Boundary Controllability of a Hybrid System Consisting in Two Flexible Beams Connected by a Point...

BOUNDARY CONTROLLABILITY OF A HYBRID SYSTEMCONSISTING IN TWO FLEXIBLE BEAMS CONNECTED BY A

POINT MASS

CARLOS CASTRO∗ AND ENRIQUE ZUAZUA∗

Abstract. We consider a hybrid system consisting on two flexible beams connected by a pointmass. The constant of rotational inertia is assumed not to be zero. In a previous paper we haveproved that, in the presence of the point mass, the system is well-posed in asymmetric spaces inwhich solutions have one more degree of regularity to one side of the mass.

We are interested in the problem of controllability when the control acts on the free extremeof one of the beams. We prove that when the control time is large enough the system is exactlycontrollable in an asymmetric space. This result is sharp. The proofs combine classical techniquesfrom asymptotic analysis and the theory of non-harmonic Fourier series.

Key words. Flexible beams, point mass, asymmetric spaces, Fourier series, controllability.

AMS subject classifications. 35L30, 35P15, 42A55

1. Introduction. In this paper we study the boundary controllability of a linearsystem modelling the vibrations of two flexible beams connected by a point mass.

We assume that the beams occupy the intervals (−1, 0) and (0, 1) and that thepoint mass is located at x = 0. By means of the scalar function u = u(x, t) definedfor x ∈ (−1, 1) and t > 0 we describe the vertical displacements of the beams and thepoint mass. The linear equations describing the small vibrations of this system canbe written as follows

γ∂2utt − utt − ∂4u = 0, for x ∈ (−1, 0), t > 0

γ∂2utt − utt − ∂4u = 0, for x ∈ (0, 1), t > 0

[u] (0, t) = [∂u](0, t) = 0, for t > 0

Mutt(0, t) + [∂3u](0, t) = 0, for t > 0

Mγ0∂utt(0, t)− [∂2u](0, t) = 0 for t > 0.

(1)

where ∂ denotes partial derivation with respect to x and the index t derivation withrespect to time. [u](0) = u(0+) − u(0−) denotes the jump of the function u at thepoint x = 0 where the mass is located. Assuming that the beams are hinged at theirextremes, system (1) has to be completed with the following boundary conditions:

u(±1, t) = ∂2u(±1, t) = 0, for t > 0.(2)

In (1) the dynamic of the beams is described by the Rayleigh beam equation whereγ ≥ 0 represents the constant of rotational inertia. The third equation guaranteesthat u and ∂u are continuous across x = 0 while the last two equations describe thevibrations of the point mass at x = 0. M is the total mass concentrated in x = 0 andγ0 the rotational inertia at this point.

∗ Departamento de Matematica Aplicada, Universidad Complutense de Madrid, 28040 Madrid,Spain ([email protected]) ([email protected]). This work was done while thefirst author was supported by a doctoral fellowship of the “Universidad Complutense de Madrid”.The authors were also partially supported by grants PB93-1203 of the DGICYT (Spain) and CHRX-CT94-0471 of the European Union.

1

2 C. CASTRO AND E. ZUAZUA

To simplify the exposition we assume M = 1 and γ = γ0 though the analysis isvalid for other values of the parameters.

It is worth noting that, in the particular case in which the constant γ of rotationalinertia vanishes (γ = 0) ∂2u is continuous across x = 0 too. This implies that theeffect of the mass point is weaker on the behavior of the system when γ = 0 thanwhen γ > 0. Thus the properties of system (1.1)-(1.2) when γ = 0 are much closerto the case in which the point mass is not present. We refer to [5] and [9] for precisestatements and the details of the proofs.

All along this paper we assume that γ > 0.System (1)-(2) has to be completed with suitable initial conditions for u(x, t),

u(0, t) and ∂u(0, t). The last two quantities will be denoted by y and z respectively,i.e.

u(0, t) = y(t); ∂u(0, t) = z(t).(3)

The initial conditions are then:

u(x, 0) = u0(x) in (−1, 0) ∪ (0, 1) ; y(0) = y0, z(0) = z0

ut(x, 0) = u1(x) in (−1, 0) ∪ (0, 1); yt(0) = y1, zt(0) = z1.(4)

System (1)-(2) has been studied in [4] where it is proved that, with appropriateregularity and compatibility conditions, on the initial data it admits a unique solutionin a suitable class. On the other hand, its energy

E(t) =∫ 1

−1

[∣∣∂2u(x, t)∣∣2 + γ |∂ut(x, t)|2 + |ut(x, t)|2

]dx + |ut(0, t)|2 + γ |∂ut(0, t)|2

(5)is constant along trajectories.

In this paper we assume that a control function q = q(t) acts on the systemthrough the extreme x = 1 on the quantity ∂2u(1, t). Then the boundary conditionsin (2) have to be replaced by

u(±1, t) = ∂2u(−1, t) = 0; ∂2u(1, t) = q(t) for t > 0.(6)

The problem of exact controllability can be formulated as follows: Given T > 0, findthe class H of initial conditions for which there exists a control q, say, in L2(0, T )such that the solution of (1), (3) with boundary conditions (6) is at rest at time t = T ,i.e. it satisfies

u(x, T ) = 0 for x ∈ (−1, 0) ∪ (0, 1), y(T ) = 0, z(T ) = 0

ut(x, T ) = 0 for x ∈ (−1, 0) ∪ (0, 1), yt(T ) = 0, zt(T ) = 0.(7)

In this formulation of the control problem we have chosen the control to belongto L2(0, T ). This is not, of course, the unique choice but it is the one that comesmore naturally when studying the problem of controllability by means of J.-L. Lions’HUM method (see [8]).

It turns that the space H of controllable initial data, can not be found among thefamily of energy spaces in which system (1)-(2) is well-posed. Indeed, all the energyspaces have in common the fact that solutions in those classes have the same regularityto both sides of the point mass. For instance, the energy E in (5) corresponds to

BOUNDARY CONTROLLABILITY OF A HYBRID SYSTEM... 3

solutions u in H2 to both sides of x = 0 and such that ut belongs to H1(−1, 1).However the space of controllable data turns out to be asymmetric in the sense thatits elemets have one more degree of regularity to the left of x = 0.

The same phenomena was observed in [6] in the case of two flexible strings con-nected by a point mass. In [6] this was proved by using the explicit formula forsolutions of the one-dimensional wave equation in terms of its initial data and it wasseen that this is a consequence of the fact that solutions gain one derivative whencrossing the mass. In [6] it was also observed that the spectral gap of the wave equa-tion vanishes in the presence of a point mass and it was conjectured these two facts(i.e. the asymmetry of the controllable space and the lack of the spectral gap) to beclosely related. Later on, in [3], it was proved that these two properties are equivalent(see also [1]).

For the fourth order system that we are considering here it has been observedthat, in presence of the point mass, the spectral gap vanishes too (see [4]). UsingFourier developments of solutions it is also proved in [4] that system (1)-(2) is wellposed in asymmetric spaces in which the solutions have one more degree of regularityto one side of x = 0. This result applies only when γ > 0 since, as we said above,when γ = 0 the presence of the mass has a much weaker effect on the behavior of thesystem. In this case (γ = 0) system (1)-(2) is not well-posed in asymmetric spaces ofthis kind.

