Sharp Rellich-Leray inequality for axisymmetric divergence-free vector fields

Abstract

In this paper, we show the N-dimensional Rellich-Leray inequality with optimal constant for axisymmetric and divergence-free vector fields. This is a second-order differential version of the former work by Costin-Maz’ya (Calc Var Partial Differ Equ 32(4):523–532, 2008) on sharp Hardy–Leray inequality for such vector fields. In the proof of our main theorem, we show the vanishing of azimuthal components of axisymmetric vector fields for $N\ge 4$, from which we also find a partial modification of the best constant derived in Costin-Maz’ya (Calc Var Partial Differ Equ 32(4):523–532, 2008).

Appendix

In this appendix, we list up some technical lemmas with proofs.

Lemma 10

Let $N\ge 2$. Let $C_0^\infty ([0,\pi ])=\big \{\psi \in C^\infty ([0,\pi ])\ ;\ \psi (0)=\psi (\pi )=0 \big \}$ and let $T=\partial _{{\theta }}D_{{\theta }}=\partial _{\theta }(\partial _ {\theta }+(N-2)\cot {\theta })$ be the second-rank derivative operator on $C_0^\infty ([0,\pi ])$. Then the set of eigenvalues of $-T$ is given by

$$\begin{aligned} Spec(-T)=\big \{\alpha _{\nu }=\nu (\nu +N-2)\ ; \ \nu \in {\mathbb {N}}\big \}. \end{aligned}$$

In addition, for every $\nu \in {\mathbb {N}}$ there exists a polynomial $P_{\nu -1}$ of degree $\nu -1$ such that the function

$$\begin{aligned} \psi _\nu (\theta )=P_{\nu -1}(-\cos \theta )\sin \theta \end{aligned}$$

satisfies the $\nu$-th eigenequation $-T\psi _\nu =\alpha _\nu \psi _\nu$. Moreover, the sequence of such functions $\{\psi _\nu (\theta )\}_{\nu \in {\mathbb {N}}}$ spans a complete basis of the Hilbert space $L^2\big ([0,\pi ],(\sin \theta )^{N-2}d\theta \big )$.

Proof

We first show that $Spec(-T)\subset \{\alpha _\nu \ ;\ \nu \in {\mathbb {N}}\}$. To do so, let $\psi \in C_0^\infty ([0,\pi ])$ and put $\psi (\theta )=\phi (\theta )\sin \theta$. Then we can see the function $\phi :[0,\pi ]\rightarrow {\mathbb {R}}$ is smooth in $(0,\pi )$ and continuous on $[0,\pi ]$, that is,

$$\begin{aligned}\phi \in C^\infty ((0,\pi ))\cap C([0,\pi ]).\end{aligned}$$

Also, by direct calculation we have

$$\begin{aligned} \begin{aligned} T\psi&=\partial _\theta \big (\partial _{\theta }+(N-2)\cot {\theta } \big )(\phi \sin {\theta })\\&=(\sin {\theta })\big ((\partial _\theta +N\cot \theta )\partial _ \theta -(N-1)\big )\phi . \end{aligned} \end{aligned}$$

Then we find that the eigenequation $-T\psi =\alpha \psi$ for $\alpha \in {\mathbb {R}}$ is equivalent to

$$\begin{aligned} - (\partial _\theta +N\cot \theta )\partial _\theta \phi =(\alpha -N+1)\phi . \end{aligned}$$

(47)

This is also equivalent to the eigenequation $-\Delta _{{\mathbb {S}}^{N+1}}\phi =(\alpha -N+1)\phi$ for the Laplace-Beltrami operator $\Delta _{{\mathbb {S}}^{N+1}}$ on ${\mathbb {S}}^{N+1}$ in the spherical polar coordinates $(\rho ,\theta _1,\ldots ,\theta _{N+1})$ with $\theta _1=\theta$. By the well-known relation $Spec(-\Delta _{{\mathbb {S}}^{N+1}})=\big \{\nu (\nu +N)\ ;\ \nu \in {\mathbb {N}}\cup \{0\}\big \}$, it then follows that

$$\begin{aligned} \begin{aligned} Spec(-T)&\subset \big \{\alpha \ ;\ \alpha -N+1\in Spec(-\Delta _{{\mathbb {S}}^{N+1}})\big \}\\&=\big \{\nu (\nu +N)+N-1\ ;\ \nu \in {\mathbb {N}}\cup \{0\}\big \}\\&=\big \{(\nu +1)(\nu +N-1)\ ;\ \nu \in {\mathbb {N}}\cup \{0\}\big \}=\{\alpha _\nu \ ;\ \nu \in {\mathbb {N}}\}. \end{aligned} \end{aligned}$$

