Lyapunov-type inequality and solution for a fractional differential equation

Abstract

In this paper, we consider the linear fractional differential equation

By obtaining the Green’s function we derive the Lyapunov-type inequality for such a boundary value problem. Furthermore, we use the contraction mapping theorem to study the existence of a unique solution for the corresponding nonlinear problem.

1 Introduction

In 1907, Lyapunov [1] stated the following outstanding result.

Theorem 1.1

([1])

If the boundary value problem (BVP)

$$ \textstyle\begin{cases} y''(t)+q(t)y(t)=0,\quad a< t< b, \\ y(a)=y(b)=0, \end{cases} $$

has a nontrivial solution, then we have the following Lyapunov inequality:

$$ \int _{a}^{b} \bigl\vert q(s) \bigr\vert \,ds >\frac{4}{b-a}. $$

(1)

Inequality (1) is very useful in various problems related to differential equations. Since the appearance of Lyapunov’s fundamental paper [1], many improvements and generalizations of inequality (1) for integer-order (second- and higher-order) BVPs have appeared in the literature; we refer the reader to the summary reference by Tiryaki [2].

Recently, the studies on Lyapunov’s inequality for fractional boundary value problem (FBVP) have begun, in which fractional derivatives (Riemann–Liouville derivative or Caputo derivative ) are used instead of the classical ordinary derivative. Such a work was initiated by Ferreira [3] in 2013, who obtained a Lyapunov inequality for the following differential equation with Riemann–Liouville fractional derivative:

(2)

subject to the boundary value condition

$$ y(a)=y(b)=0. $$

(3)

Next, in 2014, Ferreira [4] obtained a Lyapunov inequality for the following differential equation with Caputo fractional derivative:

(4)

subject to boundary value condition (3).

After [3] and [4], many results appeared in the literature; we refer the reader to [5–10], where Lyapunov or Lyapunov-type inequalities are obtained for fractional differential equation subject various boundary value conditions such as

$$\begin{aligned}& y'(a)=y'(b)=y(c)=0,\quad a< b, c\in [a,b]; \\& y(a)=y'(a)=0, \qquad y'(b)=\beta y'(\xi ); \\& y(a)=y'(a)=y''(a)=y''(b)=0; \\& y(a)=y'(a)=y(b)=0. \end{aligned}$$

Inspired by the works mentioned, in this paper, we aim to investigate the Lyapunov-type inequality for the following fractional differential equations:

(5)

where δ and γ are real numbers, and $q(t)\in L(0,1)$ is not identically zero on any compact subinterval of $(0,1)$. Furthermore, we obtain the existence of a solution for the corresponding nonlinear problem:

(6)

BVP (6) was recently studied in [11], but we should point out that only the case of $\delta >1$ and $0<\gamma <1$ was considered in [11]. In this paper, we give a comprehensive discussion on parameters δ and γ.

2 Preliminaries and lemmas

For convenience, we present some definitions and lemmas from fractional calculus theory in the sense of Riemann–Liouville and Caputo.

Definition 2.1

([12])

Let $\varGamma (v)=\int _{0}^{\infty }t^{v-1}e^{-t}\,dt$, $v>0$, be the gamma function. Then the Riemann–Liouville fractional integral of order v for $y(t)$ is defined as

Definition 2.2

([12])

Let $v>0$ and $n=[v]+1$, where $[v]$ denotes the integer part of a number v. Then the Caputo fractional derivative of order v for $y(t)$ is defined as

By Definitions 2.1 and 2.2 we have

(7)

Lemma 2.1

A function$u(t)$is a solution of the boundary value problem (5) if and only if$u(t)$satisfies

$$ u(t)= \int _{0}^{1}G(t,s)q(s)u(s)\,ds, $$

where

$$ G(t,s)=\frac{1}{\varGamma (\nu )}\textstyle\begin{cases} (1-\nu )\frac{\delta \gamma (1-t)+\gamma t}{(1-\gamma )(1-\delta )}(1-s)^{ \nu -2}-\frac{\delta }{1-\delta }(1-s)^{\nu -1}-(t-s)^{\nu -1}, \\ \quad 0\leq s \leq t\leq 1 , \\ (1-\nu )\frac{\delta \gamma (1-t)+\gamma t}{(1-\gamma )(1-\delta )}(1-s)^{ \nu -2}-\frac{\delta }{1-\delta }(1-s)^{\nu -1}, \\ \quad 0\leq t\leq s\leq 1. \end{cases} $$

(8)

Proof

Let $u(t)$ be a solution of (5). Then

By (7) we obtain

$$ u(t)=C_{1}+C_{2}t- \frac{1}{\varGamma (\nu )} \int _{0}^{t}(t-s)^{\nu -1}q(s)u(s) \,ds.$$

(9)

Considering $u(0)=\delta u(1)$, we have

$$ C_{1}=\delta C_{1}+\delta C_{2}- \frac{\delta }{\varGamma (\nu )} \int _{0}^{1}(1-s)^{ \nu -1}q(s)u(s) \,ds; $$

considering $u'(0)=\gamma u'(1)$, we have

$$ C_{2}=\frac{\gamma (\nu -1)}{\varGamma (\nu )(\gamma -1)} \int _{0}^{1}(1-s)^{ \nu -2}q(s)u(s) \,ds, $$

