ABSTRACT

In this paper, we present Lyapunov-type inequality for conformable BVP

$T_{\alpha }^{a}y\left(t\right)+q\left(t\right)y\left(t\right)=0$

with the conformable fractional derivative of order $1<\alpha \le 2$ and $2<\alpha \le 3$ with corresponding boundary conditions. We obtain the Lyapunov-type inequality by a construction Green’s function and get its corresponding maximum value. Application to the corresponding eigenvalue problem is also discussed.

Keywords:

Lyapunov-Type Inequalities, Conformable Fractional Derivative, Green’s Function, Eigenvalue

1. Introduction

Lyapunov-type inequality is an important and useful tool for studying differential equations. The classical Lyapunov-type inequality for differential equations was studied in [1] :

$\{\begin{array}{l}{y}^{″}\left(t\right)=-q\left(t\right)y\left(t\right),a<t<b,\\ y\left(a\right)=y\left(b\right)=0,\end{array}$ (1.1)

if (1.1) has a nontrivial solution, then

${\displaystyle {\int }_a^b}\left|q\left(s\right)\right|\text{d}s>\frac{4}{b-a}.$ (1.2)

Furthermore, the constant 4 in (1.2) is sharp.

More authors paid attention to study Lyapunov-type inequality for differential equations and got many results. In recent years, a series of achievements have been made in the Lyapunov-type inequalities of fractional differential equations. We refer to [2] - [12] . In [3] , Ferreira studied the following equations:

$\{\begin{array}{l}{}_a{}^{C}D{}^{\alpha }y\left(t\right)+q\left(t\right)y\left(t\right)=0,a<t<b,1<\alpha \le 2\\ y\left(a\right)=0=y\left(b\right),\end{array}$ (1.3)

if (1.3) has a nontrivial solution, then

${\displaystyle {\int }_a^b}\left|q\left(s\right)\right|\text{d}s>\frac{\Gamma \left(\alpha \right){\alpha }^{\alpha }}{{\left[\left(\alpha -1\right)\left(b-a\right)\right]}^{\alpha -1}}.$

In [7] , Abdeljanad and Baleanu obtained a Lyapunov-type inequality for ABR fractional boundary value problem

$\{\begin{array}{l}\left({}_a{}^{ABR}D{}^{\alpha }y\right)\left(t\right)+q\left(t\right)y\left(t\right)=0,a<t<b,2<\alpha \le 3,\\ y\left(a\right)=y\left(b\right)=0,\end{array}$ (1.4)

if (1.4) has a nontrivial solution, then

${\displaystyle {\int }_a^b}T\left(s\right)\text{d}s>\frac{4}{b-a},$

where

$T\left(s\right)=\left[\frac{3-\alpha }{B\left(\alpha -2\right)}\left|q\left(t\right)\right|+\frac{\alpha -2}{B\left(\alpha -2\right)}\left({}_{a}I{}^{\alpha -2}\left|q\left(s\right)\right|\right)\left(t\right)\right].$

In [10] , Abdeljawad studied a generalized Lyapunov-type inequalities for conformable BVP

$\{\begin{array}{l}{T}_{\alpha }^{c}x\left(t\right)+r\left(t\right)x\left(t\right)=0,c<t<d,1<\alpha \le 2,\\ x\left(c\right)=x\left(d\right)=0,\end{array}$ (1.5)

if (1.5) has a nontrivial solution, then

${\displaystyle {\int }_c^d}\left|r\left(s\right)\right|\text{d}s>\frac{{\alpha }^{\alpha }}{{\left(\alpha -1\right)}^{\alpha -1}{\left(d-c\right)}^{\alpha -1}}.$

Furthermore, Abdeljawad proved a Lyapunov-type inequalitiy for a sequential conformable BVP

