武汉数字工程研究所，湖北 武汉

收稿日期：2023年3月18日；录用日期：2023年4月19日；发布日期：2023年4月26日

摘要

本文利用打靶法研究了一类分数阶非线性微分方程边值问题

$\{\begin{array}{l}{}_{0}D{}_t^{\alpha }y\left(t\right)=f\left(t,y\left(t\right),{}_{0}D{}_t^{\theta }y\left(t\right)\right),t\in \left[0,1\right],\\ y\left(0\right)=y\left(1\right)=0\end{array}$

的可解性。其中 $f:\left[0,1\right]\times R^2\to R$ 连续， $1<\alpha \le 2$ ， $0<\theta \le \alpha -1$ ， ${}_{0}D{}_t^{\alpha }y\left(t\right)$ 表示标准的Riemann-Liouvile型导数。当 $f:\left[0,1\right]\times R^2\to R$ 在假设下，满足初始条件的解的唯一性和全局存在性时，则相应的边值问题也至少存在一个解。

关键词

分数阶微分方程，边值问题，初值问题，打靶法

Some Applications of Shooting Method in Boundary Value Problem of Fractional Differential Equation

Chen Chen

Wuhan Institute of Digital Engineering, Wuhan Hubei

Received: Mar. 18^th, 2023; accepted: Apr. 19^th, 2023; published: Apr. 26^th, 2023

ABSTRACT

In this paper, we use the shooting method to study the solvability of boundary value problem for fractional differential equation.

$\{\begin{array}{l}{}_{0}D{}_t^{\alpha }y\left(t\right)=f\left(t,y\left(t\right),{}_{0}D{}_t^{\theta }y\left(t\right)\right),t\in \left[0,1\right],\\ y\left(0\right)=y\left(1\right)=0\end{array}$

where $f:\left[0,1\right]\times R^2\to R$ is continuous, $1<\alpha \le 2$ , $0<\theta \le \alpha -1$ , ${}_{0}D{}_t^{\alpha }y\left(t\right)$ is the Riemann-Liouvile fractional derivative of $y\left(t\right)$ . If the uniqueness and global existence of solutions satisfying the initial conditions are assumed, then there is at least one solution to the corresponding boundary value problem.

Keywords:Fractional Differential Equation, Boundary Value Problems, Initial Value Problem, Shooting Method

Copyright © 2023 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

大量事实证明，分数阶微分方程模型在描述复杂的物理、生物问题时更具有实际意义 [1] [2] [3] 。因此，近年来越来越多的专家学者对分数阶微分方程问题给予讨论，见文献 [4] [5] [6] [7] [8] 。其中常用的研究方法有，锥上的不动点定理、上下解方法等见文献 [9] - [13] 。而打靶法 [14] 是微分方程数值解中的常用方法，其基本思想是将微分方程边值问题转化为初值问题进行求解，文献 [15] 运用压缩映射原理，研究了分数阶两点边值问题解的存在唯一性，并设计打靶法给出了数值解。而本文试图结合分数阶微分方程初值问题的相关理论和打靶法的基本思想，研究下面边值问题

$\{\begin{array}{l}{}_{0}D{}_t^{\alpha }y\left(t\right)=f\left(t,y\left(t\right),{}_{0}D{}_t^{\theta }y\left(t\right)\right),t\in \left[0,1\right],\\ y\left(0\right)=y\left(1\right)=0\end{array}$ (1)

的可解性。其中 $f:\left[0,1\right]\times R^2\to R$ 连续， $1<\alpha \le 2$ ， $0<\theta \le \alpha -1$ ， ${}_{0}D{}_t^{\alpha }y\left(t\right)$ 表示标准的Riemann-Liouvile型导数。文章的主要结构为：在第二部分中给出分数阶微分方程的相关概念；最后在第三部分将会给出主要定理的证明。

2. 预备知识

定义2.1 [15] 函数 $f:\left(a,\infty \right)\times R^2\to R$ 的 $\alpha$ 阶Riemann-Liouville型分数阶积分为：

