm is bounded by a constant C = C(ν, ϕ, T) depending on ν, ϕ, T only. It follows from Lemma 21 (see estimates (5) and (7)) for the first and the second norms. A boundedness of the third norm follows from mapping boundedness u (see Lemma 21 item (2)) and the estimate from item (4) (see Lemma 21) . Therefore, u 2 ≤ C. A boundedness of vector field u (see Lemma 21 item (2)) gives a uniform estimate u p ≤ C whenever p ≥ 2. Then, any mixed norm u p, q is finite whenever p, q from lemma condition.
Proof. Let initial data ϕ ∈ C ∞ 6/5, 3/2 . Function f from Lemma 27 is represented by integral For ϕ ∈ C ∞ 6/5, 3/2 , there is true Lemma 34. Therefore, the mapping f and any of its derivatives have a finite mixed norm · p, q . By Lemma 25 and Lemma 29, the vector field u is a weak solution of problems (1) and (2) with a finite mixed norm u p, q whenever p, q ≥ 2. Then, from Lemma 27, we conclude that u is a solution of integral Equation (44). From Lemma 28, we obtain a finiteness of mixed norms for the second derivatives D α u p 1 , q 1 , where p 1 = p/3, q 1 = q/3, |α| = 2, j = 0. Let p = 18/5, q = 12. Then, we have the statement of the lemma.
Proof. Note that weak solutions satisfy conditions: From the first equality of Lemma 22, we have: Integrals commutation is possible since the integral over R 6 is a finite. It follows from Tonnelli's theorem, boundedness and summability of u, ∇u with any exponent not less than two and Lemma A1. Here, I 1 is the Riesz potential. The interior integral in (46) is equal to zero since for any radius r. Let us prove the second equality from (45). The second equality of Lemma 22 implies: Integrals commutation we prove in the same way. There is inequality: The right-hand side is a finite because u ∈ L p , 2 ≤ p ≤ 6 (see Lemma 21 item (5), Lemma 26 item (1), Lemma A5). In addition, I 1 (|u| 2 ) ∈ L p , p > 3/2 by Lemma A1. To interior integral in (47) we apply the Stokes formula. Then, A product u i u k belongs to the space W 1 p (R 3 ) whenever p > 1. Then, the integral over surface tends to zero as r → ∞ (to apply Lemma A4 with exponent α = 1 and a mean p, close to unit). Hence, and from (47) we have: In the iterated integral |y|<r u k (t, y) we change integration order because the double integral is finite (see above). Hence, we get: The interior integral in the right-hand side of (49) is uniformly bounded with respect to r > 1. This follows from a boundedness of the Riesz potential I 2 (| u|). It is proved in the same way as Lemma 9 with applications Lemma 26 item (1) and Lemma 21 item (5). Furthermore, we use Lebesgue's theorem. Then, (48) and (49) give the equality of iterated integrals: The mapping u ∈ J 2 0 (R 3 ) (norm defined by (15)). Lemma A1 shows that Poisson's formula is true for elements of the space J 2 0 (R 3 ). Then, I 2 ( u) = −u. Therefore, we have J 2 = −J 2 from (50). The second equality from (45) is proved.
Let us show that vector field u satisfies the equality a.e. on [0, T]. We have the equality of iterated integrals: A finiteness of double integral follows from a boundedness ∇u (see Lemma 26,item (2)) and properties of the Riesz potential I 1 (| u|). Let r → ∞. The interior integral on the left-hand side of (52) tends to 4πu k (t, x) in the space L 6 (R 3 ) for almost every t. (See Lemma A1 and equality (A2), which is true for elements of the space J 2 0 (R 3 ) ). The norm u 6/5 is finite a.e. by Lemma 30. In (52), we make the passage to the limit. The interior integral on the right-hand side of (52) is replaced by application of Lemma 22. Then, we get (51). To finish the proof, we are helped with the following steps. Every equality from Lemma 25 we multiply by function u k . Thereupon, we add together them and integrate over space R 3 . From (45) and (49), we have Hence, we get the required equality.
