## Generalized Lyapunov Inequalities for a Higher-Order Sequential Fractional Differential Equation with Half-Linear Terms

Ma Dexiang,1, Özbekler Abdullah,2

AbdullahÖzbekler,E-mail:aozbekler@gmail.com , E-mail：aozbekler@gmail.com

Abstract

In this paper, we establish some new Lyapunov-type inequalities and Hartman-type inequalities for a higher-order sequential Riemann-Liouville fractional boundary value problem with some half-linear terms and a forcing term. We generalize some existing results in literature.

Keywords： Fractional derivative ; Green's function ; Lyapunov-type inequality ; Hartman-type inequality

Ma Dexiang, Özbekler Abdullah. Generalized Lyapunov Inequalities for a Higher-Order Sequential Fractional Differential Equation with Half-Linear Terms. Acta Mathematica Scientia[J], 2020, 40(6): 1537-1551 doi:

## 1 引言

$$$\label{gg}x"(t)+\nu(t)x(t)=0,$$$

$$$\label{wenti}\left\{\begin{array}{llll}(-1)^{n+1}(_a {\cal D}^{ [\alpha_n]} x)(t)+p(t)\Phi_{\beta}(x(t))=f(t), \qquad t\in(a, b), \\(_a {\cal D}^{ [\alpha_i]} x)(a)=(_a {\cal D}^{ [\alpha_i]} x)(b)=0, \quad \qquad i=0, 1, 2, \cdots , n-1.\end{array}\right.$$$

(i) $\alpha_i\in(1, 2](i=1, 2, \cdots , n, n\in {\Bbb N}$);

(ii) $\beta>0$, $\Phi_\beta(x)=|x|^{\beta-1}x$, $x\in{\Bbb R}$;

(iii) 权因子项$p(t)$和强迫项$f(t)$$[a, b]上连续; (iv) 对i=1, 2, \cdots , n, $$\label{hb} (_a {\cal D}^{[\alpha_i]}x)(t)=(_a {\cal D}^{\alpha_i}(_a {\cal D}^{[\alpha_{i-1}]}x))(t), \qquad i=1, 2, \cdots , n,$$ 其中(_a {\cal D}^{ [\alpha_0]} x)(t)=x(t); 对i=1, 2, \cdots , n,$$_a {\cal D}^{\alpha_i}$$\alpha_i阶黎曼-刘维尔分数阶导数. 注 1.1 在研究方程(1.9)之前, 我们指出文献[31]中的几点问题. (i) 在文献[31]的不等式(2.11)中, 可以证明对任意的A>0$$B>0,$若取$z_0=\big(\frac{\alpha B}{2A}\big)^{\frac{1}{2-\alpha}},$则不等式中的等号都是成立的, 并非如原文献中所述:不等式(2.11)中的等号成立当且仅当$B=z=0.$换句话说, 当$A>0, B>0$时, 不等式(2.11)在$(0, +\infty)$上不是严格不等式.

(ii) 文献[31]中的不等式(2.11)是非严格不等式, 而应该表述如下

$$$\label{9} (P_0+Q_0)x^2(c)-x(c)+\mu_0P_0+\gamma_0Q_0+F_0\geq0.$$$

(iii) 为了得到$(P_0+Q_0)(\mu_0P_0+\gamma_0Q_0+F_0)\geq\frac{1}{4}$ (文献[31]中为$(P_0+Q_0)(\mu_0P_0+\gamma_0Q_0+F_0)>\frac{1}{4}$), 条件$P_0+Q_0>0$是必要的.为此文献[31]中必须增加条件$p^+(t)\not\equiv 0$$q^+(t)\not\equiv 0.事实上, p^+(t)\not\equiv 0$$q^+(t)\not\equiv 0$在证明文献[31]中所有定理和推论时都是必须的.

(i) $(_a {\cal D}^{\alpha}g)=g^{(\alpha)}$;

(ii) 对任意$\beta\geq 0$, $(_a {\cal D}^{\alpha+\beta}g)=(_a {\cal D}^{\beta}g)^{(\alpha)}$.

