对于方程

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

有如下结论:如果$a$和$ b$ ($a<b$)是方程的非零解$x(t)$的两个连续零点, 那么必有

(1.2) $\begin{equation}\label{w2}\int_a^b|\nu(t)|{\rm d}t>\frac{4}{b-a}.\end{equation}$

上述结果参见文献[1]. (1.2)式被称为Lyapunov不等式.在文献[2]中, Wintner将(1.2)式改进如下

(1.3) $\begin{equation}\label{v08.}\int_a^b\nu^+(t){\rm d}t>\frac{4}{b-a}, \end{equation}$

其中$\nu^+(t)=\max\{\nu(t), 0\}$.称(1.3)式为Lyapunov型不等式.研究表明, Lyapunov型不等式(1.3)中的常数$4$不可能被更大的数代替(参见文献[1]和[3, 定理5.1]).后来, 对于方程(1.1), Hartman在文献[3]中得到下述Hartman型不等式

(1.4) $\begin{equation}\label{v08..} \int_a^b(b - t)(t - a)\nu^+(t){\rm d}t>{b-a}. \end{equation} $

由于对任意$t\in (a, b), $都有$(b-t)(t-a)\leq (b-a)^2/4, $因此由不等式(1.4)可得不等式(1.3).

微分方程Lyapunov不等式及广义Lyapunov不等式(如Lyapunov型不等式和Hartman型不等式)被广泛应用在研究振荡理论、渐近理论、非共轭区间的估计以及特征值估计.鉴于其广泛应用, 该研究热点被推广到各种不同的微分方程, 例如对带线性项的偶数阶线性微分方程的研究, 参见文献[4-10]; 对带线性项的奇数阶线性微分方程的研究, 参见文献[5, 11-12]; 对带半线性项的偶数阶线性微分方程的研究, 参见文献[13-14]; 对带半线性项的二阶半线性微分方程的研究, 参见文献[14-19]; 对带线性项的高阶线性微分方程的研究和带半线性项的高阶半线性微分方程的研究, 分别参见文献[20-21]和[22-23].

随着分数阶微积分的发展, 对Lyapunov不等式的研究已经推广到分数阶微分方程

(1.5) $\begin{equation}\label{f1} (_a {\cal D}^{v}x)(t)+\nu(t)x(t)=0, \qquad a<t<b. \end{equation}$

显然, 如果取$v=n\in {\Bbb N}$, 则$(_a D^{v}x)=x^{(n)}$, 从而分数阶微分方程(1.5)退化为整数阶微分方程, 因此对分数阶微分方程Lyapunov不等式的研究是对整数阶微分方程相应研究内容的推广.关于分数阶微分方程(1.5)的Lyapunov型不等式的研究结果, 当$v\in(1, 2]$时参见文献[24-27]; 当$v\in(2, 3]$时参见文献[28-29]; 当$v\in(3, 4]$时参见文献[30].最近, Agarwal等^[31]将Lyapunov不等式的研究推广到下面带强迫项的混合半线性分数阶微分方程

(1.6) $\begin{equation} \label{f2} (_a {\cal D}^{v}x)(t)+p(t)\Phi_{\mu}(x(t))+q(t)\Phi_{\gamma}(x(t))=f(t), \qquad a<t<b, \:v\in(1, 2], \end{equation} $

其中$ 0<\mu<1<\gamma<2, $$\Phi_{*}(s)=|s|^{*-1}s$.文献[31]中非常重要的改进是将(1.5)式中的线性项$\nu(t)x(t)$推广到(1.6)式中的混合半线性项$p(t)\Phi_{\mu}(x(t))+q(t)\Phi_{\gamma}(x(t))$.

显然, 当$\gamma\rightarrow 1^+$和$\mu\rightarrow{1^-}$时, (1.6)式退化成

$(_a {\cal D}^{v}x)(t)+\nu(t)x(t)=f(t), \qquad a<t<b, \:v\in(1, 2], $

其中$\nu(t)=p(t)+q(t)$.因此, 半线性问题(1.6)是线性问题(1.5)的推广.

考虑方程(1.6)的两种特殊情况.其一是次线性项的权因子$p(t)=0$, 则微分方程退化为

(1.7) $\begin{equation} \label{f2.} (_a {\cal D}^{v}x)(t)+q(t)\Phi_{\gamma}(x(t))=f(t), \qquad a<t<b, \:v\in(1, 2], \:1<\gamma<2; \end{equation} $

其二是超线性项的权因子$q(t)=0$, 则微分方程退化为

(1.8) $ \begin{equation} \label{f2..} (_a {\cal D}^{v}x)(t)+p(t)\Phi_{\mu}(x(t))=f(t), \qquad a<t<b, \:v\in(1, 2], \:0<\mu<1. \end{equation}$

截至目前, 还没有文献讨论当方程(1.7), (1.8)和(1.6)中的参数$\mu$和$\gamma$取任意正数时的结果.正如文献[31]中提到, 讨论微分方程

$(_a {\cal D}^{v}x)(t)+q(t)\Phi_{\gamma}(x(t))=f(t), \qquad a<t<b, \quad v\in(1, 2]$

在参数$\gamma\geq2$时的Lyapunov不等式是非常有意义的.本文正是基于此, 讨论一类更广泛的带强迫项的高阶半线性分数阶微分方程的广义Lyapunov不等式.本文研究结果统一并推广了许多现有文献的结果.特别的, 上述提出的问题刚好是本文的特殊情况.

