0 引言

近年来，分数阶微分方程由于在各个领域的广泛应用而备受关注^[1]，关于分数阶微分方程解的存在性和唯一性已经有了丰硕的成果^[2-7].文献[8]讨论了一类分数阶微分方程边值问题：

$ \left\{ {\begin{array}{*{20}{l}} {D_0^\alpha + u\left( t \right) = f\left( {t,u\left( t \right)} \right),t \in J: = \left[ {0,\infty } \right),}\\ {u\left( 0 \right) = 0,D_{{0^ + }}^{\alpha - 1}u\left( \infty \right) = {u_\infty },} \end{array}} \right. $

(1)

其中, D0+α是Riemann-Liouville型分数阶导数.假设f满足线性增长限制条件，应用Darbo不动点定理，得到边值问题(1) 解的存在性.受以上启发，本文主要研究以下Banach空间(E，‖·‖)中分数阶微分方程的边值问题：

$ \left\{ \begin{array}{l} D_0^\alpha + u\left( t \right) = f\left( {t,u\left( t \right),D_{{0^ + }}^{\alpha - 1}u\left( t \right)} \right),t \in J: = \left[ {0,\infty } \right)\\ u\left( 0 \right) = 0,D_{{0^ + }}^{\alpha - 1}u\left( \infty \right) = {u_\infty }, \end{array} \right. $

(2)

其中: 1＜α≤2;f∈C(J×E×E, E); u∞∈E; D0+α-1u(∞):=$\mathop {\lim }\limits_{t \to \infty } $ D0+α-1u(t); D0+α、D0+α-1是Riemann-Liouville型分数阶导数.不同于[8]，本文中讨论的函数f与u(t)的分数阶导数D0+α-1u(t)有关，假设f满足次线性和超线性增长限制条件，通过构造特殊的Banach空间，应用Darbo不动点定理得到方程(2) 解的存在性.

1 预备知识

在下文中，用D^α、D^α-1表示α及α-1阶的Riemann-Liouville型分数阶导数，关于其定义和性质可参见文献[2].

记I∈J为紧区间，C(I, E)为Banach空间上的连续函数y:I→E，‖y‖_C=$\mathop {\sup }\limits_{t \in J} $‖y(t)‖. α，α_C，α_X，α_Y分别为空间E，C(I, E)，X和Y上的Kuratowski非紧性测度.

定义空间

$ X = \left\{ {u\left( t \right) \in C\left( {J,E} \right):\mathop {\sup }\limits_{t \in J} \frac{{{{\left\| {u\left( t \right)} \right\|}^{\rho 1}}}}{{{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}}} < \infty } \right\}, $

$ Y = \left\{ {u\left( t \right) \in X:{D^{\alpha - 1}}u\left( t \right) \in C\left( {J,E} \right),\mathop {\sup }\limits_{t \in j} {{\left\| {{D^{\alpha - 1}}u\left( t \right)} \right\|}^{\rho 2}} < \infty } \right\}. $

空间上的范数是

$ {\left\| u \right\|_X} = \mathop {\sup }\limits_{t \in J} \frac{{{{\left\| {u\left( t \right)} \right\|}^{\rho 1}}}}{{{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}}},{\left\| u \right\|_Y} = \max \left\{ {{{\left\| u \right\|}_X},\mathop {\sup }\limits_{t \in J} {{\left\| {{D^{\alpha - 1}}u\left( t \right)} \right\|}^{\rho 2}}} \right\}, $

其中: 0＜ρ_i＜1(i=1, 2)，易证，(X, ‖·‖ _X)和(Y, ‖·‖ _Y)是Banach空间.

引理1^[3]   若H⊂C(I, E)有界且等度连续，则α(H(t))在I上连续且

$ {\alpha _C}\left( H \right) = \mathop {\max }\limits_{t \in I} \alpha \left( {H\left( t \right)} \right),\alpha \left( {\left\{ {\int_I {x\left( t \right){\rm{d}}t:x \in H} } \right\}} \right) \le \int_I \alpha \left( {H\left( t \right)} \right){\rm{d}}t, $

其中: H(t)={x(t):x∈H}, t∈I.

引理2^[3]  设D是Banach空间E上有界闭凸子集，若算子T:D→D是严格集压缩，则T在D内有不动点.

2 主要结果

为应用Darbo不动点定理建立边值问题(2) 解的存在性结果，假设f满足如下条件：

(H₁)存在非负函数a(t), b(t), c(t)∈C(J)，0＜ρ_i＜1(i=1, 2) 使得

$ \left\| {f\left( {t,x,y} \right)} \right\| \le a\left( t \right)\;\;{\left\| x \right\|^{\rho 1}} + b\left( t \right)\;\;{\left\| y \right\|^{\rho 2}} + c\left( t \right),t \in J,x \in E, $

$ \int_0^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]{\rm{d}}t < \frac{{\Gamma \left( \alpha \right)}}{2},} \int_0^\infty {c\left( t \right){\rm{d}}t < + \infty .} $

(H₂)对于r₁, r₂>0，I⊂J，f(t, x, y)在I×B_E(θ, r₁)×B_E(θ, r₂)上一致连续，θ是E中零元，

$ {B_E}\left( {\theta ,{r_1}} \right) = \left\{ {x \in E,\left\| x \right\| \le {r_1}} \right\},{B_E}\left( {\theta ,{r_2}} \right) = \left\{ {y \in E,\left\| y \right\| \le {r_2}} \right\}. $

