Abstract

For a fourth-order differential equation, we will establish some new Lyapunov-type inequalities, which give lower bounds of the distance between zeros of a nontrivial solution and also lower bounds of the distance between zeros of a solution and/or its derivatives. The main results will be proved by making use of Hardy’s inequality and some generalizations of Opial-Wirtinger-type inequalities involving higher-order derivatives. Some examples are considered to illustrate the main results.

1. Introduction

In this paper, we are concerned with the lower bounds of the distance between zeros of a nontrivial solution and also lower bounds of the distance between zeros of a solution and/or its derivatives for the fourth-order differential equation𝑥𝑟(𝑡)(𝑡)𝛾+𝑞(𝑡)𝑥𝛾(𝑡)=0,𝑡∈𝕀,(1.1) where 𝛾≥1,𝑟,𝑞∶𝕀→ℝ are continuous measurable functions and 𝕀 is a nontrivial interval of reals. By a solution of (1.1) on the interval 𝐽⊆𝕀, we mean a nontrivial real-valued function 𝑥∈𝐶3(𝐽), which has the property that 𝑟(𝑡)(𝑥(𝑡))𝛾∈𝐶1(𝐽) and satisfies (1.1) on 𝐽. We assume that (1.1) possesses such a nontrivial solution on 𝕀.

The nontrivial solution 𝑥(𝑡) of (1.1) is said to be oscillate or to be oscillatory if it has arbitrarily large zeros. Equation (1.1) is oscillatory if one of its nontrivial solutions is oscillatory. Equation (1.1) is said to be (𝑖,𝑗)-disconjugate if 𝑖 and 𝑗 are positive integers such that 𝑖+𝑗=4 and no solution of (1.1) has an (𝑖,𝑗)-distribution of zeros, that is, no nontrivial solution has a pair of zeros of multiplicities 𝑖 and 𝑗, respectively. In general, the differential equation of 𝑛th-order𝑥(𝑛)(𝑡)+𝑞(𝑡)𝑥(𝑡)=0(1.2) is said to be (𝑘,𝑛−𝑘)-disconjugate on an interval 𝕀 in case no nontrivial solution has a zero of order 𝑘 followed by a zero of order 𝑛−𝑘. This means that, for every pair of points 𝛼,𝛽∈𝕀,𝛼<𝛽, a nontrivial solution of (1.1) that satisfies𝑥(𝑖)𝑥(𝛼)=0,𝑖=0,…,𝑘−1,(𝑗)(𝛽)=0,𝑗=0,…,𝑛−𝑘−1(1.3) does not exist.

The least value of 𝛽 such that there exists a nontrivial solution which satisfies (1.3) is called the (𝑘,𝑛−𝑘)-conjugate point of 𝛼. The differential equation (1.2) is said to disconjugate on an interval 𝕀 if one of its nontrivial solutions has at most 𝑛−1 zeros. For our case, if no nontrivial solution of (1.1) has more than three zeros, the equation is termed disconjugate. Together with (𝑘,𝑛−𝑘)-disconjugacy, we consider the related concept, which is (𝑘,𝑛−𝑘)-disfocality. The differential equation (1.2) is said to be disfocal on an interval 𝕀 if for every nontrivial solution 𝑥 at least one of the functions 𝑥,𝑥,…,𝑥(𝑛−1) does not vanish on 𝕀. If the equation is not disfocal on 𝕀, then there exists an integer 𝑘 (1≤𝑘≤𝑛−1), a pair of points 𝛼,𝛽∈𝕀,𝛼<𝛽 and a nontrivial solution 𝑥 such that 𝑘 of the functions 𝑥,𝑥,…,𝑥(𝑛−1) vanishes at 𝛼 and the remaining 𝑛−𝑘 functions at 𝛽, that is,𝑥(𝑖)𝑥(𝛼)=0,𝑖=0,…,𝑘−1,(𝑗)(𝛽)=0,𝑗=𝑘,…,𝑛−1.(1.4) The equation (1.1) is said to be (2,2)-disconjugate on [𝛼,𝛽] if there is no nontrivial solution 𝑥(𝑡) and 𝑐,𝑑∈[𝛼,𝛽],𝑐<𝑑 such that 𝑥(𝑐)=𝑥(𝑐)=𝑥(𝑑)=𝑥(𝑑)=0. Equation (1.1) is said to be (𝑘,4−𝑘)-disfocal on an interval 𝕀 for some 1≤𝑘≤3 in case there does not exist a solution 𝑥 with a zero of order 𝑘 followed by a zero of 𝑥∗(𝑗) of order 4−𝑘, where 𝑥∗(𝑗)=𝑥(𝑗) for 𝑗=0,1,2,3 and 𝑥∗(4)=(𝑟(𝑥)𝛾).

For 𝑛th-order differential equations, (𝑘,𝑛−𝑘)-disconjugacy and disfocality are connected by the result of Nehari [1], which states that, if (1.2) is (𝑘,𝑛−𝑘)-disfocal on (𝛼,𝛽) it is disconjugate on (𝛼,𝛽). For more details about disconjugacy and disfocality and the relation between them, we refer the reader to the paper [2]. For related results to the present paper, we refer the reader to the papers [3–14] and the references cited therein.

In [4, 15], the authors established some new Lyapunov-type inequalities for higher-order differential equations. In the following, we present some of some special cases of their results for fourth-order differential equations that serve and motivate the contents of this paper. In [15], it is proved that if 𝑥(𝑡) is a solution of the fourth-order differential equation𝑥(4)(𝑡)+𝑞(𝑡)𝑥(𝑡)=0,(1.5) which satisfies 𝑥(𝛼)=𝑥(𝛼)=𝑥(𝛽)=𝑥(𝛽)=0, then𝛽𝛼||||𝑞(𝑡)𝑑𝑡≥192(𝛽−𝛼)3,(1.6) and if 𝑥(𝑡) satisfies 𝑥(𝛼)=𝑥(𝛼)=𝑥(𝛽)=𝑥(𝛽)=0, then 𝛽𝛼||||4𝑞(𝑡)𝑑𝑡≥(𝛽−𝛼)2.(1.7) In [4], the author proved that if 𝑥(𝑡) is a solution of (1.5), which satisfies 𝑥(𝛼)=𝑥(𝛽)=𝑥(𝛼)=𝑥(𝛽)=0, then𝛽𝛼||||𝑞(𝑡)𝑑𝑡≥16(𝛽−𝛼)3.(1.8) In this paper, we are concerned with the following problems for the general equation (1.1):(i)obtain lower bounds for the spacing 𝛽−𝛼, where 𝑥 is a solution of (1.1) that satisfies 𝑥(𝑖)(𝛼)=0 for 𝑖=0,1,2 and 𝑥(𝛽)=0,(ii)obtain lower bounds for the spacing 𝛽−𝛼, where 𝑥 is a solution of (1.1) that satisfies 𝑥(𝑖)(𝛽)=0 for 𝑖=0,1,2 and 𝑥(𝛼)=0,(iii)obtain lower bounds for the spacing 𝛽−𝛼, where 𝑥 is a solution of (1.1) that satisfies 𝑥(𝑖)(𝛼)=0=𝑥(𝑖)(𝛽) for 𝑖=0,1,2.

The main results will be proved in Section 2 by making use of Hardy’s inequality and some generalizations of Opial-Wirtinger-type inequalities involving higher-order derivatives. The results yield conditions for disfocality and disconjugacy. In Section 3, we will discuss some special cases of our results to derive some new results for (1.5) and give some illustrative examples. To the best of the author knowledge, this technique has not been employed before on (1.1). Of particular interest in this paper is when 𝑞 is oscillatory and 𝑟 is a negative function.

2. Main Results

In this section, we will prove the main results by making use of Hardy’s inequality and some Opial-Wirtinger-type inequalities. Throughout the paper, all the functions are assumed to be measurable functions and all the integrals that will appear in the inequalities are finite.

The Hardy inequality [16, 17] of the differential form that we will need in this paper states that, if 𝑦 is absolutely continuous on (𝛼,𝛽), then the following inequality holds𝛽𝛼||||𝑞(𝑡)𝑦(𝑡)𝑛𝑑𝑡1/𝑛≤𝐶𝛽𝛼||𝑦𝑟(𝑡)||(𝑡)𝑚𝑑𝑡1/𝑚,(2.1) where 𝑞,𝑟 the weighted functions are measurable positive functions in the interval (𝛼,𝛽) and 𝑚,𝑛are real parameters that satisfy 0<𝑛≤∞ and 1≤𝑚≤∞. The constant 𝐶 satisfies𝐶≤𝑘(𝑚,𝑛)𝛼(𝛼,𝛽),for1<𝑚≤𝑛,(2.2) where𝛼(𝛼,𝛽)∶=sup𝛼<𝑡<𝛽𝛽𝑡𝑞(𝑡)𝑑𝑡1/𝑛𝑡𝛼𝑟1−𝑚∗(𝑠)𝑑𝑠1/𝑚∗𝛼if𝑦(𝛼)=0,(𝛼,𝛽)∶=sup𝛼<𝑡<𝛽𝑡𝛼𝑞(𝑡)𝑑𝑡1/𝑛𝛽𝑡𝑟1−𝑚∗(𝑠)𝑑𝑠1/𝑚∗if𝑦(𝛽)=0,(2.3) and 𝑚∗=𝑚/(𝑚−1). Note that the inequality (2.1) has an immediate application to the case when 𝑦(𝛼)=𝑦(𝛽)=0. In this case, we see that (2.1) is satisfied if and only if𝛼(𝛼,𝛽)=sup(𝑐,𝑑)⊂(𝛼,𝛽)𝑑𝑐𝑞(𝑡)𝑑𝑡1/𝑛×min𝑐𝛼𝑟1−𝑚∗(𝑠)𝑑𝑠1/𝑚∗,𝛽𝑑𝑟1−𝑚∗(𝑠)𝑑𝑠1/𝑚∗(2.4) exists and is finite. The constant 𝑘(𝑚,𝑛) in (2.2) appears in various forms. For example,𝑘(𝑚,𝑛)∶=𝑚1/𝑚𝑚∗1/𝑚∗,𝑛𝑘(𝑚,𝑛)∶=1+𝑚∗1/𝑛𝑚1+∗𝑛1/𝑚∗,𝑢∗=𝑢𝑢−1𝑘(𝑚,𝑛)∶=Γ(𝑛/𝑠)Γ(1+(1/𝑠))Γ((𝑛−1)/𝑠)𝑠/𝑛𝑛,𝑠=.𝑚−1(2.5) In the following, we present the Opial-Wirtinger-type inequalities that we will need in the proof of the main results.

