微分包含是微分方程与集值分析的交叉学科, 在力学、工程学以及优化与控制理论中有着广泛的应用[1-2].在描述物理、力学、工程、微观经济学等系统时一般都用确定的微分方程模型, 但在实际生活及科学实践中, 确定的模型通常不适合描述某些动态系统.例如, 通常假定微分方程x′(t)=f(t, x(t))的右端为连续函数, 但实际中往往难以保证, 若将f(t, x(t))嵌入到集值映射F(t, x(t))中, 可将其转化为研究微分包含x′(t)∈F(t, x(t)).如文献[2]研究了一阶微分包含周期边值问题
$ \left\{ \begin{array}{l} y'\left( t \right) \in F\left( {t,y\left( t \right)} \right),\;\;\;\;\;t \in J = \left[ {0,T} \right],\\ y\left( 0 \right) = y\left( T \right) \end{array} \right. $ |
的可解性.
分数阶微分方程被广泛应用于解决各领域的工程问题, 如光学、流变学、新材料力学系统等[3-5].此外, 在生物学、最优控制等领域亦通过建立微分包含模型对一些实际问题进行理论分析和研究.分数阶微分包含越来越受国内外学者的关注, 并取得了一些很好的成果[6].文献[7]介绍了一种新的局部分数阶导数的定义, 称为一致分数阶导数, 较Riemann-Liouville和Caputo这2种分数阶导数具有更好的性质.文献[8]证明了一致分数阶导数的链式法则、Gronwall不等式以及Laplace变换.文献[9]给出了分数阶导数的一些有意义的计算方法.文献[10]通过定义一个tube解并运用Schauder不动点定理研究了分数阶Cauchy问题
$ \left\{ \begin{array}{l} {x^{\left( \alpha \right)}}\left( t \right) = f\left( {t,x\left( t \right)} \right),\;\;\;t \in \left[ {a,b} \right],a > 0,\\ x\left( a \right) = {x_0} \end{array} \right. $ |
的可解性.
受以上研究的启发, 在新的一致分数阶导数的定义下, 运用集值映射的不动点定理研究了微分包含问题
$ \left\{ \begin{array}{l} {x^{\left( \alpha \right)}}\left( t \right) \in F\left( {t,x\left( t \right)} \right),t \in J = \left[ {a,b} \right],a > 0,\\ x\left( a \right) = {x_0} \end{array} \right. $ | (1) |
的可解性,并通过定义问题(1) 的上下解得到了该问题解的最佳逼近.
1 预备知识记J=[a, b], C(J)为定义在J上的连续实值函数构成的Banach空间, 其范数为‖x‖∞=sup{|x(t)|t∈J}.设Ck(J)为k次连续可微实值函数构成的Banach空间, 其范数为‖x‖ck=max{‖x‖∞, …, ‖x(k)‖∞}.设L1(J, R)为定义在J上满足
设(X, |·|)是Banach空间, 若对任意的x∈X, G(x)是凸(闭)值的, 则集值映射G:X→2R是凸(闭)值的.
设任意的B
设对任意的x0∈X, G(x0)是X的非空闭子集, 且对每个包含G(x0)的N
若存在x∈X使得x∈G(x), 则G有一个不动点.
以下是一致分数阶导数和微分包含的一些相关结论, 详见文献[6-9, 11-15].
定义1(一致分数阶导数) 设α∈(0, 1]且f:[0, ∞)→R, f的一致分数阶导数可定义为
$ {T_\alpha }\left( f \right)\left( t \right): = \mathop {\lim }\limits_{\varepsilon \to 0} \frac{{f\left( {t + \varepsilon {t^{1 - \alpha }}} \right) - f\left( t \right)}}{\varepsilon },t > 0, $ |
常用fα表示.如果Tα(f)存在, 则定义
$ {f^\alpha }\left( 0 \right): = \mathop {\lim }\limits_{t \to {0^ + }} {f^\alpha }\left( t \right). $ |
定义2(一致分数次积分) 设α∈(0, 1]且f:[0, ∞)→R,f在[a, t]上的α次分数积分可定义为
$ I_\alpha ^af\left( t \right): = \int_a^t {\frac{{f\left( \tau \right)}}{{{\tau ^{1 - \alpha }}}}{\rm{d}}\tau } . $ |
命题1 设0 < a < b, 用αJab[f]表示
命题2 设f∈L1([a, b], R), 0 < a < b, 则|αJab[f]|≤αJab[|f|].
