On some applications of the boundary control method to spectral estimation and inverse problems Текст научной статьи по специальности «Математика»

ON SOME APPLICATIONS OF THE BOUNDARY CONTROL METHOD TO SPECTRAL ESTIMATION AND INVERSE PROBLEMS

S. A. Avdonin1, A. S. Mikhaylov2'3, V. S. Mikhaylov2'4

department of Mathematics and Statistics, University of Alaska, Fairbanks, USA 2St. Petersburg Department of V.A. Steklov Institute of Mathematics of the Russian Academy

of Sciences, St. Petersburg, Russia 3St. Petersburg State University, Faculty of Mathematics and Mechanics, St. Petersburg, Russia 4St. Petersburg State University, Faculty of Physics, St. Petersburg, Russia s.avdonin@alaska.edu, mikhaylov@pdmi.ras.ru, vsmikhaylov@pdmi.ras.ru

PACS 02.30 Zz, 02.30 Yy, 02.30 Nw DOI 10.17586/2220-8054-2015-6-1-63-78

We consider applications of the Boundary Control (BC) method to generalized spectral estimation problems and to inverse source problems. We derive the equations of the BC method for these problems and show that the solvability of these equations crucially depends on the controllability properties of the corresponding dynamical system and properties of the corresponding families of exponentials.

Keywords: spectral estimation problem, boundary control method, identification problem, inverse problem, Schrodinger equation, hyperbolic system. Received: 5 December 2014

1. Introduction

The classical spectral estimation problem consists of recovering the coefficients an, Ak, k = 1 ,...,N, N e N, of a signal

N

s(t) = ^ akeXkt, t ^ 0

n=1

from the given observations s(j), j = 0,..., 2N — 1, where the coefficients ak, Ak may be arbitrary complex numbers. The literature describing various methods for solving the spectral estimation problem is very extensive: see for example the list of references in [1,2]. In these papers a new approach to this problem was proposed: a signal s(t) was treated as a kernel of a certain convolution operator corresponding to an input-output map for some linear discrete-time dynamical system. While the system realized from the input-output map is not unique, the coefficients an and An can be determined uniquely using the non-selfadjoint version of the boundary control method [3].

In [4, 8], this approach has been generalized to the infinite-dimensional case: more precisely, the problem of the recovering the coefficients ak ,Ak e C, k e N, of the given signal:

<x

S(t) = ^ ak(t)eXkt, t e (0, 2T), (1.1)

k=1

from the given data S e L2(0, 2T) was considered. In [4], the case ak e C has been treated, in [8] the case when for each k, ak (t) = ^k-1 a\t are polynomials of the order Lk — 1 with complex valued coefficients ak was studied.

Recently, it was observed [9,15] that the results of [4,8] are closely related to the dynamical inverse source problem: let H be a Hilbert space, A be an operator in H with the domain D(A), Y be another Hilbert space, O : H d D(O) m Y be an observation operator (see [18]). Given the dynamical system in H:

u( — Au = 0, t > 0,

t- , , (I-2)

u(0) = a,

we denote by ua its solution, and by y(t) := (Oua)(t) the observation (output of this system). The operator that realizes the correspondence a m (Oua)(t) is called the observation operator OT : H m L2(0,T; Y). We fix some T > 0 and assume that y(t) e L2(0, T; Y). One can pose the following questions: what information on the operator A could be recovered from the observation y(t)? We mention works on the multidimensional inverse problems for the Schrodinger, heat and wave equations by one measurement, concerning this subject. Some of the results (for the Schrodinger equation) are given in [9,10,16]. To answer this question in the abstract setting, in [15] the authors derived the version of the BC-method equations under the condition that A is self-adjoint and Y = R. In the present paper, we address the same question without the assumption about selfajointness of A . The possibility of recovering the spectral data from the dynamical one is well-known for the dynamical system with a boundary control [11,12]. We extend these ideas to the case of the dual (observation) system.

The solvability of the BC-method equations for the spectral estimation problem critically depends on the properties of corresponding exponential family. The solvability of the BC-method equations for system (1.2) depends on the controllability properties of the dual system. We point out the close relation between these two problems: they both leads to essentially the same equations (see section 4 for applications), and conditions for the solvability of these equations are the same (on the connections between the controllability of a dynamical systems and properties of exponential families see [5]).

