1

Scanning control for the string equationGerald Tenenbaum, Marius Tucsnak

I. INTRODUCTION

It is well known that for pointwise control problemswe generally have a lack of robustness with respect tothe location of the actuator. More precisely, any opensubset of the considered domain ([0, π] in our case)contains points for which controllability fails, see, forinstance, [1] and references therein. A remedy whichhas been proposed in Berggren [2] is to consider anactuator which moves according to a prescribed law.To describe the problem introduced in [2], let α, βand ω be positive numbers and define

%(t) := α+ β sin(ωt) (t ∈ R). (1.1)

We consider the initial and boundary value problem

∂2w

∂t2=∂2w

∂x2+ u(t)δ%(t) (0 < x < π, t > 0), (1.2)

w(0, t) = w(π, t) = 0 (t > 0), (1.3)

w(x, 0) = 0, w(x, 0) = 0 (0 < x < π), (1.4)

where, for every real a, δa stands for the Dirac measureat a. The above system describes the linear vibrationsof an elastic string with a pointwise scanning actuator.This means that, for very t > 0 the actuator ispositioned at %(t) at instant t. It is easy to see thatif α

π ∈ Q and β = 0 (i.e., for some fixed actuators)the above system is not approximately controllable.

Based on numerical experiments, it has been conjec-tured in [2] that the system (1.2)-(1.4) is controllable(in some sense) for every α ∈ (0, π) and β, ω 6= 0.

From a theoretical point of view, related questionshave been studied in Khapalov [3], [4]. More precisely,[3] gives an algorithm defining in an implicit a man-ner a class of functions % ensuring controllability of(1.2)-(1.4) and of its generalization to several spacedimensions. In the one dimensional case, it has beenshown in [4] that for a simpler function % (but still dif-ferent of ours), approximate controllability and weakstabilizability of (1.2)-(1.4) holds. We mention that

the stabilizability of systems related (1.2)-(1.4) waspreviously studied in [5] and [6].

A related field which is not tackled in this work isthe controllability of heat equations. In this case wecannot expect exact controllability so that the naturalquestions are the approximate and null-controllability.The approximate controllability for the heat equationswith pointwise moving control has been studied in [7].As far as we know, establishing the corresponding null-controllability properties is still an open question.

Throughout this work we explicitly assume thatβω < 1, or we make assumptions implying thisinequality. This is required in our approach inasmuchwe use the fact that the map t 7→ t+ %(t) is a smoothdiffeomorphism from R onto R.

Our first main result shows that the system (1.2)-(1.4) is well posed in the natural energy space.

Theorem 1.1: Let ω, α, β denote positive numberssuch that βω < 1 and assume that α, α ± β ∈]0, π[.Then, for every w0 ∈ H1

0 ([0, π]), w1 ∈ L2([0, π]) andu ∈ L2([0, τ ]), the initial and boundary value problem(1.2)–(1.4) admits a unique solution

w ∈ C([0, τ ];H10 ([0, π]) ∩ C1([0, τ ];L2([0, π])).

The second main result asserts that the consideredsystem is controllable in a quite weak sense.

Theorem 1.2: There exists an absolute constant Asuch that, for all ε ∈]0, 1

2 [, β > 0, ω > 0, T > 0satisfying

βω <ε

A, T >

1β4ω4

,

the system (1.2)–(1.4) is approximately controllablein infinite time (in the sense that for every w0 ∈H1

0 (0, π), w1 ∈ L2(0, π) there exists u ∈ L2(0,∞)such that limτ→∞ w(·, τ) = w0, limτ→∞ w(·, τ) =w1).

The remaining part of this work is organized asfollows. In Section II we prove some trigonometricalinequalities which play a central role in this work.Section III is devoted to the proof of the main results.

17th Mediterranean Conference on Control & AutomationMakedonia Palace, Thessaloniki, GreeceJune 24 - 26, 2009

978-1-4244-4685-8/09/$25.00 ©2009 IEEE 504

II. SOME TRIGONOMETRICAL INEQUALITIES

Let α, β, ω and T be positive numbers. For t ∈ Rand ak∞k∈Z ∈ `2(Z,C), we define the function

f(t) :=∑k∈Z

akeikt sin (k%(t)) (t ∈ [−T, T ]). (2.1)

We will show, in particular, that if f vanishes inL2([−T, T ]) then all the terms of the sequence (ak)vanish.

