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The transcendental meromorphic solutions of a certain type of nonlinear differential equations

机译:一类非线性微分方程的先验亚纯解

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In this paper, we study the differential equations of the following form w(2) + R(z)(w((k)))(2) = Q(z), where R(z). Q(z) are nonzero rational functions. We proved the following three conclusions: (1) If either P(z) or Q(z) is a nonconstant polynornial or k is an even integer, then the differential equation w(2) + P(z)(2)(w((k)))(2) = Q(z) has no transcendental rnerornorphic solution; if P(z), Q(z) are constants and k is an odd integer, then the differential equation has only transcendental meromorphic solutions of the form f (z) = a CoS(bz + c). (2) If either P(z) or Q(z) is a nonconstant polynomial or k > 1, then the differential equation w(2) + (z - z(0)) P(z)(2)(w((k)))(2) = Q(z) has no transcendental meromorphic, solution, furthermore the differential equation w(2) + A(z - z(0))(w ')(2) = B, where A, B are nonzero constants, has only transcendental meromorphic solutions of the form f (z) a cos b root z - z(0), where a, b are constants such that Ab(2) = 1, a(2) = B. (3) If the differential equation w(2) + 1/P(z)(2) (w((k)))(2)= Q(z), where P is a nonconstant polynomial T, (,,-)-2 and Q is a nonzero rational function, has a transcendental meromorphic solution, then k is an odd integer and Q is a polynomial. Furthermore, if k = 1, then Q(z) C (constant) and the solution is of the form f(z) = B cos q(z), where B is a constant such that B-2 = C and q '(z) = +/- P(z). (c) 2007 Elsevier Inc. All rights reserved.
机译:在本文中,我们研究以下形式的微分方程w(2)+ R(z)(w((k)))(2)= Q(z),其中R(z)。 Q(z)是非零有理函数。我们证明了以下三个结论:(1)如果P(z)或Q(z)是非恒定多项式或k是偶数整数,则微分方程w(2)+ P(z)(2)(w ((k)))(2)= Q(z)没有先验的神经形态学解;如果P(z),Q(z)是常数且k是奇整数,则微分方程仅具有形式为f(z)= CoS(bz + c)的先验亚纯解。 (2)如果P(z)或Q(z)是非恒定多项式或k> 1,则微分方程w(2)+(z-z(0))P(z)(2)(w( (k)))(2)= Q(z)没有先验亚纯解,而且微分方程w(2)+ A(z-z(0))(w')(2)= B,其中A ,B是非零常数,仅具有形式为f(z)a cos b根z-z(0)的先验亚纯解,其中a,b是使得Ab(2)= 1,a(2)= B的常数。(3)如果微分方程w(2)+ 1 / P(z)(2)(w((k)))(2)= Q(z),其中P是一个非常数多项式T,(,, -)-2并且Q是一个非零有理函数,具有先验亚纯解,则k是一个奇数整数,而Q是一个多项式。此外,如果k = 1,则Q(z)C(常数),并且解的形式为f(z)= B cos q(z),其中B是常数,使得B-2 = C并且q' (z)= +/- P(z)。 (c)2007 Elsevier Inc.保留所有权利。

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