# Lyapunov's Type Inequalities for Fourth-Order Differential Equations

Authors: Samir H. Saker
Journal: Abstract and Applied Analysis
 (2012)

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## 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.


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## Body

## 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 equation(1.1)(r(t)(x′′′(t))γ)′+q(t)xγ(t)=0,t∈I,
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 solutionx(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-order(1.2)x(n)(t)+q(t)x(t)=0
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 satisfies(1.3)x(i)(α)=0,i=0,…,k-1,x(j)(β)=0,j=0,…,n-k-1
does not exist.The least value ofβ 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,(1.4)x(i)(α)=0,i=0,…,k-1,x(j)(β)=0,j=k,…,n-1.
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)′.Fornth-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 equation(1.5)x(4)(t)+q(t)x(t)=0,
which satisfies x(α)=x′(α)=x(β)=x′(β)=0, then(1.6)∫αβ|q(t)|dt≥192(β-α)3,
and if x(t) satisfies x(α)=x′′(α)=x(β)=x′′(β)=0, then
(1.7)∫αβ|q(t)|dt≥4(β-α)2.
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, then(1.8)∫αβ|q(t)|dt≥16(β-α)3.
In this paper, we are concerned with the following problems for the general equation (1.1):(i)
obtain lower bounds for the spacingβ-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 spacingβ-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 spacingβ-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 Section2 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 holds(2.1)(∫αβq(t)|y(t)|ndt)1/n≤C(∫αβr(t)|y′(t)|mdt)1/m,
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 satisfies(2.2)C≤k(m,n)α(α,β),for1<m≤n,
where(2.3)α(α,β):=sup⁡α<t<β(∫tβq(t)dt)1/n(∫αtr1-m*(s)ds)1/m*ify(α)=0,α(α,β):=sup⁡α<t<β(∫αtq(t)dt)1/n(∫tβr1-m*(s)ds)1/m*ify(β)=0,
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 if(2.4)α(α,β)=sup⁡(c,d)⊂(α,β)(∫cdq(t)dt)1/n×min⁡{(∫αcr1-m*(s)ds)1/m*,(∫dβr1-m*(s)ds)1/m*}
exists and is finite. The constant k(m,n) in (2.2) appears in various forms. For example,(2.5)k(m,n):=m1/m(m*)1/m*,k(m,n):=(1+nm*)1/n(1+m*n)1/m*,(u*=uu-1)k(m,n):=[Γ(n/s)Γ(1+(1/s))Γ((n-1)/s)]s/n,s=nm-1.
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 functionsϑ 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
(2.6)∫αβϕ(t)|x(k)(t)|l|x(n)(t)|mdt≤K1(α,β)[∫αβϑ(t)|x(n)(t)|μdt](l+m)/μ,
where
(2.7)K1(α,β)=(m/(l+m))m/μ(K!)l[∫αβ(ϕμ(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):=∫αt(t-s)(n-k-1)μ/(μ-1)(ϑ(s))-1/(μ-1)ds.
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
(2.8)K2(α,β)=(m/(l+m))m/μ(K!)l[∫αβ(ϕμ(t)ϑm(t))1/(μ-m)(P2,k(t))l(μ-1)/(μ-m)dt](μ-m)/μ,
where
(2.9)P2,k(t):=∫tβ(s-t)(n-k-1)μ/(μ-1)(ϑ(s))-1/(μ-1)ds.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 that(2.10)∫ατ(τ-s)(n-k-1)μ/(μ-1)(ϑ(s))-1/(μ-1)ds=∫τβ(s-τ)(n-k-1)μ/(μ-1)(ϑ(s))-1/(μ-1)ds,
and we denote by  P(α,β). This gives us the following theorem.Theorem 2.2.
Assume that the functionsϑ  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
(2.11)∫αβϕ(t)|x(k)(t)|l|x(n)(t)|mdt≤K(α,β)[∫αβϑ(t)|x(n)(t)|μdt](l+m)/μ,
where K(α,β) is defined by
(2.12)K(α,β)=(ml+m)m/μ[P(α,β)]l(μ-1)/μ(K!)l[∫αβ(ϕμ(t)ϑm(t))1/(μ-m)dt](μ-m)/μ.Theorem 2.3 ([18, Theorem  3.9.2]).
