IDENTIFICATION OF UNKNOWN PARAMETERS OF A LANDAU

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A nonlinear function which determines relations between strains and ... 1 uxuxt = g;. (t; x) 2 QT: (1). 1Institut f ur Angewandte Mathematik und Statistik der Technischen Universit at M unchen, Dachauer ...... f(x)=1+0:5 sin 5x; g(x)=1+0:5 cos 5x;.
IDENTIFICATION OF UNKNOWN PARAMETERS OF A LANDAU-GINZBURG MODEL FOR STRUCTURAL PHASE TRANSITIONS IN SHAPE MEMORY ALLOYS N.D.Botkin1

Abstract.

An algorithm for identi cation of unknown parameters of a LandauGinzburg model describing phase transitions in shape memory alloys is proposed. Distributions of the displacement and of the temperature observed with an error are input data of the algorithm. Finite element approximations of the input signal and of the model are used. The algorithm utilizes the idea of the direct minimization of the residual of equations written in an appropriate variational form. The convergence of the algorithm output to the set of all parameters compatible with the exact solution is proved. The question of the convergence under local measurements of input data is discussed and sucient conditions for the convergence are proposed. The paper is illustrated by computer simulations.

Introduction

This paper deals with the model that was studied in [1] from the point of view of numerical methods for nding solutions with the use of nite element methods. The model is characterized by heavy nonlinearities and by the forth order of derivatives with respect to spatial variables. A nonlinear function which determines relations between strains and stresses lies in the base of the model. We consider the problem of identi cation of this function and propose a numerical algorithm. The main idea of the algorithm consists in a nite-dimensional approximation of input data and the direct minimization of the defect of equations written in an appropriate variational form. Such a method was proposed in [2] for recovering unknown parameters of ordinary di erential equations. The idea of the direct minimization of residuals was applied in [3] to identi cation of heat conductivity coecients of nonlinear parabolic equations. The paper presented is closely related to the paper [3] and can be considered as its extension. It should be noted that the algorithm proposed has common features with methods developed in [4] and [5]. In [4] a method based on the direct solution of the original equation with respect to the unknown heat conductivity coecient was proposed. In [5] an approach related to direct methods and based on stabilization theory was developed.

Formulation of the problem We consider the model of structural phase transitions in shape memory alloys reported in [1]. We denote = (0; 1), ? = @ , I = (0; T ) and QT = I  . The model is described by the following system:

utt ? p(t; ; ux)x + uxxxx = f; c0t ? k0xx ? 1uxuxt = g;

(t; x) 2 QT ; (t; x) 2 QT :

(1)

1 Institut f ur Angewandte Mathematik und Statistik der Technischen Universitat Munchen, Dachauer Str. 9a, D-80335 Munchen

1

The boundary and initial conditions look as follows:

uj? = u?; uxj? = u1?; j? = ? ujt=0 = u0; utjt=0 = u1; jt=0 = 0: The functions u and  represent the displacement and the absolute temperature; e = ux is the shear strain. The function p(t; ; e) is the derivative of the free energy with respect to e. This function gives the value of the stress depending on the temperature and the strain. We consider p to be unknown. Our objective is to construct an algorithm for nding this function on the basis of solutions of (1) observed with an error. Speaking more de nitely, we have functions u ;  such that ku ? u kL1(QT )  ; k ?  kL1(QT )  ; where u;  is an exact solution of (1) and  is an error of measurements. We are to construct an algorithm that processes the functions u ;  and gives on the output a function p such that p ! p as  ! 0. We assume that p(t; ; e) is de ned on the region [0; T ]  [; ]  [e; e], where the symbols supplied with asterisks denote maximal and minimal values of  and e. We suppose that p(t; ; e) is continuous with respect to t and Lipschitzian with respect to  and e with the Lipschitz constant L. We suppose that the function p admits the following parametric representation: p(t; ; e) = P (; e; (t)); where P is a given function and is a parameter belonging to a compact subset A of a Banach space X . For example: P  (; e), A = f = ( ; :::; ; :::) : j j  M=ig  `2: Here 1) P (; e; ) = 1 1 i i i=1 i i i(; e) are uniformly bounded and Lipschitzian functions de ned on [; ]  [e; e]; M is a constant; `2 is the discrete Hilbert space. 2) P (; e; ) = ; and A is the set of all de ned on [; ]  [e; e] Lipschitzian functions, whose Lipschitz constant is less then L. In this case, A is a compact subset of the corresponding space of continuous functions. Therefore, we reduce the original problem to the approximate computation of the function ().

