为了使用单调迭代方法,我们引入了一对上下解的定义。
定义3.1
A函数\(u(t)\在C_{r}^{\alpha}[0,t]\中)被称为问题的下解(1.1)如果它满足
$$\textstyle\bbegin{cases}D_{0^{+}}^{\beta}}(0),\overline{D_{0^{+}}^{\alpha}u}(t))\geq 0。\结束{cases}$$
(3.1)
A函数\(v(t)\在C_{r}^{\alpha}[0,t]\中)被称为问题的上解(1.1)如果它满足
$$\textstyle\begin{cases}D_{0^{+}}^{\beta}(\phi_{p}(D_{0 ^{+{}}^{\alpha}v(t)))\gef(t,v(t ^{+}}^{\alpha}v}(0),\overline{D_{0^{+{}}^}\alpha}v}(t))\leq 0。\结束{cases}$$
(3.2)
我们的主要结果需要以下假设。
- \((H_{1})\)以下为:
假设\(C_{r}^{alpha}[0,T]\中的u_{0},v_{0{\)是问题的上下解决方案(1.1)分别为和\(u_{0}(t)\le v_{0neneneep(t),t\in(0,t]\)。
- \((H_{2})\)以下为:
存在一个常量M(M)这样的话
$$f\bigl(t,u(t),D_{0^{+}}^{\alpha}u(t u(t)\较大)\较大]$$
对于\(u_{0}(t)\leu(t)\tev(t。
- \((H_{3})\)以下为:
存在常量\(\lambda{1}>0,\lambda{2}\ge0\)这样的话
$$g(x{1},y{1})-g(x}2},y{2})\le\lambda{1}(x_{2} -x个_{1} )-\lambda{2}(y_{2} -年_{1}) $$
对于\(\波浪号{u}_{0}(0)\le x_{1}\le x_{2}\le波浪{v}(v)_{0}(0)\)和\(\波浪号{u}_{0}(T)\le y_{1}\le y_{2}\le \波浪{v}(v)_{0}(T)\)。
- \((H_{4})\)以下为:
存在常量\(\mu{1}>0,\mu{2}\ge0)这样的话
$$h(x{1},y{1}.)-h(x{2},y{2}.)\le\mu_1}(x_{2} -x个_{1} )-\mu{2}(y_{2} -年 _{1}) $$
对于\(\overline{D_{0^{+}}^{\alpha}u_{0}}(0)\lex_{1}\lex_2}\le\overline{D_}0^{+/}}^}^{\ alpha}v_{0{}},(0)\)和\(\overline{D_{0^{+}}^{\alpha}u_{0}}。
定理3.1
假设和\((H_{1})\)–\((H_{4})\)持有。然后存在序列\({u_{n}(t)\},\{v_{nneneneep(t)\}子集C_{r}^{alpha}[0,t]\)这样的话\(lim{n\to\infty}u{n}=x,lim{n\to\infty}v{n}=y)在\((0,T]\)和\(x,y)区间上的最小解和最大解\([u{0},v{0}]\)问题的关键(1.1),分别地,哪里
$$[u_{0},v_{0{]=\bigl\{u\在C_{r}^{\alpha}[0,T]中:u_{0:}(T)\leu(T)\ lev_{0/}(T),T\in(0,T],\ tilde{u}_{0}(0)\le\颚化符{u}(O)\le\tilde{v}(v)_{0}(0)\bigr\}$$
那就是,任何解决方案\(u\在[u{0},v{0}]\中),
$$u_{0}\leu_{1}\le\cdots\leu_}n}\le\tots\lex\leu\ley\le\tods\lev_{n}\le\cdots\lev_}1}\le v_{0{0}$$
和
$$\开始{对齐}D_{0^{+}}^{\alpha}u_{0}&\le D_{0 ^{+{}}^}\alpha}u_{1}\le\cdots\le D_}{0 ^}+}}}y\le\cdots\leD_{0^{+}}^{\alpha}v_{n}\le\cdots\\&\leD_}0^{+/}}^}\alpha{v_{1}\leD_A{0^}+}}^{\alpha}v_0}。\结束{对齐}$$
证明
让\(F(u(t)):=F(t,u(t。对于\(n=1,2,\ldots\) , 我们定义
