Observability of square Membranes by Fourier series methods Текст научной статьи по специальности «Математика»

MSC 42C99, 93B07, 35L05

DOI: 10.14529/mmp 150308

OBSERVABILITY OF SQUARE MEMBRANES BY FOURIER SERIES METHODS

V. Komornik, University of Strasbourg, Strasbourg, France, [email protected],

P. Loreti, Sapienza University of Rome, Rome, Italy, [email protected]

Fourier series methods have been successfully applied in control theory for a long time. Some theorems, however, resisted this approach. Some years ago, Mehrenberger succeeded in establishing the boundary observability of vibrating rectangular membranes (and of analogous higher dimensional problems) by developing an ingenious generalization of Ingham's classical theorem on nonharmonic Fourier series. His method turn out to be useful for other applications as well. We improve Mehrenberger's approach by a shorter proof, and we improve and generalize some earlier applications.

Keywords: observability; nonharmonic Fourier series; Ingham's theorem; wave equation.

Dedicated to the memory of Professor Alfredo Lorenzi

Introduction

Let us consider the wave equation in a bounded domain Q C Rra with boundary r

'u" - Au = 0 in R x Q,

u = 0 on R x r,

u(0) = u0 in Q,

V(0) = u1 in Q.

(1)

We recall that if u0 E ^¿(Q^d ui E L2(Q), then the problem has a unique "finite-

energy" solution satisfying

u E C(R; H1(Q)) П C 1(R; L2(Q)),

and the "energy"

E(t) := 1 \Vu(t,x)\2 + \u'(t,x)\2 dx, t E R

of the solution is in fact independent of t.

It was discovered by Lasiecka and Triggiani [1] that the normal derivative of finite-energy solutions is well defined as an element of L20C (R; L2(r)), and for each bounded interval I there exists a constant c such that all finite-energy solutions satisfy the estimate

I г

du

dv

dr dt< cE.

2

Lions [2] gave a simpler proof by using the multiplier method.

Subsequently Ho [3] discovered that the inverse inequality also holds if the interval I is long enough. This was improved and generalized by Lions [4,5], who established estimates of the type

du

E < c

dv

dr dt (2)

I S

where S is some part of the boundary T.

Both Ho and Lions applied the multiplier method. By using deeper tools of microlocal analysis, Bardos, Lebeau and Eauch [6] obtained rather precise necessary conditions and

S

I

Considering the special case where Q C R2 is a rectangle and S is the union of two

I

twice the diagonal D of the rectangle. (In the original proof the optimal condition on I was achieved by a somewhat deeper compactness-uniqueness argument; this was replaced by a short elementary proof in [7].)

Many attempts have been made to recover the last result by Fourier series techniques. Eventually, Mehrenberger [8] succeeded in devising such a proof, although under a stronger condition on I: he needed \I\ > 4v^2|D| instead of \I\ > 2|D|. His main tool was the following clever generalization of a classical theorem of Ingham on nonharmonic Fourier series:

Theorem 1. Let (wk)k=-oo a sequence of real numbers, satisfying for some nonnegative integer n and for some positive real number y the following partial gap condition:

— wk\ > ^ — k^f whenever max{|k'|, |k|} > n.

Then the following inequality holds for all square summable sequences (xk)k=-ix °f complex numbers:

R

—R

y^ Xke

k=—oo

2

i„k t ^ 4R I ^ ■ - ^

dt > n IE X |2 — ^2-2 E X f

' \k\>n k=-<x

We remark that the estimate remains valid by translation invariance if we change the integration interval (—R, R) to any other interval of length 2R.

This inequality is obvious if R < n because then the right side is nonpositive. For R > however, the result is delicate: for n = 0 this reduces to the deeper part of Ingham's theorem [9].

Theorem 1 proved to be useful for many other control problems as well [10,11]. The plan of this paper is the following:

• For the reader's convenience, in Section 1 we briefly reproduce Mehrenberger's proof of Theorem 1, based on Ingham's first method.1

R

lWe recall that Ingham proved his theorem in two different ways.

