On Controllability of an Elastic Ring

On Controllability of an Elastic Ring Sergei A. Avdonin1 Department of Mathematical Sciences University of Alaska Fairbanks, AK 99775-6660, USA http://www.dms.uaf.edu/avdonin E-mail [email protected] Boris P. Belinskiy2 Department of Mathematics University of Tennessee at Chattanooga Chattanooga, TN 37403–2598, USA http://www.utc.edu/Faculty/Boris-Belinskiy/ E-mail [email protected] Sergei A. Ivanov3 Institute of Laser Physics St. Petersburg State University St. Petersburg, 198904, Russia E-mail [email protected] Abstract We study the exact controllability problem for a ring under stretching tension that varies in time. We are looking for a couple of forces, which drive the state solution to rest. We show that applying two forces is necessary for controllability and the ring is controllable in the time interval greater than the optical length of the string. We also explain why one force would not be enough to control the ring. We use the method of moments to reduce the controllability problem to a moment problem for the controlling forces. The solution of that problem is based on an auxiliary basis property result. Both method of moments and proof of the basis property are developed for the model with time–dependent parameters. Keywords: Distributed parameter systems, wave equation, control, bases of functions, Sobolev spaces AMS Subject Classification numbers: 93B05, 35L05, 42C15, 46N20, 47N20, 46E35 1. Introduction. An elastic ring represents an important element of many engineering constructions. Google shows about 3,000 papers and patents related to elastic rings in the area of engineering. We mention here only few technical applications of the model. Vibrations of an elastic ring are considered in [27] (see also an extensive literature review therein.) As an example of a practical application, the authors mention elastic wheels for train wagons, which 1 S.A.’s research was supported in part by the NSF, grant ARC–0724860 B.B.’s research was supported in part by University of Tennessee at Chattanooga Faculty Research Grant. B. B. is the corresponding author; ph. (423)425-4748; fax (423)425-4586 3 S.I.’s research was supported in part by the Russian Foundation for Basic Research, grants 08-01-00595a and 08-01-00676a 2 1 2 could replace the usual steel wheels. We give a slightly simplified version of the governing equations for the radial w(θ, t) and circumferential u(θ, t) displacements, E EF w − (Iusss − F us) = 0, 2 R R  I  E ρF utt + αut − E F + 2 uss + (Iwsss − F ws) = 0. R R Here s = Rθ, θ is the polar angle, F is the cross-sectional area of the ring, EI is the ring-bending stiffness, α is the damping coefficient, and E is Young’s modulus. In general, this is the system of two coupled equations. But if the mechanical parameters satisfy the conditions ρF wtt + αwt + EIwssss + EI EF << 1, 2 << 1, R4 R the equations decouple and may be considered separately, ρF wtt + σwt + EIwssss = 0,  I  ρF utt + σut − E F + 2 uss = 0. R In particular, the equation for the circumferential displacement u(θ, t) has the form of the wave equation, i.e. formally coincides with the equation describing transverse vibrations of a string but with periodical boundary conditions (because the ring is closed). We further note that an elastic ring may be viewed as a short cylindrical shell (see, e.g. [34];) vibrations of an elastic shell are of permanent interest to engineers. The classical theory describing free oscillations of a thin elastic shell is well–known. The equations of motion are given by a system of PDE (see, e.g. [36]), Uzz + 1 1+σ σ 1−σ Uss − 2 Utt + Vzs + Wz = 0; 2 c 2 R 1+σ 1−σ 1 1 Uzs + Vzz + Vss − 2 Vtt + Ws = 0; 2 2 c R σ 1 1 Uz + Vs + β 2 R2 (∂zz + ∂ss)2 W + 2 Wtt = 0, R R c where U (z, s, t), V (z, s, t), and W (z, s, t) are the displacements of the points of the middle surface of a shell in the longitudinal, peripheral, and radial directions; c is the speed of the longitudinal waves in a thin plate made of the material of the shell; R is the radius of its middle surface; h is its thickness; β 2 = h2 /(12R2); z and s are the longitudinal and peripheral coordinates. It is also known that for oscillations independent on the longitudinal coordinates (∂z ≡ 0), the first equation (for longitudinal displacement) may be separated from others, 1−σ 1 Uss − 2 Utt = 0. 2 c We observe that the last equation formally coincides with the equation describing a string but with periodical boundary conditions. Two other equations may be studied separately. We note here that in both models above, one of the governing equations (or their system) is of the order higher than two (with respect to coordinate.) It is well–known from Control Theory that the second order equations are more interesting and actually more complex. The authors consider the equation (or the system of equations) of the higher order subject to periodic boundary conditions as their next goal. It should be immediately noted that the techniques used and developed in the current paper may be applied in this case as well. 3 As the last example leading to a string with periodic boundary conditions, we mention an infinite string under the periodic load with the period l. We may reduce the problem of oscillations (and hence the problem of control) to one period. From practical point of view, it is possible to slightly vary parameters of an elastic ring in time. We assume that the stretching tension of the ring varies in time. Several examples of mechanical systems with parameters that vary in time may be found in [26]. In contrast to our previous projects on control of the systems with time dependent parameters (see [2], [3]), we consider a closed oscillating object (ring) here. It is known from the general Control Theory on graphs (see, e.g. [23], [5, Sec. VII.1]) that the presence of a loop changes the controllability of a system dramatically. Indeed, we will observe that in this paper. Hence, we study the exact controllability of a ring under stretching tension, which is the sum of two terms; one is constant and another one is small and varies slowly. We say that the ring is controllable if, for any initial data by suitable manipulation of the exterior forces, the ring goes to the given regime. Our model is linear and allows inverting time. Hence, we may assume that the ring goes to rest. We are looking for a couple of forces f1,2(t), which drives the state solution to rest and explain why one force would not be enough. The controllability problem is reduced to a moment problem for the controlling forces f1,2 (t). The proof of solvability for the moment problem is based on an auxiliary basis property result. This idea has been widely used in Control Theory of distributed parameter systems since the classical papers of H.O. Fattorini and D.L. Russell (see [30], the survey [31], and book [5] for the history of the subject and extended list of references.) Yet, we believe that the necessity to use two forces to control a ring does not follow directly from the previous Control Theory results. The method of [30], as well as many other subsequent papers, is based on the properties of exponential families (usually in the space L2 (0, T )), the most important of which for Control Theory are