Thanks to the existence of the asymmetric spaces in which system (1)-(2) is wellposed and by means of the theory of non-harmonic Fourier series and more precisely,of some results by D. Ulrich [10], we prove sharp observability results. These resultsstablish the equivalence between a suitable asymmetric norm of the initial data and thequantity

∫ T

0|∂u(1, t)|2 dt which measures the amount of energy concentrated at x = 1

during the time interval t ∈ (0, T ). This result requires the time T to be sufficientlylarge and, more precisely, T ≥ 4

√γ. This is due to the finite speed of propagation

underlying in system (1) when γ > 0, and therefore it is a natural restriction to theobservability to hold.

By means of HUM and as a direct consequence of this observability result weprove the exact controllability of the system. We show, roughly, that when γ > 0the space of controllable data coincides with the subspace of H2(−1, 1) ×H1(−1, 1)of those elements that restricted to (−1, 0) have one more degree of regularity, i.e.belong to H3(−1, 0)×H2(−1, 0).

It is worth mentionning that, in the absence of mass, the space of controllabledata coincides with H2(−1, 1)×H1(−1, 1) (see [7]). Thus, as in the context of flexiblestrings with a point mass (see [6]), the presence of the point mass reduces the spaceof controllable data by one derivative on the opposite side of the mass with respectto the extreme in which the control is located.

The rest of the paper is organized as follows. In section 2 we recall withoutproofs some basic analytical results of the uncontrolled system (1)-(2) given in [4].In particular we state the well-posedness of the system in asymmetric spaces usingFourier series. In section 3 we prove suitable observability properties. In section 4we solve system (1) with non-homogeneous boundary conditions to give sense to thesolutions of the controlled problem where we have to consider the boundary condition(6). Finally, in section 6 we obtain the main controllability results.

2. Preliminary results. In this section we set some analytical properties of thesolutions of system (1) which will be used along this work. The proofs of the resultscan be seen in [4].


2.1. Spectral analysis. When decomposing solutions of (1)-(2) in Fourier seriesone is led to consider solutions in separated variables u = eiλtϕ(x). In this class ofsolutions system (1.1)-(1.2) becomes:

∂4ϕ = λ2ϕ− γλ2∂2ϕ , for x ∈ (−1, 0)

∂4ϕ = λ2ϕ− γλ2∂2ϕ , for x ∈ (0, 1)

[ϕ](0) = [∂ϕ](0) = 0

[∂2ϕ](0) = −γλ2∂ϕ(0)

[∂3ϕ](0) = λ2ϕ(0)

ϕ(±1) = ∂2ϕ(±1) = 0.

(8)

We introduce the operator (I − γ∂2)−1 : L2(−1, 1) → H2 ∩H10 (−1, 1) such that

u = (I − γ∂2)−1F if and only if u ∈ H2 ∩H10 (−1, 1) and satisfies u− γ∂2u = F .

If we define the vector valued eigenfunction φ = (ϕ,ϕ(0), ∂ϕ(0)), system (8) canbe written as

{Kφ = λ2φ∂2ϕ(±1) = 0(9)

where K is the linear operator given by

K =

(I − γ∂2)−1∂4(−1,1)/{0} 0 0

∂3+ − ∂3

− 0 0− 1

γ

(∂2+ − ∂2

−)

0 0

.

Here ∂4(−1,1)/{0} represents the operator which assigns to each function, not necessarily

continuous in x = 0, the fourth order derivative to both sides of x = 0 and ∂k±

represents the distribution which assigns to a function u the value ∂ku(0±).The following result is proved in [4]:Proposition 2.1. The eigenvalues {λk}k∈IN of system (9) are simple and con-

stitute a sequence of positive real numbers:

0 < λ1 < λ2 < · · · < λk < · · · → ∞.

Moreover, the corresponding eigenfunctions {φk}k∈IN may be normalized to forman orthonormal basis of the space

H ={φ = (ϕ,ϕ(0), ∂ϕ(0)) ∈ H2 ∩H1

0 (−1, 1)× IR× IR}

with the norm

‖φ‖H =[∫ 1

−1

|∂2ϕ|2dx

]1/2

.

Remark 1. All the results of the above Proposition but the simplicity of theeigenvalues can be proved using classical theory on compact selfadjoint operators. Thesimplicity of the eigenvalues requires a detailed analysis of the system under consid-eration.


From Proposition 2.1 the space H can also be written as follows

H =

u : u =

∑

k∈INakφk, ‖ u ‖2H=

∑

k∈IN| ak |2< ∞

.(10)

We can also define the following fractional Hilbert spaces (Hα, ‖ · ‖α)α∈IR:

Hα =

u =

∑

k∈INakφk :‖ u ‖2α=

∑

k∈IN| ak |2 λ4α

k < ∞ .(11)

We will denote by < ·, · >α the scalar product in Hα.Clearly H0 = H and ‖ · ‖H=‖ · ‖0.Observe that, if u =

∑k∈IN akφk, then Ku =

∑k∈IN

ak

λ2k

φk. Clearly K is an

isomorphism from Hα into Hα+1. We can also write explicitely K−1:

K−1u =∑

k∈INλ2

kakφk

which is continuous from Hα+1 into Hα.We need to identify the spaces Hα for some values of the parameter α ∈ IR. To do

that we denote by Hs ((−1, 1)\ {0})∩H2∩H10 (−1, 1) the subspace of H2∩H1

0 (−1, 1)constituted by the elements such that its restrictions to (−1, 0) and (0, 1) belong toHs.

We have the following characterizations of the fractional spaces Hα:Proposition 2.2. (a) H1/2 coincides algebraically and topologically with the

subspace of

H3 ((−1, 1)\ {0}) ∩H2 ∩H10 (−1, 1)× IR× IR

constituted by the elements (u, y, z) such that

∂2u(±1) = 0, u(0) = y, ∂u(0) = z.(12)

(b) H−1/2 coincides with the subspace of H10 (−1, 1)×IR×IR constituted by the elements

(u, y, z) such that u(0) = y.Moreover

‖ (u, y, z) ‖2−1/2=∫ 1

−1

[γ | ∂u |2 + | u |2] dx+ | y |2 +γ | z |2 .(13)

(c) H−1 coincides algebraically and topologically with the quotient space of L2(−1, 1)×IR × IR constituted by the classes (u, y, z) characterized in the following way: Twoelements (u1, y1, z1) and (u2, y2, z2) belong to the same class if and only if

(u1 − u2, y1 − y2, z1 − z2) = α(m,−1, 0) + β(n, 0, γ−1)

where α, β ∈ IR and m and n are the functions

m(x) =

sinh( 1+x√γ

)

2√

γ cosh( 1√γ)

if x ∈ [−1, 0]sinh( 1−x√

γ)

2√

γ cosh( 1√γ)

if x ∈ [0, 1], n(x) =

sinh( 1+x√γ

)

2γ sinh( 1√γ)

if x ∈ [−1, 0]− sinh( 1−x√

γ)

2γ sinh( 1√γ)

if x ∈ [0, 1].