${}_{0}I{}_t^{\alpha }f\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_a^t{\left(t-s\right)}^{\alpha -1}f\left(s\right)\text{d}s}$

其中 $\alpha >0$ ， $\Gamma (·)$ 为Gamma函数。

定义2.2 [15] 函数 $f:\left(a,\infty \right)\times R^2\to R$ 的 $\alpha$ 阶Riemann-Liouville型导数

${}_{a}D{}_t^{\alpha }f\left(t\right)=\frac{1}{\Gamma \left(n-\alpha \right)}{\left(\frac{\text{d}}{\text{d}t}\right)}^n{\displaystyle {\int }_a^t{\left(t-s\right)}^{n-\alpha -1}f\left(s\right)\text{d}s}$

其中 $\alpha >0$ ， $\Gamma (·)$ 为Gamma函数， $n=\left[\alpha \right]+1$ 。

引理2.1 [15] 设 $n-1<\alpha \le n$ ， $f\left(t\right)$ 在 $\left[a,b\right]$ 上的n阶导数连续，则有：

${}_{a}D{}_t^{\alpha }{}_{a}I{}_t^{\alpha }f\left(t\right)=f\left(t\right),\forall \alpha >0.$

引理2.2 [15] 令 $\alpha >0$ ， $n-1<\alpha \le n$ 。 $f\left(t\right)$ 在 $\left[a,b\right]$ 上的n阶导数连续，则有：

${}_{a}I{}_t^{\alpha }{}_{a}D{}_t^{\alpha }f\left(t\right)=f\left(t\right)-{\displaystyle \underset{j=1}{\overset{n}{\sum }}{\left[{}_{a}D{}_t^{\alpha -j}f\left(t\right)\right]}_{t=a}}\frac{{\left(t-a\right)}^{\alpha -j}}{\Gamma \left(\alpha -j+1\right)}.$

引理2.3 [4] (Banach压缩映射原理)设 $\left(X,\rho \right)$ 是完备的度量空间。 $T:X\to X$ 为压缩映射，那么映射T在X内有且只有一个不动点 $X^{\ast }$ 。

3. 主要结果及其证明

首先我们考察初值问题

$\{\begin{array}{l}{}_{0}D{}_t^{\alpha }y\left(t\right)=f\left(t,y\left(t\right),{}_{0}D{}_t^{\theta }y\left(t\right)\right),t\in \left[0,1\right],\\ {{}_{0}D{}_t^{\alpha -1}y\left(t\right)|}_{t=0}=\zeta ,{{}_{0}D{}_t^{\alpha -\text{2}}y\left(t\right)|}_{t=0}=0,\end{array}$ (2)

其中 $f:\left[0,1\right]\times R^2\to R$ 连续， $1<\alpha \le 2$ ， $0<\theta \le \alpha -1$ 。

定理3.1 (解的存在唯一性)设 $f:\left[0,1\right]\times R^2\to R$ 连续，并且满足Lipschitz条件

$\left|f\left(t,x_1,y_1\right)-f\left(t,x_2,y_2\right)\right|\le K\left|x_1-x_2\right|+L\left|y_1-y_2\right|$

则当 $L+K<\Gamma \left(\alpha -\theta +1\right)$ 时，初值问题(2)有唯一解。

证明：定义算子 $T:P\to P$ 如下：

$Ty\left(t\right)=\frac{\zeta }{\Gamma \left(\alpha \right)}{t}^{\alpha -1}+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,y\left(s\right),{}_{0}D{}_S^{\theta }y\left(s\right)\right)\text{d}s.}$

