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ON WELL-POSEDNESS OF INTEGRO-DIFFERENTIAL EQUATIONS IN WEIGHTED L2?SPACES by J.A. Burns and K. Ito January, 1990

CAMS #90-3

Center for Applied Mathematical Sciences

University of Southern California Denney Research Building 318 University of Southern California Los Angeles, CA 90089-1113 Telephone: (213) 743-8977

On Well-posedness of Integro-di erential Equations in Weighted L2-spaces John A. Burns1 Interdisplinary Center for Applied Mathematics Department of Mathematics Virginia Polytechnic Institute and State University Blacksburg, VA 24061 and Center for Applied Mathematical Sciences University of Southern California Los Angeles, CA 90089-1113 Kazufumi Ito2 Center for Applied Mathematical Sciences Department of Mathematics University of Southern California Los Angeles, CA 90089-1113

Abstract In this paper we consider the problem of constructing a well-posed state space model for a class of singular integro-di erential equations of neutral type. The work is motivated by the need to develop a framework for the analysis of numerical methods for designing control laws for aeroelastic and other uid/structure systems. Semigroup theory is used to establish existence and well-posedness results for initial data in weighted L2?spaces. It is shown that these spaces lead naturally to the dissipativeness of the basic dynamic operator. The dissipativeness of the solution generator combined with the Hilbert space structure of these weighted spaces make this choice of a state space more suitable for use in the design of computational methods for control than

previously used product spaces.

1This research was supported in part by the Air Force Oce of Scienti c Research under

grant AFOSR 89-0001 and the Defense Advanced Research Projects Agency under contracts F49620-87-C-0116 and N00014-88-K-0721. 2This research was supported in part by the Air Force Oce of Scienti c Research under grant AFOSR 90-0091 and the National Science Foundation under grant DMS-8818530.

1

1. Introduction.

The development of an appropriate state space model for the analysis and design of control laws is often a crucial step in the overall process of constructing computational methods for control. In particular, the form of the feedback control law that results from typical linear quadratic regulator design is completely determined by the choice of state space. During the past ten years considerable e ort has been devoted to the problem of constructing well-posed state space models for neutral functional di erential equations with non-atomic D operator. This work was primarily motivated by the problem of developing computational algorithms for active utter control (see [BCH], [BHT]). General well-posedness results were obtained in [BHS], [BHT] and [KZ] using product spaces of the form X p = R  Lp(?r; 0; R); 1 < p < 1: However when the results in [BHT] are applied to the utter problem, the parameter p must be chosen so that p > 2; i.e. X p is not a Hilbert space. Given that the ultimate goal of the work was to develop practical computation methods for control, the fact that the space X is not a Hilbert space complicates the problem of constructing and analyzing numerical algorithms. Therefore, one is lead naturally to the problem of nding a state space model that has the following three properties: (i) The state space is a Hilbert space Z: (ii) The problem of developing nite dimensional approximating spaces Z N is reasonable. (iii) The relationship between the states z 2 Z and the solutions to the governing functional di erential equation is as simple as possible. In this paper we show that certain weighted L2?spaces can be used as state spaces for such systems. We use linear semigroup theory [P] to establish wellposedness. Numerical schemes based on this framework will appear in [IT]. We consider several questions concerning the well-posedness of the initial value problem d Z 0 g(s)x(t + s)ds = Z 0 d(s)x(t + s) + f (t); t > 0 (1:1) dt ?r ?r with initial data (1:2) x(s) = (s); ?r < s < 0: We are interested in the case where 0 < r  +1 and g is non-negative, non-decreasing and weakly singular. 2

We assume that g() 2 L1(?r; 0) and weakly singular at s = 0. The measure () is assumed to be of the form

(s) = 0(s) ?

0

Z

s

a( )d

where 0() is of bounded variation on (?r; 0) and a() is locally absolutely continuous on (?r; 0). Additional restrictions will be placed on 0 () and a() in order to ensure the well-posedness of (1.1) in an appropriate space of initial data. Although the results in this paper apply to very general systems, it is helpful to think of g() as an Abel kernel, i.e. g(s) = jsj? for 0 < < 1 on a nite interval [?1; 0]. Moreover, equations of the form (1.1) with Abel kernels g() are found to have applications to several problems in aerodynamics and, in particular, are useful in modeling certain aeroelastic control systems (see [BCH]). The initial value problem (1:1) ? (1:2) can be written as (1:3)

Z

0

?r

g(s)x(t + s) ds =

Z

0

?r

g(s) (s) ds +

t Z 0

Z

0 0

?r



Z

t

d(s) x( + s) d + f ( )d 0

provided that the functions t ! f (t); t !R ?r d(s) x(t + s) are integrable, g()() belongs to L1(?r; 0) and the function t ! ?0r g(s) x(t + s) ds is absolutely continuous. There are cases of practical interest when the solution of the problem R

(1:4)

Z

0

?r

g(s) x(t + s) ds =  +

Z

t Z 0

0

?r



d(s) x( + s) d +

Z

t

0

f ( )d

with  6= ?0r g(s)(s)ds is also important. Thus, we shall consider both problems. It is helpful to introduce the operators D and L by setting D = R?0r g(s)(s)ds and L = R?0r d(s)(s); respectively. Using rather standard notation (1.4) can be written as R

t

Z

Dxt =  + fLx + f ( )gd

(1:5)

0

where xt : (?r; 0) ! R is de ned by xt(s) = x(t + s): Note that (1.5) is equivalent to (1.3) if D = : Moreover, since g() 2 L1(?r; 0); it follows that D : L2g ! R is a bounded linear functional satisfying

jDj 

Z

0

jg(s)j ds ?r

1=2 Z 0



?r

3

j(s)j2g(s) ds



1=2

:

The well-posedness of (1.5) has been studied by several authors. In [BHT] it was shown that under rather general conditions on D and L; (1.5) is well-posed in the sense that the solution exists and depends continuously on the initial data (; ()) in R  Lp(?r; 0) and forcing function f in Lp(0;  ) for  > 0: Most of these studies have been based on an explicit representation of the solution to Abel's equation. Moreover, as noted above when these results were applied to certain aeroelastic control problems it was shown that (1.5) is well-posed in RLp(?r; 0) only if p > 2 (see [BHS,BHT,KZ] for details). We shall investigate the well-posedness of (1.5) in the (weighted) Hilbert space L2g = L2(?r; 0; gds): In order to introduce the basic idea, we indicate how (1.1) can be (formally) written as an evolution equation on the state space L2g : Given x : (?r; +1) ! R; we de ne  : [0; +1)  (?r; 0) !R R by (t; s) = x(t + s). If x is locally absolutely continuous and the function t ! ?0r g(s) x(t + s)ds is (locally) absolutely continuous, then (1.1) can be written as the evolution equation in L2g (1:6) with the constraint that Z

(1:7)

0

?r

@ (t; s) = @ (t; s) @t @s

@ (t; s)ds = 0 d(s)(t; s) + f (t): g(s) @s Z

?r

Equation (1.7) is obtained by formally passing the time derivative inside the integral and then using (1.6). The system (1.6){(1.7) can be re-written in terms of a dynamical system on L2g by de ning an appropriate system operator. We use LAC (a; b) to denote those functions : (a; b) ! R that are locally absolutely continuous on (a; b): If (a; b) = (?r; 0), then we abbreviate and write LAC: Let A denote the operator with domain (1:8)

D(A) = f 2 L2g = 2 LAC; 0 2 L2g ; D0 ? L = 0g

and for  2 D(A) (1:9)

d : A = 0 = ds De ne the operator B : R ! L2g by

(1:10)

[Bf ](s) = b1(s)f  bf;

where 1(s) is the constant function 1 anf b = 1=L(1()): 4

We are interested in the Cauchy problem in L2g de ned by (1:11)

z_ (t) = A(z(t) + Bf (t)); t > 0

with initial data (1:12)

z(0) = :

L(1) 6= 0 is assumed in the de nition (1.10) of B: The case when L(1) = 0 is discussed in Theorem 3.2. As it will be shown in Corollary 3.3. hABf; iL2 = (0) for all  2 dom (A) where A is the adjoint operator of A (see, Lemma 3.4). Hence in this sense ABf is the delta distribution at s = 0: The goal of this paper is to show that, under mild assumptions on g and ; A generates a C0 -semigroup on L2g and solutions of (1.1){(1.2) are equivalent to solutions at (1.11){(1.12). We shall also consider the case described by (1.4) of inconsistent initial data. The paper is organized as follows. We rst consider the simple case where 0 = 0 and L = 0. Here we can establish the dissipativeness of A. This argument is then extended to certain problems for which L 6= 0. Consideration of the non-homogeneous problem follows and more general measure  are allowed. The case of inconsistent initial data is discussed in the last theorem. We shall use the following de nition of solutions to (1.1){(1.2). This de nition is analogous to the de nition of a generalized solution given in [BHT]. g

De nition 1.1 A solution to (1.1) - (1.2) is a measurable function x : (?r; 1) ! R satisfying (i) x(s) = (s) a.e. on (?r; 0) (ii) xt() 2 L2g for t  0 (iii) x satis es (1.3) on 0  t < +1.

5

2. Special Cases

We shall assume that 0 < r  +1 and the functions a(s); g(s) satisfy (H1) g(s) > 0 a.e. on (?r; 0) (H2) g() 2 L1(?r; 0) and g(s) " 1 as s ! 0? 1 (?r; 0) with g 0 (s)  0 a.e. on (?r; 0) (H3) g() 2 HLOC h i (H4) ga(()) 2 L2g .

Note that we require that g() be L1(?r; 0), but g() can have a weak singularity at s = 0. Conditions (H1){(H3) can be relaxed for certain special problems.

REMARK 2.1 Conditions (H1){(H3) imply that there exist constants

r0; 0 < r0 < r, and 0 > 0 such that (2:1) g(s)  0 > 0 a.e. on [?r0; 0): Therefore it is straight forward to show the existence of a constant K > 0 such that Z

(2:2)

T

0

jx(s)j2 ds  K

(

sup

Z

0

0tT ?r

jxt(s)j2 g(s) ds

)

for all functions x : (?r; T ) ! R satisfying x 2 L2(0; T ) and xt 2 L2g for 0  t  T .

