让我们介绍以下假设:
-
(H1)
让\(n\in\mathbb{n}\)是一个整数,让\({\mathcal{P}}=\{J{1}:=[0,T_{1}],J{2}:=(T_{1{,T_}2],J{3}:=是区间的一个分区J型,并让\(u(t):J\右箭头(1,2]\)是关于的分段常数函数\({\mathcal{P}}\)也就是说,
$$u(t)=\sum_{\ell=1}^{n} u个_{ell}I{ell}(t)=\textstyle\begin{cases}u{1}&\text{for}t在J{1}中,\\u{2}&\text在J{2}中代表}t,\\vdots\\u{n}&\ttext{for}t}在J{n}中$$
哪里\(1<u{ell}\leq2\)是常数,并且\(I_{\ell}\)是间隔的指示器\(J_{\ell}:=(T_{\ell-1},T_{\el}]\),\(\ell=1,2,\ldot,n\)(带有\(T_{0}=0\),\(T_{n}=T\)),因此
$$I{\ell}(t)=\textstyle\begin{cases}1&\text{for}t\inJ{\ell{,\\0&\text}others}。\结束{cases}$$
-
(H2)
让\(t^{\delta}f_{1}:J\乘以X\乘以X\rightarrow X\)是连续函数\(J}}|(u(t))|中的(0\leq\delta\leq\ min_{t\).存在常量\(K,L>0)使得
$$t^{delta}\bigl\Vert f_{1}_{1} -年_{2} \垂直+L\垂直z_{1} -z(-z)_{2} X\text{和J中的}t表示所有}y_{1}、y_{2}、z_{1{、z_2}$$
备注3.1
根据的评论[30]在第20页,我们可以很容易地证明条件(H2)和不等式
$$\zeta\bigl(t^{\delta}\bigl\Vert f_{1}(t,B_{1{,B_2})\bigr\Vert\bigr)\leq K\zeta(B_{1')+L\zeta$$
对于任何有界集都是等价的\(B_{1},B_{2}\子集X\)和\(单位:J\).
此外,对于给定的集合U型函数的\(u:J\至X\),让我们表示
$$U(t)=\bigl\{U(t),U\在U\bigr\}中,\quad t\在J中$$
和
$$U(J)=\bigl\{U(t):v\在U中,t\在J\bigr\}中$$
现在让我们证明BVP解的存在性(1)通过MNCK和DFPT的概念。
对于\(\ell\在\{1,2,\ldot,n\}\中),由\(E_{ell}=C(J_{ell},X)我们表示连续函数的Banach空间\(x:J_{ell}\到x\)符合规范
$$\Vertx\Vert_{E_ell}}=\sup_{t\in J_{ell}}\bigl\Vertx(t)\bigr\Vert$$
首先,我们分析了BVP(1).
由(三)BVP方程(1)可以表示为
$$\frac{d^{2}}{dt^{2{}}\int_{0}^{t}\frac}(t-s)^{1-u(t)}}{\Gamma(2-u(t$$
(4)
拿\((H1)\)考虑到,方程式(4)在间隔中\(J_{\ell},\ell=1,2,\ldot,n\),可以写为
$$开始{对齐}和\frac{d_{2}}{dt^{2}{biggl(int_{0}^{T_{1}}\frac}(T-s)^{1-u{1}{{Gamma(2-u{1{)}x(s)_{\ell})}x(s)\,ds\biggr)\\&\quad=f_{1}\bigl(T,x(T),I^{u_{ell}_{0^{+}}x。\结束{对齐}$$
(5)
现在我们向BVP介绍解决方案(1).
定义3.1
BVP公司(1)如果有函数,则有解决方案\(x_{\ell},\ell=1,2,\ldot,n\),因此\(C([0,T_{ell}],x)中的x_{ell}\)满足等式(5)和\(x{\ell}(0)=0=x{\el}(T_{\ell{)\).
根据上述观察,BVP(1)可以表示为任何\(在J_{l}中为t),\(l=1,2,点,n),作为(5).
