70.4. THE NAVIER−STOKES EQUATIONS 2411

Lemma 70.4.4 Let u0 have values in H and be F measurable, and let un be a solution to70.4.6. Then for each ω, the estimate 70.4.8 holds. Also there is a subsequence, still calledun such that the convergence for 70.4.9 - 70.4.11 are valid. For all ω , the function u(·,ω)is a solution to 70.4.13 - 70.4.14 and satisfies

u(·,ω) ∈ L∞ ([0,T ] ;H)∩L2 ([0,T ] ;W ) ,u′ ∈ L2 ([0,T ] ;V ′) .This solution is also weakly continuous into H for each ω .

Proof: All that remains to show is the last claim about weak continuity into H. Theequation 70.4.13 shows that u(·,ω) is continuous into V ′. However, the weak convergenceand the estimate 70.4.8 show that u(·,ω) is bounded in H. It follows from density of V inH that t→ u(t,ω) is weakly continuous into H.

From 70.4.13, 70.4.14, the following integral equation for a path solution holds:

u(t,ω)−u0(ω)+∫ t

0A(u(s,ω))ds+

∫ t

0N̂ (u(s,ω))ds =

∫ t

0f(s,ω)ds.

We apply Theorem 70.2.8 to prove the above solution could be taken product measurable.

Theorem 70.4.5 Let f(t,ω), q(t,ω) be product measurable and u0 be measurable, suchthat for each ω ∈ Ω, f(·,ω) ∈ L2 ([0,T ] ;W ′), q(·,ω) ∈ C ([0,T ] ;V ) with q(0) = 0, andu0(ω) ∈ H. Then there exists a global solution to the integral equation

u(t,ω)−u0(ω)+∫ t

0A(u(s,ω))ds+

∫ t

0N̂ (u(s,ω))ds =

∫ t

0f(s,ω)ds.

Proof: Letting un be a solution to 70.4.6, we verify the conditions of Theorem 70.2.8for un.

The assumption in this theorem that the un are bounded follows from the above esti-mate 70.4.8. Then it was shown in the above lemma that whenever a sequence satisfies theestimate 70.4.8, it has a subsequence which converges as in 70.4.9 - 70.4.11 to a weaklycontinuous u(·,ω). Therefore, by Theorem 70.2.8 there is a subsequence un(ω)(·,ω) con-verging weakly to u(·,ω), such that (t,ω)→ u(t,ω) is a product measurable function intoH. Then a further subsequence converges to a path solution to the above integral equation,which must be the same function because when a sequence converges, all subsequencesconverge to the same thing. In addition to this, u is also product measurable into W . Thisfollows from the above estimate 70.4.8. For φ ∈ H,(t,ω)→ (φ ,u(t,ω)) is product mea-surable. However, H is dense in W ′ and so if ψ ∈W ′, there is a sequence {φ n} in H suchthat φ n→ ψ . Then

⟨ψ,u⟩= limn→∞

(φ n,u) ,

so by the Pettis theorem [127], u is product measurable into W also.This shows much of the following theorem which is the main result.

Theorem 70.4.6 Let f(t,ω), q(t,ω) be product measurable and u0 be measurable, suchthat for each ω ∈ Ω, f(·,ω) ∈ L2 ([0,T ] ;W ′), q(·,ω) ∈ C ([0,T ] ;V ) with q(0) = 0, andu0(ω) ∈ H. Then there exists a global solution to the integral equation

u(t,ω)−u0 (ω)+∫ t

0A(u(s,ω))ds+

∫ t

0N(u(s,ω))ds =

∫ t

0f(s,ω)ds+q(t,ω) .