Pressure drop matrix for a bifuration of an artery with defects

Department of Mathematics

Pressure drop matrix for a bifuration of

an artery with defects

Vladimir Kozlov, Sergei Nazarov and German Zavorokhin

Department of Mathematics

Link¨

oping University

Vladimir Kozlov

a

, Sergei Nazarov

a,b

and German Zavorokhin

c

Abstra t. We onsider abifur ationof anartery. The inuen e of defe tsof the vessel's wallnearthe

bifur ation point on the pressure drop matrix is analyzed. The elements of this matrix are in luded

in the modied Kir hho transmission onditions, whi h were introdu ed earlier in [1℄, [2℄, and whi h

des ribeadequately the total pressure lossat the bifur ationpoint of the ow passedthrough it.

Keywords and phrases: Stokes' ow, bifur ation of a blood vessel, modied Kir hho onditions,

pressure drop matrix, total pressure loss.

1 Introdu tion

Themainobje tiveofthispaperistostudytheinuen eofdefe tsinthevesselwallsnearthebifur ation

pointonthepressuredropmatrix

Q

[3℄. We al ulatethematerialderivativeinthe aseofoblongplaques oraneurysms (see Fig.1,aand b)andthe topologi alderivativeinthe ase oflo alizedones(see Fig.1,

and d). The pressure drop matrix was introdu ed in[3℄ as an integral hara teristi of a jun tion of

a

b

c

d

Figure1: Variationsinthe shape of abifur ationnode: oblong(a)and sa ular ( ) aneurysms, oblong

parietal(b) and lo alizednodular (d) holesterol plaques.

several pipeswith absolutelyrigidwalls. It appearsthat theelementsofthis matrixare in luded inthe

modied Kir hho transmission onditions, whi h des ribe more adequately the total pressure loss at

wallswas developed. In parti ular, a new transmission ondition at the bifur ationpointwas derived,

whi h an be onsidered as a modi ation of the lassi al Kir hho ondition. Clearly, the total ux

atthe bifur ationpointis zerobut ontinuity of the pressure isnot so obvious. In uidme hani s,one

uses the total pressure lossin the ow passing the bifur ation point, see [4℄. An appropriate obje t to

des ribe this pressure loss is the pressure drop matrix, elements of whi h are involved in the modied

Kir hho onditions. Thismodi ationimprovesthe modelinseveral dire tions. First,thedis repan y

ofthe approximationof three-dimensional modelby the one-dimensional one is

O(e

−

ρ

h

)

, where

h

isthe thi kness of the vessel and

ρ

is a positive onstant. We remind that the appli ation of the lassi al Kir hho onditions brings the dis repan y

O(h

3

₎

for the velo ities and

O(h)

for the pressure. This dieren e is essential if we deal with a large system with many bifur ations. Se ond, the modied

transmission onditions depend onthe geometryof the bifur ationregion.

The pressure drop matrix

Q

is the symmetri (

2 × 2

) matrix. So it has three parameters (the diagonal elements

Q

++

, Q

−−

and the o-diagonal ones

Q

+−

= Q

−+

). The inuen e of

Q

on the transmission onditions an be taken into a ount also by the small variations in the lengths of the

edges in ident to the bifur ation point and by introdu ing ee tive lengths

L

α

(h),

α = 0, ±

, of one-dimensional images of blood vessels whilst keeping the lassi Kir hho transmission onditions and

exponentialsmallapproximation errors,see [2℄. Sin ethe number of hannelsisalsothree the ee tive

lengths

L

α

(h)

an be isomorphi ally determined by the entries of

Q

. By [2℄, the in rements of lengths

hl

α

, α = 0, ±,

l

0

= −B

0

Q

+−

= −B

0

Q

−+

,

l

±

= B

±

(Q

±∓

− Q

±±

),

B

α

=

πr

4

α

8ν

,

(1)

where

ν

isthe vis osity oftheuid and

r

α

istheradiusofthe vessel,weintrodu eperturbed edgeswith the ee tivelengths

L

α

(h) = L

α

+ hl

α

,

(2) where

L

α

are initial lengths of the edges. The ee tive lengths (2) are the attributes of the vessels themselvesand preserve when you hange the dire tionof bloodow through the node.

