Variational analysis of the planar $L_p$ dual Minkowski problem

Abstract

The $L_p$ dual curvature measures were recently introduced by Lutwak et al. (Adv Math 329:85-132, 2018) to unify the classical theory of mixed volumes and the newer theory of dual mixed volumes of convex bodies. However, the associated $L_p$ dual Minkowski problems for many important special cases remain open problems. They are analytically equivalent to a class of nonlinear problems with indices p and q. In most previous studies, conventional geometric inequalities and Aleksandrov’s variational formula for convex bodies were used to study these problems. In this paper, by using a new investigation method via directly studying the nonlinear problems characterizing the planar $L_p$ dual Minkowski problem in Sobolev spaces, several sharp functional inequalities associated with the $L_p$ dual curvature measures are established to generalize the classical inequalities, such as the Wirtinger’s inequality and the Blaschke-Santaló inequality. Based on these new sharp functional inequalities, various existence results for the planar $L_p$ dual Minkowski problem are obtained for integrable data and general $q\ge 2$. Compared with the uniqueness results of $\pi -$periodic solutions for $p=0$ and $q=2$ (in Dohmen and Giga, Proc Japan Acad Ser A Math Sci, 1994; Andrews, J Amer Math Soc, 2003), the non-uniqueness and multiple $\pi -$periodic solutions are also obtained in this paper for $p=0$ and all $q>4$ through variational analysis.

Appendix

Lemma 6.1

Let $p<0$, and $\{u_n\}$ is a sequence of positive continuous functions and $\lim \limits _{n\rightarrow +\infty }u_n(\theta )=|\sin (\theta )|$ uniformly on ${\mathbb {S}}$. Let $\varepsilon _n=\min \limits _{\theta }u_n(\theta )>0$ for each $n\in {\mathbb {N}}$. As $\varepsilon _n\rightarrow 0^+$, we have

$$\begin{aligned} \frac{3^{p}}{-2(p+1)}\varepsilon _n^{p+1}\le & {} \frac{1}{4}\int _{{\mathbb {S}}}{u_n^{p}(\theta )}d\theta \nonumber \\\le & {} \frac{3^{-p}\pi }{-2(p+1)}\varepsilon _n^{p+1} \text {holds for } p\in (-\infty ,-1); \end{aligned}$$

(6.1)

$$\begin{aligned} \frac{1}{3}\ln \left( \frac{\pi /2+\varepsilon _n}{\varepsilon _n}\right)\le & {} \frac{1}{4}\int _{{\mathbb {S}}}{u_n^{p}(\theta )}d\theta \le \frac{3\pi }{2}\ln \left( \frac{1+\varepsilon _n}{\varepsilon _n}\right) \text { holds for } p=-1;\qquad \end{aligned}$$

(6.2)

and

$$\begin{aligned} \frac{3^p}{2(p+1)}(\pi /2+\varepsilon _n)^{p+1}\le & {} \frac{1}{4}\int _{{\mathbb {S}}}{u_n^{p}(\theta )}d\theta \nonumber \\\le & {} \frac{ 3^{-p}\pi }{2(p+1)}(1+\varepsilon _n)^{p+1}\text { holds for}p\in (-1,0). \end{aligned}$$

(6.3)

Proof

For large n, it follows from the assumption that

$$\begin{aligned} \frac{1}{3}\le \frac{u_n(\theta )}{|\sin (\theta )|+\varepsilon _n} =\frac{1}{|\sin (\theta )|/u_n(\theta )+\varepsilon _n/u_n(\theta )} \le 3, \text {for all }\theta \in {\mathbb {S}}. \end{aligned}$$

We get that $\frac{|\sin (\theta )|+\varepsilon _n}{3}\le u_n(\theta )\le 3(|\sin (\theta )|+\varepsilon _n)$ for all $\theta \in {\mathbb {S}}$ and large n. Since $p<0$, we deduce that

$$\begin{aligned} {3^{p}} \int _{0}^{\pi /2}{(\sin (\theta )+\varepsilon _n)^{p}}d\theta \le \frac{1}{4}\int _{{\mathbb {S}}}{u_n^{p}(\theta )}d\theta \le {3^{-p}} \int _{0}^{\pi /2}{(\sin (\theta )+\varepsilon _n)^{p}}d\theta \qquad \end{aligned}$$

