Hadamard's inequality in the mean

arXiv (Cornell University)(2023)

引用 0|浏览1
暂无评分
摘要
Let $Q$ be a Lipschitz domain in $\mathbb{R}^n$ and let $f \in L^{\infty}(Q)$. We investigate conditions under which the functional $$I_n(\varphi)=\int_Q |\nabla \varphi|^n+ f(x)\,\mathrm{det} \nabla \varphi\, \mathrm{d}x $$ obeys $I_n \geq 0$ for all $\varphi \in W_0^{1,n}(Q,\mathbb{R}^n)$, an inequality that we refer to as Hadamard-in-the-mean, or (HIM). We prove that there are piecewise constant $f$ such that (HIM) holds and is strictly stronger than the best possible inequality that can be derived using the Hadamard inequality $n^{\frac{n}{2}}|\det A|\leq |A|^n$ alone. When $f$ takes just two values, we find that (HIM) holds if and only if the variation of $f$ in $Q$ is at most $2n^{\frac{n}{2}}$. For more general $f$, we show that (i) it is both the geometry of the `jump sets' as well as the sizes of the `jumps' that determine whether (HIM) holds and (ii) the variation of $f$ can be made to exceed $2n^{\frac{n}{2}}$, provided $f$ is suitably chosen. Specifically, in the planar case $n=2$ we divide $Q$ into three regions $\{f=0\}$ and $\{f=\pm c\}$, and prove that as long as $\{f=0\}$ `insulates' $\{f= c\}$ from $\{f= -c\}$ sufficiently, there is $c>2$ such that (HIM) holds. Perhaps surprisingly, (HIM) can hold even when the insulation region $\{f=0\}$ enables the sets $\{f=\pm c\}$ to meet in a point. As part of our analysis, and in the spirit of the work of Mielke and Sprenger (1998), we give new examples of functions that are quasiconvex at the boundary.
更多
查看译文
关键词
inequality,mean
AI 理解论文
溯源树
样例
生成溯源树,研究论文发展脉络
Chat Paper
正在生成论文摘要