Traction, deformation and velocity of deformation in a viscoelastic string

In this paper we consider a viscoelastic string whose deformation is controlled at one end. We study the relations and the controllability of the couples traction/velocity and traction/deformation and we show that the first couple behaves very like as in the purely elastic case, while new phenomena appears when studying the couple of the traction and the deformation. Namely, while traction and velocity are independent (for large time), traction and deformation are related at each time but the relation is not so strict. In fact we prove that an arbitrary number of"Fourier"components of the traction and, independently, of the deformation can be assigned at any time.


Introduction
Let a viscoelastic string be in eqilibrium on the interval [0, π] of the x-axis. When a vertical displacement is applied to the boundary point x = 0, while the boundary point x = π is kept fixed, the dynamic of the string (at rest for negative times) is described by with initial and boundary conditions w(x, 0) = 0, w t (x, 0) = 0 w(0, t) = f (t) ∈ L 2 loc (0, +∞) , w(π, t) = 0 . (2) Here w(x, t) denotes the vertical displacement and w t (x, t) the vertical velocity of the point in position x at time t. The vertical components of the stress at position x and time t is (the usual minus sign in front of the right hand side has no interest for the following and we drop it). We shall assume that the real kernel M(t) is of class H 2 loc (0, +∞) and we note that in general there will be a certain (positive) coefficient in front of the laplacian w xx , which has been taken equal to 1 for simplicity.
In order to understand the problem that we are going to study in this paper, let us first consider the purely elastic case, i.e. the case M(t) ≡ 0. In this case Eq. (1) is the string equation and it is known that: • for every T > 0 we have: w(t) ∈ C(0, T ; L 2 (0, π)) (and f → w is continuous from f ∈ L 2 (0, T )); w t (t) ∈ C(0, T ; H −1 (0, π)) (and f → w t is continuous from f ∈ L 2 (0, T )); • for every target ξ ∈ L 2 (0, π), η ∈ H −1 (0, π) it is possible to find a control f ∈ L 2 (0, T ) such that Note that we suppress the dependence on the state variable x, unless needed for clarity, so that w(t) denotes w(x, t). Moreover, w does depend on f but this is not explicitly indicated.
Consequently, we have also w x (t) ∈ C(0, T ; H −1 (0, π)) for every T > 0 and the fact that ξ and η are arbitrary shows that also the stress/velocity pair (w x (T ), w t (T )) ∈ H −1 (0, π)×H −1 (0, π) can be arbitrarily assigned under the action of the boundary control, when T ≥ 2π. This observation can be interpreted both as a controllability property of the wave equation and as the fact that stress and velocity at a certain time T (large enough) are independent (which seems to us the most enlightening interpretation). Instead the pair (w(T ), w x (T )) cannot be controlled: in fact the displacement identifies the stress (Hooke Law). Now we consider the viscoelastic string. We proved in [15] that the pair (w(t), w t (t)) has the same property as in the purely elastic case: it belongs to C(0, +∞; L 2 (0, π) × H −1 (0, π)) and its value at a certain time T can be assigned at will in L 2 (0, π) × H −1 (0, π) provided that T is large enough (T ≥ 2π). In this paper, using results in [3,4], we first prove that, in spite of the memory term, also the pair (σ(x, T ), w t (x, T )) can be arbitrarily assigned for T ≥ 2π.
Finally, in section 5 we shall study the pair deformation/stress and we shall see that these functions are not independent, i.e. there is no controllability of that pair, and that a kind of "Hooke Law" holds asymptotically, for short wavelength components. But, we shall also see that, unlike the purely elastic case, the long wavelength components are independent. The precise statement is in section 5.

Comments on previous references
Controllability properties of viscoelastic materials have been studied by several authors in past years, using different methods. See for example [5,6,11,12,13,14,16,17]. A constructive approach to the steering control (in the case of the heat equation with memory), based on moment methods, has been introduced in [18] and then developped in subsequent papers [3,4,15,19,20].
The key idea of these papers has been applied to a different class of problems in [1].
The papers [3,4] shows an interpretation of controllability of pairs of variables as independence of that controlled variables. This approach we push further in this paper. We relay on the moment methods techniques introduced in the papers just cited, and in particular we shall use some results proved in [3,4,19].
When studying distributed systems with memory, we might get the feeling that they are "perturbations" of heiter heat or wave equations, and behave much in the same way. This conjecture is disproved both from the results in Section 5 and the negative results in [8,9,10].