## 2 引理

$$$\label{lemeq1}z^m+(m-\delta)\delta^{\delta/(m-\delta)}m^{-m/(m-\delta)}\geq z^\delta, \qquad z\in(0, +\infty).$$$

(2.1)式中的等号成立当且仅当$z_0=\big(\frac{\delta}{m}\big)^{\frac{1}{m-\delta}}.$

对任意$m>\delta$, 定义函数${\cal F}(z)=z^m-z^\delta, \ z\in(0, +\infty).$容易知道定义在$(0, \infty)$上的函数${\cal F}$在点$z_0=\big(\delta/m\big)^{1/(m-\delta)}$取得最小值且最小值点唯一, 同时

$$$\label{jian1}\left\{ \begin{array}{lll}(_{a} {\cal D}^{\alpha}x)(t)=-q(t)x(t), \qquad a<t\leq b, \quad \alpha\in(1, 2], \\x(a)=x(b)=0, \end{array}\right.$$$

$[a, b]$上的连续解当且仅当其满足积分方程

$$$\label{green1}G^\alpha(t, s)=\frac{(b-a)^{1-\alpha}}{\Gamma(\alpha)}\left\{\begin{array}{ll}[(t-a)(b-s)]^{\alpha-1}-[(b-a)(t-s)]^{\alpha-1}, &a\leq s\leq t\leq b\\{}[(t-a)(b-s)]^{\alpha-1}, &a\leq t\leq s\leq b\end{array}\right..$$$

(i) 对任意$t, s\in[a, b]$,

(ii) 对任意$s\in [a, b]$,

(a) 对任意$t, s\in[a, b]$和任意$\alpha_i (i=1, 2, \cdots , n)$,

$$$\label{m}G^{\alpha_i}(t, s)\geq 0;$$$

(b) 对任意$s\in[a, b]$和任意$\alpha_i (i=1, 2, \cdots , n)$,

$$$\label{d}\max\limits_{t\in [a, b]}G^{\alpha_i}(t, s)=G^{\alpha_i}(s, s)=\frac{\left[(s-a)(b-s)\right]^{{\alpha_i}-1}}{\Gamma({\alpha_i})(b-a)^{{\alpha_i}-1}}\leq \frac{(b-a)^{\alpha_i-1}}{4^{\alpha_i-1}\Gamma(\alpha_i)};$$$

(c) 对任意$\alpha_i (i=1, 2, \cdots , n)$,

$$$\label{x}\max\limits_{t\in [a, b]}\int_a^bG^{\alpha_i}(t, s){\rm d}s=\frac{(b-a)^{{\alpha_i}}({\alpha_i}-1)^{{\alpha_i}-1}}{{\alpha_i}^{{\alpha_i}}\Gamma({\alpha_i}+1)}:={G_0^{\alpha_i}}.$$$

由(2.6)式可知

## 3 主要结果

$\label {0con1}{\mathcal G}=G_0^{\alpha_1}\cdots{G_0^{\alpha_{n-2}}}{G_0^{\alpha_{n-1}}}>0;\end {equation}$

$\label{con2}\gamma=(m-\beta)\beta^{\beta/(m-\beta)}m^{-m/(m-\beta)};\end {equation}$

$\label {con1}m=\max\{n_0, 2\}, \textrm{而}\ n_0\in {\Bbb N}\ \textrm{满足}\ n_0-1\leq\beta< n_0.\end {equation}$

$\begin{eqnarray}\label{ma1}&&\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\big(\gamma p^+(s)+f^-(s)\big){\rm d}s\bigg)^{m-1}\nonumber\\&&\times\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}p^+(s){\rm d}s\bigg)\geq\bigg(\frac{{\Gamma(\alpha_n)}(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{1-\alpha_n}}\bigg)^m, \end{eqnarray}$

$$$\label{ma2}\bigg(\int_a^b\big(\gamma p^+(s)+f^-(s)\big){\rm d}s\bigg)^{m-1}\bigg(\int_a^bp^+(s){\rm d}s\bigg)>\bigg(\frac{4^{\alpha_n-1}\Gamma(\alpha_n)(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{\alpha_n-1}}\bigg)^m;$$$

