1 Introduction

Consider the boundary value problem with Dirichlet conditions

$$\begin{aligned} {\left\{ \begin{array}{ll} x''(t)+q(t)x(t)=0,\,\,a<t<b,\\ x(a)=x(b)=0, \end{array}\right. } \end{aligned}$$
(1)

where \(q:[a,b]\rightarrow \mathbb {R}\) is a continuous function. Lyapunov in [1] proved that if Problem (1) has a nontrivial solution then

$$\begin{aligned} \int _{a}^{b}|q(s)|\,ds>\frac{4}{b-a}. \end{aligned}$$

In [2], Hartman and Wintner proved that if Problem (1) has a nontrivial solution then

$$\begin{aligned} \int _{a}^{b}(b-s)(s-a)q^{+}(s)\,ds>b-a, \end{aligned}$$

where \(q^{+}(s)=\max \{q(s),0\}\).

Inequalities of this type have appeared in the literature for other classes of boundary value problems and we refer the reader to [3,4,5,6,7] and the references therein for more details.

Recently, some Lyapunov-type inequalities have been obtained by some authors for different fractional boundary value problems (see [8,9,10,11,12], for example).

In this paper, we are concerned with the problem of finding some Lyapunov-type inequalities for the following fractional boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -^{C}\!D_{a}^{\alpha }u(t)=y(t),\quad a<t<b,\\ u'(a)=0,\,\beta \,^{C}\!D_{a}^{\alpha -1}u(b)+u(\eta )=0, \end{array}\right. } \end{aligned}$$
(2)

where \(^{C}\!D_{a}^{\alpha }\) denotes the Caputo fractional derivative of order \(\alpha \), \(1<\alpha \le 2\), \(\beta >0\) and \(a\le \eta \le b\).

As an application of our results, we obtain a lower bound for the eigenvalues of the cor-respondig problem.

The above mentioned fractional boundary value problem can be considered as the fractional version of the nonlocal boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -u''(t)=y(t),\quad 0<t<1,\\ u'(0)=0,\quad \beta u'(1)+u(\eta )=0, \end{array}\right. } \end{aligned}$$

with \(0\le \eta \le 1\) which has been studied in the special case with \(\eta =0\) in [13] and this problem models a thermostat insulated at \(t=0\) with a controller dissipating heat at \(t=1\) depending on the temperature detected by a sensor at \(t=\eta \).

2 Background

In this section, we present the basic results about fractional calculus theory which be used later. For more details, we refer the reader to [14, 15].

Definition 1

Let \(f:[a,b]\longrightarrow \mathbb {R}\) be a given function. For \(\alpha >0\), the Riemann-Liouville fractional integral of order \(\alpha \) of f is defined by

$$\begin{aligned} (I_{a}^{\alpha }f)(t)=\frac{1}{\Gamma (\alpha )}\int _{a}^{t}(t-s)^{\alpha -1}f(s)\,ds, \end{aligned}$$

where \(\Gamma (\alpha )\) denotes the classical gamma function.

Definition 2

Let \(f:[a,b]\longrightarrow \mathbb {R}\) be a given function. For \(\alpha >0\), the Caputo derivative of fractional order \(\alpha >0\) of f is given by

$$\begin{aligned} (^{C}\!D_{a}^{\alpha }f)(t)=\frac{1}{\Gamma (n-\alpha )}\int _{a}^{t}(t-s)^{n-\alpha -1}f^{(n)}(s)\,ds, \end{aligned}$$

where \(n=[\alpha ]+1\) and \([\alpha ]\) denotes the integer part of \(\alpha \).

Lemma 1

Suppose that \(f\in C(a,b)\cap L^{1}(a,b)\) with a fractional derivative of order \(\alpha >0\) belonging to \(C(a,b)\cap L^{1}(a,b)\). Then

$$\begin{aligned} I_{a}^{\alpha }(^{C}\!D_{a}^{\alpha }f)(t)=f(t)+c_{0}+c_{1}(t-a)+c_{2}(t-a)^{2}+\cdots +c_{n-1}(t-a)^{n-1}, \end{aligned}$$

for \(t\in [a,b]\), where \(c_{i}\in \mathbb {R}\,\,(i=0,1,\ldots ,n-1)\) and \(n=[\alpha ]+1\).

Lemma 2

Suposse \(f\in L^{1}(a,b)\) and \(\alpha >0\), \(\beta >0\). Then

  1. 1.

