Views
5 years ago

SN~ (~6) lff 2It 3k_~ , 5 ",,x_J {b)

SN~ (~6) lff 2It 3k_~ , 5 ",,x_J {b)

126 This ends our

126 This ends our analysis of the unperturbed semigroup. Next we define the bounded operator B from X into X ®* by Be = ((C,¢),0)= (C, ¢)r% where r ®" -- (I, 0). The following lemma is the key step in proving the equivalence between the variatlon-of-constants formula (2.1) and the RI;'DE. Lemma 3.7 Let g: R+ ~ X be a norm continuous function, then f~ ToO'(t - r)Bg(r) dr = f;"~(°"+'}(C, g(r)) dr. Proof. /o tT°O*(t- r)Bg(r) dr = fo' (((,g(r)), (i,g(r))H(t - r + .)) dr, where H is the Heavlside function defined by H(~) S 0 for t < 0 1 for t>0. l Using the definition of the weak * integral, Fubini's theorem as well as the identification of X with its embedding X ®® in X ®" we can rewrite the last integral as [' (

and the abstract integral equation (AIE) Here g : X + Rn is, say, continuous, h : R+ + Rn is, say, and r@* = (I, 0). Theorem 4.1 There is a one-to-one correspondence between solutions of (4.1) and (4.2) given by u(2) = xt and x(t) = Here (6 E X@ is the functional which assigns to an element of X@* the Rn component; so, considered as an element of X*, 6 is indeed the Dirac 6 at zero) Proof. Let x be a solution of (4.1). Define u(t) = xt. Then { +(t + 0) t t o i o u(t)(a) = "(t + a) = +(O) + S,~+"(~(U(T)) + h(7.)) dr t + a > 0 (l kmE?"3.7 (To(t)+)(u) + ~:*(t - I)@* (g(u(r)) t h(r)) dr) (a). We thus have verified that (4.2) holds. Let now u be a solution of (4.2). Define Using Lemma 3.7 once more we find or, in other words, u(t) = xt. Similarly, there is a one-to-one correspondence between the FDE Here T is the semigroup generated by A$*+B, where, as in the previous section, B : X + x@* is defined by B4 = ((C, #), 0) = r@*(C, 4). (4.5) ~7

Page 1 Page 2 D B 5 6 D MOTOR Mo~, lmotowon T R T ...
B E R L I N C H A U S S E E S T R A S S E 5 7 - 6 1 B ... - Glen Leddy
Page 1 7 6 5 | 4 L 3 | 2 | 1 CK T ll ll ll )l KLÖ'ZOJ T ll ll ll ll klözo) A B ...
1 5 . T T O - F O R U M 1 2 . T T O - F O R U M O c t o b e r 1 6 ...
D e se m b e r 0 5, V olum e 0 6, Issue 11 w w w .ta n go n o tica s.co m
P P P NP NP 12 3 4 5 6 7 8 W 9 10 11 12 13 14 W 15 16 0 17 A C B ...
DVAA: Flat Galaxies (by R.A.) A B C D E F G H I J K L 1 2 3 4 5 6 7 8 ...
H O T E L - R E S TA U R A N T- B A R S e e s t r a s s e 2 2 5 6 3 ...
Reading grade 6 2.A.5.b - mdk12
a 1 2 a b b a 3 4 a b b a 5 6 a b b a 7 8 a b b a 9 10 ... - CodeMath.com
Lösung 6 Aufgabe 1 a) b) c) Aufgabe 2 Aufgabe 3 a) 5 4 2 6 4 20 9 2 ...
8 * 5 * 4 * 3 * 2 * 1 A 8 A 7 A 6 A 5 A 4 A 3 A 2 B 8 B 7 B 6 B 5 B 4 B 3 ...
5 b 5 6 10 c 12 a 5 8 A a 13 b Classroom Exercises - flip@mrflip.com
EAS 540 - STABLE ISOTOPE GEOCHEMISTRY Thrs., BUS B-5; 6 ...
Solutions to Quiz 5 Sample B - Loyola University Chicago
Panel B 4:45-5:30 Round 2 Panel A 5:30-6:15 | Panel B 6:15-7 ...
¡' &) (10 , 2 354 6&7 98@ ) AC B$ E DG FH I 6 &Q ... - Université Lille 1
TEST ANSWERS Version A 1. B 2. B 3. E 4. E 5. D 6. B 7. C 8. B 9. A ...
V 5 1 5 B 6 L 4 X P T S F
(0 12 3¡) 4(5 6 3 798 9 %@4a7(6 b cedgfhfhiqps rti u)vxwtwty
Friday Warm-up A 5:00-5:30 PM, Warm-up B 5:30-6:00 PM Starts at ...
1. NC 2. L 3. NP 4. D 5. F 6. A 7. M a) b) b) 8. T 9 ... - Bangladesh Bank
1. Nyelvismereti feladatsor 1. C 2. B 3. D 4. D 5. A 6. B 7. B 8. C ... - Itk
1 2 (a) 2 (b) 3 4 5 6 7 8 9 10 11 - Travel Insurance
1. 2. 3. 4. 5. 6. 7. 8. Meeting A. M B. Pl Approval Awards a Public He ...
1 2 (a) 2 (b) 3 4 5 6 7 8 9 10 11 - Travel Insurance
A. LISTENING. 1-5 B. VOCABULARY 6-15 C. GRAMMAR. 16-25
1 2 3 4 5 6 A B C D Last Name FirstName Major Advisor ... - Alumni