一类非线性分数阶微分方程耦合系统正解的存在性

薛益民; 彭钟琪

doi:10.6054/j.jscnun.2020032

一类非线性分数阶微分方程耦合系统正解的存在性

薛益民^,,
彭钟琪

徐州工程学院数学与物理科学学院，徐州 221018

基金项目:

国家自然科学基金项目 11526177

江苏省自然科学基金项目 BK20151160

徐州工程学院培育项目 XKY2017113

详细信息

通讯作者:
薛益民，副教授，Email:xueym@xzit.edu.cn

中图分类号: O211.6
计量
- 文章访问数: 1695
- HTML全文浏览量: 711
- PDF下载量: 54
出版历程
- 收稿日期: 2019-07-08
- 网络出版日期: 2021-03-21
- 刊出日期: 2020-04-24

On the Existence of Positive Solutions to the Coupled System of a Class of Nonlinear Fractional Differential Equations

XUE Yimin^,,
PENG Zhongqi

School of Mathematics and Physical Science, Xuzhou University of Technology, Xuzhou 221018, China

摘要

摘要: 利用Guo-Krasnoselskii不动点定理、Schauder不动点定理和格林函数的性质，研究一类非线性Riemann-Liouville型分数阶微分方程耦合系统正解的存在性，得到了该耦合系统正解的存在性定理，并举例说明了定理的有效性.
- 分数阶微分方程 /
- 耦合系统 /
- 边值问题 /
- 不动点定理
Abstract: The Guo-Krasnoselskii's fixed point theorem, the Schauder fixed point theorem and the properties of the associated Green's function are used to study the existence of positive solutions to the coupled system of a class of nonlinear Riemann-Liouville fractional differential equations. Two theorems about the existence of positive solutions are obtained, and two examples are given to illustrate the advantages of the theorems.
- fractional differential equations /
- coupled system /
- boundary value problem /
- fixed point theorem

HTML全文

随着非线性问题研究的深入，学者们建立了比整数阶微分方程模型更为精细的分数阶微分方程模型，以更好地解决复杂的实际问题，而其中的很多问题可以化为非线性分数阶微分方程边值问题.关于非线性分数阶微分方程边值问题正解的研究也已有许多重要成果^[1-10].

BAI和LU^[10]利用锥上的不动点定理，得到了非线性分数阶边值问题

$\left\{ {\begin{array}{*{20}{l}} {D_{0 + }^\alpha u(t) + f(t,u(t)) = 0\;\;\;\;(0 < t < 1),}\\ {u(0) = u(1) = 0} \end{array}} \right.$

正解的存在性和多重性结果，其中，D₀₊^α (1 < α≤2)表示Riemann-Liouville分数阶导数，f∈C([0, 1]×[0, ∞), [0, ∞)).

受文献[10]的启发，本文利用Guo-Krasnoselskii不动点定理、Schauder不动点定理和格林函数的性质，得到如下的非线性Riemann-Liouville型分数阶微分方程耦合系统边值问题

$\left\{ \begin{array}{l} {D^\alpha }u(t) + f(t,v(t)) = 0\;\;\;\;(0 < t < 1),\\ {D^\beta }v(t) + g(t,u(t)) = 0\;\;\;\;(0 < t < 1),\\ u(0) = {D^\gamma }u(0) = {D^\gamma }u(1) = 0,\\ v(0) = {D^\delta }v(0) = {D^\delta }v(1) = 0 \end{array} \right.$

(1)

正解的存在性的充分条件，其中，2 < α, β≤3, 1 < γ≤2, 1 < δ≤2, 1+γ≤α, 1+δ≤β, f, g∈C([0, 1]×[0, ∞), [0, ∞)), D^λ表示λ阶Riemann-Liouville分数阶导数，λ∈{α, β, γ, δ}，并举例说明了定理的有效性.

1. 预备知识

定义1^[11] 函数f: $\mathbb{R}$ ⁺→ $\mathbb{R}$ 的α>0阶Riemann-Liouville积分为

${I^\alpha }f(t) = \frac{1}{{\Gamma (\alpha )}}\int_0^t {{{(t - s)}^{\alpha - 1}}} f(s){\rm{d}}s,$

其中，等式右边在 $\mathbb{R}$ ⁺上逐点定义.

定义2^[11] 函数f: $\mathbb{R}$ ⁺→ $\mathbb{R}$ 的α>0阶Riemann-Liouville导数为

${D^\alpha }f(t) = \frac{1}{{\Gamma (n - \alpha )}}{\left( {\frac{{\rm{d}}}{{{\rm{d}}t}}} \right)^n}\int_0^t {\frac{{f(s)}}{{{{(t - s)}^{\alpha - n + 1}}}}{\rm{d}}s} ,$

其中，n=[α]+1, [α]表示实数α的整数部分，等式右边在 $\mathbb{R}$ ⁺上逐点定义.

