IJDE International Journal of Differential Equations 1687-9651 1687-9643 Hindawi Publishing Corporation 724837 10.1155/2014/724837 724837 Research Article Global and Blow-Up Solutions for Nonlinear Hyperbolic Equations with Initial-Boundary Conditions http://orcid.org/0000-0002-5656-3924 Dinlemez Ülkü 1 http://orcid.org/0000-0002-7662-881X Aktaş Esra 2 Ganji D. D. 1 Department of Mathematics Faculty of Science Gazi University Teknikokullar Ankara Turkey gazi.edu.tr 2 Incirli Mahallesi Karaelmas Sokak Yunusemre Caddesi 51/18 İncirli Ankara Turkey 2014 1342014 2014 24 12 2013 07 03 2014 20 03 2014 13 4 2014 2014 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.

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 |u|αu: (1)utt+ut=(σ(|ux|2)ux)x+|u|αu,hhhhhhhhh(x,t)(0,1)×[0,T], where 1<α, σ(s) is a smooth function for 0s with the initial conditions (2)u(x,0)=u0(x),ut(x,0)=u1(x),x[0,1], and boundary conditions (3)u(1,t)=0,t(0,T),σ(|ux(0,t)|2ux(0,t))-ut(0,t)=2ϕ(t),t[0,T],σ(|ux(0,t)|2ux(0,t))+ut(0,t)=2ψ(t),t[0,T]. The problem (1)–(3) can be regarded as modelling a nonlinear string with vertical displacement function u(x,t) in . And this problem has nonlinear mechanical damping of the form |u|αu. The right end of the string makes it steady. The input ϕ(t) function and the output ψ(t) function are applied on the left.

Wu and Li  studied the motion for a nonlinear beam model with nonlinear damping a|ϕt|m-1ϕt and external forcing b|ϕ|p-1ϕ terms. They showed that this model has a unique global solution and blow-up solution under the same conditions. Levine et al.  and Levine and Serrin  studied abstract version. Georgiev and Todorova  studied nonlinear wave equations involving the nonlinear damping term |ut|m-1ut and source term of type |ut|p-1ut. They proved global existence theorem with large initial data for 1<pm. Hao and Li  studied the global solutions for a nonlinear string with boundary input and output. Dinlemez  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 .

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 σ(s), input ϕ(t), and the output ψ(t). Finally we establish a blow-up result for solutions with a negative initial energy. Our approach is similar to the one in .

2. Main Results

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

Lemma 1.

Let 1<α and u(x,t) be a solution of the problem (1)–(3). Then the energy equation of the problem (1)–(3) is (4)E(t)=12ut22-1α+2uα+2α+2+12010|ux|2σ(ξ)dξdx,(5)ddtE(t)=ϕ2(t)-ψ2(t)-ut22.

Proof.

Multiplying (1) with ut and integrating over (0,1), then we get (6)ddt{12ut22-1α+2uα+2α+2}=01(σ(|ux|2)ux)xutdx-ut22.

Applying integration by parts in the right hand side of (6), we find (7)01(σ(|ux|2)ux)xutdx=-σ(|ux(0,t)|2)ux(0,t)ut(0,t)-12ddt010|ux|2σ(ξ)dξdx.

And using boundary conditions in equality (7), we obtain (8)ddt{12ut22-1α+2uα+2α+2+12010|ux|2σ(ξ)dξdx}=ϕ2(t)-ψ2(t)-ut22. Hence the proof is completed.

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

Theorem 2.

Assume that u(x,t) is a solution of the problem (1)–(3) with 1<α and

σ(s) satisfies the following condition: (9)|s|ασ(s),fors+{0},

the input and the output functions satisfy (10)ϕ2(t)ψ2(t).

Then the solution u(x,t) is global in time.

Proof.

Let (11)G(t):=E(t)+2α+2uα+2α+2=12ut22+12010|ux|2σ(ξ)dξdx+1α+2uα+2α+2. Differentiating G(t) with respect to t and using (5), we get (12)ddtG(t)=ϕ2(t)-ψ2(t)-ut22+201|u|αuutdx. Using the Cauchy-Schwarz inequality in the last term of (12), we obtain (13)201|u|αuutdx201|u|α+1|ut|dxu2(α+1)2(α+1)+ut22, and it follows from (12), (13), and (10) that we have (14)ddtG(t)u2(α+1)2(α+1)+ut22. By assumption (9) and integrating over (0,|ux|2) and (0,1), respectively, we yield (15)1α+1ux2(α+1)2(α+1)010|ux|2σ(ξ)dξdx. Furthermore, we have (16)|u(x,t)|2(α+1)=|x1uξ(ξ,t)dξ|2(α+1)x1|uξ(ξ,t)|2(α+1)dξ01|ux(x,t)|2(α+1)dx=ux(x,t)2(α+1)2(α+1), and then (17)u(x,t)2(α+1)2(α+1)ux(x,t)2(α+1)2(α+1). Combining (11), (14), (15), and (17), we get (18)dG(t)dt1ξ1G(t), where ξ1=min{1/2,1/(α+1)}. Using Gronwall’s inequality, we have (19)G(t)G(0)e(1/ξ1)t. Therefore together with the continuation principle and the definition of G(t) 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 u(x,t) be a solution of the problem (1)–(3) with 1<α. Assume that

there exists 1<ɛ<(α+2)/2 such that the function σ(s) satisfies (20)σ(s)sɛ20sσ(ζ)dζforsϵ+{0},

the initial values satisfy     (21)E(0)0,0<01u0(x)u1(x)dx,

the input and output functions satisfy (22)ϕ2(t)ψ2(t),(ψ(t)+ϕ(t))(0t(ψ(s)-ϕ(s))ds+u0(0))0,

u(x,t) satisfies 1u.

