76.7. OTHER EXAMPLES, INCLUSIONS 2591
Theorem 76.7.2 In the situation of Corollary 76.4.9 where V is a closed subspace ofW σ ,p (U) ,σ > 1 and W is as described above for U a bounded open set, u0 ∈ L2 (Ω,W ) , u0F0 measurable. Suppose λ I +A(t,u,ω) satisfies
⟨λ I +A(t,u,ω)− (λ I +A(t,v,ω)) ,u− v⟩ ≥ δ2 ∥u− v∥p
V
for all λ large enough. Also assume Φ ∈ L∞([0,T ]×Ω,L2
(Q1/2U,H
))with Φ = BΨ
whereΨ ∈ L∞
([0,T ]×Ω,L2
(Q1/2U,W
))and progressively measurable. Then there exists a unique solution to the integral equation
Bu(t)−Bu0 +∫ t
0A(s,u)ds+
∫ t
0h(s,ω)ζ (s,ω)ds =
∫ t
0ΦdW (76.7.70)
where for a.e. s, h(s,ω)ζ (s,ω) ∈ ∂u (h(s,ω)φ (u(s))) where φ (r)≡ |r|. The symbol ∂u isthe subgradient of φ (u). Written in terms of inclusions, there exists a set of measure zerosuch that off this set,(
B(
u−∫ (·)
0ΦdW
))′+A(t,u) ∈ ∂u (h(t,ω)φ (u(s))) a.e. t
u(0) = u0
Note that one can replace
Φ ∈ L∞
([0,T ]×Ω,L2
(Q1/2U,H
))with Φ ∈ L2
([0,T ]×Ω,L2
(Q1/2U,H
))along with an assumption that t → Φ(t,ω) is
continuous. This can be done by defining a stopping time
τn ≡ inf{t : ∥Φ(t)∥> n}
Then from the above example, there exists a solution to the integral equation off a set ofmeasure zero
Bun (t)−Bu0 +∫ t
0A(s,un)ds+
∫ t
0h(s,ω)ζ n (s,ω)ds =
∫ t∧τn
0ΦdW
Since Φ is a continuous process, τn = ∞ for all n large enough. Hence, one can replace theabove with the desired integral equation. Of course the size of n depends on ω, but we candefine
u(t,ω) = limn→∞
un (t,ω)
because by uniqueness which comes from monotonicity, if for a particular ω, both n,kare sufficiently large, then un = uk. Thus u is progressively measurable and is the desiredsolution.
Next we show that the above theory can also be used as a starting point for some secondorder in time problems. Consider a beam which has a point mass of mass m attached to