Letrk,0≤k≤n-1(n≥1) be nonnegative numbers such that σ=∑k=0n-1rk>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
(2.13)∫αβϕ(t)∏k=0n|x(k)(t)|rkdt≤K1*(α,β)[∫αβϑ(t)|x(n)(t)|rdt](σ+rn)/r,
where
(2.14)K1*(α,β)=1Ω(rnσ+rn)rn/r[∫αβ(ϕr(t)ϑrn(t))1/(r-rn)∏k=0n-1(P1,k*(t))rk(r-1)/(r-rn)dt](r-rn)/r,Ω=∏k=0n-1((n-k-1)!)rk,P1,k*(t):=∫αt(t-s)(n-k-1)r/(r-1)(ϑ(s))-1/(r-1)ds.
If we replace x(i)(α)=0 by x(i)(β)=0,0≤i≤n-1(n≥1), then (2.13) holds where K1* is replaced by K2*,  which is given by
(2.15)K2*(α,β)=1Ω(rnσ+rn)rn/r[∫αβ(ϕr(t)ϑrn(t))1/(r-rn)∏k=0n-1(P2,k*(t))rk(r-1)/(r-rn)dt](r-rn)/r,
where
(2.16)P2,k*(t):=∫tβ(s-t)(n-k-1)r/(r-1)(ϑ(s))-1/(r-1)ds.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 that(2.17)∏k=0n-1(P1,k*(τ))rk(r-1)/(r-rn)=∏k=0n-1(P2,k*(τ))rk(r-1)/(r-rn),
denoted by P*(α,β). In this case the inequality (2.13) is satisfied but the constant K1*(α,β) is replaced by K*(α,β), which is defined by(2.18)K*(α,β)=(rnσ+rn)rn/r[P*(α,β)](r-rn)/rΩ[∫αβ(ϕr(t)ϑrn(t))1/(r-rn)dt](r-rn)/r.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.
For𝕀=[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
(2.19)∫αβλγ+1(t)|λ′(t)|γ|y′(t)|γ+1dt≥1(γ+1)γ+1∫αβ|λ′(t)||y(t)|γ+1dt,
for any y∈C1(𝕀) with y(α)=0=y(β).Remark 2.5.
It is clear that Theorem2.4 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.
For𝕀=[α,β] and γ≥1 being a positive integer, then we have
(2.20)∫αβ|r(t)||x′′′(t)|γ+1dt≥1(γ+1)γ+1∫αβ|p(t)||x′′(t)|γ+1dt,
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
(2.21)(r(t)(λ′(t))γ)′-(γ+1)p(t)λγ(t)=0,
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 that(2.22)∫0πt2cos⁡2tdt≃5.9531>0.39270≃14∫0πsin⁡2tdt.
Note also that (2.21) holds if one chooses r(t)=p(t)=1, where in this case(2.23)λ(t)=exp⁡(γ+1γ)1/(γ+1)t.
Now, we are ready to state and prove the main results when r(t)>0. For simplicity, we introduce the following notations:(2.24)Φ1(Q,r,P1,0):=2-γΛ[∫αβ|Q(t)|(γ+1)/γr1/γ(t)P1,0γ(t)dt]γ/(γ+1),
where Λ=(1/(γ+1))1/(γ+1),P1,0(t)=∫αt(t-s)2(γ+1)/γr-1/γ(s)ds,(2.25)Φ2(Q,r,P2,0):=2-γΛ[∫αβ|Q(t)|(γ+1)/γr1/γ(t)P2,0γ(t)dt]γ/(γ+1),
where P2,0(t)=∫tβ(s-t)2(γ+1)/γr-1/γ(s)ds,(2.26)Ψ1(Q,r,P1,0*P1,1*):=Λ[∫αβ|Q(t)|(γ+1)/γr1/γ(t)(P1,0*(t))γ-1P1,1*(t)dt]γ/(γ+1),
where P1,0*(t)=∫αt(t-s)(γ+1)/γr-1/γ(s)ds,P1,1*(t)=∫αtr-1/γ(s)ds, and(2.27)Ψ2(Q,r,P2,0*P2,1*):=Λ[∫αβ|Q(t)|(γ+1)/γr1/γ(t)(P2,0*(t))γ-1P2,1*(t)dt]γ/(γ+1),
where P2,0*(t)=∫tβ(s-t)(γ+1)/γr-1/γ(s)ds,P2,1*(t)=∫tβr-1/γ(s)ds.Remark 2.7.
