Observability of string vibrations

Transversal vibrations u = u(x, t) of a string of length l under three essential boundary conditions are studied, where u is governed by the Klein Gordon equation: utt(x, t) = a uxx(x, t)− cu(x, t), (x, t) ∈ [0, l]× R; 0 < a, c ∈ R. Su cient conditions are obtained that guarantee the unique solvability of a general observation problem with the given state functions f, g ∈ D(0, l), s ∈ R at two distinct instants of time −∞ < t1 < t2 <∞: A1u|t=t1 +B1ut|t=t1 = f, |A1|+ |B1| > 0, A1B1 ≥ 0, A2u|t=t2 +B2ut|t=t2 = g, |A2|+ |B2| > 0, A2B2 ≤ 0. Here s is arbitrary, the space D(0, l) (see [2] and [13]) is some subspace of the Sobolev space H(0, l). The essential condition of the solvability is that (t2− t1)a/l is a rational number. In fact, this result is a consequence of a general observability result related to the vibration u = u(x, t) governed by the equation utt = (p(x)ux)x − q(x)u, (x, t) ∈ [0, l]× R, 0 < p, q ∈ C∞([0, l]), subject to some initial data and linear boundary conditions (see in Proposition 1 below). This time the main restrictions are some Diophantine conditions and asymptotic properties of the eigenfrequencies ωn as n → ∞. Some other results without these restrictions are also presented.


INTRODUCTION
In control theory which is closely related to the subject of this paper, a number of monographs and articles dealt with the accessability of a given complete nal state u| t=T = f, u t | t=T = g 1 Corresponding author EJQTDE, 2013 No. 77, p. 1 of oscillations (in particular, string oscillations) in the time interval 0 ≤ t ≤ T < ∞; see, e.g., in [110].
In observation problems for the oscillations u = u(x, t) one looks for the initial functions u| t=0 = ϕ, u t | t=0 = ψ only from the partial states, e.g., so the aim is the same (roughly speaking) as in the control theory: to nd ϕ and ψ occurring above.Although, only the short communication [12] dealt with observability of the string oscillations u on [0, l] × [0, T ] governed by u tt = a 2 u xx ; and another paper [13] where u is described by the KleinGordon equation (1) u tt (x, t) = a 2 u xx (x, t) − cu(x, t), (x, t) ∈ Ω := [0, l] × R; 0 < a, c ∈ R, with the initial conditions (2) and boundary conditions (3) u| x=0 = u| x=l = 0, with u, ϕ, ψ from the corresponding generalized function spaces D s , s ∈ R. In [12] these functions ϕ and ψ were constructed only for small t 1 and t 2 : 0 ≤ t 1 < t 2 ≤ 2l/a under the additional assumption that the initial functions ϕ and ψ in question are known on some subinterval [h 1 , h 2 ] ⊂ [0, l].In [13] we dealt with the case of arbitrary t 1 , t 2 ∈ R, −∞ < t 1 < t 2 < ∞, under the restriction that (t 2 − t 1 )a/l is a rational number, and we considered the following observation conditions: (a) Now, our goal is to prove similar results related to the problem (1)(2) with the following general observation conditions: EJQTDE, 2013 No. 77, p. 2 under the following boundary conditions: either or the SturmLiouville boundary conditions Thus, we deal with observation problems related to the mixed problems (1), ( 2), ( 4); (1), ( 2), ( 5) and (1), ( 2), ( 6) respectively for the specic properties for all three mixed problems.We emphasize some common ideas that may be useful even for the study of some other problems (see Propositions 1, 2 and Lemma 1 below), e.g., for the class of the following mixed problems: where U 1 , U 2 are independent, self-adjoint (see, e. g., [14]) linear expressions (they may be dierent from (4), ( 5), ( 6)), where u, ϕ, ψ are from the generalized function spaces D s , and the conditions (e) and (9) are understood in the sense of generalized functions.
It can be supposed without loss of the generality, that the ordinary homogeneous BVP for has only the trivial solution.The ordinary dierential operator L is formally selfadjoint, therefore the operator We use the denition of the spaces D s (S), s ∈ R given in [2] and [13].Let the system {X n (x)} ∞ n=0 be a complete orthonormal basis in L 2 (S).Given arbitrary real number s, we consider on the linear span D of the functions X n (x), n ∈ N 0 = N {0}, x ∈ S, the following Euclidean norm: We use the notation S = (0, l) associated with string vibrations.

