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.
During the last century, the Navier–Stokes equations attracted very much attention. The first essential steps in this way were offered by C. Oseen , F. K. G. Oldquist , J. Leray [3,4,5], and E. Hopf . 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 ), O.A. Ladyzhenskaya wrote in her review  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 of solenoidal vector fields vanishing at infinity. Here, outer forces are trivial. Then, the class is invariant (Theorem 2). This is a new result.
The second invariant is a special parameter (see (68)) which is connected with a velocity changing of where E is a kinetic energy of a fluid flow. If or kinetic energy at a special moment is not less any mean depending on for ( i.e., changing of at moment 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.
Finally, the other parameters , (see formula (87)) and , or may also be very useful (see formula (69), Lemma 50). The first of them is a dissipation coefficient of kinetic energy . The last parameter holds a time interval of a solution regularity. These three numerical characteristics are invariant with respect to the scaling procedure.
By the way, the first attempts to estimate invariant norms were implicitly undertaken in [12,16].
An introduction of a special invariant class of vector fields and invariant parameters gives the main idea for the proof of basic results. The first step is connected with a change of the construction offered in . These changes concern solution approximations. The special kind of them gives many uniform a priori estimates. Approximations of a velocity function are built on a fundamental system with a condition for Laplacians of approximative solutions. They must be a finite part of the Fourier series. Simultaneously, approximations of a pressure function are being built. Jointly with a hydrodynamical potential, these approximations give the following facts and properties of local solutions:
solutions are bounded with respect to a uniform norm and therefore it belongs to any class ;
there is a universal time interval where bounded solutions exist;
more exact necessary conditions of a hypothetical turbulence phenomenon if it is;
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 (Section 3 and Section 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 , , are very useful.
The third part (Section 5 and Section 6) contains the proof of main statements (Theorems 7–9), which are based on properties of invariant parameters , , .
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 and, after that, it is done in class . In particular, it is shown that there is a bad solution of a class 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 . Precisely, this step distinguishes from classical way for the second time (see ).
In the considered problem, a boundedness of solutions depends on a smoothness of initial data. At least, initial data from the Sobolev class 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 :
where u is a velocity of flow, P is a pressure function, symbols
indicate a partial differentiation or differentiation in distributions, Δ is the Laplace operator, and is a positive constant (viscosity coefficient). A mapping has all derivatives and satisfies conditions of averaged growth: , . The other derivatives belong to classes for any . Furthermore, this class is denoted by symbol . A class is the class of infinitely smooth mappings with a compact support. A norm in a space is defined by formula:
A mixed norm is defined by equality:
A symbol denotes a partial differentiation or distributions with respect to a multi–index An order of the derivative is indicated by Jacobi matrix of a mapping v with respect to spatial variables is denoted by . Its modulus is
Functions’ properties from the Sobolev classes 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 and is the Euler gamma–function. The properties of these potentials can be found in .
The agreement about summation. Everywhere in this article, the repeated indices give a summation if it is not done reservation specially. For example,
Furthermore, A number we define by formula:
We apply the definition of a weak solution given in  everywhere.
2. 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.
Let be a number from formula (5) and a mapping Then, on the set , there exist weak solutions u and P of problems (1) and (2) with the following properties:
mappings u and P uniformly continuous and bounded on the set for every number ;
the solution u belongs to Sobolev classes and for every number , moreover, all norms
are uniformly bounded in spaces by a constant depending on and T only, in addition ;
gradients are bounded on the set for every number ;
the solution P satisfies uniform estimates:
for all numbers q,, and , with constants C depending on and q only;
solutions u and P are classical solutions that is for any they belong to the class .
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 ). 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 with the finite length. To the studying of the smoothness property for weak solutions, the mixed norms were applied by O. Ladyzhenskaya in  (see, also ). They were applied by other authors (see, for example, [8,10,14]). Item (5) is a particular case from . However, from this theorem, a deeper result follows (see Theorem 7).
2.1. A Priori Estimates of Gradients’ Norms
Suppose that a mapping belongs to a class and If, for every Laplacian supports are subsets of some ball with a fixed radius and , then for all mappings w satisfying condition:
The maximal mean on the right-hand side is Therefore, integrating the inequality
over the interval , we get:
Furthermore, we take a number from Corollary A4 and obtain the required estimate. □
Let be a constant from Lemma 1. Assume a mapping belongs to a class and , . Suppose that, for every t, there are fulfilled conditions:
Laplacian supports , are subsets of a ball with a fixed radius;
with constants , l the inequalities hold:
is true. Then, for every segment, where the estimate holds with a constant which depends on only.
The integral on the right-hand side in formula (8) we rewrite with two integrals and . Applying Corollary A4, we make estimates for every integral. In integral , a triple of mappings is the triple . In integral , a required triple is the triple . Therefore, condition (3) yields estimates:
Let Then, formula (9) can be transformed to the formula:
Let us integrate over segment this inequality. For the next step, we apply to each term the Hölder inequality for three and two factors, respectively getting quantities and . Hence, from condition (3), we obtain:
where Let M be a maximal mean of the function
where . Then, the last estimates give . From the definition of function g, we have . □
2.2. A Priori Estimates of Laplacian Norms
Let w, be a mapping and a number from Lemma 1. Then, for every number there exists a constant such that
for all .
We transform inequality (7) applying the estimate from Lemma 1. Then,
This inequality we integrate over the segment . Then, we estimate the right-hand side applying the Hölder inequality and underlining the integral with the term . If , then we get
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 will be obvious. If , then the estimate is acceptable. If then we have:
Hence, it follows the lemma. □
Let w be a mapping from Lemma 2 and a number from Lemma 1. Then, for every number T, , there exists a constant such that
for all .
For the mapping w, inequality (9) is fulfilled. Its right-hand side we estimate relying on Lemma 2. Then,
This inequality we integrate over segment and its right-hand side we estimate applying the Hölder inequality and underlining terms with norms . 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 , then, from
we obtain the lemma inequality. If then the estimate is acceptable. □
2.3. Basic Space of Solenoidal Vector Fields and Orthogonal Systems
Let us consider solenoidal vector fields from class with a compact support of . A closure of this class is defined by the norm:
We denote its by . From Lemmas A1 and A2, it follows that elements are represented by the Riesz potentials; moreover, . Otherwise, each element is defined uniquely by its Laplacian. The class is a separable space as a subspace of the Sobolev classes . Therefore, there exists a countable system of infinite smooth vector fields satisfying conditions:
supports of are compact sets;
the closure of a linear span in norm (11) coincides with the space .
