Theory of hyper-singular integrals and its application to the Navier-Stokes problem

TL;DR

This paper develops a theory for hyper-singular integrals, especially at the critical value , and applies it to demonstrate that the Navier-Stokes equations lack solutions, challenging their validity in fluid mechanics.

Contribution

It introduces a new framework for hyper-singular integrals and proves non-existence of solutions for Navier-Stokes equations in  case, revealing fundamental issues.

Findings

Proves existence and uniqueness of solutions for  integral equations.

Shows Navier-Stokes equations have no solutions under certain conditions.

Demonstrates a paradox where initial data implies zero velocity at initial time.

Abstract

In this paper the convolution integrals $\int_0^t(t-s)^{\lambda -1}b(s)ds$ with hyper-singular kernels are considered, where $\lambda\le 0$ and $b$ is a smooth or $b$ is in $L^1(\mathbb{R}_+)$. For such $\lambda$ these integrals diverge classically even for smooth $b$. These convolution integrals are defined in this paper for $\lambda\le 0$, $\lambda\neq 0,-1,-2,...$. Integral equations and inequalities are considered with the hyper-singular kernels $(t-s)^{\lambda -1}_+$ for $\lambda\le 0$, where $t^\lambda_+:=0$ for $t<0$. In particular, one is interested in the value $\lambda=-\frac 14$ because it is important for the Navier-Stokes problem (NSP). Integral equations of the type $b(t)=b_0(t)+ \int_0^t(t-s)^{\lambda-1}b(s)ds$, $\lambda\le 0$, are studied. The solution of these equations is investigated, existence and uniqueness of the solution is proved for $\lambda=-\frac 1 4$. This special value of $\lambda$ is of basic importance for a study of the Navier-Stokes problem (NSP). The above results are applied to the analysis of the NSP in the space $\mathbb{R}^3$ without boundaries. It is proved that the NSP is contradictory in the following sense: even if one assumes that the initial data $v_0(x):=v(x,0)\not\equiv 0$, $\nabla \cdot v_0(x)=0$ 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 it proves that the NSP does not have a solution.

Equations

$$J:={\int }_0^{\infty }{t}^{\lambda -1}\varphi (t)dt,$$

$$J={\int }_0^1{t}^{\lambda -1}\varphi (t)dt+{\int }_1^{\infty }{t}^{\lambda -1}\varphi (t)dt:=j_1+j_2.$$

$$j_1={\int }_0^1{t}^{\lambda -1}(\varphi (t)-\varphi (0))dt+\varphi (0)\frac{t^{\lambda }{|}_0^1}{\lambda }={\int }_0^1{t}^{\lambda -1}(\varphi (t)-\varphi (0))dt+\varphi (0)/\lambda .$$

$$I(t):={\Phi }_{\lambda }\star b:={\int }_0^t{\Phi }_{\lambda }(t-s)b(s)ds.$$

$$L(b):={\int }_0^{\infty }{e}^{-pt}b(t)dt.$$

$$L(t^{\lambda -1})={\int }_0^{\infty }{e}^{-pt}{t}^{\lambda -1}dt={\int }_0^{\infty }{e}^{-s}{s}^{\lambda -1}ds{p}^{-\lambda }=\frac{\Gamma (\lambda )}{p^{\lambda }}.$$

$$L({\Phi }_{\lambda })=p^{-\lambda }.$$

$$\Gamma (z+1)=z\Gamma (z),\Gamma (z)\Gamma (1-z)=\frac{\pi }{\sin (\pi z)},2^{2z-1}\Gamma (z)\Gamma (z+\frac{1}{2})={\pi }^{1/2}\Gamma (2z).$$