Ouraimwiththis arti leisto al ulateasymptoti sof thepressure dropmatrix andhen ethe total

in rements

h

P

α

l

α

, namely,

h

X

α

l

α

=

X

α

L

α

(h) −

X

α

L

α

= h (Q

+−

(B

+

+ B

−

− B

0

) − Q

++

B

+

− Q

−−

B

−

) ,

(3) of the ee tive lengths of the vessels taking into a ount the inuen e of perturbations (e.g., plaques,

aneurysms)arisingnearthebifur ationnodeofthearteryinthethree-dimensionalproblem. Asaresult,

we al ulate the total in rements of the ee tive lengths, and even determine their signs. Changes in

theee tive lengthsof the vessels orrespond tothe presen e of some defe ts inthe vessel walls. Sowe

an lo alize themby examiningthe pro ess of bloodow through a bifur ationnode.

InSe t. 2we onsider theStokessysteminanunbounded domainwith ylindri aloutletstoinnity

(see, e.g., [5, 6, 7, 8, 9℄)and provethe unique solvabilityof the problem. Forobtainingthe asymptoti

behaviorof the solutionwe exploit spe ial homogeneoussolutions tothe Stokes problemwith non-zero

uxandwithalineargrowthinthepressureatinnity(see[3℄). Asa onsequen e,weobtainadenition

ofthesymmetri pressure dropmatrix

Q

,whi hplaysa ru ialroleinthefun tioningofthe bifur ation node.

node and lose toit onthe matrix

Q

. Usingasymptoti analysis ofellipti boundaryvalueproblems in regularly(orsingularly)[10, 11℄ perturbed domainswend the in rementsof the pressure drop matrix

andalsodeterminetheirsigns. Invirtueofformulae(1)we al ulatethetotalin rementsoftheee tive

lengthsof one-dimensional imagesof the bloodvessels.

In Appendix it will be explained why the modi ation of the se ond Kir hho's law by means of

thepressure drop matrix unexpe tedly deeply in reases the a ura yof approa h forthree-dimensional

uidowina systemof thin hannelsby theone-dimensional Reynolds-Poiseuillemodel. Also,we give

proofs of supporting assertions of Se t. 2, 3. Note that onsidered in 4.2 the Cau hy problem for the

homogeneousStokes system supplemented by the Neumann ondition onthe part ofthe boundaryitis

alsoof independent interest.

2 Statement of the problem

2.1 Domains with ylindri al outlets to innity and fun tional spa es

We introdu e the domain

Ω

with three ylindri al outlets to innity (see Fig.2). Let

Ω

be an open unbounded domain with Lips hitz boundary

∂Ω

admittingthe representation

Ω = Ω

′

∪ Ω

0

∪ Ω

+

∪ Ω

−

,

where

Ω

α

_{∩ Ω}

β

_{= ∅}

for

α 6= β

,

α, β = 0, ±.

(4) Here

Ω

α

_{= {x}

α

_{= (y}

α

_{, z}

α

_{) : y}

α

_{∈ ω}

α

, z

α

> L

α

}

in a ertain Cartesian oordinate system

x

α

_{= (y}

α

_{, z}

α

₎

in

R

3

, where

y

α

are the variables inthe ross-se tion of the outlet

Ω

α

,

z

α

is the variable along the axis

of

Ω

α

and

ω

α

is a bounded domain in

R

2

. The bounded domain

Ω

′

is given by

Ω

′

_{= {x ∈ Ω : z}

α

_{< L}}

for ertain

L

,

L > max

α

L

α

. Hen eforth

x = (x

1

, x

2

, x

3

)

is a global oordinate system in

R

3

related to

the wholedomain

Ω

. We dene

L

2,β

(Ω)

asthe spa e of measurable fun tions in

Ω

with anite norm

x

x

₂

3

z

z

z

0

+

||u||

L

2,β

(Ω)

=

Z

Ω

′

|u(x)|

2

dx +

X

α=0,±

Z

Ω

α

|z

α

|

2β

|u(y

α

, z

α

)|

2

dy

α

dz

α

!

