Kenneth Kuttler

34.4. THE IMPLICIT CASE 1187

Proof: By Lemma 34.3.1, there exists a sequence of partitions{

tnk

}mnk=0 = Pn,Pn ⊆

Pn+1, of [0,T ] such that the lengths of the sub intervals converge uniformly to 0 as n→ ∞

and the step functions

mn−1

∑k=0

u(tnk )X(tn

k ,tnk+1]

(t) ≡ ul (t)

mn−1

∑k=0

u(tnk+1)X(tn

k ,tnk+1]

(t) ≡ ur (t)

converge to u in Lp (0,T ;V )≡ K. We assume that all of these partition points have emptyintersection with the set of measure zero where Bu(t) ̸= B(u(t)). Thus, at every partitionpoint, Bu(tk) = B(u(tk)). As just mentioned, Lp (0,T ;V )≡ K, Lp′ (0,T ;V ′) = K′.

Lemma 34.4.4 Let s < t. Then for u,Y satisfying 34.4.22

⟨Bu(t) ,u(t)⟩= ⟨Bu(s) ,u(s)⟩

+2∫ t

s⟨Y (r) ,u(t)⟩dr−⟨Bu(t)−Bu(s) ,u(t)−u(s)⟩ (34.4.23)

Proof: It follows from the following computations

Bu(t)−Bu(s) =∫ t

sY (r)dr

and so

2∫ t

s⟨Y (r) ,u(t)⟩dr−⟨Bu(t)−Bu(s) ,u(t)−u(s)⟩

= 2⟨∫ t

sY (r)dr,u(t)

⟩−⟨Bu(t)−Bu(s) ,u(t)−u(s)⟩

= 2⟨Bu(t)−Bu(s) ,u(t)⟩−⟨Bu(t)−Bu(s) ,u(t)−u(s)⟩

= 2⟨Bu(t) ,u(t)⟩−2⟨Bu(s) ,u(t)⟩−⟨Bu(t) ,u(t)⟩+2⟨Bu(s) ,u(t)⟩−⟨Bu(s) ,u(s)⟩

= ⟨Bu(t) ,u(t)⟩−⟨Bu(s) ,u(s)⟩

Thus⟨Bu(t) ,u(t)⟩−⟨Bu(s) ,u(s)⟩

= 2∫ t

s⟨Y (r) ,u(t)⟩dr−⟨Bu(t)−Bu(s) ,u(t)−u(s)⟩

Note that in case s = 0, you can simply write Bu(0) = Bu0 and the same argument appearsto work.

34.4. THE IMPLICIT CASE 1187Proof: By Lemma 34.3.1, there exists a sequence of partitions {1/) 7") = Pn, Pn SPx, of [0,7] such that the lengths of the sub intervals converge uniformly to 0 as n — ooand the step functionsmn—1Y w(t!) Zee (0 = w(t)k=0mn—1Y W (tes) Ze ap,,) 0) = u'(t)k=0converge to wu in L? (0,T;V) = K. We assume that all of these partition points have emptyintersection with the set of measure zero where Bu (t) 4 B(u(t)). Thus, at every partitionpoint, Bu (t,) = B(u(t,)). As just mentioned, L’ (0,7;V) = K, L” (0,T7;V') = K’Lemma 34.4.4 Lets <t. Then for u,Y satisfying 34.4.22(Bu (t) ,u(t)) = (Bu(s) ,u(s))+42 [oe )) dr — (Bu(t) — Bu(s) u(t) —u(s)) (34.4.23)Proof: It follows from the following computationstBu(t) —Bu(s) = / Y (r)drAY2 [wer t)) dr— (Bu(t) — Bu(s) ,u(t)—u(s))and so= (Bu(t) ,u(t)) — (Bu(s) ,u(s))Thus(Bu (t) ,u(t)) — (Bu(s) ,u(s))=2 fer t)) dr—(Bu(t)—Bu(s),u(t)—u(s))Note that in case s = 0, you can simply write Bu (0) = Bug and the same argument appearsto work.