Decay and Vanishing of some D-Solutions of the Navier–Stokes Equations

Abstract

An old problem since Leray (J Math Pure Appl (French) 9:1–82, 1933) asks whether homogeneous D-solutions of the 3 dimensional Navier–Stokes equation in ${\mathbb {R}}^3$ or some noncompact domains are 0. In this paper, we give a positive solution to the problem in two cases: (1) the full 3 dimensional slab case ${\mathbb {R}}^2 \times [0, 1]$ with Dirichlet boundary condition (Theorem 1.1); (2) when the solution is axially symmetric and periodic in the vertical variable (Theorem 1.3). Also, for the slab case, we prove that even if the Dirichlet integral has some growth, axially symmetric solutions with Dirichlet boundary condition must be swirl free, namely $u^\theta =0$, thus reducing the problem to essentially a “2 dimensional” problem. In addition, a general D-solution (without the axial symmetry assumption) vanishes in $\mathbb {R}^3$ if, in spherical coordinates, the positive radial component of the velocity decays at order -1 of the distance. The paper is self contained comparing with (Carrillo et al. in Funct Anal, 2020. https://doi.org/10.1016/j.jfa.2020.108504) although the general idea is related.

Appendix

In this Appendix, we are devoted to proving the Lemma 5.2 which is equivalent to the following estimates:

$$\begin{aligned} |G(x,y)|\lesssim \left\{ \begin{aligned} \begin{array}{ll} \frac{1}{|x'-y'|+|x_3-y_3|}&{}|x'-y'|<1,\\ \frac{1}{|x'-y'|^2} &{} |x'-y'|>1, \end{array} \end{aligned} \right. \end{aligned}$$

(A.14)

and

$$\begin{aligned} |\partial _{x',y'}G(x,y)|\lesssim \left\{ \begin{aligned} \begin{array}{ll} \frac{1}{|x'-y'|^2+|x_3-y_3|^2} &{} |x'-y'|<1,\\ \frac{1}{|x'-y'|^3} &{} |x'-y'|>1. \end{array} \end{aligned} \right. \end{aligned}$$

(A.15)

The proof is by direct computation. Alternatively, one can also first consider the heat kernel with Dirichlet boundary condition on ${\mathbb {R}}^2 \times [0, 1]$, which has fast decay in time. Then one can integrate out the time variable to obtain the Green’s function estimate. This was the route taken in [6]. We only show the proof of (A.14), and the proof of (A.15) will be essentially the same and we omit the details.

For simplification of notation and without cause of confusion, we denote

$$\begin{aligned} k_{n,-,+}= & {} \sqrt{|x'-y'|^2+|x_3-y_3+2n|^2},\quad k_{n,+,-}{=}\sqrt{|x'-y'|^2{+}|x_3+y_3{-}2n|^2},\\ k_{n,-,-}= & {} \sqrt{|x'-y'|^2+|x_3-y_3-2n|^2},\quad k_{n,+,+}{=}\sqrt{|x'-y'|^2{+}|x_3+y_3{+}2n|^2}. \end{aligned}$$

Case 1:$\varvec{|x'-y'|<1}$

$$\begin{aligned} G(x,y)= & {} \frac{1}{4\pi }\sum \limits ^{+\infty }_{n=-\infty }\bigg \{\frac{1}{k_{n,-,+}} -\frac{1}{k_{n,+,-}}\bigg \}\\= & {} \frac{1}{4\pi }\underbrace{\bigg \{\frac{1}{k_{0,-,+}}-\frac{1}{k_{0,+,-}} \bigg \}}_{I_1}+\frac{1}{4\pi }\underbrace{\bigg \{\frac{1}{k_{1,-,+}}-\frac{1}{k_{1,+,-}} \bigg \}}_{I_2}\\&+\frac{1}{4\pi }\sum \limits ^{n\ne 0,1}_{n\in {\mathbb {Z}}}\underbrace{\bigg \{\frac{1}{k_{n,-,+}} -\frac{1}{k_{n,+,-}}\bigg \}}_{I_{3,n}}. \end{aligned}$$

It is easy to see that

$$\begin{aligned} |I_1|+|I_2|\lesssim \frac{1}{\sqrt{|x'-y'|^2+|x_3-y_3|^2}}. \end{aligned}$$

(A.16)

We compute $I_3$ as follows:

$$\begin{aligned} |I_{3,n}|\lesssim & {} \frac{|k_{n,+,-}-k_{n,-,+}|}{k_{n,-,+}k_{n,+,-}}\nonumber \\= & {} \frac{|k^2_{n,+,-}-k^2_{n,-,+}|}{k_{n,-,+}k_{n,+,-}(k_{n,-,+}+k_{n,+,-})}\nonumber \\= & {} \frac{|4x_3(y_3-2n)|}{k_{n,-,+}k_{n,+,-}(k_{n,-,+}+k_{n,+,-})}. \end{aligned}$$