1/2

.

If

β = 0

we willuse the usual notation

L

2

(Ω)

for this spa e. ByusingtheSobolevspa e

H

1

_(Ω)

together with

L

2,1

(Ω)

weintrodu ethespa eofreal-valuedve tor fun tionsin

Ω

,

H(Ω) =

u = (u

1

, u

2

, u

3

) ∈ (H

1

(Ω))

3

| div u ∈ L

2,1

(Ω)

(5)

withthe norm given by

||u||

2

H(Ω)

=

Z

Ω

(|∇u(x)|

2

_{+ |u(x)|}

2

_{)dx +}

X

α=0,±

Z

Ω

α

|z

α

_|

2

_{|div u(y}

α

_{, z}

α

_)|

2

_dy

α

_dz

α

_.

(6)

Letalso

H

0

(Ω)

bethe subspa ein

H(Ω)

onsistingofve torfun tionsequalzeroon

∂Ω

. Thedualspa e of

H

0

(Ω)

is denoted by

(H

0

(Ω))

∗

.

2.2 Formulation of the problem

Consider the Diri hlet problem forthe stationaryStokes system with nonzero divergen e

−ν∆u(x) + ∇p(x) = F (x),

−div u(x) = G(x),

x ∈ Ω,

(7)

u(x) = 0,

x ∈ ∂Ω.

(8)

Here,

u(x) = (u

1

(x), u

2

(x), u

3

(x))

is the velo ity eld and

p(x)

isthe pressure,

ν > 0

isthe vis osity of uid,whi h isassumed tobe onstant.

In order todene aweak solution of the problem(7), (8), we introdu ea bilinearform on

H(Ω)

:

a(u, w) =

3

X

j=1

Z

Ω

∇u

j

∇w

j

dx.

Soif (u,p) isa lassi alsolutionof (7), (8), then multiplyingthe rst equation in(7)by

w ∈ H

0

(Ω)

and integrating over

Ω

, we obtain

νa(u, w) −

Z

Ω

p div w dx =

Z

Ω

F w dx

for any

w ∈ H

0

(Ω).

(9)

Weak solutionofthe problem(7),(8)is alledapair

(u, p) ∈ H

0

(Ω) ×L

2,−1

(Ω)

satisfyingtheintegral identity (9)for all

w ∈ H

0

(Ω)

and theequation

−div u = G

in

Ω

,where

F ∈ (H

0

(Ω))

∗

and

G ∈ L

2,1

(Ω)

are given.

Lemma 2.1. For arbitrary

g ∈ L

2,1

(Ω)

subje t to

Z

Ω

g(x)dx = 0

(10)

there exists a ve tor fun tion

u ∈ H

0

(Ω)

su h that

−div u = g

in

Ω

, and

||u||

H(Ω)

≤ c||g||

L

2,1

(Ω)

.

(11) Here, is a onstant independentof g.

Lemma'sproof is presented in Appendix.

Thefollowingtheoremonexisten eanduniquenessofweaksolutionstotheboundaryvalueproblem

(7)-(8)is quitestandard and we present it here for readers onvenien e.

Theorem 2.1. Suppose that

F ∈ (H

0

(Ω))

∗

and

G ∈ L

2,1

(Ω)

is su h that

Z

Ω

G(x)dx = 0.

(12)

Then there exists a weak solution

(u, p) ∈ H

0

(Ω) × L

2,−1

(Ω)

of the problem (7), (8) satisfying the estimate

||u||

H(Ω)

+ ||p||

L

2,−1

(Ω)

≤ c ||F ||

(H

0

(Ω))

∗

+ ||G||

L

2,1

(Ω)

.

(13)

Here, is a onstant independent of

F

and

G

. This solution is dened up to an additive onstant in the pressure

p

.