引理1^[11] 若α, β>0, f(x)∈L(0, 1), 则：

(ⅰ) D^βI^αf(t)=I^α－βf(t) (α>β)；

(ⅱ) D^αI^αf(t)=f(t)；

(ⅲ) ${I^\alpha }{D^\alpha }f(t) = f(t) + \sum\limits_{i = 1}^n {{c_i}} {t^{\alpha - i}}$ (c_i∈ $\mathbb{R}$ , i=1, 2, …, n, n－1 < α≤n, D^αf(t)∈L(0, 1))；

(ⅳ) $D^{\alpha} t^{\gamma}=\frac{\Gamma(\gamma+1)}{\Gamma(\gamma+1-\alpha)} t^{\gamma-\alpha} \quad(\gamma>-1, \gamma>\alpha-1, t>0)$ .

下面给出本文定理证明时所需要的引理.

引理2^[12] (Guo-Krasnoselskii不动点定理)假设P在Banach空间E中是一个锥, Ω₁和Ω₂是E的有界开子集，且0∈Ω₁, Ω₁⊂Ω₂.如果A:P∩(Ω₂\Ω₁)→P是一个全连续算子，且下列条件之一成立:

(ⅰ) ‖Ax‖≤ ‖x‖(x∈P∩∂Ω₁)且‖Ax‖≥ ‖x‖(x∈P∩∂Ω₂)；

(ⅱ)‖Ax‖≥ ‖x‖(x∈P∩∂Ω₁)且‖Ax‖≤‖x‖(x∈P∩∂Ω₂),

则A在P∩(Ω₂\Ω₁)上有一个不动点.

引理3^[13] (Schauder不动点定理)假设U是Banach空间X的非空有界闭凸子集，T是U到其自身的全连续映射，则至少存在一个x∈U，使得Tx=x.

引理4^[14] ∀y(t)∈C[0, 1], 2 < α≤3, 1 < γ≤2, 1+γ≤α，分数阶微分方程边值问题

$\left\{ {\begin{array}{*{20}{l}} {{D^\alpha }u(t) + y(t) = 0,0 < t < 1}\\ {u(0) = {D^\gamma }u(0) = {D^\gamma }u(1) = 0} \end{array}} \right.$

有唯一解 $u(t)=\int_{0}^{1} G_{\alpha}(t, s) y(s) \mathrm{d} s$ , 其中

${G_\alpha }(t,s) = \left\{ {\begin{array}{*{20}{l}} {\frac{{{t^{\alpha - 1}}{{(1 - s)}^{\alpha - \gamma - 1}} - {{(t - s)}^{\alpha - 1}}}}{{\Gamma (\alpha )}}\quad (0 \le s \le t \le 1),}\\ {\frac{{{t^{\alpha - 1}}{{(1 - s)}^{\alpha - \gamma - 1}}}}{{\Gamma (\alpha )}}\quad (0 \le t \le s \le 1).} \end{array}} \right.$

类似可得

${G_\beta }(t,s) = \left\{ {\begin{array}{*{20}{l}} {\frac{{{t^{\beta - 1}}{{(1 - s)}^{\beta - \delta - 1}} - {{(t - s)}^{\beta - 1}}}}{{\Gamma (\beta )}}\;\;\;\;\;(0 \le s \le t \le 1)}\\ {\frac{{{t^{\beta - 1}}{{(1 - s)}^{\beta - \delta - 1}}}}{{\Gamma (\beta )}}\;\;\;\;\;(0 \le t \le s \le 1)} \end{array}} \right.$

引理5^[15] 假设G(t, s)=(G_α(t, s), G_β(t, s))，则G(t, s)满足：

(ⅰ) ∀t, s∈[0, 1]，有G(t, s)∈C([0, 1]×[0, 1])；

(ⅱ) ∀t, s∈[0, 1]，有G(t, s)≥0，且∀t, s∈(0, 1)，有G(t, s)>0；

(ⅲ) ∀s∈[0, 1]，有 ${\max\limits_{t \in [0, 1]}}G(t, s) = G(1, s)$ ；

(ⅳ) ∀s∈[0, 1]，有

$\mathop {\min }\limits_{t \in [1/2,1]} G(t,s) \ge \mu \mathop {\max }\limits_{t \in [0,1]} G(t,s) = \mu G(1,s),$

其中μ=min{μ_α=(1/2)^α－1, μ_β=(1/2)^β－1}.