Approximation of input data. Algorithm and convergence Through this section we assume that values of u ;  are available in the whole domain QT . We rewrite system (1) in the following variational form:

R [u v + P (; u ; (t))v + u v ? fv] = 0; tt x x xx xx QT R [c   + k   ? u u  ? g ] = 0; QT

0 t

0 x x

1

x xt

8v 2 L2(I ; H02( )); 8 2 L2(I ; H01( )): 2

(2)

It is easy to verify that the rst equation of (2) is equivalent to the following one:

ZT where

0

2Z 32 max 4 (utt ' + P (; ux; (t))'x + uxx'xx ? f')dx5 dt = 0;

'2B02

(3)



B02 = f' 2 H02( ) : k'kH ( )  1g: Therefore, for almost all t 2 I , (t) is the minimizing point of the left-hand-side of (3). Namely, 2 0

hZ

(t) = arg min max 2 A '2B02

i2

(utt ' + P (; ux; )'x + uxx'xx ? f')dx :

(4)



The last relation is the basis of our algorithm, but the following steps should additionally be done to enable numerical implementations: 1. Replacement of utt; ux; uxx, and  by nite-dimensional approximations constructed on the basis of u and  (it is desirable that these approximations were piece-wise constant with respect to t). 2. Replacement of the set B02 by the unit ball of an appropriate nite-dimensional space, nding the max over ' in (4), and approximation of the set A. The way of realization of these steps depends on the smoothness of the exact solution u;  of (1). To be close to numerical implementations, we assume that u 2 H q+1(QT ); ; f 2 H q (QT ); q > d=2; (5) where d is dimension of the region QT (d = 2 in the case considered and d = 3 in some forthcoming examples). 1. Let 0 = x0 < x1 < x2::: < xn < xn+1 = 1 be the equidistant partition of with the step h = 1=(n ? 1). We construct the spaces of nite elements Vh  H02( ); h  H01( ) The basis of Vh consist of the following functions: 8 0; < x?xi? x 62 (xi?1; xi+1); !i(x) = : N1 ( h ); x 2 (xi?1; xi); i = 1; :::; n; N2 ( x?hxi ); x 2 (xi; xi+1) 8 0; x 62 (xi?1; xi+1); < x ? x i ? ! i(x) = : hN3 ( h ); x 2 (xi?1; xi); i = 1; :::; n; x ? x i hN4 ( h ); x 2 (xi; xi+1) where N1; N2; N3; N4 are cubic polynomials of the form: N1 ( ) = 3 2 ? 2 3; N2 ( ) = 1 ? 3 2 + 2 3; N3 ( ) = ? 2 +  3; N4 ( ) =  ? 2 2 +  3: 1

1

3

The space h has the following basis:

8 0; < x?xi? x 62 (xi?1; xi+1); ri(x) = : h ; x 2 (xi?1; xi); i = 1; :::; n: x ? x i 1 ? h ; x 2 (xi; xi+1) 1

Additionally we introduce functions which will provide the boundary conditions.

 0;



x 62 (xn; xn+1) ; !0(x) = N ( x?x ); xx 262 ((xx0;; xx1)) ; !n+1(x) = 0; x ? x n N1( h ); x 2 (xn; xn+1) 0 1 2   0; h 0; x 62 (xn; xn+1) ; ! 0(x) = hN ( x?x ); xx 262 ((xx0;; xx1)) ; !n+1(x) = hN x ? x n ); x 2 (xn; xn+1) 0 1 3( 4  0; h x 62 (x ; x )  0; hx 62 (x ; x ) r0(x) = 1 ? x?x ; x 2 (x0; x1) ; rn+1(x) = x?xn ; x 2 (xn; xn+1) : 0 1 n n+1 h h 0

0

0

Let 0 = t0 < t1 ::: < tN = T be the equidistant partition of the time interval I with the step  . For m 2 1; N and t 2 (tm?1; tm], we put

f ;h = ;;h = u;;h =

where

um = u1?(tm; 0); 0

n+1 X

n+1 X

i=0 n+1

X i=0

f (tm; xi)ri;  (tm; xi)ri;