Proof. Existen e. Let

w ∈ H

0

(Ω)

bea solutionto the problem

−div w(x) = G(x),

x ∈ Ω,

w(x) = 0,

x ∈ ∂Ω

(14) satisfying estimate (11). Su h solution exists due to Lemma 2.1. Then the ve tor fun tion

V (x) =

u(x) − w(x)

solvesthe following Stokes problem

−ν∆V (x) + ∇p(x) = ˆ

F , −div V (x) = 0,

x ∈ Ω,

(15)

V (x) = 0,

x ∈ ∂Ω,

(16)

where

F (x) = F (x)+ν∆w(x) ∈ (H

ˆ

0

(Ω))

∗

. Introdu ethespa e

H

div

0

(Ω) = {W ∈ H

0

1

(Ω) : div W = 0

in

Ω} .

Thenthe ve tor fun tion

V ∈ H

div

0

(Ω)

isfound fromthe equality

νa(V, W ) =

Z

Ω

ˆ

F W dx

for any

W ∈ H

div

0

(Ω).

(17)

Bythe Riesz theorem su hsolution exists and satises

||V ||

H(Ω)

≤ c|| ˆ

F ||

(H

0

(Ω))

∗

≤ C(||F ||

(H

0

(Ω))

∗

+ ||G||

L

2,1

(Ω)

).

To nd

p

we pro eed as follows. By Lemma (2.1), for any

g ∈ L

2,1

(Ω)

subje t to (10) there exists a ve tor fun tion

v

g

∈ H

0

(Ω)

su h that

−div v

g

= g

in

Ω

, and

Moreover the orresponden e

g → v

g

islinear. We onsider the fun tional

G(g) =

Z

Ω

ˆ

F v

g

dx − νa(V, v

g

)

(18) on

L

2,1

(Ω) = {g ∈ L

2,1

(Ω) :

R

Ω

g(x)dx = 0}

. In virtue of

|G(g)| ≤ c



|| ˆ

F ||

(H

0

(Ω))

∗

+ ||V ||

H(Ω)



||v

g

||

H(Ω)

≤ c|| ˆ

F ||

(H

0

(Ω))

∗

||g||

L

2,1

(Ω)

thelinearfun tional

G(g)

is ontinuouson

L

2,1

(Ω)

. Therefore there existanelement

p

in

L

2,−1

(Ω)

su h that

G(g) =

Z

Ω

pg dx

for all

g ∈ L

2,1

(Ω)

and

||p||

L

2,−1

(Ω)

≤ c ||F ||

(H

0

(Ω))

∗

+ ||G||

L

2,1

(Ω)

.

Clearly, the pair

(u, p)

is the required weak solution.

Uniqueness. If

F = 0

and

G = 0

then from the denition of the weak solution it follows that

a(u, u) = 0

and hen e

u = 0

. This implies that

R

Ω

p divwdx = 0

for all

w ∈ H

0

(Ω)

. Using Lemma 2.1, we on lude that

p

is onstant.

The theorem is proved.

Remark2.1. Considera non-homogeneousDiri hlet problem for Stokessystem, i.e. equations(7)are

supplied with the boundary ondition

u(x) = H,

x ∈ ∂Ω,

(19)

where

H ∈ H(Ω)

and instead (10) we require

Z

Ω

G(x)dx +

Z

∂Ω

H(x) · ndΓ = 0,

(20)

where

n

is the unit, outward normal to

∂Ω

. Substituting

u(x) = v(x) + H(x)

into (7), (19) we obtain

−ν∆v(x) + ∇p(x) = f (x), −div v(x) = g(x),

x ∈ Ω,

(21)

v(x) = 0,

x ∈ ∂Ω,

(22)

where

f (x) = F (x) + ν∆H(x) ∈ (H

0

(Ω))

∗

and

g(x) = G(x) + div H(x) ∈ L

2,1

(Ω)

veries (10). Now appli ationoftheprevioustheoremgivestheexisten eofapair

(v, p) ∈ H

0

(Ω)×L

2,−1

(Ω)

solvingproblem (7), (19) and satisfying the estimate

||v||

H(Ω)

+ ||p||

L

2,−1

(Ω)

≤ c



||f ||

(H

0

(Ω))

∗

+ ||g||

L

2,1

(Ω)

+ ||H||

H(Ω)



.