Now, we apply the Sonin–Shmidt orthogonalization to the fundamental system and construct a countable system of mappings , which would be with the orthogonality property of Laplacians in the space . That is, the scalar product
where is Kronecker’s symbol. Then, every mapping is a finite linear combination of mappings . Therefore, a support of is a compact set. Let
The system is complete for the space that is, the following proposition is true.
If in the space for a some vector field the scalar product for every then .
From chosen mappings the equality for each element of the fundamental system follows. The Stokes theorem gives
The integral over the sphere vanishes as . Actually, from Corollary A2 of Lemma A2, we have:
Furthermore, we apply Lemma A4 taking into consideration a continuity of u. The passage to the limit yields the equality or . We take a sequence of finite and smooth mappings
which converges to the vector field u in the space . Hence, . The summability of u in the space proves lemma equality. □
To the fundamental system of mappings we can adjoin any solenoidal vector field or any vector field from the class as the first element of this system.
2.4. Successive Approximations of Solutions and Its Estimates: Velocity
Let be an orthonormal system of mappings in the space constructed above with the completeness property in and conditions (12) and (13). Moreover, for all n and where a vector field is initial data in problems (1) and (2).
For successive approximations we define changing Ladyzhenskaya’s construction in (, p. 197). Set
Then, an approximative solution is built as a hydrodynamical potential
Functions are solutions of a system of differential equations:
with initial data: , where is Kronecker’s delta. Hence,
Now, we find an existence interval of a smooth solution in system (16). For every equation from (16), we multiply by functions and sum them. As a result, we have:
From Corollary A3, we get:
From Lemmas A1 and A2 vector fields , . Equalities (17) and (18) are conditions of Lemma 1 for mappings . Therefore, in system (16), an existence of smooth solutions on some interval is guaranteed by well-known theorems for ordinary differential equations. By Lemma 1 (see estimates), these solutions can be extended on the interval where is the constant in Lemma 1 (see also (5)). Thus, we proved the following statement.
Let be an interval from Lemma 1. Then, for every approximations constructed by formulas (14) and (16) satisfy conditions:
where a constant A from Lemma A1.
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. □
Let be a constant of Lemma 1. Then, for every segment approximations , which are constructed by formulae (14)–(16), satisfy inequalities:
, where a number depends on only;
with the constant A from Lemma A1.
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 and add together in final. As a result, we have
A support of is a compact set. The Stokes theorem and Corollary A3 give:
From Lemmas A1 and A2, we have
By Lemma 3, the vector field satisfies the inequality:
Then, mappings satisfy Lemma 2. This implies:
with some constant .
Let us estimate the right-hand side of (20). In (16), we take . Then, we multiply them by numbers respectively and add them together. As a result, formula (17) gives
We move derivatives with the factor in (21) using Corollary A3 and a finiteness of mapping . Then, we obtain
Apply Cauchy–Bunyakovskii’s inequality. Hence, we get the required estimate
Let be a constant of Lemma 1. Then, on every segment approximations from formulae (14)–(16) satisfy conditions:
where constants C and depend on only.
Condition (1) follows from (18). We apply the Cauchy–Bunyakovskii inequality and estimates (1) of Lemmas 6 and 7 to the scalar product . Then, . The right-hand side from (18) is estimated by applying Corollary A4, where we take the triple . From (18), we have
Apply again estimate (1) of Lemma 6. Then,
This implies condition (1). Vector fields satisfy Lemma 2 (see the proof of Lemma 6). Then, Lemma 4 gives estimate (2). □
Let be a constant of Lemma 1. Then, approximations from (14)–(16) are bounded by a constant C on the set for every where a constant C depends on only.
For approximation , we use integral representation (A2). One should replace integration over whole space by integrations over ball and its complement. Then, . Every term is estimated by application of Hölder’s inequality. We have
where are universal constants. The norm 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 , we can apply inequality (1) from Lemma 6. Hence, we get a boundedness of all vector fields by a general constant. □
Let be a constant of Lemma 1. Then, for every exponent and every segment approximations from formulae (14)–(16) satisfy the inequality with a constant C depending on only.
If , then the statement follows from Lemma 9 and estimates by item (2) of Lemma 6 and item (1) of Lemma 8. If , then we apply Hölder’s inequality. Hence, we have . Estimates of Lemma 6 and Lemma 8 prove the lemma for this exponent. An intermediate exponents is verified by Lemma A5. □
Let be a constant of Lemma 1. Then, for all approximations from formulae (18)–(20) satisfy inequalities
, with a universal constant M.
The statement of lemma is the corollary well-known results about integral differentiation with a weak singularity (see ). From the second representation of Lemma A2, we obtain two equalities: , where are some constants, are singular integral operators. Its boundedness in the space gives the required estimates. □
Let be a constant of Lemma 1. Then, for every segment for all exponents and each triple approximations from (14)–(16) satisfy inequalities:
where constants C depend on only.
Apply Hölder’s inequality. Then,
Denote . An exponent . Then, the first factor in (23) is estimated by Lemma 1 with an assumption . Uniform estimate (1) follows from Lemma 6 and Lemma 8. In the same way taking a pair we get estimate (2). Now, denote . To the first factor from the right-hand side of (23) we apply Lemma A5 relying on . The norm 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. □
2.5. Successive Approximations of Solutions and Its Estimates: Pressure
Let be an approximation from formulae (14)–(16). Fix where is the constant from Lemma 1. Consider a hydrodynamical potential
A product . This follows from estimates of Lemma 6, Lemma 8 and Hölder’s inequality. By Lemma A4 on every segment we have:
Lemma A1 implies a uniform estimate with respect to t and n:
for any exponent , where .
Let us decompose integral in (24) by two integrals and : over ball and over its exterior. Every integral we estimate by Hölder’s inequality or a simple estimation. Then,
Thus, with some constant on the set for all n, we obtain:
Function has derivatives in distributions:
The differentiation of the integral from (24), the summation and a simple estimation give:
By Lemma A1 for exponents and where , we have:
The right-hand side of (29) is bounded upper by a constant . Here, we apply inequalities from Lemma 6, Lemma 8 and Lemma A5. Therefore,
Thus, for any exponent . By Lemma A1, we obtain:
Consider two cases: and .
Let . Then, the right-hand side of (31) is bounded by a constant . This follows from estimate 4 of Lemma 12.
Let . Then, the exponent . Applying Hölder’s inequality, we get
The first factor is estimated uniformly by a some constant . 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, , we get:
Applying the integral representation for derivative in the same way we prove another uniform estimate for every exponent .