2. 主要结论

令X={u(t) |u(t)∈C([0, 1], [0, ∞))}. ∀u∈X，定义范数 $\|u\|=\max \limits_{t \in[0, 1]}|u(t)|$ ，则(X, ‖·‖)是Banach空间.令Y={v(t)|v(t)∈C([0, 1], [0, ∞))}. ∀v∈Y，定义范数 $\|v\|=\max\limits _{t \in[0, 1]}|v(t)|$ . ∀(u, v)∈X×Y，定义范数‖(u, v)‖ = ‖u‖+‖v‖，则(X×Y, ‖(u, v)‖)也是Banach空间.定义锥U⊂X×Y为

$U = \{ (u(t),v(t)) \in X \times Y:u(t) \ge 0,v(t) \ge 0,t \in [0,1]\} .$

∀(u, v)∈X×Y，定义算子T:X×Y→X×Y为

$\begin{array}{l} T(u,v)(t) = \left( {{T_\alpha }v(t),{T_\beta }u(t)} \right) = \\ \;\;\;\left( {\int_0^1 {{G_\alpha }} (t,s)f(s,v(s)){\rm{d}}s,\int_0^1 {{G_\beta }} (t,s)g(s,u(s)){\rm{d}}s} \right). \end{array}$

(2)

由引理4知T的不动点即为耦合系统(1)的解.

引理6^[14] 假设f, g∈C([0, 1]×[0, ∞), [0, ∞))，则算子T:U→U为全连续的.

为叙述简洁，记

$L = \min \left\{ {{L_\alpha } = {{\left( {\int_0^1 {{G_\alpha }} (1,s){\rm{d}}s} \right)}^{ - 1}},{L_\beta } = {{\left( {\int_0^1 {{G_\beta }} (1,s){\rm{d}}s} \right)}^{ - 1}}} \right\},$

$\Delta = \max \left\{ {{\Delta _\alpha } = {{\left( {\int_{1/2}^1 {{{(1/2)}^{\alpha - 1}}} {G_\alpha }(1,s){\rm{d}}s} \right)}^{ - 1}}} \right.,$

$\left. {{\Delta _\beta } = {{\left( {\int_{1/2}^1 {{{(1/2)}^{\beta - 1}}} {G_\beta }(1,s){\rm{d}}s} \right)}^{ - 1}}} \right\},$

其中，μ_α和μ_β由引理5的(ⅳ)给出.

定理1 假设f, g∈C([0, 1]×[0, ∞), [0, ∞))，若存在常数R_i>r_i>0 (i=1, 2)，使得下列不等式成立：

$\left( {{{\rm{H}}_1}} \right)f(t,v) \le {M_1}{R_1}\;\;\;\;\left( {(t,v) \in [0,1] \times \left[ {0,{R_1}} \right]} \right);$

$\left( {{{\rm{H}}_2}} \right)f(t,v) \ge {M_1}{r_1}\;\;\;\;\left( {(t,v) \in [0,1] \times \left[ {0,{r_1}} \right]} \right);$

$\left( {{{\rm{H}}_3}} \right)g(t,u) \le {M_2}{R_2}\;\;\;\;\left( {(t,u) \in [0,1] \times \left[ {0,{R_2}} \right]} \right);$

$\left( {{{\rm{H}}_4}} \right)g(t,u) \ge {N_2}{r_2}\;\;\;\;\left( {(t,u) \in [0,1] \times \left[ {0,{r_2}} \right]} \right);$

$\left( {{{\rm{H}}_5}} \right)0 \le {M_1},{M_2} \le L,{N_1},{N_2} \ge \Delta ,$

则耦合系统(1)至少有一个正解.

证明由引理6可知算子T:U→U是全连续的.令

$\begin{array}{l} {\mathit{\Omega }_R} = \{ (u(t),v(t))|(u(t),v(t)) \in X \times Y,\\ \;\;\;\;\;\;\;\;\left. {\left\| {(u(t),v(t))} \right\| < R = {R_1} + {R_2},t \in [0,1]} \right\}. \end{array}$

∀(u, v)∈U∩∂Ω_R，有‖(u, v)‖ =R. ∀t∈[0, 1]，由(H₁)、(H₅)和引理5的(ⅱ)、(ⅲ)，有

$\begin{array}{l} {T_\alpha }v(t) = \int_0^1 {{G_\alpha }} (t,s)f(s,v(s)){\rm{d}}s \le {M_1}{R_1}\int_0^1 {{G_\alpha }} (1,s){\rm{d}}s \le \\ \;\;\;\;\;{L_\alpha }{R_1}\int_0^1 {{G_\alpha }} (1,s){\rm{d}}s = {R_1}, \end{array}$

