Differential Equations and Their Applications Part 9 pps

150 Chapter 6. Method of Continuation where Ao, Co > 0 are undetermined constants. We first check that this E B~(Fo; [0,T]). In fact, (3.8) c(0) = -Co < 0, a(T) = Ao > O, i~(t) = -A~e A~ < O, t E [0, T], e(t) = -C~e C~ < 0, t E [0, T]. Thus, by Proposition 3.1, we see that 9 E B~(Fo; [0, T]). Next, we show that 9 E B~(F; [0, T]) for suitable choice of Ao and Co. To this end, we let L be the common Lipschitz constant for b, a, h and g. We note that (3.8) implies (1.7). Thus, it is enough to further have (3.9) a(T) + L2c(T) >_ 5, and (3.10) a(t)lx - ~12 + e(t)ly - ~12 + c(t)lz - .212 + 2a(t) (x - ~,b(t,x) - b(t,5) ) +a(t)la(t,x ) - a(t,5)] 2 + 2c(t) ( y - ~, h(t, x, y, z) - h(t, 3, ~,-2) ) < -5{1~ -~1 ~ + ly-~l 2 + Iz-~12}, VtE [0,T], x,~ E ~ n, y,~ E IR '~, z,.2 E ~ re• a.s. Let us first look at (3.10). We note that Left side of (3.10) _< a(t)lx - 512 + d(t)ly - ~]2 + c(t)]z - -212 + 2a(t)Lix - 512 + a(t)L2ix - 5} 2 (3.11) + 21c(t)lLly-~l{ix-~l+lY-Vl+lz 21} < {&(t) + 2a(t)L + a(t)L 2 + Ic(t)]L}lx - 512 + {d(t)+ 3]c(t)]L + 2L2ic(t)i}iY- ~]2 + ~_lz _ .212. Hence, to have (3.10), it suffices to have the following: { h(t) + (2L+ L2)a(t) + Lic(t)[ < -6, (3.12) d(t) + (3L + 2L2)lc(t)[ ~ -5, Vt E [0,T]. c(t) < -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~ <__ -Co <_ -25~ Vt E [0, T]. w Some solvable FBSDEs 151 These two are possible if Co > 0 is large enough. Next, for this fixed Co > 0, we choose Ao > 0 as follows. We want a(T) + c(T)L 2 = Aoe A~ - CoL2e C~ > Ao - CoL2e C~ > 5, (3.15) and a(t) + (2L + L2)a(t) + LIc(t)l (3.16) = -Ao(Ao - 2L - L2)e A~ + LCoe c~ <_ -Ao(Ao - 2L - L 2) + LCoe C~ ~_ -~. These are also possible by choosing A0 > 0 large enough. Hence, (3.9) and (3.12) hold and 9 E/~(F; [0,T]). [] From the above, we obtain that any decoupled FBSDE is solvable. In particular, any BSDE is solvable. Moreover, from Lemma 2.2, we see that the adapted solutions to such equations have the continuous dependence on the data. The above proposition also tells us that decoupled FBSDEs are very "close" to the trivial FBSDE since they can be linked by some direct strong bridges of F0. w FBSDEs with monotonicity conditions In this subsection, we are going to consider coupled FBSDEs which satisfy certain kind of monotonicity conditions. Let F = (b, a, h, g) E HI0, T]. We introduce the following conditions: (M) Let m _> n. There exists a matrix B E IR mx'~ such that for some /~ > 0, it holds that (3.17) (B(x-5),g(x)-g(5)) _>Plx-~l 2, Vx,~E~ ~, a.s. (3.18) ( B T [h(t, 0) - h(t, 9)],x - 5) + ( B [b(t, 9) - b(t, 9)], y - ~) + (B[~(t, 0) - o(t,~)], z - ~) < -~lx - ~12, Vt E [0, T], 9,9 E M, a.s. (M)' Let m < n. There exists a matrix B E ~,mxn such that for some /~ > 0, it holds that (3.17)' (B(x-x),g(x)-g(x)) ~ O, Vx,xE ]R n, a.s. (3.18)' ( B T [h(t, 9) - h(t, 9)],x - -2) + ( B [b(t, 0) - b(t, 9)], y - ~) + (B [~(t, 0) - ~(t,~)], z - ~) _< -~(lY - ~l 2 + Iz - ~12), Vte [0, T], 9,9 E M, a.s. Condition (3.17) means that the function x F-+ BTg(x) is uniformly monotone on IR ~, and condition (3.18) implies that