1. Assume that initial data ϕ ∈ C ∞ 6/5, 3/2 and its Laplacian support is a compact set. Let T 0 be a constant from Lemma 1. Then, item (3) follows from Lemma 26, and items (1) and (4) we get from Lemma 21. Let us prove estimates of item (2). A uniform boundedness with respect to t of norms u 6 , ∇u 6 , D t u 6 we obtain from Lemma 21 (item (3)). An uniform boundedness of norms ∇u 2 , ∇D t u 2 , u 2 , ∇P 2 follows from Lemma 21 (see items (4), (5), (7)). The estimate of norm u 2 follows from Lemma 31. A uniform boundedness of norms u , ij 6 we get by Lemma 26. A uniform boundedness for norm D t u 2 is the corollary of Lemma 25 because D t u is the finite linear combination of terms with uniform bounded norms in the space L 2 (R 3 ). Uniform estimates of norms in spaces L p (R 3 ), 2 < p < 6 we take from Lemma A5. The occurrence of vector field u in spaces W 1 2 (S T ) and W 1 6 (S T ) we get from the uniform estimates proved above. By Lemma 25 and Lemma 21 (see items (5) and (7)), we obtain Hence, it follows a finiteness of norm D 2 tt u 2 since u and ∇u are bounded. Therefore, u ∈ W 2 2 (S T ) . Let us prove item (5) using mixed norms (see [8,26,27]). Weak solutions u and P belong to class C(S T ) (see item 1) of this theorem). In Lemma 28, we put p = q assuming it is very large. Now, we fix an order of derivatives: m > 1. Then, by Lemma 27 and Lemma 28, derivative norms D α D j t u , |α| ≤ m are bounded in the space L r (S T ) where an exponent r ≥ 6 is an arbitrary but fixed. A boundedness of weak solution u and its summability in L 2 (S T ) imply the belonging u ∈ L r (S T ), r ≥ 2. Exponents means r, p = q we choose by large numbers so that the next conditions are fulfilled: (1) for any ball lying in S T , all conditions of Sobolev's embedding theorem in a space of continuous functions are certainly valid ( [23], p. 64); (2) at least, all derivatives of the order up to m − 1 satisfy also all conditions Sobolev's theorem from above.
Since an integer number m is an arbitrary, then a weak solution u belongs to the class C ∞ ((0, T 0 ) × R 3 ). A smoothness of function P we obtain from Lemma 25 and the smoothness of vector field u. The continuity is proved in item 1.

Direct calculations yield:
Without the second term in the first integral, the rest of the integrals of all terms in the right-hand tend to zero as r → ∞. This is guaranteed by a test-function η and a boundedness of the second derivatives Q , jk . The last follows from representation of function Q by Poisson's integral and definition of the class C ∞ 6/5, 3/2 . In this case, we have two equalities: where c kj are universal constants, and T kj are singular integral operators. Therefore, as r → ∞, then A vector field Φ r ∈ C ∞ 6/5, 3/2 . A summability of the vector field and its derivatives follows from (53) and (54), the equality . Laplacians supports Φ r are compact sets. Therefore, there exist solutions u r and P r with an initial data u r (0, x) = Φ r (x) satisfying theorem with the number From (55), we have T 0 (r) → T 0 as r → ∞. Fix a number T < T 0 . From the remark at the beginning of the proof, we conclude all estimates of the theorem for solutions u r , P r . They are uniform with respect to r for r > r 0 . Hence, sets of mappings (u r ) r>r 0 , (P r ) r>r 0 are bounded in spaces W 1 2 (S T ) and W 1 6 (S T ). Extract subsequences (u r k ) k=1,... , (P r k ) k=1,... , which converge weakly. Let u and P be its weak limits, respectively. These limits satisfy the next properties: (1) Lemma 21 is true for them (this is verified in the same way as the proof of Lemma 21 for subsequences); (2) Lemma 25 is true for them; (3) Lemma 26 is fulfilled for them. Thus, u and P are weak solutions of problems (1) and (2). Lemma 27 and Lemma 29 are true for vector field u. Conditions of growth for a mapping ϕ ∈ C ∞ 6/5, 3/2 show correctness of Lemma 28 for weak solutions from above.
