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Lemma B.3.18 Fo r u 0 K · H and f
S( 0 ,T) , and any h, k > 0, one has
C E(h,T )
E(k,T ) ,
U h (t)
U k (t)
I( 0 ,T )
+
(B.74)
with E(h,T ) as in Lemma B.3.14 . In particular ,
{ U h (t) } h> 0 is Cauchy in I( 0 ,T)
and there exists
=
u
lim
h 0
U h (t)
I( 0 ,T).
Proof We choose in ( B.73 ) v
U h (t) for some h> 0, and then exchange in the
resulting inequality the roles of k and h . Adding the resulting two inequalities for
the difference w(t)
=
U h (t) , we ge t a n inequality of the type considered in
Lemma B.3.6 , with s(t) replaced by s(t)
:=
U k (t)
f(t) : To determine r(t) in
( B.40 ), we estimate the last term in the bound ( B.73 )asfollows:using0
+
f k (t
+
k)
k (t)
1
and
U k (t),
kU k (t)
k (t)
V , V
0 ,
(B.75)
we have
A
V , V
kU k (t)
0
r(t)
:=
u k (t
+
k)
f k (t
+
k),
β
H
kU k (t)
u k (t
+
k)
V +
f k (t
+
k)
V
.
Hence, in ( B.42 ),
T
R(T )
=
r(t) d t
0
Const
S( 0 ,T + k) ·
kU k (t)
u k (t
+
k)
I( 0 ,T ) +
f k (t)
I( 0 ,T ) .
From ( B.56 ), we get
R(T )
Const
S( 0 ,T + k)
kU k (t)
w 0 H +
f k (t)
I( 0 ,T ) ,
and ( B.61 )gives
Const
S( 0 ,T + k) E(k,T ).
R(T )
w 0 H +
f k (t)
(B.76)
To estimate the value of
w( 0 )
H
in ( B.76 ) and in ( B.43 ), we use that, by
Lemma B.3.2 ,
w( 0 )
H =
(u k, 1 P
u 0 )
(u h, 1 P
u 0 )
H
E(k,T )
+
E(h,T ),
which, inserted into Lemma B.3.6 , implies the assertion.
We can now give the proof of Theorem B.2.2 (i). Lemma B.3.18 established that
u h } h> 0 is Cauchy in I( 0 ,T) , hence in particular in L 2 (J
) and in L (J
{
; V
; H
) .
C 0 (J
Th erefore, u(t)
; H
) and u( 0 )
=
lim k 0 U k ( 0 )
=
lim k 0 u k, 1 = P
u 0
· H , which is the third line in ( B.6 ).
To show that u(t) is a solution of the PVI, pick in ( B.73 ) v(t) satisfying ( B.7b ),
and pass in ( B.73 ) to the limit k
K
0, implying the second line of ( B.6 ); since
K
is
u in L (J
closed in
) ,wealsohavethefirstlineof( B.6 ).
The uniqueness and Theorem B.2 (ii) will follow from
H
and U h
; H
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