Research Article | Open Access

# Global and Blow-Up Solutions for Nonlinear Hyperbolic Equations with Initial-Boundary Conditions

**Academic Editor:**D. D. Ganji

#### Abstract

We consider an initial-boundary value problem to a nonlinear string equations with linear damping term. It is proved that under suitable conditions the solution is global in time and the solution with a negative initial energy blows up in finite time.

#### 1. Introduction

We study the damped nonlinear string equation with source term : where , is a smooth function for with the initial conditions and boundary conditions The problem (1)–(3) can be regarded as modelling a nonlinear string with vertical displacement function in . And this problem has nonlinear mechanical damping of the form . The right end of the string makes it steady. The input function and the output function are applied on the left.

Wu and Li [1] studied the motion for a nonlinear beam model with nonlinear damping and external forcing terms. They showed that this model has a unique global solution and blow-up solution under the same conditions. Levine et al. [2] and Levine and Serrin [3] studied abstract version. Georgiev and Todorova [4] studied nonlinear wave equations involving the nonlinear damping term and source term of type . They proved global existence theorem with large initial data for . Hao and Li [5] studied the global solutions for a nonlinear string with boundary input and output. Dinlemez [6] proved the global existence and uniqueness of weak solutions for the initial-boundary value problem for a nonlinear wave equation with strong structural damping and nonlinear source terms in . A lot of papers in connection with blow-up, global solutions and existence of weak solutions were studied in [7–15].

In this paper we first find energy equation for the problem (1)–(3). Then we prove the solutions of the problem (1)–(3) are global in time under some conditions on the function , input , and the output . Finally we establish a blow-up result for solutions with a negative initial energy. Our approach is similar to the one in [5].

#### 2. Main Results

Now we give the following lemma for energy equation for the problem (1)–(3).

Lemma 1. *Let and be a solution of the problem (1)–(3). Then the energy equation of the problem (1)–(3) is
*

*Proof. *Multiplying (1) with and integrating over , then we get

Applying integration by parts in the right hand side of (6), we find

And using boundary conditions in equality (7), we obtain
Hence the proof is completed.

Next we give the following theorem for global solutions in time.

Theorem 2. *Assume that is a solution of the problem (1)–(3) with and*(i)* satisfies the following condition:
*(ii)*the input and the output functions satisfy
** Then the solution is global in time.*

*Proof. *Let
Differentiating with respect to and using (5), we get
Using the Cauchy-Schwarz inequality in the last term of (12), we obtain
and it follows from (12), (13), and (10) that we have
By assumption (9) and integrating over and , respectively, we yield
Furthermore, we have
and then
Combining (11), (14), (15), and (17), we get
where . Using Gronwall’s inequality, we have
Therefore together with the continuation principle and the definition of we complete the proof of Theorem 2.

Then we give the following theorem for the blow-up solutions of the problem (1)–(3).

Theorem 3. *Let be a solution of the problem (1)–(3) with . Assume that*(i)*there exists such that the function satisfies
*(ii)*the initial values satisfy *(iii)*the input and output functions satisfy
*(iv)* satisfies .**Then the solution blows up in finite time , and
**
where is some positive constant independent of the initial value and are given by (25).*

*Proof. *We define
By virtue of (5), (21), (22), and (24), we get
Taking a derivative of (25) and using (26), we have
Multiplying (1) by and integrating over the interval and then using boundary conditions (3), we obtain
From the definition of we yield

Combining (29) and (30) in (28), we get
Using (22) in (31), we obtain
Thanks to Young's inequality,
for with and , and then we get
From embedding for and using (iv), we have and putting (34) in (32) we have
From (20), we get
Choosing and , we obtain
Thanks to (21) and (27), we yield
Now we estimate . From Holder’s inequality,
then using Young's inequality again we get
where and with . And so we have
Choosing , , we obtain
Therefore we yield
where depends on and . From (37) and (43), we have
where . Integrating (44) over , then we get
Hence blows up in finite time . is given by the inequality as below:
Consequently the solution blows up in finite time. And the proof of Theorem 3 is now finished.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

The authors would like to thank the referees for the careful reading of this paper and for the valuable suggestions to improve the presentation and style of the paper.

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#### Copyright

Copyright © 2014 Ülkü Dinlemez and Esra Aktaş. 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.