Note on a Lyapunov-type inequality for a fractional boundary value problem with Caputo-Fabrizio derivative

TL;DR

This paper introduces a corrected Lyapunov-type inequality for a fractional boundary value problem involving the Caputo-Fabrizio derivative, addressing inaccuracies in previous results.

Contribution

It provides a revised Lyapunov-type inequality for fractional boundary value problems with Caputo-Fabrizio derivatives, correcting prior inaccuracies.

Findings

Corrected Lyapunov-type inequality established

Addresses errors in previous literature

Enhances understanding of fractional boundary value problems

Abstract

In this short note, we present a Lyapunov-type inequality that corrects the recently obtained result in [M. Kirane, B. T. Torebek: A Lyapunov-type inequality for a fractional boundary value problem with Caputo-Fabrizio derivative, J. Math. Inequal. 12, 4(2018), 1005-1012].

Equations

$$\left\{\begin{array}[]{cc}{}^{CF}D_{a}^{\alpha}u(t)+q(t)u(t)=0,&0\leq a<t<b,\\ u(a)=u(b)=0,&\end{array}\right.$$

$${\int }_a^b|q(t)|ds>\frac{4(\alpha -1)(b-a)}{[(\alpha -1)(b-a)-2+\alpha {]}^2}.$$

$$u(s)={\int }_a^{b}G(t,s)q(s)u(s)ds,$$

$$G(t,s)=\left\{\begin{array}[]{lll}g_{1}(t,s)=\frac{b-t}{b-a}[(\alpha-1)(s-a)-2+\alpha],&&a\leq s\leq t\leq b,\\ &&\\ g_{2}(t,s)=\frac{t-a}{b-a}[(\alpha-1)(b-s)+2-\alpha],&&a\leq t\leq s\leq b.\end{array}\right.$$

$$t=s=\frac{1}{2}\left(b+a+\frac{2-\alpha }{\alpha -1}\right):=s^{\ast },$$

$$\displaystyle\begin{array}[]{lll}b-a<\frac{2-\alpha}{\alpha-1}&\Longleftrightarrow&2b<b+a+\frac{2-\alpha}{\alpha-1}\end{array}$$

$$\displaystyle\begin{array}[]{lll}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }&\Longleftrightarrow&b<\frac{1}{2}\left(b+a+\frac{2-\alpha}{\alpha-1}\right)\end{array}$$

$$\displaystyle\begin{array}[]{lll}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }&\Longleftrightarrow&b<s^{\ast},\end{array}$$

$$\displaystyle\begin{array}[]{ccc}b-a<\frac{2-\alpha}{\alpha-1}&\Longleftrightarrow&b<a+\frac{2-\alpha}{\alpha-1}\end{array}$$

$$\displaystyle\begin{array}[]{ccc}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }&\Longrightarrow&s<a+\frac{2-\alpha}{\alpha-1}\end{array}$$

$$\displaystyle\begin{array}[]{ccc}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }&\Longrightarrow&(\alpha-1)(s-a)-2+\alpha<0,\end{array}$$

$$\frac{b-t}{b-a}\le \frac{b-s}{b-a}.$$

$$g_1(t,s)\ge \frac{b-s}{b-a}[(\alpha -1)(s-a)-2+\alpha ],a\le s\le b<a+\frac{2-\alpha }{\alpha -1}.$$

$$h_1(s)\le g_1(t,s)\le 0,\text{ for }a\le s\le t\le a+\frac{2-\alpha }{\alpha -1},$$

$$h_1(s)=g_1(s,s)=\frac{b-s}{b-a}[(\alpha -1)(s-a)-2+\alpha ],s\in [a,b]$$

$$h_1^{\prime }(s)=-\frac{2(\alpha -1)}{b-a}s+\frac{(\alpha -1)\left(b+a\right)+2-\alpha }{b-a}.$$

$$\underset{a\le s\le t\le b<a+\frac{2-\alpha }{\alpha -1}}{\max }\left|g_1(t,s)\right|$$

$$0\le g_2(t,s)\le \frac{s-a}{b-a}[(\alpha -1)(b-s)+2-\alpha ].$$

$$h_2(s)=g_2(s,s)=\frac{s-a}{b-a}[(\alpha -1)(b-s)+2-\alpha ],a\le s\le b.$$

$$h_2^{\prime }(s)=-\frac{2(\alpha -1)}{b-a}s+\frac{(\alpha -1)\left(a+b\right)+\left(2-\alpha \right)}{b-a},$$