Furthermore, we realize the proof from the first part (see item (1) above). Therefore, the theorem is true also in this case. Theorem 1 is proved.
Proof of Theorem 2 (it is given below) is relied on for the next simple properties of mappings v ∈ C ∞ 6/5, 3/2 . For every vector field v and its derivatives of the first order, these are true for both (A1) and representation (Riesz's formula): The second equality we obtain by application of the Stokes theorem to the integral from (56) calculating over a spherical layer ε ≤ |y − x| ≤ r. From Lemma A4, . Then, the passage to limit as r → ∞, ε → 0 implies the second equality (56). The first equality is proved in the same way. We where I 1 is the Riesz potential from (4). Hardy-Littlewood-Sobolev's inequality (see Lemma A1) implies where 1 q = 1 p − 1 3 , 1 < p < q. Consider only p ∈ (1, 3). Two last estimates yield ∇v ∈ L q (R 3 ) for every q > 3/2. Analogously with the above, we show for the mapping v and a number q ∈ [3/2, 3) the belonging v ∈ L r (R 3 ) whenever r ≥ 3. The logarithmic convex of norm v p and Lemma A5 yield norm finiteness v p for p ≥ 6/5. Thus, we proved the next statement.

Remark 5.
Write Poisson's formula (the representation by Riesz's integral I 2 ) for mappings v, ∇v and D α v. Then, we have a boundedness of every vector field v ∈ C ∞ 6/5, 3/2 and its derivatives. Let (the repeated index gives summation).

Lemma 33.
Let v ∈ C ∞ 6/5, 3/2 and div v = 0. Then, the function P and all its derivatives belong to the space L r (R 3 ) whenever r > 1.
Proof. The integral from (57) we integrate by parts twice over a spherical layer ε ≤ |y − x| ≤ r. Lemma A4 and the passage to the limit as r → ∞, ε → 0 imply: where T ij is a singular integral operator with a kernel Lemma A1 and well-known Calderon-Zygmund's theorem give a summation of function P for any finite exponent r > 1 . Since then, analogously with the above, we get: Hence, we obtain a summability of ∇P whenever finite p > 1. A summability of the other derivatives follows from equalities:
Proof. Apply the second representation from Lemma 34 and make the commutation of integrals. Then, (see the first equality of Lemma A4). Changing of integration order is possible because the integral is a finite. Really, we have Then, a finiteness follows from a summability of the Riesz potential I 1 (|∇v|) with exponent 3 (see A1) and the summability of w with exponent 3/2. The first equality is proved. To prove the second formula, we observe a finiteness of integrals |w(x)|| P(y)|dxdy |x − y| 2 , Thereupon, we have: Proof of Theorem 2. Items (1), (2) and (3) from Theorem 1, Lemma 27 and Lemma 28 yield a finiteness of mixed norms D α u p 1 , q 1 where p 1 = p |α|+1 , q 1 = q |α|+1 , whenever p, q ≥ 2, 3 p + 2 q ≤ 1. For derivatives of the second order, in particular, we have a finiteness of norm u 6/5, 4 (see 30). Integrate (1) over segment [0, t] where t ≤ T < T 0 . The solution P is represented by (57). Then, from (59), we get Estimate norms in L 6/5 of every term in (61) in the usual way. We apply Hölder's inequality to interior and exterior integrals. Then, The singular integral operator T ik is bounded. Hence, from (61)-(63) and item (2) of Theorem 1, we obtain a uniform estimate of norm u 6/5 with respect to t ∈ [0, T].
The solution P is represented by (57) with replacing v by u. A summability follows from Lemma 33 whenever r > 1 . Equalities (58) and (59) and a uniform boundedness derivatives norms of vector field u prove a uniform boundedness of norms D β P r where r > 1. From (1) and proved uniform estimates from above, we have necessary statement for derivative D t u. Theorem 2 is proved.

