Electronic Journal of Differential Equations, Vol. 2007(2007), No. 113, pp. 1–7.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)
A BLOW-UP RESULT FOR A VISCOELASTIC SYSTEM IN Rn
MOHAMMAD KAFINI, SALIM A. MESSAOUDI
Abstract. In this paper we consider a coupled system of nonlinear viscoelas- tic equations. Under suitable conditions on the initial data and the relaxation functions, we prove a finite-time blow-up result.
1. Introduction
In [7], Messaudi considered the following initial-boundary value problem utt−∆u+
Z t
0
g(t−τ)∆u(τ)dτ+ut|ut|m−2=u|u|p−2, in Ω×(0,∞) u(x, t) = 0, x∈∂Ω, t≥0
u(x,0) =u0(x), ut(x,0) =u1(x), x∈Ω,
(1.1)
where Ω is a bounded domain of Rn (n≥1) with a smooth boundary∂Ω,p >2, m≥1, and g:R+→R+ is a positive non-increasing function. He showed, under suitable conditions ong, that solutions with initial negative energy blow up in finite time if p > mand continue to exist if m≥p. This result has been later pushed, by the same author [11], to certain solutions with positive initial energy. A similar result been also obtained by Wu [15] using a different method.
In the absence of the viscoelastic term (g= 0), problem (1.1) has been extensively studied and many results concerning global existence and nonexistence have been proved. For instance, for the equation
utt−∆u+aut|ut|m=b|u|γu, in Ω×(0,∞) (1.2) m, γ ≥0, it is well known that, fora= 0, the source term bu|u|γ, (γ >0) causes finite time blow up of solutions with negative initial energy (see [1]). The interaction between the damping and the source terms was first considered by Levine [4], and in [5] the linear damping case (m= 0). He showed that solutions with negative initial energy blow up in finite time. Georgiev and Todorova [2] extended Levine’s result to the nonlinear damping case (m > 0). In their work, the authors introduced a different method and showed that solutions with negative energy continue to exist globally “in time” if m ≥ γ and blow up in finite time if γ > m and the initial energy is sufficiently negative. This last blow-up result has been extended
2000Mathematics Subject Classification. 35B05, 35L05, 35L15, 35L55.
Key words and phrases. Blow up; coupled system; relaxation function; viscoelastic.
c
2007 Texas State University - San Marcos.
Submitted February 14, 2007. Published August 18, 2007.
Supported by Project SABIC-2006/01 from KFUPM.
1
to solutions with negative initial energy by Messaoudi [8] and others. For results of same nature, we refer the reader to Levine and Serrin [3], and Vitillaro [12], Messaoudi and Said-Houari [10].
For problem (1.2) inRn, we mention, among others, the work of Levine Serrin and Park [6], Todorova [12, 13], Messaoudi [9], and Zhou [16].
In this work, we are concerned with the Cauchy problem utt−∆u+
Z t
0
g(t−s)∆u(x, s)ds=f1(u, v), inRn×(0,∞) vtt−∆v+
Z t
0
h(t−s)∆v(x, s)ds=f2(u, v), inRn×(0,∞) u(x,0) =u0(x), ut(x,0) =u1(x), x∈Rn
v(x,0) =v0(x), vt(x,0) =v1(x), x∈Rn
(1.3)
where g, h, u0, u1, v0, v1 are functions to be specified later. This type of prob- lems arises in viscoelasticity and in systems governing the longitudinal motion of a viscoelastic configuration obeying a nonlinear Boltzmann’s model. Our aim is to extend the result of [16], established for the wave equation, to our problem. To achieve this goal some conditions have to be imposed on the relaxation functionsg andh.
2. Preliminaries
In this section we present some material needed in the proof of our main result.
So, we make the following assumption
(G1) g, h:R+→R+ are nonincreasing differentiable functions satisfying 1−
Z ∞
0
g(s)ds=l >0, g0(t)≤0, t≥0.
1− Z ∞
0
h(s)ds=k >0, h0(t)≤0, t≥0.
(G2) There exists a functionI(u, v)≥0 such that
∂I
∂u =f1(u, v), ∂I
∂v =f2(u, v).
(G3) There exists a constantρ >2 such that Z
Rn
[uf1(u, v) +vf2(u, v)−ρI(u, v)]dx≥0.
(G4) There exists a constantd >0 such that
|f1(ξ, ς)| ≤d(|ξ|β1+|ς|β2), ∀(ξ, ς)∈R2,
|f2(ξ, ς)| ≤d(|ξ|β3+|ς|β4), ∀(ξ, ς)∈R2, where
βi≥1, (n−2)βi≤n, i= 1,2,3,4.
