On controllability of a rotating string

J. Math. Anal. Appl. 321 (2006) 198–212 www.elsevier.com/locate/jmaa On controllability of a rotating string Sergei A. Avdonin a,1 , Boris P. Belinskiy b,∗,2 a University of Alaska Fairbanks, Fairbanks, AK 99775-6660, USA b University of Tennessee at Chattanooga, 615 McCallie Avenue, Chattanooga, TN 37403-2598, USA Received 19 April 2005 Available online 8 September 2005 Submitted by M.C. Nucci Abstract We consider the problem of controllability of a rotating string. We say that a set of initial data of the string is controllable if, for any initial data of this set by suitable manipulation of the exterior force, the string goes to rest. The main result of the paper is a description of controllable sets of initial data. The equation is not strongly hyperbolic. To get exact controllability of such equation in the sharp time interval, we use control functions from Sobolev spaces with noninteger indices. To prove our results, we apply the method of moments that has been widely used in control theory of distributed parameter systems since the classical papers of H.O. Fattorini and D.L. Russell. We use recent results about exponential bases in Sobolev spaces with noninteger indices.  2005 Elsevier Inc. All rights reserved. Keywords: Distributed parameter systems; Control; Nonharmonic Fourier series; Sobolev spaces with noninteger indices 1. Introduction A disk in a computer is rotating at a very high speed. It also performs small transverse oscillations. As a consequence of those oscillations, disk might hit surrounding surfaces and lose * Corresponding author. E-mail addresses: [email protected] (S.A. Avdonin), [email protected] (B.P. Belinskiy). URLs: http://www.dms.uaf.edu/avdonin (S.A. Avdonin), http://www.utc.edu/Faculty/Boris-Belinskiy/ (B.P. Belinskiy). 1 Research of the author was supported in part by the NSF grant no. OPP-0414128. 2 Research of the author was supported in part by the University of Tennessee at Chattanooga Faculty Research Grant and the Center of Excellence for Computer Applications Scholarship. 0022-247X/$ – see front matter  2005 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2005.08.025 S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 199 the information that it carries. Therefore it would be helpful to minimize transverse oscillations. Another important mechanical model of the similar type is a rotating propeller of a helicopter; its transversal oscillations may cause crash of helicopter. In this paper, we consider the simplest distributed rotating system, namely a rotating string. We say that a set of initial data of the string is controllable if, for any initial data of this set by suitable manipulation of the exterior force, the string goes to rest. The main result of the paper is a description of controllable sets of initial data. The equation describing a rotating string is not strongly hyperbolic: one of its leading coefficients tends to zero at the end of the string. If we restrict ourselves by controls from L2 (0, T ), then, typically, equations with singular coefficients lack exact controllability in the sharp time interval, however, can preserve approximate controllability [4]. To get exact controllability of such equations in the sharp time interval, we use control functions from Sobolev spaces with noninteger indices H s (0, T ) (see, e.g., [19] for the definition and properties of these spaces). Controls from H s (0, T ) with noninteger s were used for the first time in [6]. In the present paper we extend this approach for a new class of systems. To prove our results, we apply the method of moments that has been widely used in control theory of distributed parameter systems since the classical papers of H.O. Fattorini and D.L. Russell in the late 60s to early 70s (see the survey [21] and the book [3] for the