Global Well-Posedness of the 3D Generalized Navier-Stokes Equations with Fractional Partial Dissipation

Abstract

The existence of a global solution to the 3D incompressible Navier-Stokes equations is an outstanding open problem. However, the 3D incompressible Navier-Stokes equations with hyperdissipation $(-\Delta )^{\alpha }u$ always possesses a global smooth solution if $\alpha \geq \frac{5}{4}$. Yang-Jiu-Wu (The 3D incompressible Navier-Stokes equations with partial hyperdissipation, Math. Nachr. 292(8):1823–1836, 2019) reduced the hyperdissipation $(-\Delta )^{\alpha }u$ to $((\Lambda _{1}^{\frac{5}{2}}, \Lambda _{2}^{\frac{5}{2}})u_{1}, (\Lambda _{2}^{ \frac{5}{2}}, \Lambda _{3}^{\frac{5}{2}})u_{2}, ( \Lambda _{3}^{\frac{5}{2}}, \Lambda _{1}^{\frac{5}{2}})u_{3})^{\top }$ and obtained the global existence and uniqueness in $H^{1}$ space. However, the higher-order derivative estimate of the solution is a nontrivial question. In this paper, we prove that this solution is a global solution in $H^{s}$ with $s>\frac{5}{2}$ applying a single directional commutator estimate.

Appendix A

In this section, we present the Littlewood-Paley theory, the definition of Besov space and the proof of the Proposition 2.1.

Choose two nonnegative radial functions $\chi$, $\varphi \in C_{c}^{\infty }(\mathbb{R}^{3})$ supported in $\mathcal{B} =\{\xi \in \mathbb{R}^{3}, |\xi |\leq \frac{4}{3}\}$ and $\mathcal{C}=\{\xi \in \mathbb{R}^{3}, \frac{3}{4}\leq |\xi |\leq \frac{8}{3}\}$ respectively, such that

$$\begin{aligned} \chi (\xi )+\sum _{j\geq 0}\varphi (2^{-j}\xi )=1, \quad \xi \in \mathbb{R}^{3}, \\ \sum _{j\in \mathbb{Z}}\varphi (2^{-j}\xi )=1, \quad \xi \in \mathbb{R}^{3}\setminus \{0\}. \end{aligned}$$

Let $h=\mathcal{F}^{-1}\varphi$ and $\widetilde{h}=\mathcal{F}^{-1}\chi$, the frequency localization operators are defined by

$$\begin{aligned} \Delta _{j}f=2^{3j}\int _{\mathbb{R}^{3}}h(2^{j}y)f(x-y)\mathrm{d}y , \\ S_{j}f=2^{3j}\int _{\mathbb{R}^{3}}\widetilde{h}(2^{j}y)f(x-y) \mathrm{d}y. \end{aligned}$$

Formally, $\Delta _{j}$ is a frequency projection to the annulus $\{|\xi | \approx 2^{j}\}$, $S_{j}$ is a frequency projection to the ball $\{|\xi | \leq 2^{j}\}$. Observe that $\Delta _{j}=S_{j+1}-S_{j}$.

Let $s\in \mathbb{R}$, $1\leq p$, $q\leq \infty$, the inhomogeneous Besov space ${B}^{s}_{p,q}$ is defined by

$$\begin{aligned} {B}^{s}_{p,q}=\{f\in \mathcal{S}'(\mathbb{R}^{3})\big| \|f\|_{{B}^{s}_{p,q}}< \infty \}, \end{aligned}$$

where

$$\begin{aligned} \|f\|_{{B}^{s}_{p,q}}= \textstyle\begin{cases} \bigg(\sum _{j\geq -1}2^{jsq}\|\triangle _{j}f\|_{p}^{q}\bigg)^{ \frac{1}{q}}, \ if\ q< \infty , \\ \mathrm{sup}_{j\geq -1} 2^{js}\|\triangle _{j}f\|_{L^{p}}, \ if \ q= \infty . \end{cases}\displaystyle \end{aligned}$$

In particular, $\|f\|_{B^{s}_{2,2}}\thickapprox \|f\|_{H^{s}}$.

