# Unboundedness of Solutions of Timoshenko Beam Equations with Damping and Forcing Terms

Authors: Kusuo Kobayashi; Norio Yoshida
Journal: International Journal of Differential Equations
 (2013)

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## Abstract

Timoshenko beam equations with external  damping and internal damping terms and forcing terms are investigated, and boundary conditions (end conditions) to be considered are hinged ends (pinned ends), hinged-sliding ends, and sliding ends. Unboundedness of solutions of boundary value problems for Timoshenko beam equations is studied, and it is shown that the magnitude of the displacement of the beam grows up to ∞ ast→∞ under some assumptions on the forcing term. Our approach is to reduce the multidimensional problems to one-dimensional problems for fourth-order ordinary differential inequalities.


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## Body

## 1. Introduction

The most fundamental beam equations are of the following form:(1)ρA∂2u∂t2+EI∂4u∂x4=0(0<x<L,t>0)
with the length L, the mass density ρ, the cross-sectional area A, the modulus of elasticity E, and the flexural rigidity EI (see [1, page 416]). Taking account of the rotary inertia and the deflection due to shear, we obtain the following fourth-order beam equation for the transverse vibrations of prismatic beams on elastic foundations:
(2)ρ2Ik1G∂4u∂t4+ρA∂2u∂t2-ρ(I+EIk1G)∂4u∂t2∂x2+EI∂4u∂x4=q(x,t)
(see  [1, page 433] and Wang and Stephens [2, page 150]).Dividing the above equation byρ2I/(k1G), letting α=ρA/EI,β=k1G/ρ,γ=E/ρ,f(x,t)=(k1G/ρ2I)q(x,t), and taking into account the nonlinear term φ(u), the external damping term
(3)δ∂u∂t,
and the internal damping terms
(4)-ϵ∂3u∂t∂x2,σ∂5u∂t∂x4,
we obtain the Timoshenko beam equation
(5)∂4u∂t4+αβγ∂2u∂t2-(β+γ)∂4u∂t2∂x2+βγ∂4u∂x4+δ∂u∂t-ϵ∂3u∂t∂x2+σ∂5u∂t∂x4+φ(u)=f(x,t),(x,t)∈(0,L)×(0,∞),
where α,β,γ,δ,ϵ,σ,  and L are positive constants.LetJ=(0,L),J-=[0,L] and we assume throughout this paper that(H1)
φ(ξ) is a real-valued continuous function in ℝ;(H2)
φ(ξ)>0  and φ(-ξ)=-φ(ξ)  for  ξ>0;(H3)
φ(ξ) is a nondecreasing function in ℝ;(H4)
f(x,t) is a real-valued continuous function on J¯×[0,∞).Definition 1.
By asolution of (5), one means a function u∈C(J¯×[0,∞)) such that the partial derivatives
(6)∂4u∂t4,∂4u∂t2∂x2,∂4u∂x4,∂5u∂t∂x4
exist and are continuous on J-×[0,∞).Oscillations of beam equations have been treated by numerous authors; see, for example, Feireisl and Herrmann [3], Herrmann [4], Kopáčková [5], Kusano and Yoshida [6], Yoshida [7–10], and the references therein. In particular, we mention the paper [4] by Herrmann which deals with the Euler-Bernoulli beam equations that is similar to (5). We note that the oscillation of (5) was studied by Yoshida [10]. We refer to Ball [11], Fitzgibbon [12], and Narazaki [13] for stability and existence results for beam equations.However, there appears to be no known unboundedness results for beam equations. The objective of this paper is to provide unboundedness results for (5) by reducing the multi-dimensional problems to one-dimensional problems for ordinary differential inequalities of fourth-order.In Section2 we treat the hinged ends and reduce unboundedness problem for (5) to that for ordinary differential inequalities. Sections 3 and 4 are devoted to the hinged-sliding ends and sliding ends, respectively. In Section 5, we study fourth-order differential inequalities, and we derive unboundedness results for (5) in Section 6.