(B) 如果在$(a, b)$$x(t)恒负即x(t)<0, 那么有 \begin{eqnarray}\label{d1}&&\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\big(\gamma p^+(s)+f^+(s)\big){\rm d}s\bigg)^{m-1}\nonumber\\&&\times\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}p^+(s){\rm d}s\bigg)\geq\bigg(\frac{{\Gamma(\alpha_n)}(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{1-\alpha_n}}\bigg)^m, \end{eqnarray} $$\label{d2}\bigg(\int_a^b\big(\gamma p^+(s)+f^+(s)\big){\rm d}s\bigg)^{m-1}\bigg(\int_a^bp^+(s){\rm d}s\bigg)>\bigg(\frac{4^{\alpha_n-1}\Gamma(\alpha_n)(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{\alpha_n-1}}\bigg)^m;$$ (C) 如果在(a, b)$$x(t)$取值正负不定, 那么有

$\begin{eqnarray}\label{r1}&&\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\big(\gamma p^+(s)+|f(s)|\big){\rm d}s\bigg)^{m-1}\nonumber\\&&\times\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}p^+(s){\rm d}s\bigg)\geq\bigg(\frac{{\Gamma(\alpha_n)}(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{1-\alpha_n}}\bigg)^m, \end{eqnarray}$

$$$\label{r2}\bigg(\int_a^b\big(\gamma p^+(s)+|f(s)|\big){\rm d}s\bigg)^{m-1}\bigg(\int_a^bp^+(s){\rm d}s\bigg)>\bigg(\frac{4^{\alpha_n-1}\Gamma(\alpha_n)(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{\alpha_n-1}}\bigg)^m.$$$