(A.17)

When $n\in {\mathbb {Z}},n\ne 0,1$, we have

$$\begin{aligned} |x_3+y_3-2n|\gtrsim |n|,\quad |x_3-y_3+2n|\gtrsim |x_3-y_3|+|n|, \end{aligned}$$

which indicates that

$$\begin{aligned} k_{n,-,+}\gtrsim |x'-y'|+|x_3-y_3|+|n|,\quad k_{n,+,-}\gtrsim |n|. \end{aligned}$$

Inserting the above inequality into (A.17), we have

$$\begin{aligned} |I_{3,n}|\lesssim & {} \frac{1}{|n|\sqrt{|x'-y'|^2+|x_3-y_3|^2+n^2}}, \end{aligned}$$

so

$$\begin{aligned} \sum \limits ^{n\ne 0,1}_{n\in {\mathbb {Z}}}|I_{3,n}|\lesssim & {} \sum \limits ^{n\ne 0,1}_{n\in {\mathbb {Z}}}\frac{1}{|n|\sqrt{|x'-y'|^2+|x_3-y_3|^2+n^2}}\nonumber \\\lesssim & {} \int ^\infty _1 \frac{1}{s\sqrt{|x'-y'|^2+|x_3-y_3|^2+s^2}}ds\nonumber \\\lesssim & {} \frac{1}{\sqrt{|x'-y'|^2+|x_3-y_3|^2}}. \end{aligned}$$

(A.18)

Then (A.16) and (A.18) together indicates the first part of (A.14).

Case 2: $\varvec{|x'-y'|>1}$

Since in this situation, we need to get one more order decay with respect to $|x'-y'|$, we need to compute the sum more carefully:

$$\begin{aligned} G(x,y)= & {} \frac{1}{4\pi }\sum \limits ^{+\infty }_{n=-\infty }\bigg \{\frac{1}{k_{n,-,+}} -\frac{1}{k_{n,+,-}}\bigg \}\\= & {} \frac{1}{4\pi }\underbrace{\bigg \{\frac{1}{k_{0,-,+}}-\frac{1}{k_{0,+,-}} +\frac{1}{k_{1,-,+}}-\frac{1}{k_{1,+,-}}+\frac{1}{k_{-1,-,+}}-\frac{1}{k_{-1,+,-}} \bigg \}}_{J_1}\\&+\frac{1}{4\pi }\sum \limits _{n\geqq 2}\underbrace{\bigg \{\frac{1}{k_{n,-,+}}-\frac{1}{k_{n,+,-}} +\frac{1}{k_{n,-,-}}-\frac{1}{k_{n,+,+}}\bigg \}}_{J_{2,n}}. \end{aligned}$$

Now we will compute Js term by term. For $J_1$,

$$\begin{aligned} \Big |\frac{1}{k_{0,-,+}}-\frac{1}{k_{0,+,-}}\Big |= & {} \frac{|k^2_{0,-,+}-k^2_{0,+,-}|}{{k_{0,-,+}}{k_{0,+,-}}({k_{0,-,+}}+{k_{0,+,-}})}\nonumber \\= & {} \frac{|4x_3y_3|}{{k_{0,-,+}}{k_{0,+,-}}({k_{0,-,+}}+{k_{0,+,-}})}\nonumber \\\lesssim & {} \frac{1}{|x'-y'|^3}. \end{aligned}$$

(A.19)

By the same techniques, we can prove that

$$\begin{aligned} \Big |\frac{1}{k_{1,-,+}}-\frac{1}{k_{1,+,-}}\Big |+\Big | \frac{1}{k_{-1,-,+}}-\frac{1}{k_{-1,+,-}}\Big |\lesssim \frac{1}{|x'-y'|^3}. \end{aligned}$$

(A.20)

We see that when $|x'-y'|>1$ and $n\geqq 2$,

$$\begin{aligned} k_{n,\alpha ,\beta }\approx |x'-y'|+n. \quad \alpha ,\beta \in \{+,-\}. \end{aligned}$$

(A.21)