We first introduce some notation. First recall thatthe Fourier transform of ϕ ∈ L1(R) is given by theformula

ϕ(ξ) :=∫

Rϕ(x)eixξ dx.

The above definition is extended in the usual way forϕ ∈ L2(R). We also set

g(t) :=∑k∈Z∗

akeikt (t ∈ R),

so that, for every t ∈ R we have

f(t) :=∑k∈Z

akeikt sin(k%(t)) =g(t2)− g(t1)

2i,

where we have put tj = tj(t) := t + (−1)j%(t) forj ∈ 1, 2. For the sake of notational simplicity, weomit, here and in the remaining part of this work, thevariable t in the symbols t1 and t2.

For t ∈ R and t1, t2 defined as above, we put

Jk(T ) :=∫ T

−T|g(tk)|2 dt (k = 1, 2, T > 0).

We need the following technical result:Proposition 2.1: For every bounded interval I ⊂ R

there exist two continuous functions µ±I ∈ L1(R) suchthat

µ−I (x) 6 1lI(x) 6 µ+I (x) (x ∈ R), (2.2)

µ±I (ξ) =

|I| ± 2π if ξ = 0,0 if |ξ| > 1

(ξ ∈ R), (2.3)

where 1lI is the characteristic function of I .Proof. Following Beurling [8], we defined in [9] a

pair of functions B(x), b(x) such that

b(x) 6 sgn (x) 6 B(x) (x ∈ R)∫RB(x)− sgn (x) dx =

∫Rsgn (x)− b(x) dx = 1.

For normalisation purposes, we set here B∗(x) :=B(x/2π), b∗(x) := b(x/2π). If I := [s, t] is a realinterval, we put µ−I (x) := 1

2b∗(t − x) + b∗(x − s),

µ+I (x) := 1

2B∗(t − x) + B∗(x − s). This immedi-

ately yields (2.2). Moreover, in view of the vanishingproperties of the involved Fourier transforms recalledin [9], we see that (2.3) also holds.

We next show that, for large T , the integrals J1(T )and J2(T ) behave like 2T .

Proposition 2.2: With the above notation, assumethat β, T > 0 and that

0 6 ω < min(

1β,π

T

).

Then

|J1(T )− 2T | 6 8π1− %′(T )

=8π

1− βω cos(ωT ), (2.4)

|J2(T )− 2T | 6 8π1 + %′(T )

=8π

1 + βω cos(ωT )· (2.5)

Proof. Define

ϕ1(t) :=∫ t

0

|g(s−%(s))|21−%′(s) ds (t ∈ R),

(2.6)ψ1(t) := ϕ1(t)− t1 − α (t ∈ R), (2.7)

so that

|g(t1)|2 = 1 +ψ′1(t)

1− %′(t)(t ∈ R). (2.8)

From (2.6) and (2.2), it follows that, for all real,positive t, we have

ϕ1(t) =∫ t1

−α|g(w)|2 dw >

∫R|g(w)|2µ−[−α,t1](w) dw

=∑j,k∈Z

aj ak µ−[−α,t1](j − k),

and therefore, in view of (2.3),

ϕ1(t) > t1 + α− 2π (t > 0).

Similar calculations using µ+[−α,t1] instead of µ−[−α,t1]

yieldϕ1(t) 6 t1 + α+ 2π (t > 0).

505

Applying an analogous treatment for negative t, weobtain that the last two estimates are actually valid forall real t. From (2.8), we may therefore deduce that

|ψ1(t)| 6 2π (t ∈ R). (2.9)

Now, it follows from (2.8) that

J1(T ) = 2T +∫ T

−T

ψ′1(t)1− %′(t)

dt

= 2T +[

ψ1(t)1− %′(t)

]T−T

+∫ T

−T

ψ1(t)%′′(t)1− %′(t)2

dt.

Using (2.9) we obtain that

J1(T ) 6 2T +4π

1− %′(T )+ 2π

∫ T

−T

%′′(t)1− %′(t)2

dt

(2.10)Since βT 6 π, we have %′′(t) = −βω2 sin(ωt) 6 0for t ∈ [−π, π] and so, using (2.9),∫ T

−T

|ψ1(t)%′′(t)|1− %′(t)2

dt 6 −2π∫ T

−T

%′′(t)1− %′(t)2

dt

=4π

1− %′(T )·

This and (2.10) imply (2.4). Estimate (2.5) is derivedsimilarly.