$\{\begin{array}{l}{T}_{\alpha }^a\cdot T_{\alpha }^{a}x\left(t\right)+r\left(t\right)x\left(t\right)=0,a<t<b,\frac{1}{2}<\alpha \le 1,\\ x\left(c\right)=x\left(d\right)=0,\end{array}$ (1.6)

if (1.6) has a nontrivial solution, then

${\displaystyle {\int }_c^d}\left|r\left(t\right)\right|\text{d}s>\frac{3\alpha -1}{{\left(d-c\right)}^{2\alpha -1}}{\left(\frac{3\alpha -1}{2\alpha -1}\right)}^{\frac{2\alpha -1}{\alpha }}.$

In this paper, we establish a Lyapunov-type inequalities for conformable BVP

$\{\begin{array}{l}{T}_{\alpha }^{a}y\left(t\right)+q\left(t\right)y\left(t\right)=0,a<t<b,1<\alpha \le 2,\\ y\left(a\right)=y^{\prime }\left(b\right)=0\end{array}$ (1.7)

and

$\{\begin{array}{l}{T}_{\alpha }^{a}f\left(t\right)+p\left(t\right)f\left(t\right)=0,a<t<b,2<\alpha \le 3,\\ f\left(a\right)=f^{\prime }\left(a\right)=f^{\prime }\left(b\right)=0,\end{array}$ (1.8)

where $T_{\alpha }^a$ is conformable fractional derivative starting at a of order $\alpha$ , and $p\mathrm{,}q$ are real-valued continuous. The introduction and background of conformable fractional are given in [2] [10] . Then, we give the definition and lemma about conformable fractional derivative in the following.

Definition 1.1. [4] Let $n<\alpha \le n+1$ . Then

$\left(I_{\alpha }^{c}g\right)\left(t\right)=\frac{1}{n!}{\displaystyle {\int }_c^t}{\left(t-s\right)}^n{\left(s-c\right)}^{\alpha -n-1}g\left(s\right)\text{d}s$

is called the left conformable fractional derivative starting at c of order $\alpha$ .

Lemma 1.1. [4] Let $f\mathrm{<mo>\; :\; </mo>}\left[c\mathrm{,}\infty \right)\to R$ be $\left(n+1\right)$ times differentiable for $t>c$ , $n<\alpha \le n+1$ . Then, we have the following result:

$\left(I_{\alpha }^c\cdot T_{\alpha }^{c}f\right)\left(t\right)=f\left(t\right)-{\displaystyle {\sum }_{k=0}^n\frac{f^n\left(c\right){\left(t-c\right)}^k}{k!}}.$

2. A Lyapunov-Type Inequality for Conformable Fractional Derivative of $1<\alpha \le 2$

Theorem 2.1. $y\in C\left[a\mathrm{,}b\right]$ is a solution of the BVP (1.7) if and only if y satisfies the integral equation

$y\left(t\right)={\displaystyle {\int }_a^b}G\left(t,s\right)q\left(s\right)y\left(s\right)\text{d}s.$ (2.1)

where $G\left(t\mathrm{,}s\right)$ is the Green’s function defined as

$G\left(t,s\right)=\{\begin{array}{l}\left(t-a\right){\left(s-a\right)}^{\alpha -2},a\le t\le s\le b,\\ {\left(s-a\right)}^{\alpha -1},a\le s\le t\le b.\end{array}$ (2.2)

Proof. Applying the integral $I_a^{\alpha }$ in the (1.7), we have

$I_{\alpha }^a\cdot T_{\alpha }^{a}y\left(t\right)=-I_{\alpha }^a\left(q\left(t\right)y\left(t\right)\right).$

Then, using definition 1.1 and lemma 1.1, we obtain

$y\left(t\right)=c_0+c_1\left(t-a\right)-{\displaystyle {\int }_a^t}\left(t-s\right){\left(s-a\right)}^{\alpha -2}q\left(s\right)y\left(s\right)\text{d}s.$ (2.3)

Since $y\left(a\right)=0$ , we get immediately that $c_0=0$ .

