On controllability of a rotating string · respect to a system of nonharmonic exponential functions. In Section 4 we prove some basis properties results for these functions. In Section

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, USAb 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 thestring is controllable if, for any initial data of this set by suitable manipulation of the exterior force, thestring goes to rest. The main result of the paper is a description of controllable sets of initial data. Theequation 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 themethod of moments that has been widely used in control theory of distributed parameter systems sincethe classical papers of H.O. Fattorini and D.L. Russell. We use recent results about exponential bases inSobolev 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 os-cillations. 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 simplestdistributed rotating system, namely a rotating string. We say that a set of initial data of the stringis 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 initialdata.

The equation describing a rotating string is not strongly hyperbolic: one of its leading coeffi-cients 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 timeinterval, however, can preserve approximate controllability [4]. To get exact controllability ofsuch equations in the sharp time interval, we use control functions from Sobolev spaces withnoninteger indices Hs(0, T ) (see, e.g., [19] for the definition and properties of these spaces).Controls from Hs(0, T ) with noninteger s were used for the first time in [6]. In the present paperwe 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 controltheory of distributed parameter systems since the classical papers of H.O. Fattorini and D.L. Rus-sell in the late 60s to early 70s (see the survey [21] and the book [3] for the history of the subjectand complete references). The method is based on properties of exponential families (usually inthe space L2(0, T )), the most important of which for control theory are minimality, the Rieszbasis property, and also, as in the current paper, the L-basis property. The latter is defined as aRiesz basis property in the closure of the linear span of the family. Recent investigations intonew classes of distributed systems such as hybrid systems, systems with parameters that vary intime, and damped systems as well as problems of simultaneous control have raised a number ofnew difficult problems in the theory of exponential families (see, e.g., [1–10,14]).

In the present paper the family of nonharmonic exponentials e±iant 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 notalso a Riesz basis in L2(0, T ) for any other T . However, using recent results about exponentialbases in Sobolev spaces with noninteger indices [5], we can show that this family forms (afternormalization) 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 stronghyperbolicity. For the string with fixed end points we would get an ∼ n + 1 and for the stringwith one free and one fixed end an ∼ n + 1/2. The asymptotic formula an ∼ n + 3/4 impliesthe Riesz basis property of the family of nonharmonic exponentials in the Sobolev space withnoninteger 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 bend-ing 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 doesthe same. If the weight of the string is neglected, then it is located horizontally in the equilibriumposition. If we hit it, it starts oscillating. The oscillations are described by the wave equation. Toderive 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 fromthe elementary mechanics

dH = −ρω2x 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 endx = l of the string is free and hence

H(l) = 0. (1.2)

Integrating (1.1) and using (1.2) yields finally

H(x) = ω2

l∫x

ρx dx. (1.3)

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

Fe(x, t) = g(x)f (t), (1.5)

where f (t) is considered as a control and the distribution of the force along the string g(x) issupposed to be given. It is assumed that g ∈ L2(0, l).

For simplicity of the exposition we suppose that ρ = const > 0. The controllability resultsthat we prove are valid also for x-dependent ρ since they are based on the L-basis property of(normalized) exponentials e±iant . This property, in turn, depends on asymptotic representationof eigenfrequencies an, and that representation can be easily determined for x-dependent ρ.

For a constant ρ, (1.3) implies

H(x) = ρω2 l2 − x2

2. (1.6)

The paper is organized as follows. In Section 2 we discuss the statement of the problem, introducethe functional spaces, give the basic definitions related to controllability, and formulate the mainresults on controllability. In Section 3 we derive the moment problem for the force f (t) withrespect to a system of nonharmonic exponential functions. In Section 4 we prove some basisproperties results for these functions. In Section 5 we solve the moment problem and prove themain theorem about controllability, i.e., give the conditions of exact controllability of the rotatingstring.

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 hyperbolictype PDE:

ρytt = ρω2

2

((l2 − x2)yx

)x

+ g(x)f (t), (x, t) ∈ QT = (0, l) × (0, T ), (2.1)

y(0, t) = 0; y(x, t), yx(x, t) < ∞ as x → l−, (2.2)

y(0, x) = y0(x), yt (x,0) = y1(x). (2.3)

Here y0 and y1 are the initial displacement and velocity. This problem describes the transverseoscillations 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 goalis to find the conditions on the system under consideration that guarantee the existence of thecontrolling force f (t) such that the string goes to rest in a given time T and construct this force.


