# A Lyapunov-Type Inequality for a Fractional Differential Equation under a Robin Boundary Condition

Authors: Mohamed Jleli; Lakhdar Ragoub; Bessem Samet
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We establish a new Lyapunov-type inequality for a class of fractional differential equations under Robin boundary conditions. The obtained inequality is used to obtain an interval where a linear combination of certain Mittag-Leffler functions has no real zeros.


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## Body

## 1. Introduction and Preliminaries

Let the followingχ:[a,b]→R be a continuous function. The well-known Lyapunov inequality [1] states that a necessary condition for the boundary value problem (1)u′′t+χtut=0,a<t<bua=ub=0to have nontrivial solutions is that(2)∫ab‍χtdt>4b-a·This result found many practical applications in differential and difference equations (oscillation theory, disconjugacy, eigenvalue problems, etc.); see, for instance, [2–7]. On the other hand, many improvements of (2) have been carried out, and similar inequalities have been obtained for other types of differential equations; compare to the Pachpatte monograph [8]. The search for Lyapunov-type inequalities in which the starting differential equation is constructed via fractional differential operators has begun very recently. For example, in [9], a Lyapunov-type inequality was obtained for differential equations depending on the Riemann-Liouville fractional derivative; that is, for the boundary value problem(3)Dαaut+χtut=0,a<t<b,1<α≤2,ua=ub=0,where aDα denotes the Riemann-Liouville fractional derivative of order α. Precisely, the author proved that if (3) has a nontrivial solution, then we have(4)∫ab‍χtdt>Γα4b-aα-1.Clearly, if we let α=2 in (4), one obtains Lyapunov’s classical inequality (2).In [10], a Lyapunov-type inequality was obtained for the Caputo fractional boundary value problem(5)DaCαut+χtut=0,a<t<b,1<α≤2,ua=ub=0,where DaCα is the Caputo fractional derivative of order α. It was proved that if (5) has a nontrivial solution, then we have(6)∫ab‍χtdt>Γαααα-1b-aα-1·Similarly, if we let α=2 in (6), one obtains Lyapunov’s classical inequality (2).Motivated by the above works, we consider in this paper a Caputo fractional differential equation under Robin boundary conditions. More precisely, we consider the boundary value problem(7)DaCαut+χtut=0,a<t<b,1<α≤2,ua-u′a=ub+u′b=0,and we get a corresponding Lyapunov-type inequality. This result is then used to obtain a real interval where a linear combination of certain Mittag-Leffler functions has no (real) zeros.Before presenting the main results, let us start by recalling the concepts of the Riemann-Liouville fractional integral and the Caputo fractional derivative of orderα≥0. For more details, we refer to [11].Definition 1.
Letα≥0 and let f be a real function defined on [a,b]. The Riemann-Liouville fractional integral of order α is defined by (Ia0f)≡f and(8)Iaαft=1Γα∫at‍t-sα-1fsds,α>0,t∈a,b.Definition 2.
The Caputo fractional derivative of orderα≥0 is defined by (DaC0f)≡f and (DaCαf)(t)=(Iam-αDmf)(t) for α>0, where m is the smallest integer greater than or equal to α.The following result is standard within the fractional calculus theory involving the Caputo differential operator (see [7]).Lemma 3.
One has thatu∈C[a,b] is a solution to (7) if and only if (9)ut=c0+c1t-a-1Γα∫at‍t-sα-1χsusds,where c0 and c1 are some real constants.Now, we are ready to state and prove our main results.