In the second section, we outline the solution for the spectral estimation problem in infinite dimensional spaces (see [8] for details). In the third section, we derive the equations of the BC-method for problem (1.2), extending the results of [15] to the case of non self-adjoint operator. Also, we answer the question on the extension of the observation y(t) = (Oa)(t). The last section is devoted to the applications to inverse problem by one measurement of the Schrodinger equation on the interval and to the problem of extending the inverse data for the first order hyperbolic system on the interval, see also [4,7-9].

2. The spectral estimation problem in infinite dimensional spaces

The problem is set up in the following way: given the signal (1.1), S e L2(0,2T), for T > 0, to recover the coefficients ak (t), Ak, k e N. Below, we outline the procedure of recovering unknown parameters, for the details see [8].

We consider the dynamical systems in a complex Hilbert space H:

x(t) = Ax(t) + bf (t), t e (0,T), x(0) = 0. (2.1)

y(t) = A*y(t) + dg(t), t e (0,T), y(0) = 0, (2.2)

here b,d e H, f,g e L2(0,T), and we assume that the spectrum of the operator A, {Akis not simple. We denote the algebraic multiplicity of Ak by Lk, k e N, and also assume that the set of all root vectors {0k}, i = 1,... ,Lk, k e N, forms a Riesz basis in H. Here, the vectors from the chain {<p%k}Lki, k e N, satisfy the equations:

(A - Ak) <k = 0, (A - Ak) <k = <k-1, 2 ^ i ^ Lk.

The spectrum of A* is (Ak}^=1 and the root vectors }, i = 1,..., Lk, k e N, also form a Riesz basis in H and satisfy the equations:

(A* — Ak) ^Lk = 0, (A* — Ak) ^k = ^k+1, 1 ^ i ^ Lk — 1. Moreover, the root vectors of A and A* are normalized, in accordance with the following:

> = 0, if k = l or i = j;

<0k,^k> = 1, i = 1,...,Lk,k e N.

We consider f and g as the inputs of the systems (2.1) and (2.2) and define the outputs z and w by the formulas:

z(t) = (x(t),d), w(t) = (y(t),b).

We assume that b = k=1 bk0k, d = k=1 SL=1 dk^k. While searching for the solution to (2.1) in the form x(t) = k=1 SL=1 ck(t)0k, we arrive at the following representation for the output:

k Lk 1 z(t) = (x(t),d) = EE ck(t)dk = r(t — T)f (T) dT,

k=1 i=1

where the response function r(t) is defined as:

eAfc i

r(t) =

k=1

with ak being defined as:

Lk

tLfc-2 r, tLfc

ak + akt + ak^ + ... + afck -^ + afck

1

2 ..... k (Lk - 2)! ' k (Lk - 1)!

(2.3)

ak = E bkdj, j = 1,..., Lk, k e N. (2.4)

i=j

It is important to note that r(t) has the form of the series in (1.1). Analogously, looking for the solution of (2.2) in the form:

k Lk

y(t) = EE hk (t)^k, k=1 i=1

we arrive at:

k Lk 1

w(t) = (y(t), b) = EE hk (t)bk = r(F^g(T) dT. k=1 i=1 0

We introduce the connecting operator CT : L2(0,T) m L2(0,T), defined through its bilinear form by the formula:

<CTf,g> = (x(T),y(T)). In [8], the representation for CT was obtained: Lemma 1. The connecting operator CT has a representation

T

(CTf)(t) = / r(2T — t — t)f (t) dT. 0

2

We assume that the systems (2.1), (2.2) are spectrally controllable in time T. This means that for any i e {1,... ,Lk}, and any k e N, there exist fk, gk e H0(0,T), such that xfk (T) = <k, y9k (T) = ^k. Using ideas of the BC method [13], we are able to extract the spectral data, {Ak,a'k) , j = 1,...,Lk, k e N, from the dynamical one, r(t), t e (0, 2T), (see [4,8] for more details):

Proposition 1. The set Ak, fk, i = 1,... ,Lk, k e N, are eigenvalues and root vectors of the following generalized eigenvalue problem in L2(0, T):

T

j (r'(2T - t - t) - Ar(2T - t - t)) f (t) dT = 0. (2.5)

o

The set Ak, gk, k = 1,... i = 1,... ,Lk are eigenvalues and root vectors of the generalized eigenvalue problem in L2(0,T):

T

j (V(2T - t - T) - Ar(2T - t - T)) g(T) dT = 0. (2.6)

o

Now, we describe the algorithm of recovering ak,... a^k, k e N (see the representation (2.3)). We normalize the solutions to (2.5), (2.6) by the rule:

(cTfk ,9k ) = 1, (2.7)

and define:

T

bl = (y^ (T), b) = j r(T - T)9k (t) dT, (2.8)

o

T

d9k = (xfk(T),d) = j r(T - T fk(t) dT. (2.9)

0

Then (see (2.4))

Lk

ak = Y^9kd9k. (2.10)

i=1

We denote by d and I the differentiation operator and the identity operator in L2(0,T). We normalize the solutions to (2.5), (2.6) (for i > I) by the following rule:

([CT (d - AkI)}1 fk,gk~l) = 1, (2.11)

we define blk, dlk by (2.8), (2.9) and evaluate:

alk = £fk fk~l+1, I = 2,..., Lk. (2.12)

i=l

We conclude this section with the algorithm for solving the spectral estimation problem: suppose that we are given with the function r e L2(0, 2T) of the form (2.3) and the family Ur=1{eAk(,... ,tLk-1eXk(} is minimal in L2(0,T). Then, to recover Ak, Lk and coefficients of polynomials, one should utilize the following:

On some applications of the Boundary Control method to spectral estimation and . . . 67 Algorithm

a) solve generalized eigenvalue problems (2.5), (2.6) to find Ak, Lk and non-normalized controls. _ _ _

b) Normalize fk, gk by (2.7), define bk, dk by (2.8), (2.9) to recover ak by (2.10).

c) Normalize fk, by (2.11), define &k, dk by (2.8), (2.9) to recover ak by (2.12),

l = 2,... ,Lk — 1.

3. Equations of the BC method

Let us denote by A* the operator adjoint to A and B := O*, B : Y m H. Along with system (1.2), we consider the following dynamical control system:

vt + A*v = Bf, t < T,

v(T) = 0, ( )

and denote its solution by vf. The reason we consider the system (3.1) reverse in time is that it is adjoint to (1.2) (see [5,15]).

For every 0 ^ s < T, we introduce the control operator by Wsf := vf (s). It is easy to verify that —W° is adjoint to OT. Indeed, taking f e L2(0,T; Y), a e H we show [15] that:

T

J (f, Oa)y = — (W°f,a)n , (3.2)

0

here Oa = (Oua) (t). Due to the arbitrariness of f and a, the last equality is equivalent to

(OT )* = —W °.

We assume that the operator A satisfies the following assumptions:

Assumption 1. a) The spectrum of the operator A, {Ak }k=1 consists of the eigenvalues Ak with algebraic multiplicity Lk, k e N, and the set of all root vectors {0k}, i = 1,..., Lk, k e N, form a Riesz basis in H. Here, the vectors from the chain {0k}L=1, k e N, satisfy the equations

(A — Ak) 0k = 0, (A — Ak) 0k = 0k-1, 2 ^ i ^ Lk.

The root vectors of A*, {-0k}, i = 1,..., Lk, k e N, form a Riesz basis in H and satisfy:

(A* — Ak) 0Lk = 0, (A* — Ak) 0k = 0k+1, 1 ^ i ^ Lk — 1.

b) The system (3.1) is spectrally controllable in time T: i.e. there exists the controls fk e H° (0, T; Y) such that W°fk = 0k, for i = 1,..., Lk, k e N.

We say that the vector a is generic if its Fourier representation in the basis {0k}k=1, a = k=1 L=1 ak0k, is such that ak = 0 for all k, i. We assume that the controls from the Assumption 1 are extended by zero outside the interval (0,T). Now, we are ready to formulate.

Theorem 1. If A satisfies Assumption 1, Y = R, and source a is generic, then the spectrum of A and controls fk are the spectrum and the root vectors of the following generalized spectral problem:

2T

J ((Oa)(t) — Ak(Oa)(t),fk(t — T + tdt = 0, 0 < t < T. (3.3)

°

Here, by dot, we denote the differentiation with respect to t.

Proof. We denote by {fk} the set of controls which satisfy W°fk = fk. By {fk} we denote the set of shifted controls: fk (t) = fk (t — T). Thus, the control fk acts on the time interval (T, 2T). Let us fix some i e l,...,Lk, k e N,t e (0, T ) and consider W0 f (• + t )) :

W0 fk(• + t)) = vfk(+)(0) = vfkG+t)(0) = (Bfk(• + t)) (0) — A*vfk(^t)(0). (3.4)

Since fk e H0(T, 2T,Y ), (Bfk(• + t )) (0) = 0. The second term on the right side of (3.4) could be evaluated using the following reasons. The function vf solves:

vfl(+T) + A*vfik(^+T) = 0, 0 ^ t ^ T — t, vfk(+)(T — t ) = fk.