The differential operator (with respect to x) that appears in (2.1) is the Legendre operator (see,e.g., [15])

Lz := −((l2 − x2)z′)′

, 0 < x < l, z(0) = 0; z(x), z′(x) < ∞ as x → l−

so that its eigenfunctions (that are the solutions of Lz = λz) and eigenvalues have the form

zn(x) = P2n+1

(x

l

), λn = (2n + 1)(2n + 2), n � 0. (2.4)

The Legendre polynomials are orthogonal

l∫0

P2n+1

(x

l

)P2m+1

(x

l

)dx = l

4n + 3δmn (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 :

Ws ={

a(x) =∞∑

n=0

anψn(x): ‖a‖2s :=

∞∑n=0

|an|2λsn < ∞

}.

Evidently, W0 = L2(0, l). 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 definethese spaces, let us represent the function g(x) as a series

g(x) = ρ∑n�0

gnψn(x), (2.6)

where the factor ρ is introduced for convenience.We introduce the space W

−gs , s ∈ R,

W−gs =

{a(x) =

∞∑n=0

anψn(x): an = 0 for gn = 0, ‖a‖2s :=

∑n:gn =0

|an|2|gn|−2λsn < ∞

}.

The space Wg−s , dual to W

−gs , is defined as follows:

Wg−s =

{a(x) =

∞∑n=0

anψn(x): ‖a‖2s :=

∞∑n=0

|an|2|gn|2λ−sn < ∞

}. (2.7)

A solution of the initial boundary value problem (2.1)–(2.3) is a sum of the solutions of twoproblems. 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

ρytt = ρω2

2

((l2 − x2)yx

)x, (x, t) ∈ QT = (0, l) × (0, T ), (2.9)

with the boundary conditions (2.2) and initial conditions (2.3).


Proposition 1. The solution of the initial boundary value problem (2.9), (2.2), (2.3) exists, isunique and has the same regularity as the initial data. In particular, if y0 ∈ Ws+1, y1 ∈ Ws, theny ∈ C([0, T ];Ws+1) and yt ∈ C([0, T ];Ws). If y0 ∈ W

−g

s+1, y1 ∈ W−gs , then y ∈ C([0, T ];W−g

s+1)

and yt ∈ C([0, T ];W−gs ).

For the initial boundary value problem (2.1)–(2.3) with nonsmooth f and g we prove thefollowing result.

Theorem 1. Let T > 0, g ∈ L2(0, l).

(a) Let f ∈ L2(0, T ) and y0 ∈ W−g

1 , y1 ∈ W−g

0 . Then there is a unique solution to the initial

boundary value problem (2.1)–(2.3) such that y ∈ C([0, T ];W−g

1 ) and yt ∈ C([0, T ];W−g

0 ).

(b) Let f ∈ H1/200 (0, T ) and y0 ∈ W

−g

3/2 , y1 ∈ W−g

1/2. Then there is a unique solution to the initial

boundary value problem (2.1)–(2.3) such that y ∈ C([0, T ];W−g

3/2) and yt ∈ C([0, T ];W−g

1/2).

Here the Banach space H1/200 (0, T ) is defined as follows:

H1/200 (0, T ) = {

u ∈ H 1/2(0, T ):[t (T − t)

]−1/2u(t) ∈ L2(0, T )

}.

The discussion of the properties of the space H1/200 (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 com-pact support in [0, T ]. The space (H

1/200 (0, T ))′, dual to H

1/200 (0, T ), plays an important role in

our construction. The following result from [19, Section I.11] allows to better understand thestructure of the dual space. Any element of it admits the representation

f ∈ (H

1/200 (0, T )

)′ ⇐⇒ f = f0 + f1,

f0 ∈ H−1/2(0, T ),[t (T − t)

]1/2f1 ∈ L2(0, T ).