To prove the inverse relation $Spec(-T)\supset \{\alpha _\nu \ ;\ \nu \in {\mathbb {N}}\}$, it is enough to show the existence of a non-trivial solution to equation (47) for each $\alpha =\alpha _\nu$, $\nu \in {\mathbb {N}}$. To do this, we transform $\theta$ to $x=-\cos \theta \in [-1,1]$, whose differential obeys the chain rule $\partial _\theta =(\sin \theta )\partial _x$. Then the derivative operator in the left-hand side of (47) can be written as

$$\begin{aligned} \begin{aligned} \big (\partial _\theta +N\cot \theta \big )\partial _\theta&= \big (\partial _\theta +N\cot \theta \big )(\sin \theta )\partial _x\\&=(\cos \theta )\partial _x+(\sin \theta )\partial _\theta \partial _x +N(\cos \theta )\partial _x\\&=(1-x^2)\partial _x^2-(N+1)x\partial _x. \end{aligned} \end{aligned}$$

Hence Eq. (47) is transformed into

$$\begin{aligned} (1-x^2)\partial _x^2\phi -(N+1)x\partial _x\phi +(\alpha -N+1)\phi =0. \end{aligned}$$

(48)

On the other hand, the p-dimensional Legendre Polynomial of degree $\nu \in {\mathbb {N}}\cup \{0\}$ is given by the Rodrigues formula

$$\begin{aligned} P_\nu (x)=(1-x^2)^{(3-p)/2}\Big (\frac{d}{dx}\Big )^\nu (1-x^2)^{\nu +(p-3)/2}, \end{aligned}$$

(with the normalizing constant omitted,) which satisfies the equation

$$\begin{aligned} (1-x^2)P_\nu ''+(1-p)xP_\nu '+\nu (\nu +p-2)P_\nu =0. \end{aligned}$$

(See, e.g. [4].) Then it is easy to check for $p=N+2$ and $\nu \in \{0\}\cup {\mathbb {N}}$ that

$$\begin{aligned} \phi =P_\nu (x)=P_\nu (-\cos \theta ) \end{aligned}$$

is a solution to equation (48) or (47) with $\alpha = \alpha _{\nu +1}$. Therefore, this shows that the function $\psi _\nu (\theta )=P_{\nu -1}(-\cos \theta )\sin \theta$ for each $\nu \in {\mathbb {N}}$ satisfies the eigenequation $-T\psi _\nu =\alpha _{\nu }\psi _\nu$.

Finally, we show the completeness of $\{\psi _\nu (\theta )\}_{\nu \in {\mathbb {N}}}$. The Weierstrass approximation theorem ensures that the sequence $\{P_{\nu -1}(x)\}_{\nu \in {\mathbb {N}}}$ spans a dense subspace of $C\big ([-1,1]\big )$ in the topology of uniform convergence. Then it immediately follows that $\{\phi _\nu =P_{\nu -1}(-\cos \theta )\}_{\nu \in {\mathbb {N}}}$ also spans a dense subspace of $C([0,\pi ])$. Moreover, this implies $\{\psi _\nu =\phi _\nu (\theta )\sin \theta \}_{\nu \in {\mathbb {N}}}$ spans a dense subspace of $C_0^\infty ([0,\pi ])$ in the topology of uniform convergence, because any $\psi \in C_0^\infty ([0,\pi ])$ can be expressed as $\psi (\theta )=\phi (\theta )\sin \theta$ for some $\phi \in C([0,\pi ])$. Therefore, we see that the transitive relations

$$\begin{aligned} \mathrm{span}\{\psi _\nu \}\underset{\mathrm{dense}}{\subset } C_0^\infty \big ([0,\pi ]\big )\underset{\mathrm{dense}}{\subset } L^2\big ([0,\pi ]\big ) \end{aligned}$$

hold in the topology of $L^2\big ([0,\pi ],d\theta \big )$, and hence in that of $L^2\big ([0,\pi ],(\sin \theta )^{N-2} d\theta \big )$, which concludes the lemma. $\square$