By the boundary condition $y^{\prime }\left(b\right)=0$ , we obtain

$c_1={\displaystyle {\int }_a^b}{\left(s-a\right)}^{\alpha -2}q\left(s\right)y\left(s\right)\text{d}s.$

Hence, equation (2.3) becomes

$y\left(t\right)=\left(t-a\right){\displaystyle {\int }_a^b}{\left(s-a\right)}^{\alpha -2}q\left(s\right)y\left(s\right)\text{d}s-{\displaystyle {\int }_a^t}\left(t-s\right){\left(s-a\right)}^{\alpha -2}q\left(s\right)y\left(s\right)\text{d}s.$ (2.4)

Then, equation (2.4) can be written in the form of (2.1), where the Green’s function is defined in (2.2).

The proof is completed.

Corollary 2.1. The function G defined in Theorem 2.1 satisfied the following property:

$\underset{s\in \left[a,b\right]}{\max }G\left(t,s\right)=G\left(t,t\right)={\left(t-a\right)}^{\alpha -1}.$ (2.5)

Proof. We define the function

$g_1\left(t,s\right)=\left(t-a\right){\left(s-a\right)}^{\alpha -2}$

and

$g_2\left(t,s\right)={\left(s-a\right)}^{\alpha -1}.$

For $a\le t\le s\le b$ , differentiating $g_1\left(t\mathrm{,}s\right)$ with respect to s, we get

${g^{\prime }}_1\left(t,s\right)=\left(t-a\right)\left(\alpha -2\right){\left(s-a\right)}^{\alpha -3}<0.$ (2.6)

While for $a\le s\le t\le b$ , differentiating $g_2\left(t,s\right)$ with respect to s, we get

${g^{\prime }}_2\left(t,s\right)=\left(\alpha -1\right){\left(s-a\right)}^{\alpha -2}>0.$ (2.7)

Hence, $g_1\left(t\mathrm{,}s\right)$ is a decreasing function, $g_2\left(t,s\right)$ is an increasing function in s. Consequently, G (t, s) gets the maximum at s = t, we obtain (2.5).

Corollary 2.2. If (1.7) has a nontrivial continuous solution, then

${\displaystyle {\int }_a^b}{\left(t-a\right)}^{\alpha -1}\left|q\left(s\right)\right|\text{d}s\ge 1.$ (2.8)

Proof. Let $y\in C\left[a\mathrm{,}b\right]$ be a nontrivial solution of the BVP (1.7), where the norm

$\Vert y\Vert =\underset{t\in \left[a,b\right]}{sup}\left\{\left|y\left(t\right)\right|\right\}\mathrm{.}$

Form (2.1), we have

$\begin{array}{c}\left|y\left(t\right)\right|\le {\displaystyle {\int }_a^b}\left|G\left(t,s\right)\right|\left|q\left(s\right)\right|\left|y\left(s\right)\right|\text{d}s\\ \le {\displaystyle {\int }_a^b}\underset{s\in \left[a,b\right]}{\max }G\left(t,s\right)\left|q\left(s\right)\right|\left|y\left(s\right)\right|\text{d}s\\ \le {\displaystyle {\int }_a^b}{\left(t-a\right)}^{\alpha -1}\left|q\left(s\right)\right|\left|y\left(s\right)\right|\text{d}s.\end{array}$ (2.9)

Taking the norm leads to

$\Vert y\Vert \le \left({\displaystyle {\int }_a^b}{\left(t-a\right)}^{\alpha -1}\left|q\left(s\right)\right|\text{d}s\right)\Vert y\Vert \mathrm{.}$

Then,

${\displaystyle {\int }_a^b}{\left(t-a\right)}^{\alpha -1}\left|q\left(s\right)\right|\text{d}s\ge 1.$

This completes the proof.