$$\textstyle\bbegin{cases}D_{0^{+}}^{\beta}(\phi{p}(D_{0^{+}^{\alpha}u{n}(t))+M\phi{p}(D_{0^{+}^{\alpha}u{n}(t))=F(u{n-1}(t))+M\phi{p}(D_{0^{+}^{\alpha}u{n-1}(t)),\\quad t\in(0,t],\\\波浪号{u}_{n} (0)=\ tilde{u}_{n-1}(0)+\frac{1}{\lambda{1}}g(\波浪线{u}_{n-1}(0),\波浪号{u}_{n-1}(T)),\\\\overline{D_{0^{+}}^{\alpha}u_{n}}(0)=\overline{D_{0^{+}}^{\alpha}u _{n-1}}(0)+\frac{1}{\mu_{1}}h(\overline{D_ 0^{+}}^{\alpha}u _{n-1}}(T)),\end{cases}$$
(3.3)
和
$$\textstyle\开始{cases}D_{0^{+}}^{beta}(t)),在(0,t],波浪号{v}(v)_{n} (0)=\ tilde{v}(v)_{n-1}(0)+\frac{1}{\lambda{1}}g(\波浪线{v}(v)_{n-1}(0),\波浪号{v}(v)_{n-1}(T)),\\\\overline{D_{0^{+}}^{\alpha}v_{n}}(0)=\overline{D_{0^{+}}^{\alpha}v_{n-1}}(0)+\frac{1}{\mu_{1}}h(\overline{D_ 0^{+}^{\alpha}v_{n-1}})(0),\overline{D_ 0^{+}}^{\alpha}v_{n-1}}}(T))。\结束{cases}$$
(3.4)
发件人\(C_{r}^{alpha}[0,T]\中的u_{0},v_{0{\),我们有\(C_{r}[0,t]\中的D_{0^{+}}^{\alpha}u_{0}和\(F(u_{0}(t))+\phi_{p}(D_{0^{+}}^{\alpha}u_{0:}.鉴于引理2.3、功能\(u{1},v{1}\)在空间中有很好的定义\(C_{r}^{\alpha}[0,T]\)通过归纳,我们可以推断出\(u{n},v{n}\)在空间中有很好的定义\(C_{r}^{\alpha}[0,T]\)。
首先,我们证明\(u{0}(t)\leu{1}和\(D_{0^{+}}^{\alpha}u_{0}(t)\leD_{0对于\(在(0,t]\)中。
让\(δ(t):=φ{p}(D_{0^{+}}^{\alpha}u{1}(t)).定义\(u{1}\)假设\(u{0}\)较低的解决方案是否意味着
$$D_{0^{+}}^{\beta}\delta(t)+M\delta$$
和\(\波浪号{u}_{1} (0)-\波浪号{u}_{0}(0)=\frac{1}{\lambda{1}}g(\波浪线{u}_{0}(0),\波浪线{u}_{0}(T))\ge 0\),\(吨^{r} D类_{0^{+}}^{\alpha}u{1}(0)-t^{r} D类_{0^{+}}^{\alpha}u{0}(0)=\frac{1}{\mu{1}}h(t^{r} D类{0^{+}}^{\alpha}u{0}(0),t^{r} D类_{0^{+}}^{\alpha}u_{0}(T))\ge 0\),因此我们\(D_{0^{+}}^{\alpha}u_{0}(t)\le D_{0和\(u_{1}(t)\ge u_{0}(t),t\in(0,t]\)通过引理2.4。
使用类似的方法,我们可以证明\(v{1}(t)\le v{0}(t)\)和\(D_{0^{+}}^{\alpha}v_{1}(t)\le D_{0为所有人\(在(0,t]\)中现在,我们把\(xi(t)=\phi{p}(D_{0^{+}}^{\alpha}v{1}(t.来自(3.3), (3.4)和\((H_{2})\),我们有
$$D_{0^{+}}^{\beta}\xi(t)+M\xi}(t)\biger)\bigr]\ge 0$$
(3.5)
我们发现,通过\((H_{3})\)和\((H_{1})\),
