Navier-Stokes equations paradox

It is proved that if the Navier-Stokes problem is considered in ℝ³, then the initial velocity has to be zero. This is the paradox. The consequences of this paradox are: a) The NSP is contradictory physically and mathematically; it is not a correct formulation of the problem of motion of the incompressible viscous fluid. b) The solution to the NSP does not exist on [0, ∞). This solves one of the millennium problems.

Introduction

It is of general interest to find out whether the equations describing the fluid are correct. What does it mean that the equations are correct? It means that: a) their solution is unique (in a suitable space), b) the solution does exist (in this space), c) there are no consequences of these equations that contradict physical assumptions. Existence of such consequences we call paradoxes.

The Navier-Stokes (NS) equations (NS Problem-NSP) describe the non-compressible viscous fluid. It is proved that the solution to NSP is unique (in various spaces) [3, 5, 7, 8]. The aim of this paper is to describe a paradox in the NSP and its consequences: the NSP is contradictory and the solution to the NSP does not exists on [0, ∞), see also [5, 11, 12].

First we outline the theory. The NSP in ℝ³ consists of solving the equations:

$$v^{\prime }+\left(v,\nabla \right)v=f-\nabla p+v\Delta v,\nabla \cdot v=0,v\left(x,0\right)=v_0\left(x\right).$$

Here v = v(x, t) is the fluid velocity, v’ := dv/dt, f = f (x, t) is the exterior force, p = p (x, t) is the pressure, the fluid density ρ = 1, v = const > 0 is the viscosity coefficient, x ℝ³, t > 0. The f, v₀ are known, v, p are unknown. It is assumed that f, v₀ are smooth and rapidly decaying functions of their arguments, v, p decay

at infinity. It is proved in [5], that Eqs. (1) are equivalent to

$$v=F-{\int }_0^{t}ds\underset{{ℝ}^3}{\int }G\left(x-y,t-s\right)\left(v-\nabla \right)vdy,$$

$$F={\int }_0^{t}ds\underset{{ℝ}^3}{\int }G\left(x-y,t-s\right)f\left(y,s\right)dy+\underset{{ℝ}^3}{\int }g\left(x-y,t-s\right)v_0\left(y\right)dy,$$

$$g\left(x,t\right)=\frac{e^{-\frac{|x{|}^2}{4vt}}}{{\left(4v\pi t\right)}^{3/2}},g\left(x,0\right)=\delta \left(x\right);g\left(x,t\right)=0\text{if}t<0.$$

Here δ(x) is the delta-function and G is the tensor solving the equations:

$$G^{\prime }-v\Delta G=\delta \left(x\right)\delta \left(t\right){\delta }_{jm}-\nabla pm,\nabla \cdot G=0,G=0\text{if}t<0,$$

where δ_jm is the Kronecker symbol, p_m = p(x, t)e_m, e_j, 1 ≤ j ≤ 3, is an orthonormal basis in ℝ³, p = 0 if t < 0, ∇ p_m = $\frac{\partial p}{\partial x_j}$, e_j e_m is a tensor. Tensor G is calculated in [5] explicitly and the equivalence of (1) and (2) is proved there. Uniqueness of the solution to (2) in various functional spaces is proved in [5, 8]. Let us outline some ideas of our proofs. Fourier transform (2) with respect to x and get

$$\tilde{v}=\tilde{F}-{\int }_0^{t}ds\tilde{G}\left(\xi ,t-s\right)\tilde{v}*i\xi \tilde{v},$$

where * is the convolution in ℝ³, $\tilde{v}={\left(2\pi \right)}^{-3}{\int }_{{ℝ}^3}{e}^{-i\xi \cdot x}v\left(x,t\right)dx,$

$$\tilde{G}={\left(2\pi \right)}^{-3}\left({\delta }_{jm}-{\xi }_j{\xi }_m{\left({\xi }^2\right)}^{-1}\right)e^{-v{\xi }^{2}t},$$

see [5, p. 11]. One has $\left|v*i\xi \tilde{v}\right|\le \Vert \tilde{v}\Vert \Vert \left|\xi \right|\tilde{v}\Vert ,where\Vert \cdot \Vert ={\Vert \cdot \Vert }_{L^2\left({ℝ}^3\right)}.$ Also, $\left|\tilde{G}\right|\le c{e}^{-v{\xi }^{2}t}$ where c > 0 stands for various constants independent of t and ξ, $\Vert \tilde{G}\left(\xi ,t-s\right)\Vert \le c{\left(t-s\right)}^{-\frac{3}{4}},$

$$\Vert \left|\xi \right|\tilde{G}\left(\xi ,t-s\right)\Vert \le c{\left(t-s\right)}^{-\frac{5}{4}}.$$

