Differential Equations and Their Applications Part 3 pptx

... Proof. Theorems 3. 2 and 3. 3 tell us that (3. 17) and (3. 19) are necessary. We now prove the sufficiency. First of all, for any g E H, by (3. 17), we can find y E IR m, such that (3. 14) holds (note ... (3. 54) r~ = ~(s)dW(s). Then, from (3. 49) and (3. 53) , we have o = E(v,~ZI = E(~,(O,~) (X(r)~ ~, Y(T) ) ) (3. 55) r w Solvability of linear FBSDEs 33 Theorem 2....

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... (4 .33 ) 0 <_ VS'~(s,x,O~(s,x)) <_ cCo, V5 <_ (f(E). Then, by (3. 28), (3. 39) (with 5 = 0) and (4 .33 ), we obtain 0 < < + c0 _< + + Io (o,x)l) + Co. Now, we let 5 + 0 and ... 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

Ngày tải lên: 10/08/2014, 20:20

... 3- 540-65960-9 Mathematics Subject Classification (1991): Primary: 60H10, 15, 20, 30 ; 93E 03; Secondary: 35 K15, 20, 45, 65; 65M06, 12, 15, 25; 65U05; 90A09, 10, 12, 16 ISSN 0075-8 434 ISBN 3- 540-65960-9 ... formula 231 w An American Game Option 232 Chapter 9. Numerical Methods for FBSDEs 235 w Formulation of the Problem 235 w Numerical Approximation of the Quasilinear PDEs 237...

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... (3. 7) follows. [] We note that (3. 4) holds if both g~ and a are bounded, and (3. 6) holds if both b and a are bounded. 14 Chapter 1. Introduction An interesting corollary of Proposition 3. 2 ... for FBSDEs. Corollary 3. 3. Suppose 3 is continuous in (t, x, y, z) and uniformly Lips- chitz continuous in (x, y, z). Suppose there exists an ~ > O, such that (3. 13) {3( O,x...

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... similar to (3. 36), we have (3. 36) J~ Z ~~ >_ V~"~(s,x,y) - Eo. Combing (3. 31), (3. 32) and (3. 36), one has 0 < V$'~(s, x, y) -V~ x, y) < 2~0, which shows that (3. 37) V$'~(s,x,y)$V~ ... 0/> c (3. 34) 1 1 >V ~,~(s,x,y)-C ~-~. Combining (3. 31), (3. 32) and (3. 34), we obtain (note So > 0 is arbitrary) 0 < V~'~(s,x,y)- V~'~(s...

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... multi-index. If /3 = ( /31 ,'",/3n) is another multi-index, by /3 _< a, we mean that/3i _< ai for each i = 1, , n, and by /3 < a, we mean /3 < a and at least for one i, one has/3i < ... Zw)e~]dZ a(x,9)h(x) - (h(x)9 - 1) f3 ay(x,9 + j3(O- 9))48 (3. 33) = a(x,e) 10~ I [a(x, O)by(x, 0 + ~(~- 0)) + ay(x, 130 + t3('0- O) )b(x,O)] dl3 >_ ~_....

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... solution to (3. 32) is unique (for any f, q and uo sat- isfying (3. 31)). As a matter of fact, if ~ is another solution to (3. 32), then u - ~ is a solution of (3. 32) with f, q and uo all being ... Then, we see that u E g~(0, T; V) N C~-([0, T]; H) is a solution of (3. 32). We now combining (3. 27)- (3. 28) and (3. 34) (3. 35) to obtain the following: lu(t)]~ = Iv(t)l~ +...

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... (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 (f,g) and (f,~) ... 2.2 and 2 .3 hold for (6.1). Throughout this section, we assume that the parabolicity condition (1.6), the symmetry condition (2.2) and (H),~ (for some m _> 1) hold for (...

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... -25, Now, we take a(t) and c(t) as in (3. 7) and we require (3. 13) d(t)+(3L + 2L2)lc(t)l = -Co(Co - 3L - 2L2)e C~ -Co(Co - 3L - 2L 2) ~ -5, Vt E [0,T], and (3. 14) c(t) = -Coe C~ <__ ... T]). Again the left hand side of (3. 30) can be controlled by (3. 23) for some constant K > 0. Now, we require it(t) + Ka(t) + KIc(t)l = -hA2e A~ + 5KAoe A~ + Khe t (3. 31)...

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