Theorem 2.1 ([18, Theorem  3.9.1]). Assume that the functions 𝜗 and 𝜙 are nonnegative and measurable on the interval (𝛼,𝛽),𝑚,𝑛 are real numbers such that 𝜇/𝑚>1,𝑥(𝑡)∈𝐶(𝑛−1)[𝛼,𝛽] is such that 𝑥(𝑖)(𝛼)=0, 0≤𝑘≤𝑖≤𝑛−1(𝑛≥1), and 𝑥(𝑛−1)(𝑡) absolutely continuous on (𝛼,𝛽). Then 𝛽𝛼||𝑥𝜙(𝑡)(𝑘)||(𝑡)𝑙||𝑥(𝑛)||(𝑡)𝑚𝑑𝑡≤𝐾1(𝛼,𝛽)𝛽𝛼||𝑥𝜗(𝑡)(𝑛)||(𝑡)𝜇𝑑𝑡(𝑙+𝑚)/𝜇,(2.6) where 𝐾1(𝛼,𝛽)=(𝑚/(𝑙+𝑚))𝑚/𝜇(𝐾!)𝑙𝛽𝛼𝜙𝜇(𝑡)𝜗𝑚(𝑡)1/(𝜇−𝑚)𝑃1,𝑘(𝑡)𝑙(𝜇−1)/(𝜇−𝑚)𝑑𝑡(𝜇−𝑚)/𝜇,𝐾=(𝑛−𝑘−1),𝑃1,𝑘(𝑡)∶=𝑡𝛼(𝑡−𝑠)(𝑛−𝑘−1)𝜇/(𝜇−1)(𝜗(𝑠))−1/(𝜇−1)𝑑𝑠.(2.7) If we replace 𝑥(𝑖)(𝛼)=0 by 𝑥(𝑖)(𝛽)=0,0≤𝑘≤𝑖≤𝑛−1(𝑛≥1), then (2.6) holds where 𝐾1 is replaced by 𝐾2, which is given by 𝐾2(𝛼,𝛽)=(𝑚/(𝑙+𝑚))𝑚/𝜇(𝐾!)𝑙𝛽𝛼𝜙𝜇(𝑡)𝜗𝑚(𝑡)1/(𝜇−𝑚)𝑃2,𝑘(𝑡)𝑙(𝜇−1)/(𝜇−𝑚)𝑑𝑡(𝜇−𝑚)/𝜇,(2.8) where 𝑃2,𝑘(𝑡)∶=𝛽𝑡(𝑠−𝑡)(𝑛−𝑘−1)𝜇/(𝜇−1)(𝜗(𝑠))−1/(𝜇−1)𝑑𝑠.(2.9)

Note that the inequality (2.6) has an immediate application to the case when 𝑥(𝑖)(𝛼)=𝑥(𝑖)(𝛽)=0 for 0≤𝑖≤𝑛−1. In this case, we will assume that there exists 𝜏∈(𝛼,𝛽) such that𝜏𝛼(𝜏−𝑠)(𝑛−𝑘−1)𝜇/(𝜇−1)(𝜗(𝑠))−1/(𝜇−1)𝑑𝑠=𝛽𝜏(𝑠−𝜏)(𝑛−𝑘−1)𝜇/(𝜇−1)(𝜗(𝑠))−1/(𝜇−1)𝑑𝑠,(2.10) and we denote by 𝑃(𝛼,𝛽). This gives us the following theorem.

Theorem 2.2. Assume that the functions 𝜗 and 𝜙 are nonnegative and measurable on the interval (𝛼,𝛽),𝑚,𝑛 are real numbers such that 𝜇/𝑚>1,𝑥(𝑡)∈𝐶(𝑛−1)[𝛼,𝛽] is such that 𝑥(𝑖)(𝛼)=𝑥(𝑖)(𝛽)=0,0≤𝑘≤𝑖≤𝑛−1(𝑛≥1), and 𝑥(𝑛−1)(𝑡) absolutely continuous on (𝛼,𝛽). Then 𝛽𝛼||𝑥𝜙(𝑡)(𝑘)||(𝑡)𝑙||𝑥(𝑛)||(𝑡)𝑚𝑑𝑡≤𝐾(𝛼,𝛽)𝛽𝛼||𝑥𝜗(𝑡)(𝑛)||(𝑡)𝜇𝑑𝑡(𝑙+𝑚)/𝜇,(2.11) where 𝐾(𝛼,𝛽) is defined by 𝑚𝐾(𝛼,𝛽)=𝑙+𝑚𝑚/𝜇[]𝑃(𝛼,𝛽)𝑙(𝜇−1)/𝜇(𝐾!)𝑙𝛽𝛼𝜙𝜇(𝑡)𝜗𝑚(𝑡)1/(𝜇−𝑚)𝑑𝑡(𝜇−𝑚)/𝜇.(2.12)

Theorem 2.3 ([18, Theorem  3.9.2]). Let 𝑟𝑘,0≤𝑘≤𝑛−1(𝑛≥1) be nonnegative numbers such that ∑𝜎=𝑛−1𝑘=0𝑟𝑘>0  and 𝜗 and 𝜙 are nonnegative and measurable on the interval (𝛼,𝛽). Further, let 𝑥(𝑡)∈𝐶(𝑛−1)[𝛼,𝛽] be such that 𝑥(𝑖)(𝛼)=0,0≤𝑖≤𝑛−1(𝑛≥1) and 𝑥(𝑛−1)(𝑡) absolutely continuous on (𝛼,𝛽). Then 𝛽𝛼𝜙(𝑡)𝑛𝑘=0||𝑥(𝑘)||(𝑡)𝑟𝑘𝑑𝑡≤𝐾∗1(𝛼,𝛽)𝛽𝛼||𝑥𝜗(𝑡)(𝑛)||(𝑡)𝑟𝑑𝑡(𝜎+𝑟𝑛)/𝑟,(2.13) where 𝐾∗11(𝛼,𝛽)=Ω𝑟𝑛𝜎+𝑟𝑛𝑟𝑛/𝑟𝛽𝛼𝜙𝑟(𝑡)𝜗𝑟𝑛(𝑡)1/(𝑟−𝑟𝑛)𝑛−1𝑘=0𝑃∗1,𝑘(𝑡)𝑟𝑘(𝑟−1)/(𝑟−𝑟𝑛)𝑑𝑡(𝑟−𝑟𝑛)/𝑟,Ω=𝑛−1𝑘=0((𝑛−𝑘−1)!)𝑟𝑘,𝑃∗1,𝑘(𝑡)∶=𝑡𝛼(𝑡−𝑠)(𝑛−𝑘−1)𝑟/(𝑟−1)(𝜗(𝑠))−1/(𝑟−1)𝑑𝑠.(2.14) If we replace 𝑥(𝑖)(𝛼)=0 by 𝑥(𝑖)(𝛽)=0,0≤𝑖≤𝑛−1(𝑛≥1), then (2.13) holds where 𝐾∗1 is replaced by 𝐾∗2,  which is given by 𝐾∗21(𝛼,𝛽)=Ω𝑟𝑛𝜎+𝑟𝑛𝑟𝑛/𝑟𝛽𝛼𝜙𝑟(𝑡)𝜗𝑟𝑛(𝑡)1/(𝑟−𝑟𝑛)𝑛−1𝑘=0𝑃∗2,𝑘(𝑡)𝑟𝑘(𝑟−1)/(𝑟−𝑟𝑛)𝑑𝑡(𝑟−𝑟𝑛)/𝑟,(2.15) where 𝑃∗2,𝑘(𝑡)∶=𝛽𝑡(𝑠−𝑡)(𝑛−𝑘−1)𝑟/(𝑟−1)(𝜗(𝑠))−1/(𝑟−1)𝑑𝑠.(2.16)

Note that the inequality (2.13) has an immediate application to the case when 𝑥(𝑖)(𝛼)=𝑥(𝑖)(𝛽)=0 for 0≤𝑖≤𝑛−1. In this case, we will assume that there exists 𝜏∈(𝛼,𝛽) such that𝑛−1𝑘=0𝑃∗1,𝑘(𝜏)𝑟𝑘(𝑟−1)/(𝑟−𝑟𝑛)=𝑛−1𝑘=0𝑃∗2,𝑘(𝜏)𝑟𝑘(𝑟−1)/(𝑟−𝑟𝑛),(2.17) denoted by 𝑃∗(𝛼,𝛽). In this case the inequality (2.13) is satisfied but the constant 𝐾∗1(𝛼,𝛽) is replaced by 𝐾∗(𝛼,𝛽), which is defined by𝐾∗𝑟(𝛼,𝛽)=𝑛𝜎+𝑟𝑛𝑟𝑛/𝑟𝑃∗(𝛼,𝛽)(𝑟−𝑟𝑛)/𝑟Ω𝛽𝛼𝜙𝑟(𝑡)𝜗𝑟𝑛(𝑡)1/(𝑟−𝑟𝑛)𝑑𝑡(𝑟−𝑟𝑛)/𝑟.(2.18)

The Wirtinger-type inequality and its general forms have been studied in the literature in various modifications both in the continuous and in the discrete setting. It has an extensive applications on partial differential and difference equations, harmonic analysis, approximations, number theory, optimization, convex geometry, spectral theory of differential and difference operators, and others (see [19]).