(H₃)存在非负函数h(t)∈L¹(J)，使得α(f(t, S, D^α-1S))≤h(t)α(S), t∈J，其中S是E的有界子集，

$ \int_0^\infty {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}h\left( t \right){\rm{d}}t < \Gamma \left( \alpha \right)} . $

引理3  若条件H₁成立，则问题(2) 等价于积分方程

$ u\left( t \right) = \frac{{\left[ {{u_\infty } - \int_0^\infty {f\left( {t,u\left( t \right),{D^{\alpha - 1}}u\left( t \right)} \right){\rm{d}}t} } \right]}}{{\Gamma \left( \alpha \right)}}{t^{\alpha - 1}} + I_{{0^ + }}^\alpha + f\left( {t,u\left( t \right),D_{{0^ + }}^{\alpha - 1}u\left( t \right)} \right). $

证明过程参见文献[9]中引理3.1.

定义算子T：

$ Tu\left( t \right) = \frac{{\left[ {{u_\infty } - \int_0^\infty {f\left( {t,u\left( t \right),{D^{\alpha - 1}}u\left( t \right)} \right){\rm{d}}t} } \right]}}{{\Gamma \left( \alpha \right)}}{t^{\alpha - 1}} + \frac{1}{{\Gamma \left( \alpha \right)}}\int_0^t {{{\left( {t - s} \right)}^{\alpha - 1}}f\left( {s,u\left( s \right),D_{0^ + }^{\alpha - 1}u\left( s \right)} \right){\rm{d}}s.} $

则求解问题(2) 转化为求算子的不动点问题.

引理4  若条件H₁和H₂成立，则算子T:Y→Y有界连续.

证明  第一步：T:Y→Y.对于∀u(t)∈Y，则u(t)∈X，注意到，

$ {\int_0^\infty {f\left( {\mathit{t},\mathit{u}\left( t \right)} \right),{D^{\alpha - 1}}u\left( t \right){\rm{d}}t \le \left\| u \right\|} _Y}\int_0^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]} {\rm{d}}t + \int_0^\infty {c\left( t \right){\rm{d}}t} < \infty , $

类似于文献[8]中引理3.2的证明，可得Tu(t)∈X.又

$ {D^{\alpha - 1}}Tu\left( t \right) = {u_\infty } - \int_0^\infty {f\left( {\mathit{t},\mathit{u}\left( t \right)} \right),{D^{\alpha - 1}}u\left( t \right){\rm{d}}t + \int_0^t {f\left( {s,u\left( s \right)} \right),} } {D^{\alpha - 1}}u\left( s \right){\rm{d}}s, $

由f的连续性可得D^α-1Tu(t)∈C(J, E)，且有

$ {\left\| {{D^{\alpha - 1}}u\left( t \right)} \right\|^{\rho 2}} \le {\left( {\left\| {{u_\infty }} \right\| + 2\left\{ {{{\left\| u \right\|}_Y}\int_0^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]} {\rm{d}}t + \int_0^\infty {c\left( t \right){\rm{d}}t} } \right\}} \right)^{\rho 2}}. $

第二步：证明T:Y→Y有界.注意到上式以及

$ {\left\| {\frac{{Tu\left( t \right)}}{{1 + {t^{^{\alpha - 1}}}}}} \right\|^{\rho 1}} \le {\left( {\frac{{{{\left\| u \right\|}_\infty }}}{{\Gamma \left( \alpha \right)}} + \frac{1}{{\Gamma \left( \alpha \right)}}\left\{ {\left[ {{{\left\| u \right\|}_Y}\int_0^\infty {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} } \right]{\rm{d}}t + \int_0^\infty {c\left( t \right){\rm{d}}t} } \right\}} \right)^{\rho 1}}. $

因此‖Tu(t)‖_Y=max{$\mathop {\sup }\limits_{t \in J}\frac{{{\left\| Tu\left( t \right) \right\|}^{{{\rho }_{1}}}}}{\left( 1+{{t}^{\alpha -1}} \right)}$, $\mathop {\sup }\limits_{t \in J} $‖D^α-1u(t)‖^ρ₂}＜∞.

第三步：算子T:Y→Y连续.取{u_n}_n=1^∞ ⊂Y，u∈Y使得u_n→u(n→∞)，则{u_n}_n=1^∞是Y的有界集，即存在M>0，使得‖u_n‖_Y≤M，‖u‖_Y≤M，则

$ \begin{array}{l} {\left\| {\frac{{T{u_n}\left( t \right)}}{{1 + {t^{^{\alpha - 1}}}}} - \frac{{Tu\left( t \right)}}{{1 + {t^{^{\alpha - 1}}}}}} \right\|^{\rho 1}} \\\le \left( {\frac{2}{{\Gamma \left( \alpha \right)}}\int_0^\infty {\left\| {f\left( {t,{u_n}\left( t \right),{\mathit{D}^{\alpha - 1}}{u_n}\left( t \right)} \right) - f\left( {t,u\left( t \right),{D^{\alpha - 1}}u\left( t \right)} \right)} \right\|{\rm{d}}t} } \right) ^{\rho 1} \le \\{\left( {\frac{{4M}}{{\Gamma \left( \alpha \right)}}\int_0^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]} {\rm{d}}t + \frac{4}{{\Gamma \left( \alpha \right)}}\int_0^\infty {c\left( t \right){\rm{d}}t} } \right)^{\rho 1}}, \end{array} $

$ {\left\| {{D^{\alpha - 1}}T{u_n}\left( t \right) - {D^{\alpha - 1}}Tu\left( t \right)} \right\|^{\rho 2}} \le 4M\int_0^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]} {\rm{d}}t + 4\int_0^\infty {c\left( t \right){\rm{d}}t} . $

由Lebesgue控制收敛定理可得算子T的连续性.