Note that whenγ=1, then Ψ1(Q,r,P1,0*P1,1*) and Ψ2(Q,r,P2,0*P2,1*) become Ψ1(Q,r,P1,1*) and Ψ2(Q,r,P2,1*).Theorem 2.8.
Suppose thatx is a nontrivial solution of (1.1). If x(i)(α)=0, for i=0,1,2  and x′′(β)=0, then
(2.28)Φ1(Q,r,P1,0)+γ(γ+1)γ+1Ψ1(Q,r,P1,0*P1,1*)≥1,
where Q(t)=∫tβq(s)ds. If x(i)(β)=0, for i=0,1,2 and x′′(α)=0, then
(2.29)Φ2(Q,r,P2,0)+γ(γ+1)γ+1Ψ2(Q,r,P2,0*P2,1*)≥1,
where Q(t)=∫αtq(s)ds.Proof.
We prove (2.28). Multiplying (1.1) by x′′(t) and integrating by parts, we have
(2.30)∫αβ(r(t)(x′′′(t))γ)′x′′(t)dt=r(t)(x′′′(t))γx′′(t)|αβ-∫αβr(t)(x′′′(t))γ+1dt=-∫αβq(t)x′′(t)xγ(t)dt.
Using the boundary conditions x′′(α)=x′′(β)=0 and the assumption Q(t)=∫tβq(s)ds, we have
(2.31)∫αβr(t)(x′′′(t))γ+1dt=∫αβq(t)x′′(t)xγ(t)dt=-∫αβQ′(t)x′′(t)xγ(t)dt.
Integrating by parts the right-hand side, we see that
(2.32)∫αβQ′(t)x′′(t)xγ(t)dt=Q(t)x′′(t)xγ(t)|αβ-γ∫αβQ(t)xγ-1(t)x′(t)x′′(t)dt-∫αβQ(t)xγ(t)x′′′(t)dt.
Using the boundary conditions x′′(β)=x′′(α)=0, we have
(2.33)∫αβQ′(t)x′′(t)xγ(t)dt=-γ∫αβQ(t)xγ-1(t)x′(t)x′′(t)dt-∫αβQ(t)xγ(t)x′′′(t)dt.
Substituting (2.33) into (2.31), we obtain
(2.34)∫αβr(t)|x′′′(t)|γ+1dt≤γ∫αβ|Q(t)||x(t)|γ-1|x′(t)||x′′(t)|dt+∫αβ|Q(t)||x(t)|γ|x′′′(t)|dt.
Applying the inequality (2.6) on the integral
(2.35)∫αβ|Q(t)||x(t)|γ|x′′′(t)|dt,
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
(2.36)∫αβ|Q(t)||x(t)|γ|x′′′(t)|dt≤Φ1(Q(t),r,P1,0)∫αβr(t)|x′′′(t)|γ+1dt,
where Φ1(Q,r,P1,0) is defined as in (2.24). Applying the inequality (2.13) on the integral
(2.37)∫αβ|Q(t)||x(t)|γ-1|x′(t)||x′′(t)|dt,
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
(2.38)∫αβ|Q(t)||x(t)|γ-1|x′(t)||x′′(t)|dt≤Ψ1*(Q,r,P1,0*P1,1*)∫αβr(t)|x′′(t)|γ+1dt,
where Ψ1(Q,r,P1,0*P1,1*) is defined as in (2.26). Applying the Wirtinger inequality (2.20) on the integral
(2.39)∫αβr(t)|x′′(t)|γ+1dt,
where x′′(α)=0=x′′(β),we see that
(2.40)∫αβr(t)|x′′(t)|γ+1dt≤(γ+1)γ+1∫αβr(t)|x′′′(t)|γ+1dt,
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
(2.41)∫αβ|Q(t)||x(t)|γ-1|x′(t)||x′′(t)|dt≤Ψ1(Q,r,P1,0*P1,1*)(γ+1)-(γ+1)∫αβr(t)|x′′′(t)|γ+1dt.
