Well-Posedness and Ill-Posedness Problems of the Stationary Navier–Stokes Equations in Scaling Invariant Besov Spaces

Abstract

We consider the stationary Navier–Stokes equations in $\mathbb {R}^n$ for $n\geqq 3$ in the scaling invariant Besov spaces. It is proved that if $n<p\leqq \infty$ and $1\leqq q\leqq \infty$, or $p=n$ and $2<q\leqq \infty$, then some sequence of external forces converging to zero in $\dot{B}^{-3+\frac{n}{p}}_{p,q}$ can admit a sequence of solutions which never converges to zero in $\dot{B}^{-1}_{\infty ,\infty }$, especially in $\dot{B}^{-1+\frac{n}{p}}_{p,q}$. Our result may be regarded as showing the borderline case between ill-posedness and well-posedness, the latter of which Kaneko–Kozono–Shimizu proved when $1\leqq p<n$ and $1\leqq q\leqq \infty$.

Appendix A

Here let us prove $PI_1 \not \equiv 0$ in Step 1 of the proof of Lemma 3.3, that is,

$$\begin{aligned} P(e_2 \Phi _1 + e_3 \Phi _2) \not \equiv 0, \end{aligned}$$

(A.1)

with $\Phi _1$ and $\Phi _2$ in (3.6), and $\psi$ characterized by (3.1). We can assume here that $\psi$ is radial symmetric, i.e.,

$$\begin{aligned} \psi (x)=\psi (y), x,y\in \mathbb {R}^n\ \mathrm{with}\ |x|=|y|, \end{aligned}$$

which implies that we can write $\psi =\psi (r)$, $r=r(x)=|x|$.

Let us show (A.1) with a proof by contradiction. Suppose that

$$\begin{aligned} P(e_2 \Phi _1 + e_3 \Phi _2) \equiv 0, \end{aligned}$$

(A.2)

which yields that there exists some distribution F such that $e_2 \Phi _1 + e_3 \Phi _2 = \nabla F$. Hence it must hold that $\nabla \times (e_2 \Phi _1 + e_3 \Phi _2) \equiv 0$. Therefore, all of its components should vanish, and especially we have $\frac{\partial \Phi _1}{\partial x_1}\equiv 0.$ Actually, this yields $\Phi _1\equiv 0$, i.e.,

$$\begin{aligned} \psi _{x_3}\psi _{x_2x_3}\equiv \psi _{x_2}\psi _{x^2_3}. \end{aligned}$$

(A.3)

Indeed, if not, there exists a point $x^*\in \mathbb {R}^n$ such that $\Phi _1(x^*)\ne 0$ Then by $\frac{\partial \Phi _1}{\partial x_1}\equiv 0$, we see that $\Phi _1(x)=\Phi _1(x^*)$ on the line $\{x\in \mathbb {R}^n; x_j=x^*_j, \ \forall j=2,3,\ldots ,n\}$, which contradicts $\Phi _1 \in \mathcal {S}$. By a chain rule of differentiation, we can rewrite (A.3) as

$$\begin{aligned} \psi _r r_{x_3}(\psi _{r^2} r_{x_2}r_{x_3}+\psi _{r} r_{x_2x_3}) \equiv \psi _r r_{x_2}(\psi _{r^2} r^2_{x_3}+\psi _{r} r_{x^3_3}), \end{aligned}$$

where $\psi _{r^\alpha }\equiv \frac{\partial ^\alpha \psi }{\partial r^\alpha }$ and $r_{x_2^\alpha x_3^\beta }\equiv \frac{\partial ^{(\alpha +\beta )}}{\partial x_2^\alpha x_3^\beta }r$. Since $\psi \in \mathcal {S}\backslash \{0\}$, we have

$$\begin{aligned} r_{x_3}r_{x_2x_3}=\frac{x_3}{r}\left( -\frac{x_2x_3}{r^3}\right) \equiv r_{x_2} r_{x^3_3} = \frac{x_2}{r}\left( \frac{1}{r}-\frac{x_3^2}{r^3}\right) , \end{aligned}$$

which yields $\frac{x_2}{r^2}= 0$ for any $x\in \mathbb {R}^n$ and which is not true. Therefore, we see that the assumption (A.2) is false and hence (A.1) holds.