minimality and the Riesz basis property. Recent investigations into new classes of distributed systems such as hybrid systems and damped systems, as well as problems of simultaneous control, have raised a number of new and difficult problems in the theory of exponential families (see, e.g. [1], [6], [8], [9], [10].) The paper represents a continuation of the previous authors’ papers [2] and [3] on exact controllability of an elastic string under stretching tension. In [2], the tension is supposed to be the sum of a constant term and a small, slowly variable, term. In [3], the assumption that the tension varies slowly was removed. Though philosophically [2] and the current papers are similar (in particular, an auxiliary basis property result is similar), the problem under consideration is more complex because of necessity to also use the second force to exactly control the system. We believe that the approach of [3] could allow removing the assumption that the tension varies slowly but we do not do that in the current paper. The paper is organized as follows. In Section 2 we discuss the statement of the problem and introduce some assumptions. We include an auxiliary Section 3 to discuss the same model for the case of constant parameters. That consideration, though being simple, allows one a better understanding of the construction we use for the general case in the following sections. In Section 4 and further, we are back to the variable tension case. In Section 4 we formulate the results and outline the proof of the main result. In Section 5 we derive the moment problem for the forces f1,2 (t) with respect to a system of time dependent functions. In Section 6 we prove some basis properties results for these functions. In Section 7 we solve the moment problem and prove the main theorem about controllability, i.e. give the conditions of exact controllability of the ring under the variable in time tension. We also explain why the ring may not be controlled by one force g(x)f(t) only. In Section 8 we consider some specific distributions gj (x) that are typical for piezoelectric control. In Appendix (Section 9) we prove an auxiliary result from the theory of exponential families, which is used to prove the main result on controllability of our system. 4 2. Statement of the initial boundary value problem. For any T > 0, we consider the initial boundary value problem for the hyperbolic type PDE ρ(x)ytt = P (t)yxx + ρ(x) 2 X j=1 subject to periodic boundary conditions gj (x)fj (t), (x, t) ∈ QT = (0, l) × (0, T ) (2.1) y(0, t) = y(l, t), yx (0, t) = yx (l, t) (2.2) y(0, x) = y0 (x), yt (x, 0) = y1 (x). (2.3) and initial conditions Here ρ(x) is the density of the ring, l is its length, P (t) ≡ P0 + P1(t) is the tension with the constant component P0 and the variable component P1(t), and y0 and y1 are the initial displacement and velocity. The functions gj (x)fj (t), j = 1, 2 are two independent exterior forces with the given distributions gj (x) along the ring and unknown controls fj (t); the factor ρ(x) in the right hand side is introduced for convenience. This problem describes the small (linear) transverse oscillations of the ring. It is well-known ([22]) that the initial boundary value problem (2.1)–(2.3) has a unique solution if the functions y0 , y1 , gj , and fj are smooth enough. Our goal is to find the conditions on the system under consideration that guarantee the existence of the controlling forces fj (t) such that the ring goes to rest in a given time T and construct these forces. Similarly to the problems of control for a string, we need to introduce the optical length of the ring as the (unique) positive number T∗ satisfying the relation Z lp Z T∗ p ρ(x) dx = P (t) dt. (2.4) 0 0 We proceed further under the following Assumptions 1–7. 1. ρ = const > 0. 2. P (t) varies “slowly enough”, specifically, max 0≤t≤T √ P̈ 4π2 6D 5Ṗ 2 − < + C(T ). 4P 2 16P 3 l2 ρ (2.5) 3. maxt |P1(t)| < P0. 4. P1(t) (and hence, P (t)) is real–valued. 5. yk (x), k = 0, 1 and gj (x), fj (t), j = 1, 2 are real–valued. 6. gj (x) ∈ L2 (0, l); fj (t) ∈ L2 (0, T ), j = 1, 2. 1 7. y0 (x) ∈ Hper (0, l) := {φ ∈ H 1(0, l) : φ(0) = φ(l)}, y1 (x) ∈ L2(0, l). ⋄ Remarks. We now discuss our assumptions. Assumption 1. (i) It may be removed but sufficiently simplifies the presentation. (ii) We prefer keeping all physical parameters in our derivations so that we can (a) trace the dimension in each formula and (b) present the real–world formulas for the time of control and the form of control to engineers. Assumption 2. The expression under the symbol of the absolute value in the left-hand side of (2.5) appears regularly when studying asymptotic properties of the second order ODE (see [14].) The physical meaning of the assumption is that the tension P (t) varies slowly enough in time. This assumption contains an absolute (positive) constant D and a (positive) constant C(T ) that depends 5 on time of the control only; they will be defined below (it will be explained below that the time of control T is greater than T∗ .) As it is common in Control Theory, the proof of controllability will be reduced to the proof of the basis property of a system of (non-orthogonal) functions denoted below as F. This proof (Section 6 below) is rather complex; constants D and C(T ) will naturally appear in it. More precisely, the constant D is related to the so-called basis constant of the family F c (s) = {sin nt, cos nt}n∈N ∪ {1} ∪ {t} (2.6) (see Theorem 3 in Section 6.) The constant C(T ) is defined in (2.9) below; it naturally arises in the proof of the fact that the aforementioned family F, which appears in the controllability problem, is quadratically close to the family F c (s) (see the proof of Lemma 1 in Section 6.) Assumptions 1 and 2 certainly put some restriction on the parameters of the model. They are similar to the assumptions introduced in [2] in connection with the control problem for an elastic string. Yet, they have the technical character. In [3], they were omitted (for the model of a string) with the help of a technique that is different from one used in this paper. We will have to discuss different forms of Assumption 2 below in Sections 2 and 6. We note that Assumption 2 is formulated in terms of the tension P (t), yet technically, it will be more convenient to us reformulating it in more appropriate but not physically obvious terms. We finally note that Assumption 2 implies the inclusion P (t) ∈ C 2(0, T ). Assumption 3 guarantees the strong hyperbolicity of the PDE (2.1). Assumptions 4 and 5 are natural from the physical point of view. Assumptions 6 and 7 are standard in Control Theory. We formulate further assumptions below. Under Assumption 1, the identity (2.4) has the form s Z T∗ P (t) . (2.7) ω(t) dt = 2π where ω(t) := 2π l2 ρ 0 Assumption 2 now obtains the form Assumption 2a. √ ω̈(t) 3ω̇(t)2 − 3 < 6D C(T ) max 4 0≤t≤T 4ω (t) 2ω (t) for T such that Z (2.8) T τ := 0 ω(t) dt ∈ (2π, 2π + 1] and C(T ) := q 2 τ − 2π π2 6