• Using the result of Section 2, in Section 3 we weaken the assumption \I\ > 4y/2\D\ for the boundary observability of square membranes to \I\ > 4, 0441 \D\.

•

We illustrate this in Section 4 by generalizing the last result of [11], by observing a vibrating rectangular membrane on a finite number of horizontal and vertical lines.

1. Proof of Theorem 1

We may assume by scaling that y > 2 and R = n/2.

Following Ingham, we introduce the function h : R ^ R and its Fourier transform H : R ^ R by the formulas

hit) :

{

cost if \t\< n/2, 0 if \t\> n/2

and

H(x) := / h(t)e dt, x E R.

If x = ±1, then we have

те n/2 n/2

-txt _ / cos t cos xt dt = 2 cos t cos xt dt

H(x) = h(t)eixt dt

n/2

-n/2

0

= cos(x + 1)t + cos(x — 1)t dt

sin(x + 1)t + sin(x — 1)t

x + 1

x1

n/2

sin(x + 1)2 + sin(x — 1)2 cos nxX cos nxX —2 cos xn

x + 1

x1

x + 1 x — 1 x2 — 1

Proof of Theorem 1. Since 0 < h < 1, we have

n/2

-n/2

E

k=—oo

ak e

iwk t

dt > h(t)

E

k=

ake

iuk t

dt = H (uk' — Uk )a'k ak+

max{\k'\,\k\}>n

+ hit)

E

\k\<n

ak e

iUkt

dt > H(uk' — Uk)akak = H(0) ^ \ak\2+

max{\k'\,\k\}>n \k\>n

+ ^ H(Uk' — Uk)aa > H(0) ^ \ak\2 — ^ \H{uw — Uk

max{\k' \,\k\}>n k'=k

\k\>n

max{\k'\,\k\}>n k'=k

Since the function H is even, the last expression does not change if we change the last fraction to \ak|2. Distinguishing in the last sum the cases \k\ > n and \k\ < n, it follows

0

2

2

2

that

n/2 _ 2

-n/2

oo

£

k=-X

ak e™kt

dt

^ ^\2 ( H(0) — ^ \H(wy — wk)| J -

\k\>n \ k'=k )

>

\k\>n \ k'=k

— y^ a|2 h(wk' — wk)|.

\k\<n \k'\>n

The proof will be completed if we show the following three relations:

H (0) = 2;

g

J2H(wk' — wk)\< -2 if \k\> n;

k'=k Y

g

H(wk' — wk)\ < if \k\ <n.

\k'\>n Y

The first equality follows at once from the explicit expression of H(x). The second and third relations also follow from this expression and from the gap condition y > 2. Indeed, we have

2

H (x)\ < -— x2 — 1

if |x| > 1. Since y > 2, in case \k\ > n we have

£ W w — wk" < £ iw^kF^r < Ik W^Wf^-i < Ik W—WYF^

g ^ 1 = g ( 2 = g ( ( 1 1 ) g 41 y - k\2 _ 1 = Y2 ^ 4m2 _ 1 = y2

k =k

The proof of the third relation is identical.

□

2. A Variant of Mehrenberger's Ingham Type Theorem

Given two positive numbers y and R> we introduce four continuous even functions H,G,h,g : R ^ R by the following formulas:

Y2 , 4\k' — k\2 — 1 y2 4m2 — 1 y2 2m — 1 2m +1) y2' k' k —1 —1

H(x):^°OS ™ ifx < Y/2 \0 ifx > y/2; G := R2(H * H) + H' * H';

h(t) ■■=!H ^dx;

-(

(

g(t) := [ G(x)eitx dx.