## 2. Hinged Ends

In this section, we treat the case where the ends of the beam are hinged, so that solutionsu=u(x,t) are required to satisfy the boundary condition
(HE)u(0,t)=∂2u∂x2(0,t)=u(L,t)=∂2u∂x2(L,t)=0.Theorem 2.
Every solutionu of  (5) satisfying  (HE) is unbounded on J¯×[0,∞) if for any constant M~>0, all solutions y(t) of the fourth-order differential inequalities
(7)y(4)(t)+(αβγ+(β+γ)(πL)2)y′′(t)+(δ+ϵ(πL)2+σ(πL)4)y′(t)+βγ(πL)4y(t)-2Lπφ(M~)≤F(t),(8)y(4)(t)+(αβγ+(β+γ)(πL)2)y′′(t)+(δ+ϵ(πL)2+σ(πL)4)y′(t)+βγ(πL)4y(t)-2Lπφ(M~)≤-F(t)
are not bounded from below, where
(9)F(t)=∫0L‍f(x,t)sinπLxdx.Proof.
Suppose to the contrary that there exists a solutionu of the boundary value problem (5), (HE) which is bounded on J-×[0,∞). Then, there exists a constant M>0 such that
(10)|u(x,t)|≤M,(x,t)∈J-×[0,∞),
that is,
(11)-M≤u(x,t)≤M,(x,t)∈J-×[0,∞).
First we consider the case where -M≤u(x,t) on J¯×[0,∞). It follows from the hypotheses (H2) and (H3) that
(12)φ(u)≥φ(-M)=-φ(M)
and therefore we see from (5) that
(13)∂4u∂t4+αβγ∂2u∂t2-(β+γ)∂4u∂t2∂x2+βγ∂4u∂x4+δ∂u∂t-ϵ∂3u∂t∂x2+σ∂5u∂t∂x4-φ(M)≤f(x,t)
on J-×[0,∞). Multiplying (13) by ψ(x)=sin(πx/L) and then integrating over [0,L], we derive
(14)d4dt4∫0L‍uψ(x)dx+αβγd2dt2∫0L‍uψ(x)dx-(β+γ)d2dt2∫0L∂2u∂x2ψ(x)dx+βγ∫0L∂4u∂x4ψ(x)dx+δddt∫0Luψ(x)dx-ϵddt∫0L∂2u∂x2ψ(x)dx+σddt∫0L∂4u∂x4ψ(x)dx-2Lπφ(M)≤∫0Lf(x,t)ψ(x)dx,t≥0.
Integrating by parts and using (HE), we obtain
(15)∫0L∂2u∂x2ψ(x)dx=-(πL)2∫0L‍uψ(x)dx,∫0L∂4u∂x4ψ(x)dx=(πL)4∫0L‍uψ(x)dx.
Combining (14) and (15) yields
(16)U(4)(t)+(αβγ+(β+γ)(πL)2)U′′(t)+(δ+ϵ(πL)2+σ(πL)4)U′(t)+βγ(πL)4U(t)-2Lπφ(M)≤F(t),
where U(t)=∫0L‍uψ(x)dx.  It is easy to check that
(17)U(t)=∫0L‍uψ(x)dx≥-2LπM,
that is, U(t) is bounded from below.  Hence, we conclude that U(t) is a solution of (7) which is bounded from below. This contradicts the hypothesis. In the case where u(x,t)≤M on J¯×[0,∞), v:=-u satisfies
(18)∂4v∂t4+αβγ∂2v∂t2-(β+γ)∂4v∂t2∂x2+βγ∂4v∂x4+δ∂v∂t-ϵ∂3v∂t∂x2+σ∂5v∂t∂x4+φ(v)=-f(x,t)
in view of the hypothesis (H2). It is readily verified that v≥-M and φ(v)≥φ(-M)=-φ(M). By the same arguments as in the case where u(x,t)≥-M, we conclude that V(t)=∫0L‍vψ(x)dx is a solution of (8) which is bounded from below in light of V(t)≥-(2L/π)M. This contradicts the hypothesis and the proof is complete.


## 3. Hinged-Sliding Ends

In this section, we deal with the case of hinged-sliding ends, for which the boundary condition takes the form(HSE)u(0,t)=∂2u∂x2(0,t)=∂u∂x(L,t)=∂3u∂x3(L,t)=0.Theorem 3.
Every solutionu of  (5) satisfying  (HSE) is unbounded on J¯×[0,∞) if for any constant M~>0, all solutions y(t) of the fourth-order differential inequalities
(19)y(4)(t)+(αβγ+(β+γ)(π2L)2)y′′(t)+(δ+ϵ(π2L)2+σ(π2L)4)y′(t)+βγ(π2L)4y(t)-2Lπφ(M~)≤G(t),(20)y(4)(t)+(αβγ+(β+γ)(π2L)2)y′′(t)+(δ+ϵ(π2L)2+σ(π2L)4)y′(t)+βγ(π2L)4y(t)-2Lπφ(M~)≤-G(t)
are not bounded from below, where
(21)G(t)=∫0Lf(x,t)sinπ2Lxdx.Proof.
Suppose that there is a solutionu of the boundary value problem (5), (HSE) which is bounded on J¯×[0,∞). Let |u(x,t)|≤M on J¯×[0,∞) for some M>0. First, consider the case where -M≤u(x,t) on J¯×[0,∞). Proceeding as in the proof of Theorem 2, we find that the inequality (13) holds. Multiplying (13) by ω(x)=sin(πx/(2L)) and then integrating over [0,L], we obtain the inequality (14) with ψ(x) replaced by ω(x). Integration by parts yields
(22)∫0L∂2u∂x2ω(x)dx=-(π2L)2∫0L‍uω(x)dx,∫0L∂4u∂x4ω(x)dx=(π2L)4∫0L‍uω(x)dx.
As in the proof of Theorem 2, we observe that ∫0L‍uω(x)dx is a solution of (19) which is bounded from below. This contradicts the hypothesis. The case where u(x,t)≤M on J¯×[0,∞) can be treated similarly, and we find that ∫0L‍(-u)ω(x)dx is a solution of (20) which is bounded from below. The contradiction establishes the theorem.