$$L(I(t))=L({\Phi }_{\lambda })L(b)=\frac{L(b)}{p^{\lambda }}$$

$$I(t)=L^{-1}\big{(}L(b)p^{-\lambda}\big{)},$$

$${\Phi }_{\lambda }\star {\Phi }_{\mu }={\Phi }_{\lambda +\mu }.$$

$${\Phi }_0(t)=\delta (t),$$

$$L({\Phi }_{\lambda }\star {\Phi }_{\mu })=\frac{1}{p^{\lambda +\mu }}.$$

$$L^{-1}\Big{(}\frac{1}{p^{\lambda+\mu}}\Big{)}=\Phi_{\lambda+\mu}.$$

$$p^{-(\lambda +\mu )}=1,L^{-1}1=\delta (t).$$

$${\Phi }_{\lambda }\star {\Phi }_{\mu }=\frac{1}{\Gamma (\lambda )\Gamma (\mu )}{\int }_0^t(t-s{)}^{\lambda -1}{s}^{\mu -1}ds=\frac{t_{+}^{\lambda +\mu -1}}{\Gamma (\lambda )\Gamma (\mu )}{\int }_0^1(1-u{)}^{\lambda -1}{u}^{\mu -1}du=\frac{t_{+}^{\lambda +\mu -1}}{\Gamma (\lambda +\mu )},$$

$$B(\lambda ,\mu ):={\int }_0^1{u}^{\lambda -1}(1-u{)}^{\mu -1}du=\frac{\Gamma (\lambda )\Gamma (\mu )}{\Gamma (\lambda +\mu )}.$$

$$b(t)=b_0(t)+c{\int }_0^t(t-s{)}^{\lambda -1}b(s)ds,$$

$$|{\int }_0^{\infty }{t}_{+}^{\lambda }{\varphi }_n(t)dt|\le (a^{\lambda }+b^{\lambda }){\int }_a^b|{\varphi }_n(t)|dt.$$

$$h:={\int }_0^t(t-s{)}^{-\frac{5}{4}}b(s)ds:=t_{+}^{-\frac{5}{4}}\star b$$

$$L(t_{+}^{-\frac{5}{4}}\star b)=L(t_{+}^{-\frac{5}{4}})L(b).$$

$$\Gamma (-\frac{1}{4})=-4\Gamma (3/4):=-c_1,c_1>0.$$

$$L(h)=-c_1{p}^{\frac{1}{4}}L(b),$$

$$L(b)=-c_1^{-1}{p}^{-\frac{1}{4}}L(h).$$

$$b(t)=b_0(t)-c{c}_1{\Phi }_{\lambda }\star b,$$

$$c_1=|\Gamma (-\frac{1}{4})|,\lambda =-\frac{1}{4}.$$

$$b_{n+1}=-(c{c}_1{)}^{-1}{\Phi }_{1/4}\star b_n+(c{c}_1{)}^{-1}{\Phi }_{1/4}\star b_0,b_{n=0}=(c{c}_1{)}^{-1}{\Phi }_{1/4}\star b_0,b=\underset{n\to \infty }{\lim }{b}_n.$$

$${\Phi }_{1/4}\star b={\Phi }_{1/4}\star b_0-c{c}_{1}b,$$

$$b=-(c{c}_1{)}^{-1}{\Phi }_{1/4}\star b+(c{c}_1{)}^{-1}{\Phi }_{1/4}\star b_0.$$

$$f_{n+1}=A{f}_n+g,f_0=g;\underset{n\to \infty }{\lim }{f}_n=f,$$

Full text

In this paper the convolution integrals $\int_{0}^{t}(t-s)^{\lambda-1}b(s)ds$ with hyper-singular kernels are considered, where $\lambda\leq 0$ and $b$ is a smooth or $b$ is in $L^{1}({\mathbb{R}}_{+})$. For such $\lambda$ these integrals diverge classically even for smooth $b$. These convolution integrals are defined in this paper for $\lambda\leq 0$, $\lambda\neq 0,-1,-2,...$.

Integral equations and inequalities are considered with the hyper-singular kernels $(t-s)^{\lambda-1}_{+}$ for $\lambda\leq 0$, where $t^{\lambda}_{+}:=0$ for $t<0$.

In particular, one is interested in the value $\lambda=-\frac{1}{4}$ because it is important for the Navier-Stokes problem (NSP).

Integral equations of the type $b(t)=b_{0}(t)+\int_{0}^{t}(t-s)^{\lambda-1}b(s)ds$, $\lambda\leq 0$, are studied.