本文讨论下述带强迫项的高阶半线性分数阶微分方程的广义Lyapunov不等式

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

为方便起见, 关于(1.9)式, 做如下说明

(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$f(t)$在$[a, b]$上连续;

(iv) 对$i=1, 2, \cdots , n, $记

(1.10) $ \begin{equation}\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, \end{equation} $

其中$(_a {\cal D}^{ [\alpha_0]} x)(t)=x(t)$; 对$i=1, 2, \cdots , n, $$_a {\cal D}^{\alpha_i}$是$\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$A>0$和$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

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

因为利用其引理2.1中结论只能得到

$\label{35}x^{\mu}(c)\leq x^2(c)+\mu_0, ~~ x^{\gamma}(c)\leq x^2(c)+\gamma_0, $

而非原文中如下的严格不等式

$x^{\mu}(c)< x^2(c)+\mu_0, ~~ x^{\gamma}(c)< x^2(c)+\gamma_0.$

由此由(1.11)式只可得到

$(P_0+Q_0)(\mu_0P_0+\gamma_0Q_0+F_0)\geq\frac{1}{4}, $

而非原文中的

$(P_0+Q_0)(\mu_0P_0+\gamma_0Q_0+F_0)>\frac{1}{4}.$

从而, 文献[31]中不等式(2.3), (2.12)和(2.16), 定理2.9(i)中不等式, 定理2.10(i)中不等式, 以及不等式(2.18)都是非严格不等式.本文接下来的陈述中凡提到上述不等式(例如注3.1, 注3.2, 注3.3以及注4.2), 均指非严格不等式.

(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$p^+(t)\not\equiv 0$或$q^+(t)\not\equiv 0.$事实上, $p^+(t)\not\equiv 0$或$q^+(t)\not\equiv 0$在证明文献[31]中所有定理和推论时都是必须的.

定义 1.1  称$x(t)$是方程(1.9)在$[a, b]$上的非零解, 指$x(t)\in C[a, b]$满足方程(1.9), 且对任意$t\in(a, b), $都有$x(t)\neq 0.$

定义 1.2^[32]  $g$是定义在$[a, b]$上的实函数, $\alpha\geq 0.$函数$g$的$\alpha$阶黎曼-刘维尔分数阶积分定义如下

$ (_a {\cal I}^{\alpha}g)(t) =\left\{ \begin{array}{ll} \displaystyle\frac{1}{\Gamma(\alpha)}\int_a^t{(t-s)^{\alpha-1}g(s)}{\rm d}s, &\quad \alpha>0, \\ g(t), &\quad \alpha=0, \end{array} \right. $

其中$t\in [a, b]$, $\Gamma(\alpha)=\int_0^\infty t^{\alpha-1}e^{-t}{\rm d}t$是Gamma函数.

定义 1.3^[32]  $g$是定义在$[a, b]$上的实函数, $\alpha\geq 0.$函数$g$的$\alpha$阶黎曼-刘维尔分数阶导数定义如下

$(_a {\cal D}^{\alpha}g)(t) =\left\{ \begin{array}{ll} (_a {\cal I}^{m-\alpha}g)^{(m)}(t), &\quad \alpha>0, \\ g(t), &\quad \alpha=0, \end{array} \right.$

其中$t\in [a, b], $$m$是大于等于$\alpha$的最小正整数.

注 1.2  对任意$\alpha\in{\Bbb N}$且$\alpha\geq0$, 有

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

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

接下来, 本文在第2部分将给出一些引理, 主要结果在第3部分给出, 第4部分中会把主要结果推广到更一般的问题.最后给出例子作为理论结果的应用.

引理 2.1  设$\delta>0$是固定常数.则对任意$m>\delta$, 都有

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

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

证  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

${\cal F}_{\min}=-(m-\delta)\delta^{\delta/(m-\delta)}m^{-m/(m-\delta)}, $

即(2.1)式成立, 且知(2.1)式中的等号成立当且仅当$z_0=\big(\delta/m\big)^{1/(m-\delta)}.$证毕.