(14)


(d) H−3/2 coincides with the quotient space of H−1(−1, 1) × IR × IR constituted bythe classes (u, y, z) characterized in the following way: Two elements (u1, y1, z1) and(u2, y2, z2) belong to the same class if and only if

(u1 − u2, y1 − y2, z1 − z2) = α(m,−1, 0) + β(n, 0, γ−1)

where α, β ∈ IR, m and n are the functions given in (14).Let us recall now how solutions of (1)-(2) can be developed in Fourier series.Consider de energy space H = H0 × H−1/2 and define φk = (φk, iλkφk) where

λ−k = −λk and φ−k = φk. The set (φk)k∈ZZ constitutes an orthonormal basis inH. Then, for any initial data ((u0, y0, z0), (u1, y1, z1)) ∈ H we can find a sequence ofcoefficients (ak) such that

((u0, y0, z0), (u1, y1, z1)) =∑

k∈ZZ

akφk

and the vector valued solution U = ((u, y, z), (ut, yt, zt)) of (1), (2), (3) and (4) isgiven by

U(t) =∑

k∈ZZ

akeiλktφk.(15)

The conservation of the energy E in (5) is equivalent to the fact that system(1)-(2) generates a group of isometries in H. More precisely

E(t) =‖ U(t) ‖2H=∑

k∈ZZ

∣∣akeiλktφk

∣∣2 =∑

k∈ZZ

| ak |2=‖ U(0) ‖2H= E(0).(16)

Obviously, one can also obtain developments in Fourier series of the form (15) forsolutions of (1)-(2) in other classes Hα = Hα ×Hα−1/2.

2.2. Asymptotics of the spectrum. In this section we recall the main resultsconcerning the asymptotic behavior of the eigenvalues and eigenfunctions of (9) thatwe will need later to prove the observability inequalities.

Proposition 2.3. We have

λ2k−1 =kπ − π/2√

γ− c1(γ)√

γ (kπ − π/2)+ O(k−2), as k →∞(17)

λ2k =kπ − π/2√

γ− c2(γ)√

γ (kπ − π/2)+ O(k−2), as k →∞(18)

where c1(γ) =(2γ +

√γ tanh

(γ−1/2

))−1+(2γ)−1 and c2(γ) = γ−1/2 coth γ−1/2−2+

(2γ)−1.Moreover,

λ2k − λ2k−1 =C(γ)

(kπ − π/2) γ1/2+ O(k−2), as k →∞(19)

where C(γ) = c1(γ)− c2(γ) > 0, for all γ > 0.Remark 2. In the absence of mass the asymptotic behavior of eigenvalues is as

follows:

λk =kπ

2√

γ+ O(k−2), as k →∞.


This shows that the first term of the asymptotic expansion of λ2k is affected by thepresence of the mass but not the first term of λ2k−1.

In view of (19) the asymptotic gap λ2k − λ2k−1 decays like 1/k as k →∞ in thepresence of the mass. Note however that in the absence of the mass the eigenvaluesare uniformly separated and the gap is of the order of π/(2

√γ) for all k.

Concerning the eigenfunctions we have:Proposition 2.4. The eigenfunctions of (9) normalized in H2 ∩H1

0 (−1, 1) are

ϕ2k−1(x) =(20)

ρ2k−1

(µ+2k−1)2

[sin

(µ+

2k−1(1− |x|))− µ+

2k−1 cosµ+2k−1

µ−2k−1 cosh µ−2k−1

sinh(µ−2k−1(1− |x|)

)]

ϕ2k(x) = − ρ2k

(µ+2k)2

[sin

(µ+

2k(x

|x| − x))− sin µ+

2k

sinhµ−2k

sinh(

µ−2k(x

|x| − x))]

(21)

where ρk = 1 +O(k−1) and

µ+k = λk

√γ

√√√√12

+12

√1 +

4γ2λ2

k

, µ−k =1

√γ

√12 + 1

2

√1 + 4

γ2λ2k

.(22)

Moreover, ∂ϕk(1) 6= 0.Remark 3. We observe that the eigenfunctions ϕ2k−1 are even while ϕ2k are

odd. This fact is due to the symmetry of the problem with respect to x = 0. On theother hand, as the mass only affects the first term in the asymptotic expansion ofthe eigenvalues corresponding to odd eigenfuntions, these are the only eigenfunctionswhich are affected in a significant way by the point mass.

2.3. Asymmetric spaces. In this section we are going to introduce and charac-terize some asymmetric spaces. It is easy to see that these spaces are stable under theflow generated by system (1)-(2) and they are natural spaces to solve the boundarycontrol problem.

With the notations of section 2.1 we set

Yα =

U =

∑

k∈ZZ\{0}akφk ∈ H :(23)

‖ U ‖2Yα=

∑

k∈ZZ\{0}

( | a2k−σk|2

δ4αk

+| a2k − a2k−σk

|2δ4α+2k

)< ∞

where δk = λ2k − λ2k−σk, σk = sgn k, i.e. σk = 1 if k > 0 and σk = −1 if k < 0.

Clearly Yα endowed with the norm ‖ · ‖Yα is a Hilbert space. On the hand, itis clear that if all the δk were uniformly positive and bounded above, then Yα wouldcoincide algebraically and topologically with one of the energy spaces Hα.

Since, in view of Proposition 2.3, δk = O(k−1) = O(λ−1k ) → 0 as k → ∞ we

deduce that Yα is a strict subspace of Hα.We have the following result:Proposition 2.5. Let U0 =

((u0, y0, z0), (u1, y1, z1)

)be an element of Yα. Then,

the solution U(t) = ((u(t), y(t), z(t)) , (ut(t), yt(t), zt(t))) of (1)-(2) with initial data


U0 belongs to Yα for every t > 0 and α ∈ IR. Furthermore, for any T > 0 there existsa constant C(T ) > 0 such that

‖ U(t) ‖Yα≤ C(T ) ‖ U0 ‖Yα

, ∀0 ≤ t ≤ T,∀U0 ∈ Yα.(24)

The following theorem provides a precise characterization of the spaces Y0 andY−1:

Theorem 2.6. (a) Y0 is the subspace of elements U0 =((u0, y0, z0), (u1, y1, z1)

)of H such that the restriction of (u0, u1) to (0,1) belongs to H3(0, 1)×H2(0, 1) and,in addition to the compatibility conditions of H (

u0(0) = y0, ∂u0(0) = z0, u1(0) = y1),

the following hold:

∂u1(0+) = z1, ∂2u0(1) = 0.(25)

Furthermore, the norm ‖ · ‖Y0 is equivalent to

[‖ U ‖2H +

∥∥(u0 |(0,1), u

1 |(0,1)

)∥∥2

H3×H2(0,1)

]1/2

.

(b) Y−1 is the subspace of elements U0 =((u0, y0, z0), (u1, y1, z1)

)of H−1 such that

the restriction of (u0, u1) to (0,1) belongs to H1(0, 1)× L2(0, 1) and verify:

u0(0+) = y1, u0(1) = 0.(26)

Remark 4. The spaces Yα are asymmetric in the sense that its elements haveone more degree of regularity to the left of x = 0.

The characterization of Yα as asymmetric spaces can be explained as follows: Thevectors pk = δ2α

k (φ2k−1 + φ2k)/2, qk = δ2α+1k (φ2k−1 − φ2k)/2 constitute a Riesz basis

of Yα. We observe that pk and qk are constituted by the functions (ϕ2k−1 + ϕ2k)/2,(ϕ2k−1 − ϕ2k)/2 weighted differently. As it can be seen in Figure 1 for k = 1, dueto the presence of the mass, the profiles of (ϕ2k−1 + ϕ2k)/2 and (ϕ2k−1 − ϕ2k)/2 areessentially one reflection of the other with respect to x = 0.