In the following, we present a special case of the Wirtinger-type inequality that has been proved by Agarwal et al. in [20] and will be need in the proof the main results.

Theorem 2.4. For 𝕀=[𝛼,𝛽],𝛾≥1 is a positive integer and a positive function 𝜆∈𝐶1(𝕀) with either 𝜆(𝑡)>0 or 𝜆(𝑡)<0 on 𝕀; we have 𝛽𝛼𝜆𝛾+1(𝑡)||𝜆||(𝑡)𝛾||𝑦||(𝑡)𝛾+11𝑑𝑡≥(𝛾+1)𝛾+1𝛽𝛼||𝜆||||||(𝑡)𝑦(𝑡)𝛾+1𝑑𝑡,(2.19) for any 𝑦∈𝐶1(𝕀) with 𝑦(𝛼)=0=𝑦(𝛽).

Remark 2.5. It is clear that Theorem 2.4 is satisfied for any function 𝑦 that satisfies the assumptions of theorem. So if 𝑦(𝑡)=𝑥(𝑡) with 𝑥(𝛼)=0=𝜆(𝛽), 𝑥(𝛽)=0=𝜆(𝛼), or 𝑥(𝛼)=0=𝑥(𝛽) and 𝑝(𝑡)=𝜆(𝑡), we have the following inequality, which gives a relation between 𝑥(𝑡) and 𝑥(𝑡) on the interval [𝛼,𝛽].

Corollary 2.6. For 𝕀=[𝛼,𝛽] and 𝛾≥1 being a positive integer, then we have 𝛽𝛼||||||𝑥𝑟(𝑡)||(𝑡)𝛾+11𝑑𝑡≥(𝛾+1)𝛾+1𝛽𝛼||||||𝑥𝑝(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.20) for any 𝑥∈𝐶3(𝕀) with 𝑥(𝛼)=0=𝑟(𝛽),𝑥(𝛽)=0=𝑟(𝛼), or 𝑥(𝛼)=0=𝑥(𝛽), where 𝑟(𝑡) and 𝑝(𝑡) satisfy the equation 𝑟𝜆(𝑡)(𝑡)𝛾−(𝛾+1)𝑝(𝑡)𝜆𝛾(𝑡)=0,(2.21) for any function 𝜆(𝑡) satisfing 𝜆(𝑡)≠0.

For illustration, we apply the inequality (2.20) with 𝑥(𝑡)=sin𝑡 in the interval [0,𝜋]. If 𝑝(𝑡)=1 and 𝛾=1 and by choosing 𝑟(𝑡)=𝑡2, we see that (2.21) is satisfied when 𝜆(𝑡)=𝑡. So one can see that𝜋0𝑡2cos21𝑡𝑑𝑡≃5.9531>0.39270≃4𝜋0sin2𝑡𝑑𝑡.(2.22) Note also that (2.21) holds if one chooses 𝑟(𝑡)=𝑝(𝑡)=1, where in this case𝜆(𝑡)=exp𝛾+1𝛾1/(𝛾+1)𝑡.(2.23) Now, we are ready to state and prove the main results when 𝑟(𝑡)>0. For simplicity, we introduce the following notations:Φ1𝑄,𝑟,𝑃1,0∶=2−𝛾Λ𝛽𝛼||||𝑄(𝑡)(𝛾+1)/𝛾𝑟1/𝛾𝑃(𝑡)𝛾1,0(𝑡)𝑑𝑡𝛾/(𝛾+1),(2.24) where Λ=(1/(𝛾+1))1/(𝛾+1),𝑃1,0∫(𝑡)=𝑡𝛼(𝑡−𝑠)2(𝛾+1)/𝛾𝑟−1/𝛾(𝑠)𝑑𝑠,Φ2𝑄,𝑟,𝑃2,0∶=2−𝛾Λ𝛽𝛼||||𝑄(𝑡)(𝛾+1)/𝛾𝑟1/𝛾𝑃(𝑡)𝛾2,0(𝑡)𝑑𝑡𝛾/(𝛾+1),(2.25) where 𝑃2,0∫(𝑡)=𝛽𝑡(𝑠−𝑡)2(𝛾+1)/𝛾𝑟−1/𝛾(𝑠)𝑑𝑠,Ψ1𝑄,𝑟,𝑃∗1,0𝑃∗1,1∶=Λ𝛽𝛼||||𝑄(𝑡)(𝛾+1)/𝛾𝑟1/𝛾𝑃(𝑡)∗1,0(𝑡)𝛾−1𝑃∗1,1(𝑡)𝑑𝑡𝛾/(𝛾+1),(2.26) where 𝑃∗1,0∫(𝑡)=𝑡𝛼(𝑡−𝑠)(𝛾+1)/𝛾𝑟−1/𝛾(𝑠)𝑑𝑠,𝑃∗1,1∫(𝑡)=𝑡𝛼𝑟−1/𝛾(𝑠)𝑑𝑠, andΨ2𝑄,𝑟,𝑃∗2,0𝑃∗2,1∶=Λ𝛽𝛼||||𝑄(𝑡)(𝛾+1)/𝛾𝑟1/𝛾𝑃(𝑡)∗2,0(𝑡)𝛾−1𝑃∗2,1(𝑡)𝑑𝑡𝛾/(𝛾+1),(2.27) where 𝑃∗2,0∫(𝑡)=𝛽𝑡(𝑠−𝑡)(𝛾+1)/𝛾𝑟−1/𝛾(𝑠)𝑑𝑠,𝑃∗2,1∫(𝑡)=𝛽𝑡𝑟−1/𝛾(𝑠)𝑑𝑠.

Remark 2.7. Note that when 𝛾=1, then Ψ1(𝑄,𝑟,𝑃∗1,0𝑃∗1,1) and Ψ2(𝑄,𝑟,𝑃∗2,0𝑃∗2,1) become Ψ1(𝑄,𝑟,𝑃∗1,1) and Ψ2(𝑄,𝑟,𝑃∗2,1).

Theorem 2.8. Suppose that 𝑥 is a nontrivial solution of (1.1). If 𝑥(𝑖)(𝛼)=0, for 𝑖=0,1,2  and 𝑥(𝛽)=0, then Φ1𝑄,𝑟,𝑃1,0+𝛾(𝛾+1)𝛾+1Ψ1𝑄,𝑟,𝑃∗1,0𝑃∗1,1≥1,(2.28) where ∫𝑄(𝑡)=𝛽𝑡𝑞(𝑠)𝑑𝑠. If 𝑥(𝑖)(𝛽)=0, for 𝑖=0,1,2 and 𝑥(𝛼)=0, then Φ2𝑄,𝑟,𝑃2,0+𝛾(𝛾+1)𝛾+1Ψ2𝑄,𝑟,𝑃∗2,0𝑃∗2,1≥1,(2.29) where ∫𝑄(𝑡)=𝑡𝛼𝑞(𝑠)𝑑𝑠.