Then the solution u(x,t) blows up in finite time Tmax, and (23)Tmax(α+4αη)N-α/(α+4)(0), where η is some positive constant independent of the initial value α and N(t) are given by (25).

Proof.

We define (24)M(t):=-E(t),γ:=α2(α+2),(25)N(t):=M1-γ(t)+01u(x,t)ut(x,t)dx. By virtue of (5), (21), (22), and (24), we get (26)dM(t)dt=ut22+ψ2(t)-ϕ2(t)0,(27)0M(0)M(t),for0t. Taking a derivative of (25) and using (26), we have (28)dN(t)dt=(1-γ)M-γ(t)M(t)+01ut2dx+01uuttdx=(1-γ)M-γ(t)(ut22+ψ2(t)-ϕ2(t))+ut22+01uuttdx. Multiplying (1) by u and integrating over the interval [0,1] and then using boundary conditions (3), we obtain (29)01uuttdx=uα+2α+2-01uutdx-(ψ(t)+ϕ(t))×(0t(ψ(s)-ϕ(s))ds+u0(0))-01σ(|ux|2)ux2dx. From the definition of M(t) we yield (30)0=ɛM(t)+ɛ2ut2+ɛ2010|ux|2σ(ξ)dξdx-ɛα+2uα+2α+2.

Combining (29) and (30) in (28), we get (31)dN(t)dt=(1-γ)M-γ(t)(ut22+ψ2(t)-ϕ2(t))+ut22+uα+2α+2-01uutdx-(ψ(t)+ϕ(t))×(0t(ψ(s)-ϕ(s))ds+u0(0))-01σ(|ux|2)ux2dx+ɛM(t)+ɛ2ut22+ɛ2010|ux|2σ(ξ)dξdx-ɛα+2uα+2α+2. Using (22) in (31), we obtain (32)(1+ɛ2)ut22+(12-ɛα+2)uα+2α+2+12uα+2α+2+01(ɛ20|ux|2σ(ξ)dξ-σ(|ux|2)ux2)dx+ɛM(t)-01uutdxdN(t)dt. Thanks to Young's inequality, (33)ABδppAp+δ-qqBq,0A,B0<δ,1p+1q=1, for 01uutdx with p=q=2 and γ=2, and then we get (34)01uutdx01|uut|dxut2+14u2. From embedding for Lp(0,1) and using (iv), we have u22uα+2α+2 and putting (34) in (32) we have (35)(1+ɛ2)ut22+(12-ɛα+2)uα+2α+2+12u22+01(ɛ20|ux|2σ(ξ)dξ-σ(|ux|2)ux2)dx+ɛM(t)-ut22-14u22dN(t)dt. From (20), we get (36)ɛM(t)+ɛ2ut22+(12-ɛα+2)uα+2α+2+14u22dN(t)dt. Choosing ɛ and κ=min{ɛ/2,(1/2-ɛ/(α+2)),1/4}, we obtain (37)κ{M(t)+ut22+uα+2α+2+u22}dN(t)dt. Thanks to (21) and (27), we yield (38)0<N(0)N(t),0<t. Now we estimate [N(t)]1/(1-γ). From Holder’s inequality, (39)|01uutdx|u2ut2uα+2ut2; then using Young's inequality again we get (40)|01uutdx|δ2(1-γ)2(1-γ)ut22(1-γ)+1-2γ2(1-γ)δ-2(1-γ)/(1-2γ)uα+22(1-γ)/(1-2γ), where 0<δ and 1/p+1/q=1 with p=2(1-γ). And so we have (41)|01uutdx|1/(1-γ)21/(1-γ)((1-2γ2(1-γ))1/(1-γ)δ2(2(1-γ))1/(1-γ)ut22hhhhhhhhhh+(1-2γ2(1-γ))1/(1-γ)δ-2/(1-2γ)uα+22/(1-2γ)). Choosing β=max{δ2/(1-γ)1/(1-γ), ((1-2γ)/(1-γ))1/(1-γ)δ-2/(1-2γ)}, we obtain (42)|01uutdx|1/(1-γ)β(ut22+uα+2α+2). Therefore we yield (43)(N(t))1/(1-γ)=(M1-γ(t)+01u(x,t)ut(x,t)dx)1/(1-γ)21/(1-γ)(M(t)+|01u(x,t)ut(x,t)dx|1/(1-γ))C(M(t)+ut22+uα+2α+2+u22), where C depends on δ and α. From (37) and (43), we have (44)η(N(t))1/(1-γ)dN(t)dt, where η=κ/C. Integrating (44) over (0,t), then we get (45)1(N(0))-α/(α+4)-(α/(α+4))ηt(N(t))α/(α+4). Hence N(t) blows up in finite time Tmax. Tmax is given by the inequality as below: (46)Tmaxα+4αη(N(0))-α/(α+4). 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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