Differential Equations and Their Applications Part 9 pps

... (5.30) and (5.32), we must have (5. 29) . Hence, we obtain (5.25). This shows that a strong bridge ~(t) has been constructed with K, 5 and s being given by (5. 19) and (5.28), respectively, and ... Proof. Let F = (b,a,h,g) and F' = (b',a',h',g'). Suppose 9 6 B~(F; [0, T]). Then, for some K, 5 > 0, (1.7)-(1 .9) and (1.8)'-(1 .9) ' hold....

Ngày tải lên: 10/08/2014, 20:20

... target: (1.3) T = {(x,g(x)) I x 9 ]RU}. Problem (C). For any x 9 IR n, find an y 9 IR m and a control Z(.) 9 Z[0, T], such that (1.4) (X(T), Y(T)) 9 T, a.s. Problem (C) having a solution ... I~(t,0,0,0)l, I~(t,x,y,O)l <_ L, Vt 9 T], x,~ 9 n, y,~ 9 z,g 9 Under the above (H1), we see that for any (x, y) E]R '~ • I~ TM, and Z(.) 9 Z[0, T], (1.1) admits a unique...

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... _~ 9 BS(Fo; [O,T]) if and only if B C(0) < 0, A(T) > O, (3.2) ~(t) < 0, Yt e [0, T]. Proof. By Definition 1.3, we know that 9 9 BS(F0; [0,T]) if and only if (1.7)-(1 .9) and ... (2.11) and working on (v,p) for the transformed equations. [] Our main comparison result is the following. Theorem 6.2. Let (1.6), (2.2) and (H),~ hold for (6.2) and (6.3). Let (...

Ngày tải lên: 10/08/2014, 20:20

... differential equations and their applications / Jin Ma ; Jiongmin Yong. - Berlin, Heidelberg ; New York ; Barcelona ; Hong Kong ; London ; Milan, Paris ; Singapore, Tokyo : Springer, 199 9 ... 3-540-6 596 0 -9 Mathematics Subject Classification ( 199 1): Primary: 60H10, 15, 20, 30; 93 E03; Secondary: 35K15, 20, 45, 65; 65M06, 12, 15, 25; 65U05; 90 A 09, 10, 12, 16 ISSN 0075...

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... X(.) is in lRn; and b, a, h, 3 and g are some suit- able functions which satisfy the following Standing Assumptions: denoting M=~nxlR m x~,onehas b 9 L2~(O,T;WI'~176 cr 9 L~(O,T;WI'~215 ... for ordinary differential equations do not necessarily admit solutions. On the other hand, an FBSDE can be viewed as a two-point boundary value problem for stochastic differenti...

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... g C H if and only if (3.17) and (3. 19) hold. In this case, the adapted solution to (2.12) is unique (for any given g E H). Proof. Theorems 3.2 and 3.3 tell us that (3.17) and (3. 19) are necessary. ... fl(s)dW(s), t 9 [0, T], and take g = ~(T)~? 9 L 2 (f~;lR "~) C H. Suppose (2.12) admits an .T'~0+E adapted solution (X, ]I, Z) 9 A4[0, T] for this g. Then, for so...

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... "~ x R m• Now, we see that (2.6) and (2.8) follow from (A1) and (A2); (2.7) follows from (A1), (2.2) and (2.11); and (2 .9) -(2.10) follow from (A1) and (2.2). Therefore, by Lemma 2.1 there ... of the coefficients of the system (2.1) (i.e., b, h, a, and 3(z) - z) and their derivatives. Therefore using assumptions (H1) and (H4), and noting that supt IZ(k)(t)l <...

Ngày tải lên: 10/08/2014, 20:20

... following: A 9 L~(0, T; C~+I (]R~; Sn)), L~~ T" C m+l B 9 ~, , b (~n;~n• (2.1) ~ 9 L~(0, T; C~(~; ~)), b 9 L~(0, T; C~n(R~;~d)), c 9 L~(0, T; C~(Rn)). w Infinite horizon case 93 (3.21) ... ~?.o)Ox]dl~)w where ~o 1 c(x) = h(x) - [ay(x,O § ~w) h(x )9 - 1 - b(x,O )9~ a(x, e) + by(x, e + Zw)e~]dZ a(x ,9) h(x) - (h(x )9 - 1) f3 ay(x ,9 + j3(O- 9) )48 (3.33) = a(...

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... continuous, such that (3.20) [ v 9 52(0, T; V), ( b 9 L2(0, T; V'). Then v 9 C([O,T];H) and (3.21) d[v(t)to 2 = 2 (9( t),v(t))o, a.e.t 9 [0,T]. Let A 9 s V') be symmetric satisYying ... (3.25) Let (3.26) M(t) = fot Then, M 9 Cj:([O,T]; V) and v 9 L~(0, T;V), ~3 9 L2(0, T; V'), q 9 L2~(O, T; V) d. (q(s),dW(s) ), t 9 [0, T]. (3.27) fot fo'...

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... (A1), we see that a and fl are bounded, adapted pro- cesses. Therefore, subtracting (2 .9) from (2.8), using (2.10) and (2.11), and noting the facts that u(t, x) = qo(t, x) and u(r, X(r)) _< ... deterministic; (ii) The domains { (92 (', ")} are of the form (9( s,w) = (92 (s, Xt'X(s,w)), (s,w) C [t,T] • IR n, where (92 (t,x) = (L(t,x),U(t,x)), where L(.,.) a...

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