Differential Equations and Their Applications Part 1 doc

... mathematics ; 17 02) ISBN 3-540-65960-9 Mathematics Subject Classification (19 91) : Primary: 60H10, 15 , 20, 30; 93E03; Secondary: 35K15, 20, 45, 65; 65M06, 12 , 15 , 25; 65U05; 90A09, 10 , 12 , 16 ISSN ... feature seen in (1. 18). Again, once a solution of (1. 19) can be determined, then U(c) = Y(0) defines a unitiliy function. An interesting variation of (1. 18) and (1....

Ngày tải lên: 10/08/2014, 20:20

... GDb}dt (1. 18) + {(A1 - GA1)X + (/ ~1 - GB1)Y + (C1 - GC1)Z + D]~ - GDla}dW(t), Y(T) = Fg. Denote i ~ AA (1. 19) = (~ - GA)X + (B - GB)Y + (C - GC)Z + Db - GDb, (A1 - GA1)X § (B1 - GB1)Y. We ... of (1. 1). Then it holds (1. 21) [A1 - GA1 + (/ ~1 - GB,)G]X(T) + (B1 - GB1)Fg E T~( 01 - ac1), If, in addition, the following holds: (1. 22) { 7~(A + BG) + 7~(BF) C_ T~(D)...

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... (5 .15 ) IO'~ql 2 <__ 21BTD(O~u) + OC'ql 2 + 21BTD(a~u)l 2. Using the parabolicity condition (1. 6) and the definition of Ot (see (5 .14 )), we have (5 .16 ) IqlLl ~ C((]}/ -1 -1- lull). ... '~) by (-,-)0, and the inner product and the norm in L2(IR ~) by 11 0 Chapter 5. Linear, Degenerate BSPDEs Theorem 2.3. Let m > 1 and (H)m hold for {A, B, a,...

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... Control Left., 14 (19 90), 55- 61. [2] Backward stochastic differential equations and quasilinear parabolic partial differential equations, Lecture Notes in CIS 17 6, 200- 217 , Springer 19 92. [3] ... Stochastics, 3 (19 79), 12 7 -16 7. [2] Backward stochastic differential equations and applications, Proc. ICM, 19 94. Pardoux, E., and Peng, S., [1] Adapted...

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... + { [A1 -GA1 + (B1 -aB1)G]X -}- (gl -GB1)Y + (C1 - GCI)Z + b~3 - GD~a~dW. J Define (2.2) =[A1 - GA1 + (B1 - GB1)G]X + (BI - GB1)7 + (C1 - GC1)Z + DI~ - aDla. Since (C1 - GC1) is invertible, ... (1. 29) tiT[A1 GA1 + (B1 - GB1)G]X(T) + ~?T(B1 GB1)Fg = O, a.s. Thus, (1. 21) follows. In the case (1. 22) holds, for any x E ~'~ and g E ]R m (deterministic), by some choice of b...

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... continuous, and (H2) holds, then (1. 18) implies (1. 13). Proof. That condition (1. 13) implies (1. 18) is obvious. We need only prove the converse. Let us first assume that V is continuous and (1. 22) ... < L (1 + Ixl), (1. 22) (h(t,x,y,z),y) > -L (1 + Ixl lYl + lY12), (1. 23) (h(t,x,y,z),y) > -L(I+ [y[2), Ig(x)l <_ L. Proposition 1. 5. Let (HI) hold. Then (...

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... (1. 6), we still have (2 .11 ). Also, conditions (2 .16 ) and (2 .17 ) hold, which will lead to the existence and uniqueness of classical solutions of (2 .14 ) or (2 .13 ). Next, applying Theorem 1. 1, ... (1. 8)- (1. 10) is an adapted solution to (1. 1). Moreover, if h is also uniformly Lipschitz continuous in (x, y, z), a is bounded, and there exists a constant/3 E (0, 1)...

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... of (1. 1). Further, if (1. 11) holds, then (u, q) is an adapted classicaI solution of (1. 1). Proof. Let (u, q) be an adapted weak solution of (1. 1) such that (1. 13) holds. Then, from (1. 15), ... Dq] 1 (1. 5) - (b,q) -f }dt + ( q, dW(t) ), (t,x) G [O,T] x ]R n, ~tlt=T : g, Since (1. 1) and (1. 5) are equivalent, all the results for (1. 1) can be automatically carried over...

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... case, we still have (2.9), (2 .11 ) and (2 .12 ). Further, we have inequalities similar to (2 .13 ) and (2 .15 ) with I)~(T)I 2 and I.~(t)l 2 replaced by 0 and IY(t)l 2 + 12 (t)] 2, respectively. Thus, ... continuation 14 7 (2 .16 )-(2 .17 ) and note (2 .11 ), we have EIJ~(T)I 2 + E ".IT IX(t)l 2dt /o/ < ce {1~ 12 + El~ol 2 + E {INo(t)l 2 + I~o(t)l 2 + I'~o(t...

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... = (hi, ai, hi, gi) 6 g[o, T] (i = 1, 2), set Ilrl - r 211 o(t) = lib1 - b 211 o(t) + I1Ol - a 211 o(t) (4.8) + Ilhl - h21lo(t) + Ilgl - g 211 o. Note that I1 IIo(t) is just a family of semi-norms ... not hard to see that under (3 .17 )-(3 .18 ), (3. 21) implies (1. 8) and (3.22) implies (1. 7) and (1. 9)' (Note (1. 8) implies (1. 8)'). We see that the left hand side o...

Ngày tải lên: 10/08/2014, 20:20