## Positive Solutions for a High Order Riemann-Liouville Type Fractional Impulsive Differential Equation Integral Boundary Value Problem

Xu Jiafa,*, Yang Zhichun,

School of Mathematical Sciences, Chongqing Normal University, Chongqing 401331

 基金资助: 国家自然科学基金项目.  11971081重庆市教育委员会科技研究重大项目.  KJZD-M202000502

 Fund supported: The NSFC.  11971081Science and Technology Research Program of Chongqing Municipal Education Commission.  KJZD-M202000502

Abstract

In this paper, we study a high order Riemann-Liouville type fractional impulsive differential equation integral boundary value problem involving semipositone the nonlinear and impulsive terms. By virtue of the fixed point index, we obtain the positive solutions theorems under some appropriate superlinear and sublinear growth conditions. The results here extend the existing study.

Keywords： Fractional differential equations ; Boundary value problems ; Impulse ; Fixed point index ; Positive solutions

Xu Jiafa, Yang Zhichun. Positive Solutions for a High Order Riemann-Liouville Type Fractional Impulsive Differential Equation Integral Boundary Value Problem. Acta Mathematica Scientia[J], 2023, 43(1): 53-68 doi:

## 1 引言

$$$\label{pro} \left\{\begin{array}{l} -D_{0+}^{\alpha} u(t)=f(t, u(t)), \ t \in(0,1) \backslash\left\{t_{k}\right\}_{k=1}^{m}, \\ \Delta u\left(t_{k}\right)=I_{k}\left(u\left(t_{k}\right)\right), \ t=t_{k}, k=1,2,\cdots,m,\\ u(0)=u^{\prime}(0)=\cdots=u^{(n-2)}(0), \ u^{\prime}(1)=\int_{0}^{1} u(t) {\rm d}\eta(t), \end{array}\right.$$$

(H$\eta$) $\eta$$[0,1]上的有界变差函数, 且\int_0^1 t^{\alpha-1}{\rm d}\eta(t)\in (0,\alpha-1). 注1.1\int_{0}^{1} u(t) {\rm d}\eta(t)被称为Riemann-Stieltjes积分, 其在研究各类边值问题中起着重要作用. 若\eta可导, 则 \int_{0}^{1} u(t) {\rm d}\eta(t)=\int_{0}^{1} u(t)\eta'(t){\rm d}t, 此为Riemann积分; 若存在\{\xi_i\}_{i=1}^{m-2}满足: 0<\xi_1<\xi_2<\cdots<\xi_{m-2}<1, m是正整数且m\ge 3使得 u'(1)=\sum\limits_{i=1}^{m-2}a_iu(\xi_i), 此为多点边值问题. 若令 \eta(t)=\left\{\begin{array}{ll}0,t\in [t\in [\xi_{m-2},1],\end{array}\right. \sum\limits_{i=1}^{m-2}a_iu(\xi_i)=\int_0^1 u(t){\rm d}\eta(t), 即多点边值问题亦可用Riemann-Stieltjes积分表示. 分数阶微积分已有300多年历史, 但在发展的早期并没有引起足够的兴趣. 然而最近几十年, 研究者们发现它们在工程和物理学、力学、化学、经济学和生物学等学科中发挥着不可替代的作用, 对该类问题的研究愈发受到广泛关注, 涌现出一大批优秀的研究成果[1-29]. 众所周知, 非线性泛函分析中的拓扑方法是研究微分方程最有力的工具, 运用该方法处理的各类问题中人们普遍使用一个基本的限制条件: 方程的非线性项是非负的. 这显然局限了研究的视角和创新. 然而在刻画自然现象时, 许多不确定的物理变量、参数以及扰动因素等常常会影响到整个系统的稳定性, 非负性显然不足以描述此类问题. 例如上世纪70年代荷兰化学家Aris在研究化学反应时发现一些惰性材料、 催化剂等常常对整个反应起到加速或抑制的作用, 这样在刻画反应的微分方程中必含有某个扰动项, 即非线性项具有形式 f(t, x) \geq-M, M>0, 这就是数学模型下的半正问题. 近年来亦有很多文献研究分数阶半正问题[1-11], 例如在文献[1]中作者研究了如下抽象的 HIV-1 人口受感染的分数阶动力学模型 \left\{\begin{array}{l} D^{\alpha} u(t)+\lambda f\left(t, u(t), D^{\beta} u(t), v(t)\right)=0, \\ D^{\gamma} v(t)+\lambda g(t, u(t))=0, \\ D^{\beta} u(0)=D^{\beta+1} u(0)=0, D^{\beta} u(1)=\int_{0}^{1} D^{\beta} u(s) {\rm d} A(s), \\ v(0)=v^{\prime}(0)=0, v(1)=\int_{0}^{1} v(s) {\rm d} B(s), \end{array}\right. 其中 2<\alpha, \beta \leq 3,1<\beta<1, u 是末受感染的 {\rm CD} 4+{\rm T} 细胞, v 是已感染的 {\rm CD} 4+{\rm T} 细胞, f$$g$ 允许变号的, 且满足半正型条件, 作者借助锥上的不动点定理, 在一定的参数范围内获得该问题正解的存在性.