We are looking for a solution in the form vf ( +T)(t) = j=i ck (t)fk, then ck satisfies boundary conditions Ck(0) = Sij and equation:

l/k + Ik Ck = 0, d —

-Ck + \kCk + Ck-1 = 0, j = 2,..., Lk.

Solving this system, we obtain the following expansion:

(j — i)!

Lk (T — t — t)j-i ^

vf)(t) = £ (T (._,t) eAk(T-T-t)fj. (3.5)

3=i

Evaluating A*vfk( +T)(0), making use of (3.5) and properties of the root vectors, we arrive at:

A*vfik ) (0) = \k vfL )(0),

A*vfk(+)(0) = \kvfk(+T)(0) + vfk+ (•+T)(0), i < Lk. Then, continuing (3.4), we obtain:

W0 fLk(• + t)) = -A*vfLkk(+T)(0) = -AkWfLk, (3.6)

w0 (fk(• + t)) = -AkW0fk - AkW0fk+1, i < Lk. (3.7)

Integrating by parts and taking into account that fk(0) = fk(T) = 0 for i = 1,... ,Lk,

we get:

2T 2T

fk(

(Oa)(t), fk (t + t ))y dt = — i ((Oa)(t),fl (t + t ))y dt

00

2T

+ ((Oa)(t + t),fk(t^[=2 = — / ((Oa)(t),fk(t + t))y dt (3.8)

t=0

On some applications of the Boundary Control method to spectral estimation and . . . 69 Conversely, using the duality between W° and OT and (3.6), (3.7), we have for i = Lk:

2T

/ ((Oa)(t), fL (t + t))y dt = — (a, Wf (■ + t= (a, AkWf (■ + t=

0

2T

/0 ^Lk !

(Afca,W0/fcLk(■ + t))^ = - / (Afc(Oa)(t),/fcLk(t + t))y dt (3.9)

H

0

and for i < Lk :

2T

((Oa)(t), /(t + t))y dt = (a, AfcW0/(■ + t) + W0/k+1( + t

2T 2T

-Ak / ((Oa)(t), /k(t + t))y dt - ((Oa)(t),/k+1(t + t))y dt (3.10)

0

In what follows, we assume that elements with index i = Lk + 1 or i = 0 are zero. Combining (3.8) and (3.9), (3.10), we see that the pair Ak, /k satisfies on 0 < t < T, i = 1,...,Lk :

((Oa)(t) - Ak(Oa)(t),/k(t + t))y dt = i ((Oa)(t), /k+1 (t + t))y dt. (3.11)

2T 2T

>a)(t) — Ak(Oa)(t),fk(t + tdt = J ((Oa)

°°

Now we prove the converse; solving the generalized eigenvalue problem:

2T

((Oa)(t) - A(Oa)(t), /(t + t)) dt = 0 (3.12)

0

yields {Ak}k=1 eigenvalues of A and controls {fk}, i = 1,..., Lk, k e N.

Let the functions {f1,..., fL} satisfying (3.11) constitute the chain for (3.12) for some A. Then, as it follows from the proof that for t e (0, T):

(a, W0/(t + t+ A (a, W0/i(t + t))h = - (a, W0/m(t + t))h

h

which is equivalent to

— (a, A*vfi(t+T)(0))H + A (a,vfi(t+T)(0))H = — (a,v/i+1 (t+T)(0))H , t e (0,T). (3.13) First, we consider case i = L. We rewrite the last equality (using the notation f = fL) as:

(a,A*v/(t+T)(0) — Av/(t+T)(0))H = 0, t e (0,T). (3.14)

We assume that vf (t+T)(T — t) = ^ hen ck0k. Then, developing vf in the Fourier series as we did in (3.5), we arrive at:

kk

z=l,...,Lk

f (t+T)(0) = £ ck £ ^T^ir e"k^. (3.15)

keN j=1 (j )!

i=1,...,Lk

Applying operator A* and using the property Л*ф3к = Xk+ , we obtain:

A*vf(t+T>(0)= £ ck £ jPT> tik + ^k+1) . (3.16)

j=i (j )!

keN

Introducing the notation:

д(т):= A*vf(t+T>(0) - Xvf(t+T>(0) = £ gk(т, (3.17)

keN

relation (3.14) yields:

0 = (a,д)н = £ algk(т), т E (0,T). (3.18)

m

ken -l,...,Lk

The functions gk (t) are combination of products of eXk(T T) and polynomials ( aT) . Then, we can rewrite (3.18) as follows:

0= £ bk^T—^eXk(T-T>, т E (0,T).