As the final result from (26), (27), (30), (31), we obtain the following statement.
Let be a constant from Lemma 1. Let be a function defined by (24). Then, on every segment , with some constants there are fulfilled uniform estimates with respect to and n:
for all ;
for every ;
for every ;
, , for every .
Suppose that is the constant from Lemma 1. Let be a function defined by (24). Then, on every segment , with some constants there are fulfilled uniform estimates with respect to and n:
for every and every pair of numbers .
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. □
Let be a constant of Lemma 1. Let be a function defined by (24). Then, on every segment , with some constant there are fulfilled uniform estimates with respect to and n: for every , .
It is sufficient to repeat the proof of Lemma 11 with the application of formula (24). □
Let be a constant from Lemma 1. Supposing that is the function defined by (24), then
This follows from proposition A3. □
2.6. Estimates of Uniform Continuity of Approximations in Spaces and
Now, we estimate the integral continuity modulus of gradients and Laplacians for approximations following . Let be a constant from Lemma 1. Let T, be arbitrary numbers such that . Assume . Equations (16) we write by the following form:
Every equality we multiply by difference respectively and add together them. Setting we have
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 . Then,
Every factor from the right-hand side of these inequalities is bounded by a constant uniformly with respect to . This follows from estimates of Lemmas 6–8, definition of z and the choice of means . Since
then we get inequalities:
Integrating it over segments if and if in any case we have: . Thus, the following statement is proved.
Let be a constant from Lemma 1. Let be arbitrary but fixed numbers. Then, there exists a constant such that, for all approximations there is a fulfilled inequality:
Let be a constant from 1. Let be arbitrary but fixed numbers. Then, there exists a constant such that for all approximations there is fulfilled inequality:
where . Every equality we multiply by factor 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, 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 . A uniform boundedness of integrals follows from Corollary A4. For the verification, we take mappings triples and respectively. Finally, applying estimates from Lemma 6, Lemma 8 and Lemma 17, we obtain:
where constants depend on only.
We integrate (34) over segments if and if . Assume without restriction of the generality. Then, from (35) after Hölder’s inequality application and inequality (2) of Lemma 8, we get:
where is a new constant. From (36) in the same way, we obtain another estimate:
Integrating (34) and, gathering last estimates, we get lemma inequality. □
Let be a number from Lemma 1. Let be arbitrary but fixed numbers. Then, there exists a constant such that for all approximations and there are fulfilled inequalities:
We have . Therefore, one should find uniform estimates for every modulus on the right-hand side considering mappings .
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 only.
Let us estimate the second modulus applying Poisson’s formula (see (A1)). Then,
From the inequality,
with some constant we obtain:
Previous estimates and Lemma 8 (estimate (1)) yield:
where a constant C depends on 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 (formula (24)). The norm , which appears after applying Holder’s inequality, we must estimate by Lemma A5. Then,
Furthermore, Lemma 6 (estimates (1), (2)) and lemma 8 (estimate (1)) yield the inequality , where is some universal constant. Then, it follows:
A difference 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 and its exterior. Let . Then,
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 is some constant depending on only. Uniform estimates (39) and (40) prove the second inequality of lemma. □
2.7. Weak Limits Properties of Approximation Sequences
Let be a number from Lemma 1 and be a positive number. Then, the sequence of mappings defined by (14)–(16) is bounded in the space and the sequence constructed by formula (24) is bounded in spaces .
Estimate (2) from Lemma 6 and estimate (1) from Lemma 8 yield inequality . It is fulfilled with some constant C whenever n and . For all mappings , 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 , , such that , . All norms are uniformly bounded with respect to t. Hence, the sequence is bounded in .
Uniform boundedness of these norms , , with respect to t and n follows from Lemma 13. Therefore, the sequence is bounded in spaces . □
The spaces , are reflexive. Hence, every bounded set from it is a weakly compact set (see ). Then, by Lemma 20, sequences , 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 . This follows from Arzela’s theorem and Lemma 19.
mappings u and P are uniformly continuous on a set , moreover, ;
mappings u and P are bounded on a set ;
the mapping and there exists a constant such that following inequalities are true: , , whenever ;
, whenever , where vector field from (22), a constant ;
u has distributions of the second and third orders: , in addition, for all there are fulfilled inequalities: , where constants from Lemma 8;
the function for every , in this case, there exists a constant such that, for all estimates , are true;
there exist constants such that for every and for every ;
the function P has distributions of the second and third orders: , in addition, there exists a number such that, for all the following inequalities hold:
for every and for every
Property (1) follows from Remark 2. A uniform continuity follows from Lemma 19 and a uniform convergence of subsequences and on compact subsets of .
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 (see Lemma 8 and Lemma 11) and norms boundedness in the space (see Lemma 8 and Lemma 11) guarantee an existence of distributions , . Estimates of its norms follow from a semicontinuity of a weak limit norm.
Properties (5)–(8) are proved in the same way. For the verification, we apply Lemmas 13–15. □
The first equality is fulfilled for mappings and . The sequence is bounded in the space . In addition, estimates of norms , , are uniform with respect to t and n (see Lemmas 6–8 and 11). Apply Sobolev–Kondrashov’s embedding theorem (see , pp. 83–94) to the sequence . As a bounded set, it is embedded in the space for every ball . An exponent q satisfies condition
In this case, a dimension of spatial domain . Thus, we can assume that a subsequence converges strongly to a mapping in the space , , for every ball . Denote the integral from the first equality of the lemma by . Let . From equality
Multiply this inequality by where an arbitrary test–function. Thereupon, integrate over the set and change integration order. Then,
where is the Riesz potential, is the interior integral calculating over ball , and 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 . Then, . Integral 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 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 weakly because a function is an arbitrary. The first equality is proved. The second equality is proved in the same way. Consider the difference where is the integral of the second equality. In the integral we replace the variable by . Thereupon, we multiply the equality by a test–function and integrate its over set . Change integration order and carry over Laplace operator to function . Then,
Replace variables in the interior integral by and change integration order. Hence, we get:
Integration with respect to y we make separately over ball 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 on compact sets (see Remark 2). □
2.8. Weak Solutions and Gradients Boundedness
Let u and P be weak limits from (41). Then, for every solenoidal vector field and almost everywhere there is fulfilled integral identity:
Equalities (16) multiply by a test–function and integrate its over segment . If a subsequence converges weakly, then, for all , 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 to the mapping . It is given by support compactness of a vector field , by uniform boundedness with respect to t and n of norms and a uniform convergence of subsequence on compact subsets of A function is an arbitrary. Therefore, from (42), we obtain
It is already fulfilled for every natural number q. The construction of vector fields permits this integral identity to extend on elements of the fundamental system (see (12) and (13)), i.e.,
We show that identity (43) is true for every solenoidal vector field . Let be a sequence of a finite linear combinations of mappings , which converges to a vector field in the space . Then,
and equality (43) for mappings is true. Mappings belong to the space for a.e. t. Then,
a.e. as . Let us show
as the same condition is. Consider the equality of scalar products
and note that the right side tends to (see Lemma 21 item (4)). On the other side, . Condition (43) is true for an arbitrary . From , we have the lemma. □
(see (, pp. 41–44), see also .) Let be an arbitrary ball. Then, a space of any vector fields has a decomposition by a direct sum of orthogonal subspaces. A subspace is the space of gradients where is locally square–integrable function with a finite norm . A space is the closure with respect to the norm of all solenoidal vector fields from the class .