The result below gives conditions guaranteeing thatthe right-hand side of (2.1) does indeed define afunction in L2([−T, T ]).

Proposition 2.3: Let T, ω, α, β denotepositive numbers such that βω < 1 and letak∞k∈Z ∈ `2(Z,C). Then the series (2.1) convergesin L2([−T, T ]) to a function f satisfying

12T

∫ T

−T|f(t)|2 dt

6

(1 +

4πT1− (βω cosωT )2

) ∑k∈Z∗

|ak|2.

Proof. With no loss of generality we may assumethat

∑k∈Z |ak|2 = 1. We clearly have∫ T

−T|f(t)|2 dt = 1

4 J1(T ) + J2(T )−Q(T ) ,(2.11)

where

Jk(T ) :=∫ T

−T|g(tk)|2 dt (k = 1, 2),(2.12)

Q(T ) := 2<e∫ T

−Tg(t1)g(t2) dt. (2.13)

The above formulas clearly yield∫ T

−T|f(t)|2 dt 6 1

2

J1(T ) + J2(T )

.

The stated conclusion now follows from (2.4) and(2.5).

The main result of this section is the following.Proposition 2.4: Let M > 0, ε > 0 and letak∞k∈Z ∈ `2(Z,C) satisfy∑

k∈Zk2|ak|2 6 M.

Then there exists A = AM > 0 such that if βω < ε/Aand if T > 1/β4ω4 we have(

12 − ε

) ∑k∈Z∗

|ak|2

61

2T

∫ T

−T

∣∣∣∣∑k∈Z

akeikt sin (k%(t))∣∣∣∣2 dt. (2.14)

Proof. As in the proof of Theorem 2.3, we assumethat

∑k∈Z |ak|2 = 1 and use formula (2.11). We

appeal to (2.4) and (2.5) to bound J1(T ) and J2(T )from above or from below and so only need to considerQ(T ). Expanding g(t1) and inverting summations, weget

Q(T ) =∑k∈Z

ak

∫ T

−Teikt1 g(t2) dt. (2.15)

For j = 1, 2 and s ∈ R we write ϑj for the inversefunction of t 7→ tj . In other words, ϑj(s) is defined bythe formula ϑj + (−1)j%(ϑj) = s. Setting Tj := T +(−1)jβ sin(ωT ), we hence see that ϑ1(s) ∈ [−T, T ]whenever s ∈ [−T1 − α, T1 − α]. It follows that∫ T

−Teikt1g(t2) dt =

∫ T1+α

−T1−αeiksH(s) ds, (2.16)

where

H(s) :=g(t2(ϑ1(s)))1− %′(ϑ1(s))

(−T1 6 s+ α 6 T1).

(2.17)Now, let us assume that T1 = 2qπ for some q ∈ N.Then H , considered as initially defined on ] − T1 −α, T1−α], may be extended to a 4πq-periodic functionon R and consequently written as a Fourier series

H(s) =∑k∈Z

hkeiks/(2q) (s ∈ R).

506

This series converges in L2(I) for every boundedinterval I . Using (2.16) and Parseval’s formula weobtain that∑k∈Z

∣∣∣∣∣∫ T

−Teikt1g(t2) dt

∣∣∣∣∣2

= 4T 21

∑k∈Z|h2kq|2. (2.18)

It is well-known that the h2kq are the Fourier coeffi-cients of the Riemann sum

Hq(s) :=12q

∑−q6j<q

H(s+ 2πj). (2.19)

Indeed

Hq(s) =12q

∑k∈Z

hk eiks/(2q)∑

06j<2q

eiπjk/q

=∑k∈Z

h2kqeiks. (2.20)

This and (2.18) imply that

∑k∈Z

∣∣∣∣∣∫ T

−Teikt1g(t2) dt

∣∣∣∣∣2

= 2T1

∫ T1−α

−T1−α|Hq(s)|2 ds,

which, combined with (2.15), yields

|Q(T )|2 6 2T1

∫ T1−α

−T1−α|Hq(s)|2 ds.

Using the change of variables s = t1 = t1(t), weobtain that

|Q(T )|2 6 2T1

∫ T

−T|Hq(t1)|21− %′(t)dt. (2.21)

For t ∈ [−T, T ], j ∈ Z and tk = tk(t) (k = 1, 2)defined as before, we put

ξt(j) := t2 (ϑ1 (t1 + 2πj + 4πεmjq))− t2,

where m is the unique integer in [−q, q[ such that2mπ < t1 + α < 2(m+ 1)π and

εmj :=

−1 if q −m 6 j < q,

0 if − q −m 6 j < q −m,1 if − q 6 j < −q −m.

