# Hartman-Type and Lyapunov-Type Inequalities for a Fractional Differential Equation with Fractional Boundary Conditions

Authors: Imed Bachar; Hassan Eltayeb
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

We prove Hartman-type and Lyapunov-type inequalities for a class of Riemann–Liouville fractional boundary value problems with fractional boundary conditions. Some applications including a lower bound for the corresponding eigenvalue problem are obtained.


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## Body

## 1. Introduction

In [1], Lyapunov established the following striking inequality:Theorem 1.
Letq∈Ca,b,ℝ. Assume that the problem(1)ω″+qxω=0,x∈a,b,ωa=ωb=0,has a solution ω∈Ca,b,ℝ such that ωx≠0 for x∈a,b. Then,(2)b−a∫abqzdz>4,and constant 4 is the best possible largest number.It has been shown that this result serves as a good tool in the study of several properties of solutions of differential equations (such as eigenvalue problems and eigenvalue inequalities) (see, for example, [2–5] and the references therein). Many authors have worked on generalizations of classical inequalities (see, for instance, [4–16] and the references therein).In [11], the authors use the Hahn integral operator to prove a description of new generalization of Minkowski’s inequality.In [5], the authors improve inequality in (2) by proving the following Hartman–Winter inequality:(3)∫abb−zz−aq+zdz>b−a,where q+z=maxqz,0 is the nonnegative part of qz.Inequality (3) is also known as the best Lyapunov inequality.In [17], Ferreira considered the following fractional differential problem:(4)Da+αω+qxω=0,a<x<b,1<α≤2,ωa=ωb=0,where q∈Ca,b,ℝ and Da+α denotes the Riemann–Liouville fractional derivative of order α (see Definition 2 in the following).The author established the following Lyapunov-type inequality for problem (4).Theorem 2. (see [17]).
Assume that problem (4) has a solution ω∈Ca,b,ℝ such that ωx≠0 for x∈a,b. Then,(5)1Γα∫abqzdz>4b−aα−1.Remark 1.
Note that if we letα=2 in (5), one obtains Lyapunov’s classical inequality (2).
For the convenience of the reader, we recall the concept of fractional integral and derivative of orderγ≥0.Definition 1.
(see [18, 19]). The Riemann–Liouville fractional integral of order γ≥0 for a real-valued function ω is defined by Ia+0ωx=ωx and(6)Ia+γωx≔1Γγ∫axx−zγ−1ωzdz,γ>0,x∈a,b,where Γγ is the Euler gamma function.Definition 2.
(see [18, 19]). The Riemann–Liouville fractional derivative of order γ≥0 for function ω is defined by Da+0ωx=ωx and(7)Da+γωx≔ddxnIa+n−γωx,forγ>0,where n=γ+1 with γ the integer part of γ.
The new development in fractional calculus has attracted the attention of researchers of various disciplines. Different mathematical procedures have been considered by several authors through different research-oriented aspects of fractional differential equations (see, for instance, [20–22] and the references therein).
Our goal in this paper is to establish Hartman-type and Lyapunov-type inequalities for the following problem:(8)Da+αω+qxω=0,x∈a,b,ωa=Da+α−3ωa=Da+α−2ωa=ω″b=0,where α∈3,4 and q∈Ca,b,ℝ. Some applications are given to illustrate our result.
The organization of the paper is as follows. In Section2, we derive the explicit expression of the Green function corresponding to problem (8) and we establish some properties on it. This allows us to prove Hartman-type and Lyapunov-type inequalities for problem (8). In Section 3, we present some applications including a lower bound for the corresponding eigenvalue problem.