MAIN RESULTS
We will need the following lemma for certain estimates below.
Lemma 1.Let the sequence r n be such that 0 < M/n < |r n | → 0 with a positive constant M and let x 0 > 0 be a rational number.Then for any xed d ∈ R, there exists a number N such that Proof.Since x 0 is rational, it can be written as x 0 = p/q; p, q ∈ N, and then sin (nπx 0 + d) can assume at most q dierent values.Let N 1 and N 2 be two disjoint sets of numbers, if n ∈ N 2 .We will prove Lemma 1 according to these two cases.
First, if n ∈ N 1 , then we have for all large n ∈ N 1 since r n → 0 as n → ∞.
Combining the two cases we get the statement of Lemma 1.
Remark 2. According to condition (13), we can use Lemma 1 for the estimation of which means that sin(ω n (t 2 − t 1 ) + γ n − δ n ) = 0 automatically holds for all large n, say when n is greater than a threshold number N .
EJQTDE, 2013 No. 77, p. 5 Proof of Proposition 1.Clearly, the initial functions can be uniquely expanded into Fourier series with respect to the system {X n } with yet unknown coecients Since the solution u of the problem (7), ( 8), ( 9) has the representation (11) with some coecients α n , β n ; n ∈ N 0 , the observation problem can be reduced to the problem of nding the appropriate choices of α n and β n such that (e) is satised.For this reason, we substitute t 1 and t 2 into (11), and use both conditions in (e).As a result, we get the following necessary conditions for α n and β n : Since f ∈ D s+2 and g ∈ D s+2 , the coecients of the Fourier expansions (with respect to the system {X n }) of the functions f (x), g(x) are unambiguously determined, and comparing these Fourier series with (17) and (18), we get the following conditions for α n and β n , n ∈ N 0 : where By substituting γ n and δ n into (19) and using trigonometric identities, we get that EJQTDE, 2013 No. 77, p. 6 This linear system can be uniquely solved for the unknown coecients α n and β n , due to the fact that sin(ω n (t (20) .
Thus, the unknown initial functions ϕ and ψ are uniquely determined in the form of (15) and (16).It remains to show that ϕ, ψ are from the classes D s+1 , D s , respectively, i.e., to show that the following inequality holds: We note that condition (13) involves that |ω n | = O(n), so by using Lemma 1 and (20), we get that there is a constant M such that which means that for a suitable constant c 1 .Obviously, the inequality (22) (and thus condition (12)) can be improved when B 1 = 0 or B 2 = 0 (see Remark 4 below).
By using (22), we get that where and (f, g) ∈ D s+2 × D s+2 according to the denition of the s + 2 norm.
Remark 3. The above examinations and estimates show that the operator A, which assigns to the partial states (f, g) at t 1 and t 2 the couple (ϕ, ψ), is a continuous (bounded) EJQTDE, 2013 No. 77, p. 7 On the other hand, we get from the system of equations (19) that where and obviously Based on the behaviour of the Fourier coecients α n , β n as n → ∞ we get that which indicates a slightly stronger smoothness for the couple (ϕ, ψ) = A(f, g).
This suggests to prove a slightly sharper statement than the one in Proposition 1.To this eect, let us introduce the subspaces D s 0 ⊂ D s that contain the functions f ∈ D s whose Fourier coecients f n for n ∈ N 1 (the set N 1 depends on the problem, see its denition in Lemma 1) have the following property: Certainly this involves that D s+1 ⊂ D s 0 .
Using inequality (22) for the rst sum, we get which is nite by the denition of the spaces D s+1 0 . For Remark 4. As we mentioned before, the inequality (22) can be improved when B 1 = 0 or B 2 = 0.More precisely, if B 1 = 0 then f ∈ D s+1 ; while if B 2 = 0 then g ∈ D s+1 is sucient in condition (12).
In the special case when one of A 1 , B 1 equals zero and one of A 2 , B 2 equals zero, the observation condition (e) simplies to one of the condition (a), (b), (c), (d) (after suitably dividing with A 1 , A 2 , B 1 , or B 2 ).In this case, we have γ n ≡ 0 or γ n ≡ π/2; δ n ≡ 0 or δ n ≡ π/2, and conditions (13) and ( 14) are simplied accordingly.These terms correspond to our previous results in [13].
Remark 5. Condition (13) is suitable for the usage of Lemma 1 to have the estimation However condition (13) of Proposition 1 is not necessary, it can be replaced, e.g., with the assumption that f n = g n = 0 for all n ∈ K( ), where for the arbitrary small xed > 0, K( ) ⊂ N is the collection of all n such that EJQTDE, 2013 No. 77, p. 9 Now, we will show three applications of Proposition 1 in the following Sections 35.