(10)

and thus we get

$$\begin{aligned} C_{1} =& \frac{\delta \gamma (\nu -1)}{\varGamma (\nu )(1-\delta )(\gamma -1)} \int _{0}^{1}(1-s)^{\nu -2}q(s)u(s)\,ds \\ &{}- \frac{\delta }{\varGamma (\nu )(1-\delta )} \int _{0}^{1}(1-s)^{\nu -1}q(s)u(s) \,ds. \end{aligned}$$

(11)

Substituting (10) and (11) into (9), we obtain

$$\begin{aligned} u(t) =& \frac{\delta \gamma (\nu -1)}{\varGamma (\nu )(1-\delta )(\gamma -1)} \int _{0}^{1}(1-s)^{\nu -2}q(s)u(s)\,ds \\ &{}- \frac{\delta }{\varGamma (\nu )(1-\delta )} \int _{0}^{1}(1-s)^{\nu -1}q(s)u(s)\,ds \\ &{} + \frac{\gamma (\nu -1)t}{\varGamma (\nu )(\gamma -1)} \int _{0}^{1}(1-s)^{ \nu -2}q(s)u(s) \,ds-\frac{1}{\varGamma (\nu )} \int _{0}^{t}(t-s)^{\nu -1}q(s)u(s)\,ds \\ =& \int _{0}^{1}G(t,s)q(s)u(s)\,ds, \end{aligned}$$

where $G(t,s)$ is the Green’s function:

$$ G(t,s)=\frac{1}{\varGamma (\nu )}\textstyle\begin{cases} (1-\nu )\frac{\delta \gamma (1-t)+\gamma t}{(1-\gamma )(1-\delta )}(1-s)^{ \nu -2}-\frac{\delta }{1-\delta }(1-s)^{\nu -1}-(t-s)^{\nu -1}, \\ \quad 0\leq s \leq t\leq 1 , \\ (1-\nu )\frac{\delta \gamma (1-t)+\gamma t}{(1-\gamma )(1-\delta )}(1-s)^{ \nu -2}-\frac{\delta }{1-\delta }(1-s)^{\nu -1}, \\ \quad 0\leq t\leq s\leq 1. \end{cases} $$

 □

Lemma 2.2

When$\delta \in (0,1)$and$\gamma \in (0,1)$, Green’s function G(t,s) satisfies the following properties:

• (i) $G(t,s)\leq 0$, $(t,s)\in [0,1]\times [0,1]$;

• (ii) $\max_{0\leq t \leq 1} \vert G(t,s) \vert =-G(1,s)= \frac{(1-s)^{\nu -2}}{\varGamma (\nu )(1-\delta )(1-\gamma )}[\gamma ( \nu -1)+(1-\gamma )(1-s)]$for$s\in [0,1]$,

• (iii) $\int _{0}^{1} \vert G(t,s) \vert \,ds\leq \frac{\gamma (\nu -1)+1}{\varGamma (\nu +1)(1-\delta )(1-\gamma )}$.

Proof

(i) $G(t,s)\leq 0$ is obvious since $\delta \in (0,1)$ and $\gamma \in (0,1)$.

(ii) For $s\in [0,1]$ and $t\in [s,1]$, we have

$$ G_{t}^{\prime }(t,s)=\frac{1-\nu }{\varGamma (\nu )} \biggl[ \frac{\gamma }{1-\gamma }(1-s)^{\nu -2}+(t-s)^{\nu -2} \biggr]\leq 0, $$

which means

$$ G(1,s)\leq G(t,s)\leq G(s,s)\leq 0, \quad s\leq t\leq 1; $$

(12)

for $t\in [0,s]$, we have

$$ G_{t}^{\prime }(t,s)=\frac{\gamma (1-\nu )}{\varGamma (\nu )(1-\gamma )}(1-s)^{ \nu -2} \leq 0, $$

which means

$$ G(s,s)\leq G(t,s)\leq G(0,s)\leq 0, \quad 0\leq t\leq s. $$

(13)

Inequalities (12) and (13) show that, for $s\in [0,1]$,

$$ G(1,s)\leq G(t,s)\leq G(0,s)\leq 0,\quad 0\leq t\leq 1. $$

Therefore, for $s\in [0,1]$,

$$\begin{aligned} \max_{0\leq t\leq 1} \bigl\vert G(t,s) \bigr\vert =& -G(1,s) \\ =& \frac{1}{\varGamma (\nu )} \biggl[ \frac{(\nu -1)\gamma }{(1-\delta )(1-\gamma )}(1-s)^{\nu -2}+ \frac{(1-s)^{\nu -1}}{1-\delta } \biggr] \\ =& \frac{(1-s)^{\nu -2}}{\varGamma (\nu )(1-\delta )(1-\gamma )}\bigl[\gamma ( \nu -1)+(1-\gamma ) (1-s)\bigr]. \end{aligned}$$

(iii) By (ii) we have

$$\begin{aligned} \int _{0}^{1} \bigl\vert G(t,s) \bigr\vert \,ds \leq & \int _{0}^{1}-G(1,s)\,ds \\ =& \int _{0}^{1} \frac{(1-s)^{\nu -2}}{\varGamma (\nu )(1-\delta )(1-\gamma )}\bigl[\gamma ( \nu -1)+(1-\gamma ) (1-s)\bigr]\,ds \\ =&\frac{\gamma (\nu -1)+1}{\varGamma (\nu +1)(1-\delta )(1-\gamma )}. \end{aligned}$$

 □

Lemma 2.3

When$\delta \in (1,+\infty )$and$\gamma \in (0,1)$, Green’s function$G(t,s)$satisfies the following properties:

• (i) $G(t,s)\geq 0$, $(t,s)\in [0,1]\times [0,1]$;

• (ii) $\max_{0\leq t \leq 1}|G(t,s)|=G(0,s)= \frac{\delta (1-s)^{\nu -2}}{\varGamma (\nu )(\delta -1)(1-\gamma )}[ \gamma (\nu -1)+(1-\gamma )(1-s)]$for$s\in [0,1]$,

• (iii) $\int _{0}^{1}|G(t,s)|\,ds\leq \frac{\delta (\gamma {\nu }+1-\gamma )}{\varGamma (\nu +1)(\delta -1)(1-\gamma )}$.