'k (- 1); eAk -T), t e (0,T). (3.19)

keN ( )!

i=\,...Lk

If Y = R, the controllability of the dynamical system (3.1) imply [5] the minimality of the family (J£==1{eJk(,teJk(... ,tLk-1eJk(} in L2(0,T) in L2(0,T), so we have bk = 0 for all k,i. However, as follows from (3.15), (3.16):

bL = ckAkal - Ackak = Then, since a is generic, either A = Ak or ck = 0.

Let A = Ak, so c1 = 0. Then, for b^k-1, we have:

bL-1 = ckAk ak- Ack ak = 0 ^om which the equality c| = 0 follows. Repeating this procedure for b^k-i, i ^ 2, we obtain:

If A = Ak, then Ck = 0, i =1,..., Lk. (3.20)

We consider the second option; let A = Ak. Then, from (3.15) and (3.16):

bLk-1 = ck = 0, bLk-2 = ck ak = 0,...,b1 = cLk-1aLk = 0.

So, we arrive at the following:

If A = Ak, then dk = 0, i = 1,... ,Lk - 1, and c^k could be arbitrary. (3.21)

Finally (3.20), (3.21) imply that A = Ak> and f = ck> fL, ck = 0, for some k'.

Thus, on the first step we already obtained that A = Ak> for some k' and fL = ck> fk, The second vector f in the Jordan chain satisfies

2T 2T

L

У

(Oa)(t) - Xk'(Oa)(t),f (t + тdt = ((Oa)(t),Ck' ffr' (t + тdt.

0 0 We rewrite (3.13) in our case:

- (a,A*vf(t+T>(0))H + Xk' (a,vf(t+T>(0))H = - (a,Ck'vfyk'(i+T>(0^ , т G (0,T). (3.22)

k

On some applications of the Boundary Control method to spectral estimation and . . . 71 In this case, g, introduced in (3.17), has the form:

g(T) = £ 4 £ ^^(T T) ((Ak - Ak,) + «+')

and rewrite (3.22) as:

(j - i)!

keN j=1 ^ '

(a, g)H = (a,v/L'k' = cfc, a^' eAk' (t-t) (3.23)

otation

for (3.23) to get:

H

Using the same notations in (3.18) and (3.19), we transcribe the equalities for coefficients bk

bk = Ck' aL,h', bk = 0, k = k', In the case k = k', we repeat the arguments used above and find that:

ck = 0, i = 1,...,Lk.

When k = k', we have:

b^ = 0, b^ -1 = ck' aL,h' = 0, b^-2 = ck' aL,h' = 0,

bk' = cL'h aL'h = 0, bk' = cL'h aL'h = ck'aL'h .

So, we find:

ck' = 0, i < Lk' — 1, cLh'-1 = ck', cL'h' is arbitrary. So, finally we arrive at for some cL-1 :

f = fL-1 = ck'fL h' 1 + cL-1fkL h'

Arguing in the same fashion, we obtain that:

fi = ck' fkL h'-i + cf h', 1 ^ i < Lk' — 1.

So, we have shown that the elements of the Jordan chain for (3.3) which correspond to eigenvalue Ak' are the linear combination of corresponding controls and eigenvector (i.e. the control that generate the eigenvector of A*). □

Remark 1. The solution to (3.3) yields {Ak}k=1 eigenvalues of A and (non-normalized) root vectors {fk}, fk = ck fk + ck fL k e N, i = 1,..., Lk, cLh = 0.

For the dynamical system (1.2), under the conditions on A, Y, formulated in Theorem 1, there is the possibility to extend the observation y(t) = (Oua) (t) defined for t e (0, 2T) to t e R+. To this end, we show that for an observation having the form:

L h bj tL h -j

Oa = £ e^ ht £ y, (3.24)

(Lk - j)!'

keN j=1 ^ JJ

we can recover the coefficients bk.