THE VIBRATING STRING WITH FIXED LEFT AND FREE RIGHT ENDS
Let Ω = {(x, t) : 0 < x < l, t ∈ R}.Consider the problem of the vibrating [0, l] string with a xed left end and a free right end under the assumption that there is an elastic withdrawing force proportional to the transversal deection u(x, t) of the point x of the string at the instant denoted by t.This phenomenon is described by the KleinGordon equation as follows: (1) with the initial conditions and the homogeneous boundary conditions Some of the results of [2] (see Section 1.11.3) and [10] say that for arbitrary s ∈ R with (ϕ, ψ) ∈ D s+1 × D s , the solution of the mixed problem (1), ( 2), ( 4) satises (10), and (11); and using q(x) ≡ c > 0 we also have Here thus we have Let us consider the following observation conditions: Due to the facts that So, the condition (13) is fullled with Consequently, we can apply Proposition 1 to the vibrating string with xed left end and free right end as follows.
and suppose that sin(( Then the observation problem posed for (1), ( 2), (4) under the observation conditions has a unique solution for (ϕ, ψ) ∈ D s+1 × D s provided the elapsed time between t 1 and t 2 is a rational multiple of l/a.

THE VIBRATING STRING WITH FREE ENDS
Let Ω = {(x, t) : 0 < x < l, t ∈ R}.Consider the problem of the vibrating [0, l] string with free ends when there is an elastic withdrawing force proportional to the transversal deection u(x, t) of the point x of the string at the instant denoted by t.
This phenomenon is described by the KleinGordon equation as follows: (1) with the initial conditions and the homogeneous boundary conditions of the second kind From the same results of [2] and [10] as in Section 3 it follows that for arbitrary s ∈ R with (ϕ, ψ) ∈ D s+1 × D s the solution of the mixed problem (1), ( 2), ( 5) satises (10), (11), and due to the fact that q(x) ≡ c > 0, we have [α n cos (ω n t) + β n sin (ω n t)] X n (x), (x, t) ∈ Ω. Here If A 1 B 1 ≥ 0 and A 2 B 2 ≤ 0, then the same arguments for γ n , δ n as in Section 3 give that condition (13) is fullled with Consequently, we can apply Proposition 1 to the vibrating string with free ends as follows.
Theorem 2. Let and suppose that sin(( Then the observation problem posed for (1), ( 2), (5) under observation conditions string when there exists a varying withdrawing force proportional to the transversal deection u(x, t) of the point x of the string at the instant denoted by t.This phenomenon is described by the KleinGordon equation as follows: (1 ) and the boundary conditions The conditions for the sign of cot α and cot β in (6) are sucient to ensure the conservation of energy, hence guaranteeing the uniqueness of the solution of the mixed problem (1 ), ( 2), ( 6) in the classical case.Moreover, this condition also ensures that the (later introduced) constant γ is strictly positive.
It follows again from [2] and [10] that for arbitrary s ∈ R with (ϕ, ψ) ∈ D s+1 × D s , the solution of the mixed problem (1 ), ( 2), ( 6) satises (10) and (11).The orthonormal basis {X n }, n ∈ N 0 , in (11) can be derived from [11] as follows: Here we have where ω n can be written in the form of: EJQTDE, 2013 No. 77, p. 13 If A 1 B 1 ≥ 0 and A 2 B 2 ≤ 0, with the same arguments for γ n , δ n as in Section 3, we get that condition (13) is fullled with Consequently, we can apply Proposition 1 to the vibrating string with xing (6) as follows. Theorem and suppose that sin(( Then the observation problem posed for (1 ), ( 2), ( 6) under the observation conditions has a unique solution for (ϕ, ψ) ∈ D s+1 × D s provided the elapsed time between t 1 and t 2 is a rational multiple of l/a.

