论文标题
Zaremba问题的反向Faber-Krahn不平等
Reverse Faber-Krahn inequalities for Zaremba problems
论文作者
论文摘要
让$ω$为$ \ MATHBB {r}^n $（$ n \ geq 2 $）的多重连接域的形式的形式$ω=ω_ {\ text {out}} \ setMinus \ bar \ bar {ω___________ $ω_ {\ text {out}} $或$ω_ {\ text {in}} $。对于$ p \ in（1，\ infty），$ q \ in [1，p]，$ let $τ_{1，q}（ω）$是第一个特征 \ begin {equination*}-Δ_Pu =τ\左（\int_Ω| u |^q \ text {d} x \ right）^{\ frac {p-q} {q} {q} {q}}} | u |^{q-2 {q-2 {q-2} u = 0 \; \ text {on} \; \partialΩ_d，\ frac {\ partial u} {\ partialη} = 0 \; \; \ text {on} \; \部分ω\ setMinus \ partialω_d。 \ end {equation*} 在假设$ω_d$是凸的假设下，我们建立以下反向Faber-krahn不平等$$τ_{1，q}（ω）\ leq leq fous_ {1，q}（q}（ω^\ bigStar） $ \ mathbb {r}^n $具有相同的lebesgue度量与$ω$，并且（i）（i）（当$ω_d=ω_ {\ text {out}} $）$ w_1（ω_d）=ω__________________________nr^{n-1} $，and $（yi（ω^\ bigstar）美元 这里$ w_ {i}（ω_d）$是$ i^{\ text {th}} $ $ $ quermassIntegral $ of $ω_d。$，我们还建立了sz。 Nagy的类型不等式用于$ \ Mathbb {r}^n $（$ n \ geq 3 $）中的凸域的并行集。
Let $Ω$ be a multiply-connected domain in $\mathbb{R}^n$ ($n\geq 2$) of the form $Ω=Ω_{\text{out}}\setminus \bar{Ω_{\text{in}}}.$ Set $Ω_D$ to be either $Ω_{\text{out}}$ or $Ω_{\text{in}}$. For $p\in (1,\infty),$ and $q\in [1,p],$ let $τ_{1,q}(Ω)$ be the first eigenvalue of \begin{equation*} -Δ_p u =τ\left(\int_Ω|u|^q \text{d}x \right)^{\frac{p-q}{q}} |u|^{q-2}u\;\text{in} \;Ω,\; u =0\;\text{on}\;\partialΩ_D, \frac{\partial u}{\partial η}=0\;\text{on}\; \partial Ω\setminus \partial Ω_D. \end{equation*} Under the assumption that $Ω_D$ is convex, we establish the following reverse Faber-Krahn inequality $$τ_{1,q}(Ω)\leq τ_{1,q}(Ω^\bigstar),$$ where $Ω^\bigstar=B_R\setminus \bar{B_r}$ is a concentric annular region in $\mathbb{R}^n$ having the same Lebesgue measure as $Ω$ and such that (i) (when $Ω_D=Ω_{\text{out}}$) $W_1(Ω_D)= ω_n R^{n-1}$, and $(Ω^\bigstar)_D=B_R$, (ii) (when $Ω_D=Ω_{\text{in}}$) $W_{n-1}(Ω_D)=ω_nr$, and $(Ω^\bigstar)_D=B_r$. Here $W_{i}(Ω_D)$ is the $i^{\text{th}}$ $quermassintegral$ of $Ω_D.$ We also establish Sz. Nagy's type inequalities for parallel sets of a convex domain in $\mathbb{R}^n$ ($n\geq 3$) for our proof.
