In this paper, we establish new Lyapunov-type inequalities for a Hadamard fractional differential equation under Sturm-Liouville boundary conditions. Our conclusions cover many results in the literature.
Citation: Youyu Wang, Lu Zhang, Yang Zhang. Lyapunov-type inequalities for Hadamard fractional differential equation under Sturm-Liouville boundary conditions[J]. AIMS Mathematics, 2021, 6(3): 2981-2995. doi: 10.3934/math.2021181
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In this paper, we establish new Lyapunov-type inequalities for a Hadamard fractional differential equation under Sturm-Liouville boundary conditions. Our conclusions cover many results in the literature.
For the second order Hill's equation,
u″(t)+r(t)u(t)=0,t∈(a,b), | (1.1) |
where r(t) is a positive continuous function on [a,b], Lyapunov [1] obtain the interesting following result.
Theorem 1.1. If Eq (1.1) has a nontrivial solution u(t) satisfying u(a)=u(b)=0, then
∫bar(t)dt>4b−a. | (1.2) |
Because of the importance of Lyapunov inequality (1.2) in application, Lyapunov inequality has been extended in many directions. In the last few decades, with the increasing enthusiasm for the study of fractional differential equations, a large number of Lyapunov inequalities for fractional differential equations have appeared. For example, Ferreira [2] first obtained a Lyapunov-type inequality for Riemann-Liouville fractional differential equation, in 2014, Ferreira [3] developed a Lyapunov-type inequality for the Caputo fractional differential equation. For more details on Lyapunov-type inequalities and their applications, we refer [4,5,6,7,8,9,10,11,12,13,14,15] and the references therein.
Another kind of fractional derivatives that appears side by side to Riemann-Liouville and Caputo derivatives in the literature is the fractional derivative due to Hadamard [16]. Hadamard-type integrals arise in the formulation of many problems in mechanics such as in fracture analysis. For details and applications of Hadamard fractional derivative and integral, we refer the reader to the works in [17,18,19,20,21,22]. It must be noted that there are few papers studied Lyapunov-type inequality for Hadamard fractional differential equation. For instance, Ma et al. [23] developed a Lyapunov-type inequality for the Hadamard fractional boundary value problem in 2017.
Theorem 1.2. If the Hadamard fractional boundary value problem
(HDα1+u)(t)+q(t)u(t)=0,1<t<e, 1<α≤2, | (1.3) |
u(1)=0=u(e), | (1.4) |
has a nontrivial solution, where q:[1,e]→R is a continuous function, then
∫e1|q(s)|ds>Γ(α)λ1−α(1−λ)1−αeλ, | (1.5) |
where λ=2α−1−√(2α−2)2+12 and HDαa+ denotes the Hadamard fractional derivative of order α.
Recently, Dhar [24] and Laadjal et al. [25] generalized the Lyapunov-type inequality in Theorem 1.2 by replacing the interval [1,e] with a general interval [a,b](1≤a<b),.
Theorem 1.3. If the Hadamard fractional boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,1≤a<t<b, 1<α≤2, | (1.6) |
u(a)=0=u(b), | (1.7) |
has a nontrivial solution, where q:[a,b]→R is a continuous function, then
∫ba|q(s)|ds>4α−1Γ(α)a(lnba)1−α. | (1.8) |
Theorem 1.4. If the Hadamard fractional boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,1≤a<t<b, 1<α≤2, | (1.9) |
u(a)=0=u(b), | (1.10) |
has a nontrivial solution, where q:[a,b]→R is a continuous function, then
∫ba|q(s)|ds>Γ(α)ξ1(lnξ1alnbξ1lnba)1−α, | (1.11) |
where
ξ1=exp(12[[2(α−1)+lnba]−√4(α−1)2+ln2ba]). |
Very recently, J. Jonnalagadda and B. Debananda [26] obtained Lyapunov-type inequalities for Hadamard fractional boundary value problems associated with different sets of boundary conditions, the main results are as follows.