(23)

Letthe right-handsides in(7), (8) satisfy

Z

Ω

′

|F (x)|

2

dx +

X

α

Z

Ω

α

|F (x

α

)|

2

e

2az

α

dx

α

< ∞

(24) and

Z

Ω

′

|G(x)|

2

dx +

X

α

Z

Ω

α

|G(x

α

)|

2

e

2az

α

dx

α

< ∞,

(25)

where

a

is a positive number. Let also

G

be subje t to (12). Then a ording to Theorem 2.1 the problem(7), (8) has a solution

(u, p) ∈ H

0

(Ω) × L

2,−1

(Ω)

. We an on lude that this solution satises the following asymptoti representation atinnity

(u, p) =

X

α=0,±

χ

α

c

α

(0, 1) + (˜

v, ˜

p),

(26)

where

χ

α

= χ

α

(z

α

₎

are smooth fun tions equal

1

for

z

α

_{> L}

α

+ 1

and

0

for

z

α

_{< L}

α

,

(˜

v, ˜

p)

are exponentially de aying termsat

z

α

_{→ ∞}

and

c

α

are real onstants. Sin e this solutionis dened up to anadditive onstant we an (and will)assume

c

0

= 0

. Then the solution isunique.

The remainingpartof this se tion isdevoted tondingformulas for evaluationof onstants

c

+

and

c

−

. For this purpose we need solutions of homogeneous problem (7), (8), whi h have a linear growth at innity, namely we introdu e two linear independent solutions

(V

±

_{, P}

±

₎

whi h have the following

asymptoti representations(see [3℄)

(V

±

, P

±

) = −χ

0

(V

0

, P

0

) + χ

±

(V

±

, P

±

) + (v

±

, p

±

),

(27) where

(V

α

_{, P}

α

₎

is the Poiseuille ow inthe ylinder

Ω

α

, i.e.

V

α

y

α

i

(x) = 0

,

i = 1, 2

,

P

α

_{(x) = −C}

α

z

α

and

V

α

z

α

= V

_z

α

α

(y

α

)

solves the followingDiri hlet problemin

ω

α

∆V

_z

α

α

= −C

_α

in

ω

α

,

V

α

z

α

= 0

on

∂ω

α

. The normalizing onstant

C

α

is hoosing to satisfy

Z

ω

α

V

_z

α

α

(y

α

)dy

α

= 1.

(28)

Inthe most important ase of the ir ular ylinder, i.e.

ω

α

= {y

α

_{: |y}

α

_{| < r}

α

}

,

V

_z

α

α

(x) =

2(r

2

α

− |y

α

|

2

)

πr

4

α

,

P

α

(x) =

−8ν

πr

4

α

z

α

.

The remainder term

(v

±

_{, p}

±

₎

in(27) satises the problem

−ν∆v

±

+ ∇p

±

= f

±

, −divv

±

= g

±

in

Ω,

(29)

f

±

:= ν∆(χ

0

V

0

− χ

±

V

±

) − ∇(χ

0

P

0

− χ

±

P

±

)

(31) and

g

±

:= div(χ

0

V

0

− χ

±

V

±

)

(32) have ompa t supports. Toverify ondition (20) inTheorem (2.1) for

g

±

, weapply the Gausstheorem

forthe domain

Ω

R

= {x ∈ Ω : z

a

_{< R}}

,where

R

is a su iently largenumber, and obtain

Z

Ω

g

±

dx = lim

R→∞

Z

Ω

R

div(χ

0

V

0

− χ

±

V

±

)dx =

Z

ω

0

R

V

_z

0

0

(y

0

)dΣ −

Z

ω

±

_R

V

_z

±

±

(y

±

)dΣ = 0.