Corollary 2.3. If the BVP (1.7) has a nontrivial continuous solution, then

${\displaystyle {\int }_a^b}\left|q\left(s\right)\right|\text{d}s\ge {\left(b-a\right)}^{1-\alpha }\mathrm{.}$ (2.10)

Proof. In (2.8), let

$f\left(t\right)={\left(t-a\right)}^{\alpha -1}\mathrm{,}t\in \left(a\mathrm{,}b\right)\mathrm{.}$

Differentiating $f\left(t\right)$ on $\left(a\mathrm{,}b\right)$ , we have

$f^{\prime }\left(t\right)=\left(\alpha -1\right){\left(t-a\right)}^{\alpha -2}>0,$

hence, $f\left(t\right)$ is a increasing function, we have

$\underset{t\in \left[a,b\right]}{max}f\left(t\right)\le f\left(b\right)={\left(b-a\right)}^{\alpha -1}\mathrm{.}$

Then,

${\displaystyle {\int }_a^b}{\left(b-a\right)}^{\alpha -1}\left|q\left(s\right)\right|\text{d}s\ge 1.$

Hence, we get the inequality (2.10). The proof is complete.

Example 2.1. If the BVP

$\{\begin{array}{l}{T}_{\alpha }^{a}y\left(t\right)+\lambda y\left(t\right)=0,0<t<1,1<\alpha \le 2,\\ y\left(0\right)=y^{\prime }\left(1\right)=0\end{array}$

has a nontrivial solution, then

$\left|\lambda \right|\ge 1.$ (2.11)

Proof. Assume that $\lambda$ is an eigenvalue of (1.7). By using Corollary 2.3, we have

${\displaystyle {\int }_0^1}\left|\lambda \right|\text{d}s=\left|\lambda \right|\ge 1.$

Hence, we get the desired result (2.11). The proof is complete.

3. A Lyapunov-Type Inequality for Conformable Fractional Derivative of $2<\alpha \le 3$

Theorem 3.1. $f\in C\left[a\mathrm{,}b\right]$ is a solution of the BVP (1.8) if and only if f satisfies the integral equation

$f\left(t\right)={\displaystyle {\int }_a^b}H\left(t,s\right)p\left(s\right)f\left(s\right)\text{d}s.$ (3.12)

where $H\left(t\mathrm{,}s\right)$ is the Green’s function defined as

$H\left(t,s\right)=\{\begin{array}{l}\frac{{\left(t-a\right)}^2\left(b-s\right){\left(s-a\right)}^{\alpha -3}}{2\left(b-a\right)},a\le t\le s\le b,\\ \left[\frac{\left(b-s\right){\left(t-a\right)}^2}{2\left(b-a\right)}-\frac{{\left(t-s\right)}^2}{2}\right]{\left(s-a\right)}^{\alpha -3},a\le s\le t\le b.\end{array}$ (3.13)

Proof. Applying the integral $I_a^{\alpha }$ in the (1.8), we have

$I_{\alpha }^a\cdot T_{\alpha }^{a}f\left(t\right)=-I_{\alpha }^a\left(p\left(t\right)f\left(t\right)\right)\mathrm{.}$

Then, using definition 1.1 and lemma 1.1, we obtain

$f\left(t\right)=a_0+a_1\left(t-a\right)+a_2{\left(t-a\right)}^2-\frac{1}{2}{\displaystyle {\int }_a^t}{\left(t-s\right)}^2{\left(s-a\right)}^{\alpha -3}p\left(s\right)f\left(s\right)\text{d}s.$ (3.14)

Since $f\left(a\right)=f^{\prime }\left(a\right)=0$ , we get immediately that $a_0=a_1=0$ .

By the boundary condition $f^{\prime }\left(b\right)=0$ , we obtain

$a_2=\frac{1}{2\left(b-a\right)}{\displaystyle {\int }_a^b}\left(b-s\right){\left(s-a\right)}^{\alpha -3}p\left(s\right)f\left(s\right)\text{d}s.$

Hence, equation (3.14) becomes

$\begin{array}{c}f\left(t\right)=\frac{{\left(t-a\right)}^2}{2\left(b-a\right)}{\displaystyle {\int }_a^b}\left(b-s\right){\left(s-a\right)}^{\alpha -3}p\left(s\right)f\left(s\right)\text{d}s\\ -\frac{1}{2}{\displaystyle {\int }_a^t}{\left(t-s\right)}^2{\left(s-a\right)}^{\alpha -3}p\left(s\right)f\left(s\right)\text{d}s.\end{array}$ (3.15)

Then equation (3.15) can be written in the form of (3.12), where the Green’s function is defined in (3.13). The proof is completed.