Theorem 1.5. If the following fractional boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,0<a<t<b, 1<α≤2, | (1.12) |
l(HI2−αa+u)(a)−m(HDα−1a+u)(a)=0, | (1.13) |
nu(b)+p(HDα−1a+u)(b)=0, | (1.14) |
has a nontrivial solution, then
∫ba(lnsa)α−2|q(s)|ds>AΓ(α)[n(lnba)α−1+pΓ(α)][l(lnba)+m(α−1)], | (1.15) |
where l,p≥0;m,n>0 and
A=ln(lnba)α−1+mn(α−1)(lnba)α−2+lpΓ(α)>0. |
Theorem 1.6. If the following fractional boundary value problem
(HDα1+u)(t)+q(t)u(t)=0,1<t<T, 1<α≤2, | (1.16) |
(HI2−α1+u)(1)+(HI2−α1+u)(T)=0, | (1.17) |
(HDα−11+u)(1)+(HDα−11+u)(T)=0, | (1.18) |
has a nontrivial solution, then
∫T1(lnsa)α−2|q(s)|ds>4Γ(α)(3−α)lnT. | (1.19) |
Inspired by papers [24,25,26], in this paper, we establish a few Lyapunov-type inequalities for Hadamard fractional differential equations
(HDαa+u)(t)+q(t)u(t)=0,0<a<t<b, 1<α≤2, | (1.20) |
with the following Sturm-Liouville multi-point and integral boundary conditions,
u(a)=0,γu(b)+δu′(b)=m−2∑i=1βiu(ξi), | (1.21) |
u(a)=0,γu(b)+δu′(b)=λ∫bah(s)u(s)ds,λ≥0, | (1.22) |
where γ≥0,δ≥0,γδ>0, a<ξ1<ξ2<⋯<ξm−2<b, βi≥0(i=1,2,⋯,m−2) and h:[a,b]→[0,∞) with h∈L1(a,b).
The main difficulty of this paper is to express the solution for boundary value problems (1.20) and (1.21) with Green's function. We solve this problem by properly decomposing coefficients. For convenience, we shall adopt the following notations and assumptions
σ=√(α−1)2+(ln√ba)2,ρ1=γ(lnba)α−1+(α−1)δb(lnba)α−2>0,ρ2=γ(lnba)α−1+(α−1)δb(lnba)α−2−m−2∑i=1βi(lnξia)α−1>0,ρ3=γ(lnba)α−1+(α−1)δb(lnba)α−2−λ∫ba(lnta)α−1h(t)dt>0. |
In this paper, we shall use the following notations, definitions and some lemmas from the theory of fractional calculus in the sense of Hadamard. For more details, we refer to [27].
Definition 2.1. [27] The Hadamard fractional integral of order α∈R+ for a continuous function f:[a,b]→R is defined by
(HIαa+f)(t)=1Γ(α)∫ta(lnts)α−1f(s)dss,α>0,t∈[a,b]. |
Definition 2.2. [27] The Hadamard fractional derivative of order α∈R+ for a continuous function f:[a,b]→R is defined by
(HDαa+f)(t)=1Γ(n−α)(tddt)n∫ta(lnts)n−α−1f(s)dss,t∈[a,b], |
where n−1<α<n,n=[α]+1.
Lemma 2.3. [27] Let α>0,n=[α]+1 and 0<a<b<∞. if u∈L1(a,b) and (HIn−αa+u)(t)∈ACnδ[a,b], then
(HIαa+HDαa+u)(t)=u(t)−n∑k=1(δ(n−k)(HIn−αa+u))(a)Γ(α−k+1)(lnta)α−k. |
where ACnδ[a,b]={φ:[a,b]→C:δ(n−1)φ∈AC[a,b],δ=tddt}.