## 2. Main Results

### 2.1. A Lyapunov-Type Inequality

At first, we consider the following notations:(10)γ=b-a+2,h1t,s=1+t-a-γt-sα-1b-sα-1+α-1b-sα-2,h2t,s=1+t-a.Now, let us write problem (7) in its equivalent integral form.Lemma 4.
One has thatu∈C[a,b] is a solution to (7) if and only if u satisfies the integral equation (11)ut=∫ab‍Gt,sχsusds,where(12)Gt,s=b-sα-2b-s+α-1γΓαHt,s,(13)Ht,s=h1t,s,a≤s≤t≤b,h2t,s,a≤t≤s≤b.Proof.
From Lemma3, we have(14)ut=c0+c1t-a-1Γα∫at‍t-sα-1χsusds,where c0 and c1 are some real constants. Then, (15)u′t=c1-1Γα∫at‍α-1t-sα-2χsusds.Since u(a)-u′a=0, we obtain(16)c0=c1.From the boundary condition u(b)+u′b=0, we get(17)c1=1γΓα∫ab‍α-1b-sα-2+b-sα-1χsusds.Using (14), (16), and (17), we obtain the desired result.Lemma 5.
For all(t,s)∈[a,b]×(a,b), one has (18)Ht,s≤max1+b-a,2-αα-1b-a-1·Proof.
It is easy to see that, fora≤t≤s≤b, we have (19)0≤h2t,s≤1+b-a.On the other hand, for a≤s<t≤b, we have(20)∂h1∂tt,s=1-α-1γt-sα-2b-sα-1+α-1b-sα-2·Hence,(21)limt→s+∂h1∂tt,s∈-∞,0.Now, for fixed s in (a,b), we want to study the variation of the function t↦h1(t,s) for t in [s,b]. First, we have (22)∂h1∂tt,st=b=1-α-1γα-1+b-s·Let (23)a∗=b+1-γα-1.We distinguish two eventual cases according to the value of a∗.Case  1. If a∗≤a, in this case, we have(24)∂h1∂tt,st=b≤0,s∈a,b.From (20), (21), and (24), we deduce (25)∂h1∂tt,s≤0,s<t.This yields(26)h1b,s=1+b-a-γb-sb-s+α-1≤h1t,s≤h1s,s≤1+b-a.Observe that, in this case, we have(27)2-αα-1≤1b-a·Using the above inequality, we obtain(28)1+b-ab-s+α-11+b-a-γb-s=s-b+α-11+b-a≥b-aα-2+α-1≥0,which implies that(29)h1b,s≥0.From (26) and (29), we deduce (30)0≤h1t,s≤1+b-a.Case  2. If a<a∗≤b, in this case, we have two possibilities.(i) Ifa∗≤s<b, in this case, we have (31)∂h1∂tt,st=b≤0.Therefore, we conclude that(32)h1b,s≤h1t,s≤h1s,s≤1+b-a.On the other hand, we have(33)1+b-ab-s+α-1-γb-s=1+b-a-γb-s+α-11+b-a=s-b+α-11+b-a≥2-γ+b-aα-1≥0,which implies that(34)h1b,s≥0.From (32) and (34), we deduce(35)0≤h1t,s≤1+b-a.(ii) Ifa<s<a∗, in this case, we have (36)∂h1∂tt,st=b>0.Hence, there would existt∗∈(s,b) such that(37)∂h1∂tt,st=t∗=0.As mentioned above, it is easy to verify thath1(s,s)≥0 and h1(b,s)≤0. This yields (38)h1t∗,s≤h1b,s≤h1s,s≤rp+b-a.Then,(39)h1t,s≤max-h1t∗,s,1+b-a.Observe that∂h1/∂tt,st=t∗=0 is equivalent to (40)α-1γt∗-sα-2=b-sα-1+α-1b-sα-2.Therefore, we get(41)h1t∗,s=α-2α-1t∗+sα-1+1-a≥α-2α-1b+aα-1+1-a=b-aα-2α-1+1.Finally, using the above inequality and (39), we obtain(42)h1t,s≤max1+b-a,2-αα-1b-a-1,which makes end to the proof.Our first main result is as follows.Theorem 6.
If (7) admits a nontrivial continuous solution, then(43)∫ab‍b-sα-2b-s+α-1χsds≥γΓαmaxb-a+1,2-α/α-1b-a-1·Proof.
LetB=C[a,b] be the Banach space endowed with norm (44)x∞=maxa≤t≤bxt,x∈B.From Lemma 4, for all t∈[a,b], we have (45)ut=1γΓα×∫ab‍b-sα-2b-s+α-1Ht,sχsusds.Now, an application of Lemma 3 yields (46)u∞≤u∞max⁡b-a+1,2-α/α-1b-a-1γΓα×∫ab‍b-sα-2b-s+α-1χsds,from which the above theorem follows.