Corollary 3.1. The function H defined in Theorem 3.1 satisfied the following property:

$\underset{t\in \left[a,b\right]}{\max }H\left(t,s\right)=H\left(b,s\right)=\frac{\left(b-a\right)\left(b-s\right){\left(s-a\right)}^{\alpha -3}}{2},$

$\underset{s\in \left[a\mathrm{,}b\right]}{max}H\left(b\mathrm{,}s\right)\le \frac{\left(b-a\right){\left(s-a\right)}^{\alpha -3}}{2}\mathrm{.}$

Proof. We define the function

$h_1\left(t,s\right)=\frac{{\left(t-a\right)}^2\left(b-s\right){\left(s-a\right)}^{\alpha -3}}{2\left(b-a\right)}$

and

$h_2\left(t,s\right)=\left[\frac{\left(b-s\right){\left(t-a\right)}^2}{2\left(b-a\right)}-\frac{{\left(t-s\right)}^2}{2}\right]{\left(s-a\right)}^{\alpha -3}.$

For $a\le t\le s\le b$ , differentiating $h_1\left(t\mathrm{,}s\right)$ with respect to t, we get

${h^{\prime }}_1\left(t,s\right)=\frac{\left(t-a\right)\left(b-s\right){\left(s-a\right)}^{\alpha -3}}{b-a}\ge 0.$ (3.16)

Hence, $h_1\left(t\mathrm{,}s\right)$ is an increasing function in t.

While for $a\le s\le t\le b$ , differentiating $h_2\left(t\mathrm{,}s\right)$ with respect to t, we get

${h^{\prime }}_2\left(t,s\right)=\left[\frac{\left(b-s\right)\left(t-a\right)}{b-a}-\left(t-s\right)\right]{\left(s-a\right)}^{\alpha -3}.$

Let

$g\left(t\right)=\frac{\left(b-s\right)\left(t-a\right)}{b-a}-\left(t-s\right)=\left(\frac{b-s}{b-a}-1\right)t-\left(\frac{b-s}{b-a}\right)a+s,$

then, we have

$g^{\prime }\left(t\right)=\frac{b-s}{b-a}-1<0.$

Hence,

$g\left(t\right)\ge g\left(b\right)=0.$

That we obtain $h_2\left(t\mathrm{,}s\right)$ is an increasing function in t. Consequently, $H\left(t\mathrm{,}s\right)$ gets the maximum at $t=b$ . We have

$h_1\left(b,s\right)=\frac{{\left(b-a\right)}^2\left(b-s\right){\left(s-a\right)}^{\alpha -3}}{2\left(b-a\right)}=\frac{\left(b-a\right)\left(b-s\right){\left(s-a\right)}^{\alpha -3}}{2}$

and

$h_2\left(b,s\right)=\left[\frac{\left(b-s\right){\left(b-a\right)}^2}{2\left(b-a\right)}-\frac{{\left(b-s\right)}^2}{2}\right]{\left(s-a\right)}^{\alpha -3}=\frac{\left(b-s\right){\left(s-a\right)}^{\alpha -2}}{2}.$

Hence, $h_1>h_2$ , we obtain

$\underset{t\in \left[a,b\right]}{\max }H\left(t,s\right)=H\left(b,s\right)=h_1\left(b,s\right)=\frac{\left(b-a\right)\left(b-s\right){\left(s-a\right)}^{\alpha -3}}{2}.$

Furthermore, we have

$H\left(b,s\right)=\frac{\left(b-a\right)\left(b-s\right){\left(s-a\right)}^{\alpha -3}}{2}\le \frac{{\left(b-a\right)}^2{\left(s-a\right)}^{\alpha -3}}{2}.$

Hence,

$\underset{s\in \left[a\mathrm{,}b\right]}{max}H\left(b\mathrm{,}s\right)\le \frac{\left(b-a\right){\left(s-a\right)}^{\alpha -3}}{2}\mathrm{.}$

The proof is completed.