引理5  若条件H₁成立，B是Y的有界子集，则

(a) $\left\{ \frac{TB\left( t \right)}{1+{{t}^{\alpha -1}}} \right\}$和{D^α-1TB(t)}在任意紧区间上是等度连续的；

(b)对∀ε>0，∃N>0使得对t₁, t₂≥N, u(t)∈B，有

$ {\left\| {\frac{{Tu\left( {{t_1}} \right)}}{{1 + {t_1}^{\alpha - 1}}} - \frac{{Tu\left( {{t_2}} \right)}}{{1 + {t_2}^{\alpha - 1}}}} \right\|^{\rho 1}} < \varepsilon ,{\left\| {{D^{\alpha - 1}}Tu\left( {{t_1}} \right) - {D^{\alpha - 1}}Tu\left( {{t_2}} \right)} \right\|^{\rho 2}} < \varepsilon . $

证明   (a):由H₁及B的有界性知存在K>0，使得对u∈B，

$ \int_0^\infty {\left\| {f\left( {t,u\left( t \right),{D^{\alpha - 1}}u\left( t \right)} \right)} \right\|{\rm{d}}t \le K} , $

(3)

取R使得‖u(t)‖_Y≤R, ∀u∈B，令I=[a, b] ⊂J为紧区间，t₁, t₂∈[a, b]，t₁＜t₂则

$ \begin{array}{l} {\left\| {\frac{{Tu\left( {{t_2}} \right)}}{{1 + {t_2}^{\alpha - 1}}} - \frac{{Tu\left( {{t_1}} \right)}}{{1 + {t_1}^{\alpha - 1}}}} \right\|^{\rho 1}} \le \left\{ {\frac{{\left\| {{u_\infty }} \right\| + K}}{{\Gamma \left( \alpha \right)}}} \right.\left| {\frac{{t_2^{\alpha - 1}}}{{1 + t_2^{\alpha - 1}}} - \frac{{t_1^{\alpha - 1}}}{{1 + t_1^{\alpha - 1}}}} \right| + \frac{1}{{\Gamma \left( \alpha \right)}}\int_{{t_1}}^{{t_2}} {c\left( t \right){\rm{d}}t + } \\ \frac{R}{{\Gamma \left( \alpha \right)}}\int_{{t_1}}^{{t_2}} {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]{\rm{d}}t} + \\ {\left. {\frac{{R\left[ {{{\left( {1 + {b^{\alpha - 1}}} \right)}^{\rho 1}}{A_1} + {A_2}} \right] + {A_3}}}{{\Gamma \left( {\alpha + 1} \right)}}\left[ {\frac{{t_2^{\alpha - 1}}}{{1 + t_2^{\alpha - 1}}} - \frac{{t_1^{\alpha - 1}}}{{1 + t_1^{\alpha - 1}}} - \frac{{{{\left( {{t_2} - {t_1}} \right)}^\alpha }}}{{1 + {b^{\alpha - 1}}}}} \right]} \right\}^{^{\rho 1}}}, \end{array} $

其中: A₁=$\mathop {\max }\limits_{t \in \left[ {a,b} \right]}$a(t)；A₂=$\mathop {\max }\limits_{t \in \left[ {a,b} \right]}$b(t)；A₃=$\mathop {\max }\limits_{t \in \left[ {a,b} \right]}$c(t).

$ {\left\| {{D^{\alpha - 1}}Tu\left( {{t_2}} \right) - {D^{\alpha - 1}}Tu\left( {{t_1}} \right)} \right\|^{\rho 2}} \le {\left( {R\int_{{t_1}}^{{t_2}} {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]{\rm{d}}t + \int_{{t_1}}^{{t_2}} {c\left( t \right){\rm{d}}t} } } \right)^{\rho 2}}, $

故$\left\{ \frac{TB\left( t \right)}{1+{{t}^{\alpha -1}}} \right\}$和{D^α-1TB(t)}在[a, b]上等度连续性得证.

(b):${\left\| {\frac{{Tu\left( {{t_2}} \right)}}{{1 + t_2^{\alpha - 1}}} - \frac{{Tu\left( {{t_1}} \right)}}{{1 + t_1^{\alpha - 1}}}} \right\|^{{\rho _1}}} \le \frac{{\left\| {{u_\infty } - \int_0^\infty {f\left( {t,u\left( t \right),{D^{\alpha - 1}}u\left( t \right)} \right){\rm{d}}t} } \right\|}}{{\Gamma \left( \alpha \right)}}\left| {\frac{{t_2^{\alpha - 1}}}{{1 + t_2^{\alpha - 1}}} - \frac{{t_1^{\alpha - 1}}}{{1 + t_1^{\alpha - 1}}}} \right| + $${\left( {\frac{1}{{\Gamma \left( \alpha \right)}}\left\| {\int_0^{{t_2}} {\frac{{{{\left( {{t_2} - s} \right)}^{\alpha - 1}}}}{{1 + t_2^{\alpha - 1}}}f\left( {s,u\left( s \right),{D^{\alpha - 1}}u\left( s \right)} \right){\rm{d}}s - \int_0^{{t_1}} {\frac{{{{\left( {{t_1} - s} \right)}^{\alpha - 1}}}}{{1 + t_1^{\alpha - 1}}}f\left( {s,u\left( s \right),{D^{\alpha - 1}}u\left( s \right)} \right){\rm{d}}s} } } \right\|} \right)^{{\rho _1}}},$