Lemma 2.4. Assume that r∈C[a,b]. The Sturm-Liouville Hadamard fractional boundary value problem
{(HDαa+u)(t)+r(t)=0,0<a<t<b, 1<α≤2,u(a)=0,γu(b)+δu′(b)=∑m−2i=1βiu(ξi), |
has the unique solution
u(t)=∫baG(t,s)r(s)ds+1ρ2(lnta)α−1m−2∑i=1βi∫baG(ξi,s)r(s)ds, |
where G(t,s) is given by
G(t,s)=1sρ1Γ(α){(lnta)α−1(lnbs)α−2[γlnbs+(α−1)δb]−ρ1(lnts)α−1,a≤s≤t≤b,(lnta)α−1(lnbs)α−2[γlnbs+(α−1)δb],a≤t≤s≤b. |
Proof. Using Lemma 2.3, we have
u(t)=c1(lnta)α−1+c2(lnta)α−2−1Γ(α)∫ta(lnts)α−1r(s)dss, |
for some c1,c2∈R. Applying the boundary condition u(a)=0, we have c2=0, hence,
u(t)=c1(lnta)α−1−1Γ(α)∫ta(lnts)α−1r(s)dss, |
it is easy to obtain
u′(t)=c1(α−1)1t(lnta)α−2−1Γ(α)∫ta(α−1)1t(lnts)α−2r(s)dss=α−1t[c1(lnta)α−2−1Γ(α)∫ta(lnts)α−2r(s)dss], |
the boundary condition γu(b)+δu′(b)=∑m−2i=1βiu(ξi) imply that,
γ[c1(lnba)α−1−1Γ(α)∫ba(lnbs)α−1r(s)dss]+(α−1)δb[c1(lnba)α−2−1Γ(α)∫ba(lnbs)α−2r(s)dss]=m−2∑i=1βi[c1(lnξia)α−1−1Γ(α)∫ξia(lnξis)α−1r(s)dss], |
we obtain
c1=1ρ2Γ(α)∫ba(lnbs)α−2[γlnbs+(α−1)δb]r(s)dss−1ρ2Γ(α)m−2∑i=1βi∫ξia(lnξis)α−1r(s)dss, |
by the relation
1ρ2=1ρ1+∑m−2i=1βi(lnξia)α−1ρ1ρ2, |
we have
c1=1Γ(α)(1ρ1+∑m−2i=1βi(lnξia)α−1ρ1ρ2)∫ba(lnbs)α−2[γlnbs+(α−1)δb]r(s)dss−1ρ2Γ(α)m−2∑i=1βi∫ξia(lnξis)α−1r(s)dss=1ρ1Γ(α)∫ba(lnbs)α−2[γlnbs+(α−1)δb]r(s)dss+1ρ1ρ2Γ(α)m−2∑i=1βi∫ba(lnξia)α−1(lnbs)α−2[γlnbs+(α−1)δb]r(s)dss−1ρ2Γ(α)m−2∑i=1βi∫ξia(lnξis)α−1r(s)dss=1ρ1Γ(α)∫ba(lnbs)α−2[γlnbs+(α−1)δb]r(s)dss+1ρ2m−2∑i=1βi∫baG(ξi,s)r(s)ds, |
therefore,
u(t)=c1(lnta)α−1−1Γ(α)∫ta(lnts)α−1r(s)dss=1ρ1Γ(α)∫ba(lnta)α−1(lnbs)α−2[γlnbs+(α−1)δb]r(s)dss |
+1ρ2(lnta)α−1m−2∑i=1βi∫baG(ξi,s)h(s)ds−1Γ(α)∫ta(lnts)α−1r(s)dss=∫baG(t,s)h(s)ds+1ρ2(lnta)α−1m−2∑i=1βi∫baG(ξi,s)r(s)ds. |
The proof is complete.
Lemma 2.5. Assume that g∈C[a,b]. The Sturm-Liouville Hadamard fractional boundary value problem
{(HDαa+u)(t)+g(t)=0,0<a<t<b, 1<α≤2,u(a)=0,γu(b)+δu′(b)=λ∫bah(s)u(s)ds,λ≥0, |
has the unique solution
u(t)=∫baG(t,s)g(s)ds+λρ3(lnta)α−1∫ba(∫baG(t,s)g(s)ds)h(t)dt, |
where h:[a,b]→[0,∞) with h∈L1(a,b), G(t,s) is defined in Lemma 2.4.