If u and P are weak limits (41), then there are fulfilled equalities:
a.e. on a set for any .
Denote , , . Every vector field belongs to the space (see Lemma 21). Mappings norms are bounded by constants independent of . From the first equality of Lemma 22, we gather , where is an arbitrary. We assume the mapping H and its generators belong to the class . Otherwise, we take averages with a kernel from for them. For averages, the equality and the equality of Lemma 23 are kept. This follows from behind an arbitrary choice of a smooth function g and a field . Then, . Moreover, a smoothness H and the equality of Lemma 23 imply . From Lemma 24 on every ball , we have . A function h is infinitely smooth. This is given by smoothness . Then, . On the other hand, . Therefore, the function h is a harmonic function. Hence, and from above, there is . By Lemma A7, we have . Making an average parameter tending to zero, we obtain this equality in the general case. □
Let u and P be weak limits from (41). Then, there exists a number such that, for almost everywhere, following conditions are fulfilled:
From Lemma 25, we conclude that Laplacian is the linear combination of three vector fields , , . Coordinates are bounded on the set by Lemma 21 item (2). Then, from Lemma 21 (see estimates (3) and (6)), it follows the first part of the lemma.
Gradients boundness we obtain from the second integral representation of Lemma 22 and estimate . In the next step, we repeat the proof of Lemma 9.
Gradients boundedness we get from the first integral representation of Lemma 22 and gradients boundedness with repeating of the proof from Lemma 9. □
2.9. Weak Solutions, Integral Equations and Energetic Inequality
Let be a Weierstrass kernel. Furthermore, we consider mixed norms for mappings defined on the set .
(See , Theorem 2.1.) A vector field with a finite mixed norm is a weak solution of problems (1) and (2) if and only if when u is a solution of integral equation
where B is a some nonlinear integral operator,
(See , Theorem 3.4.) Let u be a solution of integral Equation (44) with a finite mixed norm where . Let k be a positive integer such that . If mixed norms of derivatives
with exponents are finite whenever , then also mixed norms of
are finite for the same means , .
The proof of this result relies on Calderon–Zygmund’s theorem and a boundedness of singular integral operators of parabolic type (see ).
Norms are bounded by a constant that depends on exponents p, q, derivative order and the mixed norm . It follows directly from the proof of the theorem in .
If u is a weak limit from (41), then there exists a number such that whenever .
Let be a positive arbitrary number. Integrate the equality of Lemma 25 over segment where . Then, continuity and absolute continuity on lines of mapping u give:
Every integrable term has finite norms
In addition, every norm is bounded by a constant depending on 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, . A boundedness of vector field u (see Lemma 21 item (2)) gives a uniform estimate whenever . Then, any mixed norm is finite whenever from lemma condition. □
If u is a weak limit from (41), then a mixed norm .
Let initial data . Function f from Lemma 27 is represented by integral
For there is true Lemma 34. Therefore, the mapping f and any of its derivatives have a finite mixed norm . By Lemma 25 and Lemma 29, the vector field u is a weak solution of problems (1) and (2) with a finite mixed norm whenever . 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 , where , . Let . Then, we have the statement of the lemma. □
(Energetic condition.) Let u and P be weak limits from (41). Then,
for every where from Lemma 1.
Note that weak solutions satisfy conditions:
From the first equality of Lemma 22, we have:
Integrals commutation is possible since the integral over is a finite. It follows from
Tonnelli’s theorem, boundedness and summability of with any exponent not less than two and Lemma A1. Here, 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 (see Lemma 21 item (5), Lemma 26 item (1), Lemma A5). In addition, by Lemma A1. To interior integral in (47) we apply the Stokes formula. Then,
A product belongs to the space whenever . Then, the integral over surface tends to zero as (to apply Lemma A4 with exponent and a mean p, close to unit). Hence, and from (47) we have:
In the iterated integral
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 . This follows from a boundedness of the Riesz potential . 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 (norm defined by (15)). Lemma A1 shows that Poisson’s formula is true for elements of the space . Then, . Therefore, we have from (50). The second equality from (45) is proved.
Let us show that vector field u satisfies the equality
a.e. on . We have the equality of iterated integrals:
A finiteness of double integral follows from a boundedness (see Lemma 26, item (2)) and properties of the Riesz potential . Let . The interior integral on the left-hand side of (52) tends to in the space for almost every t. (See Lemma A1 and equality (A2), which is true for elements of the space ). The norm 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 . Thereupon, we add together them and integrate over space . From (45) and (49), we have
Hence, we get the required equality. □
2.10. Proof of Theorem 1
Observe that all estimates in proved lemmas above depend on norms , (see (22)), or only and don’t depend on a diameter of Laplacian support .
If , then, by Lemma A4 integrals,
tend to zero as Therefore, equalities from Lemma A2 are true for mappings of the class . In addition, we have summability with any exponent and with any exponent (see Lemma 32).
1. Assume that initial data and its Laplacian support is a compact set. Let 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
we obtain from Lemma 21 (item (3)). An uniform boundedness of norms
follows from Lemma 21 (see items (4), (5), (7)). The estimate of norm follows from Lemma 31. A uniform boundedness of norms we get by Lemma 26. A uniform boundedness for norm is the corollary of Lemma 25 because is the finite linear combination of terms with uniform bounded norms in the space . Uniform estimates of norms in spaces , we take from Lemma A5. The occurrence of vector field u in spaces and 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 since u and are bounded. Therefore, .
