Open Journal of Mathematical Analysis

Vol. 4 (2020), Issue 2, pp. 89 – 92
ISSN: 2616-8111 (Online) 2616-8103 (Print)
DOI: 10.30538/psrp-oma2020.0066

Concerning the Navier-Stokes problem

Alexander G. Ramm
Department of Mathematics, Kansas State University, Manhattan, KS 66506, USA.; ramm@math.ksu.edu

Copyright © 2020 Alexander G. Ramm. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Received: June 16, 2020 – Accepted: August 27, 2020 – Published: August 31, 2020

Abstract

The problem discussed is the Navier-Stokes problem (NSP) in $\mathbb{R}^3$. Uniqueness of its solution is proved in a suitable space $X$. No smallness assumptions are used in the proof. Existence of the solution in $X$ is proved for $t\in [0,T]$, where $T>0$ is sufficiently small. Existence of the solution in $X$ is proved for $t\in [0,\infty)$ if some a priori estimate of the solution holds.

Keywords:

Navier-Stokes equations, uniqueness of the solutions.

1. Introduction

There is a large literature on the Navier-Stokes problem (NSP) in $\mathbb{R}^3$ ( see [1], Chapter 5) and references therein). The global existence and uniqueness of a solution in $\mathbb{R}^3$ was not proved. The goal of this paper is to prove uniqueness of the solution to NSP in a suitable functional space. No smallness assumptions are used in our proof. The NS problem in $\mathbb{R}^3$ consists of solving the equations

\begin{equation}\label{e1} v'+(v, \nabla)v=-\nabla p +\nu\Delta v +f, \quad x\in \mathbb{R}^3,\,\, t\ge 0,\quad \nabla \cdot v=0,\quad v(x,0)=v_0(x). \end{equation}

(1)

Vector-functions $v=v(x,t)$, $f=f(x,t)$ and the scalar function $p=p(x,t)$ decay as $|x|\to \infty$ uniformly with respect to $t\in \mathbb{R}_+:=[0, \infty)$, $v':=v_t$, $\nu=const>0$ is the viscosity coefficient, the velocity $v$ and the pressure $p$ are unknown, $v_0$ and $f$ are known, $ \nabla \cdot v_0=0$. Equations (1) describe viscous incompressible fluid with density $\rho=1$. We use the integral equation for $v$:

\begin{equation}\label{e2} v(x,t)=F- \int_0^tds \int_{\mathbb{R}^3} G(x-y,t-s)(v,\nabla)v dy. \end{equation}

(2)

Equation (2) is equivalent to (1), see [2]. Formula for the tensor $G$ is derived in [2], see also [1], p.41. The term $F=F(x,t)$ depends only on the data $f$ and $v_0$ (see equation (18) in [2] or formula (5.42) in [1]):

\begin{equation}\label{e3} F:=\int_{\mathbb{R}^3}g(x-y)v_0(y)dy + \int_0^tds\int_{\mathbb{R}^3}G(x-y,t-s)f(y,s)dy. \end{equation}

(3)

We assume throughout that $f$ and $v_0$ are such that $F$ is bounded in all the norms we use. Let $X$ be the Banach space of continuous functions with respect to $t$ with the norm

\begin{equation}\label{e3a} \|\tilde{v}\|:=\int_{\mathbb{R}^3}|\tilde{v}(\xi,t)|(1+|\xi|)d\xi, \end{equation}

(4)

where $t>0$, and $\tilde{v}:=(2\pi)^{-3}\int_{\mathbb{R}^3}v(x,t)e^{-i\xi \cdot x}dx$. Taking the Fourier transform of (2) yields

\begin{equation}\label{e4} \tilde{v}=\tilde{F}-\int_0^tds \tilde{G} \tilde{v}\star i\xi \tilde{v}:=B(\tilde{v}), \end{equation}

(5)

where $\star$ denotes the convolution in $\mathbb{R}^3$ and for brevity we omitted the tensorial indices: instead of $\tilde{G}_{mp}\tilde{v}_j\star (i\xi_j)\tilde{v}_p$, where one sums up over the repeated indices, we wrote $ \tilde{G}(\xi,t-s) \tilde{v}\star (i\xi \tilde{v})$. From formula (5.9) in [1] it follows that

\begin{equation}\label{e5} |\tilde{G}|\le ce^{-\nu \xi^2(t-s)}. \end{equation}

(6)

By $c>0$ we denote various constants independent of $t$ and $\xi$. Let $S(\mathbb{R}^3\times \mathbb{R}_{+})$ and $S(\mathbb{R}^3)$ be the L.Schwartz spaces. Our results are:

Theorem 1. Assume that $f$ and $v_0$ are in $S(\mathbb{R}^3\times \mathbb{R}_{+})$ and $S(\mathbb{R}^3)$ respectively. Then there is at most one solution to NSP in $X$.

Theorem 2. The solution to NSP in $X$ exists for $t\in [0, T]$ if $T>0$ is sufficiently small.

Theorem 3. The solution $v(x,t)$ to NSP in $X$ exists for all $t\ge 0$ if an a priori estimate $\sup_{t\ge 0}\|\tilde{v}(\xi,t)\|< c_a$ holds, where $c_a>0$ is a constant depending only on the data.