history of the subject and complete references). The method is based on properties of exponential families (usually in the space L2 (0, T )), the most important of which for control theory are minimality, the Riesz basis property, and also, as in the current paper, the L-basis property. The latter is defined as a Riesz basis property in the closure of the linear span of the family. Recent investigations into new classes of distributed systems such as hybrid systems, systems with parameters that vary in time, and damped systems as well as problems of simultaneous control have raised a number of new difficult problems in the theory of exponential families (see, e.g., [1–10,14]). In the present paper the family of nonharmonic exponentials e±ian t arises with an ∼ n + 3/4, n = 0, 1, . . . (up to a scaling). Such a family does not form a Riesz basis in L2 (0, 2π); it is not also a Riesz basis in L2 (0, T ) for any other T . However, using recent results about exponential bases in Sobolev spaces with noninteger indices [5], we can show that this family forms (after normalization) a Riesz basis in some Sobolev space and that, together with the duality arguments [12,18], enables studying controllability of the system in the time interval [0, 2π]. Note, the asymptotic formula an ∼ n + 3/4 is the consequence of the absence of the strong hyperbolicity. For the string with fixed end points we would get an ∼ n + 1 and for the string with one free and one fixed end an ∼ n + 1/2. The asymptotic formula an ∼ n + 3/4 implies the Riesz basis property of the family of nonharmonic exponentials in the Sobolev space with noninteger index, which, in turn, makes natural involving such a control space on the “critical” time interval. We now briefly derive our physical model. A homogeneous string with no resistance to bending is attached to the vertical axis at one end and free at the other end. In the equilibrium position, it is located vertically. If the axis is rotating with the constant angular velocity ω, the string does the same. If the weight of the string is neglected, then it is located horizontally in the equilibrium position. If we hit it, it starts oscillating. The oscillations are described by the wave equation. To derive it, we first need to find the tension of the string. Let the tension at the point x be H (x). Then dH (x) is the centripetal force acting on the element (x, x + dx) of the string. We find from the elementary mechanics dH = −ρω2 x dx. (1.1) 200 S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 Here ρ(x) is the density of the string (which is supposed to be constant in this paper). The end x = l of the string is free and hence H (l) = 0. (1.2) Integrating (1.1) and using (1.2) yields finally H (x) = ω 2 l (1.3) ρx dx. x Hence the equation of the small transverse oscillations is ρ(x)ytt = (Hyx )x + Fe (x, t). (1.4) Here Fe (x, t) is an exterior force. We suppose that (1.5) Fe (x, t) = g(x)f (t), where f (t) is considered as a control and the distribution of the force along the string g(x) is supposed to be given. It is assumed that g ∈ L2 (0, l). For simplicity of the exposition we suppose that ρ = const > 0. The controllability results that we prove are valid also for x-dependent ρ since they are based on the L-basis property of (normalized) exponentials e±ian t . This property, in turn, depends on asymptotic representation of eigenfrequencies an , and that representation can be easily determined for x-dependent ρ. For a constant ρ, (1.3) implies l2 − x2 . (1.6) 2 The paper is organized as follows. In Section 2 we discuss the statement of the problem, introduce the functional spaces, give the basic definitions related to controllability, and formulate the main results on controllability. In Section 3 we derive the moment problem for the force f (t) with respect to a system of nonharmonic