Take i G |1,...,Lk} and search for the solution to (1.2) with a = 0k in the form u = Y1 L=1 cl(t)0k, we arrive at the system (here cLk+1 = 0):

d

—ci(t) - Akc (t) = ci+1 (t), l = 1,..., Lk, dt

c (0) =

whose solution is:

ti-1

= )H eAh' l « i-

ci(t) = 0, l > i.

Thus,

(i — l)!

For the initial state a = Y1 keN Stka ak0k, we obtain:

1 ti-i

= £ TTTTTTeAh'0k- (3.25)

l.1

Lh

Lh

tLh-j

«" = £ eA"'£ aLh-j+l0

(Lk — j)!

keN j=1 v k ' l=1

So, for observation (Oa)(t) = (Oua) (t), we derive the representation (3.24) with coefficients bk, defined by:

j

bk :=£aLh-j+lO0k, k e N, j = 1,...,Lk. (3.26)

j__^ „Lh -j+lnAl

1=1

Making use of Theorem 1 (see also Remark 1), we have:

Wfk = ck0k + ck0Lh, k e N, i = 1,..., Lk, cLh =0. (3.27)

Counting (3.2), we write:

T

(W°fk,a)ff = — [ Ouf dt.

°

We plug a = 0k into the last equality and use (3.27) to get:

T

+ ck0k , 0k _

ck = (ck 0k + ck 0Lh, 0k) H = —J Ou^h fk dt. (3.28)

°

We evaluate the right side of (3.28) for all i. For i =1 we get (see (3.25)):

T

ck = —O01 y eAhf dt. °

Or equivalently:

T

c,

O0k

— eAhtf1 dt. (3.29)

Evaluating (3.28) for i = 2, counting (3.25), we obtain:

T T

ck = —O0k / eAhtf2 dt — O0k / teAhtf2 dt.

°

On some applications of the Boundary Control method to spectral estimation and . . . 73 We divide this equality by ck and plug (3.29) in to find:

J eXk(ft dt f eXk(f dt

= - f0-0I--(3.30)

k I exk(f dt - f texk(ft dt

Of

0 0

Suppose we already found —^ for l = 1,...,i — 1. To find this quantity for l = i, we evaluate (3.28), plugging the expression for v^ (3.25):

T

i f fi—i ck = — £ Of[J eXktft df.

i=i o

We divide last equality by ck to find:

feXktft df ck0

Off i—iT ^ / N -1

H 1 + E J Ä eXkf *(

(3.31)

Observe that in the right side of (3.31) in view of (3.30), we know all terms. To evaluate a%k, we use, see (3.27):

t

ak = (a, fk )H = (a, W0f - 4 fLkk) - = - i Ouaf dt- - aLkk ^ (3.32)

V / H ck J ck ck

0

We multiply (3.27) by and obtain for i <Lk:

T

4 = (W0fk,fLkk)H = — fk(f) (Ovfkk) (f) df

i=i o

0

T,O4>k I jjf—l)\eXk%fk(f) df.

k t

Lk f +Lk-l

Dividing the last equality by ck, we obtain:

L -1 t

cl- ±fe)7eXktfk(f)df, ^. (3.33)

Ck J (Lk — l)! 'k

0

Notice that in view of (3.31), we know all terms in the right hand side in (3.33). Now, we multiply (3.32) by ck:

T

ak ck = — i Ova fk df — aLk ck -.

ck

0

Since cLf = 0, we have for i = L

k = o, we nave ioi i = Lk-

T

aLk ck = —J fL (f) (Ova) (f) df, 0

and finally:

T T

ak ck = —J fk (t) (Oua) (t) dt + / t/Lh (t) (Oua) (t) dtci. (3.34)

° °

In view of (3.33), we know all terms on the right side of (3.34). Now, we rewrite formula for b, (3.26):

b, := £ {aLh-j+lck} (^) k e N, j = 1,..., Lk, (3.35)

l=1 ck

and observe that the first term in each summand is given by (3.34), while the second term by (3.31). So, we know right hand side in (3.35).

After we recovered all j by (3.35), we can extend the observation (Oa)(t) by formula (3.24) for t > 2T.

4. Application to inverse problems

Here, we provide two applications of the above-developed theory to inverse problems. Other applications of the BC approach to the spectral estimation problem can be found in [1,2,4,7-9,15].