Substituting (2.36) and (2.41) into (2.34) and cancelling the term ∫αβr(t)|x′′′(t)|γ+1dt, we have
(2.42)Φ1(Q,r,P1,0)+γ(γ+1)γ+1Ψ1(Q,r,P1,0*P1,1*)≥1,
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,P1,0*P1,1*) is replaced by Ψ2(Q,r,P2,0*P2,1*), which is defined as in (2.27). The proof is complete.In the following, we apply the Hardy inequality (2.1) on the term(2.43)∫αβr(t)|x′′(t)|γ+1dt,
by replacing y(t) by x′′(t) and use the assumption x′′(α)=0=x′′(β). In this case, we see that(2.44)∫αβr(t)|x′′(t)|γ+1dt≤CΓα(α,β,r)∫αβr(t)|x′′′(t)|γ+1dt,
where(2.45)CΓ∶=Γ1/γ(γ)Γ-1/γ(1+γγ+1)Γ-1/γ(γ2γ+1),(2.46)α(α,β,r):=sup⁡(c,d)⊂(α,β)(∫cdr(t)dt)1/(γ+1)×min⁡{(∫αcdsr1/γ(s))γ/(γ+1),(∫dβdsr1/γ(s))γ/(γ+1)}.
This implies that(2.47)∫αβ|Q(t)||x(t)|γ-1|x′(t)||x′′(t)|dt≤CΓα(α,β,r)Ψ1(Q,r,P1,0*P1,1*)×∫αβr(t)|x′′′(t)|γ+1dt.
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 thatQ′(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
(2.48)Φ1(Q,r,P1,0)+γCΓα(α,β,r)Ψ1(Q,r,P1,0*P1,1*)≥1,
where Q(t)=∫tβq(s)ds. If x(i)(β)=0, for i=0,1,2 and x′′(α)=0, then
(2.49)Φ2(Q,r,P2,0)+γCΓα(α,β,r)Ψ2(Q,r,P2,0*(t)P2,1*)≥1,
where Q(t)=∫αtq(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]).
Letrk,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/s1*=1,1/s2+1/s2*=1  and μ a constant such that μ>s2. Further assume that
(2.50)σ=∑k=0n-1rk>0,Ϝ(ϕ,ϑ):=(∫0β(1ϑ(t))s2/μdt)σ/s2*(∫0βϕs1*(t)dt)1/s1*.
Then the following inequality holds
(2.51)∫0βϕ(t)∏k=0n-1|x(k)(t)|rkdt≤Cβλ[∫0βϑ(t)|x(n)(t)|μdt]σ/μ,
where λ=∑k=0n-1(n-k-1)rk+σδ+1/s1,δ=(μ-s2)/μs2 and
(2.52)C:=Ϝ(ϕ,ϑ)∏k=0n-1[K!]-rk[(n-k-1/δ)+1]-rkδ[∑k=0n-1Krks1+σs1δ+1]1/s1.Now, by applying the inequality (2.51) on the term(2.53)γ∫0β|Q(t)||xγ-1(t)||x′(t)||x′′(t)|dt,
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 obtain(2.54)γ∫0β|Q(t)||xγ-1(t)||x′(t)||x′′(t)|dt≤γLγϜ(r,Q)β2γ-1+2/γ∫0βr(t)|x′′′(t)|γ+1dt,
where(2.55)Lγ:=12γ-1[γ4γ+1]1/γ[2γ(γ+1)+1]-(γ-1)/γ(γ+1)[γ(γ+1)+1]1/γ(γ+1),Ϝ(r,Q):=(∫0β(1r(t))γ/(γ+1)dt)(γ+1)(γ-1)/γ(∫0β|Q(t)|γ/γ-1dt)(γ-1)/γ.
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 thatQ′(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
(2.56)Φ1(Q,r,P1,0)+γLγϜ(r,Q)β2γ-1+2/γ≥1,
where Q(t)=∫tβq(s)ds and Lγ,Ϝ(r,Q) are defined as in (2.55).Remark 2.12.
Note that in the proof of Theorem2.11, 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 that(2.57)∫ατ(τ-s)2(γ+1)/γr1/γ(s)ds=∫τβ(s-τ)2(γ+1)/γr1/γ(s)ds,
denoted by P(α,β). In this case, we see that(2.58)Φ(Q,r):=Φ1(Q,r,P(t))=Φ2(Q,r,P(t)),
where in this case Φ(Q,r) is given by(2.59)Φ(Q,r):=Λ2γ[∫αβ|Q(t)|(γ+1)/γr1/γ(t)Pγ(α,β)dt]γ/(γ+1).
Second, we assume that (2.17) holds and there exists τ∈(α,β) such that(2.60)(P1,0*(α,τ))γ-1P1,1*(α,τ)=(P2,0*(τ,β))γ-1P2,1*(τ,β),
denoted by P*(α,β). In this case, we get that(2.61)Ψ(r,Q)=Ψ1*(r,Q)=Ψ2*(r,Q),
where Ψ(r,Q) is given by(2.62)Ψ(r,Q):=Λ[∫αβ|Q(t)|(γ+1)/γr1/γ(t)P*(α,β)dt]γ/(γ+1).
