Boundary Controllability of Thermoelastic Plates ... - Semantic Scholar

Boundary Controllability of Thermoelastic Plates with Free Boundary Conditions George Avalos

Irena Lasieckay May 27, 1998 Abstract

Controllability properties of a partial dierential equation (PDE) model describing a thermoelastic plate are studied. The PDE is comprised of a Kircho plate equation coupled to a heat equation on a bounded domain, with the coupling taking place on the interior and boundary of the domain. The coupling in this PDE is parameterized by  > 0. Boundary control is exerted through the (two) free boundary conditions of the plate equation, and through the Robin boundary condition of the temperature. These controls have the physical interpretation, respectively, of inserted forces and moments, and prescribed temperature, all of which act on the edges of the plate. The main result here is that under such boundary control, and with initial data in the basic space of wellposedness, one can simultaneously control the displacement of the plate exactly, and the temperature approximately. Moreover, the thermal control may be taken to be arbitrarily smooth in time and space, and the thermal control region may be any nonempty subset of the boundary. This controllability holds for arbitrary values of the coupling parameter .

1 Introduction

1.1 Statement of the Problem

Throughout, will be a bounded open subset of R2 with suciently smooth boundary ? = ?0 [ ?1 , with both ?0 and ?1 being open and nonempty, and satisfying ?0 \ ?1 = ?. Furthermore, ?2 will be any open and nonempty subset of ?. With this geometry, we shall consider here the following thermoelastic system on nite time (0; T ):

8 !tt ?
!tt +
2 ! + 
 = 0 on (0; T )  ; > > t ? 
 +  ? 
!t = 0 > > > > @! = 0 on (0; T )  ? ; > ! = > 0 @ > > > < (
! + (1 ? )B1 ! +  = u1 @
! + (1 ? ) @B2 ! ? @!tt +  @ = u on (0; T )  ?1 ; > > @ @ @ @ 2 > > > > @ +  =  u3 on (0; T )  ?2 >   0; > 0 on (0; T )  ?n?2 @ > > > : !(t = 0) = !0 ; !t (t = 0) = !1 ; (t = 0) = 0 on :

(1)

Here, , ; and  are positive constants and   0. The positive constant is proportional to the thickness of the plate and assumed to be small with 0 <  M: The boundary operators Bi are given by

 @2 ! @2!  2! 2! 2! 2! @ @ @ @ 2 2 2 2 B1 !  21 2 @x@y ? 1 @y2 ? 2 @x2 ; and B2!  (1 ? 2 ) @x@y + 1 2 @y2 ? @x2 :  Department y Department

of Mathematics and Statistics, Texas Tech University, Lubbock, Texas 79409 of Applied Mathematics, Thornton Hall, University of Virginia, Charlottesville, Va 22903

1

(2)

The constant  2 (0; 12 ) is the familiar Poisson's ratio, and  = [1 ; 2 ] denotes the outward unit normal to the boundary. Here we shall also make the following geometric assumption on the (uncontrolled) portion of the boundary ?0:

9 fx0 ; y0 g 2 R2 such that h(x; y)
  0 on ?0,

(3)

where h(x; y)  [x ? x0; y ? y0 ]. The PDE model (1), with boundary functions u1 = u2 = 0, and u3 = 0, mathematically describes an uncontrolled Kircho plate subjected to a thermal damping, with the displacement of the plate represented by the function !(t; x; y), and the temperature given by the function (t; x; y) (see [11] for a derivation of this model). The given control variables u1(t; x) and u2 (t; x) are de ned on the portion of the boundary (0; T )  ?1; the control u3 (t; x) is de ned on (0; T )  ?2. Making the denotations

8 L2( ) if  +  > 0 < L2+ ( ) = : L20( ) if  +  = 0  R (where L20( ) = f 2 L2( ) : f = 0 ), and ) ( j $ @ k k H?0 ( )  $ 2 H ( ) : @ j = 0 for j = 0; :::; k ? 1 ; ?0

(4)

(5)

we will throughout take the initial data [!0 ; !1 ; 0 ] to be in H?20 ( )  H?10 ( )  L2+ ( ). For initial data in these spaces, and controls u1 = u2 = 0, and u3 = 0, one can show the wellposedness of (1) with the corresponding solution [!; !t ; ] being in C ([0; T ]; H?20 ( )  H?10 ( )  L2+ ( )) (see e.g., [11] and [2]). In this paper, we will study controllability properties of solutions of (1) under the in uence of boundary control functions in the following spaces: [u1; u2 ; u3 ] 2 L2(0; T ; L2 (?1)  H ?1 (?1))  C r (2;T ); where r > 0 and 2;T = (0; T )  ?2:

(6)

For arbitrary [u1; u2 ; u3 ] of such smoothness, the corresponding solution [!; !t ; ] will be in the \large" space C ([0; T ]; [D(A )]0) (see the de nition of D(A ) in (52)). In particular, we intend to address, on the nite time interval [0; T ], the question of exact{approximate  controllability (this term being originally coined in [6]). To wit, for given data [!0 ; !1 ; 0 ] (initial) and !0T ; !1T ; 0T (terminal) in H?20 ( )  H?10 ( )  L2+( ), and arbitrary  > 0, is there a suitable control triple [u1; u2 ; u3 ] 2 L2 (0; T ; L2 (?1)  H ?1 (?1))  C r (2;T ) such that the corresponding solution [!; !t ; ] of (1) satis es the following steering property at terminal time T :





[!(T ); !t (T )] = !0T ; !1T ; and (T ) ? 0T L2+ ( )  ?

In regards to the literature on this particular problem, the most relevant work is that of J. Lagnese in [12] (indeed, this present paper is principally motivated by [12]). Therein, Lagnese shows that if the coupling parameter  is small enough and the boundary ? is \star{shaped", then the boundary controlled system (1) is (partially) exactly controllable with respect to the displacement !. Also in [20], a boundary{controlled system of thermoelasticity similar to (1) is studied, with a coupling parameter  likewise present therein, and a result of partial exact controllability (again for the displacement) for this PDE is cited. This controllability result is quoted to be valid for all sizes of ; however, in the private communication [21], the author of [20] has acknowledged a aw in the controllability proof, the correction of which will necessitate a smallness criterion on , akin to the situation in [12]. The chief contribution of the present paper is to remove restrictions on the size of the coupling parameter (see Theorem 1.2 below), at the expense of adding the control u3 in the thermal component. For a 1{D version of (1), S. Hansen and B. Zhang in [8], via a moment problem approach, show the system's exact null controllability with boundary control in either the plate or thermal component. Other controllability results for the thermoelastic system which do not assume any \smallness" condition on the coupling parameters deal with distributed/internal controls. These include that in [6], in which interior control is placed in the Kircho plate component subject to clamped boundary conditions; with such control, one obtains exact controllability for the displacement !, and approximate controllability for the temperature . Alternatively in [5], interior control is placed in the heat equation of (1) so as to obtain exact controllability for both components ! and . In addition, the work in [17] deals with obtaining a result of null controllability for both the displacement and the temperature, in the case that interior control is inserted in the Kircho plate component of (1). 2

So again, the main contribution and novelty of this paper is that we consider boundary controls acting via the higher order free boundary conditions, and we do not assume any size restriction on the coupling parameter . Moreover, we do not impose any geometric \star{shaped" conditions on the controlled portion of the geometry. It should be noted that the particular type of boundary conditions imposed on the mechanical variables greatly aects the analysis of the problem, even in the case of internal control. Indeed, in the case of all boundary conditions, save for the free case, it is known that the thermoelastic plate semigroup can be decomposed into a damped Kircho plate semigroup and a compact perturbation (see also [18] for a spectral decomposition of the thermoelastic system under lower order boundary conditions). Since controllability estimates are invariant with respect to compact perturbations (at least in the case of approximately controllable systems, which we are dealing with here), the aforesaid decomposition, valid for the case of lower order boundary conditions, reduces the problem of exact controllability for the mechanical variable to that of uncoupled Kircho plates. Thus, the case of lower order boundary conditions allows a reduction of the coupled problem into one which has been much studied in the past. This strategy, while successfully employed in the case of clamped or hinged boundary conditions (see [6]), is not applicable here. Indeed, in our present case of free boundary conditions, there is no decomposition with a compact part, as in the lower order case (see [16]); moreover, the controllability operator corresponding to the given boundary controls is not bounded on the natural energy space. This latter complication is due to the fact that the Shapiro{Lopatinski conditions are not satis ed for the Kircho model under free boundary conditions. The strategy adopted in this paper consists of the following steps: Initially, a suitable transformation of variables is made and applied to the equation (1); subsequently, a multiplier method is invoked with respect to the transformed equation. The mulitiplers employed here are the dierential multipliers used in the study of exact controllability for the Kircho plate model (inspired by [11]), together with the nonlocal ( DO) multipliers used in the study of thermoelastic plates in [3] and [4]. This multiplier method allows the attainment of preliminary estimates for the energy of the system. However, these estimates are \polluted" by certain boundary terms which are not majorized by the energy. To cope with these, we use the sharp trace estimates established in [15] for Kircho plates. The use of this PDE result introduces lower order terms into the energy estimate, which are eventually eliminated with the help of a new unique continuation result in [10]. It is only at the level of invoking this uniqueness result that the thermal control u3 on ?2 must be introduced. We post our main result here on controllability. Theorem 1.1 Let the assumption (3) stand. There is then a T  > 0 so that for T > T  the following controllability property holds true: For given initial data [!0 ; !1 ; 0 ] and terminal data !0T ; !1T ; 0T in the space H?20 ( )  H?10 ( )  L2+ ( ), and arbitrary  > 0, one can nd control functions [u1 ; u2 ; u3 ] 2 L2 (0; T ; L2 (?1)  H ?1(?1))  C r (2;T ) (where given r > 0) such that the corresponding solution [! ; !t ;  ] to (1) satis es at terminal time T ,   [! (T ); !t(T )] = !0T ; !1T ;

(T ) ? T

0 L2

+ ( )

< .

Theorem 1.1 is almost a corollary from the following controllability result for the displacement only, this comprising the bulk of our eort here. Theorem 1.2 With the coupling parameter  in (1) being arbitrary and the assumption (3) in place, there is then a T  > 0 so that for T > T  , the following property holds true: For all initial data [!0 ; !1 ; 0 ] 2 H?20 ( )  H?10 ( )   T 2 T 2 L+ ( ) and terminal data !0 ; !1 2 H?0 ( )  H?10 ( ), there exists [u1; u2 ; u3 ] 2 L2(0; T ; L2 (?1)  H ?1 (?1))  H !sT(; !2T;T)., where arbitrary s  0, such that the corresponding solution [!; !t ; ] to (1) satis es [!(T ); !t (T )] = 0 1 Remark 1.3 The presence of the control u3 in Theorem 1.1 is owing solely to the need to invoke the aforementioned uniqueness result of Isakov in the proof below; it plays no part whatsoever in obtaining the preliminary (lower order term{tainted) estimate on the energy. Consequently, we have the freedom to prescribe the thermal control region to be as small as we wish, and the control u3 to be as smooth in time and space as desired. Moreover, in the course of R the proof it is seen that if the parameters  =  = 0, the control u3 may be taken to satisfy ?2 u3 (t)d?2 = 0 for all t 2 [0; T ].

1.2 Operator Theoretic Formulation and Analysis 1.2.1 Preliminary De nitions

In obtaining our controllability result Theorem 1.1, it will be useful to consider the PDE system (1) as an abstract evolution equation in a certain Hilbert space, for which end we introduce the following de nitions and notations. 3

?A
! L2 ( ) to be A=
2 ; with domain  With H?k0 ( ) as de ned in (5), we de ne  A: L2( )  D   D ( A) = ! 2 H 4 ( ) \ H?20 ( ) :
! + (1? )B1 ! = 0 on ?1 and @
! + (1 ? ) @B2 ! = 0 on ? : 1 @ @  A is then positive de nite, self{adjoint, and consequently from [7] we have the characterizations D ( A 141 ) = H?10 ( ): 2 D ( A 32 ) = H ?0 ( );  D ( A 4 ) = ! 2 H 3 ( ) \ H?20 ( ) :
! + (1 ? )B1! = 0 on ?1 : Moreover, using the Green's formula in [11], we have that for !, !b \smooth enough", Z Z  @
! @B2 !  !b d? ? Z [
! + (1 ? )B !] @ !b d?; + (1 ?  ) (
2!)!b d = a (!; !b ) + 1 @ @ ?

? @ where a(
;
) is de ned by

a (!; !b) 

Z

[!xx !bxx + !yy !byy +  (!xx !byy + !yy !bxx ) + 2(1 ? )!xy !bxy ] d :

(7)

(8)

(9)

(10)

In particular, this formula and the second characterization in (8) give that for all !, !b 2 D( A 21 ),

A!; !b h



 12 !;  A 21 !b  = A A

1 1 i0 D( 2 ) D( 2 )

 A

and in addition



L2 ( )

= a (!; !b )L2 ( ) ;

(11)



2 A 12 !

L2 ( ) = a (!; !) : k!k2D(A 12 ) =



(12)

 We de ne AD : L2 ( )  D (AD ) ! L2 ( ) to be AD = ?
; with Dirichlet boundary conditions, viz. D(AD ) = H 2 ( ) \ H01 ( ):

(13)

AD is also positive de nite, self{adjoint, and by [7] 1

D(AD2 ) = H01 ( ):

(14)

 We denote the operator AR : L2( )  D (AR ) ! L2 ( ) by the following second order elliptic operator:   + # = 0 (15) AR = ?
+  I, with D(AR) = # 2 H 2 ( ) : @# @ AR is self{adjoint, positive semide nite on L2 ( ), with its fractional powers therefore being well{de ned. In particular, we have again by [7]

1

D(AR2 ) = H 1 ( ) \ L2+ ( ); with

 e #; #

H 1 ( )\L2+





= AR2 #; AR2 #e ( ) 1

1



 



= r#; r#e L2 ( )

(16)

+  #; #e L2 ( )





(note that we are using implicitly the fact that the bilinear form r#; r#e L20 ( )). 4

 

+  #; #e L2 (?) 

