Spectral analysis and stabilization of a chain of serially connected Euler-Bernoulli beams and strings

We consider $N$ Euler-Bernoulli beams and $N$ strings alternatively connected to one another and forming a particular network which is a chain beginning with a string. We study two stabilization problems on the same network and the spectrum of the corresponding conservative system: the characteristic equation as well as its asymptotic behavior are given. We prove that the energy of the solutions of the first dissipative system tends to zero when the time tends to infinity under some irrationality assumptions of the length of the strings and beams. On another hand we prove a polynomial decay result of the energy of the second system, independently of the length of the strings and beams, for all regular initial data. Our technique is based on a frequency domain method and combines a contradiction argument with the multiplier technique to carry out a special analysis for the resolvent.

Models of the transient behavior of some or all of the state variables describing the motion of flexible structures have been of great interest in recent years, for details about physical motivation for the models, see [11], [14], [16] and the references therein.
Mathematical analysis of transmission partial differential equations is detailed in [16].
Let us first introduce some notation and definitions which will be used throughout the rest of the paper, in particular some which are linked to the notion of C ν -networks, ν ∈ N (as introduced in [13] and recalled in [19]).
Let Γ be a connected topological graph embedded in R 2 , with 2N edges (N ∈ N * ). Let K = {k j : 1 ≤ j ≤ 2N } be the set of the edges of Γ. Each edge k j is a Jordan curve in R 2 and is assumed to be parametrized by its arc length x j such that the parametrization π j : [0, l j ] → k j : x j → π j (x j ) is ν-times differentiable, i.e. π j ∈ C ν ([0, l j ], R 2 ) for all 1 ≤ j ≤ 2N . The length of the edge k j is l j > 0. The C ν -network G associated with Γ is then defined as the union We study two feedback stabilization problems for a string-beam network, see [1]- [8], [16] and [27]- [28]. In the following, only chains will be considered as mathematically described in Section 5 of [20]. See also [21] and Figure 1. Following Ammari/Jellouli/Mehrenberger ( [9]), we study a linear system modelling the vibrations of a chain of alternated Euler-Bernoulli beams and strings but with N beams and N strings (instead of one string-one beam). For each edge k j (representing a string if j is odd and a beam if j is even), the scalar function u j (x, t) for x ∈ G and t > 0 contains the information on the vertical displacement of the string if j is odd and of the beam if j is even (1 ≤ j ≤ 2N ).
Our aim is to study the spectrum of the conservative spatial operator which is defined in Section 3 and to obtain stability results for (P 1 ) and (P 2 ).
We define the natural energy E(t) of a solution u = (u 1 , ..., u 2N ) of (P 1 ) or (P 2 ) by We can easily check that every sufficiently smooth solution of (P 1 ) satisfies the following dissipation law and therefore, the energy is a nonincreasing function of the time variable t.
The first result concerns the well-posedness of the solutions of (P 1 ) and the decay of the energy E(t) of the solutions of (P 1 ). We also study the spectrum of the corresponding conservative system. We give, in particular, the characteristic equation and the asymptotic behavior of the eigenvalues of the corresponding conservative system. We deduce that the generalized gap condition holds: if we denote by (λ n ) n∈N * the sequence of eigenvalues counted with their multiplicities, then Contrary to [9], it seems that the (simple) gap condition fails in general (for any N ≥ 2).
Therefore we do not succeed to obtain an observability inequality (and then to deduce stability results for (P 1 )) directly by the study of the spectrum and the eigenvectors (see, for instance, [22]). In fact, the difficulties are to locate precisely the type of eigenvalues in the packets.
However, we prove that the energy E(t) of the solutions of (P 1 ) tends to zero when t → + ∞ in an appropriate energy space (described later), under some assumptions about the irrationality properties of the length of the strings and beams. For that, we use a result from [10].
In this case, we are able to prove more interesting stability results for system (P 2 ) and to give the explicit decay rate of the energy of the solutions of (P 2 ) in an appropriate space.
In the same manner as previously and with the same energy E(t) (defined by (1.1)), every sufficiently smooth solution of (P 2 ) satisfies the following dissipation law 4) and therefore, the energy is a nonincreasing function of the time variable t.
The main result of this paper then concerns the precise asymptotic behavior of the solutions of (P 2 ). As it was shown in [9] in the case of one string and one beam connected together (i.e. N = 1), we can not except to obtain an exponential decay rate of the solutions of (P 2 ). However we are able to prove that the decay rate to zero of the energy is ln 4 (t)/t 2 , independently of the length of the strings and beams and by taking more regular initial data in an appropriate space. Our technique is based on a frequency domain method from [17] and combines a contradiction argument with the multiplier technique to carry out a special analysis for the resolvent.
This paper is organized as follows: In Section 2, we give the proper functional setting for systems (P 1 ) and (P 2 ) and prove that these two systems are well-posed. In Section 3, we study the spectrum of the corresponding conservative system and we give the asymptotic behavior of the eigenvalues. We then show that the energies of systems (P 1 ) and (P 2 ) tend to zero. Finally, in Section 4, we study the stabilization result for (P 2 ) by the frequency domain technique and give the explicit decay rate of the energy of the solutions of (P 2 ).