$\left\{ \begin{array}{ll} -D_{0+}^{\mu}\left(\varphi_{p}\left(-D_{0+}^{\alpha} u(t)\right)\right)=f(t, u(t)), \ 0<t<1, \\ u(0)=u^{\prime}(0)=\cdots=u^{(n-2)}(0)=0, \ D_{0+}^{\alpha} u(0)=D_{0+}^{\alpha} u(1)=0, \\ D_{0+}^{\beta} u(1)=\lambda \int_{0}^{\eta} D_{0+}^{\gamma} u(t) {\rm d} A(t), \end{array}\right.$

$\left\{\begin{array}{l} { }^{C} D^{\alpha} u(t)=f(t, u(t)), \quad 1<\alpha \leq 2, t \in J^{\prime}, \\ \Delta u\left(t_{k}\right)=I_{k}\left(u\left(t_{k}\right)\right), \quad \Delta u^{\prime}\left(t_{k}\right)=I_{k}^{*}\left(u\left(t_{k}\right)\right), \quad k=1,2, \cdots, p, \\ T u^{\prime}(0)=-a u(0)-b u(T), \quad T u^{\prime}(T)=c u(0)+{\rm d} u(T), \quad a, b, c, d \in \Bbb R, \end{array}\right.$

$u(t)=\left\{\begin{array}{ll} \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,u(s)){\rm d}s+\frac{(b+d) T+(a d-b c) t}{\Lambda T} \int_{t_{p}}^{T} \frac{(T-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,u(s)){\rm d}s\\ -\frac{(b+1) T+(a+b) t}{\Lambda} \int_{t_{p}}^{T} \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,u(s)){\rm d}s+{\cal A}, \quad t \in J_{0};\\ \int_{t_{k}}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,u(s)){\rm d}s+\frac{(b+d) T+(a d-b c) t}{\Lambda T} \int_{t_{p}}^{T} \frac{(T-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,u(s)){\rm d}s\\ -\frac{(b+1) T+(a+b) t}{\Lambda} \int_{t_{p}}^{T} \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,u(s)) {\rm d} s\\ +\sum_{i=1}^{k}\left[\int_{t_{i-1}}^{t_{i}} \frac{\left(t_{i}-s\right)^{\alpha-1}}{\Gamma(\alpha)} f(s,u(s)) {\rm d} s+I_{i}\left(u\left(t_{i}\right)\right)\right]\\ +\sum_{i=1}^{k-1}\left(t_{k}-t_{i}\right)\left[\int_{t_{i-1}}^{t_{i}} \frac{\left(t_{i}-s\right)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,u(s)){\rm d} s+I_{i}^{*}\left(u\left(t_{i}\right)\right)\right]\\ +\sum_{i=1}^{k}\left(t-t_{k}\right)\left[\int_{t_{i-1}}^{t_{i}} \frac{\left(t_{i}-s\right)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,u(s)) {\rm d} s+I_{i}^{*}\left(u\left(t_{i}\right)\right)\right]+{\cal A}, \\ \qquad t \in J_{k}, k=1,2, \cdots, p,\end{array}\right.$

$\begin{eqnarray*} {\cal A}&=& \frac{(b+d) T+(a d-b c) t}{\Lambda T} \sum_{i=1}^{p}\left[\int_{t_{i-1}}^{t_{i}} \frac{\left(t_{i}-s\right)^{\alpha-1}}{\Gamma(\alpha)} f(s,u(s)) {\rm d} s+I_{i}\left(u\left(t_{i}\right)\right)\right] \\ &&+\frac{(b+d) T+(a d-b c) t}{\Lambda T} \sum_{i=1}^{p-1}\left(t_{p}-t_{i}\right)\left[\int_{t_{i-1}}^{t_{i}} \frac{\left(t_{i}-s\right)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,u(s)) {\rm d} s+I_{i}^{*}\left(u\left(t_{i}\right)\right)\right] \\ &&-\sum_{i=1}^{p}\left[\frac{(1-d)(T+a t)+b(c+1) t}{\Lambda}+\frac{[(b+d) T+(a d-b c) t] t_{p}}{\Lambda T}\right]\\ &&\times \left[\int_{t_{i-1}}^{t_{i}} \frac{\left(t_{i}-s\right)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,u(s)){\rm d} s+I_{i}^{*}\left(u\left(t_{i}\right)\right)\right]. \end{eqnarray*}$