对于\(0\leq t\leq t_{\ell-1}\),采取\(x(t)等于0),我们可以写(5)作为
$$D^{u_{\ell}}_{T_{\ell-1}^{+}}x(T)=f_{1}\bigl(T,x(T$$
我们将处理以下BVP:
$$\textstyle\开始{cases}D^{u_{ell}}_{T_{ell-1}^{+}}x(T)=f_{1}(T,x(T,I^{u{ell}_{T_{ell-1}^{+}}x(T。\结束{cases}$$
(6)
为了我们的目的,以下引理将是(6).
引理3.1
一个函数 \(x\在E_{ell}\中) 形成解决方案(6)当且仅当 x 满足积分方程
$$开始{对齐}x(t)=&-(t_{\ell}-t_{\ell-1}})^{1-u{\ell{}}{\ell-1}^{+}}x(t_{\ell})\bigr)\\&{}+I^{u{\ell{}}_{t_{\ll-1}^}+}}f_{1}\bigl。\结束{对齐}$$
(7)
证明
让\(x\在E_{\ell}\中)成为问题的解决方案(6). 应用运算符\(I^{u_{ell}}_{T_{ell-1}^{+}}\)到的两侧(6),来自引理2.1我们发现
$$开始{对齐}x(t)=&\omega_1}(t-t_{{ell-1}})^{u{ell}-1}+\omega_2}(t-t_{ell-1{})在J{ell}中,ell}-1}f{1}\bigl(s,x(s),I^{u{\ell}}_{t_{\ell-1}^{+}}x[s)\bigr)\,ds,\quart-t\。\结束{对齐}$$
由于对功能的假设\(f{1}\)随着\(x(T_{ell-1})=0),我们得出结论\(ω{2}=0\).
让x满足\(x(T_{ell})=0).请注意
$$\omega_{1}=-(T_{\ell}-T_{\ell-1}})^{1-u_{\el}}I^{u_{ell}}_{T_{\el-1}^{+}}f_{1{\bigl(T_{\ ell},x(T_{ell}),I^{u_\ell}}●●●●$$
然后我们发现
$$开始{对齐}x(t)=&-(t_{\ell}-t_{\ell-1}})^{1-u{\ell{}}{\ell-1}^{+}}x(t_{\ell})\bigr)\\&{}+I^{u{\ell{}}_{t_{\el-1}^}+}}f_{1}\bigl。\结束{对齐}$$
反过来,让\(x\在E_{ell}\中)是积分方程的解(7),关于函数的连续性\(t^{\delta}f{1}\)和引理2.1,我们推断x是问题的解决方案(6). □
我们的第一个存在性结果基于定理2.1.
定理3.1
假设条件(H(H)1)和(H(H)2)保持并
$$\压裂{2(T_{\ell}-T_{\ell-1})^{u_{\el}-1}\ell}+1)}\biggr)<1$$
(8)
然后是问题(6)在上至少拥有一个解决方案 J型.
证明
我们构造运算符
$$W:E_{\ell}\rightarrow E_{\el}$$
如下:
$$\begin{aligned}Wx(t)=&-(t_{\ell}-t_{\ell-1}})^{1-u_{\el}}{\ell-1}^{+}}x(t_{\ell})\biger)\\&{}+\frac{1}{\Gamma(u_{\ell{)}\int_{t_{ell-1}}^{t}}J_{\ell}中的x(s)\bigr)\,ds,\quad t。\结束{对齐}$$
(9)
它源于分数积分的性质和函数的连续性\(t^{\delta}f{1}\)操作员W公司定义明确。
让
$$R{\ell}\geq\frac{2f^{\star}(T_{\ell{-T{\ell-1})}){(1-\delta)\Gamma(u{\ell})}(K+L\frac{$$
具有
$$f^{\star}=\sup_{t\in J_{\ell}}\bigl\Vert f_{1}(t,0,0)\bigr\Vert$$
我们考虑设置
$$B_{R_{\ell}}=\bigl\{x\在E_{\ell}中,\Vert x\Vert_{E_{\el}}\leq R_{ell}\bigr\}$$
显然,\(B_{R_{ell}}\)是非空的、闭合的、凸的和有界的。
现在我们演示一下W公司满足定理的假设2.1我们将分四个阶段进行证明。
步骤1:\(W(B_{R_{\ell}})\子结构.
