Explicit div-curl inequalities in bounded and unbounded domains of ${{\mathbb {R}}}^3$

Abstract

We give a constructive proof of some functional inequalities related to the div and curl operators in bounded and unbounded domains of ${{\mathbb {R}}}^3$. Our new innovation consists in giving explicit constants in several geometric configurations. These inequalities are of a first use in solving div-curl systems and vector potential problems arising in physics.

Appendix

1.1 Proof of Lemma 3

The proof we give here is inspired by [2, 10] and [52]. We detail it in the case ${\varOmega }={{\mathbb {R}}}^3$. The case ${\varOmega }={{\mathbb {R}}}_+^3$ can be treated exactly in the same manner. By density of ${{\mathscr {D}}}({{\mathbb {R}}}^3)$ in $H^1_{\theta }({{\mathbb {R}}}^3)$, we can consider $v \in {{\mathscr {D}}}({{\mathbb {R}}}^3)$. We have

$$\begin{aligned} \int _{{{\mathbb {R}}}^3} \varrho ^{2\theta } |v|^2 dx = \int _{{{\mathbb {S}}^2}}\left( \int _{0}^{+\infty } r^2 (r^2+\epsilon ^2)^{{\theta }} |v(r, \sigma )|^2 dr\right) d\sigma , \end{aligned}$$

where ${{\mathbb {S}}}^2= \{ {{{x}}}\in {{\mathbb {R}}}^3 \,|\,|{{{x}}}|=1\}$. Let

$$\begin{aligned} \varrho (r) = r^2 (r^2+\epsilon ^2)^{{\theta }+1/2}. \end{aligned}$$

Then,

$$\begin{aligned} \frac{\rho '(r)}{r^2 (r^2+\epsilon ^2)^{{\theta }}} = 2 \frac{\sqrt{r^2+\epsilon ^2}}{r} + (2 \theta + 1) \frac{r}{\sqrt{r^2+\epsilon ^2}} \ge k_1, \end{aligned}$$

with

$$\begin{aligned} k_1 = \min _{ x\ge 1} 2 x + \frac{(2 \theta + 1)}{x} = \left\{ \begin{array}{ll} 2 \theta + 3 &{} \text{ if } \theta \le \displaystyle { \frac{1}{2}}, \; \\ \displaystyle { 4 \sqrt{ \theta + \frac{1}{2}}} &{} \text{ if } \theta \ge \displaystyle { \frac{1}{2}}. \end{array} \right. \end{aligned}$$

For $\sigma \in {{\mathbb {S}}}^2$, integration by parts and Cauchy–Schwarz inequality give

$$\begin{aligned} \displaystyle { \int _{0}^{+\infty } \varrho '(r) |v(r, \sigma )|^2 dr}= & {} - 2 \displaystyle { \int _{0}^{+\infty } \varrho (r) v(r, \sigma ) \frac{\partial v}{\partial r}(r, \sigma ) dr} \\\le & {} 2 \displaystyle { \left( \int _{0}^{+\infty } r^2 (r^2+\epsilon ^2)^{{\theta }} |v(r, \sigma )|^2 dr \right) ^{1/2}} \\&\times \displaystyle { \left( \int _{0}^{+\infty } r^2 (r^2+\epsilon ^2)^{{\theta +1}} \left| \frac{\partial v}{\partial r}(r, \sigma )\right| ^2 dr \right) ^{1/2}}, \end{aligned}$$

(since $\varrho (0)=0$). It follows that

$$\begin{aligned} \int _{0}^{+\infty } r^2 (r^2+\epsilon ^2)^{{\theta }} |v(r, \sigma )|^2 dr \le \frac{4}{k_1^2} \int _{0}^{+\infty } r^2 (r^2+\epsilon ^2)^{{\theta }+1} |\frac{\partial v}{\partial r}(r, \sigma )|^2 dr, \end{aligned}$$

which ends the proof of (61). $\square$