其中 $P=\left\{y\left(t\right)|y\left(t\right)\in C\left[0,1\right],{}_{0}D{}_t^{\alpha }y\left(t\right)\in C\left[0,1\right]\right\}$ ，范数 $\Vert y\left(t\right)\Vert =\underset{0\le t\le 1}{\max }\left\{K\left|y\left(t\right)\right|+L\left|{}_{0}D{}_t^{\theta }y\left(t\right)\right|\right\}$ 。由引理2.2可知，分数阶微分方程初值问题(2)的解就等价于算子T的不动点。

下面证明 $T:P\to P$ 为压缩映射。对 $\forall x\left(t\right)$ ， $y\left(t\right)\in P$ ，有

$\begin{array}{c}\left|Ty\left(t\right)-Tx\left(t\right)\right|=\frac{1}{\Gamma \left(\alpha \right)}\left|{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)\text{d}s}-{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,x\left(s\right),{}_{0}D{}_s^{\theta }x\left(s\right)\right)\text{d}s}\right|\\ \le \frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}\left|f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)-f\left(s,x\left(s\right),{}_{0}D{}_s^{\theta }x\left(s\right)\right)\right|\text{d}s}\\ \le \frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}}\left(K\left|y\left(s\right)-x\left(s\right)\right|+L\left|{}_{0}D{}_t^{\theta }y\left(s\right)-{}_{0}D{}_t^{\theta }x\left(s\right)\right|\right)\text{d}s\\ \le \frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}\text{d}s\Vert y\left(t\right)-x\left(t\right)\Vert }\\ \le \frac{1}{\Gamma \left(\alpha +1\right)}\Vert y\left(t\right)-x\left(t\right)\Vert ,\end{array}$

$\begin{array}{l}\left|{}_{0}D{}_t^{\theta }Ty\left(t\right)-{}_{0}D{}_t^{\theta }Tx\left(t\right)\right|\\ =\left|{}_{0}D{}_t^{\theta }\left\{\frac{1}{\Gamma \left(\alpha \right)}\left({\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)\text{d}s}-{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,x\left(s\right),{}_{0}D{}_s^{\theta }x\left(s\right)\right)\text{d}s}\right)\right\}\right|\\ =\left|{}_{0}D{}_t^{\theta }{}_{0}I{}_t^{\alpha }\left\{f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)-f\left(s,x\left(s\right),{}_{0}D{}_s^{\theta }x\left(s\right)\right)\right\}\right|\\ =\left|{}_{0}I{}_t^{\alpha -\theta }\left\{f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)-f\left(s,x\left(s\right),{}_{0}D{}_s^{\theta }x\left(s\right)\right)\right\}\right|\end{array}$

$\begin{array}{l}=\left|\frac{1}{\Gamma \left(\alpha -\theta \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -\theta -1}\left\{f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)-f\left(s,x\left(s\right),{}_{0}D{}_s^{\theta }x\left(s\right)\right)\right\}\text{d}s}\right|\\ \le \frac{1}{\Gamma \left(\alpha -\theta \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -\theta -1}}\left|f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)-f\left(s,x\left(s\right),{}_{0}D{}_s^{\theta }x\left(s\right)\right)\right|\text{d}s\\ \le \frac{1}{\Gamma \left(\alpha -\theta \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -\theta -1}\text{d}s\Vert y\left(t\right)-x\left(t\right)\Vert }\\ \le \frac{1}{\Gamma \left(\alpha -\theta +1\right)}\Vert y\left(t\right)-x\left(t\right)\Vert ,\end{array}$

从而

$\begin{array}{c}\Vert Ty\left(t\right)-Tx\left(t\right)\Vert =\underset{0\le t\le 1}{\max }\left\{K\left|Ty\left(t\right)-Tx\left(t\right)\right|+L\left|{}_{0}D{}_t^{\theta }Ty\left(t\right)-{}_{0}D{}_t^{\theta }Tx\left(t\right)\right|\right\}\\ \le \left(\frac{K}{\Gamma \left(\alpha +1\right)}+\frac{L}{\Gamma \left(\alpha -\theta +1\right)}\right)\Vert y\left(t\right)-x\left(t\right)\Vert \le \frac{K+L}{\Gamma (\alpha -\theta +1)}\Vert y\left(t\right)-x\left(t\right)\Vert .\end{array}$