Represent the solution y(x, t) as a Fourier–Legendre series

y(x, t) =∑n�0

cn(t)ψn(x) (2.10)

with some unknown coefficients cn(t). Also represent the initial data (2.3) as the similar series

yj (x) =∑n�0

cjnψn(x), j = 0,1, (2.11)

with some (known) coefficients c0n, c

1n. 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 forthe system of (independent) ODEs

cn + 2ω2N2ncn = gnf (t), (2.12)

cn(0) = c0n, cn(0) = c1

n, n � 0, (2.13)

where

Nn =√(

n + 1

2

)(n + 1). (2.14)


It is easy to check that Nn+1 − Nn > 1 and, therefore, the set {Nn} is uniformly discrete orseparated:

infm =n

|Nm − Nn| > 0. (2.15)

Introduce the (positive) number T ∗:

T ∗ = π√

2

ω. (2.16)

To consider controllability problem for the initial boundary value problem (2.1)–(2.3), we needthe following

Definition. Let a moment T > 0 be given. The initial data (y0, y1) ∈ W−g

1 ×W−g

0 of the string aresaid to be controllable in the time interval [0, T ] by L2 controls if there exist a control functionf ∈ L2(0, T ) such that the solution of the problem (2.1)–(2.3) satisfies the additional conditionsat the moment t = T :

y(x,T ) = yt (x, T ) = 0, x ∈ [0, l]. (2.17)

The set of all such states is called the controllable set (in the time interval [0, T ]) and is denotedby G(T ).

(The similar notion will be applied for controls f ∈ H1/200 (0, T ) and initial data from W

−g

3/2 ×W

−g

1/2.)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 control-lable 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.(d) If T > T ∗, then G(T ) = W

−g

1 × W−g

0 . The space of control functions f (t) that bring thesystem from rest to rest on the time interval [0, T ], T > T ∗, is infinite-dimensional.

(e) If T � T ∗, G(T ) belongs to W−g

1 × W−g

0 but not equal to this space.

Similar investigations can be done for control functions from H1/200 (0, T ). The main advantage

of such controls is the fact of exact controllability with respect to the “natural” state space W−g

3/2 ×W

−g

1/2 in the “critical” time interval [0, T ∗].

Theorem 3. Consider the system (2.1)–(2.3). Let f ∈ H1/200 (0, T ), g ∈ L2(0, l). Then

G(T ) = W−g × W

−g for T � T∗.
3/2 1/2


3. The problem of moments

The (unique) solution of the Cauchy problem (2.12)–(2.13) may be found elementary

cn(t) = c0n cos

(ω

√2Nnt

) + c1n

sin(ω√

2Nnt)

ω√

2Nn

+ gn

t∫0

sin(ω√

2Nn(t − τ))

ω√

2Nn

f (τ) dτ.

(3.1)

The exact controllability conditions (2.17) along with orthogonality condition (2.5) lead to theconditions

cn(T ) = c′n(T ) = 0 for all n � 0. (3.2)

We find

gn

T∫0

sin(ω√

2Nn(T − τ))

ω√

2Nn

f (τ) dτ = −c0n cos

(ω

√2NnT

) − c1n

sin(ω√

2NnT )

ω√

2Nn

, (3.3)

gn

T∫0

cos(ω

√2Nn(T − τ)

)f (τ) dτ

= ω√

2Nnc0n sin

(ω

√2NnT

) − c1n cos

(ω

√2NnT

), n � 0. (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∫0

e∓iω√

2Nnτ f (τ ) dτ = −(±iω√

2Nnc0n + c1

n

), n � 0. (3.5)

Introducing the system of functions G = {Γ ±n (t)},

Γ ±n (t) = e±iωnt , ωn := ω

√2Nn, n � 0, (3.6)

the sequence

ζ±n = −(±iωnc

0n + c1

n

), n � 0, (3.7)

and denoting the inner product in L2(0, T ) as (,) yields instead of (3.5)

gn

(f,Γ ±

n

) = ζ±n for all Γ ±

n ∈ G. (3.8)

In Section 5, we will need an expression for ±iωncn(τ ) + cn(τ ). Using representation (3.1) andnotations (3.6) yields after some simplifications

±iωncn(τ ) + cn(τ ) = (±iωnc0n + c1

n

)e±iωnτ + gn

(f,Γ ±

n

)L2(0,τ )

e±iωnτ (3.9)

for any τ � 0.


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 ofelements (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 thelinear span of all the remaining elements.

If family E is minimal in H, there exists a family E ′ = {e′i} ⊂ H, which is said to be biorthog-

onal to E , such that (ej , e′i ) = δji .

Family E is said to be a Riesz basis in H, if E is an image of an isomorphic mapping V ofsome orthonormal basis F = {fj }j ⊂ H, i.e., ej = Vfj , j ∈ N, where V is a bounded linear andboundedly invertible operator. Definition and properties of Riesz bases can be found in [3,13]. Inparticular, the following inequalities hold for any f ∈ H:∑

j

∣∣(f, ej )∣∣2 � ‖f ‖2 (4.1)

and also∑j

|αj |2 � ‖f ‖2 if f =∑j

αj ej , (4.2)

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:

Nn = n + 3

4+O

(1

n

)as n → ∞. (4.3)

Lemma 1. The family G = {Γ ±n (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 thisspace.