令$||x||=\max\limits_{t\in [a, b]}|x(t)|.$

(A) 记$x(t)$是方程(1.9)在$[a, b]$上的非零解, 且在$(a, b)$$x(t)>0.则有 $$\label{0c2}(-1)^{n}(_a {\cal D}^{[\alpha_n]} x)(t)=p(t)\Phi_{\beta}(x(t))-f(t)=p(t)x^\beta(t)-f(t)\leq p^+(t)x^{\beta}(t)+f^-(t),$$ $$(_a {\cal D}^{ [\alpha_i]} x)(a)=(_a {\cal D}^{ [\alpha_i]} x)(b)=0, \quad \qquad i=0, 1, 2, \cdots , n-1. \label{c2}$$ i=1, 2, \cdots , n, 令y_{i-1}(t)=(_a {\cal D}^{[\alpha_{i-1}]}x)(t).那么由(1.10)式和(3.11)式可得 $$\label{hb1}\left\{\begin{array}{lll}(_a {\cal D}^{\alpha_{i}}y_{i-1})(t)=(_a {\cal D}^{\alpha_i}(_a {\cal D}^{[\alpha_{i-1}]}x))(t)=(_a {\cal D}^{[\alpha_i]}x)(t), \\y_{i-1}(a)=y_{i-1}(b)=0.\end{array}\right.$$ 利用引理2.2, 由(3.12)式得 $$y_{i-1}(t)=-\int_a^bG^{\alpha_i}(t, s_i)(_a {\cal D}^{[\alpha_i]}x)(s_i){\rm d}s_i,$$ $$\label{ou2g}(_a {\cal D}^{[\alpha_{i-1}]}x)(t)=-\int_a^bG^{\alpha_i}(t, s_i)(_a {\cal D}^{[\alpha_i]}x)(s_i){\rm d}s_i, \quad i=1, 2, \cdots , n.$$ 在(3.14)式中取i=1 $$\label{oua1}x(t)=(_a {\cal D}^{ [\alpha_{0}]} x)(t)=-\int_a^bG^{\alpha_{1}}(t, s_{1})(_a {\cal D}^{[\alpha_1]}x)(s_{1}){\rm d}s_{1}.$$ 在(3.14)式中取i=2并将所得结果带入(3.15)式得 \begin{eqnarray}\label{oua2}x(t)&=&-\int_a^bG^{\alpha_{1}}(t, s_{1})\bigg[-\int_a^bG^{\alpha_{2}}(s_1, s_2)(_a {\cal D}^{ [\alpha_2]} x)(s_2){\rm d}s_2\bigg]{\rm d}s_{1}\nonumber\\&=&(-1)^2\int_{[a, b]^2}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_1, s_2)(_a {\cal D}^{[\alpha_2]}x)(s_2){\rm d}s_2{\rm d}s_{1}.\end{eqnarray} 在(3.14)式中取i=3并将所得结果带入(3.16)式得 依此类推, 重复上述过程n次, 最终可得 \begin{eqnarray}\label{ou3}x(t)=(-1)^n\int_{[a, b]^n}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_n}(s_{n-1}, s_n)(_a {\cal D}^{[\alpha_n]} x)(s_n){\rm d}s_n\cdots{\rm d}s_1.\end{eqnarray} 将(3.10)式带入(3.17)式得 $$\label{oou3}x(t)\leq\int_{[a, b]^n}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_n}(s_{n-1}, s_n)[p^+(s_n)x^{\beta}(s_n)+f^-(s_n)]{\rm d}s_n\cdots{\rm d}s_1.$$ 另一方面, 由(2.5)式可知 $$\label{gm1}G^{\alpha_n}(s_{n-1}, s_n)\leq G^{\alpha_n}(s_{n}, s_n).$$ 将(3.19)式带入(3.18)式并且运用引理2.5得到 \begin{eqnarray}\label{ou3o}x(t)&\leq&\int_{[a, b]^n}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_n}(s_{n}, s_n)\left[p^+(s_n)x^{\beta}(s_n)+f^-(s_n)\right]{\rm d}s_n\cdots{\rm d}s_1\nonumber\\&=&\int_{[a, b]^{n-1}}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_{n-1}}(s_{n-2}, s_{n-1}){\rm d}s_{n-1}\cdots{\rm d}s_1\nonumber\\&&\times\int_a^bG^{\alpha_n}(s, s)\left[p^+(s)x^{\beta}(s)+f^-(s)\right]{\rm d}s\nonumber\\&\leq&{\mathcal G}\int_a^bG^{\alpha_n}(s, s)\left[p^+(s)x^{\beta}(s)+f^-(s)\right]{\rm d}s\nonumber\\&\leq&{\mathcal G}\int_a^bG^{\alpha_n}(s, s)\left[p^+(s)\|x\|^{\beta}+f^-(s)\right]{\rm d}s.\end{eqnarray} 显然, 由(3.20)式知 $$\label{ou4}||x||\leq{\mathcal G}\int_a^bG^{\alpha_n}(s, s)p^+(s)||x||^{\beta}{\rm d}s+{\mathcal G}\int_a^bG^{\alpha_n}(s, s)f^-(s){\rm d}s.$$ 注意到(3.3)式中定义的常数m满足m>\beta, 因此运用引理2.1得 $$\label{ou5}||x||^{\beta}\leq||x||^m+\gamma,$$ 其中\gamma是(3.21)式中定义的常数. 将(3.22)式带入(3.21)式得 其中 p^+(t)\not\equiv0可知{\cal A}_0>0$${\cal C}_0>0.$考虑到$z=||x||>0$为变量, 因此

$$$\label{mm}{\left[(s-a)(b-s)\right]^{{\alpha_n}-1}}\leq\frac{(b-a)^{2(\alpha_n-1)}}{4^{\alpha_n-1}}.$$$