对于\(x\在B_{R_{ell}}\中),通过(H2),我们得到:
$$\begin{aligned}\bigl\Vert Wx(t)\bigr\Vert\leq&\frac{(t_{\ell}-t_{\ell-1})^{1-u_{\el}}(t-t_{\el-1}ell}-1}\bigl\Vert f_{1}\bigle(s,x(s),I^{u{\ell}}_{t_{\ell-1}^{+}}x(s}\int_{T_{\ell-1}}^{T}^{T_{\ell}}{\Gamma(u_{\ell})}\int_{T_{\ell-1}}^{T_{ell}-s \\&{}+\frac{2}{\Gamma(u_{\ell})}\int_{T_{\ell-1}}^{T_{cell}-s)^{u_{ell}-1}\bigl\Vert f_1}(s,0,0)\bigr\Vert\,ds\\leq&\frac}2}{\Gamma(u_{\ell})}\int_{T_{\ell-1}}^{T_{cell}-s+\frac{2f^{star}(T_{ell}-T_{ell-1})-1} }{\Gamma(u_{\ell})}\int_{T_{\ell-1}}^{T_{\ ell}}s^{-\delta}\biggl(K+L\frac{(T_{\el}-T_{\el-1})^{\star}(T_{\ell}-T_{\ell-1})-T_{\ell-1}^{1-\delta})}{(1-\delta)\Gamma(u{\ell}){\biggl(K+L\frac{})^{u{\ell}}{\Gamma(u{\el})}\\leq&R{\ell},\end{aligned}$$
也就是说\(W(B_{R_{\ell}}).
步骤2:W公司是连续的。
让一个序列\((x{n})\)汇聚到x在里面\(E_{\ell}\),并让\(在J_{\ell}\中)。那么
$$\begin{aligned}&&bigl\Vert(Wx_{n})(t)-(Wx)(t{\ell}-1}\bigl\Vert f_{1}\bigl(s,x_{n}(s),I^{u_{\ell}}_{t_{\ell-1}^{+}}x_{n}(s)\bigr),I^{u_{\ell}}_{T_{\ell-1}^{+}}x(s)\bigr\Vert\,ds\\&\qquad{}+\frac{1}{\Gamma(u_{\ ell})}\int_{T_}}^{T}(T-s){\ell}}_{T_{\ell-1}^{+}}x_{n}(s)\bigr)-f_{1}\bigl(s,x(s),I^{u_{ell}_{{\ell-1}^{++}x(s)^{1-u_{ell}(T_{ell}-T_{ell-1}){T_{\ell-1}^{+}}x_{n}(s)\bigr)\\&\qquad{}-f_{1}\bigl(s,x(s),I^{u_{\ell}}_{T_{\ ell-1}^{+{}x(s}\nint_{T_{cell-1}}^{T_{ell}}(T-s)^{u_{ell}-1}\bigl\Vert f_{1}\bigr \ell-1}^{+}}x(s)\biger)\bigr\Vert\,ds\\&\quad\leq\frac{2}{\Gamma(u_{ell})}\int_{T_{ell-1}}^{T_{el}}(T_{ell}-s)^{{ell}-1}\bigl\Vert f_{1}\bigl(s,x_{n}(s),I^{u_{\ell}}_{T_{\ell-1}^{+}}x_{n}\bigr)-f_{1}\bigl(s,x(s)、I^{u_{\ ell}}_{T_{\ ell-1}^{+{}x(s,s)\bigr\Vert\,ds\\quad\leq\frac{2}\Gamma(u_{\\ell})}\nint_{T_{ell-1}}^{T_{cell}}s^{-\delta}(T_{ell}-s)^{u_{ell}-1}\bigl(K\bigl\Vertx_{n}(s)-x(s) -x(s)\biger)\|)\,ds\\&\quad\leq\frac{2K}{\Gamma(u_{ell})}\Vert x_{无}-x\Vert_{E_{ell}}\int_{T_{ell-1}}^{T_{el}s^{-\delta}(T_{ell}-s)^{u_{ell}-1}\,ds\\&\qquad{}+\frac{2L}{\Gamma(u_{el})}\bigl\VertI^{u{ell}{{ell}_{T_ell-1}^{+}}(x_{n} -x个)\bigr\Vert _{E_{\ell}}\int _{T_{\ell-1}}^{T_{\ell}s ^{-\delta}(T_{\ell}-s)^{u_{\ell}-1}\,ds\\&&quad\leq\frac{2K}{\Gamma(u_{\ell})}\Vert x_{无}-x\Vert_{E_{ell}}\int_{T_{ell-1}}^{T_{EL}}s^{-\delta}(T_{ell}-s)^{u_{ell}-1}\,ds\\&\qquad{}+\frac{2L(T_{el}-T{ell-1{)1)}\垂直x_{n} -x个\Vert_{E_{ell}}\int_{T_{ell-1}}^{T_{el}s^{-\delta}(T_{ell}-s)^{u{ell}-1}\,ds\\&\quad\leq\biggl(\frac{2K}{\Gamma{\伽玛射线(u_{\ell})\Gamma(u_}\ell}+1)}\biggr)\Vert x_{n} -x\Vert_{E_{ell}}\int_{T_{ell-1}}^{T_{el}s^{-\delta}(T_{ell}-s)^{u_{ell}-1}\,ds\\&\quad\leq\frac{}^{1-\delta})}{(1-\delta)\Gamma(u{\ell}){\biggl(2K+\frac{2L(T_{\ell{-T{\ell-1})_{n} -x个\Vert_{E_\ell}},\end{aligned}$$
也就是说,
$$\bigl\Vert(Wx_{n})-(Wx)\bigr\Vert_{E_{ell}}\rightarrow0\quad\text{as}n\rightarror\infty$$
因此,操作员W公司持续打开\(E_{\ell}\).
步骤3个:W公司有界且等度连续。
从第2步开始,我们有\(W(B_{R_{ell}})={W(x):x\在B_{R_{ell{}}}\}\子集B_{R{ell}}\中),因此,对于每个\(x\在B_{R_{ell}}\中),我们有\(W(x){E_{ell}}\leq R{ell}\),这意味着\(W(B_{R_{ell}})\)有界。还有待检查\(W(B_{R_{ell}})\)是等连续的。
对于\(J{ell}中的t_{1},t_{2}),\(t{1}<t{2})、和\(x\在B_{R_{ell}}\中),我们有:
$$\开始{aligned}&\bigl\Vert(Wx)(t_{2})-(Wx,(t_{1})\bigr\Vert\\&\quad=\biggl\Vert-\frac{(t_{ell}-t_{ell-1})^{1-u_{ell}}(t_{2} -吨_{\ell-1})^{u{\ell}-1}}{\Gamma(u{\el})}\\&\qquad{}\times\int_{T_{\ell-1}}^{T_{\ ell}}(T_{\ell}-s))\biger)\,ds\\&\qquad{}+\frac{1}{\Gamma(u{\ell})}\int_{T_{\ell-1}}^{T_{2}}(T_{2} -秒)^{u{\ell}-1}f{1}\bigl(s,x(s)_{1} -T型_{\ell-1})^{u{\ell}-1}}{\Gamma(u{\el})}\\&\qquad{}\times\int_{T_{\ell-1}}^{T_{\ ell}}(T_{\ell}-s))\大)\,ds\\&\qquad{}-\frac{1}{\Gamma(u{\ell})}\int_{T_{\ell-1}}^{T_{1}}(T_{1} -秒)^{u_{\ell}-1}f_{1}\bigl(s,x(s),I^{u_\ell}}_{T_{\ell-1}^{+}}x(s_{2} -T型_{\ell-1})^{u{\ell}-1}-(t_{1} -T型_{\ell-1})^{u_{\ell}-1}\bigr)\\\\qquad{}\times\int _{T_{\ell-1}}^{T_{\ell}}}(T_{\ell}-s)^{u_{\ell}-1}\bigl\Vert f_{1}\bigl(s,x(s),I^{u_{\ell}}_{T_{\ell-1}^{+}}x(s)\bigr \Vert,ds\\\\qquad{}+\frac{1}{\Gamma(u_{\ell})}\int _{T_{\ell-1}}}^{T_{1}}\bigl((T_{2} -秒)^{u{ell}-1}-(t_{1} -秒)^{u_{\ell}-1}\bigr)\bigl\Vert f_{1}\bigl(s,x(s),I^{u_\ell}}_{T_{\ell-1}^{+}}x(s_{2} -秒)^{u_{\ell}-1}\bigl\Vert