Let us prove item (5) using mixed norms (see [8,26,27]). Weak solutions u and P belong to class (see item 1) of this theorem). In Lemma 28, we put assuming it is very large. Now, we fix an order of derivatives: . Then, by Lemma 27 and Lemma 28, derivative norms are bounded in the space where an exponent is an arbitrary but fixed. A boundedness of weak solution u and its summability in imply the belonging . Exponents means we choose by large numbers so that the next conditions are fulfilled:
for any ball lying in , all conditions of Sobolev’s embedding theorem in a space of continuous functions are certainly valid (, p. 64);
at least, all derivatives of the order up to 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 . A smoothness of function P we obtain from Lemma 25 and the smoothness of vector field u. The continuity is proved in item 1.
2. Let initial data . We take a test-function such that if and if . Consider a solenoidal vector field
Then, . A function Q is Poisson’s integral
Hence, we have:
where is the Riesz potential, M is the maximal mean of . From Lemma A1, we obtain
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 . This is guaranteed by a test-function and a boundedness of the second derivatives . The last follows from representation of function Q by Poisson’s integral and definition of the class . In this case, we have two equalities:
where are universal constants, and are singular integral operators. Therefore, as , then
A vector field A summability of the vector field and its derivatives follows from (53) and (54), the equality
and Lemma A1. In addition, in the space in the space . Laplacians supports are compact sets. Therefore, there exist solutions and with an initial data satisfying theorem with the number
From (55), we have as . Fix a number . From the remark at the beginning of the proof, we conclude all estimates of the theorem for solutions , . They are uniform with respect to r for . Hence, sets of mappings , are bounded in spaces and . Extract subsequences , , which converge weakly. Let u and P be its weak limits, respectively. These limits satisfy the next properties:
Lemma 21 is true for them (this is verified in the same way as the proof of Lemma 21 for subsequences);
Lemma 25 is true for them;
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 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.
3. Homotopic Property of Cauchy Problem Solutions in Class
If initial data then the Cauchy problem solutions from Theorem 1 have the next homotopic property.
Let u and P be solutions of problems (1) and (2) from Theorem 1. Then, for every fixed mean (see (5)) mappings . Moreover, all norms
if , , , are uniformly bounded on every segment where .
Proof of Theorem 2 (it is given below) is relied on for the next simple properties of mappings . 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 . From Lemma A4,
as since . Then, the passage to limit as , implies the second equality (56). The first equality is proved in the same way.
where is the Riesz potential from (4). Hardy–Littlewood–Sobolev’s inequality (see Lemma A1) implies
where . Consider only . Two last estimates yield for every . Analogously with the above, we show for the mapping v and a number the belonging whenever . The logarithmic convex of norm and Lemma A5 yield norm finiteness for . Thus, we proved the next statement.
Let . Then, , whenever .
Write Poisson’s formula (the representation by Riesz’s integral ) for mappings v, and . Then, we have a boundedness of every vector field and its derivatives.
(the repeated index gives summation).
Let and . Then, the function P and all its derivatives belong to the space whenever .
The integral from (57) we integrate by parts twice over a spherical layer . Lemma A4 and the passage to the limit as , imply:
where 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 . Since
then, analogously with the above, we get:
Hence, we obtain a summability of whenever finite A summability of the other derivatives follows from equalities:
If then Poisson’s and Riesz’s formulae are true:
Suppose a function P and all its derivatives are summaable in space whenever . Let . Then,
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 with exponent 3 (see A1) and the summability of with exponent . The first equality is proved. To prove the second formula, we observe a finiteness of integrals
Thereupon, we have:
Items (1), (2) and (3) from Theorem 1, Lemma 27 and Lemma 28 yield a finiteness of mixed norms where , whenever . For derivatives of the second order, in particular, we have a finiteness of norm (see 30). Integrate (1) over segment where . The solution P is represented by (57). Then, from (59), we get
Estimate norms in of every term in (61) in the usual way. We apply Hölder’s inequality to interior and exterior integrals. Then,
The singular integral operator is bounded. Hence, from (61)–(63) and item (2) of Theorem 1, we obtain a uniform estimate of norm with respect to .
In the same way, we prove a summability of gradient with any exponent . From Lemma 27 and Lemma 28, whenever , we get a finiteness of mixed norms for derivatives of the third order where since . In particular, we have a finiteness of norm .
Let us differentiate (1) with respect to . Thereupon, we integrate its over where . Formulae (57) and (60) yield
Hence, for exponent , we obtain estimates, which are similar estimates (62) and (63). A boundedness u, uniform estimates of norm (see item (2) from Theorem 1) on segment , a finiteness of mixed norm give a uniform boundedness of norms .
Fix an exponent . Choose numbers . Then, we have a finiteness of the mixed norm . It follows
Terms in derivative without coefficients have a form: where . Then,
To the right-hand side, we apply Hölder’s inequality with exponents and . Therefore,
where , . All these mixed norms are bounded. This follows from Lemma 27 and Lemma 28. Hence, formulae (65)–(67) and a boundedness of a singular integral operator give uniform boundedness of all norms with respect to .
The solution P is represented by (57) with replacing v by u. A summability follows from Lemma 33 whenever . Equalities (58) and (59) and a uniform boundedness derivatives norms of vector field u prove a uniform boundedness of norms where . From (1) and proved uniform estimates from above, we have necessary statement for derivative . Theorem 2 is proved. □
4. 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 . A functional and the first parameter we define by
where the constant from Corollary A4. By Theorem 2, the solution of the Cauchy problem with condition 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 . In other words, the class is kept. If is the maximal interval of solution existence, then the second parameter is defined by:
Let u be a solution of problems (1) and (2) from Theorem 2. Then, functions
are continuous functions on the interval .
Let Fix where the function from Lemma 45. Choose a segment assuming . Denote . Take equalities (1) with time argument . Thereupon, we multiply them by getting the scalar product and integrate over . The derivative for all . It follows from Theorem 2. Then, by Lemma 35 (the scalar product in we write as ), we have:
Here, the right-hand side is bounded uniformly (see Theorem 2). Then, as . Triangle inequality implies the continuity of function
We write equality (1) for time arguments t, and subtract it. Thereupon, the difference we multiply by 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
Function continuity of follows also from uniform estimates of Theorem 2. Difference is considered as the sum of three integrals with combinations:
Every integral we estimate by Hölder’s inequality so that there appear norms:
The first of these norms is estimated through the second norm by the inequality from Lemma A1 with application of the second representation in Lemma 34. Therefore, on every segment with some constants , we have:
Hence, the first statement follows. Let us prove function continuity of . The estimate
was called above. The logarithmic convex inequality where and Theorem 2 (see item (2)) about uniform boundedness of norms) give the statement of the lemma. Here, it is enough to take . □
Let and , where is defined by (68), the number from Corollary A4. Then, solution u of problems (1) and (2) from Theorem 1 satisfies inequality: .
