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

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 ∞ 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: 𝜕𝜌𝐴²𝑢𝜕𝑡²𝜕+𝐸𝐼⁴𝑢𝜕𝑥⁴=0(0<𝑥<𝐿,𝑡>0) (1) 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: 𝜌²𝐼𝑘₁𝐺𝜕⁴𝑢𝜕𝑡⁴𝜕+𝜌𝐴²𝑢𝜕𝑡²−𝜌𝐼+𝐸𝐼𝑘₁𝐺𝜕⁴𝑢𝜕𝑡²𝜕𝑥²𝜕+𝐸𝐼⁴𝑢𝜕𝑥⁴=𝑞(𝑥,𝑡) (2) (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 𝛿𝜕𝑢,𝜕𝑡 (3) and the internal damping terms 𝜕−𝜖³𝑢𝜕𝑡𝜕𝑥²𝜕,𝜎⁵𝑢𝜕𝑡𝜕𝑥⁴, (4) we obtain the Timoshenko beam equation 𝜕⁴𝑢𝜕𝑡⁴𝜕+𝛼𝛽𝛾²𝑢𝜕𝑡²−𝜕𝛽+𝛾⁴𝑢𝜕𝑡²𝜕𝑥²𝜕+𝛽𝛾⁴𝑢𝜕𝑥⁴+𝛿𝜕𝑢𝜕𝜕𝑡−𝜖³𝑢𝜕𝑡𝜕𝑥²𝜕+𝜎⁵𝑢𝜕𝑡𝜕𝑥⁴(𝑢)(),()∈()×(),+𝜑=𝑓𝑥,𝑡𝑥,𝑡0,𝐿0,∞ (5) 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(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 a solution of (5), one means a function 𝑢∈𝐶(𝐽×[0,∞)) such that the partial derivatives 𝜕⁴𝑢𝜕𝑡⁴,𝜕⁴𝑢𝜕𝑡²𝜕𝑥²,𝜕⁴𝑢𝜕𝑥⁴,𝜕⁵𝑢𝜕𝑡𝜕𝑥⁴ (6) 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 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 𝑢()=𝜕0,𝑡²𝑢𝜕𝑥²()()=𝜕0,𝑡=𝑢𝐿,𝑡²𝑢𝜕𝑥²(𝐿,𝑡)=0. (HE)

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 𝑦⁽⁴⁾(𝑡)+𝜋𝛼𝛽𝛾+𝛽+𝛾𝐿²𝑦^′′(𝑡)+𝜋𝛿+𝜖𝐿²𝜋+𝜎𝐿⁴𝑦^′(𝑡)𝜋+𝛽𝛾𝐿⁴𝑦(𝑡)−2𝐿𝜋𝜑𝑀≤𝐹(𝑡),𝑦⁽⁴⁾(𝑡)+𝜋𝛼𝛽𝛾+𝛽+𝛾𝐿²𝑦^′′(𝑡)+𝜋𝛿+𝜖𝐿²𝜋+𝜎𝐿⁴𝑦^′(𝑡)𝜋+𝛽𝛾𝐿⁴𝑦(𝑡)−2𝐿𝜋𝜑𝑀≤−𝐹(𝑡) (7)(8) are not bounded from below, where 𝐹(𝑡)=^𝐿₀𝑓()𝜋𝑥,𝑡sin𝐿𝑥𝑑𝑥. (9)

