34.5. THE IMPLICIT CASE, B = B(t) 1197

This involves an operator B(t) ∈L (W,W ′) and B(t) satisfies

⟨B(t)x,x⟩ ≥ 0, ⟨B(t)x,y⟩= ⟨B(t)y,x⟩

forV ⊆W,W ′ ⊆V ′

Where we assume t→ B(t) is in C1 ([0,T ] ;L (W,W ′)) and V is dense in the Banach spaceW .

Then the main result in this section is the following integration by parts theorem.

Theorem 34.5.1 Let V ⊆W,W ′⊆V ′ be separable Banach spaces, and let Y ∈Lp′ (0,T ;V ′)and

Bu(t) = Bu0 +∫ t

0Y (s)ds in V ′, u0 ∈W,Bu(t) = B(t)(u(t)) for a.e. t (34.5.28)

As indicated, Bu is the name of a function satisfying the above equation which satisfiesBu(t) = B(t)(u(t)) for a.e. t. Thus Y = (Bu)′ as a weak derivative in the sense of V ′

valued distributions. Suppose that u ∈ Lp ([0,T ] ,V ) and (s, t)→ B′ (s)u(t) is bounded inV ′ in case p < 2. (If B(t) is constant in t this is obvious.) In the case where p ≥ 2, it isenough to assume B′ ∈C1 ([0,T ] ;L (W,W ′)). Then t→ Bu(t) is continuous into W ′ for toff a set of measure zero N and also there exists a continuous function t→ ⟨Bu,u⟩(t) suchthat for all t /∈ N,⟨Bu,u⟩(t) = ⟨B(u(t)) ,u(t)⟩ ,Bu(t) = B(t)(u(t)) , and for all t,

12⟨Bu,u⟩(t)+ 1

2

∫ t

0

⟨B′u,u

⟩ds =

12⟨Bu0,u0⟩+

∫ t

0⟨Y (s) ,u(s)⟩ds

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) ≡ uln (t)

mn−1

∑k=0

u(tnk+1)X(tn

k ,tnk+1]

(t) ≡ urn (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(t)(u(t)). Thus, at every partitionpoint, Bu(tk) = B(tk)(u(tk)). As just mentioned, Lp (0,T ;V )≡ K, Lp′ (0,T ;V ′) = K′.

Taking a subsequence, we can have∥∥∥uln−u

∥∥∥K+∥ur

n−u∥K +∥∥∥Bul

n−Bu∥∥∥

K′+∥Bur

n−Bu∥K′

+∥∥B′ur

n−B′u∥∥

L2([0,T ],W ′)+∥∥∥B′ul

n−B′u∥∥∥

L2([0,T ],W ′)< 2−n (34.5.29)

and so, we can assume that a.e. convergence also takes place for Buln,Bur

n,B′ul

n,B′ur

n,urn,u

ln.

Is Bu(0) = B(0)u0? The integral equation gives this it seems. To save notation, B(0)u0will be written as Bu0. This is not inconsistent because t → B(t)u0 is continuous and itsvalue at 0 is B(0)u0.