The Symmetric Mixed Isoperimetric Inequality Of Two Planar Convex Domains

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.