$$\开始{aligned}\波浪线{v}(v)_{1} (0)-\颚化符{u}_{1} (0)&=\波浪线{v}(v)_{0}(0)+\frac{1}{\lambda{1}}g\bigl(\波浪线{v}_{0}(0),\波浪线{v}(v)_{0}(T)\bigr)-\biggl[u_{0}(0)+\frac{1}{\lambda _{1}}g\bigl(\tilde{u}_{0}(0),\波浪线{u}_{0}(T)\bigr)\biggr]\\&=\frac{1}{\lambda_{1}}\bigl[\lambda \bigl(\波浪线{v}(v)_{0}(0)-\波浪号{u}_{0}(0)\bigr)+g\bigl(\波浪线{v}(v)_{0}(0),\波浪线{v}(v)_{0}(T)\bigr)-g\bigl(\波浪线{u}_{0}(0),\波浪号{u}_{0}(T)\biger)\bigr]\\&\ge\frac{1}{\lambda_{1}}\bigl[\lambda _{1{\bigl(\波浪线{v}(v)_{0}(0)-\波浪号{u}_{0}(0)\bigr)-\lambda_{1}\bigl(\tilde{v}(v)_{0}(0)-\波浪线{u}_{0}(0)\大)+\lambda_{2}(\波浪线{v}(v)_{0}(T)-\波浪线{u}_{0}(T)\bigr]\\&=\frac{\lambda_{2}}{\lampda_{1}}\bigl(\波浪线{v}(v)_{0}(T)-\波浪线{u}_{0}(T)\biger)\ge 0。\结束{对齐}$$
(3.6)
同样,
$$\overline{D_{0^{+}}^{\alpha}v_1}}(0)-\overline{D_}0^{+/}}^}\alpha{u_1}}(0)\geq\frac{\mu_2}}{\mu_1}}\bigl}u_{0}}(T)\bigr)\ge 0$$
(3.7)
它源自(3.5)–(3.7)和引理2.4那个\(D_{0^{+}}^{\alpha}v_1}(t)和\(v_{1}(t)\ge u_{1neneneep(t),t\in(0,t]\)。
接下来,我们展示一下\(u{1},v{1}\)是问题的上下解(1.1)分别是。
发件人(3.3)和条件\((H_{2})\)–\((H_{4})\),我们有
$$\开始{aligned}D_{0^{+}}^{\beta}\bigl(\phi_{p}\bigle(D_{0 ^{+{}}^}\alpha}u_{1}(t)\bigr)={}&F\bigl-(u_{0}(t)\biger)-F\bigl-u_{1\bigr{p}\bigl(D_{0^{+}}^{\alpha}u_{1}(t)\bigr)-\phi_{p}\ bigl(t) \bigr)-\phi_{p}\bigl(D_{0^{+}}^{\alpha}u_{0}+F\bigl(u_{1}(t)\bigr)\\={}&F\bigle(u_1}(t)\biger)\end{对齐}$$
和
$$\开始{aligned}0={}&g\bigl(\tilde{u}_{0}(0),\波浪号{u}_{0}(T)\bigr)-g\bigl(\波浪线{u}_{1} (0),\波浪形{u}_{1}(T)\bigr)+g\bigl(\波浪形{u}_{1} (0),\ tilde{u}_{1} (T)\bigr)-\lambda _{1}\bigl[\tilde{u}_{1}(0)-\tilde{u}_{0}(0)\bigr]\\le{}&g\bigl(\波浪线{u}_{1} (0),\ tilde{u}_{1} (T)\biger)-\lambda_{2}\bigl(\波浪线{u}_{1} (T)-\颚化符{u}_{0}(T)\bigr),\\0={}&h\bigl(\overline{D_{0^{+}}^{\alpha}u_{0}}(0),\overline{D_0^{+{}}^}(T)\biger)-h\bigl-(\overrine{D_0_0^{+/}}^\alpha{u_{1}}{\alpha}u_{1}}(T)\bigr)\\&{}+h\bigl(\overline{D_{0^{+}}^{\alpha}u_1}}}^{\alpha}u_{1}}(0)-\上划线{D_{0^{+}}^{\alfa}u_0}}行{D_{0^{+}}^{\alpha}u_{1}}(T)-\上一行{D_(0^{+)}^{\ alpha}u(T)\biger)。\结束{对齐}$$
自\(\波浪号{u}_{1} (T)\ge\tilde{u}_{0}(T),T^{r} D类_{0^{+}}^{\alpha}u_{1}(T)\geq T^{r} D类_{0^{+}}^{\alpha}u_{0}(T)\),上述不等式意味着
$$g\bigl(\波浪线{u}_{1} (0),\ tilde{u}_{1} (T)\biger)\ge 0,\qquad h\bigl$$
这证明了\(u{1}\)是问题的较低解决方案(1.1). 同样,我们可以证明\(v{1}\)是的上解(1.1).
