2430 CHAPTER 71. STOCHASTIC O.D.E. ONE SPACE

In particular, this holds for t = T and so, letting z ∈ L2 (Ω,C ([0,T ] ,Rn)) , {T nz} is aCauchy sequence in this space because a high enough power is a contraction map, so itconverges to a unique fixed point u. Each T nz is progressively measurable and so thefixed point is also. In L2 (Ω,C ([0,T ] ,Rn)) , you get 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)dW, (71.2.10)

Thus off a set of measure zero, the equation holds for all t and u is progressively measur-able. The place where u0 ∈ L2 (Ω) is needed is in having T z ∈ L2 (Ω,C ([0,T ] ;Rn)). Oneuses a similar procedure involving the Ito formula, the growth condition

(N(t,u,v,w,ω) ,u)≥−C (t,ω)−µ

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

).

and the Burkholder Davis Gundy inequality to verify this. Since T depends on u0, itappears that the set of measure zero, off which the integral equation holds, will also dependon u0. It appears that this ultimately results from the need to take an expectation in order todeal with the stochastic integral. If this integral could be generalized in such a way that itmade sense for each ω as in the usual Riemann Stieltjes integral, then likely this restrictioncould be removed. It is a problem because the Wiener process is not of bounded variation.

Theorem 71.2.2 Suppose the weak monotonicity condition 71.2.5 and the growth estimate71.2.3. Also assume N(t,u,v,w,ω) ∈ Rd for u,v,w ∈ Rd , t ∈ [0,T ] and (t,u,v,w,ω)→N(t,u,v,w,ω) is progressively measurable with respect to the normal filtration Ft de-termined by a given Wiener process W (t). Also suppose (t,u,v,w)→ N(t,u,v,w,ω) iscontinuous. Let f ∈ L2 (Ω,C ([0,T ] ,Rn)) and (t,u,ω)→ σ (t,u,ω) is progressively mea-surable and satisfies the linear growth condition 71.2.6 and the Lipschitz condition 71.2.9.Also suppose u0 is F0 measurable and in L2 (Ω,Rn). Then there exists a progressivelymeasurable solution u to 71.2.10. If û is another such solution, then there is a set of mea-sure zero N such that for ω /∈ N, û(t) = u(t) for all t.

Proof: It only remains to verify the uniqueness assertion. This happens because thefixed point is unique in L2 (Ω,C ([0,T ] ,Rn)). Therefore, off a set of measure zero the twosolutions are equal for all t.

71.3 Stochastic Differential EquationsIn this section, ordinary differential equations in Hilbert space which are of the form

du+N (u)dt = f dt +σ (u)dW

are considered under Lipschitz assumptions on N and σ . A very satisfactory theorem canbe proved.

The assumptions made are as follows.

∥σ (t,u,ω)−σ (t, û,ω)∥L2(Q1/2U,H) ≤ K |u−û|H , (71.3.11)