the function 9 ~-~ 152 Chapter 6. Method of Continuation -(BTh(t,O),Bb(t,O),Ba(t,O)) is monotone on the space M. The mean- ing of (3.17)' and (3.18)' are similar. Here, we should point out that (3.17) implies m _> n and (3.17)' implies m ~ n. Hence, (M) and (M)' overlaps only for the case m = n. We now prove the following. Proposition 3.4. Let T > 0 and F - (b, or, h, g) r H[0, T] satisfy (M) (resp. (M)'). Then, (3.6) holds. Consequently, F E S[0, T]. Proof. First, we assume (M) holds. Take {~(t)= (A(t) B(t) T'~ \ B(t) C(t) J (3.19) A(t) = a(t)I - 5eT-tI, t E [0, T], B(t) B, C(t) = c(t)I =_ -25CoeC~ I, with 5, Co > 0 being undetermined. Since (3.20) C(0) = -25Coi < O, A(T) = 5I > O, r = (-Se~-~, o ) 0 _2~C~eCo t < O, h(t)]xI 2 + c(t)iy] 2 + c(t)]z] 2 + 2La(t)lxl(]xI + lY] + Izl) (3.22) + 2LIc(t)I ]yi(]xI + ly] + IzI) + L2a(t)(ixI + lYl + Izl) 2 (2t3 - 5)Ix] 2 - 5(lyl 2 + Izl2), V(t,O) C [0, T] • M. It is 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 of (3.22) can be controlled by the following: {/L(t) + Ka(t) + Kic(t)]}ixI 2 + {~(t)+ Kic(t)I + Ka(t)}ly[ 2 (3.23) + {c(_~_~ +Ka(t)}iz]2, for some constant K > 0. Then, for this fixed K > 0, we now choose 5 and Co. First of all, we require (3.24) c(t) + Ka(t) = -SCoe C~ + KSe T-t ~_ -5Co + K(~e T ~ -5, (3.21) and a(T) + 23 + c(T)L 2 > 5, by Proposition 3.1, we see that 9 E BS(F0;[0, T]). Next, we prove 9 C Bs(F; [0, T]) for suitable choice of 5 and Co. Again, we let L be the common Lipschitz constant for b, a, h and g. We will choose 5 and Co so that w Some solvable FBSDEs 153 and (3.25) d(t) + KIc(t)l + Ka(t) = -25C2e C~ + 2KCohe C~ + Khe T-t < • - K) + Khe T < -6. These two can be achieved by choosing Co > 0 large enough (independent of 5 > 0). Next, we require it(t) + Ka(t) + KIc(t)l = -he T-t A- Khe T-t + 25KCoe C~ (3.26) < -6 + Khe T + 25KCoe c~ <_ 2~ - 6, and (3.27) a(T) + 2~ + c(T)L 2 = 5 + 2~ - 25CoeC~ L 2 > 6. Since/~ > 0, (3.26) and (3.27) can be achieved by letting 5 > 0 be small enough (note again that the choice of Co is independent of 5 > 0). Hence, we have (3.21) and (3.22), which proves 9 e BS(F; [0, T]). Now, we assume (M)' holds. Take (compare (3.19)) A(t) B(t)T'~ ( B(t) c(t) )' O(t) (3.28) A(t) a(t)I = 5AoeA~ Vt e [0, T], B(t) B, c(t) c(t)I - -he~• with 6, Ao > 0 being undetermined. Note that c(0) = -6I < 0, A(T) = AoI > O, Thus, by Proposition 3.1, we have 9 C /~S(Fo; [0, T]). We now choose the constants 5 and Ao. In the present case, we will still require (3.21) and the following instead of (3.22): it(t)lxl 2 + d(t)lY] 2 + c(t)lzl 2 + 2La(t)lxl(Ixl + lYl + N) (3.30) + 2LIc(t)l[Y](Ixl + lyl + Izl) + L2a(t)(]xl + lyl + IzlY <_ -51xl 2 + (2/~ - 5){lyl 2 + Iz12}, v(t,o) e [0,T] • M. These two will imply the conclusion 9 E BS(F; [0, 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) < -hAo(Ao - K) + 5Ke T < -6, 154 Chapter 6. Method of Contim~ation and a(T) + c(T)L 2 = 5Aoe A~ - 5L2 e t (3.32) > 5(Ao - L2e T) > (~. We can choose Ao > 0 large enough (independent of 5 > 0) to achieve the above two. Next, we require e(t) T + Ka(t) <_ Ka(t) <__ 5KAoe A~ ~_ 2~ - 6, (3.33) and (3.34) d(t) + Klc(t)l + Ka(t) = -Se t + KSe t + KAoSe A~ <_ 5(Ke T + KAoe A~ <_ 2~ - 5. These two can be achieved by choosing 5 > 0 small enough. Hence, we obtain (3.21) and (3.30), which gives 9 9 gS(F; [0,T]). It should be pointed out that the above FBSDEs with monotonicity conditions do not cover the decoupled case. Here is a simple example. Let n = m = 1. Consider the following decoupled FBSDE: dX(t) = X(t)dt + dW(t), (3.35) dY(t) = X(t)dt + Z(t)dW(t), X(O) = x, Y(T) = X(T). We can easily check that neither (M) nor (M)' holds. But, (3.35) is uniquely solvable over any finite time duration [0, T]. Remark 3.5. From the above, we see that decoupled FBSDEs and the FBSDEs with monotonicity conditions are two different classes of solvable FBSDEs. None of them includes the other. On the other hand, however, these two classes are proved to be linked by direct bridges to the trivial FBSDE (the one associated with Fo = (0,0, 0, 0)). Thus, in some sense, these classes of FBSDEs are very "closer" to the trivial FBSDE. w Properties of the Bridges In order to find some more solvable FBSDEs with the aid of bridges, we need to explore some useful properties that bridges enjoy. Proposition 4.1. Let T > 0. (i) For any F E H[0, T], the set BI(F; [0, T]) is a convex cone whenever it is nonempty. Moreover, (4.1) BI(F; [0, T]) : BI(F + 3'; [0, T]), V3' 9 7-/[0, T]. (ii) For any F1, F2 9 H[0, T], it holds (4.2) t31(F1;[O,T])NBx(F2;[O,T]) C_ N Bl(arl +•F2;[0,T]). c~,f~>0 w Properties of the bridges 155 Proof. (i) The convexity of BI(F; [0, T]) is clear since (1.7)-(1.9) are lin- ear inequalities in q~. Conclusion (4.1) also follows easily from the definition of the bridge. (ii) The proof follows from (2.13), (2.15) and the fact that Be(F; [0, T]) is a convex cone. [] It is clear that the same conclusions as Proposition 4.1 hold for BII(F; [0, T]) and BS(P; [0, T]).' As a consequence of (3.2), we see that if rl, F2 E HI0, T], then (4.3) Br(aF1 +/3F2; [0, T]) = r for some ct,/3 > 0, Be(F1; [0, T]) f')Bx(P2; [0, T]) = r This means that for such a case, F1 and F2 are not linked by a direct bridge (of type (I)). Let us look at a concrete example. Let Fi = (bi, ai, hi, gi) E H[0, T], i = 1,2,3, with (4.4) bl 0 b2 ~ 1 (hi)-~ (21 ~ ) (y) ( ) ( ) (Y) 1] ' h2 0 p ' (b3) ( 0 ~)(y) Crl =~ : ~ : 0, h3 1 ' gl = 92 = g3 = -x, with A, v E ~. Clearly, it holds (4.5) F3 = F1 + F2. By the remark right after Corollary 2.4, we know that B(F3; [0, T]) = r Thus, it follows from (3.5) and (4.3) that F1 and F2 are not linked by a direct bridge. However, we see that the FBSDE associated with F~ is decoupled and thus it is uniquely solvable (see Chapter 1). In w we will show that for suitable choice of s and v, F2 E S[0, T]. Hence, we find two elements in S[0, T] that are not linked by a direct bridge. This means F 1 and F2 are not very "close". Next, for any hi, b2 E L~(O, T; WI,~176 IRn)), we define (4.6) IIb~ - b211o(t) esssup sup wCf~ 0,OEM Ibl(t,O;w) - hi(t,0; w) - b2(t,O;w) + b2(t,O;w)l l0 - 0l We define [Ihl- h2llo(t) and Ilal -a2llo(t) similarly. For gl,g2 e L2r (f~; Wl'~(~n; IRm)), we define (4.7) Ilal - g2110 = esssup sup wE~ x,~ER ~ I g I(x; a)) 91 (X; CO) g2(X; W) -~ g2(X; W) I i x - ~l 156 Chapter 6. Method of Continuation Then, for any Fi = (hi, ai, hi, gi) 6 g[o, T] (i = 1, 2), set Ilrl - r211o(t) = lib1 - b211o(t) + I1Ol - a211o(t) (4.8) + Ilhl - h21lo(t) + Ilgl - g211o. Note that I1 IIo(t) is just a family of semi-norms (parameterized by t E [0, T]). As a matter of fact, lit1 - r211o(t) = o for all t E [0, T] if and only if (4.9) F2 = F1 + if, for some "y 6 7-/[0, T]. Theorem 4.2. Let T > 0 and F e H[0,T]. Let 9 C Bs(F; [0, T]). Then, there exists an e > 0, such that for any F' C H[0, T] with (4.10) IIr- r'llo(t) <~, vte [0,T], we have ~' C Bs(r'; [0,T]). 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. Now, we denote (for any 0, 0 E M) [~=x-~, 0"= 0- 0, l ~ = b(t, O) - b(t,-0), 3 = a(t, O) - a(t,-0), (4.11) I. ~ h = h(t,O) - h(t,O), "~ = g(x). - g(g), , IN,= b'(t,O) - b'(t,O), "d' = ~ (t,O) - a (t,-O), I I h' = h'(t,O) - h'(t,0), ~' = g'(x) - g'(g). Then one has (4.12) IV - ~l = Ig'(x) - g'(e) - g(x) + g(e)l -< IIg' - gllol~l. Similarly, we have (4.13) { I t' -'bl -< lib'/bllo(t)10"l, Ih' - hi -< Ilh - hllo(t)101. w Properties of the bridges Hence, it follows that 157 (4.14) > ~1~1 = + 2 < B(T)~, ~' - ~> + < C(T)('f + ~), ~' - ~) _ _ + ' gllo}l~l 2- >{8 21B(T)IIIg' gllo-IC(T)lllY gllollg ~t 5 2 provided IIg' -g]lo is small enough. Similarly, we have the following: A ^ -3) (4.15) +(r a'O I'( 0 )) _< -al~l 2 + 2 (A(t)~ + B(t)r~,~ ' - ~) + 2 (B(t)~+ C(t)~,h' -h> + 2 (B(t)T~,~ ' ~) + (A(t)(~' + "~),~' - ~) < { - 5 + 2(IA(t)l + ]B(t)])llb' - bllo(t) + 2(IB(t)l + Ic(t)l)llh' - hllo(t ) + 21B(t)lll# - ~llo(t) + IA(t)lIla' + allo(t)lJ# - ~llo(t)}lol 2. Then, our assertion follows. [] The above result tells us that if the equation associated with F is solv- able and F admits a strong bridge, then all the equations "nearby" are solvable. This is a kind of stability result. Remark 4.3. We see from (4.14) and (4.15) that the condition (4.10) can 158 Chapter 6. Method of Continuation be replaced by 2(IB(T)I + IC(T)IIIg' + gli0)iig'- glIo < 6, sup {2(IA(t)l + IB(t)l)tl b'- blio(t) (4.16) tE[O,T] + 2(IB(t)] + ]C(t)I)lih'-hilo(t) + [2[B(t)l + [A(t)I[IW + al]o(t)] I]W- allo(t)} < 5, where 6 > 0 is the one appeared in the definition of the bridge (see Defi- nition 1.3). Actually, (4.16) can further be replaced by the following even weaker conditions: 2 ( B(T)s - ~) + ( C(T)(~ + ~),~' - ~> > 2, Vx,5 E ~'~, sup {2 (A(t)~+ B(t)T~,b ' (4.17) tc[O,T] I + 2 ( B(t)~ + C(t)~,h' - h) +2 (B(t)TF, 8' - 8 > +(A(t)(8'+8),8'-~)} V0,0EM. The above means that if the perturbation is made not necessarily small but in the right direction, the solvability will be kept. This observation will be useful later. To conclude this section, we present the following simple proposition. Proposition 4.4. Let T > O, F - (b,a,h,g) 6 H[0, T] and 9 6 Bi(r; [0,T]). Let f E R and f ~(t) = e2Zto(t), t E [0, T], (4.18) = (b -flx, a,h - fly, g) 6 H[0,T]. Then, ~2 9 BI(F; [0, T]). The proof is immediate. Clearly, the similar conclusion holds if we replace BI(F; [0, T]) by Bn(F; [0, T]), B(F; [0, T]) or