The role of the variational formulation in the dimensionally-heterogeneous modelling of the human cardiovascular system - Modeling of Physiological Flows - page 255

Let us adopt γ = 1. All terms associated with the factor ( 1 − γ ) in Eq. (9.24)

drop. From the mechanical point of view this means that over Γ a all contributions

to the virtual power in the sense of the kinematics characterized by

Q 2 are not con-
sidered. With the introduction of decomposition (9.15), and using (9.19) to write
R 2 ( u 2 ) | Γ a = R 21 ( u 2 ) | Γ a + R 2 f ( u 2 ) | Γ a , Eq. (9.24) delivers the following natural con-

ditions over Γ a

) ,

, ( − ) T Γ a ( Q 1 ) ×T Γ a ( Q 1 ) = ∀ ∈T Γ a ( Q

s 1 u 1 u 21 0 s 1 (9.25) 1

t 1 + R ( u 1 ) | Γ a , v 1 T Γ a ( Q 1 ) ×T Γ a ( Q 1 ) = 0 ∀ v 1 ∈T Γ a ( Q ) , (9.26) 1 1

−R ( ) | Γ a , T Γ a ( Q 1 ) ×T Γ a ( Q 1 ) = ∀ ∈T Γ a ( Q ) ,

t 1 u 2 v 21 0 v 21 (9.27)

21 1

v 2 f ∈ ( T Γ a ( Q 1 ) ) ⊥ .

R 2 f ( u 2 ) | Γ a , v 2 f T Γ a ( Q 2 ) ×T Γ a ( Q 2 ) = 0 ∀ (9.28)

From Eq. (9.25) we have
u 1 =

u 21 in the sense of T Γ a ( Q 1 ) ,

(9.29)

u 2 f = 0 obtained after solving (9.24) ,

that, in short, means

u 1 = u 2 in the sense of T Γ a ( Q 1 ) . (9.30)

It is important to remark that Eq. (9.29) does not characterize the component u 2 f ,
but it just states that its value is different from zero, and that it can be determined
after solving (9.24).
Hence, due to the different underlying kinematics, the continuity of the primal
variables is not satisfied in the sense required by the original problem (Problem 9.1).
In fact, the fluctuation u 2 f is not necessarily null (see (9.29) 2 ) and is obtained af-
ter solving the variational equation (9.24). On the other hand from (9.26)-(9.28) we
obtain

t 1 = −R 1 (

T Γ a ( Q 1 ) ,

u 1 ) | Γ a = R 21 ( u 2 ) | Γ a in the sense of

(9.31)

) , R 2 f ( u 2 ) | Γ a = 0 in the sense of T Γ a ( Q

2

which is

T Γ a ( Q 2 ) . −R 1 ( u 1 ) | Γ a = R 2 ( u 2 ) | Γ a in the sense of (9.32)

0. Then the natural boundary conditions over
Γ a are given by (we also make use of decompositions (9.15) and (9.19))

For simplicity, let us take now γ =

s 2 ∈T Γ a ( Q 2 ) ,

s 2 , ( u 1 − u 2 ) T Γ a ( Q 2 ) ×T Γ a ( Q 2 ) = 0 ∀ (9.33)

t 21 + R 1 ( u 1 ) | Γ a , v 1 T Γ a ( Q 1 ) ×T Γ a ( Q 1 ) = 0 ∀ v 1 ∈T Γ a ( Q 1 ) , (9.34)

t 2 −R 2 ( u 2 ) | Γ a , v 2 T Γ a ( Q 2 ) ×T Γ a ( Q 2 ) = 0 ∀ v 2 ∈T Γ a ( Q 2 ) . (9.35)