# On Asymptotic Behavior and Blow-Up of Solutions for a Nonlinear Viscoelastic Petrovsky Equation with Positive Initial Energy

Authors: Gang Li; Yun Sun; Wenjun Liu
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

This paper deals with the initial boundary value problem for the nonlinear viscoelastic Petrovsky equationutt+Δ2u−∫0tgt−τΔ2ux,τdτ−Δut−Δutt+utm−1ut=up−1u. Under certain conditions on g and the assumption that m<p, we establish some asymptotic behavior and blow-up results for solutions with positive initial energy.


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## Body

## 1. Introduction

In this work, we investigate the following nonlinear viscoelastic Petrovsky problem:(1)utt+Δ2u-∫0t‍g(t-τ)Δ2u(x,τ)dτ-Δut-Δutt+|ut|m-1ut=|u|p-1u,x∈Ω,t≥0,u(x,t)=∂u(x,t)∂ν=0,x∈∂Ω,t≥0,u(x,0)=u0(x),ut(x,0)=u1(x),x∈Ω,
where Ω is a bounded domain in ℝn (n≥1) with a smooth boundary ∂Ω, m,p>1; ν is the unit outer normal on ∂Ω; and g is a nonnegative memory term.In the absence of viscoelastic term (i.e.,g=0), Guesmia [1] considered the following equation:
(2)utt+Δ2u+q(x)u(x,t)+g(ut)=0,x∈Ω,t>0,
where q:Ω→ℝ+ is a bounded function and g is continuous and increasing function satisfying g(0)=0. Under suitable growth conditions on g, the author established global existence, uniqueness, and decay results by using the semigroup method. Messaoudi [2] investigated a nonlinearly damped semilinear Petrovsky equation
(3)utt+Δ2u+a|ut|m-2ut=b|u|p-2u,x∈Ω,t>0,
where a,b>0 and p,m>2 and proved that the solution is global when m≥p while the solution blows up in finite time with negative initial energy when m<p. Later, this blow-up result was improved by Chen and Zhou [3] with positive initial energy. For a related study, we may see the work of Wu and Tsai [4]. In [5], Amroun and Benaissa studied (3) by generalizing the damping term into the form of g(ut) and obtained the global existence of the solutions by means of the stable set method combined with the Faedo-Galerkin procedure. Very recently, in the presence of the strong damping, Li et al. [6] considered the following Petrovsky equation:
(4)utt+Δ2u-Δut+|ut|m-1ut=|u|p-1u,x∈Ω,t≥0.
Without any interaction between m and p, the authors obtained the global existence and uniform decay of solutions when the initial data are in some stable set. And a blow-up result was established when m<p and the initial energy is less than the potential well depth.In the presence of the viscoelastic term (i.e.,g≠0), Muñoz Rivera et al. [7] studied the following equation:
(5)utt+Δ2u-∫0t‍g(t-τ)Δ2u(x,τ)dτ-γΔutt=0,x∈Ω,t≥0.
They proved that the memory effect produces strong dissipation capable of making uniform rate of decay for the energy. Later, in the presence of strong damping term, M. M. Cavalcanti et al. [8] considered
(6)|ut|ρutt-Δu+∫0t‍g(t-τ)Δu(x,τ)dτ-γΔut-Δutt=0,x∈Ω,t≥0,
and obtained a global existence for γ≥0 and uniform exponential decay for γ>0. This work was extended by Messaoudi and Tatar [9] to a situation where a nonlinear source term is competing with the damping induced by -γΔut and the integral term. Then in the case of γ=0, the same authors [10] showed that the damping induced by the viscoelastic term is enough to ensure global existence and uniform decay of solutions provided that the initial data are in some stable set by introducing a new functional and using the potential well method. Recently, Wu [11] improved [10] by considering the nonlinear equation:
(7)|ut|ρutt-Δu-Δutt+∫0t‍g(t-τ)Δu(x,τ)dτ+|ut|mut=|u|pu,
and a general decay result was obtained. In the presence of strong damping term Δut and dispersive term Δutt, Xu et al. [12] considered the initial boundary value problem for the following viscoelastic wave equation:
(8)utt-Δu+∫0t‍g(t-τ)Δu(x,τ)dτ-Δut-Δutt+ut=|u|p-1u.
By introducing a family of potential wells, the authors not only obtained the invariant sets, but also proved the existence and nonexistence of global weak solution under some conditions with low initial energy. Furthermore, they established a blow-up result for certain solutions with arbitrary positive initial energy (high energy case). Very recently, Tahamtani and Peyravi [13] considered problem (1) and obtained the exponential decay of the energy under some assumptions on g without any interaction between source term and damping term. Under an appropriate restriction on g, they also proved that the Lp+1 norm of any solution grows as an exponential function if m<p and the initial energy is negative. For other related works, we refer the readers to [14–24] and the references therein.Motivated by the above works, in this paper, we intend to consider problem (1) and establish some asymptotic behavior and blow-up results for solutions with positive initial energy. For our purpose, we use the functional L(t)=∥u∥22+∥∇u∥22 and give a modified manner to estimate the term |∫Ω‍u|ut|m-1utdx| so that the appearance of the form like γ=RH-σ(t) (for constants γ, R, and σ) which has been used in many earlier works (e.g., in [3, 19, 25]) can be avoided.The paper is organized as follows. In Section2 we present some assumptions and known results and state the main results. Section 3 is devoted to proof the the blow-up result—Theorem 4.


