Lyapunov’s type inequalities for hybrid fractional differential equations

Abstract

We investigate new results about Lyapunov-type inequalities by considering hybrid fractional boundary value problems. We give necessary conditions for the existence of nontrivial solutions for a class of hybrid boundary value problems involving Riemann-Liouville fractional derivative of order $2<\alpha\le3$. The investigation is based on a construction of Green’s functions and on finding its corresponding maximum value. In order to illustrate the results, we provide numerical examples.

1 Introduction and preliminaries

It is well known that various type integral inequalities play a dominant role in the study of quantitative properties of solutions of differential and integral equations. One of them is Lyapunov-type inequality which has been proved to be very useful in studying the zeros of solutions of differential equations. The well-known Lyapunov result [1] states that if the boundary value problem

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

(1.1)

has a nontrivial solution, where q is a real and continuous function, then

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

(1.2)

This result found many practical applications in differential equations (oscillation theory, disconjugacy, eigenvalue problems, etc.); see, for instance, [2–7] and references therein.

The search for Lyapunov-type inequalities in which the starting differential equation is constructed via fractional differential operators has begun very recently. The first work in this direction is due to Ferreira [8], where he derived a Lyapunov-type inequality for Riemann-Liouville fractional boundary value problem

$$ \textstyle\begin{cases} D^{\alpha}y(t)+q(t)y(t)=0, \quad a< t< b, \\ y(a)=y(b)=0, \end{cases} $$

(1.3)

where $D^{\alpha}$ is the Riemann-Liouville fractional derivative of order $1<\alpha\le2$ and $q: [a,b]\to{\mathbb{R}}$ is a continuous function. It has been proved that if (1.3) has a nontrivial solution then

$$ \int_{a}^{b}\bigl\vert q(s)\bigr\vert \,ds> \Gamma(\alpha) \biggl(\frac{4}{b-a} \biggr)^{\alpha-1}. $$

(1.4)

Clearly, if we let $\alpha=2$ in the above inequality, one obtains Lyapunov’s standard inequality. Ferreira also in [9], was obtained a Lyapunov-type inequality for the Caputo fractional boundary value problem

$$ \textstyle\begin{cases} {}^{\mathrm{C}}D^{\alpha}y(t)+q(t)y(t)=0,\quad a< t< b, \\ y(a)=y(b)=0, \end{cases} $$

(1.5)

where ${}^{\mathrm{C}}D^{\alpha}$ is the Caputo fractional derivative of order $1<\alpha\le2$. It has been proved in [9] that if (1.5) has a nontrivial solution then

$$ \int_{a}^{b}\bigl\vert q(s)\bigr\vert \,ds> \frac{\Gamma(\alpha)\alpha^{\alpha}}{[(\alpha -1)(b-a)]^{\alpha-1}} . $$

(1.6)

Similarly if we let $\alpha=2$ in (1.6), one obtains Lyapunov’s classical inequality (1.2).

In [10], Jleli and Samet considered the fractional differential equation

$$ {}^{\mathrm{C}}D^{\alpha}y(t)+q(t)y(t)=0,\quad a< t< b, $$

(1.7)

with the mixed boundary conditions

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

(1.8)

or

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

(1.9)

For boundary conditions (1.8) and (1.9), two Lyapunov-type inequalities were established, respectively, as follows:

$$ \int_{a}^{b}(b-s)^{\alpha-2}\bigl\vert q(s) \bigr\vert \,ds>\frac{\Gamma(\alpha)}{\max\{ \alpha-1, 2-\alpha\}(b-a)} $$

(1.10)

and

$$ \int_{a}^{b}(b-s)^{\alpha-1}\bigl\vert q(s) \bigr\vert \,ds> \Gamma(\alpha). $$

(1.11)

Recently Rong and Bai [11] considered (1.7) under boundary condition

$$ y(a)=0,\qquad {}^{\mathrm{C}}D^{\beta}y(t)=0, \quad 0< \beta\le1, $$

(1.12)

and established the following Lyapunov-type inequality:

$$ \int_{a}^{b}(b-s)^{\alpha-\beta-1}\bigl\vert q(s) \bigr\vert \,ds> \frac{(b-a)^{-\beta}}{\max \{\frac{1}{\Gamma(\alpha)}-\frac{\Gamma(2-\beta)}{\Gamma (\alpha-\beta)}, \frac{\Gamma(2-\beta)}{\Gamma(\alpha-\beta)}, \frac{2-\alpha}{\alpha-1}\cdot\frac{\Gamma(2-\beta)}{\Gamma (\alpha-\beta)} \}}. $$

(1.13)

For other work on Lyapunov-type inequalities for fractional boundary value problems we refer the reader to [12–14].