We are going to prove the following variant of Theorem 1:

Theorem 2. Let (uk)k=-oo a sequence of real numbers, satisfying for some nonnegative integer n and for some positive real number 7 the partial gap condition

\uk' — uk\>\k' — k\j whenever max{|k'|, |k|} > n. (3)

Then the following inequality holds for all square summable sequences (xk)k=-<x> of complex numbers:

R

У^ Xk e

k=-œ

2

-R

We notice that

dt > -л I G(0) V \xk|2 - max\xk|2 I . (4)

max о I '

v k=-œ \k\<n

a := max Id = max q < 00 and d := max IGI = max IGI <00.

[-R,R] [-7,7]'

The first relation follows from the formula q(t) = (R2 — t2)h(t)2 implying that q > 0 in [—R, R] and q < 0 outside [-R, R]. The second one holds because G vanishes outside

Since ^d G do not vanish identically, we have also a,ft> 0. Proof. We write

<x

x(t) := £ ak'

for brevity.

Using the Fourier inversion formula

(X

—itx

g(t)e x dt = 2nG(x)

we have

R 00

aj \x(t)\2 dt > j g(t)\x(t)\2 dt = 2n J] G(uk — ш«)xkxk. (5)

-R -ж k,k'=-<x

Since G = 0 outside [—7,7], applying (3) the last sum is equal to

те

2

G(0) £ \xk\2 + G(uk ^k' %)xk xk' ■

k=-<x \k\<n

\k'\<n k=k'

Remarking that

£ G(шk — ш„)xkxk7 > — в £ \xk\2 +2 |xk'\2 = —в E \xk\2,

\ k\ <n \ k\ <n \ k\ <n

\k'\<n \k'\<n

k=k'

we deduce from (5) the inequality

f \x(t)\2 dt > 2n ( G(0) j \xk\2 - P j \xk\

\ k=—^ \k\<n

i.e., (4).

□

Remark 1. For the purposes of the next section we determine explicitly the function G.

For y = n an easy computation leads to the expressions2

, 4(R2 - t2) 2 nt g(t)= 9(t) :=(1+ t)2(1 - t)2 C°S "2

and

(R2 + 1) sin x + (R2 — 1)(n — x) c°s x, if x < n

(

2G(x) = G(x) :-- .

0, if x > n.

The general case hence follows by a change of variable.3 Setting c := n/y we have H (x) = H(cx),

h(t) = f H(x)eitx dx = f H(cx)el(t/c)(cx) dx = c-1 f H(cx)ei(t/c)(cx) d(cx) = c-1 h(t/c);

G(x) = R2(H * H)(x) + (H' * H')(x) = R2 J H(y)H(x — y) dy+

+ J H'(y)H'(x — y) dy = R2 J H(cy)H(cx — cy) dy + c2 J H' (cy)H' (cx — cy) dy 12

c R J H(z)H(cx — z) dz + cj H'(z)H'(cx — z) dz c(R(H * H) + (H' * H')^J (cx);

g(t) = J G(x)eitx dx = c-1R2 J (H * H)(cx)eUx dx + c J (H' * H')(cx)eitx dx = = c-1R2 J(iH * Jff)(cx)ei(t/c)(cx) dx + c j(H' * H')(cx)ei(t/c)(cx) dx = = c-2R2 J(^ * H)(z)ei(t/c)z dz + J(H' * H')(z)ei(t/c)z dz =

= c-2R2 f ( H * H)(z)ei(t/c)z dz — c-2t2 f ( H * H)(z)ei(t/c)z dz = c-2(R2 — t2)h2(t/c).

2See [12, pp. 62 and 64.J.

3 All integrals will be taken over R.

3. Boundary Observability of Square Membranes I

We investigate the validity of the inverse inequality (2) for the solutions of (1) when Q is a square membrane with diagonal D. (The computations of this section may be easily adapted to general rectangular domains but the results are less elegant.) We are going to prove the following

Proposition 1. If \I\ > 4,0441 \D\, then the finite energy solutions of (1) satisfy the estimate (2).