OBSERVABILITY RESULTS FOR ARBITRARY t 1 < t 2
Considering the observation problem posed in Proposition 1 without conditions (13), ( 14) and for randomly chosen t 1 < t 2 , we have to investigate the necessity of the description of all possibilities of the observability.We shall give an alternative treatment, and for this reason we introduce the following notations: for every n ∈ N 0 , let d n denote the distance of h n = ω n (t 2 − t 1 ) + γ n − δ n to the nearest zero of the sine function, i.e., The promised alternative is the following: 1.If d n > 0 for all n ∈ N 0 then the observation problem is (formally) uniquely solvable, since the coecients α n , β n for the denition of ϕ, ψ can be uniquely nded from the linear system (19) for every n ∈ N 0 .Although, for the estimates of α n , β n (and for the belonging of ϕ, ψ to the corresponding spaces D s+1 , D s , respectively), some EJQTDE, 2013 No. 77, p. 14 special tools are needed (e.g., we used Diophantine ones in Sections 15).A general result can be the following: The formal series: have the following properties: (24) Moreover, the relations in (24) also hold in the case if we replace d n by d n , where with arbitrarily small xed δ and arbitrary |b n | ≤ d n .
We emphasize, that all remarks of the present section are also valid for the observation problems posed for the case c = 0 of the KleinGordon equation, i.e., for the standard vibrating string.

1 2 .
EJQTDE, 2013 No. 77, p. 3 Completing D with respect to this norm, we obtain a Hilbert space denoted by D s .

Proposition 2 .
If the conditions of Proposition 1 hold and f, g ∈ D s+1 0 (instead of D s+2 ), then the observation problem posed for (7), (8), (9) under condition (e) has a unique solution for (ϕ, ψ) ∈ D s+1 × D s .In other terms, the operator A : A(f, g) := (ϕ, ψ) maps D s+1 0 ×D s+1 0 into D s+1 ×D s as a continuous (bounded) operator.Proof of Proposition 2. We can do the same steps as in the proof of Proposition 1.It remains only to show that (21) also holds for (f, g) ∈ D s+1 0 × D s+1 0 .Indeed, we have has a unique solution for (ϕ, ψ) ∈ D s+1 × D s provided the elapsed time between t 1 and t 2 is a rational multiple of l/a.EJQTDE, 2013 No. 77, p. 125.THE VIBRATING STRING WITH BOUNDARY CONDITION OF THE THIRD KIND (STURMLIOUVILLE BOUNDARY CONDITIONS)Let Ω = {(x, t) : 0 < x < l, t ∈ R}.Consider the problem of the vibrating [0, l] EJQTDE, 2013 No. 77, p. 4 Second, if n ∈ N 2 , then this means that there is a constant d 1 > 0 such that | sin (nπx 0 + d) | > d 1 for all n ∈ N 2 .Due to the uniform continuity of the sine function, n ∈ N 2 we can improve inequality (22).Namely, for these values of n (if n is large enough) we have sin(ω n (t 2 − t 1 ) + γ n − δ n ) > d 1 /2 as it was shown in the proof of Lemma 1, whence it follows that max{|α n |, |β n |} < c 2 max{|f n |, |g n |} ∀n ∈ N 2 with a suitable constant c 2 .So, we get, that 2013No.77, p. 10 and let us suppose that A 1 B 1 ≥ 0 and A 2 B 2 ≤ 0. Without loss of the generality, we can take A 1 , B 1 , B 2 ≥ 0 and A 2 ≤ 0. Hence it follows that sin γ n