即

$\left\| {{T_\alpha }v(t)} \right\| \le {R_1}.$

(3)

由(H₃)、(H₅)和引理5的(ⅱ)、(ⅲ)，有

$\begin{array}{l} {T_\beta }u(t) = \int_0^1 {{G_\beta }} (t,s)g(s,u(s)){\rm{d}}s \le {M_2}{R_2}\int_0^1 {{G_\beta }} (1,s){\rm{d}}s \le \\ \;\;\;\;\;{L_\beta }{R_2}\int_0^1 {{G_\beta }} (1,s){\rm{d}}s = {R_2}, \end{array}$

即

$\left\| {{T_\beta }u(t)} \right\| \le {R_2}.$

(4)

由式(3)、(4)，可得

$\left\| {T(u,v)} \right\| = \left\| {{T_\alpha }v(t)} \right\| + \left\| {{T_\beta }u(t)} \right\| \le {R_1} + {R_2} = R = \left\| {(u,v)} \right\|,$

即

$\left\| {T(u,v)} \right\| \le \left\| {(u,v)} \right\|\;\;\;\;\left( {(u,v) \in U \cap \partial {\mathit{\Omega }_R}} \right).$

令Ω_r={(u(t), v(t))|(u(t), v(t))∈X×Y, ‖(u(t), v(t))‖ < r=r₁+r₂, t∈[0, 1]}. ∀(u, v)∈U∩∂Ω_r, 有‖(u, v)‖ =r. ∀t∈[0, 1]，由(H₂)、(H₅)和引理5的(ⅱ)、(ⅳ)，有

$\begin{array}{l} {T_\alpha }v(t) = \int_0^1 {{G_\alpha }} (t,s)f(s,v(s)){\rm{d}}s \ge {N_1}{r_1}\int_0^1 {{G_\alpha }} (t,s){\rm{d}}s = \\ \;\;\;\;\;\;\;\;{N_1}{r_1}\left( {\int_0^{1/2} {{G_\alpha }} (t,s){\rm{d}}s + \int_{1/2}^1 {{G_\alpha }} (t,s){\rm{d}}s} \right) \ge \\ \;\;\;\;\;\;\;\;{N_1}{r_1}\int_{1/2}^1 {{G_\alpha }} (t,s){\rm{d}}s \ge {N_1}{r_1}\int_{1/2}^1 {\mathop {\min }\limits_{t \in [1/2,1]} } {G_\alpha }(t,s){\rm{d}}s \ge \\ \;\;\;\;\;\;\;\;{\Delta _\alpha }{r_1}\int_{1/2}^1 {{{\left( {1/2} \right)}^{\alpha - 1}}{G_\alpha }\left( {1,s} \right){\rm{d}}s} = {r_1}, \end{array}$

即

$\left\| {{T_\alpha }v(t)} \right\| \ge {r_1}.$

(5)

由(H₄)、(H₅)和引理5的(ⅱ)、(ⅳ)，有

$\begin{array}{l} {T_\beta }u(t) = \int_0^1 {{G_\beta }} (t,s)g(s,u(s)){\rm{d}}s \ge {N_2}{r_2}\int_0^1 {{G_\beta }} (t,s){\rm{d}}s = \\ \;\;\;\;\;{N_2}{r_2}\left( {\int_0^{1/2} {{G_\beta }} (t,s){\rm{d}}s + \int_{1/2}^1 {{G_\beta }} (t,s){\rm{d}}s} \right) \ge \\ \;\;\;\;\;{N_2}{r_2}\int_{1/2}^1 {{G_\beta }} (t,s){\rm{d}}s \ge {N_2}{r_2}\int_{1/2}^1 {\mathop {\min }\limits_{t \in [1/2,1]} } {G_\beta }(t,s){\rm{d}}s \ge \\ \;\;\;\;\;{\Delta _\beta }{r_2}\int_{1/2}^1 {{{\left( {1/2} \right)}^{\beta - 1}}} {G_\beta }(1,s){\rm{d}}s = {r_2}, \end{array}$

即

$\left\| {{T_\beta }u(t)} \right\| \ge {r_2}.$

(6)

由式(5)、(6)，可得

$\left\| {T(u,v)} \right\| = \left\| {{T_\alpha }v(t)} \right\| + \left\| {{T_\beta }u(t)} \right\| \ge {r_1} + {r_2} = r = \left\| {(u,v)} \right\|,$

即

$\left\| {T(u,v)} \right\| \ge \left\| {(u,v)} \right\|\;\;\;\;\left( {(u,v) \in U \cap \partial {\mathit{\Omega }_{{r^ * }}}} \right).$

由引理2，算子T至少有一个不动点(u, v)∈U∩(Ω_R\Ω_r)，即耦合系统(1)至少有一个正解.证毕.

