2614 CHAPTER 77. STOCHASTIC INCLUSIONS
Then there is a continuous function t→ ⟨Bu,v⟩(t) such that
⟨Bu,v⟩(t) = ⟨B(u(t)) ,v(t)⟩
a.e. t such thatsup
t∈[0,T ]|⟨Bu,v⟩(t)| ≤C∥u∥X ∥v∥X
and if K : X → X ′
⟨Ku,v⟩ ≡∫ T
0⟨Lu,v⟩ds+ ⟨Bu,v⟩(0)
Then K is continuous and linear and
⟨Ku,u⟩= 12[⟨Bu,u⟩(T )+ ⟨Bu,u⟩(0)]
If u ∈ X and Bu(0) = 0 then there exists a sequence {un} such that ∥un−u∥X → 0 butun (t) = 0 for all t close to 0.
77.4 Measurable Approximate SolutionsThe main result in this section is the following theorem. Its proof follows a method due toBrezis and Lions [91] adapted to the case considered here where the operator is set valued.In this theorem, we let F : V →V ′ be the duality map ⟨Fu,u⟩= ∥u∥p ,∥Fu∥= ∥u∥p−1 forp > 1.
As above, Lu = (Bu)′ . In addition to this, define Λ to be the restriction of L to thoseu ∈ X which have Bu(0) = 0. Thus
D(Λ) = {u ∈ X : Bu(0) = 0}
Then one can show that Λ∗ is monotone. It is not hard to see that this should be the case.Let v ∈ D(Λ∗) and suppose it is smooth. Then∫ T
0⟨Λu,v⟩dt = ⟨Bu(T ) ,v(T )⟩−
∫ T
0
⟨Bu,v′
⟩dt
and so, if∣∣∣∫ T
0 ⟨Λu,v⟩dt∣∣∣≤C∥u∥V , then we need to have v(T ) = 0 and Λ∗v =−Bv′. Now
it is just a matter of doing the computations to verify that
⟨Λ∗v,v⟩ ≥ 0.
Lemma 77.4.1 Let K and L be as in Proposition 77.3.7. Then for each f ∈ V ′ and u0 ∈W,there exists a unique u ∈ X such that
⟨Ku,v⟩+Fu = ⟨ f ,v⟩+ ⟨Bv(0) ,u0⟩ (77.4.9)
for all v ∈ X . Also, the mapping which takes ( f ,u0) to this solution is demicontinuous inthe sense that if fn→ f strongly in V ′ and u0n→ u0 in W, then un→ u weakly in V .