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《中国科学:数学》2015,(2):212-215
<正>Objective triangle functors RINGEL Claus MichaelZHANG Pu Abstract An additive functor F:A→B between additive categories is said to be objective,provided any morphism f in A with F(f)=0 factors through an object K with F(K)=0.We concentrate on triangle functors between triangulated categories.The first aim of this paper is to characterize objective triangle functors F in several ways.Second,we are interested in the corresponding Verdier quotient functors VF:A→A/Ker F,in particular we want to know under what conditions VF is full.The third question to be considered concerns  相似文献   

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In this paper, using the exp-function method we obtain some new exact solutions for (1+1)-dimensional and (2+1)-dimensional Kaup–Kupershmidt (KK) equations. We show figures of some of the new solutions obtained here. We conclude that the exp-function method presents a wider applicability for handling nonlinear partial differential equations.  相似文献   

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试题研讨(10)     
题1 (2003年元月湖北省八校高三第一次联考)设P是直线x-y+9=0上的一点,过P点的椭圆以双曲线4x2-5y2=20的焦点为焦点,试求P点在什么位置时,所求椭圆的长轴最短,并写出这个具有最短长轴的椭圆方程.  相似文献   

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<正>On the lower bounds of the curvatures in a bounded domain LU Qi Keng Abstract Let KD(z,z)be the Bergman kernel of a bounded domain D in Cnand Sect D(z,ξ)and Ricci D(z,ξ)be the holomorphic sectional curvature and Ricci curvature of the Bergman metric ds2=TD(z,z)dzαdzβrespectivelyαβat the point z∈D with tangent vectorξ.It is proved by constructing suitable minimal functions that  相似文献   

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116 设任意△HBC中,D、E、F分别是△ABC的边BC、CA、AB上的内点,△DEF、△AEF、△BDF、△CED 的周长分别记为m0、m1、m2。m3, =min{A、B、C},则 上述命题可向平面n边形推广,另猜测,在任意△ABC中,有 (吴善和.1999,4) 117 如果△ABC内的三个圆都与三角形的内切圆相切,并且每个圆与△ABC的两边相切,设r、ra.rb、rc分别为内切圆及其余三个圆的半径,则 (赵长健.1999,4) 118 在交叉四边形 ABCD中,a、b、c/及S分别表示其边长和面…  相似文献   

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For initial value problems in ordinary second-order differential equations of the special form y″ = f(x, y), mew explicit, direct Runge-Kutta-Nyström formula-pairs of order 8(7), 9(8), 10(9) and 11(10) are presented using the mode of Bettis, Dormand and Prince. Two numerical examples demonstrate the efficiency of the new formula-pairs.  相似文献   

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解方程 sin5x=sin4x。解法一:因为与a有相同的正弦值的弧度数x的集合是{x|x=kπ (-1)~ka,k∈Z},所以原方程可以化成 5x=kπ (-1)4x (k∈Z) 解之得:x=kπ/5 (-1)~(k 1) 所以原方程的解集是{x|2= 解法二:原方程等价为sin4x=sin5x,m同解法一得:4x=kπ (-1)~k5x 解之得:x=kπ/4 (-1)~(k 1)5(k∈z)  相似文献   

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