# Lyapunov-Type Inequalities for Second-Order Boundary Value Problems with a Parameter

Authors: Haidong Liu
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this paper, we will establish some new Lyapunov-type inequalities for a class of second-order boundary value problems with a parameter. The inequalities generalize some early results in the literature.


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## Body

## 1. Introduction

Up until now, integral inequalities have attracted the attention of many researchers, due to its wide applications in the research of qualitative and quantitative properties such as global existence, boundedness, and stability of differential and integral equations (see [1–26] and the references therein). Among these inequalities, one important kind is the Lyapunov-type inequality, which was originally presented by Lyapunov in [27] as follows.Ifut is a solution of(1)u′′+qtu=0,satisfying ua=ub=0a<b and ut≠0 for t∈a,b, then(2)∫abqtdt>4b−a,and afterwards by Wintner [28] as(3)∫abq+tdt>4b−a,where q+t:=maxqt,0.Following Lyapunov’s landmark work, there have been plenty of references focused on the Lyapunov-type inequality and its generalizations which are widely used in various problems such as asymptotic theory, disconjugacy, and eigenvalue problems of differential equations and difference equations (see [29–41] and the references therein).For example, in 2003, Yang [29] obtained the following result for the second-order half-linear equation:(4)rtu′tp−1u′t′+qtutp−1ut=0,ua=ub=0,ut≠0,t∈a,b,where q,r∈Ca,b,ℝ such that rt>0, for t∈a,b, and p>0.Theorem 1.
(see [29]). Assume boundary value problem (4) has a solution ut; then, the following inequality holds:(5)∫abq+tdt≥2p+1∫abr−1/ptdtp,where q+t:=maxqt,0.
In 2012, Tiryaki et al. [34] established an inequality for boundary value problem of the form(6)rtu′tα−2u′t′+qtutα∗−2,ut=0,ua=ub=0,ut≠0,t∈a,b,where α>1 and α∗=α/α−1. Their result is as follows.Theorem 2.
(see [34]). Assume boundary value problem (6) has a solution ut; then, the following inequality holds:(7)∫abq+tha1−αt+hb1−αtdt≥1,where hat=∫atr1−α∗sds, hbt=∫tbr1−α∗sds, and q+t:=maxqt,0.In 2015, Agarwal et al. [36] established a Lyapunov-type inequality for the second-order forced boundary value problem of the form:(8)rtu′tβ−1u′t′+qtutγ−1,ut=ft,ua=ub=0,ut≠0,t∈a,b,in the subhalf-linear 0<γ<β and the super-half-linear 0<β<γ<2β cases, where rt and qt are integrable on a,b with rt>0 on a,b. Their result is as follows.Theorem 3.
(see [36]). Suppose that a,b, a<b, are consecutive zeros of a nontrivial solution of the first part of equation (8), then the inequality(9)2Γγβ∫abq+tdt+∫abftdt>2β+1Γγβ∫abr−1/βtdt−βholds, where γ∈0,2β and Γγβ=2β−γγγ/2β−γβ−2β/2β−γ2−2β/2β−γ>0.Agarwal and Özbekler [36] also established a Lyapunov-type inequality for the second-order forced boundary value problem with mixed nonlinearities:(10)u′′t+ptutα−1ut+qtutγ−1ut=ft,ua=ub=0,ut≠0,t∈a,b,where 0<γ<1<α<2. The result is as follows.Theorem 4.
(see [36]). Suppose that a,b, a<b, are consecutive zeros of a nontrivial solution of the first part of equation (10), then the inequality(11)∫abp++q+tdt∫abα0p+t+γ0q+t+ftdt>4b−a2,holds, where α0=2−ααα/2−α22/α−2>0 and γ0=2−γγγ/2−γ22/γ−2>0.We find that in [36], the authors studied the case 0<γ<1<α<2 of equation (10). It will be interesting to prove Lyapunov-type inequalities for equation (10) or other equation when α and γ have other relation. Motivated by [36], in this paper, we will establish a Lyapunov-type inequality for the nonlinear second-order boundary value problem of the form(12)ptu′α−2u′′−λqtuβ−2u+λrtuγ−2u+htuα−2u=0,t∈a,b,(13)ua=ub=0,ut≠0,t∈a,b,where p,q,r,h∈Ca,b,ℝ such that pt>0, qt>0, rt>0 for t∈a,b, 1<α<γ<β, and λ≥0 is a real parameter, and(14)ptu′α−2u′′+λqtu′β−2u′′+λrtuγ−2u+htuα−2u=0,t∈a,b,with the boundary condition (13), where p,q,r,h∈Ca,b,ℝ such that pt>0, qt>0, rt>0 for t∈a,b, 1<α<γ<β, and λ≥0 is a real parameter. Our results extend and compliment the results of [29, 36].


