让\(E=C[0,1]\)显然,空间电子是一个Banach空间,如果它被赋予如下规范:
$$\Vert x\Vert=\max_{t\in[0,1]}\bigl\Vert x(t)\bigr\Vert$$
定义运算符\(T:E\右箭头E\)作为
$$开始{对齐}(Tx)(t)=&\frac{1}{\Gamma(\alpha+\beta)}\int_{0}^{t}(t-s)^{\alpha+/\beta-1}f\bigl(s,x(s)\bigr)\,ds-\frac}t^{\alpha}{\alpha+\beta-1}f\bigl(s,x(s)\bigr)\,ds+\frac{\lambda t^{\alba}(2t-\alpha-1)}{(1-\alpha)\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}x(s)\,ds\\&{}-\frac{t^{\alpha}(1-t)}{(1-\alpha)\Gamma(\alpha+1)\Garma(\beta)}\int_{0}^{1}}\int_{0}^{1} x个(s) \,ds\\&{}+\frac{\t^{alpha}(2t-\alpha-1)}{1-\alpha}\int_{0}^{1} x个(s) \,ds-\frac{\lambda}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}x(s)\,ds,\quad t\in[0,1]。\结束{对齐}$$
(3)
很容易看出(1)等于T型.
引理3.1
-
(i)
\(在[0,1]}|t^{alpha}(2t-\alpha-1)|=max\{(\frac{\alpha}{2})^{\alfa},1-\alpha\});
-
(ii)
\([0,1]}t^{alpha}(1-t)=\frac{alpha^{alpha}}{(1+alpha)^{1+\alpha}}).
证明
(i) 让\(g(t)=t^{α}(2t-α-1),\(0\leq t\leq 1\),然后\(g'(t)=(\alpha+1)t^{\alpha-1}(2t-\alpha),\(g(0)=0),\(g(1)=1-α>0).
什么时候?\(0\leq t<\frac{\alpha}{2}\),\(g’(t)\leq 0); 什么时候\(压裂{\alpha}{2}<t\leq 1\),\(g’(t)\geq 0)总之,\(|g(t)|{max}=\max\{|g(\frac{alpha}{2})|,g(1).
(ii)出租\(g(t)=t^{α}(1-t)),\(0\leq t\leq 1\),然后\(g'(t)=t^{\alpha-1}[\alpha-(\alpha+1)t]\).
什么时候?\(0\leq t<\frac{\alpha}{1+\alpha{),\(g’(t)\geq 0); 什么时候\(压裂{\alpha}{1+\alpha{<t\leq 1\),\(g’(t)\leq 0).所以\(g(t)_{\max}=g(\frac{\alpha}{1+\alpha})=\frac{\alpha ^{\alpha}}{(1+\alpha)^{1+\alpha}}\).
证明已完成。□
让\(eta=\frac{1}{1-\alpha}\max\{(\frac}\alpha{2})^{alpha},1-\alfa\}\).
引理3.2
\(T:E\右箭头E\)
是完全连续的.