因此当 $K+L\le \Gamma \left(\alpha -\theta +1\right)$ 时 $\frac{K+L}{\Gamma \left(\alpha -\theta +1\right)}<1$ 。即T为压缩映射。从而由引理2.3可知，算子T存在唯一不动点 $y^{\ast }\in P$ ，此不动点即为分数阶微分方程初值问题(2)的局部唯一解。下面将证这个唯一解可延拓到整个 $\left[0,1\right]$ 区间。

定理3.2 (解的延展性) 设定理3.1的条件成立，且存在常数 $M>0$ ，使得

$\left|f\left(t,x,y\right)\right|<M,\forall \left(t,x,y\right)\in \left[0,1\right]\times R^2.$

则初值问题(2)的解的极大存在区间为 $\left[0,1\right]$ 。

证明 不妨设 $y\left(t\right)$ 的极大存在区间是 $I=\left[0,T\right]$ ， $T<1$ ，由定理3.1可知：

$y\left(t\right)=\frac{\zeta }{\Gamma \left(\alpha \right)}{t}^{\alpha -1}+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,y\left(s\right),{}_{0}D{}_S^{\theta }y\left(s\right)\right)\text{d}s.}$

则对 $\forall \epsilon$ ， $\exists \delta =\frac{\epsilon }{M}$ ，对 $\forall t_1,t_2\in \left(T-\delta ,T\right)$ ，令 $t_1<t_2$

$\begin{array}{c}{}_{0}D{}_t^{\alpha -1}y\left(t\right)={}_{0}D{}_t^{\alpha -1}\left\{\frac{\zeta }{\Gamma \left(\alpha \right)}{t}^{\alpha -1}+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,y\left(s\right),{}_{0}D{}_S^{\theta }y\left(s\right)\right)\text{d}s}\right\}\\ =\frac{\zeta }{\Gamma \left(\alpha \right)}{}_{0}D{}_t^{\alpha -1}{t}^{\alpha -1}+{}_{0}D{}_t^{\alpha -1}{}_{0}I{}_t^{\alpha }f\left(t,y\left(t\right),{}_{0}D{}_t^{\theta }y\left(t\right)\right)\\ =\frac{\zeta }{\Gamma \left(\alpha +\theta \right)}+{}_{0}I{}_t^{1}f\left(t,y\left(t\right),{}_{0}D{}_t^{\theta }y\left(t\right)\right)\\ =\frac{\zeta }{\Gamma \left(\alpha +\theta \right)}+{\displaystyle {\int }_0^{t}f\left(s,y\left(s\right),{}_{0}D{}_S^{\theta }y\left(s\right)\right)\text{d}s},\end{array}$

$\begin{array}{c}\left|{}_{0}D{}_t^{\alpha -1}y\left(t_2\right)-{}_{0}D{}_t^{\alpha -1}y\left(t_1\right)\right|=\left|{\displaystyle {\int }_0^{t_2}f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)\text{d}s}-{\displaystyle {\int }_0^{t_1}f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)\text{d}s}\right|\\ =\left|{\displaystyle {\int }_{t_1}^{t_2}f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)\text{d}s}\right|\le {\displaystyle {\int }_{t_1}^{t_2}\left|f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)\right|\text{d}s}\\ \le M\left|t_2-t_1\right|\le \epsilon .\end{array}$

因此 $\underset{t\to T^{-}}{\lim }{}_{0}D{}_t^{\alpha -1}y\left(t\right)$ 存在，记为 ${\xi }^{·}$ 。

再由定理3.1可知，初值问题

$\{\begin{array}{l}{}_{0}D{}_t^{\alpha }x\left(t\right)=f\left(t,x\left(t\right),{}_{0}D{}_t^{\theta }x\left(t\right)\right),t\in \left[0,1\right],\\ {{}_{0}D{}_t^{\alpha -1}x\left(t\right)|}_{t=T}={\xi }^{·},{{}_{0}D{}_t^{\alpha -\text{2}}y\left(t\right)|}_{t=T}=0\end{array}$