引理 2.2^[24]  $x(t)$是分数阶微分方程边值问题

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

在$[a, b]$上的连续解当且仅当其满足积分方程

$x(t)=\int^b_aG^\alpha(t, s)q(s)x(s){\rm d}s, \qquad t\in [a, b], $

其中$G^\alpha(t, s)$是问题(2.2)的Green函数且

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

引理 2.3^[24]  (2.3)式中定义的Green函数$G^\alpha(t, s)$满足下述性质

(i) 对任意$t, s\in[a, b]$,

$G^\alpha(t, s)\geq0;$

(ii) 对任意$s\in [a, b]$,

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

引理 2.4^[33]  (2.3)式中定义的Green函数$G^\alpha(t, s)$满足

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

为方便, 对$i=1, 2, \cdots , n$, 若记

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

那么, 由引理2.3和引理2.4可得

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

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

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

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

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

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

引理 2.5  对任意$t\in [a, b]$,

$\int_{[a, b]^{n-1}}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_{n-1}}(s_{n-2}, s_{n-1}){\rm d}s_{n-1}\cdots{\rm d}s_1\leq G_0^{\alpha_1}\cdots{G_0^{\alpha_{n-2}}}{G_0^{\alpha_{n-1}}}.$

证  由(2.6)式可知

$\int_a^bG^{\alpha_i}(t, s){\rm d}s\leq{G_0^{\alpha_i}}, \quad \forall t\in[a, b], \quad i=1, 2, \cdots , n.$

因此对任意$t\in [a, b]$有

$\begin{eqnarray*}&&\int_{[a, b]^{n-1}}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_{n-1}}(s_{n-2}, s_{n-1}){\rm d}s_{n-1}\cdots{\rm d}s_1\nonumber\\&=&\int_{[a, b]^{n-2}}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_{n-2}}(s_{n-3}, s_{n-2})\nonumber\\&&\times\bigg\{\int_a^bG^{\alpha_{n-1}}(s_{n-2}, s_{n-1}){\rm d}s_{n-1}\bigg\}{\rm d}s_{n-2}\cdots{\rm d}s_1\nonumber\\&\leq &\int_{[a, b]^{n-2}}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_{n-2}}(s_{n-3}, s_{n-2}){\rm d}s_{n-2}\cdots{\rm d}s_1G_0^{\alpha_{n-1}}\nonumber\\&=&\int_{[a, b]^{n-3}}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_{n-3}}(s_{n-4}, s_{n-3})\nonumber\\&&\times\bigg\{\int_a^bG^{\alpha_{n-2}}(s_{n-3}, s_{n-2}){\rm d}s_{n-2}\bigg\}{\rm d}s_{n-3}\cdots{\rm d}s_1G_0^{\alpha_{n-1}}\nonumber\\&\leq &\int_{[a, b]^{n-3}}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_{n-3}}(s_{n-4}, s_{n-3}){\rm d}s_{n-3}\cdots{\rm d}s_1G_0^{\alpha_{n-2}}G_0^{\alpha_{n-1}}, \nonumber\label{p1}\end{eqnarray*}$

依此类推得

$\int_{[a, b]^{n-1}}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_{n-1}}(s_{n-2}, s_{n-1}){\rm d}s_{n-1}\cdots{\rm d}s_1 \leq G_0^{\alpha_1}\cdots{G_0^{\alpha_{n-2}}}{G_0^{\alpha_{n-1}}}.$

引理2.5证毕.

为方便起见, 记$u^\pm=\max\{\pm u, 0\};$引入常数

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

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

其中(3.2)式中$m$定义如下

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

定理 3.1  假设在$[a, b]$上$p^+(t)\not\equiv0$且方程(1.9)在$[a, b]$上有非零解$x(t)$, 那么有下面结论:

(A) 如果在$(a, b)$上$x(t)$恒正即$x(t)>0$, 那么有

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

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

(B) 如果在$(a, b)$上$x(t)$恒负即$x(t)<0$, 那么有

(3.6) $\begin{eqnarray}\label{d1}&&\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\big(\gamma p^+(s)+f^+(s)\big){\rm d}s\bigg)^{m-1}\nonumber\\&&\times\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}p^+(s){\rm d}s\bigg)\geq\bigg(\frac{{\Gamma(\alpha_n)}(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{1-\alpha_n}}\bigg)^m, \end{eqnarray}$

(3.7) $\begin{equation}\label{d2}\bigg(\int_a^b\big(\gamma p^+(s)+f^+(s)\big){\rm d}s\bigg)^{m-1}\bigg(\int_a^bp^+(s){\rm d}s\bigg)>\bigg(\frac{4^{\alpha_n-1}\Gamma(\alpha_n)(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{\alpha_n-1}}\bigg)^m;\end{equation}$

(C) 如果在$(a, b)$上$x(t)$取值正负不定, 那么有

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

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

证  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$(a, b)$上$x(t)>0$.则有

(3.10) $\begin{equation}\label{0c2}(-1)^{n}(_a {\cal D}^{[\alpha_n]} x)(t)=p(t)\Phi_{\beta}(x(t))-f(t)=p(t)x^\beta(t)-f(t)\leq p^+(t)x^{\beta}(t)+f^-(t), \end{equation}$