${\left[(s-a)(b-s)\right]^{{\alpha_n}-1}}\not\equiv\frac{(b-a)^{2(\alpha_n-1)}}{4^{\alpha_n-1}}$$p^+(t)\not\equiv0, 可得 $$\label{mm1}\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}p^+(s){\rm d}s<\frac{(b-a)^{2(\alpha_n-1)}}{4^{\alpha_n-1}}\int_a^bp^+(s){\rm d}s,$$ $$\label{mm2}\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\left[\gamma p^+(s)+f^-(s)\right]{\rm d}s <\frac{(b-a)^{2(\alpha_n-1)}}{4^{\alpha_n-1}}\int_a^b\left[\gamma p^+(s)+f^-(s)\right]{\rm d}s.$$ 由(3.24)式和(3.25)式, 结合(3.4)式易得(3.5)式. (B) 记x(t)是方程(1.9)在[a, b]上的非零解, 且在(a, b)$$x(t)<0$.$z(t)=-x(t)$必是下述问题

$$$\label{t1}\left\{\begin{array}{lll}(-1)^{n+1}(_a {\cal D}^{ [\alpha_n]} z)(t)+p(t)\Phi_{\beta}(z(t))=-f(t), \qquad t\in(a, b), \\(_a {\cal D}^{ [\alpha_i]} z)(a)=(_a {\cal D}^{ [\alpha_i]} z)(b)=0, \qquad\qquad i=0, 1, 2, \cdots , n-1\end{array}\right.$$$

$[a, b]$上的非零解, 且在$(a, b)$$z(t)>0.由(A)中结论可得 \begin{eqnarray}\label{tt2}&&\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\left[\gamma p^+(s)+(-f)^-(s)\right]{\rm d}s\bigg)^{m-1}\nonumber\\&&\times\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}p^+(s){\rm d}s\bigg)\geq\bigg(\frac{{\Gamma(\alpha_n)}(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{1-\alpha_n}}\bigg)^m;\end{eqnarray} $$\label{tt3}\bigg(\int_a^b\left[\gamma p^+(s)+(-f)^-(s)\right]{\rm d}s\bigg)^{m-1}\bigg(\int_a^bp^+(s){\rm d}s\bigg)>\bigg(\frac{4^{\alpha_n-1}\Gamma(\alpha_n)(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{\alpha_n-1}}\bigg)^m.$$ 考虑到(-f)^-=f^+, 则不等式(3.27)和(3.28)分别为(3.6)和(3.7)式. (C) 记x(t)是方程(1.9)在[a, b]上的非零解.那么必有x(t)>0, t\in(a, b)$$x(t)<0, t\in(a, b)$.考虑到$f^\pm(t)\leq|f(t)|$, 由(3.4)和(3.6)式易得(3.8)式; 由(3.5)和(3.7)式易得(3.9)式.定理3.1证毕.

$n=1$时, 方程(1.9)退化为

$$$\label{wentik}\left\{\begin{array}{lll} (_a {\cal D}^{ \alpha} x)(t)+p(t)\Phi_{\beta}(x(t))=f(t), \qquad t\in(a, b), \quad\alpha\in (1, 2], \quad\beta>0, \\ x(a)=x(b)=0. \end{array}\right.$$$

$$$\label{p11} \frac{2^{3n-3}}{(b-a)^{2n-3}}\leq \left\{ \begin{array}{ll} \displaystyle\int_a^b(s-a)(b-s)p^+(s){\rm d}s, &\quad n=2r+1\\[4mm] \displaystyle\int_a^b(s-a)(b-s)p^-(s){\rm d}s, &\quad n=2r+2 \end{array}, r=0, 1, 2, 3\cdots; \right.$$$

$$$\label{p12} \frac{2^{3n-1}}{(b-a)^{2n-1}}< \left\{ \begin{array}{ll} \displaystyle\int_a^bp^+(s){\rm d}s, &\quad n=2r+1\\ [4mm]\displaystyle\int_a^bp^-(s){\rm d}s, &\quad n=2r+2 \end{array}, r=0, 1, 2, 3\cdots;\right.$$$

$$$\label{p120} \frac{2^{3n-1}}{(b-a)^{2n-1}}< \displaystyle\int_a^b|p(s)|{\rm d}s, n=1, 2, 3\cdots.$$$