定理2 假设f, g∈C([0, 1]×[0, ∞), [0, ∞))，若存在常数a_i, b_i>0 (i=1, 2)，使得下列不等式成立：

(I₁) f(t, v)≤a₁+b₁v^ρ₁, ∀(t, v)∈[0, 1]×[0, ∞), ρ₁∈(0, 1)；

(I₂) g(t, u)≤a₂+b₂u^ρ₂, ∀(t, u)∈[0, 1]×[0, ∞), ρ₂∈(0, 1)，

则耦合系统(1)至少有一个正解.

证明取定常数r^*，使得

${r^ * } \ge \max \left\{ {4{a_1}{\mathit{\Lambda }_1},{{\left( {4{b_1}{\mathit{\Lambda }_1}} \right)}^{\frac{1}{{1 - {\rho _1}}}}},4{a_2}{\mathit{\Lambda }_2},{{\left( {4{b_2}{\mathit{\Lambda }_2}} \right)}^{\frac{1}{{1 - {\rho _2}}}}}} \right\},$

(7)

其中， $\mathit{\Lambda}_{1}=\int_{0}^{1} G_{\alpha}(1, s) \mathrm{d} s, \mathit{\Lambda}_{2}=\int_{0}^{1} G_{\beta}(1, s) \mathrm{d} s$ .定义

$\begin{array}{l} {\mathit{\Omega }_{{r^ * }}} = \left\{ {(u,v)(t)|(u,v)(t) \in X \times Y,\left\| {(u,v)(t)} \right\| \le {r^*},} \right.\\ \;\;\;\;\;\;t \in [0,1]\} , \end{array}$

(8)

则Ω_r^*是Banach空间X×Y的非空有界闭凸子集，下面证明T:Ω_r^*→Ω_r^*. ∀(u, v)∈Ω_r^*，由式(2)、(7)、(8)和(I₁)，有

$\begin{array}{l} {T_\alpha }v(t) = \int_0^1 {{G_\alpha }} (t,s)f(s,v(s)){\rm{d}}s \le \\ \;\;\;\;\;\;\int_0^1 {{G_\alpha }} (1,s)\left( {{a_1} + {b_1}{v^{{\rho _1}}}} \right){\rm{d}}s \le \left( {{a_1} + {b_1}{r^{*{\rho _1}}}} \right){\mathit{\Lambda }_1} \le \\ \;\;\;\;\;\;{r^*}/4 + {r^*}/4 = {r^*}/2, \end{array}$

即

$\left\| {{T_\alpha }v(t)} \right\| \le {r^*}/2.$

(9)

由式(2)、(7)、(8)和(I₂)，有

$\begin{array}{l} {T_\beta }u(t) = \int_0^1 {{G_\beta }} (t,s)g(s,u(s)){\rm{d}}s \le \\ \;\;\;\;\;\;\;\int_0^1 {{G_\beta }} (1,s)\left( {{a_2} + {b_2}{u^{{\rho _2}}}} \right){\rm{d}}s \le \left( {{a_2} + {b_2}{r^{*{\rho _2}}}} \right){\mathit{\Lambda }_2} \le \\ \;\;\;\;\;\;\;{r^*}/4 + {r^*}/4 = {r^*}/2, \end{array}$

即

$\left\| {{T_\beta }u(t)} \right\| \le {r^*}/2.$

(10)

由式(9)、(10)，可得

$\begin{array}{l} \left\| {T(u,v)} \right\| = \left\| {{T_\alpha }v(t)} \right\| + \left\| {{T_\beta }u(t)} \right\| \le {r^*}\\ \;\;\;\left( {\forall (u,v) \in {\mathit{\Omega }_{{r^*}}}} \right), \end{array}$

即T：Ω_r^*→Ω_r^*.

由引理6，知算子T：Ω_r^*→Ω_r^*是全连续的.由引理3，耦合系统(1)至少有一个正解.证毕.

3. 应用举例

本节给出2个例子以验证定理的有效性.