(3.11) $\begin{equation}(_a {\cal D}^{ [\alpha_i]} x)(a)=(_a {\cal D}^{ [\alpha_i]} x)(b)=0, \quad \qquad i=0, 1, 2, \cdots , n-1. \label{c2}\end{equation}$

对$i=1, 2, \cdots , n$, 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

(3.12) $\begin{equation}\label{hb1}\left\{\begin{array}{lll}(_a {\cal D}^{\alpha_{i}}y_{i-1})(t)=(_a {\cal D}^{\alpha_i}(_a {\cal D}^{[\alpha_{i-1}]}x))(t)=(_a {\cal D}^{[\alpha_i]}x)(t), \\y_{i-1}(a)=y_{i-1}(b)=0.\end{array}\right.\end{equation}$

利用引理2.2, 由(3.12)式得

(3.13) $\begin{equation}y_{i-1}(t)=-\int_a^bG^{\alpha_i}(t, s_i)(_a {\cal D}^{[\alpha_i]}x)(s_i){\rm d}s_i, \end{equation}$

即

(3.14) $\begin{equation}\label{ou2g}(_a {\cal D}^{[\alpha_{i-1}]}x)(t)=-\int_a^bG^{\alpha_i}(t, s_i)(_a {\cal D}^{[\alpha_i]}x)(s_i){\rm d}s_i, \quad i=1, 2, \cdots , n.\end{equation}$

在(3.14)式中取$i=1$得

(3.15) $\begin{equation}\label{oua1}x(t)=(_a {\cal D}^{ [\alpha_{0}]} x)(t)=-\int_a^bG^{\alpha_{1}}(t, s_{1})(_a {\cal D}^{[\alpha_1]}x)(s_{1}){\rm d}s_{1}.\end{equation}$

在(3.14)式中取$i=2$并将所得结果带入(3.15)式得

(3.16) $\begin{eqnarray}\label{oua2}x(t)&=&-\int_a^bG^{\alpha_{1}}(t, s_{1})\bigg[-\int_a^bG^{\alpha_{2}}(s_1, s_2)(_a {\cal D}^{ [\alpha_2]} x)(s_2){\rm d}s_2\bigg]{\rm d}s_{1}\nonumber\\&=&(-1)^2\int_{[a, b]^2}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_1, s_2)(_a {\cal D}^{[\alpha_2]}x)(s_2){\rm d}s_2{\rm d}s_{1}.\end{eqnarray}$

在(3.14)式中取$i=3$并将所得结果带入(3.16)式得

$\begin{eqnarray*}x(t)&=&(-1)^2\int_{[a, b]^2}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_1, s_2)\bigg[-\int_a^bG^{\alpha_3}(s_2, s_3)(_a {\cal D}^{[\alpha_3]} x)(s_3){\rm d}s_3\bigg]{\rm d}s_2{\rm d}s_{1}\\&=&(-1)^3\int_{[a, b]^3}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_1, s_2)G^{\alpha_{3}}(s_2, s_3)(_a {\cal D}^{ [\alpha_3]} x)(s_3){\rm d}s_3{\rm d}s_2{\rm d}s_{1}.\end{eqnarray*}$

依此类推, 重复上述过程$n$次, 最终可得

(3.17) $\begin{eqnarray}\label{ou3}x(t)=(-1)^n\int_{[a, b]^n}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_n}(s_{n-1}, s_n)(_a {\cal D}^{[\alpha_n]} x)(s_n){\rm d}s_n\cdots{\rm d}s_1.\end{eqnarray}$

将(3.10)式带入(3.17)式得

(3.18) $\begin{equation}\label{oou3}x(t)\leq\int_{[a, b]^n}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_n}(s_{n-1}, s_n)[p^+(s_n)x^{\beta}(s_n)+f^-(s_n)]{\rm d}s_n\cdots{\rm d}s_1.\end{equation}$

另一方面, 由(2.5)式可知

(3.19) $\begin{equation}\label{gm1}G^{\alpha_n}(s_{n-1}, s_n)\leq G^{\alpha_n}(s_{n}, s_n).\end{equation}$

将(3.19)式带入(3.18)式并且运用引理2.5得到

(3.20) $\begin{eqnarray}\label{ou3o}x(t)&\leq&\int_{[a, b]^n}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_n}(s_{n}, s_n)\left[p^+(s_n)x^{\beta}(s_n)+f^-(s_n)\right]{\rm d}s_n\cdots{\rm d}s_1\nonumber\\&=&\int_{[a, b]^{n-1}}G^{\alpha_{1}}(t, s_{1})G^{\alpha_{2}}(s_{1}, s_{2})\cdots G^{\alpha_{n-1}}(s_{n-2}, s_{n-1}){\rm d}s_{n-1}\cdots{\rm d}s_1\nonumber\\&&\times\int_a^bG^{\alpha_n}(s, s)\left[p^+(s)x^{\beta}(s)+f^-(s)\right]{\rm d}s\nonumber\\&\leq&{\mathcal G}\int_a^bG^{\alpha_n}(s, s)\left[p^+(s)x^{\beta}(s)+f^-(s)\right]{\rm d}s\nonumber\\&\leq&{\mathcal G}\int_a^bG^{\alpha_n}(s, s)\left[p^+(s)\|x\|^{\beta}+f^-(s)\right]{\rm d}s.\end{eqnarray}$