2. Proofs

Proof. [Proof of Theorem 1] Let $\tilde{v}$ and $\tilde{w}$ belong to $X$ and solve equation (5). Denote $z:=\tilde{v}- \tilde{w}$. Then (5) implies

\begin{equation}\label{e6} z=-\int_0^tds \tilde{G}(z\star i\xi \tilde{v}+\tilde{w}\star i\xi z). \end{equation}

(7)

Let $\|z(\xi,t)\|:=u(t)$ and $\int_{\mathbb{R}^3}:=\int$. From (7) and (6) one gets

\begin{equation}\label{e7} u(t)\le c\int_0^tds \int d\xi e^{-\nu \xi^2(t-s)}(1+|\xi|) \left[\int|z(\xi-\zeta,s)||\zeta| |\tilde{v}(\zeta,s)|d\zeta\\ +\int |\tilde{w}(\xi-\zeta,s)||\zeta||z(\zeta,s)|d\zeta\right]. \end{equation}

(8)

Let $\eta:=\xi-\zeta$. One has:

\begin{equation}\label{e8} \int d\zeta |\zeta||\tilde{v}|\int d\xi (1+|\xi|)|z(\xi-\zeta,s)|e^{-\nu \xi^2(t-s)}\le \|\tilde{v}\|u(s)\max_{\zeta\in \mathbb{R}^3} \left\{ e^{-\nu |\eta+\zeta|^2 (t-s)}\frac{1+|\eta+\zeta|}{1+|\eta|}\right\}. \end{equation}

(9)

Furthermore,

\begin{equation}\label{e9} \max_{\zeta\in \mathbb{R}^3}\left\{ e^{-\nu |\eta+\zeta|^2 (t-s)}\frac{1+|\eta+\zeta|}{1+|\eta|}\right\}=(1+|\eta|)^{-1} \max_{p\in \mathbb{R}_+}\{(1+p)e^{-\nu p^2(t-s)}\} \le\\ 1+\frac{c_{\nu}}{(t-s)^{1/2}}, \end{equation}

(10)

where $c_{\nu}=c\nu^{-0.5}$. Indeed, if $h(r)=(1+r)e^{-\nu(t-s)r^2}$, then $\max_{r>0}h(r)=h(R)\le 1+\frac {c_\nu}{(t-s)^{1/2}}$, where $R=-\frac 1 2+(\frac 1 4+ \frac 1 {2\nu (t-s)})^{1/2}$ and $h'(R)=0$. A similar estimate holds for the second integral in (8):

\begin{equation}\label{e10} \begin{split} \int d\zeta (1+|\zeta|) |z(\zeta,s)|\int d\xi e^{-\nu \xi^2(t-s)}(1+|\xi|)|\tilde{w}(\xi-\zeta,s)|\le u(s)\max_{\zeta \in \mathbb{R}^3} \int dp |\tilde{w}(p,s)|(1+|p+\zeta|)e^{-\nu \xi^2(t-s)}. \end{split} \end{equation}

(11)

The right side of \eqref{e10} is $u(s)J$, where

\begin{equation}\label{e11} \begin{split} J=\int dp |\tilde{w}(p,s)|(1+p) \max_{\zeta \in \mathbb{R}^3}\left(\frac{1+|p+\zeta|}{1+p} e^{-\nu \xi^2(t-s)}\right)\le \|\tilde{w}\|\left(1+ \frac {c_\nu}{(t-s)^{1/2}}\right). \end{split} \end{equation}

(12)

From (7)--(12) one gets

\begin{equation}\label{e12} u(t)\le C(t)\int_0^t \left(1+\frac {c_\nu}{(t-s)^{1/2}}\right) u(s)ds, \quad C(t)=c\max_{0\le s \le t}(\|\tilde{v}(p,s)\|+ \|\tilde{w}(p,s)\|), \end{equation}

(13)

where $C(t)>0$ is a continuous function and $u(t)\ge 0$. Note that $C(t)$ is a continuous function of $t$ for all $t\ge 0$ because we assume that the solutions $\tilde{v}$ and $\tilde{w}$ belong to $X$ and $C(t)$ is the sum of the norms of the two elements of $X$. The Volterra inequality (13) has only the trivial solution $u(t)=0$, as follows from Lemma 1, proved below. Theorem 1 is proved.

Lemma 1. Inequality (13) has only the trivial non-negative solution $u(t)=0$.

Proof. [Proof of Lemma 1] Denote $\frac {u(t)}{C(t)}=q(t)$. Then

\begin{equation}\label{e13} q(t)\le \int_0^t \left(1+\frac {c_\nu}{(t-s)^{1/2}}\right) C(s)q(s)ds:= \int_0^t K(t,s)q(s)ds. \end{equation}

(14)

The kernel $K(t,s)>0$ is weakly singular. Any solution $q\ge 0$ to (14) satisfies the estimate $0\le q\le Q$, where $Q\ge 0$ solves the Volterra equation

\begin{equation}\label{e14} Q(t)=\int_0^t K(t,s)Q(s)ds. \end{equation}

(15)

This equation has only the trivial solution $Q=0$. Lemma 1 is proved.