### 2.2. Nonexistence Result of Real Zeros for a Linear Combination of Certain Mittag-Leffler Functions

Letα,β>0 be fixed. The complex function(47)Eα,βz=∑k=0∞‍zkΓkα+β,α>0,β>0,z∈Cis analytic in the whole complex plane; it will be referred to [12–14] as the Mittag-Leffler function with parameters (α,β).At this stage, using the above Lyapunov-type inequality, we give an interval where a linear combination of Mittag-Leffler functions has no real zeros. More precisely, we proveTheorem 7.
Let1<α≤2. Then Eα,2z+Eα,1z+zEα,αz has no real zeros for (48)z∈-3αΓα1+αmax2,2-α/α-1-1,0.Proof.
Let(a,b)=(0,1), and consider the fractional Sturm-Liouville eigenvalue problem(49)D0Cαut+λut=0,0<t<1,ua-u′a=ub+u′b=0.
The real values ofλ, for which there exists a non-trivial solution to (49), are called eigenvalues of (49); and the corresponding solutions are called eigenfunctions.
As established in [15], the eigenvalues of (49) must be positive; moreover, the positive number λ is an eigenvalue of (49) if and only if (50)Eα,2-λ+Eα,1-λ-λEα,α-λ=0.Thanks to Theorem 6, if a positive real eigenvalue λ of (49) exists, then (51)λ∫01‍1-sα-2α-sds≥3Γαmax2,2-α/α-1-1·Hence,(52)λα+1α≥3Γαmax2,2-α/α-1-1,which concludes the proof.