Proof

(i) When $0\leq t\leq s\leq 1$,

$$ G(t,s)=(\nu -1) \frac{\delta \gamma (1-t)+\gamma t}{\varGamma (\nu )(\delta -1)(1-\gamma )}(1-s)^{ \nu -2}+ \frac{\delta }{\varGamma (\nu )(\delta -1)}(1-s)^{\nu -1}\geq 0. $$

When $0\leq s \leq t\leq 1$,

$$\begin{aligned} G(t,s) =&(\nu -1) \frac{\delta \gamma (1-t)+\gamma t}{\varGamma (\nu )(\delta -1)(1-\gamma )}(1-s)^{ \nu -2}+ \frac{(1-s)^{\nu -1}}{\varGamma (\nu )} \biggl[ \frac{1}{\delta -1}+1-\biggl(\frac{t-s}{1-s} \biggr)^{\nu -1} \biggr] \\ \geq& 0. \end{aligned}$$

(ii) For $s\in [0,1]$ and $t\in [s,1]$, we have

$$ G_{t}^{\prime }(t,s)=\frac{1-\nu }{\varGamma (\nu )} \biggl[ \frac{\gamma }{1-\gamma }(1-s)^{\nu -2}+(t-s)^{\nu -2} \biggr]\leq 0, $$

which means

$$ 0\leq G(1,s)\leq G(t,s)\leq G(s,s), \quad s\leq t \leq 1. $$

(14)

For $t\in [0,s]$, we have

$$ G_{t}^{\prime }(t,s)= \frac{\gamma (1-\nu )(1-s)^{\nu -2}}{(1-\gamma )\varGamma (\nu )}\leq 0, $$

which means

$$ G(s,s)\leq G(t,s)\leq G(0,s), \quad 0\leq t\leq s. $$

(15)

Inequalities (14) and (15) show us that, for $s\in [0,1]$,

$$ \max_{0\leq t \leq 1} \bigl\vert G(t,s) \bigr\vert =G(0,s)= \frac{\delta (1-s)^{\nu -2}}{\varGamma (\nu )(\delta -1)(1-\gamma )}\bigl[ \gamma (\nu -1)+(1-\gamma ) (1-s)\bigr]. $$

(iii) From (ii) we have

$$\begin{aligned} \int ^{1}_{0} \bigl\vert G(t,s) \bigr\vert \,ds \leq & \int ^{1}_{0}G(0,s)\,ds \\ =& \int ^{1}_{0} \frac{\delta (1-s)^{\nu -2}}{\varGamma (\nu )(\delta -1)(1-\gamma )}\bigl[ \gamma (\nu -1)+(1-\gamma ) (1-s)\bigr]\,ds \\ =&\frac{\delta (\gamma \nu +1-\gamma )}{\varGamma (\nu +1)(\delta -1)(1-\gamma )}. \end{aligned}$$

 □

Lemma 2.4

When$\delta \in (0,1)$and$\gamma \in (1,1+\frac{(\nu -1)\delta }{2-\nu }]$, Green’s function$G(t,s)$satisfies the following properties:

• (i) $G(t,s)\geq 0$, $(t,s)\in [0,1]\times [0,1]$;

• (ii) $\max_{0\leq t\leq 1}|G(t,s)|=G(1,s)= \frac{(1-s)^{\nu -2}}{\varGamma (\nu )(1-\delta )(\gamma -1)} [ \gamma (\nu -1)-(\gamma -1)(1-s) ]$for$s\in [0,1]$,

• (iii) $\int ^{1}_{0}|G(t,s)|\,ds\leq \frac{\gamma (\nu -1)+1}{\varGamma (\nu +1)(1-\delta )(\gamma -1)}$.

Proof

We first prove (i) and (ii). For $s\in [0,1]$, when $t\in [0,s]$,

$$ G_{t}^{\prime }(t,s)= \frac{\gamma (\nu -1)(1-s)^{\nu -2}}{\varGamma (\nu )(\gamma -1)}\geq 0, $$

which means that, for $s\in [0,1]$,

$$ G(0,s)\leq G(t,s)\leq G(s,s), \quad t\in [0,s].$$

(16)

When $t\in [s,1]$,

$$\begin{aligned}& \begin{aligned}&G_{t}^{\prime }(t,s)=\frac{(\nu -1)}{\varGamma (\nu )} \biggl[ \frac{\gamma }{\gamma -1}(1-s)^{\nu -2}-(t-s)^{\nu -2} \biggr], \\ & G_{tt}^{\prime \prime }(t,s)=\frac{1}{\varGamma (\nu )}( \nu -1) (2-\nu ) (t-s)^{\nu -3} \geq 0. \end{aligned} \end{aligned}$$

(17)

Letting $G_{t}^{\prime }(t,s)=0$, we get ${t^{*}=(\frac{\gamma }{\gamma -1})^{\frac{1}{\nu -2}}(1-s)+s}\in [s,1]$. Combining with (17), for $s\in [0,1]$, we have

$$\begin{aligned}& G\bigl(t^{*},s\bigr)\leq G(t,s) \leq G(s,s), \quad t\in \bigl[s,t^{*}\bigr], \\ \end{aligned}$$