证明
由于(f),\(T:E\右箭头E\)是连续的。对于任何有界集\(D\子集E\),存在\(K>0\)这样的话\(对于D\中的所有x\),\(\|x\|\leq K\).存在一个常量\(L_{1}>0\)这样的话\(|f(t,x)|\leq L_{1}\)对于任何\(在[0,1]\中)和\(x\英寸[-K,K]\).然后\(对于D\中的所有x\),因此
$$开始{aligned}\bigl\vert(Tx)(t)\bigr\vert\leq&\frac{1}{\Gamma(\alpha+\beta)}\int_{0}^{t}(t-s)^{\alpha+\beta-1}\bigle\vertf\bigl ^{\alpha+\beta-1}\bigl\vert f\bigl(s,x(s)\biger)\bigr\vert\,ds\\&{}+\frac{\lambda\eta}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}\bigl\vert x vert\,ds\\&{}+\frac{\lambda\mu\alpha^{\alpha}}{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1)}\int_{0}^{1}\bigl\vert x(s)\bigr\vert \,ds+\mu\eta\int_{0}^{1}\bigl\ vert x int_{0}^{t}(t-s)^{\alpha+\beta-1}\,ds+\frac{L_{1}\eta}{\Gamma(\alpha+/beta)}\int_{0}^{1}(1-s)^{\alpha+\beta-1}\,ds\\&{}+\frac{\eta\lambda\Vertx\Vert}{\Gamma(\alpha ^{1}(1-s)^{\beta-1}\,ds\\&{}+\frac{\lambda\mu\alpha^{\alpha}\Vertx\Vert}{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1)}\\&{}+\eta\mu\Vert x\Vert+\frac{\lambda\Vert x \Vert}{\Garma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\,ds\\=&\frac{L_{1} t吨^{\alpha+\beta}}{\Gamma(\alpha+/\beta+1 \beta+1)}\\&{}+\frac{\lambda\mu\alpha^{\alpha}\Vert x\Vert}{(1+\alpha)^{1+\alpha}(1-\alpha)\伽马(\alpha+1)}+\eta\mu\Vert x\Vert+\frac{\lambda\Vert x \Vert t^{\alpha}}{\Gamma(\ alpha+1{1}}{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1)\Garma(\beta+1)}\\&{}+\frac{\lambda\mu\alpha^{\alpha}\Vert x\Vert}{(1+\alpha)^{1+\alpha}^{\alpha}L_{1}}{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1)\Gamma(\beta+1)}\\&{}+\frac{\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\lang1033\$$
这意味着技术总监一致有界。
此外,对于\(D\中的x\),\(0\leq t_{1}<t_{2}\leq 1\),我们有
$$开始{对齐}和\bigl\vert(Tx)(t_{2})-(Tx_{2} -秒)^{\alpha+\beta-1}f\bigl(s,x(s)\bigr)\,ds-\frac{t{2}^{\alfa}(2t_{2}-\α-1)}{(1-\alpha)\Gamma(\alpha+\beta)}\\&\qquad{}\times\int_{0}^{1}(1-s)^{\alpha+/\beta-1}f\bigl(s,x(s)\bigr)\,ds+\frac{\lambda t_{2}^{\alpha}(2t_{2}-\α-1)}{(1-\α)\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}x)\biger)\,ds+\frac{\lambda\mu t_{2}^{\alpha}(1-t{2})}{(1-\alpha)\Gamma(\alpha+1)}\int_{0}^{1} x个(s) \,ds\\&\qquad{}+\frac{\mu t_{2}^{\alpha}(2t_{2}-\α-1)}{1-\alpha}\int_{0}^{1} x个(s) \,ds-\frac{\lambda}{\Gamma(\alpha)}\int_{0}^{t{2}}(t_{2} -秒)^{\alpha-1}x(s)\,ds\\&\qquad{}-\frac{1}{\Gamma(\alpha+\beta)}\int_{0}^{t_{1}}(t_{1} -秒)^{\alpha+\beta-1}f\bigl(s,x(s)\bigr)\,ds\\&\qquad{}+\frac{t{1}^{\alfa}(2t_{1}-\α-1)}{(1-\alpha)\Gamma(\alpha+\beta)}\int_{0}^{1}(1-s)^{\alpha+/\beta-1}f\bigl(s,x(s)\bigr)\,ds\\&\qquad{}-\frac{\lambda t_{1}^{\alpha}(2t_{1}-\α-1)}{(1-\α)\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}x)\biger)\,ds\\&\qquad{}-\frac{\lambda\mu