Proof. Using Lemma 2.3, we have
u(t)=c1(lnta)α−1+c2(lnta)α−2−1Γ(α)∫ta(lnts)α−1g(s)dss, |
for some c1,c2∈R. Using the boundary condition u(a)=0, we have c2=0, therefore,
u(t)=c1(lnta)α−1−1Γ(α)∫ta(lnts)α−1g(s)dss, |
and
u′(t)=α−1t[c1(lnta)α−2−1Γ(α)∫ta(lnts)α−2g(s)dss], |
the boundary condition γu(b)+δu′(b)=λ∫bah(t)u(t)dt imply that,
γ[c1(lnba)α−1−1Γ(α)∫ba(lnbs)α−1g(s)dss]+(α−1)δb[c1(lnba)α−2−1Γ(α)∫ba(lnbs)α−2g(s)dss]=λ∫bah(t)u(t)dt, |
we obtain
c1=1ρ1Γ(α)∫ba(lnbs)α−2[γlnbs+(α−1)δb]g(s)dss+λρ1∫bah(t)u(t)dt, |
therefore, the solution of the boundary value problem is
u(t)= c1(lnta)α−1−1Γ(α)∫ta(lnts)α−1g(s)dss=1ρ1Γ(α)∫ba(lnta)α−1(lnbs)α−2[γlnbs+(α−1)δb]h(s)dss+λρ1(lnta)α−1∫bah(t)u(t)dt−1Γ(α)∫ta(lnts)α−1g(s)dss=∫baG(t,s)g(s)ds+λρ1(lnta)α−1∫bah(t)u(t)dt. |
Multiplying both side of above equation by h(t) and integrating from a to b, we obtain
∫bah(t)u(t)dt=∫ba(∫baG(t,s)g(s)ds)h(t)dt+λρ1∫bah(t)u(t)dt⋅∫ba(lnta)α−1h(t)dt, |
and
∫bah(t)u(t)dt=ρ1ρ3∫ba(∫baG(t,s)g(s)ds)h(t)dt, |
thus
u(t)=∫baG(t,s)g(s)ds+λρ3(lnta)α−1∫ba(∫baG(t,s)g(s)ds)h(t)dt, |
which concludes the proof.
Lemma 2.6. The function G defined in Lemma 2.4 satisfies the following properties:
1).G(t,s)≥0on[a,b]×[a,b],2).maxt∈[a,b]G(t,s)=G(s,s)=1sρ1Γ(α)(lnsa)α−1(lnbs)α−2[γlnbs+(α−1)δb]. |
Proof. (1). Firstly, we define two functions as follows
g1(t,s)=1sρ1Γ(α)[(lnta)α−1(lnbs)α−2[γlnbs+(α−1)δb]−ρ1(lnts)α−1],a≤s≤t≤b,g2(t,s)=1sρ1Γ(α)(lnta)α−1(lnbs)α−2[γlnbs+(α−1)δb],a≤t≤s≤b. |
Obviously, g2(t,s)≥0. Next, we consider function g1(t,s), differentiating g1(t,s) with respect to t, we get
∂g1(t,s)∂t=α−1stρ1Γ(α)[(lnta)α−2(lnbs)α−2[γlnbs+(α−1)δb]−ρ1(lnts)α−2]=α−1stρ1Γ(α){γ[(lnta)α−2(lnbs)α−1−(lnts)α−2(lnba)α−1]+(α−1)δb[(lnta)α−2(lnbs)α−2−(lnts)α−2(lnba)α−2]}, |
since a≤s≤t≤b, we have
(lnta)α−2<(lnts)α−2,(lnbs)α−1<(lnba)α−1, |
which imply that
(lnta)α−2(lnbs)α−1−(lnts)α−2(lnba)α−1<0. |
The inequality
lntalnbs−lntslnba=lntalnbs−lnts(lnbs+lnsa)=lnbs(lnta−lnts)−lnsalnts=lnbslnsa−lnsalnts=lnsalnbt≥0, |
implying that
(lnta)α−2(lnbs)α−2−(lnts)α−2(lnba)α−2<0, |
so we have ∂g1(t,s)∂t≤0, this means that g1(b,s)≤g1(t,s)≤g1(s,s). On the other hand,
sρ1Γ(α)g1(b,s)=(lnba)α−1(lnbs)α−2[γlnbs+(α−1)δb]−ρ1(lnbs)α−1=(α−1)δb[(lnba)α−1(lnbs)α−2−(lnbs)α−1(lnba)α−2]=(α−1)δb(lnba)α−2(lnbs)α−2[lnba−lnbs]=(α−1)δb(lnba)α−2(lnbs)α−2lnsa≥0, |
hence, g1(t,s)≥g1(b,s)≥0 and G(t,s)≥0.