The solution of these equations is investigated, existence and uniqueness of the solution is proved for $\lambda=-\frac{1}{4}$. This special value of $\lambda$ is of basic importance for a study of the Navier-Stokes problem (NSP).

The above results are applied to the analysis of the NSP in the space ${\mathbb{R}}^{3}$ without boundaries. It is proved that the NSP is contradictory in the following sense: even if one assumes that the initial data $v(x,0)>0$ one proves that the solution $v(x,t)$ to the NSP has the property $v(x,0)=0$, in general. This paradox shows that the NSP is not a correct description of the fluid mechanics problem and it proves that the NSP does not have a solution, in general.

1 Introduction

1.1. In this paper the new definition of the convolution $\int_{0}^{t}(t-s)^{\lambda-1}b(s)ds$ with hyper-singular functions is given. We compare this definition with the one, based on the distribution theory, [2]. The $b(s)$ is assumed locally integrable function on ${\mathbb{R}}_{+}:=[0,\infty)$. This assumption is satisfied in the Navier-Stokes problem (NSP), see [5], Chapter 5, where the integral equations of the type $b(t)=b_{0}(t)+\int_{0}^{t}(t-s)^{\lambda-1}b(s)ds$ with $\lambda=-\frac{1}{4}$ are of interest. Classically these integral equations do not make sense because the integrals diverge if $\lambda\leq 0$. In Sections 1–4 of this paper a new definition of such integrals is given, the solution to the integral equation with hyper-singular kernel is investigated. These results are used in Section 5 of this paper, where the basic results concerning the NSP are obtained.

We analyze the NSP and prove that the NSP is physically not a correct description of the fluid mechanics problem and that the NSP does not have a solution, in general. The words ”in general” mean that if the initial velocity $v_{0}=0$ and the force $f=0$, then the NSP has the solution $v(x,t)=0$ which exists for all $t\geq 0$. This meaning is valid in Theorem 5.2, below.

For future use we define $\Phi_{\lambda}(t):=\frac{t^{\lambda-1}}{\Gamma(\lambda)}$ and the convolution $\Phi_{\lambda}\star b:=\int_{0}^{t}\Phi_{\lambda}(t-s)b(s)ds$. Here and below $t:=t_{+}$, that is, $t=0$ if $t<0$ and $t=t$ if $t\geq 0$.

1.2. Let us give the standard definition of the singular integral used in the distribution theory. Let

[TABLE]

where the test function $\phi(t)\in C^{\infty}_{0}({\mathbb{R}})$.

Integral (1) diverges classically (that is, in the classical sense) if $\lambda\leq 0$. It is defined in distribution theory (for example, in [2]) as follows:

[TABLE]

The integral $j_{2}$ converges classically for any complex $\lambda\in{\mathbb{C}}$ and is analytic with respect to $\lambda$. The integral $j_{1}$ for $\lambda>0$ converges classically and can be written as

[TABLE]

The right side of (3) admits analytic continuation with respect to $\lambda$ from Re$\lambda>0$ to the region Re$\lambda>-1$. Thus, formulas (2) and (3) together define integral (1) for Re$\lambda>-1$. The singular integral $J$ has a simple pole at $\lambda=0$, diverges classically for $-1<\lambda<0$, but is defined in this region by formulas (2) and (3) by analytic continuation with respect to $\lambda$. This procedure can be continued and $J$ can be defined for an arbitrary large fixed negative $\lambda$, $\lambda\neq 0,-1,-2,...$.

1.3. Let us define the convolution

[TABLE]

We assume that $b(t)\in L^{1}({\mathbb{R}}_{+})$ and the Laplace transform of $b$ is defined for Re$p\geq 0$ by the formula

[TABLE]

Let us define $L(t^{\lambda-1})$ not using the distribution theory. For $\lambda>0$ one has:

[TABLE]

It follows from (6) and from the definition of $\Phi_{\lambda}=\frac{t^{\lambda-1}}{\Gamma(\lambda)}$ that

[TABLE]

The gamma function $\Gamma(\lambda)$ is analytic in $\lambda\in{\mathbb{C}}$ except for the simple poles at $\lambda=-n$, $n=0,1,2....$, with the residue at $\lambda=-n$ equal to $\frac{(-1)^{n}}{n!}$. It is known that

[TABLE]

The function $\frac{1}{\Gamma(\lambda)}$ is an entire function of $\lambda$. These properties of $\Gamma(z)$ can be found, for example, in [4].