Lemma 11

Let $N\ge 3$ and $\nu \in {\mathbb {N}}$. For any $h\in C^\infty ({\mathbb {R}}_{\ge 0})$, let ${\varvec{u}}:{\mathbb {R}}^N\rightarrow {\mathbb {R}}^N$ be an axisymmetric vector field whose polar components for $\rho >0$ are given by

$$\begin{aligned} \begin{aligned}&u_\rho (\rho ,\theta _1)=-h(\rho )D_{\theta _1}\psi _\nu (\theta _1)\ ,\quad \ u_{\theta _1}(\rho ,\theta _1)=\rho D_\rho h(\rho )\psi _\nu (\theta _1),\\&\quad \mathrm{and}\qquad u_{\theta _2}=\cdots =u_{\theta _{N-1}}=0\quad \mathrm{identically}. \end{aligned} \end{aligned}$$

(49)

Here $\psi _\nu$ denotes the $\nu$-th eigenfunction of $-\partial _{\theta _1}D_{\theta _1}$. Then ${\varvec{u}}$ is smooth and divergence-free in ${\mathbb {R}}^N\backslash \{{\varvec{0}}\}$. In particular, let

$$\begin{aligned} {\varvec{w}}=-{\varvec{e}}_\rho \rho ^{\nu -1}D_{\theta _1}\psi _\nu (\theta _1)+{\varvec{e}}_{\theta _1}\rho D_\rho \rho ^{\nu -1}\psi _\nu (\theta _1). \end{aligned}$$

Then ${\varvec{w}}$ is divergence-free and harmonic on ${\mathbb {R}}^N$.

Proof

First it is clear from (49) and Lemma 10 that ${\varvec{u}}$ satisfies (26) and (27), and hence ${\mathrm{div}}\,{\varvec{u}}=0$. Next we show that $\Delta {\varvec{w}}={\varvec{0}}$ in ${\mathbb {R}}^N\backslash \{{\varvec{0}}\}$. By (15) and (16), we have

$$\begin{aligned} \begin{aligned} \rho ^2 \Delta {\varvec{w}}&=\rho ^2 D_\rho \partial _\rho {\varvec{w}}+\Delta _\sigma {\varvec{w}}=\alpha _{\nu -1}{\varvec{w}}+\Delta _\sigma {\varvec{w}}. \end{aligned} \end{aligned}$$

To show the vanishing of the right-hand side, abbreviating as $\psi =\psi _\nu$ and $T=\partial _{\theta _1}D_{\theta _1}$, we check by Lemma 5 that

$$\begin{aligned} \begin{aligned} -\Delta _\sigma ({\varvec{e}}_\rho D_{\theta _1}\psi )&=-2 {\varvec{e}}_{\theta _1}T\psi +{\varvec{e}}_\rho \big (-D_{\theta _1}T\psi +(N-1)D_{\theta _1}\psi \big )\\ {}&=2\alpha _{\nu } {\varvec{e}}_{\theta _1}\psi +{\varvec{e}}_\rho (\alpha _{\nu }+N-1)D_{\theta _1}\psi ,\\ \Delta _\sigma ({\varvec{e}}_{\theta _1}\psi )&=-2 {\varvec{e}}_\rho D_{\theta _1}\psi +{\varvec{e}}_{\theta _1}\big (N-3+T\big ) \psi \\&=-2 {\varvec{e}}_{\rho }D_{\theta _1}\psi +(N-\alpha _{\nu }-3){\varvec{e}}_{\theta _1}\psi . \end{aligned} \end{aligned}$$