Since our assumption is between \I\ > 4v^2\D\ and the optimal condition \I\ > 2\D\, our result is weaker than the result obtained by the multiplier method, but stronger than the first theorem found in [8] by Fourier series techniques.

Remark 2. As in [8], the proof of the proposition, given below, may be adapted to higher dimensions when Q is an ^-dimensional interval, and S is the union of the sides of Q having a common vertex. Furthermore, we may consider various other (mixed) boundary conditions such that the corresponding eigenfunctions are still products of sine and cosine functions.

For the proof we assume by scaling that Q = (0,^) x (0,^) and

S := {(x1,x2) E [0,^] x [0,^] : x1 = 0 or x2 = 0}

is the union of two adjacent sides.

We recall that the solutions of (1) are given by the series

те те

u(t,x1,x2) = ^^2 aеЦЩ + bk e-m) sin(kixi) sintta), (6)

u

ki = 1 k2 = 1

with the usual notation \k\ := \Jkf + , and that the complex coefficients satisfy the equality (see, e.g., [11, (3.7)])

k1 = 1 k2 = 1

Turning to the study of the boundary integral first we observe that

[ ^^ (t,x1, 0) dx1 dt

J dx2

2

We need the following lemma:

Lemma 1. Fix an integer N > 2 an d N — 1 intege rs k1,... ,kN -1 > 1. If kN ,k'N are two positive integers satisfying

max{kN, kN} > max{k1;... , kN-1},

then

y/k2 + ••• + kN-1 + kN — ^ k2 + ••• + kN-1 + (kN)

> (vN — Vn — i)kn — kn\

This lemma improves a result in [8] by increasing the constant 1/2VN to VN — VN — 1. For example, we get y/2 — 1 & 0, 41 instead of 1/2y/2 & 0, 35 if N =F2:oof.

Assuming by symmetry that kN > k'N and setting s := k2 + • • • + kN-1 for brevity, we have s < (N — 1)k%, and

Ws + kN — Vs + (kN)2\ _ y/s + kN — y/s + (kN)

\kN — kN \

kN — kN

kN + kN

>

kN + kN

v^s+kN + V s + (kN )2 yNkN + y/(N — 1)kN + (kN )2'

Setting x :— kN/kN £ (0,1) for brevity, the last expression is equal to

f (x)

1 + x

vn + V n — 1 + x2

The lemma follows because f (x) > f (0) = VN — VN — 1 for a 11 x £ (0,1). f

f '(x)

( ^ N — 1 — x ,

VN + , | > 0

(vN + VN — 1 + x2)2 V VN — 1 + x2

for all x £ (0,1).

□

N=2

applying Theorem 2 with n = k1 and 7 = V~2 — 1 for each fixed k1. This yields the following inequality:

R n

a

n2

-R 0

du

dx2

(t,x1, 0)

dx1 dt >

> G(0) £ £ kl (\ak\2 + \bk\2) — (ak\2 + \bk\2)

>

ki = 1 k2 = 1

00 00

ki = 1 k2<ki 00

> G(0) £ £ k2 (\ak\2 + \bk n — 2 £ E \k\2 (\ak\2 + \bk\2) .

ki = 1 k2 = 1 ki = 1 k2<ki

2

2

1

2

By symmetry we also have

R n

a ~~2

П

-R 0

du

dx1

(t, 0,X2)

dx2 dt >

> g(O^ Y.kii2+\bkn - Il E iki2i2+\bkn.

fcl = l &2 = 1

Adding the two inequalities we conclude that

*2 = 1 kl <*2

R

-R S

du

dv

П

dr dt > —

a

(G(0) - f) EE \k\2 (Ы2 + \bk |2) . ^ ' k1=i k2=i

Taking (7) into account we obtain our final estimate:

R

-R S

du

dv

Г ät> 4G(0) - 2в E.

a

It remains to choose R > 0 so as to satisfy 4G(0) — > 0, i.e.,

max G < 2G(0).