With the above notation, we have

Hq(t1) = Gq(t)

:=12q

∑−q6j<q

g(t2 + ξt(j))1− %′(t2 + ξt(j))

(−T < t 6 T )

so that, from (2.21),

|Q(T )|2 6 2T1

∫ T

−T|Gq(t)|21− %′(t) dt.

The 1-periodic function g has zero mean-value and%′ is small when ωT is small. Our strategy is to showthat, for each t, the sequence ξt(j)q−1

j=−q is well-distributed in the torus, so as to infer that Gq(t) issmall in size. To this end, we introduce the Weyl sums

σh(t) :=12q

∑−q6j<q

eihξt(j) (h ∈ Z).

We set v := j + 2εmjq, so that |t1 + 2πv| 6 T forevery t ∈ [−T, T ]. It is easily seen that

4πβω cos(ωT )1− βω cos(ωT )

6 ξ′t(j)− 2π

=4π%′(ϑ1(t1 + 2πv))

1− %′(ϑ1(t1 + 2πv))6

4πβω1− βω

·

Using the Kusmin–Landau inequality (see, for in-stance, [10], Theorem I.6.7) it follows that

|σh(t)| 6 3πqβωh

=6

T1βωh, (2.22)

provided that

ωT 6π

3, βω 6

15, 1 6 4h+ 1 6

1βω·

Write

gK(s) :=∑k6K

ak cos(ks), rK(s) := g(s)− gK(s).

Then

Gq(t) := PK(t) +RK(t) + V (t), (2.23)

where

PK(t) :=12q

∑−q6j<q

gK(t2 + ξt(j)), (2.24)

RK(t) :=12q

∑−q6j<q

rK(t2 + ξt(j)),(2.25)

(2.26)

V (t)

:=12q

∑−q6j<q

g(t2 + ξt(j))%′(t2 + ξt(j))1− %′(t2 + ξt(j))

·

507

Let K > 0 be an integer such that 4K + 1 6 1/βω.Inverting summations yields

PK(t) =∑

16|k|6K

akσk(t)eikt2 ,

and so, using (2.22),

|PK(t)|2 662

T 21 β

2ω2

∑16|k|6K

|ak|2∑

16|k|6K

1k2

612π2

T 21 β

2ω2·

It follows that∫ T

−T|PK(t)|2 dt 6

24π2T

T 21 β

2ω2· (2.27)

On the other hand, we derive from (2.25) and (2.5)that∫ T

−T|RK(t)|2 dt

6 (2T + 8π)∑|k|>K

|ak|2 62M(T + 4π)

K2· (2.28)

Finally, we plainly infer from (2.26) that, for a suitableabsolute constant C, we have∫ T

−T|V (t)|2dt 6 Cβ2ω2T. (2.29)

Choosing K equal to the integer part of 1/(4βω), T >K4 and βω/ε small enough, it follows from (2.23)combined with (2.27)–(2.29) that Q(T ) 6 εT .

III. PROOF OF THE MAIN RESULTS

Consider the following homogeneous initial andboundary value problem associated to (1.2)-(1.4)

∂2η

∂t2(x, t) =

∂2η

∂x2(x, t) (0 < x < π, t > 0), (3.1)

η(0, t) = η(π, t) = 0 (t > 0), (3.2)

η(x, 0) = η0(x), η(x, 0) = η1(x) (0 < x < a).(3.3)

The above system models the linear vibrations of anelastic string occupying the interval [0, π] and fixedat both ends. The functions η and η stand for thetransverse displacement and the transverse velocity ofthe string, respectively.

It is well-known that for every η0 ∈ H2(0, π) ∩H1

0 (0, π) and η0 ∈ H10 (Ω), the initial and boundary

value problem (3.1)–(3.3) admits a unique solution

η ∈ C([0,∞);H2(0, π) ∩ C1([0,∞);H1

0 (0, π))).

We consider the observation problem which is basedmeasuring, at each t > 0 of the transverse velocity ηat x = %(t), where % : [0,∞[→ [0, π] is the functiondefined in (1.1), i.e. we associate to (3.1)-(3.3) theoutput law

y(t) := η(%(t), t). (3.4)

The main ingredient of the proofs of our main resultsis the following proposition.