L2 ( )

L2 ( )

(17)

is H 1 ( ){elliptic on H 1 ( ) \

 We denote the operator AN : L2 ( )  D (AN ) ! L2( ) by the following second order elliptic operator:

(

AN = ?
, with D(AN ) = # 2

H 2 ( ) :



)

= 0 : #j?0 = @# @ ?1

(18)

Once again by [7], we have 1

D(AN2 ) = H?10 ( ):

(19)

 ( 0; 1 ) will denote the classical Sobolev trace maps, which yield for f 2 C 1( )

0 f = f j? ; 1f = @f @ ? :  We de ne the elliptic operators G1 ; G2 , and D as follows: 8 2
v = 0 on

> > > > > < v = @v = 0 on ?0 ; @ G1h = v () > ( >
v + (1 ? )B1 v = h > >
v + (1 ? ) @B2 v = 0 on ?1 ; > : @@ @

Dh = v ()

(20)


v = 0 on

vj? = h on ?;

8 2
v = 0 on

> > > > > < v = @v = 0 on ?0 ; @ G2h = v () > ( >
v + (1 ? )B1 v = 0 > >
v + (1 ? ) @B2 v = h on ?1: > : @@ @

(21)

Furthermore, if we de ne the space L2+ (?2) in a manner totally analogous to that for L2+ ( ) (see (4)), we q can the construct the spaces H+ (?2) for q 2 R by setting

Hq + (?2) = H q (?2) \ L2+ (?2): Therewith, we can then properly de ne the (Robin) map R by having for all h 2 Hq + (?2),

8   I v = 0 on

> ?
+  > Rh = v () > < @ +  = h on ?2 > : @v + v = 0 on ?n? : :> @

(22)

(23)

2

The classic regularity results of [19] (p.152) then provide that for q ? 12 ;

8   > D 2 L  H q (?); H q+ 21 ( ) ; > > < R 2 L Hq + (?2); H q+ 23 ( ) \ L2+(?2) ;   > G1 2 L H q (?1); H q+ 25 ( ) ; > > : G2 2 L H q? 12 (?1); H q+3 ( ) : 5

(24)





Denoting the topological dual of Hq + (?2) as Hq + (?2) 0 (pivotal with respect to the L2{inner product), then with the elliptic operators AR and R as de ned above, one can show that for q  ? 12 , the (Banach space)  1   adjoint R AR 2 L D(AR2 ); Hq + (?2) 0 satis es

RAR # = #j?2 for all # 2 D(AR2 ). 1

(25)

Moreover, with the operators  A and Gi as de ned above, one can readily with the use of the Green's  show 1 1  i A 2 L D ( A 2 ); H ? 12 (?1) satisfy for i = 1; 2 formula (9) that 8 $ 2 D( A 2 ) the (Banach space) adjoints Gi 

G A$ = i

 (?1)i?1 $j on ? 2?i ?1 1 0

(26)

on ?0:

 With AN given by (18), we de ne the operator P : D(P )  L2 ( ) ! L2( ) by P  I + AN .

(27)

(i) With the parameter > 0; we de ne a space H?10 ; ( ) equivalent to H?10 ( ) with its inner product being ( !1 ; !2 )H?10 ; ( )  (!1 ; !2 )L2 ( ) + (r!1 ; r!2 )L2 ( ) 8 !1 ; !2 2 H?10 ( );

 1

(28)

and with its dual denoted as H??01; ( ). After recalling that H?10 ( ) = D AN2 (by (19)), two extensions by continuity will then yield that





P 2 L H?10 ; ( ); H??01; ( ) , with hP !1 ; !2 iH??01; ( )H?10; ( ) = (!1 ; !2 )H?10 ; ( ) :



(29)



Furthermore, the obvious H?10 ; ( ){ellipticity of P and Lax{Milgram give us that P 2 L H?10 ; ( ); H??01; ( ) is boundedly invertible, with





P ?1 2 L H??01; ( ); H?10 ; ( ) :

(30)

Moreover, P being positive de nite and self{adjoint as an operator P : L2( )  D(P ) ! L2( ), the square 1 1 root P 2 is consequently well{de ned with D(P 2 ) = H?10 ; ( ), by (19). It then follows from (28) and (29) that for ! and !b 2 H?10 ; ( );

12

2

P ! L2 ( ) = k!k2L2( ) + kr!k2L2 ( ) = k!k2H?10 ; ( ) ;  12 21  P !; P !b L2 ( ) = (!; !b )H?10 ; ( ) :

(31) (32)

(ii) Finally, by Green's formula we have for !, !b 2 D( A 21 ),

? AG
!; !b


+ 2 1 H??1; ( )H?1 ; ( ) =

 0@!  0
? A!b L2 (?1 ) ? (r!; r!b )L2 ( ) + @ ; !b 2 + 1 !; G2  L (?1 ) = ? (r!; r!b )L2 ( ) = ? hAN !; !b iH??01; ( )H?10 ; ( ) ;

after using (26). We thus obtain after two extensions by continuity to H?10 ; ( ) that

?








P = I ?
+  AG2 1 as elements of L H?10 ; ( ); H??01; ( ) :

(33) (34)

In obtaining the equality above, we have used implicitly the fact that for every $ 2 H??01; ( ) and $ 2 D( A 21 )

h$; $iH??01; ( )H?10 ; ( ) = h$; $i[D(A 12 )]0 D(A 21 ) : 6

(35)

 With L2+ ( ) as de ned in (4), we denote the Hilbert space H to be H  D( A 12 )  H?10 ; ( )  L2+( );

(36)

with the inner product

02 ! 3 2 !b 31 1       @4 !12 5 ; 4 !b2 5A = A 21 !1 ; A 12 !b1 2 + P 21 !2 ; P 12 !b2 2 + ; b 2 : L ( ) L ( ) L ( )  b H

(37)

 With the above de nitions, and making the denotation G1 0 +  (|)  AR ?  ? A AG2 0 ; we then set A : H  D(A ) ! H to be 0 I 0 0 10 0 I 0 1 (|) C A  @ 0 P ?1 0 A B @ ?A  0 A 0 ? AD (I ? D 0) ?  AR 0 0n I A 21 )  D(AR) \ L2+ ( ) with D(A ) = [!0 ; !1 ; 0 ] 2 D( A 12 )  D( such that  A!0 +  AG1 0 0 2 H??01; ( ) and 
!1 + 
0 2 L2+( ) .

(38)

(39)

 We make the following denotations for the space of controllability: Us = L2 (?1)  H ?1 (?1)  Hs+ (?2); ?
Us = L2 (0; T ; L2 (?1)  H ?1 (?1))  H s (0; T ; L2+ (?2)) \ L2 0; T ; Hs+ (?2) ;

(40)

where Hs+ (?2) is as de ned in (22), and s  0. We de ne the control operator B on Us by having for every u = [u1 ; u2 ; u3 ] 2 Us,

2 3 0 AG1 u1 +  AG2u2 ] 75 : Bu = 64 P ?1[  AR Ru3

 

(41)







Note that a priori, the mapping B only makes sense as an element of L Us ; D(A ) 0 , where H  D(A ) 0 . Indeed, for xed u = [u1 ; u2 ; u3 ] 2 Us one has, upon using the expression for the inverse A? 1 given in (144) below, and the de nition of the elliptic operators G1 , G2 and R in (21) and (23) above, that

2 3 2 ?G u ? G u ? A?1 (|)Ru 3 0 3 A G 1 u1 +  AG2 u2] 75 = A 4 1 1 2 2 0 5 2 D(A )0 ; Bu = A A? 1 64 P ?1[  AR Ru3

?Ru3

(42)

where (|) is as de ned in (38) (note that we are implicitly using the fact that  A?1(|) 2 L(L2( ); D( A 21 )), by Proposition 4.1 of the Appendix).  By duality, we have



?



Us = L2 (?1)  H 1 (?1)  Hs+ (?2) 0 ;   Us = L2 (0; T ; L2 (?1)  H 1 (?1))  H s (0; T ; L2+ (?2)) \ L2(0; T ; Hs+ (?2)) 0 ;




and B 2 L D(A ); Us . 7

(43)

1.2.2 Abstract Operator Formulation If we take the initial data [!0 ; !1 ; 0 ] to be in H , and control u 2 Us , where Us is as de ned in (40), then considering

the operator de nitons above, the coupled system (1) can be rewritten a fortiori as the operator theoretic model

2

3

2

2 !(0) 3 2 ! 3 4 !t(0) 5 = 4 !01 5 ;

3

d 4 !! 5 = A 4 !! 5 + Bu;

t dt t 



(0)



(44)

0

with this equation having sense in D(A ) 0 (a space strictly larger than H ). Given the operator de nitions for A and B above, the solution [!; !t ; ] to the ODE (44) (and so to the PDE (1)) is given by

2 !(
) 3 2! 3 Z (
) 4 !t (
) 5 = eA (
) 4 !01 5 + eA (
?s) Bu(s)ds; (
)

0

0





(45)



which by (42) and the convolution theorem is an element of C [0; T ]; D(A ) 0 . With this representation of the    solution [!; !t ; ] in mind, we de ne the input ! terminal state map LT 2 L Us; D(A ) 0 as

LT u =

ZT 0

eA (T ?s) Bu(s)ds:

(46)

Taken as an unbounded operator from Us into H , then LT : D(LT )  Us ! H is closed and densely de ned, with its domain of de nition D(LT ) given to be (47) D(LT ) = fu 2 Us : LT u 2 H g : Its adjoint LT : D(LT )  H ! Us , where Us is as given in (43), is likewise closed and densely de ned, with 8 2 3 9 = < 0 (48) D(LT ) = :[0 ; 1 ; 0 ] 2 H : LT 4 1 5 2 Us ; : 0

As we are concerned with obtaining1exact controllability of the displacement [!; !t ] only, we accordingly de ne the projection operator  : H ! D( A 2 )  H?10 ; ( ) by

2$ 3   0 0  4 $1 5 = $ $1 :

(49)

#0

Henceforth,1the work here will be concerned with determining the surjectivity of the closed operator LT , D(LT )  Us ! D (  A 2 )  H?10 ; ( ), with LT u = 

ZT 0

eA (T ?s) Bu(s)ds;

(50)

and with D(LT ) = D(LT ). Determining the ontoness of the operator LT for some T > 0 becomes our concern here, since it is equivalent to showing the exact controllability of the solution component [!; !t ] to (1) (Theorem 1.2). This surjectivity for LT is in turn equivalent to the existence of a certain observability 1inequality pertaining to the range of the adjoint LT  (the inequality (66) below), where LT  : D(LT )  D( A 2 )  H?10 ; ( ) ! H  is likewise a closed densely de ned operator (as LT is), with its domain given by

o

n

D(LT ) = [0 ; 1 ] 2 D( A 12 )  H?10 ( ) : [0 ; 1 ; 0] 2 D(LT ) :

(51)

It is the injectivity condition (66) which we intend to directly verify. In order to rewrite this abstract inequality in \PDE form" (i.e., as the inequality (67) below), we need the following two propositions, the rst of which is proved in the Appendix below. 8

Proposition 1.4 The Hilbert space adjoint A of A , as de ned in (39), is given to be 0 I 0 0 10 0 ?I 0 1 A  0 ?(|) C A = @ 0 P ?1 0 A B @  A; 0 AD (I ? D 0) ? AR 0 0 I n A 12 )  D(AR) \ L2+ ( ) with D(A ) = [0 ; 1 ; 0 ] 2 D( A 21 )  D( such that  A0 +  AG1 0 0 2 H??01; ( ) and such that ? 
1 + 
0 2 L2+ ( ) (above, (|) is the same denotation made in (38)).

(52)

n o Remark 1.5 Using the semigroup eA t t0 generated by A , then for terminal data [0 ; 1 ; 0 ] 2 H , 2 (t) 3 2 3 4 t(t) 5 = eA (T ?t) 4 01 5 2 C ([0; T ]; H )

(t) 0 is the solution to the following backwards problem: 8   ?
 +
2  + 
= 0 tt tt > > > t + 
?  ? 
t = 0 on (0; 1)  ;

> > > >  = @ = 0 on (0; 1)  ?0 ; > @ > > < (
 + (1 ? )B1  +  = 0
 + (1 ? ) @B2  ? @tt +  @ = 0 > > @@ @ @ @ > > > @ +  = 0 on (0; 1)  ?;   0; > > > @ > > :

on (0; 1)  ?1 ;

(53)

(54)

[(T ); t(T ); (T )] = [0; 1 ; 0 ] .

Remark 1.6 For terminal data [0 ; 1 ; 0 ] in D(A ), the two equations of (54) may be written pointwise as P tt = ? A ?  AG1 0 +  AG2 0 ? 
in H??01; ( ); (55) 2 t = ?
+  +
t in L ( ); (56) [(T ); t(T ); (T )] = [0 ; 1 ; 0 ] : (57) Remark 1.7 Since ?0 \ ?1 = ;, and ? is smooth, we can assume throughout that D(A ) is dense in the graph topology of D(LT ). Proposition 1.8 The adjoint LT 2 D (LT )  H ! Us of LT is computed to be 2 3 2 3 " # 0 0 t  (58) ; ?  LT 4 1 5 = @ t j?1 ;  j?2 for all 4 1 5 2 D (LT ) , @ ? 1 0 0 where

  @t ;  j ; j ?2 are boundary \traces" of the solution [; t ; ] to the coupled system (54). @ ?1 t ?1

Proof: By Remark 1.7, it is enough to show the characterization in (58) for [0 ; 1 ; 0 ] 2 D(A ). With this in

mind, one has readily the classic representation

2 3 2 3 2 3 0 0 0 ?
LT 4 1 5 = BeA (T ?t) 4 1 5 for every 4 1 5 2 D A ; 0

0

0

9

(59)

? ?





where again, B 2 L D A ; Us is the adjoint of B. We must show that the right hand of this equality ?
may be written explicitly in \PDE form" as (58). To this end, for every [u1 ; u2 ; u3 ] 2 Us, and [0 ; 1 ; 0 ] 2 D A , we have

* 2 u1 3 2 0 3+ LT 4 u2 5 ; 4 1 5

=

u3

[D(A )]0 D(A )

0

*Z T

=

0

2 u (s) 3 2  3+ 0 1 eA (T ?s) B 4 u2(s) 5 ds; 4 1 5 u3(s)

2 u (s) 3 2  3+ 0 1 eA (T ?s) A A? 1B 4 u2 (s) 5 ; 4 1 5

Z T*

u3 (s)

0

[D(A )]0 D(A )

0

ds

0

 0



[D(A )] D(A ) 3 1 2 2 3 0 ZT 0 u1(s)  = @A? 1 B 4 u2(s) 5 ; eA (T ?s) A 4 1 5A ds

u3(s)

0

H

0

2  31 Z T 02 ?G1u1 (s) ? G2u2(s) ? A?1(|)Ru3 3 0  5 ; A eA (T ?s) 4 1 5A ds: 0 = @4 ?Ru3

0

0

H

(60)

2 (t) 3 2 3 0 Noting that 4 t(t) 5  eA (T ?t) 4 1 5 gives the solution to the backward problem (54), we then use this relation,