Preliminaries
The following computations make sense for smooth boundary inputs f and are then extended by continuity to f ∈ L 2 (0, T ). Let (note that 1 is the coefficient of the laplacian in (1)). Then, integrating both the sides we can write (1) in the form We introduce This shows a relation with the first order systems studied in [3,4].
In particular, this lemma shows that the stress, as an element of H −1 (0, π), can be computed at each time t.
The multiplicative transformation is innocuous since, with M α (t) = e 2αt M(t), we have Projecting the solutions of Eq. (6) on the spaces generated by 2/π sin nx in L 2 (0, π) we find the following representation/definition for the solutions of Eq. (6) (see [18]): where θ n (0) = 0 and θ n (t) solves Let us introduce the solutions z n (t) of the problem Then we have (we rename f (t) the function e 2αt f (t)) So, the quantities of our interest are: where Convergence of the previous series in the appropriate spaces, C(0, T ; L 2 (0, π)) for the first and C(0, T ; H −1 (0, π)) is known, see [3,15,18,19].
It is clear from these formulas that control problems are easily reduced to moment problems. So, before we proceed, we present some background information on moment problems and Riesz sequences.

Preliminaries: moment problems, Riesz bases and Riesz sequences
Let H be a Hilbert space and {h n } a (fixed) sequence in H. Let us consider the infinite set of equations u, h n = c n (13) wher ·, · is the inner product in H and {c n } is a sequence of complex number. Under the heading "moment problem" is intended the problem to caracterize those sequences {h n } such that a solution u of the equations ( If furthermore the sequence {h n } is complete, then it is a Riesz basis, and conversely. Let {h n } be a sequence in H. A Paley-Wiener theorem, adapted to Hilbert spaces and orthonormal bases, states that if {e n } is an orthonormal basis of H and h n − e n 2 < 1 then {h n } is a Riesz basis. A corollary which will be used is as follows: We stress that the sequence {e n } in Corollary 5 need not be an orthonormal basis. Condition (15) does not imply that {h n } is a Riesz sequence but Theorem 6 (Bari Theorem) If both the condition (15) and the condition (16) below hold then {h n } is a Riesz sequence.

The additional condition (16) is called ω-independence and it is
The convergence of the series is in H so that, as noted in Corollary 5, the convergence of the series in (16) implies {α n } ∈ l 2 . Finally, we state the following lemma. For completeness, we give a proof in Appendix A.

The stress and the velocity
In this section we consider the pair stress/velocity and we prove Theorem 1. We proceed in several steps. Lemma 7 shows that every pair (ξ, η) ∈ H −1 (0, π) × H −1 (0, π) can be represented as and conversely. Hence, given T > 0, the pair (ξ, η) in H −1 (0, π) is reachable at time T by the pair (w t (·, T ), σ(·, T )) if the following moment problem is solvable (see (11). We ignore the inessential factor (π/2)e 2αT ): So, our goal is the proof that this moment problem is solvable for arbitrary sequences {(ξ n , η n )} in l 2 × l 2 , i.e. arbitrary {ξ n + iη n } in l 2 C , the l 2 -space of complex valued sequences (and n is natural, n > 0). Even more, we prove that the solution f (t) ∈ L 2 (0, T ) depends continuously on {(ξ n , η n )}.
We introduce γ n = ξ n + iη n and so that the moment problem (17) takes the form So, the moment problem (17) is solvable and the solution f (t) depends continuously on the l 2 sequences {ξ n } and {η n } if and only if the sequence {Z n (t)} is a Riesz sequence in L 2 (0, T ). This we are going to prove now, and we shall see that any T ≥ 2π will do.

