Logarithmic stabilization of the Euler–Bernoulli transmission plate equation with locally distributed Kelvin–Voigt damping

Abstract

In this paper we will study the asymptotic behaviour of the energy decay of a transmission plate equation with locally distributed Kelvin-Voigt feedback. Precisly, we shall prove that the energy decay at least logarithmically over the time. The originality of this method comes from the fact that using a Carleman estimate for a transmission second order system which will be derived from the plate equation to establish a resolvent estimate which provide, by the famous Burq's result [Bur98], the kind of decay mentionned above.

Stabilization for the Wave Equation with Singular Kelvin–Voigt Damping

“…According to [7], [11] or [12] we can find four weight functions ϕ 1,1 , ϕ 1,2 , ϕ 2,1 and ϕ 2,2 , a finite number of points x i j,k where B(x i j,k , 2ε) ⊂ Ω j for all j, k = 1, 2 and i = 1, . .…”

Exponential stability for a coupled system of damped–undamped plate equations

“…In fact, due to the invertibility of the Dirichlet Laplacian in Ω, the norms ∆u L 2 (Ω) and u H 2 (Ω) are equivalent on the space H 2 (Ω) ∩ H 1 0 (Ω). As H 2 0 (Ω) is a closed subspace of H 2 (Ω) ∩ H 1 0 (Ω), these norms are also equivalent on H 2 0 (Ω), and now the assertion follows from part a) (see also [11], Proposition 2.1 and Proposition 2.2).…”

“…Concerning the higher-order transmission conditions (2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14) and (2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14)(15), note that for all (u 1 , u 2 ) ∈ H 4 (Ω 1 )×H 4 (Ω 2 ) satisfying (2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12) and (2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) all tangential derivatives of u 1 −u 2 and ∂ ν u 1 −∂ ν u 2 along Γ disappear. Therefore, for such u the transmission conditions (2-14)-(2-15) are equivalent to the conditions…”

“…We will show that (u 1 , u 2 ) = 0. We multiply (2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14)(15)(16) and (2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14)(15)(16)(17) with u 1 and u 2 , respectively. Summing up and performing an integration by parts yields…”

“…Because of (2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14)(15)(16)(17)(18)(19), the trivial extension u 2 by zero to R n belongs to H 4 (R n ) and satisfies ∆ 2 u 2 = λ 2 u 2 in R n . As ∆ 2 in L 2 (R n ) has no eigenvalues, this implies u 2 = 0 and therefore u 2 = 0.…”