【作者】 岳双珊；

【导师】 周家足；

【作者基本信息】 西南大学 ， 基础数学， 2011， 硕士

【摘要】 本文的主要工作是拓展包含经典的等周不等式,Bonnesen等周不等式在内的几何不等式.第一部分：著名的平面等周不等式是最早用基本的几何不变量来刻画平面几何图形的几何不等式.即平面上固定周长的简单闭曲线中,圆所围的面积最大.换句话来说,平面上固定面积的区域中,圆盘的周长最小.用数学语言表述为(等周不等式)欧氏平面R2中域K的面积A,周长P满足不等式P2-4πA≥0, (1)等号成立的充要条件是K为圆盘.加强的等周不等式是以下是著名的Bonnesen等周不等式(Bonnesen等周不等式)欧氏平面R2中域K的面积A,周长P,包含于K内的最大内接圆半径为ri,包含K的最小外接圆半径为re,则有P2-4πA≥π2(re-ri)2, (2)等号成立的充要条件是K为圆盘.J.Zhou用积分几何方法揭示了一系列的类似Bonnesen不等式并给出了统一的证明.设K为平面上面积为A,周长为P的域,ri和re分别为K的最大内接圆半径和最小外接圆半径,设ri≤r≤re,则下列不等式成立P2-4πA≥(P2-2πr)2, P2-4πA≥(P2-2A/r)2, P2-4πA≥(A/r-πr)2,等号成立的充要条件是K为圆盘.我们主要研究平面上两个凸域的对称混合等周不等式(the symmetric mixed isoperimetric inequality)设Kk(k=i,j)为欧氏平面R2中面积为Ak,周长为Pk的域,它们的对称混合等周亏格(the symmetric mixed isoperimetric deficit)被J.Zhou定义为σ(Ki,Kj)=Pi2Pj2-16πAiAj. (4)根据周家足的思想方法和积分几何包含测度理论,我们利用欧氏平面上一个域包含另一个域的包含测度把平面上的Bonnesen等周不等式和Bonnesen不等式拓广,得到Bonnesen对称混合等周不等式及加强形式和Bonnesen对称混合不等式,主要得到以下定理定理1.设Kk(k=i,j)为欧氏平面R2中面积为Ak的凸域,其边界(?)Kk是简单闭曲线,周长为Pk,则我们有定理2.设Kk(k=i,j)为欧氏平面R2中面积为Ak的凸域,其边界(?)Kk是简单闭曲线,周长为Pk,则我们有Pi2Pj2-16π2AiAj≥4π2Ai2(tM-tm)2+(2πAitm+2πAitM-PiPj)2. (6)当Ki取为单位圆盘时即为P2-4πA≥π2(re-ri)2+(πre+πri-P)2. (7)这是著名的Bonnesen等周不等式(2)的加强.第二部分：记△=P2-4πA,则△为等周亏格,它的上界是几何中一个十分有趣的问题,欧氏平面上等周亏格的上界估计有下面的定理设K为欧氏平面上面积为A,周长为P的域,ρm和ρM分别为(?)K的最小曲率半径和最大曲率半径,则下列不等式成立P2-4πA≤π2 (ρM-ρm)2, (8)等号成立的充要条件为ρm=ρM,即K为圆盘.以上结果由Bottema于1933年得到.1955年Pleijel得到Bottema不等式(8)的加强,即P2-4πA≤πr(4-π)(ρM-ρm)2 (9)等号成立的充要条件为ρm=ρM,即K为圆盘.本文我们得到了欧氏平面上的对称混合等周亏格的上界.定理3.设Kk(k=i,j)为欧氏平面R2中面积为Ak的凸域,其边界(?)Kk是简单闭曲线,周长为Pk,则我们有Pi2Pj2-16π2AiAh≤4π2PiPj(tMRi2-tmri2). (10)其中tm=max{t:g(tki)(?)Kj,g∈G},tM=min{t:g(tKi)(?)Kj,g∈G}等号成立的充分必要条件是Ki和Kj均为圆盘.更多还原

【Abstract】 In this paper, we extend the classical isoperimetric inequality and the Bonnesen isoperimetric inequality in the Euclidean plane.Part I:isoperimetric inequality is an old problem, it says that given a domain K in the Euclidean plane, if the area is fixed, then the circle has the minimum length; or if the length is fixed, the circle has the maximum area. And expressed in mathematic language is that:(Isoperimetric inequality) The area A and the length P of any domain K in the plane satisfies the inequality P2-4πA≥0, (1) with the equality hold if and only if K is a disk.The sharping isoperimetric inequality is the Bonnesen inequality. The following is the famous Bonnesen isoperimetric inequality:(Bonnesen isoperimetric inequality) Let K be a domain of perimeter P and area A. Denote ri the radius of the maximal circle contained in K and by re the radius of the minimal circle containing K. Then the following inequality hold P2-4πAA≥π2(re-ri)2, (2) with the equality hold if and only if K is a disk.J. Zhou [33] obtain some Bonnesen inequalities by using the integral geometry method.We investigate the symmetric mixed isoperimetric inequalities of two planar convex domains. Let Kk(k= i,j) be the domain of the area of Ak, and of the perimeter Pk,respectively.The symmetric mixed isoperimetric deficit of Ki and Kj is defined by J.Zhou asσ(Ki,Kj)=Pi2Pj2-16πAiAj. (4)In this thesis the problem we solved is that we use the method of integral geometry and the condition of one domain to contain another domain in the plane, get the Bonnesen symmetric mixed isoperimetric inequality and Bonnesen mixed inequality.We get the following the results.Theorem 1.Let Kk(k=i,j)be the convex domain of the area of Ak,and of the perimeter Pk,respectively.Then Pi2Pj2-16π2AiAj≥(PiPj-4πAit)2, Pi2Pj2-16π2AiAj≥(PiPj-4πAi/t)2, (5) Pi2Pj2-16π2AiAj≥4π2(Aj/t-Ait)2Theorem 2.Let Kk(k=i,j)be the convex domain of the area of Ak,and of the perimeter Pk,respectively.Then Pi2Pj2-16π2AiAj≥4π2Ai2(tM-tm)2+(2πAitm+2πAitM-PiPj)2. (6)When Ki is the unit disk,we have P2-4πA≥π2(re-ri)2+(πre+πri-P)2. (7) This the sharping Bonnesen isoperimetric inequality.PartⅡ:LetΔ2(K)=P2-4πA,be the isoperimetric deficit of K.Its upper limit is an interesting problem.For upper limit of the isoperimetric deficit,We have the following theorems:Let K be a domain of perimeter P and area A in the plane.Assume that the boundary (?)K of the convex set K has a continuous radius of curvatureρ.Letρm andρM be the smallest and greatest values,respectively,ofρ.We have P2-4πA≤π2(ρM-ρm)2, (8) with the equality hold if and only ifρm=ρM,that is,K is a disk. This inequality is due to Bottema in 1933. In 1955,Pleijel obtained an im-provement as follow. P2-4πA≤π(4-π)(ρM-ρm)2, (9) with the equality hold if and only ifρm=ρM,that is,K is a disk.We obtain the following symmetric mixed isoperimetric upper limits.Theorem 3.Let Kk(k=i,j)be the convex domain of the area of Ak,and the perimeter Pk,respectively.Let tm=max{t:9(tKi)(?)Kj,g∈G},tM=min{t: g(tKi)(?)Kj,g∈G},then Pi2Pj2-16π2AiAj≤4π2PiPj(tMRi2-tmri2). (10) The equality holds if and only if Ki and Kj are discs.更多还原