## 2. Main Results

### 2.1. Green’s Function

First, we recall the following well-known properties (see, for example, [18, 19]).Lemma 1.
Letα∈3,4 and ω∈Ca,b,ℝ∩L1a,b. Then,(i)
For0<γ<α, Da+γIa+αω=Ia+α−γω and Da+αIa+αω=ω(ii)
Da+αωx=0 if and only if ωx=∑i=14cix−aα−i, where ci∈ℝ, for i∈1,2,3,4(iii)
Assume thatDa+αω∈Ca,b,ℝ∩L1a,b; then,(9)Ia+αDa+αωx=ωx+∑i=14cix−aα−i,where ci∈ℝ, for i∈1,2,3,4.Lemma 2.
Letω∈Ca,b be a solution of problem (8). Then,(10)ωx=∫abGαx,yqyωydy,where Gαx,y is Green’s function of problem (8) given by(11)Gαx,y=1Γαb−yb−aα−3x−aα−1−x−yα−1,a≤y≤x≤b,b−yb−aα−3x−aα−1,a≤x≤y≤b.Proof.
Letω be such solution. By Lemma 1, we have(12)ωx=∑i=14cix−aα−i−Ia+αqωx.
Using the fact thatωa=Da+α−3ωa=Da+α−2ωa=ω″b=0, we obtain c2=c3=c4=0 and b−aα−3Γαc1=∫abb−yα−3qyωydy.
Therefore,(13)ωx=1Γα∫abb−yb−aα−3x−aα−1qyωydy−1Γα∫axx−yα−1qyωydy=∫abGαx,yqyωydy.
This ends the proof.To get a quick perspective, in Figure1, we have the representation of Green’s function G7/2x,y with the contours and some projections.One can see from Figure1 that Green’s function G7/2x,y≥0 and it is nondecreasing with respect to the first variable. This important observation will be proved for Gαx,y with α∈3,4.Figure 1
Gαx,y for α=7/2. (a) Gα (x, y) and contours. (b) Projection on xz. (c) Projection on yz.
(a)(b)(c)Definition 3.
Letf,g:a,b×a,b⟶ℝ with f,g≥0. We say that(14)fx,y≈gx,yona,b×a,b,if there exists c>0 such that 1/cgx,y≤fx,y≤cgx,y for all x,y∈a,b×a,b.Remark 2.
Letτ>0 and x,z∈0,1. Then,(15)min1,τ1−zx≤1−zxτ≤max1,τ1−zx.
Next, we establish some properties on Green’s functionGαx,y given by (11).Proposition 1.
(i)
Ona,b×a,b,(16)Gαx,y≈x−aα−2b−yα−3minx−a,y−a.(ii)
Ona,b×a,b,(17)∂∂xGαx,y≈x−aα−3b−yα−3minx−a,y−a.(iii)
The functionGα satisfies the following property:(18)0≤Gαx,y≤Gαb,y,x,y∈a,b×a,b.Proof.
(i)
From Lemma2, for x,y∈a,b, we have(19)Gαx,y=1Γαb−yb−aα−3x−aα−11−b−yb−a2b−ax−y+x−ab−yα−1,wherex−y+=maxx−y,0.Now, sinceb−ax−y+/x−ab−y∈0,1, for x,y∈a,b, then by using Remark 2, with τ=α−1 and z=b−y2/b−a2∈0,1, we obtain(20)Gαx,y≈b−yα−3x−aα−2b−ax−a−b−yx−y+.Hence, inequalities in (16) follow by observing that(21)b−ax−a−b−yx−y+≈minx−a,y−a.(ii)
We have(22)∂∂xGαx,y=α−1Γαb−yb−aα−3x−aα−2−x−yα−2,a≤y≤x≤b,b−yb−aα−3x−aα−2,a≤x≤y≤b.Similar to casei, by using the fact that(23)∂∂xGαx,y=α−1Γαb−yb−aα−3x−aα−21−b−yb−ab−ax−y+x−ab−yα−2and applying Remark2 with τ=α−2 and z=b−y/b−a∈0,1, we obtain the required result.(iii)
Lety∈a,b. Since the function x⟶∂/∂xGαx,y is nondecreasing on a,b, we deduce that(24)0=Gαa,y≤Gαx,y≤Gαb,y.
This completes the proof.