使用数学归纳法,我们知道
$$开始{对齐}开始{校准}和u{0}{+}}^{\alpha}u_{1}\le\cdots\le D_{0^{+}{^{\alpha}u_{n}\leD_{0 ^{+{}}^}\alpha{u_{n+1}\\&\幻影{D_{0 ^{+}}\勒D_{0^{+}}^{\alpha}v_{n}\le\cdots\le D_{0 ^{+{}}^{\alpha}v_{1}\le D_}0^{+}}^}\alpha{v_{0},\end{aligned}\end{arigned}$$
(3.8)
对于\(在(0,t]\)中和\(n=1,2,3,\ldots\) .
序列\({t^{1-\alpha}u{n}\}\)和\(\{t^{r} 天_{0^{+}}^{\alpha}u{n}\}\)一致有界且等距[14]. 类似地,我们可以证明序列\({t^{1-\alpha}v{n}\}\)和\(\{t^{r} D类_{0^{+}}^{\alpha}v{n}\}\)一致有界且等距。Arzela–Ascoli定理保证\({t^{1-\alpha}u{n}\}\)和\({t^{1-\alpha}v_{n}\}\)汇聚到\(t^{1-\alpha}x(t)\)和\(t^{1-\alpha}y(t)\)一致开启\([0,T]\)分别为;\(\{t^{r} D类_{0^{+}}^{\alpha}u{n}\}\)和\(\{t^{r} D类_{0^{+}}^{\alpha}v{n}\}\)汇聚到\(\{t^{r} D类_{0^{+}}^{\alpha}x(t)\}\)和\(\{t^{r} D类_{0^{+}}^{\alpha}y(t)\}\)一致开启\([0,T]\)分别是。因此\(\|u_{n} -x个\|_{C_{r}^{\alpha}}\rightarrow 0,\|v_{n} -年\|_{C_{r}^{alpha}}\右箭头0\(n\rightarrow\infty)\)。
通过积分表示(2.2)对于线性分式问题,解\(u{n}(t)\)问题的(3.3)可以表示为
$$开始{对齐}u_{n}(t)={}&t^{α-1}k_{n-1}+\frac{1}{\varGamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\phi_{q}\biggl[\varGamma(\beta)\phi_}(h_{n-1)s^{\beta-1}E_{beta,\beta}\bigl(-Ms^{beta}大)\\&{}+\int_{0}^{s}(s-\tau)^{\beta-1}E_{\beta,\beta}\bigl(-M(s-\tau)^{\ beta}\bigr)\eta_{n-1}(\tau,\biggr]\,d\tau,\四边形t\in(0,t],\end{对齐}$$
哪里\(k{n-1}=widetilde{u}_{n-1}(0)+\frac{1}{\lambda}g(\widetilde{u}_{n-1}(0),\widetilde{u}_{n-1}(T)),\(eta_{n-1}(s)=F(u_{n-1}(s))+M\phi_{p}(D^{alpha}_{0^{+}}u{n-1)和\(h{n-1}=\overline{D_{0^{+}}^{\alpha}u_{n-1}}}(0)+\frac{1}{\mau{1}}h(\overline{D_ 0^{+}^{\alpha}u_{n-1}}(T))\)。
假设(f)应用支配收敛定理,\(x(t)\)满足以下积分方程:
$$\开始{对齐}x(t)={}&t^{\alpha-1}\widetilde{x}(0)+\frac{1}{\varGamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\ phi_{q}\biggl[\varGamma(\beta)\phi_{p}(h{0})s^{\beta-1}E_{beta,\beta}\bigl(-Ms^{beta}\big r)\\&{}+\int_{0}^{s}(s-\tau)^{\beta-1}E_{\beta,\beta}\bigl(-M(s-\tau)^{\ beta}\ bigr)\eta(\tau,\biggr]\,d\tau,\quad t\in(0,t],\end{对齐}$$