Note that (G1) is necessary to guarantee the hyperbolicity of the system (1.3). As an example of functions satisfying (G2)-(G4), we have
I(u, v) =a
ρ|u−v|ρ, ρ >2, (n−2)ρ≤2(n−1).
Condition (G4) is necessary for the existence of a local solution to (1.3).
We introduce the “modified” energy functional E(t) :=1
2kutk22+1
2kvtk22+1 2(1−
Z t
0
g(s)ds)k∇uk22+1 2(1−
Z t
0
h(s)ds)k∇vk22 +1
2(g◦ ∇u) +1
2(h◦ ∇v)− Z
Rn
I(u, v)dx,
(2.1) where
(g◦ ∇u)(t) = Z t
0
g(t−τ)||u(t)− ∇u(τ)||22dτ.
(h◦ ∇v)(t) = Z t
0
h(t−τ)||∇v(t)− ∇v(τ)||22dτ.
(2.2)
3. Blow up results In this section we state and prove our main result.
Theorem 3.1. Assume that (G1)–(G4)hold and that maxnZ +∞
0
g(s)ds, Z +∞
0
h(s)dso
≤ ρ(ρ−2)
1 +ρ(ρ−2). (3.1) Then for initial data (u0, v0),(u1, v1)∈H1(RN)×L2(RN), with compact support, satisfying
E(0) = 1
2ku1k22+1
2k∇u0k22+1
2kv1k22+1
2k∇v0k22− Z
Rn
I(u0, v0)dx <0, (3.2) the corresponding solution (of (1.3))blows up in finite time.
Proof. Multiplying (1.3) byutandvtrespectively, integrating over Rn, using inte- gration by parts, and repeating the same computations as in [7], we obtain
E0(t) = 1
2(g0◦ ∇u) +1
2(h0◦ ∇v)−1
2g(s)k∇uk22−1
2h(s)k∇vk22≤0. (3.3) Hence,
E(t)≤E(0)<0. (3.4)
We then define
F(t) =1 2
Z
Rn
[|u(x, t)|2+|v(x, t)|2]dx+1
2β(t+t0)2, (3.5) fort0>0 andβ >0 to be chosen later. By differentiatingF twice we get
F0(t) = Z
Rn
(utu+vtv)dx+β(t+t0), (3.6) F00(t) =
Z
Rn
(uttu+vttv)dx+ Z
Rn
(|ut|2+|vt|2)dx+β. (3.7)
To estimate the termR
Rn(uttu+vttv)dxin (3.7), we multiply the equations in (1.3) byuandv respectively and integrate by parts overRn to get
Z
Rn
(uutt+vvtt)dx=− Z
Rn
(|∇u|2+|∇v|2)dx+ Z
Rn
[uf1(u, v) +vf2(u, v)]dx +
Z t
0
g(t−s) Z
Rn
∇u(x, t).∇u(x, s)dx ds +
Z t
0
h(t−s) Z
Rn
∇v(x, t).∇v(x, s)dx ds.
Using Young’s inequality and (G3) we arrive at Z
Rn
(uutt+vvtt)dx≥h
−1−δ+ Z t
0
g(s)dsi
k∇uk22+ρ Z
Rn
I(u, v)dx
− 1 4δ
Z t
0
g(s)ds
(g◦ ∇u) +h
−1−δ+ Z t
0
h(s)dsi k∇vk22
− 1 4δ
Z t
0
h(s)ds
(h◦ ∇v) + Z
Rn
|ut|2+|vt|2 dx,
(3.8) we then insert (3.8) in (3.7) to obtain
F00(t)≥(−1−δ+ Z t
0
g(s)ds)k∇uk22− 1 4δ(
Z t
0
g(s)ds)(g◦ ∇u) + (−1−δ+
Z t
0
h(s)ds)k∇vk22− 1 4δ(
Z t
0
h(s)ds)(h◦ ∇v) +ρ
Z
Rn
I(u, v)dx+ 2 Z
Rn
(|ut|2+|vt|2)dx+β.
