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1.
奇摄动非线性系统Robin边值问题   总被引:3,自引:0,他引:3  
莫嘉琪 《应用数学》1998,11(2):113-115
本文研究了非线性系统奇摄动问题:ε2y"-(x,y,y)=0,0<x<1,0<ε≤1,y(0)-py'(0)=A,p>0,y(1)=B,其中y,f,A,B为n维向量.在相应的假设下,利用代数型边界层函数,证明了该问题存在一个解y(x,ε),并利用微分不等式方法得到了其解的渐近估计.  相似文献   

2.
在处理一类椭圆C:x^2+y^2/a^2+b^2=1(a〉0,b〉0,a≠b)与直线l:y=kx+h的有关问题时,若能根据题意令x/a=x′,y/b=y′,即可把椭圆C、直线l分别变成圆C′:x′^2+y′^2=1、直线l′:by′=kax′+h,从而把椭圆与直线的位置关系问题转化为圆与直线的位置关系问题.如果需要还可以利用公式x/a=x′、y/b=y′将所得结果再转化回来.此法新颖、别致、简捷、实用,下面举例说明.  相似文献   

3.
谢淳  罗治国 《大学数学》2011,27(4):57-62
利用拓扑度理论和上下解方法讨论了一类三阶微分方程组{x′′′(t)+f1(t,y(t),x′(t),x″(t))=0,0≤t≤1,y′′′(t)+f2(t,x(t),y′(t),y″(t))=0,0≤t≤1在适当的条件下解的存在性.  相似文献   

4.
本文旨在通过实例,归纳总结出形如y=x+和y=x的最小值问题的统一解法及一般结果对能直接利用基本不等式a+b≥2,a+b+c≥3求解的情形,本文将略去.第一类的最小值问题情形1例1求的最小值,这里x(O,π).以上两个木等号中的等式同时成立,当且仅例2求函数y的值域.解函数的定义域为一1≤X≤1,于是令t=(1-x2)+,于是只需求出t的值域,即可得到y的值域.以上两个不等号中的等式同时成立,当且仅。。M。。。。。。4.例3(一般情形)求y一x十上的最小值,其中,0<X<b,户是一个正常数,且产)矿.上述两个不等号中的等式同时成立…  相似文献   

5.
本文研究kolmogorov捕食系统{(dx/dt)=x(ψ(x)-φ(y) (dx/dt)=y(bx^m-d) 得到了极限环存在唯一的条件,从而推广了前人相关的结果.其中:ψ(x)=a0+a1x+a2x^2+…+a(a-1)x^(n-1) -anx^n;n≥m≥1(n,m∈N),φ(0)=0,φ(y)〉ε〉0(y〉0).  相似文献   

6.
题1已知函数y=kx与y=x^2+2(x≥0)的图象相交于不同两点A(x1,Y1),B(x2,y2),l1,l2分别是y=x^2+2(x≥0)的图象在A,B两点的切线,M,N分别是l1,l2与x轴的交点,P为l1与l2的交点.  相似文献   

7.
第31届西班牙数学奥林匹克第2题为 命题1如果(x+√x^2+1)(y+√y^2+1)=1,则x+y=0.文[1]给出下面推广: 命题2如果m〉0,x,y∈[m,+∞)或x,y∈(-∞,+m]且(x+√x^2-m^2)(y+√y^2-m^2)=m^2,那么x=Y. 文[1]采用换元法证明了命题2,仔细研读后笔者给出命题2的另一种简洁证法。  相似文献   

8.
记f(x,y)=Ax^2+Bxy+Cy^2+Dx+Ey+F. 设点P(m,n)是圆锥曲线C:f(x,y)=0的一条弦AB的中点,C′是C关于点P对称的曲线(如图1),则曲线C上点A(B)关于点P(m,n)的对称点,B(A)在曲线C′上,故A,B是两曲线C,C′的交点。  相似文献   

9.
题29 已知函数f(x)对任意的实数x、Y都有厂(x+Y)=f(x)+f(y)-1,且当X〉0时,f(x)〉1.  相似文献   

10.
方差的计算公式为S2人教版初中《代数》第三册),它又可化为9一上【】X~上(,IZI方差具有非负性,即5270,当且仅当xl—12一…一xu,SZ=0,利用方差公式及其非负性,可以将不少数学问题转化为方差问题来解决.例1已知a,heR”,a+b—1,求证a’+bZk且2”证考虑a,b二项的方差例3已知实数x,y,z满足x-y=6,cy=zZ十9,求证:X一y·证工,y的方差为Y一言【x“十y“一二(x+y广」”亏以x+y)“一zry一青(x+y广」一一子【一子X十一月.t+o)1一一,J>几4=O,于是X一y·例4已知0<6<。,求函数y=/厂工面F肩而十/而百…  相似文献   