显然, 由(3.20)式知

(3.21) $\begin{equation}\label{ou4}||x||\leq{\mathcal G}\int_a^bG^{\alpha_n}(s, s)p^+(s)||x||^{\beta}{\rm d}s+{\mathcal G}\int_a^bG^{\alpha_n}(s, s)f^-(s){\rm d}s.\end{equation}$

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

(3.22) $\begin{equation}\label{ou5}||x||^{\beta}\leq||x||^m+\gamma, \end{equation}$

其中$\gamma$是(3.21)式中定义的常数.

将(3.22)式带入(3.21)式得

${\cal A}_0||x||^m-||x||+{\cal C}_0\geq 0, $

其中

${\cal A}_0={\mathcal G}\int_a^bG^{\alpha_n}(s, s)p^+(s){\rm d}s, \quad{\cal C}_0={\mathcal G}\int_a^bG^{\alpha_n}(s, s)[\gamma p^+(s)+f^-(s)]{\rm d}s.$

由$p^+(t)\not\equiv0$可知${\cal A}_0>0$且${\cal C}_0>0.$考虑到$z=||x||>0$为变量, 因此

$\begin{eqnarray*}\min\limits_{0<z<\infty}\left\{{\cal A}_0 z^m- z+{\cal C}_0\right\}&=&\left\{{\cal A}_0 z^m- z+{\cal C}_0\right\}|_{z=(m{\cal A}_0)^{\frac{1}{1-m}}}\\&=&{\cal C}_0-(m-1){\cal A}_0^{1/(1-m)}m^{m/(1-m)}\geq 0, \end{eqnarray*}$

即

${\cal C}_0\geq(m-1){\cal A}_0^{1/(1-m)}m^{m/(1-m)}, $

此式即为(3.4)式.

考虑到

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

且${\left[(s-a)(b-s)\right]^{{\alpha_n}-1}}\not\equiv\frac{(b-a)^{2(\alpha_n-1)}}{4^{\alpha_n-1}}$和$p^+(t)\not\equiv0, $可得

(3.24) $\begin{equation}\label{mm1}\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}p^+(s){\rm d}s<\frac{(b-a)^{2(\alpha_n-1)}}{4^{\alpha_n-1}}\int_a^bp^+(s){\rm d}s, \end{equation}$

(3.25) $\begin{equation}\label{mm2}\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\left[\gamma p^+(s)+f^-(s)\right]{\rm d}s <\frac{(b-a)^{2(\alpha_n-1)}}{4^{\alpha_n-1}}\int_a^b\left[\gamma p^+(s)+f^-(s)\right]{\rm d}s.\end{equation}$

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$(a, b)$上$x(t)<0$.则$z(t)=-x(t)$必是下述问题

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

在$[a, b]$上的非零解, 且在$(a, b)$上$z(t)>0$.由(A)中结论可得

(3.27) $\begin{eqnarray}\label{tt2}&&\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\left[\gamma p^+(s)+(-f)^-(s)\right]{\rm d}s\bigg)^{m-1}\nonumber\\&&\times\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}p^+(s){\rm d}s\bigg)\geq\bigg(\frac{{\Gamma(\alpha_n)}(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{1-\alpha_n}}\bigg)^m;\end{eqnarray}$

(3.28) $\begin{equation}\label{tt3}\bigg(\int_a^b\left[\gamma p^+(s)+(-f)^-(s)\right]{\rm d}s\bigg)^{m-1}\bigg(\int_a^bp^+(s){\rm d}s\bigg)>\bigg(\frac{4^{\alpha_n-1}\Gamma(\alpha_n)(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{\alpha_n-1}}\bigg)^m.\end{equation}$

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$x(t)>0, t\in(a, b)$或$x(t)<0, t\in(a, b)$.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.

当$n=1$时, 方程(1.9)退化为

(3.29) $\begin{equation}\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.\end{equation} $

此时由定理3.1(C)得下面推论.