命题3 设r∈Cα([a, b], R), 0 < a < b, 在{t∈[a, b]:r(t)>0}上有rα(t) < 0.若r(a)≤0, 则对任意t∈[a, b], 有r(t)≤0.
定义3 集值映射F:J×R→2R若满足:
(ⅰ)对任意的x∈R, 有t→F(t, x)是可测的;
(ⅱ)对几乎处处的t∈J, 有x→F(t, x)是上半连续的;
(ⅲ)对任意的k>0, 存在hk∈L1(J, R+), 使得‖F(t, x)‖=sup{|v|:v∈F(t, x)}≤hk(t), 其中|x|≤k, t∈J;
则其是L1-Carathéodary函数.
引理1[10](Bohenblust-Karlin不动点定理) 设X是Banach空间, D是X的非空有界闭凸子集, 设H:D→BCC(X), 使得H(D)
引理2[12] 设I是紧的实区间,且X是Banach空间, 令F:I×X→CC(X).对任意的y∈X, (t, x)→F(t, x)关于t可测, 且对几乎处处的t∈I, (t, x)→F(t, x)关于x是上半连续的.对于每个不动点x∈C[J, R], 定义SF, x={v∈L1(J, R):v(t)∈F(t, x(t)), t∈J}≠
引理3[16] 若g∈L1(J, R), 则函数x:J→R,
$ x\left( t \right): = {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha J_a^t\left[ {\frac{{g\left( s \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right) $ |
是
$ \left\{ \begin{array}{l} {x^\alpha }\left( t \right) + \frac{1}{{{a^\alpha }}}x\left( t \right) = g\left( t \right),\;\;t \in \left[ {a,b} \right],\;\;\;a > 0,\\ x\left( a \right) = {x_0} \end{array} \right. $ | (2) |
的解.
定义4 函数x∈AC(J, R)是式(1) 的解, 如果存在v∈F(t, x(t)), 对任意的t∈J, 使得xα(t)=v(t)且x(a)=x0.
2 主要结果及其证明定义5 设φ∈AC(J, R), 若存在v1∈L1(J, R),满足
$ \left\{ \begin{array}{l} {v_1}\left( t \right) \in F\left( {t,\varphi \left( t \right)} \right),\;\;\;t \in J,a > 0,\\ {\varphi ^\alpha }\left( t \right) \le {v_1}\left( t \right),\;\;\;\;\varphi \left( a \right) \le {x_0}, \end{array} \right. $ |
则称φ(t)是式(1) 的下解.
定义6 设ψ∈AC(J, R),若存在v2∈L1(J, R), 满足
$ \left\{ \begin{array}{l} {v_2}\left( t \right) \in F\left( {t,\psi \left( t \right)} \right),\;\;\;t \in J,a > 0,\\ {\psi ^\alpha }\left( t \right) \geqslant {v_2}\left( t \right),\;\;\;\;\psi \left( a \right) \geqslant {x_0}, \end{array} \right. $ |
则称ψ(t)是式(1) 的上解.
定理1 设F:J×R→CC(R)是L1-Carathéodary集值映射, 若存在φ, ψ∈AC(J, R), 分别是式(1) 的下解与上解, 其中φ≤ψ.令E={(t, x)∈J×R|φ(t)≤x(t)≤ψ(t)}, 则对任意的t∈J, 式(1) 至少有1个解x∈E满足φ(t)≤x(t)≤ψ(t).