(6)

u (tm; xi)!i + umi! i;

i=0 m un+1 = u1?(tm; 1), and  (tm ; xi+1) ? u (tm ; xi?1) u m ui = 2h

for i 2 1; n. We de ne the discussed approximations as follows (the approximations of  and f are already de ned in (6) ): u;;h = u;;h 2 x if t 2 (tm?1; tm] for some m 2 1; N ;





u;;h =  ?2 u;;h (tm; ) ? 2u;;h (tm?1; ) + u;;h (tm?2; ) ; 1 2 ;;h 1 ;;h u;;h = 61 u;;h xx (tm ; ) + 3 uxx (tm?1; ) + 6 uxx (tm?2; ) 3

if t 2 (tm?1; tm] for some m 2 2; N ; and





= 2 ?2 u;;h (t1 ; ) ? u;;h (t0 ; ) ? uh1 () ; u;;h 1 4

(7)

2 ;;h  h = 31 u;;h u;;h xx (t1 ; ) + 3 uxx (t0 ; ) + 6 u1 xx () 3

if t 2 [t0; t1]. Here

uh 1

=

n+1 X i=0

u1(xi)!i + u1i!i;

where u1 0 = u1?(0; 0); u1 n+1 = u1?(0; 1), and u1i = u1(xi+1) 2?h u1(xi?1) for i 2 1; n. PROPOSITION 1. Let  and h have comparable magnitude (there exist 1 and 2 such that 1 < =h < 2 ). Then the following estimates hold under assumptions (5):

kf ;h ? f kL (QT ) k;;h ? kL (QT ) ku;;h ? uttkL (QT ) 1 ku;;h ? uxkL (QT ) 2 ku;;h ? uxxkL (QT ) 3 2

2

2

2

2

    

M  kf kH q (QT )  max(; h);  M  kkH q (QT )  max(; h) + meas ( )  ;  M  kukH q (QT )  max(; h) + meas ( ) = 2 ; M  kukH q (QT )  max(; h) + meas ( ) =h ;  M  kukH q (QT )  max( ; h) + meas ( ) =h2 : 1 2

+1

1 2

+1

1 2

+1

1 2

(8)

1 2

Moreover, if ; ux 2 L1(QT ), then

ku;;h 2 kL1 (QT )  M (1 + =h); k;;h kL1(QT )  M (1 + ):

Here, M is an independent from ;  , and h constant. The proof of this proposition is based on properties of the standard interpolation operators related to the spaces Vh ; h, see [6, 7, 8]. We give a sketch of the proof in the appendix. For the sake of simplicity, denote the collection (; ; h) by ". We shall write u"1; u"2; u"3; ", ;;h ;;h ;;h , and f ;h. We say that " ! 0 i ; ; h ! 0; and f " instead of u;;h 1 ; u2 ; u3 ;  = 2; =h2 ! 0; and there exist constants 1; 2 such that 1 < =h < 2. 2. Let us replace the set B02 by the unit ball Sh of the space Vh . Let A be a family of nite dimensional sets such that A  A whenever 2  1, and cl [>0 A = A. To nd max in (4) (we consider the approximations are already substituted into (4)) over the set Sh , we de ne a function t; ;" 2 Vh by the condition: 2

(

t; ;" ; ') 2 = H0 ( )

Z

1

(u"1' + P ("; u"2; )'x + u"3 'xx ? f "')dx; 8' 2 Vh:



5

(9)

This relation can be considered as the representation of the continuous functional de ned on the Hilbert space Vh via the inner product. Recall that the functions u"1; u"2; u"3; 3", and f " are piece-wise constant w.r.t. t and, hence, the function t; ;" is constant w.r.t. t on each interval (tm?1; tm], m 2 1; N . Using the function t; ;", we came to the following procedure:

" (t) = arg min 2A

Z

(u"1

t; ;" + P ("; u" ; ) t; ;" + u" t; ;" ? f " t; ;" )dx 2 x 3 xx

(10)



if t 2 (tm?1; tm] for some m 2 1; N . The following assertion is true.

THEOREM 1. Let U be the set of all functions () satisfying (3) (the set of all parameters compatible with the exact solution u; ). Then for any q  1, distLq (I ;X )( " (); U ) ! 0 as " ! 0 and  ! 0.