(t) 0 the de nition of the adjoint A in (52), and Proposition 4.1 of the Appendix to obtain

* 2 u1(s) 3 2 0 3+ LT 4 u2(s) 5 ; 4 1 5 u3(s)

 0



[D(A )] D(A ) 02 2 31 Z T B ?G1u1 (s) ? G2 u2(s) ? A?1 (|)Ru3 3 6 ?1 ?t ?1 5 ; 4  P A ? P (|) 75CA ds = 0 @4 0 ?Ru3 AD (I ? D 0)t ? AR H  Z T  1    1 1 1

= =

0

 A 2 t A 2 G1u1 ; 

0

Z T "

0

A 2 G2u2;  A 2 t + 

L2 ( )

+  (Ru3 ; AR )L2 ( ) dt

#

  At ? 3 dt + u1 ; G1  At H ? 21 (?1 )H 21 (?1 ) + u2; G2  3 H 2 (?1 )H 2 (?1 )

Z T " @t  = u1 ; 0

L2 ( )

@

#

ZT 0

 (u3 ; )L2 (?2 ) dt

? hu2 ; t iH?1 (?1)H1(?1 ) dt

L2 (?1 )

+ hu3; iH s(0;T ;L2+ (?2 ))\L2 (0;T ;Hs+ (?2 ))[H s (0;T ;L2+ (?2 ))\L2 (0;T ;Hs+ (?2 ))]0 ;

(61)

thereby completing the proof of Proposition 1.8.  Immediately, we have

Corollary 1.9 The adjoint operator LT  : D(LT )  D( A 21 )  H?10 ; ( ) ! Us is given by



#  " @ t 1 = @ ?1 ; ?tj?1 ;  j?2 #

L  0 T

"



for all [0 ; 1 ] 2 D(LT ), where @t ; t j?1 ; @

?1

(62)

j?2 are boundary traces of the solution [; t ; ] to the following

10

(backwards) system:

8   ?
 +
2  + 
= 0 > tt tt on (0; 1)  ; > > t + 
?  ? 
t = 0 > > > > = 0 on (0; 1)  ?0 ;  = @ > @ > > < (
 + (1 ? )B1  +  = 0 @
 + (1 ? ) @B2  ? @tt +  @ = 0 on (0; 1)  ?1 ; > > > @ @ @ @ > > > @ +  = 0 on (0; 1)  ?;   0; > > @ > > :

(63)

[(T ); t(T ); (T )] = [0; 1 ; 0] .

We conclude this section with a regularity result for the thermal component of the solution [; t ; ] to (54), this being originally derived in [11] and [2] for the forward problem (1). Assuming terminal data [0 ; 1 ; 0 ] 2 D(A ), we have, by using (53), the equality

2 (t) 3

2 0 2 (t) 3 2 (t) 31

d 4 @ 4 54 5 5A : dt

t((tt))

= ?2 A t((tt)) ; t((tt)) H

H

(64)

Integrating this equation from 0 to T , performing computations similar to those performed for the proof of Proposition 1.4, recalling the characterization (16) and (17), and subsequently invoking a density argument, we have the

following:

Proposition 1.10 With terminal data [0; 1 ; 0 ] 2 H , we have that the component of the solution of (54) is 1  an element of L2 0; 1; D(AR2 ) . Indeed, we have the following relation valid for all T > 0:

2  3

2

2 (0) 3

2

4 0 5

?

4  (0) 5

= 2 Z T

A 21

2 dt:

R L2 ( )

1



t(0)

0 0 H H

(65)

2 Proof of Theorem 1.2

2.1 The Necessary Inequality

As stated above, showing the partial exact controllability of the system (1) for some time T > 0 is equivalent to 1  2 showing the surjectivity of the operator LT : D(LT )  Us ! D(A )  H?10 ; ( ), where LT is as de ned in (50), and with D(LT ) as de ned in (47). By the classical functional analysis (see e.g. [9], Lemma 3.8.18), the ontoness of LT for some time T > 0 is tantamount to the existence of a constant CT > 0 such that following inequality satis ed for all [0 ; 1 ] 2 D(LT ), where D(LT ) is as de ned in (51):

2 

L 4 0

T 01

3

 

5  CT

0 1

Us



1 D ( A 2 )H 1

?0 ; ( )

:

(66)

Corollary 1.9 then gives that this abstract inequality above may be rewritten as having for all [0 ; 1 ] 2 D(LT ), # Z T "

@t

2 2



+ ktkH 1 (?1 ) dt +  k k2[H s(0;T ;L2+ (?2 ))\L2 (0;T ;Hs+ (?2 ))]0 @ 0 L2 (?1 )

  

2 ; (67)  CT

01

1 1 D ( A 2 )H?0 ; ( ) 11

"



#

t where @ @ ?1 ; tj?1 ; j?2 are traces of the solution [; t ; ] to the backwards system (63) (this being \adjoint"

with respect to (1)). So to prove the partial exact controllability of the system (Theorem 1.2), it will hence suce to establish the inequality (67), for T > 0 large enough . In this connection, the bulk of the work will entail the derivation of the following estimate: Theorem 2.1 For T > 0 large enough, the solution [; t ; ] to (54) with terminal data [0 ; 1 ; 0 ] 2 D(LT ) satis es the following inequality:

Z T 

1

2

1 2  1

2

1 2

1 2  A 2 0 L2 ( ) +

P 2 1

L2 ( ) A 2  L2 ( ) +

P 2 t

L2 ( ) +

AR2

L2 ( ) dt +



0 #

@

2 Z T" 2 t 2 2 1

 CT k 0 kL2 ( ) + kt kH 1 (?1 ) + @ + k kH 2 + ( ) dt 0 L2 (?1 )  2 2 2

+ kkL1 (0;T ;H 32 + ( )) + kt kL1 (0;T ;H 12 + ( )) + k kL1 (0;T ;H ? 12 + ( )) : (68) This theorem will follow from a chain of results. Given the density of D(A ) in D (LT ) (see Remark 1.7), and the fact that the solution of (54) has the representation 2 (t) 3 2 3  4 t (t) 5 = eA (T ?t) 4 01 5 ; (69) (t) 0 it will be to terminal data in enough to show the inequality   2 (68) for solutions [; t ; ] to (54) corresponding   2  2 D A . Taking [0 ; 1 ; 0 ] 2 D A , we then have that [; t ; ] is an element of C ([0; T ]; H )\ C 1([0; T ]; D(A ))\     C [0; T ]; D A 2 , and as such has the additional regularity (see [3] [Theorem 2], and also [12]):

 2 C ([0; T ]; H 4 ( )); t 2 C ([0; T ]; H 3 ( )); tt 2 C ([0; T ]; D( A 21 )); 2 t 2 C ([0; T ]; D (AR ) \ L+ ( ));  ? G2 1tt + G1 0 ? G2 0 2 C ([0; T ]; D( A)): (70) This extra regularity of [; t ; ], corresponding to smooth initial data, will justify the computations to be done

below.

2.2 Proof of Theorem 2.1

  

As mentioned above, the terminal data [0; 1 ; 0 ] will be considered to be in D A 2 ; accordingly the corresponding solution [; t ; ] of (54) will be a classical one, with the regularity posted in (70). With the end in mind of deriving the estimate (67), we start by making the substitution b(t) = e?t (t); and b(t) = e?t (t); (71)

h

i

where parameter  2 R is to be determined. Necessarily then, b; bt ; b solves the coupled (backwards) system

882   2b b b  2b b b b > <   + 2   +  ?


  +2 t + tt +
 + 
b = 0 t tt >   on (0; 1)  ; > > :  b + bt + 
b ?  b ? 
 b + bt = 0 > > > > b > > b = @@ = 0 on (0; 1)  ?0 ; > > > > : @
b + (1 ? ) @B2b ? @  2 b + 2 bt + btt +  @ b = 0 on (0; 1)  ?1 ; > > @ @ @ @ > > > @ b +  b = 0 on (0; 1)  ?;   0; > > @ > > > h > : b(T ); bt(T ); b(T )i = e?T 0 ; ?e?T 0 + e?T 1 ; e?T 0 . 12

(72)

  

h

i

Since [0 ; 1 ; 0 ] 2 D A 2 , the extra regularity in (70) gives that b; bt ; b is a classical (not just weak) solution of (72); accordingly, we can rewrite (72) abstractly as (see Remark 1.6 and (34))

 2b









  + 2 bt + btt ?
 2 b + 2 bt + btt ?  AG2 1  2b + 2 bt + btt + Ab +  AG1 0 b ?  AG2 0 b + 
b = 0 in H??01; ( );











 b + bt + 
b ?  b ? 
 b + bt = 0 in L2( );

(73) (74)

i   (T ); bt(T ); b(T ) = e?T 0 ; ?e?T 0 + e?T 1 ; e?T 0 : (75) Now multiplying the heat equation (74) by  and adding it to the Kircho plate (73), and subsequently taking 2 the parameter  to be   2  , we obtain the single equation hb





btt ?
btt +  Ab ?  AG2 1  2b + 2 bt + btt +  AG1 0 b ?  A G2 0 b b b b b b h b = bc0 + cb1 t +i c2 ?+Tc3 t + c4?
T ; ?T ?T  (T ); t(T ); (T ) = e 0 ; ?e 0 + e 1 ; e 0 ;

(76) (77)

3

2 , and c = ? 4 (note that the particular where the constants c0 = 2  2 ?  , c1 =   , c2 = ? 4 242 , c3 = ?  4 4 2 b choice of  made here eliminates the higher order term
t ). The system (76){(77) may be rewritten in PDE form as the Kircho plate equation

8b > tt ?
btt +
2 b = c0 b + c1 bt + c2 b + c3 bt + c4
b on (0; 1)  ; > > > b > > b = @@ = 0 on (0; 1)  ?0 ; > > > @
b + (1 ? ) @B2b = @  2 b + 2 b + b  ?  @ b on (0; 1)  ?1 ; > : > t tt > @ @ @ @ > > i h > : b(T ); bt(T ); b(T ) = e?T 0 ; ?e?T 0 + e?T 1; e?T 0 .  2b b b  b b

(78)

b  hAsb b?i G2 1   + 2 t + tt + G1 0 ? G2 0 2 C ([0; T ]; D(A)) (using the last containment in (70)), then ; t is a classical solution of (78).

We note at this point that one can readily i the trace estimate Lemma 4.5 (of the Appendix below) for h b bderive b b the plate component
 ? of the solution ; t ; of (72). The proof of this is relegated to the Appendix since it 0 is entirely analogous to that shown for the forward problem in [3] and [4]. This estimate will be critical in the proof of the following lemma.

h

i

Lemma 2.2 (a) The solution b; bt ; b to (72) satis es the following relation for all s and  2 [0; T ]:



1

2

1 2 t= A 2 b(t) L2 ( ) +

P 2 bt (t)

L2 ( ) = F (s;  );

 t=s

(79)

where F (
;
) is a function (de ned below in (97)) which obeys the following estimate for all s and  2 [0; T ] and

13

 > 0:

0Z 2 3 1

b

2



2



2



2 T 2 @ 



t + b H 21 + ( ) + b H 23 + ( ) + bt H 21 + ( ) 5 dtA F (s;  )  C @ 4 bt H 12 (? ) +

@

1 0 2 L (? ) Z T 1

1

2

12

2 

21

2



2 b b 2  +  + P  +CT b L1(0;T ;H ? 12 + ( )) +  A

L2 ( ) t L2 ( ) AR b L2 ( ) dt 0  1

2

1

2

1 2

1 2  +

 A 2 b(s) L2 ( ) +

P 2 bt (s)

L2 ( ) +

 A 2 b( ) L2 ( ) +

P 2 bt ( )

L2 ( ) : h b b bi

(b) For  > 0 small enough, the solution ; t ;

(80)

to (72) satis es the following estimate for all s and  2 [0; T ]:

 1

2

1

2

21

2

21

2  1 + 

 b b b 2 2  A A  (  )   (  ) +  ( s ) + P

L2 ( ) t L2( ) 1 ? 

L2( ) P bt(s) L2 ( ) 3 1 0Z 2

b

2



T 2 2 2 2 @ 

+

b

H 21 + ( ) +

b

H 23 + ( ) +

bt H 21 + ( ) 5 dtA +C @ 4

bt

H 21 (? ) +

@t

1 0 L2 (?1 ) Z T 

1

2



2

21

2 

21

2  b b b  2 +CT L1 (0;T ;H ? 12 + ( )) + 1 ? 

A  L2 ( ) + P t L2 ( ) + AR b L2 ( ) dt; 0

(81)

where the constant CT above depends upon time, but C does not.

Proof: We take the duality pairing of the abstract equation (76) with bt and integrate in time and space so as

to get

# E D E Ab; bt [D(A 21 )]0 D(A 12 ) dt btt ?
btt ?  AG2 1 btt; bt H ?1( )H 1 ( ) +  s ?0 ?0 Z D   E =

 AG2 1  2 b + 2 bt ?  AG2 0 b; bt [D(A 21 )]0D(A 21 ) dt AG1 0 b +  sZ   + c0 b + c1 bt + c2 b + c3 bt + c4
b; bt 2 dt Z  "D

L ( )

s

(82)

(note that here we are using implicitly the fact that the terminal data [0 ; 1 ; 0 ] being in D(A ) implies that   Ab +  AG2 1  2b + 2 bt + btt ?  AG1 0 b +  AG2 0 b is an element of C ([0; T ]; H??01 ( ))). Secondly, denoting A?D1 to be the inverse of the elliptic operator de ned in (13), we multiply the pde (78) by ? c 1 A?D1 b, and subsequently integrate in time and space so as to get

Z  s

 btt ?
btt +
2 b ? hc0 b + c1 bt + c2 b + c3 bt + c4
bi ; ? c1 A?D1 b

L2 ( )

dt:

(83)

(A1) Rewriting the equation (82). Using the equality (34) and the characterizations in (26), we have upon the taking of adjoints that (82) may be rewritten as

t= Z    Z  b b  b b b b b = c0 + c2  + c3 t + c4
; t dt + c1 t; t dt s s t=s 2 3 ! ! Z b b b 5 dt : +  b; @@t ? 4  2 @@ + 2  @@t +  0 b; bt (84) 

1

P 12 b (t)

2 +

 1 b 2 2 (t) A

t 2 2 L2 ( ) L ( ) s

L2 (?1 )

L2 (?1 )

(A2) Rewriting the equation (83).