类似于文献[8]中引理3.3的证明易知：∃N₁>0，对t₁, t₂≥N₁,

$ \left\| {\int_0^{{t_2}} {\frac{{{{\left( {{t_2} - s} \right)}^{\alpha - 1}}}}{{1 + {t_2}^{\alpha - 1}}}f\left( {s,u\left( s \right),{D^{\alpha - 1}}u\left( s \right)} \right){\rm{d}}s - \int_0^{{t_1}} {\frac{{{{\left( {{t_1} - s} \right)}^{\alpha - 1}}}}{{1 + {t_1}^{\alpha - 1}}}f\left( {s,u(s} \right),{D^{\alpha - 1}}u\left( s \right){\rm{d}}s} } } \right\| < \frac{{\Gamma \left( \alpha \right)\varepsilon }}{2}. $

又因为$\mathop {\lim }\limits_{t \to \infty } {\mkern 1mu} \frac{{{t^{\alpha - 1}}}}{{1 + {t^{\alpha - 1}}}}$=1，∃N₂>N₁使得对∀t₁, t₂≥N₂有

$ \begin{array}{l} \left| {\frac{{t_2^{\alpha - 1}}}{{1 + t_2^{\alpha - 1}}} - \frac{{t_1^{\alpha - 1}}}{{1 + t_1^{\alpha - 1}}}} \right| \le \left( {1 - \frac{{t_2^{\alpha - 1}}}{{1 + t_2^{\alpha - 1}}}} \right) + \left( {1 - \frac{{t_1^{\alpha - 1}}}{{1 + t_1^{\alpha - 1}}}} \right) \le \\ \frac{{\Gamma \left( \alpha \right)}}{{\left\| {{u_\infty } - \int_0^\infty {f\left( {t,u\left( t \right),{D^{\alpha - 1}}u\left( t \right)} \right){\rm{d}}t} } \right\|}} \cdot \frac{\varepsilon }{2}, \end{array} $

因此, 对∀ε>0，当t₁, t₂≥N₂时，${{\left\| \frac{Tu\left( {{t}_{2}} \right)}{1+t_{2}^{\alpha -1}}-\frac{Tu\left( {{t}_{1}} \right)}{1+t_{1}^{\alpha -1}} \right\|}^{{{\rho }_{1}}}}$＜ε，且由(3) 式知∃N₃>0，使得对∀u∈B有$\int_{{N_3}}^\infty {} $‖f(t, u(t), D^α-1u(t))‖dt≤$\frac{\varepsilon }{2}$.当t₁, t₂≥N₃有

$ \begin{array}{l} {\left\| {{D^{\alpha - 1}}Tu\left( {{t_2}} \right) - {D^{\alpha - 1}}Tu\left( {{t_1}} \right)} \right\|^{\rho 2}} \le {\left\| {\int_0^{{t_2}} {f\left( {s,u\left( s \right),\\ {D^{\alpha - 1}}u\left( s \right)} \right)} {\rm{d}}s - \int_0^{{t_1}} {f\left( {s,u\left( s \right),{D^{\alpha - 1}}u\left( s \right)} \right)} {\rm{d}}s} \right\|^{\rho 2}} \le \\ {\left( {\int_{{N_3}}^{{t_2}} {\left\| {f\left( {s,u\left( s \right),{D^{\alpha - 1}}u\left( s \right)} \right)} \right\|} \;{\rm{d}}s + \int_{{N_3}}^{{t_1}} {\left\| {f\left( {s,u\left( s \right),{D^{\alpha - 1}}u\left( s \right)} \right)} \right\|} \;{\rm{ds}}} \right)^{\rho 2}} \le \\ {\left( {2\int_{{N_3}}^\infty {\left\| {f\left( {t,u\left( t \right),{D^{\alpha - 1}}u\left( t \right)} \right)} \right\|} \;{\rm{dt}}} \right)^{\rho 2}} < \varepsilon . \end{array} $

故对∀ε>0，取N=max{N₂, N₃}时, (b)得证.

引理6  若条件H₁成立，B是Y的有界子集，则α_Y(TB)=$\mathop {\sup }\limits_{t \in J} $α(‖TB(t)‖_Y).

证明  对∀ε>0，存在划分TB=$\bigcup\limits_{i = 1}^n {} $TB_i，使得diam_Y(TB_i)＜α_Y(TB)+ε，所以对∀t∈J，u₁, u₂∈B_i，${\left\| {\frac{{T{u_1}\left( t \right)}}{{1 + {t^{\alpha - 1}}}} - \frac{{T{u_2}\left( t \right)}}{{1 + {t^{\alpha - 1}}}}} \right\|^{{\rho _1}}}$≤‖Tu₁-Tu₂‖_Y＜α_Y(TB)+ε.

由TB(t)=$\bigcup\limits_{i = 1}^n {} $TB_i(t)及ε的任意性可得$\mathop {\sup }\limits_{t \in J} \alpha \left( {\frac{{{{\left\| {TB\left( t \right)} \right\|}^{{\rho _1}}}}}{{{{\left( {1 + t_1^{\alpha - 1}} \right)}^{{\rho _1}}}}}} \right)$≤α_Y(TB).