The aim of this manuscript is to establish some Lyapunov’s type inequalities for hybrid fractional boundary value problem

$$ \textstyle\begin{cases} D_{a}^{\alpha} [ \frac {y(t)}{f(t,y(t))}- \sum_{i=1}^{n} I_{a}^{\beta }h_{i}(t,y(t)) ]+g(t)y(t)=0, \quad t\in(a,b), \\ y(a)=y'(a)=y(b)=0, \end{cases} $$

(1.14)

where $D_{a}^{\alpha}$ denotes the Riemann-Liouville fractional derivative of order $\alpha\in(2,3]$ starting from a point a, the functions $y\in C([a,b], \mathbb{R})$, $g\in L^{1}((a,b], \mathbb{R})$, $f\in C^{1}([a,b]\times\mathbb{R}, \mathbb{R}\setminus\{0\})$, $h_{i}\in C([a,b]\times\mathbb{R}, \mathbb{R})$, $\forall i=1,2,\ldots ,n$, and $I_{a}^{\beta}$ is the Riemann-Liouville fractional integral of order $\beta\geq\alpha$ with the lower limit at a point a.

We recall the basic definitions, [15–17].

Definition 1.1

The fractional integral of order q with the lower limit a for a function f is defined as

$$I_{a}^{q} f(t)= \frac{1}{\Gamma(q)} \int_{a}^{t}\frac{f(s)}{(t-s)^{1-q}}\,ds,\quad t>a, q>0, $$

provided the right-hand side is point-wise defined on $[a,\infty)$, where $\Gamma(\cdot)$ is the gamma function, which is defined by $\Gamma(q)=\int_{0}^{\infty}t^{q-1}e^{-t}\,dt$.

Definition 1.2

The Riemann-Liouville fractional derivative with the lower limit a of order $q>0$, $n-1< q< n$, $n\in N$, is defined as

$$ D_{a}^{q}f(t)=\frac{1}{\Gamma(n-q)} \biggl(\frac{d}{dt} \biggr)^{n} \int_{a}^{t} (t-s)^{n-q-1}f(s)\,ds, $$

where the function f has absolutely continuous derivative up to order $(n-1)$.

2 Main results

We consider two cases: (I) $h_{i}=0$, $i=1,2,\ldots,n$, and (II) $h_{i}\ne 0$, $i=1,2,\ldots,n$.

2.1 Case I: $h_{i}=0$, $i=1,2,\ldots,n$

We consider problem (1.14) with $h_{i}(t,\cdot)=0$ for all $t\in[a,b]$. For $\alpha\in(2,3]$, we first construct a Green’s function for the following boundary value problem:

$$ \textstyle\begin{cases} D_{a}^{\alpha}[ \frac{y(t)}{f(t,y(t))} ]+g(t)y(t)=0, \quad t\in(a,b), \\ y(a)=y'(a)=y(b)=0, \end{cases} $$

(2.1)

with the assumption that f is continuously differentiable and $f(t,y(t))\neq0$ for all $t\in[a,b]$.

Lemma 2.1

Let $y\in AC([a,b],\mathbb{R})$ be a solution of problem (2.1). Then the function y satisfies the following integral equation:

$$ y=f(t,y) \int_{a}^{b}G(t,s)g(s)y(s)\,ds, $$

(2.2)

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

$$ G(t,s)= \textstyle\begin{cases} \frac{(b-s)^{\alpha-1}(t-a)^{\alpha-1}}{\Gamma(\alpha )(b-a)^{\alpha-1}}, &a\leq t\leq s\leq b, \\ \frac{(b-s)^{\alpha-1}(t-a)^{\alpha-1}}{\Gamma(\alpha )(b-a)^{\alpha-1}}-\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}, &a\leq s \leq t \leq b. \end{cases} $$

(2.3)

Proof

Taking the Riemann-Liouville fractional integral of order α from a to t of both sides of (2.1), we obtain

$$ \frac{y(t)}{f(t,y(t))}=c_{0}(t-a)^{\alpha -1}+c_{1}(t-a)^{\alpha-2}+c_{2}(t-a)^{\alpha-3}-I_{a}^{\alpha}g(t)y(t). $$

(2.4)

Putting $t=a$ in (2.4), we get a constant $c_{2}=0$. Differentiating both sides of equation (2.4) with respect to t, we have

$$ \frac{f(t,y(t))y'(t)-y(t)f_{t}(t,y(t))}{f^{2}(t,y(t))}=c_{0}(\alpha -1) (t-a)^{\alpha-2}+c_{1}( \alpha-2) (t-a)^{\alpha-3}-I_{a}^{\alpha-1}g(t)y(t). $$

