2550 CHAPTER 76. IMPLICIT STOCHASTIC EQUATIONS

It follows from the uniqueness assumption 76.3.17 that for each ω off a set of measurezero, there exists a unique solution to

(Bu)′ (·,ω)+ Ā(·,u(·,ω) ,ω) = f (·,ω) in V ′ω ,

B(u(·,ω))(0) = 0

You can consider the function of two variables u(t,ω) . Is this function progressively mea-surable? Right now, this is not clear because we have done nothing more than solve aproblem for each ω .

However, we can at least say that uh is progressively measurable because uh ∈V . Recallalso that

1h(I− τh)Buh + Ā(ω)uh = f , uh ∈ V

Next we show that because of uniqueness, one can assume that u is progressively measur-able. To do this, we show that the sequence for which convergence holds in the above canbe chosen independent of ω .

Claim: A single sequence h→ 0 works for all ω off a set of measure zero.Proof: Since there is only one solution to the above initial value problem for ω /∈ N,

then letting h→ 0 be a single sequence, one can conclude that uh (·,ω)⇀ u(·,ω) in Vω =Lp (0,T,V ). Otherwise, from the above argument, one could obtain another subsequencewhich converges to a solution different than u(·,ω) which would violate uniqueness.

From the coercivity condition, it follows that there exists a constant C ( f ) depending onf such that for all h,

∥uh∥V ≤C ( f )

Therefore, there is a further subsequence still denoted by h such that

uh ⇀ ū in Lp ([0,T ]×Ω;V ) (76.3.21)

where the measurable sets are just the product measurable sets B ([0,T ])×FT . Then itfollows from Lemma 76.3.4 that u(·,ω) = ū(·,ω) in Vω for all ω off a set of measure zero.It follows that in all of the above, we could substitute ū for u at least for ω off a single setof measure zero. Thus u can be assumed progressively measurable.

Note the importance of path uniqueness in obtaining the result on progressive measur-ability of the solutions.

We will write u rather than ū to save notation. Now with this lemma, it is easy to obtainthe following proposition.

Proposition 76.3.6 Let q ∈ V such that t → q(t,ω) is continuous and q(0,ω) = 0, andlet the conditions 76.3.14 - 76.3.17 be valid. Also let u0 ∈ L2 (Ω,V ) such that u0 is F0measurable. Let f ∈ V ′ be given. Then for each ω off a set of measure zero, there existsu(·,ω) ∈ Vω such that (Bu)′ (·,ω) ∈ V ′ω and

Bu(0,ω) = Bu0

and also the following equation holds in V ′ω for a.e. ω

(Bu−Bq)′ (·,ω)+A(·,u(·,ω) ,ω) = f (·,ω)