证明 考虑辅助问题
$ \left\{ \begin{array}{l} {x^\alpha }\left( t \right) + \frac{1}{{{a^\alpha }}}x\left( t \right) \in F\left( {t,x\left( t \right)} \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {t,x\left( t \right)} \right),\;\;\;t \in J,\\ x\left( a \right) = {x_0},\;\;\;a > 0, \end{array} \right. $ | (3) |
其中,γ:J×R→R定义为
$ \gamma \left( {t,x\left( t \right)} \right) = \left\{ \begin{array}{l} \varphi \left( t \right),\;\;\;\;\varphi \left( t \right) > x\left( t \right),\\ x\left( t \right),\;\;\;\;\;\varphi \left( t \right) \le x\left( t \right) \le \psi \left( t \right),\\ \psi \left( t \right),\;\;\;\;\;\psi \left( t \right) < x\left( t \right). \end{array} \right. $ |
由F是有非空闭凸值的L1-Carathéodary集值映射, 可得存在Φ∈L1(J, R+), 使得
$ \begin{array}{l} Tx\left( t \right): = \left\{ {h \in C\left( {J,R} \right):h\left( t \right) = {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + } \right.} \right.\\ \;\;\;\;\;\;\;\;\;\;\;\left. {\left. {\alpha \mathscr{J}_a^t\left[ {\frac{{v\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {t,x} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right)} \right\}. \end{array} $ |
只需证明算子T满足引理1的假设条件, 从而可得T有不动点即式(2) 的解.
下面分两部分证明:
第1部分:证明上述定义的算子T有不动点.
i)对每个x∈C(J), T(x)是凸的.设h, h∈T(x), 则存在v∈SF, x且v∈SF, x, 使得
$ \begin{array}{l} h\left( t \right) = {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{v\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {t,x} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right),\\ \overline {h\left( t \right)} = {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{\overline {v\left( s \right)} + \frac{1}{{{a^\alpha }}}\gamma \left( {t,x} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right). \end{array} $ |
设0≤λ≤1, 则对每一个t∈
$ \begin{array}{l} \left[ {\lambda h + \left( {1 - \lambda } \right)\bar h} \right]\left( t \right) = \\ \;\;\;\;\;\;\;\lambda {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{v\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {t,x} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right) + \\ \;\;\;\;\;\;\;\left( {1 - \lambda } \right){{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}} \times \\ \;\;\;\;\;\;\;\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{\overline {v\left( s \right)} + \frac{1}{{{a^\alpha }}}\gamma \left( {t,x} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right) + \\ \;\;\;\;\;\;\;\alpha \mathscr{J}_a^t\left[ {\frac{{\left( {1 - \lambda } \right)\left( {\overline {v\left( s \right)} + \frac{1}{{{a^\alpha }}}\gamma \left( {t,x} \right)} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right] = \\ \;\;\;\;\;\;\;{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{\lambda \left( {v\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {t,x} \right)} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}} + } \right.} \right.\\ \;\;\;\;\;\;\;\left. {\left. {\frac{{\left( {1 - \lambda } \right)\left( {\overline {v\left( s \right)} + \frac{1}{{{a^\alpha }}}\gamma \left( {t,x} \right)} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right). \end{array} $ |
因为F有凸值, 所以SF, x是凸的, 从而可得λh+(1-λ)h∈T(x).