RT

1=q

k (t)kqX dt

We recall that k ()kLq(I ;X ) = : For brevity we introduce the following 0 notations: Z `" (t; ; ') = (u"1' + P ("; u"2; )'x + u"3 'xx ? f "')dx

Z

`(t; ; ) = (utt + P (; ux; ) x + uxx

xx ? f

)dx:



It is convenient to extend these function setting them to be equal to zero for t  0. The proof of Theorem 1 is based on the following proposition.

PROPOSITION 2.

ZT 0

max j` (t; ; 2 A "

t; ;") ? max `2 (t; ; 2B02

)jdt ! 0

as " ! 0. The proof is given in the Appendix. PROOF OF THEOREM 1. The proof is exactly the same as in [3]: We embed the functions " () into the set M = f(j)g of Young measures (functions mapping I in the set of regular probability measures de ned on A). Namely, we set " (tj) =  " (t), where the last symbol denotes the Dirac measure concentrated at the point " (t). The set M endowed with the weak* topology of L1(I ; C (A)) is compact. If the conclusion of Theorem 1 is not ful lled, we can nd a sequence "k k () converging to some measure ~(j) (that is, "k k ! ~) and strictly separated from U . By the de nition of "k k , we have

ZT Z 0 A

`"k (t; ;

t; ;"k )"k k (tjd )dt =

ZT 0

6

min ` (t; ; 2Ak "k

t; ;"k )dt:

Using Proposition 2, it is easily to prove that

ZT Z 0 A

and

`"k (t; ;

ZT 0

t; ;"k )"k k (tjd )dt !

min ` (t; ; 2Ak "k

t; ;"k )dt !

ZT 0

ZT Z 0 A

max `2 (t; ; ')~(tjd )dt

'2B02

min max `2 (t; ; ')dt = 0 2 A '2B02

(the last equality is valid due to (3)). So, we conclude that

ZT Z 0 A

max `2 (t; ; ')~(tjd )dt = 0;

'2B02

which implies that ~(j) \is concentrated" on the set-valued function





Q(t) = 2 A : max ` t; ; ') = 0 : '2B02

2(

This means that ~(tjA n Q(t)) = 0 at almost all t 2 I . Since U is the set of all selectors of the set-valued mapping Q, we obtain, using some result of [2], that "kk () ! U in Lq (I ; X ). This is a contradiction. Let us consider now numerical implementation of (10). If we gather all things involved into the right-hand-side of (10), we obtain the following equality:

`" (t; ; t; ;") = (q"m( ) ? bm")T S"?1(q"m( ) ? bm") if t 2 (tm?1; tm] for some m 2 1; N . Here

0 (! ; ! ) : : : (! ; ! ) (! ; ! ) : : : (! ; ! ) 1 BB 1xx 1xx : : : nxx 1xx 1xx 1xx : : : nxx 1xx C C BB (!1xx; !nxx) : : : (!nxx; !nxx) (!1xx; !nxx) : : : (!nxx; !nxx) C C ; S" = B C BB (! ; ! ) : : : (! ; ! ) (! ; ! ) : : : (! ; ! ) C C B@ 1xx 1xx : : : nxx 1xx 1xx 1xx : : : nxx 1xx C A (!1xx; ! nxx) : : : (!nxx; ! nxx) (!1xx; ! nxx) : : : (!nxx; ! nxx)

0 (f "; ! ) ? (u" ; ! ) ? (u" ; ! ) 1 1 1 1 3 1xx C BB ::: C " " " C B ( f ; ! ) ?  ( u ; ! ) ?

( u ; ! ) n n nxx m 1 3 ; b" = B " " " C BB (f ; ! 1) ? (u1; ! 1) ? (u3; !1xx) C C A @ ::: (f "; ! n) ? (u"1; ! n) ? (u"3 ; ! nxx) 7

(11)

0 (P ("; u" ; ); ! ) 1 1x 2 B C : : : B C " " B C; ( P (  ; u ; ) ; ! ) nx m 2 B q" ( ) = B (P ("; u" ; ); ! ) C 1x C 2 B C @ A :::

(P ("; u"2; ); ! nx) where parentheses denote integration over the region . The following generalization will play a special part in the next section, where local measurements of solutions will be discussed. Let k(") 2 0; N be some integer-valued function. We put 1 " (t) = arg min  2 A k(") + 1

m X

(q"i ( ) ? bi")T S"?1(q"i ( ) ? bi" )

i=m?k(") some m 2 k(") + 1; N and " (t)  " (tk("))

(12)

if t 2 (tm?1; tm] for if t 2 [0; tk(")]. THEOREM 2. Let " () be de ned by (12) with k(") ! 0 as " ! 0. Then the conclusion of Theorem 1 holds. The proof is based on the next proposition which di ers from Proposition 2 in the presence of the small time delay k(") involved into the approximating functions.