同理可证$\mathop {\sup }\limits_{t \in J} $ α(‖D^α-1TB(t)‖^ρ₂)≤α_Y(TB).故$\mathop {\sup }\limits_{t \in J} $α(‖TB(t)‖_Y)≤α_Y(TB).

由引理5可知∀ε>0，∃N>0, 使得对t₁, t₂≥N，u∈B有${\left\| {\frac{{Tu\left( {{t_1}} \right)}}{{1 + t_1^{\alpha - 1}}} - \frac{{Tu\left( {{t_2}} \right)}}{{1 + t_2^{\alpha - 1}}}} \right\|^{{\rho _1}}}$＜ε₁，则

$ \left\| {\frac{{Tu\left( {{t_1}} \right)}}{{1 + {t_1}^{\alpha - 1}}} - \frac{{Tu\left( {{t_2}} \right)}}{{1 + {t_2}^{\alpha - 1}}}} \right\| < \varepsilon . $

(4)

定义TB在[0，N]上的限制TB_{[0, N]}，由引理1得

$ \begin{array}{l} {\alpha _Y}\left( {TB|{\;_{\left[ {0,N} \right]}}} \right) \le \max \left\{ {{\alpha _X}\left( {TB|{\;_{\left[ {0,N} \right]}}} \right),{\alpha _C}\left( {{D^{\alpha - 1}}TB|{\;_{\left[ {0,N} \right]}}} \right)} \right\} \le \\ \mathop {\sup }\limits_{t \in J} \left\{ {\alpha \left( {\frac{{{{\left\| {TB\left( t \right)} \right\|}^{\rho 1}}}}{{{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}}}} \right),\alpha {{\left\| {{D^{\alpha - 1}}TB\left( t \right)} \right\|}^{\rho 2}}} \right\}. \end{array} $

又注意到存在B的划分，使得B=$\bigcup\limits_{i = 1}^n {} $B_i，TB_{[0, N]}=$\bigcup\limits_{i = 1}^n {} $TB_{i[0, N]}且

$ dia{m_X}\left( {T{B_i}|{\;_{\left[ {0,N} \right]}}} \right) < \mathop {\sup }\limits_{t \in J} \alpha \left( {\frac{{{{\left\| {TB\left( t \right)} \right\|}^{\rho 1}}}}{{{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}}}} \right) + \varepsilon ,i = 1,2, \cdots ,n, $

(5)

$ diam\left( {\left( {{D^{\alpha - 1}}T{B_i}} \right)|{\;_{\left[ {0,N} \right]}}} \right) < \mathop {\sup }\limits_{t \in J} \alpha \left( {{{\left\| {{D^{\alpha - 1}}TB\left( t \right)} \right\|}^{\rho 2}}} \right) + \varepsilon ,i = 1,2, \cdots ,n. $

(6)

则对∀Tu₁, Tu₂∈TB_i, t≥N，应用(4) 和(5) 式有

$ \begin{array}{l} {\left\| {\frac{{T{u_1}\left( t \right)}}{{1 + {t^{\alpha - 1}}}} - \frac{{T{u_2}\left( t \right)}}{{1 + {t^{\alpha - 1}}}}} \right\|^{\rho 1}} \le \left( {\left\| {\frac{{T{u_1}\left( t \right)}}{{1 + {t^{\alpha - 1}}}} - \frac{{T{u_1}\left( N \right)}}{{1 + {N^{\alpha - 1}}}}} \right\| + \left\| {\frac{{T{u_2}\left( N \right)}}{{1 + {N^{\alpha - 1}}}} - \frac{{T{u_2}\left( N \right)}}{{1 + {N^{\alpha - 1}}}}} \right\| + } \right.\\ {\left. {\left\| {\frac{{T{u_2}\left( N \right)}}{{1 + {N^{\alpha - 1}}}} - \frac{{T{u_2}\left( t \right)}}{{1 + {t^{\alpha - 1}}}}} \right\|} \right)^{\rho 1}} \le {\left( {\left\| {\frac{{T{u_1}\left( N \right)}}{{1 + {N^{\alpha - 1}}}} - \frac{{T{u_2}\left( N \right)}}{{1 + {N^{\alpha - 1}}}}} \right\| + 2\varepsilon } \right)^{\rho 1}}. \end{array} $

同理可得‖D^α-1Tu₁(t)-D^α-1Tu₂(t)‖^ρ₂≤(‖D^α-1Tu₁(N)-D^α-1Tu₂(N)‖+2ε)^ρ₂.

又因TB=$\bigcup\limits_{i = 1}^n {} $TB_i，由(5) 和(6) 式及ε的任意性可知

$ {\alpha _Y}\left( {TB} \right) \le \mathop {\sup }\limits_{t \in J} \left\{ \alpha {\left( {\frac{{{{\left\| {TB\left( t \right)} \right\|}^{\rho 1}}}}{{{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}}}} \right),\alpha \left( {{{\left\| {{D^{\alpha - 1}}TB\left( t \right)} \right\|}^{\rho 2}}} \right)} \right\}. $

故$\mathop {\sup }\limits_{t \in J} $α(‖TB(t)‖_Y)≥α_Y(TB).

定理1  若条件H₁~H₃成立，则问题(2) 至少存在一个解.