在 $\left[T,T+h\right]$ 上有唯一解，则初值问题(2)的解 $y\left(t\right)$ 可以延展到 $\left[0,T+h\right]$ 上，这与 $y\left(t\right)$ 的极大存在区间为 $\left[0,T\right]$ 矛盾。故 $y\left(t\right)$ 的极大存在区间是 $\left[0,1\right]$ 。

定理3.3 (解对初值的连续依赖性)设定理3.1的条件成立， $y\left(t,\zeta \right)$ 是初值问题(2)的解，它在区间 $\left[0,1\right]$ 上有定义，那么对 $\forall \epsilon >0,\exists \delta$ ，使得当

$\left|\zeta -{\zeta }_1\right|<\delta$

时，问题(2)满足 ${{}_{0}D{}_t^{\alpha -1}|}_{t=0}={\zeta }_1$ 的解 $y\left(t,{\zeta }_1\right)$ 在区间 $\left[0,1\right]$ 上有也定义，并且

$\Vert y\left(t,\zeta \right)-y\left(t,{\zeta }_1\right)\Vert <\epsilon ,0\le t\le 1.$

证明 对 $\forall \epsilon >0$ ， $\exists \delta =\frac{\epsilon }{G}$ ，其中 $G=\frac{\frac{K}{\Gamma \left(\alpha \right)}+\frac{L}{\Gamma \left(\alpha +\theta \right)}}{1-\frac{K}{\Gamma \left(\alpha +1\right)}-\frac{L}{\Gamma \left(\alpha -\theta +1\right)}}$ ，则当 $\left|\zeta -{\zeta }_1\right|<\delta$ 时有：

$\begin{array}{l}\left|y\left(t,\zeta \right)-y\left(t,{\zeta }_1\right)\right|\\ =\left|\frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)}\left(\zeta -{\zeta }_1\right)+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}\left\{f\left(s,y\left(s,\zeta \right),{}_{0}D{}_s^{\theta }y\left(s,\zeta \right)\right)-f\left(s,y\left(s,{\zeta }_1\right),{}_{0}D{}_s^{\theta }y\left(s,{\zeta }_1\right)\right)\right\}\text{d}s}\right|\\ \le \frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)}\left|\zeta -{\zeta }_1\right|+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}\left|f\left(s,y\left(s,\zeta \right),{}_{0}D{}_s^{\theta }y\left(s,\zeta \right)\right)-f\left(s,y\left(s,{\zeta }_1\right),{}_{0}D{}_s^{\theta }y\left(s,{\zeta }_1\right)\right)\right|\text{d}s}\\ \le \frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)}\left|\zeta -{\zeta }_1\right|+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}\left(K\left|y\left(s,\zeta \right)-y\left(s,{\zeta }_1\right)\right|+L\left|{}_{0}D{}_s^{\theta }y\left(s,\zeta \right)-{}_{0}D{}_s^{\theta }y\left(s,{\zeta }_1\right)\right|\right)\text{d}s}\\ \le \frac{1}{\Gamma \left(\alpha \right)}\left|\zeta -{\zeta }_1\right|+\frac{1}{\Gamma \left(\alpha +1\right)}\Vert y\left(t,\zeta \right)-y\left(t,{\zeta }_1\right)\Vert ,\end{array}$