PROPOSITION 3. ZT 1 X k(") max `" (t ? i; ; 2A k(") + 1 i =0 0

t?i; ;") ? max `2 (t; ; ) dt ! 0 2B 2 0

as " ! 0. The proof is given in the Appendix. The proof 2 is the same as the one of Theorem 1. We only need to use the P ("of) `Theorem t?i; ;") instead of `" (t; ; t; ;"). ( t ? i; ; term k("1)+1 ik=0 "

REMARK 1. If the set U contains at least one constant with respect to t function, then

the requirement k(") ! 0 can be omitted. Such constant functions can be approximated by elements of X found from the relation

"

N X 1 (q"i ( ) ? bi")T S"?1(q"i ( ) ? bi"): = arg min 2A N + 1 i=0

(13)

THEOREM 3. Let " is de ned by (13). Then distX ( " ; U c ) ! 0 as " ! 0 and  ! 0. Here U c is the set of all elements of X which satisfy (3), being substituted in place of (t). 8

PROOF OF THEOREM 3. From (11), we have N 1 X i i T ?1 i i

N + 1 i=0 (q" ( ) ? b") S" (q" ( ) ? b") =

ZT

`" (t; ;

t; ;")dt:

0

Hence, by Proposition 2, we obtain

Z N 1 X 2 (t; ; )dt i i T ? 1 i i ` ( q ( ) ? b ) S ( q ( ) ? b ) ! max " " " " " N + 1 i=0 2B T

2 0

0

(14)

as " ! 0 uniformly with respect to 2 A (note that N also depends on the generalized parameter "). Let "k k is a sequence of constant with respect to t elements of M constructed like that in the proof of Theorem 1. So, we have

Z ZT A 0

`"k (t; ;

t; ;"k )dt"k k (d ) =

min

2Ak

ZT

`"k (t; ;

t; ;"k )dt:

(15)

0

Because of the compactness of M, we suppose that "k k converge to an element ~ 2 M. One can prove that ~ is also independent from t. Due to Proposition 2, the rst and the second terms of (15) converge to

Z ZT A 0

max ` t; ; )dt~(d ) 2B02

2(

ZT

min max `2 (t; ; )dt; 2 A

and

0

2B02

respectively. From the assumption about the existence of a constant function in U , we conclude (see (3)) that

ZT

max `2 (t; ; )dt = 0; min 2 A and, hence,

0

Z ZT A 0

2B02

max `2 (t; ; )dt~(d ) = 0: 2B02

Now we can nish the proof like that of Theorem 1, considering the following constant with respect to t multivalued function

9 8 ZT = < Q(t)  U c = : 2 A : max `2 (t; ; ') = 0; : '2 B 0 2 0

REMARK 2. If the ifunction ! P (; ; ) is linear with respect to , then q"i ( ) = i 

Q" ; 2 A , where Q" is the corresponding matrix consisting of 2n rows and of dim(A ) columns. In this case, the right-hand-side of (12) is a quadratic form with a positive semide nite matrix. 9

REMARK 3. If the function ! P (; ; ) is linear with respect to and the set A

is convex, then the set U is a convex closed subset of Lq (I ; X ) (for arbitrary q  1). Additionally, if X is a Hilbert space, then there exists an element 0() 2 U with the minimum norm of L2(I ; X ). The next theorem shows how to provide convergence to the function 0(). The formulation of this theorem refers to a function whose existence is declared by the next proposition. function such that k(") ! 0 as PROPOSITION 4. Let k(") 2 0; N be an integer-valued R T " ! 0, then there exists a function (t; ";  ) such that 0 (t; ";  )dt ! 0 as ";  ! 0 and for any function () 2 U , m X 1 i i T ?1 i i k(") + 1 i=m?k(")(q" (  (t)) ? b") S" (q" (  (t)) ? b")  (t; ";  )