. ⋄ + 2 τ5 (2.9) We will further use the eigenfunctions associated with the homogeneous PDE (2.1) subject to the boundary conditions (2.2). Separating variables yields the Sturm-Liouville problem, e′′(x) + λe(x) = 0, 0 < x < l, e(0) = e(l), e′ (0) = e′ (l), (2.10) the solution to which is given by λn =  2πn l 2 , en (x) = e 2iπnx l √ ,n∈Z l (2.11) (see [21].) It is well-known that the system {en (x), n ∈ Z} forms an orthonormal basis in L2 (0, l). We note that all eigenvalues λn , n 6= 0 have the multiplicity 2. Actually, this is the mathematical 6 reason why two controls are required to exactly control our model. In turn, the presence of two controls reduces the time of control. It is known (see, e.g. [2], [3], [30], [31]) that a string with Dirichlet, Neumann, or Robin boundary conditions is controllable by one control f(t) (for the right choice of the control profile g(x)) in the time interval [0, T ], T > 2T∗ (or T ≥ 2T∗ as in the case of Dirichlet boundary conditions.) In the present paper, we prove that the ring is controllable (for the right choice of the control profiles g1,2(x)) with the help of two controls f1,2(t) in the time interval [0, T ], T > T∗ (where T∗ is defined in (2.4).) Controllability of a model means that we may drive a “wide” class of the initial conditions to 1 rest. We describe this class precisely later but note here that this class is dense in Hper (0, l)×L2 (0, l). In particular, a model is called spectrally controllable in the time interval [0, T ] if all initial data of the form (y0 , y1 ) = (em , en ); m, n ∈ Z, belong to the controllable (on [0, T ]) set. We note that the corresponding classes of the initial conditions for a string subject to singular boundary conditions are described in [4]. Let us remind some notions from the Hilbert spaces theory. Let E = {ej }j∈N be a family of elements (vectors) ej of a Hilbert space H. Family E is called minimal in H if any element ej does not belong to the closure of the linear span of all the remaining elements. If a family E is minimal in H, there exists a family E ′ = {e′i } ⊂ H, which is said to be biorthogonal to E, such that (ej , e′i ) = δji . A family E is said to be a Riesz basis in H, if E is an image of an isomorphic mapping V of some orthonormal basis F = {fj }j ⊂ H, i.e. ej = V fj , j ∈ N, where V is a bounded linear and boundedly invertible operator. Definition and properties of Riesz bases can be found in [5, 16]. In particular, the following inequalities hold for any f ∈ H X 2 |(f, ej | ≍ ||f||2 (2.12) j and also X j 2 |αj | ≍ ||f||2 if f = X αj ej (2.13) j where ≍ stands for the two side inequality with constants independent of f. The inequalities (2.12) and (2.13) are used below in Section 5. We note that a family biorthogonal to a Riesz basis also forms a Riesz basis in H. A family E is said to be an L-basis in H if it forms a Riesz basis in the closure of its linear span. If a family E is an L-basis and complete in H, it clearly forms a Riesz basis in H. 3. Controllability in the case of constant parameters. In this section, we consider the initial boundary value problem similar to (2.1)–(2.3) in the case of constant parameters. That consideration, though being simple, allows one to better understand the construction we use in the following sections for the general case of non-constant parameters. Because we will cite the results of this section below, we sometimes use the upper index c here to identify the results for constant parameters. Assuming P (t) ≡ P0 and hence, by (2.7), s P0 ω(t) = 2π 2 ≡ ω0, l ρ we get according to (2.7), ω0T∗ = 2π. 7 The initial boundary value problem has the form ρytt = P0 yxx + ρ 2 X j=1 gj (x) fj (t), (x, t) ∈ QT = (0, l) × (0, T ), y(0, t) = y(l, t) yx (0, t) = yx (l, t), 0 (3.1) (3.2) 1 y(x, 0) = y (x), yt (x, 0) = y (x). (3.3) Assumptions 1–4 are satisfied automatically for the model under consideration. The results on existence, uniqueness and regularity of the solution for the initial value problem (3.1)–(3.3) are described by the following Theorem 1c . Let Assumptions 5–7 be satisfied. Then, there exists a unique (real–valued) solution for the initial boundary value problem (3.1)–(3.3) satisfying the following inclusions     1 y ∈ C [0, T ]; Hper (0, l) , yt ∈ C [0, T ]; L2(0, l) . (3.4) We formulate and discuss a similar result for non-constant parameters below in Section 4. l−periodicity of the problem and Assumptions 5–7 for the terminal data yk (x) and the space distributions of the controls gj (x) allow representing them as the Fourier series with respect to the basis {en (x), n ∈ Z} (see (2.11)) with the (known) coefficients ynj and gnj correspondingly, X yk (x) = ynk en (x), k = 0, 1, (3.5) X gj (x) = gnj en (x), j = 1, 2. (3.6) Here and below the absence of the summation index means that it is carried over all n ∈ Z. Assumption 6 implies X ||gj ||2 = |gnj |2 < ∞. (3.7) j k = gnj for all indices. Assumption 5 implies that y−n = ynk , g−n We proceed further under the following assumptions on the functions gj (x). Assumption 8. 2 1 8a. △n ≡ gn1 g−n − gn2 g−n 6= 0 ∀n ∈ Z′. |g01| |g02 | 8b. + 6= 0. ⋄ Here and below we use the following set of integer numbers Z′ ≡ Z \ {0}. (3.8) (3.9) (3.10) The meaning of the assumption is that we actually consider the controllability of the system in the “general position”, as it is rather common in Physics. The specific examples considered below in Section 8 demonstrate that. Specifically, the space distribution of the controls there has the form of the Dirac delta (or its derivative) at two points x1, x2, such that (x2 − x1 )/π is irrational. Note that the identity △−n = −△n, n ∈ Z′ holds. The main result of this section concerning controllability of the system (3.1)–(3.3) is given by Theorem 2c. Let T > T∗ and Assumptions 5–8 be satisfied. If j X g−n nyn0 △n ′ n∈Z 2 j X g−n yn1 < ∞ and △n ′ n∈Z 2 < ∞, j = 1, 2, (3.11) then, there exist the functions f1,2 ∈ L2 (0, T ), such that y(x, T ) = yt (x, T ) = 0, x ∈ (0, l). (3.12) 8 Proof. Theorem 1c allows representing the solution y(x, t) as the Fourier series with respect to the basis {en (x), n ∈ Z} with the unknown coefficients An (t) satisfying Än + ω02 n2 An = 2 X j=1 We conclude from (3.13) that An (T ) = yn0 cos ω0 |n|T + gnj fj (t), An (0) = yn0 , Ȧn (0) = yn1 , n ∈ Z, sin ω0 |n|T yn1 ω0 |n| + Ȧn (T ) = −ω0 |n|yn0 sin ω0|n|T + yn1 cos ω0 |n|T + Z Z T 0 2 sin ω0|n|(T − t) X j gn fj (t)dt, ω0|n| (3.14) j=1 T 0 (3.13) cos ω0|n|(T − t) 2 X j=1 gnj fj (t)dt, n ∈ Z; (3.15) for n = 0, this representation should be understood as the limit as n → 0. Conditions of exact controllability (3.12) may be reformulated in terms of coefficients An, An (T ) = Ȧn (T ) = 0, n ∈ Z. The last conditions, for n 6= 0, may be rewritten as follows,     Z T 2 X cos ω0|n|t ˆ −yn1 j f (t)dt = W|n| gn W|n| , n ∈ Z′ . sin ω0 |n|t j ω0 |n|yn0 0 (3.16) (3.17) j=1 Here the matrices Wn are defined by  sin ω0 nT Wn = cos ω0nT − cos ω0 nT sin ω0 nT  , n∈N and are invertible. Hence, the equalities (3.17) may be rewritten as follows,    Z T  2 X cos ω0|n|t ˆ −yn1 j gn f (t)dt = , n ∈ Z′ . sin ω0 |n|t j ω0 |n|yn0 0 (3.18) (3.19) j=1 It is convenient to rewrite Equations (3.19) in another form, for n ∈ N only,     Z T     Z T f1 (t) −yn1 f1 (t) ω0nyn0 cos ω0 nt dt = , Gn sin ω0 nt dt = , n ∈ N. (3.20) Gn 1 0 f2 (t) −y−n f2 (t) ω0ny−n 0 0 Here   gn1 gn2 Gn = , n ∈ N. 1 2 g−n g−n Equations (3.20) may be reduced to the separate moment problems for controls f1 and f2 , Z T Z T cos ω0 nt fj (t)dt = Cnj , sin ω0nt fj (t)dt = Snj , n ∈ N, j = 1, 2 0 (3.21) (3.22) 0 with the uniquely determined Cnj and Snj (see Assumption 8a),    1   1  1 ω0nyn0 Sn −yn Cn −1 −1 , n ∈ N. = G , = G 0 1 n n ω0ny−n Sn2 −y−n Cn2 (3.23) Here  2  1 g−n −gn2 , n ∈ N. 1 gn1 △n −g−n It is easy to check that inequalities (3.11) imply the inclusions {Snj }, {Cnj } ∈ l2. Gn−1 = (3.24) 9 The moment equations look differently for n = 0, Z T Z g01 (T − t)f1 (t)dt + g02 0 g01 Z T f1 (t)dt + 0 g02 Assumption 8b allows finding the moments Z T Z tfj (t)dt and 0 Z T (T − t)f2 (t)dt = y00 , (3.25) f2 (t)dt = y01 . (3.26) fj (t)dt, j = 1, 2 (3.27) 0 T 0 T 0 by y00 and y01 from (3.25)–(3.26) but not uniquely. Having found all moments, we reduce the problem of controllability to two separate moment problems (3.22), (3.25), and (3.26) for controls f1 and f2 with respect to the system of functions F c, 2 F c ≡ {sin ω0nt, cos ω0nt}n∈N ∪ {1} ∪ {ω0t} (3.28) in L (0, T ). To be more precise, let c c c (t) ≡ cos ω0nt, F2n−1 (t) ≡ sin ω0 nt, n ≥ 1. ≡ ω0t, F0c ≡ 1, F2n F c ≡ {Fnc }n≥−1, where F−1 (3.29) Then the moment problem has the form,
