Remark on the strong solvability of the Navier–Stokes equations in the weak $L^n$ space

Abstract

The initial value problem of the incompressible Navier–Stokes equations in $L^{n,\infty }({\mathbb {R}}^n)$ is investigated. Introducing the real interpolation estimates for the Duhamel terms, we construct global and local in time mild (and strong) solutions in $BC\bigl ((0,T)\,;\,L^{n,\infty }({\mathbb {R}}^n)\bigr )$ for external forces with non-divergence form in the scale invariant class. We observe that a mild solution becomes the strong solution, i.e., it satisfies the differential equation in the critical topology of $L^{n,\infty }({\mathbb {R}}^n)$ with an additional condition only on the external force, even though the Stokes semigroup in not strongly continuous on $L^{n,\infty }({\mathbb {R}}^n)$. Furthermore, via the existence and the uniqueness of local in time solutions, we extend the uniqueness theorem within the solution class $BC\bigl ([0,T)\,;\,L^{n,\infty }({\mathbb {R}}^n)\bigr )$.

Inclusion between \(X_\sigma ^{n,\infty }\) and \({\widetilde{L}}_{\sigma }^{n,\infty }({\mathbb {R}}^n)\)

We show, in general, that $X_{\sigma }^{n,\infty } \ne {\widetilde{L}}_{\sigma }^{n,\infty }({\mathbb {R}}^n)$.

For simplicity, we consider $n=1$ and forget the solenoidal condition for a while. Actually, we give an example that $a\not \in \overline{D(\Delta )}^{\Vert \cdot \Vert _{1,\infty }}$ and $a \in {\widetilde{L}}^{1,\infty }({\mathbb {R}})$. Here, we note that $L^{1,\infty }({\mathbb {R}})$ is a quasi Banach space but the following contains an essential idea.

Let

$$\begin{aligned} R(x)= {\left\{ \begin{array}{ll} |x|, &{} 0\le |x| \le 1, \\ r_k(x), &{} k < |x| \le k+1, k=1,2,\dots , \end{array}\right. } \end{aligned}$$

where $r_k(x) = \mathrm {sgn} (\sin (2^k\pi x))$ denote the Rademacher functions.

Suppose the following function:

$$\begin{aligned} a(x) := \frac{R(x)}{|x|} \quad \text {for }x \in {\mathbb {R}}. \end{aligned}$$

We note that $|R(x)|\le 1$ and $|a(x)|={\left\{ \begin{array}{ll}1, &{} |x|\le 1, \\ 1/|x|, &{} |x|\ge 1. \end{array}\right. }$

Then, we shall show that $|e^{t\Delta }a(x)|=o(1/|x|)$ as $|x|\rightarrow \infty .$ Take an arbitrary small $\varepsilon >0$. Then we can choose $N_{\varepsilon } \in {\mathbb {N}}$ so that

$$\begin{aligned} \int _{|y| \ge N_\varepsilon } E_1(y)dy<\varepsilon \quad \text {and }\quad \int _{|y| \ge N_\varepsilon } 2|y| E_1(y)dy <\varepsilon , \end{aligned}$$

where $E_t(x)=(4\pi t)^{-1/2}\exp (-|x|^2/(4t))$, $t>0$ is the heat kernel.

For a while, we assume $|x| \ge 2N_\varepsilon$ and $t\le 1$. Put

$$\begin{aligned} \begin{aligned} e^{t\Delta }a(x)&=\int _{|y|<|x|/2} E_t(y)a(x-y)dy+ \int _{|y|\ge |x|/2} E_t (y) a(x-y)dy \\&=:I_1+ I_2. \end{aligned} \end{aligned}$$

Firstly, we estimate $I_2$ as follows:

$$\begin{aligned} \begin{aligned} |I_2|&\le \int _{|y|\ge |x|/2} E_t(y)dy = \int _{|y|\ge |x|/2} E_t(y) \frac{|x|}{|x|}dy \\&\le \frac{1}{|x|}\int _{|y|\ge N_\varepsilon } 2|y|E_t(y)dy \le \frac{1}{|x|}\sqrt{t}\int _{|y|\ge N_\varepsilon } 2|y|E_1(y)dy \\&\le \frac{1}{|x|}\sqrt{t}\varepsilon \le \frac{\varepsilon }{|x|}. \end{aligned} \end{aligned}$$