证明   ① 取R≥max{$\frac{{\left\| {{u_\infty }} \right\| + 2\int_0^\infty {c\left( t \right){\rm{d}}t} }}{{\Gamma \left( \alpha \right) - 2\int_0^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{{\rho _1}}}\alpha \left( t \right) + b\left( t \right)} \right]{\rm{d}}t} }}$, $\frac{1}{{2\int_0^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{{\rho _1}}}\alpha \left( t \right) + b\left( t \right)} \right]{\rm{d}}t} }}$}.令B={u∈Y:‖u‖_Y≤R}，下证T:B→B.对∀u(t)∈B，有

$ \begin{array}{l} \frac{{{{\left\| {Tu\left( t \right)} \right\|}^{\rho 1}}}}{{{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}}} \le {\left( {\frac{1}{{\Gamma \left( \alpha \right)}}\left[ {\left\| {{u_\infty }} \right\| + \int_0^\infty {\left\| {f\left( {t,u\left( t \right),{D^{\alpha - 1}}u\left( t \right)} \right)} \right\|{\rm{d}}t} } \right]} \right)^{\rho 1}} < \\ {\left( {\frac{{\left\| {{u_\infty }} \right\|}}{{\Gamma \left( \alpha \right)}} + \frac{2}{{\Gamma \left( \alpha \right)}}\left\{ {{{\left\| u \right\|}_Y}\int_0^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]} \;{\rm{d}}t + \int_0^\infty {c\left( t \right){\rm{d}}t} } \right\}} \right)^{^{\rho 1}}} \le \\ \frac{{\left\| {{u_\infty }} \right\|}}{{\Gamma \left( \alpha \right)}} + \frac{2}{{\Gamma \left( \alpha \right)}}\left\{ {{{\left\| u \right\|}_Y}\int_0^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]} \;{\rm{d}}t + \int_0^\infty {c\left( t \right){\rm{d}}t} } \right\} \le R. \end{array} $

$ {\left\| {{D^{\alpha - 1}}Tu\left( t \right)} \right\|^{\rho 2}} \le {\left( {\left\| {{u_\infty }} \right\| + 2\left\{ {{{\left\| u \right\|}_Y}\int_0^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]{\rm{d}}t + \int_0^\infty {} } } \right\}c\left( t \right){\rm{d}}t} \right)^{\rho 2}} \\ \le R. $

② 取D=$\overline {co} $(TB)，则D是B的有界闭凸子集，下证T:D→D是严格集压缩.定义

$ \begin{array}{l} {T_n}u\left( t \right): = \frac{{{u_\infty }}}{{\Gamma \left( \alpha \right)}}{t^{\alpha - 1}} - \frac{1}{{\Gamma \left( \alpha \right)}}\int_0^t {\left[ {{t^{\alpha - 1}} - {{\left( {t - s} \right)}^{\alpha - 1}}} \right]} f\left( {s,u\left( s \right),{D^{\alpha - 1}}u\left( s \right)} \right){\rm{d}}s - \\ \frac{1}{{\Gamma \left( \alpha \right)}}\int_t^n {{t^{\alpha - 1}}} f\left( {s,u\left( s \right),{D^{\alpha - 1}}u\left( s \right)} \right){\rm{d}}s, \end{array} $

$ {\left\| {\frac{{{T_n}u\left( t \right)}}{{1 + {t^{\alpha - 1}}}} - \frac{{{T_n}u\left( t \right)}}{{1 + {t^{\alpha - 1}}}}} \right\|^{\rho 1}} \le {\left( {\frac{1}{{\Gamma \left( \alpha \right)}}\left[ {R\int_n^\infty {\left[ {{{\left( {1 + {t^{^{\alpha - 1}}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]{\rm{d}}t + \int_n^\infty {c\left( t \right){\rm{d}}t} } } \right]} \right)^{\rho 1}}, $

$ {\left\| {{D^{\alpha - 1}}T{u_m}\left( t \right) - {D^{\alpha - 1}}Tu\left( t \right)} \right\|^{\rho 2}} \le {\left( {R\int_n^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]} {\rm{d}}t + \int_n^\infty {c\left( t \right){\rm{d}}t} } \right)^{\rho 2}}. $

当n→∞时，d(‖T_nS(t)‖_Y, ‖TS(t)‖_Y)→0，由非紧性测度性质得：

$ \mathop {\lim }\limits_{n \to \infty } \alpha \left( {{{\left\| {{T_n}S\left( t \right)} \right\|}_{\;Y}}} \right) = \alpha \left( {\left\| {TS\left( t \right)} \right\|{\;_Y}} \right),t \in J. $

由引理5知$\left\{ {\frac{{S\left( t \right)}}{{1 + {t^{\alpha - 1}}}}} \right\}$，{D^α-1S(t)}在J的任意紧区间等度连续，由条件H₁和H₂中f一致连续性得{f(t, u(t), D^α-1u(t)):u(t)∈S}在区间[0, n]上有界且等度连续，利用引理1、引理6和H₃得：