哪里\(h_{0}=\上划线{D_{0^{+}}^{\alpha}x}(0),eta(s)=F(x(s))+M\phi_{p}.通过引理2.3,我们知道\(x(t)\)是问题的解决方案(1.1). 用同样的方法,我们可以证明\(y(t)\)也是问题的解决方案(1.1),并满足\(u{0}\lex\ley\lev{0}\)在\((0,T]\)。
为了证明这一点\(x(t),y(t)\)是的极值解(1.1),让\(u\在[u{0},v{0}]\中)是这个问题的任何解决方案(1.1). 我们认为\(u_{n}\leu\lev_{n{,t\in(0,t]\)对一些人来说n个.让\(\zeta(t)=\phi{p}(D_{0^{+}}^{\alpha}u(t))-\phi{p}(D_{0^{+}}^{\alpha}u{n+1}(t)),\varrho(t)=\phi{p}(D_{0^{+}^{\alpha}v{n+1}(t))-\phi p}(D_{0^{+}^{\alpha}u(t)))然后,根据条件\((H_{2})\),我们看到了
$$D_{0^{+}}^{\beta}\zeta(t)+M\zeta第0页$$
和
$$D_{0^{+}}^{\beta}\varrho(t)+M\varrho(t)=F\bigl(v_{n}(t)\bigr)-F\bigl(u(t)\bigr)+M\bigl[\phi_{p}\bigl(D_{0^{+}}^{\alpha}v_{n}\bigr)-\phi_{p}\bigl(D_{0^{+}}^{\alpha}v\bigr)\bigr]\ge 0$$
此外,根据条件\((H_{3})\),我们有
$$\开始{aligned}\波浪线{u}(0)-\波浪线{u}_{n+1}(0)&=\tilde{u}(O)+\frac{1}{\lambda_{1}}g\bigl{u}_{n} (0)+\frac{1}{\lambda{1}}g\bigl(\波浪线{u}_{n} (0),\ tilde{u}_{n} (T)\bigr)\biggr]\\&=\frac{1}{\lambda{1}}\bigl[\lambda{1}\tilde{u}(0)+g\bigl{u}_{n} (0)+g\bigl(\tilde{u}_{n} (0),\波浪号{u}_{n} (T)\biger)\bigr]\&\ge\frac{\lambda_{2}}{\lampda_{1}}\bigl(\波浪线{u}(T)-\波浪线{u}_{n} (T)\biger)\ge 0\end{对齐}$$
和
$$\开始{aligned}\波浪线{v}(v)_{n+1}(0)-\波浪线{u}(O)&=\波浪线{v}(v)_{n} (0)+\frac{1}{\lambda{1}}g\bigl(\波浪线{v}(v)_{n} (0),\ tilde{v}(v)_{n} (T)\bigr)-\biggl[\tilde{u}(0)+\frac{1}{\lambda-1}g\bigl(\tilde}u}{v}(v)_{n} (0)+g\bigl(\tilde{u}(0),\tilde{u}(T)\bigr)-(\lambda _{1}\tilde{u}(0)+g\bigl(\tilde{u}_{n} (0),\tilde{u}_{n}(T)\bigr)\biger]\\&\ge\frac{\lambda_{2}}{\lampda_{1}}\bigl(\tillde{v}_{n}。\结束{对齐}$$
按条件\((H_{4})\),我们有
$$开始{对齐}\上划线{D_{0^{+}}^{\alpha}u}(0),上划线{D_{0^{+}}^{\alpha}u}u_{n}}(0),\overline{D_{0^{+}}^{\alpha}u_{n}(T)\biger)\biggr]\\={}&\frac{1}{\mu_1}}\bigl[\mu_{1}\overline{D_0^{+{}}^\alpha{u}\overline{D_{0^{+}}^{\alpha}u}(T)\biger)\\&{}-\bigl(\mu_{1}\overline{D_}0^{++}}^}\alpha}u_{n}}(0)+h\ bigl,\overline{D_{0^{+}}^{\alpha}u_{n}}(T)\biger)\bigr]\\ge{}&\frac{\mu_{2}}{\mu_1}}\bigl{对齐}$$