(b) The family G forms an L-basis in L2(0, T ) for T > T ∗. Moreover, there exits an infinitefamily 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 twosubfamilies, G = G0 ∪ G1, where G0 forms a Riesz basis in L2(0, T ) and G1 is an infinitefamily.

(d) The family {ω1/2n Γ ±

n (t)} is a Riesz basis in [H 1/200 (0, T ∗)]′.

Here [H 1/200 (0, T ∗)]′ is the space dual to H

1/200 (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 formsan L-basis we show that it can be extended to a Riesz basis. Then, as a subset of a Riesz basis itis, by definition, an L-basis.


We denote Ω := {±ωn}n�0, and

n+(r) := supx∈R

#{Ω ∩ [x, x + r)

}, (4.4)

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

D+(Ω) := limr→∞

n+(r)

r.

The limit exists due to the sub-additivity of n+(r). Similarly, one can introduce the function

n−(r) := infx∈R

#{Ω ∩ [x, x + r)

}and the lower uniform density of Ω :

D−(Ω) := limr→∞

n−(r)

r

(function n−(r) is super-additive, and so, the limit exists).It is easy to prove, using (4.3), that

D+(Ω) = D−(Ω) = 1

ω√

2. (4.5)

Statements (b) and (c) follow now from the results of Seip [22]:

For any T > 2πD+(Ω) (= π√

2ω

= T ∗) the family G can be extended to a family of exponen-tials that forms a Riesz basis in L2(0, T ) by adding an infinite family of exponentials.

For any T < 2πD−(Ω) (= π√

2ω

= T ∗) the family G can be turned into a Riesz basis inL2(0, T ) by extracting an infinite family of exponentials.

(a) Let us denote by F(z) the generating function of the family G:

F(z) =∞∏

n=0

(1 − z2

ω2n

)(4.6)

and let

Φ(z) =∞∏

n=0

(1 − z2

μ2n

), (4.7)

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:

For any real h = 0,∣∣∣∣F(x + ih)

Φ(x + ih)

∣∣∣∣ � exp�{−

∑|μn|�2|x|

(δn

μn

+ δn

x + ih − μn

)}, (4.8)

where � stands for the two sides inequality with some positive constants independent ofx ∈ R.


From (4.3) it follows that

δn = ω√

2

4+O

(1

n

). (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 anyreal h∣∣F(x + ih)

∣∣ � (1 + |x|)−1/2

, x ∈ R. (4.10)

Hence+∞∫

−∞

|F(x)|21 + x2

dx < ∞,

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

L2(0, T ∗) [17].In the theory of exponential Riesz bases, an important role is played by the Muckenhoupt (A2)

condition

sup

(1

|I |∫I

w(x)dx · 1

|I |∫I

1

w(x)dx

)< ∞,

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 meansthat the norms of the family biorthogonal to G are uniformly bounded. This fact can be provedsimilarly to the proof of [3, Theorem 3(v)] where a family with the same (as F ) behavior ofa 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, whosezero set {ξk} lies in a strip parallel to the real axis and satisfies the separation conditioninfk =j |ξk − ξj | > 0. If for some real h and s

c(1 + |x|)s �

∣∣F(x + ih)∣∣ � C

(1 + |x|)s

, x ∈ R,

with positive constants c and C independent of x ∈ R, then the family {eiξkt /(1+|ξk|)s} formsa Riesz basis in Hs(0, a) for s /∈ {−N + 1

2 } and in [Hn−1/200 (0, a)]′ for s = −n + 1/2, n ∈ N.

Here [Hn−1/200 (0, a)]′ is the dual space to H

n−1/200 (0, a) relative to L2(0, a) (see [19, Sec-

tions I.11, I.12] for detailed description of these spaces; we do not discuss this descriptionbecause we actually do not use that space below).

