Simultaneous observability of networks of beams and strings

In this paper we investigate a finite system of vibrating beams and strings. We obtain results on simultaneous observability by observing a common endpoint.

(1) Assume that we can measure the total force f := N j=1 u j,x (0, •) exerced on the beams at the common endpoint during some time.Investigating the observability of the problem, our question is whether this information is sufficient in order to identify all initial data?This problem was first studied for vibrating strings in the case N = 2 in [7] and for arbitrary N in [2] , and for beams in [2] in the case N = 2.
Recall (see e.g. [10]) that for every initial data (u j0 , u j1 ) in the natural energy space, there exists a unique solution satisfying (well-posedness) and that (hidden regurality).See, e.g., Lasiecka and Triggiani [8] and [9] for results of such type.Moreover, the linear maps are continuous with respect to these topologies.It follows that for every bounded interval I there exists a constant c such that for all initial data.Now our question is whether the linear map is one-to-one?If yes, we can ask whether the inverse linear map is also bounded, that is, whether the inverse inequality to (2) holds true.It would mean that there exists another constant c such that for all initial data.

Statement of the theorem and starting idea of the proof
Let us begin with a simple but important observation: if there exist two beams with commeasurable lengths, then the map (3) is not one-to-one for any interval I. Indeed, if for example l 1 l 2 = p q with two positive integers p and q, then the formulae , u j (x, t) := 0, j = 3, . . ., N, define a nonzero solution of (1) with suitable initial data for which f vanishes identically on R. Thus we cannot hope positive results unless is irrational for all j = k. ( Remark 1 The set of excluded N-tuples (l 1 , ..., l N ), where at least one of the fractions lj l k is rational, has zero measure.Consequently, the complement set of admissible N-tuples is dense in (0, ∞) N .
The following result can be obtained.
Theorem 1 Let I be an arbitrarily short bounded interval and s < 1.Then for almost all N -tuples (l 1 , . . ., l N ) of positive real numbers statisfying (5) there exists for all initial data.
The starting idea of the proof is the following.The solution of ( 1) is given by the formulas with suitable complex coefficients depending on the initial data, and by rearranging the exponents k |k| π 2 l −2 j into an increasing sequence (λ n ) and denoting the corresponding coefficients kπb jk l −1 j by b n .It follows from (5) that A straightforward computation shows that the estimate ( 6) is equivalent to the inequality In the following section we present some results concerning this type of estimates.

Preliminary results
Let (λ n ) be a strictly increasing sequence of real numbers satisfying the following uniform gap condition: We have the following theorem due to Ingham [6] .
Theorem 2 Assume (9).If I is a bounded interval of length |I| > 2π/γ, then there exist two positive constants c 1 and c 2 such that for all square summable sequences of complex numbers b n .
An optimal condition for the length of I satisfying the above inequalities was given by Beurling.This is expressed by the so-called upper density of the sequence (λ n ) , a notion due to Pólya (see [11] ).
Definition 3.1 Let us denote by n + (r) the maximal possible number of elements of (λ n ) contained in an interval of length r > 0. Then the limit exists and is equal to We call D + the upper density of the sequence (λ n ).
The result of Beurling [4] is as follows.In our original problem on the observability of beams we have Indeed, for each fixed j, an interval of length r contains at least √ rl 2 j /π 2 − 1 and at most √ rl 2 j /π 2 + 1 elements of the sequence k |k| π 2 /l 2 j k∈Z .Hence It is thus tempting to apply Beurling's theorem which would also yield the condition But there is a serious obstacle in our case: the uniform gap condition ( 9) is not satisfied if N ≥ 2. Therefore we have to generalize our condition.Let (λ n ) +∞ n=−∞ again be a strictly increasing sequence of real numbers.The following result in [3] establishes a connection between the assumptions of Ingham and Beurling.
Fixing such a γ and M , we introduce the divided differences of the close exponential functions.
Definition 3.2 Fix a number 0 < γ ≤ γ .We say that λ m , . . ., λ m+k−1 is a chain of close exponents belonging to γ if It follows from the property (11) and from the choice of γ that k ≤ M, and that every λ n belongs to a unique chain.
For each chain λ m , . . ., λ m+k−1 let us denote by e m (t) , . . ., e m+k−1 (t) the divided differences of the exponential functions exp (iλ m t) , . . ., exp (iλ m+k−1 t), defined by the formula for n = m, . . ., m + k − 1.In particular we have e m (t) = exp (iλ m t).If λ m , . . ., λ n are pairwise distinct, then we have the more familiar expressions We recall the following result of [3] .11), then there exist two constants c 1 and c 2 such that, using the above notation, we have for every sequence (a n ) +∞ n=−∞ of complex numbers.