f_{1}\bigle(s,x(s),I^{u_\ell}}_{T_{\ell-1}^{+}}x(s_{2} -T型_{\ell-1})^{u{\ell}-1}-(t_{1} -T型_{\ell-1})^{u_{\ell}-1}\bigr)\\&\qquad{}\times\int_{T_{\ell-1}}^{T_{cell}(T_{ell}-s)^}u_{ell}-1{\bigl\Vert f_{1}\bigl{1}(s,0,0)\bigr\Vert\,ds\\&\qquad{}+\frac{(T_{ell}-T_{ell-1})^{1-u_{ell}}{\Gamma(u_{el})}\bigl((T_{2} -T型_{\ell-1})^{u{\ell}-1}-(t_{1} -T型_{\ell-1})^{u_{\ell}-1}\bigr)int_{T_{\ell-1}}^{T_{ell}-s ^{T{1}}\bigl((T_{2} -秒)^{u{ell}-1}-(t_{1} -秒)^{u_{\ell}-1}\bigr)\bigl\Vert f_{1}\bigl(s,x(s),I^{u_\ell}}_{T_{\ell-1}^{+}}x(s T{1}}\bigl((T_{2} -秒)^{u{ell}-1}-(t_{1} -秒)^{u_{ell}-1}\bigr)\bigl\Vert f_{1}(s,0,0)\bigr\Vert\,ds\\&\qquad{}+\frac{1}{\Gamma(u_{el})}\int_{t_1}}^{t_2}}(t_{2} -秒)^{u_{\ell}-1}\bigl\Vert f_{1}\bigle(s,x(s),I^{u_\ell}}_{T_{\ell-1}^{+}}x(s(吨)_{2} -秒)^{u_{ell}-1}\bigl\Vert f_{1}(s,0,0)\bigr\Vert\,ds.\\&\quad\leq\frac{(T_{ell}-T_{ell-1})^{1-u{ell}}{\Gamma(u_{cell})}\bigle((T_{2} -吨_{\ell-1})^{u{\ell}-1}-(t_{1} -吨_{\ell-1})^{u_{\ell}-1}\bigr)\\&\qquad{}\times\int_{T_{\ell-1}}^{T_{\ ell}}(T_{\el}-s)^{u_\ell}-1}s^{-\delta}\bigl(K\bigl\Vert x(s)\bigr\Vert+L\bigl\ Vert I^{u_{\ell{}}}x(s)\bigr\Vert\bigr)\,ds\\&\qquad{}+\frac{f^{star}(T_{ell}-T_{ell-1})^{1-u{ell}}{\Gamma(u{ell{)}\bigl((T_{2} -T型_{\ell-1})^{u{\ell}-1}-(t_{1} -T型_{\ell-1})^{u_{\ell}-1}\biger)\int_{T_{\ell-1}}^{T_{\ ell}(T_{\el}-s)_{2} -秒)^{u{ell}-1}-(t_{1} -秒)^{u_{\ell}-1}\bigr)\bigl(K\bigl\Vert x(s)\bigr\Vert+L\bigl\Vert I^{u_{\ell}}_{T_{\ell-1}^{+}}x(s)\bigr\Vert\bigr)\,ds\\\\qquad{}+\frac{f^{\star}}{\Gamma(u_{\ell})}\int _{\ell-1}}^{T_{1}}}\bigl(T_{2} -秒)^{u{ell}-1}-(t_{1} -秒)^{u{\ell}-1}\biger)\,ds\\&\qquad{}+\frac{1}{\Gamma(u{\el})}\int_{t{1}}^{t{2}}s^{-\delta}(t_{2} -秒)^{u_{\ell}-1}\bigl(K\bigl\Vert x(s)\bigr\Vert+L\bigl\ Vert I^{u_}\ell}}_{T_{\ell-1}^{+}}x(s T吨_{2} -秒)^{u{\ell}-1}\,ds\\&\quad\leq\frac{1}{\Gamma(u{\el})}\bigl((t_{2} -T型_{\ell-1})^{u{\ell}-1}-(t_{1} -T型_{\ell-1})^{u_{\ell}-1}\bigr qquad{}+\frac{f^{star}(T_{ell}-T{ell-1})}{Gamma(u{ell}+1)}\bigl((T_{2} -T型_{\ell-1})^{u{\ell}-1}-(t_{1} -T型_{\ell-1})^{u_{\ell}-1}\bigr)\\&\qquad{}+\frac{1}{\Gamma(u_{\ ell})}\bigl(K\Vert x\Vert_{E_{\ell{}}+L\bigl\Vert I^{u_-1}}^{T{1}}s^{-\delta}\bigl((T_{2} -吨_{1} )^{u{\ell}-1}\bigr)\,ds\\&\qquad{}+\frac{f^{\star}}{\Gamma(u{\el})}\biggl(\frac}(t_{2} -T型_{\ell-1})^{u{\ell}}{u{\ ell}}-\frac{(t_{2} -吨_{1} )^{u{\ell}}{{u{\ ell}}}-\frac{(t_{1} -T型_{\ell-1})^{u{\ell}}{{u{\ ell}}}\biggr)\\&\qquad{}+\frac{(t_{2} -吨_{1} )^{u_{\ell}-1}}{\Gamma(u_{\tell})}\bigl(K\Vert x\Vert_{E_{ell}}+L\bigl\Vert I^{u{\ell{}}_{T_{\ell-1}^{+}}x\bigr\Vert_1}}\bigr裂缝{f^{\star}}{\Gamma(u{\ell})}\frac{(T_{2} -吨_{1} )_{2} -吨_{\ell-1})^{u{\ell}-1}-(t_{1} -T型_{\ell-1})^{u_{\ell}-1}\bigr)\\&\qquad{}\times\biggl(K\Vert x\Vert_{E_{ell}}+L\frac{(T_{ell}-T_{ell-1}裂缝{f^{star}(T_{ell}-T_{ell-1})}{Gamma(u{ell}+1)}\bigl((T_{2} -T型_{\ell-1})^{u{\ell}-1}-(t_{1} -T型_{\ell-1})^{u_{\ell}-1}\bigr)\\&\qquad{}+\biggl(\frac{({t_{1}}^{1-\delta}-{t_{\ell-1}}^}{1-\delta})(t_{2} -吨_{1} )^{u{\ell}-1}}{(1-\delta)\Gamma(u_{\ell})}\biggr)\biggl(K\Vert x\Vert _{E_{\ell}}+L\frac{_{\ell}+1)}\bigl((T_{2} -T型_{\ell-1})^{u{\ell}}-(t_{2} -吨_{1} )^{u_{ell}}-(t_{1} -吨_{\ell-1})^{u_{\ell}}\biger)\\&\qquad{}+\frac{({t{2}}^{1-\delta}-{t{1}}^}{1-\delta})(t_{2} -吨_{1} )^{u{\ell}-1}}{(1-\delta)\Gamma(u{\el})}\biggl(K\Vert x\Vert_{E_{ell}}+L\frac{(T_{ell}-T_{ell-1})}(T_{2} -吨_{1} )^{u{\ell}}{\Gamma(u{\el}+1)}\\&\quad\leq\biggl}}{\Gamma(u_{\ell}+1)}\biggr)\Vert x\Vert_{E_ell}}+\frac{f^{\star}\倍\bigl((t_{2} -T型_{\ell-1})^{u{\ell}-1}-(t_{1} -T型_{\ell-1})^{u_{\ell}-1}\bigr)\\&\qquad{}+\biggl(\frac{{t_{2}}^{1-\delta}-{t_{\ell-1}}^{1-\delta}}{(1-\delta)\Gamma{\Gamma(u_{\ell}+1)}\biggr)\Vert x\Vert_{E_ell}}\bigr)(t_{2} -吨_{1} )^{u{\ell}-1}\\&\qquad{}+\frac{f^{\star}}{\Gamma(u{\el}+1)}\bigl((t_{2} -T型_{\ell-1})^{u{\ell}}-(t_{1} -吨_{\ell-1})^{u{\ell}}\bigr)。\结束{对齐}$$
因此\((Wx)(t_{2})-(Wx作为\(|t_{2} -吨_{1} |\右箭头0\),这意味着\(T(B_{R_{ell}})\)是等连续的。
步骤4:W公司是一个k个-设置收缩。
对于\(B_{R_{ell}}中的U)和\(在J_{\ell}\中),我们有:
$$\begin{aligned}\zeta\bigl(W(U)(t)\bigr)=&\zeta\ bigl{\ell-1}}^{t_{\ell}}{\Gamma(u_{\ell})}\int_{T_{\ell-1}}^{T}。\结束{对齐}$$
