Weak stability of a laminated beam

In this paper, we consider the stability of a laminated beam equation, derived by Liu, Trogdon, and Yong [ 6 ], subject to viscous or Kelvin-Voigt damping. The model is a coupled system of two wave equations and one Euler-Bernoulli beam equation, which describes the longitudinal motion of the top and bottom layers of the beam and the transverse motion of the beam. We first show that the system is unstable if one damping is only imposed on the beam equation. On the other hand, it is easy to see that the system is exponentially stable if direct damping are imposed on all three equations. Hence, we investigate the system stability when two of the three equations are directly damped. There are a total of seven cases from the combination of damping locations and types. Polynomial stability of different orders and their optimality are proved. Several interesting properties are revealed.

(1. 10) which can be simplified into a six-order PDE for w. When the extensional motion of the bottom and top layers is neglected, we obtain the model proposed by Hansen and Spies [5].
Investigation on the qualitative properties of these models, such as stability and regularity of the solution, subject to certain damping mechanism started in the 1990's. In [6], exponential stability was proved for the Mead-Makrus model (1.9)-(1.10) when the shear stress τ and shear strain γ relation is assumed to be viscoelastic of Boltzmann type. When this relationship is of Kelvin-Voigt type, analyticity of the associated semigroup was proved by Hansen and Liu [4], which was further extended to the corresponding multi-layers beam and plate model by Allen and Hansen in [2,1]. For the Rao-Nakra model (1.6)-(1.8), exponential stability was obtained when standard boundary damping is imposed on one end of the beam for all three displacements [9]. For the model in [5], exponential stability was proved [12] when structural damping and boundary damping are added, or when viscous damping are added to all three equations [11].
In this paper, we are interested in the stability of the Rao-Nakra model (1.6)-(1.8) with weaker damping. It is clear that exponential stability holds if all three displacements are damped by viscous or Kelvin-Voigt damping distributed over the spatial domain. On the other hand, the system could be unstable if only one displacement is damped, which will be proved in next section. Therefore, we consider the case when two of the displacements are damped, i.e., (1.12) where a i , b i , c i ≥ 0, i = 1, 2 and only two of them not in the same equation are positive. At the boundary, the beam is hinged and the bottom and top layers satisfy Neumann boundary conditions. That is The initial condition is This paper is organized as follows. In Section 2, we present the semigroup setting of the system for well-posedness. Section 3 is devoted to show the polynomial stability of the system by the frequency domain method for seven different cases of damping locations and types. We continue to prove the optimality of the polynomial stability order in Section 4 by spectral analysis.
2. Preliminary and main results. Notice that (u 1 , u 3 , w) = (C, C, 0) is a static solution to system (1.11)-(1.16) for any constant C. In order to have a unique static solution, we shall choose a proper state space. Let for Y = (y 1 , · · · , y 6 ) T , Z = (z 1 , · · · , z 6 ) T ∈ H. Hereafter, we use ·, · and · to denote the usual L 2 inner product and norm. Let v 1 = u 1 t , v 3 = u 3 t , y = w t , and U = (u 1 , v 1 , u 3 , v 3 , w, y) T . Define an operator A : D(A) → H by 792 YANFANG LI, ZHUANGYI LIU AND YANG WANG We now rewrite system (1.11)-(1.17) as a first order evolution equation in H The energy associated with (2.4) is A straight forward calculation leads to Theorem 2.1. The operator A defined above is the infinitesimal generator of a C 0 -semigroup e At of contractions in the Hilbert space H.
Proof. It is obvious that D(A) is dense in H. From (2.6), the system is dissipative. Moreover, assume that 0 ∈ σ(A). Then there is a unit sequence U n = (u 1 n , v 1 n , u 3 n , v 3 n , w, y) T ∈ D(A) such that AU n H = o(1).
we conclude that U n H = o(1), which contradicts our assumption. Hence, 0 ∈ ρ(A). By Theorem 1.2.4 in [7], A is the infinitesimal generator of a C 0 -semigroup e At of contractions in the Hilbert space H.
We first present the following conclusion.
Substituting (2.7) into system (2.4), we obtain that This implies that there are infinitely many eigenvalues ±iβ n on the imaginary axis.