ii)对任意常数r>0, 令Br={x∈C(J):‖x‖≤r}, 则Br为C(J)上的有界闭凸集.下证存在r>0, 使得T(Br)
假设存在xr∈Br, hr∈T(xr)与v∈SF, x, 使得
$ h\left( t \right) = {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{\pi }{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{v\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {t,x} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right). $ |
对任意的t∈J, 有
$ \begin{array}{l} \left| {{h_r}\left( t \right)} \right| = \\ \;\;\;\;\left| {{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{v\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {t,y} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right)} \right| \le \\ \;\;\;\;{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}\left| {{x_0}} \right| + {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\alpha \mathscr{J}_a^t \times } \right.\\ \;\;\;\;\left. {\left[ {\frac{{\left| {v\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {t,y} \right)} \right|}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right) \le \frac{K}{C}\left[ {\mathit{\Phi }\left( t \right) } \right.+\\ \;\;\;\;\left. {\max \frac{1}{{{a^\alpha }}}\left( {\sup \left| {\varphi \left( t \right)} \right|,\sup \left| {\psi \left( t \right)} \right|} \right)} \right]\frac{{{b^\alpha } - {a^\alpha }}}{\alpha } + \\ \;\;\;\;K{{\rm{e}}^{\frac{1}{\alpha }}}\left| {{x_0}} \right|. \end{array} $ |
其中,
$ \begin{array}{l} 令\;\;r = K{{\rm{e}}^{\frac{1}{\alpha }}}\left| {{x_0}} \right| + \frac{K}{C}\left[ {\mathit{\Phi }\left( t \right) } \right.+ \\ \;\;\;\;\;\;\;\;\;\left. {\max \frac{1}{{{a^\alpha }}}\left( {\sup \left| {\varphi \left( t \right)} \right|,\sup \left| {\psi \left( t \right)} \right|} \right)} \right]\frac{{{b^\alpha } - {a^\alpha }}}{\alpha }, \end{array} $ |
则‖hr‖≤r, 即存在r>0, 使得T(Br)
iii)T(Br)是等度连续的.设t1, t2∈J, t1≠t2, 设x∈Br且h∈T(x), 则存在v∈SF, x, 使得对任意t∈J, 有
$ h\left( t \right) = {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{v\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {t,x} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right). $ |
因此当t2→t1时,有
$ \begin{array}{l} \left| {h\left( {{t_2}} \right) - h\left( {{t_1}} \right)} \right| \le \left| {{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{{{t_2}}}{a}} \right)}^\alpha }}}{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} - {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{{{t_1}}}{a}} \right)}^\alpha }}}{{\rm{e}}^{\frac{1}{\alpha }}}{x_0}} \right| + \\ \;\;\;\;\;\;\;\left| {{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{{{t_2}}}{a}} \right)}^\alpha }}}\alpha \mathscr{J}_a^{{t_2}}\left[ {\frac{{v\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {{t_2},x} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right] - } \right.\\ \;\;\;\;\;\;\;\left. {{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{{{t_1}}}{a}} \right)}^\alpha }}}\mathscr{J}_a^{{t_1}}\left[ {\frac{{v\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {{t_1},x} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right| \le \\ \;\;\;\;\;\;\;{{\rm{e}}^{\frac{1}{\alpha }}}\left| {{x_0}} \right|\left| {{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{{{t_2}}}{a}} \right)}^\alpha }}} - {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{{{t_1}}}{a}} \right)}^\alpha }}}} \right| + \\ \frac{K}{C}\left[ {\mathit{\Phi }\left( t \right) + \max \frac{1}{{{a^\alpha }}}\left( {\sup \left| {\varphi \left( t \right)} \right|,\sup \left| {\psi \left( t \right)} \right|} \right)} \right] \times \\ \frac{{\left| {t_2^\alpha - t_1^\alpha } \right|}}{\alpha } \to 0. \end{array} $ |
由i)~iii)知T是紧值映射.
iv)T有闭图像.设xn→x*, hn∈T(xn), 且hn→h*, 下证h*∈T(x*).