Hereweusedthenormalization ondition(28). Therefore,

(v

±

_{, p}

±

₎

admitstheasymptoti representation

(26),where

c

0

= 0

and

(˜

v, ˜

p)

exponentially tends tozero when

z

α

_{→ ∞}

.

Now we an present formulas for al ulation of oe ients in(26)

Theorem 2.2. Let the fun tions

F

and

G

satisfy (24)-(25) and let the asymptoti formula (26) be valid with

c

0

= 0

. Then

c

±

=

Z

Ω

F V

±

+ GP

±

dx

(33)

Proof. Let

Ω

R

bethe samedomainasbefore. Multiplyingequations(7), (8)by

(V

±

_{, P}

±

₎

,

integrat-ingover

Ω

R

and using Green's formula, we obtain

Z

Ω

R

(−ν∆v + ∇p) V

±

_{− divvP}

±

_dx

=

X

α

Z

ω

α



− ν V

±

∂

z

α

v

_z

α

− v∂

_z

α

V

±

z

α

+ pV

_z

±

α

− P

±

v

_z

α







z

α

_=R

dy

α

_.

Takinghere limitand using asymptoti formulas for

(V

±

_{, P}

±

₎

and (26) we arrive at(33).

Applyingformula(33)tothesolution

(v

±

_{, p}

±

₎

oftheproblem(7), (8)withtherighthandsidesgiven

by (31), (32), we obtainthe representations

(v

±

, p

±

) = χ

±

Q

±±

(0, 1) + χ

∓

Q

±∓

(0, 1) + (˜

v

±

, ˜

p

±

),

(34) wherethe oe ients are evaluateda ording to

Q

γτ

=

Z

Ω

(f

γ

V

τ

+ g

γ

P

τ

) dx,

γ, τ = ±.

(35) From (27) and (34) we get the followingrepresentations

(V

±

, P

±

) = −χ

0

(V

0

, P

0

) + χ

±

(V

±

, P

±

)

+χ

±

Q

±±

(0, 1) + χ

∓

Q

±∓

(0, 1) + (˜

v

±

, ˜

p

±

),

(36) withtheremainders

(˜

v

±

, ˜

p

±

)

exponentiallyde ayingatinnity. Notethatastraightforward al ulation gives the equality

Q

γτ

= Q

τ γ

. The oe ients

Q

γτ

in the expansion (36) of the pressure at innity in

Ω

±

form the symmetri

(2 × 2)

 matrix

Q

alled the pressure drop matrix. Another approa h to introdu ingthe matrix

Q

was presented in[3℄.

3.1 Regular perturbation of the boundary of

Ω

We assume that the boundary

∂Ω

is su iently smooth. We introdu e oordinates

(n, τ )

in a neigh-borhoodof theboundaryasfollows:

n

isthe oriented distan e to

∂Ω

(

n > 0

outside

Ω

¯

) and

τ

isalo al oordinateon

∂Ω

. Let

ϕ = ϕ(τ )

be a smooth fun tion (positive or negative) with a ompa t support on

∂Ω

. Nowdene the surfa e

Γ

ε

asthe perturbation of the surfa e

∂Ω

as

Γ

ε

= {x : n = εϕ(τ )} ,

(37) where

ε > 0

isa smallparameter. Let

Ω

ε

is domainwith the boundary

Γ

ε

.

Forthe perturbed domain

Ω

ε

we have analog of formula(36)

(V

_ε

±

, P

_ε

±

) = −χ

0

(V

0

, P

0

) + χ

±

(V

±

, P

±

) + (v

ε

±

, p

ε

±

),

(38)

(v

±

ε

, p

ε

±

) = χ

±

Q

±±

(ε)(0, 1) + χ

∓

Q

±∓

(ε)(0, 1) + (˜

v

±

ε

, ˜

p

ε

±

).