- 1 1 - 1 1

Fig. 1. (ϕ1 − ϕ2)/2 and (ϕ1 + ϕ2)/2

This explains the asymmetric structure of Yα.

3. Observability. As we mentioned in the introduction, using HUM [8], thecontrollability of system (1) with controls of the form (6) can be reduced to the


obtention of suitable observability estimates for the system in the absence of control:

γ∂2utt − utt − ∂4u = 0, for x ∈ (−1, 0), 0 < t < Tγ∂2utt − utt − ∂4u = 0, for x ∈ (0, 1), 0 < t < T[u] (0, t) = [∂u] (0, t) = 0, for 0 < t < Tutt(0, t) +

[∂3u

](0, t) = 0, for 0 < t < T

γ∂utt(0, t)− [∂2u

](0, t) = 0, for 0 < t < T

u(±1, t) = ∂2u(±1, t) = 0, for 0 < t < T(u(x, 0), u(0, 0), ∂u(0, 0)) =

(u0, y0, z0

),

(ut(x, 0), ut(0, 0), ∂ut(0, 0)) =(u1, y1, z1

).

(27)

As in previous sections, we identify the solution u of (1) with the vector valuedunknown U . The following holds:

Lemma 3.1. For any T > 0 there exists C(T ) > 0 such that∫ T

0

|∂u(1, t)|2 dt ≤ C ‖ U0 ‖2Y−1, ∀U0 ∈ Y−1.(28)

Moreover, if T ≥ 4√

γ there exists C(T ) > 0 such that

‖ U0 ‖2Y−1≤ C

∫ T

0

|∂u(1, t)|2 dt, ∀U0 ∈ Y−1.(29)

Remark 5. The first estimate (28) of this lemma establishes a hidden regularityresult since, in view of Proposition 2.5 and Theorem 2.6, the fact that U0 ∈ Y−1

implies u∣∣(0,1) ∈ C

([0, T ];H1(0, 1)

) ∩ C1([0, T ];L2(0, 1)

)but this is not sufficient to

guarantee that ∂u(1, t) ∈ L2(0, T ).The second estimate (29) of the lemma guarantees that the norm of the initial data

in Y−1 can be observed continuously in terms of the L2(0, T )-norm of ∂u(1, t). In viewof Proposition 2.5 this implies that the norm of the solution U in C([0, T ], Y−1) canbe observed too. Due to the finite speed of propagation of the system the time requiredfor the uniform observability (29) has to be large enough. The lower bound 4

√γ is

sharp.In order to prove Lemma 3.1 we need the following result on non-harmonic Fourier

series due to Ulrich [10].Theorem 3.2. Let (σn)n∈ZZ and (τn)n∈ZZ be two sequences of distinct complex

numbers such that σn 6= τn for all n ∈ ZZ and

lim|n|→∞

|σn − n| = lim|n|→∞

|τn − n| = 0.(30)

Then{eiσnt

}n∈ZZ

forms a Riesz basis of L2(0, 2π) and moreover

{eiσnt

}n∈ZZ

∪{

eiσnt − eiτnt

σn − τn

}

n∈ZZ

(31)

forms a Riesz basis of L2(0, 4π).Remark 6. We refer to [3] and [1] for a generalization of this result.As an immediate consequence of this result the following holds:Corollary 3.3. Let γ > 0. Then, if

{λ2

k

}k∈ZZ

denote the eigenvalues of system(27),

{eiλ2kt

}k∈ZZ

∪{

eiλ2kt − eiλ2k−σkt

λ2k − λ2k−σk

}

k∈ZZ

(32)


form a Riesz basis of L2(0, 4

√γ).

Proof. We introduce the change of variables s = tπ/√

γ that transforms thefunctions in (32) into:

e−is/2{

ei(λ2k√

γ/π+1/2)s}

k∈ZZ∪e−is/2

{ei(λ2k

√γ/π+1/2)s − ei(λ2k−σk

√γ/π+1/2)s

δk

}

k∈ZZ

and the interval t ∈ (0, 4

√γ)

into s ∈ (0, 4π). Obviously the common multiplicativefactor e−is/2 does not affect the fact that these functions constitute a Riesz basis ofL2(0, 4π) or not.

By setting

τk = λ2k

√γ

π+

12, σk = λ2k−σk

√γ

π+

12

we are in the conditions of Theorem 3.2 in view of the asymptotic form of the eigen-values proved in Propositions 2.3.

Undoing the change of variables we deduce that Corollary 3.3 holds.The following characterization of Riesz basis (which can be seen in Young [11])

will also be used:Theorem 3.4. Let H be a separable Hilbert space. The following two properties

are equivalent:(a) {en}n∈ZZ forms a Riesz basis of H;(b) {en}n∈ZZ is a complete sequence in H and there exists two positive constants

A,B > 0 such that

A

n∑

i=1

| ci |2≤∥∥∥∥∥

n∑

i=1

ciei

∥∥∥∥∥

2

H

≤ B

n∑

i=1

| ci |2

for any n ∈ IN and ci, i = 1, · · · , n.We are now ready to prove Lemma 3.1.

Proof of Lemma 3.1 Recall that the solution u of (27) can be written as

(u, u(0), ∂u(0)) =∑

k∈ZZ\{0}akeiλktφk(x)

where φk = (ϕk, ϕk(0), ∂ϕk(0)) are the eigenfunctions of the eigenvalue problem (8)with ϕk normalized in H2 ∩H1

0 (−1, 1) such that (−1)k∂ϕk(1) > 0.They are of the form

ϕ2k−σk(x) =

ρ2k−σk(µ+

2k−σk

)2

[sin

(µ+

2k−σk(1− x)

)

− µ+2k−σk

cos µ+2k−σk

µ−2k−σkcoshµ−2k−σk

sinh(µ−2k−σk

(1− x))]

ϕ2k(x) = − ρ2k(µ+

2k

)2

[sin

(µ+

2k(1− x))− sin

(µ+

2k

)

sinh(µ−2k

) sinh(µ−2k(1− x)

)]

in the interval (0, 1).


We introduce the following change in the Fourier coefficients of the solutions:ak = akλk |∂ϕk(1)|. Then

u(x, t) =∑

k∈ZZ\{0}

ak

λk | ∂ϕk(1) |eiλktϕk(x).

Observe that {ak} ∈ `2 if and only if {ak} ∈ `2. For that it is sufficient to seethat there exist positive constants A,B > 0 such that A ≤| λk || ∂ϕk(1) |≤ B for allk. To do that we observe that

|λ2k−σk∂ϕ2k−σk

(1)| = λ2k−σkρ2k−σk(

µ+2k−σk

)2

(µ+

2k−σk− µ+

2k−σk

cosµ+2k−σk

cosh µ−2k−σk

)(33)

=1√γ

+ O(k−1),

|λ2k∂ϕ2k(1)| = λ2kρ2k(µ+

2k

)2

(µ+

2k −sin

(µ+

2k

)µ−2k

sinh(µ−2k

))

=1√γ

+ O(k−1)(34)

in view of (22) and the asymptotic results of section 2.2. Thus, | λk∂ϕk(1) |→ 1/√

γas | k |→ ∞ and, on the other hand, | λk∂ϕk(1) |6= 0 for all k 6= 0. Therefore,constants A,B > 0 exist.