Now, we prove the Proposition 2.1. Taking the inner product with $u$ in $\mathbb{R}^{3}\times (0, T)$ on (1.3)₁-(1.3)₃, respectively, using integration by part and divergence free condition of $u$, one has

$$\begin{aligned} &\|u\|^{2}_{2}+2\int _{0}^{T} \bigg(\| (\Lambda _{1} ^{\frac{5}{ 4}}, \Lambda _{2}^{\frac{5}{ 4}}) u_{1} \|^{2}_{2} +\| ( \Lambda _{2}^{ \frac{5}{ 4}}, \Lambda _{3}^{\frac{5}{ 4}})u_{2}\|^{2}_{2} +\| ( \Lambda _{3}^{\frac{5}{ 4}} , \Lambda _{1}^{\frac{5}{ 4}} )u_{3}\|^{2}_{2} \bigg)\text{d}t\le \|u_{0}\|^{2}_{2}. \end{aligned}$$

(4.1)

In the following, we establish the $\dot{H}^{1}$-estimate. Putting the operator $\nabla$ to the equations (1.3)₁-(1.3)₃ and taking the $L^{2}$ inner product with $\nabla u_{i}$ for $i=1,2,3$, respectively, we have

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \|\nabla u\|^{2}_{2}+ \| (\Lambda _{1} ^{ \frac{5}{ 4}}, \Lambda _{2}^{\frac{5}{ 4}}) \nabla u_{1} \|^{2}_{2} + \| ( \Lambda _{2}^{\frac{5}{ 4}}, \Lambda _{3}^{\frac{5}{ 4}})\nabla u_{2}\|^{2}_{2} +\| (\Lambda _{3} ^{\frac{5}{ 4}} , \Lambda _{1} ^{\frac{5}{ 4}} )\nabla u_{3}\|^{2}_{2} \\ =&-\sum _{i,j,k=1}^{3}\int _{\mathbb{R}^{3}}\partial _{j} u_{k} \partial _{k} u_{i} \partial _{j} u_{i} dx\triangleq I. \end{aligned}$$

(4.2)

In the following, we bound the term $I$.

$$\begin{aligned} I =&-\sum _{i,k=1}^{3}\int _{\mathbb{R}^{3}}\partial _{1} u_{k} \partial _{k} u_{i} \partial _{1} u_{i} \text{d}x -\sum _{i,k=1}^{3} \int _{\mathbb{R}^{3}}\partial _{2} u_{k} \partial _{k} u_{i} \partial _{2} u_{i} \text{d}x -\sum _{i,k=1}^{3}\int _{\mathbb{R}^{3}} \partial _{3} u_{k} \partial _{k} u_{i} \partial _{3} u_{i} \text{d}x \\ \triangleq &I_{1}+I_{2}+I_{3}. \end{aligned}$$

We estimate $I_{1}$ in detail.

$$\begin{aligned} I_{1} =& -\int _{\mathbb{R}^{3}}\partial _{1} u_{1} \partial _{1}u_{1} \partial _{1} u_{1} \text{d}x-\int _{\mathbb{R}^{3}}\partial _{1} u_{1} \partial _{1} u_{2} \partial _{1} u_{2} \text{d}x-\int _{\mathbb{R}^{3}} \partial _{1} u_{1} \partial _{1} u_{3} \partial _{1} u_{3} \text{d}x \\ -&\int _{\mathbb{R}^{3}}\partial _{1} u_{2} \partial _{2} u_{1} \partial _{1} u_{1} \text{d}x-\int _{\mathbb{R}^{3}}\partial _{1} u_{2} \partial _{2} u_{2} \partial _{1} u_{2} \text{d}x-\int _{\mathbb{R}^{3}} \partial _{1} u_{2} \partial _{2} u_{3} \partial _{1} u_{3} \text{d}x \\ -&\int _{\mathbb{R}^{3}}\partial _{1} u_{3} \partial _{3} u_{1} \partial _{1} u_{1} \text{d}x-\int _{\mathbb{R}^{3}}\partial _{1} u_{3} \partial _{3} u_{2} \partial _{1} u_{2}\text{d}x-\int _{\mathbb{R}^{3}} \partial _{1} u_{3} \partial _{3}u_{3} \partial _{1} u_{3} \text{d}x\triangleq \sum _{k=1}^{9}I_{1k}. \end{aligned}$$