t_{1}^{\alpha}(1-t{1})}{(1-\alpha)\Gamma(\alpha+1)}\int_{0}^{1} x个(s) \,ds\\&\qquad{}-\frac{\mut_{1}^{\alpha}(2t_{1}-\alpha-1)}{1-\alpha}\int_{0}^{1} x个(s) \,ds+\frac{\lambda}{\Gamma(\alpha)}\int_{0}^{t_{1}}(t_{1} -秒)^{\alpha-1}x(s)\,ds\\&\quad=\biggl\vert\int_{0}^{t_{1}}\frac{f(s,x(s_{2} -秒)^{\α+\β-1}-(t_{1} -秒)^{\alpha+\beta-1}\bigr]\,ds+\int_{t{1}}^{t{2}}\frac{f(s,x(s))}{\Gamma(\alpha+/beta)}(t_{2} -秒)^{\alpha+\beta-1}\,ds\\&\qquad{}+\frac{(\alpha+1)(t{2}^{\alalpha}-t{1}^{\ alpha})+2(t{1}^{\alpha+1}-t{2}^}\alpha+1})}{(1-\alpha)\Gamma(s)\大)\,ds\\&\qquad{}+\frac{\lambda(\alpha+1))}{(1-\α)\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}x(s)\,ds\\&\qquad{}+\frac{_{0}^{1}(1-s)^{\beta-1}f\bigl(s,x(s)\bigr,ds\\&\qquad{}+\frac{\lambda\mu^{\阿尔法}-t{1}^{\alpha})}{(1-\alpha)\Gamma(\alpha+1)}\int_{0}^{1} x个(s) \,ds\\&\qquad{}+\frac{2\mu(t_{2}^{\alpha+1}^{\ alpha+1{)+\mu(\alpha+1)(t_{1}^{\alpha}-t{2}^{\alpha})}{1-\ alpha}\int_{0}^{1} x个(s) \,ds\\&\qquad{}+\int_{0}^{t_{1}}\frac{\lambda x(s)}{\Gamma(\alpha)}\bigl[(t_{1} -秒)^{\α-1}-(t_{2} -秒)^{\alpha-1}\bigr]\,ds-\int_{t{1}}^{t{2}}\frac{\lambdax(s)}{\Gamma(\alpha)}(t_{2} -秒)^{α-1},ds\biggr\vert\&\quad\leq\frac{L_{1}\vert t_{2}^{\alpha+\beta}-t_{1{^{\alpha+\beta}\vert}{\Gamma(\alpha+/beta+1)}+\frac}L_1}\verts(\alba+1)-t{2}^{\alpha+1})\vert}{(1-\alpha)\Gamma(\alpha+\beta+1)}\\&\qquad{}+\frac{\vert\lambda(\alfa+1)(t_{1}^{\alpha}-t{2}^{\alba})+2\lambda(t_{2}^{\alpha+1}-t_{1}^{\alpha+1})\vert\Vertx\vert}{(1-\alpha)\Gamma(\alpha+1)}\\&\qquad{}+\frac{L_1}\vert}-t{2}^{\alpha})\vert}{(1-\alpha)\Gamma(\alpha+1)\Garma(\beta+1)}+\frac{\vert\lambda\mu(t_{1}^{\ alpha+1}-t{2}^{α+1})+\lambda\mu ^{\alpha})\vert\vert x\vert}{1-\alpha{+\frac{\lambda\vert x\ vert\vert t_{1}^{\alpha}-t_{2}^{\ alpha}\vert}{\Gamma(\alpha+1)}\\&\quad\leq\frac{L_{1}}{\Gamma(\ alpha+\beta+1)}\bigl \Gamma(\alpha+1)}\&\qquad{}+\frac{L_{1}}{(1-\alpha)\Gamma(\alfa+1)\Gamma(\beta+1)}+\frac{\lambda\mu K}{(1-\alpha)\Gamma(\alpha+1)}+\frac{2\mu K}{1-\alpha}\biggr]\bigl{(1-\alpha)\Gamma(\alpha+1)}+\frac{L_{1}}{(1-\ alpha+\frac{\lambda\mu K}{(1-\alpha)\Gamma(\alpha+1)}+\frac{\mu K(\alfa+1)}{1-\alpha}+\frac{\lambda K}{\Gamma(\alba+1)}\biggr]\bigl(t_{2}^{\alpha{-t_{1}^{\ alpha}\bigr),\end对齐}$$
这意味着技术总监是等连续的。因此,通过阿尔泽拉–阿斯科利定理,\(T:E\右箭头E\)是完全连续的。
证明已完成。□
定理3.1
假设
(f)
满足Lipschitz条件
$$\bigl\vert f(t,x)-f(t,y)\bigr\vert\leq L\vert x-y\vert,\quad\对于[0,1]中的所有t,R中的x,y\$$
和
$$开始{对齐}A=&\frac{(1+\eta)L}{\Gamma(\alpha+\beta+1)}+\frac}(1+/eta)\lambda}{\Gamma(\ alpha+1)}+\frac:\alpha^{\alpha}L}{{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1)}+\eta\mu<1。\结束{对齐}$$
然后(1)有独特的解决方案.