Consequently

A∑

k∈ZZ\{0}| ak |2≤

∑

k∈ZZ\{0}| ak |2≤ B

∑

k∈ZZ\{0}| ak |2 .(35)

On the other hand, the norm in the asymmetric space Y−1 can also be written in anequivalent form in terms of the coefficients ak. Indeed,

A2

∑

k∈ZZ\{0}δ4k |a2k−σk

|2 +∑

k∈ZZ\{0}δ2k | a2k − a2k−σk

|2(36)

≤∑

k∈ZZ\{0}δ4k |a2k−σk

|2 +∑

k∈ZZ\{0}δ2k| a2k − a2k−σk

|2

≤ B2

∑

k∈ZZ\{0}δ4k | a2k−σk

|2 +∑


|2 .

We start with the second inequality:∑


|2(37)

=∑

k∈ZZ\{0}δ2k| a2kλ2k | ∂ϕ2k(1) | −a2k−σk

λ2k−σk| ∂ϕ2k−σk

(1) ||2

≤ 4∑

k∈ZZ\{0}

[| a2k |2 δ4

k

∣∣∣∣λ2k | ∂ϕ2k(1) | −1/

√γ

δk

∣∣∣∣2

+ | a2k−σk|2 δ4

k

∣∣∣∣λ2k−σk

| ∂φ2k−σk(1) | −1/

√γ

δk

∣∣∣∣2]

+2γ

∑


|2


≤ C

∑

k∈ZZ

δ4k |a2k−σk

|2 +∑

k∈ZZ\{0}δ4k | a2k |2

+

2γ

∑


|2

since, in view of (33)-(34), | λ2k | ∂ϕ2k(1) | −1/√

γ |, | λ2k−σk| ∂ϕ2k−σk

(1) | −1/√

γ |and δk are of the order of 1/k.

Clearly, the last term in (37) can be bounded in terms of∑

k∈ZZ\{0}δ4k | a2k−σk

|2 +∑

k∈ZZ\{0}δ2k | a2k − a2k−σk

|2 .

The first inequality in (36) can be proved in a similar way.Now, taking into account that ∂u(1, t) =

∑k∈ZZ\{0} ak∂ϕk(1)eiλkt we deduce that

∫ T

0

|∂u(1, t)|2 dt =∫ T

0

∣∣∣∣∣∣∑

k∈ZZ\{0}ak∂ϕk(1)eiλkt

∣∣∣∣∣∣

2

dt =∫ T

0

∣∣∣∣∣∣∑

k∈ZZ\{0}(−1)k ak

λkeiλkt

∣∣∣∣∣∣

2

dt

=∫ T

0

∣∣∣∣∣∣∑

k∈ZZ\{0}

[(a2k

λ2k− a2k−σk

λ2k−σk

)eiλ2kt + δk

(a2k−σk

λ2k−σk

)eiλ2kt − eiλ2k−σkt

δk

]∣∣∣∣∣∣

2

dt

≤ C∑

k∈ZZ\{0}

(∣∣∣∣a2k

λ2k− a2k−σk

λ2k−σk

∣∣∣∣2

+ δ2k

∣∣∣∣a2k−σk

λ2k−σk

∣∣∣∣2)

≤ C∑

k∈ZZ\{0}

(|a2k − a2k−σk

|2λ2

2k

+ |a2k−σk|2

∣∣∣∣1

λ2k− 1

λ2k−σk

∣∣∣∣2

+δ2k

λ2k−σk

|a2k−σk|2

)

≤ C∑

k∈ZZ\{0}

(δ2k |a2k − a2k−σk

|2 + δ4k |a2k−σk

|2)≤ C ‖ U0 ‖2Y

in view of Corollary 3.3, inequalities (36) and the fact that λ2k ∼ 1/δk and λ2k−σk∼

1/δk.A similar computation shows that (29) holds too.Inequalities (36) imply the statement of Lemma 3.1 in view of Theorem 3.4.

4. On the solvability of the system with non-homogeneous data. Inthis section we analyze the existence, uniqueness and regularity of non-homogeneousboundary value problems that appear when addressing the control problem.

In order to state the main results of this section it is convenient to introduce thefollowing asymmetric spaces:

V +0 =

{(ϕ, η, ξ) ∈ H−3/2 : ϕ|(0,1) ∈ L2(0, 1)

};(38)

V +1 =

{(ϕ, η, ξ) ∈ H−1 : ϕ|(0,1) ∈ H1(0, 1), ϕ(0+) = η, ϕ(1) = 0

}.(39)

In these notations the superscripts + indicate that the elements of these spaces aremore regular to the right of x = 0, while the subscripts 0 (resp. 1) indicates that themaximal regularity is L2 (resp. H1).

Remark 7. In view of the characterization of Y−1 given in Theorem 2.6 weobserve that Y−1 = V +

1 × V +0 .

This section is divided in two parts. In the first one we analyze systems with non-zero right hand side terms. In the second one we address non-homogeneous boundaryvalue problems by transposition.


4.1. Systems with non-zero right hand side. Let us consider the non ho-mogeneous system:

γ∂2utt − utt − ∂4u = f, x ∈ (−1, 0), 0 < t < Tγ∂2utt − utt − ∂4u = f, x ∈ (0, 1), 0 < t < T[u] (0, t) = [∂u] (0, t) = 0, 0 < t < Tutt(0, t) +

[∂3u

](0, t) = g, 0 < t < T

γ∂utt(0, t)− [∂2u

](0, t) = h, 0 < t < T

u(±1, t) = ∂2u(±1, t) = 0, 0 < t < T(u(x, 0), u(0, 0), ∂u(0, 0)) =

(u0, y0, z0

),

(ut(x, 0), ut(0, 0), ∂ut(0, 0)) = (u1, y1, z1).

(40)

Observe that the boundary conditions at x = ±1 vanish.We have the following result:Theorem 4.1. Assume that U0 =

((u0, y0, z0), (u1, y1, z1)

) ∈ H−1/2 ×H−1 and

((1− γ∂2)−1f, g, h) ∈ L2 (0, T ;H−1) .(41)

Then, there exists an unique solution U ∈ C([0, T ]; H−1/2 ×H−1

)of (40).

Moreover, there exists C(T ) > 0 such that

∫ T

0

|∂u(1, t)|2 dt(42)

≤ C(‖ ((1− γ∂2)−1f, g, h) ‖2L1(0,T ;H0(−1,1)) + ‖ U0 ‖2H−1/2×H−1

)

for all U0 and (f, g, h) as above.Remark 8. The first result of this theorem is classical and provides the existence

of solutions with values in H−1/2 ×H−1, which is a natural symmetric energy spacefor solving (40).

Inequality (42) extends the “hidden regularity” result of Lemma 3.1 to the solu-tions of the non-homogeneous system. However, (42) is not sharp since it requiresthe same degree of regularity at both sides of x = 0 while in Lemma 3.1 this degree ofregularity is only required on (0, 1). The next theorem provides a sharp result.

Theorem 4.2. Assume that U0 ∈ Y−1 and ((1 − γ∂2)−1f, g, h) ∈ L1(0, T ;V +0 ).

Then, there exists an unique solution U ∈ C ([0, T ];Y−1) of (40).Moreover, there exists C(T ) > 0 such that

∫ T

0

|∂u(1, t)|2 dt ≤ C[‖ ((1− γ∂2)−1f, g, h) ‖2

L1(0,T ;V +0 )

+ ‖ U0 ‖2Y−1

],(43)

for every U0 and (f, g, h) as above.The proof of both theorems is rather similar. For simplicity we only prove Theo-

rem 4.2.