(3.30) fj , Fnc = Bnj , n ≥ −1, j = 1, 2 with some known coefficients Bnj . c 2 We remind that 2π ω0 = T∗ . The family F does not form a basis in L (0, T∗ ) (indeed, if we remove the function ω0 t, we get an orthogonal basis in L2 (0, T∗)), yet, F c forms an L-basis in L2(0, T ) for any T > T∗ (see [33], [5, Sec. II.4].) The L-basis property of F c implies controllability stated in Theorem 2c [5, Secs. I.2, III.3] • It is important to note that if the initial data contain any harmonic with n 6= 0, then two controls are necessary; but if they contain only one harmonic with n = 0, then one control is enough. If T ≤ T∗ , the family F c contains a proper subfamily which forms a Riesz basis in L2 (0, T ). For T = T∗ this is obvious and for T < T∗ it follows from the results [33], [5, Sec. II.4]. Therefore the family F c is not minimal in L2 (0, T ) for T ≤ T∗ . Hence, system (2.1), (2.2) is not spectrally controllable in this case [5, Sec. III.2]. The system may be approximately controllable in this case if the coefficients gn1,2 exponentially decrease as n → ∞ (see [5, Sec. II.6].) However, this controllability is unstable with respect to perturbation of the parameters of the system. 4. The main results. We now return to the general case of the variable parameters of the model. For any T > 0, we consider the initial boundary value problem (2.1)–(2.3) and proceed under Assumptions 1–8. The existence, uniqueness and regularity of the solution for the initial value problem (2.1)–(2.3) are described by the following Theorem 1. Let Assumptions 1–8 be satisfied. Then, there exists a unique solution for the initial value problem (2.1)–(2.3) satisfying the following inclusions     1 y ∈ C [0, T ]; Hper (0, l) , yt ∈ C [0, T ]; L2(0, l) . (4.1) Though this result is used below, it has an auxiliary character and may be proved by different means, e.g. similarly to [2]. 10 The main result of the paper is given by Theorem 2. Let Assumptions 1–8 be satisfied. If j X g−n nyn0 △n ′ 2 n∈Z j X g−n yn1 < ∞ and △n ′ n∈Z 2 < ∞, j = 1, 2, and T > T∗ , then, there exist the functions f1,2 ∈ L2 (0, T ), such that y(x, T ) = yt (x, T ) = 0, x ∈ (0, l). (4.2) (4.3) We note that the inequalities (4.2) represent some restrictions on both initial data and space distributions of the controlling forces. Plan of the proof. Our proof is rather cumbersome but similar to one in Section 3 for the case of constant parameters. The proof contains several steps, which we indicate below along with the corresponding sections. (1) We first apply the Fourier method (separation of variables) to reduce the problem of control for the initial boundary value problem to the moment problem. For that, we obtain the ODEs for the Fourier coefficients of the solution y(x, t). The solution of these ODEs may be represented in terms of the system {Φn, Ψn } of fundamental solutions of an auxiliary second order ODE with the time dependent coefficient P (t) (Section 5.) (2) We prove that the system of functions {Φn, Ψn} forms an L-basis in L2 (0, T ) for T > T∗ (Section 6 and Appendix.) (3) Based on the basis property of the system {Φn, Ψn} we solve the moment problem and hence, solve the problem of control (Section 7.) The rest of the paper is devoted to the realization of the program outlined above. Also, in Section 8, we consider the case of special distributions gj (x) that are typical for the piezoelectric control. 5. The moment problem. For a linear dynamical system, it is standard in Control Theory to reduce the problem of control to a moment problem. Yet, our model is described by Equation (2.1) with time–dependent coefficients, and hence, requires developing the previous methods further. We first proceed as in the proof of Theorem 2c by looking for a solution of the initial boundary value problem (2.1)–(2.3) in the form of the Fourier series with respect to the basis {en(x), n ∈ Z} with the unknown coefficients An (t) satisfying the Cauchy problem similar to (3.13), 2 ρÄn + X 4π2n2 P (t) gnj fj (t), An(0) = yn0 , Ȧn(0) = yn1 , n ∈ Z. An = ρ 2 l (5.1) j=1 The ODEs (5.1) may be rewritten as Än + ω2 (t)n2 An = 2 X j=1 gnj fj (t), n ∈ Z. (5.2) We observe that the ODEs (5.1) will produce the moment problem of a new type. Unlike the moment problem in many previous Control Theory papers (see, e.g. [30], [31], [5], [7], [10], [18], where P (t) = const), it is based on a system of functions that differs from the system of non– harmonic exponential functions and even may not be found explicitly because of the variability of the coefficient P (t) (except for very specific P (t).) Yet, we are able to prove the solvability of the moment problem. We mostly follow the scheme of our paper [2] where the exact controllability of an elastic string with tension that varies in time was proved. Our main tool is the Liouville transformation that is well known in the asymptotic analysis of the ODEs and allows reducing the 11 moment problem to another form, for which we can prove an auxiliary basis property result. Hence, we claim that our basis property result is asymptotic in nature. We now describe the Liouville transformation. We introduce the new variable Z t s= ω(ξ)dξ. (5.3) 0 We notice that s(T∗ ) = 2π and put τ = s(T ). We suppose now that T > T∗ and note that the transformation t 7→ s of the interval [0, T ] onto the interval [0, τ ] is one–to–one and hence, (5.3) defines a function (5.4) t = t(s), t ∈ C 3 [0, τ ]. We also introduce the new unknown functions according to the formulas An(t) = ω−1/2(t)Ân (s), n ∈ Z. (5.5) Applying the transformation (5.3), (5.5) to the ODEs and initial conditions (5.1)–(5.2), we get a new Cauchy problem (where the factor ρ is canceled) Â′′n + (Q(s) + n2)Ân = 2 X gnj fˆj (s), j=1 = a1n, n ∈ Z, 1 where a0n = ω1/2(0)yn0 , a1n = ω−1/2(0)yn1 + ω−3/2(0)ω̇(0)yn0 2 and the following functions are introduced, (5.7) 3ω̇(t)2 ω̈(t) fj (t) − 3 and fˆj (s) = with t = t(s). 