REMARK 2.2 If r < +1 is nite, then condition (H1) can be relaxed provided that g(s) is bounded from below. In particular, if g(s)  ?c, then de ne g~(s) = g(s)+c  0 and write (1.1) as

d 0 g~(s) x(t + s) ds = 0 d~(s) x(t + s) + f (t) (2:3) dt ?r ?r where ~(s) = c]0;0] ? c]?r;0] + : In this case g~() will satisfy (H1)-(H4) and the results presented below will apply. Z

Z

REMARK 2.3 If one de nes g^(s) = e!sg(s) for ! > 0, then x will satisfy (1.1) if

y(t) = e?wtx(t) satis es 0 0 g^(s) y(t + s) ds = [?!g^(s) ds + e!sd(s)]y(t + s) + e?!tf (t) (2:4) dtd ?r ?r Z



Z

6

with y(s) = e?!s x(s) for ?r < s < 0. Note that g^0(s) = e!s(!g + g0): Thus, if gg0 is bounded below, then one can choose an ! > 0 such that g^0  0 on (?r; 0); i.e., (H3) holds for g^:

REMARK 2.4: If g 62 L1(?1; 0), then we may proceed as follows. Suppose x(t); t > ?r is the solution to Z

0

0

g(s)(s)ds +

?r the function y(t) = e?!tx(t);

?r

De ne

Z

g(s) x(t + s)ds =

e!t Z

0

?r

g(s)e!s

(s) ds +

Z

0

?r

Z

0

t Z 0

?r



d(s)x( + s) + f () d:

t  ?r for ! 2 R. Then y satis es g(s)e!s y(t + s) ds =

 t ! Z 0 !s ? ! e e d(s)y( + s) + e f () 0 ?r

Z

d

R Where y(s) = e?!s (s) for ?r < 0 < 0. If (t) = ?0r g(s) y(t + s) ds and 0 = R0 ?r g (s) (s) ds, then we obtain  Z Z t 0 !s ? ! ? ! ( t ? s ) ? !t 0 e d(s)y( + s) + e f () d: (t) = e  + e ?r 0

Thus, y satis es (2.5)

= ?!

d 0 g^(s)y(t + s) ds dt ?r Z

Z

0

?r !s e g(s).

g^(s)y(t + s) ds +

Z

0 !s e d(s)y(t + s) + e?!tf (t):

?r

where g^(s) = Assume that there exists a constant ! > 0 such that g^ 2 1 L (?1; 0). Then, Theorem 3.2 shows that (2.5) has the unique solution y(t + ) 2 C ([0; T ]; L2g^) provided that the initial condition = e?!s  satis es (2:6)

Z

0

?1

0 geq?!s jj2 ds g^ j j2 ds = Z

?1

and that f 2 L2(0; T ). Suppose y is the said solution and de ne x(t) = e!ty(t); t > ?1. Then, x(t); t > ?1 satis es (1.1) using exactly the same arguments as above. This implies that (1.1) has the solution x(t); t > ?1 provided (2.6) and f 2 L2(0; T ) and x(t + ) 2 C ([0; T ]; L2g) where g = geq?!s. 7

We begin by considering special cases for the form of A as de ned by (1.3){(1.9).

THEOREM 2.1 Assume that (H1)-(H4) hold. If  = 0, then A is dissipative and

generates a strongly continuous contraction semigroup on L2g . Proof: If  2 D(A), then < A;  > = RR?0r < 0(s); (s) > g(s) ds R = R?0r < 0(s); (s) ? (0) > g(s) ds + ?0r < 0(s); (0) > g(s) = ?0r < 0(s); (s) > g(s) ds; where (s) = (s) ? (0); ?r < s < 0. For each k;  satisfying ?r < ?k  ?r0  ? < 0 de ne the integral

I;k =

? < 0(s); ?k

Z

(s) > g(s) ds:

Since g 2 H 1[?k; ?], we have 1 R ? d j (s)j2g (s) ds I = ;k 2 ?k ds R (2:7) = 12 fj (?)j2g(?) ? j (?k)j2g(?k)g ? 21 ??k g0(s)j (s)j2 ds: R R Also, (?) = (0) ? ?0 0(s)ds = ? ?0 0(s) ds so it follows that 2 Z 0 0 2 (s)ds g(?)j (?)j = g(?)

?

2 Z

 g(?) ? j 0(s)j g(s) ds g(s) 4

0

q

q

 g(?) ? gds (s) Z



Z



Z

0

0 g (?)

?

0

?r

!

g(s) ds

! Z

Z

r

0

2

3 5

0

j 0(s)j2g(s)ds ?



j 0(s)j2g(s)ds

j 0(s)j2g(s)ds ! 0

as  ! 0+ . Note that we used the fact that g is non-decreasing and g(s)  0 > 0. Since, I;k < A;  >= lim !+ 

o

k!r?

8

and

I;k  2 ?0r j0(s)j2g(s)ds ? 21 j (?k)j2g(?k) + ? k j (s)j2g0(s)ds it follows that < A;  > 0 and A is dissipative. Moreover, if r is nite, 0 (2:8) < A;  > = ? 21 g(?r)j (?r)j2 ? 21 ?r g0(s)j (s)j2ds Next consider the resolvent equation for A (2:9) (s) ? 0(s) = h(s) 2 L2g and h

R

R

i

Z

(2:10)

Z

0 = D =

0

?r

g(s)0(s)ds:

Solving (2.9) we obtain the representation

(s) = es(0) +

(2:11)

Z

s

For Re > 0 R0 0 (s?)h( )j2 ds ?r g (s)j s e

R



Z



Z

0 (s?) e h()d;

0

?r

0

j

Z

s

?r < s  0:

0 (s?)Re q e g()jh()jdj2 ds

0 (s?)Re Z 0 (s?)Re g()jh()j2 dds d) e ( e Z

?r s

s

Z 0 Z s 1  Re ?r ( ?r e(s?)Reds)g()jh()j2 d Z 0 1  (Re)2 ?r g()jh()j2 Thus, the boundary condition (2.4) becomes

(2:12)

Z

0

?r

esg(s)ds



(0) =

Z

0

?r



g(s) h(s) ?