和
$$开始{对齐}\上划线{D_{0^{+}}^{阿尔法}v_{n+1}}^{\alpha}u}(T)\bigr)\ge 0。\结束{对齐}$$
因此,\(D_{0^{+}}^{\alpha}u_{n+1}(t)\leD_{0 ^{+{}}^}\alpha}u(t)\ leD__{0和\(u_{n+1}(t)\le u(t)\le v_{n+1}(t),t\in(0,t]\)此外,通过归纳法\(x(t)\leu(t)\ ley(t),D_{0^{+}}^{\alpha}x\leD_{0 ^{+{}}^}\alpha}u\leD_{0^}+}}^{\alpha}y\)在\((0,T]\)通过服用\(到英寸).证据完整。□
定理3.2
定理中的假设3.1保持不变,存在一个常数N个这样的话
$$f\bigl(t,u(t),D_{0^{+}}^{\alpha}u(t v(t)\较大)\较大]$$
(3.9)
对于\(u_{0}(t)\leu(t)\tev(t,和\(\widetilde{u}_{0}(0)=\widetilde{v}(v)_{0}(0)\),\(\overline{D_{0^{+}}^{\alpha}u_{0}}(0)=\overline{D_0^{+{}}^}\alpha{v_{0{}},(0)\)。然后是问题(1.1)在订单间隔中具有唯一的解决方案\([u{0},v{0}]\)。
证明
来自定理3.1,我们知道\(x(t)\)和\(y(t)\)是极值解\(x(t)\ le y(t),t \ in(0,t]\).足以证明\(x(t)\ge y(t),t\in(0,t]\)。
事实上,通过(3.8)和\(\overline{D_{0^{+}}^{\alpha}u_{0}}(0)=\overline{D_0^{+{}}^}\alpha{v_{0{}},(0)\),我们知道\(\overline{D_{0^{+}}^{\alpha}x}(0)=\overline{D_0^{+{}}^}\alpha{y}(O)\).让\(w(t)=φ{p}(D_{0^{+}}^{\alpha}x(t)),我们有,来自(3.9),
$$\textstyle\开始{cases}D_{0^{+}}^{\beta}w(t)=F(x(t))-F(y(t)。\结束{cases}$$
然后\(w(t)\ge 0,t\in(0,t]\),即。\(D_{0^{+}}^{alpha}x(t)。也通过(3.8)和\(\widetilde{u}_{0}(0)=\widetilde{v}(v)_{0}(0)\),我们有\(\widetilde{x}(0)=\widetelde{y}(0\)),引理2.4暗示\(x(t)\ge y(t),t\in(0,t]\)因此,我们获得\(x(t)=y(t)\).问题(1.1)有一个独特的解决方案。证据完整。□
最后,我们用一个例子来说明这个定理3.1。
示例3.1
考虑以下分数周期边值问题:
$$\textstyle\begin{cases}D_{0^{+}}^{\beta}(\phi_{p}(D_{0 ^{+{}}^{\alpha}u(t))=t^{1/2}(1-t)-2[D_{0^{+}}^}\ alpha}u(t{2\varGamma(4/3)}-\widetilde{u}(1))=0,\\(\frac{1}{2}+\overline{D_{0^{+}}^{\alpha}u}$$
(3.10)
哪里\(α=1/2,β=2/3,p=3,T=1),\(f(t,u,D_{0^{+}}^{\alpha}u)=t^{1/2}(1-t)-2[D_{0 ^{+{}}^}\alpha{u(t)]^{2}+u(t,\(g(x,y)=x(\frac{\varGamma(5/6)}{2\varGamma(4/3)}-y)\)、和\(h(x,y)=(分形{1}{2}+x)(1-y)。
