具有一阶导数依赖的半线上奇异三阶bvp的正解

Pub Date : 2021-08-01 DOI:10.2478/ausm-2021-0006
Abdelhamid Benmezaï, El-Djouher Sedkaoui
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引用次数: 1

摘要

摘要本文研究了三阶边值问题{- u′(t) +k2u′(t)=φ(t)f(t,u(t),u′(t))的正解的存在性,t >0u(0)=u′(0)=u′(+∞)=0,\左\{\矩阵{- u′′\左(t \右)+ {k′2}u′\左(t \右)= \phi \左(t \右)f\左({t,u\左(t \右),\,\,\,t >0 \hfill \cr u\左(0 \右)=u′\左(0 \右)=u′\左({+ \infty} \右)=0,\hfill \cr u\左(0 \右)=u′\左(0 \右)。其中k是一个正常数,则φ∈L1(0;+∞)是非负的,并且在(0;+∞)上相等消失,函数f: m + x(0;+∞)×(0;+∞)→m +是连续的,并且在空间变量及其导数处可能是奇异的。
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Positive solution for singular third-order BVPs on the half line with first-order derivative dependence
Abstract In this paper, we investigate the existence of a positive solution to the third-order boundary value problem { -u‴(t)+k2u′(t)=φ(t)f(t,u(t),u′(t)),   t>0u(0)=u′(0)=u′(+∞)=0, \left\{ \matrix{- u'''\left( t \right) + {k^2}u'\left( t \right) = \phi \left( t \right)f\left( {t,u\left( t \right),u'\left( t \right)} \right),\,\,\,t > 0 \hfill \cr u\left( 0 \right) = u'\left( 0 \right) = u'\left( { + \infty } \right) = 0, \hfill \cr} \right. where k is a positive constant, ϕ ∈ L1 (0;+ ∞) is nonnegative and does vanish identically on (0;+ ∞) and the function f : ℝ+ × (0;+ ∞) × (0;+ ∞) → ℝ+ is continuous and may be singular at the space variable and at its derivative.
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