$ \begin{array}{l} \alpha \left( {\frac{{{{\left\| {{T_n}S\left( t \right)} \right\|}^{\rho 1}}}}{{{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}}}} \right) \le \left\{ {\frac{1}{{\Gamma \left( \alpha \right)}}\int_0^t {\alpha \left( {\left\{ {f\left( {t,u\left( t \right),{D^{\alpha - 1}}u\left( t \right)} \right):u\left( t \right) \in S} \right\}} \right)} {\rm{d}}t + } \right.\\ \left. {\frac{1}{{\Gamma \left( \alpha \right)}}\int_t^n {\alpha \left( {\left\{ {f\left( {t,u\left( t \right),{D^{\alpha - 1}}u\left( t \right)} \right):u\left( t \right) \in S} \right\}} \right)} {\rm{d}}t} \right\} \le \\ \frac{1}{{\Gamma \left( \alpha \right)}}{\int_0^n {\left( {1 + {t^{\alpha - 1}}} \right)} ^{\rho 1}}h\left( t \right)\alpha \left( {\frac{{{{\left\| {S\left( t \right)} \right\|}^{\rho 1}}}}{{{{\left( {1 + {t^{\alpha - 1}}} \right)}^{^{\rho 1}}}}}} \right){\rm{d}}t \le \\ \frac{1}{{\Gamma \left( \alpha \right)}}{\int_0^n {\left( {1 + {t^{\alpha - 1}}} \right)} ^{\rho 1}}h\left( t \right){\rm{d}}t{\alpha _Y}\left( S \right) \le \frac{1}{{\Gamma \left( \alpha \right)}}{\int_0^\infty {\left( {1 + {t^{\alpha - 1}}} \right)} ^{\rho 1}}h\left( t \right){\rm{d}}t{\alpha _Y}\left( S \right), \end{array} $

注意到(1+t^α-1)^ρ₁>1，1/Γ(α)>1，

$ \begin{array}{l} \alpha \left( {{{\left\| {{D^{\alpha - 1}}{T_n}S\left( t \right)} \right\|}^{\rho 2}}} \right) \le \int_t^n {\alpha \left( {\left\{ {f\left( {t,u\left( t \right),{D^{\alpha - 1}}u\left( t \right)} \right):u\left( t \right) \in S} \right\}} \right)} {\rm{d}}t \le \\ \int_t^n {h\left( t \right)\alpha } \left( {{{\left\| {{D^{\alpha - 1}}S\left( t \right)} \right\|}^{\rho 2}}} \right){\rm{d}}t \le \int_0^\infty {h\left( t \right){\rm{d}}t{\alpha _Y}} \left( S \right) \le \frac{1}{{\Gamma \left( \alpha \right)}}{\int_0^\infty {\left( {1 + {t^{\alpha - 1}}} \right)} ^{\rho 1}}h\left( t \right){\rm{d}}t{\alpha _Y}\left( S \right). \end{array} $

令λ=$\frac{1}{{\Gamma \left( \alpha \right)}}\int_0^\infty {} $(1+t^α-1)^ρ₁h(t)dt，由H₃知0＜λ＜1，故$\mathop {\sup }\limits_{t \in J} $α(‖TS(t)‖_Y)≤λα_Y(S)，由引理6及Darbo不动点定理知问题(2) 在D中至少有一个解.

下面考虑问题(2) 中u_∞=0，且定理1中f的次线性条件改为超线性条件时，则有定理2.

定理2  若f满足(H₀), 存在非负函数a(t), b(t)∈C(J)，ρ_i>1(i=1, 2), 使得

$ \left\| {f\left( {t,x,y} \right)} \right\| \le a\left( t \right)\;\;{\left\| x \right\|^{\rho 1}} + b\left( t \right)\;\left\| y \right\|{\;^{\rho 2}},t \in J,x \in E, $

$ \int_0^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{\rho 1}}a\left( t \right) + b\left( t \right)} \right]{\rm{d}}t < \Gamma \left( \alpha \right)} , $

且同时满足H₂和H₃，则问题(2) 至少存在一个解.

证明过程类似前定理，只需取ρ=max{ρ₁, ρ₂}，0＜R＜${(\frac{{\Gamma \left( \alpha \right)}}{{2\int_0^\infty {\left[ {{{\left( {1 + {t^{\alpha - 1}}} \right)}^{{\rho _1}}}\alpha \left( t \right) + b\left( t \right)} \right]{\rm{d}}t} }})^{\frac{\rho }{{{\rho ^{ - 1}}}}}}$.

3 例子

定义空间E=l_∞={u=(u₁, u₂, …, u_n, …):$\mathop {\sup }\limits_n $|u_n|＜∞}，空间E上的范数为‖u‖=$\mathop {\sup }\limits_n $|u_n|，则E是Banach空间.考虑边值问题:

$ \left\{ \begin{array}{l} {D^{3/2}}{u_n}\left( t \right) = \frac{{t{{\left| {{u_n}\left( t \right)} \right|}^{1/2}}}}{{8{{\rm{e}}^{{t^2}}}{{\left( {1 + \sqrt t } \right)}^{1/2}}}} + \frac{{\sqrt {\left| {\sin {{\left( {{D^{1/2}}{u_{n + 1}}\left( t \right)} \right)}^{2/3}}} \right|} }}{{16{n^2}\left( {9 + {t^2}} \right)}},\\ {u_n}\left( 0 \right) = 0,{D^{1/2}}{u_n}\left( \infty \right) = {u_{n\infty }}, \end{array} \right. $

(7)

其中:t∈[0, ∞); α=$\frac{3}{2}$; ρ₁=$\frac{1}{2}$; ρ₂=$\frac{1}{3}$; f(t, x, y)=(f₁(t, x, y), f₂(t, x, y), …, f_n(t, x, y), …);

$ {f_n}\left( {t,x,y} \right) = \frac{{t{{\left| {{x_n}} \right|}^{1/2}}}}{{8{{\rm{e}}^{{t^2}}}{{\left( {1 + \sqrt t } \right)}^{1/2}}}} + \frac{{\sqrt {\left| {\sin y_{n + 1}^{2/3}} \right|} }}{{16{n^2}\left( {9 + {t^2}} \right)}}. $