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 that(2.63)P(α,β):=2-(3γ+2)/γγ(3γ+2)(β-α)(3γ+2)/γ.
Also, when r(t)=1, then P*(α,β) becomes(2.64)P*(α,β):=(γ2γ+1)γ-1(β-α2)((2γ+1)(γ-1)+γ)/γ.Theorem 2.13.
Assume thatQ′(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
(2.65)Φ(Q,r)+γCΓα(α,β,r)Ψ(r,Q)≥1,
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
(2.66)∫αβ(r(t)(x′′′(t))γ)′x′′(t)dt=r(t)(x′′′(t))γx′′(t)|αβ-∫αβr(t)(x′′′(t))γ+1dt=-∫αβq(t)xγ(t)x′′(t)dt.
Using the boundary conditions x′′(α)=x′′(β)=0, we get that
(2.67)∫αβr(t)(x′′′(t))γ+1dt=∫αβq(t)xγ(t)x′′(t)dt=∫αβQ′(t)xγ(t)x′′(t)dt.
Integrating by parts the right-hand side, we see that
(2.68)∫αβQ′(t)x′′(t)xγ(t)dt=Q(t)x′′(t)xγ(t)|αβ-γ∫αβQ(t)xγ-1(t)x′(t)x′′(t)dt-∫αβQ(t)xγ(t)x′′′(t)dt.
Using the boundary conditions x′′(β)=x′′(α)=0, we see that
(2.69)∫αβQ′(t)x′′(t)xγ(t)dt=-γ∫αβQ(t)xγ-1(t)x′(t)x′′(t)dt-∫αβQ(t)xγ(t)x′′′(t)dt.
Substituting (2.69) into (2.67), we have
(2.70)∫αβr(t)|x′′′(t)|γ+1dt≤γ∫αβ|Q(t)||x(t)|γ-1|x′(t)||x′′(t)|dt+∫αβ|Q(t)||x(t)|γ|x′′′(t)|dt.
Applying the inequality (2.6) on the integral
(2.71)∫αβ|Q(t)||x(t)|γ|x′′′(t)|dt,
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
(2.72)∫αβ|Q(t)||x(t)|γ|x′′′(t)|dt≤Φ(Q,r)∫αβr(t)|x′′′(t)|γ+1dt,
where Φ(Q,r) is defined as in (2.59). Applying the inequality (2.13) on the integral
(2.73)∫αβ|Q(t)||x(t)|γ-1|x′(t)||x′′(t)|dt,
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
(2.74)∫αβ|Q(t)||x(t)|γ-1|x′(t)||x′′(t)|dt≤Ψ(Q,r)∫αβr(t)|x′′(t)|γ+1dt,
where Ψ(Q,r) is defined as in (2.62). Applying the Hardy inequality (2.1) on the term
(2.75)∫αβr(t)|x′′(t)|γ+1dt
with y(t)=x′′(t) where x′′(α)=0=x′′(β), we see that
(2.76)∫αβr(t)|x′′(t)|γ+1dt≤CΓα(α,β,r)∫αβr(t)|x′′′(t)|γ+1dt,
where CΓ and α(α,β,r) are defined as in (2.45) and (2.46).Substituting (2.76) into (2.74), we have
(2.77)∫αβ|Q(t)||x(t)|γ-1|x′(t)||x′′(t)|dt≤γCΓα(α,β,r)Ψ(Q,r)∫αβr(t)|x′′′(t)|γ+1dt.
Substituting (2.72) and (2.77) into (2.70) and cancelling the term ∫αβr(t)|x′′′(t)|γ+1dt, we have
(2.78)Φ(Q,r)+γCΓα(α,β,r)Ψ(Q,r)Ψ(Q,r)≥1,
which is the desired inequality (2.28). The proof is complete.In the proof of Theorem2.13 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 thatrQ′(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
(2.79)Φ(Q,r)+γ(γ+1)γ+1Ψ(Q,r)≥1,
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 whenr(t)<0. For simplicity, we denote(2.80)α1(α,β,r2)∶=sup⁡(c,d)⊂(α,β)(∫cdr2(t)dt)1/(γ+1)×min⁡{(∫αcdsr2/γ(s))γ/(γ+1),(∫dβdsr2/γ(s))γ/(γ+1)},(2.81)K1*∶=(γγ+1)γ/(γ+1)[∫αβ|r(t)r′(t)|γ+1r2γ(t)P1,2γ(t)dt]1/(γ+1),K2*∶=(γγ+1)γ/(γ+1)[∫αβ|r(t)r′(t)|γ+1r2γ(t)P2,2γ(t)dt]1/(γ+1),
where(2.82)P1,1(t)∶=∫αt(t-s)γ+1/γ(1r2(s))1/γds,P1,2(t)∶=∫αtdsr2/γ(s),P2,1(t)∶=∫tβ(t-s)γ+1/γ(1r2(s))1/γds,P2,2(t)∶=∫tβdsr2/γ(s).Theorem 2.15.