$\left\{\begin{array}{l} { }^{C} D^{q} u(t)=f(t, u(t)), \quad \mbox{ a.e. }\ t \in[0,1], \\ \Delta u\left(t_{k}\right)=I_{k}\left(u\left(t_{k}\right)\right), \quad \Delta u^{\prime}\left(t_{k}\right)=J_{k}\left(u\left(t_{k}\right)\right), \quad k=1,2, \cdots, m, \\ a u(0)-b u(1)=0, \quad a u^{\prime}(0)-b u^{\prime}(1)=0 \end{array}\right.$

$u(t)=\int_{0}^{1} G_{1}(t, s) f(s,u(s)){\rm d}s+\sum_{i=1}^{m} G_{2}\left(t, t_{i}\right) J_{i}\left(u\left(t_{i}\right)\right)+\sum_{i=1}^{m} G_{3}\left(t, t_{i}\right) I_{i}\left(u\left(t_{i}\right)\right).$

## 2 基础知识

$D_{0+}^{\alpha} f(t)=\frac{1}{\Gamma(n-\alpha)}\left(\frac{\rm d}{{\rm d}t}\right)^{n} \int_{0}^{t} \frac{f(s)}{(t-s)^{\alpha-n+1}}{\rm d}s,$

$\begin{eqnarray*} t^{\alpha-1} s(1-s)^{\alpha-2} \left[1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right] &\le& \Gamma(\alpha) G(t, s)\\ &\le &s(1-s)^{\alpha-2} \left[1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right],\forall t,s \in [0,1]. \end{eqnarray*}$

$\kappa_1=\frac{\alpha \Gamma(\alpha-1)}{\Gamma(2 \alpha)}\left[1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right], \kappa_2=\frac{1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}}{\alpha(\alpha-1) \Gamma(\alpha)},$

$\kappa_1 \varphi(s)\le \int_0^1 G(t,s)\varphi(t){\rm d}t \le \kappa_2 \varphi(s),\forall s\in [0,1].$

$PC[0,1]=\{u\in [0,1]\to \Bbb R ^+: u$$t\not =t_k上连续, 且 u(t_k^-), u(t_k^+)存在, u(t_k^-)=u(t_k),1\le k\le m \}, 则PC[0,1] 带上范数\|u\|=\sup\limits_{t\in [0,1]}|u(t)|构成一Banach 空间. 令P=\{u\in PC[0,1]: u(t)\ge 0,t\in [0,1]\}, 则P$$PC[0,1]$上的锥. 根据引理2.2知问题(1.1)等价于如下的积分方程

$\begin{matrix}\label{equ1} u(t)&=&\int_{0}^{1} G(t, s) f(s,u(s)){\rm d}s+t^{\alpha-1} \sum_{t \leq t_{k}<1} t_{k}^{1-\alpha} I_{k}\left(u\left(t_{k}\right)\right) \\ & = &\int_{0}^{1} G(t, s) f(s,u(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) I_{k}\left(u\left(t_{k}\right)\right), \end{matrix}$

$H(t,t_k)=\left\{\begin{array}{ll}t^{\alpha-1}t_{k}^{1-\alpha}, & 0\le t \leq t_{k}<1,\\ 0, & 0<t_k< t \le 1.\end{array}\right.$

$$$\label{w} \left\{\begin{array}{l} -D_{0+}^{\alpha} w(t)=M, \ t \in(0,1) \backslash\left\{t_{k}\right\}_{k=1}^{m}, \\ \Delta w\left(t_{k}\right)=M_k, \ t=t_{k}, k=1,2,\cdots,m,\\ w(0)=w^{\prime}(0)=\cdots=w^{(n-2)}(0), \ w^{\prime}(1)=\int_{0}^{1} w(t) {\rm d}\eta(t), \end{array}\right.$$$

$$$\label{w1}w(t)=\int_{0}^{1} G(t, s) M{\rm d}s+ \sum_{k=1}^m H(t,t_k) M_k,t\in [0,1].$$$