4 4ω (t) 2ω (t) ω(t)3/2 (5.8) Ân (0) = Q(s) = a0n, Â′n (0) (5.6) We note that the properties of P (t) imply Q(s) ∈ C[0, τ ] and fˆj (s) ∈ L2 (0, τ ). (5.9) It is helpful to note at this moment that the conditions (4.2) are equivalent to j X g−n na0n △n ′ n∈Z 2 j X g−n a1n < ∞ and △n ′ n∈Z 2 < ∞, j = 1, 2. ⋄ (5.10) The solution of the Cauchy problem (5.6), (5.7) may be given in terms of two independent solutions of the corresponding homogeneous ODE wn′′ + (Q(s) + n2 )wn = 0 n = 0, 1, 2, ... (5.11) We introduce two solutions Φn (s) and Ψn (s) of (5.11) satisfying different initial conditions Φ′′n Ψ′′n + Φ′′0 + Q(s)Φ0 = 0, Φ0 (0) = 0, Φ′0 (0) = 1, (5.12) Ψ′′0 + Q(s)Ψ0 = 0, Ψ0 (0) = 1, Ψ′0 (0) = 0, (5.13) 2 + (Q(s) + n )Φn = 0, Φn(0) = 0, Φ′n(0), (Q(s) + n2 )Ψn = 0, Ψn (0) = 1, Ψ′n (0) = n ∈ N, 0, n ∈ N. (5.14) (5.15) We need to make some simple remarks about the aforementioned solutions. (a) The Wronskian of the solutions Φn(s) and Ψn(s) is equal to −1 for n = 0 and −n for n ∈ N, and hence, they are linearly independent. (b) Different structure of the spectrum of the Sturm–Liouville problem (2.10) for n = 0 and n 6= 0 forces us here and below to consider the corresponding equations for n = 0 and 12 n 6= 0 separately. (c) If the tension P (t) is independent of time, then according to (5.8), Q(s) ≡ 0 and hence, the solutions of the Cauchy problems (5.12)–(5.15) are Φ0 (s) = s, Ψ0 (s) = 1, Φn (s) = sin ns, Ψn (s) = cos ns, n ∈ N, (5.16) We note that the set {Φn, Ψn , n ∈ N ∪ {0}} is similar to the set F c (see (3.28).) We are going back to the general case. The representation for the (unique) solution of the problems (5.6), (5.7) and its derivatives has now the form       Â0(s) 1 Φ0 (s) 0 Ψ0 (s) = a0 + a0 Φ′0 (s) Ψ′0 (s) Â′0(s)    !X Z s 2 Φ0 (s) Ψ0 (s) (5.17) g0j fˆj (ξ)dξ, + Ψ0 (ξ) − Φ (ξ) 0 ′ ′ Φ (s) Ψ (s) 0 0 0 j=1 + Z       1 1 Φ|n| (s) Ân (s) 0 Ψ|n| (s) = a + a ′ n Ψ′ (s) |n| n Φ|n| (s) Â′n (s) |n|    !X 2 Φ|n| (s) Ψ|n| (s) gnj ˆ Ψ|n| (ξ) fj (ξ)dξ, n ∈ Z′ . − Φ|n|(ξ) ′ ′ Φ|n| (s) Ψ|n| (s) |n| s 0 (5.18) j=1 We are looking for the exterior forces fj (t) (or fˆj (s)) that drive the solution y(x, t) to rest in the given time T, i.e. An (T ) = Ȧn (T ) = 0, n ∈ Z. (5.19) RT According to (5.3), the value of the variable s that corresponds to t = T is given by 0 ω(ξ)dξ = τ. Hence, the requirements (5.19), in terms of the transformation (5.5) lead to the conditions Ân(τ ) = Â′n (τ ) = 0, n ∈ Z. (5.20) Substituting s = τ into (5.17)–(5.18), we get the following equations for the forces fˆj (s) that guarantee the exact controllability of the ring,    !X Z τ 2 Φ0 (τ ) Ψ0 (τ ) Ψ0 (ξ) g0j fˆj (ξ)dξ − Φ0 (ξ) ′ ′ Φ0 (τ ) Ψ0 (τ ) 0 j=1 Z τ 0     1 Φ0 (τ ) 0 Ψ0 (τ ) = −a0 − a0 , Φ′0 (τ ) Ψ′0 (τ )    !X 2 gnj ˆ Φ|n| (τ ) Ψ|n| (τ ) fj (ξ)dξ Ψ|n| (ξ) − Φ|n|(ξ) ′ ′ Φ|n| (τ ) Ψ|n| (τ ) |n| (5.21) j=1     1 1 Φ|n|(τ ) 0 Ψ|n| (τ ) = − an − an , n ∈ Z′. Φ′|n|(τ ) Ψ′|n|(τ ) |n| (5.22) These equations represent the moment problem similar to one we discussed in Section 3. In particular, it is easy to check that if we assume, as in Section 3, that Q(s) ≡ 0, then the moment problem that appears in place of (5.21)–(5.22) may be transformed to two separate moment problems (3.22), (3.25), and (3.26). Hence, we further reformulate the moment problem (5.21)–(5.22) as a problem for separated forces fˆj , j = 1, 2. We introduce matrices   Φ|n|(τ ) −Ψ|n| (τ ) Wn = , n ∈ N ∪ {0}, (5.23) Φ′|n|(τ ) −Ψ′|n| (τ ) 13 the (unknown) vectors X~n = Z and the (known) vectors τ 0   2 Ψ|n|(ξ) X j ˆ gnfj (ξ)dξ, n ∈ Z, Φ|n|(ξ) (5.24) j=1  1  1 −a0 −an ~ ; A = , n ∈ Z′. n a00 |n|a0n Using the notations (5.23)–(5.25) in the systems (5.21)–(5.22), we get ~n , n ∈ Z. WnX~n = WnA ~0 = A (5.25) (5.26) The matrices Wn are invertible because of the linear independence of Φn and Ψn for all n ∈ N ∪ {0} (the similar logic was used in Section 3.) Hence, the linear algebraic systems (5.26) have the unique solution, ~n, n ∈ Z. X~n = A (5.27) Rewriting (5.27) in terms of (5.24) and (5.25), we get (not yet separated) moment problems for the forces fˆj (s),  1  2 Z τ  −a0 Ψ0 (ξ) X j ˆ g0 fj (ξ)dξ = , (5.28) a00 Φ0 (ξ) 0 j=1 Z τ 0  1   2 −an Ψ|n| (ξ) X j ˆ gn fj (ξ)dξ = , n ∈ Z′. |n|a0n Φ|n| (ξ) (5.29) j=1 We first work with Equations (5.29) and rewrite them in another form, for n ∈ N only,  1    Z τ  1   0  Z τ fˆ (ξ) −a1n fˆ (ξ) nan Ψn (ξ) dξ = , Gn ˆ2 Φn (ξ) dξ = , n ∈ N, Gn ˆ2 1 0 −a na f (ξ) f (ξ) −n −n 0 0 (5.30) where the invertible matrix Gn is used again (see Section 3, (3.21).) Introducing the moments of controlling forces similar to (3.22), Z τ Z τ Cnj ≡ Ψn (ξ)fˆj (ξ)dξ, Snj ≡ Φn (ξ)fˆj (ξ)dξ, n ∈ Z′, j = 1, 2, (5.31) 0 0 we get, instead of (5.30), the system of equations for the moments similar to (3.22)–(3.23),  0     1  1 nan −a1n Sn Cn −1 −1 , n ∈ N. , = G = G n n na0−n −a1−n Sn2 Cn2 (5.32) As in Section 3, Assumption 8a allows finding the moments Snj , Cnj , n ∈ Z′ , j = 1, 2 uniquely from the systems (5.32) and Assumption 8b allows finding the moments S0j , C0j , not uniquely though, from (5.28). To formulate the final form of the moment problem, we let and F ≡ {Fn}n≥−1, where F2n(t) ≡ Ψn (s), F2n−1(s) ≡ Φn(s), n ∈ N ∪ {0} (5.33) j j ≡ Snj , n ∈ N ∪ {0}. ≡ Cnj , B2n−1 B2n (5.34) ˆ1 ˆ2 Then, the final form of two separate moment problems for controls f and f with respect to the system of functions F in L2 (0, τ ) is as follows,
fˆj , Fn = Bnj , n ∈ N ∪ {0} ∪ {−1}, j = 1, 2 (5.35) with the coefficients Bnj found uniquely for n ∈ N from (5.32) and not uniquely for n = 0 from (5.28). Here (·, ·) is the inner product in L2 (0, τ ). Conditions (4.2) and, equivalently, (5.10) imply 14 {Bnj } ∈ l2 . We note that the form (5.35) of the moment problem is equivalent to the form (5.32) because the functions Φn (s), Ψn (s) are real–valued. We have reduced the problem of controllability to the moment problem for the functions fj (ξ), j = 1, 2. As it is well–known from Control Theory (see [30]), the solvability of the moment problem is based on the basis property of the family F. 6. Basis properties of the family F. We prove here the L-basis property of the set F(s) of the solution of the initial value problems (5.12)–(5.15). In the case when Q(s) ≡ 0, the solutions of the initial value problems are F c(s) = {sin ns, cos ns}n∈N ∪ {1} ∪ {s}, (6.1) c c c (s) ≡ s, F0c ≡ 1, F2n (s) ≡ cos ns, F2n−1 (s) ≡ sin ns, n ≥ 1. F−1 (6.2) To be more precise, we let F c ≡ {Fnc }n≥−1, where c 2 The set of functions F forms an L-basis in L (0, τ ) for any τ > 2π, i.e. a Riesz basis in the closure L̄ of its linear span (see Sec. II.4, [6]). We introduce also a biorthogonal basis F c,0 so that c,0 (Fnc , Fm ) = δnm . (6.3) We have to be more specific here. We introduce L̄⊥ as the orthogonal complement of L̄ in L2 (0, τ ) and represent L2 (0, τ ) = L̄ ⊕ L̄⊥ . The biorthogonal basis F c,0 above is supposed to be extended by zero on L̄⊥ . We now prove that the set F = {Fn} forms an L-basis in L2 (0, τ ) for all τ > 2π, i.e. a Riesz basis in the closure L̄ of its linear span. We need two simple technical results. Proposition 1. The functions Fn(s) admit an asymptotic representation   1 c Fn(s) = Fn (s) + O as n → ∞, (6.4) n that is uniform on [0, τ ]. This result may be easily proved by using the methods from [14]. It implies that the families F and F c are quadratically close. Proposition 2. The following estimate holds, r τ3 c ∀n. (6.5) ||Fn|| ≤ 3 This result may be easily proved by elementary means. We introduce the linear operator K : L2(0, τ ) → L2 (0, τ ) according to the formula, KFnc = Fn − Fnc , ∀ Fnc ∈ F c . (6.6) The action of K on any other element of L̄ is defined by the linearity and continuity; it is supposed also to be extended by zero to the subspace L̄⊥ , orthogonal to L̄. Lemma 1. Let τ > 2π and Assumptions 2a and 3 be satisfied. Then the set F has the following properties. (a) The one-to-one correspondence between the sets of F c and F is given by the linear invertible operator I + K : L2 (0, τ ) → L2 (0, τ ) (see(6.6)), (I + K)Fnc = Fn, ∀ Fnc ∈ F c. (b) The set F forms an L-basis in L2 (0, τ ). (6.7) 15 Proof. If (a) is proven, then (b) immediately follows from the definition of an L-basis. The proof of (a) requires several steps. As we show below, this result has an asymptotic nature. We are going to prove that the one-to-one correspondence between Fnc and Fn is given by (6.7). For that, we are going to prove that ||K|| < 1 and hence, the operator I + K is invertible. We first derive from (6.7) and (6.5) r τ3 c c , ∀ n. (6.8) ||Fn|| = ||(I + K)Fn || ≤ ||I + K|| · ||Fn|| ≤ ||I + K|| 3 Further, let f be an arbitrary element of L̄. We find X f = fn Fnc (6.9) n with fn to be Fourier coefficients of f. Using (6.3), we get X (f, Fnc,0 )Fnc , f = (6.10) n so that Kf = X n (f, Fnc,0)(Fn − Fnc ). (6.11) We now estimate ||Kf||. Our goal is to (a) prove that ||Kf|| < ||f||, which will imply that ||K|| < 1, and also (b) find an explicit estimate for ||K||. We find X X ||Kf|| = || (f, Fnc,0 ) · (Fn − Fnc )|| ≤ |(f, Fnc,0)| · ||Fn − Fnc )|| n ≤ s X n n |(f, Fnc,0)|2 · s X n ||Fn − Fnc ||2 ≤ C1||f|| · s X n ||Fn − Fnc ||2 (6.12) with some positive constant C1 independent of f. The clarification of the last step is not trivial. For an orthonormal basis, it would be due to Parceval’s equality. Because the basis Fnc is not orthogonal in L2 (0, τ ), it is not easy to estimate C1 . We by-pass the problem by using the following result that is proved in Appendix. Theorem 3. There exists an absolute constant D > 0 such that for any finite complex sequence {ck } and for any ε ∈ (0, 1) the following inequality takes place X X ε2 D |cn|2 ≤ || cnFnc ||2L2(0,2π+ε). n n In the rest of this section and in the next section we show that Theorem 3 implies controllability of the system (2.1), (2.2) in the time interval [0, Tε], Tε := t(2π +ε) (see the definition of the function t(s) in (5.4).) This certainly implies controllability for all T greater than Tε . Hence, it is enough to restrict ε to any finite interval; we chose 0 < ε < 1 and put τ = 2π + ε. Let cn ≡ (f, Fnc,0). Then Theorem 3 and the representation (6.10) imply that X ε2 D |(f, Fnc,0)|2 ≤ ||f||2. n We conclude that the constant C1 in (6.12) may be estimated above by s X 1 ||K|| ≤ √ ||Fn − Fnc ||2 . ε D n 1 √ . ε D Hence, (6.13) 16 If ||K|| < 1, the operator I + K is invertible. According to (6.13), this condition holds if 1 X ||K||2 ≤ 2 ||Fn − Fnc ||2 < 1. ε D n (6.14) We now prove (6.14) by using Assumption 2a (see(2.8).) We introduce the function φn (s) = Fn(s) − Fnc (s). (6.15) According to (5.12)–(5.15)) this function is the solution of the following initial boundary value problem φ′′n + n2 φn = −Q(s)Fn , φn (0) = 0, φ′n(0) = 0, n ∈ N, (6.16) φ′′n = −Q(s)Fn , φn (0) = 0, φ′n(0) = 0, n = −1, 0. (6.17) The condition (6.14) now has the following form X 1 ||φn||2 < 1. (6.18) ||K||2 ≤ 2 ε D n∈N∪{−1}∪{0} The standard methods of the ODE theory applied to (6.16) yield Z s sin n(s − ξ) Q(ξ)Fn(ξ)dξ, n ∈ N, φn(s) = − n 0 Z s φn(s) = − (s − ξ)Q(ξ)Fn (ξ)dξ, n = −1, 0, (6.19) (6.20) 0 that, along with (6.8), allows estimating the norm of φn . We first find Z s √ |φn(s)| ≤ rn max |Q(s)| |Fn(ξ)|dξ ≤ rn max |Q(s)| s||Fn||, [0,τ ] [0,τ ] 0 where rn = Hence, we find for the norm, 1 , n ∈ N; rn = 1, n = −1, 0. n ||φn|| ≤ rn max |Q(s)| · ||Fn|| [0,τ ] We further use (6.5) and (6.8) to find Z τ 0 1/2 sds = rn max |Q(s)| · ||Fn|| [0,τ ] ||φn|| ≤ rn max |Q(s)| · ||I + K|| [0,τ ] (6.21) r τ5 , 6 r τ2 . 2 (6.22) (6.23) We now rewrite the estimate for the norm in (6.14) as follows, X τ5 τ X rn2 ||φn||2 ≤ 2 max |Q(s)|2 ||I + K||2 ||K||2 ≤ 2 ε D n 6ε D [0,τ ] n≥−1 ≤ τ5 max |Q(s)|2 (1 + ||K||)2 6ε2 D [0,τ ] X n≥1    2 τ5 1 2 2 π +2 = 2 max |Q(s)| (1 + ||K||) + 2 .(6.24) n2 6ε D [0,τ ] 6 The last inequality contains ||K|| in both sides. Solving it for ||K|| yields, q
π2 5 max[0,τ ] |Q(s)| ε√16D 6 +2 τ q ||K|| ≤
. π2 1 − max[0,τ ] |Q(s)| ε√16D + 2 τ5 6 We now are able to formulate Assumption 2 (or 2a) in its final form. (6.25) 17 Assumption 2b. Let τ ∈ (2π, 2π + 1] and √ √ τ − 2π max |Q(s)| < 6D q = 6D q
s∈[0,τ ] π2 5 2 2 6 +2 τ Combining (6.25) and (6.26), we finally get ε π2 6
. ⋄ + 2 τ5 ||K|| < 1. • (6.26) (6.27) Remark. We do not consider control on the large intervals of time, i.e. for large s. Our (important) estimate in Section 9 is found for the time interval (0, 2π + ε), 0 < ε < 1. However, if a system is controllable on the time interval [0, T1], then it is controllable on [0, T2] for any T2 > T1. 7. Solution of the moment problem. Controllability. As it was shown in Section 5, the problem of controllability may be reduced to the moment equations (5.35) with respect to the family F. According to Section 6, the family F forms an Lbasis in L2 (0, τ ), i.e. a Riesz basis in the closure of its linear span L̄ for all τ > 2π. We fix an arbitrary τ > 2π and introduce a biorthogonal basis F 0 = {Fn0} ⊂ L̄ to the Riesz basis F, so that 0 ) = δnm , ∀n, m ≥ −1. (Fn, Fm (7.1) ˆ The moment equations (5.35) mean that are the Fourier coefficients of fj with respect to the basis F. Hence, the functions fˆj may be reconstructed as follows, X X fˆj (s) = (fˆj , Fn) Fn0(s) = Bnj Fn0 (s). (7.2) {Bnj } n n This reconstruction is not unique. Indeed, an arbitrary element from the orthogonal complement to L̄ may be added to fˆj (s) and the dimension of that complement is infinite. This proves Theorem 2 which states a positive result on controllability of the system (2.1)–(2.3) for T > T∗ . Since the family F is quadratically close to F c , it is easy to show that F is not minimal in L2 (0, τ ) for τ ≤ 2π. Therefore, similarly to the system with constant coefficients, the system (2.1)–(2.3) is not spectrally controllable in the time interval [0, T ] if T ≤ T∗ . As well as the system (3.1)–(3.3), the system (2.1)–(2.3) may be approximately controllable in this case if the coefficients gn1,2 decrease exponentially as n → ∞. However, this controllability is unstable with respect to perturbation of the parameters of the system (coefficients of the equation, length of the ring, and functions g1,2 .) 8. Special case of the space distribution of the control. There is an interest to the so–called piezoelectric control in the Control Theory literature. By the piezoelectric control, we understand control with the space distribution concentrated at some point(s). For example, such a control is discussed in [12], [35] for a beam model where the space distribution is taken in the form of the derivative of the Dirac delta. We note that the beam equation is of the fourth order with respect to x, so that for the ring, we may also try the Dirac delta itself as a distribution of control. We consider two examples. For simplicity, we consider the case of constant coefficients only and let l = 2π. 1. Let gj (x) = 2π δ(x − xj ), xj ∈ (0, 2π), j = 1, 2. It is possible to check that Theorem 1 is still valid in this case. Obviously gnj = e−inxj , so that △n = e−inx1 einx2 − e−inx2 einx1 = 2i sin n(x2 − x1). We conclude that △n 6= 0 if and only if the ratio ξ := (x2 − x1)/π is irrational. Also (8.1) |g01 | + |g02 | = 2. 18 2. Similarly for gj (x) = 2π δ ′(x − xj ), we find gnj = ine−inxj , so that