Then

$$\begin{aligned} J_{2,n}= & {} \frac{k^2_{n,+,-}-k^2_{n,-,+}}{{k_{n,-,+}}{k_{n,+,-}}({k_{n,-,+}} +{k_{n,+,-}})}+\frac{k^2_{n,+,+}-k^2_{n,-,-}}{{k_{n,-,-}}{k_{n,+,+}}({k_{n,-,-}} +{k_{n,+,+}})}\\= & {} \frac{4x_3(y_3-2n)}{{k_{n,-,+}}{k_{n,+,-}}({k_{n,-,+}}+{k_{n,+,-}})} +\frac{4x_3(y_3+2n)}{{k_{n,-,-}}{k_{n,+,+}}({k_{n,-,-}}+{k_{n,+,+}})}\\= & {} 4x_3y_3\underbrace{\bigg [\frac{1}{k_{n,-,+}k_{n,+,-}(k_{n,-,+}+k_{n,+,-})} +\frac{1}{k_{n,-,-}k_{n,+,+}(k_{n,-,-}+k_{n,+,+})}\bigg ]}_{K_{n,1}}\\&-8x_3n\underbrace{\bigg [\frac{1}{k_{n,-,+}k_{n,+,-}(k_{n,-,+}+k_{n,+,-})} -\frac{1}{k_{n,-,-}k_{n,+,+}(k_{n,-,-}+k_{n,+,+})}\bigg ]}_{K_{n,2}}. \end{aligned}$$

Thus, we have

$$\begin{aligned} |J_{2,n}|\lesssim |K_{n,1}|+n|K_{n,2}|. \end{aligned}$$

Next we will show that

$$\begin{aligned} \sum \limits ^\infty _{n=2} |J_{2,n}|\lesssim & {} \sum \limits ^\infty _{n=2} |K_{n,1}|+ \sum \limits ^\infty _{n=2}n|K_{n,2}|\nonumber \\\lesssim & {} \frac{1}{|x'-y'|^2}. \end{aligned}$$

(A.22)

Then (A.19), (A.20) and (A.22) together indicate the second part of (A.14).

Using (A.21),

$$\begin{aligned} \sum \limits ^\infty _{n=2} |K_{n,1}|\lesssim & {} \sum \limits ^\infty _{n=2} \frac{1}{(|x'-y'|+n)^3}\nonumber \\\lesssim & {} \int ^\infty _0\frac{1}{(|x'-y'|+s)^3}ds\lesssim \frac{1}{|x'-y'|^2}. \end{aligned}$$

The hardest part is to estimate $K_{n,2}$ since the sum has one more increasing term n before $K_{n,2}$. When we estimate $K_{n,2}$, we need one more $\frac{1}{n}$ coming out compared with $K_{n,1}$.

Denote

$$\begin{aligned} K_{n,2,1}\equiv & {} {k_{n,-,+}\, \cdot k_{n,+,-} \cdot (k_{n,-,+}+k_{n,+,-})}\approx (|x'-y'|+n)^3,\\ K_{n,2,2}\equiv & {} {k_{n,-,-} \, \cdot k_{n,+,+} \cdot (k_{n,-,-}+k_{n,+,+})}\approx (|x'-y'|+n)^3. \end{aligned}$$

Therefore,

$$\begin{aligned} |K_{n,2}|= & {} \Big |\frac{1}{K_{n,2,1}}-\frac{1}{K_{n,2,2}}\Big |\nonumber \\= & {} \frac{|K^2_{n,2,2}-K^2_{n,2,1}|}{K_{n,2,1}K_{n,2,2}(K_{n,2,1}+K_{n,2,2})}\nonumber \\\approx & {} \frac{|K^2_{n,2,2}-K^2_{n,2,1}|}{(|x'-y'|+n)^9}. \end{aligned}$$

(A.23)

By a direct computation, we can see that

$$\begin{aligned} |K^2_{n,2,2}-K^2_{n,2,1}|\lesssim |x'-y'|^4n+|x'-y'|^2n^3+n^5. \end{aligned}$$

(A.24)

Inserting (A.24) into (A.23), we can get

$$\begin{aligned} \sum \limits ^\infty _{n=2} n|K_{n,2}|\lesssim & {} \sum \limits ^\infty _{n=2} \frac{|x'-y'|^4n^2+|x'-y'|^2n^4+n^6}{(|x'-y'|+n)^9}\nonumber \\\lesssim & {} \int ^\infty _0\frac{|x'-y'|^4s^2+|x'-y'|^2s^4+s^6}{(|x'-y'|+s)^9}ds\nonumber \\\lesssim & {} \frac{1}{|x'-y'|^2}. \end{aligned}$$

(A.25)

Combining the above, we have proved the estimate in (A.14) for G(x, y). The estimate of $\partial _{x',y'}G(x,y)$ will be essentially the same as G(x, y), since one $\frac{1}{|x-y|}$ will come out when we differentiate G(x, y) on $x',y'$, so we omit the details.