32.5 The Implicit Case, B = B
The above theorem can be generalized to the case where the formula is of the
form
∫ t
BX (t) = BX0 + Y (s)ds
0
This involves an operator B
∈ℒ and
B satisfies
〈B (t)x,x〉 ≥ 0, 〈B (t)x,y〉 = 〈B (t)y,x〉
for
Where we assume t → B
is in
C ^{1} and
V is dense in the Banach
space
W .
Then the main result in this section is the following integration by parts
theorem.
Theorem 32.5.1 Let V ⊆ W,W ^{′} ⊆ V ^{′} be separable Banach spaces, and let
Y ∈ L ^{p′
}
and
∫ t
Bu (t) = Bu0 + Y (s)ds in V ′, u0 ∈ W, Bu(t) = B (t) (u (t)) for a.e. t
0
(32.5.28)
(32.5.28)
As indicated, Bu is the name of a function satisfying the above equation which satisfies
Bu
=
B for a.e. t . Thus Y =
^{′} as a weak derivative in the sense of V ^{′}
valued distributions. Suppose that u ∈ L ^{p} and → B ^{′} u is bounded in
V ^{′} in case p < 2
. (If B is constant in t this is obvious.) In the case where p ≥ 2
, it is
enough to assume B ^{′} ∈ C ^{1} . Then t → Bu is continuous into W ^{′} for
t off a set of measure zero N and also there exists a continuous function t →
such that for all t N, =
,Bu =
B , and for all
t,
1 1 ∫ t ′ 1 ∫ t
2 〈Bu,u 〉(t)+ 2 〈B u,u〉ds = 2 〈Bu0,u0〉+ 〈Y (s),u(s)〉ds
0 0
Proof: By Lemma 32.3.1 , there exists a sequence of partitions
_{k=0} ^{mn} =
P _{n} , P _{n} ⊆P _{n+1} ,
of
such that the lengths of the sub intervals converge uniformly to 0 as
n →∞ and
the step functions
mn∑ −1 n l
u(tk)X(tnk,tnk+1](t) ≡ un (t)
k=0
mn∑ −1 ( n ) r
u tk+1 X(tnk,tnk+1](t) ≡ un (t)
k=0
converge to
u in
L ^{p} ≡ K . We assume that all of these partition points have
empty intersection with the set of measure zero where
Bu ≠ B .
Thus, at every partition point,
Bu =
B . As just mentioned,
L ^{p} ≡ K, L ^{p′
} =
K ^{′} .
Taking a subsequence, we can have
∥∥ul − u∥∥ + ∥ur− u∥ + ∥∥Bul − Bu ∥∥ + ∥Bur − Bu ∥ ′
n K n K n K′ n K
′ r ′ ∥∥ ′ l ′∥∥ −n
+ ∥B un − B u ∥L2([0,T ],W ′) + B un − B u L2([0,T ],W ′) < 2 (32.5.29)
(32.5.29)
and so, we can assume that a.e. convergence also takes place for
Bu _{n} ^{l} ,Bu _{n} ^{r} ,B ^{′} u _{n} ^{l} ,B ^{′} u _{n} ^{r} ,u _{n} ^{r} ,u _{n} ^{l} .
Is Bu
=
B u _{0} ? The integral equation gives this it seems. To save notation,
B u _{0} will be written as
Bu _{0} . This is not inconsistent because
t → B u _{0} is
continuous and its value at 0 is
B u _{0} .
Lemma 32.5.2 Let s < t. Then for u,Y satisfying 32.5.28
〈Bu (t) ,u (t)〉− 〈Bu (s),u (s)〉+ 〈(B(t)− B (s))u (s),u (t)〉
∫ t
+ 〈(B (t)− B (s))u(s),u(t) − u(s)〉 = 2 〈Y (r),u(t)〉 dr
s
− 〈B (t)u(t)− B (t)u (s),u (t)− u (s)〉 (32.5.30)
(32.5.30)
Proof: It follows from the following computations
∫ t
B (t)u (t)− B (s)u(s) = Y (r) dr
s
and so
∫
t
2 s 〈Y (r),u(t)〉dr − 〈B (t)u(t) − B(s)u(s),u(t)− u(s)〉
〈 ∫ t 〉
= 2 s Y (r)dr,u(t) − 〈B (t)u(t)− B (s) u(s),u (t)− u (s)〉
= 2 〈B (t)u(t)− B (s)u (s),u (t)〉− 〈B(t)u(t)− B (s)u (s),u (t)− u (s)〉
= 2〈B (t)u(t),u(t)〉− 2 〈B (s)u(s),u(t)〉 − 〈B (t)u(t) ,u (t)〉
+〈B (t) u(t),u (s)〉+ 〈B (s)u (s) ,u (t)〉− 〈B (s)u (s),u (s)〉
= 〈B(t)u(t),u(t)〉− 〈B (s) u(s),u(s)〉
+ [〈B (t)u (t),u(s)〉− 〈B(s)u(s),u(t)〉]
= 〈B (t)u (t),u(t)〉 − 〈B (s)u(s),u(s)〉
+ 〈(B (t)− B (s))u(s),u(t)〉
Thus
〈Bu (t),u (t)〉− 〈Bu (s),u (s)〉+ 〈(B(t)− B (s))u (s),u (t)〉
∫
t
= 2 s 〈Y (r),u(t)〉dr− 〈B(t)u(t)− B (s)u (s) ,u (t)− u (s)〉
Now consider the last term. It equals
〈B(t)u(t)− (B(s)− B (t) +B (t))u(s),u(t) − u (s)〉
= 〈B (t)u (t)− ((B(s)− B (t))u (s)+ B (t)u(s)),u (t)− u (s)〉
= 〈B (t)u (t)− B (t)u(s),u(t) − u (s)〉+ 〈(B (t)− B (s))u (s) ,u (t)− u (s)〉
It follows that
〈Bu (t),u (t)〉− 〈Bu (s),u (s)〉+ 〈(B(t)− B (s))u (s),u (t)〉
+ 〈(B (t)− B (s))u (s),u(t)− u(s)〉
∫ t
= 2 〈Y (r),u (t)〉dr− 〈B (t)u(t)− B (t)u (s),u (t)− u (s)〉
s
Of course this computation is under the assumption that neither s,t are in the
exceptional set off which B
u =
Bu . In case
s = 0 the same formula holds
except you need to replace
u with
u _{0} and
Bu with
B u _{0} =
Bu .
