# 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 [510], 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:

1. (i)

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

2. (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$$s\in [0,1]$$,

3. (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:

1. (i)

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

2. (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$$s\in [0,1]$$,

3. (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:

1. (i)

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

2. (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$$s\in [0,1]$$,

3. (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:

1. (i)

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

2. (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}
3. (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$$u(t)$$.

1. (i)

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

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

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

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

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

$$\int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)-(\gamma -1) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds> \varGamma ( \nu ) (1-\delta ) (\gamma -1);$$
4. (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

1. (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.

2. (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.

3. (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.

4. (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.

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Correspondence to Dexiang Ma.

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Ma, D., Yang, Z. Lyapunov-type inequality and solution for a fractional differential equation. J Inequal Appl 2020, 181 (2020). https://doi.org/10.1186/s13660-020-02448-z

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• DOI: https://doi.org/10.1186/s13660-020-02448-z

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### Keywords

• Fractional differential equation
• Green’s function
• Lyapunov-type inequality
• Contraction mapping theorem