证明
定义\(Q=\max_{t\in[0,1]}|f(t,0)|\)并选择\(r\geq\frac{\frac{2Q}{\Gamma(\alpha+\beta+1)}+\frac}\alpha^{\alpha}Q}{(1+\alpha)^{1+\alpha}(1-\ alpha,将闭合球定义为\(B_{r}=\{x\在E:\|x\|\leq r \}\中),然后针对\(x\在B_{r}\中),我们有
$$开始{aligned}\bigl\vert(Tx)(t)\bigr\vert\leq&\frac{1}{\Gamma(\alpha+\beta)}\int_{0}^{t}(t-s)^{\alpha+\beta-1}\bigle\vertf\bigl ^{\alpha+\beta-1}\bigl\vert f\bigl(s,x(s)\biger)\bigr\vert\,ds\\&{}+\frac{\lambda\eta}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}\bigl\vert x vert\,ds\\&{}+\frac{\lambda\mu\alpha^{\alpha}}{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1)}\int_{0}^{1}\bigl\vert x(s)\bigr\vert \,ds+\mu\eta\int_{0}^{1}\bigl\ vert x 0}^{t}(t-s)^{\alpha+\beta-1}\bigl\bigr\vert\bigr)\,ds\\&{}+\frac{\eta}{\Gamma(\alpha+\beta)}\int_{0}^{1}(1-s)^{\alpha+/beta-1}\bigl Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}\bigl\vert x(s)\bigr\vert,ds+\frac{\alfa^{\alpha}}{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1)\Garma(\beta)}\\&{}\times\int_{0}^{1}(1s)^{\beta-1}\bigl{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1)}\int_{0}^{1}\bigl\vert x(s)\bigr\vert,ds+\mu\eta\int_{0}^{1}\bigl\vert x(s)\bigr\vert c{\eta(Lr+Q)}{\Gamma(\alpha+\beta+1)}+\frac{\lambdar\eta}{\Gamma(\ alpha+1)}\\&{}+\frac{\alpha^{\ alpha}(Lr+Q)}{(1+\alpha)^{1+\alpha}(alpha+1)}\\leq&\frac{(1+\eta)(Lr+Q)}{\Gamma(alpha+\beta+1)}+\frac}(1+/eta)\lambda r}{\Gamma(\alpha+1)}+\frac{\alpha^{\ alpha}(Lr+Q)}{(1+\alpha)^{1+\alpha}}$$
这意味着\(\ | Tx \ | \leq r \)也就是说,\(T(B_{r})\子集B_{r}\)在以下内容中\(E中的x,y\),每个\(在[0,1]\中),我们可以做到
$$开始{对齐}和\bigl\vert(Tx)(t)-(Ty)(t}{\Gamma(\alpha+\beta)}\int_{0}^{1}(1-s)^{\alpha+/beta-1}\bigl\vert f\bigl(s,x(s)\bigr)-f\bigl(s,y(s)\bigr)\bigr\vert\,ds\\&\qquad{}+\frac{\lambda\eta}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}\bigl\vert x(s)-y)}\int_{0}^{1}(1-s)^{\beta-1}\bigl\vert f\bigl,ds\\&\qquad{}+\frac{\lambda\mu\alpha^{\alpha}}{(1+\alpha)^{1+\alpha}{\lambda}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\bigl\vert x(s)-y(s)\bigr\vert,ds\\&&quad\leq\frac{(1+\eta)L\Vert x-y\Vert}{\Gamma(\alpha+\beta+1)}+\frac{(1+\eta)\lambda \Vert x-y\Vert}{\Gamma(\alpha+1)}+\frac{\alpha^{\alpha}L\Vert x-y\Vert}{(1+\alpha)^{1+\alpha}\mu\alpha^{\alpha}\Vertx-y\Vert}{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1)}+\eta\mu\Vert x-y\Vert=A\Vert x-y\Vert,\end{aligned}$$
这意味着T型是一种收缩。因此,根据巴拿赫不动点定理[30],T型具有唯一的不动点(1)有一个独特的解决方案。
证明已完成。□
定理3.2
假设存在
\(M>0)
和
\(抄送0)
具有
\(\frac{(1+\eta)c}{\Gamma(\alpha+\beta+1)}+\frac}(1+/eta)\lambda}{\Gamma(\ alpha+1)}+\frac[\alpha^{\alpha}c}{(1+\alpha)^{1+\alfa}}(1-\alpha)\Gamma(\alpha+1)}+\eta\mu<1)
这样的话
\(|f(t,x)|\leq c|x|+M\)
对于
\(在[0,1]\中)
和
\(R\中的x\),然后(1)至少有一个解决方案.