## 4 结论的推广

$\left\{ {\begin{array}{*{20}{l}}{{{( - 1)}^{n + 1}}{(_a}{D^{[{\alpha _n}]}}x)(t) + \sum\limits_{j = 1}^k {{p_j}} (t){\Phi _{{\beta _j}}}(x(t)) = f(t), \qquad t \in (a, b), }\\{{(_a}{D^{[{\alpha _i}]}}x)(a) = {(_a}{D^{[{\alpha _i}]}}x)(b) = 0, \qquad \quad i = 0, 1, 2, \cdots , n - 1, }\end{array}} \right.$

$\label{con11}m=\max\{n_0, 2\}, ~~\textrm{而}\ n_0\in{\Bbb N}\ \textrm{满足}\ n_0-1\leq\max\limits_{1\leq j\leq k}\beta_j< n_0;\end {equation}$

$\label{con21}\gamma_j=(m-\beta_j)\beta_j^{\beta_j/(m-\beta_j)}m^{-m/(m-\beta_j)}, \quad j=1, 2, \cdots k.\end {equation}$

$\begin{eqnarray}\label{ch1}&&\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\bigg(\sum\limits_{j=1}^{k}\gamma_j p^+_j(s)+f^-(s)\bigg){\rm d}s\bigg)^{m-1}\nonumber\\&&\times\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\sum\limits_{j=1}^{k}p^+_j(s){\rm d}s\bigg)\geq\bigg(\frac{{\Gamma(\alpha_n)}(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{1-\alpha_n}}\bigg)^m, \end{eqnarray}$

$$$\label{ch2}\bigg(\int_a^b\bigg(\sum\limits_{j=1}^{k}\gamma_j p^+_j(s)+f^-(s)\bigg){\rm d}s\bigg)^{m-1}\int_a^b\sum\limits_{j=1}^{k}p^+_j(s){\rm d}s>\bigg(\frac{4^{\alpha_n-1}\Gamma(\alpha_n)(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{\alpha_n-1}}\bigg)^m;$$$

(B*)如果在$(a, b)$$x(t)恒负即x(t)<0, 那么有 \begin{eqnarray}\label{cd1}&&\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\bigg(\sum\limits_{j=1}^{k}\gamma_j p^+_j(s)+f^+(s)\bigg){\rm d}s\bigg)^{m-1}\nonumber\\&&\times\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\sum\limits_{j=1}^{k}p^+_j(s){\rm d}s\bigg)\geq\bigg(\frac{{\Gamma(\alpha_n)}(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{1-\alpha_n}}\bigg)^m, \end{eqnarray} $$\label{cd2}\bigg(\int_a^b\bigg(\sum\limits_{j=1}^{k}\gamma_j p^+_j(s)+f^+(s)\bigg){\rm d}s\bigg)^{m-1}\int_a^b\sum\limits_{j=1}^{k}p^+_j(s){\rm d}s>\bigg(\frac{4^{\alpha_n-1}\Gamma(\alpha_n)(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{\alpha_n-1}}\bigg)^m;$$ (C*)如果在(a, b)$$x(t)$取值正负不定, 那么有

${(\int_a^b {{{\left[ {(s - a)(b - s)} \right]}^{{\alpha _n} - 1}}} (\sum\limits_{j = 1}^k {{\gamma _j}} p_j^ + (s) + |f(s)|){\rm{d}}s)^{m - 1}} \\\times (\int_a^b {{{\left[ {(s - a)(b - s)} \right]}^{{\alpha _n} - 1}}} \sum\limits_{j = 1}^k {p_j^ + } (s){\rm{d}}s) \ge {(\frac{{\Gamma ({\alpha _n}){{(m - 1)}^{1 - 1/m}}}}{{m{\cal G}{{(b - a)}^{1 - {\alpha _n}}}}})^m},{\rm{ }}$

$$$\label{t2}\bigg(\int_a^b\bigg(\sum\limits_{j=1}^{k}\gamma_j p^+_j(s)+|f(s)|\bigg){\rm d}s\bigg)^{m-1}\int_a^b\sum\limits_{j=1}^{k}p^+_j(s){\rm d}s>\bigg(\frac{4^{\alpha_n-1}\Gamma(\alpha_n)(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{\alpha_n-1}}\bigg)^m.$$$