Then we have

$$\begin{aligned} \begin{aligned} \Delta _\sigma {\varvec{w}}&=\Delta _\sigma \big (-{\varvec{e}}_\rho \rho ^{\nu -1}D_{\theta _1}\psi +{\varvec{e}}_{\theta _1}(\nu +N-2)\rho ^{\nu -1}\psi \big )\\&=\rho ^{\nu -1}\Big (-\Delta _\sigma ({\varvec{e}}_\rho D_{\theta _1}\psi )+(\nu +N-2)\Delta _\sigma ({\varvec{e}}_{\theta _1}\psi )\Big )\\&=\rho ^{\nu -1}\Big (\big (2\alpha _{\nu }+({\nu }+N-2) (N-\alpha _{\nu }-3)\big ){\varvec{e}}_{\theta _1}\psi \\&\quad +\big (\alpha _{\nu }+N-1-2({\nu }+N-2)\big ){\varvec{e}}_{\rho }D_{\theta _1}\psi \Big )\\&=-\alpha _{\nu -1}{\varvec{w}}. \end{aligned} \end{aligned}$$

Therefore, we obtain $\rho ^2\Delta {\varvec{w}}={\varvec{0}}$, which implies the harmonicity of ${\varvec{w}}$ in ${\mathbb {R}}^N\backslash \{{\varvec{0}}\}$. Since ${\varvec{w}}$ is bounded near ${\varvec{0}}$, the (well-known) removal of singularity shows that $\Delta {\varvec{w}}={\varvec{0}}$ on ${\mathbb {R}}^N$. $\square$

Corollary 12

Choose $h\in C^\infty ({\mathbb {R}}_{\ge 0})$ to be such that $h(\rho )= {\left\{ \begin{array}{ll} \rho ^{\nu -1} &{}\mathrm{for}\ 0\le \rho \le 1\\ 0&{} \mathrm{for}\ 2\le \rho \end{array}\right. }$. Then (49) satisfies $\int _{{\mathbb {R}}^N}|\Delta {\varvec{u}}|^2|{\varvec{x}}|^{2\gamma }dx<\infty$ and $\int _{{\mathbb {R}}^N}|{\varvec{u}}|^{2}|{\varvec{x}}|^{2\gamma -4}dx=\infty$ if $\gamma \le 3-\frac{N}{2}-\nu$.

Lemma 13

Let $({M}, \mu )$ be a measure space and let $\xi _1, \xi _2: {M} \rightarrow {\mathbb {R}}$ be $\mu$-integrable functions such that $\xi _1 > 0$ $\mu$-a.e.. Then we have

$$\begin{aligned} \frac{\int _{M} \xi _2 d\mu }{\int _{M} \xi _1 d\mu } \ge \mathrm{ess} \inf _{x \in {M}} \frac{\xi _2(x)}{\xi _1(x)}, \end{aligned}$$

where the equality holds if and only if $\displaystyle -\infty <\mathrm{ess}\inf _{y\in {M}}\frac{\xi _2(y)}{\xi _1(y)}=\dfrac{\xi _2(x)}{\xi _1(x)}$$\mu$-a.e. x.

Proof

Let $I = \mathrm{ess} \, \inf _{x \in {M}} \frac{\xi _2(x)}{\xi _1(x)}$. Then $\frac{\xi _2}{\xi _1} \ge I$ $\mu$-a.e. Multiplying both sides by $\xi _1> 0$ , we have $\xi _2\ge I \xi _1$ $\mu$-a.e.. Integrating over M, we obtain

$$\begin{aligned} \int _{M} \xi _2 d\mu \ge I \int _{M} \xi _1 d\mu . \end{aligned}$$

Since $\int _{M}\xi _1d\mu >0$, this leads the result:

$$\begin{aligned} \frac{\int _{M}\xi _2d\mu }{\int _{M}\xi _1d\mu }\ge I. \end{aligned}$$

Here the equality holds if $-\infty < I=\frac{\xi _2}{\xi _1}$ $\mu$-a.e., because the same above argument with every “$\ge$” replaced by “$=$” also holds. Conversely if $\frac{\int _{M}\xi _2d\mu }{\int _{M}\xi _1d\mu }= I$, then $-\infty <I$ and

$$\begin{aligned} \int _{M}(\xi _2-I\xi _1) d\mu =0. \end{aligned}$$

Since $\xi _2-I\xi _1\ge 0$$\mu$-a.e., the integrand must vanish: $\xi _2-I\xi _1=0$ $\mu$-a.e., which concludes the lemma. $\square$