Proof of Theorem 4.2. Taking into account that the system is linear and in viewof Lemma 3.1 it is sufficient to prove it when U0 ≡ 0.

To simplify the notation we identify (I − γ∂2)−1f and the vector valued function((I − γ∂2)−1f, g, h

).

We observe that (I − γ∂2)−1f ∈ L1(0, T ;V +0 ) and then, in view of Remark 7, we

have(0, (I − γ∂2)−1f

) ∈ L1(0, T ; Y−1)


On the other hand, composing system (27) with the operator (I − γ∂2)−1 andidentifying the unknown vector (u, u(0), ∂u(0)) with u, it can be written as:

utt + Au = (I − γ∂2)−1f

where A = K−1 is the underlying elliptic operator.Since U0 ≡ 0, by the variation of constants formula u(t) =

∫ t

0v(t− s; s)ds where

v(·, ·; s) satisfies{

vtt + Av = 0v(0; s) = 0, vt(0; s) = (I − γ∂2)−1f(s).(44)

In view of Proposition 2.5 it is easy to see that u or, more precisely, its correspondingvector-valued solution U , belongs to C([0, T ]; Y−1).

On the other hand, in view of Lemma 3.1 we have

∫ T

0

|∂v(1, t; s)|2 dt ≤ C∥∥(

0,[(I − γ∂2)−1f

](s)

)∥∥2

Y−1

= C ‖ [(I − γ∂2)−1f

](s) ‖2

V +0

, ∀s ∈ [0, T ].

By Minkowski’s inequality we deduce that

‖ ∂u(1, t) ‖L2(0,T )=∥∥∥∥∫ t

0

∂v(1, t− s; s)ds

∥∥∥∥L2(0,T )

≤ C∥∥(I − γ∂2)−1f

∥∥L1(0,T ;V +

0 ).

The following result is also needed:Theorem 4.3. Assume that U0 ≡ 0 and (f, g, h) = ∂t(F,G, H) satisfying ((I −

γ∂2)−1F,G, H) ∈ L1(0, T ;V +

1

). Then the solution of (40) verifies U ∈ C([0, T ];Y−1).

Moreover, there exists C(T ) > 0 such that

∫ T

0

|∂u(1, t)|2 dt ≤ C ‖ ((I − γ∂2)−1F,G, H) ‖2L1(0,T ;V +

1 )(45)

for every (F,G, H) as above.Proof of Theorem 4.3. As in Theorem 4.2 above we identify (I − γ∂2)−1F (resp.

(I − γ∂2)−1Ft) with ((I − γ∂2)1F, G, H) (resp. (I − γ∂2)−1Ft, Gt,Ht)) to simplifythe notation.

On the other hand, u = vt where v, which is also identified with the unknownvector (v, v(0), ∂v(0)), is the solution of

{vtt + Av = (I − γ∂2)−1Fv(0) = vt(0) = 0.

(46)

Observe that we have taken null initial data for v. This is due to the fact that inthis proof we may assume ((1−γ∂2)−1F, G, H) to be of compact support in time. In-deed, if Theorem 4.3 is proved for those ((1−γ∂2)−1F, G,H), it can then be extendedby density to all ((1− γ∂2)−1F, G,H) ∈ L1(0, T ;V +

1 ).With this in mind we see that the appropriate initial conditions for v are as

follows:

v(0) = ut(0) = 0; vt(0) = utt(0) = (I − γ∂2)−1F (0)−Au(0) = 0.


To complete the proof of Theorem 4.3 it is sufficient to prove that the followinglemma holds:

Lemma 4.4. Assume that U0 = 0 and ((1−γ∂2)−1f, g, h) ∈ L1(0, T ; V +1 ). Then,

there exists C > 0 such that the solution of (40) satisfies

∫ T

0

|∂ut(1, t)|2 dt ≤ C ‖ ((1− γ∂2)−1f, g, h) ‖2L1(0,T ;V +

1 )(47)

for all (f, g, h) as above.Proof of Lemma 4.4. First of all we observe that in view of the characterization

of the asymmetric space Y in Theorem 2.6 it is easy to see that ((I − γ∂2)−1f, 0) ∈L1(0, T ; Y−1). Note that we identify (I−γ∂2)−1f with the vector ((I−γ∂2)−1f, g, h)to simplify the notation.

As in Theorem 4.2 above u =∫ t

0v(x, t− s; s)ds where v solves (44). Then ω = vt

verifies{

ωtt + Aω = 0ω(0; s) = (I − γ∂2)−1f(s), ωt(0; s) = 0.

(48)

In view of Lemma 3.2 we have∫ T

0

|∂ω(1, t)|2 dt ≤ C ‖ ((I − γ∂2)−1f, 0) ‖2Y−1= C ‖ (I − γ∂2)−1f ‖2

V +1

.(49)

On the other hand

∂ut =∫ t

0

∂vt(t− s; s)ds =∫ t

0

∂ω(t− s; s)ds

and therefore∫ T

0

|∂ut(1, t)|2 dt =∥∥∥∥∫ t

0

∂ω(1, t− s; s)ds

∥∥∥∥2

L2t (0,T )

.(50)

Now, by Minkowski’s inequality and (49) we deduce that∥∥∥∥∫ t

0

∂ω(1, t− s; s)ds

∥∥∥∥L2

t (0,T )

≤ C ‖ (I − γ∂2)−1f ‖L1(0,T ;V +1 ) .(51)

Combining (50)-(51) we deduce that (47) holds.This concludes the proof of Lemma 4.4 and Theorem 4.3.

4.2. Non-homogeneous boundary conditions. Let us consider now the sys-tem

γ∂2utt − utt − ∂4u = 0, x ∈ (−1, 0), 0 < t < Tγ∂2utt − utt − ∂4u = 0, x ∈ (0, 1), 0 < t < T[u] (0, t) = [∂u] (0, t) = 0, 0 < t < Tutt(0, t) +

[∂3u

](0, t) = 0, 0 < t < T

∂utt(0, t)− [∂2u

](0, t) = 0, 0 < t < T

u(±1, t) = ∂2u(−1, t) = 0, ∂2u(1, t) = q(t), 0 < t < T(u(x, 0), u(0, 0), ∂u(0, 0)) =

(u0, y0, z0

),

(ut(x, 0), ut(0, 0), ∂ut(0, 0)) =(u1, y1, z1

).

(52)


Observe that the boundary conditions in (52) are non-homogeneous since theboundary condition ∂2u(1, t) takes the value q.

We assume that

q ∈ L2(0, T )(53)

and

U0 ∈ Y − ={U0 ∈ H0 ×H−1/2 : u0|(−1,0) ∈ H3(−1, 0), ∂2u0(−1) = 0,(54)

u1|(−1,0) ∈ H2(−1, 0), ∂u1(0−) = z1}

.

Observe that Y − is the reflection of the space Y0 (characterized in Theorem 2.6) withrespect to x = 0. In other words, U0 ∈ Y − if and only if V 0(x) = U0(−x) ∈ Y0.

