Comments on the Navier–Stokes Problem

: The aim of this paper is to explain for broad audience the author’s result concerning the Navier–Stokes problem (NSP) in R 3 without boundaries. It is proved that the NSP is contradictory in the following sense: if one assumes that the initial data v ( x ,0 ) (cid:54)≡ 0, ∇ · v ( x ,0 ) = 0 and the solution to the NSP exists for all t ≥ 0, then one proves that the solution v ( x , t ) to the NSP has the property v ( x ,0 ) = 0. This paradox shows that the NSP is not a correct description of the ﬂuid mechanics problem and the NSP does not have a solution. In the exceptional case, when the data are equal to zero, the solution v ( x , t ) to the NSP exists for all t ≥ 0 and is equal to zero, v ( x , t ) ≡ 0. Thus, one of the millennium problems is solved.


Introduction
The results of this paper are proved in detail in the monograph [1]. In the author's papers, listed in the References (see [2][3][4][5]), some preliminary results are obtained. In paper [6] some of the results are summarized. These results are stated in the abstract. The aim of this paper is to explain for broad audience the author's result concerning the Navier-Stokes problem (NSP) in R 3 without boundaries. The result, proven in detail in the book [1], can now be stated: If the exterior force f (x, t) = 0, the initial velocity v 0 (x) := v(x, 0) ≡ 0, ∇ · v 0 (x) = 0 and the solution v(x, t) of the NSP exists for all t ≥ 0, then v 0 (x) = 0.
This result, that we call the NSP paradox (or just paradox), shows that The NSP is not a correct statement of the problem of motion of viscous incompressible fluid. The NSP is neither physically nor mathematically correct statement of the dynamics of incompressible viscous fluid.
Let us explain the steps of our proof. The NSP consists of solving the equations: (1) see, for example, books [1,7]. Here v = v(x, t) is the velocity of incompressible viscous fluid, The data v 0 and f are given, the v and p are to be found. The fluid's density ρ = 1.
(a) First we reduce the NSP to an equivalent integral equation. v where F = F(x, t) depends only on the data f and v 0 , see [1]. We assume that f = 0. This is done for simplicity only. Under this assumption one has (see [1]): where The tensor G = G(x, t) = G jm (x, t) is calculated explicitly in [1]: Let us define the Fourier transform: Take the Fourier transform of Equation (2) and get the integral equation: where denotes the convolution in R 3 . The following inequality, that follows from the Cauchy inequality, is useful: One proves a priori estimate (see [1]): By c here and throughout the paper various positive constants, independent of t, are denoted. We denote by c 1 := |Γ(− 1 4 )| > 0 the special constant from Equation (27), see below.
We denote v j,m := ∂v j ∂x m , := R 3 , write Equation (1) as where over the repeated indices summation is understood, 1 ≤ j ≤ 3. We assume that Here · is L 2 (R 3 ) norm. Multiply Equation (10) by v j , integrate over R 3 and sum up over j to get where z ,t := ∂z ∂t . In deriving inequality (12) we have used integration by parts: This and our assumption (11) imply estimate sup t≥0 v . By Parseval equality the desired estimate (9) follows. Estimate (9) is proved.
Secondly, we prove that any solution to Equation (7) satisfies integral inequality (15), see below. The integral in this inequality is a convolution with the kernel that is hyper-singular; this integral diverges classically, that is, from the classical point of view. This inequality is derived in Section 2 (see also [1]): The norm here and below is L 2 (R 3 ) norm. Since the convolution integral in (15) diverges classically, we give a new definition to this integral in Section 3 and estimate the solution b(t) to integral inequality (15) by the solution q(t) to an integral equation with the same hyper-singular kernel: Namely, we prove the following inequality The term b 0 (t) depends on the data only (on v 0 since we assume f = 0) and we may assume that b 0 (t) is smooth and rapidly decaying as t → ∞.
(c) We prove a priori estimate part of which is inequality (9). One has by the Parseval equality. We prove that Equation (17) has a solution in the space C(R + ), sup t≥0 q(t) < c, provided that the data v 0 (x) is smooth and rapidly decaying at infinity. Moreover, this solution is unique and (d) We prove that any solution b(t) ≥ 0 of inequality (15) with b 0 (t) a smooth rapidly decaying function satisfies inequality (18). Since q(0) = 0 and 0 ≤ b(t) ≤ q(t) it follows that b(0) = 0. This yields the NSP paradox mentioned at the beginning of this section. Indeed, the initial data v 0 (x) ≡ 0, so b(0) > 0 and we prove that b(0) = 0.
The NSP paradox impies the conclusions we have made: The NSP is physically not a correct description of motion of incompressible viscous fluid in R 3 and the NSP does not have a solution on the interval [0, ∞) unless the data are equal to zero. In this case the solution to the NSP does exist on the whole inteval [0, ∞) and is identically equal to zero.
The uniqueness of the solution to NSP is proved in Section 4, see Theorem 3.
In this paper, by c various constants, independent of t, are denoted. Multiply inequality (22) by |ξ|, take the norm · of both sides of the resulting inequality and get inequality (15). In this calculation one uses the formulas which are easy to derive. The c are different in these formulas.
To study integral Equation (17) and integral inequality (15) we need to define the hyper-singular integral in this equation.
To do this, one needs some auxiliary material. Let us define the function where Γ(λ) is the gamma function. Here and throughout t = t + , that is, t = 0 for t < 0, t := t for t ≥ 0. It is known (see [8] ) that Γ(λ) is an analytic function of λ ∈ C except for the points λ = 0, −1, −2, ...., at which it has simple poles. The function 1 Γ(λ) is entire function of λ.
Consider the convolution operator One has where denotes the convolution on R + and c 1 := |Γ(− 1 4 )|. Inequality (15) can be written as Consider also the corresponding integral equation:

