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Data Matrix 2d barcode for .NET = C|(X +BU )1 ( A|X ) 1 B + D. in .NET Integrated barcode data matrix in .NET = C|(X +BU )1 ( A|X ) 1 B + D.

= C. use none none writer toinsert none on nonedata matrix creating asp.net (X +BU )1 ( A Barcode FAQs X ) 1 B + D. Compatible and regular systems (iii) The alternative formula for C (X +BU )1 follows from (4.7.3). For each w X 1 , w = 0 and C (X +BU )1 w = C&D w w = C&D = Cw, w 0 so C (X +BU )1 is a (compa tible) extension of C B(X 1 ; Y ). By (5.1.

2) and (5.2.2), the corresponding feedthrough operator is D( ) C.

(X +BU )1 ( A X ) 1 B = D. A&B Corollary 5.2.7 L none none et S = C&D be a boundary control node on (Y, X, U ), with main operator A, control operator B, observation operator C, and transfer function D.

Let D B(U ; Y ), (A), and de ne C. (X +BU )1 as in Theorem 5.2.6. The the following claims are true. (i) The transfer function D is given by D(z) = C (X +BU )1 (z A X ) 1 B + D, z (A) none for none . (ii) If S is a system node, then we can replace the equation (4.7.

4) in Lemma 4.7.8 by x (t) A.

(X +BU )1 B = y(t) C (X +BU )1 D x(t) , u( none none t) t s, x(s) = xs . (5.2.

4). (iii) If S is an L p Reg-well-posed system none for none node then we can replace equations (4.6.6) and (4.

6.7) in Theorem 4.6.

5 by x (t) A. (X +BU )1 B = y(t) C (X +BU )1 D x(t) , u( none for none t) t R, (5.2.5).

and we can replace fo rmulas (4.6.13) and (4.

6.14) in Theorem 4.6.

11 by (5.2.4).

Proof This follows from Lemma 5.1.4, Corollary 5.

1.7, and Theorem 5.2.

6. In the above corollary we only used a small part of Theorem 5.2.

6, namely the fact that C. (X +BU )1 D is a comp none none atible extension of C&D. We can also use the operators and to get a different representation formula for the state trajectory x in Lemma 4.7.

8 and Theorems 4.6.5 and 4.

6.11..

A&B Corollary 5.2.8 L et S = C&D be a system node on (Y, X, U ) which is also a boundary control node, with semigroup generator A, control operator B, observation operator C, and transfer function D.

Let D B(U ; Y ), (A), and de ne , , and C. (X +BU )1 as in Theorem 5.2.6. Then the following claims are true. 5.2 Boundary control systems (i) The state traject none for none ory x and output function y in Lemma 4.7.8 satisfy x (t) = x(t), x(t) = u(t), t s, t s, (5.

2.6) y(t) = C. (X +BU )1 x(t) + Du(t), x(s) = xs . (ii) If S is L Reg-well-posed then the state trajectory x and output function y in Theorem 4.6.5 satisfy x (t) =. x(t),. x(t) = u(t),. t R, t R,. y(t) = C (X +BU )1 x(t) + Du(t),. (5.2.7).

and the state traject none none ory x and output function y in Theorem 4.6.11 satisfy (5.

2.6). Proof This follows from Theorem 5.

2.6 and Corollary 5.2.

7. Remark 5.2.

9 Especially in the L -well-posed and Reg-well-posed cases it may be confusing that we have several different extensions of C, namely the unique extension to an operator C. X B(X ; Y ) given b y Theorem 4.4.2(ii), and the extensions given in Theorem 5.

2.6 to a family of operators in B((X + BU )1 ; Y ) which are parametrized by the operator D. It is not true, in general, that C.

(X +BU )1 B((X + BU )1 ; Y ) in Theorem 5.2.6 is a restriction of C X to (X + BU )1 : thi s is true if and only if we choose D in Theorem 5.2.6 to be the feedthrough operator D de ned in Theorem 4.

5.2. The same comment applies to the case where the system is regular in the sense of De nition 5.

6.3. Remark 5.

2.10 Continuing the preceding remark, although the operator B((X + BU )1 ; X ) is an extension of the generator A B(X 1 ; X ), it is not the restriction to (X + BU )1 of the extension A. X of A to B(X ; X 1 none none ) that we constructed in Section 3.6. This follows from the fact that ( A) maps X 1 onto X for (A), whereas ( ) maps all of (X + BU )1 into X .

In particular, ( A) is injective but ( ) is not. Remark 5.2.

11 The operators and in Theorem 5.2.6 can be interpreted in the following way.

The operator is an abstract partial differential operator with an insuf cient set of boundary conditions, so that does not itself generate a semigroup, i.e., the solution x of (5.

2.6) is not unique if we remove the abstract boundary condition x(t) = u(t). Often is referred to as a trace operator .

It follows from (5.2.2) that x := ( A.

X ) 1 Bu is the uniqu e (static) solution in (X + BU )1 of the abstract elliptic problem x = x, x = u. (5.2.

8). Boundary control syst none none ems arise in at least two different ways: in realization theory (see Example 5.2.4), and in the theory of partial differential equations.

In.
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