Proof. We prove (2.28). Multiplying (1.1) by 𝑥(𝑡) and integrating by parts, we have 𝛽𝛼𝑥𝑟(𝑡)(𝑡)𝛾𝑥𝑥(𝑡)𝑑𝑡=𝑟(𝑡)(𝑡)𝛾𝑥||(𝑡)𝛽𝛼−𝛽𝛼𝑥𝑟(𝑡)(𝑡)𝛾+1𝑑𝑡=−𝛽𝛼𝑞(𝑡)𝑥(𝑡)𝑥𝛾(𝑡)𝑑𝑡.(2.30) Using the boundary conditions 𝑥(𝛼)=𝑥(𝛽)=0 and the assumption ∫𝑄(𝑡)=𝛽𝑡𝑞(𝑠)𝑑𝑠, we have 𝛽𝛼𝑥𝑟(𝑡)(𝑡)𝛾+1𝑑𝑡=𝛽𝛼𝑞(𝑡)𝑥(𝑡)𝑥𝛾(𝑡)𝑑𝑡=−𝛽𝛼𝑄(𝑡)𝑥(𝑡)𝑥𝛾(𝑡)𝑑𝑡.(2.31) Integrating by parts the right-hand side, we see that 𝛽𝛼𝑄(𝑡)𝑥(𝑡)𝑥𝛾(𝑡)𝑑𝑡=𝑄(𝑡)𝑥(𝑡)𝑥𝛾||(𝑡)𝛽𝛼−𝛾𝛽𝛼𝑄(𝑡)𝑥𝛾−1(𝑡)𝑥(𝑡)𝑥−(𝑡)𝑑𝑡𝛽𝛼𝑄(𝑡)𝑥𝛾(𝑡)𝑥(𝑡)𝑑𝑡.(2.32) Using the boundary conditions 𝑥(𝛽)=𝑥(𝛼)=0, we have 𝛽𝛼𝑄(𝑡)𝑥(𝑡)𝑥𝛾(𝑡)𝑑𝑡=−𝛾𝛽𝛼𝑄(𝑡)𝑥𝛾−1(𝑡)𝑥(𝑡)𝑥(𝑡)𝑑𝑡−𝛽𝛼𝑄(𝑡)𝑥𝛾(𝑡)𝑥(𝑡)𝑑𝑡.(2.33) Substituting (2.33) into (2.31), we obtain 𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡≤𝛾𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾−1||𝑥||||𝑥(𝑡)||+(𝑡)𝑑𝑡𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾||𝑥(||𝑡)𝑑𝑡.(2.34) Applying the inequality (2.6) on the integral 𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾||𝑥||(𝑡)𝑑𝑡,(2.35) with 𝜙(𝑡)=|𝑄(𝑡)|,𝜗(𝑡)=𝑟(𝑡),𝑚=1,𝑘=0,𝑙=𝛾,𝑛=3, and 𝜇=𝛾+1, we get (note that 𝑥(𝑖)(𝛼)=0, for 𝑖=0,1,2) that 𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾||𝑥||(𝑡)𝑑𝑡≤Φ1𝑄(𝑡),𝑟,𝑃1,0𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.36) where Φ1(𝑄,𝑟,𝑃1,0) is defined as in (2.24). Applying the inequality (2.13) on the integral 𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾−1||𝑥||||𝑥(𝑡)||(𝑡)𝑑𝑡,(2.37) with 𝜙(𝑡)=|𝑄(𝑡)|,𝜗(𝑡)=𝑟(𝑡),𝑛=2,𝑟0=𝛾−1,𝑟1=1,𝑟2=1,𝜎+𝑟2=𝛾+1, and 𝑟=𝛾+1, we see that 𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾−1||𝑥||||𝑥(𝑡)||(𝑡)𝑑𝑡≤Ψ∗1𝑄,𝑟,𝑃∗1,0𝑃∗1,1𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.38) where Ψ1(𝑄,𝑟,𝑃∗1,0𝑃∗1,1) is defined as in (2.26). Applying the Wirtinger inequality (2.20) on the integral 𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.39) where 𝑥(𝛼)=0=𝑥(𝛽),we see that 𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡≤(𝛾+1)𝛾+1𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.40) where 𝑟(𝑡) satisfies (2.21) for any positive function 𝜆(𝑡) and 𝑝(𝑡) is replaced by 𝑟(𝑡). Substituting (2.40) into (2.38), we have 𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾−1||𝑥||||𝑥(𝑡)||Ψ(𝑡)𝑑𝑡≤1𝑄,𝑟,𝑃∗1,0𝑃∗1,1(𝛾+1)−(𝛾+1)𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡.(2.41) Substituting (2.36) and (2.41) into (2.34) and cancelling the term ∫𝛽𝛼𝑟(𝑡)|𝑥(𝑡)|𝛾+1𝑑𝑡, we have Φ1𝑄,𝑟,𝑃1,0+𝛾(𝛾+1)𝛾+1Ψ1𝑄,𝑟,𝑃∗1,0𝑃∗1,1≥1,(2.42) which is the desired inequality (2.28). The proof of (2.29) is similar by using the integration by parts and Φ1(𝑄,𝑟,𝑃1,0) is replaced by Φ2(𝑄,𝑟,𝑃2,0), which is defined as in (2.25), and Ψ1(𝑄,𝑟,𝑃∗1,0𝑃∗1,1) is replaced by Ψ2(𝑄,𝑟,𝑃∗2,0𝑃∗2,1), which is defined as in (2.27). The proof is complete.

In the following, we apply the Hardy inequality (2.1) on the term𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.43) by replacing 𝑦(𝑡) by 𝑥(𝑡) and use the assumption 𝑥(𝛼)=0=𝑥(𝛽). In this case, we see that𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡≤𝐶Γ𝛼(𝛼,𝛽,𝑟)𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.44) where𝐶Γ∶=Γ1/𝛾(𝛾)Γ−1/𝛾𝛾1+Γ𝛾+1−1/𝛾𝛾2,𝛾+1(2.45)𝛼(𝛼,𝛽,𝑟)∶=sup(𝑐,𝑑)⊂(𝛼,𝛽)𝑑𝑐𝑟(𝑡)𝑑𝑡1/(𝛾+1)×min𝑐𝛼𝑑𝑠𝑟1/𝛾(𝑠)𝛾/(𝛾+1),𝛽𝑑𝑑𝑠𝑟1/𝛾(𝑠)𝛾/(𝛾+1).(2.46) This implies that𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾−1||𝑥||||𝑥(𝑡)||(𝑡)𝑑𝑡≤𝐶Γ𝛼(𝛼,𝛽,𝑟)Ψ1𝑄,𝑟,𝑃∗1,0𝑃∗1,1×𝛽𝛼||𝑥𝑟(𝑡)(||𝑡)𝛾+1𝑑𝑡.(2.47) Proceeding as in the proof of Theorem 2.8 and using (2.47) instead of (2.41), we have the following result.

Theorem 2.9. Assume that 𝑄(𝑡)=𝑞(𝑡) and 𝑥 is a nontrivial solution of (1.1). If 𝑥(𝑖)(𝛼)=0, for 𝑖=0,1,2 and 𝑥(𝛽)=0, then Φ1𝑄,𝑟,𝑃1,0+𝛾𝐶Γ𝛼(𝛼,𝛽,𝑟)Ψ1𝑄,𝑟,𝑃∗1,0𝑃∗1,1≥1,(2.48) where ∫𝑄(𝑡)=𝛽𝑡𝑞(𝑠)𝑑𝑠. If 𝑥(𝑖)(𝛽)=0, for 𝑖=0,1,2 and 𝑥(𝛼)=0, then Φ2𝑄,𝑟,𝑃2,0+𝛾𝐶Γ𝛼(𝛼,𝛽,𝑟)Ψ2𝑄,𝑟,𝑃∗2,0(𝑡)𝑃∗2,1≥1,(2.49) where ∫𝑄(𝑡)=𝑡𝛼𝑞(𝑠)𝑑𝑠.

In the following, we will apply a new inequality to establish a new result but on the interval [0,𝛽]. The inequality that we will apply is given in the following theorem.

Theorem 2.10 ([18, Theorem  3.7.4]). Let 𝑟𝑘,0≤𝑘≤𝑛−1(𝑛≥1) be nonnegative numbers and 𝜗 and 𝜙 nonnegative and measurable on the interval (0,𝛽).   Let 𝑥(𝑡)∈𝐶(𝑛−1)[0,𝛽] be such that 𝑥(𝑖)(0)=0,0≤𝑖≤𝑛−1(𝑛≥1),𝑥(𝑛−1)(𝑡), is absolutely continuous on (0,𝛽), and let 𝑠1, 𝑠2 be constants greater than 1,1/𝑠1+1/𝑠∗1=1,1/𝑠2+1/𝑠∗2=1  and 𝜇 a constant such that 𝜇>𝑠2. Further assume that 𝜎=𝑛−1𝑘=0𝑟𝑘>0,ϝ(𝜙,𝜗)∶=𝛽01𝜗(𝑡)𝑠2/𝜇𝑑𝑡𝜎/𝑠∗2𝛽0𝜙𝑠∗1(𝑡)𝑑𝑡1/𝑠∗1.(2.50) Then the following inequality holds 𝛽0𝜙(𝑡)𝑛−1𝑘=0||𝑥(𝑘)||(𝑡)𝑟𝑘𝑑𝑡≤𝐶𝛽𝜆𝛽0||𝑥𝜗(𝑡)(𝑛)||(𝑡)𝜇𝑑𝑡𝜎/𝜇,(2.51) where ∑𝜆=𝑛−1𝑘=0(𝑛−𝑘−1)𝑟𝑘+𝜎𝛿+1/𝑠1,𝛿=(𝜇−𝑠2)/𝜇𝑠2 and 𝐶∶=ϝ(𝜙,𝜗)𝑛−1𝑘=0[]𝐾!−𝑟𝑘[](𝑛−𝑘−1/𝛿)+1−𝑟𝑘𝛿∑𝑛−1𝑘=0𝐾𝑟𝑘𝑠1+𝜎𝑠1𝛿+11/𝑠1.(2.52)

Now, by applying the inequality (2.51) on the term𝛾𝛽0||||||𝑥𝑄(𝑡)𝛾−1||||𝑥(𝑡)||||𝑥(𝑡)||(𝑡)𝑑𝑡,(2.53) with 𝜙(𝑡)=|𝑄(𝑡)|,𝜗(𝑡)=𝑟(𝑡),𝑛=3,𝑟0=𝛾−1,𝑟1=1,𝑟2=1,𝜎=𝛾+1,𝜇=𝛾+1,𝑠1=𝑠2=𝛾, and 𝛼=0, we obtain𝛾𝛽0||||||𝑥𝑄(𝑡)𝛾−1||||𝑥(𝑡)||||𝑥(𝑡)||(𝑡)𝑑𝑡≤𝛾𝐿𝛾ϝ(𝑟,𝑄)𝛽2𝛾−1+2/𝛾𝛽0||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.54) where𝐿𝛾1∶=2𝛾−1𝛾4𝛾+11/𝛾[]2𝛾(𝛾+1)+1−(𝛾−1)/𝛾(𝛾+1)[]𝛾(𝛾+1)+11/𝛾(𝛾+1),ϝ(𝑟,𝑄)∶=𝛽01𝑟(𝑡)𝛾/(𝛾+1)𝑑𝑡(𝛾+1)(𝛾−1)/𝛾𝛽0||||𝑄(𝑡)𝛾/𝛾−1𝑑𝑡(𝛾−1)/𝛾.(2.55) Proceeding as in the proof of Theorem 2.8 by using (2.54) instead of (2.38), we get the following result.