(i) An integration by parts, the use of the heat equation (74), and the fact that AR b = ?
b +
D 0 b +  b =

14

AD (I?D 0) b +  b yield

 c  Z  c1     c1 Z   b ?1 b  1 b ? 1 ? 1 b b b ? tt; AD L2 ( ) dt = ? t; AD L2 ( ) + t ; AD t L2 ( ) dt s s      s c  Z  c   1 1 b ?1 b ?1 b  b b 

t; ?  AD + (I ? D 0) ? t; AD L2 ( ) + s s  Z    dt: bt ; (I ? D 0) b +  (I ? D 0) bt ? c 1 s L2 ( )

=

L2 ( )

dt

(85)

(ii) An integration by parts and employment of Green's Theorem yield

Z  Z    ? 1 b b c1
tt; AD L2 ( ) dt = ? c1 rbtt ; rA?D1 b L2 ( ) dt s   b ?1 b  Zs   b ?1 b  = ?c1 rt ; rAD L2 ( ) + c1 rt ; rAD t L2 ( ) dt s s  Z    Z   ?1 bt ! @A D = ?c1 rbt ; rA?D1 b L2 ( ) + c1 bt ; AD A?D1 bt L2 ( ) dt + c1 bt; @ dt s s s L2 (?1 )  Z     Z     @A?1 b  @ (I ? D 0) b !  D b ?1 b b b b = ?c1 rt ; rAD

?c1

Z

L2 ( ) s

+ c1

t ;

s

t

L2 ( )

@ (I ? D 0)b +  @ (I ? D 0)bt bt;  @ @

s

!

dt + c1

L2 (?1 )

s

t ; ? 

@ +

@

L2 (?1 )

dt:

dt

(86)

(iii) Through the use of the Green's Theorem (9) and the boundary conditions in (78), we obtain

?

Z  c1  s



2 b ?1 b


; AD

L2 ( )

dt

Z   c1 Z  b @A?D1 b c 1 ? 1 b b =? a ; AD dt ? ; @ s s

!

Z ?1 b c 1 dt +
b; @A@D s L2 (?1 )

Jointly then, the equalities (83) and (85){(87) give the relation

! L2 (?0 )

dt:

(87)

 Z     c  1 ? 1 b b b 0 = ?c1 dt ? dt t; ?  AD + (I ? D 0) L2 ( ) s s L2 ( )  Z     @A?1 b  @ (I ? D 0) b ! Z    c  1 b b b + dt ? c1 bt ; ?  @D + dt t; (I ? D 0)  + (I ? D 0) t @ 2 s s L ( ) L2 (?1 ) ! Z @ ( I ? D  0 )b  @ (I ? D 0 )bt b dt + +c1 t; @ @ s L2 (?1 ) Z  Z  c1 Z  b @A?D1 b ! ?1 b ! c c @A 1 1 ? 1 b b b dt ? dt + a ; AD dt + ; @
; @D s s s L2 (?1 ) L2 (?0 )   Z    b ?1 b  1 c 1 ?1 b ? 1 b b b b b b b dt + c1 rt ; rAD L2 ( ) + t ; AD L2 ( ) :(88) ? c0 + c1 t + c2 + c3 t + c4
; AD s s L2 ( ) Z 

 bt; bt

15

Summing the relations (84) and (88), we obtain



1

P 12 b (t)

2 +

 1 b 2 2 (t) A

t 2 2 L2 ( ) L ( )

t= Z   =



c0 b + c2b + c3 bt + c4
b; bt ? c 1 A?D1 b dt

s t=s ! ! 3 Z  2 @ b b b @  @  t t 5 dt +  b; @ ? 4  2 @ + 2  @ +  0 b; bt s 2 2 L (?1 ) L (?1 )      Z     dt + c1 rbt ; rA?D1 b L2 ( ) bt ;  ?  A?D1 b +  (I ? D 0) b ? c 1 s s L2 ( ) 2 ! ! 3   Z  bt  ? 1 b b 5 dt +c1 4 ; (I ? D 0) b +  (I ? D 0) bt ? bt;  ?  @A@D +  @ (I ?@D 0) s 2 2 L (?1 ) !L ( ) Z Z    b b @ (I ? D 0) +  @ (I ? D 0)t dt + c 1 a b; A?D1 b L2 ( ) dt +c1 bt ;  @ @ s s L2 (?1 ) 2 3 ! !  Z   ?1 b ?1 b 5 dt + c1 bt; A?D1 b 2 (89) ?
b; @AD + c1 4  b; @AD

s

@

L2 (?1 )

@

R 

L2 (?0 )



L ( ) s

(note the cancellation of the high order term s bt; bt L2 ( ) dt). We now proceed to estimate the right hand side of this relation. In so doing, we will be using implicitly in (B1){(B7) below the inequality ab  a2 + C b2 . (B1) We have by trace theory,

! ! 3 Z  2 @ b b b @  @  5 dt ? 4  2 @ + 2  @t +  0 b; bt +  b; @t s L2 (?1 ) L2 (?1 )

b

2 3 Z T 2



2



2



2 5 dt:  C 4 b H 32 + ( ) + bt H 21 + ( ) + b H 21 + ( ) +

@@t

0 L2 (?1 )

(90)

(B2) As A?D1 is a bounded operator, we have

Z 

 c 1 ?1 b b b b b b c0 + c2  + c3 t + c4
; t ? AD dt s Z T 1

2 Z T 



2

2 

2 A 2 b L2 ( ) dt:  C b L2( ) +

bt

L2 ( ) +

b

L2 ( ) dt + 6



0 0

(91)

(B3) As D 0 2 L(H s( )) for s > 21 (by standard elliptic theory), and A?D1 2 L(L2( ); D(AD)), we then have in

conjunction with trace theory

  Z     Z    dt dt + c 1 bt ; ?  A?D1 b +  (I ? D 0) b bt; (I ? D 0) b +  (I ? D 0) bt s s L2 ( ) L2 ( ) Z  Z  @A?1 b ! Z     @A?1 b !  c c 1 1 b; D dt + a b; A?D1 b dt + ?c1 bt; ?  @D @ L2 (?1 ) dt s s s 2 L (?1 )   Z T

1

2 Z T 



2



2



2 b b b (92)  C + +  

H 12 + ( ) t H 21 + ( ) H 21 + ( ) dt + 6 0 A 2 b L2 ( ) dt: 0 b @ (B4) Using the fact that D 0 2 L(H s ( )) for s > 21 , and @ (t) = ? b(t) ? , we have along with trace theory ? ? c 1

16

that

c1

Z s

b @ (I ? D 0)b +  @ bt bt ; ?  @@ +  @ @

!

dt

b

2 Z T 2



2



2



2  C 4 b H 23 + ( ) dt + bt H 12 + ( ) + b H 21 + ( ) +

@@t

0 L2 (?1 )

L2 (?1 )

3 5 dt:

(93)

(B5) By [1] (p. 311, Theoreme 3) and trace theory we deduce that @@ D 0 2 L(H 1 ( ); H ? 21 (?)), and so accordingly we have c1

Z s

0 b  @D 0 bt bt;  @D @ ? @

!

L2 (?1 )

Z T 

1

2 Z T



2

12

2   2 b b dt  C t H 12 (? ) dt + 6

AR L2 ( ) + P bt L2 ( ) dt: 1

(94)

0

0

(B6) As A?D1 2 L(H ?1( ); H01 ( )), by the characterizations of elliptic operators given in [7], we then have for

all t 2 [0; T ],

2

1 2



 b  rt(t); rA?D1 b(t) L2 ( )  C

rbt (t)

L2 ( )

rA?D1 b(t)

L2 ( )  6

P 2 bt (t)

L2 ( ) + C

b

L1(0;T ;H ?1 ( )) : We thus have





t=  



t=

c1 rbt(t); rA?D1 b L2 ( ) + c 1 bt(t); A?D1 b(t) L2 ( ) t=s



2

12

2

12 t=s

2   b b b  6 P t ( ) L2 ( ) + 6 P t(s) L2 ( ) + C L1 (0;T ;H ?1 ( )) :

(95)

(B7) Finally, we can use the trace result Lemma 4.5 and the fact that A?D1 2 L(H ? 12 + ( ); H 23 + ( )) (again

by [7])) to have

Z ?1 b ? c 1
b; @A@D s

!

L2 (?0 )

dt

Z 



2 Z T



2 Z 





 b b b  C
 L2 (?0 ) H ? 12 + ( ) dt  e
 L2 (?0 ) dt + C b H ? 12 + ( ) dt 6C s 0 s (where the constant Ce above is the very same as that in (155)) Z T 

1

2

21

2   b 2   6

A  L2 ( ) + P bt L2 ( ) dt 0

12

2

21

2 

1

2

1

2  b b b 2 2   + 6 A (s) L2 ( ) + P t(s) L2 ( ) + A ( ) L2 ( ) + P bt( ) L2 ( ) Z T 2

2 +C

b

H 12 + ( ) dt + CT;

b

L1 (0;T ;H ? 21 + ( )) : 0

(96)

Therefore, if we de ne F (s;  ) to be the estimates (90){(96) then give

F (s;  )  2  (Right Hand Side of (89)) ;

(97)

3 1

b

2



2 2 2 @  +

t

+

b

H 21 + ( ) +

b

H 23 + ( ) +

bt

H 21 + ( ) 5 dtA H (?1 ) @ 2 0 L (?1 ) Z

21

2 

21

2 T  1 2

2 b b 2 +CT k kL1(0;T ;H ? 21 + ( )) + 

A  L2 ( ) + P t L2 ( ) + AR b L2 ( ) dt 0 

1

2

21

2

12

2 

1

2 b b b 2 2   + A (s) L2 ( ) + P t (s) L2 ( ) + A ( ) L2 ( ) + P bt( ) L2 ( ) ; (98)

0Z 2 T 2 F (s;  )  C @ 4

bt

12

17

where the constant CT above depends upon time, but the constant C does not (we have not noted the noncrucial dependence of C and CT on ). This and the equality (89) prove (a). To prove (b), we combine (79) and (80) and subsequently take  > 0 small enough. The proof of Lemma 2.2 is concluded.  Taking an arbitrary radial vector eld h 2 R2 of the form h(x; y)  [x ? x0 ; y ? y0 ] , (99) where (x; y) 2 and (x0; y0 ) 2 R2 is xed, one has the following relation which is essentially demonstrated in [12], the complete proof of which is carried out in the Appendix below (Proposition 4.6). i h Proposition 2.3 With the vector eld h as de ned in (99), the solution b; bt ; b to (78), corresponding to terminal    data [0 ; 1 ; 0 ] 2 D A 2 , satis es the following equality for arbitrary 0 2 [0; T ):









Z T ?0

21

2 1 Z T ?0

 1 b 2 b 2 dt c0 b + c2 b + c3bt + c4
b; h
rb ? 21 b

A  L2 ( ) + P t L2 ( ) dt =  2 0 L2 ( ) 0   Z T ?0  Z T ?0 Z 2  1 1 2 b b b b ?c1 ; h
rt ? 2 t h
 t + rbt d?dt dt + 2 0 ?1 0 L2 ( ) Z T ?0



2 Z T ?0 Z  2 1 b ? t L2 ( ) dt + 2 h

b d?dt

0 ?0 0 ! 3  Z T ?0 2  @    b 1 @ @ 1 5 dt 4 b; h
rb ? b + @  2 b + 2 bt ?  @ ; h
rb ? 2 b ? @ 2 2 0 L (?1 ) L2 (? ) t=T ?0    ? (bt; h
rb)L2( ) + rbt; r h
rb L2 ( ) ? 12 (bt; b)L2 ( ) t=0 "  #t=T ?0    + 2 rbt ; rb L2 ( ) + c1 b; h
rb ? 12 b L2 ( ) t=0 2 !2 3 ! ! ! 2 b! Z T ?0 Z h
 @ 2 b 2 @ 2 b 2 2 2 b b 4 2 + 2 + 2 @ 2 @ 2 + 2(1 ? ) @  5 dtd?: ? 2 @x @y @x @y @x@y ?1 0 h b b bi 

Lemma 2.4 For all 0 2 (0; T ), the solution ; t ;

1

(100)

to (78) satis es

 2 Z T ?0 

1

2

1 2   A 2 b L2 ( ) +

P 2 bt

L2 ( ) dt  CT

b

L1 (0;T ;H ? 21 + ( ))

0 31

b

2 Z T 2



2



2 2 2 @ 





+ 4 bt H 1 (? ) +

@t

+ b H 21 + ( ) + b H 23 + ( ) + bt H 21 + ( ) dt5A 1 0 L2 (?1 ) 

1

2

2 

1

2

2

1

1 +C  A 2 b(T ? 0 ) L2 ( ) +

P 2 bt(T ? 0 )

L2 ( ) +

 A 2 b(0 ) L2 ( ) +

P 2 bt(0 )

L2 ( ) ;

(101)

where the constant CT depends on T , but the constant C does not. Proof: We proceed to majorize the right hand side of (100). To this end, we now specify (x0; y0 ) 2 R2, so that h(x; y)  [x ? x0; y ? y0 ] of (99) satis es h
  0 on ?0: (102)

(A.1) First o, we have by Cauchy Schwarz and the inequality ab  a2 + C b2 ,

Z T ?0 

 1 b b b b b b c0 + c2  + c3 t + c4
; h
r ? 2  dt 0 L2 ( ) Z T 



2



2



2  ! Z T ?0

1

2  C

A 2 b L2 ( ) dt:

b L2 ( ) dt + b H 1 ( ) dt + bt L2( ) dt +  0

0

18

(103)

(A.2) Likewise by ab  a2 + C b2 and Sobolev trace theory,



2 # + bt L2 ( ) dt c1 ? 0 L2 ( ) ! ! 3 Z T ?0 2  2b b  @ b b @  @  1 b 4?   + 2 t ?  ; h
rb ? b 5 + @ @ 2 L2 (? ) + 2 ; @ L2 (? ) dt 0 1 1 Z T ?0

1

2 Z T 2

@ bt

2



2 3



2



2  C 4

@

dt + b H 23 + ( ) + bt L2 ( ) + b H 12 + 5 dt + 

P 2 bt L2 ( ) dt: (104)  0 0 L2 (?1 ) Z T ?0 " 

b; h
rbt ? 1 bt 2

(A.3) Using (102), we have



1 Z T ?0 Z h
 
b2  0: 2 0 ?0

(105)

(A.4) As h
rb ? 12 b 2 H?10 ; ( ), we have for all t 2 [0; T ]



   b(t); h
rb(t) ? 1 b(t) b(t); h
rb(t) ? 1 b(t) = 2 2 L2 ( ) H??01; ( )H?10 ; ( ) !

2



2

1

2 !