△2n e−inx1 einx2 − e−inx2 einx1 = 2in2 sin n(x2 − x1). (8.2) To describe further results, we need some concepts from the theory of Diophantine approximation. We denote by S the set of all irrational numbers θ such that, if [a0, a1, . . . , an, . . . ] is the expansion of θ as a continued fraction, then the sequence of partial quotients (an ) is bounded. This is the set of “badly approximable numbers”. We note that S is obviously uncountable and, by classical results on Diophantine approximation (cf. [11, p. 120]), its Lebesgue measure is equal to zero. We will also use the fact (see, e.g. [24]) that if θ ∈ S, then there exists Cθ > 0 such that inf |θn − m| ≥ m∈N Cθ n ∀n ∈ N. (8.3) On the other hand (see [11]), for any ε > 0 there exists a set Bε ⊂ (0, 1) of Lebesgue measure equal to 1 such that for any θ ∈ Bε , there exists Cθ > 0 such that inf |θn − m| ≥ m∈N Cθ . n1+ε (8.4) √ Theorem 1 is not applicable for gj (x) = 2π δ(x−xj ). However, it is easy to check (see, e.g. [5, Secs. III.2, V.1]) that the solution to the initial value problem (2.1)–(2.3) satisfies the same inclusions     1 y ∈ C [0, T ]; Hper (0, 2π) , yt ∈ C [0, T ]; L2(0, 2π) (8.5) 1 if (y0 , y1 ) ∈ Hper (0, 2π) × L2 (0, 2π). Theorem 4. Let gj (x) = 2π δ(x−xj ), j = 1, 2 and ξ = (x2 −x1 )/π. Then for any T > T∗ and ε > 0, the following statements are valid. 2 (a) If ξ ∈ S, the controllability set of the system (2.1)–(2.2) contains the space Hper (0, 2π) × 1 Hper (0, 2π). 2+ε (b) For almost all ξ, the controllability set of the system (2.1)–(2.2) contains the space Hper (0, 2π)× 1+ε Hper (0, 2π). √ One can check ([5, Secs. III.2, V.1]) that for gj (x) = 2π δ ′(x − xj ) the solution to the initial value problem (2.1)–(2.3) satisfies the inclusions   
′  1 (8.6) y ∈ C [0, T ]; L2(0, 2π) , yt ∈ C [0, T ]; Hper (0, 2π)
′
′ 1 1 1 if (y0 , y1) ∈ L2 (0, 2π) × Hper (0, 2π) . Here Hper (0, 2π) is the space dual to Hper (0, 2π). j ′ Theorem 5. Let g (x) = 2π δ (x−xj ), j = 1, 2 and ξ = (x2 −x1 )/π. Then for any T > T∗ and ε > 0, the following statements are valid. (a) If ξ ∈ S, the controllability set of the system (2.1)–(2.2) contains the space L2 (0, 2π) × 1 ((Hper (0, 2π))′. (b) For almost all ξ, the controllability set of the system (2.1)–(2.2) contains the space H ε (0, 2π)×
′ 1−ε Hper (0, 2π) . Proof. We prove the statement of Theorem 4a. It follows from (8.3) that in this case | △n | ≥ C/|n|, |gnj | ≍ 1, and inequalities (4.2) are valid if X X 2 2 (8.7) nyn1 < ∞. n2yn0 < ∞ and Inequalities (8.7) are equivalent to the inclusion 2 1 (y0 , y1 ) ∈ Hper (0, 2π) × Hper (0, 2π). 19 Other statements of Theorems 4 and 5 can be proved in a similar way using the estimates (8.3) and (8.4). • 9. Appendix. We introduce the family E = {fk }k∈Z ∪ {f00 }, where fk (t) = eikt, k ∈ Z, and f00 = t. Let 0 < ε < 1 and γ = 2π + ε. We prove the following result. Theorem 3A. There exists an absolute constant D0 > 0 such that for any finite complex sequence {ck }, the following inequality takes place, " # 2 X X 2 2 2 I(c) := (9.1) |ck | + |c00| . ck fk + c00f00 ≥ ε D0 k k L2 (0,γ) Comment. Theorem 3A immediately implies the following result that we used in Section 6 for the basis F c . Theorem 3. There exists an absolute constant D(= D0 /2) such that for any finite complex sequence {ck } and for any ε ∈ (0, 1) the following inequality takes place, X X ε2 D |cn|2 ≤ || cnFnc ||2L2(0,2π+ε). n n The proof of Theorem 3A and the analytical expression of the lower estimate of D0 follow. For ε ≤ 1 a calculation gives that the numerical value of D0 is more than 0.000017. We note that we do not find the sharp estimate of the value of this basis constant. Hence, our estimate is not optimal. The idea of the proof is the following. (i) We add a simple exponential family to E, such that the joint family forms a basis in L2 (0, γ). The lower basis constant of the new system is not less than the basis constant to the primer system. (ii) We consider the system E α := e−αt E that corresponds to a shift of the set Z of exponents of E to the upper half plane. (iii) We consider the system E α in L2 (0, ∞). The Fourier transform of this family gives the family of partial fractions. We study the basis properties of family of these fractions in the Hardy space 2 H+ and estimate the basis constant of this family. (iv) We estimate the norm of the operator that is inverse to the projector from the span of the exponentials in L2 (0, ∞) onto L2 (0, γ). The estimate is based on the explicit form of the generating function for the joint system. Proof. Take α > 0 and consider the family E α := e−αtE. Evidently, ! 2 2 X X −αt −αt αt −αt −αt e ck fk + c00e f00 I(c) = e ≥ ck e fk + c00e f00 . k L2 (0,γ) k L2 (0,γ) Now the elements of E α may be treated as the projections of elements from L2 (0, ∞) onto L2 (0, γ). We introduce the entire function of the exponential type γ z − iα F (z) = sin[πε(z − iβ)] sin[π(z − iα)] , (9.2) z − iβ S where β > 0, α 6= β. The set Λ of zeros of F is Λα Λb , n Λα = {n + iα}n∈Z , Λβ = { + iβ}n∈Z, n6=0 ε 20 and z = iα is a double root of F . With Λ we connect the family EF = {eiλt}λ∈Λ ∪ te−at . We introduce also the families of fractions Xαβ = Xα ∪ Xβ , where Xα = {xλ }λ∈Λα ∪ {x0αi}, Xβ = {xλ}λ∈Λβ with xλ = r   √ ℑλ 1 = F −i 2ℑλ e−iλ̄t , x0αi = π z − λ̄   √ = F 2 α3 te−αtχ[0,∞) . r 2α3 1 π (z + iα)2 2 Here F is the inverse Fourier transform. We note that these functions are normalized in H+ (for 2 the definition and properties of the Hardy space H+ see, e.g. [20], [15], [28].) The function F given by (9.2) is a sine type function with the width of the indicator diagram equal to γ and Λ is a separated set of roots of F, i. e. inf λ6=µ; λ,µ∈Λ |λ − µ| > 0. Then, by the Levin–Golovin theorem [25, 17] (see also [19, 5]), the exponential family EF corresponding to Xαβ forms a Riesz basis in L2 (0, γ). In fact, in the Levin–Golovin theorem only the simple zeros are considered. The roots of F are simple except the double zero at z = iα. The double root gives the term te−at in the family (the exact statement for multiply zeros may be found in [32], but we do not use this fact here and present it only to make the situation more clear.) Our next goal is to establish a connection between the exponential series in L2 (0, ∞) and L2 (0, γ). We introduce the Blaschke products b = bαbβ , bα (z) = sin π(z − iα) z − iα sin πε(z − iβ) z + iβ , bβ (z) = . sin π(z + iα) z + iα sin πε(z + iβ) z − iβ These Blaschke products are generated by the families Xα and Xβ correspondingly. We introduce the subspaces of the Hardy space connected with the introduced inner functions, 2 2 2 2 2 2 Kα := H+ ⊖ bαH+ , Kβ := H+ ⊖ bβ H+ , Kb = H+ ⊖ bH+ . It is well-known [20], [15], [28] that Kα := We introduce also _ Xα, Kβ := _ Xβ . 2 2 Kγ := H+ ⊖ eiγz H+ . By the Paley–Wiener theorem, this subspace is the Fourier transform of the space L2 (0, γ). 