exponential functions. In Section 4 we prove some basis properties results for these functions. In Section 5 we solve the moment problem and prove the main theorem about controllability, i.e., give the conditions of exact controllability of the rotating string. H (x) = ρω2 2. Statement of the initial boundary value problem and of the main results For any T > 0, we consider the following initial boundary value problem for the hyperbolic type PDE:   ρω2  2 l − x 2 yx x + g(x)f (t), (x, t) ∈ QT = (0, l) × (0, T ), 2 y(0, t) = 0; y(x, t), yx (x, t) < ∞ as x → l − , ρytt = y(0, x) = y0 (x), yt (x, 0) = y1 (x). (2.1) (2.2) (2.3) Here y0 and y1 are the initial displacement and velocity. This problem describes the transverse oscillations of the rotating string. It is well known [16] that the initial boundary value problem (2.1)–(2.3) has a unique solution if the functions y0 , y1 , g, and f are smooth enough. Our goal is to find the conditions on the system under consideration that guarantee the existence of the controlling force f (t) such that the string goes to rest in a given time T and construct this force. S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 201 The differential operator (with respect to x) that appears in (2.1) is the Legendre operator (see, e.g., [15])   ′ z(x), z′ (x) < ∞ as x → l − Lz := − l 2 − x 2 z′ , 0 < x < l, z(0) = 0; so that its eigenfunctions (that are the solutions of Lz = λz) and eigenvalues have the form   x zn (x) = P2n+1 (2.4) , λn = (2n + 1)(2n + 2), n  0. l The Legendre polynomials are orthogonal l P2n+1 0     x x l P2m+1 dx = δm l l 4n + 3 n (2.5) √ and the functions ψn (x) = (4n + 3)/ l P2n+1 (x/ l) form an orthonormal basis in L2 (0, l). For any real s we introduce the space Ws :  ∞ ∞   2 |an |2 λsn < ∞ . an ψn (x): a s := Ws = a(x) = n=0 n=0 L2 (0, l). Evidently, W0 = For s > 0 the space Ws is the domain of the corresponding (namely, s/2) power of the Legendre operator. We need also the weight spaces depending on the function g(x) that appears in (1.5). To define these spaces, let us represent the function g(x) as a series  g(x) = ρ gn ψn (x), (2.6) n0 where the factor ρ is introduced for convenience. −g We introduce the space Ws , s ∈ R,  ∞  −g Ws = a(x) = an ψn (x): an = 0 for gn = 0, a n=0 2 s :=  n:gn =0 |an |2 |gn |−2 λsn < ∞ . −g g The space W−s , dual to Ws , is defined as follows:  ∞ ∞   g an ψn (x): a 2s := W−s = a(x) = |an |2 |gn |2 λ−s n <∞ . n=0 (2.7) n=0 A solution of the initial boundary value problem (2.1)–(2.3) is a sum of the solutions of two problems. The first one is Eq. (2.1) with boundary conditions (2.2) and zero initial conditions y(0, x) = 0, yt (x, 0) = 0. (2.8) The second initial boundary value problem consists of the equation   ρω2  2 l − x 2 yx x , (x, t) ∈ QT = (0, l) × (0, T ), 2 with the boundary conditions (2.2) and initial conditions (2.3). ρytt = (2.9) 202 S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 Proposition 1. The solution of the initial boundary value problem (2.9), (2.2), (2.3) exists, is unique and has the same regularity as the initial data. In particular, if y0 ∈ Ws+1 , y1 ∈ Ws , then −g −g −g y ∈ C([0, T ]; Ws+1 ) and yt ∈ C([0, T ]; Ws ). If y0 ∈ Ws+1 , y1 ∈ Ws , then y ∈ C([0, T ]; Ws+1 ) −g and yt ∈ C([0, T ]; Ws ). For the initial boundary value problem (2.1)–(2.3) with nonsmooth f and g we prove the following result. Theorem 1. Let T > 0, g ∈ L2 (0, l). −g −g (a) Let f ∈ L2 (0, T ) and y0 ∈ W1 , y1 ∈ W0 . Then there is a unique solution to the initial −g −g boundary value problem (2.1)–(2.3) such that y ∈ C([0, T ]; W1 ) and yt ∈ C([0, T ]; W0 ). 1/2 −g −g (b) Let f ∈ H00 (0, T ) and y0 ∈ W3/2 , y1 ∈ W1/2 . Then there is a unique solution to the initial −g −g boundary value problem (2.1)–(2.3) such that y ∈ C([0, T ]; W3/2 ) and yt ∈ C([0, T ]; W1/2 ). 