$$\underset{a\le t\le s\le b<a+\frac{2-\alpha }{\alpha -1}}{\max }\left|g_2(t,s)\right|$$

$$\underset{a\le t,s\le b<a+\frac{2-\alpha }{\alpha -1}}{\max }\left|G(t,s)\right|=2-\alpha .$$

$$\left\{\begin{array}[]{l}0\leq g_{1}(t,s)\leq h_{1}(s),\text{ \ }a+\frac{2-\alpha}{\alpha-1}\leq s\leq b.\\ \\ h_{1}(s)\leq g_{1}(t,s)\leq 0,\text{ \ }a\leq s\leq a+\frac{2-\alpha}{\alpha-1}<s^{\ast}.\end{array}\right.$$

$$\underset{a\le s\le t\le a+\frac{2-\alpha }{\alpha -1}\le b}{\max }\left|g_1(t,s)\right|$$

$$\underset{a\le t\le s\le a+\frac{2-\alpha }{\alpha -1}\le b}{\max }\left|g_2(t,s)\right|$$

$$\underset{a\le t,s\le a+\frac{2-\alpha }{\alpha -1}\le b}{\max }\left|G(t,s)\right|$$

$$\underset{a\le t,s\le a+\frac{2-\alpha }{\alpha -1}\le b}{\max }\left|G(t,s)\right|=\frac{[(\alpha -1)(b-a)+\left(2-\alpha \right){]}^2}{4(\alpha -1)(b-a)}.$$

$$\underset{(t,s)\in [a,b]\times [a,b]}{\max }\left|G(t,s)\right|=2-\alpha ,$$

$$\underset{(t,s)\in [a,b]\times [a,b]}{\max }\left|G(t,s)\right|=\frac{[(\alpha -1)(b-a)+\left(2-\alpha \right){]}^2}{4(\alpha -1)(b-a)}.$$

$$\int_{a}^{b}|q(t)|ds\geq\left\{\begin{array}[]{l}\frac{1}{2-\alpha.},\text{ \ if \ }b-a<\frac{2-\alpha}{\alpha-1},\\ \\ \frac{4(\alpha-1)(b-a)}{[(\alpha-1)(b-a)+\left(2-\alpha\right)]^{2}},\text{ \ if \ }b-a\geq\frac{2-\alpha}{\alpha-1}.\end{array}\right.$$

Full text

**Keywords: **Caputo-Fabrizio derivative, Lyapunov-type inequality, boundary value problem.

MSC (2010): 34A08, 34A40, 26A33.

1 Introduction

Recently, in [1] the authors discussed a Lyapunov-type inequality for the following linear fractional boundary value problem:

[TABLE]

where ${}^{CF}D_{a}^{\alpha}$ denotes the Caputo-Fabrizio derivative [2, 3] of order $\alpha,(1<\alpha\leq 2),$ $q:[a,b]\rightarrow\mathbb{R}$ is a continuous function. And they included the following result:

Theorem 1** ([1])**

If the fractional boundary value problem (1) has a nontrivial solution, then

[TABLE]

We have noticed that, the denominator in the inequality (2) is equal zero when $b-a=\frac{2-\alpha}{\alpha-1}.$ So a mistake has been occured during the previous result and some other results (Corrolary 3.4 and Corrolary 3.5 in [1]) are also incorrect. These mistakes come from the main wrong in (Lemma 3.2 in [1] related to the calculations for the maximum value of the Green’s function of the problem (1).

This work aims to show these mistakes and present the correct version of them.

We will also refer the interested reader in studying the Lyapunop-type inequalities for a fractional boundary value problems to the works compiled in chapter in the book [4], as well as some other papers published recently, for example see [5]–[10] and the references cited therein.

2 Main results

The fractional boundary value problem (1) is equivalent to the integral equation

[TABLE]

where $G(t,s)$ is called the Green’s function of the problem (1) and it’s difened by

[TABLE]

See [1] for more details, (note that this function in (Lemma 3.1, [1]) is written in wrong way. Although its proof is true).

The mistake alluded to in ([1], Section 3) is that the authors concluded that the maximum value of the function $G(t,s)$ is obtained at the point

[TABLE]

where $s^{\ast}$ is defined by (equality (3.5) in [1]). However, this is wrong for $(t,s)\in[a,b]\times[a,b]$ with $1<\alpha\leq 2$, as we will show nextly.