(4.3)

$I_{1}$ can be divided into two classes $I_{11}$, $I_{13}$, $I_{14}$, $I_{17}$, $I_{18 }$, $I_{19}$ and $I_{12}$, $I_{15}$, $I_{16}$. Using Hölder, Gagliardo-Nirenberg, Minkowski inequalities, Lemma 2.1, Lemma 2.2 and Young inequality, we have

$$\begin{aligned} I_{14} \le &\|\partial _{1} u_{2}\|_{L^{2}}\|\partial _{2} u_{1}\|_{L^{4}_{x_{1}}L^{2}_{x_{2}}L^{ \infty }_{x_{3}}} \|\partial _{1} u_{1}\|_{L^{4}_{x_{1}}L^{\infty }_{x_{2}} L^{2}_{x_{3}}} \\ \le & \|\partial _{1} u_{2}\|_{L^{2}}\|\Lambda _{1}^{\frac{1}{4}} \partial _{2} u_{1}\|_{L^{2}_{x_{1}}L^{2}_{x_{2}}L^{\infty }_{x_{3}}} \|\Lambda _{1}^{\frac{1}{4}}\partial _{1} u_{1}\|_{L^{2}_{x_{1}}L^{ \infty }_{x_{2}} L^{2}_{x_{3}}} \\ \le & \|\partial _{1} u_{2}\|_{L^{2}}\|\Lambda _{1}^{\frac{1}{4}} \partial _{2} u_{1}\|_{L^{\infty }_{x_{3}}L^{2}_{x_{1}}L^{2}_{x_{2}}} \|\Lambda _{1}^{\frac{1}{4}}\partial _{1} u_{1}\|_{L^{\infty }_{x_{2}}L^{2}_{x_{1}} L^{2}_{x_{3}}} \\ \le & \|\partial _{1} u_{2}\|_{L^{2}}\|\Lambda _{1}^{\frac{1}{4}} \partial _{2} u_{1}\|^{\frac{1}{2}}_{L^{2}}\|\Lambda _{1}^{\frac{1}{4}} \partial _{3}\partial _{2} u_{1}\|^{\frac{1}{2}}_{L^{2}} \|\Lambda _{1}^{ \frac{1}{4}}\partial _{1} u_{1}\|^{\frac{1}{2}}_{L^{2}}\|\Lambda _{1}^{ \frac{1}{4}}\partial _{2}\partial _{1} u_{1}\|^{\frac{1}{2}}_{L^{2}} \\ \le & \|\partial _{1} u_{2}\|_{L^{2}}\|(\Lambda _{1}^{\frac{5}{4}}, \Lambda _{2}^{\frac{5}{4}}) u_{1}\|^{\frac{1}{2}}_{L^{2}}\|(\Lambda _{1}^{ \frac{5}{4}},\Lambda _{2}^{\frac{5}{4}}) \partial _{3} u_{1}\|^{\frac{1}{2}}_{L^{2}} \|\Lambda _{1}^{\frac{5}{4}} u_{1}\|^{\frac{1}{2}}_{L^{2}}\|\Lambda _{1}^{ \frac{5}{4}}\partial _{2} u_{1}\|^{\frac{1}{2}}_{L^{2}} \\ \le & \frac{1}{16}\bigg(\|(\Lambda _{1}^{\frac{5}{4}},\Lambda _{2}^{ \frac{5}{4}}) \partial _{3} u_{1}\|^{2}_{L^{2}}+\|\Lambda _{1}^{\frac{5}{4}} \partial _{2} u_{1}\|^{2}_{L^{2}}\bigg) \\ & +C \|\partial _{1} u_{2}\|^{2}_{L^{2}} \bigg(\|(\Lambda _{1}^{\frac{5}{4}},\Lambda _{2}^{\frac{5}{4}}) u_{1}\|^{2}_{L^{2}}+ \|\Lambda _{1}^{\frac{5}{4}} u_{1}\|^{2}_{L^{2}}\bigg). \end{aligned}$$