证明
首先我们分析方程解的先验界\(x=σTx)对一些人来说\([0,1]\中的σ).
如果\(x=σTx)对一些人来说\(σ[0,1]\),\(E\中的x\),然后我们得到
$$\begin{aligned}\对于[0,1]中的所有t,\quad\bigl\vert x(t)\bigr\vert=&\bigl\vert\sigma Tx(t,ds\\&{}+\frac{\eta}{\Gamma(\alpha+\beta)}\int_{0}^{1}(1-s)^{\alpha+/beta-1}\bigl\vert f\bigl(s,x(s)\bigr)\biger\vert\,ds\\&{}+\frac{\lambda\eta}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}\bigl\ vert x(s \Gamma(\beta)}\int_{0}^{1}(1-s)^{\beta-1}\bigl\vert f\bigl(s,x(s)\bigr)\biger\vert\,ds\\&{}+\frac{\lambda\mu\alpha^{\alpha}}{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1 ^{t}(t-s)^{\alpha-1}\bigl\vert x(s)\bigr\vert\,ds\\leq&\frac{(1+\eta)2(c\vert x\vert+M)}{\Gamma(\alpha+\beta+1)}+\frac{(1+\eta)\lambda\Vertx\Vert}{\Garma(\alpha+1 \Vert x\Vert}{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1)}+\eta\mu\Vert x \Vert。\结束{对齐}$$
所以
$$开始{对齐}\Vert x\Vert\leq&\frac{(1+\eta)(c\Vert x\Vert+M)}{\Gamma(\alpha+\beta+1)}+\frac}(1+/eta)\lambda\Vert x \Vert}{\Gamma+1)\Gamma(\beta+1)}\\&{}+\frac{\lambda\mu\alpha^{\alpha}\Vert x\Vert}{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1)}+\eta\mu\Vert x\Vert。\结束{对齐}$$
这意味着
$$开始{对齐}\Vert x \Vert\leq&\frac{\frac}2M}{\Gamma)\lambda}{\Gamma(\alpha+1)}+\frac{\alpha^{\alha}c}{(1+\alpha)^{1+\alpha}\伽马(\beta+1)}+\frac{\lambda\mu\alpha^{\alpha}}{(1+\alpha)^{1+\alpha}(1-\alpha)\Gamma(\alpha+1)}+\eta\mu)}\\:=&B.结束{对齐}$$
让\(伽马=B+1).设置\(P_{\gamma}=\{x\在E:\parallel x\parallel<\gamma\}\中),然后\(对于所有x\ in \ partial P_{\gamma}\),\(\parallel x\parallel=\gamma>B\)现在,我们考虑\(T:\上划线{P_{\gamma}}\右箭头E\)通过以上分析,可以得出以下结论\(x\neq\西格玛Tx\)对于\(对于部分P_{gamma}中的所有x,对于[0,1]\中的所有σ).
定义运算符\(H_{\sigma}:E\右箭头E\)(\([0,1]\中的所有\σ\)),作为\(H_{\sigma}(x)=x-\sigma Tx\)。很容易看出
$$\对于[0,1]中的所有\sigma\,\对于部分P_{\gamma}中的所有x\,\quad H_{\sigma}(x)=x-\sigma-Tx\neq 0$$
根据引理3.2,T型是完全连续的。这就产生了\([0,1]\中的所有\σ\),\(H_{\西格玛}\)是一个完全连续的场。
因此,通过Leray–Schauder度的同伦不变性,我们知道
$$\操作员名称{deg}(H_{\sigma},P_{\gamma},0)=\操作员名称{deg}(H_{1},P_{\gamma},0)=\操作员名称{deg}(H_{0},P_{\gamma},0)=\操作员名称{deg}(I,P_{\gamma},0)=1\neq 0$$
根据Leray–Schauder度的非零性质\(H_{1}(x)=x-Tx=0\)中至少有一个解决方案\(P_{\gamma}\)这就是问题所在(1)至少有一个解决方案。证明已完成。□