BS(F; [0, T]). w Construction of Bridges In this section, we are going to present some more results on the solvability of FBSDEs by constructing certain bridges. w A general consideration Let us start with the following linear FBSDE: { d {x(t) ={~4 X(t) bo(t) {ao(t) dW(t), ) ( )( )} (5.1) \ Y(t) Y(t) + ho(t) dt + t 9 [0, T], X(O) = x, Y(T) = GX(T) + go, w Construction of bridges 159 where ,A E IR (n+m)• G E IR mxn, ")' - (bo,~o, ho,go) C 7/[0, T] (see (1.3)) and x E IRn. We have the following result. Lemma 5.1. Let T > O, Then, the two-point boundary value problem (5.1) is uniquely solvable for all V E 7/[0, T] if and only if Proof. Let \,(t)/ I) f x(t) k Y(t) / Then we have the linear FBSDE for (~, 7) as follows: d r + ( bo(t) ho(t)Gbo(t) ) }dt (5.3) + z(t) -a~o(t) [(0) = x, ~(T) = go, Clearly, the solvability of (5.3) is equivalent to that of (5.1). By Theorem 3.7 of Chapter 2, we obtain that (5.3) is solvable for all V E 7/[0, T] if and only if (3.16) and (3.19) of Chapter 2 hold. In the present case, these two conditions are the same as (5.2). This proves the result. [] Now, let us relate the above result to the notion of bridge. From Theorem 2.1, we know that if F1 and F2 are linked by a bridge, then F1 and F2 have the same solvability. On the other hand, for any given F, Corollary 2.4 tells us that if F admits a bridge, then, the FBSDE associated with F admits at most one adapted solution. The existence, however, is not claimed. The following result tells us something concerning the existence. This result will be useful below. Proposition 5.2. Let To > 0 and F = (b, 0, h, g) with (:) (5.4) ~ h(t,O) = A , g(x) = Gx, V(t, 0) 6 [0, To] x M. Then F E S[0,T] for all T E (0, To] if/3(F; [0, T]) r r for all T E (0, T0]. Proof. Since B(F; [0, T]) ~ r by Corollary 2.4, (5.1) admits at most one solution. By taking V = (bo,ao,ho,go) = 0 and x = 0, we see that the resulting homogeneous equation only admits the zero solution. This is equivalent to that (5.1) with the nonhomogeneous terms being zero only [...]... (5.21) C(t) _ e Let us now look at these requirements separately First of all, it is clear true that A(t) > 0 for all t 9 [0, T] N ext, C(t) e+ (e T M + 2e at + 3) A2 + #2 z~ A2 t = f ( t ) , t 9 [o, T] Since f " ( t ) >_ 0 for all t E [0, o0), the function f ( t ) is convex Thus, (5.22) holds if and only if (5.23)... under (5.15) and (5.28), the following holds (5. 29) F ( T ) - g2 f(O) >_ O In fact, by the choice of 6 and by (5.27), (5.30) F(O) - g2 f(O) = ff'(O) = O On the other hand, (5.31) 1 ( 1 + ~ ) 2 e 2 ~ T + ~-f ( l + g # ~ e ~ T g# F ' ( T ) = -~ ~] g2(52+# 2) s2 w Construction of bridges 165 Thus, by (5.15), it follows that Then, by F " ( T ) > 0, together with (5.30) and (5.32), we must have (5. 29) Hence,... 