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 |𝑢(𝑥,𝑡)|≤𝑀,(𝑥,𝑡)∈𝐽×[0,∞), (10) that is, −𝑀≤𝑢(𝑥,𝑡)≤𝑀,(𝑥,𝑡)∈𝐽×[0,∞). (11) First we consider the case where −𝑀≤𝑢(𝑥,𝑡) on 𝐽×[0,∞). It follows from the hypotheses (H2) and (H3) that 𝜑(𝑢)≥𝜑(−𝑀)=−𝜑(𝑀) (12) and therefore we see from (5) that 𝜕⁴𝑢𝜕𝑡⁴𝜕+𝛼𝛽𝛾²𝑢𝜕𝑡²−𝜕𝛽+𝛾⁴𝑢𝜕𝑡²𝜕𝑥²𝜕+𝛽𝛾⁴𝑢𝜕𝑥⁴+𝛿𝜕𝑢𝜕𝜕𝑡−𝜖³𝑢𝜕𝑡𝜕𝑥²𝜕+𝜎⁵𝑢𝜕𝑡𝜕𝑥⁴−𝜑(𝑀)≤𝑓(𝑥,𝑡) (13) on 𝐽×[0,∞). Multiplying (13) by 𝜓(𝑥)=sin(𝜋𝑥/𝐿) and then integrating over [0,𝐿], we derive 𝑑⁴𝑑𝑡^4𝐿₀(𝑥)𝑑𝑢𝜓𝑑𝑥+𝛼𝛽𝛾²𝑑𝑡^2𝐿₀(𝑥)−𝑑𝑢𝜓𝑑𝑥𝛽+𝛾²𝑑𝑡^2𝐿₀𝜕²𝑢𝜕𝑥²𝜓(𝑥)𝑑𝑥+𝛽𝛾^𝐿₀𝜕⁴𝑢𝜕𝑥⁴𝜓(𝑥)𝑑𝑑𝑥+𝛿𝑑𝑡^𝐿₀(𝑥)𝑑𝑢𝜓𝑑𝑥−𝜖𝑑𝑡^𝐿₀𝜕²𝑢𝜕𝑥²𝜓(𝑥)𝑑𝑑𝑥+𝜎𝑑𝑡^𝐿₀𝜕⁴𝑢𝜕𝑥⁴𝜓(𝑥)𝑑𝑥−2𝐿𝜋𝜑(𝑀)≤^𝐿₀𝑓(𝑥,𝑡)𝜓(𝑥)𝑑𝑥,𝑡≥0. (14) Integrating by parts and using (HE), we obtain ^𝐿₀𝜕²𝑢𝜕𝑥²𝜓(𝑥)𝜋𝑑𝑥=−𝐿^2𝐿₀𝑢𝜓(𝑥)𝑑𝑥,_𝐿0𝜕⁴𝑢𝜕𝑥⁴𝜓(𝑥)𝜋𝑑𝑥=𝐿^4𝐿₀𝑢𝜓(𝑥)𝑑𝑥. (15) Combining (14) and (15) yields 𝑈⁽⁴⁾(𝑡)+𝜋𝛼𝛽𝛾+𝛽+𝛾𝐿²𝑈^′′(𝑡)+𝜋𝛿+𝜖𝐿²𝜋+𝜎𝐿⁴𝑈^′(𝑡)𝜋+𝛽𝛾𝐿⁴𝑈(𝑡)−2𝐿𝜋𝜑(𝑀)≤𝐹(𝑡), (16) where 𝑈(𝑡)=∫^𝐿₀𝑢𝜓(𝑥)𝑑𝑥. It is easy to check that 𝑈(𝑡)=^𝐿₀𝑢𝜓(𝑥)𝑑𝑥≥−2𝐿𝜋𝑀, (17) 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 𝜕⁴𝑣𝜕𝑡⁴𝜕+𝛼𝛽𝛾²𝑣𝜕𝑡²−𝜕𝛽+𝛾⁴𝑣𝜕𝑡²𝜕𝑥²𝜕+𝛽𝛾⁴𝑣𝜕𝑥⁴+𝛿𝜕𝑣𝜕𝜕𝑡−𝜖³𝑣𝜕𝑡𝜕𝑥²𝜕+𝜎⁵𝑣𝜕𝑡𝜕𝑥⁴+𝜑(𝑣)=−𝑓(𝑥,𝑡) (18) 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)=∫^𝐿₀𝑣𝜓(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 𝑢()=𝜕0,𝑡²𝑢𝜕𝑥²(0,𝑡)=𝜕𝑢()=𝜕𝜕𝑥𝐿,𝑡³𝑢𝜕𝑥³(𝐿,𝑡)=0. (HSE)

