71.3. STOCHASTIC DIFFERENTIAL EQUATIONS 2431

|N (t,u1,v1,w1,ω)−N (t,u2,v2,w2,ω)| ≤ K (|u1−u2|+ |v1− v2|+ |w1−w2|) (71.3.12)

where the norms |·| refer here to the Hilbert space H. Assume N,σ are both progressivelymeasurable. From the Lipschitz condition given above,

|N (t,u,v,w,ω)−N (t,0,0,0,ω)| ≤ K (|u|+ |v|+ |w|)

and it is assumed thatt→ N (t,0,0,0,ω) (71.3.13)

is in L2 (Ω,C ([0,T ] ;H)). Also consider the growth condition which is implied by the abovecondition and the Lipschitz assumption.

(N (t,u,v,w,ω) ,u)≥−C (t,ω)−µ

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

)(71.3.14)

where C ∈ L1 ([0,T ]×Ω) and the linear growth condition for σ ,

∥σ (t,u,ω)∥ ≤ a+b |u|H (71.3.15)

71.3.1 The Lipschitz CaseTheorem 71.3.1 Suppose 71.3.11, 71.3.13, 71.3.15, 71.3.12 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. (71.3.16)

71.3.16 where u ∈ L2 (Ω,C ([0,T ] ;H)) ,u0 ∈ L2 (Ω) ,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 v ∈ L2 (Ω;C ([0,T ] ;H)) where v is also progressively measurable. Then letu be given by

u(t,ω)−u0(ω)+∫ t

0N (s,v(s,ω),v(s−h,ω) ,w(s,ω) ,ω)ds

=∫ t

0f (s,ω)ds+

∫ t

0σ (s,v,ω)dW. (71.3.17)

The Lipschitz condition 71.3.12, the assumption 71.3.13, and the linear growth assertion71.3.15, implies that u is also in L2 (Ω;C ([0,T ] ;H)) . The proof of this involves the same