In view of Proposition 2.5 and taking into account that system (52) is symmetricwith respect to x = 0 when q = 0 the following holds:

Lemma 4.5. When q ≡ 0 and U0 ∈ Y − system (6.16) admits an uniquesolution U ∈ C([0, T ];Y −). Moreover, there exists C(T ) > 0 such that

‖ U ‖L∞(0,T ;Y −)≤ C(T ) ‖ U0 ‖Y − , ∀U0 ∈ Y −.(55)

As a consequence of this Lemma and since system (52) is linear it is sufficient toanalyze solutions of (52) when U0 ≡ 0 and q is as in (53). Therefore, in the sequel weassume that U0 ≡ 0.

Solutions of (52) can be understood in the sense of transposition. To make precisethis notion we consider the adjoint system:

γ∂2ϕtt − ϕtt − ∂4ϕ = f, x ∈ (−1, 0), 0 < t < Tγ∂2ϕtt − ϕtt − ∂4ϕ = f, x ∈ (0, 1), 0 < t < T[ϕ] (0, t) = [∂ϕ] (0, t) = 0, 0 < t < Tϕtt(0, t) +

[∂3ϕ

](0, t) = g(t), 0 < t < T

γ∂ϕtt(0, t)− [∂2ϕ

](0, t) = h(t), 0 < t < T

ϕ(±1, t) = ∂2ϕ(±1, t) = 0, 0 < t < T(ϕ(x, T ), ϕ(0, T ), ∂ϕ(0, T )) = (ϕt(x, T ), ϕt(0, T ), ∂ϕt(0, T )) ≡ 0.

(56)

Given ((I − γ∂2)−1f, g, h) ∈ L1(0, T ; V +0 ) and taking into account that system

(56) is time-reversible, in view of Theorem 4.2 we deduce that (56) admits an uniquesolution Φ ∈ C([0, T ]; Y−1) (by Φ we denote the vector-valued unknown associated toϕ).

On the other hand, ∂ϕt(1, t) ∈ L2(0, T ).Multiplying in (52) by ϕ and integrating by parts we get, at least formally, the

following identity:

∫ T

0

< (I − γ∂2)u, (I − γ∂2)−1f >H10 ,H−1 dt +

∫ T

0

y(t)g(t)dt +∫ T

0

z(t)h(t)dt(57)

=∫ T

0

∂ϕ(1, t)q(t)dt,

where y(t) = u(0, t), z(t) = ∂u(0, t) and <,>H10 ,H−1 represents the duality product

between H10 (−1, 1) and H−1.


Note that we have used the selfadjointness of (I − γ∂2)−1 in the identity∫ 1

−1

uf =∫ 1

−1

(I − γ∂2)u(I − γ∂2)−1f =< (I − γ∂2)u, (I − γ∂2)−1f >H10 ,H−1 .

We adopt (57) as definition of solution of (52) when U0 ≡ 0.Definition. We say that U ∈ C(0, T ; Y −) is a solution of (52) with U0 ≡ 0 in thesense of transposition if (57) holds for any ((I − γ∂2)−1f, g, h) ∈ L1(0, T ; V +

0 ).Remark 9. As we will see below, solutions in the sense of transposition are more

regular on the left hand side of x = 0. They satisfy

u|(−1,0) ∈ C([0, T ]; H3(−1, 0)

) ∩ C1([0, T ];H2(−1, 0)

)(58)

and the compatibility conditions

∂ut(0−, t) = zt(t), ∂2u(−1) = 0.(59)

Observe that the initial condition U0 ≡ 0 is implicit in (57).Theorem 4.6. When U0 ≡ 0 and q is as in (53), system (52) admits an unique

solution in the sense of transposition.Moreover, there exists C(T ) > 0 such that

‖ U ‖C([0,T ];Y −)≤ C(T ) ‖ q ‖L2(0,T ),(60)

for every q as above.

Proof of Theorem 4.6. In view of Theorem 4.2 the right hand side of (57) de-fines a linear continuous operator in L1(0, T ; V +

0 ). Therefore, by duality, we deducethat there exists an unique (u, y, z) ∈ L∞

(0, T ; (V +

0 )′)

solution of (57) where(V +

0

)′denotes the dual of V +

0 . Moreover, there exists C > 0 such that

‖ ((I − γ∂2)u, y, z) ‖L∞(0,T ;(V +0 )′)≤ C ‖ q ‖L2(0,T ) .(61)

Furthermore, as a consequence of Theorem 4.2 we deduce that (ut, yt, zt) ∈L∞

(0, T ;

(V +

1

)′)and the estimate

∥∥((I − γ∂2)ut, yt, zt

)∥∥L∞(0,T ;(V +

1 )′) ≤ C ‖ q ‖L2(0,T )(62)

holds for every q as in (53).Observe now that the duals of V +

0 ,(V +

0

)′, and V +

1 , coincide respectively withthe spaces

V −1 =

{(v, y, z) ∈ L2(−1, 1)× IR× IR : v|(−1,0) ∈ H1(−1, 0), v(−1) = 0,

[(I − γ∂2)−1v](0) = y, [∂(I − γ∂2)−1v](0) = z}

V −0 =

{(v, y, z) ∈ H−1(−1, 1)× IR× IR : v|(−1,0) ∈ L2(−1, 0),

[(1− γ∂2)−1v](0) = y, [∂(1− γ∂2)−1v](0−) = z}

.

On the other hand, V −1 × V −

0 is the image of the space Y − by the operator Ldefined as

L((u0, y0, z0), (u1, y1, z1)) = (((I − γ∂2)u0, y0, z0), ((I − γ∂2)u1, y1, z1)).(63)


We observe that L is in fact an isomorphism from Y − to V −1 × V −

0 .With the above considerations and (61)-(62) we deduce

‖ U ‖L∞(0,T ;Y −)≤ C(T ) ‖ q ‖L2(0,T ) .

Now by density it follows that (60) hold. To see this it is sufficient to observethat when q is smooth enough and of compact support, solutions of (52) belongto C

([0, T ];H1/2 ×H0

).

5. Controllability. In this section we prove the main controllability result forthe system

γ∂2utt − utt − ∂4u = 0, x ∈ (−1, 0), 0 < t < Tγ∂2utt − utt − ∂4u = 0, x ∈ (0, 1), 0 < t < T[u] (0, t) = [∂u] (0, t) = 0, 0 < t < Tutt(0, t) +

[∂3u

](0, t) = 0, 0 < t < T

γ∂utt(0, t)− [∂3u

](0, t) = 0, 0 < t < T

u(±1, t) = ∂2u(−1, t) = 0, ∂2u(1, t) = q(t), 0 < t < T(u(x, 0), u(0, 0), ∂u(0, 0)) =

(u0, y0, z0

),

(ut(x, 0), ut(0, 0), ∂ut(0, 0)) =(u1, y1, z1

).

(64)

The following holds:Theorem 5.1. Assume that T ≥ 4

√γ. Then for every

((u0, y0, z0

),(u1, y1, z1

))∈ Y − there exists a control q ∈ L2(0, T ) such that the solution of (64) in the sense oftransposition satisfies

((u(x, T ), u(0, T ), ∂u(0, T )), (ut(x, T ), ut(0, T ), ∂ut(0, T ))) ≡ 0.(65)

Moreover, there exists C > 0 such that

‖ q ‖L2(0,T )≤ C ‖ U0 ‖Y − , ∀U0 ∈ Y −.(66)

Remark 10. Theorem 5.1 states the exact controllability of (64) in the space Y −

with controls in L2(0, T ) provided T ≥ 4√

γ.The functional frame we have chosen for the control problem (U0 ∈ Y − and

q ∈ L2(0, T )) is not unique. A similar result holds for U0 in Y −−1 where

Y −−1 =

{U0 ∈ H−1 ×H−3/2 : u0|(−1,0) ∈ H1(−1, 0),

u1|(−1,0) ∈ L2(−1, 0), u0(0−) = y0, u0(−1) = 0}

with controls q ∈ H−2(0, T ). In this case exact controllability holds at time T > 4√

γbecause it is convenient to take controls q of compact support in order to avoid furthersingularities in the solutions at t = 0 and t = T .