Suppose thatx is a nontrivial solution of (1.1). If x(i)(α)=0, for i=0,1,2 and x′′(β)=0, then
(2.83)Φ1(Q1,r2,P1,0)+γCΓα1(α,β,r2)Ψ1(Q1,r2,P1,0*P1,1*)+K1*≥1,
where Q1(t)=∫tβr(s)q(s)ds. If x(i)(β)=0, for i=0,1,2 and x′′(α)=0, then
(2.84)Φ2(Q1,r2,P2,0)+γCΓα1(α,β,r2)Ψ2(Q1,r2,P2,0*P2,1*)+K2*≥1,
where Q1(t)=∫αtr(s)q(s)ds.Proof.
We prove (2.83). Multiplying (1.1) by r(t)x′′(t) and integrating by parts, we have
(2.85)∫αβ(r(t)(x′′′(t))γ)′r(t)x′′(t)dt=r2(t)(x′′′(t))γx′′(t)|αβ-∫αβr2(t)(x′′′(t))γ+1dt-∫αβr(t)r′(t)x′′(t)(x′′′(t))γdt=-∫αβr(t)q(t)x′′(t)xγ(t)dt.
Using the boundary conditions x′′(α)=x′′(β)=0 and the assumption Q1(t)=∫tβr(s)q(s)ds, we have
(2.86)∫αβr2(t)(x′′′(t))γ+1dt=-∫αβr(t)r′(t)x′′(t)(x′′′(t))γdt+∫αβq(t)x(t)x′′(t)dt=-∫αβr(t)r′(t)x′′(t)(x′′′(t))γdt-∫αβQ1′(t)xγ(t)x′′(t)dt.
Integrating by parts the last term in the right-hand side, we see that
(2.87)∫αβQ1′(t)x′′(t)xγ(t)dt=Q1(t)x′′(t)xγ(t)|αβ-γ∫αβQ1(t)xγ-1(t)x′(t)x′′(t)dt-∫αβQ1(t)xγ(t)x′′′(t)dt.
Using the boundary conditions x′′(β)=x′′(α)=0, we see that
(2.88)∫αβQ1′(t)x(t)x′′(t)dt=-γ∫αβQ1(t)xγ-1(t)x′(t)x′′(t)dt-∫αβQ1(t)xγ(t)x′′′(t)dt.
Substituting (2.88) into (2.86), we have
(2.89)∫αβr2(t)|x′′′(t)|γ+1dt≤∫αβ|Q1(t)||x(t)|γ|x′′′(t)|dt+γ∫αβ|Q1(t)||x(t)|γ-1|x′(t)||x′′(t)|dt+∫αβ|r(t)r′(t)||x′′(t)||x′′′(t)|γdt.
Applying the inequality (2.6) on the integral
(2.90)∫αβ|Q1(t)||x(t)|γ|x′′′(t)|dt,
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
(2.91)∫αβ|Q1(t)||x(t)|γ|x′′′(t)|dt≤Φ1(|Q1(t)|,r2,P1,0)[∫αβr2(t)|x′′′(t)|2dt],
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
(2.92)∫αβ|Q1(t)||x(t)|γ-1|x′(t)||x′′(t)|dt,
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
(2.93)∫αβ|Q1(t)||x(t)|γ-1|x′(t)||x′′(t)|dt≤Ψ1(Q1,r2,P1,0*P1,1*)∫αβr2(t)|x′′(t)|γ+1dt,
where Ψ1(Q1,r2,P1,0*P1,1*) is defined as in (2.26) after replacing Q by Q1 and r by r2. Applying the Hardy inequality (2.1) on the term
(2.94)∫αβr2(t)|x′′(t)|γ+1dt,
with y(t)=x′′(t) where x′′(α)=0=x′′(β), we see that
(2.95)∫αβr2(t)|x′′(t)|γ+1dt≤CΓα1(α,β,r2)∫αβr2(t)|x′′′(t)|γ+1dt,
where CΓ and α1(α,β,r2) are defined as in (2.45) and (2.80).Substituting (2.95) into (2.93), we have
(2.96)∫αβ|Q(t)||x(t)|γ-1|x′(t)||x′′(t)|dt≤CΓα1(α,β,r2)Ψ1(Q,r2,P1,0*P1,1*)×∫αβr(t)|x′′′(t)|γ+1dt.