Denote $\Vert \left|\xi \right|\tilde{v}\Vert :=b\left(t\right)\ge 0,\Vert \left|\xi \right|\tilde{F}\Vert :=b_0\left(t\right)\ge 0$. Multiply (6) by $\left|\xi \right|$, take the norm $\Vert \cdot \Vert$ of both sides of (6) and use (8) to get

$$b\left(t\right)\le b_0\left(t\right)+c{\int }_0^t{\left(t-s\right)}^{-\frac{5}{4}}b\left(s\right)ds.$$

The integral ${\int }_0^t{\left(t-s\right)}^{\lambda -1}b\left(s\right)ds:=t^{\lambda -1}*b$ is the convolution of the distribution t^λ-1 and b(t) in ℝ₊ = [0, ∞). We assume that b = b(t) is continuous on ℝ₊ and t^λ = 0 for t < 0. The integral t^λ-1 *b diverges classically if λ ≤ 0. Integral (9) with λ = -1/4 diverges classically. In [5, 9, 10, 11, 12] such integrals are defined in a new way. Let us define it for λ ≤ 0 by the formula

$$L\left(t^{\lambda -1}*b\right)=L\left(t^{\lambda -1}\right)L\left(b\right).$$

Here L is the Laplace transform, $L\left(b\right)={\int }_0^{\infty }{e}^{-pt}b\left(t\right)dt$. One has

$$L\left(t^{\lambda -1}\right)=\frac{\Gamma \left(\lambda \right)}{p^{\lambda }},\forall \lambda \ne 0,-1,-2\dots ,$$

see [5, 1]. Here Γ(λ) is the gamma-function, which is analytic with respect to λ ∈ ℂ except for λ = -n, n = 0, 1, 2,…. At the point λ = -n the Γ(λ) has a simple pole with the residue (-1)ⁿ/n!, 1/Γ(λ) is an entire function, Γ(z + 1) = zΓ(z), Γ(z)Γ(1 - z) = π/sin(πz), see [4]. Let Φ_λ:= t^λ-1/Γ(A). Then L(Φ_λ) = p^−λ for all λ ∈ ℂ. For λ ≤ 0 one has

$$L\left({\Phi }_{\lambda }*b\right)=L\left(b\right)/p^{\lambda },{\Phi }_{\lambda }*b=L^{-1}\left(L\left(b\right)/p^{\lambda }\right).$$

This formula defines the convolution for λ < 0, that is, for the values of λ for which the convolution is not defined classically, in particular, for $\lambda =-\frac{1}{4}$. We avoided the usage of distribution theory, cf. [2].

Lemma 1. One has

$${\Phi }_{\lambda }*{\Phi }_{\mu }={\Phi }_{\lambda +\mu },\underset{\lambda \to +0}{lim}{\Phi }_{\lambda }=\delta \left(t\right),$$

where δ(t) is the Dirac distribution.

Proof: Since L(Φ_λ * Φ_μ) = L(Φ_λ)L(Φ_μ) = 1/p^λ+μ and L⁻¹(1/p^λ+μ) = Φ_λ+μ, the first part of Lemma 1 is proved. To prove the second part we use the formula

$${\Phi }_{\lambda }*{\Phi }_{\mu }=\frac{1}{\Gamma \left(\lambda \right)\Gamma \left(\mu \right)}{\int }_0^t{\left(t-s\right)}^{\lambda -1}{s}^{\mu -1}ds=\frac{t^{\lambda +\mu -1}}{\Gamma \left(\lambda \right)\Gamma \left(\mu \right)}B\left(\lambda ,\mu \right),$$

where B(x, y) is the beta-function,

$$B\left(x,y\right)=\frac{\Gamma \left(x\right)\Gamma \left(y\right)}{\Gamma \left(x+y\right)}={\int }_0^1{\left(1-u\right)}^{x-1}{u}^{y-1}du,$$

see [4]. Formula (14) gives an alternative derivation of the first formula (13) because the right side of (14) equals to Φ_λ+μ(t).

Let ϕ ∈ C∞(ℝ₊) be an arbitrary function vanishing near infinity. Note that 1/Γ(λ) ~ λ as λ → 0. Therefore

$$\underset{\lambda \to +0}{lim}{\int }_0^{\infty }{\Phi }_{\lambda }\left(t\right)\varphi \left(t\right)dt=\underset{\lambda \to +0}{lim}\lambda {\int }_0^{\infty }{t}^{\lambda -1}\varphi \left(t\right)dt=\varphi \left(0\right).$$

Lemma 1 is proved. See [5, p. 24] for a different proof of Lemma 1.