4.1. Reconstructing the potential for the 1D Schrodinger equation from boundary measurements

Let the real potential q e L1(0,1) and a e H°(0,1) be fixed, we consider the boundary value problem:

iut(x, t) — uxx (x, t) + q(x)u(x, t) = 0 t > 0, 0 < x < 1 u(0,t) = u(1, t) = 0 t> 0, (4.1)

u(x, 0) = a(x) 0 < x < 1.

Assuming that the initial datum a is generic (but unknown), the inverse problem we are interested in is to determine the potential q from the trace of the derivative of the solution u to (4.1) on the boundary:

{r°(t),n(t)} := {ux(0,t),ux(1,t)}, t e (0, 2T), It is well known that the self-adjoint operator A defined on L2(0,1) by:

A0 = —0'' + q0, D(A) := H2(0,1) n H°(0,1), (4.2)

admits a family of eigenfunctions {0k}k=1 forming a orthonormal basis in L2(0, 1), and associated sequence of eigenvalues Ak m Using the Fourier method, we can represent the solution of (4.1) in the form:

k

u(x,t) = £ akeiAht0k(x), ak = (a, 0k)L2(°,1) (4.3) k=1

The inverse data admits the representation:

£ akeiAht0k(0), £ akeiAht0k(1) . (4.4)

k=1 k=1 J

One can prove that r°,r1 e L2(0,T). Using the method from the first section, we recover the eigenvalues Ak of A and the products 0k(0)ak and 0k(1)ak. So (as a is generic) we recovered the spectral data consisting of:

{Ak« L • (45)

Now from D we construct the spectral function associated with A. Given A e C, we denote by y(-, A) the solution to:

—y''(x, A) + q(x)y(x, A) = Ay(x, A), 0 < x < 1, y(0, A) = 0, y'(0, A) = 1.

Then, the eigenvalues of the Dirichlet problem of A are exactly the zeros of the function y(1, A), while a family of normalized corresponding eigenfunctions is given by 0k(x) = y(x,Ak)

Thus, we can rewrite the second components in D in the following way:

0k (1) y'(1,Ak)

, Ak

0k (0) y'(0,Ak)

y'(1,Ak)=: Ak. (4.6)

Let us denote by dot the derivative with respect to A and An be an eigenvalue of A. We borrowed the following fact from [17, p. 30]:

Ak)||L2 = y' (1,Ak )y(1, Ak), A„ — A

y(1,A) = II

n2n2

y(1,Ak) = — ^ n ^^ =: Bk.

ra^1,ra=k

Notice that the set of pairs {Ak, ||y(-, Ak)||L2}k=1 =: D is "classical" spectral data. Using the above relations, we come to D = {Ak, AkBk}k=1. Let := ||y(-, Ak)|L2 = AkBk, we introduce the spectral function associated with A:

' — E % A« a > 0,

°<Ah^A h

which is a monotonously increasing function having jumps at the points of the Dirichlet spectra. The regularized spectral function is introduced by:

„(A) i p(A) — (A) A»0. p (A) ^ 1 A> 0 ff(a)n p(A) a < P°(A)=4:sa (Ok)2 A> 0.

where p° is the spectral Unction associated with the operator A with q = 0. The potential can thus be recovered from a(A) by Gelfand-Levitan, Krein or the BC method (see [6,14]). Once the potential has been found, we can recover the eigenfunctions 0k, the traces 0k (0) and Fourier coefficients ak, k = 1,... to. Thus, the initial state can be recovered via its Fourier series.

4.2. Extension of the inverse data

We fix p, G C 1([0,1]; C), di, d2 G L2(0,1; C) and consider on interval (0,1) the initial boundary value problem:

dt

dx

0 1 1 0

P11 P12

i P21 P22

0, t > 0,

t > 0,

(4.7)

0 € x € 1.

< u(0,t) = u(1,t) = 0, u(x, 0 m / d1(x) v(x, 0W ld2(x)

We fix some T > 0 and define R(t) := {v(0,t),v(1,t)}, 0 ^ t ^ T. Here, we focus on the problem of the continuation of the inverse data: we assume that R(t) is known on the interval (0,T), T > 2, and recover it on the whole real axis. The problem of recovering unknown coefficients p, and initial state c1,2 has been considered in [19,20], where the authors established the uniqueness result, having the response R(t) on the interval (—T, T) for large enough T.

