In this paper, we study the initial-boundary value problem for a system of nonlinear viscoelastic Petrovsky equations. Introducing suitable perturbed energy functionals and using the potential well method we prove uniform decay of solution energy under some restrictions on the initial data and the relaxation functions. Moreover, we establish a growth result for certain solutions with positive initial energy. | Turk J Math (2014) 38: 87 – 109 ¨ ITAK ˙ c TUB ⃝ Turkish Journal of Mathematics doi: Research Article Global existence, uniform decay, and exponential growth of solutions for a system of viscoelastic Petrovsky equations Faramarz TAHAMTANI∗, Amir PEYRAVI Department of Mathematics, College of Sciences, Shiraz University, Shiraz, 71454, Iran Received: • • Accepted: Published Online: • Printed: Abstract: In this paper, we study the initial-boundary value problem for a system of nonlinear viscoelastic Petrovsky equations. Introducing suitable perturbed energy functionals and using the potential well method we prove uniform decay of solution energy under some restrictions on the initial data and the relaxation functions. Moreover, we establish a growth result for certain solutions with positive initial energy. Key words: Global existence, uniform decay, exponential growth, viscoelastic Petrovsky equation 1. Introduction In this paper, we investigate the following initial-boundary value problem: |ut |ρ utt + ∆2 u − ∆utt − (g1 ∗ ∆2 u)(t) − ∆ut + |ut |p−1 ut = f1 (u, v), (x, t) ∈ Ω × [0, T ), ρ 2 2 q−1 |vt | vtt + ∆ v − ∆vtt − (g2 ∗ ∆ v)(t) − ∆vt + |vt | vt = f2 (u, v), (x, t), ∈ Ω × [0, T ), u(x, 0) = u0 (x), ut (x, 0) = u1 (x), x ∈ Ω, v(x, 0) = v (x), v (x, 0) = v (x), x ∈ Ω, 0 t 1 u(x, t) = ∂ u(x, t) = 0, v(x, t) = ∂ν v(x, t) = 0, (x, t) ∈ ∂Ω × [0, T ), ν () where ρ > 0 , p, q ≥ 1, T > 0, Ω is a bounded domain of Rn (n = 1, 2, 3) with a smooth boundary ∂Ω so that the divergence theorem can be applied, ν denotes the outward normal derivative, g1 and g2 are positive functions satisfying some conditions to be specified later, and ∫ t (gi ∗ ϕ)(t) = gi (t − τ )ϕ(τ )dτ, i = 1, 2. 0 By taking [ ] r−3 r+1 f1 (u, v) = (r + 1) a|u + v|r−1 (u + v) + b|u| 2 |v| 2 u , [ ] r−3 r+1 f2 (u, v) = (r + 1) a|u + v|r−1 (u + v) + b|v| 2 |u| 2