2

1 b

b b

b A 2 b(t) L2 ( ) ;  C L1 (0;T ;H ?1 ( )) +   C (t) H ?1 ( ) + h
r(t) ? 2 (t) 1 ?0 ; ?0 ; H ( ) ?0 ;

accordingly, there is the estimate

t=T ?0  b  b 1 b b b b ? (t; h
r)L2 ( ) + rt ; r h
r L2 ( ) ? 2 (t; )L2 ( ) t=0 "  # t = T ?    0 

1

2  b(T ? 0 )

2 2 2 A + 2 rbt ; rb L2 ( ) + c1 b; h
rb ? 12 b  C

b

L1 (0;T ;H ? 21 + ( )) +

 L ( ) L2 ( ) t=0 

1

2

2

1

1

2 (106) : +

P 2 bt(T ? 0)

L2 ( ) +

 A 2 b(0 ) L2 ( ) +

P 2 bt(0)

L2 ( ) :  @  R T ?0 b b h
r : Noting that (A.5) Handling the term ?  ; 

0

@

@ h
rb =  b +  (x ? x ) b +  (y ? y ) b +  (x ? x ) b +  b +  (y ? y ) b ; 1 x 1 0 xx 1 0 xy 2 0 xy 2 y 2 0 yy @

(107) we then have by Cauchy Schwarz, the trace estimate (172) for Kircho plates in the Appendix below, the use of the boundary data in (78) and standard Sobolev trace theory that

Z T ?0  @    b; @ h
rb dt  Z0T ?0 



2



2



2



2



2



2 b b b b b b C

H 21 + ( ) + xx L2 (?1 ) + yy L2 (?1 ) + 2 xy L2 (?1 ) + x H 12 + ( ) + y H 21 + ( ) dt 0 2 3

2 b

2

2 b

2

2 b

2 Z T ?0



2



2 2 @  @  @ 





b

b 4 b 12 + +

2

5 C @ L2 (?1 ) +

@ 2

L2 (?1 ) + 2

@@

L2 (?1 ) + x H 12 + ( ) + y H 21 + ( ) dt H ( ) 0

2 #

@  Z T "

2



2  2 CT

c0 b + c1 bt + c2 b + c3bt + c4
b [H 23 ? ( )]0 + b H 21 + ( ) +

@  b + 2 bt

2 dt 0 L (?1 )  ! Z T 



2



2



2 + + rbt L2 (? ) + b H 32 + ( ) + bt H 21 + ( ) dt 1 0  Z T 



2



2



2



2



2 b b b b b b (108) CT

t H 1 (?1 ) + c1 t + c4
 [H 32 ? ( )]0 +  H 32 + ( ) + t H 21 + ( ) + H 21 + ( ) dt 0 ?

 =





19

R

2

To handle the term 0T

c1 bt + c4
b

[H 32 ? ( )]0 dt, we use Proposition 4.4 in the Appendix below, and the fact that bt = ? b(t) + e?t t (t) and b(t) = e?t(t) to have

Z T

Z T

2

2 b b c1 t + c4
 [H 23 ?( )]0 dt = ?c1 b(t) + c1 e?t t (t) + c4e?t
(t)

[H 32 ? ( )]0 dt

0 0

@

2 # Z T "



2 2 2 2 b 2 + kkH 23 + ( ) + k kH 12 + ( ) + kt kH 12 + ( ) +

t

 C

@ L2 (?1 ) dt L ( ) 0

b

2 3 Z T 2



2



2



2 5 dt:  CT 4 b H 32 + ( ) + b H 21 + ( ) + bt H 21 + ( ) +

@@t

0 L2 (?1 )

(109)

Collectively, the estimates (108) and (109) then give

0Z 2

b

2 3  Z T ?0  @  T  2

5 dt  b; @ h
rb dt  CT @ 4 bt H 1 (? ) +

@@t

? 1 0 0 2 L (?1 )  ! Z T 



2



2



2 +

b H 23 + ( ) + bt H 21 + ( ) + b H 21 + ( ) dt : 0

(110)

(A.6) In the same way as in (A.5) we have,

! 2 b! !2 3 Z T ?0 Z h
 2 @ 2 b !2 @ 2 b!2 2 2 b b 4 2 + 2 + 2 @ 2 @ 2 + 2(1 ? ) @  5 dtd? ? 2 @x @y @x @y @x@y ?1 0 3 2

b

2 Z T



2



2 2 2 +

b

H 23 + ( ) +

bt

H 21 + ( ) +

b

H 21 + ( ) 5 dt:  CT 4 bt H 1 (? ) +

@@t

1 0 L2 (?1 )

(A.7) Finally,

1 Z T ?0 Z 2

0

?1

2

b

2  Z T 2 2

2 h
 bt + rbt d?dt  C 4 bt H 1(? ) +

@@t

1 0 

L2 (?1 )

3 5 dt:

(111)

(112)

The estimate (101) now comes about by stringing together (100), (103){(106), (110){(112), and taking  > 0 small enough. 

i

h

Lemma 2.5 For T > 0 large enough, the solution b; bt ; b of (78) satis es the following estimate:

Z T 

1

2

21

2

1

21

2 

1

2

2 b b 2  A 2 b(T ) L2 ( ) +

P 2 bt (T )

L2 ( )

A  L2 ( ) + P t L2 ( ) + AR b L2 ( ) dt +  0 3 0 2

b

2 Z



T 2 2 2 2 @  +

b

H 12 + ( ) +

b

H 23 + ( ) +

bt

H 21 + ( ) 5 dt  CT @k 0 k2L2 ( ) + 4

bt

H 1(? ) +

@t

1 0 L2 (?1 ) 



2 + b L1 (0;T ;H ? 12 + ( )) : (113)

20

Proof: We have for any 0 2 (0; T ),

Z T 

1

2

1 2   A 2 b L2 ( ) +

P 2 bt

L2( ) dt

Z0 0 

1

2

21

2  Z T 

1

2

21

2  b b b 2 2   + + P   = A dt + A 

L2 ( ) t L2 ( )

L2 ( ) P bt L2 ( ) dt 0 T ?0 Z T ?0 

1

2

12

2  b 2  +

A  L2 ( ) + P bt L2 ( ) dt 0 

2

2 

12 2  0 (1 + )  12 b 2 b  1 ?  A (T ) L2 ( ) + P t(T ) L2 ( ) + CT

b

L1 (0;T ;H ? 21 + ( )) 3

b

2 Z T 2



2



2 2 2 +

b

H 21 + ( ) +

b

H 23 + ( ) +

bt

H 21 + ( ) 5 dt +C 4 bt H 12 (? ) +

@@t

1 0 L2 (?1 )    Z

T

A 12 b

2 +

P 21 bt

2 +

AR21 b

2 dt + Z T ?0

A 21 b

2 +

P 12 bt

2 + 20 



dt 1? 0 L2 ( ) L2 ( ) L2 ( ) L2 ( ) L2 ( ) 0 (after applying Lemma 2.2(b) twice)  1 2 Z T 

1

2

1 2 

1

2 

1 2   201(1?+ )

 A 2 b(T )

L2 ( ) +

P 2 bt(T )

L2 ( ) + 12?0  A 2 b L2 ( ) +

P 2 bt

L2 ( ) +

AR2 b

L2 ( ) dt

0 3 1 0 2

b

2 Z T



2

2

2

2

2 +CT @ 4 bt H 1 (? ) +

@@t

+

b

H 21 + ( ) +

b

H 32 + ( ) +

bt

H 21 + ( ) 5 dt +

b

L1 (0;T ;H ? 21 + ( )) A 1 0 L2 (? ) 

1

21 1

1

2

2

2 

1

2 +C  A 2 b(T ? 0 ) 2 + P bt (T ? 0 )

2 +

 (114) A 2 b(0 ) 2 +

P 2 bt (0 )

2 ; L ( )

L ( )

L ( )

L ( )

after applying Lemma 2.4. Applying Lemma 2.2(b) twice more to the right hand side of (114) yields



1 2 Z T 

1

2

2 

21

2 

21 1 + 

b b b b

A 2  L2 ( ) + P t L2 ( ) dt  C0 1 ?  A 2 (T ) L2 ( ) + P t(T ) L2 ( ) 0 0 3

b

2 Z T 2



2



2 2 2 +

b

H 21 + ( ) +

b

H 23 + ( ) +

bt

H 21 + ( ) 5 dt +CT @ 4 bt H 1 (? ) +

@@t

1 0 L2 (?1 ) Z

21

2 

21

2

 C0  T 

 1 b 2 b 2 1 + + k k2 1 A + dt;  + P 

AR b

t ? + L (0;T ;H

2

( ))

1?

L2 ( )

L2 ( )

0

L2 ( )

(115)

where C0 = 2 (0 + C ). Moreover, we can multiply the equation (74) by b, integrate in time and space, use the characterization (17), and the fact that b(T ) = e?T 0 to have



2 t=T Z T   Z T

1

2  

2 b b b b b  AR L2 ( ) dt = 2 L2 ( ) +  ? 
; L2 ( ) dt 0 0 t=0 2 ! 3 ZT   b 5 dt +  4 rbt ; r b L2 ( ) ? @@t ; b 0 L2 (?1 ) 2 ! Z     T b

@  2 t ? T  2 e 0 L2 ( ) + 4 b ? 
b; b L2 ( ) + rbt ; r b L2 ( ) ?  @ ; b 0

L2 (?1 )

Majorizing this expression results in

3 2 0 Z Z T

1

2

T @ b 2 2 +

b

H 21 + ( )) 5 dt  AR2 b L2 ( ) dt  C @k 0k2L2 ( ) + 4

@t

0 0 L2 (?1 ) Z T 

1

2

21

2  ! Z T

21

2 +

A 2 b L2 ( ) + P bt L2 ( ) dt + e AR b L2 ( ) dt; e

0

0

21

3 5 dt: (116)

and taking e > 0 small enough above, this becomes

0 2 Z T

1

2 Z T @ b 2

AR2 b L2 ( ) dt  C1 @k 0 k2L2 ( ) + 4

@t

0 0

L2 (?1 )

1

2



2

12

2 3 1 A 2 b L2 ( ) + P bt L2 ( ) 5 dtA ; + b H 12 + ( )) + 

(117)

where C1 = C?e e . Combining (115) and (117) then, we have

Z T 

1

2

12

2 

21

2 b b 2 

A  L2 ( ) + P t L2 ( ) + AR b L2 ( ) dt 0 



2 

2 

12

( C 1 + 1) C0 (1 + )  12 b 2 2 b  C1 k 0kL2 ( ) +

A (T ) L2 ( ) + P t (T ) L2 ( ) + CT b L1 (0;T ;H ? 21 + ( )) 1 ?  3 1

b

2 Z T 2



2



2 2 2 @  +

b

H 21 + ( ) +

b

H 23 + ( ) +

bt

H 21 + ( ) 5 dtA + 4 bt H 1(? ) +

@t

1 0 L2 (?1 )  Z

21

2 

21

2 T 1 b 2 C

0 b 2 A  dt: (118) + AR b + (C1 + 1) + P t



1? 0 L2 ( ) Taking  > 0 small enough above will then yield

L2 ( )

L2 ( )

Z T 

1

2

21

2 

12

2 b b

A 2  L2 ( ) + P t L2 ( ) + AR b L2 ( ) dt 0 



2

1

2

2 

12 2 b b  C k 0 kL2 ( ) +  A 2 (T ) L2 ( ) + P t(T ) L2 ( ) + CT b L1 (0;T ;H ? 21 + ( )) 3 1

b

2 Z T 2



2



2 2 2 4 bt 1 +

@ t

+

b

H 12 + ( ) +

b

H 23 + ( ) +

bt

H 21 + ( ) 5 dtA ; @ H (? ) 1 0 L2 (?1 )

(119)

where CT is a constant which depends on T , but the constant C does not. In addition, we have by (79) that

 1

2 Z T 

1

2

2  Z T

1 2 

1  A 2 b(T ) L2 ( ) +

P 2 bt (T )

L2( ) + F (T; t)dt; A 2 b(t) L2 ( ) +

P 2 bt (t)

L2 ( ) dt = T



0 0

(120)

where the function F is as de ned in (97). Combining (119) and (120) then yields

Z T  1

1

2

12

2

21

2  1 b b b 2  2 A (t) L2 ( ) + 2 P t(t) L2 ( ) + AR L2 ( ) dt 0  1

2

2  Z T

1 + T2

 A 2 b(T ) L2 ( ) +

P 2 bt (T )

L2 ( ) + 21 F (T; t)dt 0   2 



2 2 1 1  C k 0 k2L2 ( ) +

 A 2 b(T )

L2 ( ) +

P 2 bt(T )

L2 ( ) + CT

b

L1(0;T ;H ? 21 + ( )) 3 1

b

2 Z T 2



2



2



2



2 @ 

t + b H 21 + ( ) + b H 23 + ( ) + bt H 21 + ( ) 5 dtA : + 4 bt H 1 (? ) +

@

1 0 L2 (?1 )

22

(121)

Applying the estimate (80) to (121) then gives

Z T  1

1

2



b(t) 2 + 1

P 21 bt (t)

2 2 +

AR21 b

2 2 dt 2  A

L ( ) L ( ) 0  2

1

2L ( )

2 1

2 T + 2  A 2 b(T ) L2 ( ) + P 2 bt(T )

L2 ( ) 



2 

1

2

2 

21 2 b b  C k 0 kL2 ( ) + (C + T)  A 2 (T ) L2 ( ) + P t (T ) L2 ( ) + CT b L1 (0;T ;H ? 12 + ( )) 3 1

b

2 Z T 2



2



2 2 2 4 bt 1 +

@ t

+

b

H 21 + ( ) +

b

H 23 + ( ) +

bt

H 21 + ( ) 5 dtA @ H (? ) 1 0 L2 (? ) Z T 

1

2 1

1

2

1 2   + (T + ) A 2 b(t) L2 ( ) + P 2 bt(t) L2 ( ) +

AR2 b(t)

L2 ( ) dt:

0

(122)

Invoking the estimate (119) to the right hand side of this expression nally yields

Z T  1

1

2

21

2 

21

2 1 b b

A 2 (t) L2 ( ) + 2 P t (t) L2 ( ) + AR b L2 ( ) dt 2 0 

1

2

12

2  T b b 2 + 2  A (T ) L2 ( ) + P t (T ) L2 ( ) 

1

2

12

2  2 b b 2  C (T +  + 1) k 0 kL2 ( ) + (T (C + 1) + C (1 + ))  A (T ) L2 ( ) + P t (T ) L2 ( ) 3 0Z 2

b

2



T 2 2 2 2 +

b

H 12 + ( ) +

b

H 23 + ( ) +

bt

H 21 + ( ) 5 dt +CT @ 4

bt

H 1(? ) +

@@t

1 0 L2 (?1 ) 



2 + b 1 : ? 1 + L (0;T ;H

2

( ))

(123)

From this, one has the asserted inequality (113) for T > 1 C (1 + ) (where  < 2(C1+1) ).  2 ? (C + 1) Conclusion of the Proof of Theorem 2.1: Assume initially that [0 ; 1 ; 0 ] 2 D(A ). Through the change i h 2 of variable b(t) = e?t(t) and b(t) = e?t (t), where again b; bt ; b solves (78) and   2  > 0, we have for

23

T > 0 large enough,

Z T 

1

2

12

2

1

2

21

2 

1

2 2  A (t) L2 ( ) + P t (t) L2 ( ) + AR (t) L2 ( ) dt +  A 2 (T ) L2 ( ) +

P 2 t (T )

L2 ( )

Z0 T 

1

2 

1

1

2  2  = A 2 et b(t)

L2 ( ) +

P 2 et bt (t) + et b(t)

L2( ) +

AR2 et b(t)

L2 ( ) dt

0

1 T

2

1  2 +

 A 2 e b(T ) L2 ( ) +

P 2 eT bt (T ) + eT b(T )

L2 ( ) 0 2 3

b

2 Z



T 2 2 2  CT @k 0 k2L2 ( ) + 4

bt

H 1 (? ) +

@@t

+

b

H 12 + ( ) +

bt

H 21 + ( )5 dt 1 0 L2 (?1 ) 



2



2 b b +  L1 (0;T ;H 32 + ( )) + L1(0;T ;H ? 12 + ( )) (after using the estimate " (113))

@ ?