(2). By the above discussion, for function g1(t,s) satisfy 0≤g1(t,s)≤g1(s,s)=G(s,s). It is easy to see that 0≤g2(t,s)≤g2(s,s)=g1(s,s)=G(s,s) for a≤t≤s≤b.
Lemma 2.7. Assume 0<a≤s≤b and 1<α<2, then
0≤1s(lnsalnbs)α−1≤1√ab⋅(α−1)α−1(lnba)2(α−1)eσ(2e)α−1(α−1+σ)α−1. |
Proof. Let
f(s)=1s(lnsalnbs)α−1, s∈[a,b]. |
Clearly f(a)=f(b)=0 and f(s)>0 on (a,b). By Rolle's Theorem, there exists s∗∈(a,b) such that f(s∗)=maxf(s) on (a,b), i.e., f′(s∗)=0. Note that
f′(s)=1s2(lnsalnbs)α−1[(α−1)lnbs−lnsalnsalnbs−1], |
denote x=lnbs, y=lnsa, let f′(s)=0, we obtain (α−1)(x−y)=xy and x+y=lnba, by these two equalities, we get
(xy)2=(α−1)2(x−y)2=(α−1)2[(x+y)2−4xy]=(α−1)2[(lnba)2−4xy], |
note the fact xy>0, we have
xy=(α−1)[−2(α−1)+√4(α−1)2+(lnba)2]=(α−1)(lnba)22(α−1)+√4(α−1)2+(lnba)2, |
and
y=12(lnba−xyα−1)=12(lnba+2(α−1)−√4(α−1)2+(lnba)2)=ln√ba+(α−1)−√(α−1)2+(ln√ba)2, |
so,
s∗=aey=√abeα−1e√(α−1)2+(ln√ba)2, |
it is easy to show that s∗∈(a,b), thus
maxf(s)=f(s∗)=1s∗(lns∗alnbs∗)α−1=1√ab⋅(α−1)α−1(lnba)2(α−1)e√(α−1)2+(ln√ba)2(2e)α−1(α−1+√(α−1)2+(ln√ba)2)α−1. |
which concludes the proof.
We now present Lyapunov-type inequalities for the Sturm-Liouville Hadamard fractional boundary value problems (1.20) and (1.21).
Theorem 3.1. If a nontrivial continuous solution of the Sturm-Liouville Hadamard fractional boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,0<a<t<b, 1<α≤2,u(a)=0,γu(b)+δu′(b)=m−2∑i=1βiu(ξi), |
exists, where q:[a,b]→R is a continuous function, and γ≥0,δ≥0,γδ>0, a<ξ1<ξ2<⋯<ξm−2<b, βi≥0(i=1,2,⋯,m−2), then we have
∫ba1s(lnsa)α−1(lnbs)α−2[γlnbs+(α−1)δb]|q(s)|ds≥ρ1ρ2ρ2+∑m−2i=1βi(lnba)α−1Γ(α). | (3.1) |
Proof. Let B=C[a,b] be the Banach space endowed with norm ‖x‖=supt∈[a,b]|x(t)|. From Lemma 2.4, for all t∈[a,b], we have
u(t)=∫baG(t,s)q(s)u(s)ds+1ρ2(lnta)α−1m−2∑i=1βi∫baG(ξi,s)q(s)u(s)ds, |
Now, an application of Lemma 2.6 yields
‖u‖≤‖u‖(∫baG(s,s)|q(s)|ds+1ρ2m−2∑i=1βi(lnba)α−1∫baG(s,s)|q(s)|ds)=‖u‖(1+1ρ2m−2∑i=1βi(lnba)α−1)∫baG(s,s)|q(s)|ds, |
Since u is non trivial, then ‖u‖≠0, so
∫ba1s(lnsa)α−1(lnbs)α−2[γlnbs+(α−1)δb]|q(s)|ds≥ρ1ρ2ρ2+∑m−2i=1βi(lnba)α−1Γ(α). |
from which inequality in (3.1) follows.