(4.4)

Arguing similarly as the above argument, we obtain

$$\begin{aligned} &I_{11}+I_{13}+I_{17}+I_{18}+I_{19} \\ \le &\frac{1}{8}\bigg(\|(\Lambda _{1}^{\frac{5}{4}},\Lambda _{2}^{ \frac{5}{4}}) \nabla u_{1}\|^{2}_{L^{2}} +\|(\Lambda _{2}^{\frac{5}{4}}, \Lambda _{3}^{\frac{5}{4}}) \nabla u_{2}\|^{2}_{L^{2}}+\|(\Lambda _{1}^{ \frac{5}{4}}, \Lambda _{3}^{\frac{5}{4}})\nabla u_{3}\|^{2}_{L^{2}}\bigg) \\ & +C \|\nabla u\|^{2}_{2}\bigg(\|(\Lambda _{1}^{\frac{5}{4}},\Lambda _{2}^{ \frac{5}{4}}) u_{1}\|^{2}_{L^{2}} +\|(\Lambda _{2}^{\frac{5}{4}},\Lambda _{3}^{ \frac{5}{4}}) u_{2}\|^{2}_{L^{2}}+\|(\Lambda _{1}^{\frac{5}{4}}, \Lambda _{3}^{ \frac{5}{4}}) u_{3}\|^{2}_{L^{2}}\bigg). \end{aligned}$$

(4.5)

Because the form of $I_{12}$, $I_{15}$, $I_{16}$ is different from the others in $I_{1}$, we use a different method to estimate the three terms. By divergence free condition of $u$, integrating by part, and Hölder, Gagliardo-Nirenberg, Minkowski inequalities, Lemma 2.1, Lemma 2.2 and Young inequality, one has

$$\begin{aligned} &I_{12}+I_{15}=\int _{\mathbb{R}^{3}}\partial _{3}u_{3} \partial _{1} u_{2} \partial _{1} u_{2} \text{d}x =-2\int _{\mathbb{R}^{3}}u_{3} \partial _{1} u_{2} \partial _{13} u_{2} \text{d}x \\ \le & C\|u_{3}\|_{L^{2}_{x_{3}}L^{2}_{x_{2}}L^{\infty }_{x_{1}}} \| \partial _{1} u_{2}\|_{L^{4}_{x_{3}}L^{\infty }_{x_{2}}L^{2}_{x_{1}}} \|\partial _{13} u_{2}\|_{L^{4}_{x_{3}}L^{2}_{x_{2}}L^{2}_{x_{1}}} \\ \le & C\|u_{3}\|^{\frac{1}{2}}_{L^{2}}\|\partial _{1} u_{3}\|^{\frac{1}{2}}_{L^{2}} \|\Lambda _{3}^{\frac{1}{4}}\partial _{1} u_{2}\|_{L^{2}}^{\frac{1}{2}} \| \Lambda _{3}^{\frac{1}{4}}\partial _{12} u_{2}\|_{L^{2}}^{\frac{1}{2}} \| \Lambda ^{\frac{5}{4}}_{3}\partial _{1} u_{2}\|_{L^{2}} \\ \le & C\|u_{3}\|^{\frac{3}{5}}_{L^{2}}\|\Lambda ^{\frac{5}{4}}_{1} u_{3}\|^{ \frac{2}{5}}_{L^{2}} \|\partial _{1} u_{2}\|_{L^{2}}^{\frac{2}{5}} \| \Lambda _{3}^{\frac{5}{4}}\partial _{1} u_{2}\|_{L^{2}}^{\frac{1}{10}} \|( \Lambda _{2}^{\frac{5}{4}}, \Lambda ^{\frac{5}{4}}_{3})\partial _{1} u_{2}\|^{ \frac{3}{2}}_{L^{2}} \\ \le & \frac{1}{8} \|(\Lambda _{2}^{\frac{5}{4}}, \Lambda ^{\frac{5}{4}}_{3}) \partial _{1} u_{2}\|^{2}_{L^{2}} +C\|\partial _{1} u_{2}\|_{L^{2}}^{2} \|\Lambda ^{\frac{5}{4}}_{1} u_{3}\|^{2}_{L^{2}} \|u_{3}\|^{3}_{L^{2}}. \end{aligned}$$