The right side of (6) is analytic with respect to $\lambda\in{\mathbb{C}}$ except for $\lambda=0,-1,-2,....$ and, therefore, defines $L(t^{\lambda-1})$ for all these $\lambda$ by analytic continuation with respect to $\lambda$ without using the distribution theory.

Let us define the convolution $I(t)$ using its Laplace transform

[TABLE]

and its inverse:

[TABLE]

where $L^{-1}$ is the inverse of the Laplace transform. Since the null-space of $L$ is trivial, that is, the zero element, the inverse $L^{-1}$ is well defined on the range of $L$.

For $-1<\lambda<0$, in particular for $\lambda=-\frac{1}{4}$, formula (10), can be interpreted as a generalized Fourier integral. The value $\lambda=-\frac{1}{4}$ is very important in NSP, see monograph [5], Chapter 5, and Section 5 below. We return to this question later, when we discuss the integral equations with hyper-singular integrals.

1.4. Let us now prove the following result that will be used later.

Theorem 1.1. One has

[TABLE]

for any $\lambda,\mu\in{\mathbb{C}}$. If $\lambda+\mu=0$ then

[TABLE]

*where $\delta(t)$ is the Dirac distribution. *

Proof. By formulas (7) and (9) with $b(t)=\Phi_{\mu}(t)$ one gets

[TABLE]

By formula (7) one has

[TABLE]

This proves formula (11).

If $\lambda+\mu=0$ then

[TABLE]

This proves formula (12).

Theorem 1.1 is proved. $\Box$

Remark 1.1. Let us give an alternative proof of formula (11).

For Re$\lambda>0$, Re$\mu>0$ one has

[TABLE]

where the right side of (16) is equal to $\Phi_{\lambda+\mu}$ and we have used the known formula for the beta function:

[TABLE]

Analytic properties of beta function follow from these of the gamma function.

Remark 1.2. Theorem 1.1 is proved in [2], pp.150–151. Our proof differs from the proof in [2]. It is not clear how the proof in [2] is related to the definition of regularized hyper-singular integrals used in [2].

2 Preparation for investigation of integral equations

with hyper-singular kernels

In this section we start an investigation of equations of the following type

[TABLE]

where $b_{0}$ is a smooth functions rapidly decaying with all its derivatives as $t\to\infty$, $b_{0}(t)=0$ if $t<0$. We are especially interested in the value $\lambda=-\frac{1}{4}$, because of its importance for the Navier-Stokes theory, [5], Chapter 5, [8], [9].

The integral in (18) diverges in the classical sense for $\lambda\leq 0$. Our aim is to define this hyper-singular integral.

There is a regularization method to define singular integrals $J:=\int_{{\mathbb{R}}}t_{+}^{\lambda-1}\phi(t)dt$, $\lambda\leq 0$, in the distribution theory, see the Introduction, Section 1.2. The integral in (18) is a convolution, which is defined in [2], p.135, as a direct product of two distributions.

This definition is not suitable for our purposes because $t_{+}^{\lambda-1}$ for any $\lambda\leq 0$, $\lambda\neq 0,-1,-2,...$ is a distribution on the space $\mathcal{K}:=C^{\infty}_{0}(\mathbb{R}_{+})$ of the test functions, but it is not a distribution in the space of the test functions $K:=C^{\infty}_{0}({\mathbb{R}})$ used in [2].

Indeed, one can find $\phi\in K$ such that $\lim_{n\to\infty}\phi_{n}=\phi$ in $K$, but $\lim_{n\to\infty}\int_{{\mathbb{R}}}t_{+}^{\lambda-1}\phi_{n}(t)dt=\infty$ for $\lambda\leq 0$, so that $t_{+}^{\lambda-1}$ is not a linear bounded functional in $K$, i.e., not a distribution.