2 # ZT


2 ? t ? t ? t ? t

e t(t) ? e (t) H 1 (?1 ) + @ e t(t) ? e (t)

dt  CT k 0 kL2 ( ) + 0 L2 (?1 ) ! Z T h i

2 2 2 2 ? t ? t ? t

e (t) H 12 + ( ) + e t (t) ? e (t) H 21 + ( ) dt + kkL1 (0;T ;H 32 + ( )) + k kL1(0;T ;H ? 21 + ( )) + 0 #

@ (t)

2 Z T" t 2 1 2 2

 CT k 0 kL2 ( ) + kt (t)kH 1(?1 ) + @ 2 + k kH 2 + ( ) dt 0 L (?1 )  2 2 2 2

+ kkL1 (0;T ;H 23 + ( )) + ktkL1 (0;T ;H 12 + ( )) + k kL1 (0;T ;H ? 12 + ( )) :

(124)

This gives the desired inequality (68). 

2.3 Conclusion of the Proof of Theorem 1.2

For [0 ; 1 ] 2 D(LT ), we immediately have from Theorem 2.1 the corollary:

Corollary 2.6 For [0 ; 1 ] 2 D(LT ), the corresponding solution [; t ; ] of (63) satis es the following inequality:

Z T 

1

2

1 2

12

2

21

2 

1

2 2  A 2 0 L2 ( ) +

P 2 1

L2( ) A (t) L2 ( ) + P t (t) L2 ( ) + AR (t) L2 ( ) dt + 

0 #

@ (t)

2 Z T" t 2 2  CT kt(t)kH 1 (?1 ) +

@

2 + k kH 12 + ( ) dt 0 L (?1 )  2 2 2 + kkL1(0;T ;H 32 + ( )) + ktkL1(0;T ;H 12 + ( )) + k kL1 (0;T ;H ? 12 + ( )) :

(125)

We will have the desired inequality (67) by now eliminating the tainting lower order terms in (125). To this end, we invoke a (by now) classical compactness{uniqueness argument (see e.g. [13] and [2]) which makes crucial use of the Holmgren's{type uniqueness result for the thermoelastic system posted in the introduction. It is at this point that the boundary trace j?2 , corresponding to the control u3, comes into play.

Lemma 2.7 For T > 0 large enough, the existence of the inequality (125) implies that for initial data [0 ; 1 ] 2 D(LT ), there exists a CT such that the following estimate holds true: kk2L1 (0;T ;H 23 + ( )) + ktk2L1 (0;T ;H 12 + ( )) + k k2L1(0;T ;H ? 12 + ( )) +  CT

Z T" 0

ZT

k k2H 21 + ( ) dt

!

@

2 # t 2 kt kH 1 (?1 ) +

@

2 dt + k k[H s(0;T ;L2+ (?2 ))\L2 (0;T ;Hs+ (?2 ))]0 : L (?1 ) 2

24

0

(126)

nh

Proof: If the proposition is false, there then exists a sequence (0n) ; (1n) nh io1 solution sequence (n) ; (tn) ; (n) n=1 to (63) which satis es

io1 n=1

 D(LT ), and a corresponding

(n)

2

2

2

 L1 (0;T ;H 23 + ( )) +

(tn)

L1 (0;T ;H 21 + ( )) +

(n)

L1 (0;T ;H ? 21 + ( )) Z T 2 (127) +

(n)

H 12 + ( ) dt = 1 8 n; 2Z 20 3 3

(n)

2

(n)

2 T 2 @t



( n ) 4 4 5 +  lim dt +



[H s(0;T ;L2 (?2 ))\L2 (0;T ;H s (?2 ))]0 5 = 0. (128) t H 1 (? ) @ n!1 0 1 + + 2 L (?1 )

(125) and (127){(128) then imply the boundedness of the sequence

8Z 2  91 < T (n) (t) 

2

21 (n)

2 3

" (0n) #

2 =

(n)

1 4 5 + A (129) : dt + ( t )



(n) R :0 ;

L2 ( ) t (t) D(A 2 )H?1 ; ( ) 1  1 ( ) 1  2 )  H D ( A 0 ?0 ; n=1 nh (n) (n) io1 he e i 1 1

There thus exists a subsequence, still denoted here as that

n=1

, and 0; 1 2 D( A 2 )  H?0 ; ( ), such

(0n) ! e0 in D( A 12 ) weakly; (1n) ! e1 in H?10 ; ( ) weakly.

i

h

0 ; 1

(130) (131)

i

h

If we further denote e; et ; e as the solution to (63), corresponding to initial data e0 ; e1 ; 0 , then a fortiori,

h

(n) ; (tn) ;

i h e e ei ! ; t ; in L1 (0; T ; H ) weak star.  h 1 i0  n (n) o1 ?1 1 (n)

From Proposition 4.3 we have that tt

( h

(0n) ; (1n) i

1 D ( A 2 )H 1 (n)

t

)1

?0 ; ( )

n=1

n=1

is bounded in L

0; T ; D( A 2 P )

(132) , inasmuch as

is bounded in D( A 21 )  H?10 ; ( ). Also, from Proposition 4.4, we have that

2 L2 (0; T ; [H 32 ? ( )]0) for all n, with the estimate

Z T

0



(n) 2 t [H 23 ? ( )]0 dt

Z T 2

 C 4 0 n

(n)

2

2 2

@t



( n ) (n) + + 

1 + 1 + t 2 2 H ( ) H ( ) @

L2 (?1 )

o1

3 5 dt,

(133)

and this combined with (127) yields that t(n) n=1 is bounded in L2(0; T ; [H 32 ? ( )]0). This boundedness of nh (n) (n) io1 tt ; t n=1 , and that for the sequence posted in (129), alllows us to deduce through a compactness result of Simon's in [22] that

(n) ! e strongly in L1(0; T ; H 23 + ( )); (tn) ! et strongly in L1 (0; T ; H 21 + ( )); (n) ! e strongly in L2 (0; T ; H 21 + ( )); (n) ! e strongly in L1 (0; T ; H ? 12 + ( )).

These convergences and (127) thus give

2

2



2

e L1(0;T ;H 32 + ( )) +

et

L1 (0;T ;H 21 + ( )) +

e

L1 (0;T ;H ? 21 + ( )) Z T 2 +

e

H 12 + ( ) dt = 1. 0 25

(134)

Moreover, representation of LT  in (62), and the convergences posted in (128) and (130){(132) give h e e i the explicit that 0 ; 1 2 D(LT ), with



2 Z T 2



2 4 et 1 +

@ et

@ H (?1 ) 0

L2 (?1 )

Now if we make the change of variable

3

5 dt +

e

2

[H s (0;T ;L2+ (?2 ))\L2 (0;T ;Hs + (?2 ))]0

= 0.

(135)

$ = et ; # = et ;

then using (135), [$; #] solve the system

8  $ ?
$ +
2 $ + 
# = 0 > tt tt on (0; 1)  ; > # > t + 
# ? # ? 
$t = 0 > > > > = 0 on (0; 1)  ?; $ = @$ > @ > > < (
$ + (1 ? )B1$ + # = 0 @
$ + (1 ? ) @B2 $ ? @$tt +  @# = 0 on (0; 1)  ?1 ; > > > @ @ @ @ > > > @# + # = 0 on (0; 1)  ?; > > @ > > :

(136)

# = 0 on (0; 1)  ?2 :

Consequently, by Isakov's uniqueness theorem in [10], we have that for T > 0 large enough [$; #] = [0; 0], and so e is a constant and e is a constant. From the essential boundary condition on ?0 in (63), weh theni have e = 0 on (0; T )  . In turn, the free boundary conditions on ?1 give that e = 0 on (0; T )  . Thus e; e = [0; 0], which contradicts the equality given in (134). This concludes the proof of the lemma. 

Corollary 2.6 and Lemma 2.7 in combination give the inequality (67),1 the establishment of which veri es the surjectivity of the control to partial state map LT : D(LT )  Us ! D( A 2 )  H?10; ( ). This completes the proof of Theorem 1.2.

3 The Proof of Theorem 1.1

Given the space C r (2;T ), we consider the system (1) under the in uence of boundary controls in Ur+1 , as de ned in (40). The controlled PDE is then approximately controllable for T large enough. Indeed, if we take arbitrary [0 ; 1 ; 0 ] from the null space of LT , then using the form of this operator given in (58), we have necessarily that t tj?1 = @ @ ?1 = 0, and j?2 = 0, where [; t ; ] is the solution to (54). We can then use the uniqueness theorem of Isakov, in a fashion similar to that employed in Lemma 2.7, to show that [; t ; ] = [0; 0; 0] on (0; T )  , and in particular [0 ; 1 ; 0 ] = [0; 0; 0]. With the designated control space Ur+1 we then take T > 0 large enough so as to ensure both the approximate controllability of the entire system (1), and the exact controllability with respect to the displacement (see Theorem 1.2). In this event, we have the observability inequality (67), and therewith one can show in a manner identical to that done in [14] (Appendix B) that LT LT  is boundedly invertible as an element of L(D(LT ); [D(LT )]0), 1 where the projection  onto D( A 2 )  H?10 ; ( ) is as de ned in (49). Subsequently, by the Closed Graph Theorem, we will have that LT LT  (LT LT )?1  2 L(H ).   Step 1. For arbitrary h > 0 we select a u1 2 iD(LT )  Ur+1 , so that for any terminal state !0T ; !1T ; 0T 2 H , the corresponding solution !(1) (t); !t(1) (t); (1) (t) to (1) with u  u1 satis es

2 !(1) (T ) ? !T 3 2! 0 0

4 (1) T 5 + eA T 4 !1 t (T ) ? !1

!(1) T   (T ) ? 0

0

3

 5

<

;   

H 1 + (I?  )LT LT  (LT LT )?1 

L(H )

26

(137)

where I : H ! H denotes the Identity). Step 2. We h now select u2 2 D(LT ) to i be the \minimal norm steering control" with respect to the partial terminal state !0T ? !(1) (T ); !1T ? !t(1) (T ) . That is to say, u2 satis es

2! 3  0 T ? ! (1) (T )  ! 0 A T

5 4 !1 = T (1) ; LT u2 + e

(138)

!1 ? !t (T )

0

and minimizes the functional 12 kukUs , over all u 2 Us which satis es LT u = 2

 !T ? !(1) (T )  0

!1T ? !t(1) (T )

2! 3 0 ? eA T 4 !1 5 0

(by Theorem 1.2, we know there exists at least one such u). By convex optimization theory and Lax{Milgram, the minimizer u2 can be given explicitly by

2 ! 31 02 !T ? !(1) (T ) 3 0 0 u2 = LT  (LT LT )?1  @4 !1T ? !t(1) (T ) 5 ? eA T 4 !1 5A T 0

?

0

(1) (T )

(139)

(see (B.20) of [14], p. 288). With this representation, we then have from (137) the norm bound

kLT u2 kH 



 

LT LT  (LT LT )?1 



1 +

(I? )LT LT  (LT LT )?1 

L(H

:

(140)

)

Step 3. Set the control u = u1 + u2 . Consequently, there is the equality 2 2! 3 2 ! 3 2 !T 3 2 ! 33 0 0 0 0 LT u + eA T 4 !1 5 = LT u1 + LT u2 + eA T 4 !1 5 = 4 !1T 5 + (I? ) 4LT u2 + eA T 4 !1 55 : 0 0 0 (141) (1) (T ) Letting [!; !t ;  ]denote the solution of (1) corresponding to the chosen control u , we then have from (141) that [! (T ); !t (T )] = !0T ; !1T . Moreover, from (141), (137) and (140) we obtain the estimate

2 2 2 ! 33

3 0 0

(T ) ? T

) 4eA T 4 !1 55

+ k(I? )LT u2 k

5 4 0 + ( I ?   2 0 L+ ( ) H

(1) (T ) ? T 0 H 0 

< + k(I? )LT u2kH < : (142) ? 1      1 + (I?  )LT LT  (LT LT  )  L(H )

Thus, the constructed control u = [u1 ; u2 ; u3 ] 2 Ur+1 satis es the desired exact{approximate controllability property. Moreover, the Sobolev Embedding Theorem gives that u3 2 C r (2;T ). This concludes the proof of Theorem 1.1.

4 Appendix

AG2 0? AG1 0 is an element of L(L2( ); [D( A 21 )]0 ) and Proposition 4.1 The operator AR ?  +    AR ?  +  AG2 0 ?  AG1 0 = AD (I ? D 0) as elements of L(D( A 21 ); L2( )).