## 2. Main Results

Lemma 1.
Ifu is differential on [a, b] satisfying ua=ub=0 and ut≠0 for t∈a,b, then(15)supa≤t≤but≤12∫abu′tdt.Proof.
Sinceu is differential on [a, b] satisfying ua=ub=0, then we have(16)ut=12∫atu′sds−12∫tbu′sds,t∈a,b.
So,(17)ut≤12∫atu′sds+12∫tbu′sds=12∫abu′sds,t∈a,b.Therefore, (15) holds.Lemma 2.
(see [13]). Let A>0,B>0, and 1<α<γ<β be given. Then, for each x≥0,(18)Axγ−Bxβ≤Aβ−γβ−αβ−αBγ−αAγ−α/γ−βxα,holds.Theorem 5.
Assumeu is a solution of equation (12) satisfying the boundary conditions (13). Then,(19)λ2αγ−αβ−αγ−α/β−γβ−γβ−α∫abrβ−αtqα−γt1/β−γdt+12α∫abh+tdt≥∫abp1/1−αtdt1−α,where h+t:=maxht,0.Proof.
Multiplying (12) by ut and integrating over a,b, yields:(20)∫abptu′tα−2u′t′utdt−λ∫abqtutβdt+λ∫abrtutγdt+∫abhtutαdt=0.
Using integration by parts to the first integral on the left-hand side of (20) and from (13), we have(21)−∫abptu′tαdt−λ∫abqtutβdt+λ∫abrtutγdt+∫abhtutαdt=0.
Then, we obtain(22)−∫abptu′tαdt−λ∫abqtutβdt+λ∫abrtutγdt+∫abh+tutαdt≥0,i.e.,(23)∫abptu′tαdt≤λ∫abrtutγdt−λ∫abqtutβdt+∫abh+tutαdt.
By using Hölder’s inequality with indices1/τ+1/ρ=1,(24)∫abftgtdt≤∫abftτdt1/τ∫abgtρdt1/ρ,with ft=p1/αtu′t,gt=p−1/αt,τ=α,andρ=α/α−1, we obtain that(25)∫abu′tdt≤∫abptu′tαdt1/α∫abp1/1−αtdtα−1/α,i.e.,(26)∫abu′tdtα∫abp1/1−αtdtα−1≤∫abptu′tαdt.
On the contrary, from Lemma1, we obtain(27)∫abh+tutαdt≤supa≤t≤butα∫abh+tdt≤12α∫abh+tdt∫abu′tdtα.
Then, from (23), (26), and (27), we obtain(28)∫abu′tdtα∫abp1/1−αtdtα−1≤∫abptu′tαdt=λ∫abrtutγdt−∫abqtutβdt+∫abh+tut2dt≤λ∫abrtutγ−qtutβdt+12α∫abh+tdt∫ab|u′t|dtα.
For the first integral on the right-hand side of (28), inequality (18) in Lemma 2 with A=rt, B=qt, and x=ut≥0 for t∈a,b implies that(29)rtutγ−qtutβ≤rβ−α/β−γtqα−γ/β−γtγ−αβ−αγ−α/β−γβ−γβ−αutα.
By (28) and (29), we obtain(30)∫abu′tdtα∫abp1/1−αtdtα−1≤λ∫abrβ−αtqα−γt1/β−γγ−αβ−αγ−α/β−γβ−γβ−αutαdt+12α∫abh+tdt∫abu′tdtα.
From Lemma1, we have(31)supa≤t≤butα≤12α∫abu′tdtα.
In view of (30) and (31), we obtain that(32)∫abu′tdtα∫abp1/1−αtdtα−1≤λ2α∫abrβ−αtqα−γt1/β−γγ−αβ−αγ−α/β−γβ−γβ−αdt∫abu′tdtα+12α∫abh+tdt∫abu′tdtα.
Since∫abu′tdt>0 (in fact, if ∫abu′tdt=0, we have u′t=0 for t∈a,b. By condition (13), we obtain ut=0 for t∈a,b, which contradicts to ut≠0, t∈a,b), dividing both sides of (32) by ∫abu′tdtα, we obtain(33)λ2α∫abrβ−αtqα−γt1/β−γγ−αβ−αγ−α/β−γβ−γβ−αdt+12α∫abh+tdt≥∫abp1/1−αtdt1−α,which also leads to (19). The proof is complete.