Well-posedness of the systems
In order to study systems (P 1 ) and (P 2 ) we need a proper functional setting. We define the following space Note the following lemma: Lemma 2.1. We have that 0 is an eigenvalue associated to (P 1 ) and (P 2 ) of multiplicity N − 1, i.e. there exists a subspace of V of dimension N − 1 such that any φ in this subspace satisfies Proof. Let φ be a non-trivial solution of (EP 0 ). By the two first equations of (EP 0 ), for j ∈ {1, · · · , N }, φ 2j−1 is a first order polynomial and φ 2j is a third order polynomial.
It is well-known that system (P 1 ) may be rewritten as the first order evolution equation where U is the vector U = (u, ∂ t u) t and the operator A 1 : with satisfies (2.7) to (2.10) hereafter} , It is clear that < . , . > V does not define a norm for V but only a semi-norm since, for all u ∈ V , we have < u, u > V = 0 if and only if u satisfies (EP 0 ). In order to get a Hilbert space we define by E 0 , the eigenspace of A 1 associated to the eigenvalue 0, i.e.
where γ is a simple closed curve enclosing only the eigenvalue 0 (see Theorem III-6.17 of [15]). Now let H 1 the Hilbert space defined by where Then H 1 is a Hilbert space, equipped with the usual inner product From now on we consider the operator A 1 restricted to the space H 1 ∩ Y 1 with value in H 1 , since A 1 commutes with P 0,1 . By abuse of notation, this operator will be always denoted by A 1 and D(A 1 ) will be its domain, i.e. Therefore Moreover the norm on D(A 1 ) is defined by Note that, with all these notation, problem (P 1 ) is rewritten in an abstract way as: find Now we can prove the well-posedness of system (P 1 ) and that the solution of (P 1 ) satisfies the dissipation law (1.2).
(ii) The solution u of (P 1 ) with initial datum in D(A 1 ) satisfies (1.2). Therefore the energy is decreasing.
Proof. (i) By Lumer-Phillips' theorem (see [24,26]), it suffices to show that A 1 is dissipative and maximal. We By integration by parts, we have .7) and (2.8), and by the continuity of v at the interior nodes, we obtain by (2.9), (2.10) and since v ∈ V . Therefore This shows the dissipativeness of A 1 .
Let us now prove that A 1 is maximal, i.e. that λI − A 1 is surjective for some λ > 0.
Suppose that we have found u with the appropriate regularity. Then for all j ∈ It remains to find u. By (2.15) and (2.16), u j must satisfy, for all j = 1, ..., N , and Multiplying these identities by a test function φ, integrating in space and using integration by parts, we obtain This problem has a unique solution u ∈ V by Lax-Milgram's lemma, because the lefthand side of (2.17) is coercive on V equipped with the inner product defined by . Coming back to (2.17) and by integrating by parts, we find Consequently, by taking particular test functions φ, we obtain since the resolvent of A 1 commutes with P 0,1 (see [15]).
In summary we have found (u, v) t ∈ D(A 1 ) satisfying (2.14), which finishes the proof of (i).
(ii) To prove (ii), it suffices to derivate the energy (1.1) for regular solutions and to use system (P 1 ). The calculations are analogous to those of the proof of the dissipativeness of A 1 in (i), and then, are left to the reader.
We see, in the same manner, that problem (P 2 ) can be rewritten in an abstract way as: satisfies (2.7), (2.9), (2.10) and (2.18) hereafter} , Then we define the Hilbert space H 2 by (with γ is a simple closed curve enclosing only the eigenvalue 0), and Then The following proposition holds: (ii) The solution u of (P 2 ) with initial datum in D(A 2 ) satisfies (1.4). Therefore the energy is decreasing.
Proof. The proof of (i) and (ii) is the same as the proof of Proposition 2.2, and therefore is left to the reader.