(6.4)

holds for large n. Noting that the inequality $\frac{2}{\pi }\le \frac{\sin (\theta )}{\theta }<1$ holds for all $\theta \in (0,\frac{\pi }{2}]$, we have

$$\begin{aligned} \int _{0}^{\pi /2}{(\theta +\varepsilon _n)^{p}}d\theta \le \int _{0}^{\pi /2}(\sin (\theta )+\varepsilon _n)^{p}d\theta \le \int _{0}^{\pi /2}{(2\theta /\pi +\varepsilon _n)^{p}}d\theta . \end{aligned}$$

(6.5)

And it is easy to get that

$$\begin{aligned} \int _{0}^{\pi /2}{(2\theta /\pi +\varepsilon _n)^{p}}d\theta =\left\{ \begin{array}{ll} \frac{\pi }{2(1+p)}\left( {(1+\varepsilon _n)^{p+1}}-{\varepsilon _n^{p+1}}\right) ,\ p\not =-1,\\ \frac{\pi }{2}\ln \frac{1+\varepsilon _n}{\varepsilon _n},\ p=-1. \end{array}\right. \end{aligned}$$

(6.6)

and

$$\begin{aligned} \int _{0}^{\pi /2}{(\theta +\varepsilon _n)^{p}}d\theta =\left\{ \begin{array}{ll} \frac{1}{1+p}\left( {(\pi /2+\varepsilon _n)^{p+1}}-{\varepsilon _n^{p+1}}\right) ,\ p\not =-1,\\ \ln \left( \frac{\pi /2+\varepsilon _n}{\varepsilon _n}\right) ,\ p=-1. \end{array}\right. \end{aligned}$$

(6.7)

Then, (6.1)-(6.3) follow from (6.4)-(6.7). $\square$

Lemma 6.2

Let $q\ge 2$, $f(\theta )=|\sin (\theta )|$, denote $f_0(\theta )=|f(2\theta /3)|$ for $\theta \in [0,3\pi /2]$; and $f_0(\theta )=0$ for $\theta \in (3\pi /2, 2\pi )$. Then $f_0\in H^{1,q}({\mathbb {S}})$, and there exists $\tau _0>0$ such that

$$\begin{aligned} \int _{{\mathbb {S}}}\left( f^q_0-\int _0^1\frac{(f_0^2+s^2f'^2_0)^{q/2}-f_0^q}{s^2}ds\right) d\theta =\tau _0>0. \end{aligned}$$

Proof

By Theorem 3.3, we get that

$$\begin{aligned} \int _{0}^{\pi } f^q d\theta -\int _{0}^{\pi }\int _0^1\frac{(f^2+s^2f'^2)^{q/2}-f^q}{s^2}dsd\theta =0. \end{aligned}$$

(6.8)

A simple calculation together with (6.8) gives that

$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {S}}} f^q_0 d\theta -\int _{{\mathbb {S}}}\int _0^1\frac{(f_0^2+s^2f'^2_0)^{q/2}-f_0^q}{s^2}dsd\theta \\&\quad =\int _{0}^{3\pi /2} f^q_0 d\theta -\int _{0}^{3\pi /2}\int _0^1\frac{(f_0^2+s^2f'^2_0)^{q/2}-f_0^q}{s^2}dsd\theta \\&\quad =\frac{3}{2}\left( \int _{0}^{\pi } f^q d\theta -\int _{0}^{\pi }\int _0^1\frac{(f^2+4s^2f'^2/9)^{q/2}-f^q}{s^2}dsd\theta \right) \\&\quad =\frac{3}{2}\int _{0}^{\pi }\int _0^1\frac{(f^2+s^2f'^2)^{q/2}-(f^2+4s^2f'^2/9)^{q/2}}{s^2}dsd\theta :=\tau _0>0. \end{aligned} \end{aligned}$$

$\square$

Lemma 6.3

Let $q>1$, $t>0$, $\tau \not =0$, $\varsigma \in {\mathbb {R}}$, $A_1(\tau ,\varsigma )$ and $B_1(\tau ,\varsigma )$ be given by (2.14) with $t=1$. Then,

$$\begin{aligned} \frac{\partial A_1(\tau ,\varsigma )}{\partial \tau }= & {} -(q-2)\tau \varsigma \int _0^1(\tau ^2+s^2\varsigma ^2)^{{q}/{2}-2}ds=\frac{\partial B_1(\tau ,\varsigma )}{\partial \varsigma }, \end{aligned}$$

(6.9)

$$\begin{aligned}&\frac{\partial A_1(\tau ,\varsigma )}{\partial \varsigma }=-(\tau ^2+\varsigma ^2)^{{q}/{2}-1}, \end{aligned}$$

