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t = At x0 B e u(0) + Bt v 1, p in .NET Insert gs1 datamatrix barcode in .NET t = At x0 B e u(0) + Bt v 1, p




How to generate, print barcode using .NET, Java sdk library control with example project source code free download:
t = At x0 B e u(0) + Bt v 1, p generate, create 2d data matrix barcode none in .net projects ASP.NET Web Application Framework = At (x0 Be u(0)) + Bt v = x1 (t) + x2 (t). Recall that, by Theorem 4.2.1(iii), Be u 0 = ( A X ) 1 Bu 0 for all u 0 U . The assumptions x0 X and A X x0 + Bu(0) X imply ( A X )x0 Bu(0) X , hence x0 ( A X ) 1 Bu(0) = x0 Be u(0) X 1 . 4.3 Differential representations Thus, x1 (t) = At (x0 .net vs 2010 2d Data Matrix barcode Be u(0)) is continuously differentiable in X with x1 = Ax1 . By Theorem 4.

3.6, also x2 is continuously differentiable in X with x2 = A. X x2 + Bu. Thus x is continuously differentiable in X and x = A X x + Bu (in X ). Corol DataMatrix for .NET lary 4.

3.8 Let U and X be Banach spaces, let A be a C0 semigroup on X , and let B be a L p . Reg-well-posed input ma p for A with input space U . Let s R, xs X , and u s U . (i) If B is L p -well-posed for some p < , then the following conditions are equivalent: (a) A.

X xs + Bu s X ; 1, p visual .net datamatrix 2d barcode (b) for each u Wloc ([s, ); U ) satisfying u(s) = u s the strong solution x of (4.3.

2) is continuously differentiable in X ; 1, p (c) for some u Wloc ([s, ); U ) satisfying u(s) = u s the strong solution x of (4.3.2) is continuously differentiable in X .

(ii) If B is L -well-posed or Reg-well-posed, then the following conditions are equivalent: (a) A. X xs + Bu s X ; (b) f or each u Reg1 ([s, ); U ) satisfying u(s) = u s the strong loc solution x of (4.3.2) is continuously differentiable in X ; (c) for some u Reg1 ([s, ); U ) satisfying u(s) = u s the strong loc solution x of (4.

3.2) is continuously differentiable in X . Proof The proof is the same in both cases.

By Theorem 4.3.7, (a) (b), and obviously (b) (c).

If the strong solution x of (4.3.2) is continuously differentiable in X , then x (s) = A.

X x(s) + Bu(s) = A X xs + Bu s X , so (c ) (a). Occasionally we shall need a version of Corollary 4.3.

8 where we do not know that B is L p . Reg-well-posed, but onl Data Matrix for .NET y that the control operator B satis es B B(U ; X 1 ). As the following theorem shows, the conclusion of Corollary 4.

3.8 remains valid in this case, too, provided we increase the regularity of the input function slightly. Theorem 4.

3.9 Let U and X be Banach spaces, let A be a C0 semigroup on X , and let B B(U ; X 1 ). Let s R, xs X , and u s U .

Then the following conditions are equivalent: (i) A. X xs + Bu s X ; 2,1 ( ii) for each u Wloc ([s, ); U ) satisfying u(s) = u s the strong solution x of (4.3.2) is continuously differentiable in X ; 2,1 (iii) for some u Wloc ([s, ); U ) satisfying u(s) = u s the strong solution x of (4.

3.2) is continuously differentiable in X ..

The generators Proof That (i) (ii) f ollows from Theorem 3.8.3.

Obviously, (ii) (iii). If (iii) holds, then x (s) = A. X x(s) + Bu(s) = A X xs + Bu s X ; hence (iii) (i). It is possible to reformulate parts of the conclusions of Theorems 4.3.

6, 4.3.7 and 4.

3.9 by introducing some additional spaces, namely D (A&B) = [ w ] u. A. X w + Bu X ,. (4.3.5).

(X + BU )1 = {w X A. X w + Bu X for some u U }. (4.3.

6) The former of the above spaces becomes important in Section 4.6 in connection with the de nition of a system node. The latter is important in the theory of compatible systems and boundary control systems presented in 5.

Lemma 4.3.10 Let A be a densely de ned operator on X with a nonempty resolvent set, and let B B(U ; X 1 ).

De ne D (A&B) by (4.3.5).

Then D (A&B) X X is dense in U , and the restriction A&B of A. X B : U X 1 to X D ( .net vs 2010 ECC200 A&B) is closed (as an unbounded operator from U to X ). The domain D (A&B) of A&B is a Banach space with the graph norm [w] u.

D(A&B) 2 = . A. X w + Bu 2 + . w. 2 + . u. U X X 1/2. (4.3.7).

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