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 𝑦⁽⁴⁾(𝑡)+𝜋𝛼𝛽𝛾+𝛽+𝛾2𝐿²𝑦^′′(𝑡)+𝜋𝛿+𝜖2𝐿²𝜋+𝜎2𝐿⁴𝑦^′(𝑡)𝜋+𝛽𝛾2𝐿⁴𝑦(𝑡)−2𝐿𝜋𝜑𝑀≤𝐺(𝑡),𝑦⁽⁴⁾(𝑡)+𝜋𝛼𝛽𝛾+𝛽+𝛾2𝐿²𝑦^′′(𝑡)+𝜋𝛿+𝜖2𝐿²𝜋+𝜎2𝐿⁴𝑦^′(𝑡)𝜋+𝛽𝛾2𝐿⁴𝑦(𝑡)−2𝐿𝜋𝜑𝑀≤−𝐺(𝑡) (19)(20) are not bounded from below, where 𝐺(𝑡)=^𝐿₀𝑓()𝜋𝑥,𝑡sin2𝐿𝑥𝑑𝑥. (21)

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 −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 ^𝐿₀𝜕²𝑢𝜕𝑥²𝜔(𝑥)𝜋𝑑𝑥=−2𝐿^2𝐿₀𝑢𝜔(𝑥)𝑑𝑥,_𝐿0𝜕⁴𝑢𝜕𝑥⁴𝜔(𝑥)𝜋𝑑𝑥=2𝐿^4𝐿₀𝑢𝜔(𝑥)𝑑𝑥. (22) As in the proof of Theorem 2, we observe that ∫^𝐿₀𝑢𝜔(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 ∫^𝐿₀(−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 𝜕𝑢()=𝜕𝜕𝑥0,𝑡³𝑢𝜕𝑥³(0,𝑡)=𝜕𝑢()=𝜕𝜕𝑥𝐿,𝑡³𝑢𝜕𝑥³(𝐿,𝑡)=0. (SE)

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 𝑦⁽⁴⁾(𝑡)+𝛼𝛽𝛾𝑦^′′(𝑡)+𝛿𝑦^′(𝑡)−𝐿𝜑𝑀≤𝐻(𝑡),𝑦⁽⁴⁾(𝑡)+𝛼𝛽𝛾𝑦^′′(𝑡)+𝛿𝑦^′(𝑡)−𝐿𝜑𝑀≤−𝐻(𝑡) (23)(24) are not bounded from below, where 𝐻(𝑡)=^𝐿₀𝑓(𝑥,𝑡)𝑑𝑥. (25)

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 ^𝐿₀𝜕²𝑢𝜕𝑥²𝑑𝑥=𝜕𝑢()−𝜕𝑥𝐿,𝑡𝜕𝑢()𝜕𝑥0,𝑡=0,_𝐿0𝜕⁴𝑢𝜕𝑥⁴𝜕𝑑𝑥=³𝑢𝜕𝑥³()−𝜕𝐿,𝑡³𝑢𝜕𝑥³(0,𝑡)=0. (26) Hence, we have the inequality 𝑈⁽⁴⁾(𝑡)+𝛼𝛽𝛾𝑈^′′(𝑡)+𝛿𝑈^′(𝑡)−𝐿𝜑(𝑀)≤𝐻(𝑡),𝑡≥0, (27) where 𝑈(𝑡)=∫^L₀𝑢𝑑𝑥. 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 ∫^𝐿₀(−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 𝑦⁽⁴⁾(𝑡)+𝑘𝑦^′′(𝑡)+ℓ𝑦^′(𝑡)+𝑚𝑦(𝑡)−𝜇≤𝑔(𝑡),𝑡≥0, (28) 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 solution 𝑦(𝑡) of (28) is not bounded from below if liminf_𝑡→∞1𝑡^𝑡₀𝜉1−𝑡³𝑔(𝜉)𝑑𝜉=−∞. (29)

