共查询到20条相似文献,搜索用时 78 毫秒
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本文研究了高阶复线性微分方程解在角域上的增长性问题.利用Nevanlinna理论和共形变换的方法,获得了一些使得方程非平凡解在角域上有快速增长的系数条件,这些结果丰富了复方程解在角域上增长性的研究. 相似文献
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应用角域Nevanlinna理论和Ahlfors覆盖曲面理论, 研究了二阶微分方程f’’+A(z)f=0的解的零点分布. 证明了在复平面上至少存在一条半直线, 使得二阶微分方程解在该直线上的零点的径向收敛指数为无穷. 用新的方法证明了伍胜健在文献[5]中的一个定理. 相似文献
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应用角域Nevanlinna理论,研究了二阶亚纯系数微分方程f′′+A(z)f=0的解的零点聚值线和Borel方向之间的关系.推广了文献[5]中的一个定理. 相似文献
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研究了一类亚纯函数系数的高阶线性微分方程的解的不动点问题,应用值分布的理论和方法,得到了复域微分方程亚纯解的不动点性质. 相似文献
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本文通过对一类非线性微分方程解的性质研究,给出了测度值分枝扩散过程在球域上负荷概率的精确估计 相似文献
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高阶线性微分方程的解及其解的导数的不动点 总被引:2,自引:0,他引:2
研究了复域齐次和非齐次线性微分方程的解及其解的导数的不动点与超级问题,得到了整函数系数的齐次和非齐次线性微分方程的解及其解的导数的不动点的两个结果,所得结果推广了一些相关结果. 相似文献
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研究了亚纯函数系数的二阶线性微分方程解的不动点及超级问题,得到了有关复域微分方程亚纯解的不动点性质,并且由于受到微分方程的制约,其性质与一般亚纯函数的不动点性质相比,显得十分有趣. 相似文献
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研究了有限迭代级整系数的复线性微分方程,应用Nevalinna和Ahlfors的角域理论,得到了有关解的迭代级,零点迭代收敛指数以及角域中的零点分布的结果. 相似文献
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通过把两个奇异端点的边界条件加以分离,利用微分方程的解(实参数解或复参数解)给出了实系数对称微分算子最大算子域的一种新的分解.进而应用这些解统一对其自共轭域进行描述,给出了自共轭域的完全刻画. 相似文献
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In this paper we determine the minimum and maximum values of the sum of squares of degrees of bipartite graphs with a given number of vertices and edges. 相似文献
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Jogi Henna 《Annals of the Institute of Statistical Mathematics》2005,57(4):655-664
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
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Nguyêñ Quoôć Thǎ;ńg 《代数通讯》2013,41(3):1097-1110
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
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Janusz Sokó? 《Journal of Mathematical Analysis and Applications》2008,344(2):869-875
We consider the classes of analytic functions introduced recently by K.I. Noor which are defined by conditions joining ideas of close-to-convex and of bounded boundary rotation functions. We investigate coefficients estimates and radii of convexity. 相似文献
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Piotr Je¸drzejewicz 《代数通讯》2013,41(4):1500-1508
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
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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
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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
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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