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

Authors: Ülkü Dinlemez; Esra Aktaş
Journal: International Journal of Differential Equations
 (2014)

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## 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.


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## Body

## 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 0≤s 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 ℝ. 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 [1] 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. [2] and Levine and Serrin [3] studied abstract version. Georgiev and Todorova [4] 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<p≤m. 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 ℝ. 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 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 [5].


## 2. Main Results

Now we give the following lemma for energy equation for the problem (1)–(3).Lemma 1.
Let1<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)=12∥ut∥22-1α+2∥u∥α+2α+2+12∫01∫0|ux|2σ(ξ)dξdx,(5)ddtE(t)=ϕ2(t)-ψ2(t)-∥ut∥22.Proof.
Multiplying (1) with ut and integrating over (0,1), then we get
(6)ddt{12∥ut∥22-1α+2∥u∥α+2α+2}=∫01(σ(|ux|2)ux)xutdx-∥ut∥22.
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)-12ddt∫01‍∫0|ux|2σ(ξ)dξdx.
And using boundary conditions in equality (7), we obtain
(8)ddt{12∥ut∥22-1α+2∥u∥α+2α+2+12∫01∫0|ux|2σ(ξ)dξdx}=ϕ2(t)-ψ2(t)-∥ut∥22.
Hence the proof is completed.Next we give the following theorem for global solutions in time.Theorem 2.
Assume thatu(x,t) is a solution of the problem (1)–(3) with 1<α and(i)
σ(s) satisfies the following condition:
(9)|s|α≤σ(s),fors∈ℝ+∪{0},(ii)
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+2∥u∥α+2α+2=12∥ut∥22+12∫01∫0|ux|2σ(ξ)dξdx+1α+2∥u∥α+2α+2.
Differentiating G(t) with respect to t and using (5), we get
(12)ddtG(t)=ϕ2(t)-ψ2(t)-∥ut∥22+2∫01|u|αuutdx.
Using the Cauchy-Schwarz inequality in the last term of (12), we obtain
(13)2∫01|u|αuutdx≤2∫01|u|α+1|ut|dx≤∥u∥2(α+1)2(α+1)+∥ut∥22,
and it follows from (12), (13), and (10) that we have
(14)ddtG(t)≤∥u∥2(α+1)2(α+1)+∥ut∥22.
By assumption (9) and integrating over (0,|ux|2) and (0,1), respectively, we yield
(15)1α+1∥ux∥2(α+1)2(α+1)≤∫01∫0|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)dξ≤∫01|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)dt≤1ξ1G(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.
Letu(x,t) be a solution of the problem (1)–(3) with 1<α. Assume that(i)
there exists1<ɛ<(α+2)/2 such that the function σ(s) satisfies
(20)σ(s)s≤ɛ2∫0sσ(ζ)dζforsϵℝ+∪{0},(ii)
the initial values satisfy(21)E(0)≤0,0<∫01u0(x)u1(x)dx,(iii)
the input and output functions satisfy(22)ϕ2(t)≤ψ2(t),(ψ(t)+ϕ(t))(∫0t(ψ(s)-ϕ(s))ds+u0(0))≤0,(iv)
u(x,t) satisfies 1≤∥u∥.
Then the solutionu(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=∥ut∥22+ψ2(t)-ϕ2(t)≥0,(27)0≤M(0)≤M(t),for0≤t.
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)(∥ut∥22+ψ2(t)-ϕ2(t))+∥ut∥22+∫01uuttdx.
Multiplying (1) by u and integrating over the interval [0,1] and then using boundary conditions (3), we obtain
(29)∫01uuttdx=∥u∥α+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)+ɛ2∥ut∥2+ɛ2∫01∫0|ux|2σ(ξ)dξdx-ɛα+2∥u∥α+2α+2.
Combining (29) and (30) in (28), we get
(31)dN(t)dt=(1-γ)M-γ(t)(∥ut∥22+ψ2(t)-ϕ2(t))+∥ut∥22+∥u∥α+2α+2-∫01uutdx-(ψ(t)+ϕ(t))×(∫0t(ψ(s)-ϕ(s))ds+u0(0))-∫01σ(|ux|2)ux2dx+ɛM(t)+ɛ2∥ut∥22+ɛ2∫01∫0|ux|2σ(ξ)dξdx-ɛα+2∥u∥α+2α+2.
Using (22) in (31), we obtain
(32)(1+ɛ2)∥ut∥22+(12-ɛα+2)∥u∥α+2α+2+12∥u∥α+2α+2+∫01(ɛ2∫0|ux|2‍σ(ξ)dξ-σ(|ux|2)ux2)dx+ɛM(t)-∫01uutdx≤dN(t)dt.
Thanks to Young's inequality,
(33)AB≤δppAp+δ-qqBq,0≤A,B∀0<δ,1p+1q=1,
for ∫01‍uutdx with p=q=2 and γ=2, and then we get
(34)∫01uutdx≤∫01|uut|dx≤∥ut∥2+14∥u∥2.
From embedding for Lp(0,1) and using (iv), we have ∥u∥22≤∥u∥α+2α+2 and putting (34) in (32) we have
(35)(1+ɛ2)∥ut∥22+(12-ɛα+2)∥u∥α+2α+2+12∥u∥22+∫01(ɛ2∫0|ux|2σ(ξ)dξ-σ(|ux|2)ux2)dx+ɛM(t)-∥ut∥22-14∥u∥22≤dN(t)dt.
From (20), we get
(36)ɛM(t)+ɛ2∥ut∥22+(12-ɛα+2)∥u∥α+2α+2+14∥u∥22≤dN(t)dt.
Choosing ɛ and κ=min⁡{ɛ/2,(1/2-ɛ/(α+2)),1/4}, we obtain
(37)κ{M(t)+∥ut∥22+∥u∥α+2α+2+∥u∥22}≤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|≤∥u∥2∥ut∥2≤∥u∥α+2∥ut∥2;
then using Young's inequality again we get
(40)|∫01uutdx|≤δ2(1-γ)2(1-γ)∥ut∥22(1-γ)+1-2γ2(1-γ)δ-2(1-γ)/(1-2γ)∥u∥α+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)∥ut∥22hhhhhhhhhh+(1-2γ2(1-γ))1/(1-γ)δ-2/(1-2γ)∥u∥α+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(∥ut∥22+∥u∥α+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)+∥ut∥22+∥u∥α+2α+2+∥u∥22),
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)Tmax⁡≤α+4αη(N(0))-α/(α+4).
Consequently the solution blows up in finite time. And the proof of Theorem 3 is now finished.



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*Source: 724837-2014-04-13.xml*