例1 考虑如下耦合系统边值问题

$\left\{ {\begin{array}{*{20}{l}} {{D^{5/2}}u(t) + \frac{{{t^2}}}{3} + \frac{t}{{2 + t}}\ln (1 + v(t)) = 0\;\;\;\;\;(0 < t < 1)}\\ {{D^{7/3}}v(t) + \frac{{{t^2}\sin t}}{4} + \frac{{\sqrt t u(t)}}{{\left( {1 + {{\rm{e}}^t}} \right)(1 + u(t))}} = 0\;\;\;\;(0 < t < 1)}\\ {u(0) = {D^{3/2}}u(0) = {D^{3/2}}u(1) = 0}\\ {v(0) = {D^{5/4}}v(0) = {D^{5/4}}v(1) = 0} \end{array}} \right.$

(11)

其中，2 < α=5/2, β=7/3≤3, 1 < γ=3/2, δ=5/4≤2，满足1+γ≤α, 1+δ≤β，而且

$f(t,v(t)) = \frac{v}{2} + \frac{{{{\sin }^2}t}}{{10}} + 1,g(t,u(t)) = \frac{{{v^2}}}{{2 + t}} + \frac{t}{3} + 1,$

易知f, g∈C([0, 1]×[0, ∞), [0, ∞)).经计算，可得

$\begin{array}{l} \int_0^1 {{G_\alpha }} (1,s){\rm{d}}s = \frac{1}{{\Gamma (\alpha )}}\int_0^1 {\left( {{{(1 - s)}^{\alpha - \gamma - 1}} - {{(1 - s)}^{\alpha - 1}}} \right)} {\rm{d}}s = \\ \frac{1}{{\Gamma (2.5)}}\left( {\int_0^1 {{{(1 - s)}^0}} {\rm{d}}s - \int_0^1 {{{(1 - s)}^{3/2}}} {\rm{d}}s} \right) \approx \frac{1}{{5.538918}}, \end{array}$

$\begin{array}{l} \int_0^1 {{G_\beta }} (1,s){\rm{d}}s = \frac{1}{{\Gamma (\beta )}}\int_0^1 {\left( {{{(1 - s)}^{\beta - \lambda - 1}} - {{(1 - s)}^{\beta - 1}}} \right)} {\rm{d}}s = \\ \frac{1}{{\Gamma (7/3)}}\left( {\int_0^1 {{{(1 - s)}^{1/12}}} {\rm{d}}s - \int_0^1 {{{(1 - s)}^{4/3}}} {\rm{d}}s} \right) \approx \frac{1}{{5.691837}}. \end{array}$

因此

$\begin{array}{l} M = \min \left\{ {{M_\alpha } = {{\left( {\int_0^1 {{G_\alpha }} (1,s){\rm{d}}s} \right)}^{ - 1}},} \right.\\ \;\;\;\;\;\;\;\;\left. {{M_\beta } = {{\left( {\int_0^1 {{G_\beta }} (1,s){\rm{d}}s} \right)}^{ - 1}}} \right\} \approx 5.538918. \end{array}$

由于

$\begin{array}{l} \int_{1/2}^1 {{{(1/2)}^{\alpha - 1}}} {G_\alpha }(1,s){\rm{d}}s = \\ \;\;\;\;\;\;\frac{{{{(1/2)}^{\alpha - 1}}}}{{\Gamma (\alpha )}}\int_{1/2}^1 {\left( {{{(1 - s)}^{\alpha - \gamma - 1}} - {{(1 - s)}^{\alpha - 1}}} \right)} {\rm{d}}s = \\ \;\;\;\;\;\;\frac{{{{(1/2)}^{3/2}}}}{{\Gamma (2.5)}}\int_{1/2}^1 {\left( {{{(1 - s)}^0} - {{(1 - s)}^{3/2}}} \right)} {\rm{d}}s \approx \frac{1}{{21.896336}}, \end{array}$

$\begin{array}{l} \int_{1/2}^1 {{{(1/2)}^{\beta - 1}}} {G_\beta }(1,s){\rm{d}}s = \\ \;\;\;\;\;\frac{{{{(1/2)}^{\beta - 1}}}}{{\Gamma (\beta )}}\int_{1/2}^1 {\left( {{{(1 - s)}^{\beta - \lambda - 1}} - {{(1 - s)}^{\beta - 1}}} \right)} {\rm{d}}s = \\ \;\;\;\;\;\frac{{{{(1/2)}^{4/3}}}}{{\Gamma (2.5)}}\int_{1/2}^1 {\left( {{{(1 - s)}^{1/12}} - {{(1 - s)}^{4/3}}} \right)} {\rm{d}}s \approx \frac{1}{{23.886016}}. \end{array}$