Applying the inequality (2.6) on the integral
(2.97)∫αβ|r(t)r′(t)||x′′(t)||x′′′(t)|γdt,
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
(2.98)∫αβ|r(t)r′(t)||x′′(t)||x′′′(t)|γdt≤K1*[∫αβr2(t)|x′′′(t)|γ+1dt],
where K1* is defined as in (2.81). Substituting (2.91), (2.96), and (2.98) into (2.89) and cancelling the term ∫αβr2(t)|x′′′(t)|γ+1dt, we have
(2.99)Φ1(Q1,r2,P1,0)+γCΓα1(α,β,r2)Ψ1(Q1,r2,P1,0*P1,1*)+K1*≥1,
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,P1,0*P1,1*);K1* are replaced by Φ2(Q1,r2,P2,0),Ψ2(Q1,r2,P2,0*P2,1*);K2* 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 existsτ∈(α,β) such that(2.100)∫αtdsr2/γ(s)=∫tβdsr2/γ(s),
denoted by Pαβ. In this case, we denote(2.101)K*(r2)=K1*=K2*,
where(2.102)K*(r2):=(γPαβγ+1)γ/(γ+1)[∫αβ|r(t)r′(t)|γ+1r2γ(t)dt]1/(γ+1).
We also assume that there exists τ∈(α,β) such that(2.103)∫ατ(τ-s)2(γ+1)/γr2/γ(s)ds=∫τβ(s-τ)2(γ+1)/γr2/γ(s)ds,
denoted by Pr2(α,β). By using r2 instead of r in Φ(Q,r) and Ψ(Q,r), we have(2.104)Φ(Q,r2):=Λ2γ[∫αβ|Q(t)|(γ+1)/γr2/γ(t)Pr2γ(α,β)dt]γ/(γ+1),Ψ(Q,r2):=Λ[∫αβ|Q(t)|(γ+1)/γr1/γ(t)Pr2*(α,β)dt]γ/(γ+1),
where Pr2*(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 thatQ′(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
(2.105)Φ(Q,r2)+γCΓα1(α,β,r2)Ψ(Q,r2)+K*(r2)≥1,
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 Theorems2.13 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 thatQ′(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
(2.106)Φ(Q,r2)+γ(γ+1)γ+1Ψ(Q,r2)+K*(r2)≥1,
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 thatQ′(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
(2.107)Φ(Q,r2)+γ(γ+1)γ+1Ψ(Q,r2)+K*(r2)≥1,
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 Section2 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(3.1)((x′′′(t)γ)′+q(t)xγ(t)=0,t∈[α,β].
When r(t)=1, we see that(3.2)Φ1(Q,1,P1,0)≤12γ(1γ+1)1/(γ+1)[γ(β-α)3γ+3(3γ+3)(3γ+2)]γ/(γ+1)max⁡t∈[α,β]Q(t),Ψ1(Q,1,P1,0*P1,1*)≤max⁡t∈[α,β]|Q(t)|(1γ+1)1/(γ+1)(γ2γ+1)γ(γ-1)/(γ+1)×(γ(2γ+1)(γ-1)+2γ)γ/(γ+1)(β-α)((2γ+1)(γ-1)+2γ)/(γ+1),
where Q(t)=∫tβq(s)ds. The same will be for Φ2(Q,1,P1,0) and Ψ2(Q,1,P1,0*P1,1*), but in this case we assume that Q(t)=∫αtq(s)ds. This gives us the following result for (3.1).Theorem 3.1.