■
It is good to emphasize part of the above.
〈B (t)u (t)− B (t)u (s),u (t)− u (s)〉− 〈B (t)u (t)− B (s)u(s),u(t) − u (s)〉
= 〈(B (s)− B (t))u(s),u(t)− u(s)〉
Lemma 32.5.3 Let the partitions P _{k} be as above such that 32.5.29 , P _{k} =
_{j=0} ^{mk} .
Then for any m ≤ m _{k} ,
m∑− 1
〈B(tkj+1)u(tkj+1)− B (tkj+1)u (tkj),u (tkj+1)− u(tkj)〉−
j=0
m −1
∑ 〈B (tk )u(tk )− B (tk)u (tk) ,u (tk )− u (tk)〉 = εm (k)
j=0 j+1 j+1 j j j+1 j
where lim_{k→∞} ε ^{m}
= 0
. Here
m∑−1〈( ( ) ( )) ( ) ( ) ( )〉
εm(k) = B tkj − B tkj+1 u tkj ,u tkj+1 − u tkj
j=0
Proof: From the above lemma, the absolute value of the left side is no larger
than
m∑−1
||〈(B (tkj) − B (tkj+1))u(tkj),u(tkj+1) − u(tkj)〉||
j=0
m∑−1∫ tkj+1 ∥ ( )∥ ∥ ( ) ( )∥
≤ ∥B′(τ)u tkj ∥W′ dτ ∥u tkj+1 − u tkj∥W (32.5.31)
j=0 tkj
(32.5.31)
In case p ≥ 2 then for C ≥ max_{s}
_{ℒ(W,W′)
} ,
m∑− 1∫ tkj+1∥ l ∥ ∥ r l ∥
≤ C k ∥u (τ)∥W ∥uk(τ)− uk(τ)∥W dτ
j=0 tj
m∑− 1∫ tkm ∥ l ∥ ∥ r l ∥
= C X[tkj,tkj+1](τ)∥uk (τ)∥W ∥uk (τ) − uk(τ)∥W dτ
j=0 0
∫ tkm m∑− 1 ∥ l ∥ ∥ r l ∥
= C X[tkj,tkj+1](τ)∥uk (τ)∥W ∥uk (τ) − uk(τ)∥W dτ
0 j=0
∫ tkm ∥ l ∥ ∥ r l ∥
= C ∥u (τ)∥W ∥uk(τ)− uk(τ)∥W dτ
∥∥0l∥∥ ∥∥ r l ∥∥
≤ C uk Lp([0,T],V) u k(τ)− uk(τ) Lp([0,T],V)
≤ Cˆ(2)2−k
by
32.5.29 . In case
p < 2
, then from assumption and
32.5.31 , the absolute value of the left
side is no larger than
m∑−1
C (tkj+1 − tkj)∥∥u (tkj+1)− u (tkj)∥∥
j=0 W
m∑ −1∫ tkj+1
= C X kk (s)∥∥urk(s) − ulk(s)∥∥
j=0 tkj [tj,tj+1] W
∫ tkm
= C ∥∥urk(s) − ulk(s)∥∥
0 W
which converges to 0 as
k →∞ thanks to
32.5.29 .
■
Lemma 32.5.4 In the above situation,
∫
T ′
st∈uNpC 〈Bu (t),u (t)〉+ 0 〈B u,u〉ds ≤ C(∥Y ∥K′ ,∥u∥K)
Also, t → Bu
is weakly continuous with values in W ^{′} on N ^{C} where N is a set of
measure zero including the set where Bu ≠ B .
Proof: From the above formula of Lemma 32.5.2 applied to the k ^{th} partition of
described above,
m− 1
∑
〈Bu (tm ),u(tm)〉− 〈Bu0,u0〉+ j=0 〈(B (tj+1)− B (tj))u (tj),u(tj+1)〉
∗
m∑−1
+ 〈(B (tj+1)− B (tj))u(tj),u (tj+1)− u(tj)〉
j=0
∫
m∑−1 tj+1
= 2 tj 〈Y (r),u(tj+1)〉dr − 〈B (tj+1)u (tj+1)− B (tj+1)u (tj),u(tj+1)− u (tj)〉
j=0
(32.5.32)
(32.5.32)
Consider the third term on the left,
m∑n−1〈( (n ) ( n)) ( n) ( n )〉
B tj+1 − B tj u tj ,u tj+1
j=0 〈 ( ) ( ) 〉
∫ tmn mn∑ −1 B tnj+1 − B tnj l r
= X(tnj,tnj+1](t)----tn---−-tn----un (t),un(t) dt
0 j=0 j+1 j
Using a simple approximate identity argument and the assumption that
t → B is in
C ^{1} ,
m −1 ( ) ( )
∑n n n B--tnj+1--−-B-tnj- ′
X(tj,tj+1](t) tnj+1 − tnj → B (t)
j=0
uniformly on (0,T ]. Then ∑
_{j=0} ^{mn−1} X _{(t
jn,tj+1n]}
n n
B(tjt+n1)−−-Btn(tj)-
j+1 j