因此

$\begin{array}{l} N = \max \left\{ {{N_\alpha } = {{\left( {\int_{1/2}^1 {{{(1/2)}^{\alpha - 1}}} {G_\alpha }(1,s){\rm{d}}s} \right)}^{ - 1}},} \right.\\ \;\;\;\;\;\;\left. {{N_\beta } = {{\left( {\int_{1/2}^1 {{{(1/2)}^{\beta - 1}}} {G_\beta }(1,s){\rm{d}}s} \right)}^{ - 1}}} \right\} \approx 23.886016. \end{array}$

选取R₁=4, r₁=1/12，R₂=3, r₂=1/9，则有

$\begin{array}{*{20}{c}} {f(t,v(t)) = v + \frac{{{{\sin }^2}t}}{{1 + {{\rm{e}}^t}}} + 2 \le 7 \le 2{R_1} = {M_1}{R_1}}\\ {\left( {(t,v) \in [0,1] \times \left[ {0,{R_1}} \right]} \right),} \end{array}$

$\begin{array}{*{20}{c}} {f(t,v(t)) = v + \frac{{{{\sin }^2}t}}{{1 + {{\rm{e}}^t}}} + 2 \ge 2 = 24{r_1} = {N_1}{r_1}}\\ {\left( {(t,v) \in [0,1] \times \left[ {0,{r_1}} \right]} \right),} \end{array}$

$\begin{array}{*{20}{c}} {g(t,u(t)) = \frac{{{v^2}}}{{2 + t}} + \frac{t}{3} + 3 \le 11 \le 4{R_2} = {M_2}{R_2}}\\ {\left( {(t,u) \in [0,1] \times \left[ {0,{R_2}} \right]} \right),} \end{array}$

$\begin{array}{*{20}{c}} {g(t,u(t)) = \frac{{{v^2}}}{{2 + t}} + \frac{t}{3} + 3 \ge 3 \ge 25{r_2} = {N_2}{r_2}}\\ {\left( {(t,u) \in [0,1] \times \left[ {0,{r_2}} \right]} \right),} \end{array}$

且满足0 < M₁, M₂≤M，N₁, N₂≥N.定理1的条件均被满足，所以耦合系统(11)至少存在1个正解.

例2 考虑如下耦合系统边值问题

$\left\{ \begin{array}{l} ^c{D^{5/2}}u(t) + \frac{{\left( {1 + {{\rm{e}}^{ - t}}} \right)v}}{{\left( {2 + {{\cos }^2}t} \right)\left( {1 + {v^{4/7}}} \right)}} + \frac{6}{7} = 0,\\ ^c{D^{7/3}}v(t) + \frac{{\left( {1 + {{\rm{e}}^{3t}}} \right)\ln (1 + u)}}{{\left( {1 + {{\rm{e}}^{2t}}\sin t} \right)\left( {2 + {u^{1/3}}} \right)}} + \frac{{11}}{7} = 0,\\ u(0) = {D^{3/2}}u(0) = {D^{3/2}}u(1) = 0,\\ v(0) = {D^{5/4}}v(0) = {D^{5/4}}v(1) = 0, \end{array} \right.$

(12)

其中，0 < t < 1，n=2，2 < α=5/2, β=7/3 < 3，满足1+γ≤α, 1+ δ≤β，而且

$f(t,v) = \frac{{\left( {1 + {{\rm{e}}^{ - t}}} \right)v}}{{\left( {2 + {{\cos }^2}t} \right)\left( {1 + {v^{4/7}}} \right)}} + \frac{6}{7}\;\;\;\;((t,v) \in [0,1] \times [0,\infty )),$

$\begin{array}{l} g(t,u) = \frac{{\left( {1 + {{\rm{e}}^{3t}}} \right)\ln (1 + u)}}{{\left( {1 + {{\rm{e}}^{2t}}\sin t} \right)\left( {2 + {u^{1/3}}} \right)}} + \frac{{11}}{7}\;\;\;\;((t,v) \in [0,1] \times \\ \;\;\;\;\;\;\;\;\left. {[0,\infty )} \right). \end{array}$

易知f, g∈C([0, 1]×[0, ∞), [0, ∞))，且

$|f(t,v)| \le 0.683940{v^{3/7}} + 0.857143,$

$|g(t,u)| \le 21.085537{u^{2/3}} + 1.571429.$

定理2的条件均被满足，故耦合系统(12)在[0, 1]上至少有1个正解.