    \(^{C}\!D_{a}^{\alpha }I_{a}^{\alpha }f(t)=f(t)\)

  2. 2.

    \(I_{a}^{\alpha }(I_{a}^{\beta })f(t)=(I_{a}^{\alpha +\beta }f)(t)\)

3 Main results

Our starting point in this section is the following lemma which gives us an expression for the Green’s function of the boundary value problem (2). The case for \(a=0\) and \(b=1\) appears in [16, Lemma 2.4].

Lemma 3

Suppose \(y\in C[a,b]\). A function \(u\in C[a,b]\) is a solution of Problem (2) if and only if it satisfies the integral equation

$$\begin{aligned} u(t)=\int _{a}^{b}G(t,s)y(s)\,ds, \end{aligned}$$

where G(t, s) is the Green’s function given by

$$\begin{aligned} G(t,s)=\beta +H_{\eta }(s)-H_{t}(s), \end{aligned}$$

where for \(r\in [a,b]\), \(H_{r}:[a,b]\rightarrow \mathbb {R}\) is the function defined as

$$\begin{aligned} H_{r}(s)= {\left\{ \begin{array}{ll} \frac{(r-s)^{\alpha -1}}{\Gamma (\alpha )},&{}\quad \text {for}\quad a\le s\le r\le b,\\ 0, &{} \quad \text {for}\quad a\le r<s\le b. \end{array}\right. } \end{aligned}$$

Proof

Using Lemma 2, we have

$$\begin{aligned} u(t)=-I_{a}^{\alpha }y(t)+c_{0}+c_{1}(t-a) =-\frac{1}{\Gamma (\alpha )}\int _{a}^{t}(t-s)^{\alpha -1}y(s)\,ds+c_{0}+c_{1}(t-a), \end{aligned}$$

for some constants \(c_{0},c_{1}\in \mathbb {R}\).

This gives us

$$\begin{aligned} u'(t)=-\frac{1}{\Gamma (\alpha )}\int _{a}^{t}(\alpha -1)(t-s)^{\alpha -2}y(s)\,ds+c_{1}. \end{aligned}$$

From the boundary condition \(u'(a)=0\), we get \(c_{1}=0\).

This gives us

$$\begin{aligned} u(t)=-\frac{1}{\Gamma (\alpha )}\int _{a}^{t}(t-s)^{\alpha -1}y(s)\,ds+c_{0}. \end{aligned}$$

By using the fact that \(^{C}\!D_{a}^{\alpha -1}c_{0}=0\), and Lemma 2, we have

$$\begin{aligned} ^{C}\!D_{a}^{\alpha -1}u(t)=-^{C}\!D_{a}^{\alpha -1}I_{a}^{\alpha }y(t)=-^{C}\!D_{a}^{\alpha -1}I_{a}^{\alpha -1}I_{a}y(t) =-I_{a}y(t)=-\int _{a}^{t}y(s)\,ds. \end{aligned}$$

This gives us

$$\begin{aligned} \beta \,^{C}\!D_{a}^{\alpha -1}u(b)=-\beta \int _{a}^{b}y(s)\,ds. \end{aligned}$$

Taking into account the boundary condition

$$\begin{aligned} \beta \,^{C}\!D_{a}^{\alpha -1}u(b)+u(\eta )=0, \end{aligned}$$

we have

$$\begin{aligned} 0=-\beta \int _{a}^{b}y(s)\,ds-\frac{1}{\Gamma (\alpha )}\int _{a}^{\eta }(\eta -s)^{\alpha -1}y(s)\,ds+c_{0} \end{aligned}$$

and, from this, it follows

$$\begin{aligned} c_{0}=\beta \int _{a}^{b}y(s)\,ds+\frac{1}{\Gamma (\alpha )}\int _{a}^{\eta }(\eta -s)^{\alpha -1}y(s)\,ds. \end{aligned}$$

Consequently,

$$\begin{aligned} u(t)=-\frac{1}{\Gamma (\alpha )}\int _{a}^{t}(t-s)^{\alpha -1}y(s)\,ds +\beta \int _{a}^{b}y(s)\,ds+\frac{1}{\Gamma (\alpha )}\int _{a}^{\eta }(\eta -s)^{\alpha -1}y(s)\,ds. \end{aligned}$$