Statement (d) follows now from this theorem (with a = T ∗, s = −1/2) and (4.10). �


5. Controllability: Proof of the main results

Proof of Proposition 1 and Theorem 1(a). From the definition of spaces Ws and asymptoticrepresentation of ωn (see (3.6), (4.3)) it is clear that∥∥{

y(·, τ ), yτ (·, τ )}∥∥2

W1×W0�

∑n�0

(ω2

n

∣∣cn(τ )∣∣2 + ∣∣cn(τ )

∣∣2) (5.1)

and ∥∥{y(·, τ ), yτ (·, τ )

}∥∥2W

−g1 ×W

−g0

�∑

n:gn =0

(ω2

n

∣∣cn(τ )∣∣2 + ∣∣cn(τ )

∣∣2)|gn|−2 (5.2)

(these relations mean the equivalence of norms of functions and their Fourier representations inthe corresponding spaces).

In Section 3 we have shown that

±iωncn(τ ) + cn(τ ) = e±iωnτ(±iωnc

0n + c1

n + gn

(f,Γ ±

n

)L2(0,τ )

). (5.3)

So, if f = 0,

ω2n

∣∣cn(τ )∣∣2 + ∣∣cn(τ )

∣∣2 = ω2n

∣∣c0n

∣∣2 + ∣∣c1n

∣∣2, 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∑n:gn =0

(ω2

n

∣∣cn(τ )∣∣2 + ∣∣cn(τ )

∣∣2)|gn|−2 �∑n,±

∣∣(f,Γ ±n

)L2(0,τ )

∣∣2. (5.5)

On the other hand, the inequality

∑n,±

∣∣(f,Γ ±n

)L2(0,τ )

∣∣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.

From (5.2), (5.5), and (5.6) it follows that {y(·, τ ), yτ (·, τ )} ∈ W−g

1 × W−g

0 .

Using the relations (5.1), (5.2), (5.4)–(5.6), one can easily check that the sequence

YN(t) :={

N∑n=0

cn(t)ψn,

N∑n=0

cn(t)ψn

}converges to

{y(·, t), yt (·, t)

}

in W−g

1 × W−g

0 norm uniformly on any finite interval t ∈ [0, T ]. Therefore {y(·, t), yt (·, t)}depends continuously on t in W

−g

1 × W−g

0 norm, which proves Proposition 1 and Theo-rem 1(a). �

The proof of the statement (b) of Theorem 1 will be provided below together with the proofof Theorem 3.


Proof of Theorem 2. (a) The first assertion of the theorem is a well-know result in control theorythat we include for the completeness purposes only. Let for some integer m � 0,

gm = 0,∣∣c0

m

∣∣ + ∣∣c1m

∣∣ = 0.

It follows directly from (3.7), (3.8) that, for any T > 0, the controllable set G(T ) is orthogonalto 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) isequivalent to minimality of {gnΓ

±n } in L2(0, T ). In view of Lemma 1(c), for T < T ∗ the family

{Γ ±n } is not minimal. Therefore, the family {gnΓ

±n } is also not minimal even in the case when

gn = 0 for all n.

(c) Due to Lemma 1((a) or (b)) the family {Γ ±n } is minimal in L2(0, T ∗). If gn = 0 for all n,

the same is true for the family {gnΓ±n }. 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

L2(0, T ). Introduce the basis {γ ±n } that is biorthogonal to the basis {Γ ±

n }, i.e.,(Γ +

n , γ −m

) = (Γ −

n , γ +m

) = 0,(Γ +

n , γ +m

) = (Γ −

n , γ −m

) = δmn for any n,m � 0.

Therefore, a formal solution of the moment equalities (3.8) can be written as

f (t) =∑n,±

ζ±n

gn

γ ±n (s). (5.7)

From the definitions of the coefficients ζ±n and spaces W

−gs it follows that this formula actually

gives a function f from L2(0, T ) if and only if (y0, y1) ∈ W−g

1 × W−g

0 .

Note, control is not unique in this case. Indeed, because the exponential family is not completefor T > T ∗, we can always add a function that is orthogonal to all exponentials; such a controldrives the system from rest to rest. The dimension of the space of control sequences such that(

f,Γ ±n

) = 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)).

(e) Inclusion G(T ) ⊂ W−g

1 × W−g

0 for T � T ∗, was really obtained in the proof of Theo-

rem 1(a). If the equality G(T ∗) = W−g

1 × W−g

0 was true, then due to Theorem III.3.10(a) of [3]the family {Γ ±

n } would form Riesz basis or an L-basis in L2(0, T ∗). This is impossible becauseof the statement (a) of Lemma 1. �Remark 2. It is interesting to note that if gn = 0 for all n and |gn| < c1 exp(−c2n) (where c1 andc2 are positive constants independent of n), then the system (2.1), (2.2) is approximately control-lable 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):

ρutt = ρω2

2

((l2 − x2)ux

)x, (x, t) ∈ QT = (0, l) × (

0, T ∗), (5.8)

u(0, t) = 0; u(x, t), ux(x, t) < ∞ as x → l−, (5.9)

u(x,T ∗) = u0(x), ut

(x,T ∗) = u1(x) (5.10)


with the observation

O{u0, u1} = z(t) := (u(·, t), g)

L2(0,l). (5.11)

Proposition 2. Operator O is an isomorphism between Wg

−1/2 × Wg

−3/2 and [H 1/200 (0, T ∗)]′.