For example, the integral $\int_{0}^{\infty}t^{\lambda-1}\phi(t)dt$ is not a bounded linear functional on $K$: take a $\phi$ which is vanishing for $t>1$, positive near $t=0$ and non-negative on $[0,1]$. Then this integral diverges at such $\phi$ and is not a bounded linear functional on $K$.

On the other hand, one can check that $t_{+}^{\lambda-1}$ for any $\lambda\in R$ is a distribution (a bounded linear functional) in the space $\mathcal{K}=C^{\infty}_{0}({\mathbb{R}}_{+})$ with the convergence $\phi_{n}\to\phi$ in $\mathcal{K}$ defined by the following requirements:

a) the supports of all $\phi_{n}$ belong to an interval $[a,b]$, $0<a\leq b<\infty$,

b) $\phi_{n}^{(j)}\to\phi^{(j)}$ in $C([a,b])$ for all $j=0,1,2,....$.

Indeed, the functional $\int_{0}^{\infty}t_{+}^{\lambda}\phi(t)dt$ is linear and bounded in $\mathcal{K}$:

[TABLE]

A similar estimate holds for all the derivatives of $\phi_{n}$.

Although $t_{+}^{-\frac{5}{4}}$ is a distribution in $\mathcal{K}$, the convolution

[TABLE]

*cannot be defined similarly to the definition in the book [2] because the function

$\int_{0}^{\infty}\phi(u+s)b(s)ds$ does not, in general, belong to $\mathcal{K}$ even if $\phi\in\mathcal{K}$.*

Let us define the convolution $h$ using the Laplace transform (6). Laplace transform of distributions is studied in [1]. There one finds a definition of the Laplace transform of distributions, the Laplace transform of convolutions, tables of the Laplace transforms of distributions, in particular, formula (6) and other information.

One has

[TABLE]

To define $L(t^{\lambda-1})$ for $\lambda\leq 0$, note that for Re$\lambda>0$ the classical definition (6) holds. The right side of (6) admits analytic continuation to the complex plane of $\lambda$, $\lambda\neq 0,-1,-2,....$. This allows one to define integral (6) for any $\lambda\neq 0,-1,-2,...$. It is known that $\Gamma(z+1)=z\Gamma(z)$, so

[TABLE]

Therefore, we define $h$ by the formula $h=L^{-1}(Lh)$ and defining $L(h)$ as follows:

[TABLE]

where formula (6) with $\lambda=-\frac{1}{4}$ was used and we assume that $b$ is such that $L(b)$ can be defined. That $L(b)$ is well defined in the Navier-Stokes theory follows from the a priori estimates proved in [5], Chapter 5 and in Section 5 below, see Theorem 5.1.

From (23) one gets

[TABLE]

3 Integral equation

Consider equation (18). It can be rewritten as

[TABLE]

where

[TABLE]

Theorem 3.1. Equation (25)-(26) has a unique solution in $C(0,T)$ for any $T>0$ if $b_{0}$ is sufficiently smooth and rapidly decaying as $t$ grows. This solution can be obtained by iterations:

[TABLE]

Proof. Applying to equation(25) the operator $\Phi_{1/4}\star$ and using equation (12) one gets a Volterra-type equation

[TABLE]

or

[TABLE]

The operator $\Phi_{\lambda}\star$ with $\lambda>0$ is a Volterra-type operator. Therefore equation (29) can be solved for $b$ by iterations, see Lemma 3.1 below and [5], p.53, Lemmas 5.10 and 5.11.

If $b_{0}\geq 0$ and $cc_{1}$ is sufficiently large, then the solution to (25) is non-negative, $b\geq 0$, see Reamark 3.1 below.

Theorem 3.1 is proved. $\Box$

For convenience of the reader let us prove the result about solving equation (29) by iterations, mentioned above.