## 2. Preliminaries and Main Results

In this section, we first present some assumptions and known results which will be used throughout this work.For the relaxation functiong, we give the following assumptions:(G1)
g(t):ℝ+→ℝ+ is a C1 function such that
(9)g(0)>0,1-∫0∞‍g(s)ds=l>0,(G2)
g′(t)≤0, for all t≥0.Lemma 1 (Sobolev-Poincaré inequality).
Letq be a number with 2≤q<∞(n=1,2,3,4) or 2≤q≤2n/(n-4)(n≥5); then for u∈H02(Ω) there exists a positive C*=C(Ω,q) such that
(10)∥u∥q≤C*∥Δu∥2.We assume thatm,p satisfy
(11)1<p<∞(n=1,2,3,4)or1<p≤nn-4(n≥5),1<m<∞(n=1,2,3,4)or1<m≤n+4n-4(n≥5).
We state a local existence theorem that can be established by adopting the arguments of [2, 5, 22].Theorem 2.
Assume that (11), (G1), and (G2) hold. Let (u0,u1)∈H02(Ω)×H01(Ω) be given. Then there exists a unique weak solution u(t) such that
(12)u∈C([0,T];H02(Ω))∩C1([0,T];L2(Ω)),ut∈L2([0,T];H01(Ω))∩Lm+1(Ω×(0,T)),
for T>0 small enough.Next, we define the following functionals:(13)I(t)≔I(u(t))=(1-∫0t‍g(τ)dτ)∥Δu∥22+(g∘Δu)(t)-∥u∥p+1p+1,J(t)≔J(u(t))=12(1-∫0t‍g(τ)dτ)∥Δu∥22+12(g∘Δu)(t)-1p+1∥u∥p+1p+1,E(t):=E(u(t))=12∥ut∥22+12∥∇ut∥22+J(t),
where
(14)(g∘v)(t)=∫0t‍∫Ω‍g(t-τ)|v(t)-v(τ)|2dxdτ.Remark 3.
A multiplication of (1) by ut and integration over Ω easily yield
(15)E′(t)=-12g(t)∥Δu∥22-∥∇ut∥22+12(g′∘Δu)(t)-∥ut∥m+1m+1≤-∥∇ut∥22-∥ut∥m+1m+1≤0,
since g′(t)≤0.
Our main results read as follows.Theorem 4.
Suppose that (11), (G1), and (G2) hold and m<p. Let u be the unique solution to problem (1) and denote d1=((p-1)/2(p+1))(l/C*2)(p+1)/(p-1). Then, for any fixed 0<θ<1, E(0)=θd1<d1, I(0)<0, and g satisfies
(16)∫0∞‍g(τ)dτ<(p-1)(1-θ)p-1+(1/(p+1-θ(p-1))),
either there exists some T* such that the solution of problem (1) blows up in T* in the sense of limt→T*-E(t)=-∞, or one has limt→+∞E(t)=-∞.