Z

s

0

es? h()d



ds:

These imply that  2 (A) = P(A) satisfying Re  0 if and only if ?0r esg(s)ds = R0 0. RSince  ! ?r esg(s)ds is continuous on [0; 1), there exists a 0 > 0 such that 0 ?0r e0sg(s)ds > 0. Moreover, from (2.10) and (2.11) there exists M > 0 such that j(0I ? A)?1hjL2  M jhjL2 for all h 2 L2g R

g

g

9

Hence one can conclude by the Lumer-Phillips theorem (e.g. Theorem 4.3 and 4.6 in [P, Chap. 1] ) that if (H 1) ? (H 3) are satis ed, then A generates a C0-semigroup on L2g . The dissipative argument in the proof of Theorem 2.1 can now be applied to a certain case where RHS 6= 0.

COROLLARY 2.2 Assume that r is nite, 0(s) = a0]0;0] + a1]?r;0] and g satis es: (H 3) g0  c g on (?r; 0) for some c > 0 Then, if 2 0 a (2:13) a0 + a1 + 2g(?1 r) ? 2c ( g(s)ds) < 0; ?r then there exists !  0 such that A ? !I is dissipative. Proof: Note that for  2 dom(A) Z

< A;  >=

Z

0

?r

< (s); (s) > g(s)ds + (0)(a0(0) + a1(?r) +

Z

0

?r

a(s)(s)ds)

Thus, from (2.8) Z 0 = ? 21 g(?r)j (?r)j2 ? 21 g0(s)j (s)j2ds

?r



+(0) (a0 + a1)(0) ? a1 (?r) + Here for  > 0 Z

(

0

?r

g(s)ds)j(0)j2 =

Z

0

?r

Z

0 ?r

a(s)(s)ds



g(s)j(s) ? (s)j2ds

 (1 + 1 )

Z

0

g (s)j(s)j2 + (1 + ) ?r

Z

0

g (s)j (s)j2ds ?r

Z 0 Z 0 1 +  1 +  2   ?r g(s)j(s)j + c ?r g0(s)j (s)j2ds

Hence,

2 0 < A;  > a0 + a1 + 2g(a?1 r) ? 2(1 c+ ) g(s)ds j(0)j2 ?r Z

10

!

1=2 2 a ( g )ds j(0)j jjL2 + 2c jj2L2 + ?r If (2.13) is satis ed, then there exists !  0 such that < A;  > !jj2L2 for all  2 dom(A). Next, we consider the case with non-zero right hand side. To this end, rst we study the equation of the form d 0 g(s) x(t + s)ds = ?x(t) + f (t): (2:14) dt ?r In this case, < A;  > ?j(0)j2 for  2 dom(A). Using exactly the same arguments as in the proof of Theorem 2.1, one thus can show that A associated with (2.14) generates a C0-semigroup S (t); t  0 on L2g . Z

0

!

g

g

g



Z

THEOREM 2.3: For  2 L2g and f = 0, there exists a unique solution x(t); t  0

to (2.14) and we have

(S (t)) (s) = x(t + s) on (?r; 0) for t  0:

Proof: Suppose  2 dom(A2); z(t) = (t; ) = S (t) 2 dom(A2); t  0 and it satis es

@ (t; s) = @ (t; s) in (?r; 0) @t @s 0 @ (t; s)ds = ?(t; 0): g (s) @s ?r Since z(t) 2 C 2(0; T ; L2g ) for any T  0, 8 > > > > >
Z > > > > :

(2:15)

0 (t; 0) = ? dtd g(s)(t; s)ds 2 H 1(0; T ): Z

?r

Thus, (t; s); t  0 satis es the initial-boundary value problem @t@  = @s@  in (?r; 0) and (t; 0) = \given H 1-function"; so that there exists a x 2 H 1[?r; T ] such that

(t; s) = x(t + s); t  0 and ? r  s  0: From (2.15), (2:16)

d ( 0 g(s) x(t + s)ds) = ?x(t); t  0: dt ?r Z

11

Since dom(A2) is dense in L2g , there exists a sequence fng in dom(A2) such that jn ? jL2 ! 0 as n ! 1. Suppose xn(t + s) = n(t; s) = (S (t)n) (s); t  0, then from (2.16) Z t Z 0 Z 0 g(s) xn(t + s)ds = g(s)n(s)ds ? xn(s)ds: g

0

?r

?r n jS (t) ? S (t)jL2g

! 0 as n ! 1, uniformly on [0; T ], there exists a unique Since 2 limit x(t + ) 2 C (0; T ; Lg) that satis es (S (t))(s) = x(t + s) (?r; 0); t  0 and

Z

0 ?r

g(s)x(t + s)ds =

Z

0

?r

g(s)(s)ds ?

Z

0

t

x(s)ds:

Conversely, suppose x(t) is a solution to (2.14). De ne z(t) = x(t + ) 2 L2g : Then z(t); t  0 satis es t

Z

z(t) =  + A 0 z(s)ds; i.e. z(t) is the weak-solution to (1.11). In fact, if  = 0t z()d : i.e.,

(2:17)

R

Z

t

(s) = 0 x( + s)d s 2 [?r; 0] then  is locally absolutely continuous with d  = x(t + s) ? x(s) 2 L2 : g ds Moreover, since x(t) satis es (2.14), we have 0 d ds = ? t x()d = ?(0): g(s) ds 0 ?r Hence,  2 dom(A) and A = z(t) ?  which implies z satis es (2.17). Since the weak solution to (1.11) is unique, (2.14) has a unique solution. Now, we consider the case when  2 L2g and f 2 L2loc (0; 1). Z

Z

THEOREM 2.4: For any  2 L2g and f 2 L2loc (0; 1) there exists a unique solution x(t); t  0 to (2.14) and it satis es 

Z

t



jx(t + )j2L2  jj2L2 + 0 jf (s)j2ds ; t  0 g

g

12

If we de ne the function z(t) = x(t + s) 2 L2g , then

z(t) = S (t) + A

t

Z

0

S (t ? s) f (s)ds

1 (0; 1), then z (t) is the strong and moreover if  + f (0) 2 dom(A) and f 2 Hloc solution to d z(t) = A (z(t) + f (t)) ; t > 0 dt where = ?1(constant) 2 L2g .

