脈沖微分方程非局部奇異邊值問題

苗春梅,葛渭高

(1. 長春大學(xué) 理學(xué)院,長春 130022;2. 吉林大學(xué) 數(shù)學(xué)學(xué)院,長春 130012;3. 北京理工大學(xué) 數(shù)學(xué)學(xué)院,北京 100081)

0 引 言

脈沖微分方程應(yīng)用廣泛,關(guān)于其穩(wěn)定性、 振動性、 邊值問題等研究目前已有許多結(jié)果[1-12]. 但關(guān)于脈沖微分方程非局部奇異邊值問題的研究結(jié)果較少,多數(shù)都是兩點(diǎn)邊值問題[13-16].

Dai等[14]運(yùn)用上下解方法研究了如下奇異Emden-Fowler邊值問題:

其中:λ,m,a,b,c,d≥0;p(t),q(t)在t=0和t=1處具有奇性;非線性項(xiàng)f(t,x)=p(t)xλ+q(t)x-m在x=0處具有奇性,但該問題中的非線性項(xiàng)是具體的多項(xiàng)式函數(shù),不具有一般性.

在非局部邊值問題中,帶有積分邊界條件的問題在熱傳導(dǎo)問題[17]、 水力問題[18]和半導(dǎo)體問題[19]中應(yīng)用廣泛,兩點(diǎn)邊界條件實(shí)質(zhì)上是積分邊界條件的特殊情形.

基于此,本文研究如下帶有積分邊界條件的脈沖微分方程奇異邊值問題:

(1)

其中: 0

假設(shè)如下條件成立:

1 預(yù)備知識

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

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

定義1如果函數(shù)u∈PC1[J]∩C2[J′]滿足式(1)且u(t)>0,t∈(0,1),則稱u為邊值問題(1)的正解.

定義2[4]如果對任意的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,則稱集合S?PC[J]是擬等度連續(xù)的.

定義3[4]如果對任意的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,則稱集合S?PC1[J]是擬等度連續(xù)的.

引理1[4]集合S?PC[J](S?PC1[J])在PC[J](PC1[J])上相對緊當(dāng)且僅當(dāng)S是有界的且擬等度連續(xù).

引理2對常數(shù)ak,bk≥0(k=1,2,…,p)和y∈L1[J],邊值問題

(2)

存在唯一解:

證明: 由式(2)知

u″(t)=-y(t),t∈J′.

(4)

(5)

再對式(5)從0到t積分得

(6)

將式(6)兩邊同時(shí)乘以g(t),再從0到1積分,由邊界條件可得

結(jié)合式(6)可得

從而,對任意的t∈[0,1],有

證畢.

為研究奇異邊值問題(1)解的存在性,先考慮如下邊值問題:

(7)

定義算子T:PC[J] →PC[J]為

引理3算子T:PC[J] →PC[J]全連續(xù).

因此,T(B)是一致有界的.

對任意的ε>0,t,s∈Jk(k=0,1,…,p),

引理4假設(shè)存在與λ無關(guān)的常數(shù)R>a≥0,使得對任意的λ∈(0,1),邊值問題

(9)

的解u(t)都有‖u‖≠R,則當(dāng)λ=1時(shí),邊值問題(9)至少有一個(gè)解u∈PC1[J]∩C2[J′],且‖u‖≤R.

證明: 對任意的λ∈[0,1],u∈PC[J],定義

由引理3知,Nλ:PC[J] →PC[J]全連續(xù). 易證u(t)是邊值問題(9)的解當(dāng)且僅當(dāng)u是Nλ在PC[J]中的不動點(diǎn). 令Ω={u∈PC[J]|‖u‖a,因此(I-N0)u(t)=u(t)-N0u(t)=u(t)-a≠0,t∈J. 對λ∈(0,1),若存在u∈?Ω,使得(I-Nλ)u(t)=0,t∈J,則u(t)是問題(9)的解,由引理的條件可得‖u‖≠R,與u∈?Ω矛盾. 因此,對任意的u∈?Ω和λ∈[0,1],有Nλu≠u. 又由Leray-Schauder度的同倫不變性得,Deg{I-N1,Ω,θ}=Deg{I-N0,Ω,θ}=1. 因此,N1在Ω中存在不動點(diǎn)u,即λ=1時(shí),邊值問題(9)至少存在一個(gè)解u∈PC1[J]∩C2[J′],且‖u‖≤R. 證畢.

引理5如果u(t)是邊值問題(7)的解,則:

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 主要結(jié)果

定理1設(shè)(H1)成立,再假設(shè)下列條件成立,則邊值問題(1)至少存在一個(gè)正解:

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

證明: 由(H4)知,存在M>0和0<ε<(1-σ)M,使得

(11)

(12)

有解.

為了證明對任意的m∈N0,邊值問題(12)都有解,先考慮邊值問題:

(13)

其中:

下面應(yīng)用引理4證明邊值問題(13)有解. 為此,先考慮如下一族邊值問題:

(14)

由(H2)知,對任意的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)

對式(15)從t(t∈J0)到1積分并由u′(t)的單調(diào)性可得

(16)

將式(16)兩邊同時(shí)除以f1(u(t)),再由0到1積分可得

進(jìn)而有

結(jié)合式(11)可得‖u‖=u(1)≠M(fèi). 又由引理4知,對任意固定的m∈N0,邊值問題(13)至少有一個(gè)解um∈PC1[J]∩C2[J′],滿足‖um‖≤M. 由引理5知,um(t)≥1/m>0,從而

因此um(t)也是邊值問題(12)的解.

下面將獲得um(t)(?m∈N0)的擬下界,即存在常數(shù)L>0,L0≥0(與m無關(guān)),使得

um(t)≥Lt+L0,t∈J.

(17)

由于

0<1/m≤um(t)≤M,t∈J.

(18)

故由(H3)知,存在連續(xù)函數(shù)ψM: [0,1] → (0,∞),使得:f(t,um(t))≥ψM(t),t∈J′. 又由引理2知,

最后證明{um(t)}m∈N0在J上是一致有界且擬等度連續(xù)的. 由式(18)知{um(t)}m∈N0在J上是一致有界的. 下面證明其在J上是擬等度連續(xù)的.

因?yàn)閡m(t)是式(12)的解,因此對x∈J′,有

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

(20)

將式(20)從t(t∈J0)到1積分并由u′(t)的單調(diào)性可得

(21)

因此{(lán)um(t)}m∈N0在J上是擬等度連續(xù)的.

又由于

在式(22)中,令m→ ∞,m∈N*,由Lebesgue控制收斂定理可得

因此u(t)是邊值問題(1)的正解,且u(t)≥Lt+L0,t∈J,‖u‖≤M. 證畢.

3 應(yīng)用實(shí)例

例1考慮邊值問題:

(23)

邊值問題(23)至少有一個(gè)正解.

從而(H4)成立,因此,由定理1可知,邊值問題(23)至少存在一個(gè)正解. 證畢.

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