论文标题
牛顿的最小阻力问题中的鼻子拉伸方法
Method of nose stretching in Newton's problem of minimal resistance
论文作者
论文摘要
我们考虑问题$ \ inf \ big \ {\ int \！\！\int_Ω（1 + | \ nabla u（x，x，y）|^2）^{ - 1} dx dy：\ text {function} fartion} fortial} u function} u：ω\ to \ mathbb {r} \ r} \ r} \ text text { }（x，y）\ inω= \ {（x，y）：x^2 + y^2 \ le 1 \} \，\ big \} $（牛顿的问题）及其概括。在纸上\ cite {brfk}中，证明，如果解决方案$ u $为$ c^2 $，则在开放式套装中$ \ mathcal {u} \ subsetω$，则$ \ det d^2u = 0 $ in $ \ nathcal {u} $。遵循该图$（u）\ rfloor_ \ mathcal {u} $不包含$ u $的子图的极端点。 在本文中，我们证明结果有所更强。也就是说，存在具有以下属性的解决方案$ u $。如果$ u $是$ c^1 $在打开集中的$ \ mathcal {u} \subsetΩ$，则Graph $（u \ rfloor_ \ Mathcal {u}）$不包含凸面的极端点$ C_U = \ = \ {（x，y，z） \} $。结果，我们有$ c_u = \ text {\ rm cons}（\ overline {\ text {\ rm sing $ c_u $}}）$，其中sing $ c_u $表示$ \ partial c_u $的单数点。我们证明了广义牛顿问题的类似结果。
We consider the problem $\inf\big\{ \int\!\!\int_Ω(1 + |\nabla u(x,y)|^2)^{-1} dx dy : \text{ the function } u : Ω\to \mathbb{R} \text{ is concave and } 0 \le u(x,y) \le M \text{ for all } (x,y) \in Ω=\{ (x,y): x^2 + y^2 \le 1 \} \, \big\}$ (Newton's problem) and its generalizations. In the paper \cite{BrFK} it is proved that if a solution $u$ is $C^2$ in an open set $\mathcal{U} \subset Ω$ then $\det D^2u = 0$ in $\mathcal{U}$. It follows that graph$(u)\rfloor_\mathcal{U}$ does not contain extreme points of the subgraph of $u$. In this paper we prove a somewhat stronger result. Namely, there exists a solution $u$ possessing the following property. If $u$ is $C^1$ in an open set $\mathcal{U} \subset Ω$ then graph$(u\rfloor_\mathcal{U})$ does not contain extreme points of the convex body $C_u = \{ (x,y,z) :\, (x,y) \in Ω,\ 0 \le z \le u(x,y) \}$. As a consequence, we have $C_u = \text{\rm Conv} (\overline{\text{\rm Sing$C_u$}})$, where Sing$C_u$ denotes the set of singular points of $\partial C_u$. We prove a similar result for a generalized Newton's problem.