## 2.1. A Lyapunov-Type Inequality

At first, we consider the following notations:(10)γ=b-a+2,h1t,s=1+t-a-γt-sα-1b-sα-1+α-1b-sα-2,h2t,s=1+t-a.Now, let us write problem (7) in its equivalent integral form.Lemma 4.
One has thatu∈C[a,b] is a solution to (7) if and only if u satisfies the integral equation (11)ut=∫ab‍Gt,sχsusds,where(12)Gt,s=b-sα-2b-s+α-1γΓαHt,s,(13)Ht,s=h1t,s,a≤s≤t≤b,h2t,s,a≤t≤s≤b.Proof.
From Lemma3, we have(14)ut=c0+c1t-a-1Γα∫at‍t-sα-1χsusds,where c0 and c1 are some real constants. Then, (15)u′t=c1-1Γα∫at‍α-1t-sα-2χsusds.Since u(a)-u′a=0, we obtain(16)c0=c1.From the boundary condition u(b)+u′b=0, we get(17)c1=1γΓα∫ab‍α-1b-sα-2+b-sα-1χsusds.Using (14), (16), and (17), we obtain the desired result.Lemma 5.
For all(t,s)∈[a,b]×(a,b), one has (18)Ht,s≤max1+b-a,2-αα-1b-a-1·Proof.
It is easy to see that, fora≤t≤s≤b, we have (19)0≤h2t,s≤1+b-a.On the other hand, for a≤s<t≤b, we have(20)∂h1∂tt,s=1-α-1γt-sα-2b-sα-1+α-1b-sα-2·Hence,(21)limt→s+∂h1∂tt,s∈-∞,0.Now, for fixed s in (a,b), we want to study the variation of the function t↦h1(t,s) for t in [s,b]. First, we have (22)∂h1∂tt,st=b=1-α-1γα-1+b-s·Let (23)a∗=b+1-γα-1.We distinguish two eventual cases according to the value of a∗.Case  1. If a∗≤a, in this case, we have(24)∂h1∂tt,st=b≤0,s∈a,b.From (20), (21), and (24), we deduce (25)∂h1∂tt,s≤0,s<t.This yields(26)h1b,s=1+b-a-γb-sb-s+α-1≤h1t,s≤h1s,s≤1+b-a.Observe that, in this case, we have(27)2-αα-1≤1b-a·Using the above inequality, we obtain(28)1+b-ab-s+α-11+b-a-γb-s=s-b+α-11+b-a≥b-aα-2+α-1≥0,which implies that(29)h1b,s≥0.From (26) and (29), we deduce (30)0≤h1t,s≤1+b-a.Case  2. If a<a∗≤b, in this case, we have two possibilities.(i) Ifa∗≤s<b, in this case, we have (31)∂h1∂tt,st=b≤0.Therefore, we conclude that(32)h1b,s≤h1t,s≤h1s,s≤1+b-a.On the other hand, we have(33)1+b-ab-s+α-1-γb-s=1+b-a-γb-s+α-11+b-a=s-b+α-11+b-a≥2-γ+b-aα-1≥0,which implies that(34)h1b,s≥0.From (32) and (34), we deduce(35)0≤h1t,s≤1+b-a.(ii) Ifa<s<a∗, in this case, we have (36)∂h1∂tt,st=b>0.Hence, there would existt∗∈(s,b) such that(37)∂h1∂tt,st=t∗=0.As mentioned above, it is easy to verify thath1(s,s)≥0 and h1(b,s)≤0. This yields (38)h1t∗,s≤h1b,s≤h1s,s≤rp+b-a.Then,(39)h1t,s≤max-h1t∗,s,1+b-a.Observe that∂h1/∂tt,st=t∗=0 is equivalent to (40)α-1γt∗-sα-2=b-sα-1+α-1b-sα-2.Therefore, we get(41)h1t∗,s=α-2α-1t∗+sα-1+1-a≥α-2α-1b+aα-1+1-a=b-aα-2α-1+1.Finally, using the above inequality and (39), we obtain(42)h1t,s≤max1+b-a,2-αα-1b-a-1,which makes end to the proof.Our first main result is as follows.Theorem 6.
If (7) admits a nontrivial continuous solution, then(43)∫ab‍b-sα-2b-s+α-1χsds≥γΓαmaxb-a+1,2-α/α-1b-a-1·Proof.
LetB=C[a,b] be the Banach space endowed with norm (44)x∞=maxa≤t≤bxt,x∈B.From Lemma 4, for all t∈[a,b], we have (45)ut=1γΓα×∫ab‍b-sα-2b-s+α-1Ht,sχsusds.Now, an application of Lemma 3 yields (46)u∞≤u∞max⁡b-a+1,2-α/α-1b-a-1γΓα×∫ab‍b-sα-2b-s+α-1χsds,from which the above theorem follows.


## 2.2. Nonexistence Result of Real Zeros for a Linear Combination of Certain Mittag-Leffler Functions

Letα,β>0 be fixed. The complex function(47)Eα,βz=∑k=0∞‍zkΓkα+β,α>0,β>0,z∈Cis analytic in the whole complex plane; it will be referred to [12–14] as the Mittag-Leffler function with parameters (α,β).At this stage, using the above Lyapunov-type inequality, we give an interval where a linear combination of Mittag-Leffler functions has no real zeros. More precisely, we proveTheorem 7.
Let1<α≤2. Then Eα,2z+Eα,1z+zEα,αz has no real zeros for (48)z∈-3αΓα1+αmax2,2-α/α-1-1,0.Proof.
Let(a,b)=(0,1), and consider the fractional Sturm-Liouville eigenvalue problem(49)D0Cαut+λut=0,0<t<1,ua-u′a=ub+u′b=0.
The real values ofλ, for which there exists a non-trivial solution to (49), are called eigenvalues of (49); and the corresponding solutions are called eigenfunctions.
As established in [15], the eigenvalues of (49) must be positive; moreover, the positive number λ is an eigenvalue of (49) if and only if (50)Eα,2-λ+Eα,1-λ-λEα,α-λ=0.Thanks to Theorem 6, if a positive real eigenvalue λ of (49) exists, then (51)λ∫01‍1-sα-2α-sds≥3Γαmax2,2-α/α-1-1·Hence,(52)λα+1α≥3Γαmax2,2-α/α-1-1,which concludes the proof.



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