### 2.2. Statements and Proofs of Main Results

Theorem 3. (Hartman–Winter-type inequality).
Letq∈Ca,b,ℝ. Assume that problem (8) has a solution ω∈Ca,b,ℝ such that ωx≠0 for x∈a,b. Then,(25)∫abb−zα−3z−a2b−a−zq+zdz≥Γα,where q+z=maxqz,0.Proof.
From Lemma2, we know that(26)ωx=∫abGαx,zqzωzdz,x∈a,b.
Without loss of generality, we may assume thatωx>0 for x∈a,b.
Using (26), Proposition 1 (iii) and the fact that qz≤q+z, we deduce that(27)ωx≤∫abGαx,zq+zωzdz≤∫abGαb,zq+zωzdz.
Hence,(28)ω≤∫abGαb,zq+zωdz,or equivalently,(29)1≤∫abGαb,zq+zdz.
Therefore,(30)1≤1Γα∫abb−zα−3z−a2b−a−zq+zdz,from which inequality (25) follows.Remark 3.
Letq∈Ca,b,ℝ. Under the same conditions as in Theorem 3, we have(31)∫abb−zα−3z−a2b−a−zqzdz≥Γα.
By applying the previous theorem withα=4, we obtain the following:Corollary 1.
Letq∈Ca,b,ℝ. Assume that the problems(32)ω4+qxω=0,x∈a,b,ωa=ω′a=ω″a=ω″b=0,admit a solution ω∈Ca,b,ℝ such that ωx≠0 for x∈a,b. Then,(33)∫abz−ab−z2b−a−zq+zdz≥6.
In particular,(34)∫abz−ab−zq+zdz≥3b−a.Corollary 2.
(Lyapunov−type inequality)
Under the same conditions as in Theorem3, we have(35)∫abq+zdz≥Γαα−1α−22b−aα−1α−1α−3α−3.Proof.
By Theorem3, we have(36)∫abfzq+zdz≥Γα.where fz≔b−zα−3z−a2b−a−z≥0.
Forz∈a,b, we have(37)f′z=b−zα−42b−z2−α−3z−a2b−a−z.
Note that(38)f′z=0ona,bif and only ifz=z∗≔1α−1α−1b−b−aα−1α−3.
Furthermore,f′z>0 on a,z∗ and f′z<0 on z∗,b.
Hence,(39)supz∈a,bfz=fz∗=2b−aα−1α−1α−2α−1α−3α−3.
So Lyapunov-type inequality (35) follows from (36) and (39).Corollary 3.
Letq∈Ca,b,ℝ. Assume that the problems(40)ω4+qxω=0,x∈a,b,ωa=ω′a=ω″a=ω″b=0,admit a solution ω∈Ca,b,ℝ such that ωx≠0 for x∈a,b. Then,(41)∫abq+zdz≥93b−a3.Proof.
Inequality (41) follows from (39) with α=4.