(6.10)

and

$$\begin{aligned}&\frac{\partial B(\tau ,\varsigma )}{\partial \tau } =(q-1)|\tau |^{q-2}-\int _0^1\frac{(\tau ^2+s^2\varsigma ^2)^{{q}/{2}-1}-|\tau |^{q-2}}{s^2}ds\nonumber \\&\quad +(2-q)\tau ^2\int _0^1\frac{(\tau ^2+s^2\varsigma ^2)^{{q}/{2}-2}-|\tau |^{q-4}}{s^2}ds. \end{aligned}$$

(6.11)

If $q\ge 2$, there exists $C>0$ depending only on q such that

$$\begin{aligned} \left| \frac{\partial A_1(\tau ,\varsigma )}{\partial \tau }\right| + \left| \frac{\partial B_1(\tau ,\varsigma )}{\partial \varsigma }\right| \le C [\tau ^2+\varsigma ^2]^{\frac{q}{2}-1}. \end{aligned}$$

(6.12)

If $1<q<2$, further assume that $|\tau |>\alpha >0$, then there exists $C>0$ depending only on q and $\alpha$ such that

$$\begin{aligned} \left| \frac{\partial A_1(\tau ,\varsigma )}{\partial \tau }\right| + \left| \frac{\partial B_1(\tau ,\varsigma )}{\partial \varsigma }\right|= & {} 2(2-q)\left| \tau \varsigma \int _0^1(\tau ^2+s^2\varsigma ^2)^{{q}/{2}-2}ds\right| \nonumber \\\le & {} C|\tau |. \end{aligned}$$

(6.13)

Proof

By calculation directly, we get (6.9), (6.11) and

$$\begin{aligned} \frac{\partial A_1(\tau ,\varsigma )}{\partial \varsigma }= & {} -\int _0^1[\tau ^2+s^2\varsigma ^2]^{\frac{q}{2}-1}ds-(q-2)\varsigma ^2\int _0^1[\tau ^2+s^2\varsigma ^2]^{\frac{q}{2}-2}s^2ds\\= & {} -[\tau ^2+\varsigma ^2]^{\frac{q}{2}-1}. \end{aligned}$$

If $2< q<4$, then ${q}/{2}-2\in (-1,0)$, and it follows from the Hölder inequality that $(\tau ^2+s^2\varsigma ^2)^{{q}/{2}-2}\le \left( {|\tau s\varsigma |}{2}\right) ^{{q}/{2}-2}$. For $2< q<4$, we deduce that

$$\begin{aligned} \left| \tau \varsigma \int _0^1(\tau ^2+s^2\varsigma ^2)^{{q}/{2}-2}ds\right| \le \frac{1}{2}|2\tau \varsigma |^{{q}/{2}-1}\int _0^1\frac{1}{s^{2-q/2}}ds\le \frac{1}{q-2}(\tau ^2+\varsigma ^2)^{{q}/{2}-1}. \end{aligned}$$

If $q\ge 4$, it is easy to see that

$$\begin{aligned} \left| \tau \varsigma \int _0^1[\tau ^2+s^2\varsigma ^2]^{\frac{q}{2}-2}ds\right| \le \left| \tau \varsigma \int _0^1(\tau ^2+\varsigma ^2)^{{q}/{2}-2}ds\right| \le \frac{1}{2}(\tau ^2+\varsigma ^2)^{{q}/{2}-1}. \end{aligned}$$

It follows from $q>2$ that

$$\begin{aligned} \left| \tau \varsigma \int _0^1(\tau ^2+s^2\varsigma ^2)^{{q}/{2}-2}ds\right| \le \left( \frac{1}{q-2}+ \frac{1}{2}\right) (\tau ^2+\varsigma ^2)^{{q}/{2}-1}. \end{aligned}$$

(6.14)

By (6.14), there exists $C>0$ depending on q such that

$$\begin{aligned} \begin{aligned} \left| \frac{\partial B_1(\tau ,\varsigma )}{\partial \varsigma }\right| =\left| \frac{\partial A_1(\tau ,\varsigma )}{\partial \tau }\right|&=\left| -(q-2)\tau \varsigma \int _0^1(\tau ^2+s^2\varsigma ^2)^{{q}/{2}-2}ds\right| \\&\le C (\tau ^2+\varsigma ^2)^{{q}/{2}-1}. \end{aligned} \end{aligned}$$

which is (6.12). If $q=2$, it is trivial to get (6.12). For the case $1<q<2$, we see that $\int _0^{+\infty }(\tau ^2+s^2)^{{q}/{2}-2}ds<\int _0^{+\infty }(\alpha ^2+s^2)^{{q}/{2}-2}ds<+\infty$. It follows that (6.13) holds for some $C>0$, which depends only on q and $\alpha$. $\square$