Usefull estimates
The sequence {z n (t)} has been studied in previous papers, in particular in [18,19,20], where we proved the following representation formula. Computing a second derivative of both the sides of (9) we see that i.e. (using N ′ α (0) = 0, see (7)) and so This equality holds with the possible exception of one index n 0 : the exceptional index exists if there exists a natural number n 0 such that α 2 = n 2 0 , i.e. β n 0 = 0. In this case we have to replace the previous representation formula with the expression in [19, formula (18)]. We don't insist on this rather exceptional case here and we assume β n = 0 for every n.
We integrate by parts the last integral and we rewrite formula (22) as follows: Here, given by We note that L(t) has the same regularity as N ′ α (t) and L(0) = 0. Then we have the following equality: We shall use the following result from [3, formulas (2.14) and (2.27)]: Lemma 8 For every T > 0 there exists a number M such that for every n we have: (we can replace β n with n in the previous formulas, since β n ≍ n). In particular, the sequence {z n (t)} is bounded on bounded intervals.
Furthermore, using the representation (23) and we see: Proof. In this proof, {M n (t)} denotes a sequence of functions which is bounded on [0, T ] (not the same functions at every occurrence). We use equality (23) and boundedness on [0, T ] of the sequence {z n (t)} to see that Using L(0) = 0 and differentiability of F (t) and L(t), two integrations by parts in the last integral shows that it can be absorbed in M n (t)/β n . We integrate by parts the first integral in the right hand side and we use (n/β n ) − 1 ≍ 1/n 2 to see that If F ′′ (t) ∈ L 2 a further integration by parts shows that the last line is M n (t)/β n . Otherwise we note that {e iβnt } is a Riesz sequence in L 2 (0, T ) for every T ≥ 2π (see [3, Appendix 5.1]). Hence, the sequence {sin β n t} is Riesz on every interval L 2 (0, T ), T ≥ π (the proof is similar to the corresponding proof for the cosine sequence given in [7]). We fix T 0 = max{π, T } and we note that for every fixed t ∈ [0, T ] we have where H(t) denotes the Heavisede function. Hence, for every fixed t, these integrals are the "Fourier" coefficients of [H(t − τ )F ′ (t − τ )e ατ ] in the expansion in terms of the biorthogonal of {sin β n t} and this gives (for a suitable constant M) A further integration from 0 to T gives the result.

The proof of Theorem 1
Statements 1 and 2 of Theorem 1 are in Lemma 2. In order to prove the statement 3 we must prove that the sequence {Z n (t)} in (18) is a Riesz sequence in L 2 (0, T ), provided that T ≥ 2π. This is the bulk of the proof, which requires several steps. It is convenient to introduce the following notations: Z ′ = Z − {0} and, for n < 0: β −n = β n , γ n = γ −n .
Here {γ n } ∈ l 2 C (Z ′ ) (we shall denote l 2 C (Z ′ ) simply as l 2 ). So, both z n (t) and Z n (t) are defined also for n < 0 and (since the memory kernels are real) and the moment problem (17) is equivalent to We are going to prove that {Z n (t)} n∈Z ′ , is a Riesz sequence in L 2 (0, T ), T ≥ 2π.
The value of T ≥ 2π is now fixed so that, using (18) and Lemma 9 with F (t) = K(t) (and using K(0) = 1) we get: As we noted, the sequence {e (α+iβn)t } is a Riesz sequence in L 2 (0, T ) when T ≥ 2π so that condition (31) implies the existence of N such that {Z n (t)} |n|>N is a Riesz sequence and so, using Bari Theorem combined with (31), We recall that that the series here has to converge in L 2 (0, T ) and this is the case if and only if {α n } ∈ l 2 . We proceed in several steps to prove that {Z n (t)} is ω-independent.
Step 1: an equation for Z n (t). Using (12) we see that Hence, using (9), We compute the derivative of both the sides, using (21). We get Collecting corresponding terms, we see that Z n (t) solves the following integrodifferential equation: (the definitions of H(t) and K(t) are in (12)) and Z n (0) = 1. Note that this is similar to [3, formula (2.22)].
Step 2: The sequence {Z n (t)} is linearly independent in L 2 (0, T ) for every T > 0. The proof is by contradiction. If it is linearly dependent then there exist N > 0 and, corresponding to it, an index −K < 0, and coefficients α n ∈ C such that Remark 10 We recall that the indices are from Z ′ ; i.e. n = 0 is excluded. We can also include n = 0 in the sums, but then α 0 = 0.
We choose N > 0 to be the first index which corresponds to the minimum value of the numbers K > 0. Then we have also The first series in the right hand side is zero (use the first equality in (35)). Computing with t = 0 and using the second equality in (35), we get Using that N α (t) is differentiable with N α (0) = 1, we see that This can be combined with (35) to see that an equality of the form (35) holds for N replaced by N − 1, without increasing K. In fact we get This contradicts the definition of N.
Step 3: the sequence {Z n (t)} ω-independent, hence it is Riesz, in L 2 (0, T ) if T ≥ 2π We need the following Lemma, whose proof is in Appendix B.
Lemma 11 There exists a sequence {M n (t)} n∈Z ′ of H 2 functions, for which the following properties hold: • the following series converge in L 2 (0, T ) for every sequence {α n } n∈Z ′ ∈ l 2 and for every T > 0: We recall that, in order to prove that {Z n (t)} is a Riesz sequence in L 2 (0, T ), we must prove that it is ω-independent. We proceed as follows: we assume that a sequence {α n } satisfies n =0 α n Z n (t) = 0 (38) in L 2 (0, T ) (so that necessarily {α n } ∈ l 2 ) and we prove {α n } = 0. Relaying on Lemma 11, we first prove the following additional "regularity" of the sequence {α n }.
Lemma 12 Let (38) hold. Then, there exists {γ n } ∈ l 2 such that Proof. In this proof we use the fact that {e iβnt } n∈Z ′ is a Riesz sequence in L 2 (0, T ) for every T ≥ 2π of deficiency 1, and we get a Riesz basis if we add β 0 = 0 and we consider {e iβnt } n∈Z , see the appendices in the papers [3,18]. Consequently, from [21, Theorem 1], if we add a further exponential e ict with c = β n for every n ∈ Z we get a Riesz basis of H 1 (0, T ), whose elements are the functions 1, e ct and (1/β n )e iβnt (here n = 0). As we noted, convergence of the series in (38) implies that {α n } ∈ l 2 . We multiply both the sides of (38) with e −αt and we use the representation (37) we see that − n =0 Both the series converge in L 2 (0, T ) for every T and Lemma 11 asserts that the series n =0 α n M ′ n (t) converges in L 2 (0, T ) too. Hence, it represents an H 1 (0, T ) functions, which can be expanded in series of 1, e ict and (1/β n )e iβnt . So we have also Equating the corresponding coefficients we see that δ 0 = 0, δ c = 0 and Now we compute the derivatives of both the sides of (41) and we get: Our assumption now is that the series converge in L 2 (0, T ) and so the right hand side of (42) belongs to H 1 (0, T ). As above, being T ≥ 2π, we have We use this lemma as follows. Equality (39) implies convergence of the following series, which then have to converge to 0: As we noted, the first series on the right side vanishes. Equality (39) shows convergence of the last series and also it shows that we can compute both the sides with t = 0. We get We can combine (38) and (43) so to get a new series Note that if n = 1 then α (1) n = 0 if and only if α n = 0. We repeate this procedure till we remove N − 1 positive and N − 1 negative coefficients and we end up with the equality |n|≥Nα n Z n (t) = 0 and so {α n } = 0 since {Z n (t)} |n|≥N is a Riesz sequence. Asα n = 0 if and only if α n = 0, we see that the series (38) is a finite sum, and so we have also α n = 0 if |n| < N since the sequence {Z n (t)} is linearly independent.
Remark 13 Finally we note that Lemma 11 can be interpreted as a weaker version of the asymptotic estimates in [3,Lemma 5.3], which required one more derivative of N α (t).