Applying the conditions of problem (2.1), the constant $c_{1}$ is vanished. Replacing t by b with $c_{1}=c_{2}=0$ in (2.4) and using the last condition of (2.1), the constant $c_{0}$ is obtained as follows:

$$ c_{0}=\frac{I_{a}^{\alpha}g(b)y(b)}{(b-a)^{\alpha-1}}. $$

Hence a solution y of problem (2.1) satisfies the following integral equation:

$$ y(t)=f\bigl(t,y(t)\bigr) \biggl[ \int_{a}^{b}\frac{(b-s)^{\alpha-1}}{\Gamma(\alpha )}\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}g(s)y(s) \,ds- \int_{a}^{t}\frac {(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s)y(s)\,ds \biggr]. $$

(2.5)

By the definition of the Green’s function as in (2.3), equation (2.5) can be written in the form of (2.2). The proof is completed. □

Lemma 2.2

The Green’s function defined in (2.3) satisfies:

• (i) $G(t,s)\ge0$, $\forall t,s\in[a,b]$.

• (ii) $G(t,s)\le H(s):= \frac{(b-s)^{\alpha -1}}{\Gamma(\alpha-1)}$.

• (iii) $\max_{s\in[a,b]} H(s) =\frac {(b-a)^{\alpha-1}}{\Gamma(\alpha-1)}$.

Proof

First of all, we define the following two functions:

$$\begin{aligned}& g_{1}(t,s) = \frac{(b-s)^{\alpha-1}(t-a)^{\alpha -1}}{(b-a)^{\alpha-1}}, \quad a\leq t\leq s\leq b, \\& g_{2}(t,s) = \frac{(b-s)^{\alpha-1}(t-a)^{\alpha -1}}{(b-a)^{\alpha-1}}- (t-s)^{\alpha-1}, \quad a\leq s \leq t\leq b. \end{aligned}$$

(i) It is obvious that $g_{1}(t,s)\ge0$. To show that $g_{2}(t,s)\ge0$, we use the following observation of Ferreira in [8]:

$$a+\frac{(s-a)(b-a)}{t-a}\ge s \quad \mbox{is equivalent to } s\ge a. $$

Then we have

$$\begin{aligned} (t-s)^{\alpha-1} =&(t-a+a-s)^{\alpha-1} \\ =& \biggl[(t-a) \biggl(1+\frac {a-s}{t-a} \biggr) \biggr]^{\alpha-1} \\ =& \biggl[(b-a) \biggl(1+\frac{a-s}{t-a} \biggr) \biggr]^{\alpha-1} \frac {(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}} \\ =& \biggl[b- \biggl(a+\frac{(b-a)(s-a)}{t-a} \biggr) \biggr]^{\alpha -1} \frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}, \end{aligned}$$

(2.6)

which leads to

$$\begin{aligned} g_{2}(t,s) =& \frac{(b-s)^{\alpha-1}(t-a)^{\alpha -1}}{(b-a)^{\alpha-1}}- (t-s)^{\alpha-1} \\ =& \frac{(b-s)^{\alpha-1}(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}- \biggl[b- \biggl(a+\frac{(b-a)(s-a)}{t-a} \biggr) \biggr]^{\alpha-1}\frac {(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}} \\ \ge& \frac{(b-s)^{\alpha-1}(t-a)^{\alpha-1}}{(b-a)^{\alpha -1}}-(b-s)^{\alpha-1}\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}} \\ =& 0. \end{aligned}$$

Therefore, part (i) is proved.

(ii) For $a\le s\le t\le b$, we have

$$\begin{aligned} g_{2}(t,s) =& \biggl[\frac{(b-s)(t-a)}{b-a} \biggr]^{\alpha -1}-(t-s)^{\alpha-1} \\ =&(\alpha-1) \int_{t-s}^{[(b-s)(t-a)]/(b-a)}x^{\alpha-2}\,dx \\ \le&(\alpha-1) \biggl[\frac{(b-s)(t-a)}{b-a} \biggr]^{\alpha-2} \biggl[ \frac{(b-s)(t-a)}{b-a}-(t-s) \biggr] \\ \le&(\alpha-1)\frac{(b-s)^{\alpha-1}(t-a)^{\alpha -1}}{(b-a)^{\alpha-1}} \\ \le&(\alpha-1) (b-s)^{\alpha-1}, \end{aligned}$$

and consequently

$$ \frac{g_{2}(t,s)}{\Gamma(\alpha)}\le\frac{(b-s)^{\alpha-1}}{\Gamma (\alpha-1)}. $$