$\widetilde{f}(t,u)=\left\{\begin{array}{ll}f(t,u)+M,& u\ge 0,\\ f(t,0)+M, & u<0,\end{array}\right. \widetilde{I}_k(u)=\left\{\begin{array}{ll}I_k(u)+M_k,& u\ge 0,\\ I_k(0)+M_k, & u<0,\end{array}\right.$

$$$\label{equ2}u(t)=\int_{0}^{1} G(t, s) \widetilde{f}(s,u(s)-w(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) \widetilde{I}_{k}\left(u\left(t_{k}\right)-w(t_k)\right),\forall t\in [0,1].$$$

(ii) 若$u^*$是方程(2.4)的解且$u^*(t)\ge w(t),t\in [0,1]$, 则$u^*-w$是方程(2.1)的解.

$\begin{eqnarray*} u^*(t)+w(t)& =&\int_{0}^{1} G(t, s) \widetilde{f}(s,u^*(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) \widetilde{I}_{k}\left(u^*\left(t_{k}\right)\right)\\ & =&\int_{0}^{1} G(t, s) [{f}(s,u^*(s))+M]{\rm d}s+ \sum_{k=1}^m H(t,t_k) [{I}_{k}\left(u^*\left(t_{k}\right)\right)+M_k]\\ & = &\int_{0}^{1} G(t, s) {f}(s,u^*(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) {I}_{k}\left(u^*\left(t_{k}\right)\right)\\ && + \int_{0}^{1} G(t, s) M {\rm d}s+ \sum_{k=1}^m H(t,t_k) M_k. \end{eqnarray*}$

$u^*(t)-w(t) =\int_{0}^{1} G(t, s) f(s,u^*(s)-w(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) I_{k}\left(u^*\left(t_{k}\right)-w(t_k)\right),$

$\begin{eqnarray*} u^*(t)& =&\int_{0}^{1} G(t, s) f(s,u^*(s)-w(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) I_{k}\left(u^*\left(t_{k}\right)-w(t_k)\right) +w(t) \\ & =&\int_{0}^{1} G(t, s) [f(s,u^*(s)-w(s)) +M]{\rm d}s+ \sum_{k=1}^m H(t,t_k) [I_{k}\left(u^*\left(t_{k}\right)-w(t_k)\right)+M_k]\\ & =&\int_{0}^{1} G(t, s) \widetilde{f}(s,u^*(s)-w(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) \widetilde{I}_{k}\left(u^*\left(t_{k}\right)-w(t_k)\right). \end{eqnarray*}$

$$$\label{opA}(Au)(t)=\int_{0}^{1} G(t, s) \widetilde{f}(s,u(s)-w(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) \widetilde{I}_{k}\left(u\left(t_{k}\right)-w(t_k)\right),\forall t\in [0,1].$$$

(H2) 存在$Q(t):[0,1]\to \Bbb R ^+(\not \equiv 0)$$N_k>0\ (k=1,2,\cdots,m)使得 \widetilde{f}(t,u)\le \frac{\Gamma(\alpha)}{1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}}\frac{\Theta_{\kappa_1,\kappa_2,M,M_k} Q(t)}{3\int_0^1 Q(t)\varphi(t){\rm d}t}, \widetilde{I}_k(u)\le \frac{\Theta_{\kappa_1,\kappa_2,M,M_k} N_k}{2\sum\limits_{k=1}^m t_k^{1-\alpha}N_k}, \forall t\in [0,1], u\in [0,\Theta_{\kappa_1,\kappa_2,M,M_k}], k=1,2,\cdots,m; (H3) 存在d\ge 0,$$e_k\ge 0(\not \equiv 0)$, $k=1,2,\cdots,m$使得

$d<\kappa_2^{-1}\Longrightarrow \frac{\omega_{\kappa_1,\kappa_2}(1-d\kappa_2)\Gamma(\alpha+1)\Gamma(\alpha-1)}{\Gamma(2\alpha)}- \sum\limits_{k=1}^m t_k^{1-\alpha} e_k \int_0^{t_k}t^\alpha (1-t)^{\alpha-2}{\rm d}t>0,$
$\limsup_{u\to +\infty}\frac{I_k(u)}{u}\le e_k, k=1,2,\cdots,m, \ \limsup_{u\to +\infty}\frac{f(t,u)}{u}\le d,$