if t 2 (tm?1; tm] for some m 2 k(") + 1; N . Here  (t) is the projection of (t) onto the set A . The proof is given in the Appendix. THEOREM 4. Let the conditions of Remark 3 be ful lled, and" let k(") 2 0; N be an integer-valued function such that k(") ! 0 as " ! 0. Let () be found from the following program: " (t) = arg min k kX 2A subject to m X 1 i ( ) ? bi )T S ?1(q i ( ) ? bi )   (t; ";  ) ( q (16) " " " " " k(") + 1 i=m?k(")

if t 2 (tm?1; tm] for some m 2 k(") + 1; N , and " (t)  " (tk(") ) if t 2 [0; tk(")]. Then " () ! 0() in L2(I ; X ) as ";  ! 0. PROOF OF THEOREM 4. Let us prove rst that distL (I ;X )( " (); U ) ! 0: 2

To do this, we use arguments similar to the arguments of the proof of theorem 1. Assume that a sequence "k k 2 M is constructed like that in the proof of Theorem 1 and "k k ! ~. Then, taking into account that m X

(q"i ( ) ? bi")T S"?1(q"i ( ) ? bi") =

i=m?k(")

we have

ZT Z 0 A

k(") X i=0

`" (t ? i; ;

t?i; ;");

t 2 (tm?1; tm]; m 2 1; N;

("k ) 1 kX t?ik; ;"k )"k k (tjd )dt  k("k ) + 1 i=0 `"k (t ? ik ; ;

ZT

 (t; "k ; k )dt ! 0: 0

10

Using Proposition 3, we obtain

ZT Z

0 A

Z ("k ) 1 kX t ? i ; ;" "  k k k k ) (tjd )dt ! k("k ) + 1 i=0 `"k (t ? ik ; ;

T

Z

0 A

and, hence,

ZT Z 0 A

max `2 (t; ; )~(tjd )dt; 2B02

max `2 (t; ; )~(tjd )dt = 0: 2B02

The rest of the proof goes in the same way as in Theorem 1. Assume now that "kk () 6! 0() in L2(I; X ). Then on can nd a sequence "kk () such that k "k k () ? 0()kL (I ;X )  r > 0 and k "kk () ? uk ()kL (I ;X )  1=k for some sequence uk () 2 U . Since U is compact w.r.t. the weak topology of L2(I; X ), we may assume that uk () ! u0() 2 U weakly. Hence, "kk () ! u0() weakly. By the de nition of "kk (), for any u() 2 U , we have k "kk (t)kX  kuk (t)kX ; t 2 I; where uk (t) is the projection of u(t) onto the set Ak . Hence, lim sup k "kk ()kL (I ;X )  lim sup kuk ()kL (I ;X ) = ku()kL (I ;X ): (17) 2

2

k!1

2

k!1

2

2

Applying (17) to the function u0(), we conclude that "kk () ! u0() strongly in L2(I ; X ). With (17) this implies ku0()kL (I ;X )  ku()kL (I ;X ) for any u() 2 U . Therefore, u0() = 0(). This contradiction proves the theorem . REMARK 4. Let the set U contains at least one constant w.r.t t function, then the set c U of all such functions is compact (not necessarily convex). Suppose there exists a unique element 0 2 U c with the minimum norm. For nding this element, one can use the next theorem where the requirement k(") ! 0 is omitted (k(") = N is set). The proof of this theorem is based on the following proposition. 2

2

PROPOSITION 4c. Let the set U contains at least one constant w.r.t t function. Then c there exists a function (";  ) such that (";  ) ! 0 as ";  ! 0 and for any 2 U ,

N 1 X i ( ) ? bi )T S ?1(q i ( ) ? bi )   (";  ): ( q  " " " "  " N + 1 i=0 Here  is the projection of onto the set A . The proof is given in the Appendix.

THEOREM 4c. Let the conditions of Remark 4 be ful lled and the set U contains at " least one constant w.r.t. t function. Let be found from the following program: " = arg min k kX 2A subject to

N 1 X i ( ) ? bi )T S ?1(q i ( ) ? bi )   (";  ): ( q " " " " " N + 1 i=0

11

Then " ! 0 in X as ";  ! 0.