(A*)证明过程与定理3.1(A)类似, 关键不同在于$m$的取值.同定理3.1(A)的证明, 从开始直到得到(3.21)式, 这里我们得到的是

$$$\label{ou41}\|x\|\leq{\mathcal G}\int_a^bG^{\alpha_n}(s, s)\sum\limits_{j=1}^{k} p_j^+(s)\|x\|^{\beta_j}{\rm d}s+{\mathcal G}\int_a^bG^{\alpha_n}(s, s)f^-(s) {\rm d}s.$$$

$$$\label{ou51}\|x\|^{\beta_j}\leq \|x\|^m+\gamma_j;\qquad j=1, 2, \cdots , k,$$$

$p^+_{j_0}(t)\not\equiv0$可知$\overline{{\cal A}_0}>0, \overline{{\cal C}_0}>0.$

$$$\label{1c}\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\sum\limits_{j=1}^{k}p^+_j(s){\rm d}s\bigg)<\frac{(b-a)^{2(\alpha_n-1)}}{4^{\alpha_n-1}}\int_a^b\sum\limits_{j=1}^{k}p^+_j(s){\rm d}s,$$$

$\begin{eqnarray} &&\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\bigg(\sum\limits_{j=1}^{k}\gamma_j p^+_j(s)+f^-(s)\bigg){\rm d}s\\&<& \frac{(b-a)^{2(\alpha_n-1)}}{4^{\alpha_n-1}}\int_a^b\bigg[\sum\limits_{j=1}^{k}\gamma_j p^+_j(s)+f^-(s)\bigg]{\rm d}s, \end{eqnarray}$

(B*)和(C*)的证明由(A*)即得, 此处略.

$n=1$$k=2时, 方程(4.1)退化为 $$\label{wenti8}\left\{\begin{array}{lll}(_a {\cal D}^{ \alpha} x)(t)+p_1(t)\Phi_{\mu}(x(t))+p_2(t)\Phi_{\gamma}(x(t))=f(t), \quad t\in(a, b), \alpha\in (1, 2], \\ x(a)= x(b)=0, \end{array}\right.$$ 其中, \mu>0, \gamma>0. 注 4.2 文献[31]中研究了方程(4.14)的Lyapunov型不等式.本文中, 我们去掉了文献[31]中的条件0<\mu<1<\gamma<2, 而允许\mu$$\gamma$为任意正数.当然, 若$0<\mu<1<\gamma<2$, 那么按本文中(4.2)式可得$m=2$, 此时易验证定理4.1(A*)与文献[31]中定理2.1和定理2.5一致, 定理4.1(B*)与文献[31]中定理2.2和2.6一致, 定理4.1(C*)与文献[31]中定理2.3和定理2.7一致.

$f_0=0$, $\alpha_i=2(i=1, 2, \cdots n)$时, 方程(4.1)退化为

$$$\label {wenti0}\left\{\begin{array}{lll} x^{(2n)}(t)+(-1)^{n-1}\sum\limits_{j=1}^{k}f_j(t)\Phi_{\beta_j}(x(t))=0, \qquad t\in(a, b), \\[3mm] x^{(2i)}(a)=x^{(2i)}(b)=0, \qquad i=0, 1, 2, \cdots, n-1. \end{array}\right.$$$

$$$\label{ex1}\left\{\begin{array}{lll}-(_a {\cal D}^{\frac{7}{2}}x)(t)+p_1(t)|x(t)|^{-\frac{1}{2}}x(t)+p_2(t)|x(t)|^{\frac{1}{2}}x(t)+p_3(t)|x(t)|^{\frac{3}{2}}x(t)=f(t), \ t\in(a, b), \\x(a)=x(b)=(_a {\cal D}^{\frac{3}{2}}x)(a)=(_a {\cal D}^{\frac{3}{2}}x)(b)=0, \end{array}\right.$$$

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