11.
An estimator of the number of components of a finite mixture ofk-dimensional distributions is given on the basis of a one-dimensional independent random sample obtained by a transformation of ak-dimensional independent random sample. A consistency of the estimator is shown. Some simulation results are given in a case of finite mixtures of two-dimensional normal distributions.  相似文献   

12.
Let A be a UFD of characteristic p > 0, let 𝒵 be a set of some eigenvectors of a derivation of A. We prove, under some additional assumptions, a necessary and sufficient condition for 𝒵 to be a p-basis of the minimal ring of constants containing 𝒵. The main preparatory result is the unique decomposition theorem with respect to a factor from a given subalgebra containing Ap.  相似文献   

13.
N/Kbe a Galois extension of number fields with finite Galois group G.We describe a new approach for constructing invariants of the G-module structure of the K groups of the ring of integers of N in the Grothendieck group of finitely generated projective Z[G]modules. In various cases we can relate these classes, and their function field counterparts, to the root number class of Fröhlich and Cassou-Noguès.  相似文献   

14.
有资格限制的指派问题的求解方法   总被引:3,自引:0,他引:3  
在实际的指派工作中,常会遇到某个人有没有资格去承担某项工作的问题,因此,本建立了有资格限制的指派问题的数学模型。在此数学模型中,将效益矩阵转化为判定矩阵,由此给出了判定此种指派问题是否有解的方法;在有解的情况下,进一步将效益矩阵转化为求解矩阵,从而将有资格限制的指派问题化为传统的指派问题来求解。最后给出了一个数值例子来说明这样的处理方法是有效的。  相似文献   

15.
Tai Keun Kwak  Yang Lee 《代数通讯》2013,41(9):4033-4046
We study the nilpotency of the sums of all coefficients of some sorts of products of polynomials over reversible, IFP, and NI rings, and introduce an SCN ring as a generalization. We characterize SCN rings in relation with related ring properties, and also provide several useful properties and ring extensions of SCN rings.  相似文献   

16.
It is a well-known result of M. Brodmann that if is an ideal of a commutative Noetherian ring , then the set of associated primes of the -th power of is constant for all large . This paper is concerned with the following question: given a prime ideal of which is known to be in for all large integers , can one identify a term of the sequence beyond which will subsequently be an ever-present? This paper presents some results about convergence of sequences of sets of associated primes of graded components of finitely generated graded modules over a standard positively graded commutative Noetherian ring; those results are then applied to the above question.

  相似文献   


17.
Let L be the Euclidean functional with p-th power-weighted edges. Examples include the sum of the p-th power-weighted lengths of the edges in minimal spanning trees, traveling salesman tours, and minimal matchings. Motivated by the works of Steele, Redmond and Yukich (Ann. Appl. Probab. 4, 1057–1073, 1994, Stoch. Process. Appl. 61, 289–304, 1996) have shown that for n i.i.d. sample points {X 1,…,X n } from [0,1] d , L({X 1,…,X n })/n (dp)/d converges a.s. to a finite constant. Here we bound the rate of convergence of EL({X 1,…,X n })/n (dp)/d . Y. Koo supported by the BK21 project of the Department of Mathematics, Sungkyunkwan University. S. Lee supported by the BK21 project of the Department of Mathematics, Yonsei University.  相似文献   

18.
19.
We present a unified approach to compute the number of connected components in the group of real points of adjoint almost simple real algebraic groups.  相似文献   

20.
Let G be a graph and let Pm(G) denote the number of perfect matchings of G.We denote the path with m vertices by Pm and the Cartesian product of graphs G and H by G×H. In this paper, as the continuance of our paper [W. Yan, F. Zhang, Enumeration of perfect matchings of graphs with reflective symmetry by Pfaffians, Adv. Appl. Math. 32 (2004) 175-188], we enumerate perfect matchings in a type of Cartesian products of graphs by the Pfaffian method, which was discovered by Kasteleyn. Here are some of our results:1. Let T be a tree and let Cn denote the cycle with n vertices. Then Pm(C4×T)=∏(2+α2), where the product ranges over all eigenvalues α of T. Moreover, we prove that Pm(C4×T) is always a square or double a square.2. Let T be a tree. Then Pm(P4×T)=∏(1+3α2+α4), where the product ranges over all non-negative eigenvalues α of T.3. Let T be a tree with a perfect matching. Then Pm(P3×T)=∏(2+α2), where the product ranges over all positive eigenvalues α of T. Moreover, we prove that Pm(C4×T)=[Pm(P3×T)]2.  相似文献   

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