(H4) 存在$t_0\in (0,1]$, $\widetilde{Q}(t):[0,1]\to \Bbb R ^+(\not \equiv 0)$$\widetilde{N}_k>0\ (k=1,2,\cdots,m)使得 \widetilde{f}(t,u)\ge \frac{\widetilde{Q}(s)\Theta_{\kappa_1,\kappa_2,M,M_k}\Gamma(\alpha)}{t_0^{\alpha-1}\left[1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right]\int_0^1 \varphi(s)\widetilde{Q}(s){\rm d}s }, \widetilde{I}_k(u)\ge \frac{\Theta_{\kappa_1,\kappa_2,M,M_k} \widetilde{N}_k}{t_0^{\alpha-1}\sum\limits_{t \leq t_{k}＜1} t_k^{1-\alpha} \widetilde{N}_k }, \forall t\in [0,1], u\in [0,\Theta_{\kappa_1,\kappa_2,M,M_k}], k=1,2,\cdots,m. 定义B_\rho=\{u\in P: \|u\|<\rho\}, \rho>0, 则B_\rho$$P$中的开球, 且

$\overline{B_\rho}=\{u\in P: \|u\|\le\rho\}, \partial B_\rho=\{u\in P: \|u\|=\rho\}.$

$$$\label{que}u\not = Au+\lambda \phi, \lambda\ge 0, u\in \partial B_{R_1}\cap P,$$$

$u=\lambda Au,$

$\begin{eqnarray*} u(t)& \le &(Au)(t)\\ & \le &\int_{0}^{1} \frac{\varphi( s)}{\Gamma(\alpha)} \left[1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right] \frac{\Gamma(\alpha)}{1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}}\frac{\Theta_{\kappa_1,\kappa_2,M,M_k} Q(s)}{3\int_0^1 Q(t)\varphi(t){\rm d}t}{\rm d}s+ \sum_{k=1}^m t_k^{1-\alpha} \frac{\Theta_{\kappa_1,\kappa_2,M,M_k} N_k}{2\sum\limits_{k=1}^m t_k^{1-\alpha}N_k} \\ & =&\frac{5}{6} \Theta_{\kappa_1,\kappa_2,M,M_k}, \end{eqnarray*}$

$$$\label{idex1}i(A, B_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\cap P, P)=1.$$$

$\begin{eqnarray*} i\left(A,\left(B_{R_1} \backslash \overline{B}_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\right) \cap P, P\right) &=&i(A, B_{R_1}\cap P, P)- i(A, B_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\cap P, P)\\ &=&0-1=-1. \end{eqnarray*}$

$\liminf\limits_{u\to +\infty}\frac{f(t,u)}{u}\ge b$

$n=3,\alpha=2.5,\eta(t)=t$, 则

$\delta_{\alpha,\eta}=1.5-\int_0^1 t^{1.5}{\rm d}t=1.1, \kappa_1=\frac{2.5 \Gamma(1.5)}{\Gamma(5)}\left[1+\frac{\int_0^1 \tau^{1.5}{\rm d}\tau}{1.1}\right]=0.126,$
$\kappa_2=\frac{1+\frac{\int_0^1 {\rm d}\tau}{1.1}}{2.5\times 1.5 \Gamma(2.5)}=0.383, \omega_{\kappa_1,\kappa_2}=\frac{0.126\Gamma(5)}{2.5^2\times 1.5^2\times 0.383 \Gamma^2(1.5)}=0.715.$

$\int_0^{t_1} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t=\int_0^{0.5} t^{2.5}(1-t)^{0.5}{\rm d}t=0.02,$
$\int_0^{t_2} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t=\int_0^{0.6} t^{2.5}(1-t)^{0.5}{\rm d}t=0.035,$
$\Theta_{\kappa_1,\kappa_2,M,M_k}=0.715^{-1}\left[\frac{M}{1.5\Gamma(2.5)}+ t_1^{-1.5} M_1+t_2^{-1.5} M_2\right]=0.701 M+3.956M_1+3.01M_2.$

(H1) 存在$b\ge 0$, $l_1,l_2\ge 0(\not \equiv 0)$使得

$\mbox{当$b<\kappa_1^{-1}$时:}\ \frac{0.126 b-1}{3.75} + 0.0143l_1+0.025l_2 >0,$

$\liminf_{u\to +\infty}\frac{I_1(u)}{u}\ge l_1,\ \liminf_{u\to +\infty}\frac{I_2(u)}{u}\ge l_2, \ \liminf_{u\to +\infty}\frac{f(t,u)}{u}\ge b,$