Equality (1) we multiply by getting the scalar product and integrating over the whole space. Then, from Theorem 2 and Lemma 35, we have:
where from Lemma 32. Now, we show that the function is negative. Note . Suppose the opposite. Then,
From Corollary A4 (it is extended on the class by Lemma 34), we have estimate:
(it follows from Lemma 35), then the last two inequalities imply . We have a contradiction.
Let be a maximal interval where function . Suppose . Continuity condition (see Lemma 32 and Theorem 2) gives .
Repeating arguments from above, we obtain estimate:
With the other hand, function from Lemma 32 is a decreasing function on interval . Therefore, . Since then . Compare this inequality with (70). Then, we have a contradiction. □
where is defined by (68), numbers from Corollary A4,
Then, for solution u of problems (1) and (2) from Theorem 1, there exists a number such that
Suppose the opposite. Then, on interval the inequality holds:
Integrate it over this interval. Since
Take out a nonpositive term on the right-hand side and input the mean from (5). Then,
We have a contradiction with the condition. □
Let and where is defined by (68) and the number from Corollary A4. Then, problems (1) and (2) have unique solution u and solution P such that are defined on the set . In addition, these solutions have properties (1)–(5) from Theorem 1 on every fixed segment and satisfy Theorem 2. Moreover, the norm , as a function of argument t, is a decreasing function on the set .
If then the statement of theorem follows from Theorems 1, 2 and Lemma 37. A finiteness of mixed norms we get from a boundedness of the vector field u and estimates . Solution uniqueness in the class , is proved in [7,8,26] (see also ).
Norm monotonicity as a function on time argument t follows from condition (see proof of Lemma 37).
Let be an interval of the maximal length such that there exist solutions with the estimates of Theorem 2.
Suppose . Let and . By Theorem 2 mapping, belongs to class . 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 where
We have from condition , for means . Hence, the solution u is extended with the half-interval on an interval of more length . We have a contradiction. □
where is defined by (68), number from Corollary A4. Then, problems (1) and (2) have a unique solution u and a solution P that are defined on the set . In addition, these solutions have properties (1)–(5) from Theorem 1 on every fix segment and satisfy Theorem 2. The norm , as a function of t, is not decreasing function on the set , where constant from (5).
Let u and P be solutions of problems (1) and (2) from Theorem 2. The proof proceeds from induction with respect to number m from Lemma 38. Let . By Lemma 38, there exists a number such that
By Theorems 1–3, there exists a global solution w of problems (1) and (2) with changed initial data . 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 such that
Now, we take initial data such that
By Lemma 38, there exists satisfying
By Theorem 2 and the induction hypothesis, there exists a global solution w of problems (1) and (2) with a new initial data . 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
as . □
4.2. Critical 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 and Theorem 1.
The solution u of problems (1) and (2) from Theorem 1 satisfies estimate:
Let and i.e.,
Let u be a solution of problems (1) and (2) from Theorem 1 If
for a some number , then solution u can be extended by a global solution with properties (1)–(5) from Theorem 1 and estimates from Theorem 2.
We construct the extension in the same way as in the proof of Theorem 4. □
Let and parameter (see (68)). If u is the solution of problems (1) and (2) from Theorem 1, then on interval the inequality holds:
We integrate the inequality of Lemma 39 over the segment . Since
then, applying Newton–Leibnitz’s formula, we obtain the statement. □
Let and parameter (see (68)). Let u be a solution of problems (1) and (2) from Theorem 1. Suppose, on the interval , there is fulfilled estimate:
If , then there exists a number such that
Both parts we raise to the second power and integrate over the interval . From (72), we get:
where the number from Theorem 1. Since , 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 (see Lemma 36), it is fulfilled:
Integrate (72) over the interval . 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 . Since , 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 . The last statement of lemma we prove from the opposite in the same way. □
Let initial data and parameter . There doesn’t exist solution u of problems (1) and (2) satisfying (73). It is always true inequality .
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 then there exists a constant c such that
for any . Hence, we have for each pair and , respectively. From Lemma A6, it follows —a contradiction. The lemma is proved. □
Let initial data and parameter . If u is the solution of problems (1) and (2), then
Suppose the opposite. Then, we have the equality in Lemma 41. It implies the second equality from (73). Repeating the proof of Lemma 43, we obtain a contradiction. □
Let u be a solution of problems (1) and (2) with initial data . Then, a function
and a function
with condition are not decreasing functions on the interval where constant from Corollary A4.
From Theorem 2, we have safety of class for every if u satisfies lemma conditions. The both functions are continuous (see Lemma 36 and Theorem 2). Inequality (84) (see below) is true for any mean . Rewrite its in another form:
and integrate its over the segment . Simple transformations give:
Hence, and from lemma condition, it follows for all . Furthermore, we use inequality and get the monotonicity of the second function. For the monotonicity, the first function follows from inequality (84) because, in this case, . □
4.3. Solutions Extension in Global with Condition
: Necessary Conditions For Hypothetical Turbulence Solutions
Let . Suppose that and parameter λ from (68). If or and the solution u of problems (1) and (2) from Theorem 2 satisfies
Then, there exists a number such that inequality is fulfilled:
where constant from Corollary A4.
Assume . Then, the statement follows from Lemmas 42 and 43. Let . Suppose the opposite. Then, we have:
Make the passage to the limit in (80) as and compare the new estimate with the inequality from lemma condition. Taking (5), (68) and (80), we conclude Consider a function It vanishes at boundary points of ; moreover, (see (80)). Let be an interval, where the function vanishes at boundary points and on its interior. Then, there exists a point , where . Hence, from (72), we get:
In particular, by Lemma 36, . 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 be an interval of the maximal length, where solutions u and P of problems (1) and (2) have properties from Theorem 2. Then, and (see (69)). Hence, . Therefore, J. Leray’s estimate from  can be given in invariant form in the following statement.
Let and Suppose that 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:
A function is unbounded in some left neighborhood of the point . Suppose the opposite. Then, for every point , there exists solution v of problems (1) and (2) with initial data which satisfy Theorems 1 and 2.