Basic Parameters and Extension of the Cauchy Problem Solutions
Now, we define two from three basic parameters. They have a key part for an extension of the Cauchy problem solutions as solutions with initial data from the class C ∞ 6/5, 3/2 . A functional l(ϕ) and the first parameter λ we define by where the constant a 1 from Corollary A4. By Theorem 2, the solution of the Cauchy problem with condition ϕ ∈ C ∞ 6/5, 3/2 can be extended as the solution in any time t. Moreover, extended solutions keep uniform estimates of all norms from Theorem 2 on extended segments [0, T] ⊂ [0, T * ). In other words, the class C ∞ 6/5, 3/2 is kept. If [0, T * ) is the maximal interval of solution existence, then the second parameter is defined by: where T 0 from (5). The third parameter ε is defined below by (87). Proof. Let s ∈ (0, T * ). Fix t ∈ (s, τ 1 (s) where the function τ 1 from Lemma 45. Choose a segment [T, T 1 ] ⊂ (s, τ 1 (s)) assuming t, t + h ∈ [T, T 1 ]. Denote z = z(t, h, x) = u(t + h, x) − u(t, x). Take equalities (1) with time argument t + h. Thereupon, we multiply them by z getting the scalar product and integrate over R 3 . The derivative D t u ∈ C ∞ 6/5, 3/2 for all t ∈ [0, T * ). It follows from Theorem 2. Then, by Lemma 35 (the scalar product in L 2 we write as ( f , g)), we have:

Solutions Extension in Global with
Here, the right-hand side is bounded uniformly (see Theorem 2). Then, ∇z 2 2 = O(h) as h → 0. Triangle inequality implies the continuity of function η 1 . We write equality (1) for time arguments t, t + h and subtract it. Thereupon, the difference we multiply by D t u(t + h, x) getting the scalar product and integrating over the whole space. As a result, we have Uniform estimates from Theorem 2 and an integrability for any exponent a.e. imply the equality: Then, we have the continuity of function η 2 . Function continuity of η 3 follows also from uniform estimates of Theorem 2. Difference η 3 (t) − η 3 (t 0 ) is considered as the sum of three integrals with combinations: Every integral we estimate by Hölder's inequality so that there appear norms: u i (t, ·) − u i (t 0 , ·) 6 , u k, i (t, ·) − u k, i (t 0 , ·)) 2 , u k (t, ·) − u k (t 0 , ·)) 2 .
Proof. Suppose the opposite. Then, on interval [0, T 0 ), the inequality holds: Integrate it over this interval. Since Take out a nonpositive term on the right-hand side and input the mean T 0 from (5). Then, We have a contradiction with the condition.
Proof. If T < T 0 , then the statement of theorem follows from Theorems 1, 2 and Lemma 37. A finiteness of mixed norms u p, q we get from a boundedness of the vector field u and estimates u 2 ≤ ϕ 2 . Solution uniqueness in the class L p, q , 3 p + 2 q ≤ 1 is proved in [7,8,26] (see also [13]). Norm monotonicity ∇u 2 as a function on time argument t follows from condition η < 0 (see proof of Lemma 37). Let [0, T * ) be an interval of the maximal length such that there exist solutions with the estimates of Theorem 2. Suppose T * < ∞. Let t 0 < T * and T * − t 0 < 0, 5T * . By Theorem 2 mapping, u(t 0, ·) belongs to class C ∞ 6/5, 3/2 . Therefore, by Theorem 1 with this initial data, there is the unique solution w of the Cauchy problem that can be built that can be considered as the extension of solution u (see Lemma 36 and Theorem 1). Extension of u is the unique solution of problems (1) and (2) that satisfies the theorem, at least, on the interval [0, t 0 + T 2 ), where We have T 2 ≥ T 0 from condition ∇u 2 ≤ |∇ϕ 2 , w(t, x) = u(t 0 + t, x) for means t < T * − t 0 . Hence, the solution u is extended with the half-interval [0, T * ) on an interval of more length [0, T * + 0, 5T 0 ). We have a contradiction. By Theorems 1-3, there exists a global solution w of problems (1) and (2) with changed initial data w(0, x) = u(t 0 , x). This is the unique smooth extension of solution u that satisfies the proving theorem. Assume the theorem is true for a some natural number m. That is, every solution u has a global extension with properties of the theorem if, for this u, there exists a number t 0 ∈ (0, T 0 ) such that l(u(t 0 , ·)) < q α m ν a 1 2 . Now, we take initial data ϕ such that By Lemma 38, there exists t 0 ∈ (0, T 0 ) satisfying l(u(t 0 , ·)) < q α m ν a 1 2 .