(4.6)

Similarly,

$$\begin{aligned} &I_{16}=\int _{\mathbb{R}^{3}}u_{2}\partial _{12}u_{3} \partial _{1} u_{3} \text{d}x+\int _{\mathbb{R}^{3}}u_{2} \partial _{2} u_{3} \partial _{11}u_{3} \text{d}x \\ \le & \|u_{2}\|_{L^{2}_{x_{3}}L^{2}_{x_{1}}L^{\infty }_{x_{2}}}\| \partial _{1}u_{3}\|_{L^{4}_{x_{3}}L^{\infty }_{x_{1}}L^{2}_{x_{2}}} \| \partial _{12} u_{3}\|_{L^{4}_{x_{3}}L^{2}_{x_{1}}L^{2}_{x_{2}}} \\ & +\|u_{2}\|_{L^{2}_{x_{3}}L^{2}_{x_{1}}L^{\infty }_{x_{2}}} \| \partial _{2} u_{3}\|_{L^{4}_{x_{3}}L^{\infty }_{x_{1}}L^{2}_{x_{2}}} \|\partial _{11}u_{3}\|_{L^{4}_{x_{3}}L^{2}_{x_{1}}L^{2}_{x_{2}}} \\ \le & \frac{1}{8}\|(\Lambda _{1}^{\frac{5}{4}}, \Lambda _{3}^{\frac{5}{4}}) \nabla u_{3}\|^{2}_{L^{2}}+C\|\partial _{2} u_{2}\|_{L^{2}}^{2}\|( \Lambda _{3}^{\frac{5}{4}}, \Lambda _{1}^{\frac{5}{4}})u_{3}\|^{2}_{L^{2}} \|u_{2} \|_{L^{2}}^{2} \\ &+C \|\partial _{2}u_{3}\|^{ 2}_{L^{2}} \|\Lambda _{2}^{\frac{5}{4}} u_{2} \|_{L^{2}}^{2} \|u_{2}\|_{L^{2}}^{ 3}. \end{aligned}$$

(4.7)

Adding (4.4)-(4.7), we get

$$\begin{aligned} I_{1} \le & \frac{1}{4}\bigg(\|(\Lambda _{1}^{\frac{5}{4}},\Lambda _{2}^{ \frac{5}{4}}) \nabla u_{1}\|^{2}_{L^{2}}+\|(\Lambda _{2}^{\frac{5}{4}}, \Lambda _{3}^{\frac{5}{4}})\nabla u_{2}\|^{2}_{L^{2}}+\|(\Lambda _{3}^{ \frac{5}{4}},\Lambda _{1}^{\frac{5}{4}})\nabla u_{3}\|^{2}_{L^{2}}\bigg) \\ & +C \|\nabla u\|^{2}_{2}\bigg(\|(\Lambda _{1}^{\frac{5}{4}},\Lambda _{2}^{ \frac{5}{4}}) u_{1}\|^{2}_{L^{2}} +\|(\Lambda _{2}^{\frac{5}{4}},\Lambda _{3}^{ \frac{5}{4}}) u_{2}\|^{2}_{L^{2}}+\|(\Lambda _{1}^{\frac{5}{4}}, \Lambda _{3}^{ \frac{5}{4}}) u_{3}\|^{2}_{L^{2}}\bigg) \\ &\times \bigg(1+ \|u\|^{3}_{L^{2}}+ \|u\|_{L^{2}}^{2}\bigg). \end{aligned}$$

(4.8)