This solution gives the unique extension u on the interval , where and M is the supremum of . A point is an arbitrary, therefore, solutions u and P can be extended on the interval . 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 by Theorem 2 and Lemma 35 because the solution . The integral representation from Lemma 34 permits to apply Corollary A4 and estimate of the right-hand side in the last equality. Repeating the proof of Lemma 1, we obtain
Replace in (84) s on , where and as . The passage to the limit in (85) gives estimate (82). □
Let and For finite interval of the maximal length, the parameter μ is not greater than the number , where λ is defined by (68).
In the inequality
we apply Lemma (79). From (5) and (68), we get the statement. □
Let , and λ be a parameter from (68). If the interval has the maximal finite length and solutions u, P of problems (1) and (2) have properties from Theorem 2 on this interval, then unextended solutions satisfy conditions:
Therefore, we have the second inequality in (86). □
Set and with a constant from A4. Let u be a solution of problems (1) and (2) from Theorem 2. If u satisfies condition
then problems (1) and (2) have global solutions u and P. Moreover, they have properties (1)–(5) from Theorem 1 on every segment and satisfy conditions of Theorem 2 there. As a function of argument t, the product is a decreasing function on the set , where constant from (5).
Let a number from Lemma 46. Without norm monotonicity, the statement of theorem follows from Theorem 2 and Lemma 40. The product is a decreasing function on the set . It follows from Theorem 4. Therefore, the theorem is proved. □
Now, we give one result that is connected with a local solutions’ extension. If , then we introduce the third parameter , which gives a dissipation quantity of a kinetic energy. It is defined by formula:
We observe from Lemmas 43 and 44 that the parameter satisfies strong inequalities: . This is very important for the furthest. The usefulness of this parameter is explained by the following result.
Suppose initial data and
where is defined by (68). If solution u of problems (1) and (2) from Theorem 2 satisfies (87), then this solution has an extension on the set where
This extension has properties (1)–(5) from Theorem 1 on every segment .
Now, we consider only that solutions which don’t have any global and smooth extension. Take . From theorem condition and the second inequality of Lemma 49, we obtain: . Hence, we get: . Then, the statement of the theorem follows from the definition of parameter . The theorem is proved. □
Suppose . A finite mean of parameter μ satisfies inequalities:
In the first inequality of Lemma 49, we take . Then, we get the necessary upper estimate. The strong lower estimate doesn’t follow from (71) yet. Let
Consider a function
We observe (see formula (87)). Hence, there exists a number such that . Then,
where function from Lemma 45. Since this function does not decrease then for every t, , we have . Therefore,
which holds for every t, . Integrating this inequality over interval from formula (87), we gather:
If then, in formula (86), we must have identical equalities (see (88)). It implies the identity
Take . Hence, we get . This contradicts Theorem 6, from which we obtain because . The last proves the strong lower estimate for . □
5. Main Results, Existence of Global Regular Solutions, and Sufficient Conditions
Now, we prove the basic result which is described by Theorem 7.
Let be initial data, the parameter λ from (68) and the number from (5), the vector field u from Theorem 1. If parameter or in opposite case
then the Cauchy problems (1) and (2) have global solutions and with the following properties:
mappings u and P are uniformly continuous and bounded on a set for every number ;
for every numbers , and multi-indices all norms
are uniformly bounded on the segment , moreover ;
gradients are bounded on the set for every ;
solution u has a finite mixed norm on the set for every and every pair of exponents ;
solutions u, P belong to class i.e., these solutions are classical.
If parameter , then the function is a decreasing function on the interval . If and condition (89) is fulfilled then the function is a decreasing function on the interval .
Let . Then, the statement follows from Theorem 4.
Let . 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 and condition (89) is fulfilled. Then, the statement of the theorem arises from Lemmas 45 and 46, Theorem 5. The theorem is proved. □
Let be initial data, the parameter (see (68)) a vector field u is a weak solution from the Cauchy problems (1) and (2). If on an interval an inequality
is fulfilled, then the weak solutions and are regular on interval and satisfy Theorem 5.
Let be a maximal interval where the weak solution u is regular. Suppose . Then, from (86) and (90), we have
Since solution u is regular on the interval then, by differentiation of the previous identity at point , we obtain . From Lemma 50, we have a contradiction. Therefore, . □
Does there exist weak solution u satisfying opposite inequality (90) if ? It is unknown.
6. The Cauchy Problem with Less Smoothness of Initial Data
In addition, the invariant class 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 and were shown in . Therefore, we consider the first space from them.
Let be a solenoidal vector field. Set a sequence of of finite, solenoidal and infinitely smooth vector fields, which converges to the field in the space . We observe that . Let , be sequences of solutions in the Cauchy problem for Navier–Stokes equations with the initial dates . Then, all pairs satisfy all uniform estimates of Lemma 21 on any compact set of the interval where since upper bounds in these inequalities depend on a set and . Therefore, on every fixed segment we can take these constants as common for all because in the space . Then, without loss of generality, we assume that the sequence converges weakly in the space to a field . In addition, we suppose that and converge weakly in to and . More generally, weak limits satisfy all conclusions of Lemma 21 and they are weak solutions of problems (1) and (2). From the equality (1) for couple and items (2), (4), (5), (8) differentiating (1), we obtain that distributions belong to the space for almost everywhere t. Thus, the class is invariant similar to the class . For this case, in the same way, we can define the basic parameters . After that, one should note that the statement of Lemma 36 will be true when initial data . Repeating the proof of Theorem 7, we obtain the following result.
Let be initial data, the parameter λ from (68) and the number from (5). Let be a vector field u is a weak solution of the Cauchy problems (1) and (2). If parameter or in opposite case
then the Cauchy problems (1) and (2) have global solutions and with the following properties:
mappings u and P are uniformly continuous and bounded on a set for every number ;
for every numbers , all norms
are uniformly bounded and mixed norms are finite on segment ;
solutions u, P belong to class i.e., these solutions are classical.
If parameter , then the function is a decreasing function on the interval . If and
then the function is a decreasing function on the interval .
Let be initial data in problems (1) and (2). If parameter then the solution u from Theorem 9 satisfies:
a power of norm is a convex function;
there is fulfilled:
It follows from Lemma 45 because this lemma is true for solution u from Theorem 9. □
7. Integral Identities for Solenoidal Vector Fields: Dimensions Comparison
Some review and results about integral identities for solenoidal vector fields are given by authors in [33,34]. Here, we reduce one from these identities, which shows the essential distinction for the Navier–Stokes equations between space and plane.