2 Let PKγ Kb be the orthogonal projector PKγ from H+ onto Kγ restricted on Kb . Let ϕ = PKγ ψ. Then, evidently, h i−1 kϕkH+2 ≥ PKγ Kb kψkH+2 . If ψ is represented as a series with respect to the elements of Xαβ , the last inequality has the following form, 21  P Kγ  X cλ xλ + Λα ∪Λβ  c00x0iα  2 2 H+ h ≥ P Kγ Kb i−1 −2 X 2 cλ xλ + c00x0iα Λα ∪Λβ . (9.3) 2 H+ √ √ We denote the exponentials normalized in L2 (0, ∞) by eλ = 2ℑλe−iλ̄t and e0iα = 2 α3 te−αt . In terms of the Fourier images, inequality (9.3) takes the form, X 2 cλ eλ + c00e0iα Λα ∪Λβ L2 (0,γ) h ≥ P Kγ Kb i−1 −2 X 2 cλeλ + c00e0iα Λα ∪Λβ . (9.4) L2 (0,∞) The following result is shown in [29] (see also [5, Sec.II.3.3].) Proposition 3.  PKγ Kb −1 −1 bH 2 + = sin ϕ(Kb⊥ , Kγ ) = Pexp(izγ)H 2 − −1 ≥ inf x∈R |F (x)| =: C(F ). supx∈R |F (x)| (9.5) bH 2 + Here ϕ(Kb⊥ , Kγ ) is the angle between the subspaces Kb⊥ and Kγ and Pexp(izγ)H 2 is the skew pro− 2 2 jection onto eiγz H− parallel to bH+ . As Proposition 3 shows, we need an estimate to the constant C(F ) in (9.5) to proceed further. That estimate is given by the following lemma. Lemma 2. α β C(F ) ≥ min{ , } tanh(πα) tanh(πεβ). β α Proof. By Proposition 3 and (9.2) C(F ) ≥ inf x−iα x−iβ sup x−iα x−iβ inf | sin π(x − iα) sin πε(x − iβ)| . sup | sin π(x − iα) sin πε(x − iβ)| Since an elementary calculation gives sup Thus, x − iα x − iβ 2 = x2 + α 2 , x2 + β 2 α α x − iα x − iα = , α > β; 1, a ≤ β; inf = 1, α > β; , α ≤ β. x − iβ β x − iβ β inf x−iα x−iβ sup x−iα x−iβ = min{α/β, β/α}. (9.6) We further use the following elementary estimates, sinh(ab) ≤ | sin(a(x + ib))| ≤ cosh(ab), ∀a, b, x ∈ R. These estimates and (9.6) give the assertion of the lemma. • We further need to estimate the basis constant of the family of partial fractions on the semi-axis. First, we separate two elements xiα, x0iα corresponding to the double root of F from the rest of the family Xαβ . For any elements x, y of a Hilbert space, 22 kx + yk2 ≥ kxk2 + kyk2 − 2 cos ϕkxkkyk ≥ (1 − cos ϕ)(kxk2 + kyk2 ), (9.7) where cos ϕ = |(x, y)| , ϕ ∈ [0, π/2]. kxkkyk 2 To apply this inequality to the elements xαi and x0αi in H+ , we first calculate (xαi, x0αi)H+2 α2 √ = 2 π Z∞ −∞ √ 1 1 −i 1 1 dz = iα22 2 resαi = √ . 2 2 (z + αi) z − αi (z + αi) z − αi 2 (9.8) √ Thus, cos ϕ = 1/ 2 and hence, √ kc0xiα + c00x0iαk2H 2 ≥ (1 − 1/ 2)(|c0 |2 + |c00|2 )). (9.9) + Let now ψ be the angle between Kbiα and Kebiα , where biα is the Blaschke product corresponding to iα, 2  z − iα biα(z) = z + iα and ebiα = b/biα. Then (9.7), (9.9) imply X k cλ xλ + c0 xiα + c00 x0iαk2H 2 + λ6=iα  ≥ (1 − cos ψ) k X λ6=iα cλxλ k2H 2 +  √ + (1 − 1/ 2)(|c0 |2 + |c00|2) . By Vasyunin’s theorem, which gives an estimate of the angle between two subspaces of the form 2 2 2 2 H+ ⊖ S 1 H+ and H+ ⊖ S 2 H+ with inner functions S1 , S2 (see, e.g. [28], Lect.10, n.1), we have sin ψ ≥ |ebiα(iα)|2. Further, sin π(z − iα) e biα(iα) = (z − iα) z=iα Since we may replace the function sin ψ ≥ · z + iα z − iα sin πε(z − iβ) z + iβ sin π(z + iα) z + iα sin πε(z + iβ) z − iβ sin π(z−iα) z−iα (9.10) z=iα . by π at z = iα, (9.10) gives 2πiα sin (πεi(α − β)) α + β sin(2iπα) sin (πεi(α + β)) α − β 2 = 2πα sinh (πε(α − β)) α + β sinh 2πα sinh (πε(α + β)) α − β 2 . (9.11) sin2 ψ. We have now the estimate   X X 1 1 cλxλ + c0xiα + c00x0iαk2H 2 ≥ sin2 ψ k cλ xλk2H 2 + (1 − √ )(|c0 |2 + |c00|2 ) . k + + 2 2 λ6=iα λ6=iα It is easy to show that 1 − cos ψ ≥ 1 2 23 Proposition 4. [28, Lec. 7.2] The family Xαβ \{xiα, xiα0 } forms a Riesz basis in its span in the 2 Hardy space H+ and   2  −1 X X 1  ≥ 32(1 + 2 log ) cλ xλ |cλ|2 , δ λ∈Λ\iα λ∈Λ\iα 2 H+ where δ = δ(α, β, ε) is the Carleson constant of the set Λ\iα. Remark. The spectrum (the set of poles of partial fractions of the family Xαβ ) is in a strip in C+ and it is separable. In this case the spectrum is a Carleson set (see, e.g. [28, Lecture 11]), and we may apply the Shapiro–Shields theorem (see, e.g. [28, Lecture 6].) For the expression of the basis constant via the Carleson constant, see [28, Lecture 7.2]. To proceed further, we have to find the Carleson constant of the set Λ\iα. For simplicity, we add the points iα and iβ to the set, i.e. we study the set [ e = {n + iα}n∈Z { n + iβ}n∈Z . Λ ε e We remind that for any set {µn }∞, We denote by e b the Blaschke product corresponding to Λ. 1 the corresponding Blaschke product is ∞ Y n=1 εn z − µn , z − µ̄n where the factors εn have unit absolute value and are included for convergence of the product (see [28].) In our case we have a simple expression, sin π(z − iα) sin πε(z − iβ) e . b(z) = sin π(z + iα) sin πε(z + iβ) Let x′λ be the biorthogonal element to xλ with respect to the family {xλ}λ∈Λ e . It is well-known (see, e.g. [5, II.1.12]), that Y µ − λ̄ = |ebλ (λ)|−1 kx′λk = µ−λ µ6=λ z−λ̄ with ebλ(z) = eb(z) z−λ . e We now estimate the Carleson constant δ = inf kx′λk−1 for the spectrum Λ. Lemma 3. δ ≥ min λ∈Λ  2πα sinh[πε|α − β|] 2πεβ sinh[π|α − β|] , sinh 2απ sinh[πε(α + β)] sinh 2πεβ sinh[π(α + β)]  . Proof. Take first λ + iα ∈ Λα . Then sin π(z − iα) z − n + iα sin πε(z − iβ) |ebλ(λ)| = z − n − iα sin π(z + iα) sin πε(z + iβ) = 2πα = sinh 2πα s z+iα 2πα sin[πε(n + i(α − β))] sinh 2πα sin[πε(n + i(α + β))] cos2 (πεn) sinh2 [πε(α − β)] + sin2(πεn) cosh 2[πε(α − β)] cos2 (πεn) sinh2 [πε(α + β)] + sin2(πεn) cosh 2[πε(α + β)] 24 2πα = sinh 2πα Here we use s − cos2 (πεn) + cosh2[πε(α − β)] 2πα ≥ sinh 2πα − cos2 (πεn) + cosh2[πε(α + β)] min [0,1] s cosh2 [πε(α − β)] − 1 . cosh2 [πε(α + β)] − 1 a−x a−1 = b−x b−1 for a < b. Similarly, we find the estimate for |ebλ(λ)| with λ = nε + βi. • Combining the inequality (9.4) with Proposition 3, Lemma 3, and the estimates (9.9) and (9.10), we find 2  2 X 1 2 α β 0 cλ eλ + c00eiα ≥ sin ψ min{ , } · tanh2 (πα) tanh2 (πεβ) 2 β α Λα ∪Λβ L2 (0,γ) · min     X √ 1 , (1 − 1/ 2)  |cλ|2 + |c00|2) . 32[1 + 2 log δ1 ] Λ ∪Λ α β This inequality is valid, in particular, for cλ = 0, λ ∈ Λβ . Then the left hand side is equal to I(c), and for all β > 0, β 6= α, we have 2  α β I(c) ≥ sin2 ψ min{ , } · tanh2 (πα) tanh2(πεβ) β α ! X 1 1 1 2 2 · min{ , 3 } |ck | + |c00| ) . 2α 4α 64[1 + 2 log δ1 ] k We now simplify the expressions in the right hand-side of the last inequality. By assumption, tanh x ε ≤ 1. In what follows, we also assume that β ≤ 1. Since decreases, x tanh(πεβ) ≥ tanh π. εβ We note that the presence of the term tanh2(πεβ) implies appearance of ε2 in the main estimate (9.1). We further find lower estimates of sin ψ and the Carleson constant δ (see Lemma 3), which do sinh[πε(α − β)] not depend on ε. Since the ratio is a decreasing function of ε, we may replace it by sinh[πε(α + β)] its value at ε = 1 in inequality (9.11). To estimate the Carleson constant δ (see Lemma 3), we use the inequality 2πεβ 2πβ ≥ . sinh(2πεβ) sinh(2πβ) We conclude that the inequality (9.1) has been finally proved with D0 ≥ Here sup R(α, β). α>0, β>0,α6=β 2  1 1 1 α β R(α, β) = β 2 sin2 ψ min{ , } · tanh2 (πα) · tanh2 π · min{ , 3 } β α 2α 4α 64[1 + 2 log 1δ ] 25 with δ ≥ min  2πα 2πβ , sinh 2απ sinh 2πβ  sinh[π|α − β|] sinh[π(α + β)] and sin ψ ≥ 2πα sinh (π(α − β)) α + β sinh 2πα sinh (π(α + β)) α − β 2 . • We can find the suboptimal value R on a grid numerically. Calculation gives that on the square [0, 1] × [0, 1], the maximal value of R is R = .0000171775 for α = 0.17608, β = 0.21008. Acknowledgment The authors are grateful to the anonymous referee for a very helpful criticism. 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