AN EXTENSION OF A LYAPUNOV APPROACH TO THE STABILIZATION OF SECOND ORDER COUPLED

. This paper deals with the convergence to zero of the energy of the solutions of a second order linear coupled system. It revisits some previous results on the stabilization of such systems by exhibiting Lyapunov functions. The ones used are constructed according to some scalar cases situations. These simpler situations explicitely show that the assumptions made on the operators in the coupled systems seem, ﬁrst, natural and, second, give insight on their forms.

As we previously said, in this paper, we investigate such a method in the case when A 2 = A 2 1 and C = A β 1 with β ∈ [0, 3 2 ]. The main result of this paper is Theorem 3.3 which also proves the polynomial convergence to 0 of the solution (u, v). Compared to the result in ( [4], p. 144, Prop. 5.3), the convergence that we obtain is in weaker norms, but requires less regularity on the initial data.
Let us mention that the stabilization of such systems settled in abstract form has been widely studied and many results are connected to those in our paper. The case when A 1 and A 2 are unbounded operators, and C is a bounded was considered by Alabau in [3] and by Alabau et al. in [5] where the polynomial statibility is proven. The optimal energy decay was proven in [11] in the case A 1 = A 2 = A, C = I and B = A γ where γ < 0. The indirect stabilization of abstract coupled equations is also considered in [1,2,6,7].
In order to motivate the Lyapunov function that we construct in the proof of our main result, we explain the strategy in Section 2 in the framework of a coupled scalar differential system.
In Section 3 we introduce the functional framework and an existence theorem that lead to state and prove our main result, namely Theorem 3.3.

A Lyapunov function for the scalar case
As mentioned in the preceding section, we consider the (real) scalar coupled system where λ, µ > 0, and c are such that 0 < c 2 < λµ. The damping coefficient is set to 1 for simplicity but a time scale change reduces general damping terms bu to this case. In order to shorten the formulas, let us introduce for each solution (u, v) of (2.1), its total energy Then we have for all t ≥ 0

Now we introduce
Our first result is the following Proposition 2.1. There are some constants η > 0, δ > 0 such that Proof. For all ε > 0 we define the function It is easy to check that and In fact, using Young's inequality, we get Then we deduce It is clear that for all i ∈ {1, 2, 3, 4} where C 2 is given in (2.5). A similar proof gives similar inequalities for C 1 . Let ε 1 > 0 the value of ε such that C 1 defined in (2.4) is equal to 0. Let ε ∈ (0, ε 1 ). In this case C 1 becomes positive.

Now we have
Similarly, we get and then Using Young's inequality, we can find some constants c 1 , c 2 , c 3 > 0 such that Finally we obtain Now by choosing ε ∈ (0, ε 1 ) such that 1 − (2 + c 1 + c 2 + c 3 )ε > 0, some constant C 3 > 0 can be found such that By combining this with the inequality (2.3), we get for all t ≥ 0 We conclude the proof by integrating this last inequality and using (2.3) again.
3. The case A 2 = A 2 , and C = A β with β ∈ [0, 3 2 ] This section is devoted to the proof of Theorem 3.3. In order to proceed we first introduce the functional framework and give an existence theorem.

Functional framework
Let H be a separable Hilbert space, whose norm and scalar product will be denoted · and ·, · respectively. We consider A : H → H an unbounded closed self-adjoint operator such that the injection D(A) ⊂ H is dense and compact. We assume moreover throughout the paper that there exists a > 0 such that Following for example the exposition given in [10], by denoting (λ n ) n∈N * the increasing sequence of eigenvalues of A, the largest a for which (3.1) is true is λ 1 .
Besides, let us consider (e n ) n∈N * an orthonormal basis of H constituted by eigenvectors of A. For any β > 0, u, e i 2 < ∞), and we set A β : H → H given by then (see e.g. [10]) and A β is an unbounded self-adjoint operator such that the inclusion D(A β ) ⊂ H is dense and compact. We also have for some a > 0. The largest a for which this inequality is true being λ β 1 . As usual we write A 0 = Id. In this case of course the operator A α is a continuous linear operator on H. We will denote V = D(A 1/2 ) and W = D(A). Thus V and W are Hilbert spaces whose norms · V and · W are given respectively by We have, if we identify H with its dual with dense and compact injections when the norms on the Hilbert spaces V and W are given by where ·, · V ,V denotes the action of V on V (with a similar notation for W ). Of course when u ∈ H one has Let us remark that with these definitions A maps continuously V to V and A 2 maps W to W .