If we takeα=2 and λ=1 in inequality (19), we obtain the following result.Corollary 1.
Assumeu is a solution of equation(34)ptu′′−qtuβ−2u+rtuγ−2u+htu=0,t∈a,b,satisfying boundary condition (13). Then,(35)14γ−2β−2γ−2/β−γβ−γβ−2∫abrβ−2tq2−γt1/β−γdt+14∫abh+tdt≥∫abp−1tdt−1,where h+t:=maxht,0.Theorem 6.
Assumeu is a solution of equation (14) satisfying boundary condition (13). Then,(36)λ12γβ−α/β−γγ−αβ−αγ−α/β−γβ−γβ−α,∫abq1/1−βtdt1−βγ−α/γ−β∫abrtdtβ−α/β−γ+12α∫abh+tdt≥∫abp1/1−αtdt1−α,where h+t:=maxht,0.Proof.
Multiplying (14) by ut and integrating over a,b yields(37)∫abptu′tα−2u′t′utdt+λ∫abqtu′tβ−2u′t′utdt+λ∫abrtutγdt+∫abhtutαdt=0.
Using integration by parts to the first and second integrals on the left-hand side of (37) and from (13), we have(38)−∫abptu′tαdt−λ∫abqtu′tβdt+λ∫abrtutγdt+∫abhtutαdt=0,i.e.,(39)∫abptu′tαdt=λ∫abrtutγdt−∫abqtu′tβdt+∫abhtutαdt≤λ∫abrtutγdt−∫abqtu′tβdt+∫abh+tutαdt.
By using Hölder’s inequality (24) with ft=q1/βtu′t,gt=q−1/βt,τ=β,andρ=β/β−1, we obtain that(40)∫abu′tdt≤∫abqtu′tβdt1/β∫abq1/1−βtdtβ−1/β,i.e.,(41)∫abu′tdtβ∫abq1/1−βtdtβ−1≤∫abqtu′tβdt.
From (26), (39), (41), and Lemma 1, we obtain(42)∫abu′tdtα∫abp1/1−αtdtα−1≤∫abptu′tαdt=λ∫abrtutγdt−∫abqtu′tβdt+∫abh+tutαdt≤λ∫abrtutγdt−∫abu′tdtβ∫abq1/1−βtdtβ−1+∫abh+tutαdt≤λsupa≤t≤butγ∫abrtdt−∫abu′tdtβ∫abq1/1−βtdtβ−1+∫abh+tutαdt≤λ12γ∫abu′tdtγ∫abrtdt−∫abu′tdtβ∫abq1/1−βtdtβ−1+∫abh+tutαdt.
For the right-hand side of (42), inequality (18) in Lemma 2 with A=1/2γ∫abrtdt, B=∫abq1/1−βtdt1−β, and x=∫abu′tdt>0 implies that(43)12γ∫abrtdt∫abu′tdtγ−∫abq1/1−βtdt1−β∫abu′tdtβ≤12γβ−α/β−γ∫abrtdtβ−α/β−γγ−αβ−αγ−α/β−γβ−γβ−α⋅∫abq1/1−βtdt1−βγ−α/γ−β∫abu′tdtα.
From (27), (42), and (43), we have(44)∫abu′tdtα∫abp1/1−αtdtα−1≤λ12γβ−α/β−γ∫abrtdtβ−α/β−γγ−αβ−αγ−α/β−γβ−γβ−α⋅∫abq1/1−βtdt1−βγ−α/γ−β∫abu′tdtα12α∫abh+tdt∫abu′tdtα.
Thus, dividing both sides of (29) by ∫abu′tdtα, we obtain(45)λ12γβ−α/β−γγ−αβ−αγ−α/β−γβ−γβ−α,∫abq1/1−βtdt1−βγ−α/γ−β∫abrtdtβ−α/β−γ+12α∫abh+tdt≥∫abp1/1−αtdt1−α,which also leads to (36). The proof is complete.Remark 1.
We note that whenλ=0, α=p+1, and (5) can be obtained from Theorems 5 and 6, respectively.