(3.9)
Now, we exploit (2.1) to substitute forR
RnI(u, v)dx, thus (3.9) takes the form F00(t)≥ −ρE(t) +β+ [(−1−δ+
Z t
0
g(s)ds) +ρ 2
1− Z t
0
g(s)ds ]k∇uk22 + [(−1−δ+
Z t
0
h(s)ds) +ρ 2(1−
Z t
0
h(s)ds)]k∇vk22 +hρ
2 − 1 4δ
Z t
0
g(s)dsi
(g◦ ∇u) +hρ
2 − 1 4δ
Z t
0
h(s)dsi
(h◦ ∇v) + (ρ
2 + 2)[kutk22+kvtk22].
(3.10)
At this point, we introduce
G(t) :=F−γ(t),
forγ >0 to be chosen properly. By differentiatingGtwice we arrive at G0(t) =−γF−(γ+1)(t)F0(t), G00(t) =−γF−(γ+2)(t)Q(t),
where
Q(t) =F(t)F00(t)−(γ+ 1)(F0)2(t)
≥F(t)n
−ρE(t) +β+h
(−1−δ+ρ 2)−(ρ
2 −1) Z t
0
g(s)ds)i k∇uk22 +h
(−1−δ+ρ 2)−(ρ
2 −1) Z t
0
h(s)ds)i k∇vk22 +ρ
2 − 1 4δ(
Z t
0
g(s)ds)
(g◦ ∇u) +ρ 2− 1
4δ( Z t
0
h(s)ds)
(h◦ ∇v) + (ρ
2 + 2)[kutk22+kvtk22]o
−(γ+ 1)hZ
Rn
(utu+vtv)dx+β(t+t0)i2 . (3.11) Using Young’s and Cauchy-Schwartz inequalities, we estimate the last term in (3.11) as follows:
hZ
Rn
(utu+vtv)dx+β(t+t0)i2
≤Z
Rn
(utu+vtv)dx2 + 2hε
2 Z
Rn
(utu+vtv)dx2 + 1
2εβ2(t+t0)2i
+β2(t+t0)2
≤(1 +ε)(
Z
Rn
(utu+vtv)dx)2+ (1 + 1
ε)β2(t+t0)2
≤(1 +ε)hZ
Rn
u2dx+ Z
Rn
v2dxihZ
Rn
u2tdx+ Z
Rn
v2tdxi + (1 +1
ε)β2(t+t0)2
≤2F(x)h (1 +ε)
Z
Rn
u2tdx+ Z
Rn
v2tdx
+ (1 + 1 ε)βi
.
Hence, (3.11) becomes Q(t)≥F(t)nh
(−1−δ+ρ 2)−(ρ
2 −1) Z t
0
g(s)ds)i k∇uk22 +h
(−1−δ+ρ 2)−(ρ
2 −1) Z t
0
h(s)ds)i k∇vk22 + [ρ
2− 1 4δ
Z t
0
g(s)ds](g◦ ∇u) + [ρ 2 − 1
4δ Z t
0
h(s)ds](h◦ ∇v) +ρ
2 + 2−2(γ+ 1)(1 +ε
kutk22+kvtk22
−ρE0−2(γ+ 1)(1 +1 ε)βo
, ∀ε >0.
(3.12)
We chooseε < ρ/4, 0< γ <(ρ−4ε)/(4(1 +ε)), andβ small so that
−ρE0−[2 +2
ε+γ(2 +2
ε)]β ≥0.
Next, we chooseδ >0 so that
−1−δ+ Z t
0
g(s)ds+ρ 2(1−
Z t
0
g(s)ds)≥0, ρ 2 − 1
4δ Z t
0
g(s)ds≥0, and
−1−δ+ Z t
0
h(s)ds+ρ 2 1−
Z t
0
h(s)ds
≥0, ρ 2− 1
4δ Z t
0
h(s)ds≥0.
This is, of course, possible by (3.1), we then conclude, from (3.12), thatQ(t)≥0, for allt≥0. ThereforeG00(t)≤0 for allt≥0; which implies thatG0 is decreasing.
By choosingt0large enough we get F0(0) =
Z
Rn
(u0u1+v0v1)dx+βt0>0, henceG0(0)<0. Finally Taylor expansion ofGyields
G(t)≤G(0) +tG0(0), ∀t≥0,
which shows thatG(t) vanishes at a time tm ≤ −GG(0)0(0). ConsequentlyF(t) must
become infinite at timetm.
Acknowledgments. The authors would like to express their sincere thanks to King Fahd University of Petroleum and Minerals for its support.
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Mohammad Kafini
Dammam Community College, Dammam, Saudi Arabia E-mail address:[email protected]
Salim A. Messaoudi
Mathematical Sciences Department, KFUPM, Dhahran 31261, Saudi Arabia E-mail address:[email protected]