## 3. Proof of the Main Results

We denote(17)d(t)=infu∈H02(Ω)∖{0}supλ≥0J(λu),𝒩={u∈H02(Ω)∖{0}:I(u(t))=0},
then we can prove the following lemma.Lemma 5.
Fort≥0, we have
(18)0<d1≤d(t)≤d2(u)=supλ≥0J(λu),(19)d(t)=infu∈𝒩J(u).Proof.
Obviously,(20)d(t)≤d2(u)=supλ≥0J(λu).
Straightforward computations yield
(21)J(λu)=λ22[(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)]-λp+1p+1∥u(t)∥p+1p+1,
then
(22)ddλJ(λu)=λ[(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)]-λp∥u(t)∥p+1p+1.
Let
(23)ddλJ(λu)=0,
which leads to
(24)λ1-=0,λ2-=[(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)]1/(p-1)(∥u(t)∥p+1p+1)1/(p-1).
An elementary calculation shows
(25)d2dλ2J(λ1-u)>0,d2dλ2J(λ2-u)<0.
Using (G1) and Lemma 1 we arrive at
(26)supλ≥0J(λu)=J(λ2-u)=p-12(p+1)×[(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)(∥u(t)∥p+1p+1)2/(p+1)](p+1)/(p-1)≥p-12(p+1)[l∥Δu(t)∥22(∥u(t)∥p+1p+1)2/(p+1)](p+1)/(p-1)≥p-12(p+1)(lC*2)(p+1)/(p-1)=d1>0,
which implies that d(t)≥d1.
To get (19), straightforward computations lead to
(27)(1-∫0t‍g(τ)dτ)∥Δ(λ2-u)∥22+(g∘Δ(λ2-u))(t)-∥(λ2-u)∥p+1p+1=(1-∫0t‍g(τ)dτ)×[(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)]2/(p-1)(∥u(t)∥p+1p+1)2/(p-1)×∥Δu(t)∥22+[(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)]2/(p-1)(∥u(t)∥p+1p+1)2/(p-1)×(g∘Δu)(t)-[(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)](p+1)/(p-1)(∥u(t)∥p+1p+1)(p+1)/(p-1)×∥u(t)∥p+1p+1=[(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)](p+1)/(p-1)∥u(t)∥p+12(p+1)/(p-1)×{(1-∫0t‍g(τ)dτ)∥Δu(t)∥22(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)222+(g∘Δu)(t)(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)2222222-∥u(t)∥p+1p+1∥u(t)∥p+1p+1}=0,
which implies that λ2-u∈𝒩. Also, for any u∈𝒩, we note that
(28)λ2-(u)=[(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)]1/(p-1)(∥u(t)∥p+1p+1)1/(p-1)=1.
Therefore we have λ2-(u)u=u for all u∈𝒩. Hence, we complete the proof.Lemma 6.
Suppose that (11), (G1), and  (G2) hold; E(0)<d1; and I(0)<0; then we have I(t)<0, for all t∈[0,T) and
(29)d1<p-12(p+1)[(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)]<p-12(p+1)∥u∥p+1p+1.Proof.