Let βi=0(i=1,2,⋯,m−2) in Theorem 3.1, we have
Corollary 3.2. If a nontrivial continuous solution of the Sturm-Liouville Hadamard fractional boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,0<a<t<b, 1<α≤2,u(a)=0,γu(b)+δu′(b)=0, |
exists, where q:[a,b]→R is a continuous function, and γ≥0,δ≥0,γδ>0, then we have
∫ba1s(lnsa)α−1(lnbs)α−2[γlnbs+(α−1)δb]|q(s)|ds≥Γ(α)(lnba)α−2[γlnba+(α−1)δb]. | (3.2) |
Let γ=1,δ=0 or γ=0,δ=1 in Corollary 3.2, we can obtain the following Lyapunov-type inequalities.
Corollary 3.3. If a nontrivial continuous solution of the Hadamard fractional boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,0<a<t<b, 1<α≤2,u(a)=0,u(b)=0, |
exists, where q:[a,b]→R is a continuous function, then we have
∫ba1s(lnsa)α−1(lnbs)α−1|q(s)|ds≥Γ(α)(lnba)α−1. | (3.3) |
This is Theorem 1 in [25].
Corollary 3.4. If a nontrivial continuous solution of the Hadamard fractional boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,0<a<t<b, 1<α≤2,u(a)=0,u′(b)=0, |
exists, where q:[a,b]→R is a continuous function, then we have
∫ba1s(lnsa)α−1(lnbs)α−2|q(s)|ds≥Γ(α)(lnba)α−2. | (3.4) |
Let γ=1,δ=0 in Theorem 3.1, we have the following Lyapunov-type inequality.
Corollary 3.5. If a nontrivial continuous solution of the Hadamard fractional multi-point boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,0<a<t<b, 1<α≤2,u(a)=0,u(b)=m−2∑i=1βiu(ξi), |
exists, where q:[a,b]→R is a continuous function, a<ξ1<ξ2<⋯<ξm−2<b, βi≥0(i=1,2,⋯,m−2) with 0≤∑m−2i=1βi<1, then we have
∫ba1s(lnsa)α−1(lnbs)α−1|q(s)|ds≥(lnba)α−1[(lnba)α−1−∑m−2i=1βi(lnξia)α−1)](lnba)α−1−∑m−2i=1βi(lnξia)α−1+∑m−2i=1βi(lnba)α−1Γ(α). | (3.5) |
Let γ=0,δ=1 in Theorem 3.1, we have the following Lyapunov-type inequality.
Corollary 3.6. If a nontrivial continuous solution of the Hadamard fractional multi-point boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,0<a<t<b, 1<α≤2,u(a)=0,u′(b)=m−2∑i=1βiu(ξi), |
exists, where q:[a,b]→R is a continuous function, a<ξ1<ξ2<⋯<ξm−2<b, βi≥0(i=1,2,⋯,m−2) with 0≤∑m−2i=1βi<1, then we have
∫ba1s(lnsa)α−1(lnbs)α−2|q(s)|ds≥(lnba)α−2[α−1b(lnba)α−2−∑m−2i=1βi(lnξia)α−1)]α−1b(lnba)α−2−∑m−2i=1βi(lnξia)α−1+∑m−2i=1βi(lnba)α−1Γ(α). | (3.6) |
Let β=β1≥0,β2=β3=⋯=βm−2=0,ξ=ξ1 in Corollary 3.5, we obtain three-point Lyapunov-type inequality.