Proposition 3.1: Let τ, ω, α, β denote positive num-bers with such that βω < 1 and assume that α, α±β ∈]0, π[. Then there exists a positive constant K =K(τ, ω, α, β) such that for any η0 ∈ H2(0, π) ∩H1

0 (0, π), η1 ∈ H10 (0, π) the solution η of (3.1)–(3.4)

satisfies∫ τ

0

|y(t)|2dt 6 K(‖η0‖2H1

0 (0,π) + ‖η1‖2L2[0,π]

)(3.5)

Moreover, there exists a positive constant ω0 suchthat for any ω ∈]0, ω0[, the system (3.1)–(3.4) isapproximately observable in infinite time, i.e. if y(t) =0 for every t > 0 then η0 = η1 = 0.

Proof Let X := H10 (0, π) × L2[0, π]. Is is well

known that X is a Hilbert space if endowed with thescalar product⟨[

f1g1

],

[f2g2

]⟩=∫ π

0

df1dx

(x)df2dx

(x)dx

+∫ π

0

g1(x)g2(x)dx.

For k ∈ Z∗ and x ∈ [0, π], we write ϕk(x) :=√2/π sin(kx). Since (ϕk)k∈N∗ is an orthonormal ba-

sis in L2[0, π], simple calculations yield that the family(ϕk)k∈Z∗ defined by

ϕk =1√2

[ϕk/(ik)ϕk

](k ∈ Z∗), (3.6)

is an orthonormal basis in X . Therefore, it sufficesto prove (3.5) under the assumption that there existsN0 > 0 for such that:⟨

dη0dx

,dϕkdx

⟩L2[0,π]

= 〈η1, ϕk〉L2[0,π] = 0 (k > N0).

508

In this case the solution η of (3.1)–(3.3) is given by[η(·, t)η(·, t)

]=

1√2

∑k∈Z∗

eikt(i

k

⟨dη0dx

,dϕkdx

⟩+ 〈η1, ϕk〉

)ϕk.

It follows that

η(ρ(t), t)

=∑k∈Z∗

eikt√π

(i

k

⟨dη0dx

,dϕkdx

⟩+ 〈η1, ϕk〉

)sinkρ(t).

The above formula, (3.4) and Theorem 2.3 imply (3.5).Finally, the fact that, for small enough ω0 > 0, the

unique continuation property stated holds follows fromProposition 2.4.

We are now in a position to prove our main results.Proof of Theorem 1.1 The result follows from the

first part of Proposition 3.1 by applying the transposi-tion method of Lions and Magenes.

Proof of Theorem 1.2 The result follows from thesecond part of Proposition 3.1 by applying the HUMmethod of Lions.

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[3] A. Y. Khapalov, “Controllability of the wave equation withmoving point control,” Appl. Math. Optim., vol. 31, no. 2, pp.155–175, 1995.

[4] ——, “Observability and stabilization of the vibrating stringequipped with bouncing point sensors and actuators,” Math.Methods Appl. Sci., vol. 24, no. 14, pp. 1055–1072, 2001.

[5] A. Bamberger, J. Jaffre, and J. P. Yvon, “Punctual control ofa vibrating string: numerical analysis,” Comput. Math. Appl.,vol. 4, no. 2, pp. 113–138, 1978.

[6] J. R. McLaughlin and M. Slemrod, “Scanning control of avibrating string,” Appl. Math. Optim., vol. 14, no. 1, pp. 27–47, 1986.

[7] C. Castro and E. Zuazua, “Unique continuation and controlfor the heat equation from an oscillating lower dimensionalmanifold,” SIAM J. Control Optim., vol. 43, no. 4, pp. 1400–1434 (electronic), 2004/05.

[8] A. Beurling, “Mittag-Leffler lectures on harmonic analysis,” inThe collected works of Arne Beurling. Vol. 2, ser. Contempo-rary Mathematicians. Boston, MA: Birkhauser Boston Inc.,1989, pp. xx+389, harmonic analysis, Edited by L. Carleson,P. Malliavin, J. Neuberger and J. Wermer.

[9] G. Tenenbaum and M. Tucsnak, “Fast and strongly localizedobservation for the Schrodinger equation,” Transactions of theAmerican Mathematical Society, vol. to appear, 2008. [Online].Available: http://hal.archives-ouvertes.fr/hal-00140540/en/

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