(18)

$$\begin{aligned}& G\bigl(t^{*},s\bigr)\leq G(t,s) \leq G(1,s), \quad t\in \bigl[t^{*},1\bigr]. \end{aligned}$$

(19)

Inequalities (16), (18), and (19) show us that, for $s\in [0,1]$,

$$ \max_{0\leq t \leq 1} \bigl\vert G(t,s) \bigr\vert = \max \bigl\{ \bigl\vert G(0,s) \bigr\vert , \bigl\vert G(s,s) \bigr\vert , \bigl\vert G\bigl(t^{*},s\bigr) \bigr\vert , \bigl\vert G(1,s) \bigr\vert \bigr\} . $$

Now we prove $G(0,s)$, $G(s,s)$, $G(t^{*},s)$, and $G(1,s)$ are all nonnegative.

For $G(0,s)$, we have

$$\begin{aligned}& G(0,s)=\frac{1}{\varGamma (\nu )} \biggl[ \frac{\delta \gamma (\nu -1)}{(1-\delta )(\gamma -1)}(1-s)^{\nu -2}- \frac{\delta }{1-\delta }(1-s)^{\nu -1} \biggr], \\& G_{s}^{\prime }(0,s)=\frac{\delta (\nu -1)(1-s)^{\nu -3}}{(1-\delta )\varGamma (\nu )} \biggl[(1-s)+ \frac{ \gamma (2-\nu )}{\gamma -1} \biggr]\geq 0, \end{aligned}$$

which means that $G(0, s)$ is increasing for $s\in [0,1]$. Considering that $\gamma <\frac{1}{2-\nu }$ in case of $1<\gamma \leq 1+\frac{(\nu -1)\delta }{2-\nu }$ and $0<\delta <1$, we have, for $s\in [0,1]$,

$$ G(0,s)\geq G(0,0)=\frac{\delta }{\varGamma (\nu )(1-\delta )(\gamma -1)}\bigl[1- \gamma (2-\nu ) \bigr]>0. $$

(20)

Inequalities(16) and (20) show that $G(s,s)>0$.

For $G(t^{*},s)$, we have

$$\begin{aligned} G\bigl(t^{*},s\bigr) =& \frac{(1-s)^{\nu -2}}{\varGamma (\nu )} \biggl\{ \frac{\delta \gamma (\nu -1)}{(\gamma -1)(1-\delta )}+ \biggl[(2-\nu ) \biggl( \frac{\gamma }{\gamma -1} \biggr)^{\frac{\nu -1}{\nu -2}}+ \frac{\delta }{1-\delta } \biggr](s-1) \\ &{}+ \frac{\gamma (\nu -1)}{\gamma -1}s \biggr\} \\ =& \frac{(1-s)^{\nu -2}}{\varGamma (\nu )}g(s), \end{aligned}$$

(21)

where

$$ g(s)=\frac{\delta \gamma (\nu -1)}{(\gamma -1)(1-\delta )}+ \biggl[(2- \nu ) \biggl(\frac{\gamma }{\gamma -1} \biggr)^{\frac{\nu -1}{\nu -2}}+ \frac{\delta }{1-\delta } \biggr](s-1)+ \frac{\gamma (\nu -1)}{\gamma -1}s,\quad s\in [0,1]. $$

Obviously, $g(s)$ is increasing on $[0,1]$, and thus

$$ g(s)\geq g(0)=\frac{\delta \gamma (\nu -1)}{(\gamma -1)(1-\delta )}-(2- \nu ) \biggl( \frac{\gamma }{\gamma -1} \biggr)^{\frac{\nu -1}{\nu -2}}- \frac{\delta }{1-\delta },\quad s\in [0,1]. $$

(22)

Let $k(t)=\frac{\delta (\nu -1)}{1-\delta }t-(2-\nu )t^{\frac{\nu -1}{\nu -2}}-\frac{\delta }{1-\delta }$, $t\in [1,+\infty )$. Then

$$ k'(t)=\frac{\delta (\nu -1)}{1-\delta }+(\nu -1)t^{\frac{1}{\nu -2}} \geq 0, $$

(23)

which mean that $k(t)$ is increasing in $[1,+\infty )$. Letting $k(t_{0})=0$, we get $t_{0}= \frac{(1-\delta )(2-\nu )t_{0}^{\frac{\nu -1}{\nu -2}}+\delta }{\delta (\nu -1)}$ and

$$\begin{aligned} t_{0}-1= \frac{(1-\delta )(2-\nu )t_{0}^{\frac{\nu -1}{\nu -2}}+\delta (2-\nu )}{\delta (\nu -1)}>0. \end{aligned}$$

Then

$$\begin{aligned} \frac{t_{0}}{t_{0}-1} =&1+\frac{1}{t_{0}-1} \\ =&1+ \frac{(\nu -1)\delta }{\delta (2-\nu )+(1-\delta )(2-\nu )t_{0}^{\frac{\nu -1}{\nu -2}}} \\ >&1+\frac{(\nu -1)\delta }{2-\nu }\geq \gamma , \end{aligned}$$

that is, $t_{0}<\frac{\gamma }{\gamma -1}$. By (23) we obtain $g(0)=k(\frac{\gamma }{\gamma -1})\geq k(t_{0})=0$, and thus

$$ g(s)\geq g(0)\geq 0. $$

(24)

From (21) and (24) it follows that $G(t^{*},s)\geq 0$.