The above proof also leads to This has motivated us to consider the cases where two of the equations in system (2.4) are directly damped. Our main results are summarized in the next theorem. Theorem 2.3. Let e At be the semigroup associated with system (2.4). Case (I). When c 1 = c 2 = 0, (i) a 1 = b 1 = 0, a 2 , b 2 > 0, the semigroup e At is polynomially stable of order 1 3 ; (ii) a 1 , b 1 > 0, a 2 = b 2 = 0, the semigroup e At is polynomially stable of order 1 2 ; (iii) a 1 = b 2 = 0, a 2 , b 1 > 0, the semigroup e At is polynomially stable of order 1 2 . Case (II). When a 1 = a 2 = 0, (i) b 1 = c 1 = 0, b 2 , c 2 > 0, when E1 ρ1 = E3 ρ3 , i.e., the two wave equations have the same propagation speeds, the semigroup e At is polynomially stable of order 1 2 ; when E1 ρ1 = E3 ρ3 , i.e., the two wave equations have different propagation speeds, the semigroup e At is polynomially stable of order 1 4 ; (ii) b 1 = c 2 = 0, b 2 , c 1 > 0, when E1 ρ1 = E3 ρ3 , the semigroup e At is polynomially stable of order 1 2 ; when E1 ρ1 = E3 ρ3 , the semigroup e At is polynomially stable of order 1 4 ; (iii) b 1 , c 1 > 0, b 2 = c 2 = 0, the semigroup e At is polynomially stable of order 1 4 ; (iv) b 1 , c 2 > 0, b 2 = c 1 = 0, the semigroup e At is polynomially stable of order 1 4 . Here, we say that the semigroup e At is polynomially stable of order 1 k , if ∀U 0 ∈ D(A), there is a constant C > 0 such that the solution U of (2.4) satisfies Our main tool is the frequency domain characterization of polynomial stability by Borichev and Tomilov.
Theorem 2.5. [3] Let H be a Hilbert space and A generates a bounded C 0 −semi-

YANFANG LI, ZHUANGYI LIU AND YANG WANG
Then, there exists a positive constant C > 0 such that To end this section, we give the following remark on our main results, which provides useful information to the design of laminated beam. Proof. Assume that (2.11) is false, i.e., there is a λ such that iλ ∈ σ(A). Then there exist a sequence β n → λ and a unit norm sequence U n = (u 1 i.e., The second, forth and sixth equations are in spaces L 2 * , L 2 * and L 2 , respectively. For convenience, we omit the subscript n hereafter. From (3.1) and the dissipativeness of A in (2.6), , respectively, we obtain (3.11) Then, the respective inner product of (3.9)-(3.11) with u 1 , u 3 , w yields (3.14) In what follows, we will show that the above will lead to a contraction U H = o(1).
Since β is finite, by (3.2) and (3.4), we also get Using (3.15)-(3.16) in (3.12)-(3.13), also by the boundedness of w x , it is easy to see that 17) The inner product of (3.10) with w x in L 2 leads to w x = o(1), hence by the Poincaré inequality, w = o(1) and βw = o(1) since β is finite. Therefore, from (3.6) and (3.14), Since β is finite, by (3.2) and (3.4), then By the same reasoning in case (i), we can also get It is straightforward to get u 1 by the same argument used in case (i).
Since β is finite, by (3.4) and (3.6) Repeating the analysis in (3.26)-(3.30) in the case (i), we can also get u 3 Since β is finite, by (3.4) and (3.6) Repeat the analysis in (3.26)-(3.30) in the case (i), we also get u 1 = o(1) and u 1 Since β is finite, by (3.4) and (3.6) Repeating the analysis in (3.26)-(3.30) in the case (i), we can also get u 1 = o(1) and u 1 . For all the cases above we obtain U H = o(1) which contradicts with U H = 1. Hence, we conclude that iR ⊂ ρ(A).