Therefore,

$$\begin{aligned} u(t)=\beta \int _{a}^{b}y(s)\,ds+\int _{a}^{b}H_{\eta }(s)y(s)\,ds-\int _{a}^{b}H_{t}(s)y(s)\,ds \end{aligned}$$

or, equivalently,

$$\begin{aligned} u(t)=\int _{a}^{b}\left( \beta +H_{\eta }(s)-H_{t}(s)\right) y(s)\,ds. \end{aligned}$$

This completes the proof. \(\square \)

Remark 1

Notice that the Green’s function can be expressed as

$$\begin{aligned} G(t,s)= {\left\{ \begin{array}{ll} \beta +H_{\eta }(s)-\dfrac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}, &{}\quad \text {for}\,\,a\le s\le t\le b,\\ \beta +H_{\eta }(s), &{}\quad \text {for}\,\,a\le t\le s\le b. \end{array}\right. } \end{aligned}$$

In the following proposition, we present some properties about Green’s function

Proposition 1

The Green’s function satisfies:

  1. (i)

    \(\max \{G(t,s):t,s\in [a,b]\}=\beta +\dfrac{(\eta -a)^{\alpha -1}}{\Gamma (\alpha )}\).

  2. (ii)

    \(\min \{G(t,s):t,s\in [a,b]\}=\beta -\dfrac{(b-\eta )^{\alpha -1}}{\Gamma (\alpha )}\).

Proof

  1. (i)

    Notice that for \(s\in [a,b]\) fixed, we have

    $$\begin{aligned} \frac{\partial G}{\partial t}(t,s)= {\left\{ \begin{array}{ll} 0,&{}\quad \text {for}\,\,a\le t\le s, \\ -\dfrac{(\alpha -1)(t-s)^{\alpha -2}}{\Gamma (\alpha )},&{}\quad \text {for}\,\,a\le t\le s\le b. \end{array}\right. } \end{aligned}$$

    From this, it follows that G(t, s) is a decreasing function in t, and this gives us

    $$\begin{aligned}&\max \{G(t,s):t,s\in [a,b]\}=G(a,s)\nonumber \\&\quad = {\left\{ \begin{array}{ll} \beta +\dfrac{(\eta -s)^{\alpha -1}}{\Gamma (\alpha )},&{}\quad \text {for}\,\,a\le s\le \eta \le b,\\ \beta , &{}\quad \text {for}\,\,a\le \eta \le s\le b. \end{array}\right. } \end{aligned}$$

    On the other hand, if we put \(\varphi (s)=\beta +\frac{(\eta -s)^{\alpha -1}}{\Gamma (\alpha )}\) for \(s\in [a,\eta ]\), since \(\varphi '(s)=-\frac{(\alpha -1)(\eta -s)^{\alpha -2}}{\Gamma (\alpha )}<0\), \(\varphi \) is a decreasing function and we infer that \(\max \{\varphi (s):s\in [a,\eta ]\}=\varphi (a)=\beta +\frac{(\eta -a)^{\alpha -1}}{\Gamma (\alpha )}\). Therefore,

    $$\begin{aligned} \max \{G(t,s):t,s\in [a,b]\}=\max \left\{ \beta ,\beta +\frac{(\eta -a)^{\alpha -1}}{\Gamma (\alpha )}\right\} =\beta +\frac{(\eta -a)^{\alpha -1}}{\Gamma (\alpha )} \end{aligned}$$

    and this proves (i).

  2. (ii)

    Since G(t, s) is a decreasing function in t, we have

    $$\begin{aligned}&\min \{G(t,s):t,s\in [a,b]\}=G(b,s)\nonumber \\&\quad = {\left\{ \begin{array}{ll} \beta +\dfrac{(\eta -s)^{\alpha -1}}{\Gamma (\alpha )}-\dfrac{(b-s)^{\alpha -1}}{\Gamma (\alpha )}, &{}\quad \text {for}\,\,a\le s\le \eta \le b,\\ \beta -\dfrac{(b-s)^{\alpha -1}}{\Gamma (\alpha )}, &{}\quad \text {for}\,\,a\le \eta \le s\le b. \end{array}\right. } \end{aligned}$$