Since $G(t^{*},s)\geq 0$, by (19) it follows that $G(1,s)\geq 0$

Above all, we conclude that

$$\begin{aligned} G(t,s)\geq 0, \quad (t,s)\in [0,1]\times [0,1] \end{aligned}$$

and

$$ \max_{0\leq t \leq 1} \bigl\vert G(t,s) \bigr\vert =\max \bigl\{ G(s,s),G(1,s)\bigr\} . $$

Since $\gamma <\frac{1}{2-\nu }$ in the case of $1<\gamma \leq 1+\frac{(\nu -1)\delta }{2-\nu }$ and $0<\delta <1$, we get

$$\begin{aligned} G(s,s)-G(1,s) =&\frac{(1-\nu )(1-s)^{\nu -2}}{\varGamma (\nu )} \biggl[ \frac{\delta \gamma (1-s)+\gamma s}{(1-\gamma )(1-\delta )}- \frac{\gamma }{(1-\gamma )(1-\delta )}+(1-s)^{\nu -1} \biggr] \\ =&\frac{(1-s)^{\nu -1}}{\varGamma (\nu )(\gamma -1)}\bigl[(2-\nu )\gamma -1\bigr] \leq 0, \end{aligned}$$

so

$$ \max_{0\leq t \leq 1} \bigl\vert G(t,s) \bigr\vert = G(1,s)= \frac{(1-s)^{\nu -2}}{\varGamma (\nu ){(1-\delta )(\gamma -1)}} \bigl[{ \gamma (\nu -1)-(\gamma -1) (1-s)} \bigr]. $$

(iii) By (ii) we have

$$\begin{aligned} \int _{0}^{1} \bigl\vert G(t,s) \bigr\vert \,ds \leq & \int _{0}^{1}G(1,s)\,ds \\ =& \int _{0}^{1} \frac{(1-s)^{\nu -2}}{\varGamma (\nu ){(1-\delta )(\gamma -1)}} \bigl[{ \gamma (\nu -1)-(\gamma -1) (1-s)} \bigr]\,ds \\ =&\frac{\gamma (\nu -1)+1}{\varGamma (\nu +1)(1-\delta )(\gamma -1)}. \end{aligned}$$

 □

Lemma 2.5

When$\delta \in (1,+\infty )$and$\gamma \in (1,\frac{1}{2-\nu }]$, Green’s function$G(t,s)$satisfies the following properties:

• (i) $G(t,s)\leq 0$, $(t,s)\in [0,1]\times [0,1]$;

• (ii) For$s\in [0,1]$,

  $$\begin{aligned}& \max_{0\leq t\leq 1} \bigl\vert G(t,s) \bigr\vert \\& \quad \leq \frac{(1-s)^{\nu -2}}{\varGamma (\nu )(\delta -1)(\gamma -1)} \\& \qquad {}\times \biggl\{ \delta \gamma (\nu -1)- \biggl[\delta (\gamma -1)-\gamma (2- \nu ) ( \delta -1) \biggl(\frac{\gamma -1}{\gamma }\biggr)^{\frac{1}{2-\nu }} \biggr](1-s) \biggr\} ; \end{aligned}$$

• (iii) $\int ^{1}_{0}|G(t,s)|\,ds\leq \frac{1}{\varGamma (\nu +1)(\delta -1)(\gamma -1)} \{ \delta [1+ \gamma (\nu -1)]+(\delta -1)\gamma (2-\nu )(\frac{\gamma -1}{\gamma })^{ \frac{1}{2-\nu }} \}$.

Proof

We first prove (i) and (ii). For $s\in [0,1]$ and $t\in [0,s]$,

$$\begin{aligned} G_{t}^{\prime }(t,s)= \frac{\gamma (\nu -1)(1-s)^{\nu -2}}{(\gamma -1)\varGamma (\nu )}\geq 0, \end{aligned}$$

which means

$$ G(0,s)\leq G(t,s)\leq G(s,s),\quad t\in [0,s]. $$

(25)

When $t\in [s,1]$,

$$\begin{aligned}& \begin{aligned}&G_{t}^{\prime }(t,s)=\frac{(\nu -1)}{\varGamma (\nu )} \biggl[ \frac{\gamma }{\gamma -1}(1-s)^{\nu -2}-(t-s)^{\nu -2} \biggr], \\ & G^{\prime \prime }_{tt}(t,s)=\frac{1}{\varGamma (\nu )}( \nu -1) (2-\nu ) (t-s)^{\nu -3} \geq 0. \end{aligned} \end{aligned}$$

(26)

Letting $G_{t}^{\prime }(t,s)=0$, we get $t^{*}={(\frac{\gamma }{\gamma -1})^{\frac{1}{\nu -2}}(1-s)+s}\in [s,1]$. Combining (26), for $s\in [0,1]$, we have

$$\begin{aligned}& G\bigl(t^{*},s\bigr)\leq G(t,s)\leq G(s,s),\quad t \in \bigl[s,t^{*}\bigr], \end{aligned}$$

(27)

$$\begin{aligned}& G\bigl(t^{*},s\bigr)\leq G(t,s)\leq G(1,s), \quad t\in \bigl[t^{*},1\bigr]. \end{aligned}$$

(28)

Inequalities (25), (27), and (28) show that, for $s\in [0,1]$,

$$ \max_{0\leq t \leq 1} \bigl\vert G(t,s) \bigr\vert =\max \bigl\{ \bigl\vert G(0,s) \bigr\vert , \bigl\vert G(s,s) \bigr\vert , \bigl\vert G\bigl(t^{*},s\bigr) \bigr\vert , \bigl\vert G(1,s) \bigr\vert \bigr\} . $$

We now prove that $G(0,s)$, $G(s,s)$, $G(t^{*},s)$, $G(1,s)$ are all nonpositive.

