共查询到20条相似文献,搜索用时 421 毫秒
1.
非自治时滞微分方程的扰动全局吸引性* 总被引:1,自引:1,他引:0
考虑具有扰动项的非自治时滞微分方程x>(t)=-a(t)x(t-τ)+F(t,xt),t≥0(*)其中F:[0,∞)×C[-δ,0]→R且连续,C[-δ,0]表示将[-δ,0]映射到R的所有连续函数集合.F(t,0)≡0,a(t)∈C((0,∞),(0,∞)),τ≥0.通常文献对a(t)不依赖于t即a(t)为自治情形,研究了方程(*)零解的局部或全局渐近性质[1~5,7].本文对a(t)为非自治即依赖于t之情形,获得了方程(*)零解全局吸引的充分条件,所得结论在某种意义上说是不可改进的.本文改进和推广了已有文献的相应结果,同时本文采用的方法可应用到非自治非线性扰动方程. 相似文献
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本文研究了高阶线性微分方程$$f^{(k)}(z)+A_{k-2}(z)f^{(k-2)}(z)+\cdots+A_0(z)f(z)=0,\eqno(*)$$解的线性相关性,其中$A_j(z)(j=0,2,\ldots,k-2)$是常数, $A_1$为非常数的的整周期函数,周期为$2\pi i$,且是$e^z$的有理函数.在一定条件下,我们给出了方程(*)解的表示. 相似文献
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命题若 x∈R,y>0, 则x/y(x-y)≥x-y (*) 证明 x/y(x-y)≥x-y(?)x(x-y)≥y(x -y)(y>0)(?)x2-xy≥xy-y2(?)x2-2xy+y2≥0(?)(x-y)2≥0,而此式显然成立,故(*)式成立,从证明过程易知等号成立的条件是x=y. 下面我们来看这个命题的一点应用. 例1 对任意实数a>1,b>1有不等式: 相似文献
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众所周知,当a、b为实数时有(a-b)~2≥0,而有a~2+b~2≥2ab,当且仅当a=b时等号成立。进一步引伸,不难得到: x+y/2≥(xy)~(1/2)≥2/(1/x+1/y) (*) 这里,x>0,y>0,当且仅当x=y时等号成立。不等式(*)有着广泛的运用,在很多书刊上 相似文献
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In this paper, the authors study the existence and nonexistence of multiple positive solutions for problem(*)μwhere h ∈ H-1(RN), N ≥ 3, |f(x,u)| ≤ C1up-1 + C2u with C1 > 0, C2∈ [0,1) being some constants and 2 < p < ∞. Under some assumptions on f and h, they prove that there exists a positive constant μ* <∞ such that problem (*)μ has at least one positive solution uμ if μ,∈ (0,μ*), there are no solutions for (*)μ if μ, > μ* and uμ is increasing with respect to μ∈ (0,μ*); furthermore, problem (*)μ has at least two positive solution for μ ∈ (0,μ*) and a unique positive solution for μ, =μ* if p ≤2N/N-2. 相似文献
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For Lienard's equation or its equivalent system (?)=y-F(x),(?)=-x,(F(x)=integral from n=0 to x (f(ξ)dξ)). (*) there were quite a few papers studied (*) to have some limit cycles. Although there were few papers studied (*) to have at most m limit cycles, but good results were given already. A. Lino, W. Demelo and C. C. Pugh conjectured following fact: when f(x)=sum from N (a_ix~i) (N=2n+1,2n+2),system (*) has at most n limit cycles. 相似文献
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判别式法和换元法,是求函数极值时常用的初等方法,解题过程中,往往由于忽视基本理论知识而导致错误。这里仅以求函数y=x+4+(5-x~2)~(1/2)的极值为例来阐明解题过程中应注意的两个问题。用判别式法求上述函数的极值时,先变形为y-x-4=(5-x~2)~(1/2),再两边平方整理,得 2x~2+(8-2y)x+(y~2-8y+11)=0 因为x为实数,所以其判别式△=4(4-y)~2-8(y~2-8y+11)≥0 (*) 即 y~2-8y+6≤0 解之,得 4-10~(1/2)≤y≤4+10~(1/2)。假若至此就得出 y_(maX)=4+10~(1/2),y_(mlx)=4-10~(1/2)。那将是错误的,因为事实上应为4-10~(1/2)0或△=0,并非要求两者同时成立,其次由(*)成立,并不能逆推出上一式的成立。因 相似文献
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该文考虑广义Beltrami方程组D~tf(x)H(x)Df(x)=J(x,f)~(2/n)G(x).(*)利用能量和变分方法,在矩阵H(x),G(x)∈S(n)满足一致椭圆型条件下,得到了(*)式所满足的齐次散度型椭圆方程DivA(x,Df(x))=0,并得到了(*)式的分量函数的弱单调性和Caccioppoli不等式. 相似文献
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对Linard方程作变换,得到二个方程 dz/dy=F_i(z)-y,(i=1,2)。(i) 设(*)满足解的存在唯一性条件,F_1(z)=-F_2(z),F′_1(z)连续,F′_1(0)<0。记F(z)=F_1(z),方程(1)可写为 dz/dy=F(z)-y。(3)方程(3)的过点(z_0,F(z_o))的、在特征曲线y=F(z)上方的轨线用表示,下方的用y(z)表示。针对文[3]中定义的二个状态函数, 相似文献
11.