由hn∈T(xn)可知,存在vn∈SF, xn, 使得
$ \begin{array}{l} {h_n}\left( t \right) = {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{{{t_1}}}{a}} \right)}^\alpha }}} \times \\ \;\;\;\;\;\;\;\;\;\;\;\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{{v_n}\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {t,{x_n}} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right). \end{array} $ |
需证存在v*∈SF, x*, 使得对任意的t∈J,
$ \begin{array}{l} {h_ * }\left( t \right) = {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{{{t_1}}}{a}} \right)}^\alpha }}} \times \\ \;\;\;\;\;\;\;\;\;\;\;\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{{v_*}\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {t,{x_*}} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right). \end{array} $ |
因为xn→x*, hn→h*, 且r连续, 则当n→∞时,
$ \begin{array}{l} \left\| {\left( {{h_n}\left( t \right) - {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{\gamma \left( {t,{x_n}} \right)}}{{{a^\alpha }{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right)} \right) - } \right.\\ \left. {\left( {{h_*}\left( t \right) - {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{\gamma \left( {t,{x_*}} \right)}}{{{a^\alpha }{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right)} \right)} \right\| \to 0. \end{array} $ |
考虑线性连续算子Γ:L1(J, R)→C(J, R),
$ v \to \Gamma \left( v \right)\left( t \right) = {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{v\left( s \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right). $ |
命题4 Γ
$ \begin{array}{l} \left( {{h_n}\left( t \right) - {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}}\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{v\left( n \right)}}{{{a^\alpha }{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right)} \right) \in \\ \;\;\;\;\;\;\;\Gamma \left( {{S_{F,{x_n}}}} \right). \end{array} $ |
因为yn→y*, 由命题1知, 对某些v*∈SF, x*, 有
$ \begin{array}{l} {h_*}\left( t \right) = {{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{t}{a}} \right)}^\alpha }}} \times \\ \;\;\;\;\;\;\;\;\;\;\;\left( {{{\rm{e}}^{\frac{1}{\alpha }}}{x_0} + \alpha \mathscr{J}_a^t\left[ {\frac{{{v_*}\left( s \right) + \frac{1}{{{a^\alpha }}}\gamma \left( {t,{x_*}} \right)}}{{{{\rm{e}}^{ - \frac{1}{\alpha }{{\left( {\frac{s}{a}} \right)}^\alpha }}}}}} \right]} \right). \end{array} $ |
结合i)~iv), T满足命题3的假设, 从而T有不动点,即式(3) 的解.
第2部分:证明式(3) 的任意解x(t):φ(t)≤x(t)≤ψ(t).
先证φ(t)≤x(t).反设φ(t)不全小于x(t), 则存在t∈J, 使得φ(t)>x(t).则由文献[16]可知, 存在v(t)=xα(t), v1(t)=φα(t)且v1(t)≤v(t).令h(t)=φ(t)-x(t), 即存在t0∈J, 使得
$ \begin{array}{l} {h^\alpha }\left( {{t_0}} \right) = {\left( {\varphi \left( {{t_0}} \right) - x\left( {{t_0}} \right)} \right)^\alpha } = {\varphi ^\alpha }\left( {{t_0}} \right) - {x^\alpha }\left( {{t_0}} \right) \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;{v_1}\left( t \right) - v\left( t \right) \le 0, \end{array} $ |
且h(a)=φ(a)-x(a)≤x0-x0=0, 即h(a)≤0, 由命题3, 任意t∈J, h(t)≤0, 与h(t)>0矛盾, 从而φ(t)≤x(t), t∈J.
同理可证x(t)≤ψ(t).
综上可得, 式(3) 至少有1个解x∈ACα(J, R), 使得对任意t∈J, 有φ(t)≤x(t)≤ψ(t), 而此时γ(t, x(t))=x(t), 式(3) 即为式(1), 定理得证.
推论1(次线性增长条件) 假设F:J×R→BCC(R), (t, x)→F(t, x)关于t可测、关于x上半连续, 且存在函数a(t), b(t)∈L1(J, R+), μ∈[0, 1], 使得
$ \left\| {F\left( {t,x} \right)} \right\| \le a\left( t \right){\left| x \right|^\mu } + b\left( t \right),\;\;\;\;\left( {t,x} \right) \in J \times R, $ |
则式(1) 至少有1个解.
证明 在该假设条件下, 取hk(t)=a(t)|x|μ+b(t), 由定理1即可证得.
推论2(至多线性增长条件) 假设F:J×R→BCC(R), (t, x)→F(t, x)关于t可测、关于x上半连续, 且存在函数a(t), b(t)∈L1(J, R+), 使得
$ \left\| {F\left( {t,x} \right)} \right\| \le a\left( t \right)\left| x \right| + b\left( t \right),\;\;\;\;\left( {t,x} \right) \in J \times R, $ |
则式(1) 至少有1个解.
证明 在该假设条件下, 取hk(t)=a(t)|x|+b(t), 由定理1即可证得.
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