Proof of the theorem
Let us consider the sequence (λ n ) in our problem, defined in (7).We know from the previous section that D + = 0.By Lemma 3.1, for every bounded interval there exist an integer M ≥ 1 and a positive number γ such that the sequence satisfies the "generalized uniform gap condition" (11).Let us now fix a number 0 < γ ≤ γ and a chain of close exponents, λ m , . . ., λ m+k−1 belonging to γ .Define Using the definition of the divided differences, since the elements of the sequence (λ n ) are pairwise distinct, one can show by (13) that for suitable coefficients a n , n = m, . . ., m + k − 1.Moreover, by ( 14) there exists a constant c such that Therefore, applying Theorem 4, we obtain the inequality In order to obtain the estimate (8), it suffices to show the existence of a constant for all n ∈ Z.
For this we need some classical results on Diophantine approximation (see [5] , Chapter VII., Theorem 1).
Theorem 5 Let φ (q) be a decreasing function of the integer variable q > 0 with 0 ≤ φ (q) ≤ 1/2.Then the set of inequalities (the ,,norm" is the distance from the set Z) has infinitely many integer solutions q > 0 for almost no or for almost all n-tuples (θ 1 , . . ., θ n ) of real numbers according to whether (φ (q)) n converges or diverges.
Let ε > 0, and φ (q) := q −1−ε .Then by Theorem 5 for almost all choices of the lengths l j and for all ε > 0 we can find c ε > 0 such that Thus for all ε > 0 there exists c ε such that Since all the integers belonging to the lambda's of the chain are equivalent (see the above meaning of ), we have for all n ∈ Z and for every positive ε.Taking ε = (1 − s)/(N − 1), inequality (15) and hence Theorem 1 follows.
there exists a constant c = c (|I| , α, β) such that for all initial data.
Remark 2 Similar results were announced without proof in [1] , writing that the proofs are more complicated and will be presented elsewhere.Our proof below, based on earlier results of Baiocchi et al., is short and elementary.
Proof.We proceed as above.For the solution of the system of strings we have It follows from ( 19) and (20) that and it is easy to see that we have excluded a set of measure 0 from the N -tuples (l 1 , . . ., l N ).Now, we have to prove the estimate We introduce the divided differences for the chains of close exponents, see (12) and (13), and define d n , n ∈ Z, as in (14).Applying Theorem 4, it leads again to the

Theorem 3
Assume (9) again.(a) If I is a bounded interval of length |I| > 2πD + , then the inequalities (10) hold true.(b) If |I| < 2πD + , then the first inequality of (10) does not hold true (with a constant independent of the choice of (b n )).

Lemma 3 . 1
Let x be a positive number satisfying x > 2πD + .Then there exists a real number γ > 0 and an integer M ≥ 1 such that x > 2π γ and λ n+M − λ n ≥ M γ for all n.

=k∈Z b jk sin kπx l j exp ik |k| π 2 t l 2 j
u j (x, t) = 0 =k∈Z b jk sin kπx l j exp ikπt l j , j = 1, . . ., N 1 ,and for the solution for the system of beamsu j (x, t) = 0 , j = N 1 + 1, . . ., N,as above.For the total force measured at the common endpoint we have f (b n e iλnt .