Lemma 3.1. The operator $Af:=\int_{0}^{t}(t-s)^{p}f(s)ds$ in the space $X:=C(0,T)$ for any fixed $T\in[0,\infty)$ and $p>-1$ has spectral radius $r(A)$ equal to zero, $r(A)=0$. The equation $f=Af+g$ is uniquely solvable in $X$. Its solution can be obtained by iterations

[TABLE]

for any $g\in X$ and the convergence holds in $X$.

Proof. The spectral radius of a linear operator $A$ is defined by the formula

[TABLE]

By induction one proves that

[TABLE]

From this formula and the known asymptotic of the gamma function the conclusion $r(A)=0$ follows. The convergence result (30) is analogous to the well known statement for the assumption $\|A\|<1$. A more detailed argument can be found in [5], p.53.

Lemma 3.1 is proved. $\Box$

Remark 3.1. If $c>0$ is sufficiently large, then the norm of the operator $B:=(cc_{1})^{-1}\Phi_{1/4}\star$ in $C(0,T)$ is less than one: $\|B\|<1$. In this case, $(I-B)^{-1}=\sum_{j=0}^{\infty}B^{j}$ is a positive operator.

Let us now give another approach to solving integral equation (25) with $\lambda=-\frac{1}{4}$.

Theorem 3.2. The solution to equation (25) with $\lambda=-\frac{1}{4}$ does exist, is unique, and belongs to $C({\mathbb{R}}_{+})$ provided that $b_{0}(t)\in C({\mathbb{R}}_{+})$ and $|b_{0}(t)|+|b^{\prime}(t)|\leq c(1+t)^{-2}$.

Proof. Take the Laplace transform of equation (25) with $\lambda=-\frac{1}{4}$, use formula (7) to get

[TABLE]

Thus,

[TABLE]

Therefore

[TABLE]

Let us check that

[TABLE]

From our assumptions about $b_{0}(t)$ it follows that $|L(b_{0})|\leq c(1+|p|)^{-1}$, Re$p\geq 0$. Let $p=iw$. Since $b(t)=(2\pi)^{-1}\int_{-\infty}^{\infty}e^{iwt}L(b)dw$, one gets

[TABLE]

where $c_{2}>0$ is some constant. Here we have used the inequality

[TABLE]

Recall that by $c>0$ various constants are denoted.

Let us check (37) for $w\geq 0$. For $w<0$ the argument is similar. One has $(iw)^{1/4}=e^{i\pi/8}w^{1/4}$,

[TABLE]

where $C:=cc_{1}w^{1/4}>0$. Therefore,

[TABLE]

Consequently, inequality (37) is checked.

Theorem 3.2 is proved. $\Box$

Remark 3.2. It follows from formula (29) that $b(0)=0$ because $\lim_{t\to 0}\Phi_{\frac{1}{4}}\star b_{0}=0$ and $\lim_{t\to 0}\Phi_{\frac{1}{4}}\star b=0$ holds if $b$ is a locally integrable bounded on ${\mathbb{R}}_{+}$ function $b=b(t)$.

4 Integral inequality

Consider the following inequality

[TABLE]

where $c_{1}=-\Gamma(-\frac{1}{4})$ for $\lambda=-\frac{1}{4}$

Let $f=f(t)\in L^{1}({\mathbb{R}}_{+})$ be some function. If $q\leq f$ then $\Phi_{1/4}\star q\leq\Phi_{1/4}\star f.$ Therefore, inequality (38) with $\lambda=-\frac{1}{4}$, after applying to both sides the operator $\Phi_{1/4}\star$, implies

[TABLE]

Inequality (39) for sufficiently large $c>0$ can be solved by iterations with the initial term $(cc_{1})^{-1}\Phi_{1/4}\star b_{0}$, see Remark 3.1. This yields

[TABLE]

where $b$ solves the integral equation (25). This follows from Theorem 4.1.

Theorem 4.1. Assume that $b$ solves (25), $c>0$ is sufficiently large, $b_{0}(t)$ satisfies conditions stated in Theorem 3.2 and $q\geq 0$ solves inequality (38). Then inequality (40) holds.

Proof. Denote $z:=b-q$, where

[TABLE]

Then

[TABLE]

Solving this inequality by iterations and using Remark 3.1 one obtains (40). If $c>0$ is arbitrary, then this argument yields inequality (40) for sufficiently small $t>0$ because the norm of the operator $(cc_{1})^{-1}\Phi_{1/4}\star$ tends to zero when $t\to 0$.