(39) Theorem 3.1. Let

Ω

ε

be domainwith regular perturbation of the boundary (37). Then formulae (38), (39) have the asymptoti expansion of elements of the matrix

Q

Q

γτ

(ε) = Q

γτ

+ εc

τ

γ

+ O(ε

2

),

γ, τ = ±.

(40) Here,

c

τ

γ

= ν

R

Γ

0

ϕ∂

n

V

γ

· ∂

n

V

τ

dΣ

. Proof. Let

(w

ε

, q

ε

) = (v

_±

ε

, p

ε

_±

) − (v

±

, p

±

) = (V

ε

±

, P

ε

±

) − (V

±

, P

±

).

If it is needed we extend smoothly fun tions

(v

±

, p

±

)

outside

Ω

. Then the pair

(w

ε

_{, q}

ε

₎

satises the

following problem

−ν∆w

ε

+ ∇q

ε

= 0, −divw

ε

= 0

in

Ω

ε

,

(41)

w

ε

= −V

±

on

Γ

ε

.

(42)

We take the asymptoti ansatzfor a solution(41), (42) as follows:

(w

ε

, q

ε

) = ε(v

_±

′

, p

′

_±

) + ε

2

(v

_±

′′

, p

′′

_±

) + . . .

(43) where

(v

′

±

, p

′

±

)

must satisfythe problemin

Ω

:

−ν∆v

_±

′

+ ∇p

′

_±

= 0, −divv

_±

′

= 0

in

Ω,

(44)

v

_±

′

= h

±

on

∂Ω.

(45)

Comparing(42) and (45) we see that

Finally, using that

∂

n

V

±

n

= 0

on

∂Ω

(this follows from

−divV

±

_{= 0}

) weobtain (40).

Note that the matrix

{c

τ

γ

}

γ,τ =±

in (40) is positive denite for

ϕ > 0

and is negative denite for

ϕ < 0

. Indeed,if someof the oe ients

c

τ

γ

= 0

then

∂

n

V

τ

_{= 0,}

_{τ = ±}

,onthe

supp ϕ

and by 4.2(see Appendix) the homogeneousStokes problem(7), (8) willhavetrivial solution

(u, p) = (0, 0, 0; 1)

only.

Taking into a ount the formulae (3), (40) we obtain the asymptoti expansion for the total

in re-ments of the ee tive lengths

X

α

l

α

(ε) =

X

α

l

α

+ εΨ + O(ε

2

),

(47) where

Ψ = ν

Z

Γ

0

ϕ{(B

+

+ B

−

− B

0

)∂

n

V

+

· ∂

n

V

−

− B

+





∂

_n

V

+





2

− B

−





∂

_n

V

−





2

}dΣ.

Letthe radii of the bloodvessels

r

α

, α = 0, ±

,be onne ted as follows

r

±

= δ

±

r

0

, 0 < δ

±

< 1

. We have the following theorem that gives us possibility to estimate by means of the ee tive lengths the

inuen e of hanges inthe vessel walls geometry.

Theorem 3.2. Let

δ

±

be real numbers su h that

|δ

2

+

− δ

−

2

| < 1

, then in (47) the matrix of quadrati form

Ψ

is negative denite for

ϕ > 0

and is positive denite for

ϕ < 0

.

Proof. Substitutingtheasymptoti expansion(40)into(3),usingSylvester's riterionforthematrix

ofquadrati form

Ψ

, we immediatelyprove the assertion of Theorem 3.2.

So we an on lude that if near the bifur ation of an artery the holesterol plaque (in the ase of

ϕ < 0

) islo atedthen thetotal ee tivelengthof thevessels in reases. Vas ularinjury asso iatedwith aneurysm(inthe aseof

ϕ > 0

) orresponds tothe de reasinginthe totalee tivelengthof thevessels, see Appendix.

3.2 Model problem in a half-spa e

ConsiderthehomogeneousStokessysteminthehalf-spa e

R

3

+

= {Y = (Y

′

, Y

3

) = (Y

1

, Y

2

, Y

3

) : Y

3

> 0}

:

−ν∆U(Y ) + ∇P (Y ) = 0,

−div U(Y ) = 0,

Y ∈ R

3

₊

,

(48)

U(Y

′

, 0) = 0,

Y

′

∈ R

2

.

