一類分?jǐn)?shù)階微分方程三點(diǎn)邊值問題正解的存在性與不存在性

何健堃, 賈 梅, 陳 輝

(上海理工大學(xué) 理學(xué)院, 上海 200093)

0 引 言

分?jǐn)?shù)階微積分理論在物理、 化學(xué)和工程技術(shù)等領(lǐng)域應(yīng)用廣泛, 目前已受到廣泛關(guān)注[1-3]. 關(guān)于分?jǐn)?shù)階微分方程邊值問題的研究已取得了很多成果[4-16]. Su等[7]研究了非齊次邊界條件分?jǐn)?shù)階微分方程兩點(diǎn)邊值問題:

正解的存在性和不存在性, 其中: 1<δ<2;J=(0,1]; 擾動(dòng)參數(shù)a≥0,b≥0.

本文考慮分?jǐn)?shù)階微分方程關(guān)于含擾動(dòng)參數(shù)的三點(diǎn)邊值問題:

(1)

1)f(·,u)對u∈+是可測的;

2)f(t,·)對幾乎處處t∈[0,1]是連續(xù)的;

3) 對每個(gè)r>0, 存在φr∈L1[0,1],φr(t)≥0, 使得|u|>≤r,t∈[0,1]時(shí), 滿足|f(t,tα-2u)|>≤φr(t).

文獻(xiàn)[7]研究了非齊次邊界條件下兩點(diǎn)邊值問題正解的存在性與不存在性, 通過假設(shè)非線性項(xiàng)f(t,u)是關(guān)于u的單調(diào)遞增函數(shù), 利用上下解方法、 Schauder不動(dòng)點(diǎn)定理和不動(dòng)點(diǎn)指數(shù)理論, 獲得了該問題正解存在與不存在的充分條件. 由于非局部邊值問題應(yīng)用廣泛, 因此本文考慮在邊界條件中含有分?jǐn)?shù)階導(dǎo)數(shù)的三點(diǎn)邊值問題正解的存在性與不存在性, 利用范數(shù)形式的錐拉伸與錐壓縮不動(dòng)點(diǎn)定理, 證明該問題在參數(shù)滿足不同范圍時(shí)正解的存在性與不存在性. 通過對擾動(dòng)參數(shù)的控制, 得到了該問題正解存在性與不存在性的結(jié)果. 本文結(jié)果是在非線性項(xiàng)滿足超線性與次線性條件下得出的, 因此不需要非線性項(xiàng)f(t,u)關(guān)于u單調(diào)遞增的限制.

1 定義與引理

定義1[2]函數(shù)u: (0,∞)→的α階Riemann-Liouville分?jǐn)?shù)階積分定義為

對任意的α>0, 右端積分在(0,∞)上逐點(diǎn)可積.

定義2[2]函數(shù)u: (0,∞)→的α階Riemann-Liouville分?jǐn)?shù)階導(dǎo)數(shù)定義為

對任意的α>0, 右端積分在(0,∞)上逐點(diǎn)可積,n為大于或等于[α]的最小整數(shù).

于[0,1]中幾乎處處成立, 其中ci∈(i=1,2,…,n)為任意常數(shù),n為大于或等于[α]的最小整數(shù).

引理2記ρ=(1-λξα-β-1)-1, 對任意的y∈L1[0,1], 則邊值問題

(2)

的唯一解可表示為

(3)

其中:

(6)

其中ci∈(i=1,2). 根據(jù)邊界條件得c2=a, 則

(7)

因此, 將c1的值代入式(7), 則邊值問題(2)的解為

其中G(t,s)和K(ξ,s)分別由式(4)和式(5)定義.

反之, 如果式(3)成立, 則易得u為邊值問題(2)的解. 證畢.

引理3記H(t,s)=t2-αG(t,s), 則分別由式(4),(5)定義的函數(shù)G(t,s),K(ξ,s)和函數(shù)H(t,s)滿足如下性質(zhì):

1) 對任意的t,s∈[0,1],G(t,s)是連續(xù)函數(shù), 且G(t,s)≥0;

3) 對任意的t,s∈[0,1],H(t,s)是連續(xù)函數(shù), 且H(t,s)≥0;

4) 對任意的t,s∈[0,1], 有

證明: 1) 由G(t,s)的表達(dá)式易知對任意的t,s∈[0,1],G(t,s)是連續(xù)函數(shù), 且對任意的0≤t≤s≤1, 有G(t,s)≥0; 另一方面, 當(dāng)0≤s≤t≤1, 有

因此, 結(jié)論1)成立.

因此, 結(jié)論2)成立.

3) 由結(jié)論1)易知結(jié)論3)成立.

當(dāng)0≤s≤t≤1時(shí), 有

當(dāng)0≤t≤s≤1時(shí), 有

引理4(α-β-1)(λξα-β-2-1)+(α-1)ρ-1≥0.

證明: 如果λξα-β-2-1≥0, 即λ≥ξ2-α+β, 則不等式顯然成立. 如果λξα-β-2-1<0, 即0≤λ<ξ2-α+β, 則有

當(dāng)α-β-1-(α-1)ξ≥0時(shí), 不等式顯然成立; 當(dāng)α-β-1-(α-1)ξ<0時(shí), 有

λξα-β-2(α-β-1-(α-1)ξ)+β≥α-β-1-(α-1)ξ+β=(α-1)(1-ξ)≥0.

證畢.