Theorem 4.1 is proved. $\Box$

Papers [6], [7], [10] also deal with hyper-singular integrals.

5 Application to the Navier-Stokes problem

In this Section we apply the results of Sections 1–4 to the Navier-Stokes problem. Especially the results of Sections 3 and 4 will be used.

The Navier-Stokes problem (NSP) in ${\mathbb{R}}^{3}$ is discussed in many books and papers ( see [5], Chapter 5, and references therein). The uniqueness of a solution in ${\mathbb{R}}^{3}$ was proved in [3], [5] and in [9] in different norms. The existence of the solution to the NSP is discussed in [5].

The goal of this Section is to prove that the statement of the NSP is contradictory. Therefore, the NSP is not a physically correct statement of the problem of fluid mechanics. We prove that the solution to the NSP does not exist, in general. Therefore, in this Section a negative solution to one of the millennium problems is given.

What is a physically correct statement of problems of fluid mechanics remains an open problem.

We prove the paradox in the NSP. This paradox can be described as follows:

*One can have initial velocity $v(x,0)>0$ in the NSP and, nevertheless, the solution $v(x,t)$ to this NSP must have the zero initial velocity: $v(x,0)=0$. *

This paradox proves that the statement of the NSP is contradictory, that the NSP is not a physically correct statement of the fluid mechanics problem and the solution to the NSP does not exist, in general.

The NSP in ${\mathbb{R}}^{3}$ consists of solving the equations

[TABLE]

see, for example, books [3] and [5], Chapter 5.

Vector-functions velocity $v=v(x,t)$ and the exterior force $f=f(x,t)$ and the scalar function $p=p(x,t)$, the pressure, are assumed to decay as $|x|\to\infty$ and $t\in{\mathbb{R}}_{+}:=[0,\infty)$.

The derivative with respect to time is denoted $v^{\prime}:=v_{t}$, $\nu=const>0$ is the viscosity coefficient, the velocity $v=v(x,t)$ and the pressure $p=p(x,t)$ are unknown, $v_{0}=v(x,0)$ and $f(x,t)$ are known. It is assumed that $\nabla\cdot v_{0}=0$. Equations (43) describe viscous incompressible fluid with density $\rho=1$.

Let us assume for simplicity that $f=0$. This do not change our arguments and our logic.

The solution to NSP (43) solves the integral equation:

[TABLE]

where $(v,\nabla)v=v_{j}v_{p,j}$, over the repeated indices summation is assumed and $v_{p,j}:=\frac{\partial v_{p}}{\partial x_{j}}$.

Equation (44) implies an integral inequality of the type studied in Sections 3 and 4 (see also [5], Chapter 5).

Formula for the tensor $G=G(x,t)=G_{pm}(x,t)$ is derived in [5], p.41:

[TABLE]

The term $F=F(x,t)$, in our case when $f=0$, depends only on the data $v_{0}$ (see formula (5.42) in [5]):

[TABLE]

where

[TABLE]

We assume throughout that

[TABLE]

is such that $F$ is bounded in all the norms we use.

Let us use the Fourier transform:

[TABLE]

Fourier transform equation (44) and get the integral equation:

[TABLE]

where $\bigstar$ denotes the convolution in ${\mathbb{R}}^{3}$.

For brevity we omitted the tensorial indices: instead of $\tilde{G}_{mp}\tilde{v}_{j}\bigstar(i\xi_{j})\tilde{v}_{p}$, where one sums up over the repeated indices, we wrote $\tilde{G}\tilde{v}\bigstar(i\xi\tilde{v})$.

From formula (5.9) in [5], see formula (45) one gets:

[TABLE]

One has $|\delta_{pm}-\frac{\xi_{p}\xi_{m}}{\xi^{2}}|\leq c$. Therefore,

[TABLE]

We denote by $c>0$ various constants independent of $t$ and $\xi$, by $\|\tilde{v}\|$ the norm in $L^{2}({\mathbb{R}}^{3})$ and by $(v,w)$ the inner product in $L^{2}({\mathbb{R}}^{3})$.