Proof: Without loss of generality, one can assume  = 0 First consider f (t)  f (constant). Note that (2:18)

A

t

Z

0

S (t ? s)  ds = S (t) ?  for  2 L2g :

R Let  2 L2g be the solution to boundary-value problem: A = 0 = 0 with ?0r g(s)0(s)ds+ (s) = ?f ; i.e. (s) = ?f (constant). De ne the function ^(t; ) = S (t) ?  2 L2g ; t  0:

Then ^(t; ) 2 L2g satis es

@ ^ ? @ ^ = 0 @t @s 0 @ ^(t; s)ds = ?^(t; 0) + f: g(s) @s ?r In fact, if x^(t + ) = S (t), then from Theorem 2.3 x^ satis es 8 > > > > >
Z > > > > :

Z

0

?r

g(s)^x(t + s)ds =

Z

0

?r

g(s)(s)ds ?

Z

t

0

x^(s)ds:

Thus, if we de ne x(t) = x^(t) + f; t  ?r; then x satis es d Z 0 g(s)x(t + s)ds = ?x(t) + f: dt ?r where x(t + s) = S (t) ?  2 L2g . Moreover, since 1 d j^j2  ^(0)(?^(0) + f )  1 jf j2; 2 dt L2 2 g

13

we have

j^(t; )j2L2  tjf j2; t  0: g

Next, we assume f (t) is given by

f (t) =

n

X

i=1

fi[ti?1; ti)

with 0 = t0 < t1 < : : : < tn = T . Applying the above arguments, successively on the intervals [ti?1; ti); i = 1; 2; : : : ; n, we obtain the function x de ned by (2:19)

x(t + s) =

m

" X

i=1

#

fi (S (t ? ti?1) ? S (t ? ti) ) (s)

for t 2 [tm; tm+1), satis es d Z 0 g(t)x(t + s)ds = ?x(t) + f ; t  t < t (2:20) i i?1 i dt ?r and we have

jx(t + )j2L2 

(2:21)

g

m

X

i=1

jfij2jti ? ti?1j2; t  0:

Finally, for any f 2 L2(0; T ), there exists a sequence of step function ff n g such that jf n ? f jL2 (0;T ) ! 0 as n ! 1. Let xn (t; ) be the corresponding solution to f n de ned by (2.19). Then from (2.21), t ! x(t + ) 2 L2g de nes a Cauchy sequence in C [0; T ; L2g ). Thus

jxn(t + ) ? xm(t + )jL2 ! 0 as n; m ! 1 g

uniformly in t on [0; T ]. Hence there exists a unique limit

x(t + s) = nlim xn(t + s); "1 t  0 and ? r  s  0 and x satis es Z

0

?r

and moreover,

g(s)x(t + s)ds = ?

Z

Z

t

0

t

(x(s) ? f (s)) ds; t  0

jx(t + )j2L2  0 jf (s)j2ds; t  0 g

14

when x(t + ) 2 C [0; T ; L2g). From (2.18) and (2.19) t

Z

xn(t + ) = A 0 S (t ? s) f n(s)ds Since A is closed, the above convergence implies that 0t S (t ? s) f (s)ds 2 dom(A) and R

(2:22)

x(t + ) = A

Z

0

t

S (t ? s) f (s)ds

for all f 2 L2(0; T ). Moreover, suppose f 2 H 1(0; T ), then

z(t) = x(t + ) = S (t)( + f (0)) ? f (t) +

t

Z

0

Thus, if  + f (0) 2 dom(A), then

S (t ? s) f 0(s)ds

z(t) + f (t) 2 C (0; T ; dom(A)) and

A(z(t) + f (t)) = A S (t)( + f (0)) + A

t

Z

0

S (t ? s) f 0(s)ds:

Similarly, one can show that dtd z(t) exists for all t > 0 and dtd z(t) = A(z(t) + f (t)).

15

3. The General Case

Let us now consider the general equation: d Z 0 g(s) x(t + s)ds = Z 0 d(s) x(t + s) + f (t) (3:1) dt ?r ?r R where  = 0 ? s0 a()d, 0 is of bounded variation on (?r; 0) and we assume Z 0 jd0(s)j < 1 (H 5) ?r g (s) From Remark 2.2, if r is nite, then without loss of generality one can assume that g  g(?r) > 0 on [?r; 0). Thus, in this case, (H5) is equivalent to 0 being of bounded variation on (?r; 0).