参考文献(15)

[1]	SUN Y P, ZHAO M. Positive solutions for a class of fractional differential equations with integral boundary conditions[J]. Applied Mathematics Letters, 2014, 34:17-21. doi: 10.1016/j.aml.2014.03.008
[2]	ZHANG X Q, WANG L, SUN Q. Existence of positive solutions for a class of nonlinear fractional differential equations with integral boundary conditions and a parameter[J]. Applied Mathematics and Computation, 2014, 226:708-718. doi: 10.1016/j.amc.2013.10.089
[3]	CHEN T Y, LIU W B, HU Z G. A boundary value pro-blem for fractional differential equation with P-Laplacian operator at resonance[J]. Nonlinear Analysis, 2012, 75(6):3210-3217. doi: 10.1016/j.na.2011.12.020
[4]	GOODRICH C S. On a fractional boundary value problem with fractional boundary conditions[J]. Applied Mathematics Letters, 2012, 25(8):1101-1105. doi: 10.1016/j.aml.2011.11.028
[5]	GRAEF J R, KONG L J. Positive solutions for a class of higher order boundary value problems with fractional Q-derivatives[J]. Applied Mathematics and Computation, 2012, 218(19):9682-9689. doi: 10.1016/j.amc.2012.03.006
[6]	SHAH K, KHAN R A. Existence and uniqueness of positive solutions to a coupled system of nonlinear fractional order differential equations with anti-periodic boundary conditions[J]. Differential Equations & Applications, 2015, 7(2):245-262.
[7]	JIANG W H. Solvability for a coupled system of fractional differential equations with integral boundary conditions at resonance[J]. Advances in Differential Equations, 2013, 324:1-13. doi: 10.1186/1687-1847-2013-324
[8]	AHMAD B, NTOUYAS S K. A fully Hadamard type integral boundary value problem of a coupled system of fractional differential equations[J]. Fractional Calculus and Applied Analysis, 2014, 17(2):348-360. http://www.wanfangdata.com.cn/details/detail.do?_type=perio&id=47f19bd98d1cc43baee8c8424dcbe7a8
[9]	SHAH K, KHALIL H, KHAN R A. Upper and lower solutions to a coupled system of nonlinear fractional differential equations[J]. Progress in Fractional Differentiation and Applications, 2016, 2(1):1-10. doi: 10.18576/pfda/020101
[10]	BAI Z B, LU H S. Positive solutions for boundary value problem of nonlinear fractional differential equation[J]. Journal of Mathematical Analysis and Applications, 2005, 311:495-505. doi: 10.1016/j.jmaa.2005.02.052
[11]	KILBAS A A, SRIVASTAVA H M, TRUJILLO J J. Theory and applications of fractional differential equations[M]. Amsterdam:Elsevier, 2006.
[12]	GUO D J, LAKSHMIKANTHAM V. Nonlinear problems in abstract cones[M]. San Diego:Academic Press, 1988.
[13]	GRANAS A, DUGUNDJI J. Fixed point theory[M]. New York:Springer, 2005.
[14]	薛益民, 戴振祥, 刘洁.一类Riemann-Liouville型分数阶微分方程正解的存在性[J].华南师范大学学报(自然科学版), 2019, 51(2):105-109. doi: 10.6054/j.jscnun.2019033 XUE Y M, DAI Z X, LIU J. On the existence of positive solutions to a type of Riemann-Liouville fractional diffe-rential equations[J]. Journal of South China Normal University(Natural Science Edition), 2019, 51(2):105-109. doi: 10.6054/j.jscnun.2019033
[15]	薛益民, 苏有慧, 刘洁, 等.一类分数阶微分方程耦合系统边值问题解的存在性[J].徐州工程学院学报(自然科学版), 2018, 33(1):41-47. XUE Y M, SU Y H, LIU J, et al. Existence of solutions of the boundary value problem to a coupled system of a certain fractional differential equations[J]. Journal of Xuzhou Institute of Technology(Natural Science Edition), 2018, 33(1):41-47.

施引文献(4)

期刊类型引用(4)

1.	葛月英，葛琦. 一类非线性混合分数阶微分方程系统解的稳定性. 延边大学学报(自然科学版). 2024(01): 1-12 . 百度学术
2.	戴振祥，薛益民，彭钟琪. 非线性耦合分数阶微分方程组正解的存在性. 徐州工程学院学报(自然科学版). 2024(03): 72-81 . 百度学术
3.	朱鹏程. Caputo型线性分数阶常微分方程的一种新的高阶数值方法. 科学技术创新. 2022(34): 35-39 . 百度学术
4.	薛益民，戴振祥，刘洁. 一类分数阶微分方程耦合系统正解的多重性. 徐州工程学院学报(自然科学版). 2020(03): 59-63 . 百度学术