Spectral analysis of a chain of serially connected Euler-Bernoulli beams and strings
In this section, we study the spectral analysis of the corresponding conservative system.
Let Φ be the solution of the conservative system derived from problems (P 1 ) and (P 2 ) given in the introduction, i.e. Φ is the solution of the following system where we have replaced the dissipative conditions (in bold in systems (P 1 ) and (P 2 )) by the conservative ones.
We can rewrite system (P c ) in an abstract way as: find (Φ, Ψ) t ∈ D(A c ) such that and Then we define the Hilbert space H c by with P 0,c : V × 2N j=1 L 2 (0, l j ) → E 0 the projection onto E 0 defined by with (with γ is a simple closed curve enclosing only the eigenvalue 0), and Due to (3.21), we set V c the Hilbert space defined by equipped with the inner product (2.5).
Following Section 2, it is clear that system (P c ) is well-posed in the natural energy space. If we suppose that (u 0 , u 1 ) ∈ H c = V c × 2N j=1 L 2 (0, l j ), then problem (P c ) admits a unique solution This system is obviously conservative, i.e. its energy is constant.

The characteristic equation
Let φ be a non-trivial solution of the eigenvalue problem (EP ) associated to the conservative problem (P c ) and λ 2 be the corresponding eigenvalue. That is to say, φ ∈ V c satisfies the transmission and boundary conditions (3.22)-(3.26) hereafter as well as Note that this also means that (φ, λφ) ∈ D(A c ) is an eigenvector of A c associated to the eigenvalue λ. By the definition of A c and of its domain, 0 is not an eigenvalue of A c . Moreover 0 is not an eigenvalue of A 1 and A 2 .

Following Paulsen ([23]) and Mercier ([18]
), we will rewrite this eigenvalue problem on a chain of 2N beams and strings using only square matrices of order 2 in the following way: we define, for each j ∈ {1, ..., N }, the vector functions V 2j−1 and V 2j by Define the matrices A j by with j ∈ {1, · · · , N } and with the notation c 2j = cos(l 2j · z), s 2j = sin(l 2j · z).

(3.27)
The matrix T is defined by: To finish with, the matrix M (z) is the square matrix of order 2 given by Lemma 3.1. (A few trivial but useful properties) With the notation introduced above, we have: In this basis, if we consider the two following functions d 1 , d 2 with coordinates d 1 := (−e l j z sin(l j z), e l j z cos(l j z) − 1, 0, −e l j z sin(l j z)) d 2 := (e l j z − e −l j z , 0, cos(l j z) − e −l j z , e l j z − cos(l j z), we can see that they are independent and satisfy (3.22). Consequently u j can be expressed as a linear combination of these two functions. Now, to find A j , we proceed as follows: let (α, β) t the coordinates of u j in the basis (d 1 , d 2 ). There exist two matrices M 0 , M 1 such that V j (0) = M 0 (α, β) t and V j (l j ) = M 1 (α, β) t , then A j is the matrix Moreover the transmission conditions (3.24), (3.25) and (3.26) imply the second and third equations.
The fourth one is the logical consequence of the first three applied successively for j = 1, Proof. Let φ be a non-trivial solution of the eigenvalue problem (EP ) and λ 2 be the corresponding eigenvalue, where λ = iz 2 (z ∈ R + * ).
Using the boundary conditions as well as V 2N (l 2N ) = M (z)V 1 (0), it follows: It is clear that the vector of the second part of the previous equality is non-trivial since φ is a non-trivial solution of problem (EP ). Hence the result.  Proof. In the following, the notation o(h(λ)) is used for a square matrix of order 2 such that all its terms are dominated by the function λ → h(λ) asymptotically. For any j ∈ {1, . . . , N }, which leads, after some calculations, to: Thus