定義P={u∈E:u(t)≥0,t∈(0,1]}, 顯然P是E中的錐. 定義算子T:P→E, 則

令P0={u∈E:t2-αu(t)≥βt‖u‖,t∈(0,1]}, 則P0?P為E的一個(gè)錐.

引理5算子T:P→P0是全連續(xù)算子.

證明: 由引理3和引理4, 顯然當(dāng)u∈P時(shí), 對任意的t∈(0,1], 有Tu(t)≥0, 且

另一方面, 由引理3, 有

則t2-αTu(t)≥βt‖Tu‖, 即T:P→P0.

令{un}?P,u0∈P, 且‖un-u0‖→0, 當(dāng)n→+∞時(shí), 存在常數(shù)l>0, 使得‖un‖≤l, ‖u0‖≤l. 則對任意的t∈[0,1], 有|t2-αun(t)|>≤l, |t2-αu0(t)|>≤l. 由f滿足L1-Carathéodory條件可知, 對幾乎處處s∈[0,1], 有

設(shè)Ω?P有界, 存在一個(gè)常數(shù)l>0, 使得對任意的u∈Ω, 有‖u‖≤l, 因此, 對任意的u∈Ω, 存在φl∈L1[0,1], 使得

|f(s,u(s))|>=|f(s,tα-2t2-αu(s))|>≤φl(s),s∈[0,1],

因此T(Ω)一致有界.

此外, 對任意的t,s∈[0,1], 因?yàn)镚(t,s)是連續(xù)函數(shù), 則其為一致連續(xù)函數(shù), 因此對任意的ε>0, 存在常數(shù)

使得對所有的t1,t2,s∈[0,1], 當(dāng)|t1-t2|><δ時(shí), 有

從而

因此T(Ω)是等度連續(xù)的. 由Arzela-Ascoli定理可知,T(Ω)為相對列緊集. 從而T:P→P0是全連續(xù)的. 證畢.

由引理2, 易得：

引理6邊值問題(1)存在一個(gè)正解當(dāng)且僅當(dāng)算子T在P中存在一個(gè)不動(dòng)點(diǎn). 此外, 若u是邊值問題(1)的一個(gè)正解, 則u∈P0.

1) ‖Tu‖≤‖u‖,u∈P∩?Ω1; ‖Tu‖≥‖u‖,u∈P∩?Ω2;

2) ‖Tu‖≥‖u‖,u∈P∩?Ω1; ‖Tu‖≤‖u‖,u∈P∩?Ω2.

2 邊值問題正解的存在性與不存在性

為方便, 記

定理1若f0M2, 則存在a0,b0>0, 使得當(dāng)0≤a≤a0, 0≤b≤b0時(shí), 邊值問題(1)至少存在一個(gè)正解.

證明: 首先, 由于f00, 使得對任意的t∈[0,1],u∈(0,r1], 有f(t,tα-2u)

故對任意的u∈?Ω1, ‖Tu‖≤‖u‖.

根據(jù)引理3可得

故對任意的u∈?Ω2, ‖Tu‖≥‖u‖.

定理2若f∞M2, 則存在a0,b0>0, 使得當(dāng)0≤a≤a0, 0≤b≤b0時(shí), 邊值問題(1)至少存在一個(gè)正解.

f(t,tα-2u)≤φR1(t)+(M1-ε)u.

令

Ω3={u∈P0: ‖u‖

且max{a0,b0}≤M1r3. 則對任意的u∈?Ω3, 有‖u‖=r3, 并且

故對任意的u∈?Ω1, ‖Tu‖≤‖u‖.

定理3若f∞>M2, 則存在a1,b1>0, 使得當(dāng)a>a1,b>b1時(shí), 邊值問題(1)無正解.

令

因此

由式(5)和引理3可得

從而

即‖u‖>R.

即‖u‖>‖u‖+R, 矛盾, 假設(shè)不成立. 故存在正常數(shù)a1,b1, 使得當(dāng)a>a1,b>b1時(shí), 邊值問題(1)無正解. 證畢.

3 應(yīng)用實(shí)例

例1考慮邊值問題:

(8)

可得以下結(jié)論:

1) 如果參數(shù)a∈[0,0.000 833 361 6),b∈[0,0.000 833 361 6), 則邊值問題(8)至少存在一個(gè)正解.

2) 如果參數(shù)a∈[1.134×107,+∞),b∈[1.134×107,+∞), 則邊值問題(8)不存在正解. 這里邊值問題(1)中

α=5/3,β=1/2,ξ=4/5,λ=1,f(t,u)=t2/3u2arctantu+t1/2(1+sint)u3/2.

則存在φr(t)=r2arctant2/3r+(1+sint)r3/2, 使得對所有的u≤r及t∈(0,1), 有|f(t,t-1/3u)|>≤φr(t), 易見f滿足L1-Carathéodory條件. 通過簡單計(jì)算可得:

M1=0.027 778 72,M2=264.884 7,f0=0M2.

1) 令r1=0.03, 選取max{a0,b0}

2) 令R=3.14×106, 當(dāng)u∈[3.925×105,+∞),t∈(0,1)時(shí), 有f(t,t-1/3u)>M2u. 選取min{a1,b1}>4Γ(α)R=1.134×107. 由定理3可知, 當(dāng)a,b∈[1.134×107,+∞)時(shí), 邊值問題(8)不存在正解.

例2考慮邊值問題:

(9)

這里邊值問題(1)中α=3/2,β=1/3,ξ=3/5,λ=1,f(t,u)=etq/2uq, 0M2,f∞=0

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