For $G(s,s)$, we have

$$\begin{aligned} G(s,s) =&\frac{(1-s)^{\nu -2}}{\varGamma (\nu )(\delta -1)(\gamma -1)} \bigl\{ \bigl[(2-\nu )\gamma -1\bigr]\delta + \bigl[(\nu -1)\delta \gamma -(\gamma -1) \delta -(\nu -1)\gamma \bigr]s \bigr\} \\ =&\frac{(1-s)^{\nu -2}}{\varGamma (\nu )(\delta -1)(\gamma -1)}L(s), \end{aligned}$$

where

$$ L(s)=\bigl[(2-\nu )\gamma -1\bigr]\delta +\bigl[(\nu -1)\delta \gamma -(\gamma -1) \delta -(\nu -1)\gamma \bigr]s,\quad s\in [0,1]. $$

We have

$$ L(0)=\delta \bigl[(2-\nu )\gamma -1\bigr]\leq 0, \qquad L(1)=-(\nu -1)\gamma < 0, $$

which means that $L(s)\leq 0$, $s\in [0,1]$, and therefore

$$ G(s,s)\leq 0, \quad s\in [0,1]. $$

Inequalities $G(0,s)\leq 0$ and $G(t^{*},s)\leq 0$ follow from (25), (27), and $G(s,s)\leq 0$.

For $G(1,s)$, we have

$$\begin{aligned}& G(1,s)=\frac{1}{\varGamma (\nu )} \biggl[ \frac{\gamma (\nu -1)}{(1-\delta )(\gamma -1)}(1-s)^{\nu -2}- \frac{1}{1-\delta }(1-s)^{\nu -1} \biggr], \\& G_{s}^{\prime }(1,s)=\frac{-(\nu -1)(1-s)^{\nu -3}}{\varGamma (\nu )(\delta -1)(\gamma -1)}\bigl[(2- \nu ) \gamma +(\gamma -1) (1-s)\bigr]\leq 0, \end{aligned}$$

and thus, for $s\in [0,1]$,

$$ G(1,s)\leq G(1,0)=\frac{1}{\varGamma (\nu )(\delta -1)(\gamma -1)}\bigl[ \gamma (2-\nu )-1\bigr]\leq 0. $$

Above all, we get that $G(t,s)\leq 0$ and, for $s\in [0,1]$,

$$ \max_{0\leq t\leq 1} \bigl\vert G(t,s) \bigr\vert =\max \bigl\{ -G(0,s),-G\bigl(t^{*},s\bigr)\bigr\} . $$

We can easily compute that

$$ -G\bigl(t^{*},s\bigr)=\frac{(1-s)^{\nu -2}}{\varGamma (\nu )}h_{1}(s), $$

where

$$\begin{aligned} h_{1}(s) =& \biggl\{ \frac{\delta [1-\gamma (2-\nu )]}{(\delta -1)(\gamma -1)}+(2-\nu ) \biggl( \frac{\gamma }{\gamma -1}\biggr)^{\frac{\nu -1}{\nu -2}} \biggr\} \\ &{} - \biggl\{ \frac{\delta [1-\gamma (2-\nu )]}{(\delta -1)(\gamma -1)}+(2-\nu ) \biggl( \frac{\gamma }{\gamma -1} \biggr)^{\frac{\nu -1}{\nu -2}}- \frac{\gamma (\nu -1)}{(\delta -1)(\gamma -1)} \biggr\} s, \end{aligned}$$

and

$$ -G(0,s)=\frac{(1-s)^{\nu -2}}{\varGamma (\nu )}h_{2}(s), $$

where

$$ h_{2}(s)=\frac{\delta [1-\gamma (2-\nu )]}{(\delta -1)(\gamma -1)}+ \frac{\delta }{\delta -1}s. $$

Obviously,

$$\begin{aligned}& h_{1}(0)=\frac{\delta [1-\gamma (2-\nu )]}{(\delta -1)(\gamma -1)}+(2- \nu ) \biggl( \frac{\gamma }{\gamma -1} \biggr)^{\frac{\nu -1}{\nu -2}}>\frac{\delta [1-\gamma (2-\nu )]}{(\delta -1)(\gamma -1)}=h_{2}(0)>0, \\& h_{2}(1)=\frac{\delta \gamma (\nu -1)}{(\delta -1)(\gamma -1)}>\frac{\gamma (\nu -1)}{(\delta -1)(\gamma -1)}=h_{1}(1)>0. \end{aligned}$$

