IJDE International Journal of Differential Equations 1687-9651 1687-9643 Hindawi Publishing Corporation 435456 10.1155/2013/435456 435456 Research Article Unboundedness of Solutions of Timoshenko Beam Equations with Damping and Forcing Terms Kobayashi Kusuo Yoshida Norio Jaros Jaroslav Department of Mathematics University of Toyama Toyama 930-8555 Japan u-toyama.ac.jp 2013 27 3 2013 2013 15 01 2013 21 02 2013 2013 Copyright © 2013 Kusuo Kobayashi and Norio Yoshida. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 ∞ as t 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.

1. Introduction

The most fundamental beam equations are of the following form: (1)ρA2ut2+EI4ux4=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  4ut4+ρA2ut2-ρ(I+EIk1G)4ut2x2+EI4ux4=q(x,t) (see [1, page 433] and Wang and Stephens [2, page 150]).

Dividing the above equation by ρ2I/(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)δut, and the internal damping terms (4)-ϵ3utx2,σ5utx4, we obtain the Timoshenko beam equation (5)4ut4+αβγ2ut2-(β+γ)4ut2x2+βγ4ux4+δut-ϵ3utx2+σ5utx4+φ(u)=f(x,t),(x,t)(0,L)×(0,), where α,β,γ,δ,ϵ,σ, and L are positive constants.

Let J=(0,L),J-=[0,L] and we assume throughout this paper that

φ(ξ) is a real-valued continuous function in ;

φ(ξ)>0 and φ(-ξ)=-φ(ξ) for ξ>0;

φ(ξ) is a nondecreasing function in ;

f(x,t) is a real-valued continuous function on J¯×[0,).

Definition 1.

By a solution of (5), one means a function uC(J¯×[0,)) such that the partial derivatives (6)4ut4,4ut2x2,  4ux4,5utx4 exist and are continuous on J-×[0,).

Oscillations of beam equations have been treated by numerous authors; see, for example, Feireisl and Herrmann , Herrmann , Kopáčková , Kusano and Yoshida , Yoshida , and the references therein. In particular, we mention the paper  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 . We refer to Ball , Fitzgibbon , and Narazaki  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 Section 2 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 solutions u=u(x,t) are required to satisfy the boundary condition (HE)u(0,t)=2ux2(0,t)=u(L,t)=2ux2(L,t)=0.

Theorem 2.

Every solution u 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 solution u 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)-Mu(x,t)M,(x,t)J-×[0,). First we consider the case where -Mu(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)4ut4+αβγ2ut2-(β+γ)4ut2x2+βγ4ux4+δut-ϵ3utx2+σ5utx4-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)d4dt40Luψ(x)dx+αβγd2dt20Luψ(x)dx-(β+γ)d2dt20L2ux2ψ(x)dx+βγ0L4ux4ψ(x)dx+δddt0Luψ(x)dx-ϵddt0L2ux2ψ(x)dx+σddt0L4ux4ψ(x)dx-2Lπφ(M)0Lf(x,t)ψ(x)dx,t0. Integrating by parts and using (HE), we obtain (15)0L2ux2ψ(x)dx=-(πL)20Luψ(x)dx,0L4ux4ψ(x)dx=(πL)40Luψ(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)4vt4+αβγ2vt2-(β+γ)4vt2x2+βγ4vx4+δvt-ϵ3vtx2+σ5vtx4+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)=2ux2(0,t)=ux(L,t)=3ux3(L,t)=0.

Theorem 3.

Every solution u 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 solution u 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 -Mu(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)0L2ux2ω(x)dx=-(π2L)20Luω(x)dx,0L4ux4ω(x)dx=(π2L)40Luω(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)ux(0,t)=3ux3(0,t)=ux(L,t)=3ux3(L,t)=0.

Theorem 4.

Every solution u 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 -Mu(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)0L2ux2dx=ux(L,t)-ux(0,t)=0,0L4ux4dx=3ux3(L,t)-3ux3(0,t)=0. Hence, we have the inequality (27)U(4)(t)+αβγU′′(t)+δU(t)-Lφ(M)H(t),t0, 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)+y(t)+my(t)-μg(t),t0, and derive sufficient condition for every solution y(t) of (28) to be unbounded from below. It is assumed that k,,m, and μ are nonnegative constants, and g(t) is a continuous function on [0,).

Theorem 5.

Every solution y(t) of (28) is not bounded from below if (29)liminft1t0t(1-ξt)3g(ξ)dξ=-.

Proof.