In view of the last remark of the preceding section this is equivalent to the inequality

(K + 1) sin x + (K — 1) (n — x) cos x < 2 (K — 1) n, for all 0 < x < n, where we write K := R2y2/n2 for brevity. This is equivalent to

K + 1 2n + (x — n) cos x < -—---

K- 1

sin x

for all 0 < x < n, and this is satisfied by a simple computation if K+l < 5,97, which in turn is satisfied if K > 1, 403. This is equivalent to4

R > y/1, 403n/Y « 1,18448301n/Y.

Using the value y = \f2 — 1 we arrive finally to the sufficient condition R > 2, 859594947n of the proposition.

4. Internal Observability of Square Membranes

We continue to consider sufficiently smooth solutions of (3). The following notion was introduced in [13] and generalized in [11]:

4This is better than the condition R > ^/2-k/y in [8].

2

2

2

Definition 1. A function f G L1(0, n) is pcyclic for some integer p > 2 if its 2n-periodic

odd extension satisfies the equality

% f (»?)-•

for almost all t G R.

We fix two integers p,q > 2 and we consider only smooth solutions of (3) whose initial data are pcyclic in the first variable x1 and q-cyclic in the second variable x2. We observe the solutions simultaneously on p — 1 vertical lines and q — 1 horizontal lines, given by the equations

Jn ■ 1 1 a Jn ■ 1 1

x1 — —, j — 1,...,p — 1 and x2 — —, J — 1,...,q — 1. pq

We call these solutions (p, q)-cyclic; see the figure below.

7T

7r/q

0 7v/p

Our purpose is to prove the following theorem: p — q

R >

(2 + V2)n

P

0)

then there exists a positive constant c = c(R) such that all (p,q)-cyclic solutions of (1) satisfy the inequality

R n

p-1

cE * £ J J

J-1-R 0

u'I t,—,z\ + u'\ t,z, — \

V p ) V p )

dz dt.

p = q

p-1 R I q-1

cE < / / | u'(t,jn/p,x2) | 2 dx2 dt + / / | u'(t,x1,jn/q) | 2 dx1 dt

R n

j=i

-R 0

j=i

-R 0

2

if R is sufficiently large. However, for p = q the condition (9) is replaced by a more complicated expression, difficult to evaluate (see (10) below). For the proof we start by deducing from (6) that

ro ro

u'(t, xbx2)=E (akeikt — bke-ilklt) sin(k1x1) sin(k2x2).

k1 = 1 k2 = 1

Hence for each j = 1,... ,p — 1 we have

ro ro

u'(t,jn/p,x2) = E i\k\ aem — bke-iWt) sin(fo jn/p) sintta) = k1 = 1 k2 = 1 2p ro

= Y,sin(jn/p)^ E i\k\ (ak eilklt — bk e-ikt) sin^); 1=1 k1=l k2 = 1

in the middle sum of the last line k1 runs over the set of positive integers congruent to 1 2p

Since sin(l(j + p)n/p) = (—1)1 sin(ljn/p) for j = 1,...,p, we may rewrite the last equality as

p

u' (t,jn/p,x2) = ^2sin(£jn/p)fe(t,x2)

1=1

with

ro

fi(t, x2) = i ^ J] ±\k\ (akeilklt — bke-ikt) sin^); k1=l k2 = 1

here and in the following formulas k1 runs over the set of positive integers congruent to 1 p 2p

Since sin(ljn/p) = 0 for j = p and since the matrix (sin(ljn/p))P--=1 is invertible6, it follows that

p- 1 p- 1 p- 1

01^2 \fl(t,x2)\2 <Y.\u' (t,jn/p,x2)\2 < C^ \fl(t, x2) \2

1=1 j=1 1=1

with two positive constants ci = ci(p), i = 1, 2.