So, if we make a line $H(s)$ through $(0,h_{1}(0))$ and $(1,h_{2}(1))$, that is,

$$ H(s)=\frac{\delta [1-\gamma (2-\nu )]}{(\delta -1)(\gamma -1)}+(2- \nu ) \biggl(\frac{\gamma }{\gamma -1} \biggr)^{\frac{\nu -1}{\nu -2}}+ \biggl[\frac{\delta }{\delta -1}-(2-\nu ) \biggl( \frac{\gamma }{\gamma -1}\biggr)^{ \frac{\nu -1}{\nu -2}} \biggr]s, $$

then we have

$$ 0\leq h_{1}(s), h_{2}(s)\leq H(s), \quad s\in [0,1]. $$

Therefore, for $s\in [0,1]$,

$$\begin{aligned}& \max_{0\leq t\leq 1} \bigl\vert G(t,s) \bigr\vert \\& \quad =\max \bigl\{ -G(0,s),-G\bigl(t^{*},s\bigr)\bigr\} \\& \quad = \frac{(1-s)^{\nu -2}}{\varGamma (\nu )}\max \bigl\{ h_{1}(s),h_{2}(s)) \bigr\} \\& \quad \leq \frac{(1-s)^{\nu -2}}{\varGamma (\nu )}H(s) \\& \quad =\frac{(1-s)^{\nu -2}}{\varGamma (\nu )(\delta -1)(\gamma -1)} \biggl\{ \delta \gamma (\nu -1)- \biggl[\delta ( \gamma -1)-\gamma (2-\nu ) ( \delta -1) \biggl(\frac{\gamma -1}{\gamma } \biggr)^{\frac{1}{2-\nu }} \biggr](1-s) \biggr\} . \end{aligned}$$

(iii) easily s follows by (ii):

$$\begin{aligned}& \int _{0}^{1} \bigl\vert G(t,s) \bigr\vert \,ds \\& \quad \leq \int _{0}^{1}\max_{0\leq t \leq 1} \bigl\vert G(t,s) \bigr\vert \,ds \\& \quad \leq \int _{0}^{1} \frac{(1-s)^{\nu -2}}{\varGamma (\nu )(\delta -1)(\gamma -1)} \\& \qquad {}\times \biggl\{ \delta \gamma (\nu -1)- \biggl[\delta (\gamma -1)- \gamma (2-\nu ) ( \delta -1) \biggl(\frac{\gamma -1}{\gamma }\biggr)^{\frac{1}{2-\nu }} \biggr](1-s) \biggr\} \,ds \\& \quad =\frac{1}{\varGamma (\nu +1)(\delta -1)(\gamma -1)} \biggl\{ \delta \bigl[1+ \gamma (\nu -1)\bigr]+(2- \nu ) (\delta -1)\gamma \biggl(\frac{\gamma -1}{\gamma }\biggr)^{ \frac{1}{2-\nu }} \biggr\} . \end{aligned}$$

 □

3 Main result

Theorem 3.1

Suppose the boundary value problem (5) has a nonzero solution$u(t)$.

• (i) If$\delta \in (0,1)$and$\gamma \in (0,1)$, then

  $$ \int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)+(1-\gamma ) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds> \varGamma ( \nu ) (1-\delta ) (1-\gamma ); $$

• (ii) If$\delta \in (1,+\infty )$and$\gamma \in (0,1)$, then

  $$ \int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)+(1-\gamma ) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds> \frac{\varGamma (\nu )(\delta -1)(1-\gamma )}{\delta }; $$

• (iii) If$\delta \in (0,1)$and$\gamma \in (1,1+\frac{(\nu -1)\delta }{2-\nu }]$, then

  $$ \int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)-(\gamma -1) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds> \varGamma ( \nu ) (1-\delta ) (\gamma -1); $$

• (iv) if$\delta \in (1,+\infty )$and$\gamma \in (1,\frac{1}{2-\nu }]$, then

  $$\begin{aligned}& \int _{0}^{1}(1-s)^{\nu -2} \biggl\{ \delta \gamma (\nu -1)- \biggl[ \delta (\gamma -1)-\gamma (2-\nu ) (\delta -1) \biggl( \frac{\gamma -1}{\gamma }\biggr)^{\frac{1}{2-\nu }} \biggr](1-s) \biggr\} \\& \qquad {}\times \bigl\vert q(s) \bigr\vert \,ds \\& \quad > \varGamma (\nu ) (\delta -1) (\gamma -1). \end{aligned}$$

Proof

Let $u(t)$ be a nonzero solution of the boundary value problem (5). By Lemma 2.1 we have

$$ u(t)= \int _{0}^{1}G(t,s)q(s)u(s)\,ds. $$

Let $m=\max_{t\in [0,1]} \vert u(t) \vert$. Then

$$ \bigl\vert u(t) \bigr\vert \leq \int _{0}^{1} \bigl\vert G(t,s) \bigr\vert \bigl\vert q(s) \bigr\vert \bigl\vert u(s) \bigr\vert \,ds\leq m \int _{0}^{1} \bigl\vert G(t,s) \bigr\vert \bigl\vert q(s) \bigr\vert \,ds . $$

Next, since $|G(t,s)| \vert q(s) \vert \leq \max_{0\leq t\leq 1}|G(t,s)||q(s)|$, but $|G(t,s)| \vert q(s) \vert \not \equiv \max_{0\leq t\leq 1}|G(t,s)| |q(s)|$, we have

$$ \int _{0}^{1} \bigl\vert G(t,s) \bigr\vert \bigl\vert q(s) \bigr\vert \,ds< \int _{0}^{1} \max_{0\leq t\leq 1} \bigl\vert G(t,s) \bigr\vert \bigl\vert q(s) \bigr\vert \,ds, $$

which means

$$ \bigl\vert u(t) \bigr\vert < m \int _{0}^{1}\max_{0\leq t\leq 1} \bigl\vert G(t,s) \bigr\vert \bigl\vert q(s) \bigr\vert \,ds, $$

that is,

$$ 1< \int _{0}^{1}\max_{0\leq t\leq 1} \bigl\vert G(t,s) \bigr\vert \bigl\vert q(s) \bigr\vert \,ds.$$

(29)

Substituting Lemma 2.2(ii), Lemma 2.3(ii), Lemma 2.4(ii), and Lemma 2.5(ii) into (29), we easily get statements (i), (ii), (iii), and (iv) of Theorem 3.1. □

By Theorem 3.1 we have the following conclusions.