然后备注3.1意味着,对于每个\(在J_{i}中为\),
$$开始{对齐}和\zeta\bigl(W(U)(t)\bigr)\\&\quad\leq\biggl\{\frac{(t_{\ell}-t_{\ell-1})^{1-U{\ell{}}(t-t_{\el-1}\ell}}(t_{ell}-s)^{U_{ell}-1}\biggl[K\widehat{\zeta}(U)\int_{t_{{ell-1}}^{t_{ell{}s^{-\delta}\,ds+L\frac{(t_}\ell}-T_{\ell-1})^{u_{\ell}}}{\Gamma(u_{\ell}+1)}\widehat{zeta}hat{\zeta}(u)\ int _{T_{\ell-1}}}^{t} 秒^{-\delta}\,ds\\&\qquad{}+L\frac{(T_{\ell}-T{\ell-1})^{t} 秒^{-\delta}\,ds\biggr],x\在U\biggr中^{t_{ell}}s^{-\delta}\,ds+L\frac{(t_{ell{-t_{ell-1})^{U{ell}{\Gamma(U_{ell}+1)}\widehat{\zeta}(U)\int_{t_{el-1}}}^{T_{\ell}}s ^{-\delta}\,ds\biggr]\\&\qquad{}+\frac{(T-T_{\ell-1})^{u{\ell}-1}}}{\Gamma(u{\ell})}\ int _{T_{\ell-1}}}^{T}\biggl[K\widehat{zeta}(u)\ int _{\ell-1}}}^{t} 秒^{-\delta}\,ds\\&\qquad{}+L\frac{(T_{\ell}-T{\ell-1})^{t} 秒^{-\delta}\,ds\biggr],x\在U\biggr\}\\&\quad\leq\frac{[({T_{ell}}^{1-\delta{-\delta}}^}{1-\ delta})+伽马(U{\ell})}\biggl(K+L\frac{(T_{\ell{-T_{\ell-1})^{U{\el}}{\Gamma(U{\ ell}+1)}\biggr)\widehat{\zeta}(U)\\&\quad\leq\frac{2({T_{ell}}^{1-\delta}-{T_{cell-1}}^{1-\ delta})(T_{ell{-T_{ell-1})}}{\Gamma(u_{\ell}+1)}\biggr)\widehat{\zeta}(u)。\结束{对齐}$$
因此,我们有:
$$\widehat{\zeta}(WU)\leq\frac{2 ^{u{\ell}}{\Gamma(u{\el}+1)}\biggr)\widehat{\zeta}(u)$$
因此,从(8)我们推断W公司形成一个固定的收缩。因此根据定理2.1问题(6)至少有一个解决方案\(\widetilde{x{ell}}\)在里面\(B_{R_{ell}}\).
让
$$ {x}_{\ell}=\textstyle\begin{cases}0,&t\in[0,t_{\ell-1}],\\widetilde{x}_{\ell},&t\在J_{\ellneneneep中。\结束{cases}$$
(10)
我们知道这一点\(C([0,T_{ell}],x)中的x_{ell}\)由定义(10)满足等式
$$\frac{d_{2}}{dt^{2}{biggl(int_{0}^{T_{1}}\frac{(T-s)^{1-u{1}{{Gamma(2-u{1{)}x{ell}(s)\,ds+\cdots+\int_{T_{ell-1}}^{T}\frac}{\ell})}{x_{\ell{}}(s)\,ds\biggr)=f_{1}\bigl$$
对于\(在J_{\ell}\中),这意味着\(x_{\ell}\)是的解决方案(5)带有\(x{\ell}(0)=0\)和\(x_{\ell}(T_{\ell{)=\widetilde{x}_{\ell}(T_{\ell{)=0\)。那么
$$x(t)=\textstyle\begin{cases}x_{1}(t),在J_{1{中为四t,在J_1}中为(t){x}_{2} ,在J{2}中为&t,在[0,t_{ell-1}]中为&t\{x}_{\ell},&t\在J_{\ellneneneep中,\end{cases}\displaystyle\end{cases}$$
形成BVP的解决方案(1). □