PROOF OF THEOREM 4c. We may consider U c as the subset of"A. cUsing arguments

of the proof of Theorem 3 and Theorem 4, we can prove that distX ( ; U ) ! 0: If " 6! 0, then we can nd a subsequence "k k such that k "kk ? 0kX  r > 0 and k "kk ? uk kX  1=k for some sequence uk 2 U c . Since U c is compact, we may assume that uk ! u0 2 U c in X (note that at this point we had to assume the convexity of U to have the weak compactness of this set). Obviously, "kk ! u0 in X . It follows from the construction of "kk that for any u 2 U c , k "k k kX  kuk kX ; where uk is the projection of u onto Ak . So, we have ku0kX = klim k "kk kX  klim ku k = kukX : !1 !1 k X Therefore, u0 = 0. This is a contradiction.

REMARK 5 Bearing in mind the Kuhn-Tucker theorem, the following heuristic simpli -

cation of the optimization problem of Theorem 4c can be proposed:

"

N X 1 = arg min (q"i ( ) ? bi")T S"?1(q"i ( ) ? bi") + c(";  )k k2X : 2A N + 1 i=0

(18)

Here c(";  ) is a suciently small coecient, whose value should be de ned more precisely from numerical experiments.

EXAMPLE 1. The system (1) was simulated with the following data p(t; ; e) = e ? e3 + e5; f (x) = 1 + 0:5 sin5x; g(x) = 1 + 0:5 cos 5x;  = 1; = 0:1 or 0:01; c0 = 1; k0 = 1; h = 1=21;  = h=10; T = 1: Consider rst the case where the structure of the function p is assumed to be known. That is, we know that p(t; ; e) = 1e + 2e3 + 3e5; and we have to identify unknown coecients 1; 2; 3, and the quantities  and as well. We set 1 = e; 2 = e3; 3 = e5 and apply procedure (12). Figure 1 shows the time performance of the procedure. The exact value of was 0:1 in this case. Now we set = 0:01 and consider the case where the structure of the function p is assumed to be unknown. We put j (; e) = kj e`j ; j = 1; :::; 36; 12

so that we have all terms of the polynomial of two variables of the fth degree. We apply procedure (18). Figures 2 shows the recovered function. Figure 3 depicts the di erence between the exact and recovered functions. Figures 4 and 5 stand for the same example with the di erence that the uniformly distributed random error of the amplitude  =  3 is present. Figure 4 shows the recovered function and Figure 5 depicts the di erence between the exact and recovered functions. 1.5

1; 

1

3

0.5

0 -0.5

2

-1 -1.5 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Figure 1

0.25

5.0E-05

0

-5.0E-05

-0.25

-1.5E-04

0

0 0.1



0.1

0.2

-1

-0.5

0

0.5



1

e Figure 2

13

0.2

-1

-0.5

0

e Figure 3

0.5

1

0.25

0.025

0

0.005

-0.25

-0.015

0

0 0.1



0.1

0.2

-1

0

-0.5



1

0.5

e Figure 4

0.2

-1

-0.5

0.5

0

1

e Figure 5

Local measurements Now we consider the case where the measurements are curried out in a subregion

0h  . This subregion can depend on the parameter h and get smaller when h decreases. For m  k, consider the function m X 1 (q i 0h ( ) ? bi 0h )T S ?0h1(q i 0h ( ) ? bi 0h );

h ;";k; ( ) = k + 1 i=m?k

Gm0

(19)

where the functions q i 0h ( ), the vectors bi 0h , and the matrix S 0h are de ned as follows. Let xi ; xi ; :::; xin0 be the points of the partition of which lie in the interior of 0h. We set 1

2

0 (! ; ! ) : : : (! ; ! ) (! ; ! ) : : : (! ; ! ) 1 i xx i xx in0 xx i xx C BB i xx i xx : : : in0 xx i xx ::: C BB (!i xx; !in0 xx) : : : (!in0 xx; !in0 xx) (!i xx; !in0 xx) : : : (!in0 xx; !in0 xx) C C ; S 0h = B C BB (! ; ! ) : : : (! ; ! ) (! ; ! ) : : : (! ; ! ) C C i xx i xx in0 xx i xx C B@ i xx i xx : : : in0 xx i xx A ::: 1

1

1

1

1

1

1

1

1

1

1

1

1

1

(!i xx; ! in0 xx) : : : (!in0 xx; ! in0 xx) (!i xx; ! in0 xx) : : : (!in0 xx; ! in0 xx) 1

1

0 (f "; ! ) ? (u" ; ! ) ? (u" ; ! ) 1 i 1 i 3 i xx C BB ::: C " " " C B ( f ; ! ) ?  ( u ; ! ) ?