(H2) 取$M=5,M_1=6,M_2=7$, $N_1=2.5,N_2=3.4$, $Q(t)\equiv 10, \forall t\in [0,1]$, 则$0.701 M+3.956M_1+3.01M_2=48.311,$$6.657N_1+4.303N_2=31.27, \widetilde{f}(t,u)\le \frac{\Gamma(2.5)}{1+\frac{\int_0^1 {\rm d}\tau}{1.1}}\frac{(0.701 M+3.956M_1+3.01M_2) Q(t)}{3\int_0^1 Q(t)\varphi(t){\rm d}t}=42.03, \widetilde{I}_1(u)\le \frac{(0.701 M+3.956M_1+3.01M_2)N_1 }{6.657N_1+4.303N_2}=3.862, \widetilde{I}_2(u)\le \frac{(0.701 M+3.956M_1+3.01M_2)N_2 }{6.657N_1+4.303N_2}=5.253, \forall t\in [0,1], u\in [0,48.311]. 从而令 {f}(t,u)=\sin \pi t+\frac{1}{25}u-5,I_1(u)=\frac{1}{650}u^2-6, I_2(u)=\frac{1}{2}\sqrt{u}-7, t\in [0,1],u\in \Bbb R ^+, \widetilde{f}(t,u)=\sin \pi t+\frac{1}{25}u\le 2.932, \widetilde{I}_1(u)=\frac{1}{650}u^2\le 3.59, \widetilde{I}_2(u)=\frac{1}{2}\sqrt{u}\le 3.475,t\in [0,1],$$u\in [0,48.311]$.

$\liminf_{u\to +\infty}\frac{f(t,u)}{u}=\liminf_{u\to +\infty}\frac{\sin \pi t+\frac{1}{25}u-5}{u}=\frac{1}{25}<\kappa_1^{-1},$
$\liminf_{u\to +\infty}\frac{I_1(u)}{u}=\liminf_{u\to +\infty}\frac{\frac{1}{650}u^2-6}{u}=+\infty,$
$\liminf_{u\to +\infty}\frac{I_2(u)}{u}=\liminf_{u\to +\infty}\frac{\frac{1}{2}\sqrt{u}-7}{u}=0,$

(H3) 存在$d\ge 0,$$e_k\ge 0(\not \equiv 0), k=1,2使得 d<\kappa_2^{-1}\Longrightarrow 0.088(1-0.383d)-0.057e_1-0.075e_2>0, \limsup_{u\to +\infty}\frac{I_k(u)}{u}\le e_k, k=1,2, \ \limsup_{u\to +\infty}\frac{f(t,u)}{u}\le d, 对任意的t\in [0,1]一致成立; (H4) 取t_0=0.5,M=7,M_1=8,M_2=9, \widetilde{N}_1=1.5,\widetilde{N}_2=1.7, \widetilde{Q}(t)\equiv 20,t\in [0,1], 则 \Theta_{\kappa_1,\kappa_2,M,M_k}=0.701 M+3.956M_1+3.01M_2=63.645, \widetilde{f}(t,u)\ge 642,\widetilde{I}_1(u)\ge 34.178, \widetilde{I}_2(u)\ge 38.735, \forall t\in [0,1], u\in [0,63.645], k=1,2. f(t,u)=\ln(e^{650}+u)+|\cos\pi t|-7, I_1(u)=\ln (e^{35}+u)-8,I_2(u)=\ln (e^{39}+u)-9,t\in [0,1], u\in \Bbb R ^+, \widetilde{f}(t,u)=\ln(e^{650}+u)+|\cos\pi t|\ge 650, \widetilde{I}_1(u)=\ln (e^{35}+u)\ge 35, \widetilde{I}_2(u)=\ln (e^{39}+u)\ge 39, \forall t\in [0,1],u\in [0,63.645]. 另外可计算极限 \limsup_{u\to +\infty}\frac{f(t,u)}{u}=\limsup_{u\to +\infty}\frac{\ln(e^{650}+u)+|\cos\pi t|-7}{u}=0, \limsup_{u\to +\infty}\frac{I_1(u)}{u}=\limsup_{u\to +\infty}\frac{\ln (e^{35}+u)-8}{u}=0,$$\limsup_{u\to +\infty}\frac{I_2(u)}{u}=\limsup_{u\to +\infty}\frac{\ln (e^{39}+u)-9}{u}=0,$

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