By Theorem 2 and the induction hypothesis, there exists a global solution w of problems (1) and (2) with a new initial data w(0, x) = u(t 0 , x). By a uniqueness theorem, it is the unique smooth extension of solution u that satisfies the proving theorem. By the induction principle, the theorem is proved because a −2 1 as m → ∞.

Critical λ Parameter Mean and the First Hypothetical Turbulent Solution
Furthermore, it is important in principle an invariant form of a priori estimate for the Cauchy problem solution. An invariance follows from Lemmas 1, 6, 20 and 25, Remark 2, norm semicontinuity of ∇u 2 and Theorem 1.
Proof. We construct the extension in the same way as in the proof of Theorem 4.
Proof. Both parts we raise to the second power and integrate over the interval [0, T 0 ). From (72), we get: where the number T 0 from Theorem 1. Since λ = 1, then Therefore, the limit in (74) is equal to zero because, in (74), it must be equalities. This is possible only if, on the interval [0, T 0 ) (see Lemma 36), it is fulfilled: Integrate (72) over the interval [0, T 0 ). As the result, we have: (we take into consideration in formula (74) the limit vanishes ). Apply the estimate of Lemma 39. Then, Hence, we have the inequality λ ≥ 1. Since λ = 1, then the inequality from Lemma (71) must be as the equality. The second formula of lemma is proved. The first follows from (75) and condition λ = 1. The last statement of lemma we prove from the opposite in the same way.
Proof. If such solution exists, then, from (73), we obtain Here, the identical equality is impossible because, for any solution u, the inequality (see (7)) is fulfilled: Apply estimates from the proof of Lemma 1. Then, we obtain: Compare this inequality with the identity above. Therefore, we must have the equalities for intermediate estimates of Corollary A4 and Lemma 1. Since we used Cauchy-Bunyakovskii's inequality in the Hilbert space L 2 (R 3 ), then there exists a constant c such that for any i, j = 1, 2, 3. Hence, we have u i, j = u j, i for each pair i, j and u ≡ 0, respectively. From Lemma A6, it follows u ≡ 0-a contradiction. The lemma is proved.
Proof. Assume λ = 1. Then, the statement follows from Lemmas 42 and 43. Let λ < 1. Suppose the opposite. Then, we have: Hence, and from (72), we get Let (79) Integrate (78) over segment [0, t]. Then, we obtain: Make the passage to the limit in (80) as t ↑ T 0 and compare the new estimate with the inequality from lemma condition. Taking (5), (68) and (80), we conclude lim t→T 0 u 2 It vanishes at boundary points of [0, T 0 ]; moreover, β ≤ 0 (see (80)). Let I ⊂ (0, T 0 ) be an interval, where the function β vanishes at boundary points and β < 0 on its interior. Then, there exists a point t 0 ∈ I, where β (t 0 ) = 0. Hence, from (72), we get: From (80), we have: Compare the left and right sides of this formula and, after we apply (79). Then, l(u(t 0 , ·)) ≤ The hypothesis from proof beginning gives the equality: Therefore, in (81), the inequality must be by the equality. Hence, we get β(t 0 ) = 0. This goes to a contradiction with the choice of the interval I. It implies β = 0. Hence, Multiply these equalities. From (78), we obtain: In particular, by Lemma 36, l(ϕ) ≥ 81ν 2 8 . This is impossible with the considering lemma condition. This contradiction proves the lemma. Now, we shall study properties of unextended solutions of problems (1) and (2) if such solutions exist. Let [0, T * ) be an interval of the maximal length, where solutions u and P of problems (1) and (2) have properties from Theorem 2. Then, T * ≥ T 0 and T * = µT 0 (see (69)). Hence, µ ≥ 1. Therefore, J. Leray's estimate from [3] can be given in invariant form in the following statement.