Suppose thatx is a nontrivial solution of (3.1). If x(i)(α)=0, for i=0,1,2 and x′′(β)=0, then
(3.3)Mγmax⁡t∈[α,β]|Q(t)|(β-α)γ((3γ+3)/(γ+1))+γNγmax⁡t∈[α,β]|Q(t)|(β-α)((2γ+1)(γ-1)+2γ)/(γ+1)≥1,
where Q(t)=∫tβq(s)ds, and
(3.4)Mγ∶=12γ(1γ+1)1/(γ+1)[γ(3γ+3)(3γ+2)]γ/(γ+1),Nγ∶=(γ+1)γ+1(γ2γ+1)γ(γ-1)/(γ+1)(γ(2γ+1)(γ-1)+2γ)γ/(γ+1)(1γ+1)1/(γ+1).
If x(i)(β)=0, for i=0,1,2 and x′′(α)=0, then (3.3) holds with Q(t)=∫αtq(s)ds.As a special case of Theorem3.1, if γ=1, we have the following result.Theorem 3.2.
Suppose thatx is a nontrivial solution of
(3.5)x′′′′(t)+q(t)x(t)=0,t∈[α,β].
If x(i)(α)=0, for i=0,1,2 and x′′(β)=0, then
(3.6)1415max⁡t∈[α,β]|∫tβq(s)ds|[(β-α)3+815(β-α)]≥1.
If x(i)(β)=0, for i=0,1,2 and x′′(α)=0, then
(3.7)1415max⁡t∈[α,β]|∫αtq(s)ds|[(β-α)3+815(β-α)]≥1.As a special case of Theorem2.11, if r(t)=1, we have the following result.Theorem 3.3.
Suppose thatx is a nontrivial solution of (3.1). If x(i)(0)=0, for i=0,1,2 and x′′(β)=0, then
(3.8)max⁡t∈[0,β]|∫0tq(s)ds|[1415βγ((3γ+3)/(γ+1))+1423β2γ-1+2/γ]≥1.Example 3.4.
Consider the equation(3.9)x(4)(t)+cos⁡(αt)x(t)=0,0≤t≤β,
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
(3.10)max⁡t∈[α,β]|∫0tq(t)dt|=max⁡t∈[0,β]|∫0tcos⁡(αt)dt|:=1α≥415+423β3.
That is β≥(415+423)1/3α1/3.Using the definitions of the functionsP1,0 and P2,0 and putting r(t)=1, we see after simplifications that(3.11)P1,0(t)∶=∫αt(t-s)2(γ+1)/γds=(t-α)(3γ+2)(3γ+2),P2,0(t)∶=∫tβ(s-t)2(γ+1)/γds=(β-t)(3γ+2)(3γ+2).
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 that(3.12)P(α,β)=γ2(3γ+2)/γ(3γ+2)(β-α)(3γ+2)/γ.
Also, when r(t)=1, one can get that(3.13)P*(α,β)=(γ2γ+1)γ-1(β-α2)(2γ+1)(γ-1)/γ+1.
In this case, we have that(3.14)Φ(Q,1):=Λ2γ[∫αβ|Q(t)|(γ+1)/γPγ(α,β)dt]γ/(γ+1),Ψ(Q,1):=Λ[∫αβ|Q(t)|(γ+1)/γP*(α,β)dt]γ/(γ+1).As a special case of Theorem2.15, if r(t)=1, then we have the following result.Theorem 3.5.
Assume thatr(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
(3.15)Φ(Q,1)+γ(γ+1)γ+1Ψ(Q,1)≥1.As a special case whenγ=1, we see that(3.16)Φ(Q,1)≤1805max⁡t∈[α,β]|∫tq(t)dt|×(β-α)3,Ψ(Q,1)≤12max⁡t∈[α,β]|∫tq(t)dt|(β-α).
This gives us the following result for (3.5).Theorem 3.6.
Assume thatr(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
(3.17)max⁡t∈[α,β]|∫tq(t)dt|[1805(β-α)3+2(β-α)]≥1.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(3.18)x(4)(t)+λcos⁡2(αt)x(t)=0,0≤t≤π,
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
(3.19)λmax⁡t∈[0,π]∫0tcos⁡2(αt)dt=λmax⁡t∈[0,π][12t+14αsin⁡(2αt)]×[1805π3+2π]=(λπ2+λ4α)×[1805π3+2π].
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(3.20)x(α)=x′(α)=x(β)=x′(β)=0,
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(3.21)x(α)=x′(α)=x′′(β)=x′′′(β)=0,
which correspond to a beam clamped at t=α and free at t=β.Remark 3.8.
The study of the boundary conditionsx(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].



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*Source: 795825-2012-06-04.xml*