01 10

We introduce the notations B operators A, A* acting by the rule:

P

P11 P12 1P21 P22

D

and the

d

A = BT- + P,

dx d

on (0,1),

A*é = —+ PT, on (0,1), dx

with the domains:

D(A) = D(A*) = p

P1

G H 1(0, 1; C2) | P1(0) = P1(1) = 0

The spectrum of the operator A has the following structure (see [19,20]): a(A) = E1 U E2, where E1 n E2 = 0 and there exists N1 G N such that

1) E1 consists of 2N1 — 1 eigenvalues including algebraical multiplicities

2) E2 consists of infinite number of eigenvalues of multiplicity one

3) Root vectors of A form a Riesz basis in L2(0,1; C2).

Let m denote the algebraical multiplicity of eigenvalue A, and we introduce the notations:

E1 = {A' G a(A), mj ^ 2, 1 € i € N} , E2 = {An G a(A), An is simple , n G Z} .

Let e1 := (°). The root vectors are introduced in the following way:

(A - A') 01 = 0, (A - A') 0} = 0j-i, 2 ^ j ^ mi, 0}(0) = ei, 0} G D(A), 1 ^ j ^ mi. For the adjoint operator, the following equalities are valid:

A*— A )é

0,

A* — Ajéj = éj+1, 1 € j € mi — 1, éj(0) = e1, éi G D(A*), 1 € j € mj.

d

For the simple eigenvalues, we have:

(A — A„) fn = 0, (A* — Jn) fn = 0, for n e Z, fn(0) = fn(0) = ei, fn e D(A), fn e D(A*). Moreover, the following biorthogonality conditions hold:

fn) =0, (fn,fj ) =0, (fk ,fn) = 0,

/fk) =0, if i = k or j = l,

Pj = (fj,fj) , i = !,... ,N, j = 1,..., mi,

Pn = (fn,fn), n e Z.

We represent the initial state as the series:

N mi

D = dj fj (x) + £ dnfn(x), (4.8)

jj

i=i j=i nez

and search for the solution to (4.7) in the form:

N mi

u

I (x,t) =

i=i j=i nez

(x, t) = £ £ cj (t)fj (x) + £ Cn(t)fn(x).

Using the method of moments, we can derive the system of ODe's for cj, i e {1,...,N}, j e {1,... ,mi}, cn, n e Z, solving which we obtain:

cj (t) = e

\it

t2 tmi-j

di + di t + di — + + di _-_

dj + dj+it + dj+2 2 + ... + dmi (m j)!

cn(t) = dneXnt.

Notice that the response {v(0, t), v(1, t)} has a form depicted in (1.1):

N

(0, t) = £ eAita0(t) + £ eXntdn (fn(0))2 , (4.9)

i=i nez

N

'(l,t) = £ eAita!(t) + £ eXntdn (fn(1))2 , (4.10)

eAita!(t) + £

i=i nez

where the coefficients of a0(t) = mi-1 altk are given by

1 mi

«0 = £ di (fi(0)) 2 , «1 = £ di (fi-i(0^ 2 , «2 = ^ ($-2(0)) 1=1 1=2 1=3

1 mi ( ) 1 ( ) £ di (fi-k(0))2 ,. . . <-1 = (m _1)!dmi (f 1(0))2 .

" k!^ lvvl-kK Jmi-1 (mi - 1)!

The coefficients a1(t), i = 1,... ,N are defined by the similar formulas.

We assume that the initial state D is generic. Introducing the notation U := ("), we consider the dynamical system with the boundary control f e L2(R+):

U( - AU = 0, 0 ^ x ^ 1, t> 0, u(0,t) = f (t),u(1,t) = 0, t> 0, U(x, 0) = 0.

It is not difficult to show that this system is exactly controllable in time T > 2. This implies (see [5]) that the family (JIN=1{ext(,... ,tmi-1eXH} U {eXn(}n^Z forms a Riesz basis in a closure

v

2

of its linear span in L2((0,T); C). So we can apply the method from the second sections to recover A', m', coefficients of polynomials a0,1(t) i = 1,..., N, An, n e Z. The latter allows one to extend the inverse data R(t) to all values of t e R by formulas (4.9), (4.10). This is important for the solution of the identification problem, see [20].

Acknowledgments

Sergei Avdonin was supported by the NSF grant DMS 1411564; Alexander Mikhaylov was supported by NSh-1771.2014.1, RFBR 14-01-00306, IRSES (FP7-PEOPLE-2012-IRSES Marie Curie Actions); Victor Mikhaylov was supported by RFBR 14-01-00535, RFBR 14-0131388 andNIR SPbGU 11.38.263.2014.

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