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 𝑦⁽⁴⁾(𝑡)+𝑘𝑦^′′(𝑡)+ℓ𝑦^′(𝑡)−𝑀𝑚−𝜇≤𝑔(𝑡),𝑡≥0. (30) Integrating (30) over [0,𝑡], we get 𝑦⁽³⁾(𝑡)+𝑘𝑦^′(𝑡)+ℓ𝑦(𝑡)−𝑀𝑚𝑡−𝜇𝑡≤𝑐₁+^𝑡₀𝑔(𝜉)𝑑𝜉, (31) and hence 𝑦⁽³⁾(𝑡)+𝑘𝑦^′(𝑡)−𝑀ℓ−𝑀𝑚𝑡−𝜇𝑡≤𝑐₁+^𝑡₀𝑔(𝜉)𝑑𝜉, (32) where 𝑐₁=𝑦⁽³⁾(0)+𝑘𝑦^′(0)+ℓ𝑦(0). Integration of (32) over [0,𝑡] yields 𝑦^′′(𝑡)+𝑘𝑦(𝑡)−𝑀ℓ𝑡−𝑀𝑚2𝑡²−𝜇2𝑡²≤𝑐₀+𝑐₁𝑡+^𝑡₀𝑑𝑠₁^𝑠1₀𝑔(𝜉)𝑑𝜉, (33) that is, 𝑦^′′(𝑡)−𝑀𝑘−𝑀ℓ𝑡−𝑀𝑚2𝑡²−𝜇2𝑡²≤𝑐₀+𝑐₁𝑡+^𝑡₀𝑑𝑠₁^𝑠1₀𝑔(𝜉)𝑑𝜉, (34) where 𝑐₀=𝑦^′′(0)+𝑘𝑦(0). Integrating (34) over [0,𝑡] twice, we obtain 𝑦(𝑡)−𝑀𝑘2𝑡²−𝑀ℓ6𝑡³−𝑀𝑚𝑡24⁴−𝜇𝑡24⁴≤𝑑₀+𝑑₁𝑡+𝑑₂𝑡²+𝑑₃𝑡³+^𝑡₀𝑑𝑠₃^𝑠3₀𝑑𝑠₂^𝑠2₀𝑑𝑠₁^𝑠1₀𝑔(𝜉)𝑑𝜉, (35) and therefore −𝑀−𝑀𝑘2𝑡²−𝑀ℓ6𝑡³−𝑀𝑚𝑡24⁴−𝜇𝑡24⁴≤𝑑₀+𝑑₁𝑡+𝑑₂𝑡²+𝑑₃𝑡³+16^𝑡₀(𝑡−𝜉)³𝑔(𝜉)𝑑𝜉, (36) in view of the identity ^𝑡₀𝑑𝑠₃^𝑠3₀𝑑𝑠₂^𝑠2₀𝑑𝑠₁^𝑠1₀𝑔(𝜉)1𝑑𝜉=6^𝑡₀(𝑡−𝜉)³𝑔(𝜉)𝑑𝜉, (37) where 𝑑₀=𝑦(0),𝑑₁=𝑦^′(0),𝑑₂=𝑐₀/2, and 𝑑₃=𝑐₁/6. Dividing (36) by 𝑡⁴ yields −𝑀𝑡⁴−𝑀𝑘2𝑡²−𝑀ℓ−6𝑡𝑀𝑚−𝜇24−𝑑24𝑡⁰⁴+𝑑₁𝑡³+𝑑₂𝑡²+𝑑₃𝑡≤16𝑡^𝑡₀𝜉1−𝑡³𝑔(𝜉)𝑑𝜉. (38) 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 2–4 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 liminf_𝑡→∞1𝑡^𝑡₀𝜉1−𝑡³𝐹(𝜉)𝑑𝜉=−∞,limsup_𝑡→∞1𝑡^𝑡₀𝜉1−𝑡³𝐹(𝜉)𝑑𝜉=∞. (39)(40)