27

Proof. For every # 2 D(AR) and $ 2 D( A 21 ), we have     G2 0 ? A G1 0 #; $ AR ?  + A [D( A 12 )]0 D ( A 21 ) 

 AG1 0#; $ [D(A 12 )]0 D(A 21 ) (?
#; $)L2 ( ) +  AG2 0#; $ [D(A 12 )]0 D(A 21 ) ?    ? #; G A$
?  A$
2 ; $ +  = (r#; r$)L2( ) ? @# 0 2 (?1 ) ? 0 #; G1  2 L L (?1 ) @ L2 (?1 )

(after the use of Green's formula and the taking of adjoints) ?
 = (r#; r$)L2( ) ? 0 #; G1  A$ L2 (?1 ) = (#; ?
$)L2 ( ) (after one more use of Green's Theorem and the characterization (26)) = (#; AD (I ? D 0)$)L2 ( ) : As D(AR) is dense in L2( ), this equality proves the assertion.  Lemma 4.2 The Hilbert space adjoint A of A , as de ned in (39), is given to be

0 I 0 0 10 0 ?I  B  ? 1 A 0 A = @ 0 P 0 A @  0 AD (I ? D 0) 0 0 I n 1 1 

0 1 ?(|) C A; ?  AR A 2 )  D(AR) \ L2+ ( ) with D(A ) = [0 ; 1 ; 0 ] 2 D( A 2 )  D ( such that  A0 +  AG1 0 0 2 H??01; ( ) and such that ? 
1 + 
0 2 L2+ ( )

(above, (|) is the same denotation made in (38)). Proof: We de ne S  H to be

S 

n

A 21 )  D(AR) \ L2+ ( ) [0 ; 1 ; 0 ] 2 D( A 12 )  D( such that  A!1 +  AG1 0 2 H??01; ( ) and such that ? 
1 + 
0 2 L2+ ( ) ;

h

i

and proceed to show that D(A ) = S . Indeed, if [!1 ; !2 ; ] 2 D(A ) and !e1 ; !e2 ; e 2 S we have by using (39),

0 2 ! 3 2 !e 31 1 @A 4 !12 5 ; 4 !e2 5A =  e H    1     A 2 !2 ;  A 21 !e1 L2 ( ) + P ?1(? A!1 + AR  ?   ?  A G

 +  A G

 ) ; ! e 1 0 2 0 2  H?10 ( )     e ? AD (I ? D 0)!2 ; e L2 ( ) ?  AR;  L2 ( )



 A!e2 [D(A 12 )]0 D(A 21 ) ?  (
; !e2 )L2 ( ) = !2 ;  A!e1 [D(A 12 )]0D(A 21 ) ? !1 ;      e



   ? AG1 0; !e2 [D(A 21 )]0D(A 21 ) +  AG2 0 ; !e2 [D(A 12 )]0 D(A 21 ) + 
!2 ;  2 ? ; AR e =

L ( )

(after

using  the equality posted

in (35) 

1 1 1 ? !1 ;  1 +  (r; r! !2 ;  A!e1 A!e2 e2 )L2 ( )  @ [D(A2 )]0D(A 2 )  @ !e2 [D(A 2 )]0 D(A 2 ) ? @ ; 0 !e2 2 ?  0 ; @ 2 ?  ( 0 ; 0!e2 )L2 (?1 ) L (?1 )  @!  L (?1 )      ? ; AR e ? r!2 ; re L2 ( ) +  @2 ; 0 e L2 (?1 ) L2 (?1 )

(after using Green's Theorem and (26))

28



L2 ( )

  

  + !2 ; A!e1 [D(A 21 )]0 D(A 12 ) + ; ?
!e2 L2 ( ) L2 ( ) !       e e @   e +  G1  A!2 ; 0  L2 (?1 ) ? ; AR e 2 + !2 ;
 L2 ( ) ?  0!2 ; @ L (?1 ) L2 (?1 )    1    

A!e  1  e e e  1 A 2 !1 ;  = ?  +  ! ; ? A  + ?  G A ! ;

 A 12 !e2 L2 ( ) + !2 ;  2 R 2 0 1 [D( 2 A2) A 2 )]0 D( L2 (?1 )  L2 ( )     E D + !2 ;  AG1 0 e [D(A 21 )]0D(A 21 ) + ;  AD (I ? D 0)!e2 2 ? ;  AR e 2 L ( ) L (?1 )    1  1 1 AG1 0e ?  AG2 0 e)] 2 A 2 !1 ;  A!e1 + (?AR e +  e +  = ?  A 12 !e2 L2 ( ) + P 2 !2 ; P 2 P ?1[ L ( )       + ; AD (I ? D 0)!e2 ? ; AR e 2 L2 ( ) h L (?1 ) ei (after again using (35), (26) and the fact that !e1 ; !e2 ;  2 S ) 02 ! 3 2 !e1 31 1 = @4 !2 5 ; T 4 !e2 5A ;  e H 

A 12 !1 ;  = ?  A 12 !e2

where

Thus,



0 I 0 0 10 0 ?I  B A 0 ? 1 @ A T  0 P 0 @  A (I ? D ) 0 0 0 I 0 D

0 1 ?(|) C A: ?  AR

S  D(A ) and A S = T :

(143)

To show the opposite containment: One can straightforwardly compute the inverse A? 1 2 L(H ; D(A )) as

1 0 2 ?1 ?1   ? 1 (|)A?1 ? 1P   A ( | ) A A ( I ? D ) ? A ? A ? 0

R D R C B   CC : (144) A? 1 = B I 0 0 B A @  0 ?  A?R1 ?  A?R1AD (I ? D 0) ?
In turn, one can use this quantity and Proposition 4.1 to compute the Hilbert space adjoint A ?1 of A? 1 as 0 2 ?1 ?1 1 ?1(|)A?1 ?1P ?     ? A ( | ) A A ( I ? D ) A A D 0 R R C   ?A
?1 = B B CC : (145) B ? I 0 0

@ A  ?1 ?1 ?  AR AD (I ? D 0)

0

?  AR

2! 3 0 With this quantity in hand, we then have that for arbitrary [0 ; 1 ; 0 ] 2 D(A ), and corresponding 4 !1 5 = 0 2 3 0 A 4 1 5 2 H , 0

2 2 ! 3 6 ? 2 A?1(|)A?R1AD (I ? D 0)!0 + A?1 P !1 ?  A?1(|)A?R10 2 3 0 0  4 1 5 = ?A
?1 4 !1 5 = 66  ? ! 0 4  ?1 ?1 0 0 ?  AR AD (I ? D 0)!0 ?  AR 0 29

3 77 75 : (146)

i2

h

A fortiori then, [0 ; 1 ] 2 D( A 21 ) and operator (|) in (38) gives

0

2 D(AR) \ L2+ ( ). Moreover, (146) and the de nition of the

 A0 +  AG1 01 2 H?10 ; ( ); AD (I ? D 0)1 ? AR 0 +  0 2 L2+ ( ): (147) Thus, D(A )  S , and this combined with (143) concludes Lemma 4.2.  Proposition 4.3 For arbitrary terminal data [0; 1 ; 0 ] 2 H , the solution [; t ; ] to (54) has the following additional regularity:

kttkL1(0;T ;[D(A 21 P ?1 )]0 )  C k[0 ; 1 ; 0 ]kH , de ned operator,  A 21 P ?1 : D( A 21 P ?1)  L2 ( ) ! L2 ( ), with where  A 21 P ?1 nis taken as a closed and densely o A 12 ) . D ( A 21 P ?1) = ' 2 L2 ( ) : P ?1' 2 D(

Proof: For terminal data [0 ; 1 ; 0 ] 2 D(A ), we have for all $ 2 L1(0; T ; D1( A 21 P ?1)), upon using the abstract equation (55), the characterizations in (26), the fact that P ?1$ 2 L1(0; T ; D( A 2 ) \ D(AN )) (recall the de nition of @ AN in (18) and P in (27)) and that @ ? = ? j? ,

ZT

($; tt)L2 ( ) dt =

ZT?

 A ? AG + AG ? 

dt $; P ?1 ? 1 0 2 0 L2 ( )

0  Z0 T   1  ? ? ?


1   ? 1 ? 1 ? 1 ? 1 2 2     = ? A P $; A  L2 ( ) ?  G1 AP $; 0 L2 (?1 ) +  G2 AP $; 0 L2 (?1 ) ?  P $;
L2 ( ) dt 0 " #   ZT  1  ?
@ 1 +  P ?1$; A2 = ? A 2 P ?1$;  dt ?  P ?1$;


0

= =

Z T  0

@

L2 ( )

 21 P ?1$;  ?A A 12 

? 
P ?1$; L2 ( )



?



L2 (?1 )



?
+  rP ?1$; r

A 12  ? A 12 P ?1$; 

Z T  0



L2 ( )




L2 ( )



L2 ( )

L2 ( )

dt

dt:

(148)

Estimating the farnside of this expression by using the fact that P ?1 2 L(L2( ); D(AN )), followed by the contraction o  of the semigroup eA t t0, one has the estimate

ZT 0

2  3

0 ($; tt)L2 ( ) dt  C

4 1 5

k$kL1 (0;T ;D(A 21 P ?1 )) dt:

0 H

(149)

A density argument concludes the proof. 

Proposition 4.4 If [; t ; ] denotes the solution to (54), corresponding to terminal data [0 ; 1 ; 0 ], we have the following estimates: 1. The map [0 ; 1 ; 0 ] !
 is an element of L(H ; L2 (0; T ; [H 1 ( )]0)), with the norm bound k
kL2 (0;T ;[H 1( )]0 )  C kkL2 (0;T ;H 32 + ( )) :



h

i2 

2. The map [0 ; 1 ; 0 ] ! [t ; t ] is an element of L D(LT ); L2 (0; T ; [H 32 ? ( )]0)

(150) , with the norm bound

!

@

t : k[t ; t ]k[L2 (0;T ;[H 32 ?( )]0 )]2  C k kL2 (0;T ;H 21 + ( )) + kt kL2 (0;T ;H 12 + ( )) +

@

2 2 L (0;T ;L (?1 )) (151)

30

Proof of (i): For all $ 2 L2(0; T ; H 1 ( )), we easily have

ZT 0

 C

(
; $)L2 ( ) dt = ?

Z Th 0

ZT 0

(r; r$)L2 ( ) dt +

Z T  @  0

i

@ ; $

L2 (?1 )

dt

krkL2 ( ) kr$kL2 ( ) + kkH 23 + ( ) k$kH 1 ( ) dt  C kkL2 (0;T ;H 32 + ( )) k$kL2 (0;T ;H 1 ( )) ,(152)

and this estimate gives the asserted result. Proof of (ii): If [0 ; 1 ; 0 ] 2 D(A ), then [; t ; ] 2 C ([0; T ]; D(A )) \ C 1([0; T ]; H ), 3and so in particular
t 2 L2(0; T ; L2 ( )). Taking the L2 {inner product with respect to arbitrary $ 2 L2(0; T ; H 2 ? ( )), we have upon the use of Green's Theorem, the de nition of AR in (15), and the fact that D(AR) = H 2 ( ) for 0    34 ? 2 (by [7]),

?

ZT 0

(
t ; $)L2 ( ) dt =

Z T

1 2

1 2

AR t ; AR $



dt ?

Z T  @t 

@ ; $ L2 (?1 ) dt Z T  1+  3?  Z T  @t  4 2 4 2 = AR t; AR $ L2 ( ) dt ? @ ; $ L2 (?1 ) dt 0 0 !

@

Z T

1 + 



3 ? 

t 4 2 4 2



AR t L2 ( ) AR $ L2 ( ) + @ 2 k$kH 23 ?( ) dt 0 ! L (?1 )

@

 kt kL2 (0;T ;H 12 + ( )) +

@t

k$kL2 (0;T ;H 32 ?( )) : L2 (0;T ;L2 (?1 )) 0

L2 ( )

0

Moreover, as [0 ; 1 ; 0 ] 2 D(A ), we can take the L2 {inner product of and use (56) and (153) to obtain

ZT 0

= ?1

( t; $)L2 ( ) dt = ?1

Z T 

ZT 0

AR4 + 2 ; AR4 ? 2 $ 1



3





t

with arbitrary $ 2 L2 (0; T ; H 32 ?( ))

(AR + 
t ; $)L2 ( ) dt



+ (
t ; $)L2 ( ) dt L2 ( )

!

@

t  C k kL2 (0;T ;H 21 + ( )) + ktkL2 (0;T ;H 12 + ( )) +

@

k$kL2 (0;T ;H 23 ? ( )) . L2 (0;T ;L2 (?1 )) 0

(153)

(154)

Having obtained the estimates (153) and (154) with smooth data [0 ; 1 ; 0 ], a density argument (see Remark 1.7) and a recollection of the form of the adjoint LT in (58) will allow us to obtain the norm bound (151) for all terminal data in D(LT ).  i h Lemma 4.5 Concerning the component b of the solution b; bt ; b of (72), one has that
b ? 2 L2(0; T ; L2 (?0)) 0 with the following estimate valid for all s and  2 [0; T ] :

 Z 



2 Z T 

1

2



2

12

2 b b b b e 2 


 L2 (?0 ) dt  C 0 A  L2 ( ) + P t L2 ( ) + H 21 + ( ) dt s 1

"

" #

2 #

2

2 b b  (  )  ( s )

b

A +

b t(s)

D(A 21 )H 1 ( ) +

bt ( )

D(A 12 )H 1 ( ) + L1 (0;T ;H??01; ( )) : ?0 ;

(155)

?0 ;

Proof: So as to obtain the inequality (155), we multiply the rst equation of (78) by the quantity h
rb, where   2 2 h(x; y)  [h1 (x; y); h2 (x; y)] is a C ( ) vector eld1 which satis es  ?0 h ? = [01 ; 2 ] on (156) on ?1 ; and follow this by an integration from s to  ; i.e., we will work with the equation

Z  s

1

 btt ?
btt +
2 b; h
rb

dt = L2 ( )

Z  s



c0 b + c1 bt + c2 b + c3 bt + c4
b; h
rb

Here we make use of the fact that ?0 and ?1 are separated.