Integrating the left inequality and using the orthogonality of the functions sin(k2x2) we get

p-1 n P-1 I

/ \u'(t, Ijn /p, x2) \2 dx2 > c^E / \fi(t, x2) \2 dx2 =

j=1 0 l=1 0

p-1 ro 2

01n EE

2

i=i fc2=i

J2±\k\ (akeikt — bke-i№) ki=£

5The signs ± can be precised, but it is not necessary for the sequel: it is equal to 1 if k1 = I mod 2p, and — 1 if ki = I + p mod 2p.

6See [16, Problem 277].

Integrating in time the preceding inequality, using Lemma 1 and observing that now we may apply Theorem 1 with 71 — (v^2 — 1)p (instead of 71 — (v^2 — 1)) for each fixed

k2 because of the congruence condition k1 = 1 mod p, we obtain the following estimate: p-1 R n

J J | u'(t,ljn/p,x2) | 2 dx2 dt >

j=1-R 0

p-1 x . D / 2 \

EEf E I k 12 ( I ak I2 + I bk I 2) — R^E I k I 2 ( I ak I2 + | bk |2) .

1=1 k2 = 1 ki =1 Y1 ki =1 / \ki>k2

Since for (p, q)-cyclic solutions we have ak — bk — 0 whenever k1 = 0 (see [11, Lemma 4.1]), we may change YiP—i artificially to ^P=r Then the last expression becomes

p-1 R n

J J Iu'(t,£jn/p,x2)I2 dx2 dt >

j=1-R 0

x / x 2 x \

> 2C1R £ E IkI2 (IakI2 + IbkI2) — R^E IkI2 (IafcI2 + IbkI2) k2 = 1 \ki=k2 Y1 ki = 1 /

x1 x2

— R n

£ J J Iu'(t,xj/q)\2 dx1 dt >

j=1-R 0

x / x 2 x \

> 2C1R £ E IkI2 (Iak I2 + Ibk I2) — R-2E IkI2 (Iak I2 + Ibk I2) ki = 1 \k2 =ki Y2 k2 = 1 '

with y2 — (v^2 — 1)q. Adding the two inequalities we conclude that

£ J J Wdx dt + £ J J W(t.MM dt

j=-R 0 j=-R 0

xx

> c£ EIkI2 (Iak I2 + Ibk I2) ki = 1 k2 = 1

with

/ n?2 C

r'--9 F? I min J/^-iT)! — -

/2 R2Y2

In view of (7) the theorem follows if c > 0. For p — q this is equivalent to (9).

c — 2R (min{c1(p).c1(q)}— — R|>) . (10)

Pari of this work was done during the visit of the first author at the Dipartimento di Seienze di Base e Applieate per I'Ingegneria of the Sapienza Universita di Roma in May-July 2014- The author wishes to thank the department for its hospitality.

R n R n

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Vilmos Komornik, Département de mathématique, Université de Strasbourg, 7 rue René Descartes, 67084 Strasbourg Cedex, France, [email protected].

Paola Loreti, Dipartimento di Scienze di Base e Applicate per PIngegneria, Sapienza Université di Roma, via A. Scarpa n. 16, 00161 Roma, Italia, [email protected].

Долгое время в теории управления успешно применялись методы Фурье. Однако для некоторых теорем этот подход не применим. Несколько лет назвд Мегрен-берг установил граничную наблюдаемость колебаний прямоугольной мембраны (и для аналогичных задач большей размерности), обобщив классическую теорему Ингама о негармонических рядах Фурье. Предложенный им метод оказался полезным и для других приложений. Мы совершенствуем подход Мегренберга, сократив доказательство, а также обобщаем некоторые ранее рассмотренные приложения.

Ключевые слова: наблюдаемость; негармонический ряд Фурье; теорема Ингама; волновое уравнение.

Вилмос Коморник, профессор, Страсбургский университет (г. Страсбург, Франция), [email protected].

Паола Лорети, профессор, Римский университет Ла Сапиенца (г. Рим, Италия), [email protected].

Received December 11, 2014

УДК 517.9

DOI: 10.14529/ mm р 150308

Поступила в редакцию 11 декабря 2014 г.