共查询到20条相似文献,搜索用时 0 毫秒
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Teodor M. Atanackovic Stevan Pilipovi? 《Nonlinear Analysis: Theory, Methods & Applications》2010,72(11):4101-4114
The system , where Dγ,γ∈[0,2] are operators of fractional differentiation, is investigated and the existence of a mild and classical solution is proven. Also, a necessary and sufficient condition for the existence and uniqueness of a solution to a general linear fractional differential equation , in is given. 相似文献
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Pablo Amster María Cristina Mariani 《Journal of Mathematical Analysis and Applications》2007,325(2):1133-1141
We study the existence of periodic solutions for a nonlinear fourth order ordinary differential equation. Under suitable conditions we prove the existence of at least one solution of the problem applying coincidence degree theory and the method of upper and lower solutions. 相似文献
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研究常微分方程d2u/dx2 K(x)e2u=0在(-∞, ∞)上整体解的存在性问题.此方程是熟知的在R2上预定高斯曲率方程的一个特例.本文证明了一个存在性定理. 相似文献
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František Neuman 《Aequationes Mathematicae》1993,46(1-2):38-43
Summary It is shown that a change of the basis in the solution space of a second order linear differential equation induces a covariant change in the solution space of the corresponding iterative equation. Also studied is the problem to what extent a solution of an iterative equation determines the equation.Dedicated to the memory of Alexander M. Ostrowski on the occasion of the 100th anniversary of his birth. 相似文献
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Oleg Palumbíny 《Czechoslovak Mathematical Journal》1999,49(4):779-790
The paper deals with the oscillation of a differential equation L
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y + Q(t)y 0 as well as with the structure of its fundamental system of solutions. 相似文献
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A. V. Glushak 《Mathematical Notes》2010,87(5-6):654-662
In Banach space, we consider the problem of determining the solution and a summand of a differential equation of fractional order from the initial and redundant conditions containing fractional Riemann-Liouville integrals. It is shown that the solvability of the problem under consideration depends on the distribution of zeros of the Mittag-Leffler function. 相似文献