推论 3.1  假设在$[a, b]$上$p^+(t)\not\equiv0$且问题\reff{wentik}在$[a, b]$上有非零解$x(t)$, 那么有

(3.30) $ \begin{eqnarray}\label{k3} &&\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha-1}}[\gamma p^+(s)+|f(s)|]{\rm d}s\bigg)^{m-1}\nonumber\\ &&\times\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha-1}}p^+(s){\rm d}s\bigg)\geq\bigg(\frac{{\Gamma(\alpha)}(m-1)^{1-1/m}}{m(b-a)^{1-\alpha}}\bigg)^m, \end{eqnarray}$

(3.31) $\begin{equation}\label{k4} \bigg(\int_a^b(\gamma p^+(s)+|f(s)|){\rm d}s\bigg)^{m-1}\bigg(\int_a^bp^+(s){\rm d}s\bigg)>\bigg(\frac{4^{\alpha-1}\Gamma(\alpha)(m-1)^{1-1/m}}{m(b-a)^{\alpha-1}}\bigg)^m. \end{equation}$

注 3.1  当$\beta\in(0, 1)$时, 由(3.3)式得$m=2$.此时推论3.1的结论即为文献[31]中定理2.9.

注 3.2  当$\beta\in(1, 2)$时, 由(3.3)式得$m=2$.此时推论3.1的结论即为文献[31]中定理2.10.

注 3.3  当$\beta=1$时, 分别由(3.3)和(3.2)式得$m=2$和$\gamma=1/4$.此时推论3.1的结论即为文献[31]中推论2.11.

在方程(1.9)中如果取$\beta=1$, $f=0, $$\alpha_i=2(i=1, 2, \cdots , n)$, 那么方程(1.9)退化为

(3.32) $\begin{equation}\label{wenti1}\left\{\begin{array}{lll}x^{(2n)}(t)+(-1)^{n-1}p(t)x(t)=0, \qquad t\in(a, b), \\x^{(2i)}(a)=x^{(2i)}(b)=0, \qquad i=0, 1, 2, \cdots , n-1.\end{array}\right.\end{equation}$

此时常数${\mathcal G}$, $m$, $\gamma$分别是${(b-a)^{2(n-1)}}/{8^{n-1}}$, $2$, $1/4$, 由定理3.1(C)得下面推论.

推论 3.2  假设在$[a, b]$上$p^+(t)\not\equiv 0$, $p^-(t)\not \equiv 0$, 且方程(3.32)在$[a, b]$上有非零解$x(t)$, 那么有

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

(3.34) $\begin{equation}\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. \end{equation} $

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

注 3.4  问题(3.32)的Lyapunov不等式在文献[4-10]中被广泛研究.在文献[8]的定理2和文献[9]的定理2.1中, 不等式(3.35)左侧的常数是$\frac{2^{2n}}{(b-a)^{2n-1}}$.因此, 当$n>1$时, 本文中的Lyapunov不等式(3.35)要优于文献[8-9]中结论; 当$n=1$时, 本文中的Lyapunov不等式(3.35)与文献[8-9]中结论一致.当然, 不等式(3.33)和(3.34)是全新结果.

研究如下带强迫项的混合半线性分数阶微分方程的Lyapunov型和Hartman型不等式

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

其中$n, k\in{\Bbb N}$且$n\geq 1, k\geq 2$; 对$i=1, 2, \cdots , n, $有$\alpha_i\in(1, 2];$对$j=1, 2, \cdots , k, $有$\beta_j>0$且对$i\neq j, $有$\beta_i\neq \beta_j;$对$ j=1, 2, \cdots , k, $有$p_j(t)\in C([a, b], {\Bbb R});$$f(t)\in C([a, b], {\Bbb R})$.

为方便, 记

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

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

定理 4.1  假设存在$j_0\in\{1, 2, \cdots , k\}$使得在$[a, b]$上$p_{j_0}^+(t)\not\equiv0, $且方程(4.1)在$[a, b]$上有非零解$x(t)$.

(A*)如果在$(a, b)$上$x(t)$恒正即$x(t)>0$, 那么有

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

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

(B*)如果在$(a, b)$上$x(t)$恒负即$x(t)<0$, 那么有

(4.6) $\begin{eqnarray}\label{cd1}&&\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\bigg(\sum\limits_{j=1}^{k}\gamma_j p^+_j(s)+f^+(s)\bigg){\rm d}s\bigg)^{m-1}\nonumber\\&&\times\bigg(\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\sum\limits_{j=1}^{k}p^+_j(s){\rm d}s\bigg)\geq\bigg(\frac{{\Gamma(\alpha_n)}(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{1-\alpha_n}}\bigg)^m, \end{eqnarray}$

(4.7) $\begin{equation}\label{cd2}\bigg(\int_a^b\bigg(\sum\limits_{j=1}^{k}\gamma_j p^+_j(s)+f^+(s)\bigg){\rm d}s\bigg)^{m-1}\int_a^b\sum\limits_{j=1}^{k}p^+_j(s){\rm d}s>\bigg(\frac{4^{\alpha_n-1}\Gamma(\alpha_n)(m-1)^{1-1/m}}{m{\mathcal G}(b-a)^{\alpha_n-1}}\bigg)^m;\end{equation}$