Next, we estimate $I_1$:

$$\begin{aligned} \begin{aligned} I_1&= \int _{|y|\le |x|/2} E_t(y)R(x-y)\bigg \{\frac{1}{|x-y|}-\frac{1}{|x|}\bigg \}dy + \int _{|y|\le |x|/2} E_t(y)\frac{R(x-y)}{|x|}dy\\&=: J_1 + J_2. \end{aligned} \end{aligned}$$

Since $|x-y| \ge |x|/2$ when $|y|\le |x|/2$, we have

$$\begin{aligned} |J_1|\le & {} \int _{|y|\le |x|/2} E_t(y)|R(x-y)| \frac{|y|}{|x||x-y|}dy \le \frac{1}{|x|^2}\int _{|y|\le |x|/2} 2|y|E_t(y)dy\\\le & {} \frac{1}{|x|}\frac{\sqrt{t}M}{|x|}. \end{aligned}$$

Here, we put $M=\int _{{\mathbb {R}}} 2|y|E_1(y)dy$. Therefore, if $|x|>M/\varepsilon$, then $|J_1| \le \varepsilon /|x|$.

In order to estimate $J_2$, it suffices to show that there exists $L_{\varepsilon ,t}>0$ such that if $|x| > L_{\varepsilon ,t}$ then

$$\begin{aligned} \left| \int _{|y| \le N_\varepsilon } E_t(y) R(x-y)dy\right| <\varepsilon . \end{aligned}$$

(A.1)

Since $E_t$ is uniformly continuous on $[-N_\varepsilon ,N_\varepsilon ]$, we can choose $k_0\in {\mathbb {N}}$ such that the lower Darboux sum with partition of $[-N_\varepsilon ,N_\varepsilon ]$ by partial length $2^{-k_0}$ approximates the integral $\int _{|y|\le N_{\varepsilon }}E_t(y)\,dy$. Indeed, let $x_\ell =-N_\varepsilon +\ell 2^{-k_0}$ $(0\le \ell \le 2N_\varepsilon 2^{k_0})$ and set

$$\begin{aligned} S_t(x)=\sum _{\ell =0}^{2N_\varepsilon 2^{k_0}-1} a_{\ell ,t} \chi _{[x_\ell ,x_{\ell +1}]}(x), \qquad \text { where }\quad a_{\ell ,t}=\inf _{[x_\ell ,x_{\ell +1}]} E_t(x), \end{aligned}$$

then we suppose that

$$\begin{aligned} \int _{|y|\le N_\varepsilon } (E_t(y) - S_t(y))\,dy \le \frac{\varepsilon }{2}. \end{aligned}$$

Moreover, we note that

$$\begin{aligned} \sum _{\ell =0}^{2N_\varepsilon 2^{k_0}-1} a_{\ell ,t} 2^{-k_0} \le 1. \end{aligned}$$

Take $K\in {\mathbb {N}}$ so that $2^{-K+k_0} < \varepsilon /2$. Hereafter, we assume $|x| > K +N_\varepsilon +1$ and $|y| \le N_\varepsilon$. Then $|x-y|\ge K+1$. So under $|x-y|>K+1$, we note that $R(x-y)=r_k(x-y)$ and $k\ge K+1$.