For $a\le t\le s\le b$, we have

$$\begin{aligned} g_{1}(t,s) =& \biggl[\frac{(b-s)(t-a)}{b-a} \biggr]^{\alpha-1} \\ \le&\frac{(b-s)^{\alpha-1}(s-a)^{\alpha-1}}{(b-a)^{\alpha-1}} \\ =&(\alpha-1) \int_{0}^{[(b-s)(s-a)]/(b-a)}x^{\alpha-2}\,dx \\ \le&(\alpha-1)\frac{(b-s)^{\alpha-2}(s-a)^{\alpha -2}}{(b-a)^{\alpha-2}} \biggl[\frac{(b-s)(s-a)}{b-a} \biggr] \\ =&(\alpha-1)\frac{(b-s)^{\alpha-1}(s-a)^{\alpha-1}}{(b-a)^{\alpha -1}} \\ \le&(\alpha-1) (b-s)^{\alpha-1}, \end{aligned}$$

which yields

$$ \frac{g_{1}(t,s)}{\Gamma(\alpha)}\le \frac{(b-s)^{\alpha-1}}{\Gamma (\alpha-1)}. $$

It follows that

$$ G(t,s) \leq H(s), \quad \forall s,t\in[a,b], $$

which is the proof of part (ii).

(iii) This is obvious, since $H'(s)<0$. The proof is complete. □

Theorem 2.1

The necessary condition for the existence of a nontrivial solution for the boundary value problem (2.1) is

$$ \frac{\Gamma(\alpha-1)}{\|f\|} \leq \int_{a}^{b} (b-s)^{\alpha-1}\bigl\vert g(s) \bigr\vert \,ds, $$

(2.7)

where $\|f\|= \sup_{t\in[a,b], y\in{\mathbb{R}}} |f(t,y)|$.

Proof

From Lemma 2.1, the solution of (2.1) satisfies the following integral equation:

$$ y(t)=f\bigl(t,y(t)\bigr) \biggl( \int_{a}^{b} G(t,s)g(s)y(s)\,ds \biggr). $$

The continuity of functions y and f on their compact domains yield

$$ \|y\| \leq\|f\| \biggl( \int_{a}^{b} \bigl\vert G(t,s)\bigr\vert \bigl\vert g(s)\bigr\vert \bigl\vert y(s)\bigr\vert \,ds \biggr). $$

Simplifying above inequality, we get

$$ 1\leq\|f\| \biggl( \int_{a}^{b} H(s)\bigl\vert g(s)\bigr\vert \,ds \biggr). $$

(2.8)

Applying the result in Lemma 2.2, the desired inequality in (2.7) is obtained. □

Corollary 2.1

The necessary condition for the existence of a nontrivial solution for the boundary value problem (2.1) is

$$ \frac{\Gamma(\alpha-1)}{\|f\|}(b-a)^{1-\alpha}\leq\|g\|_{L^{1}}. $$

(2.9)

Corollary 2.2

Consider the fractional Sturm-Liouville problem given by

$$ \textstyle\begin{cases} D_{0}^{\alpha}[ \frac {y(t)}{f(t,y(t))} ]+\lambda y(t)=0, \quad \alpha\in(2,3], t\in (0,1), \\ y(0)=y'(0)=y(1)=0, \end{cases} $$

(2.10)

where $f(t,y(t))\neq0$ for all $t\in[0,1]$ and $\lambda\in\mathbb {R}$. The necessary condition for the existence of a nontrivial solution for the boundary value problem (2.10) is

$$ |\lambda|\geq\frac{\alpha\Gamma(\alpha-1)}{\|f\|}. $$

(2.11)

Proof

From the Lyapunov-type inequality in Theorem 2.1 and replacing the values $a=0$, $b=1$, and $g(t)\equiv\lambda$ for $t\in[0,1]$, the inequality (2.7) becomes

$$ \frac{\Gamma(\alpha-1)}{\|f\|} \leq \int_{0}^{1} (1-s)^{\alpha -1}|\lambda|\,ds = \frac{1}{\alpha}|\lambda|. $$

This completes the proof. □

Corollary 2.3

Consider the fractional Sturm-Liouville problem of the form

$$ \textstyle\begin{cases} D_{a}^{\alpha}[ \frac{y(t)}{f(t,y(t))} ]+\lambda y(t)=0, \quad \alpha\in(2,3], t\in(a,b), \\ y(a)=y'(a)=y(b)=0, \end{cases} $$

(2.12)

where $f(t,y(t))\neq0$ for all $t\in[a,b]$ and $\lambda\in\mathbb {R}$. The necessary condition for the existence of a nontrivial solution for the boundary value problem (2.12) is

$$ |\lambda|\geq\frac{\Gamma(\alpha-1)}{(b-a)^{\alpha} \|f\|}. $$

(2.13)