THEOREM 3.1 For any T > 0,  2 L2g and f 2 L2(0; T ) (3.1) has a unique solution x(t); t  0 and there exists a K , which depends only on T , such that (3:2)

jx(t + )jL2  K g

 Z t 2 2 jjL2g + 0 jf (s)j ds



Proof: Let M be the set de ned by M = fx(t + ) 2 C (0; T ; L2g) and x(s) = (s) on (?r; 0)g: De ne the solution map T on M by y = Tx where y is de ned as the solution to d Z 0 g(s) y(t + s)ds = ?y(t) + x(t) + Z 0 d(s) x(t + s) + f (t) dt ?r ?r with y(s) = (s); s < 0. We will show that for x 2 M ,

t ! x(t) +

Z

0

?r

d(s)x(t + s) 2 L2(0; T ):

From Remark 2.1, there exists a r0 > 0 such that g  1 on [?r0; 0). Thus, x 2 M implies x 2 L2(0; T ). From (H4), for x 2 M , Z

0

?r

a(s)x(t + s)ds 2 C (0; T ):

Suppose r = 1. Then for any R > 0, Z

0

?R

d0(s)x(t + s) 2 L2(0; T ) 16

and we have T Z 0

Z

0



?R



2

d0(s)x(t + s) dt 

jd0(s)j ?R

0 0

Z

0

Z

j d0(s)j jx(t + s)j2dt ?R

0

Z

jd0 (s)j ?R

jd0 (s)j ?R

=

Z

0

+ Z



0

?R

0

Z

Z

0

jd0(s)j ?R

jd0 (s)j

"

Z

T

jx(t + s)j2dt

0

j d0 (s)j( jdg(0s()s)j ?R ?R

Z



(3:3)

TZ0

Z

Z

(T +s)^0

g()j()j2 d

s

(T +s)_0 jx()j2d) 0

Z

0

?R

jd0 (s)j jj2 2 +

Z

Lg

g(s)

0

?R

jd0(s)j jxj2 2

L (0;T )

Similarly, for T  R1 < R2 Z

?R1

?R2

Thus, (3: 4)Z

nR

0

?1

d0(s)x(t + s)

0

o

?k d0 (s)x(t + s) k1

d0(s)x(t + s)



2 L2 (0;T )



2 L2 (0;T )



Z

?R1

?R2

jd0(s)j

Z

?R1

?R2

jd0(s)j jj2 2 : g(s)

Lg

is a Cauchy sequence in L2(0; T ) and hence from (3.3)



Z

0

?1

jd0(s)j

" Z

jd0 j jj2 2 +

0

g

?1

Lg

Z

0

?1

#

jd0 j jxj2 2

L (0;T )

For r nite, we have (3:5)

Z

0

?r

d0(s)x(t + s)

2



L2 (0;T )



Z

0 ?r

jd0(s)j

2

 "

1 jj2 + jxj2 L2(0;T ) g(?r) L2

#

g

Hence it follows from Theorem 2.4 that Tx 2 M for all x 2 M . De ne the sequence fxk g in M by xk+1 = Txk where x0(t) = 0; t > 0. Then, from Theorem 2.4 (3:6)

Z

t

jx1(t + )j2L2  jj2L2 + 0 jf (s)j2ds: g

g

Now, let yk = xk+1 ? xk for k  1 then yk satis es  Z Z 0 0 d k?1 k k k ? 1 (3:7) dt ?r g(s)y (t + s)ds = ?y (t) + y (t) + ?r d(s)y (t + s) 17

#

:

with yk (s) = 0; s < 0. For t < r, it follows from Theorem 2.4 and (3.4){(3.5)

jyk (t + )jL2g

"

3 1+

jd0 (s)j ?r

0 a2(s)

Z

+3

0

Z

3 1+

Z

!

0

jd0 (s)j ?r

Z

0 Z

0 a2

!

2

t k?1 2 jy (s)j ds

tZ 0

g(s) ds 0

?t

"

2

 #Z

 #

?

g(s)jyk?1( + s)j2dsd

1 jyk?1(t + )j2 L2 g(?t) g

max jyk?1(s + )j2L2g s2[0;t]

!

g ds where we used g(s)  g(?t); s  ?t. Since g(?t) " 1 as t ! 0+ , for t0 suciently small jyk (t + )jC(0;t0;L2 )  21 jyk?1(t + )jC(0;t0;L2 ); k  2: This implies that fxk (t + )gk1 is a Cauchy sequence in C (0; t0; L2g ) and hence from (3.7) its unique limit +3t

?t

g

g

x(t + ) = lim xk (t + ) as k ! 1 in C (0; t0; L2g ) satis es (3.1) on t 2 [0; t0]. From Theorem 2.4 and (3.4){(3.6), there exists a constant M1 > 0 (independent of  2 L2g and f 2 L2loc ) such that

jy1(t + )j2C(0;t0;L2 )  x1(t) + ?0r d(s)x1(t + s) L2(0;t0)  M1 jj2L2 + 0t0 jf (s)j2ds g





R



R



g

Since we have

jx(t + )jC(0;t0;L2 )  2jy1(t + )jC(0;t0;L2 ) + jx1(t + )jC(0;t0;L2 ) g

g

 Z t 0 2 2 C (0;t0;L2g )  (2M1 + 1) jjL2g + 0 jf (s)j ds :

g



jx(t + )j2

Since (3.1) is linear and time-invariant in x, one can extend the interval of existence to arbitrary bounded interval [0; T ], employing the above construction of solutions, successively on the intervals [jt0; (j + 1)t0], j  1.

18

The following theorem proves the equivalence of solutions to the equation (3.1) and the Cauchy problem (1.11).

THEOREM 3.2: Let x(t); t  ?r be the solution to (3.1) with f  0 and de ne the solution semigroup S (t); t  0 on L2g by (S (t))(s) = x(t + s) on (?r; 0):

(3:8)

Then, S (t); t  0 forms a C0-semigroup on L2g whose in nitesimal generator A is given by dom(A) = f 2 L2g :  is locally absolutely continuous with 0 2 L2g and (3.9)

Z

0

0

Z

d(s)(s)g and A = 0 for  2 dom(A). Moreover,  2 (A) if and only if () 6= 0, where the characteristic function () is given by ?r

(3:10)

g(s)0(s)ds =

() = 

If  2 (A) \ R+ and can be represented as

Z

0

?r

g(s)esds ?