1/2 Here the Banach space H00 (0, T ) is defined as follows: 1/2 H00 (0, T ) = u ∈ H 1/2 (0, T ): t (T − t) −1/2 u(t) ∈ L2 (0, T ) . 1/2 The discussion of the properties of the space H00 (0, T ) can be found in [19, Section I.11]. It is shown there that this space is the subspace of H 1/2 (R) with the elements having a com1/2 1/2 pact support in [0, T ]. The space (H00 (0, T ))′ , dual to H00 (0, T ), plays an important role in our construction. The following result from [19, Section I.11] allows to better understand the structure of the dual space. Any element of it admits the representation  1/2 ′ f ∈ H00 (0, T ) ⇐⇒ f = f0 + f1 , f0 ∈ H −1/2 (0, T ), t (T − t) 1/2 f1 ∈ L2 (0, T ). Represent the solution y(x, t) as a Fourier–Legendre series  cn (t)ψn (x) y(x, t) = (2.10) n0 with some unknown coefficients cn (t). Also represent the initial data (2.3) as the similar series  j yj (x) = cn ψn (x), j = 0, 1, (2.11) n0 with some (known) coefficients cn0 , cn1 . Substituting the representations (2.10) and (2.11) into Eqs. (2.1) and (2.3) and using the orthogonality condition (2.5) yields the Cauchy problem for the system of (independent) ODEs c̈n + 2ω2 Nn2 cn = gn f (t), cn (0) = cn0 , ċn (0) = cn1 , (2.12) n  0, (2.13) where Nn =  n+  1 (n + 1). 2 (2.14) 203 S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 It is easy to check that Nn+1 − Nn > 1 and, therefore, the set {Nn } is uniformly discrete or separated: inf |Nm − Nn | > 0. (2.15) m=n Introduce the (positive) number T ∗ : √ π 2 ∗ . (2.16) T = ω To consider controllability problem for the initial boundary value problem (2.1)–(2.3), we need the following −g −g Definition. Let a moment T > 0 be given. The initial data (y0 , y1 ) ∈ W1 ×W0 of the string are said to be controllable in the time interval [0, T ] by L2 controls if there exist a control function f ∈ L2 (0, T ) such that the solution of the problem (2.1)–(2.3) satisfies the additional conditions at the moment t = T : y(x, T ) = yt (x, T ) = 0, (2.17) x ∈ [0, l]. The set of all such states is called the controllable set (in the time interval [0, T ]) and is denoted by G(T ). 1/2 −g (The similar notion will be applied for controls f ∈ H00 (0, T ) and initial data from W3/2 × −g W1/2 .) The system is called spectrally controllable in the time interval [0, T ] if all initial data of the form (y0 , y1 ) = (ψm , ψn ), m, n ∈ N, belong to G(T ). The system (2.1)–(2.2) is called approximately controllable in the time interval [0, T ] if G(T ) is dense in W1 × W0 . Note, the uniqueness of the solution (see Theorem 1) implies y(x, t) ≡ 0 for t  T if equalities (2.17) are true and f (t) ≡ 0 for t  T . We now formulate our main results on controllability of the rotating string. Theorem 2. Consider the system (2.1)–(2.3). Let T > 0, g ∈ L2 (0, l), f ∈ L2 (0, T ). (a) If there exists n such that gn = 0, then the system (2.1)–(2.3) is not approximately controllable for any T . (b) If T < T ∗ , the system is not spectrally controllable for any g ∈ L2 (0, l). (c) If T  T ∗ and gn = 0 for all n, then the system is spectrally controllable. −g −g (d) If T > T ∗ , then G(T ) = W1 × W0 . The space of control functions f (t) that bring the system from rest to rest on the time interval [0, T ], T > T ∗ , is infinite-dimensional. −g −g (e) If T  T ∗ , G(T ) belongs to W1 × W0 but not equal to this space. 1/2 Similar investigations can be done for control functions from H00 (0, T ). The main advantage −g of such controls is the fact of exact controllability with respect to the “natural” state space W3/2 × −g W1/2 in the “critical” time interval [0, T ∗ ]. 1/2 Theorem 3. Consider the system (2.1)–(2.3). Let f ∈ H00 (0, T ), g ∈ L2 (0, l). Then −g −g G(T ) = W3/2 × W1/2 for T  T∗ . 