Proof. The (unique) solution of (5.8)–(5.10) may be constructed as in Section 2. Let

uj (x) =∑n�0

ujnψn(x), j = 0,1, and g(x) =

∑n�0

gnψn(x).

It can be easily checked that

u(x, t) =∑n�0

(u0

n cosωn

(t − T ∗) + u1

n

sinωn(t − T ∗)ωn

)ψn(x). (5.12)

Orthogonality condition (2.5) yields the representation for the observation (5.11),

z(t) =∑n�0

gn

(u0

n cosωn

(t − T ∗) + u1

n

sinωn(t − T ∗)ωn

)

=∑n,±

gnb±n e±iωn(t−T ∗) =

∑n,±

gnb±n ω

−1/2n ω

1/2n e±iωn(t−T ∗), (5.13)

where b±n := (1/2)(u0

n ± u1n/(iωn)). Since the family {ω1/2

n e±iωnt } forms a Riesz basis in

[H 1/200 (0, T ∗)]′ (see Lemma 1(d)), so does the family {ω1/2

n e±iωn(t−T ∗)}. Hence (5.13) implies(see (4.2))

‖z‖2[H 1/2

00 (0,T ∗)]′ �∑n,±

∣∣gnb±n ω

−1/2n

∣∣2 �∑n

|gn|2(∣∣u0

nω−1/2n

∣∣2 + ∣∣u1nω

−3/2n

∣∣2). (5.14)

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, andusing (5.8)–(5.11) and (2.2) yields

ρ

l∫0

[yt

(x,T ∗)u0(x) − y

(x,T ∗)u1(x)

]dx =

T ∗∫0

f (t)z(t) dt. (5.15)

For nonsmooth data this equality is understood as an equality between linear functionals actingon elements from dual spaces and can be written as

ρ⟨Cf, {u0, u1}

⟩ = ⟨f,O{u0, u1}

⟩, (5.16)

where C is the (control) operator: Cf = {yt (·, T ∗),−y(·, T ∗)}. From (5.16), ρC = O∗ and Propo-sition 2 implies that operator C is an isomorphism of H

1/200 (0, T ∗) and W

−g

1/2 × W−g

3/2.

Basing on this result we are able now to complete the proofs of Theorems 1(b) and 3. First turnto Theorem 1(b). Due to Proposition 1 it suffice to prove its statement for zero initial conditions.


Suppose that t � T ∗. From (3.1), (3.3), (3.4), and (3.6) and using the definition of spaces W−gs

it follows that∥∥{y(·, t), yt (·, t)

}∥∥2W

−g3/2×W

−g1/2

�∑

n:gn =0

∣∣(f,Γ ±n

)L2(0,t)

∣∣2λ

1/2n

�∑

n:gn =0

∣∣(f , Γ ±n

)L2(0,T ∗)

∣∣2λ

1/2n

� ∥∥{y(·, T ∗), yt

(·, T ∗)}∥∥2W

−g3/2×W

−g1/2

.

Here

f (t) ={

f (s) if 0 � s � t,

0 if t < s � T ∗ < 0.

Since we have just proved that operator C is an isomorphism, and so{y(·, T ∗), yt

(·, T ∗)} ∈ W−g

3/2 × W−g

1/2,

it follows that{y(·, t), yt (·, t)

} ∈ W−g

3/2 × W−g

1/2.

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 forall t.

We have proved that, for all T , the reachability set R(T )—set of states reachable from zeroinitial state with the help of controls from H

1/200 (0, T )—belongs to W

−g

3/2 × W−g

1/2. From Propo-

sition 2 we know that R(T ∗) = W−g

3/2 × W−g

1/2. Therefore R(T ) = W−g

3/2 × W−g

1/2 for T � T ∗.Because of invertibility of time in Eq. (2.1), G(T ) = R(T ) for T � T ∗, that proves Theo-rem 3. �References

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