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 we may take (5.33) f(T) K = f(0) v It is interesting that the if(-) constructed in the above is not in /3(Fo; [0, T]) for any T > 0 since )~(t) > 0 On the other hand, we note that both A(t) and B(t) are independent of T However, due to the fact that K depending on T,... slightly different from those we have seen before w Forward S D E s w i t h Reflections Let (9 be a closed convex domain in ~ n Define for any x E 0 (9 the set of inward normals to O at x by (1.1) h/'x = {~: I '1 = 1, and (% x - y) < O, Vy e (9) It is clear that if the boundary 0( .9 is smooth (say, C1), then for any x E 0 (9, the set Af~ contains only one vector, that is, the unit inner normal vector at x We... functions of bounded variation; and for ~ 9 BV([O,T]; ]Rn), we denote 17/I(T) to be the total variation of ~ on [0, T] A general form of (forward) SDEs with reflection (FSDER, for short) is the following: (1.2) X(t) = x + /0 b(s, X(s))ds + /0 a(s, X(s))dW(s) + ~(t) Here the b and a are functions of (t, x, w) 9 [0, T] • ~ x f~ (with w being suppressed, as usual); and ~ 9 BVj:([O,T];]R'~), the set of... 7) 9 L~=([0, T]; ~ n ) • BV~:([0, T]; nZ~) is called a solution to the FSDER (1.2) if 1) x ( t ) 9 (9, vt 9 [0,T], a.s.; 2) rl(t) = f~ l{x(8)coo}7(s)dl~l(s), where q'(s) 9 JY'X(s), 0 < s < t < T, dlr/I-a.e.; 3) equation (1.2) is satisfied almost surely A widely used tool for solving an FSDER is the following (deterministic) function-theoretic technique known as the Skorohod Problem: Let the domain (9. .. 4)~(1+ 92 ), 3 4;~(1+ g2), t e [0,T], where f(t) is defined by (5.22) In fact, by (5.20), (5.22), (5.28) and (5.33), we have (5.35) C(t) = f(t) - f(O) V f(T) 4),(1 + g2)" Clearly, C(t) is convex Thus, (5.36) C(t) 0, only depending on )~ and. .. = GX(T) + ~(X(T)), admits a unique adapted solution 0 - (X, ]I, Z) E A4[0, T] Proof We note that if (5 .9) ~(t, 0) = eetg(t, e-z~0), V(t, 0) E [0, T] x M, then, (5.10) Ilbllo(t) = Ilbllo(t), Vt c [0, T] Similar conclusion holds for ~, h and ~ if we define ~, h and ~" similar to (5 .9) On the other hand, if O(t) _= (X(t), Y(t), Z(t)) is an adapted solution of (5.8) with /3 = 0, then ~)(t)z~ eZtO(t ) is... BS(F; [0, T]) if it is the solution to the following differential equation for some constants K, K, 5, s > 0, ( ~ ( t ) + ~ffr + r (i)(0) = ( K f~), (5.11) = -5I, t 9 [0, T], satisfying the following additional conditions: (5.12) I (I, GT)d2(T) ( G ) ~ sI On the other hand, we find that the solution to (5.11) is given by (5.13) ~(t) = e-Art ( K 0 t -:t 9 [0, T] Thus, in principle, if we can find constants... exists a unique T1 > To, such that f(T1) = f(0), and (5.38) f f(t) < f(O), f(t) > f(0), Vt E [0, 7"1], Vt C (7"1,oo) Hence, we obtain (5. 39) C(T) > f(To) - f(O) This proves the second relation in (5.34) 4A(1 + g2)" 166 Chapter 6 Method of Continuation Now, from Remark 4.3 and Theorem 5.3, we know that the following FBSDEs is solvable on [0, T] (5.40) dX(t) = {( /9 - A)X(t) + #Y(t) + b(t, X(t), Y(t), Z(t)) . (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. ~, a.s. (3.18) ( B T [h(t, 0) - h(t, 9) ],x - 5) + ( B [b(t, 9) - b(t, 9) ], y - ~) + (B[~(t, 0) - o(t,~)], z - ~) < -~lx - ~12, Vt E [0, T], 9, 9 E M, a.s. (M)' Let m < n. There. (3.18)' ( B T [h(t, 9) - h(t, 9) ],x - -2) + ( B [b(t, 0) - b(t, 9) ], y - ~) + (B [~(t, 0) - ~(t,~)], z - ~) _< -~(lY - ~l 2 + Iz - ~12), Vte [0, T], 9, 9 E M, a.s. Condition (3.17)