Theorem 3.2

• (i) when$\delta \in (0,1)$and$\gamma \in (0,1)$, if

  $$ \int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)+(1-\gamma ) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds\leq \varGamma (\nu ) (1-\delta ) (1-\gamma ), $$

  then the boundary value problem (5) has no nonzero solution.

• (ii) when$\delta \in (1,+\infty)$and$\gamma \in (0,1)$, if

  $$ \int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)+(1-\gamma ) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds\leq \frac{\varGamma (\nu )(\delta -1)(1-\gamma )}{\delta }, $$

  then the boundary value problem (5) has no nonzero solution.

• (iii) when$\delta \in (0,1)$and$\gamma \in (1,1+\frac{(\nu -1)\delta }{2-\nu }]$, if

  $$ \int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)-(\gamma -1) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds\leq \varGamma (\nu ) (1-\delta ) (\gamma -1), $$

  then the boundary value problem (5) has no nonzero solution.

• (iv) when$\delta \in (1,+\infty )$and$\gamma \in (1,\frac{1}{2-\nu }]$, if

  $$\begin{aligned}& \int _{0}^{1}(1-s)^{\nu -2} \biggl\{ \delta \gamma (\nu -1) \\& \qquad {}- \biggl[ \delta (\gamma -1)-\gamma (2-\nu ) (\delta -1) \biggl( \frac{\gamma -1}{\gamma }\biggr)^{\frac{1}{2-\nu }} \biggr](1-s) \biggr\} \bigl\vert q(s) \bigr\vert \,ds \\& \quad \leq \varGamma (\nu ) (\delta -1) (\gamma -1), \end{aligned}$$

  then the boundary value problem (5) has no nonzero solution.

Now we consider the existence of solutions to the following nonlinear boundary value problem:

(30)

Theorem 3.3

Let$f:[0,1]\times R\rightarrow R$be continuous and satisfy the following Lipschitz condition with Lipschitz constantL:

$$ \bigl\vert f(t,u_{1})-f(t,u_{2}) \bigr\vert \leq L \vert u_{1}-u_{2} \vert $$

(31)

for all$(t,u_{1}),(t,u_{2})\in [0,1]\times R$. If

$$ L< \textstyle\begin{cases} \frac{\varGamma (\nu +1)(1-\delta )(1-\gamma )}{\gamma (\nu -1)+1},\quad \delta \in (0,1),\gamma \in (0,1), \\ \frac{\varGamma (\nu +1)(\delta -1)(1-\gamma )}{\delta (\gamma \nu +1-\gamma )},\quad \delta \in (1,+\infty ),\gamma \in (0,1), \\ \frac{\varGamma (\nu +1)(1-\delta )(\gamma -1)}{\gamma (\nu -1)+1},\quad \delta \in (0,1),\gamma \in (1,1+\frac{\delta (\nu -1)}{2-\nu }], \\ \frac{\varGamma (\nu +1)(\delta -1)(\gamma -1)}{\delta [1+\gamma (\nu -1)]+(\delta -1) \gamma (2-\nu )(\frac{\gamma -1}{\gamma })^{\frac{1}{2-\nu }}},\quad \delta \in (1,+\infty ],\gamma \in (1,\frac{1}{2-\nu }], \end{cases} $$

(32)

then (30) has a unique solution.

Proof

Let E be the Banach space $C[0,1]$ with norm $\Vert u\Vert =\max_{t\in [0,1]}|u(t)|$.

By Lemma 2.1, $u\in E$ is a solution of (30) if and only if it satisfies the integral equation

$$ u(t)= \int _{0}^{1}G(t,s)f\bigl(s,u(s)\bigr)\,ds. $$

Define the operator $T:E\rightarrow E$ by

$$ Tu(t)= \int _{0}^{1}G(t,s)f\bigl(s,u(s)\bigr)\,ds. $$

Then T is completely continuous. We claim that T has a unique fixed point in E. In fact, for any $u_{1},u_{2}\in E$, we have

$$\begin{aligned} \bigl\vert Tu_{1}(t)-Tu_{2}(t) \bigr\vert \leq& \int _{0}^{1} \bigl\vert G(t,s) \bigr\vert \bigl\vert f\bigl(s,u_{1}(s)\bigr)-f\bigl(s,u_{2}(s) \bigr) \bigr\vert \,ds \\ \leq& L \int _{0}^{1} \bigl\vert G(t,s) \bigr\vert \bigl\vert u_{1}(s)-u_{2}(s) \bigr\vert \,ds \\ \leq& L \int _{0}^{1} \bigl\vert G(t,s) \bigr\vert \,ds \Vert u_{1}-u_{2} \Vert . \end{aligned}$$

(33)

Substituting of Lemma 2.2(iii), Lemma 2.3(iii), Lemma 2.4(iii), and Lemma 2.5(iii) into (33), we conclude that T is a contraction mapping and thus obtain the desired result. □

4 Conclusion

In this paper, we study a linear fractional differential equation. Firstly, by obtaining the Green’s function we derive a Lyapunov-type inequality for such a boundary value problem. Furthermore, we use the contraction mapping theorem to study the existence of a unique solution for the corresponding nonlinear problem.