胡衛(wèi)敏, 蔣達(dá)清

(1. 伊犁師范學(xué)院 數(shù)學(xué)與統(tǒng)計(jì)學(xué)院應(yīng)用數(shù)學(xué)研究所, 新疆 伊寧 835000; 2. 東北師范大學(xué) 數(shù)學(xué)與統(tǒng)計(jì)學(xué)院, 長(zhǎng)春 130024)

0 引 言

考慮如下奇異邊值問(wèn)題:

(1)

目前, 關(guān)于邊值問(wèn)題

(2)

的研究已有許多結(jié)果[1-15]. 文獻(xiàn)[5-8]給出了問(wèn)題(2)當(dāng)非線(xiàn)性項(xiàng)不具有奇性時(shí)的存在性結(jié)果; 文獻(xiàn)[1,9]給出了當(dāng)p=2時(shí)奇異邊值問(wèn)題(奇性依賴(lài)于變量)的一些存在性原則; 文獻(xiàn)[10]研究了當(dāng)p=2 時(shí)離散邊值問(wèn)題解的存在性; 文獻(xiàn)[11]研究了當(dāng)p=2時(shí)連續(xù)邊值問(wèn)題解的存在性. 而關(guān)于二階p-Laplacian方程組奇異邊值問(wèn)題解的存在性研究目前文獻(xiàn)報(bào)道較少.

若奇異邊值問(wèn)題(1)滿(mǎn)足以下條件, 則稱(chēng)(x(t),y(t))是問(wèn)題(1)的正解:

1) (x,y)∈C[0,1]×C[0,1]∩C1(0,1)×C1(0,1);

2) ?t∈(0,1), (x,y)>(0,0), 且x(0)=y(0)=A,x(1)=y(1)=B;

3)φ(x′(t)),φ(y′(t))在(0,1)中絕對(duì)連續(xù), 且滿(mǎn)足:

1 存在性原則

假設(shè)條件:

(H1)fi(t,x,y)∈C((0,1)×R2,R)(i=1,2);

(3)

(4)

(5)

注2容易驗(yàn)證條件(H2)蘊(yùn)含著

其中φ-1(t)是φ(t)的反函數(shù). 事實(shí)上,

類(lèi)似地, 有

引理1邊值問(wèn)題

(6)

證明: 由于唯一性的證明很簡(jiǎn)單, 這里只證明存在性. 對(duì)任意的0

由注2, 顯然y(t)在(0,1)中連續(xù)非增且y(0+)<0

取

(7)

則Ur是定義在(0,1)上的函數(shù), 且有

(8)

對(duì)于0

類(lèi)似地有, 當(dāng)0<ν

因此,Ur(t)在[0,1]上連續(xù),

同理, 若取

也有類(lèi)似結(jié)論.

類(lèi)似引理1的證明, 有:

引理2邊值問(wèn)題

對(duì)每個(gè)固定的(x,y)∈D, 考慮如下邊值問(wèn)題:

(9)

先考慮修正后的邊值問(wèn)題:

(10)n

其中:n≥4是自然數(shù);ηn(t)在[0,1]上連續(xù), 且滿(mǎn)足0≤ηn(t)≤1及

(12)n

和

(13)n

引理3令(ln(t),wn(t))是問(wèn)題(10)n的解, 則

(ur(t),vr(t))≤(ln(t),wn(t))=(Tn(l,w))(t)≤(Ur(t),Vr(t)), 0≤t≤1.

證明: 由于(ln(t),wn(t))≥(ur(t),vr(t))在[0,1]上成立與(ln(t),wn(t))≤(Ur(t),Vr(t))在[0,1]上成立本質(zhì)上一致, 所以本文只證明后者即可.

(14)

對(duì)式(14)兩邊關(guān)于t從t0到t∈(t0,t2)積分, 得

即

則有w(t0)≤w(t2)=0, 矛盾.

類(lèi)似地, 有wn(t)≤Vr(t), 所以?t∈[0,1], (ln(t),wn(t))≤(Ur(t),Vr(t)).

證明: 令[a,b]?(0,1)是一緊區(qū)間, 可得

(15)

其中Cn是方程

的解. 根據(jù)積分第一中值定理, 存在ξn∈[a,b], 使得

即

又由引理3知,ur(t)≤ln(t)≤Ur(t), 從而存在M=M(r,a,b)>0, 使得

(16)

(17)

成立. 由式(15)～(17), 有

類(lèi)似地, 有

根據(jù)引理4, 可以證明:

(18)

其中τ=φ(l′(1/2))是方程

(19)

的解. 綜上可得l(t)=(Tl)(t), 類(lèi)似可證w(t)=(Tw)(t). 因此, (l,w)=(T(l,w))(t)是式(9)的解.

由于D是C[0,1]×C[0,1]中的任意有界集, 于是有:

引理6T:C[0,1]×C[0,1]→C[0,1]×C[0,1]全連續(xù).

對(duì)問(wèn)題(1)利用Schauder不動(dòng)點(diǎn)定理和Leray-Schauder非線(xiàn)性抉擇定理可得更一般的存在性原則.

定理1假設(shè)(H1)和(H2)成立. 設(shè)存在常數(shù)M>A+B(不依賴(lài)于λ), 且

(20)

其中(x,y)∈C[0,1]×C[0,1]∩C1(0,1)×C1(0,1)是邊值問(wèn)題

(21)λ

的解,λ∈(0,1). 則問(wèn)題(1)存在一個(gè)解(x,y)且滿(mǎn)足‖(x,y)‖≤M.

證明: 式(21)λ等價(jià)于不動(dòng)點(diǎn)問(wèn)題

(x,y)=λT(x,y), (x,y)∈C[0,1]×C[0,1],

(22)λ

定理2假設(shè)(H1)和(H2)成立. 設(shè)存在常數(shù)M>A+B(不依賴(lài)于λ), 且式(20)成立, 其中(x,y)∈C[0,1]×C[0,1]∩C1(0,1)×C1(0,1)是邊值問(wèn)題

(23)λ

的解,λ∈(0,1). 則問(wèn)題(1)存在一個(gè)解(x,y)且滿(mǎn)足‖(x,y)‖≤M.

證明: 式(23)λ等價(jià)于不動(dòng)點(diǎn)問(wèn)題

(x,y)=(1-λ)(Q,Q)+λT(x,y),Q=A(1-t)+Bt.

(24)λ

證明: 問(wèn)題(1)等價(jià)于不動(dòng)點(diǎn)問(wèn)題(x,y)=T(x,y). 因?yàn)門(mén):C[0,1]×C[0,1]→C[0,1]×C[0,1]全連續(xù), 則利用Schauder不動(dòng)點(diǎn)定理可證得結(jié)論.

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