Proof of Theorem 5.1. In view of the observability results of Lemma 3.1 it is a directapplication of HUM (see [8]).

Given T ≥ 4√

γ, for any Φ0 =((ϕ0, ψ0, ξ0), (ϕ1, ψ0, ξ1)

) ∈ Y−1 we solve the


adjoint system

γ∂2ϕtt − ϕtt − ∂4ϕ = 0, −1 < x < 0, 0 < t < Tγ∂2ϕtt − ϕtt − ∂4ϕ = 0, 0 < x < 1, 0 < t < T[ϕ] (0, t) = [∂ϕ] (0, t) = 0, 0 < t < Tϕtt(0, t) +

[∂3ϕ

](0, t) = 0, 0 < t < T

γ∂ϕtt(0, T )− [∂2ϕ

](0, t) = 0, 0 < t < T

ϕ(±1, t) = ∂2ϕ(±1, t) = 0, 0 < t < TΦ(0) ≡ ((ϕ(x, 0), ϕ(0, 0), ∂ϕ(0, 0)) , (ϕt(x, 0), ϕt(0, 0), ∂ϕt(0, 0)))

=((ϕ0, ψ0, ξ0), (ϕ1, ψ0, ξ1)

)= Φ0.

(67)

In view of Proposition 2.5, system (67) admits an unique solution Φ ∈ C([0, T ;Y−1).Moreover, thanks to Lemma 3.1, ∂ϕ(1, t) ∈ L2(0, T ).

We then solve

γ∂2utt − utt − ∂4u = 0, −1 < x < 0, 0 < t < Tγ∂2utt − utt − ∂4u = 0, 0 < x < 1, 0 < t < T[u] (0, t) = [∂u] (0, t) = 0, 0 < t < Tutt(0, t) +

[∂3u

](0, t) = 0, 0 < t < T

γ∂utt(0, T )− [∂2u

](0, t) = 0, 0 < t < T

u(±1, t) = 0, ∂2u(−1, t) = 0, ∂2u(1, t) = ∂ϕ(1, t)(u(x, T ), u(0, T ), ∂u(0, T )) ≡ (ut(x, T ), ut(0, T ), ∂ut(0, T )) ≡ 0.

(68)

In view of Theorem 4.6 and the time-reversibility of system (68) we deduce that ithas an unique solution defined by transposition.

We define the linear map

ΛΦ0 = ((u(x, 0), u(0, 0), ∂u(0, 0)) , (ut(x, 0), ut(0, 0), ∂ut(0, 0))) .

Multiplying in (68) by ϕ and integrating by parts (this a formal computation thatmay be done rigorous by the definition of solution in the sense of transposition) itfollows that

〈LΛΦ0, Φ0〉 =∫ T

0

|∂ϕ(1, t)|2 dt, ∀Φ0 ∈ Y−1(69)

where

LΦ0 =((

(I − γ∂2)ϕ1(x), ψ1, ξ1),− (

(I − γ∂2)ϕ0(x), ψ0, ξ0)))

.

In view of identity (69) it follows that LΛ is an isomorphism from Y−1 into itsdual Y ′

−1 and therefore Λ is an isomorphism from Y−1 into L−1Y ′−1.

Now we observe that, as it was pointed out in Remark 7, Y−1 = V +1 × V +

0 andthen Y ′

−1 = V −0 × V −

1 where V −0 and V −

1 are the spaces introduced in the proof ofTheorem 4.6.

It is easy to see that L−1(V −0 × V −

1 ) ≡ Y − algebraically and topologically.This implies that Λ : Y−1 → Y − is an isomorphism. Therefore, for any U0 ∈ Y −

there exists an unique Φ0 ∈ Y−1 such that ΛΦ0 = U0. This means that the solutionU of (68) with control q = ∂ϕ(1, t) where ϕ is the solution of (67) with initial dataΦ0 = Λ−1U0 is such that U(0) ≡ U0. Therefore q is the control we were looking for.


We also have by construction and in view of identity (69) and the observabilityinequalities of Lemma 3.1 that

∫ T

0

|∂ϕ(1, t)|2 dt =∣∣〈LΛΦ0, Φ0〉

∣∣ =∣∣〈LU0, Φ0〉

∣∣

≤ C ‖ Φ0 ‖Y−1‖ U0 ‖Y −≤ C

(∫ T

0

|∂ϕ(1, t)|2 dt

)1/2

‖ U0 ‖Y − .

Therefore∫ T

0

|∂ϕ(1, t)|2 dt ≤ C ‖ U0 ‖2Y −

and consequently (66) holds.

REFERENCES

[1] C. Castro, Asymptotic analysis and control of a hybrid system composed by two vibrat-ing strings connected by a point mass, Control, Optimization and Calculus of Variations(ESAIM COCV), http://www.emath.fr/cocv/, 2 (1997), pp. 231–280.

[2] C. Castro and E. Zuazua, Analyse spectrale et controle d’un systeme hybride compose dedeux poutres connectees par une masse ponctuelle, C. R. Acad. Sci. Paris, t. 322, Serie I,(1996), pp. 351–356.

[3] C. Castro and E. Zuazua, Une remarque sur les series de Fourier non-harmoniques et sonapplication a la controlabilite des cordes avec densite singuliere, C. R. Acad. Sci. Paris, t.323, Serie I, (1996), pp. 365–370.

[4] C. Castro and E. Zuazua, A hybrid system consisting of two flexible beams con-nected by a point mass: Spectral analysis and well-posedness in asymmetricspaces; in Elasticite, Viscoelasticite et Controle Optimal; ESAIM Proceedings,http://www.emath.fr/proc/Vol.2/, (1997), pp. 17–53.

[5] C. Castro and E. Zuazua, Exact boundary controllability of two Euler-Bernoulli beams con-nected by a point mass, Mathematical & computer modelling, to appear.

[6] S. Hansen and E. Zuazua, Exact controllability and stabilization of a vibrating string with aninterior point mass, SIAM J. Cont. Optim., 33 (5) (1995), pp. 1357–1391.

[7] J. E. Lagnese and J.-L. Lions, Modelling, Analysis and Control of Thin Plates, Masson,RMA, 1988.

[8] J.-L. Lions, Controlabilite exacte, stabilisation et perturbations de systemes distribues. Tome1. Controlabilite exacte, Masson, RMA8, 1988.

[9] S. W. Taylor, Exact boundary controllability of a Beam and Mass system, in Computationaland Control IV, Birkhauser Boston, Ed. K.L. Bowers and J. Land, Series: Prog. syst.control theory, 20, 1995.

[10] D. Ulrich, Divided Differences and Systems of Nonharmonic Fourier Series, Proc. of theAmer. Math. Soc., 80 (1) (1980), pp. 47–57.

[11] R. M. Young, An introduction to Nonharmonic Fourier series, Academic Press, 1980.

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