If we takeα=2 and λ=1 in inequality (36), we obtain the following result.Corollary 2.
Assumeu is a solution of equation(46)ptu′′+qtu′β−2u′′+rtuγ−2u+htuα−2u=0,t∈a,b,satisfying boundary condition (13). Then,(47)12γβ−2/β−γγ−2β−2γ−2/β−γβ−γβ−2,∫abq1/1−βtdt1−βγ−2/γ−β∫abrtdtβ−2/β−γ+14∫abh+tdt≥∫abp−1tdt−1,where h+t:=maxht,0.Theorem 7.
Assumeu is a solution of equation (14) satisfying boundary condition (13). Then,(48)λ12γβ−α/β−γb−aβ−1γ−αqβ−αγ−α/β−γβ−γβ−α,∫abrtdtβ−α/β−γ+12α∫abh+tdt≥pb−a1−α,where p:=minpt:t∈a,b, q:=minqt:t∈a,b, and h+t:=maxht,0.Proof.
From the proof of Theorem6, we have (39) holds. By (39) and Lemma 1, we obtain(49)p∫abu′tαdt=mina≤t≤bpt⋅∫abu′tαdt≤∫abptu′tαdt=λ∫abrtutγdt−∫abqtu′tβdt+∫abh+tutαdt≤λ∫abrtutγdt−mina≤t≤bqt∫abu′tβdt+∫abh+tutαdt≤λsupa≤t≤butγ∫abrtdt−q∫abu′tβdt+∫abh+tutαdt≤λ12γ∫abu′tdtγ∫abrtdt−q∫abu′tβdt+∫abh+tutαdt.
By using Hölder’s inequality (24) with ft=u′t,gt=1,τ=α,βandρ=α/α−1,β/β−1, respectively, we obtain that(50)∫abu′tdt≤∫abu′tαdt1/αb−aα−1/α,∫abu′tdt≤∫abu′tβdt1/βb−aβ−1/β.
Therefore,(51)b−a1−α∫abu′tdtα≤=∫abu′tαdt,(52)b−a1−β∫abu′tdtβ≤∫abu′tβdt.
From (49), (51), and (52), we have(53)pb−a1−α∫abu′tdtα≤λ12γ∫abrtdt∫abu′tdtγ−qb−a1−β∫abu′tdtβ+∫abh+tutαdt.
For the right-hand side of (53), inequality (18) in Lemma 2 with A=1/2γ∫abrtdt, B=qb−a1−β, and x=∫abu′tdt>0 implies that(54)12γ∫abrtdt∫abu′tdtγ−qb−a1−β∫abu′tdtβ≤12γβ−α/β−γ∫abrtdtβ−α/β−γb−aβ−1γ−αqβ−αγ−α/β−γβ−γβ−α∫abu′tdtα.
From (53) and (54), we have(55)pb−a1−α∫abu′tdtα≤λ12γβ−α/β−γ∫abrtdtβ−α/β−γb−aβ−1γ−αqβ−αγ−α/β−γβ−γβ−α∫abu′tdtα+12α∫abh+tdt∫abu′tdtα.
Thus, dividing both sides of (55) by ∫abu′tdtα, we obtain(56)λ12γβ−α/β−γ∫abrtdtβ−α/β−γb−aβ−1γ−αqβ−αγ−α/β−γ⋅β−γβ−α+12α∫abh+tdt≥pb−a1−α,which also leads to (48). The proof is complete.Remark 2.
We note that whenλ=0, α=2, and pt≡1, classical result (3) can be obtained from Theorems 5–7, respectively.
If we takeα=2 and λ=1 in inequality (48), we obtain the following result.Corollary 3.
Assumeu is a solution of equation (46), satisfying boundary condition (13). Then,(57)12γβ−2/β−γb−aβ−1γ−2qβ−2γ−2/β−γβ−γβ−2∫abrtdtβ−2/β−γ+14∫abh+tdt≥pb−a,where p:=minpt:t∈a,b, q:=minqt:t∈a,b, and h+t:=maxht,0.



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