204 S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 3. The problem of moments The (unique) solution of the Cauchy problem (2.12)–(2.13) may be found elementary √ √ t  √  sin(ω 2Nn (t − τ )) 1 sin(ω 2Nn t) + gn f (τ ) dτ. ω 2 Nn t + cn √ √ ω 2Nn ω 2Nn cn (t) = cn0 cos 0 (3.1) The exact controllability conditions (2.17) along with orthogonality condition (2.5) lead to the conditions cn (T ) = cn′ (T ) = 0 for all n  0. (3.2) We find gn T √ √   √ sin(ω 2Nn (T − τ )) 1 sin(ω 2Nn T ) 0 f (τ ) dτ = −cn cos ω 2Nn T − cn , √ √ ω 2Nn ω 2Nn T   √ cos ω 2Nn (T − τ ) f (τ ) dτ 0 gn 0 √  √   √  = ω 2Nn cn0 sin ω 2Nn T − cn1 cos ω 2Nn T , n  0. (3.3) (3.4) Equations (3.3)–(3.4) form the moment problem for the function f (τ ). Multiplying (3.3) by √ ±iω 2Nn and adding to (3.4) yields after some simplifications gn T e∓iω 0 √ 2Nn τ √   f (τ ) dτ = − ±iω 2Nn cn0 + cn1 , n  0. Introducing the system of functions G = {Γn± (t)}, √ Γn± (t) = e±iωn t , ωn := ω 2Nn , n  0, (3.5) (3.6) the sequence   ζn± = − ±iωn cn0 + cn1 , n  0, (3.7) and denoting the inner product in L2 (0, T ) as (,) yields instead of (3.5)   gn f, Γn± = ζn± for all Γn± ∈ G. (3.8) In Section 5, we will need an expression for ±iωn cn (τ ) + ċn (τ ). Using representation (3.1) and notations (3.6) yields after some simplifications     ±iωn cn (τ ) + ċn (τ ) = ±iωn cn0 + cn1 e±iωn τ + gn f, Γn± L2 (0,τ ) e±iωn τ for any τ  0. (3.9) S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 205 4. Basis properties of the family G 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 for any j element ej does not belong to the closure of the linear span of all the remaining elements. If family E is minimal in H, there exists a family E ′ = {ei′ } ⊂ H, which is said to be biorthogonal to E, such that (ej , ei′ ) = δj i . 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 = Vfj , j ∈ N, where V is a bounded linear and boundedly invertible operator. Definition and properties of Riesz bases can be found in [3,13]. In particular, the following inequalities hold for any f ∈ H:   (f, ej )2 ≍ f 2 (4.1) j and also  j |αj |2 ≍ f 2 if f =  (4.2) αj ej , j where ≍ stands for the two sides inequality with constants independent of f. The inequalities (4.1) and (4.2) are used below in Section 5. Note that a family biorthogonal to a Riesz basis also forms a Riesz basis in H. 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 family E is an L-basis and complete in H, it clearly forms a Riesz basis in H. In this section we study minimality and the basis properties of the set of the functions G given by (3.6). The following asymptotic relation is easy to be verified:   1 3 as n → ∞. (4.3) Nn = n + + O 4 n Lemma 1. The family G = {Γn± (t)} has the following properties: (a) The family G is complete and minimal in L2 (0, T ∗ ), however, it is not a Riesz basis in this space. (b) The family G forms an L-basis in L2 (0, T ) for T > T ∗ . Moreover, there exits an infinite family of exponentials G1 such that the family G ∪ G1 forms a Riesz basis in L2 (0, T ). (c) For T < T ∗ , the family G is not minimal in L2 (0, T ). Moreover, it can be split into two subfamilies, G = G0 ∪ G1 , where G0 forms a Riesz basis in L2 (0, T ) and G1 is an infinite family. 1/2 1/2 (d) The family {ωn Γn± (t)} is a Riesz basis in [H00 (0, T ∗ )]′ . 1/2 1/2 Here [H00 (0, T ∗ )]′ is the space dual to H00 (0, T ∗ ) provided L2 (0, T ∗ ) is identified with its dual (see [19, Section I.12]). Proof. We begin with the proof of the assertions (b) and (c). To prove that the family G forms an L-basis we show that it can be extended to a Riesz basis. Then, as a subset of a Riesz basis it is, by definition, an L-basis. 