So we note that if $[x_\ell ,x_{\ell +1}]$ is contained within the region so that $R(x-\cdot )=r_k(x-\cdot )$, i.e., the case for all $y \in [x_l,x_{l+1}]$ it holds that $k<|x-y|\le k+1$, then $\int _{[x_\ell ,x_{\ell +1}]}a_{\ell ,t}r_k(x-y)dy =0$. Otherwise, $\bigl | \int _{[x_\ell ,x_{\ell +1}]} a_{\ell ,t}R(x-y)dy\bigr |\le 2\cdot 2^{-K-1} a_{\ell ,t}$. Hence,

$$\begin{aligned} \begin{aligned} \left| \int _{|y|\le N_\varepsilon } S_t(y)R(x-y)dy\right|&\le \sum _{\ell =0}^{2N_\varepsilon 2^{k_0}-1} \left| \int _{[x_\ell ,x_{\ell +1}]} a_{\ell ,t} R(x-y)dy\right| \\&\le \sum _{\ell =0}^{2N_\varepsilon 2^{k_0}-1} 2^{-K}a_{\ell ,t} \\&= 2^{-K+k_0} \sum _{\ell =0}^{2N_\varepsilon 2^{k_0}-1} a_{\ell ,t}2^{-k_0} \le \frac{\varepsilon }{2}. \end{aligned} \end{aligned}$$

Therefore, taking $L_{\varepsilon ,t}=K+N_\varepsilon +1$, we obtain (A.1).

We estimate $J_2$ for $|x|>L_{\varepsilon ,t}$. It holds that

$$\begin{aligned} \begin{aligned} |J_2|&\le \left| \int _{|y|\le N_\varepsilon } E_t(y)R(x-y)dy\right| + \left| \int _{N_\varepsilon \le |y| \le |x|/2} E_t(y)R(x-y)dy \right| \\&\le \varepsilon + \int _{|y|\ge N_\varepsilon } E_t(y)dy \le 2\varepsilon . \end{aligned} \end{aligned}$$

As a consequence, for $|x| \ge L:=\max \{2N_\varepsilon , M/\varepsilon ,L_{\varepsilon ,t}\}$ we obtain that

$$\begin{aligned} |e^{t\Delta }a(x)|\le \frac{4\varepsilon }{|x|}. \end{aligned}$$

Next, we show that $e^{t\Delta }a \not \rightarrow a$ as $t\rightarrow 0$ in $L^{1,\infty }({\mathbb {R}})$. Let $t\le 1$ be fixed. Since $\Vert f+g\Vert _{1,\infty } \le 2 (\Vert f\Vert _{1,\infty }+\Vert g\Vert _{1,\infty })$ we observe that

$$\begin{aligned} \begin{aligned} \Vert e^{t\Delta }a -a\Vert _{1,\infty }&\ge \Vert e^{t\Delta }a-a\Vert _{L^{1,\infty }(\{|x|\ge L\})} \\&\ge \frac{1}{2}\Vert a \Vert _{L^{1,\infty }(\{|x|\ge L\})} - \Vert e^{t\Delta }a \Vert _{L^{1,\infty }(\{|x|\ge L\})} \\&\ge 1-8\varepsilon . \end{aligned} \end{aligned}$$

Since $\varepsilon >0$ is arbitrary, choosing $\varepsilon <1/8$, we conclude that $e^{t\Delta }a \not \rightarrow a$ as $t\rightarrow 0$ in $L^{1,\infty }({\mathbb {R}})$, i.e., $a \not \in \overline{D(\Delta )}^{\Vert \cdot \Vert _{1,\infty }}$. On the other hand, $a \in L^{1,\infty }({\mathbb {R}})\cap L^\infty \subset {\widetilde{L}}^{1,\infty }({\mathbb {R}})$.

For $n\ge 2$, we can apply a similar argument for solenoidal functions. In 2D case, we introduce

$$\begin{aligned} a(x_1,x_2)=R(x)\left( \frac{-x_2}{|x|^2}, \frac{x_1}{|x|^2}\right) \end{aligned}$$

where R(x) is modified with the Rademacher functions on dyadic squares. Then each segment

$$\begin{aligned} \pm \chi _{Q}(x)\left( \frac{-x_2}{|x|^2},\frac{x_1}{|x|^2}\right) \in L_{\sigma }^{2,\infty }({\mathbb {R}}^2), \end{aligned}$$

where Q is a square. This implies $(a,\nabla p)=0$ for all $\nabla p\in G^{2,1}({\mathbb {R}}^2)$. Hence $a \in L_{\sigma }^{2,\infty }({\mathbb {R}}^2)$. Finally, cutting off the singularity suitably, we obtain a desired example.