Let be any triple of solenoidal vector fields from the class . Denote
(see ) For every triple of solenoidal vector fields from the class the identity holds:
Hence, it follows (one should take ):
(see .) If dimension , then every solenoidal vector satisfies the integral identity:
Obviously, it implies some interesting applications to the 2D Navier–Stokes and Euler equations (see ).
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 ).
In the case , we have formula (83) and, therefore, the norm 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 . In addition, the estimate is exact and it does not follows from Judovich’s results . This explains "the simplicity" of a motion of an ideal fluid on plane.
Let , . Then, the product is a decreasing function in any case.
If dimension , then integral from (91) may be not equal to null.
For a simple example, there is the vector field with the following coordinates:
where is the i –th vector row of the skew-symmetric matrix. Since
then simple calculations show
with a constant . A coefficient may be not equal to zero for fixed different means k and i because there is the linear independence of polynomials 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 can be taken with a large value implying a positive mean of the difference
for . It is possible because we can take a factor for initial data or diminish viscosity coefficient . This implies a growth of the norm for space. Obviously, on the plane, this phenomena does not appear.
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 or kinetic energy satisfies inequality:
No phenomena blow up on the time interval if kinetic energy satisfies inequality:
with condition .
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.
This research received no external funding.
I would like to say many thanks to all my reviewers for their remarks and advices.
Conflicts of Interest
The author declares no conflict of interest.
Appendix A.1. About the Riesz Potentials and Integral Representations
Some technical results are given.
(Hardy–Littlewood–Sobolev’s inequality (, p. 141). Let be Riesz’s potential defined by (4). Set . Then, there exists a constant where , , such that the following inequality holds:
In a special case, we give an estimate for operator norm.
The inequality is true, i.e., .
It is sufficient to verify this inequality for smooth and finite mappings. From Riesz’s formula, we have:
Multiply it by . Then, we make a simple estimate and integrate over space. Hence,
The interior integral we estimate applying Leray’s inequality
(see , also , p. 24), thereupon we use Hölder’s inequality. Then,
It gives the required estimate. □
Let us make more precise well-known integral representations as Poisson’s formula and Riesz’s formula for smooth functions with compact support.
Let be a mapping and its Laplacian has a compact support. Then, the equalities hold:
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 and two integrals over sphere . They are:
The third and the fourth integrals we transform again applying the Stokes formula and getting integrals over balls , and ,respectively. Every integral must contain Laplacian. Since support of is a compact set, then these integrals tend to zero as , .
The second integral in (A3) we denote by a symbol I. Then,
The Stokes formula application gives the equality:
The second term in (A4) we integrate by parts. Therefore,
The integral from the first term in (A4) we estimate applying the Hölder’s inequality. Then,
where —is the volume of a unit ball. From compactness of Laplacian support and lemma condition, we obtain that integral as . The first integral in (A3) tends to the mean as formula (A2) we prove by the same way. □
A mapping w from Lemma A2 satisfies inequalities: , with some constants and .
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:
A change of variables proves the second estimate. □
Let v, be mappings which satisfy conditions from Lemma A2. Then,
We apply the Stokes formula to the integral from the left side of this equality. From Corollary A2 on a sphere , we get the following formula: . A passage to the limit as gives the required equality. □
Let be a function and with some exponent and distributions for some exponent . Then, for function P, Poisson’s formula (the first equality from (A2)) is true.
For any smooth function P, we verify the integral identity the same way as in Lemma A2 with application of Lemma A4 (see below). A density of smooth functions and Lemma A1 prove the statement in a general case because there is continuity of the Riesz potentials in spaces . □
Suppose that a continuous mapping belongs to the class Then, for any point x, an exponent α, where
Hölder’s inequality implies an estimate:
Here, – is the surface measure of an unit sphere, . Let
The second integral on the right-hand side is estimated by application of Hölder’s inequality. Furthermore, we replace the integration over a ball by the integration over the whole space. Hence,
and as Therefore, from (A5), we have the statement. □
Appendix A.2. Logarithmic Convexity Inequalities and Its Corollaries
(, p. 21). A function is a logarithmic convex function. That is, for exponents , with condition where , the inequality is fulfilled.
Let u, v, be a triple of mappings satisfying conditions of Lemma A2. Then, the inequality holds:
with a constant . In addition, for a solenoidal vector field u, there is a more exact estimate:
We have estimates:
which follow from the Cauchy–Bunyakovskii’s inequality. Apply Hölder’s inequality for three factors. Then,
For each coordinate , we have where (see Corollary A1). A density of smooth functions and Lemma A1 give the required estimate in a general case.
Since (we apply the Minkovskii’s inequality with exponent 3), then . Respectively, we have and , .
From Lemma A5 with exponents and number , we obtain: . Then, from inequalities above and formula (A6), we prove the estimate with a constant .
Now, we verify the other inequality. For solenoidal vector fields, we get (see Corollary A2):
where is Kronecker’s delta. Applying Hölder’s inequality to a pair
then, from the inequality
for vector fields (see , Chapter 2 and ), we get the second part of the lemma comparing all estimates from above. □
Appendix A.3. Vanishing of Harmonic and Biharmonic Functions
If a harmonic function is represented by sum where functions , , then .
Without loss of generality, we assume that functions are smooth. Otherwise, we take its average defined by a formula
with a kernel . In the equality,
, we transform the integral applying the Stokes theorem. Let . Then,
The integral over surface since . Hence, . This integral is transformed in the same way as the integral I. From the equality,
by application of the theorem on the mean value of a harmonic function, we conclude the formula:
For chosen means , each potential , is finite (see Lemma A1). The passage to the limit in (A7) as yields the equality: . □
If a biharmonic function has a decomposition where functions , then .
Without loss of the generality, we can replace functions by its averages (see above). Then, every average . Now, we fix the averaging parameter . Let and . Let us show that function h is a harmonic function. It is sufficient to apply the theorem about the mean value of a harmonic function to and use the spherical coordinates. Then, for the average, we have:
The integral is transformed by applying three times of the Stokes theorem: twice to the integrals over volume and once to the integral over surface. As a result, we obtain:
Furthermore, we apply again the theorem about a mean value for a harmonic function to the third integral. After that, we input the mean of integral in (A8). Then, we conclude:
Here, the integral over the volume set tends to a finite mean as . The finiteness of this mean is proved in the same way as in Lemma A6. This implies
An exponent mean belongs to the interval . Hence, and from Lemma A4, we obtain . Taking assumption about a ball center, we obtain that the function is a harmonic function. Then, from Lemma A6, . Let . Then, . □
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