206 S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 We denote Ω := {±ωn }n0 , and n+ (r) := sup # Ω ∩ [x, x + r) , (4.4) x∈R where #A is the number of elements in the set A. Function n+ (r) is clearly sub-additive: n+ (r1 + r2 )  n+ (r1 ) + n+ (r2 ). Let us define in a standard way (see, e.g., [11, p. 346]) the upper uniform density of Ω to be n+ (r) . r→∞ r The limit exists due to the sub-additivity of n+ (r). Similarly, one can introduce the function D+ (Ω) := lim n− (r) := inf # Ω ∩ [x, x + r) x∈R and the lower uniform density of Ω: n− (r) r→∞ r − (function n (r) is super-additive, and so, the limit exists). It is easy to prove, using (4.3), that 1 D+ (Ω) = D− (Ω) = √ . ω 2 Statements (b) and (c) follow now from the results of Seip [22]: D− (Ω) := lim (4.5) √ For any T > 2πD+ (Ω) (= π ω 2 = T ∗ ) the family G can be extended to a family of exponen2 (0, T ) by adding an infinite family of exponentials. tials that forms a Riesz basis in L√ π 2 − For any T < 2πD (Ω) (= ω = T ∗ ) the family G can be turned into a Riesz basis in L2 (0, T ) by extracting an infinite family of exponentials. (a) Let us denote by F (z) the generating function of the family G:  ∞   z2 1− 2 F (z) = ωn (4.6) n=0 and let  ∞   z2 1− 2 , µn n=0 √ where µn := ω 2(n + 1/2). Let δn := ωn − µn . We put also µ−n = −µn , δ−n = −δn . A particular case of [1, Lemma 4] can be formulated as follows: Φ(z) = For any real h = 0,     F (x + ih)    ≍ exp ℜ −  Φ(x + ih)     δn δn , + µn x + ih − µn (4.7) (4.8) |µn |2|x| where ≍ stands for the two sides inequality with some positive constants independent of x ∈ R. S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 From (4.3) it follows that √   1 ω 2 +O . δn = 4 n 207 (4.9) √ Taking into account that Φ(z) = cos(πz/ω 2) = cos(zT ∗ /2) and, so, |Φ(x + ih)| ≍ 1, x ∈ R, we can derive from (4.8), (4.9) (the more general result proved in [1, Theorem 3]) that for any real h     F (x + ih) ≍ 1 + |x| −1/2 , x ∈ R. (4.10) Hence +∞ −∞ |F (x)|2 dx < ∞, 1 + x2 that proves minimality of G in L2 (0, T ∗ ) (see [20], [3, Section II.4]). Since the generating function F (x) does not belong to L2 (R), the family G is complete in 2 L (0, T ∗ ) [17]. In the theory of exponential Riesz bases, an important role is played by the Muckenhoupt (A2 ) condition     1 1 1 dx < ∞, w(x) dx · sup |I | |I | w(x) I I where the supremum is taken over all intervals I of the real axis. The relation (4.10) implies that |F (x + ih)|2 does not satisfy the Muckenhoupt condition on straight lines parallel to the real axis (it is the well-know fact, complete proof can be found in [3, Section II.3.4]). Therefore the family G does not form a Riesz basis in L2 (0, T ∗ ) (see, e.g., [3, Sections II.3.4, II.4.2]). Remark 1. Family G not only is minimal but also is uniformly minimal in L2 (0, T ∗ ). It means that the norms of the family biorthogonal to G are uniformly bounded. This fact can be proved similarly to the proof of [3, Theorem 3(v)] where a family with the same (as F ) behavior of a generating function was studied. To prove statement (d), we apply now the following theorem [5, Theorem 2]. Let F be an entire function of exponential type, with indicator diagram of width a, whose zero set {ξk } lies in a strip parallel to the real axis and satisfies the separation condition infk =j |ξk − ξj | > 0. If for some real h and s   s  s  c 1 + |x|  F (x + ih)  C 1 + |x| , x ∈ R, with positive constants c and C independent of x ∈ R, then the family {eiξk t /(1 + |ξk |)s } forms n−1/2 (0, a)]′ for s = −n + 1/2, n ∈ N. a Riesz basis in H s (0, a) for s ∈ / {−N + 21 } and in [H00 n−1/2 n−1/2 Here [H00 (0, a)]′ is the dual space to H00 (0, a) relative to L2 (0, a) (see [19, Sections I.11, I.12] for detailed description of these spaces; we do not discuss this description because we actually do not use that space below). Statement (d) follows now from this theorem (with a = T ∗ , s = −1/2) and (4.10). ✷ 208 S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 5. Controllability: Proof of the main results Proof of Proposition 1 and Theorem 1(a). From the definition of spaces Ws and asymptotic representation of ωn (see (3.6), (4.3)) it is clear that    2  2    y(·, τ ), yτ (·, τ ) 2 (5.1) ωn2 cn (τ ) + ċn (τ ) ≍ W ×W 1 0 n0 and    2  2     y(·, τ ), yτ (·, τ ) 2 −g −g ≍ ωn2 cn (τ ) + ċn (τ ) |gn |−2 W ×W 1 0 (5.2) n:gn =0 (these relations mean the equivalence of norms of functions and their Fourier representations in the corresponding spaces). In Section 3 we have shown that     ±iωn cn (τ ) + ċn (τ ) = e±iωn τ ±iωn cn0 + cn1 + gn f, Γn± L2 (0,τ ) . (5.3) So, if f = 0, 2 2   2  2  ωn2 cn (τ ) + ċn (τ ) = ωn2 cn0  + cn1  , n = 0, 1, . . . . (5.4) Therefore, for any τ > 0, the state {y(·, τ ), yτ (·, τ )} belongs to the same space as the initial data {y0 , y1 }. Let now y0 = y1 = 0. Then (5.3) implies     2  2 2    f, Γ ± 2  (5.5) ωn2 cn (τ ) + ċn (τ ) |gn |−2  n L (0,τ ) . n,± n:gn =0 On the other hand, the inequality    f, Γ ± n L2 (0,τ ) n,± 2  K τ 0   f (t)2 dt  K f 2 L2 (0,T )  (5.6) is valid for any positive τ , any f ∈ L2 (0, τ ), and some constant K independent of f. For τ > T ∗ inequality (5.6) follows from Lemma 1(b) and (4.1). Therefore, it is obviously true for τ  T ∗ . Indeed, it is sufficient to put f (t) = 0 for t > τ. −g −g From (5.2), (5.5), and (5.6) it follows that {y(·, τ ), yτ (·, τ )} ∈ W1 × W0 . Using the relations (5.1), (5.2), (5.4)–(5.6), one can easily check that the sequence  N N   YN (t) := cn (t)ψn , ċn (t)ψn converges to y(·, t), yt (·, t) n=0 −g n=0 −g in W1 × W0 norm uniformly on any finite interval t ∈ [0, T ]. Therefore {y(·, t), yt (·, t)} −g −g depends continuously on t in W1 × W0 norm, which proves Proposition 1 and Theorem 1(a). ✷ The proof of the statement (b) of Theorem 1 will be provided below together with the proof of Theorem 3. S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 209 Proof of Theorem 2. (a) The first assertion of the theorem is a well-know result in control theory that we include for the completeness purposes only. Let for some integer m  0,  0  1 c  + c  = 0. gm = 0, m m It follows directly from (3.7), (3.8) that, for any T > 0, the controllable set G(T ) is orthogonal to the two-dimensional subspace of W1 × W0 spanned by (ψm , 0) and (0, ψm ). (b) According to [3, Theorem III.3.10], spectral controllability of the system (2.1), (2.2) is equivalent to minimality of {gn Γn± } in L2 (0, T ). In view of Lemma 1(c), for T < T ∗ the family {Γn± } is not minimal. Therefore, the family {gn Γn± } is also not minimal even in the case when gn = 0 for all n. (c) Due to Lemma 1((a) or (b)) the family {Γn± } is minimal in L2 (0, T ∗ ). If gn = 0 for all n, the same is true for the family {gn Γn± }. Then the system is spectrally controllable in the time interval [0, T ∗ ] and hence, in any time interval [0, T ] with T > T ∗ . (d) Suppose T > T ∗ . Then, according to Lemma 1(b) the family G forms an L-basis in 2 L (0, T ). Introduce the basis {γn± } that is biorthogonal to the basis {Γn± }, i.e.,  + −  − +  + +  − − Γn , γm = Γn , γm = δnm for any n, m  0. Γn , γm = Γn , γm = 0, Therefore, a formal solution of the moment equalities (3.8) can be written as  ζ± n γn± (s). f (t) = g n n,± (5.7) −g From the definitions of the coefficients ζn± and spaces Ws it follows that this formula actually −g −g gives a function f from L2 (0, T ) if and only if (y0 , y1 ) ∈ W1 × W0 . Note, control is not unique in this case. Indeed, because the exponential family is not complete for T > T ∗ , we can always add a function that is orthogonal to all exponentials; such a control drives the system from rest to rest. The dimension of the space of control sequences such that   f, Γn± = 0 for all n  0 is equal to the codimension of the linear span of G in L2 (0, T ). For T > T ∗ it is infinite (Lemma 1(b)). −g −g (e) Inclusion G(T ) ⊂ W1 × W0 for T  T ∗ , was really obtained in the proof of Theo−g −g rem 1(a). If the equality G(T ∗ ) = W1 × W0 was true, then due to Theorem III.3.10(a) of [3] the family {Γn± } would form Riesz basis or an L-basis in L2 (0, T ∗ ). This is