Tamtam Proceedings - lamsin

122 Bouzahirwhere the linear operator D ε : B → E is defined for ε ∈ [0, +∞), by{Dε = W ε ◦ τ ε ,τ ε (ϕ)(θ) = ϕ(θ − ε), for ϕ ∈ B and θ ∈ (−∞, 0] .We impose also that (T 0 (t)) t≥0is exponentially stable, i.e.,(H3) there exist two positive constants ¯M and ω 0 such that‖T 0 (t)‖ D(A)≤ ¯Me −ω0t , t ≥ 0,and that the operator D is stable in the sense that(H4) For all T > 0, there exist positive constants a 1 , α and β such that for all g ∈C([0, +∞) , E), the integral solution of the functional equation{ Dxt = g(t), t ∈ [0, T ] ,x 0 = ψ, where ψ ∈ B and Dψ = g(0),satisfies the inequality ‖x t ‖ B≤ a 1 sup |g(s)| + βe −αt ‖ψ‖ Bfor all t ∈ [0, T ] and0≤s≤tψ ∈ B with Dψ = g(0).Consider the following system{ ∂∂t Du t = ADu t if t ≥ 0,u(θ) = ϕ (θ) if θ ∈ (−∞, 0] with ϕ ∈ B.(5)Using equality (4), we can see that a necessary condition for u : (−∞, b) → E, b > 0, tobe a solution of Eq. (5) is that it verifies the following integrated one on (−∞, b){Dut = T 0 (t)Dϕ, t ≥ 0,(6)u 0 = ϕ,whereϕ ∈ Y :={}ϕ ∈ B : Dϕ ∈ D(A) .The following fundamental inequality is essential.Lemma 1. Suppose that Conditions (H2) and (H4) are fulfilled and let ψ ∈ B andh ∈ C ([0, T ] ; E), T > 0 such that Dψ = h(0). Then, for any ε > 0 sufficiently smallsuch that K(0) ‖D 0 ‖ < 1 and M(s) < 1 for all s ∈ [0, t − ε], t ∈ [0, T ] , there exista > 0 and functions b, c, d ∈ L ∞ ([0, T ] , IR + ) such that the solution v of the equation{ Dvt = h(t), t ∈ [0, T ] ,(7)v(t) = ψ(t), t ∈ (−∞, 0]satisfies the inequality‖v t ‖ B≤ e[b(t) −a(t−ε) ‖ψ‖ B+ c(t) sup |h(s)|0≤s≤ε]+ d(t) sup |h(s)| , t ∈ [ε, T ] .ε≤s≤t(8)TAMTAM –Tunis– 2005