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 liminf_𝑡→∞1𝑡^𝑡₀𝜉1−𝑡³(−𝐹(𝜉))𝑑𝜉=−limsup_𝑡→∞1𝑡^𝑡₀𝜉1−𝑡³𝐹(𝜉)𝑑𝜉=−∞, (41) 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 liminf_𝑡→∞1𝑡^𝑡₀𝜉1−𝑡³𝐺(𝜉)𝑑𝜉=−∞,limsup_𝑡→∞1𝑡^𝑡₀𝜉1−𝑡³𝐺(𝜉)𝑑𝜉=∞. (42)

Theorem 8. Every solution 𝑢 of (5) satisfying (SE) is unbounded on 𝐽×[0,∞) if liminf_𝑡→∞1𝑡^𝑡₀𝜉1−𝑡³𝐻(𝜉)𝑑𝜉=−∞,limsup_𝑡→∞1𝑡^𝑡₀𝜉1−𝑡³𝐻(𝜉)𝑑𝜉=∞. (43)

Remark 9. If a solution 𝑢 of (5) is unbounded on 𝐽×[0,∞), then there exists a sequence {(𝑥_𝑛,𝑡_𝑛)}⊂𝐽×[0,∞) such that lim_𝑛→∞𝑢𝑥_𝑛,𝑡_𝑛=∞. (44)

Remark 10. In the case where 𝜑(𝜉)=𝑝𝜉^𝑟−1𝜉, (45)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 𝜕⁴𝑢𝜕𝑡⁴𝜕+3²𝑢𝜕𝑡²𝜕−6⁴𝑢𝜕𝑡²𝜕𝑥²𝜕+9⁴𝑢𝜕𝑥⁴+13𝜕𝑢−1𝜕𝑡3𝜕³𝑢𝜕𝑡𝜕𝑥²+13𝜕⁵𝑢𝜕𝑡𝜕𝑥⁴(),()∈()×()+𝑢=𝑓𝑥,𝑡𝑥,𝑡0,𝜋0,∞ (46) with the hinged ends (HE), where 𝑓(𝑥,𝑡)=(sin𝑥)2𝑡⁵sin𝑡+𝑡⁵𝜋sin𝑡+2+5𝑡⁴sin𝑡−70𝑡⁴𝜋sin𝑡+2+60𝑡³sin𝑡+240𝑡²𝜋sin𝑡+2+120𝑡sin𝑡. (47) Here 𝐿=𝜋,𝛼=1/3,𝛽=𝛾=3,𝛿=𝜖=𝜎=1/3, and 𝜑(𝑢)=𝑢. An easy computation shows that 𝐹(𝑡)=𝜋22𝑡⁵sin𝑡+𝑡⁵𝜋sin𝑡+2+5𝑡⁴sin𝑡−70𝑡⁴𝜋sin𝑡+2+60𝑡³sin𝑡+240𝑡²𝜋sin𝑡+2+120𝑡sin𝑡. (48) Since 1𝑡^𝑡₀𝜉1−𝑡³𝜉⁵sin𝜉+𝜃₁𝑑𝜉=6𝑡sin𝜉+𝜃₁+𝐵₁𝑡,𝜃₁,1𝑡^𝑡₀𝜉1−𝑡³𝜉^𝑁sin𝜉+𝜃₂𝑑𝜉=𝐵₂𝑡,𝜃₂(𝑁=1,2,3,4), (49) we observe that 1𝑡∫^𝑡₀𝜉1−𝑡³𝐹(𝜉)𝜋𝑑𝜉=3𝜋𝑡2sin𝑡+sin𝑡+2+𝐵₃(𝑡), (50) where 𝜃_𝑖(𝑖=1,2) are constants, 𝐵_𝑖(𝑡,𝜃_𝑖)(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.