Theorem 2.11. Assume that 𝑄(𝑡)=𝑞(𝑡) and 𝑥 is a nontrivial solution of (1.1). If 𝑥(𝑖)(0)=0, for 𝑖=0,1,2 and 𝑥(𝛽)=0, then Φ1𝑄,𝑟,𝑃1,0+𝛾𝐿𝛾ϝ(𝑟,𝑄)𝛽2𝛾−1+2/𝛾≥1,(2.56) where ∫𝑄(𝑡)=𝛽𝑡𝑞(𝑠)𝑑𝑠 and 𝐿𝛾,ϝ(𝑟,𝑄) are defined as in (2.55).

Remark 2.12. Note that in the proof of Theorem 2.11, we do not require additional inequalities like the Hardy inequality or the Wirtinger inequality. So it will be interesting to extend the proof of Theorem 2.10 to cover the boundary conditions 𝑥(𝑖)(𝛼)=𝑥(𝑖)(𝛽)=0 and replace the interval [0,𝛽] by [𝛼,𝛽].

In the following, we will assume that (2.10) and (2.17) hold. First, we assume that (2.10) holds and there exists 𝜏∈(𝛼,𝛽) such that𝜏𝛼(𝜏−𝑠)2(𝛾+1)/𝛾𝑟1/𝛾(𝑠)𝑑𝑠=𝛽𝜏(𝑠−𝜏)2(𝛾+1)/𝛾𝑟1/𝛾(𝑠)𝑑𝑠,(2.57) denoted by 𝑃(𝛼,𝛽). In this case, we see thatΦ(𝑄,𝑟)∶=Φ1(𝑄,𝑟,𝑃(𝑡))=Φ2(𝑄,𝑟,𝑃(𝑡)),(2.58) where in this case Φ(𝑄,𝑟) is given byΛΦ(𝑄,𝑟)∶=2𝛾𝛽𝛼||||𝑄(𝑡)(𝛾+1)/𝛾𝑟1/𝛾𝑃(𝑡)𝛾(𝛼,𝛽)𝑑𝑡𝛾/(𝛾+1).(2.59) Second, we assume that (2.17) holds and there exists 𝜏∈(𝛼,𝛽) such that𝑃∗1,0(𝛼,𝜏)𝛾−1𝑃∗1,1𝑃(𝛼,𝜏)=∗2,0(𝜏,𝛽)𝛾−1𝑃∗2,1(𝜏,𝛽),(2.60) denoted by 𝑃∗(𝛼,𝛽). In this case, we get thatΨ(𝑟,𝑄)=Ψ∗1(𝑟,𝑄)=Ψ∗2(𝑟,𝑄),(2.61) where Ψ(𝑟,𝑄) is given byΨ(𝑟,𝑄)∶=Λ𝛽𝛼||||𝑄(𝑡)(𝛾+1)/𝛾𝑟1/𝛾𝑃(𝑡)∗(𝛼,𝛽)𝑑𝑡𝛾/(𝛾+1).(2.62) Note that when 𝑟(𝑡)=1, we have that the condition (2.57) is satisfied when (𝜏−𝛼)(3𝛾+2)/𝛾=(𝛽−𝜏)(3𝛾+2)/𝛾. This in fact is satisfied when 𝜏=(𝛼+𝛽)/2. In this case, we see that2𝑃(𝛼,𝛽)∶=−(3𝛾+2)/𝛾𝛾(3𝛾+2)(𝛽−𝛼)(3𝛾+2)/𝛾.(2.63) Also, when 𝑟(𝑡)=1, then 𝑃∗(𝛼,𝛽) becomes𝑃∗𝛾(𝛼,𝛽)∶=2𝛾+1𝛾−1𝛽−𝛼2((2𝛾+1)(𝛾−1)+𝛾)/𝛾.(2.64)

Theorem 2.13. Assume that 𝑄(𝑡)=𝑞(𝑡) and 𝑥(𝑡) is a nontrivial solution of (1.1). If 𝑥(𝑖)(𝛼)=0=𝑥(𝑖)(𝛽), for 𝑖=0,1,2, then Φ(𝑄,𝑟)+𝛾𝐶Γ𝛼(𝛼,𝛽,𝑟)Ψ(𝑟,𝑄)≥1,(2.65) where 𝐶Γ and 𝛼(𝛼,𝛽,𝑟)  are defined as in (2.45) and (2.46) and Φ(𝑄,𝑟)  and Ψ(𝑄,𝑟) are defined as in (2.59) and (2.62).

Proof. Multiplying (1.1) by 𝑥(𝑡) and integrating by parts, we have 𝛽𝛼𝑥𝑟(𝑡)(𝑡)𝛾𝑥𝑥(𝑡)𝑑𝑡=𝑟(𝑡)(𝑡)𝛾𝑥||(𝑡)𝛽𝛼−𝛽𝛼𝑥𝑟(𝑡)(𝑡)𝛾+1𝑑𝑡=−𝛽𝛼𝑞(𝑡)𝑥𝛾(𝑡)𝑥(𝑡)𝑑𝑡.(2.66) Using the boundary conditions 𝑥(𝛼)=𝑥(𝛽)=0, we get that 𝛽𝛼𝑥𝑟(𝑡)(𝑡)𝛾+1𝑑𝑡=𝛽𝛼𝑞(𝑡)𝑥𝛾(𝑡)𝑥(𝑡)𝑑𝑡=𝛽𝛼𝑄(𝑡)𝑥𝛾(𝑡)𝑥(𝑡)𝑑𝑡.(2.67) Integrating by parts the right-hand side, we see that 𝛽𝛼𝑄(𝑡)𝑥(𝑡)𝑥𝛾(𝑡)𝑑𝑡=𝑄(𝑡)𝑥(𝑡)𝑥𝛾||(𝑡)𝛽𝛼−𝛾𝛽𝛼𝑄(𝑡)𝑥𝛾−1(𝑡)𝑥(𝑡)𝑥−(𝑡)𝑑𝑡𝛽𝛼𝑄(𝑡)𝑥𝛾(𝑡)𝑥(𝑡)𝑑𝑡.(2.68) Using the boundary conditions 𝑥(𝛽)=𝑥(𝛼)=0, we see that 𝛽𝛼𝑄(𝑡)𝑥(𝑡)𝑥𝛾(𝑡)𝑑𝑡=−𝛾𝛽𝛼𝑄(𝑡)𝑥𝛾−1(𝑡)𝑥(𝑡)𝑥−(𝑡)𝑑𝑡𝛽𝛼𝑄(𝑡)𝑥𝛾(𝑡)𝑥(𝑡)𝑑𝑡.(2.69) Substituting (2.69) into (2.67), we have 𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡≤𝛾𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾−1||𝑥||||𝑥(𝑡)||+(𝑡)𝑑𝑡𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾||𝑥(||𝑡)𝑑𝑡.(2.70) Applying the inequality (2.6) on the integral 𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾||𝑥||(𝑡)𝑑𝑡,(2.71) with 𝜙(𝑡)=|𝑄(𝑡)|,𝜗(𝑡)=𝑟(𝑡),𝑚=1,𝑘=0,𝑙=𝛾,𝑛=3, and 𝜇=𝛾+1, we get (noting that 𝑥(𝑖)(𝛼)=𝑥(𝑖)(𝛽)=0, for 𝑖=0,1,2) that 𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾||𝑥||(𝑡)𝑑𝑡≤Φ(𝑄,𝑟)𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.72) where Φ(𝑄,𝑟) is defined as in (2.59). Applying the inequality (2.13) on the integral 𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾−1||𝑥||||𝑥(𝑡)||(𝑡)𝑑𝑡,(2.73) with 𝜙(𝑡)=|𝑄(𝑡)|,𝜗(𝑡)=𝑟(𝑡),𝑛=2,𝑟0=𝛾−1,𝑟1=1,𝑟2=1,𝜎+𝑟2=𝛾+1, and 𝑟=𝛾+1, we see that 𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾−1||𝑥||||𝑥(𝑡)||(𝑡)𝑑𝑡≤Ψ(𝑄,𝑟)𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.74) where Ψ(𝑄,𝑟) is defined as in (2.62). Applying the Hardy inequality (2.1) on the term 𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡(2.75) with 𝑦(𝑡)=𝑥(𝑡) where 𝑥(𝛼)=0=𝑥(𝛽), we see that 𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡≤𝐶Γ𝛼(𝛼,𝛽,𝑟)𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.76) where 𝐶Γ and 𝛼(𝛼,𝛽,𝑟) are defined as in (2.45) and (2.46). Substituting (2.76) into (2.74), we have 𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾−1||𝑥||||𝑥(𝑡)||(𝑡)𝑑𝑡≤𝛾𝐶Γ𝛼(𝛼,𝛽,𝑟)Ψ(𝑄,𝑟)𝛽𝛼||𝑥𝑟(𝑡)||(𝑡)𝛾+1𝑑𝑡.(2.77) Substituting (2.72) and (2.77) into (2.70) and cancelling the term ∫𝛽𝛼𝑟(𝑡)|𝑥(𝑡)|𝛾+1𝑑𝑡, we have Φ(𝑄,𝑟)+𝛾𝐶Γ𝛼(𝛼,𝛽,𝑟)Ψ(𝑄,𝑟)Ψ(𝑄,𝑟)≥1,(2.78) which is the desired inequality (2.28). The proof is complete.

