Biomedical Engineering Reference
In-Depth Information
every t. Hence for " small enough,
0
2
V
"
,since
"
converges uniformly to as
" # 0.
The second pair of test functions we consider is .
1
;b/ where b belongs to
C
1
.Œ0;TI H
0
.0;L//, with
R
!
b D 0 and for a.e. t:
LJ
LJ
LJ
LJ
.0;b/
T
in B n C
Ǜ=2
R
1
D
.0;b/
T
in C
Ǜ=2
:
Since min
Œ0;T
Œ0;L
.1 C
"
/ Ǜ=2, .
1
;b/is a pair of admissible test functions for
all ".
With both types of test functions, it is easy to pass to the limit in the weak
formulation as " goes to zero. We obtain the existence of a weak solution on .0;T/
satisfying energy estimates.
Eventually, we show that we can extend the solution as long as we have
min
Œ0;T
Œ0;L
.1 C / > 0. Let us build an increasing sequence of times .T
k
/
k
1
as follows. First we choose a time T
1
>0such that there exists a weak solution
up to T
1
, with m
1
D min
Œ0;T
1
Œ0;L
.1 C / > 0. Possibly changing slightly T
1
,we
may assume that @
t
.T
1
/ 2 L
2
.0;L/ and
u
.T
1
/ 2 L
2
.
.T
1
// (since this is true for
almost every time).
Now, let k 1 and assume that we have built a solution up to some time T
k
, with
m
k
D min
Œ0;T
k
Œ0;L
.1C/ > 0. Our construction allows us to build an extension of
our solution, on some time interval starting from T
k
. Thanks to the a priori energy
estimate, we have for s T
k
1 C .s/ 1 C .T
k
/ .s T
k
/
2
C.T
k
;s/ m
k
.s T
k
/
2
C.T
k
;s/; (1.59)
with
C.T
k
;s/D C
k
u
.T
k
/k
L
2
.
0
/
; k@
t
.T
k
/k
H
0
.0;L/
;
Z
s
exp :.s
u
/.kf k
L
2
.
.
u
//
.
u
/ Ckgk
L
2
.0;L/
.
u
//d
u
;
T
k
where C is positive and nondecreasing with respect to its arguments, and C.T
k
;s/
C.0;s/. This a priori estimate shows that if we let
k
D minf1;.m
k
=2C.T
k
;T
k
C 1//
2
g;
we can build a solution starting from
u
.T
k
/, .T
k
/ and @
t
.T
k
/ up to the time T
k
C
k
(this corresponds to choosing Ǜ D m
k
=2 in the construction of the solution). The
time T
k
C
1
is chosen close to T
k
C
k
(in ŒT
k
C
k
=2;T
k
C
k
), in order to have also
@
t
.T
k
C
1
/ 2 L
2
.!/ and
u
.T
k
C
1
/ 2 L
2
.
.T
k
C
1
//.
We l e t T
sup
k
T
k
.IfT
< C1,thenm
D
D
min
Œ0;T
Œ0;L
.1 C / D 0.
Otherwise, since m
k
m
for all k,
k
minf1;.m
=2C.0;T
//
2
g >0.But
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