Proof

From Corollary 2.1, we get

$$ \frac{\Gamma(\alpha-1)}{\|f\|}(b-a)^{1-\alpha} \leq \int_{a}^{b} |\lambda| \,ds = (b-a)|\lambda|, $$

which is the inequality in (2.13). This completes the proof. □

Example 2.1

Consider the following boundary value problem of the hybrid fractional differential equation:

$$ \textstyle\begin{cases} D_{1}^{5/2} [\frac{y(t)}{(t+1)+\frac {|y(t)|+3}{|y(t)|+5}} ]+\lambda y(t)=0, \quad t \in (1,3 ), \\ y(1)=y'(1)=y(3)=0. \end{cases} $$

(2.14)

Here $\alpha=5/2$, $a=1$, $b=3$, $f(t,y)=(t+1)+(|y|+3)/(|y|+5)$. We find that $\|f\|=5$. Applying Corollary 2.3, we see that the necessary condition for the existence of a nontrivial solution for the boundary value problem (2.14) is

$$ |\lambda|\geq0.03133285342. $$

2.2 Case II: $h_{i}\ne0$, $i=1,2,\ldots, n$

In this section we will construct Lyapunov-type inequalities for the boundary value problem (1.14). We recall that f is continuously differentiable.

Lemma 2.3

Let $y\in AC[a,b]$ be a solution of problem (1.14). Then the function y can be written as

$$ y(t)=f\bigl(t,y(t)\bigr) \Biggl[ \int_{a}^{b} G(t,s)g(s)y(s)\,ds-\sum _{i=1}^{n} \int _{a}^{b}G^{*}(t,s)h_{i}\bigl(s,y(s) \bigr)\,ds \Biggr], $$

(2.15)

where $G(t,s)$ is defined as in (2.3) and $G^{*}(t,s)$ is defined by

$$ G^{*}(t,s)= \textstyle\begin{cases} \frac{(b-s)^{\beta-1}(t-a)^{\alpha-1}}{\Gamma(\beta )(b-a)^{\alpha-1}}-\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}, &a\leq s\leq t \leq b, \\ \frac{(b-s)^{\beta-1}(t-a)^{\alpha-1}}{\Gamma(\beta )(b-a)^{\alpha-1}}, &a\leq t\leq s\leq b. \end{cases} $$

(2.16)

Proof

The general solution of problem (1.14) is given by

$$\begin{aligned}& \frac{y(t)}{f(t,y(t))}-\sum_{i=1}^{n} I_{a}^{\beta}h_{i}\bigl(t,y(t)\bigr) \\& \quad = -I_{a}^{\alpha}g(t)y(t)+c_{1}(t-a)^{\alpha-1}+c_{2}(t-a)^{\alpha -2}+c_{3}(t-a)^{\alpha-3}. \end{aligned}$$

(2.17)

By condition $y(a)=0$, the constant $c_{3}=0$. Differentiating equation (2.17), we get

$$\begin{aligned}& \frac{f(t,y(t))y'(t)-y(t)f_{t}(t,y(t))}{f^{2}(t,y(t))}-\sum_{i=1}^{n} I_{a}^{\beta-1}h_{i}\bigl(t,y(t)\bigr) \\& \quad = -I_{a}^{\alpha-1}g(t)y(t)+c_{1}(\alpha-1) (t-a)^{\alpha-2}+c_{2}(\alpha -2) (t-a)^{\alpha-3}. \end{aligned}$$

Replacing t by a to the above equation, we have $c_{2}=0$. Equation (2.17) becomes

$$ y(t) = f\bigl(t,y(t)\bigr) \Biggl[-I_{a}^{\alpha}g(t)y(t)+c_{1}(t-a)^{\alpha-1}+\sum_{i=1}^{n} I_{a}^{\beta}h_{i}\bigl(t,y(t)\bigr) \Biggr]. $$

(2.18)

Since $y(b)=0$, we have

$$ c_{1}=(b-a)^{1-\alpha} \Biggl(I_{a}^{\alpha}g(b)y(b)- \sum_{i=1}^{n} I_{a}^{\beta}h_{i} \bigl(b,y(b)\bigr) \Biggr). $$

Substituting the constant $c_{1}$ into equation (2.18), the solution of problem (1.14) is in the form

