脉冲微分方程非局部奇异边值问题

苗春梅,葛渭高

(1. 长春大学 理学院,长春 130022;2. 吉林大学 数学学院,长春 130012;3. 北京理工大学 数学学院,北京 100081)

0 引 言

脉冲微分方程应用广泛,关于其稳定性、 振动性、 边值问题等研究目前已有许多结果[1-12]. 但关于脉冲微分方程非局部奇异边值问题的研究结果较少,多数都是两点边值问题[13-16].

Dai等[14]运用上下解方法研究了如下奇异Emden-Fowler边值问题:

其中:λ,m,a,b,c,d≥0;p(t),q(t)在t=0和t=1处具有奇性;非线性项f(t,x)=p(t)xλ+q(t)x-m在x=0处具有奇性,但该问题中的非线性项是具体的多项式函数,不具有一般性.

在非局部边值问题中,带有积分边界条件的问题在热传导问题[17]、 水力问题[18]和半导体问题[19]中应用广泛,两点边界条件实质上是积分边界条件的特殊情形.

基于此,本文研究如下带有积分边界条件的脉冲微分方程奇异边值问题:

(1)

其中: 0

假设如下条件成立:

1 预备知识

PC[J]={u:J→R|u(t)在J0上连续且u(tk-0)=u(tk),u(tk+0)(k=1,2,…,p)存在};

PC1[J]={u:J→R|u′(t)在J0上连续且u′(tk-0)=u′(tk),u′(tk+0)(k=1,2,…,p)存在}.

定义1如果函数u∈PC1[J]∩C2[J′]满足式(1)且u(t)>0,t∈(0,1),则称u为边值问题(1)的正解.

定义2[4]如果对任意的u∈S,ε>0,存在δ>0,使得s,t∈Jk(k=1,2,…,p)且|s-t|<δ,有|u(s)-u(t)|<ε,则称集合S⊂PC[J]是拟等度连续的.

定义3[4]如果对任意的u∈S,ε>0,存在δ>0,使得s,t∈Jk(k=1,2,…,p)且|s-t|<δ,有|u(s)-u(t)|<ε和|u′(s)-u′(t)|<ε,则称集合S⊂PC1[J]是拟等度连续的.

引理1[4]集合S⊂PC[J](S⊂PC1[J])在PC[J](PC1[J])上相对紧当且仅当S是有界的且拟等度连续.

引理2对常数ak,bk≥0(k=1,2,…,p)和y∈L1[J],边值问题

(2)

存在唯一解:

证明: 由式(2)知

u″(t)=-y(t),t∈J′.

(4)

(5)

再对式(5)从0到t积分得

(6)

将式(6)两边同时乘以g(t),再从0到1积分,由边界条件可得

结合式(6)可得

从而,对任意的t∈[0,1],有

证毕.

为研究奇异边值问题(1)解的存在性,先考虑如下边值问题:

(7)

定义算子T:PC[J] →PC[J]为

引理3算子T:PC[J] →PC[J]全连续.

因此,T(B)是一致有界的.

对任意的ε>0,t,s∈Jk(k=0,1,…,p),

引理4假设存在与λ无关的常数R>a≥0,使得对任意的λ∈(0,1),边值问题

(9)

的解u(t)都有‖u‖≠R,则当λ=1时,边值问题(9)至少有一个解u∈PC1[J]∩C2[J′],且‖u‖≤R.

证明: 对任意的λ∈[0,1],u∈PC[J],定义

由引理3知,Nλ:PC[J] →PC[J]全连续. 易证u(t)是边值问题(9)的解当且仅当u是Nλ在PC[J]中的不动点. 令Ω={u∈PC[J]|‖u‖a,因此(I-N0)u(t)=u(t)-N0u(t)=u(t)-a≠0,t∈J. 对λ∈(0,1),若存在u∈∂Ω,使得(I-Nλ)u(t)=0,t∈J,则u(t)是问题(9)的解,由引理的条件可得‖u‖≠R,与u∈∂Ω矛盾. 因此,对任意的u∈∂Ω和λ∈[0,1],有Nλu≠u. 又由Leray-Schauder度的同伦不变性得,Deg{I-N1,Ω,θ}=Deg{I-N0,Ω,θ}=1. 因此,N1在Ω中存在不动点u,即λ=1时,边值问题(9)至少存在一个解u∈PC1[J]∩C2[J′],且‖u‖≤R. 证毕.

引理5如果u(t)是边值问题(7)的解,则:

1)u(t)在Jk(k=0,1,…,p)上是凹的;

2)u′(t)≥ 0,t∈J0,u′(tk-0)≥u′(tk+0)≥0,Δu(tk)≥0,k=1,2,…,p;

3)u(t)≥a,t∈[0,1].

2 主要结果

定理1设(H1)成立,再假设下列条件成立,则边值问题(1)至少存在一个正解:

(H3) 对任意的l>0,存在函数ψl: [0,1] → (0,∞),使得f(t,u)≥ψl(t),(t,u)∈J′×(0,l];

证明: 由(H4)知,存在M>0和0<ε<(1-σ)M,使得

(11)

(12)

有解.

为了证明对任意的m∈N0,边值问题(12)都有解,先考虑边值问题:

(13)

其中:

下面应用引理4证明边值问题(13)有解. 为此,先考虑如下一族边值问题:

(14)

由(H2)知,对任意的x∈J′,

-u″(x)=λq(x)f*(x,u(x))=λq(x)f(x,u(x))≤q(x)[f1(u(x))+f2(u(x))],

(15)

对式(15)从t(t∈J0)到1积分并由u′(t)的单调性可得

(16)

将式(16)两边同时除以f1(u(t)),再由0到1积分可得

进而有

结合式(11)可得‖u‖=u(1)≠M. 又由引理4知,对任意固定的m∈N0,边值问题(13)至少有一个解um∈PC1[J]∩C2[J′],满足‖um‖≤M. 由引理5知,um(t)≥1/m>0,从而

因此um(t)也是边值问题(12)的解.

下面将获得um(t)(∀m∈N0)的拟下界,即存在常数L>0,L0≥0(与m无关),使得

um(t)≥Lt+L0,t∈J.

(17)

由于

0<1/m≤um(t)≤M,t∈J.

(18)

故由(H3)知,存在连续函数ψM: [0,1] → (0,∞),使得:f(t,um(t))≥ψM(t),t∈J′. 又由引理2知,

最后证明{um(t)}m∈N0在J上是一致有界且拟等度连续的. 由式(18)知{um(t)}m∈N0在J上是一致有界的. 下面证明其在J上是拟等度连续的.

因为um(t)是式(12)的解,因此对x∈J′,有

-um″(x)=q(x)f(x,um(x))≤q(x)[f1(um(x))+f2(um(x))],

(20)

将式(20)从t(t∈J0)到1积分并由u′(t)的单调性可得

(21)

因此{um(t)}m∈N0在J上是拟等度连续的.

又由于

在式(22)中,令m→ ∞,m∈N*,由Lebesgue控制收敛定理可得

因此u(t)是边值问题(1)的正解,且u(t)≥Lt+L0,t∈J,‖u‖≤M. 证毕.

3 应用实例

例1考虑边值问题:

(23)

边值问题(23)至少有一个正解.

从而(H4)成立,因此,由定理1可知,边值问题(23)至少存在一个正解. 证毕.

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