impossible because of the statement (a) of Lemma 1. ✷ Remark 2. It is interesting to note that if gn = 0 for all n and |gn | < c1 exp(−c2 n) (where c1 and c2 are positive constants independent of n), then the system (2.1), (2.2) is approximately controllable in any interval [0, T ]. The proof of this fact is similar to the proof of [3, Theorem V.1.3(e)]. Proof of Theorems 3 and 1(b). Let us introduce the problem which is dual to problem (2.1)– (2.3):     ρω2  2 l − x 2 ux x , (x, t) ∈ QT = (0, l) × 0, T ∗ , 2 u(0, t) = 0; u(x, t), ux (x, t) < ∞ as x → l − ,     ut x, T ∗ = u1 (x) u x, T ∗ = u0 (x), ρutt = (5.8) (5.9) (5.10) 210 S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 with the observation   O{u0 , u1 } = z(t) := u(·, t), g L2 (0,l) . (5.11) g g 1/2 Proposition 2. Operator O is an isomorphism between W−1/2 × W−3/2 and [H00 (0, T ∗ )]′ . Proof. The (unique) solution of (5.8)–(5.10) may be constructed as in Section 2. Let   j uj (x) = gn ψn (x). un ψn (x), j = 0, 1, and g(x) = n0 n0 It can be easily checked that     sin ωn (t − T ∗ ) u0n cos ωn t − T ∗ + u1n ψn (x). u(x, t) = ωn (5.12) n0 Orthogonality condition (2.5) yields the representation for the observation (5.11),      sin ωn (t − T ∗ ) z(t) = gn u0n cos ωn t − T ∗ + u1n ωn n0   ∗ ∗ −1/2 1/2 = gn bn± e±iωn (t−T ) = gn bn± ωn ωn e±iωn (t−T ) , n,± (5.13) n,± 1/2 where bn± := (1/2)(u0n ± u1n /(iωn )). Since the family {ωn e±iωn t } forms a Riesz basis in ∗ 1/2 1/2 [H00 (0, T ∗ )]′ (see Lemma 1(d)), so does the family {ωn e±iωn (t−T ) }. Hence (5.13) implies (see (4.2))      −3/2 2  −1/2 2 gn b± ωn−1/2 2 ≍ (5.14) ≍ z 2 1/2 |gn |2 u0n ωn  + u1n ωn  . n ∗ ′ [H00 (0,T )] n n,± The proposition is proved. ✷ Going back to the proof of Theorems 3 and 1(b), we suppose that in (2.3) y0 = y1 = 0 and functions f, g, u0 , and u1 are smooth enough, so the initial boundary value problems (2.1)–(2.3) and (5.8)–(5.10) have classical solutions. Multiplying (2.1) by u(x, t), integrating over QT = (0, l) × (0, T ), integrating by parts, and using (5.8)–(5.11) and (2.2) yields ρ l 0     yt x, T ∗ u0 (x) − y x, T ∗ u1 (x) dx = T ∗ f (t)z(t) dt. (5.15) 0 For nonsmooth data this equality is understood as an equality between linear functionals acting on elements from dual spaces and can be written as     (5.16) ρ Cf, {u0 , u1 } = f, O{u0 , u1 } , where C is the (control) operator: Cf = {yt (·, T ∗ ), −y(·, T ∗ )}. From (5.16), ρC = O∗ and Propo−g −g 1/2 sition 2 implies that operator C is an isomorphism of H00 (0, T ∗ ) and W1/2 × W3/2 . Basing on this result we are able now to complete the proofs of Theorems 1(b) and 3. First turn to Theorem 1(b). Due to Proposition 1 it suffice to prove its statement for zero initial conditions. 211 S.A. Avdonin, B.P. Belinskiy / J. Math. Anal. Appl. 321 (2006) 198–212 −g Suppose that t  T ∗ . From (3.1), (3.3), (3.4), and (3.6) and using the definition of spaces Ws it follows that     2 1/2   y(·, t), yt (·, t) 2 −g −g ≍   f, Γ ± 2 n L (0,t) λn W ×W 3/2 1/2 n:gn =0 ≍     f˜, Γ ± n L2 (0,T ∗ ) n:gn =0 2 1/2  λn      2 ≍  y ·, T ∗ , yt ·, T ∗ W −g ×W −g . 3/2 Here f˜(t) =  f (s) 0 1/2 if 0  s  t, if t < s  T ∗ < 0. Since we have just proved that operator C is an isomorphism, and so     −g −g y ·, T ∗ , yt ·, T ∗ ∈ W3/2 × W1/2 , it follows that −g −g y(·, t), yt (·, t) ∈ W3/2 × W1/2 . The same arguments as in proving Theorem 1(a) lead to the continuity of the state in this space. Taking into account Proposition 1, we conclude that the statement of Theorem 1(b) is valid for all t. We have proved that, for all T , the reachability set R(T )—set of states reachable from zero −g −g 1/2 initial state with the help of controls from H00 (0, T )—belongs to W3/2 × W1/2 . From Propo−g −g −g −g sition 2 we know that R(T ∗ ) = W3/2 × W1/2 . Therefore R(T ) = W3/2 × W1/2 for T  T ∗ . 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