Investigation of Integral Equations and Inequalities with Hyper-Singular Kernel
In this section, we solve Equation (28) analytically and prove estimate (18). First, let us define the hyper-singular integral ψ := Φ λ q. We are especially interested in the value λ = − 1 4 because it appears in Equation (28). For λ > 0 the convolution Φ λ q is defined classically for q ∈ L 2 (R + ) and one has L(ψ) = L(q)p −λ , where L is the Laplace transform operator defined as which is analytic in the region Rep > 0 if q ∈ L 2 (R + ). If q(t)e −at ∈ L 2 (R + ) and a = const > 0, then L(q) is an analytic function of p in the region Rep > a. The Laplace transform is injective on any domain of its definition. Therefore the inverse operator L −1 is well defined on the range of L. The inversion formula is known: where C σ is the straight line σ = const > a, p = σ + iω, ω changes from −∞ to ∞ and L(q) is a function of p. In Appendix 3 of [1] one finds information on the Laplace transform used in this paper. In particular, the following Lemmas 1-3 will be used. Their proofs can be found in Appendix 3 of [1].

Lemma 1. Example text of a Lemma.
If Q(p) is analytic in the region Rep > 0 and then q(t) = 1 2π ∞ −∞ e iωt Q(iω)dω, q(t) ∈ L 2 (R + ) and L(q) = Q(p). In Lemma 1 sufficient conditions are given for a function, analytic in the region Rep > 0 to be the Laplace transform of an L 2 (R + ) function.
Here we have used the known result (see [1] or [9]): and the known formula For λ > 0 and q smooth and decaying at infinity this formula can be understood classically. For λ < 0 it is defined by the analytic continuation with respect to λ ∈ C where L(Φ λ ) is given in formula (33). Formula (33) is valid for all λ ∈ C by the analytic continuation from the region Reλ > 0, where it is valid classically.
Let us define convolution ψ := Φ λ q by the formula: The expression under the sign L −1 is an entire function of λ. For λ > 0 the ψ(t) is well defined classically if q ∈ C(R + ) ∩ L 2 (R + ). The function L(ψ) admits analytic continuation with respect to λ to the whole complex plane C. Therefore, the convolution ψ is defined for all λ ∈ C. We are especially interested in the value λ = − 1 4 because it appears in Equation (27).
To illustrate the argument with analytic continuation, consider a simple example: where s = pt. Formula (36) is valid classically for Rez > 0, but remains valid for all z ∈ C, z = 0, −1, −2, . . . . . ., by the analytic continuation with respect to z because Γ(z) is analytic for z ∈ C, z = 0, −1, −2, . . . . . ., and p −z is an entire function of z. Formula (33) follows from (36) immediately: just divide both sides of (36) by Γ(z). The integral (36) diverges classically for Rez ≤ 0, but formula (36) is valid by analytic continuation for all z ∈ C except for z = 0, −1, −2, . . . . . .. In [10], a regularization method is described for defining divergent integrals. By this method one writes and uses analytic continuation with respect to z. The third term of the right side of Equation (37) for Rez > 0 can be written as 1 z + 0.5 1 z+1 . The first integral on the right side of (37) is analytic with respect to z in the region Rez > −2, the second integral is also analytic with respect to z in this region and the third term, 1 z + 0.5 1 z+1 , admits analytic continuation with respect to z from the region Rez > 0 to the complex plane C except for the points z = 0 and z = −1. Thus, the right side of Equation (37) admits analytic continuation with respect to z to the region Rez > −2, except for the points z = 0 and z = −1 at which it has simple poles. So, this right side is well defined at z = − 1 4 . However, the right side of (37) is much less convenient than Γ(z), the expression we use. If one deals with the integral Φ λ q, then the advantage of our definition, based on the Laplace transform, is even greater because the three terms, analogous to the terms on the right side of Equation (37), will depend on p and on z and there is no separation of z-dependence similar to the one we have in Equation (36). Furthermore, these three terms are not all the Laplace transforms. Consequently, it is wrong to use the regularization procedure from [10] in our problem.
where δ(t) is the Dirac distribution.
This proves formula (39). Lemma 4 is proved. 2 This proof is taken from [1]. Our plan is to prove that Equation (28) has a solution q(t) ∈ C(R + ) provided that b 0 (t) is smooth and rapidly decaying as t → ∞. Moreover, this solution is unique in C(R + ) and q(0) = 0. Any solution b(t) ≥ 0 to inequality (27) satisfies the relation b(t) ≤ q(t).
In particular, b(0) = 0. This is the NSP paradox because a priori b(0) = 0. To realize this plan, let us investigate Equation (28). First, let us apply to (28) the operator Φ 1/4 and use Lemma 4 to get Since L(η) ≥ 0, one concludes that 1 ≤ c(−c 1 p 1 4 + p − 1 4 Γ(1/4)). If L(η) ≡ 0, then one has a contradiction: take p → +∞, then the above inequality yields 1 ≤ −∞. This contradiction proves that L(η) = 0, so z = 0. Theorem 3 is proved. 2 Theorem 3 is not used in the derivation of our basic conclusions. This theorem is new. Earlier uniqueness theorems were proved under different assumptions on the spaces to which the solution to the NSP belongs, see [2,11].

Conclusions
From Theorems 1 and 2 the NSP paradox follows. From the NSP paradox we conclude that the NSP is physically and mathematically contradictive and is not a correct description of the dynamics of incompressible viscous fluid.
Thus, one of the millennium problems is solved.