Let us start to discuss the previous value of $s^{\ast}$.

Note that, if $b-a<\frac{2-\alpha}{\alpha-1},$ we have

[TABLE]

thus $s^{\ast}\notin[a,b].$ Then the maximum value of the function $G(t,s)$ is not at $s^{\ast}$ when $b-a<$ $\frac{2-\alpha}{\alpha-1}.$

Now, for $a\leq s\leq t\leq b,$ with $b-a<\frac{2-\alpha}{\alpha-1},$ we have

[TABLE]

on other hand, we have

[TABLE]

By the inequalities (14) and (15) we get

[TABLE]

Observe that the inequality (16) is contrary to (the inequality (3.4) in [1]).

Remark 2

Note that on a general interval $[a,b],0\leq a<b,$ we have

[TABLE]

where the function $h_{1}$ is defined by

[TABLE]

We differentiate the function $h_{1}(s)$ to get

[TABLE]

We have $s^{\ast}$ is the unique solution of the equation $h_{1}^{\prime}(s)=0,$ where $s^{\ast}$ is given by (5) but the value of $s^{\ast}$ in some cases does not belong to the interval $[a,b]$ as we have shown previously.

By the discussion above, we can conclude that the maximum value of the function $G(t,s)$ lays in the following two cases:

Case 1. $b-a<$ $\frac{2-\alpha}{\alpha-1}.$

Because $b-a<$ $\frac{2-\alpha}{\alpha-1},$ so then $s\leq b<s^{\ast}$ (here $a+\frac{2-\alpha}{\alpha-1},s^{\ast}\notin[a,b])$, we obtain $h_{1}^{\prime}(s)\geq$ [math] and $h_{1}(s)\leq 0$ for $s\leq b<a+\frac{2-\alpha}{\alpha-1}.$ So by (17) and the contunuity of the function $h_{1}$ we conclude that

[TABLE]

Next, for $a\leq t\leq s\leq b.$ Obviously,

[TABLE]

We define a function $h_{2}$ by

[TABLE]

Differentiating the function $h_{2}(s)$

[TABLE]

which implies that the function $h_{2}^{\prime}(s)$ has a unique zero, at the point $s^{\ast},$ but $s^{\ast}>b$ (i.e. $s^{\ast}\notin[a,b]$). Because $h_{2}^{\prime}(s)>0$ for all $s\in[a,b]$ and $g_{2}(t,s)\geq 0$ for $a\leq t\leq s\leq b,$ and $h_{2}(s)$ is continuous function, then

[TABLE]

By (24) and (20) we get

[TABLE]

Case 2. $b-a\geq$ $\frac{2-\alpha}{\alpha-1}.$

From the inequality $b-a\geq$ $\frac{2-\alpha}{\alpha-1}$ we get $a+\frac{2-\alpha}{\alpha-1}\in[a,b]$ and $s^{\ast}\in[a+\frac{2-\alpha}{\alpha-1},b],$ we obtain

[TABLE]

Because $h_{1}(s)$ is continuous function, and $h_{1}(a+\frac{2-\alpha}{\alpha-1})=h_{1}(b)=0,$ and $h_{1}(a)=-(2-\alpha),$ we conclude that

[TABLE]

On other hand, we have

[TABLE]

By (27) and (28) we get

[TABLE]

using the inequality $\frac{(A+B)^{2}}{4}\geq AB,$ with $A=(\alpha-1)(b-a)$ and $B=\left(2-\alpha\right),$ we obtain

[TABLE]

Thus we conclude the follwing result.

Proposition 3

*The Green’s function $G$ defined by (4), has the following properties:

$i\mathbf{).}$ If $b-a<$ $\frac{2-\alpha}{\alpha-1},$ then*

[TABLE]

$ii).$* If $b-a\geq$ $\frac{2-\alpha}{\alpha-1},$ then*

[TABLE]

Hence we have the following Lyapunov-type inequality.

Theorem 4

If the fractional boundary value problem (1) has a nontrivial solution. Then

[TABLE]

Proof. Since the proof is well-known so that the reader can easily check it on, where it’s used in [1] but in here we get into details in two cases related the properties (30) and (31).

By using Theorem 4, the reader can smoothly correct (Corrolary 3.4 and Corrolary 3.5 in [1]), but should seperate each of them in two cases $b-a<$ $\frac{2-\alpha}{\alpha-1},$ and $b-a\geq$ $\frac{2-\alpha}{\alpha-1}$.