$$\begin{aligned} y(t) =& f\bigl(t,y(t)\bigr) \Biggl[-I_{a}^{\alpha}g(t)y(t)+ \biggl(\frac {t-a}{b-a} \biggr)^{\alpha-1} \Biggl( I_{a}^{\alpha}g(b)y(b)- \sum_{i=1}^{n} I_{a}^{\beta}h_{i} \bigl(b,y(b)\bigr) \Biggr) \\ &{}+\sum_{i=1}^{n} I_{a}^{\beta}h_{i} \bigl(t,y(t)\bigr) \Biggr] \\ =& f\bigl(t,y(t)\bigr) \Biggl[- \int_{a}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)}g(s)y(s)\,ds \\ &{}+ \biggl( \frac{t-a}{b-a} \biggr)^{\alpha-1} \Biggl( \int_{a}^{b}\frac{(b-s)^{\alpha-1}}{\Gamma(\alpha)}g(s)y(s)\,ds \\ &{}-\sum_{i=1}^{n} \int_{a}^{b}\frac{(b-s)^{\beta-1}}{\Gamma(\beta )}h_{i} \bigl(s,y(s)\bigr)\,ds \Biggr)+\sum_{i=1}^{n} \int_{a}^{t}\frac{(t-s)^{\beta -1}}{\Gamma(\beta)} h_{i} \bigl(s,y(s)\bigr)\,ds \Biggr] \\ =&f\bigl(t,y(t)\bigr) \Biggl[ \int_{a}^{b} G(t,s)g(s)y(s)\,ds-\sum _{i=1}^{n} \int _{a}^{b}G^{*}(t,s)h_{i}\bigl(s,y(s) \bigr)\,ds \Biggr], \end{aligned}$$

where the Green’s functions $G(t,s)$ and $G^{*}(t,s)$ are defined by (2.3) and (2.16), respectively. The proof is completed. □

Lemma 2.4

The Green’s function $G^{*}(t,s)$, which is given by (2.16), satisfies the following inequalities:

• (i) $G^{*}(t,s)\geq0$, $\forall t,s\in[a,b]$;

• (ii) $G^{*}(t,s)\le J(s):= \frac{(\alpha -1)(b-s)^{\beta-1}}{\Gamma(\beta)}$.

Also we have

• (iii) $\max_{s\in[a,b]} J(s) =\frac{(\alpha -1)(b-a)^{\beta-1}}{\Gamma(\beta)}$.

Proof

From Lemma 2.3, we define

$$\begin{aligned}& g_{3}(t,s) = \frac{(b-s)^{\beta-1}(t-a)^{\alpha -1}}{(b-a)^{\alpha-1}}- (t-s)^{\beta-1}, \quad a\leq s \leq t\leq b, \\& g_{4}(t,s) = \frac{(b-s)^{\beta-1}(t-a)^{\alpha -1}}{(b-a)^{\alpha-1}}, \quad a\leq t\leq s\leq b. \end{aligned}$$

It is obvious that $g_{4}(t,s)\geq0$. By using (2.6) with replacing α by β, we have

$$\begin{aligned} g_{3}(t,s) = & \frac{(b-s)^{\beta-1}(t-a)^{\alpha-1}}{(b-a)^{\alpha -1}}- (t-s)^{\beta-1} \\ \geq& \frac{(b-s)^{\beta-1}(t-a)^{\alpha-1}}{(b-a)^{\alpha -1}}-\frac{(b-s)^{\beta-1}(t-a)^{\beta-1}}{(b-a)^{\beta-1}} \\ =&(b-s)^{\beta-1} \biggl[ \biggl(\frac{t-a}{b-a} \biggr)^{\alpha -1}- \biggl(\frac{t-a}{b-a} \biggr)^{\beta-1} \biggr]. \end{aligned}$$

As $\beta\geq\alpha$, we deduce that $g_{3}(t,s)\geq0$. Therefore, we have $G^{*}(t,s)\geq0$ for all $t,s\in[a,b]$.

We omit the proofs of (ii) and (iii), since these are similar to that for the Green’s function $G(t,s)$ in Lemma 2.2. □

In the following results will be used the following condition:

• (H) $|h_{i}(t,y(t))|\leq|x_{i}(t)||y(t)|$ where $x_{i}\in C([a,b],\mathbb{R})$, $i=1,2,\ldots,n$.

Theorem 2.2

Assume that the condition (H) holds with $[a,b]=[0,1]$. The necessary condition for the existence of a nontrivial solution for problem (1.14) on $[0,1]$ is

$$ \Gamma(\alpha-1) \Biggl(\frac{1}{\|f\|}-\frac{(\alpha-1)}{\Gamma (\beta+1)}\sum _{i=1}^{n}\|x_{i}\| \Biggr)\leq \int_{0}^{1}(1-s)^{\alpha-1}\bigl\vert g(s) \bigr\vert \,ds. $$

(2.19)

Proof

From Lemma 2.3, the solution of problem (1.14) on $[0,1]$ is given by

$$ y(t)=f\bigl(t,y(t)\bigr) \Biggl[ \int_{0}^{1} G(t,s)g(s)y(s)\,ds-\sum _{i=1}^{n} \int _{0}^{1}G^{*}h_{i}\bigl(s,y(s)\bigr) \,ds \Biggr]. $$