Using (15), we have E(t)<d1, for all t∈[0,T). To show that I(t)<0 on [0,T), we proceed by contradiction. Assume there exists t0∈[0,T) such that I(t0)≥0. Since I(0)<0, it follows that there exists some t~∈(0,t0] such that I(u(t~))=0. Now, we define
(30)t*=inf{t~∈(0,t0]:(1-∫0t~‍g(τ)dτ)2222×∥Δu(t~)∥22+(g∘Δu)(t~)=∥u(t~)∥p+1p+1(1-∫0t~‍g(τ)dτ)}.
Then, we have I(u(t*))=0 and
(31)(1-∫0t‍g(τ)dτ)∥Δu∥22+(g∘Δu)(t)<∥u∥p+1p+1,∀0≤t<t*.
Suppose that∥Δu(t*)∥22=0, by the regularity of u(t), we have
(32)limt→t*-∥Δu(t)∥22=0.
On the other hand, applying Lemma 1 to (31), we obtain
(33)(1-∫0t‍g(τ)dτ)∥Δu∥22+(g∘Δu)(t)<∥u∥p+1p+1≤C*p+1∥Δu∥2p+1,0≤t<t*.
From the above inequality, we can easily get ∥Δu(t)∥22≠0, for all t∈[0,t*), and
(34)lC*p+1<∥Δu∥2p-1,
which implies
(35)limt→t*-∥Δu(t)∥2>(lC*p+1)1/(p-1)
and this contradicts to (32).
Suppose that∥Δu(t*)∥22≠0. Applying Lemma 5, we see that d(t) is the infimum of J(u(t)) over all functions u in 𝒩 and J(u(t*))≥d(t)≥d1, which contradicts to J(u(t*))≤E(t*)<d1. Thus, we conclude that I(t)<0 for all t∈[0,T).
To get (29), using Lemma 5, (31) and the conclusion that I(t)<0, for all t∈[0,T), we obtain
(36)d1≤p-12(p+1)×([(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)]×[(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)]2/(p-1)×((∥u∥p+1p+1)2/(p-1))-1)<p-12(p+1)[(1-∫0t‍g(τ)dτ)∥Δu(t)∥22+(g∘Δu)(t)]<p-12(p+1)∥u∥p+1p+1,0≤t<T,
which completes the proof.Proof of Theorem4.
Suppose on the contrary that there exists a positive constantC0 such that E(t)≥-C0, for all t≥0. An integration of (15) over [0,t] yields
(37)d+C0>E(0)-E(t)≥∫0t‍(∥∇ut(τ)∥22+∥ut(τ)∥m+1m+1)dτ,∀t≥0.
Set L(t)=∥u∥22+∥∇u∥22; then we have
(38)L′(t)=2∫Ω‍uutdx+2∫Ω‍∇u·∇utdx,(39)L′′(t)=2∫Ω‍ut2dx+2∫Ω‍uuttdx+2∫Ω‍∇ut2dx+2∫Ω‍∇u·∇uttdx=2(∥ut∥22+∥∇ut∥22-∥Δu∥22)+2∫Ω‍Δu∫0t‍g(t-τ)Δu(x,τ)dτdx-2∫Ω‍∇u·∇utdx-2∫Ω‍u|ut|m-1utdx+2∥u∥p+1p+1.
We estimate (39) as follows. By using Cauchy-Schwarz inequality and Young's inequality, we get for any η>0(40)∫Ω‍∫0t‍g(t-τ)Δu(t)Δu(τ)dτdx=∫Ω‍∫0t‍g(t-τ)Δu(t)(Δu(τ)-Δu(t))dτdx+(∫0t‍g(τ)dτ)∥Δu(t)∥22≥-∫0t‍g(t-τ)∥Δu(t)∥2∥Δu(τ)-Δu(t)∥2dτ+(∫0t‍g(τ)dτ)∥Δu(t)∥22≥-η(g∘Δu)(t)+(1-14η)(∫0t‍g(τ)dτ)∥Δu(t)∥22.
Using Young's inequality, we get for any ε1>0(41)|2∫Ω‍∇u·∇utdx|≤ε1∥∇u∥22+C(ε1)∥∇ut∥22.
By Hölder and Young inequalities, we arrive at