## 2.1. Green’s Function

First, we recall the following well-known properties (see, for example, [18, 19]).Lemma 1.
Letα∈3,4 and ω∈Ca,b,ℝ∩L1a,b. Then,(i)
For0<γ<α, Da+γIa+αω=Ia+α−γω and Da+αIa+αω=ω(ii)
Da+αωx=0 if and only if ωx=∑i=14cix−aα−i, where ci∈ℝ, for i∈1,2,3,4(iii)
Assume thatDa+αω∈Ca,b,ℝ∩L1a,b; then,(9)Ia+αDa+αωx=ωx+∑i=14cix−aα−i,where ci∈ℝ, for i∈1,2,3,4.Lemma 2.
Letω∈Ca,b be a solution of problem (8). Then,(10)ωx=∫abGαx,yqyωydy,where Gαx,y is Green’s function of problem (8) given by(11)Gαx,y=1Γαb−yb−aα−3x−aα−1−x−yα−1,a≤y≤x≤b,b−yb−aα−3x−aα−1,a≤x≤y≤b.Proof.
Letω be such solution. By Lemma 1, we have(12)ωx=∑i=14cix−aα−i−Ia+αqωx.
Using the fact thatωa=Da+α−3ωa=Da+α−2ωa=ω″b=0, we obtain c2=c3=c4=0 and b−aα−3Γαc1=∫abb−yα−3qyωydy.
Therefore,(13)ωx=1Γα∫abb−yb−aα−3x−aα−1qyωydy−1Γα∫axx−yα−1qyωydy=∫abGαx,yqyωydy.
This ends the proof.To get a quick perspective, in Figure1, we have the representation of Green’s function G7/2x,y with the contours and some projections.One can see from Figure1 that Green’s function G7/2x,y≥0 and it is nondecreasing with respect to the first variable. This important observation will be proved for Gαx,y with α∈3,4.Figure 1
Gαx,y for α=7/2. (a) Gα (x, y) and contours. (b) Projection on xz. (c) Projection on yz.
(a)(b)(c)Definition 3.
Letf,g:a,b×a,b⟶ℝ with f,g≥0. We say that(14)fx,y≈gx,yona,b×a,b,if there exists c>0 such that 1/cgx,y≤fx,y≤cgx,y for all x,y∈a,b×a,b.Remark 2.
Letτ>0 and x,z∈0,1. Then,(15)min1,τ1−zx≤1−zxτ≤max1,τ1−zx.
Next, we establish some properties on Green’s functionGαx,y given by (11).Proposition 1.
(i)
Ona,b×a,b,(16)Gαx,y≈x−aα−2b−yα−3minx−a,y−a.(ii)
Ona,b×a,b,(17)∂∂xGαx,y≈x−aα−3b−yα−3minx−a,y−a.(iii)
The functionGα satisfies the following property:(18)0≤Gαx,y≤Gαb,y,x,y∈a,b×a,b.Proof.
(i)
From Lemma2, for x,y∈a,b, we have(19)Gαx,y=1Γαb−yb−aα−3x−aα−11−b−yb−a2b−ax−y+x−ab−yα−1,wherex−y+=maxx−y,0.Now, sinceb−ax−y+/x−ab−y∈0,1, for x,y∈a,b, then by using Remark 2, with τ=α−1 and z=b−y2/b−a2∈0,1, we obtain(20)Gαx,y≈b−yα−3x−aα−2b−ax−a−b−yx−y+.Hence, inequalities in (16) follow by observing that(21)b−ax−a−b−yx−y+≈minx−a,y−a.(ii)
We have(22)∂∂xGαx,y=α−1Γαb−yb−aα−3x−aα−2−x−yα−2,a≤y≤x≤b,b−yb−aα−3x−aα−2,a≤x≤y≤b.Similar to casei, by using the fact that(23)∂∂xGαx,y=α−1Γαb−yb−aα−3x−aα−21−b−yb−ab−ax−y+x−ab−yα−2and applying Remark2 with τ=α−2 and z=b−y/b−a∈0,1, we obtain the required result.(iii)
Lety∈a,b. Since the function x⟶∂/∂xGαx,y is nondecreasing on a,b, we deduce that(24)0=Gαa,y≤Gαx,y≤Gαb,y.
This completes the proof.