    Put \(\psi (s)=\beta -\frac{(b-s)^{\alpha -1}}{\Gamma (\alpha )}\) for \(s\in [\eta ,b]\). Since \(\psi '(s)=\frac{(\alpha -1)(b-s)^{\alpha -2}}{\Gamma (\alpha )}\ge 0\), \(\psi \) is a nondecreasing, and, consequently, \(\min \{\psi (s):s\in [\eta ,b]\}=\psi (\eta )=\beta -\frac{(b-\eta )^{\alpha -1}}{\Gamma (\alpha )}\). On the other hand, put \(\alpha (s)=\beta +\frac{(\eta -s)^{\alpha -1}}{\Gamma (\alpha )} -\frac{(b-s)^{\alpha -1}}{\Gamma (\alpha )}\) for \(s\in [a,\eta ]\), since \(\alpha '(s)=-\frac{(\alpha -1)(\eta -s)^{\alpha -2}}{\Gamma (\alpha )} +\frac{(\alpha -1)(b-s)^{\alpha -2}}{\Gamma (\alpha )}=\frac{(\alpha -1)}{\Gamma (\alpha )} \left[ (b-s)^{\alpha -2}-(\eta -s)^{\alpha -2}\right] \le 0\), (because \(1<\alpha \le 2\)), \(\alpha \) is decreasing on \([a,\eta ]\) and, therefore, \(\min \{\alpha (s):s\in [a,\eta ]\}=\alpha (\eta )=\beta -\frac{(b-\eta )^{\alpha -1}}{\Gamma (\alpha )}\). These facts say us that

    $$\begin{aligned} \min \{G(t,s):t,s\in [a,b]\}=\beta -\frac{(b-\eta )^{\alpha -1}}{\Gamma (\alpha )} \end{aligned}$$

    and this completes the proof.

\(\square \)

Remark 2

Notice that if \(\beta \Gamma (\alpha )\ge (b-\eta )^{\alpha -1}\) then \(G(t,s)\ge 0\).

In the case \(\beta \Gamma (\alpha )<(b-\eta )^{\alpha -1}\) then, since

$$\begin{aligned} \beta -\frac{(b-\eta )^{\alpha -1}}{\Gamma (\alpha )}\le G(t,s)\le \beta +\frac{(\eta -a)^{\alpha -1}}{\Gamma (\alpha )}, \end{aligned}$$

we have that

$$\begin{aligned} |G(t,s)|\le \max \left\{ \beta +\frac{(\eta -a)^{\alpha -1}}{\Gamma (\alpha )}, \frac{(b-\eta )^{\alpha -1}}{\Gamma (\alpha )}-\beta \right\} ,\quad \text {for}\,\,t,s\in [0,1]. \end{aligned}$$

Our main result is the following Lyapunov-type inequality.

Theorem 1

Suppose that the fractional boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -^{C}\!D_{a}^{\alpha }u(t)=q(t)u(t),\quad a<t<b,\\ u'(a)=0,\,\beta \,^{C}\!D_{a}^{\alpha -1}u(b)+u(\eta )=0, \end{array}\right. } \end{aligned}$$

with \(1<\alpha \le 2\), \(\beta >0\), \(a\le \eta \le b\) and \(\beta \ge \frac{(b-\eta )^{\alpha -1}}{\Gamma (\alpha )}\), where \(q:[a,b]\rightarrow \mathbb {R}\) is a continuous function, has a nontrivial continuous solution then

$$\begin{aligned} \int _{a}^{b}|q(s)|\,ds\ge \frac{\Gamma (\alpha )}{\beta \Gamma (\alpha )+(\eta -a)^{\alpha -1}}. \end{aligned}$$

Proof

Consider the Banach space \(C[a,b]=\{x:[a,b]\rightarrow \mathbb {R}:x\,\,\text {continuous}\}\) with the standard norm \(\Vert x\Vert _{\infty }=\max \{|x(t)|:a\le t\le b\}\), for \(x\in C[a,b]\).

By Lemma 3,

$$\begin{aligned} u(t)=\int _{a}^{b}G(t,s)q(s)u(s)\,ds,\quad \text {for}\,\,a\le t\le b, \end{aligned}$$

where G(t, s) is the Green’s function appearing in Lemma 3.