Lemma 47. Let ϕ ∈ C ∞ 6/5, 3/2 and l(ϕ) > 81ν 2 8 . Suppose that [0, T * ) is the maximal interval where solutions u and P of problems (1) and (2) have solution properties from Theorem 2. If this interval is a finite, then the following estimate holds: This solution gives the unique extension u on the interval [t 0 , t 0 + l), where l ≥ 9 4 ν 3 /(4M 4 and M is the supremum of η 1 . A point t 0 is an arbitrary, therefore, solutions u and P can be extended on the interval [0, T * + l). In addition, they have solutions' properties from Theorem 2 on this interval. This contradicts the choice of interval with the maximal length. Now, we prove estimate (82). For solution u, we have: which follows from (1). It is true for every mean t ∈ [0, T * ) by Theorem 2 and Lemma 35 because the solution u ∈ C ∞ 6/5, 3/2 . The integral representation from Lemma 34 Replace in (84) s on s m , where s m ↑ T * and η 1 (s m ) → ∞ as m → ∞. The passage to the limit in (85) gives estimate (82).

Proof. Consider a function
From (5) and (68), (72), we have: Therefore, ω(0) ≥ ω(t) ≥ ω(T * − 0). Hence, it follows the first inequality from (86). Integrate over [0, t] the inequality of Lemma (79). From (72), we obtain: Applying (5), (68) and (69), we get: Therefore, we have the second inequality in (86).  Proof. Let t 0 a number from Lemma 46. Without norm monotonicity, the statement of theorem follows from Theorem 2 and Lemma 40. The product u 2 ∇u 2 is a decreasing function on the set [t 0 , ∞). It follows from Theorem 4. Therefore, the theorem is proved. Now, we give one result that is connected with a local solutions' extension. If λ < 1, then we introduce the third parameter ε, which gives a dissipation quantity of a kinetic energy. It is defined by formula: lim We observe from Lemmas 43 and 44 that the parameter ε satisfies strong inequalities: 0 < ε < 1. This is very important for the furthest. The usefulness of this parameter is explained by the following result.

Main Results, Existence of Global Regular Solutions, and Sufficient Conditions
Now, we prove the basic result which is described by Theorem 7.
If parameter λ ≥ 1, then the function l(t) = u 2 ∇u 2 is a decreasing function on the interval [0, ∞) . If λ < 1 and condition (89) is fulfilled then the function l = l(t) is a decreasing function on the interval [T 0 , ∞).
Proof. Let λ > 1. Then, the statement follows from Theorem 4. Let λ = 1. In this case, the theorem arises from Lemmas 42, Lemma 43 and Theorem 4. The monotonicity of the function l follows from Lemma 45. Let λ < 1 and condition (89) is fulfilled. Then, the statement of the theorem arises from Lemmas 45 and 46, Theorem 5. The theorem is proved.
Proof. Let be [0, T * ) ⊆ [0, T) a maximal interval where the weak solution u is regular. Suppose T * < T. Then, from (86) and (90), we have Since solution u is regular on the interval [0, T 0 ), then, by differentiation of the previous identity at point t = 0, we obtain µ = 1. From Lemma 50, we have a contradiction. Therefore, T * = T.
Does there exist weak solution u satisfying opposite inequality (90) if t > T 0 ? It is unknown.

The Cauchy Problem with Less Smoothness of Initial Data
In addition, the invariant class C ∞ 6/5, 3/2 Sobolev space as the closure of infinitely smooth vector fields is another important invariant class, which satisfy existence condition of global solutions.