The result follows by induction.
Remark 3.4. We can note that the eigenvalues λ = iz 2 of (EP ) have 2N families of asymptotic behavior: It follows that the generalized gap condition (1.3) holds. If λ = 0 is an eigenvalue of the operator A c and E λ is the associated eigenspace, then the dimension of E λ is one.
Proof. The eigenvectors φ ∈ V c associated to the eigenvalue λ 2 (cf. problem (EP )) are entirely determined by their values at the nodes of the network (i.e. where the beams and strings are connected to one another). Due to Lemma 3.1, they are also determined (3.22)) and ∂ x φ 1 (0) may take any value in R * . Hence the result.
3.2 Strong stability of (P 1 ) and (P 2 ) We first prove the following lemma: for all eigenvectors φ ∈ V c of (EP ).
Proof. Let φ ∈ V c be an eigenvector of (EP ) associated to the eigenvalue λ 2 , where λ = iz 2 (z ∈ R + * ). Assume that (3.34) is false, i.e. that we have We use in the following the basis introduced in the proof of Lemma 3.1.
As a consequence of the previous lemma, we can prove the following proposition.
First, we show that A 1 has no eigenvalue on the imaginary axis. If it is not the case, let iω be an eigenvalue of A 1 where ω ∈ R * . Let Z ∈ D(A 1 ) be an eigenvector associated with iω. Then Z is of the form It is an immediate consequence of the identity (iωI − A 1 )Z = 0.
We now take the inner product ., . H 1 between A 1 Z and Z. By (2.13), we have Since Z is an eigenvector of A 1 associated with iω and ω = 0, we obtain Note that Z satisfies the eigenvalue problem (EP ) and Z belongs to D(A c ), since (where we use (λI − A c )( 1 λ−iω Z) = Z and where γ is a simple closed curve enclosing only 0), and thus Z = Z − P 0,c Z ∈ (I − P 0,c )(V × 2N j=1 L 2 (0, l j )) = H c . Then this contradicts (3.34). Therefore A 1 has no eigenvalue on the imaginary axis. Now, we can apply the main theorem of Arendt and Batty [10]: Since σ(A 1 ) ∩ iR is empty, we obtain (3.38).
⇒ Let us show that (3.38) implies (3.34). For that purpose we use a contradiction argument. Suppose that there exists an eigenvector φ ∈ V c of (EP ) of associated eigenvalue λ 2 (where λ = iz 2 , z ∈ R + * ) such that Let us set u(., t) = φ cos(z 2 t).
Then u is solution of (P 1 ) and satisfies This contradicts (3.38).
It suffices to use Lemma 3.6 to finish the proof.
Moreover, with the same method as previously, we are able to prove the decay to zero of the energy of solutions without restriction about the irrational properties of the lengths.
Proposition 3.8. We have lim t → +∞ E(t) = 0 for any solution of (P 2 ) with (u 0 , u 1 ) in Proof. As in the proof of Proposition 3.7, we can show that the energy of solutions of (P 2 ) tends to zero if and only if for all eigenvectors φ of (EP ). Let φ be an eigenvector of (EP ) such that (3.40) is false.
By the same proof as Lemma 3.6, this implies that φ = 0, which is impossible. Then (3.40) holds and therefore the energy decays to 0.
Remark 3.9. If we take the initial data in V × 2N j=1 L 2 (0, l j ), the energy of the solutions of (P 1 ) and (P 2 ) do not decay to 0, since u = φ, where (φ, 0) t is an eigenvector of A i (i = 1, 2) associated to the eigenvalue 0, is solution of (P 1 ) and (P 2 ) with constant energy.

Stabilization result for (P 2 )
We prove a decay result of the energy of system (P 2 ), independently of the length of the strings and beams, for all regular initial data. In [9], the authors prove that the system described by (P 2 ) is not exponentially stable in H 2 with N = 1 (i.e. with one string and one beam). Therefore, in the general case (for N ∈ N * ), we can not except to obtain an exponential decay for the energy of the solutions of (P 2 ), but only a weaker decay rate, and in this general case, we prove a polynomial decay rate. To obtain this, our technique is based on a frequency domain method and combines a contradiction argument with the multiplier technique to carry out a special analysis for the resolvent.
The following theorem is a direct generalisation of the result in [9], which we note, due to a mistake in the choice of θ, the decay rate in the following ln 4 (t) t 2 has been written ln 6 (t) t 4 (corresponding to a choice of θ = 1 and not to θ = 1/2).
Theorem 4.1. There exists a constant C > 0 such that, for all (u 0 , u 1 ) ∈ D(A 2 ), the solution of system (P 2 ) satisfies the following estimate Proof. We will employ the following frequency domain theorem for polynomial stability (see ) of a C 0 semigroup of contractions on a Hilbert space: for some constant C > 0 and for θ > 0 if where ρ(L) denotes the resolvent set of the operator L.
Then the proof of Theorem 4.1 is based on the following two lemmas. Proof. Since A 2 has compact resolvent, its spectrum σ(A 2 ) only consists of eigenvalues of A 2 . We will show that the equation with Z = (y, v) t ∈ D(A 2 ) and β = 0 has only the trivial solution.
The second lemma shows that (4.43) holds with L = A 2 and θ = 1.
Fourth step. Finally, we choose q 2j−1 and q 2j such that dq 2j−1 dx is strictly positive and dq 2j dx is strictly negative. This can be done by taking q 2j−1 (x) = e x − 1, q 2j (x) = e (l 2j −x) − 1.