## 4. Sliding Ends

We study the case of sliding ends for which the boundary condition takes the form(SE)∂u∂x(0,t)=∂3u∂x3(0,t)=∂u∂x(L,t)=∂3u∂x3(L,t)=0.Theorem 4.
Every solutionu of  (5) satisfying  (SE) is unbounded on J¯×[0,∞) if for any constant M~>0, all solutions y(t) of the fourth-order differential inequalities
(23)y(4)(t)+αβγy′′(t)+δy′(t)-Lφ(M~)≤H(t),(24)y(4)(t)+αβγy′′(t)+δy′(t)-Lφ(M~)≤-H(t)
are not bounded from below, where
(25)H(t)=∫0Lf(x,t)dx.Proof.
Suppose that the boundary value problem (5), (SE) has a solution u which is bounded on J¯×[0,∞). There exists a constant M>0 such that |u(x,t)|≤M on J¯×[0,∞). First, consider the case where -M≤u(x,t) on J¯×[0,∞). Arguing as in the proof of Theorem 2, we see that the inequality (13) holds. Integrating (13) over [0,L], we obtain the inequality (14) with -((2L)/π)φ(M) replaced by -Lφ(M) and ψ(x)≡1. It is easy to see that
(26)∫0L∂2u∂x2dx=∂u∂x(L,t)-∂u∂x(0,t)=0,∫0L∂4u∂x4dx=∂3u∂x3(L,t)-∂3u∂x3(0,t)=0.
Hence, we have the inequality
(27)U(4)(t)+αβγU′′(t)+δU′(t)-Lφ(M)≤H(t),t≥0,
where U(t)=∫0L‍udx. Therefore, we conclude that U(t) is a solution of (23) which is bounded from below. This contradicts the hypothesis. The case where u(x,t)≤M on J¯×[0,∞) can be treated analogously, and we observe that ∫0L‍(-u)dx is a solution of (24) which is bounded from below. This is a contradiction and the proof is complete.


## 5. Fourth-Order Ordinary Differential Inequalities

We deal with the ordinary differential inequality of the fourth order(28)y(4)(t)+ky′′(t)+ℓy′(t)+my(t)-μ≤g(t),t≥0,
and derive sufficient condition for every solution y(t) of (28) to be unbounded from below. It is assumed that k,ℓ,m, and μ are nonnegative constants, and g(t) is a continuous function on [0,∞).Theorem 5.
Every solutiony(t) of  (28) is not bounded from below if
(29)liminft→∞1t∫0t(1-ξt)3g(ξ)dξ=-∞.Proof.
Assume, on the contrary, that there exists a solutiony(t) of (28) which is bounded from below. Let y(t)≥-M on [0,∞) for some constant M>0. Then we obtain from (28) the inequality
(30)y(4)(t)+ky′′(t)+ℓy′(t)-Mm-μ≤g(t),t≥0.
Integrating (30) over [0,t], we get
(31)y(3)(t)+ky′(t)+ℓy(t)-Mmt-μt≤c1+∫0tg(ξ)dξ,
and hence
(32)y(3)(t)+ky′(t)-Mℓ-Mmt-μt≤c1+∫0tg(ξ)dξ,
where c1=y(3)(0)+ky′(0)+ℓy(0). Integration of (32) over [0,t] yields
(33)y′′(t)+ky(t)-Mℓt-Mm2t2-μ2t2≤c0+c1t+∫0t‍ds1∫0s1‍g(ξ)dξ,
that is,
(34)y′′(t)-Mk-Mℓt-Mm2t2-μ2t2≤c0+c1t+∫0t‍ds1∫0s1‍g(ξ)dξ,
where c0=y′′(0)+ky(0). Integrating (34) over [0,t] twice, we obtain
(35)y(t)-Mk2t2-Mℓ6t3-Mm24t4-μ24t4≤d0+d1t+d2t2+d3t3+∫0t‍ds3∫0s3‍ds2∫0s2‍ds1∫0s1‍g(ξ)dξ,
and therefore
(36)-M-Mk2t2-Mℓ6t3-Mm24t4-μ24t4≤d0+d1t+d2t2+d3t3+16∫0t‍(t-ξ)3g(ξ)dξ,
in view of the identity
(37)∫0t‍ds3∫0s3‍ds2∫0s2‍ds1∫0s1‍g(ξ)dξ=16∫0t‍(t-ξ)3g(ξ)dξ,
where d0=y(0),d1=y′(0),d2=c0/2, and d3=c1/6. Dividing (36) by t4 yields
(38)-Mt4-Mk2t2-Mℓ6t-Mm24-μ24-(d0t4+d1t3+d2t2+d3t)≤16t∫0t(1-ξt)3g(ξ)dξ.
The left hand side of (38) is bounded from below, whereas the right hand side of (38) is not bounded from below from the condition (29). This is a contradiction and the proof is complete.