## 2.2. Statements and Proofs of Main Results

Theorem 3. (Hartman–Winter-type inequality).
Letq∈Ca,b,ℝ. Assume that problem (8) has a solution ω∈Ca,b,ℝ such that ωx≠0 for x∈a,b. Then,(25)∫abb−zα−3z−a2b−a−zq+zdz≥Γα,where q+z=maxqz,0.Proof.
From Lemma2, we know that(26)ωx=∫abGαx,zqzωzdz,x∈a,b.
Without loss of generality, we may assume thatωx>0 for x∈a,b.
Using (26), Proposition 1 (iii) and the fact that qz≤q+z, we deduce that(27)ωx≤∫abGαx,zq+zωzdz≤∫abGαb,zq+zωzdz.
Hence,(28)ω≤∫abGαb,zq+zωdz,or equivalently,(29)1≤∫abGαb,zq+zdz.
Therefore,(30)1≤1Γα∫abb−zα−3z−a2b−a−zq+zdz,from which inequality (25) follows.Remark 3.
Letq∈Ca,b,ℝ. Under the same conditions as in Theorem 3, we have(31)∫abb−zα−3z−a2b−a−zqzdz≥Γα.
By applying the previous theorem withα=4, we obtain the following:Corollary 1.
Letq∈Ca,b,ℝ. Assume that the problems(32)ω4+qxω=0,x∈a,b,ωa=ω′a=ω″a=ω″b=0,admit a solution ω∈Ca,b,ℝ such that ωx≠0 for x∈a,b. Then,(33)∫abz−ab−z2b−a−zq+zdz≥6.
In particular,(34)∫abz−ab−zq+zdz≥3b−a.Corollary 2.
(Lyapunov−type inequality)
Under the same conditions as in Theorem3, we have(35)∫abq+zdz≥Γαα−1α−22b−aα−1α−1α−3α−3.Proof.
By Theorem3, we have(36)∫abfzq+zdz≥Γα.where fz≔b−zα−3z−a2b−a−z≥0.
Forz∈a,b, we have(37)f′z=b−zα−42b−z2−α−3z−a2b−a−z.
Note that(38)f′z=0ona,bif and only ifz=z∗≔1α−1α−1b−b−aα−1α−3.
Furthermore,f′z>0 on a,z∗ and f′z<0 on z∗,b.
Hence,(39)supz∈a,bfz=fz∗=2b−aα−1α−1α−2α−1α−3α−3.
So Lyapunov-type inequality (35) follows from (36) and (39).Corollary 3.
Letq∈Ca,b,ℝ. Assume that the problems(40)ω4+qxω=0,x∈a,b,ωa=ω′a=ω″a=ω″b=0,admit a solution ω∈Ca,b,ℝ such that ωx≠0 for x∈a,b. Then,(41)∫abq+zdz≥93b−a3.Proof.
Inequality (41) follows from (39) with α=4.


## 3. Applications

### 3.1. Lower Bound for the Eigenvalues

Consider the following eigenvalue problem:(42)D0+αωx+λωx=0,x∈0,1,3<α≤4,ω0=D0+α−3ω0=D0+α−2ω0=ω″1=0.Theorem 4.
Assume that eigenvalue problem (42) has a solution ω∈Ca,b,ℝ such that ωx≠0 for x∈a,b. Then,(43)λ≥α−2Γα+12.Proof.
By Remark3 (with a=0 and b=1), we have(44)λ∫011−zα−3z2−zdz≥Γα,from which inequality (43) follows by observing that(45)∫011−zα−3z2−zdz=2αα−2.


### 3.2. Nonexistence Results

Consider the following problem:(46)D0+αω+qxω=0,x∈0,1,3<α≤4,ω0=D0+α−3ω0=D0+α−2ω0=ω″1=0,where q∈C0,1,ℝ. Then, we have the following result.Theorem 5.
Assume that(47)∫01q+zdz<Γαα−1α−22α−1α−3α−3.
Then, problem (46) has no nontrivial solution.Proof.
The assertion follows from Corollary2.


## 3.1. Lower Bound for the Eigenvalues

Consider the following eigenvalue problem:(42)D0+αωx+λωx=0,x∈0,1,3<α≤4,ω0=D0+α−3ω0=D0+α−2ω0=ω″1=0.Theorem 4.
Assume that eigenvalue problem (42) has a solution ω∈Ca,b,ℝ such that ωx≠0 for x∈a,b. Then,(43)λ≥α−2Γα+12.Proof.
By Remark3 (with a=0 and b=1), we have(44)λ∫011−zα−3z2−zdz≥Γα,from which inequality (43) follows by observing that(45)∫011−zα−3z2−zdz=2αα−2.


## 3.2. Nonexistence Results

Consider the following problem:(46)D0+αω+qxω=0,x∈0,1,3<α≤4,ω0=D0+α−3ω0=D0+α−2ω0=ω″1=0,where q∈C0,1,ℝ. Then, we have the following result.Theorem 5.
Assume that(47)∫01q+zdz<Γαα−1α−22α−1α−3α−3.
Then, problem (46) has no nontrivial solution.Proof.
The assertion follows from Corollary2.



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