Biomedical Engineering Reference
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time-marching, splitting scheme, which defines an approximate solution on .0;T/
of our main FSI problem ( 2.111 )-( 2.118 ). Furthermore, the scheme is designed in
such a way that for each t > 0 the approximate FSI solution satisfies a discrete
version of an energy estimate for the continuous problem.
We would like to ultimately show that, as t ! 0, the sequence of solutions
parameterized by N (or t) converges to a weak solution of ( 2.111 )-( 2.118 ).
Furthermore, we also need to show that 1 C n
R min >0is satisfied for each
n D 0;:::;N1. In order to obtain this result, it is crucial to show that the discrete
energy of the approximate FSI solutions defined for each t is uniformlybounded,
independently of t (or N). This result is obtained by the following Lemma.
Lemma 2.1 (The uniform energy estimates). Let t > 0 and N D T=t >0 .
Furthermore, let E n C
1
2
N , E n C N , and D N be the total energy and dissipation given
by ( 2.141 ) and ( 2.142 ) , respectively.
There exists a constant C>0 independent of t (and N ) such that the following
estimates hold:
1. E n C 2
N
C; E n C 1
N
C , for all n D 0;:::;N 1;
2. P j D 1 D N C;
Z
N 1
X
1
2
.1 C n /j u n C 1
u n
2
Ckv n C 1
v n C
2
L 2 .0;1/
3.
j
k
F
L 2 . S / C;
n D 0
1
2
L 2 .0;1/ CkV n C 1
V n
Ckv n C
v n
2
2
k
k
.k@ z . n C 1
d n
N 1
X
L 2 .0;1/ C a S d n C 1
n /k
2
d n ;d n C 1
4.
C:
n D 0
In fact, C D E 0 C C
L 2 .0;T/ ,where C is the constant
from ( 2.147 ) , which depends only on the parameters in the problem.
Proof. We begin by adding the energy estimates ( 2.144 )and( 2.147 ) to obtain
2
2
kP in k
L 2 .0;T/ CkP out k
Z
1
2
2
E n C 1
N
C D n C 1
N
.1 C n / j u n C 1
u n
2
Ck v n C 1
v n C
2
L 2 .0;1/
C
j
k
F
1
2
1
2
Ck v n C
v n
2
L 2 .0;1/ Ck V n C 1
V n
2
L 2 . S / Ck @ z . n C
n / k
2
L 2 .0;1/
k
k
d n
C a S d n C 1
d n ;d n C 1
E N C Ct..P in / 2
C .P out / 2 /; n D 0;:::;N 1:
Then, we calculate the sum, on both sides, and cancel out like terms in the kinetic
energy that appear on both sides of the inequality to obtain
Z
N 1
X
N 1
X
1
2
v n C 2
E N C
D n C 1
N
.1 C n /j u n C 1
u n
2
Ckv n C 1
2
L 2 .0;1/
C
j
k
F
n D 0
n D 0
Ckv n C 2
L 2 . S / Ck@ z . n C 2
v n
2
L 2 .0;1/ CkV n C 1
V n
2
n /k
2
L 2 .0;1/
k
k
 
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