Different exceptions for solenoidal vector fields from Sobolev classes [31]. Therefore, we consider the first space from them.
Let ϕ ∈ • W 3 2 (R 3 ) be a solenoidal vector field. Set (ϕ m ) m=1,... a sequence of of finite, solenoidal and infinitely smooth vector fields, which converges to the field ϕ in the space . We observe that ϕ m ∈ C ∞ 6/5, 3/2 . Let (u m ) m=1,... , (P m ) m=1,... be sequences of solutions in the Cauchy problem for Navier-Stokes equations with the initial dates ϕ m . Then, all pairs u m , P m satisfy all uniform estimates of Lemma 21 on any compact set of the interval [0, T m 0 ) where T m 0 = 9 4 ν 3 /4 4 ∇ϕ m 4 2 since upper bounds in these inequalities depend on a set and ν, ∇ϕ m 2 , ϕ m 2 . Therefore, on every fixed segment [0, T] ⊂ [0, T 0 ], we can take these constants as common for all u m , P m because ϕ m → ϕ in the space • W 3 2 (R 3 ). Then, without loss of generality, we assume that the sequence (u m ) m=1,... converges weakly in the space W 1 6 (S T ) to a field u 0 . In addition, we suppose that ( u m ) m=1,... , (∇D t u m ) m=1,... and (∇P m ) m=1,... converge weakly in L 1 2 (S T ) to u 0 , ∇D t u 0 and ∇P 0 . More generally, weak limits u 0 , P 0 satisfy all conclusions of Lemma 21 and they are weak solutions of problems (1) and (2). From the equality (1) for couple u 0 , P 0 and items (2), (4), (5), (8) differentiating (1), we obtain that distributions u 0 , j , j = 1, 2, 3, belong to the space L 2 (R 3 ) for almost everywhere t. Thus, the class • W 3 2 (R 3 ) is invariant similar to the class C ∞ 6/5, 3/2 . For this case, in the same way, we can define the basic parameters λ, µ, ε. After that, one should note that the statement of Lemma 36 will be true when initial data ϕ ∈ • W 3 2 (R 3 ). Repeating the proof of Theorem 7, we obtain the following result.
Theorem 10. Let ϕ ∈ W 3 2 (R 3 ) be initial data in problems (1) and (2). If parameter λ > 1, then the solution u from Theorem 9 satisfies: (1) a power of norm u 4 2 is a convex function; (2) there is fulfilled: Proof. It follows from Lemma 45 because this lemma is true for solution u from Theorem 9.

Integral Identities for Solenoidal Vector Fields: Dimensions Comparison
Some review and results about integral identities for solenoidal vector fields are given by authors in [32,33]. Here, we reduce one from these identities, which shows the essential distinction for the Navier-Stokes equations between space and plane.
Let u, v, w : R n → R n be any triple of solenoidal vector fields from the class C 2 0 (R n ). Denote c ki (u) = u k, i − u k, i , k, i = 1, 2, . . . , n.
Lemma 51. (see [32]) For every triple u, v, w : R n → R n of solenoidal vector fields from the class C 2 0 (R n ), the identity holds: (w i, j + w j, i )c ki (v)c kj (u)dx = − w i (c ki (u) v k + c ki (v) u k )dx.
Hence, it follows (one should take u = v = w): u i, j c ki (u)c kj (u)dx = − u i c ki (u) u k dx.
Corollary 1. (see [33].) If dimension n = 2, then every solenoidal vector u ∈ C 2 0 (R 2 ) satisfies the integral identity: Obviously, it implies some interesting applications to the 2D Navier-Stokes and Euler equations (see [32]). (1) We deduce a priori estimate for a solution u, which is not independent of a viscosity: where f is an outer force. This improves essentially Ladyzhenskaya's estimate (see [34]). (2) In the case f = 0, we have formula (83) and, therefore, the norm ∇u 2 is a decreasing function.