In the proof of Theorem 2.13 if we apply the Wirtinger inequality (2.20) instead of the Hardy inequality (2.1), then we have the following result.

Theorem 2.14. Assume that 𝑟𝑄(𝑡)=𝑞(𝑡) and 𝑥(𝑡) is a nontrivial solution of (1.1). If 𝑥(𝑖)(𝛼)=0=𝑥(𝑖)(𝛽), for 𝑖=0,1,2, then Φ(𝑄,𝑟)+𝛾(𝛾+1)𝛾+1Ψ(𝑄,𝑟)≥1,(2.79) where Φ(𝑄,𝑟) and Ψ(𝑄,𝑟) are defined as in (2.59) and (2.62).

Next, in the following, we establish some results which allow us to consider the case when 𝑟(𝑡)<0. For simplicity, we denote𝛼1𝛼,𝛽,𝑟2∶=sup(𝑐,𝑑)⊂(𝛼,𝛽)𝑑𝑐𝑟2(𝑡)𝑑𝑡1/(𝛾+1)×min𝑐𝛼𝑑𝑠𝑟2/𝛾(𝑠)𝛾/(𝛾+1),𝛽𝑑𝑑𝑠𝑟2/𝛾(𝑠)𝛾/(𝛾+1),𝐾(2.80)∗1𝛾∶=𝛾+1𝛾/(𝛾+1)𝛽𝛼||𝑟(𝑡)𝑟||(𝑡)𝛾+1𝑟2𝛾𝑃(𝑡)𝛾1,2(𝑡)𝑑𝑡1/(𝛾+1),𝐾∗2𝛾∶=𝛾+1𝛾/(𝛾+1)𝛽𝛼||𝑟(𝑡)𝑟||(𝑡)𝛾+1𝑟2𝛾𝑃(𝑡)𝛾2,2(𝑡)𝑑𝑡1/(𝛾+1),(2.81) where𝑃1,1(𝑡)∶=𝑡𝛼(𝑡−𝑠)𝛾+1/𝛾1𝑟2(𝑠)1/𝛾𝑑𝑠,𝑃1,2(𝑡)∶=𝑡𝛼𝑑𝑠𝑟2/𝛾,𝑃(𝑠)2,1(𝑡)∶=𝛽𝑡(𝑡−𝑠)𝛾+1/𝛾1𝑟2(𝑠)1/𝛾𝑑𝑠,𝑃2,2(𝑡)∶=𝛽𝑡𝑑𝑠𝑟2/𝛾.(𝑠)(2.82)

Theorem 2.15. Suppose that 𝑥 is a nontrivial solution of (1.1). If 𝑥(𝑖)(𝛼)=0, for 𝑖=0,1,2 and 𝑥(𝛽)=0, then Φ1𝑄1,𝑟2,𝑃1,0+𝛾𝐶Γ𝛼1𝛼,𝛽,𝑟2Ψ1𝑄1,𝑟2,𝑃∗1,0𝑃∗1,1+𝐾∗1≥1,(2.83) where 𝑄1∫(𝑡)=𝛽𝑡𝑟(𝑠)𝑞(𝑠)𝑑𝑠. If 𝑥(𝑖)(𝛽)=0, for 𝑖=0,1,2 and 𝑥(𝛼)=0, then Φ2𝑄1,𝑟2,𝑃2,0+𝛾𝐶Γ𝛼1𝛼,𝛽,𝑟2Ψ2𝑄1,𝑟2,𝑃∗2,0𝑃∗2,1+𝐾∗2≥1,(2.84) where 𝑄1∫(𝑡)=𝑡𝛼𝑟(𝑠)𝑞(𝑠)𝑑𝑠.

Proof. We prove (2.83). Multiplying (1.1) by 𝑟(𝑡)𝑥(𝑡) and integrating by parts, we have 𝛽𝛼𝑥𝑟(𝑡)(𝑡)𝛾𝑟(𝑡)𝑥(𝑡)𝑑𝑡=𝑟2𝑥(𝑡)(𝑡)𝛾𝑥||(𝑡)𝛽𝛼−𝛽𝛼𝑟2𝑥(𝑡)(𝑡)𝛾+1−𝑑𝑡𝛽𝛼𝑟(𝑡)𝑟(𝑡)𝑥(𝑥𝑡)(𝑡)𝛾𝑑𝑡=−𝛽𝛼𝑟(𝑡)𝑞(𝑡)𝑥(𝑡)𝑥𝛾(𝑡)𝑑𝑡.(2.85) Using the boundary conditions 𝑥(𝛼)=𝑥(𝛽)=0 and the assumption 𝑄1∫(𝑡)=𝛽𝑡𝑟(𝑠)𝑞(𝑠)𝑑𝑠, we have 𝛽𝛼𝑟2𝑥(𝑡)(𝑡)𝛾+1𝑑𝑡=−𝛽𝛼𝑟(𝑡)𝑟(𝑡)𝑥𝑥(𝑡)(𝑡)𝛾𝑑𝑡+𝛽𝛼𝑞(𝑡)𝑥(𝑡)𝑥(𝑡)𝑑𝑡=−𝛽𝛼𝑟(𝑡)𝑟(𝑡)𝑥(𝑥𝑡)(𝑡)𝛾𝑑𝑡−𝛽𝛼𝑄1(𝑡)𝑥𝛾(𝑡)𝑥(𝑡)𝑑𝑡.(2.86) Integrating by parts the last term in the right-hand side, we see that 𝛽𝛼𝑄1(𝑡)𝑥(𝑡)𝑥𝛾(𝑡)𝑑𝑡=𝑄1(𝑡)𝑥(𝑡)𝑥𝛾||(𝑡)𝛽𝛼−𝛾𝛽𝛼𝑄1(𝑡)𝑥𝛾−1(𝑡)𝑥(𝑡)𝑥(−𝑡)𝑑𝑡𝛽𝛼𝑄1(𝑡)𝑥𝛾(𝑡)𝑥(𝑡)𝑑𝑡.(2.87) Using the boundary conditions 𝑥(𝛽)=𝑥(𝛼)=0, we see that 𝛽𝛼𝑄1(𝑡)𝑥(𝑡)𝑥(𝑡)𝑑𝑡=−𝛾𝛽𝛼𝑄1(𝑡)𝑥𝛾−1(𝑡)𝑥(𝑡)𝑥−(𝑡)𝑑𝑡𝛽𝛼𝑄1(𝑡)𝑥𝛾(𝑡)𝑥(𝑡)𝑑𝑡.(2.88) Substituting (2.88) into (2.86), we have 𝛽𝛼𝑟2||𝑥(𝑡)||(𝑡)𝛾+1𝑑𝑡≤𝛽𝛼||𝑄1||||||(𝑡)𝑥(𝑡)𝛾||𝑥||(𝑡)𝑑𝑡+𝛾𝛽𝛼||𝑄1(||||||𝑡)𝑥(𝑡)𝛾−1||𝑥(||||𝑥𝑡)(||+𝑡)𝑑𝑡𝛽𝛼||𝑟(𝑡)𝑟||||𝑥(𝑡)||||𝑥(𝑡)||(𝑡)𝛾𝑑𝑡.(2.89) Applying the inequality (2.6) on the integral 𝛽𝛼||𝑄1||||||(𝑡)𝑥(𝑡)𝛾||𝑥||(𝑡)𝑑𝑡,(2.90) with 𝜙(𝑡)=|𝑄(𝑡)|,𝜗(𝑡)=𝑟(𝑡),𝑚=1,𝑘=0,𝑙=𝛾,𝑛=3, and 𝜇=𝛾+1, we get (note that 𝑥(𝑖)(𝛼)=0, for 𝑖=0,1,2) that 𝛽𝛼||𝑄1||||||(𝑡)𝑥(𝑡)𝛾||𝑥||(𝑡)𝑑𝑡≤Φ1||𝑄1||(𝑡),𝑟2,𝑃1,0𝛽𝛼𝑟2||𝑥(𝑡)||(𝑡)2,𝑑𝑡(2.91) where Φ1(𝑄1,𝑟2,𝑃1,0) is defined as in (2.24) by replacing 𝑄 by 𝑄1 and 𝑟 by 𝑟2. Applying the inequality (2.13) on the integral 𝛽𝛼||𝑄1||||||(𝑡)𝑥(𝑡)𝛾−1||𝑥||||𝑥(𝑡)||(𝑡)𝑑𝑡,(2.92) with 𝜙(𝑡)=|𝑄1(𝑡)|,𝜗(𝑡)=𝑟2(𝑡),𝑛=2,𝑟0=𝛾−1,𝑟1=1,𝑟2=1,𝜎+𝑟2=𝛾+1, and 𝑟=𝛾+1, we see that 𝛽𝛼||𝑄1||||||(𝑡)𝑥(𝑡)𝛾−1||𝑥||||𝑥(𝑡)||(𝑡)𝑑𝑡≤Ψ1𝑄1,𝑟2,𝑃∗1,0𝑃∗1,1𝛽𝛼𝑟2||𝑥(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.93) where Ψ1(𝑄1,𝑟2,𝑃∗1,0𝑃∗1,1) is defined as in (2.26) after replacing 𝑄 by 𝑄1 and 𝑟 by 𝑟2. Applying the Hardy inequality (2.1) on the term 𝛽𝛼𝑟2||𝑥(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.94) with 𝑦(𝑡)=𝑥(𝑡) where 𝑥(𝛼)=0=𝑥(𝛽), we see that 𝛽𝛼𝑟2||𝑥(𝑡)||(𝑡)𝛾+1𝑑𝑡≤𝐶Γ𝛼1𝛼,𝛽,𝑟2𝛽𝛼𝑟2||𝑥(𝑡)||(𝑡)𝛾+1𝑑𝑡,(2.95) where 𝐶Γ and 𝛼1(𝛼,𝛽,𝑟2) are defined as in (2.45) and (2.80). Substituting (2.95) into (2.93), we have 𝛽𝛼||||||||𝑄(𝑡)𝑥(𝑡)𝛾−1||𝑥||||𝑥(𝑡)||(𝑡)𝑑𝑡≤𝐶Γ𝛼1𝛼,𝛽,𝑟2Ψ1𝑄,𝑟2,𝑃∗1,0𝑃∗1,1×𝛽𝛼||𝑥𝑟(𝑡)(||𝑡)𝛾+1𝑑𝑡.(2.96) Applying the inequality (2.6) on the integral 𝛽𝛼||𝑟(𝑡)𝑟||||𝑥(𝑡)||||𝑥(𝑡)||(𝑡)𝛾𝑑𝑡,(2.97) with 𝜙(𝑡)=|𝑟(𝑡)𝑟(𝑡)|,𝜗(𝑡)=𝑟2(𝑡),𝑚=𝛾,𝑘=2,𝑙=1,𝑛=3, and 𝜇=𝛾+1, we get (note that 𝑥(𝑖)(𝛼)=0, for 𝑖=0,1,2) that 𝛽𝛼||𝑟(𝑡)𝑟||||𝑥(𝑡)||||𝑥(𝑡)||(𝑡)𝛾𝑑𝑡≤𝐾∗1𝛽𝛼𝑟2||𝑥(𝑡)||(𝑡)𝛾+1,𝑑𝑡(2.98) where 𝐾∗1 is defined as in (2.81). Substituting (2.91), (2.96), and (2.98) into (2.89) and cancelling the term ∫𝛽𝛼𝑟2(𝑡)|𝑥(𝑡)|𝛾+1𝑑𝑡, we have Φ1𝑄1,𝑟2,𝑃1,0+𝛾𝐶Γ𝛼1𝛼,𝛽,𝑟2Ψ1𝑄1,𝑟2,𝑃∗1,0𝑃∗1,1+𝐾∗1≥1,(2.99) which is the desired inequality (2.83). The proof of (2.84) is similar to (2.83) by using the integration by parts and Φ1(𝑄1,𝑟2,𝑃1,0),Ψ1(𝑄1,𝑟2,𝑃∗1,0𝑃∗1,1);𝐾∗1 are replaced by Φ2(𝑄1,𝑟2,𝑃2,0),Ψ2(𝑄1,𝑟2,𝑃∗2,0𝑃∗2,1);𝐾∗2 are defined by (2.25), (2.27), and (2.81) by replacing 𝑟 by 𝑟2. The proof is complete.