(42)|2∫Ω‍uut|ut|m-1dx|≤2∥u∥m+1∥ut∥m+1m≤2|Ω|(p-m)/((p+1)(m+1))∥u∥p+1∥ut∥m+1m≤2|Ω|(p-m)/((p+1)(m+1))∥u∥p+11-(p+1)/(m+1)×∥u∥p+1(p+1)/(m+1)∥ut∥m+1m<2|Ω|(p-m)/((p+1)(m+1))(2(p+1)d1p-1)(1/(p+1))-(1/(m+1))×∥u∥p+1(p+1)/(m+1)∥ut∥m+1m≤Cd(ε2∥u∥p+1p+1+C(ε2)∥ut∥m+1m+1),
where Cd=2|Ω|(p-m)/((p+1)(m+1))(2(p+1)d1/(p-1))(1/(p+1))-(1/(m+1)). Combining (40)–(42), we obtain
(43)L′′(t)+C(ε1)∥∇ut∥22+CdC(ε2)∥ut∥m+1m+1>2(∥ut∥22+∥∇ut∥22)-2[1-(1-14η)(∫0t‍g(τ)dτ)]∥Δu(t)∥22-2η(g∘Δu)(t)+(2-Cdε2)∥u∥p+1p+1-ε1∥∇u∥22≥2(∥ut∥22+∥∇ut∥22)-2[1+ε1B2-(1-14η)(∫0t‍g(τ)dτ)]∥Δu(t)∥22-2η(g∘Δu)(t)+(2-Cdε2)∥u∥p+1p+1,
where B is the Poincaré constant. Since
(44)-2[1+ε1B2-(1-14η)(∫0t‍g(τ)dτ)]∥Δu(t)∥22-2η(g∘Δu)(t)+(2-Cdε2)∥u∥p+1p+1≥-2[1+ε1B2-(1-14η)(∫0t‍g(τ)dτ)]∥Δu(t)∥22-2η(g∘Δu)(t)+(2-Cdε2)∥u∥p+1p+1+σ(E(t)-E(0))≥{σ2-2-ε1B-[σ2-2(1-14η)]∫0t‍g(τ)dτ}×∥Δu(t)∥22+σ2(∥ut∥22+∥∇ut∥22)+(σ2-2η)(g∘Δu)(t)+(2-Cdε2-σp+1)∥u∥p+1p+1-σE(0),∀t≥0,
where σ>0 will be chosen later. Then, we have
(45)L′′(t)+C(ε1)∥∇ut∥22+CdC(ε2)∥ut∥m+1m+1>(2+σ2)(∥ut∥22+∥∇ut∥22)+{σ2-2-ε1B-[σ2-2(1-14η)]∫0t‍g(τ)dτ}×∥Δu(t)∥22+(σ2-2η)(g∘Δu)(t)+(2-Cdε2-σp+1)∥u∥p+1p+1-σE(0),2222222222222222222222222222222∀t≥0.
Choosing η<(σ/4)-(θ(p-1)/2) and ε1 small enough and using (16), we can get a certain constant σ∈(2θ(p-1)+4,2(p+1)), such that
(46){σ2-2-ε1B-[σ2-2(1-14η)]∫0t‍g(τ)dτ}>θ(p-1),σ2-2η>θ(p-1).
Therefore, it follows from (G1) and Lemma 6 that
(47){σ2-2-ε1B-[σ2-2(1-14η)]∫0t‍g(τ)dτ}∥Δu(t)∥22+(σ2-2η)(g∘Δu)(t)-σE(0)>θ(p-1)(∥Δu(t)∥22+(g∘Δu)(t))-σE(0)>2θ(p+1)d1-σE(0)≥2(p+1)E(0)-σE(0)>0,
since the choice of σ. Finally, we take ε2 small enough so that 2-Cdε2-(σ/(p+1)):=C1>0 and use Lemma 6 to get
(48)L′′(t)+C(ε1)∥∇ut∥22+CdC(ε2)∥ut∥m+1m+1>C1∥u∥p+1p+1>2C1(p+1)p-1d1.
Integrate twice the above inequality over [0,t] and use (37); we have
(49)L(t)>C1(p+1)d1p-1t2+[L′(0)-(C(ε1)+C(ε2)(C0+d1))]t+L(0),
which implies that L(t) grows more quickly than the linear growth for t≥0.
On the other hand, by (37) and Hölder's inequality, we have
(50)∥u(t)∥2≤∥u0∥2+∫0t∥ut(τ)∥2dτ≤∥u0∥2+C(C0+d1)1/(m+1)t(m-1)/(m+1),∥∇u∥2≤∥∇u0∥2+∫0t‍∥∇ut(τ)∥2dτ≤∥∇u0∥2+(C0+d1)1/2.
Thus,
(51)L(t)=∥u∥22+∥∇u∥22≤C2t2(m-1)/(m+1)+C3t+C4,∀t≥0,
where C2, C3, and C4 are positive constants. Obviously, (51) contradicts (49).



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