2434 CHAPTER 71. STOCHASTIC O.D.E. ONE SPACE

71.3.2 The Locally Lipschitz CaseNow replace the Lipschitz assumpton 71.3.12 with the locally Lipschitz assumption whichsays that if max(|u| , |v| , |w|)< R, then there is a constant KR such that

|N (t,u1,v1,w1,ω)−N (t,u2,v2,w2,ω)| ≤ K (R)(|u1−u2|+ |v1− v2|+ |w1−w2|)(71.3.19)

Also assume the growth condition

(N (t,u,v,w,ω) ,u)≥−C (t,ω)−µ

(|u|2 + |v|2 + |w|2

)(71.3.20)

and the linear growth condition on σ

∥σ (t,u,ω)∥ ≤ a+b |u|H

and the Lipschitz condition on σ 71.3.11. This can likely be relaxed as in the case of theLipschitz condition for N but for simplicity, we keep it.

Theorem 71.3.2 Suppose 71.3.11, 71.3.14, 71.3.15,71.3.13, 71.3.19 and let

w(t) = w0 +∫ t

0u(s)ds, w0 ∈ L2 (Ω) , w0 is F0 measurable.

Then there exists a unique progressively measurable solution u to the integral equation

u(t,ω)−u0(ω)+∫ t

0N (s,u(s,ω),u(s−h,ω) ,w(s,ω) ,ω)ds

=∫ t

0f (s,ω)ds+

∫ t

0σ (s,u,ω)dW. (71.3.21)

71.3.16 where u ∈ L2 (Ω,C ([0,T ] ;H)) ,u0 ∈ L2 (Ω,H) ,u0 is F0 measurable, f is progres-sively measurable and in L2 ([0,T ]×Ω;H) . Here there is a set of measure zero such thatif ω is not in this set, then u(·,ω) solves the above integral equation 71.3.16 and further-more, if û(·,ω) is another solution to it, then u(t,ω) = û(t,ω) for all t if ω is off some setof measure zero.

Proof: Let un be the unique solution to the integral equation

un (t,ω)−u0(ω)+∫ t

0N (s,Pnun(s,ω),Pnun (s−h,ω) ,Pnwn (s,ω) ,ω)ds

=∫ t

0f (s,ω)ds+

∫ t

0σ (s,un,ω)dW. (71.3.22)

where Pn is the projection onto B(0,9n). Thus the modified problem is in the situation ofTheorem 71.3.1 so there exists such a solution. Then let

τn = inf{t : |un (t)|+ |wn (t)|> 2n}