Proof. [Proof of Theorem 2] From (5) after multiplying by $1+|\xi|$, integrating over $\mathbb{R}^3$ and using calculations similar to the ones in equation (12), one gets

\begin{equation}\label{e15} u(t)\le b(t)+c\int_0^t\left(1+\frac {c_\nu}{(t-s)^{1/2}}\right)u^2(s)ds:=A(u), \end{equation}

(16)

where $b(t):=\int|\tilde{F}(\xi, t)|(1+|\xi|)d\xi$ and $u(t):=\|\tilde{v}(\xi,t)\|$. For sufficiently small $T$ equation $U=AU$ is uniquely solvable by iterations according to the contraction mapping principle. If $\sup_{t\in [0,T]}b(t)\le c_0$ and $T$ is sufficiently small, then a ball $\sup_{t\in [0,T]}u(t)\le c_1$, $c_1>c_0$, is mapped by the operator $A$ into itself and $A$ is a contraction mapping. The operator $A$ maps positive functions into positive functions. Thus, $u(t)\le U(t)$. Theorem 2 is proved.

Proof. [Proof of Theorem 3] Under the assumption of Theorem 3 inequality (16) implies:

\begin{equation}\label{e16} u(t)\le b(t)+cc_a\int_0^t\left(1+\frac {c_\nu}{(t-s)^{1/2}}\right)u(s)ds:= A_1(u), \end{equation}

(17)

The corresponding equation $U=A_1U$ is a linear Volterra integral equation. It has a unique solution defined for all $t\ge 0$, and $0\le u(t)\le U(t)$. Theorem 3 is proved.

Remark 1. The following a priori estimates for solutions to NSP hold:

\begin{equation}\label{e17} \|v\|_{L^2(\mathbb{R}^3)}\le c, \quad \int_0^t\|\nabla v(x,s)\|^2_{L^2(\mathbb{R}^3)}ds\le c, \end{equation}

(18)

and

\begin{equation}\label{e18} \sup_{t\in [0,T]}|\tilde{v}(\xi,t)|\le c+cT^{1/2}, \quad sup_{t\ge 0; \xi \in \mathbb{R}^3} (|\xi||\tilde{v}|)< c. \end{equation}

(19)

Proof. [Proof of (18)] First estimate (18) is well known. It remains to prove the second estimate (18). For this, multiply (1) by $v$ and integrate over $\mathbb{R}^3$ to get (see [1]): $$0.5\frac {d \int v^2 dx}{dt}+\nu \int |\nabla v|^2dx=\int fvdx.$$ Integrating over $t$ one gets: $$0.5 \int v^2dx +\int_0^tds \int |\nabla v(x,s)|^2dx \le 0.5 \int v_0^2dx + \int_0^t ds \int fvdx.$$ One has $\int_0^tds \int fvdx\le \int_0^t ds (\int |f(x,s)|^2)^{1/2} (\int|v(x,s)|^2)^{1/2}\le c$. Indeed, it is assumed that $f$ decays fast, so $\sup_{t\ge 0}\int_0^tds(\int |f|^2dx)^{1/2}\le c$. Using this and estimates (18) we get $\int_0^tds \int |fv|dx\le c.$ Thus, the second estimate (18) is proved.

Proof. [Proof of estimate (19)] From Equation (5) one gets:

\begin{equation}\label{e19} |\tilde{v}|\le |\tilde{F}|+c\int_0^t e^{-\nu \xi^2 (t-s)}|\tilde{v}|\star (|\xi||\tilde{v}|) ds:= |\tilde{F}|+I. \end{equation}

(20)

One has $\sup_{t\ge 0} |\tilde{F}|\le c$ under the assumptions of Theorem 1. By the Cauchy inequality, the first estimate (18) and Parseval's equality one gets $ |\tilde{v}|\star (|\xi||\tilde{v})\le \|\tilde{v}|\|_{L^2(\mathbb{R}^3)} \| |\xi|\tilde{v}\|_{L^2(\mathbb{R}^3)}.$ Thus, using the Cauchy inequality, and the second estimate (18), one gets

\begin{equation}\label{e20} I\le c\int_0^t e^{-\nu \xi^2 (t-s)}\| |\xi|\tilde{v}\|_{L^2(\mathbb{R}^3)}ds\le ct^{1/2} [\int_0^t \| |\xi|\tilde{v}\|^2_{L^2(\mathbb{R}^3)}]^{1/2}\le ct^{1/2}. \end{equation}

(21)

From (20) and (21) estimate (19) follows. The second estimate (19) is proved in [1], p. 50, inequality (5.39), under the assumption of Theorem 1.

Competing Interests

The author does not have any competing interests in the manuscript.

References

Ramm, A. G. (2019). Symmetry Problems. The Navier-Stokes Problem, Morgan & Claypool Publishers, San Rafael, CA.
Ramm, A. G. (2017). Global existence, uniqueness and estimates of the solution to the Navier-Stokes equations, Applied Mathematics Letters, 74, 154-160.