Let us introduce the norm

[TABLE]

One has

[TABLE]

by the Parceval equality.

Assumption A. Assume that $F(x,t)$ is a smooth function rapidly decaying together with all its derivatives. In particular,

[TABLE]

Assumption A holds throughout Section 5 and is not repeated. It is known that

[TABLE]

[TABLE]

see [5], p.52.

Theorem 5.1. *Inequalities (55)–(56) hold. *

Theorem 5.2. The NSP (43) does not have a solution, in general.

Proof of Theorem 5.1. Proof of Theorem 5.1 can be found in [5]. Because of the importance of the third inequality (56) and of its novelty, we give its proof in detail.

Let $|\tilde{v}(\xi,t)|:=u$, $|\tilde{F}|:=\mu(\xi,t):=\mu$. From equation (50) one gets:

[TABLE]

where the Parceval formula

[TABLE]

was used.

By direct calculation one derives the following inequality:

[TABLE]

It follows from this inequality and from (57) by multiplying by $|\xi|$ and taking the norm $\|\cdot\|$ of the resulting inequality that the following integral inequality holds:

[TABLE]

where

[TABLE]

The function $b_{0}(t)$ is smooth and rapidly decaying due to Assumption A.

Let $\beta$ solve the following equation:

[TABLE]

Equation (62) can be written as

[TABLE]

where $\star$ denotes the convolution of two functions on ${\mathbb{R}}_{+}$ and $c_{1}=|\Gamma(-\frac{1}{4})|$. The convolution on ${\mathbb{R}}_{+}$ was defined in the Introduction. In Section 4 the relation between the solutions to integral equation (63) and integral inequality (60) was studied and the inequality

$b(t)\leq\beta(t)$ was proved.

Taking the Laplace transform of equation (62) and using equation (6) with $\lambda=-\frac{1}{4}$, we get

[TABLE]

so

[TABLE]

Therefore,

[TABLE]

It follows from (66) that

[TABLE]

where the argument $p$ of the function $L(b_{0})$ is equal to $i\tau$, $p=i\tau$, and we have used the decay $O(|\tau|^{-1})$ of $|L(b_{0})|$ as a function of $p=i\tau$ as $|\tau|\to\infty$.

This decay follows from Assumption A and implies that the integrand in (67) belongs to $L^{1}({\mathbb{R}})$ because of the following inequality, proved at the end of Section 3:

[TABLE]

Thus, the third estimate (56) of Theorem 5.1 is proved. $\Box$

Proof of theorem 5.2. If $v_{0}(x)=v(x,0)\not\equiv 0$ and $\nabla\cdot v_{0}(x)=0$ then $b_{0}(0)>0$. Apply to equation (62) the operator $\Phi_{1/4}\star$ and use Theorem 1.1. This yields

[TABLE]

where formula (6) was used, $c_{1}=-\Gamma(-\frac{1}{4})>0$ and $\Phi_{\frac{1}{4}}\star\Phi_{-\frac{1}{4}}=\delta$, where $\delta$ is the delta-function, see formulas (11)–(12). We assume that $b_{0}(t)$ satisfies Assumption A, so it is smooth and rapidly decaying. Then equation (69) can be solved by iterations by Theorem 3.1 and the solution $\beta$ is also smooth. Therefore,

[TABLE]

Consequently, it follows from (69) that $\beta(0)=0$. Since $0\leq b(t)\leq\beta(t)$, one concludes that

[TABLE]

This result proves that the NSP problem does not have a solution, in general. Indeed, starting with an initial data which is positive we prove that the corresponding solution to the NSP must have the initial data equal to zero. This is the NSP paradox, see [11].

Of course, if the data $v_{0}(x)=v(x,0)=0$ then the solution exists for all $t\geq 0$ and is equal to zero by the uniqueness theorem, see, for example, [5], [9].

Other paradoxes of the theory of fluid mechanics are mentioned in [3].

Theorem 5.2 is proved. $\Box$