Deformation and stress
Now we examine the pair of the deformation w(·; T ) and the stress σ(·; T ). Let us go back to the series of the deformation and the series of the stress in (11) which converge respectively in L 2 (0, π) and H −1 (0, π). We shall see that the sequence of the Fourier coefficients of this series are asymptotic one to the other. Namely we shall prove: Theorem 14 Let T > 0 and let f (t) be a control which drives the deformation from the initial condition w(·, 0) = 0 to w(·, T ). Let w n be the Fourier coefficients of w(·, T ), i.e.
Let (12)). Then, there exists a number M (which depends on T and f ) such that Proof. In this theorem we are not assuming that T has to be "large". We assume only that it is positive. So, even the sole sequence {w n } in (44) will not be arbitrary. Furthermore, the sequencese {w m } and {σ n } do not uniquely identify the function f . So, let us consider one special f ∈ L 2 (0, T ) for which equalities (44) and (45) both hold. We observe that The last integral is less then (46) The function F (t) is twice differentiable and moreover F (0) = 0 so that, using Lemma 9, The result follows from here.
Up to now the results we have found parallel those of the purely elastic case. Now we can observe a difference, which might have some interest, concerning the long wavelength components. Clearly components which correspond to φ n (x) with "large" n, i.e. short wavelength, as computed by model (1) will not represent the real behavior of the system, due to unmodeled dynamics, dissipations etc., not taken into account when deriving Eq. (1) and only the first components will be (hopefully) realistic. And here we have a difference with the purely elastic case, since in the purely elastic case the generalized Fourier coefficients cannot be assigned at will for the deformation and stress, not even for a single wavelength. Instead: Theorem 15 Let N > 0 be fixed and let {c n } and {d n } be two finite sequences of real numbers, 1 ≤ n ≤ N. Then there exists a function f (t) ∈ L 2 (0, T ) which assign the "Fourier" coefficients {c n } to the deformation and {d n } to the stress.
Proof. In order to prove the theorem, it is sufficient that we show solvability of the following (finite) moment problem (the function F (t) is defined in (46)): This problem is solvable if and only if the functions Uniqueness of solution of Volterra integral equations implies N n=1 α n z n (r) = 0 and this is possible only when each coefficient α n = 0, because the sequence {z n (t)} is linearly independent, see [19].
In order to prove the second statement we show that any χ ∈ H −1 (0, π) can be represented as c n (n cos nx) (so that the sequence {n cos nx} is complete in H −1 (0, π)) and that there exist m > 0 and M such that (i.e., the proof relays on Theorem 4).
To be more precise the distributions n cos nx in this formula are distributions on (−π, π), localized to (0, π).
This shows that (2c n ) (n cos nx) . (51) and (52) in order to get