?r  = ()?1 es

Z

0

?r

t

Z

0

Proof: Since x(t) satis es (2.14) with f (t) = x(t) +

0 ?r

d(s)es:

2 L2g , then the solution x(t + ) 2 L2g to (3.1)

x(t + ) = S (t) ? (A ? I )

(3:11)

Z

S (t ? s) f (s)ds:

d(s)x(t + s) 2 L2(0; T );

it follows from Theorem 2.4, the semigroup S (t) de ned by (3.8) is strongly continuous. Let  2 dom(A) and choose  2 R+ suciently large. Then for some 2 L2g

 = (I ? A)?1

=

which implies

(s) =

Z

Z

1 ?t e S (t)

0

Z 1 1 ?t e x(t + s)dt = es e? x( )d

0

0

19

?

dt Z

s

0 (s?) e ()d

where x(t); t  ?r is the solution to (3.1) with initial data and f  0. This equality shows that  is locally absolutely continuous on (?r; 0] and 0 =  ? 2 L2g . Thus, A = 0 and Z

0 ?r

g(s)0(s)ds = = = =

1

Z

0 1

Z

0 0

Z

e?t d(s)

?r Z

e?t

0

d(s)

?r

0

Z

?r Z

g(s)x(t + s)ds ?

tZ 0

0 ?r Z

Z



0

?r



g(s) (s)ds dt



d(s)x( + s)d dt

1 Z 1

0

Z



e?tdt x( + s)ds

Z 0 1 ? e x( + s)ds = d(s)(s): ?r 0

Hence we have proved that dom(A) is contained in the set D de ned by (3.9) Conversely, suppose  2 D. For  suciently large =  ? 0 2 L2g and  = (I ? A)?1 2 dom(A); i.e., = (I ? A). Then, Z

 ? 0 =  ? 0

Z 0 Z 0 Z 0 0 0 g(s) (s)ds ? d(s)(s) = g(s)(s)ds ? d(s)(s): ?r ?r ?r ?r

0

These two equations imply

(s) = (s) ? (s) = es(0) and



()(0) = 

Since for  > 0 and " > 0,



Z

0 ?r

Z

0 ?r

g(s)esds ?

Z

0

?r

d(s)es



(0) = 0:

g(s)e?sds  g(?")(1 ? e?");

for  suciently large, () 6= 0 and therefore  =  2 dom(A). Exactly the same calculation as above shows that  2 (A) ifand only if () 6= 0. For  2 (A) \ R+, let y(t; s) = e?tx(t + s); t  0 on (?r; 0) where 20

x(t); t  ?r is the solution to (3.1). Then, y(t; ); t  0 formally satis es d y(t; ) = (A ? I )y(t; ) in (?r; 0) dt 0 @ y(t; s)ds = 0 d(s)y(t; s) + e?tf (t): g (s) @s ?r ?r Using the same arguments as in the proof of Theorem 2.4, one obtains Z

Z

y(t; ) = e?tS (t) ? (A ? I )

Z

0

t ?(t?s) e S (t ? s)

?s e f (s)ds

which means (3.11). (A ? I )  in (3.11) is the delta distribution at s = 0 in the following sense.

Corollary 3.3: For  2 dom(A) (3:12) ?h(A ? I ) ; i = (0) where



is de ned as in Theorem 3.2.

In order to prove Corollary 3.3 we need

Lemma 3.4: The adjoint operator A of A is given by d (g)(s)( (s) ? (0)) ? (s) (0)) A = ? g1 ds with domain dom(A) = f 2 L2g (?r; 0) : z(s) = g(s)( (s) ? (0)) ? (s) (0) is locally 0 absolute continuous with zg 2 L2g ; lim z() = (0) (0) and lim z() = 0g !0? !r+

Proof: Since A is densely de ned, the adjoint operator A exists and it satis es hA; i = h; yi for all  2 dom(A) where 2 dom(A) and y = A 2 L2g : Thus, Z

0 ?r

g(s)0(s)

(s)ds = 21

Z

0 ?r

g(s)(s)y(s)ds

(3.13)

= (0)(

Z

0

g()y()d) ? ?r

Z

0

Z

(

s

?r ?r

g()y()d)0 (s)ds:

R First we assume that (0) = ?0r d() 6= 0; where one can assume (r) = 0 without loss of generality. Then Z 0 (s) = (0) ? 0()d s and therefore Z

0

?r

g(s)0(s)ds

0

= (0)(0) ?

Z

= (0)(0) ?

Z

= (0)(0) ?

Z

?r

0

d(s) Z

(



?r ?r

0

?r

Z

s

0 0  ()d

d(s))0()d

(s)0(s)ds:

1 R 0 (g (s) + (s))0(s)ds and it follows from (3.13) that Thus (0) = (0) ?r  Z Z s Z 0 0  ( s ) 1 gyd + gyd 0(s) = 0 g (s)( (s) ? (0) ( gyd)) ? (0) ?r ?r ?r ?r for all 0 2 L2g (?r; 0): This implies 1 (Z 0 gyd)) ? (s) (Z 0 gyd) + Z s gyd = 0; a.e. g(s)( (s) ? (0) (0) ?r ?r ?r R Since g() " 1 as  ! 0? ; ?0r g()y()d = (0) (0) and therfore Z

0

#

"

z(s) = g(s)( (s) ? (0)) ? (s) (0) + and

Z

s

?r

g()y()d = 0

lim z(s) = (0) (0); slim z(s) = 0: !r+

s!0?