We give the new proof of the existence of a global weak solution for the Euler equations in plane in the case when an outer force f = 0. In addition, the estimate ∇u 2 ≤ ∇ϕ 2 is exact and it does not follows from Judovich's results [35]. This explains "the simplicity" of a motion of an ideal fluid on plane.
Remark 6. Let n = 2, f = 0. Then, the product ∇u 2 u 2 is a decreasing function in any case.

Remark 7.
If dimension n = 3, then integral from (91) may be not equal to null.
For a simple example, there is the vector field with the following coordinates: where l i is the i -th vector row of the skew-symmetric matrix. Since with a constant c = 0. A coefficient ∑ j l ki l ij l kj may be not equal to zero for fixed different means k and i because there is the linear independence of polynomials λ 2 i λ 4 k − λ 2 k λ 4 i , i < k, i, k = 1, 2, 3. It gives a distinct from zero of the integral when we choose a suitable skew-symmetric matrix. Respectively, the right side (see (83)) for dimensions n ≥ 3 can be taken with a large value implying a positive mean of the difference R n n ∑ i, j=1 u i u j, i u j dx − ν u 2 2 for t 0. It is possible because we can take a factor for initial data αϕ or diminish viscosity coefficient ν. This implies a growth of the norm ∇u 2 for space. Obviously, on the plane, this phenomena does not appear.

Conclusions
Briefly, the main achievements (see Theorems 7-10) have an obvious physical interpretation and, therefore, it may be interesting for applications. Nevertheless, they are connected with monitoring of blow up.
First of all, no phenomena blow up if parameter λ ≥ 1 or kinetic energy satisfies inequality: u(t, ·) 2 2 ≥ ϕ 2 2 1 − λ 2 t T 0 with condition λ < 1. Finally, we have the importance of the exact lower estimates for kinetic energy of a fluid flow. It is possible that this is one of the new ways where the interesting problem will be studied.
Funding: This research received no external funding.
we apply twice the Stokes formula removing integrals over spherical layer and derivatives of the mapping w. As the result, we have two integrals over sphere | x − y |= ε and two integrals over sphere | x − y |= r. They are: |x−y|=ε w(y)dS ε 2 , |x−y|=r w(y)dS r 2 , (A3) |x−y|=ε w , j (y)(x j − y j )dS ε 2 , |x−y|=r w , j (y)(x j − y j )dS r 2 .
The third and the fourth integrals we transform again applying the Stokes formula and getting integrals over balls | x − y |≤ ε, and | x − y |≤ r, respectively. Every integral must contain Laplacian. Since support of w is a compact set, then these integrals tend to zero as ε → 0, r → ∞. The second integral in (A3) we denote by a symbol I. Then, The Stokes formula application gives the equality: |x−y|≤r (3w(y) + w , j (y)(x j − y j ))dy.
The integral from the first term in (A4) we estimate applying the Hölder's inequality. Then, |I| ≤ 3r −3 w p (σ 3 r 3 ) 1−1/p + 1 2r |x−y|≤r | w(y)|dy, where σ 3 -is the volume of a unit ball. From compactness of Laplacian support and lemma condition, we obtain that integral I → 0 as r → ∞. The first integral in (A3) tends to the mean 4πw(x) as ε → 0. formula (A2) we prove by the same way.
Proof. The first inequality follows from the second representation of Lemma A2 and compactness of Laplacian support. The second estimate follows from the third representation of Lemma A2 because the first estimate from the corollary gives: |w(x)| ≤ C 1 4π R 3 dy |y| 2 |x − y| 2 .
A change of variables y = |x|z proves the second estimate.
Corollary A3. Let v,w : R 3 → R 3 be mappings which satisfy conditions from Lemma A2. Then, R 3 v k w k dy = − v k, j w k, j dy.
Proof. We apply the Stokes formula to the integral from the left side of this equality. From Corollary A2 on a sphere |y| = r, we get the following formula: v k w k, j = O(r −3 ). A passage to the limit as r → ∞ gives the required equality.