Using Remark 2, since \(\beta \ge \frac{(b-\eta )^{\alpha -1}}{\Gamma (\alpha )}\), \(G(t,s)\ge 0\) and, moreover \(\max \{G(t,s):t,s\in [a,b]\}=\beta +\frac{(\eta -a)^{\alpha -1}}{\Gamma (\alpha )}\), we infer, for any \(t\in [a,b]\),

$$\begin{aligned} |u(t)|\le \int _{a}^{b}G(t,s)|q(s)||u(s)|\,ds \le \Vert u\Vert _{\infty }\int _{a}^{b}\frac{\beta \Gamma (\alpha )+(\eta -a)^{\alpha -1}}{\Gamma (\alpha )}|q(s)|\,ds, \end{aligned}$$

and, this gives us

$$\begin{aligned} \Vert u\Vert _{\infty }\le \Vert u\Vert _{\infty }\frac{\beta \Gamma (\alpha )+(\eta -a)^{\alpha -1}}{\Gamma (\alpha )} \int _{a}^{b}|q(s)|\,ds. \end{aligned}$$

Since the solution u is nontrivial, we get

$$\begin{aligned} 1\le \frac{\beta \Gamma (\alpha )+(\eta -a)^{\alpha -1}}{\Gamma (\alpha )}\int _{a}^{b}|q(s)|\,ds \end{aligned}$$

and this gives us the desired result. \(\square \)

Theorem 1 gives us the following corollary.

Corollary 1

Suppose that the boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -u''(t)=q(t)u(t),\,\,a<t<b,\\ u'(a)=0,\,\beta u'(b)+u(\eta )=0, \end{array}\right. } \end{aligned}$$

where \(\beta >0\), \(a\le \eta \le b\) and \(\beta \ge (b-\eta )\) and \(q:[a,b]\rightarrow \mathbb {R}\) is a continuous function, has a nontrivial continuous solution then

$$\begin{aligned} \int _{a}^{b}|q(s)|\,ds\ge \frac{1}{\beta +(\eta -a)}. \end{aligned}$$

Proof

Apply Theorem 1 for \(\alpha =2\). \(\square \)

4 Application

In this section, we present some applications of the results obtained in Sect. 3 to eigenvalue problem.

\(\lambda \in \mathbb {R}\) is said to be an eigenvalue of the fractional boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -^{C}\!D_{a}^{\alpha }u(t)=\lambda u(t),\,\,a<t<b,\\ u'(a)=0,\,\beta \,^{C}\!D_{a}^{\alpha -1}u(b)+u(\eta )=0, \end{array}\right. } \end{aligned}$$
(3)

where \(1<\alpha \le 2\), \(\beta >0\) and \(a\le \eta \le b\) if Problem (3) has at least a nontrivial continuous solution \(x_{\lambda }\). In this case, we say that \(x_{\lambda }\) is an eigenvector associated to the eigenvalue \(\lambda \).

Corollary 2

Under assumption \(\beta \ge \frac{(b-\eta )^{\alpha -1}}{\Gamma (\alpha )}\) and suppose that \(\lambda \) is an eigenvalue of Problem (3) then

$$\begin{aligned} |\lambda |\ge \frac{\Gamma (\alpha )}{(\beta \Gamma (\alpha )+(\eta -a)^{\alpha -1})(b-a)}. \end{aligned}$$

Proof

As \(\lambda \) is an eigenvalue of Problem (3), this means that Problem (3) has a nontrivial continuous solution \(x_{\lambda }\) and, by using Theorem 1, we have

$$\begin{aligned} \int _{a}^{b}|\lambda |\,ds\ge \frac{\Gamma (\alpha )}{\beta \Gamma (\alpha )+(\eta -a)^{\alpha -1}}. \end{aligned}$$

Therefore,

$$\begin{aligned} |\lambda |\ge \frac{\Gamma (\alpha )}{(\beta \Gamma (\alpha )+(\eta -a)^{\alpha -1})(b-a)}, \end{aligned}$$

which yields the desired result. \(\square \)

Corollary 3

Suppose that \(\lambda \) is an eigenvalue of the ordinary boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -u''(t)=\lambda u(t),\,\,a<t<b,\\ u'(a)=0,\,\beta u'(b)+u(\eta )=0, \end{array}\right. } \end{aligned}$$
(4)

where \(\beta >0\), \(a\le \eta \le b\) and \(\beta \ge (b-a)\), then

$$\begin{aligned} |\lambda |\ge \frac{1}{(\beta +(\eta -a))(b-a)}. \end{aligned}$$

Proof

Since \(\lambda \) is an eigenvalue of Problem (4), this says that Problem (4) admits a nontrivial continuous solution \(x_{\lambda }\). Now, by using Corollary 1, we get

$$\begin{aligned} |\lambda |(b-a)\ge \frac{1}{\beta +(\eta -a)}. \end{aligned}$$

This gives us the desired result. \(\square \)