(C*)如果在$(a, b)$上$x(t)$取值正负不定, 那么有

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

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

证  (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

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

由(4.2)式可断定$m>\beta_j(j=1, 2, \cdots , k)$, 因此由引理2.1可得

(4.11) $\begin{equation}\label{ou51}\|x\|^{\beta_j}\leq \|x\|^m+\gamma_j;\qquad j=1, 2, \cdots , k, \end{equation}$

其中$\gamma_j (j=1, 2, \cdots , k)$是(4.3)式中定义的常数.记

$\overline{{\cal A}_0}={\mathcal G}\int_a^bG^{\alpha_n}(s, s)\sum\limits_{j=1}^{k}p_j^+(s){\rm d}s, \, \, \, \, \, \overline{{\cal C}_0}={\mathcal G}\int_a^bG^{\alpha_n}(s, s)\bigg(\sum\limits_{j=1}^{k}\gamma_j p_j^+(s)+f^-(s)\bigg){\rm d}s.$

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

将(4.11)式带入(4.10)式得

$\begin{eqnarray*}%\label{ou6}\overline{{\cal A}_0} \|x\|^m- \|x\|+\overline{{\cal C}_0}\geq 0.\end{eqnarray*}$

考虑到$\|x\|$为变量, 因此

$\begin{array}{lll}\min\limits_{0<z<\infty}\left\{\overline{{\cal A}_0} z^m-z+\overline{{\cal C}_0}\right\}&=\left\{\overline{{\cal A}_0} z^m-z+\overline{{\cal C}_0}\right\}|_{z=(m\overline{{\cal A}_0})^{\frac{1}{1-m}}}\\&=\overline{{\cal C}_0}-(m-1)\overline{{\cal A}_0}^{1/(1-m)}m^{m/(1-m)}\geq 0, \end{array}$

即

$\overline{{\cal C}_0}\geq(m-1)\overline{{\cal A}_0}^{1/(1-m)}m^{m/(1-m)}, $

此即为不等式(4.4).

接下来, 与(3.24)和(3.25)式推导过程类似, 可得

$\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}p^+_{j_0}(s){\rm d}s<\frac{(b-a)^{2(\alpha_n-1)}}{4^{\alpha_n-1}}\int_a^bp^+_{j_0}(s){\rm d}s, $

$\int_a^b{\left[(s-a)(b-s)\right]^{\alpha_n-1}}\left[\gamma p^+_{j_0}(s)+f^-(s)\right]{\rm d}s <\frac{(b-a)^{2(\alpha_n-1)}}{4^{\alpha_n-1}}\int_a^b\left[\gamma p^+_{j_0}(s)+f^-(s)\right]{\rm d}s, $

从而有

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

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

由(4.12)和(4.13)式, 结合(4.4)式即得(4.5)式.

(B*)和(C*)的证明由(A*)即得, 此处略.

推论 4.1  假设存在$j_1, j_2\in\{1, 2, \cdots , k\}$使得$p_{j_1}^+(t)\not\equiv 0$且$p_{j_2}^+(t)\not\equiv0.$如果方程(4.1)在$[a, b]$上有非零解$x(t)$, 那么不等式(4.4), (4.6)和(4.8)都是严格不等式.

证  证明过程与定理4.1类似, 从不等式(4.11)开始有所不同.对(4.11)式, 由引理2.1我们知道, 对$j=1, 2, \cdots, k, $不等式(4.11)中得等号成立当且仅当$z_0^j=\big(\frac{\beta_j}{m}\big)^{\frac{1}{m-\beta_j}}.$考虑到$\big(\frac{x}{m}\big)^{\frac{1}{m-x}}$关于$x$是单调上升的, 因此由$\beta_i\neq \beta_j$知$z_0^i\neq z_0^j$.也即对(4.11)式中的$k$个不等式, 至多有一个不等式是不严格的, 而其它$k-1$个不等式都是严格的.由条件$p_{j_1}^+(t)\not\equiv 0$且$p_{j_2}^+(t)\not\equiv 0, $可知

$\begin{eqnarray*}\|x\|&\leq&{\mathcal G}\int_a^bG^{\alpha_n}(s, s)\sum\limits_{j=1}^{k} p_j^+(s)\|x\|^{\beta_j}{\rm d}s+{\mathcal G}\int_a^bG^{\alpha_n}(s, s)f^-(s) {\rm d}s\\ &<&{\mathcal G}\int_a^bG^{\alpha_n}(s, s)\sum\limits_{j=1}^{k} p_j^+(s){\rm d}s\|x\|^{m}+{\mathcal G}\int_a^bG^{\alpha_n}(s, s)\bigg(\sum\limits_{j=1}^{k}\gamma_j p_j^+(s)+f^-(s)\bigg) {\rm d}s, \end{eqnarray*}$