( u ; ! ) i i i xx 0 0 0 m 1 3 n n n ; b 0h = B " " " C BB (f ; !i ) ? (u1; ! i ) ? (u3; !i xx) C C A @ ::: 1

1

1

1

1

1

(f "; ! in0 ) ? (u"1; ! in0 ) ? (u"3 ; ! in0 xx) 0 (P ("; u" ; ); ! ) 1 ix 2 C B : : : B "; u" ; ); !i x) C C B ( P (  m 2 n0 C : q 0h ( ) = B " " B (P ( ; u2; ); ! i x) C C B A @ ::: " " (P ( ; u2; ); ! in0x) 1

1

14

Let us assume that we have chosen an integer-valued function k(") 2 0; N and a function  (k; ") To simplify notations, we set

Gm" ( ) = Gm 0h ;";k(");(k(");") ( ):

(20)

Suppose the function Gm" ( ) has the following properties: G1. There exists a unique element m" 2 A(k(");") such that

Gm" ( m") =

min Gm" ( ):

2A(k(");")

G2. There exists q  1 such that for any 2 A(k(");"),

Gm" ( ) ? Gm" ( m")  K" (t)k ? m"kqX if t 2 (tm?1; tm] for some m 2 k(") + 1; N , where K" (t) is a strictly positive function. We put K" (t)  K"(tk(") ) if t 2 [0; tk(")]. Let us de ne

"(t) = m" if t 2 (tm?1; tm] for some m 2 k(") + 1; N and " (t)  " (tk(")) if t 2 [0; tk(")].

(21)

The next proposition is a particular case of Proposition 4. We give it to de ne a function that will be used in the formulation of the next theorem. PROPOSITION 5. Let k(") ! 0 and  (k("); ") ! 0 as " ! 0, then there exists an R T integrable w.r.t. t function R" (t) such that R" (t)  R" (tk(")) when t 2 [0; tk(")], 0 R" (t)dt ! 0 as " ! 0, and for any function () 2 U ,

jGm" (  (t)) ? Gm" ( m")j  R" (t) if t 2 (tm?1; tm] for some m 2 1; N . Here  (t) is the projection of (t) onto the set

A(k(");").

THEOREM 5. Let "() be de ned by (21). If there exist functions k(") and  (k; ") such that:

i) k(") ! 0 and  (k("); ") ! 0 as " ! 0. ii) RThe function Gm" ( ) de ned by (19),(20) has the properties G1 and G2. iii) 0T R" (t)=K"(t)dt ! 0 as " ! 0, where R" (t) is the function whose existence is declared by Proposition 5. R Then the set U consists of a single function () and 0T k (t) ? "(t)kqX dt ! 0 as " ! 0. PROOF OF THEOREM 5. Let () 2 U . Then, due to the property G2 and Proposition 5, for any t 2 (tm?1; tm]; m 2 1; N , we have

K" (t)k  (t) ? m"(t)kqX  Gm" (  (t)) ? Gm" ( m"(t))  R" (t): 15

Hence,

ZT 0

k (t) ? m"(t)kqX dt 

ZT 0

ZT

R" (t)=K" (t)dt + k (t) ?  (t)kqX dt: 0

k (t) ?  (t)kqX

Since, for any t 2 I , ! 0 and k (t) ?  (t)kqX  2 diam(A), we conclude that the last integral goes to zero as  goes to zero. This proves the theorem. REMARK 6. If it is known a priori that the set U contains at least one time-independent function, then the requirement k(") ! 0 of Proposition 5 and, hence, of Theorem 5 can be omitted. It is meaningful to set k(") = N . Let us discuss assumptions G1 and G2 in the case of linearity of the function ! P (; ; ). In this case the functions Gm" ( ) become quadratic forms, i.e.

Gm" ( ) = T D"m + dm" T + c"m; 2 A(k(");"); where

D"m =

m X 1 i T ?1 i k(") + 1 i=m?k(") Q" S 0h Q":

All matrices Qi"T S ?0h1Qi" are positive semi-de nite. Each of them may be singular but if they di er one from other, the matrix D"m may be positive de ned and well conditioned. Note that the number of terms in the right-hand-side of the last equality may increase as " ! 0. The following can be recommended in the case where the set U contains at least one time-independent function: One should set k(") = N and choose  (") so that dim A(")