$\begin{array}{l}\left|{}_{0}D{}_t^{\theta }y\left(t,\zeta \right)-{}_{0}D{}_t^{\theta }y\left(t,{\zeta }_1\right)\right|\\ =\left|\frac{\zeta -{\zeta }_1}{\Gamma \left(\alpha \right)}{}_{0}D{}_t^{\theta }{t}^{\alpha -1}+{}_{0}I{}_t^{\alpha -\theta }\left\{f\left(s,y\left(s,\zeta \right),{}_{0}D{}_s^{\theta }y\left(s,\zeta \right)\right)-f\left(s,y\left(s,{\zeta }_1\right),{}_{0}D{}_s^{\theta }y\left(s,{\zeta }_1\right)\right)\right\}\right|\\ \le \frac{\left|\zeta -{\zeta }_1\right|}{\Gamma \left(\alpha +\theta \right)}{t}^{\alpha -\theta -1}+\frac{1}{\Gamma \left(\alpha -\theta \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -\theta -1}\left|f\left(s,y\left(s,\zeta \right),{}_{0}D{}_s^{\theta }y\left(s,\zeta \right)\right)-f\left(s,y\left(s,{\zeta }_1\right),{}_{0}D{}_s^{\theta }y\left(s,{\zeta }_1\right)\right)\right|\text{d}s}\\ \le \frac{\left|\zeta -{\zeta }_1\right|}{\Gamma \left(\alpha +\theta \right)}{t}^{\alpha -\theta -1}+\frac{1}{\Gamma \left(\alpha -\theta \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -\theta -1}\left(K\left|y\left(s,\zeta \right)-y\left(s,{\zeta }_1\right)\right|+L\left|{}_{0}D{}_s^{\theta }y\left(s,\zeta \right)-{}_{0}D{}_s^{\theta }y\left(s,{\zeta }_1\right)\right|\right)\text{d}s}\\ \le \frac{\left|\zeta -{\zeta }_1\right|}{\Gamma \left(\alpha +\theta \right)}+\frac{1}{\Gamma \left(\alpha -\theta +1\right)}\Vert y\left(t,\zeta \right)-y\left(t,{\zeta }_1\right)\Vert .\end{array}$

从而

$\begin{array}{c}\Vert y\left(t,\zeta \right)-y\left(t,{\zeta }_1\right)\Vert =\underset{0\le t\le 1}{\max }\left\{K\left|y\left(t,\zeta \right)-y\left(t,{\zeta }_1\right)\right|+L\left|{}_{0}D{}_t^{\theta }y\left(t,\zeta \right)-{}_{0}D{}_t^{\theta }y\left(t,{\zeta }_1\right)\right|\right\}\\ \le \frac{K\left|\zeta -{\zeta }_1\right|}{\Gamma \left(\alpha \right)}+\frac{L\left|\zeta -{\zeta }_1\right|}{\Gamma \left(\alpha +\theta \right)}+\left(\frac{K}{\Gamma \left(\alpha +1\right)}+\frac{L}{\Gamma \left(\alpha +\theta \right)}\right)\Vert y\left(t,\zeta \right)-y\left(t,{\zeta }_1\right)\Vert \\ \le \frac{\frac{K}{\Gamma \left(\alpha \right)}+\frac{L}{\Gamma \left(\alpha +\theta \right)}}{1-\frac{K}{\Gamma \left(\alpha +1\right)}-\frac{L}{\Gamma \left(\alpha -\theta +1\right)}}\left|\zeta -{\zeta }_1\right|\\ \le \epsilon .\end{array}$

成立，所以解 $y\left(t\right)$ 关于初值 $\zeta$ 连续。

定理3.4 设定理3.1-3.2的条件成立，则边值问题(1)在P中至少存在一个解。

证明 记初值问题(1)的解为 $y\left(t,\zeta \right)$ ，根据定理3.2可知， $y\left(t,\zeta \right)$ 在 $\left[0,1\right]$ 上唯一存在。并有

$\begin{array}{c}y\left(t,\zeta \right)=\frac{\zeta }{\Gamma \left(\alpha \right)}{t}^{\alpha -1}+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)\text{d}s}\\ \ge \frac{\zeta }{\Gamma \left(\alpha \right)}{t}^{\alpha -1}-\frac{M}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}\text{d}s}\\ =\frac{\zeta }{\Gamma \left(\alpha \right)}{t}^{\alpha -1}-\frac{M}{\Gamma \left(\alpha +1\right)}{t}^{\alpha }\end{array}$