Let us write (9) as

$$b\left(t\right)\le b_0\left(t\right)-c{c}_1{\Phi }_{-\frac{1}{4}}*b,c_1:=\left|\Gamma \left(-\frac{1}{4}\right)\right|>0.$$

Apply the operator ${\Phi }_{1/4}*$ to (16) and use (13) to get

$${\Phi }_{1/4}*b\le {\Phi }_{1/4}*b_0-c{c}_{1}b,c{c}_1>0.$$

If λ > 0 and g(t) ≥ 0, then ${\Phi }_{1/4}*$ g ≥ 0. Since b ≥ 0 it follows from (17) that

$$b\left(t\right)\le \frac{1}{c{c}_1}{\Phi }_{1/4}*b_0.$$

Since b₀(t) is a smooth bounded function on ℝ₊, it follows that b(t) has these properties. We have proved the following result.

Theorem 1. If sup_t≥0 b₀(t) < ∞, then

$$\underset{t>0}{sup}b\left(t\right)<\infty ,$$

and

$$b\left(0\right)=0.$$

Proof: Formula (20) holds because lim_t→0 Φ_¼ * b₀ = 0.

Formula (20) is the NSP paradox: originally it was not assumed that v(x, 0) = v₀(x) = 0, so b(0) ≠ 0, but we have derived that b(0) = 0, so v₀(x) = 0.

Lemma 2. If b(t) ∈ C([0, T]), then

$${\int }_0^t{\left(t-s\right)}^{\frac{5}{4}}b\left(s\right)ds\sim t^{-\frac{1}{4}}b\left(0\right)\Gamma \left(\frac{1}{4}\right)/\Gamma \left(3/4\right),t\to 0.$$

where the sign ~ stands for asymptotic equivalence.

Proof: One has

$${\int }_0^t{\left(t-s\right)}^{\frac{5}{4}}b\left(s\right)ds=t^{-\frac{1}{4}}{\int }_0^t{\left(1-u\right)}^{-\frac{5}{4}}b\left(tu\right)du\sim t^{-\frac{1}{4}}b\left(0\right)B\left(-\frac{1}{4},1\right).$$

Since b(tu) → b(0) uniformly with respect to u ∈ [0, 1] and B( $-\frac{1}{4}$, 1) = Γ( $-\frac{1}{4}$)/Γ(3/4), Lemma 2 is proved.

The integral ${\int }_0^1{\left(1-u\right)}^{-\frac{5}{4}}du$ diverges classically. We define it as $B\left(-\frac{1}{4},1\right)$. The beta-function B(x, y) is well defined for all x, y ∈ ℂ except for x, y = 0, -1, -2,…, so at the values x = $-\frac{1}{4}$ = 1 the function B(x, y) is well defined.

Lemma 3. Let b(t) ≥ 0 satisfy (9) and β(t) ≥ 0 satisfy the equation

$$\beta \left(t\right)=b_0\left(t\right)+c{\int }_0^t{\left(t-s\right)}^{-\frac{5}{4}}\beta \left(s\right)ds.$$

Then

$$b\left(t\right)\beta \left(t\right).$$

Proof: Let z(t):= β(t) - b(t). Subtract from (9) (23) and get

$$0\le z\left(t\right)-c{\int }_0^t{\left(t-s\right)}^{-\frac{5}{4}}z\left(s\right)ds.$$

By formula (21) one gets from (25) that

$$0\le z\left(t\right)+c{c}_1{t}^{-\frac{1}{4}}\left(0\right)/\Gamma \left(3/4\right),t\to 0.$$

Thus, it follows from (25) that z(0) ≥ 0 as t → 0.

Similarly, if there is a point t₀ at which z(t₀) < 0, then we have a contradiction with (25). A detailed and different proof of Lemma 3 is given in [5, p. 20].

Lemma 3 is proved.

Eq. (23) can be solved analytically, see [6, 9]. Take the Laplace transform of (23) and get L(β) = L(b₀) - cc₁p¼ L(β). Thus

$$\beta \left(t\right)=L^{-1}\left(\frac{L\left(b_0\right)}{1+c{c}_1{p}^{1/4}}\right).$$

Let us state a priori estimates, proved in [5], for the solution to (1) assuming that $\left|\xi \right|\left|\tilde{F}\right|<c$:

$$\underset{t>0}{sup}\left(\Vert v\Vert +{\int }_0^{t}b\left(s\right)ds\right)<c,\underset{t>0}{sup}b\left(t\right):=\Vert \left|\xi \right|\tilde{v}\Vert <c,$$

$$\Vert \left|\xi \right|\tilde{v}\Vert <c,\left|\tilde{v}\right|\le c\left(1+t^{1/2}\right).$$