Since $y\in C([0,1],\mathbb{R})$ and $f\in C^{1}([0,1]\times\mathbb {R},\mathbb{R}\setminus\{0\})$, we get

$$\begin{aligned} \bigl\vert y(s)\bigr\vert \leq& \|f\| \Biggl[ \int_{0}^{1} \bigl\vert G(t,s)\bigr\vert \bigl\vert g(s)\bigr\vert \bigl\vert y(s)\bigr\vert \,ds + \sum _{i=1}^{n} \int_{0}^{1} \bigl\vert G^{*}(t,s)\bigr\vert \bigl\vert h_{i}\bigl(s,y(s)\bigr)\bigr\vert \,ds \Biggr] \\ \leq& \|f\| \Biggl[ \int_{0}^{1} H(s)\bigl\vert g(s)\bigr\vert \bigl\vert y(s)\bigr\vert \,ds + \sum_{i=1}^{n} \int _{0}^{1} J(s) \bigl\vert x_{i}(s) \bigr\vert \bigl\vert y(s)\bigr\vert \,ds \Biggr] \\ \leq&\|f\| \Biggl[ \int_{0}^{1}\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha -1)}\bigl\vert g(s)\bigr\vert \bigl\vert y(s)\bigr\vert \,ds \\ &{}+\frac{(\alpha-1)}{\Gamma(\beta)}\sum_{i=1}^{n} \int _{0}^{1}(1-s)^{\beta-1}\bigl\vert x_{i}(s)\bigr\vert \bigl\vert y(s)\bigr\vert \,ds\Biggr], \end{aligned}$$

which leads to

$$\begin{aligned} \|y\| \leq\|f\| \Biggl[\frac{\|y\|}{\Gamma(\alpha-1)} \int _{0}^{1}(1-s)^{\alpha-1}|g(s)|\,ds + \frac{(\alpha-1)\|y\|}{\Gamma(\beta+1)}\sum_{i=1}^{n} \|x_{i}\| \Biggr]. \end{aligned}$$

Therefore, we deduce that the inequality in (2.19) holds. □

Corollary 2.4

Assume that the condition (H) holds with $[a,b]=[0,1]$. Consider the problem

$$ \textstyle\begin{cases} D_{0}^{\alpha} [ \frac{y(t)}{f(t,y(t))}-\sum_{i=1}^{n} I_{0}^{\beta}h_{i}(t,y(t)) ]+\lambda y(t)=0, \quad \alpha \in(2,3], t\in(0,1), \\ y(0)=y'(0)=y(1)=0. \end{cases} $$

(2.20)

The necessary condition for the existence of a nontrivial solution for problem (2.20) on $[0,1]$ is

$$ \alpha\Gamma(\alpha-1) \Biggl(\frac{1}{\|f\|}- \frac{(\alpha -1)}{\Gamma(\beta+1)}\sum_{i=1}^{n} \|x_{i}\| \Biggr)\leq|\lambda|. $$

(2.21)

Proof

Setting the function $g(t)\equiv\lambda$ for $t\in[0,1]$ and applying Theorem 2.2, we obtain the following inequality:

$$ \Gamma(\alpha-1) \Biggl(\frac{1}{\|f\|}-\frac{(\alpha-1)}{\Gamma (\beta+1)}\sum _{i=1}^{n}\|x_{i}\| \Biggr)\leq \int_{0}^{1}(1-s)^{\alpha -1}|\lambda|\,ds= \frac{1}{\alpha}|\lambda|, $$

from which the result in (2.21) is proved. □

Theorem 2.3

Suppose that the condition (H) holds. The necessary condition for the existence of a nontrivial solution for problem (1.14) on $[a,b]$, is

$$ \|g\|_{L^{1}} \geq \frac{\Gamma(\alpha-1)}{(b-a)^{\alpha-1}} \Biggl( \frac{1}{\|f\|} - (\alpha-1)\frac{(b-a)^{\beta-1}}{\Gamma(\beta )}\sum _{i=1}^{n} \|x_{i}\|_{L^{1}} \Biggr). $$

(2.22)

Proof

From Lemmas 2.2 and 2.4, we have

$$\begin{aligned} \bigl\vert y(s)\bigr\vert \leq& \|f\| \Biggl[ \int_{a}^{b} \bigl\vert G(t,s)\bigr\vert \bigl\vert g(s)\bigr\vert \bigl\vert y(s)\bigr\vert \,ds + \sum _{i=1}^{n} \int_{a}^{b} \bigl\vert G^{*}(t,s)\bigr\vert \bigl\vert h_{i}\bigl(s,y(s)\bigr)\bigr\vert \,ds \Biggr] \\ \leq& \|f\| \Biggl[ \int_{a}^{b} H(s)\bigl\vert g(s)\bigr\vert \bigl\vert y(s)\bigr\vert \,ds + \sum_{i=1}^{n} \int _{a}^{b}J(s)\bigl\vert x_{i}(s) \bigr\vert \bigl\vert y(s)\bigr\vert \,ds \Biggr]. \end{aligned}$$