Corollary 3.2. If (1.8) has a nontrivial continuous solution, then

${\displaystyle {\int }_a^b}{\left(s-a\right)}^{\alpha -3}\left|p\left(s\right)\right|\text{d}s\ge \frac{2}{{\left(b-a\right)}^2}\mathrm{.}$ (3.17)

Proof. Let $f\in C\left[a\mathrm{,}b\right]$ be a nontrivial solution of the BVP (1.8), where the norm

$\Vert f\Vert =\underset{t\in \left[a\mathrm{,}b\right]}{sup}\left\{\left|f\left(t\right)\right|\right\}\mathrm{.}$

Form (3.1), we have

$\begin{array}{c}\left|f\left(t\right)\right|\le {\displaystyle {\int }_a^b}\left|H\left(t\mathrm{,}s\right)\right|\left|p\left(s\right)\right|\left|f\left(s\right)\right|\text{d}s\\ \le {\displaystyle {\int }_a^b}\underset{s\in \left[a\mathrm{,}b\right]}{max}H\left(b\mathrm{,}s\right)\left|p\left(s\right)\right|\left|f\left(s\right)\right|\text{d}s\\ \le {\displaystyle {\int }_a^b}\frac{{\left(b-a\right)}^2{\left(s-a\right)}^{\alpha -3}}{2}\left|p\left(s\right)\right|\left|f\left(s\right)\right|\text{d}s\mathrm{.}\end{array}$ (3.18)

Taking the norm leads to

$\Vert f\Vert \le \left({\displaystyle {\int }_a^b}\frac{{\left(b-a\right)}^2{\left(s-a\right)}^{\alpha -3}}{2}\left|p\left(s\right)\right|\text{d}s\right)\Vert f\Vert \mathrm{.}$

Then,

${\displaystyle {\int }_a^b}\frac{{\left(b-a\right)}^2{\left(s-a\right)}^{\alpha -3}}{2}\left|p\left(s\right)\right|\text{d}s\ge 1.$

Hence, we get the inequality in (3.17). This completes the proof.

Example 3.1. If the BVP

$\{\begin{array}{l}{T}_{\alpha }^{a}f\left(t\right)+\lambda f\left(t\right)=0,0<t<1,2<\alpha \le 3,\\ f\left(0\right)=f^{\prime }\left(0\right)=f^{\prime }\left(1\right)=0\end{array}$

has a nontrivial continuous solution, then

$\left|\lambda \right|\ge 2\left(\alpha -2\right)\mathrm{.}$ (3.19)

Proof. Assume that $\lambda$ is an eigenvalue of (1.8). By using Corollary 3.2, we have

${\displaystyle {\int }_0^1}{s}^{\alpha -3}\left|\lambda \right|\text{d}s\ge 2.$

Then, we obtain

$\left|\lambda \right|{\displaystyle {\int }_0^1}{s}^{\alpha -3}\text{d}s=\left|\lambda \right|\frac{1}{\alpha -2}\ge 2.$

We get the desired result (3.19). The proof is complete.

4. Conclusion

On the base of [10] , by changing and increasing the edge value conditions, we establish some new Lyapunov-type inequalities for conformable BVP with the conformable derivative of order $1<\alpha \le 2$ and $2<\alpha \le 3$ . In Section 2 and Section 3, by Green’s function and its corresponding maximum value, we obtain new results about Lyapunov-type inequalities for conformable BVP.

Funding

This research is supported by National Science Foundation of China (11671227) and Academic dissertation research innovation funding fund.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

Cite this paper

Wang, X. and Xu, R. (2018) Lyapunov-Type Inequalities for Conformable BVP. Journal of Applied Mathematics and Physics, 6, 1549-1557. https://doi.org/10.4236/jamp.2018.67131

References