显然f(t, x, y)∈C(J×E×E, E).注意到

$ \left\| {\;{{\left| x \right|}^{1/2}}\;} \right\| = \mathop {\sup }\limits_n \left| {\;\left| {{x_n}} \right|{\;^{1/2}}} \right| = \mathop {\sup }\limits_n \left| {{x_n}} \right|{\;^{1/2}} = {\left( {\mathop {\sup }\limits_n \left| {{x_n}} \right|\;} \right)^{^{1/2}}} = {\left\| x \right\|^{^{1/2}}}, $

$ \left\| {f\left( {t,x,y} \right)} \right\| \le \frac{t}{{8{{\rm{e}}^{{t^2}}}{{\left( {1 + \sqrt t } \right)}^{1/2}}}}{\left\| x \right\|^{^{1/2}}} + \frac{1}{{16\left( {9 + {t^2}} \right)}}{\left\| y \right\|^{1/3}}, $

且$\frac{1}{2}$Γ(2/3)≈0.443 1，$\int_0^\infty {\left[ {{{\left( {1 + \sqrt t } \right)}^{1/2}}\frac{t}{{8{{\rm{e}}^{{t^2}}}{{\left( {1 + \sqrt t } \right)}^{1/2}}}} + \frac{1}{{16\left( {9 + {t^2}} \right)}}} \right]} {\rm{d}}t = \frac{1}{{16}} + \frac{{\rm{ \mathsf{ π} }}}{{96}}$≈0.095 2.

故取a(t)=$\frac{t}{{8{{\rm{e}}^{{t^2}}}{{\left( {1 + \sqrt t } \right)}^{1/2}}}}$，b(t)=${\frac{1}{{16\left( {9 + {t^2}} \right)}}}$时条件H₁满足, 条件H₂显然满足.

对于H₃，记f=f⁽¹⁾+f⁽²⁾，其中f_n⁽¹⁾=$\frac{{t{{\left| {{x_n}} \right|}^{1/2}}}}{{8{{\rm{e}}^{{t^2}}}{{\left( {1 + \sqrt t } \right)}^{1/2}}}}$，f_n⁽²⁾=$\frac{{\sqrt {\left| {\sin y_{n + 1}^{2/3}} \right|} }}{{16{n^2}\left( {9 + {t^2}} \right)}}$.

下证对任一有界集D⊂E，α(f⁽²⁾(t, x, D))=0.取有界集{y^(m)} E，即‖y^(m)‖≤M，m=1, 2, ….其中, y^(m)=(y₁^(m), y₂^(m), …y_n^(m), …).固定某一t∈J, x∈E，

$ \left| {f_n^{\left( 2 \right)}\left( {t,x,{y^{\left( m \right)}}} \right)} \right| \le \frac{{{M^{1/3}}}}{{144{n^2}}}, $

(8)

所以{f_n⁽²⁾(t, x, y^(m))}有界.利用对角线方法，选取{y^(m_i)} {y^(m)}，使得当i→∞时，

$ f_n^{\left( 2 \right)}\left( {t,x,{y^{\left( {{m_i}} \right)}}} \right) \to {w_n}. $

(9)

由(8) 式可知

$ \left| {{w_n}} \right| \le \frac{{{M^{1/3}}}}{{144{n^2}}},n = 1,2, \cdots , $

(10)

所以w={w₁, w₂, …, w_n, …}∈E.由(8) 和(10) 式知，对∀ε₁>0，∃n>N时, 有

$ \left| {f_n^{\left( 2 \right)}\left( {t,x,{y^{\left( {{m_i}} \right)}}} \right)} \right| < {\varepsilon _1},\left| {{w_n}} \right| \le {\varepsilon _1},i = 1,2, \cdots . $

(11)

由(9) 式知, 对∀ε>0，∃K>0，使得当i>K时，

$ \left| {f_n^{\left( 2 \right)}\left( {t,x,{y^{\left( {{m_i}} \right)}}} \right) - {w_n}} \right| < \varepsilon ,n = 1,2, \cdots ,N. $

(12)

由(11) 和(12) 式易得‖f⁽²⁾(t, x, y^(m))-w‖≤ε，即当i→∞时，‖f⁽²⁾(t, x, y^(m))-w‖→0.所以对D⊂E，f⁽²⁾(t, x, D)是相对紧集.故α(f⁽²⁾(t, x, D))=0.因此对任一有界集S⊂E，有

$ \alpha \left( {f\left( {t,S,{D^{\alpha - 1}}S} \right)} \right) \le \alpha \left( {{f^{\left( 1 \right)}}\left( {t,S,{D^{\alpha - 1}}S} \right)} \right) = \frac{{\alpha \left( S \right)}}{{8{t^{ - 1}}{{\rm{e}}^{{t^2}}}{{\left( {1 + \sqrt t } \right)}^{1/2}}}}. $

又因为$\int_0^\infty {{{\left( {1 + \sqrt t } \right)}^{1/2}}} $$\frac{1}{{8{t^{ - 1}}{{\rm{e}}^{{t^2}}}{{\left( {1 + \sqrt t } \right)}^{1/2}}}}{\rm{d}}t$＜Γ(3/2)，故取h(t)=$\frac{1}{{8{t^{ - 1}}{{\rm{e}}^{{t^2}}}{{\left( {1 + \sqrt t } \right)}^{1/2}}}}$，条件H₃满足.

由定理1可知问题(7) 至少存在一个解.