Next, assume that (2.12) is false. Then there exist a sequence β → ∞ and a unit sequence i.e., The second, forth and sixth equations equal to o(1) in spaces L 2 * , L 2 * and L 2 , respectively. Again, we are going to show U H = o(1) for a contradiction. It follows from dissipation (3.8) and (3.37) that Next, take L 2 inner product of (3.45)-(3.47) with u 1 , u 3 , w, respectively. Since βu 1 , βu 3 , βw are equivalent to v 1 , v 3 , y which are bounded, we have Then, it follows from (3.38) and (3.40) that Since βw and w xx are bounded, β 1 2 w x is also bounded by interpolation. Applying (3.51)-(3.52) to (3.48) yields for k ≥ 3, we obtain i.e., Similarly, we also have In order to estimate the w term, we take the L 2 inner product of (3.45) with β  and integrating by parts, we can rewrite it as For k = 3, (3.58) becomes As w xx = O(1) and βu 1 x = o(1) which is guaranteed by (3.55) for k = 3, we conclude that β Finally, (3.50) is now reduced to This leads to βw = o(1), which further implies that y = o(1). We have arrived the promised contradiction for k = 3.
(ii) a 1 , b 1 > 0, a 2 = b 2 = 0. From dissipation (3.44), Combining it with (3.38) and (3.40) yields By the Poincaré inequality, Next, the L 2 inner product of (3.52) with β Dividing (3.65) by β and integrating by parts, we have for k ≥ 2 yields Let's consider the cases that the bottom and top layers have same or different wave speeds. When E1 ρ1 = E3 ρ3 : Taking L 2 inner product of (3.39) with u 3 , and (3.41) with u 1 gives (3.77) We subtract (3.76) from (3.77), then take the real part. This will cancel the first two terms in each of them. Thus, Take k = 2, by (3.70)-(3.72), we have When E1 ρ1 = E3 ρ3 : Taking L 2 inner product of (3.41) with u 1 and dividing by β (3.81) By (3.70)-(3.72), (3.53) becomes i.e., β  (3.81). This leads to By interpolation, Repeat the arguments in (3.76)-(3.87) in case (i). Then, (1), when E1 ρ1 = E3 ρ3 and k = 2, or when E1 ρ1 = E3 ρ3 and k = 4. We have arrived the promised contradiction. The L 2 inner product of (3.41) with u 1 gives Applying (3.91) and (3.92), the above is further reduced to for k ≥ 4, we obtain Assume that the order of polynomial stability is k + for any > 0. Then, By the resolvent identity, for |ξη k+ | < 1 2M , Thus, there is a curve on the complex plane for some ξ > 0, which divides the complex plane into two parts, one is on the right side of the curve including the curve itself. We denote it by C ξ k+ . The above resolvent identity implies that C ξ k+ ⊂ ρ(A). But, the eigenvalues we found will eventually fall into C ξ k+ as Imλ → ∞. A contraction.
For the convenience and readability, set Then ρh = 3. We denote EI by K. E 1 is left alone since it will determine whether the wave speeds of the bottom and top layers are same or not.

YANFANG LI, ZHUANGYI LIU AND YANG WANG
Now, substituting (4.27)-(4.30) into ∆(λ) = 0 again, we obtain that Then, we haveσ Therefore, in this case, one branch of eigenvalues of A has the asymptotic expression as following: whereλ 0 denotes as (4.26). When E 1 = 1, substituting (4.27)-(4.29) into ∆(λ) = 0, we get ∆(λ) where every term of H 2 (ˆ n ) is a polynomial function ofˆ n and H 2 (ˆ n )/ˆ n → 0, n → ∞. Hence,ˆ Therefore, in this case, one branch of eigenvalues of A has the asymptotic expression as following: where every term of G 2 (ˆ n ) is a polynomial function ofˆ n and G 2 (ˆ n )/ˆ n → 0, n → ∞. Hence,ˆ n =˜ n i +σ n , Therefore, in this case, one branch of eigenvalues of A has the asymptotic expression as following: Therefore, in this case, one branch of eigenvalues of A has the asymptotic expression as following:  where every term of G 2 (ˆ n ) is a polynomial function ofˆ n and G 2 (ˆ n )/ˆ n → 0, n → ∞. Hence,ˆ n =˜ n i +σ n , whereλ 0 denotes as (4.26).