In the following, we assume that there exists 𝜏∈(𝛼,𝛽) such that𝑡𝛼𝑑𝑠𝑟2/𝛾=(𝑠)𝛽𝑡𝑑𝑠r2/𝛾,(𝑠)(2.100) denoted by 𝑃𝛼𝛽. In this case, we denote𝐾∗𝑟2=𝐾∗1=𝐾∗2,(2.101) where𝐾∗𝑟2∶=𝛾𝑃𝛼𝛽𝛾+1𝛾/(𝛾+1)𝛽𝛼||𝑟(𝑡)𝑟||(𝑡)𝛾+1𝑟2𝛾(𝑡)𝑑𝑡1/(𝛾+1).(2.102) We also assume that there exists 𝜏∈(𝛼,𝛽) such that𝜏𝛼(𝜏−𝑠)2(𝛾+1)/𝛾𝑟2/𝛾(𝑠)𝑑𝑠=𝛽𝜏(𝑠−𝜏)2(𝛾+1)/𝛾𝑟2/𝛾(𝑠)𝑑𝑠,(2.103) denoted by 𝑃𝑟2(𝛼,𝛽). By using 𝑟2 instead of 𝑟 in Φ(𝑄,𝑟) and Ψ(𝑄,𝑟), we haveΦ𝑄,𝑟2Λ∶=2𝛾𝛽𝛼||||𝑄(𝑡)(𝛾+1)/𝛾𝑟2/𝛾𝑃(𝑡)𝛾𝑟2(𝛼,𝛽)𝑑𝑡𝛾/(𝛾+1),Ψ𝑄,𝑟2∶=Λ𝛽𝛼||||𝑄(𝑡)(𝛾+1)/𝛾𝑟1/𝛾𝑃(𝑡)∗𝑟2(𝛼,𝛽)𝑑𝑡𝛾/(𝛾+1),(2.104) where 𝑃∗𝑟2(𝛼,𝛽) is obtained from (2.103). Using these new values and proceeding as in the proof of Theorem 2.15, we have the following result.

Theorem 2.16. Assume that 𝑄(𝑡)=𝑞(𝑡) and 𝑥(𝑡) is a nontrivial solution of (1.1). If 𝑥(𝑖)(𝛼)=0=𝑥(𝑖)(𝛽), for 𝑖=0,1,2, then Φ𝑄,𝑟2+𝛾𝐶Γ𝛼1𝛼,𝛽,𝑟2Ψ𝑄,𝑟2+𝐾∗𝑟2≥1,(2.105) where 𝐶Γ and 𝛼1(𝛼,𝛽,𝑟) are defined as in (2.45) and (2.80) and 𝐾∗(𝑟2),Φ(𝑄,𝑟2), and Ψ(𝑟2,𝑄) are defined as in (2.102), (2.104).

In the proofs of Theorems 2.13 and 2.15 if we apply the Wirtinger inequality (2.20) instead of the Hardy inequality (2.1), then we have the following results.

Theorem 2.17. Assume that 𝑄(𝑡)=𝑞(𝑡) and 𝑥(𝑡) is a nontrivial solution of (1.1). If 𝑥(𝑖)(𝛼)=0=𝑥(𝑖)(𝛽), for 𝑖=0,1,2, then Φ𝑄,𝑟2+𝛾(𝛾+1)𝛾+1Ψ𝑄,𝑟2+𝐾∗𝑟2≥1,(2.106) where 𝐶Γ and 𝛼1(𝛼,𝛽,𝑟) are defined as in (2.45) and (2.80) and 𝐾∗(𝑟2), Φ(𝑄,𝑟2), and Ψ(𝑟2,𝑄) are defined as in (2.102), (2.104).

Theorem 2.18. Assume that 𝑄(𝑡)=𝑞(𝑡) and 𝑥(𝑡) is a nontrivial solution of (1.1). If 𝑥(𝑖)(𝛼)=0=𝑥(𝑖)(𝛽), for 𝑖=0,1,2, then Φ𝑄,𝑟2+𝛾(𝛾+1)𝛾+1Ψ𝑄,𝑟2+𝐾∗𝑟2≥1,(2.107) where 𝐾∗(𝑟2),Φ(𝑄,𝑟2),Ψ(𝑟2,𝑄), and 𝐾∗ are defined as in (2.102), (2.104).

3. Discussions and Examples

In this section, we establish some special cases of the results obtained in Section 2 and also give some illustrative examples. We begin with Theorem 2.8 and consider the case when 𝑟(𝑡)=1. In this case, (1.1) becomes(𝑥(𝑡)𝛾+𝑞(𝑡)𝑥𝛾[].(𝑡)=0,𝑡∈𝛼,𝛽(3.1) When 𝑟(𝑡)=1, we see thatΦ1𝑄,1,𝑃1,0≤12𝛾1𝛾+11/(𝛾+1)𝛾(𝛽−𝛼)3𝛾+3(3𝛾+3)(3𝛾+2)𝛾/(𝛾+1)max[]𝑡∈𝛼,𝛽Ψ𝑄(𝑡),1𝑄,1,𝑃∗1,0𝑃∗1,1≤max𝑡∈[𝛼,𝛽]||||1𝑄(𝑡)𝛾+11/(𝛾+1)𝛾2𝛾+1𝛾(𝛾−1)/(𝛾+1)×𝛾(2𝛾+1)(𝛾−1)+2𝛾𝛾/(𝛾+1)(𝛽−𝛼)((2𝛾+1)(𝛾−1)+2𝛾)/(𝛾+1),(3.2) where ∫𝑄(𝑡)=𝛽𝑡𝑞(𝑠)𝑑𝑠. The same will be for Φ2(𝑄,1,𝑃1,0) and Ψ2(𝑄,1,𝑃∗1,0𝑃∗1,1), but in this case we assume that ∫𝑄(𝑡)=𝑡𝛼𝑞(𝑠)𝑑𝑠. This gives us the following result for (3.1).