Li Dening 《数学年刊B辑(英文版)》1986,7(2):147-159
Consider the nonlinear inltial-boundary value problem for quasilinear hyperbolicsystem:Let k≥2[n/2] 6,(F,g)∈ H~k(R_ ;Ω)×H~k(R_ ;Ω),and their traces at t=0 are zeroup to the order k-1.If for u=0,the problem(*)at t=0 is a Kreiss hyperbolic system,and the boundaryconditions satisfy the uniformly Lopatinsky criteria,then there exists a T>0 such that(*)has a unique H~k soluton in(0,T).In the Appendix,for symmetric hyperbolic systems,a comparison between theuniformly Lopatinsky condition and the stable admissible condition is given. 相似文献
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高国柱 《数学物理学报(B辑英文版)》1990,(3)
Recently the authors have studied the oscillations of some neutral differential difference equations and obtained very good results (see [1—4]). In this paper we consider the oscillations of the neutral differential difference equationd/dt[x(t)+sum from i=1 px(t-τ)+sum form j=1 to n qx(t-σ)=0, t≥t,] (*)where p, τ, q and σ (i=1, 2, …, m; j=1, 2, …, n) are positive constants. Some sufficient conditions for all solutions of (*) to oscillate are obtained. And in some ease we give neeessaxy and sufficient conditions for (*) to oscillate. 相似文献
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1992年6月号问题解答 (解答由问题提供人给出) 25.证明因为不等式(*)关于x,y,z对称,所以不妨设x≤y≤z,令y=x+m,z=x+m+n(x≥0,m≥0,n≥0),代入不等式(*)两边得 x·(x+2m+n)~2+(x+m)·(x+n)~2+(x+m+n)·(x-n)~2 相似文献
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张书年 《数学年刊A辑(中文版)》1986,(5)
本文讨论下形之时滞微分方程 (t)=q(t)x(t)-p(t)x(t-r),t≥0,(*)其中g(t),p(t)为[0,∞)上实值连续函数,具公共周期ω>0,而r=kω,k为正整数。 记 我们定义了方程(*)的“度数”m(,,r),它给出了方程(*)具正实部的特征指数的个数(按重数计),它由(,,r)的值所决定。并且,我们得出了(*)任一解的结构性表达式,即任一解均可表示为m个固定的无界解的线性组合与某个有界解之和。 相似文献
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本刊刊登的文〔1〕、〔2〕、〔3〕阅来颇有收益,深受启发,联想到我们在求y=P(x)/Q(x)(P(x)、Q(x)的次数不超过2)的值域时,经常采用的判别式法,笔者依法炮制出一个与之类似的三角判别式法,现简介如下。定理:设方程asinx+bcosx+c=0(a、b不同时为零,x_0≤x0时,方程(*)有相异二实根 (2)当△=0时,方程(*)有相等二实根 (3)当△<0时,方程(*)没有实数根。 相似文献
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定理若直线lx+my+n=0(n≠0)和曲线Ax2+Bxy+Cy2+Dx+Ey+F=0有两个交点P,Q,O为坐标原点,则直线OP,OQ上的点均满足方程Ax2+Bxy+Cy2+(Dx+Ey)(-lx+nmy)+F(-lx+nmy)2=0.(*)证设点P的坐标为(x1,y1),则lx1+my1+n=0,即-lx1+nmy1=1(1)Ax12+Bx1y1+Cy12+Dx1+Ey1+F=0(2)又直线OP上的点均可表示为(tx1,ty1),其中t为任意实数.∵当x=tx1,y=ty1时,方程(*)的左端Ax2+Bxy+Cy2+(Dx+Ey)(-lx+nmy)+F(-lx+nmy)2=t2[Ax12+Bx1y1+Cy12+(Dx1+Ey1)(-lx1+nmy1)+F(-lx1+nmy1)2]=t2(Ax12+Bx1y1+Cy12+Dx1+Ey1+F)=0,∴直线OP上的点都在方程(*)表示的曲线上… 相似文献
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文[1]用两种方法证明了“一个奇妙的组合恒等式”:
n∑j=0(-1)j(n -j)nCjn=n!(n∈N+)……(*)j=0
实际上,文[2]与文[3]分别用数学归纳法和概率证法证明了比(*)更强的组合恒等式: 相似文献
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In order to study three-point BVPs for fourth-order impulsive differential equation of the form with the following boundary conditions u'(0) = u(1) = 0. u"(0) == 0 = u"(1) -φq(α)u"(η). the authors translate the fourth-order impulsive differential equations with p-Laplacian (*) into three-point BVPs for second-order differential equation without impulses and two-point BVPs for second-order impulsive differential equation by a variable transform. Based on it, existence theorems of positive solutions for the boundary value problems (*) are obtained. 相似文献
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杨芬 《数学物理学报(A辑)》2015,(2):282-287
研究如下非齐次双调和方程-△~2u+u~p+f(x)=0,x∈R~n(*)正解的存在性,其中△~2是双调和算子,p1,n≥5,f≠0.在文献[16[的基础上,得到:对f给定条件,方程(*)有一类不同于文献[16]的两种衰减的正解. 相似文献