Existence result
Let α and β two reals numbers with β ≥ 0. We recall that we consider the problem which can be rewritten as the first order system Let us first establish an existence and uniqueness result for (3.5).
We concentrate on the case β ∈ [ 1 2 , 3 2 ], the case β ∈ [0, 1 2 ) being easier. Let us consider where ·, · V ,V denotes the usual duality pairing between V and V , with similar notation for W , while ·, · denotes the scalar product on H for which it is a Hilbert space.
It is straightforward to prove that ·, · H defines a scalar product on H for which it is a Hilbert space provided that From now on we assume that (3.7) holds true.
We now consider the unbounded operator A : H −→ H defined by It is clear that A has a dense domain in H.
Let us remark that for any U = (u, v, w, z) ∈ D(A) one has and therefore AU, U = w 2 ≥ 0. Indeed Let us show that I + A is onto. For this we take (f, g, h, k) ∈ H. We want to find (u, v, w, z) ∈ D(A) such that Clearly Φ is continuous on V × W . It is also clear that Φ is coercive if we assume |α| < λ 3−2β 2 1 . By the Lax-Milgram theorem, there exists a unique (u, v) ∈ V × W such that We therefore get We have thus proven that A is maximal monotone. By classical theory, we get that

Main result of the paper
Our main result is the following 3 2 ] and α = 0. Let (u, v) be a solution of (3.5) such that (u(0), u (0), v(0), v (0)) ∈ H, then there exists a constant c > 0 such that Remark 3.4. If we replace (3.5) by where, as mentionned in the introduction, B is a bounded self-adjoint operator on H for which there exists µ > 0 such that the results of Theorem 3.1 and Remark 3.4 remain true.
Remark 3.5. If we replace (3.5) by where A 2 is a self-adjoint unbounded operator such that D(A 2 ) = D(A 2 ) and there exist ν 1 , ν 2 > 0 such that and if B is as in the remark 3.4, the result of Theorem 3.3 remains true provided |α| is small enough (depending on λ 1 , ν 1 and ν 2 ).
Remark 3.6. In the case β = 0, in order to obtain the decay of the energy, we must assume In the paper [4], the authors obtain such a decay with merely We, of course, would expect that the energy decay also holds with (u(0), v(0), u (0), v (0)) ∈ D(A) but unfortunately we are not able to prove it for the moment being.
Proof of Theorem 3.3. All the computations below will be made assuming that (u 0 , v 0 , u 1 , v 1 ) ∈ D(A) which ascertains them. By density and continuity the inequalities stated in Theorem 3.3 remain true.
Let us also recall that α satisfies |α| < λ 3−2β 2 1 . We introduce the energy of the system by Then we have Let p > 1 and ε > 0 two real numbers to be fixed later and let and a = min(0, 1 − β). We find easily First case : β ∈ [0, 1]. In this case a = 0. We have

Now since
Au, u W = 1 we get Let us remark that Thus where we choose γ > 0 such that which is equivalent to This choice is possible provided that (3.10) Now we choose p > 1 such that (3.10) is satisfied. Then we have Now let us observe first that if one considers some U ∈ W one has Second, since u ∈ V , one has Au ∈ V . Therefore according to (3.11) we have Moreover A 1/2 u ∈ H, thus, one has According to (3.13), since
By choosing ε small enough, we find a constant η = η(p, ε) > 0 such that for all t ≥ 0 For all t ≥ 0, we haveẼ ThenẼ is nonincreasing. Observe that from which we deduce that (3.14) Now since then there exists a constant γ > 0 such that for all t ≥ 0 From (3.15), assuming ε possibly smaller in order to achieve positivity of the quadratic form H ε , we get Using inequality (3.14), we obtain
we get Let us remark that we therefore get where we choose γ > 0 such that which is equivalent to This choice is possible provided that p + 1 p − 1

Examples
This section is devoted to giving examples of operators to which Theorem 3.3 applies.
Example 4.1. The first case that we consider is when H = L 2 (Ω) and where the coefficients a i,j ∈ C 1 (Ω) satisfy a ij = a ji , ∀i, j,  where ρ 1 > 0.
Here we can, as in [4] consider the case where where ρ 2 > 0.
Example 4.3. Let us remark that due to remark 3.5 and the Poincaré inequality, our result applies to the case when A 1 = A is as in example 1 and A 2 = A 2 1 + ζA 1 for any ζ > 0.