特别的当 $t=1$ 时有： $y\left(1,\zeta \right)\ge \frac{\zeta }{\Gamma \left(\alpha \right)}-\frac{M}{\Gamma \left(\alpha +1\right)}$ 。对充分大的 $\zeta ={\zeta }_1$ ，可以使得 $y\left(1,{\zeta }_1\right)>0$ 。

另一方面，

$\begin{array}{c}y\left(t,\zeta \right)=\frac{\zeta }{\Gamma \left(\alpha \right)}{t}^{\alpha -1}+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,y\left(s\right),{}_{0}D{}_s^{\theta }y\left(s\right)\right)\text{d}s}\\ \le \frac{\zeta }{\Gamma \left(\alpha \right)}{t}^{\alpha -1}+\frac{M}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}\text{d}s}\\ =\frac{\zeta }{\Gamma \left(\alpha \right)}{t}^{\alpha -1}+\frac{M}{\Gamma \left(\alpha +1\right)}{t}^{\alpha }\end{array}$

特别的当 $t=1$ 时有： $y\left(1,\zeta \right)\le \frac{\zeta }{\Gamma \left(\alpha \right)}+\frac{M}{\Gamma \left(\alpha +1\right)}$ 。对绝对值充分大的 $\zeta ={\zeta }_2$ ，可以使得 $y\left(1,{\zeta }_2\right)<0$ 。

由定理3.3可知， $y\left(t,\zeta \right)$ 是 $\zeta$ 的一个连续函数，所以至少存在一个 ${\zeta }^{\ast }$ 使 $y\left(t,{\zeta }^{\ast }\right)=0$ ，此时初值问题(1)的解 $y\left(t,{\zeta }^{\ast }\right)$ 也就是所求边值问题(1)的解。

用打靶法解分数阶两点边值问题(1)的具体计算过程如下：

步骤1：选取初值 ${\zeta }^0$ 和迭代精度 $\epsilon$ 。

步骤2：通过解分数阶初值问题(2)，其中 ${{}_{0}D{}_t^{\alpha -1}y\left(t\right)|}_{t=0}={\zeta }^k,\left(k=0,1,2,\cdots \right)$ ，得到 $y\left(t,{\zeta }^k\right)$ ，若 $y\left(b,{\zeta }^k\right)=0$ ，则 $y\left(t,{\zeta }^k\right)\ne 0$ 就是问题(1)的解。一般情况下 $y\left(b,{\zeta }^k\right)\ne 0$ ，则记 $\varphi \left(\zeta \right):=y\left(b,\zeta \right)$ ，从初值逼近 ${\zeta }^0$ 开始，得到 ${\zeta }^k$ 如下

${\zeta }^{k\text{+}1}={\zeta }^k-\frac{\varphi \left({\zeta }^k\right)}{\Delta \varphi \left({\zeta }^k\right)},k=0,1,2,\cdots .$

步骤3：重复第二步，直至 $\left|{\zeta }^{k+1}-{\zeta }^k\right|<\epsilon$ 。并将最后的 ${\zeta }^{k+1}$ 记作 ${\zeta }^{*}$ 。

步骤4：最后通过解分数阶初值问题(2)，其中 ${{}_{0}D{}_t^{\alpha -1}y\left(t\right)|}_{t=0}={\zeta }^{*},\left(k=0,1,2,\cdots \right)$ ，得到 $y\left(t,{\zeta }^{*}\right)$ 。即为边值问题(1)的解。

文章引用

陈 辰. 打靶法在分数阶微分方程边值问题中的一些应用
Some Applications of Shooting Method in Boundary Value Problem of Fractional Differential Equation[J]. 理论数学, 2023, 13(04): 895-901. https://doi.org/10.12677/PM.2023.134095

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