Theorem 3.1. Suppose that 𝑥 is a nontrivial solution of (3.1). If 𝑥(𝑖)(𝛼)=0, for 𝑖=0,1,2 and 𝑥(𝛽)=0, then 𝑀𝛾max𝑡∈[𝛼,𝛽]||||𝑄(𝑡)(𝛽−𝛼)𝛾((3𝛾+3)/(𝛾+1))+𝛾𝑁𝛾max𝑡∈[𝛼,𝛽]||||𝑄(𝑡)(𝛽−𝛼)((2𝛾+1)(𝛾−1)+2𝛾)/(𝛾+1)≥1,(3.3) where ∫𝑄(𝑡)=𝛽𝑡𝑞(𝑠)𝑑𝑠, and 𝑀𝛾1∶=2𝛾1𝛾+11/(𝛾+1)𝛾(3𝛾+3)(3𝛾+2)𝛾/(𝛾+1),𝑁𝛾∶=(𝛾+1)𝛾+1𝛾2𝛾+1𝛾(𝛾−1)/(𝛾+1)𝛾(2𝛾+1)(𝛾−1)+2𝛾𝛾/(𝛾+1)1𝛾+11/(𝛾+1).(3.4) If 𝑥(𝑖)(𝛽)=0, for 𝑖=0,1,2 and 𝑥(𝛼)=0, then (3.3) holds with ∫𝑄(𝑡)=𝑡𝛼𝑞(𝑠)𝑑𝑠.

As a special case of Theorem 3.1, if 𝛾=1, we have the following result.

Theorem 3.2. Suppose that 𝑥 is a nontrivial solution of 𝑥[].(𝑡)+𝑞(𝑡)𝑥(𝑡)=0,𝑡∈𝛼,𝛽(3.5) If 𝑥(𝑖)(𝛼)=0, for 𝑖=0,1,2 and 𝑥(𝛽)=0, then 14√15max𝑡∈[𝛼,𝛽]||||𝛽𝑡||||𝑞(𝑠)𝑑𝑠(𝛽−𝛼)3√+815(𝛽−𝛼)≥1.(3.6) If 𝑥(𝑖)(𝛽)=0, for 𝑖=0,1,2 and 𝑥(𝛼)=0, then 14√15max𝑡∈[𝛼,𝛽]||||𝑡𝛼||||𝑞(𝑠)𝑑𝑠(𝛽−𝛼)3√+815(𝛽−𝛼)≥1.(3.7)

As a special case of Theorem 2.11, if 𝑟(𝑡)=1, we have the following result.

Theorem 3.3. Suppose that 𝑥 is a nontrivial solution of (3.1). If 𝑥(𝑖)(0)=0, for 𝑖=0,1,2 and 𝑥(𝛽)=0, then max𝑡∈[0,𝛽]||||𝑡0||||1𝑞(𝑠)𝑑𝑠4√𝛽15𝛾((3𝛾+3)/(𝛾+1))+1√423𝛽2𝛾−1+2/𝛾≥1.(3.8)

Example 3.4. Consider the equation 𝑥(4)(𝑡)+cos(𝛼𝑡)𝑥(𝑡)=0,0≤𝑡≤𝛽,(3.9) where 𝜆 and 𝛼 are positive constants. Theorem 3.3 gives that if the solution of (3.9) satisfies 𝑥(𝑖)(0)=0, for 𝑖=0,1,2 and 𝑥(𝛽)=0, then max𝑡∈[𝛼,𝛽]||||𝑡0||||𝑞(𝑡)𝑑𝑡=max𝑡∈[0,𝛽]||||𝑡0||||1cos(𝛼𝑡)𝑑𝑡∶=𝛼≥4√√15+423𝛽3.(3.10) That is √𝛽≥(4√15+423)1/3𝛼1/3.

Using the definitions of the functions 𝑃1,0 and 𝑃2,0 and putting 𝑟(𝑡)=1, we see after simplifications that𝑃1,0(𝑡)∶=𝑡𝛼(𝑡−𝑠)2(𝛾+1)/𝛾𝑑𝑠=(𝑡−𝛼)(3𝛾+2),𝑃(3𝛾+2)2,0(𝑡)∶=𝛽𝑡(𝑠−𝑡)2(𝛾+1)/𝛾𝑑𝑠=(𝛽−𝑡)(3𝛾+2)(.3𝛾+2)(3.11) The condition (2.57) is satisfied when (𝜏−𝛼)(3𝛾+2)/𝛾=(𝛽−𝜏)(3𝛾+2)/𝛾. This in fact is satisfied when 𝜏=(𝛼+𝛽)/2. In this case, we see that𝑃𝛾(𝛼,𝛽)=2(3𝛾+2)/𝛾(3𝛾+2)(𝛽−𝛼)(3𝛾+2)/𝛾.(3.12) Also, when 𝑟(𝑡)=1, one can get that𝑃∗𝛾(𝛼,𝛽)=2𝛾+1𝛾−1𝛽−𝛼2(2𝛾+1)(𝛾−1)/𝛾+1.(3.13) In this case, we have thatΛΦ(𝑄,1)∶=2𝛾𝛽𝛼||||𝑄(𝑡)(𝛾+1)/𝛾𝑃𝛾(𝛼,𝛽)𝑑𝑡𝛾/(𝛾+1),Ψ(𝑄,1)∶=Λ𝛽𝛼||||𝑄(𝑡)(𝛾+1)/𝛾𝑃∗(𝛼,𝛽)𝑑𝑡𝛾/(𝛾+1).(3.14)

As a special case of Theorem 2.15, if 𝑟(𝑡)=1, then we have the following result.

Theorem 3.5. Assume that 𝑟(𝑡)>0,𝑄(𝑡)=𝑞(𝑡), and 𝑥(𝑡) is a nontrivial solution of (3.1). If 𝑥(𝑖)(𝛼)=0=𝑥(𝑖)(𝛽) for 𝑖=0,1,2, then Φ(𝑄,1)+𝛾(𝛾+1)𝛾+1Ψ(𝑄,1)≥1.(3.15)

As a special case when 𝛾=1, we see thatΦ1(𝑄,1)≤√805max𝑡∈[𝛼,𝛽]||||𝑡||||×𝑞(𝑡)𝑑𝑡(𝛽−𝛼)3,1Ψ(𝑄,1)≤2max[]𝑡∈𝛼,𝛽||||𝑡||||(𝑞(𝑡)𝑑𝑡𝛽−𝛼).(3.16) This gives us the following result for (3.5).

Theorem 3.6. Assume that 𝑟(𝑡)>0,𝑄(𝑡)=𝑞(𝑡), and 𝑥(𝑡) is a nontrivial solution of (3.5). If 𝑥(𝑖)(𝛼)=0=𝑥(𝑖)(𝛽) for 𝑖=0,1,2, then max𝑡∈[𝛼,𝛽]||||𝑡||||1𝑞(𝑡)𝑑𝑡√805(𝛽−𝛼)3+2(𝛽−𝛼)≥1.(3.17)

One can also use the rest of theorems to get some new results and due to the limited space the details are left to the reader. The following example illustrates the result.

Example 3.7. Consider the equation 𝑥(4)(𝑡)+𝜆cos2(𝛼𝑡)𝑥(𝑡)=0,0≤𝑡≤𝜋,(3.18) where 𝜆 and 𝛼 are positive constants. If 𝑥(𝑡) is a solution of (3.18), which satisfies 𝑥(𝑖)(0)=𝑥(𝑖)(𝜋)=0 for 𝑖=0,1,2, then 𝜆max𝑡∈[0,𝜋]𝑡0cos2(𝛼𝑡)𝑑𝑡=𝜆max𝑡∈[0,𝜋]121𝑡+×14𝛼sin(2𝛼𝑡)√805𝜋3=+2𝜋𝜆𝜋2+𝜆×14𝛼√805𝜋3.+2𝜋(3.19) That is, √𝜆(𝜋/2+1/4𝛼)≥1/[(1/80)5𝜋3+2𝜋].

It will be interesting to establish some new results related to some boundary value problems in bending of beams, see [21, 22].

Problem 1. In particular, one can consider the boundary conditions 𝑥(𝛼)=𝑥(𝛼)=𝑥(𝛽)=𝑥(𝛽)=0,(3.20) which correspond to a beam clamped at each end and establish some new Lyapunov’s type inequalities. The main problem in this case that has been appeared when I tried to treat it is the integral ∫(𝑥)𝛾𝑑𝑡. Note that this integral is trivial if 𝛾=1. So to complete the proof, one should give a relation between this integral and (𝑥)𝛾.

Problem 2. One can also consider the boundary conditions 𝑥(𝛼)=𝑥(𝛼)=𝑥(𝛽)=𝑥(𝛽)=0,(3.21) which correspond to a beam clamped at 𝑡=𝛼 and free at 𝑡=𝛽.

Remark 3.8. The study of the boundary conditions 𝑥(𝛽)=𝑥(𝛽)=𝑥(𝛼)=𝑥(𝛼)=0, which correspond to a beam clamped at 𝑡=𝛽 and free at 𝑡=𝛼, and the boundary conditions 𝑥(𝛼)=𝑥(𝛼)=𝑥(𝛽)=𝑥(𝛽)=0, which correspond to a beam hinged or supported at both ends will be similar to the proof of the boundary conditions (3.20)-(3.21) and will be left to the interested reader. For more discussions of boundary conditions of the bending of beams, we refer to [21, 22].

Acknowledgment

This project was supported by King Saud University, Deanship of Scientific Research, College of Science Research Centre.