Remark 16
The properties of the sine and cosine sequences, one an orthonormal basis and the second a Riesz basis, can be interchanged, since every Riesz basis is an orthonormal basis with a suitable, equivalent, Hilbert space norm.

B Appendix: the proof of Lemma 11
We denote {M n (t)} a sequence of functions with the properties 1), 2), 3) stated in the second item of Lemma 11. Unless needed for clarity, we don't distinguish among different occurrencies of these functions, so M n (t) is not the same at every occurrence.
Remark 17 Properties 1), 2) and 3) of a sequence {M n (t)} are retained under convolution with a fixed integrable kernel.
Using formula (32), we represent Z n (t) as and we prove the existence of two sequences {M n (t)}, {M n (t)} such that z n (t) = e αt cos β n t +M n (t) , n t 0 R n (t − s)z n (s) ds = e αt sin β n t +M n (t) .
We prove the first equality in (53). We relay on (23). First we prove that every term in the rows (24)-(26) has the properties of an M n (t) so that, by applying the observation in Remark 17 to the convolution in (23) we shall get z n (t) − e αt cos β n t ≤ M n (t) .
In fact, it is clear that since L ′ (t) ∈ H 1 (0, T ). As to the first function in (24), we clearly have α βn e αt sin β n t = M n (t). The second function in (24) has the first and second derivative given by so that the required properties 1), 2) and 3) are clear, using the second inequality in (27) and 1 − µ n = α/(n 2 − α 2 ). We consider row (25). We study the series n =0 α n µ n β n t 0 e α(t−r) sin β n (t − r)z n (r) dr which is uniformly convergent, since {z n (t)} is bounded on bounded intervals.
Computing the first derivatives we have two series: α n µ n t 0 e α(t−r) cos β n (t − r)z n (r) dr . L 2 (0, T )-convergence is now clear. So, properties 1) and 2) hold. In order to prove property 3) replace α n with α n /β n in the series (54), and compute the second derivative. The series that we get can be treated as above.
Finally, we consider the term (26). Properties 1) and 2) are obvious, because the derivative of the term at the line (26) is the sum of the following terms: The first term has the property of an M n (t) and the same holds for the second too, since This shows properties 1) and 2) of the second term in (55). We note now property 3). In fact, the second derivative is the sum of the following terms: The corresponding series (with coefficients α n /β n ) is seen to be L 2 -convergent, using arguments similar to the previous ones. This gives the required property of the first addendum on the right hand side of (18). The fact to be noted is that in these computations the regularity used is M ∈ H 2 . Now we consider the convolution of z n (t) with the functions H(t) and nK(t) (we recall K(0) = 1).
The convolution of H(t) with every term in the right hand side of (23) gives a sum of terms which have the properties of M n (t). This is clear, using Remark 17, for every term, a part possibly the first one t 0 H(t − s)e αs cos β n s ds , whose first and second derivatives are respectively H(0)e αt cos β n t + t 0 H ′ (s)e α(t−s) cos β n (t − s) ds , e αt αH(0) cos β n t − β n H(0) sin β n t + α The notation M n (t) to denote this term is legitimate, as seen using K ∈ H 2 and (n/β n − 1) ≍ 1/n 2 . Finally, the convolution of nK(t) with the functions at the row (24)-(26) contributes terms with the properties of {M n (t)}. The computations are similar to the previous ones, and are left to the reader. We confine ourselves to note that in these computations we encounter integrals of the form (54).