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 \(\sup_{t\ge 0}\|\tilde{v}(\xi,t)\|0\) 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.