也即

$\overline{{\cal A}_0} \|x\|^m- \|x\|+\overline{{\cal C}_0}> 0.$

考虑到$\|x\|$是一个变量, 得

$\begin{eqnarray*}\min\limits_{0<z<\infty}\left\{\overline{{\cal A}_0} z^m-z+\overline{{\cal C}_0}\right\}&=&\left\{\overline{{\cal A}_0} z^m-z+\overline{{\cal C}_0}\right\}|_{z=(m\overline{{\cal A}_0})^{\frac{1}{1-m}}}\\ &=&\overline{{\cal C}_0}-(m-1)\overline{{\cal A}_0}^{1/(1-m)}m^{m/(1-m)}> 0, \end{eqnarray*}$

即

$\overline{{\cal A}_0}\times\overline{{\cal C}_0}^{m-1}>(m-1)^{m-1}m^{-m}, $

此式说明不等式(4.4)是严格不等式.因而, (4.6)和(4.8)式都是严格不等式.

注 4.1  由推论4.1知, 必须在文献[31]的定理2.1, 2.2和2.3中增加条件$p^+(t)\not\equiv 0$且$q^+(t)\not\equiv 0, $文献[31]中的不等式(2.3), (2.12)和(2.16)才是严格不等式, 涉及其后面的推论亦如此.

当$n=1$且$k=2$时, 方程(4.1)退化为

(4.14) $\begin{equation}\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.\end{equation} $

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$\mu$和$\gamma$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.

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

(4.15) $\begin{equation}\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.\end{equation} $

注 4.3  关于方程(4.15)的Lyapunov型不等式, 文献[13]中研究了$n\geq 1$的情形, 文献[14]研究了$n=1$的情形.这里特别指出, 文献[13]和[14]中都要求$\beta_j\in(0, 2)\setminus\{1\}$, $j=1, 2, \cdots , k$, 但本文中去掉了这些限制.

例 4.1  考虑下面带强迫项的混合半线性分数阶微分方程

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

其中$p_j(j=1, 2, 3)$和$f$均在$[a, b]$上连续; $p^+_1(t)\not\equiv 0$或$p^+_2(t)\not\equiv 0$或$p^+_3(t)\not\equiv 0.$

由注  知$ (_a {\cal D}^{\frac{7}{2}}x)(t)=(_a {\cal D}^{\frac{3}{2}}x)"(t)=(_a {\cal D}^{ 2}(_a {\cal D}^{\frac{3}{2}}x))(t).$将问题(4.16)与(4.1)进行比对, 可见$n=2$, $k=3$, $\alpha_1=\frac{3}{2}$, $\alpha_2=2$, $\beta_1=\frac{1}{2}$, $\beta_2=\frac{3}{2}$以及$\beta_3=\frac{5}{2}$.按(4.2)和(4.3)式计算得$m=3$, $\gamma_1=5\times6^{-\frac{6}{5}}$, $\gamma_2=\frac{1}{4}$, $\gamma_3=5^5\times6^{-6}$以及$ {\mathcal G}=\frac{8\sqrt{3}}{27\sqrt{\pi}}(b-a)^{\frac{3}{2}}. $

由定理4.1(C*)可以断定, 如果方程(4.16)在$[a, b]$上有非零解, 则有

(4.17) $\begin{eqnarray}\label{eqex1}(b-a)^{-\frac{3}{2}}&\leq& \int_a^b(s-a)(b-s)\big(p^+_1(s)+p^+_2(s)+p^+_3(s)\big){\rm d}s\nonumber\\&&\times\bigg(\int_a^b(s-a)(b-s)\big(c_1p^+_1(s)+c_2p^+_2(s)+c_3p^+_3(s)+c_4|f(s)|\big){\rm d}s\bigg)^2, \end{eqnarray}$

(4.18) $\begin{eqnarray}\label{eqex2}(b-a)^{-\frac{15}{2}}&\leq&\int_a^b\big(p^+_1(s)+p^+_2(s)+p^+_3(s)\big){\rm d}s\nonumber\\&&\times\bigg(\int_a^b\big(k_1p^+_1(s)+k_2p^+_2(s)+k_3p^+_3(s)+k_4|f(s)|\big){\rm d}s\bigg)^2, \end{eqnarray}$

其中

$\begin{array}{lllll}&c_1\approx0.235725, \quad &c_2\approx0.101195, \quad &c_3\approx0.027111, \quad &c_4\approx0.404780;\\&k_1\approx0.029465, \quad &k_2\approx0.012649, \quad &k_3\approx0.003388, \quad &k_4\approx0.050597.\end{array}$

进一步, 由定理4.1(A*)可知, 如果在$(a, b)$上$x(t)>0$, 那么在不等式(4.17)和(4.18)中将$|f(s)|$替换为$f^-(s)$时不等式仍成立; 由定理4.1(B*)可知, 如果在$(a, b)$上$x(t)<0$, 那么在不等式(4.17)和(4.18)中将$|f(s)|$替换为$f^+(s)$时不等式亦成立.