Next we assume that (0) = 0: Then, Z

0

?r

(g(s) + (s))0(s)ds = 0

and (0) 2 R is arbitrary. It thus follows from (3.13) that Z

0

?r

g()y()d = 0 22

and

Z

s

g(s) (s) + g()y()d = c(g(s) + (s)) ?r where c is a constant. Since g(s) " 1 as s ! 0? c = (0) and we obtain g(s)( (s) ? (0)) ? (s) (0) +

Z

s

?r

g()y()d = 0:

Hence slim z(s) = 0 = (0) (0) and slim z(s) = 0: !0? !r+

We now return to the proof of Corrollary 3.3: It follows from Lemma 3.3 that for  2 dom(A); Z 0 ?h ; (A ? I )i = 41() ?r es(g(s)(s) + dsd z(s))ds where z(s) = g(s)((s) ? (0)) ? (s)(0))  Z 0 s 1 e (g(s) + (s))(0)ds ? (0)(0) = 4() ?r = (0); which means (3.12). Finally we will discuss the inconsistent initial data case; for given  = 1 2 R, consider the equation Z

0

?r

(3:12)

g(s)x(t + s)ds = 1 +

x(s)

Z

tZ 0

0 ?r

d(s)x( + s)d

= 0; s < 0:

THEOREM 3.5 Assume that there exists a  2 (A) \ R+ such that

?1(i! + ) 2 L2(0; 1) i! +  where  is the characteristic function de ned by (3.10). Then, there exists a locally measurable function x satis es (3.12) a.e., with x(t + ) 2 L2loc (0; 1; L2g ).



Proof: Consider a family of equation (3:13)

Z

0

?r

g(s)x(t + s)ds =

Z

0

t Z 0

?r

23



d(s)x( + s) + f () d

with f (s) = e? s; s > 0;  0 and x(s) = 0; s < 0. By Theorems 3.1 and 3.2, (3.12) possesses a unique solution x (t); t  ?r and Z

t

z (t) = x (t + ) = (A ? I ) 0 S (t ? s) f (s)ds 2 L2g Let y (t) = z (t)e?t 2 L2g ; t  0. Then y 2 L2(0; 1; L2g ) and its (one-sided) Laplace transformation is given by 1 y^ (z) = e?zt y (t)dt = (A ? I )(z +  ? A)?1 +  + z  0

(3:14)

Z

?1 =  z(+z + ) +  + z e(z+)s 2 L2g ; < z > 0: ?1

Note that if let y^(z) =  z+(z+) e(z+)x; Re z > 0, then y^ belongs to the HardyLebesque class H 2(0) and Z

1

jy^ (i!) ? y^(i!)j2L2 d! ! 0 as ! 1:

?1

g

Let y 2 L2(0; 1; L2g ) be the inverse Fourier transform of y^(i!):

y(t) = lim

N i!t e y^(i!)d! ?N

Z

2 L2g as N ! 1:

Then by the Fourier-Plancherel's theorem 1

Z

0

jy (t) ? y(t)j2L2 dt ! 0 as ! 1: g

Suppose z(t) = ety(t). Then for any T > 0 Z

0

Since Z

t

T

jz (t) ? z(t)j2L2 dt ! 0 as ! 1: g

Z

t

? s S (t ? s) f (s)ds = S (t)   ? A S (t ? s) e ds 0 0 ! S (t)  in L2(0; T ; L2g ) and A is closed on L2(0; T ; L2g), from (3.14) we have

? e? t

z(t) = (A ? I )S (t)  2 L2(0; T ; L2g ): 24

This implies that there exists a locally measurable function x such that z(t) = x(t + ) 2 L2g a.e. Now, since Z t f (s)ds = 1 ? e? t; t  0; 0

it follows from (3.13) that x(t + ) 2 L2(0; T ; L2g ) satis es (3.12) a.e. in (0; T ) and this completes the proof. In conclusion we note that the Hilbert space L2g (?r; 0) provides a state space in which (3.1) is well-posed. Theorem 3.2 establishes the basic equivalence between the solutions to (3.1) and the semigroup generated by A: Moreover, the space L2g (?r; 0) is simple enough to be used in the development of approximating systems and hence we have shown that L2g (?r; 0) is a state space that meets the basic requirements (i)-(iii) as given in the introduction. Numerical results based on this framework can be found in [IT].

25

4. References

[BCH] J.A. Burns, E.M. Cli and T.L. Herdman, A state space model for an aeroelastic system, Proc. 22nd IEEE Conference on Decision and Control. [BHS] J.A. Burns, T.L. Herdman and H.W. Stech, Linear functional di erential equations as semigroups on product speces, SIAM J. Math. Analysis 14 (1983), 98-116. [BHT] J.A. Burns, T.L. Herdman and J. Turi, Neutral functional integro-di erential equations with weakly singular kernels, J. Math. Anal. Appl. 145 (1990), 371-401. [IT] K. Ito and J. Turi, Numerical methods for a class of singular integro-di erential equations based on semigroup approximation, submitted. [KZ] F. Kappel and K.P. Zhang, On neutral functional di erential equations with nonatomic di erence operator, J. Math. Anal. Appl. 113 (1986), 311-343. [P] A. Pazy, \Semigroup of Linear Operators and Applications to Partial Di erential Equations," Springer-Verlag, New York, 1983.

26