Corollary 3.7. If a nontrivial continuous solution of the Hadamard fractional three-point boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,0<a<t<b, 1<α≤2,u(a)=0,u(b)=βu(ξ), |
exists, where q:[a,b]→R is a continuous function, a<ξ<b, 0≤β<1, then we have
∫ba1s(lnsa)α−1(lnbs)α−1|q(s)|ds≥(lnba)α−1[(lnba)α−1−β(lnξa)α−1)](lnba)α−1+β[(lnba)α−1−(lnξa)α−1]Γ(α). | (3.7) |
By the relation
1s(lnsa)α−1(lnbs)α−2[γlnbs+(α−1)δb]=1s(lnsa)α−1(lnbs)α−1⋅γlnbs+(α−1)δblnbs, |
and applying Theorem 3.1 and Lemma 2.7, we easily get the following results.
Theorem 3.8. If a nontrivial continuous solution of the Sturm-Liouville Hadamard fractional boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,0<a<t<b, 1<α≤2,u(a)=0,γu(b)+δu′(b)=m−2∑i=1βiu(ξi), |
exists, where q:[a,b]→R is a continuous function, and γ≥0,δ≥0,γδ>0, a<ξ1<ξ2<⋯<ξm−2<b, βi≥0(i=1,2,⋯,m−2), then we have
∫baγlnbs+(α−1)δblnbs|q(s)|ds≥√abρ1ρ2Γ(α)ρ2+∑m−2i=1βi(lnba)α−1⋅(2e)α−1(α−1+σ)α−1(α−1)α−1(lnba)2(α−1)eσ. | (3.8) |
A similar discussion can be made for Theorem 3.8. We omit the details here.
Next we give Lyapunov-type inequalities for Sturm-Liouville Hadamard fractional boundary value problems (1.20)–(1.22). The proof is essentially the same as that of Theorem 3.1 and Theorem 3.8. Therefore, we omit the proof.
Theorem 3.9. If a nontrivial continuous solution of the Sturm-Liouville Hadamard fractional boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,0<a<t<b, 1<α≤2,u(a)=0,γu(b)+δu′(b)=λ∫bah(s)u(s)ds,λ≥0, |
exists, where q:[a,b]→R is a continuous function and h:[a,b]→[0,∞) with h∈L1(a,b), γ≥0,δ≥0,γδ>0, then we have
∫ba1s(lnsa)α−1(lnbs)α−2[γlnbs+(α−1)δb]|q(s)|ds≥ρ1ρ3ρ3+λ(lnba)α−1∫bah(t)dtΓ(α). | (3.9) |
Theorem 3.10. If a nontrivial continuous solution of the Sturm-Liouville Hadamard fractional boundary value problem
(HDαa+u)(t)+q(t)u(t)=0,0<a<t<b, 1<α≤2,u(a)=0,γu(b)+δu′(b)=λ∫bah(s)u(s)ds,λ≥0, |
exists, where q:[a,b]→R is a continuous function and h:[a,b]→[0,∞) with h∈L1(a,b), γ≥0,δ≥0,γδ>0, then we have
∫baγlnbs+(α−1)δblnbs|q(s)|ds≥√abρ1ρ3Γ(α)ρ3+λ(lnba)α−1∫bah(t)dt⋅(2e)α−1(α−1+σ)α−1(α−1)α−1(lnba)2(α−1)eσ. | (3.10) |
Remark 3.11. The similar discussion can be made for Theorem 3.9 and Theorem 3.10. We omit the details here.
Remark 3.12. Comparisons with previous literatures, the conclusion of this paper has the following characteristics. Firstly, our results are new and include all the conclusions of [23,24,25]. Secondly, the result of Lemma 2.7 is better than that of Lemma 5 in [25].
The authors would like to thank the referees for their helpful comments and suggestions. This work is supported by the Tianjin Natural Science Foundation (grant no. 20JCYBJC00210).
The authors declare no conflict of interest.
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