By the same way as in deriving (4.8), one has

$$\begin{aligned} I_{2}+I_{3} \le & \frac{1}{4}\bigg(\|(\Lambda _{1}^{\frac{5}{4}}, \Lambda _{2}^{\frac{5}{4}}) \nabla u_{1}\|^{2}_{L^{2}}+\|(\Lambda _{2}^{ \frac{5}{4}},\Lambda _{3}^{\frac{5}{4}})\nabla u_{2}\|^{2}_{L^{2}}+\|( \Lambda _{3}^{\frac{5}{4}},\Lambda _{1}^{\frac{5}{4}})\nabla u_{3}\|^{2}_{L^{2}} \bigg) \\ & +C \|\nabla u\|^{2}_{2}\bigg(\|(\Lambda _{1}^{\frac{5}{4}},\Lambda _{2}^{ \frac{5}{4}}) u_{1}\|^{2}_{L^{2}} +\|(\Lambda _{2}^{\frac{5}{4}},\Lambda _{3}^{ \frac{5}{4}}) u_{2}\|^{2}_{L^{2}}+\|(\Lambda _{1}^{\frac{5}{4}}, \Lambda _{3}^{ \frac{5}{4}}) u_{3}\|^{2}_{L^{2}}\bigg) \\ &\times \bigg(1+ \|u\|^{3}_{L^{2}}+ \|u\|_{L^{2}}^{2}\bigg). \end{aligned}$$

(4.9)

Inserting (4.8), (4.9) to (4.2), we have

$$\begin{aligned} &\frac{d}{dt} \|\nabla u\|^{2}_{2} + \| (\Lambda _{1} ^{ \frac{5}{ 4}}, \Lambda _{2}^{\frac{5}{ 4}}) \nabla u_{1} \|^{2}_{2} + \| ( \Lambda _{2}^{\frac{5}{ 4}}, \Lambda _{3}^{\frac{5}{ 4}})\nabla u_{2}\|^{2}_{2} +\| (\Lambda _{3} ^{\frac{5}{ 4}} , \Lambda _{1} ^{\frac{5}{ 4}} )\nabla u_{3}\|^{2}_{2} \\ \le &C\|\nabla u\|^{2}_{L^{2}}\bigg(\|(\Lambda _{1}^{\frac{5}{4}}, \Lambda _{2}^{\frac{5}{4}}) u_{1}\|_{L^{2}}^{2}+\|(\Lambda _{2}^{ \frac{5}{4}},\Lambda _{3}^{\frac{5}{4}}) u_{2}\|_{L^{2}}^{2}+\|(\Lambda _{1}^{ \frac{5}{4}}, \Lambda _{3}^{\frac{5}{4}}) u_{3}\|_{L^{2}}^{2}\bigg) \\ &\times \bigg(1+\|u\|^{2}_{L^{2}}+\|u\|^{3}_{L^{2}}\bigg). \end{aligned}$$

(4.10)

By Gronwall’s inequality, we obtain

$$\begin{aligned} &\|\nabla u\|^{2}_{2} + \int _{0}^{T}\bigg(\| (\Lambda _{1} ^{ \frac{5}{ 4}}, \Lambda _{2}^{\frac{5}{ 4}}) \nabla u_{1} \|^{2}_{2} + \| ( \Lambda _{2}^{\frac{5}{ 4}}, \Lambda _{3}^{\frac{5}{ 4}})\nabla u_{2}\|^{2}_{2} +\| (\Lambda _{3} ^{\frac{5}{ 4}} , \Lambda _{1} ^{\frac{5}{ 4}} )\nabla u_{3}\|^{2}_{2} \bigg)\text{d}t \\ \le & \|\nabla u_{0}\|^{2}_{L^{2}}e^{C\int ^{T}_{0}\big(\|(\Lambda _{1}^{ \frac{5}{4}}, \ \Lambda _{2}^{\frac{5}{4}}) u_{1}\|_{L^{2}}^{2}+\|(\Lambda _{2}^{ \frac{5}{4}}, \ \Lambda _{3}^{\frac{5}{4}}) u_{2}\|_{L^{2}}^{2}+\|(\Lambda _{3}^{ \frac{5}{4}}, \ \Lambda _{1}^{\frac{5}{4}}) u_{3}\|_{L^{2}}^{2}\big) \big(1+ \|u\|^{3}_{L^{2}}\big)\text{d}t} \\ \le & \|\nabla u_{0}\|^{2}_{L^{2}}e^{ C\big(1+\|u_{0}\|_{L^{2}}^{5} \big)}. \end{aligned}$$

(4.11)

Noting the $L^{2}$-estimate (4.1) of $u$, the proof of the Proposition 2.1 is completed.