## 6. Unboundedness Results for Timoshenko Beam Equations

Combining Theorems2–4 with Theorem 5, we present unboundedness results for the three types of boundary value problems for (5) under consideration.Theorem 6.
Every solutionu of  (5) satisfying  (HE) is unbounded on J¯×[0,∞) if
(39)liminft→∞1t∫0t(1-ξt)3F(ξ)dξ=-∞,(40)limsupt→∞1t∫0t(1-ξt)3F(ξ)dξ=∞.Proof.
The hypothesis (39) implies that every solution y(t) of (7) is not bounded from below via Theorem 5. Since the hypothesis (40) implies that
(41)liminft→∞1t∫0t(1-ξt)3(-F(ξ))dξ=-limsupt→∞1t∫0t(1-ξt)3F(ξ)dξ=-∞,
we observe that every solution y(t) of (8) is not bounded from below. The conclusion follows from Theorem 2. The proof is complete.We combine Theorems3 and 4 with Theorem 5 to establish the following two theorems.Theorem 7.
Every solutionu of  (5) satisfying  (HSE) is unbounded on J¯×[0,∞) if
(42)liminft→∞1t∫0t(1-ξt)3G(ξ)dξ=-∞,limsupt→∞1t∫0t(1-ξt)3G(ξ)dξ=∞.Theorem 8.
Every solutionu of  (5) satisfying  (SE) is unbounded on J¯×[0,∞) if
(43)liminft→∞1t∫0t(1-ξt)3H(ξ)dξ=-∞,limsupt→∞1t∫0t(1-ξt)3H(ξ)dξ=∞.Remark 9.
If a solutionu of (5) is unbounded on J¯×[0,∞), then there exists a sequence {(xn,tn)}⊂J¯×[0,∞) such that
(44)limn→∞u(xn,tn)=∞.Remark 10.
In the case where(45)φ(ξ)=p|ξ|r-1ξ,p>0, and r>0 being constants, we see that φ(ξ) satisfies the hypotheses (H1)–(H3).Example 11.
We consider the Timoshenko beam equation(46)∂4u∂t4+3∂2u∂t2-6∂4u∂t2∂x2+9∂4u∂x4+13∂u∂t-13∂3u∂t∂x2+13∂5u∂t∂x4+u=f(x,t),(x,t)∈(0,π)×(0,∞)
with the hinged ends (HE), where
(47)f(x,t)=(sinx)[2t5sint+t5sin(t+π2)+5t4sint-70t4sin(t+π2)+60t3sint+240t2sin(t+π2)+120tsint].
Here L=π,α=1/3,β=γ=3,δ=ϵ=σ=1/3, and φ(u)=u. An easy computation shows that
(48)F(t)=π2[2t5sint+t5sin(t+π2)+5t4sint-70t4sin(t+π2)+60t3sint+240t2sin(t+π2)+120tsint].
Since
(49)1t∫0t(1-ξt)3ξ5sin(ξ+θ1)dξ=6tsin(ξ+θ1)+B1(t,θ1),1t∫0t(1-ξt)3ξNsin(ξ+θ2)dξ=B2(t,θ2)(N=1,2,3,4),
we observe that
(50)1t∫0t(1-ξt)3F(ξ)dξ=3πt(2sint+sin(t+π2))+B3(t),
where θi(i=1,2) are constants, Bi(t,θi)(i=1,2) and B3(t) are bounded functions on [0,∞). Hence, the conditions (39) and (40) are satisfied. It follows from Theorem 6 that every solution u of the problem (46), (HE) is unbounded on [0,π]×[0,∞). For example, u=(sinx)t5sint is such a solution.



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