Consequently the above inequality becomes

$$ \|y\| \leq \|f\| \Biggl[\frac{(b-a)^{\alpha-1}\|y\|}{\Gamma(\alpha -1)} \int_{a}^{b} |g(s)|\,ds + \frac{(\alpha-1)(b-a)^{\beta-1}\|y\|}{\Gamma(\beta)}\sum _{i=1}^{n} \int_{a}^{b} \bigl\vert x_{i}(s)\bigr\vert \,ds \Biggr], $$

which leads to

$$ 1 \leq\|f\| \Biggl[ \frac{(b-a)^{\alpha-1}}{\Gamma(\alpha-1)} \|g\| _{L^{1}}+\frac{(\alpha-1)(b-a)^{\beta-1}}{\Gamma(\beta)} \sum_{i=1}^{n} \|x_{i} \|_{L^{1}} \Biggr]. $$

Then the estimate in (2.22) holds. □

Corollary 2.5

Let the condition (H) holds. Consider the fractional boundary value problem given by

$$ \textstyle\begin{cases} D_{a}^{\alpha} [ \frac{y(t)}{f(t,y(t))}-\sum_{i=1}^{n} I_{a}^{\beta}h_{i}(t,y(t)) ]+\lambda y(t)=0, \quad \alpha \in(2,3], t\in(a,b), \\ y(a)=y'(a)=y(b)=0. \end{cases} $$

(2.23)

The necessary condition for the existence of a nontrivial solution for problem (2.23) on $[a,b]$ is

$$ |\lambda| \geq \frac{\Gamma(\alpha-1)}{(b-a)^{\alpha}} \Biggl(\frac{1}{\|f\|} - \frac{(\alpha-1)(b-a)^{\beta-1}}{\Gamma(\beta )}\sum_{i=1}^{n} \|x_{i}\|_{L^{1}} \Biggr). $$

(2.24)

Proof

Applying the inequality in (2.22) with $g(s)=\lambda$, $s\in[a,b]$, it follows that

$$ \int_{a}^{b}|\lambda|\,ds \geq\frac{\Gamma(\alpha-1)}{(b-a)^{\alpha -1}} \Biggl(\frac{1}{\|f\|} - \frac{(\alpha-1)(b-a)^{\beta -1}}{\Gamma(\beta)}\sum_{i=1}^{n} \|x_{i}\|_{L^{1}} \Biggr), $$

which implies the inequality in (2.24). □

Example 2.2

Consider the following boundary value problem of the hybrid fractional differential equation:

$$ \textstyle\begin{cases} D_{0}^{5/2} [\frac{y(t)}{(t+1)+\frac {2|y(t)|+1}{3|y(t)|+2}}-\sum_{i=1}^{3}I_{0}^{7/2}\frac {t^{(i+1)/(i+2)}y^{2}(t)}{1+|y(t)|} ]+\lambda y(t)=0,\quad t \in (0,\frac{1}{2} ), \\ y(0)=y'(0)=y (\frac{1}{2} )=0. \end{cases} $$

(2.25)

Here $\alpha=5/2$, $\beta=7/2$, $a=0$, $b=1/2$, $n=3$, $h_{i}(t,y)=(t^{(i+1)/(i+2)}y^{2})/(1+|y|)$, $i=1,2,3$, $f(t,y)=(t+1)+(2|y|+1)/(3|y|+2)$. We find that $\|f\|=13/6$, and $|h_{i}(t,y)|\leq|t^{(i+1)/(i+2)}||y|$. Setting $x_{i}(t)=t^{(i+1)/(i+2)}$, $i=1,2,3$, we have $\|x_{1}\| _{L^{1}}=0.1889881575$, $\|x_{2}\|_{L^{1}}=0.1698867308$, and $\|x_{3}\| _{L^{1}}=0.1595414382$. Applying Corollary 2.5, we see that the necessary condition for the existence of a nontrivial solution for the boundary value problem (2.25) is

$$ |\lambda|\geq2.106444184. $$

Remark 2.1

The boundary value problem (1.14) can be rewritten by

$$ \textstyle\begin{cases} D_{a}^{\alpha} [ \frac {y(t)}{f(t,y(t))} ]+g(t)y(t)= \sum_{i=1}^{n} I_{a}^{\beta-\alpha}h_{i}(t,y(t)), \quad t\in(a,b), \\ y(a)=y'(a)=y(b)=0, \end{cases} $$

(2.26)

which is a hybrid fractional integro-differential equation with boundary conditions. Therefore, all results can be applied.