设置
$$u_{0}(t)\equiv 0,\qquad v_{0neneneep(t)=\frac{\varGamma(5/6)}{\varGamma(4/3)}t^{1/3},\quad t\in[0,1]$$
很容易验证\(D_{0^{+}}^{1/2}u_{0}(t)等于0,D_{0对于\(t \ in(0,1]\)和
$$t^{1/6}D_{0^{+}}^{1/2}u_{0}{0^{+}}^{1/2}v{0}(t){t=1}=1$$
因此,
$$开始{对齐}和D_{0^{+}}^{2/3}\bigl埃蒂尔德{u}_{0}(0),\widetilde{u}_{0}(1)\bigr)=0,\qquad h\bigl(t^{1/6}D_{0^{+}}^{1/2}u_{0}(t)|{t=0},t^{1/1}D_{0。\结束{对齐}$$
这些表明\(u{0}\)是的较低解决方案(3.10). 我们有
$$开始{对齐}和D_{0^{+}}^{2/3}\bigl v_{0}\bigr)\\&\幻影{D_{0^{+}}^{2/3}\bigl,\\&g\bigl(\widetilde{v}(v)_{0}(0),\widetilde{v}(v)_{0}(1)\bigr)=0,h\bigl(t^{1/6}D_{0^{+}}^{1/2}v_{0}(t)|_{t=0},t^{1/16}D_0^{+{}}^}1/2}v_0}(t)。\结束{对齐}$$
这些表明\(v{0}\)是的上解(3.10)、和\(u{0}(t)\lev{0}(t)\)在\([0,1]\)。
对于\(u{0}\leu\lev\lev{0}\),我们有\(\ phi_{3}(D_{0^{+}}^{1/2}v)-\phi_{3}(D_{0^{+}}^{1/2}u)=(D_{0^{+}}^{1/2}v)^{2}-(D_{0^{+}}^{1/2}u)^{2}\)和
$$f\bigl(t,u,D_{0^{+}}^{1/2}u\bigr)+2\phi_{3}\bigle(D_{0 ^{+{}}^{1/2}u\ bigr v\le 0$$
因此,\(f(t,u,D_{0^{+}}^{1/2}u)-f(t,v,D_}0^{++}^{1/2}v)\le 2[\phi_{3}(D_{0 ^{+{}}^1/2}v。
此外,\(\frac{\partial g(x,y)}{\partial x}=\frac{\varGamma(5/6)}{2\varGamma(4/3)}-y\geq-\frac{\varGamma(5/6)}{2\varGamma(4/3)},\frac{\partial g(x,y)}{\partial y}=-x\)对于\(\widetilde{u}_{0}(0)\le x\le\widetilde{v}(v)_{0}(0),y\在[\widetilde中{u}_{0}(1),\widetilde{v}(v)_{0}(1)]=[0,\frac{\varGamma(5/6)}{\varGamma(4/3)}]\)因此,\(g(u{1},v{1})-g(u}2},v{2})\le\frac{varGamma(5/6)}{2\varGamma(4/3)}(u_{2} -u个_{1})\)对于\(\widetilde{u}_{0}(0)\leu{1}\leu}2}\le\widetilde{v}(v)_{0}(0),\widetilde{u}_{0}(1)\lev{1}\lev}2}\le\widetilde{v}(v)_{0}(1)\).以同样的方式,\(h(u{1},v{1}.)-h(u{2},v{2}.)_{2} -u个_{1})\),用于\(t^{1/6}D_{0^{+}}^{1/2}u_{0},\(t^{1/6}D_{0^{+}}^{1/2}u_{0}。
因此,条件\((H_{1})\)–\((H_{4})\)都很满意。存在两个单调迭代序列\({u{k})和\({v_{k}\}\)一致收敛于问题的最小和最大解(3.10)英寸\([u{0},v{0}]\)通过定理3.1。