Assume, on the contrary, that there exists a solution y(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)+y(t)-Mm-μg(t),t0. Integrating (30) over [0,t], we get (31)y(3)(t)+ky(t)+y(t)-Mmt-μtc1+0tg(ξ)dξ, and hence (32)y(3)(t)+ky(t)-M-Mmt-μtc1+0tg(ξ)dξ, where c1=y(3)(0)+ky(0)+y(0). Integration of (32) over [0,t] yields (33)y′′(t)+ky(t)-Mt-Mm2t2-μ2t2c0+c1t+0tds10s1g(ξ)dξ, that is, (34)y′′(t)-Mk-Mt-Mm2t2-μ2t2c0+c1t+0tds10s1g(ξ)dξ, where c0=y′′(0)+ky(0). Integrating (34) over [0,t] twice, we obtain (35)y(t)  -Mk2t2-M6t3-Mm24t4-μ24t4d0+d1t+d2t2+d3t3+0tds30s3ds20s2ds10s1g(ξ)dξ, and therefore (36)-M-Mk2t2-M6t3-Mm24t4-μ24t4d0+d1t+d2t2+d3t3+160t(t-ξ)3g(ξ)dξ, in view of the identity (37)0tds30s3ds20s2ds10s1g(ξ)dξ=160t(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-M6t-Mm24-μ24-(d0t4+d1t3+d2t2+d3t)16t0t(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 Theorems 24 with Theorem 5, we present unboundedness results for the three types of boundary value problems for (5) under consideration.

Theorem 6.

Every solution u of (5) satisfying (HE) is unbounded on J¯×[0,) if (39)liminft1t0t(1-ξt)3F(ξ)dξ=-,(40)limsupt1t0t(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)liminft1t0t(1-ξt)3(-F(ξ))dξ=-limsupt1t0t(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 Theorems 3 and 4 with Theorem 5 to establish the following two theorems.

Theorem 7.

Every solution u of (5) satisfying (HSE) is unbounded on J¯×[0,) if (42)liminft1t0t(1-ξt)3G(ξ)dξ=-,limsupt1t0t(1-ξt)3G(ξ)dξ=.

Theorem 8.

Every solution u of (5) satisfying (SE) is unbounded on J¯×[0,) if (43)liminft1t0t(1-ξt)3H(ξ)dξ=-,limsupt1t0t(1-ξt)3H(ξ)dξ=.

Remark 9.

If a solution u of (5) is unbounded on J¯×[0,), then there exists a sequence {(xn,tn)}J¯×[0,) such that (44)limnu(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)4ut4+32ut2-64ut2x2+94ux4+13ut-133utx2+135utx4+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)1t0t(1-ξt)3ξ5sin(ξ+θ1)dξ=6tsin(ξ+θ1)+B1(t,θ1),1t0t(1-ξt)3ξNsin(ξ+θ2)dξ=B2(t,θ2)(N=1,2,3,4), we observe that (50)1t0t(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.

Timoshenko S. Young D. H. Weaver W. Jr. Vibration Problems in Engineering 1974 4th New York, NY, USA John Wiley & Sons Wang T. M. Stephens J. E. Natural frequencies of Timoshenko beams on pasternak foundations Journal of Sound and Vibration 1977 51 2 149 155 2-s2.0-0017463899 Feireisl E. Herrmann L. Oscillations of a non-linearly damped extensible beam Applications of Mathematics 1992 37 6 469 478 MR1185802 Herrmann L. Vibration of the Euler-Bernoulli beam with allowance for dampings 2 Proceedings of the World Congress on Engineering 2008 London, UK Kopáčková M. On periodic solution of a nonlinear beam equation Aplikace Matematiky 1983 28 2 108 115 MR695184 Kusano T. Yoshida N. Forced oscillations of Timoshenko beams Quarterly of Applied Mathematics 1985 43 2 167 177 MR793524 Yoshida N. Forced oscillations of extensible beams SIAM Journal on Mathematical Analysis 1985 16 2 211 220 10.1137/0516016 MR777463 Yoshida N. Forced oscillations of nonlinear extensible beams Proceedings of the 10th International Conference on Nonlinear Oscillations (Varna, 1984) 1985 Sofia, Bulgaria 814 817 Yoshida N. On the zeros of solutions of beam equations Annali di Matematica Pura ed Applicata 1988 151 389 398 10.1007/BF01762806 MR964521 Yoshida N. Vibrations of Timoshenko beams with damping and forcingterms Proceedings of the IEEE International Conference on Industrial Engineering and Engineering Management of (IEEM '12) 2012 Ball J. M. Stability theory for an extensible beam Journal of Differential Equations 1973 14 399 418 MR0331921 Fitzgibbon W. E. Global existence and boundedness of solutions to the extensible beam equation SIAM Journal on Mathematical Analysis 1982 13 5 739 745 10.1137/0513050 MR668317 Narazaki T. On the time global solutions of perturbed beam equations Proceedings of the Faculty of Science of Tokai University 1981 16 51 71 MR632661