31

L2 ( )

dt:

(157)

To handle the left hand side of (157): (i) First,

 Z    b b h
r  ?  ; dt t t L2 ( ) s L2 ( ) s s Z Z b   1 Z  Z  b2  1 b = t; h
r L2 ( ) ? 2 div t h dtd + 2 b2t [h1x + h2y ] dtd

s

s

s b   1 Z  Z b2 b = t ; h
r  2 + t [h1x + h2y ] dtd ; Z 



btt ; h
rb





dt = bt; h
rb L2 ( )

2

L ( ) s

s



(158)

after making use of the divergence theorem and the fact that bt = 0 on ?0. (ii) Next,

 b  b   b b ?
tt; h
r L2 ( ) dt = rt; r h
r L2 ( ) s s Z    ? rbt ; r h
rbt L2 ( ) dt s  b  b  1 Z  Z  b 2  = rt; r h
r L2 ( ) ? 2 div rt h dtd

s

s# " Z  Z b2 h1x b2ty h2y Z Z h tx b b b b Z 

?

s



s



2

+

2

dtd ?

i

txty h2x + tx ty h1y dtd

s

Z  Z " b2 h2y b2ty h1x # tx + 2 + 2 dtd

s

 b  b  Z  Z " b2txh2y b2ty h1x b2txh1x b2ty h2y # = rt; r h
r L2 ( ) + 2 + 2 ? 2 ? 2 dtd

s

s Z Z h b b b b i

?

txty h2x + tx ty h1y dtd ;

(159)

 2  R 2 R after again using the divergence theorem and the fact that div rbt h d = ?0 rbt d?0 = 0 (as bt (t) 2 H?20 ( )).

(iii) To handle the biharmonic term, we use the Green's Theorem (9), the given boundary conditions of (78), (156), and the fact that b 2 H?20 ( ) to obtain

Z 

Z   a b; h
rb dt s s Z Z  Z Z  2b b
r  @h b d?1dt ?
b + (1 ? )B1 b @ 2 d?0dt: +



2 b; h
rb

s

?1

dt = L2 ( )

@

s

?0

@

(160)

We note at this point that we can rewrite the rst term on the right hand side of (160) as

! Z  Z T

1

2 Z Z  h i 1 2 2 2 h
r bxx + byy + 2bxx byy + 2(1 ? )bxy dtd + O a b; h
rb dt = 2

A 2 b L2 ( ) dt ; s

s

0



(161)

R T

1

2  b  2 where O 0 A  L2 ( ) dt denotes a series of terms which can be majorized by the L2 (0; T ; D( A 21 )){norm

32

of b. We consequently have by the divergence theorem that

! Z  Z T

1

2 Z Z  h i 1 2 2 2 a b; h
rb dt = 2 h
r bxx + byy + 2bxx byy + 2(1 ? )bxy dtd + O

A 2 b L2( ) dt s 0 s

! Z Z n h Z T

1

2 io 1 = 2 div h b2xx + b2yy + 2bxx byy + 2(1 ? )b2xy dtd + O

A 2 b L2 ( ) dt s

0 ! Z Z h Z T

1

2 i 1 2 2 2 = 2 bxx + byy + 2bxx byy + 2(1 ? )bxy dtd?0 + O

A 2 b L2 ( ) dt s ?0 0 ! Z  Z  2 Z T

1

2 1 = 2 (162)
b dt + O

A 2 b L2 ( ) dt ; s

?0

0

@ b b where in the last step above, we have used the fact (as reasoned in [11], Ch. 4) that  ? = @ = 0 implies 0 ?0   2 2 2 2 b b b b b b that xx + yy + 2xx yy + 2(1 ? )xy =
 on ?0 . To handle the last term on the right hand side of (160), we note that B1b = 0 on ?0 , which implies that b @ 2 b b b
 =
 + (1 ? )B1  = @ 2 on ?0 ;

(163)

we consequently have upon the insertion of (162) into (160), followed by the consideration of (163) that

Z  s

! Z 1

2 Z Z Z b b
rb 2 dt = ? 1 


b

2 2 dt +   b
@h
rb d?1dt + O T

 2 A  L2 ( ) dt : 2 s @ L ( ) L (?0 ) 0 s ?1 (164)


2 ; h

(iv) To handle the right hand side of equation (157), an integration by parts yields

Z  s

= c1





c0 b + c1 bt + c2 b + c3bt + c4
b; h
rb

b; h
rb



L2 ( ) s

? c1

Z  s

b; h
rbt

L2 ( )

dt

dt + L2 ( )

Z  s

As h
rb 2 C ([0; T ]; H?10 ; ( )), we have for all t 2 [0; T ]

b

(t); h
rb(t)



= L2 ( )

Db

(t); h
rb(t)



c0 b + c2b + c3 bt + c4
b; h
rb

E

H??01; ( )H?10 ; ( )

L2 ( )

dt:

:

Accordingly, we have

"D #  E c0 b + c1 bt + c2b + c3 bt + c4
b; h
rb L2 ( ) dt = c1 b; h
rb H ?1 ( )H 1 ( ) ?0 ; ?0 ; s Z  Z    b b b b b b b

Z  s

?c1

s

; h
rt

L2 ( )

dt +

s

c0 + c2 + c3 t + c4
; h
r

L2 ( )

dt:

(165)

To nish the proof, we rewrite (157) by collecting the relations given above in (158), (159), (164) and (165) to

33

attain

Z Z 1 Z 


b

2 @h
rb d? dt b dt = 


1 2 s @ L2 (?0 ) s ?1 ! Z T

1

2 1 Z  Z b2 [h + h ] dtd

b 2   dt + +O A

L2 ( ) 2 s t 1x 2y 0

Z  Z " b2 h2y b2ty h1x b2 h1x b2ty h2y #  tx tx b b b b b ? c0 + c2  + c3 t + c4
; h
r L2 ( ) dt + 2 + 2 ? 2 ? 2 dtd

s

s Z Z h Z  i b b b b b; h
rbt  2 dt ? tx ty h2x + txty h1y dtd + c1 L ( ) s # "s

    E D : (166) + rbt ; h
rb 2 + bt ; r h
rb 2 ? c1 b; h
rb ?1 1 Z 

L ( )

H?0 ; ( )H?0 ; ( )

L ( )

s

The desired inequality (155) now comes about by majorizing the right hand side of this expression (note that in this b
r is a \lower order term", as h = 0).  majorization we are using implicitly the fact that @h@ ?1 ?1

h

i

Proposition 4.6 With vector eld h as de ned in (99), the solution b; bt ; b to (78), corresponding to terminal  the   data [0 ; 1 ; 0 ] 2 D A 2 , satis es the equality (100) for arbitrary 0 2 [0; T ). Proof: we multiply the equation (78) by h
rb ? 21 b and subsequently integrate in time and space; i.e., we will

consider the equation

Z T ?0  0

h



i

btt ?
btt +
2 b ? c0 b + c1 bt + c2 b + c3 bt ; h
rb ? 21 b

L2 ( )

dt = 0:

(167)

the computationsperformed in [12] for the quantity  using directly R T ?First, 0 b b dt, in the case that h is a radial vector eld (see the relations (3.12)  ?
 +
2 ; h
rb ? ? 1 b tt

0

tt

2

and (3.16) of [12]), we have

L2 ( )

Z T ? 0 

 btt ?
btt +
2 ; h
rb ? ? 1 b 2 L2( ) dt 0  t=T ?0  b  b 1 b b b b = (t; h
r)L2 ( ) + rt ; r h
r L2 ( ) ? 2 (t; )L2( ) t=0 t=T ?0 1 Z T ?0 Z    2   2 b b b b ?2 ? 2 rt ; r L2 ( ) h
 t + rt d?dt ?1 0 t=0 2 ! 3 Z T ?0 1

1

2

b 2 4 P 2 bt 2 +

bt

2 ? h
rb ? 1 b; @ tt 5 + 2 2 @ L2 (? ) dt L ( ) L ( ) 0 1 #   Z T ?0 " 1 Z     2 @ 1 b b b bb +

0

2 a(; ) + 2

0

?1

?0

h

 d? ?
; @ h
r

L2 (?0 )

dt

!    3 Z T ?0 2 @
b b 1 1  @B @ 2 b b b b b b 4 5 + @ + (1 ? ) @ ; h
r ? 2  L2 (?1 ) ?
 + (1 ? )B1 ; @ h
r ? 2  L2 (?1 ) dt 0 !3 ! 2 b! Z T ?0 Z h
 2 @ 2 b !2 @ 2 b !2 2 b 2 b 2 @ @  @ 5 dtd?: 4 2 + 2 + 2 2 (168) + 2 + 2(1 ? ) 2

@x

@y

@x

34

@y

@x@y


?
= h

 , this equation becomes ?0 ?0   Z T ?0  Z T ?0 

1

2 1 2 btt ?
btt +
2 ; h
rb ? ? 1 b b b b

P t L2 ( ) + a(; ) dt 2 L2 ( ) dt = 2 0 0 t=T ?0     + (bt; h
rb)L2 ( ) + rbt; r h
rb L2 ( ) ? 12 (bt; b)L2 ( ) t=0 t=T ?0 1 Z T ?0 Z      2 ?2 ? 2 rbt ; rb L2 ( ) h
 b2t + rbt d?dt ?1 0 t=0  Z T ?0 



2 Z   2 1 b b +

t L2 ( ) ? 2 ? h

 d? dt 0 0 2 ! 3  Z T ?0  @    b @  2b + 2 b ?  @ ; h
rb ? 1 b 4 b; h
rb ? 1 b 5 + + @ t @ 2 @ 2 L2 (? ) dt 2 0 L (?1 ) 1 3 ! ! ! 2 Z T ?0 Z h
 2 @ 2 b !2 @ 2 b !2 4 2 + 2 + 2 @ 2 b2 @ 2 b2 + 2(1 ? ) @ 2 b 5 dtd?: (169) + @x @y @x @y @x@y ?1 2 0 h b b b b bi b 1b ?

@ h
r Using the boundary conditions in (78) and the fact that @

Secondly, we multiply c0 + c1 t + c2  + c3 t + c4
 by h
r ? 2 , and integrate by parts to obtain

"



Z T ?0 



#t=T ?

0 dt = c1 b; h
rb ? 12 b c0 b + c1 bt + c2 b + c3 bt + c4
b; h
rb ? 12 b 0 L2 ( ) L2 ( ) t=0    Z T ?0  Z T ?0 1 1 b b b b b b b b b ?c1 ; h
rt ? 2 t c0 + c2  + c3t + c4
; h
r ? 2  dt + dt: (170) 0 0 L2 ( ) L2 ( ) To now obtain (100), we combine the expressions (167) and (169){(170), and follow this by a rearrangement of terms and use of the characterization (11).  We nally cite here a critical trace result which is used in the derivation of Lemma 2.4. Trace Theorem ([15]): Let the function '(t; x) satisfy the following Kircho equation on an open, bounded domain  Rn, with smooth boundary ?, ? = ?0 [ ?1 , where each ?i is open and nonempty, with ?0 \ ?1 = ;: 8 2 > > 'tt ?
'tt +
' = f on (0; T )  ;

> > > < ' = @' = 0 on (0; T )  ?0 ; @ > > (
' + (1 ? )B1 ' = g1 > >
' + (1 ? ) @B2' ? @'tt = g : @@ 2 @ @

(171) on (0; T )  ?1

(here the boundary operators B1 and B2 are as given in (2)). Let 0 < 0 < T and  > 0 be arbitrary. Then the following inequality holds true for the solution ':

@ 2 '

2

@ 2 '

2 # Z T ?0 "

@ 2 '

2



2

dt +

2

+

0

 CT;0 ;

@

("Z T 0

L2 (?1 )

kf k

@

2 3 [H 2 ? ( )]0

L2 (?1 )

@@

L2 (?1 )

+ kg1 kL2 (?1 ) + k'kH 2

2 3 2 +

o

+ kg2 k2H ?1 (0;T ?1 ) :

(?1 )

#

+ kjr't jkL2 (?1 ) + k't kL2 (?1 ) dt 2

2

(172)

Remark 4.7 In the original statement of this theorem (see Theorem 2.1 in [15]), the term R0T kf k2[H 23 ? ( )]0 dt in the inequality (172) is replaced by kf k2[H q (0;T  )]0 , where q < 12 . However, if one is replacing the H ?q (0; T  ) spaces with L2 (0; T ; [H q ( )]0), the values of allowed parameters extend to q < 3=2 + . This is in line with elliptic

theory corresponding to free boundary conditions.

35

References [1] J. P. Aubin, Analyse fonctionelle appliquee Tome 2, Presses Universitaire de France (1979.) [2] G. Avalos, \The exponential stability of a coupled hyperbolic/parabolic system arising in structural acoustics", Abstract and Applied Analysis, Vol. 1, No. 2 (1996), pp. 203{217.

[3] G. Avalos and I. Lasiecka, \Exponential stability of a free thermoelastic system without mechanical dissipation", SIAM [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22]

Journal Math. Anal., Vol. 29, No. 1, (January 1998), pp. 155{182. G. Avalos and I. Lasiecka, \Exponential stability of a thermoelastic system without mechanical dissipation", Rendiconti Di Instituto Di Matematica Dell'Universita di Trieste, Supp. Vol. XXVIII, 1{28 (1997). G. Avalos, \Exact controllability of a thermoelastic system with control in the thermal component only", IMA Preprint Series #1532, and submitted to Advances in Dierential Equations. L. de Teresa and E. Zuazua, \Controllability for the linear system of thermoelastic plates", Advances in Dierential Equations, Vol. 1, Number 3 (1996), pp. 369{402. P. Grisvard, \Caracterization de quelques espaces d'interpolation", Arch. Rat.. Mech. Anal. 25 (1967), 40{63. S. Hansen and B. Zhang, \Boundary control of a linear thermoelastic beam", J. Math. Anal. Appl., 210, (1997), 182{205.

V. Hutson and J. S. Pym, Applications of functional analysis and operator theory, Academic Press, New York (1980). V. Isakov, \On the uniqueness of the continuation for a thermoelasticity system", preprint (1998). J. Lagnese, Boundary stabilization of thin plates, SIAM Stud. Appl. Math., 10 (1989). J. Lagnese, \The reachability problem for thermoelastic plates", Arch. Rational Mech. Anal., 112 (1990), pp. 223{267. I. Lasiecka and R. Triggiani, \Uniform stabilization of the wave equation with Dirichlet or Neumann feedback control without geometrical conditions", Appl. Math. Optim., 25 (1992), 189{224. I. Lasiecka and R. Triggiani, \Exact controllability of the wave equation with Neumann boundary control", Appl. Math. Optim., 19 (1989), 243{290. I. Lasiecka and R. Triggiani, \Sharp trace estimates of solutions to Kircho and Euler{Bernoulli equations", Appl. Math. Optim., 28 (1993), 277{306 I. Lasiecka and R. Triggiani,\Structural decomposition of thermoelastic semigroups with rotational forces", to appear in Semigroup Forum. G. Lebeau and E. Zuazua, \Null{controllability of a system of linear thermoelasticity", Arch. Rational Mech., Vol. 141 (1998), 297{329. G. Leugering, \A decomposition method for integro{partial dierential equations and applications", J. Math. Pures Appl., 71 (1992), 561{587. J.L. Lions and E. Magenes, Non-Homogeneous boundary value problems and applications, vol. 1. Springer{Verlag, New York, 1972. W. Liu, \Partial exact controllability and exponential stability in higher{dimensional linear thermoelasticity", ESAIM: Control, Optimisation and Calculus of Variations, Vol. 3 (1998), 23{48. W. Liu, private communication. J. Simon, Compact sets in the space Lp(0; T ; B ), Ann. Mat. Pura Appl. (4) 148 (1987), pp. 65{96.

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