(49) We are interested in solutions of (48), (49) having the form

U(x) = r

λ

_u(ω)

,

P (x) = r

λ−1

_p(ω)

, where

r = |Y |

,

ω = Y /r

and

λ

is a omplex number. Su h solutions exist only when

λ = 1, 2, , . . .

or

λ = −2, −3, . . .

. Moreover, the spa e of su h solutions is the same for

λ

and

−1 − λ

, see for example Theorem 5.2.1 in[11℄.

For

λ = 1

this problemhas the following three solutions

(V

k

, P

k

)

,

k = 1, 2, 3

,where

V

1

(Y ) = (Y

3

, 0, 0), V

2

(Y ) = (0, Y

3

, 0), V

3

= 0, P

1

= P

2

= 0

and

P

3

= 1.

Thersttwove torfun tionsare alledthe Quetteowsand the thirdoneis onstant. UsingTheorem

5.4.4[11℄,we an des ribeall solutionsfor

λ = −2

. They are given by pairs

(U, P )

,

(δ + 6)v = 3c

in

S

2

+

and

v = 0

on

∂S

2

+

. (50)

Here

δ

istheLapla e-Beltramioperatoron

S

2

. Solutionsto(50)are obtainedfromsolutionsto

∆v = 3c

in

R

3

+

with zero Diri hlet boundary onditions and

v

being se ond order polynomial. Therefore these solutionsare given as

(V

1

, P

1

)

,

(V

2

, P

2

)

and

(V

3

, P

3

)

,where

V

k

(Y ) =

ω

k

ω

3

νr

2

(ω

1

, ω

2

, ω

3

), P

k

(Y ) = 2

ω

k

ω

3

r

3

, k = 1, 2,

and

V

3

(Y ) =

ω

2

3

νr

2

(ω

1

, ω

2

, ω

3

), P

3

(Y ) = 2

ω

2

3

r

3

−

2

3r

3

.

Thesefun tions verify the following bi-orthogonality onditions

h(V

k

, P

k

), (V

j

, P

j

)i =

Z

R

3

+



− ν∆(χV

k

) + ∇(χP

k

)
· V

j

− div(χV

k

)P

j



dY

=

Z

S

2

+



− ν(∂

r

V

k

· V

j

− V

k

· ∂

r

V

j

) + P

k

ω · V

j

− ω · V

k

P

j



r

2

dS

ω

= M

k

δ

k

j

,

(51)

where

χ

is asmooth fun tionequals

0

forsmall

|Y |

and

1

for large

|Y |

,

dS

ω

spheri alarea element and

M

k

= 5

Z

S

2

+

ω

_k

2

ω

₃

2

dS

ω

for

k = 1, 2

and

M

3

=

−1

ν

Z

S

2

+

ω

₃

2

dS

ω

.

3.3 Domain lose to a half-spa e

Now let us turn to the Stokes system in a domain

Ξ

whi h oin ides with

R

3

+

outside a ball

B

2δ

(0)

given by

|Y | ≤ 2δ

and

Ξ = {Y ∈ B

δ

(0) : Y

3

> φ(Y

1

, Y

2

)}

, where

φ

isa smooth fun tion equalto

0

for

|Y

′

_{| > δ}

:

−ν∆u(Y ) + ∇p(Y ) = F (Y ),

−div u(Y ) = G(Y ),

Y ∈ Ξ,

(52)

u(Y ) = 0,

Y ∈ ∂Ξ,

(53)

Theorem 3.3. Let

F ∈ (H

1

0

(Ξ))

∗

and

G ∈ L

2

_(Ξ)

. Then the problem (52), (53) has a unique weak

solution

(u, p) ∈ H

1

0

(Ξ) × L

2

(Ξ)

. This solution satises

||u||

H

1

_(Ξ)

+ ||p||

_L

2

_(Ξ)

≤ C



||F ||

(H

1

0

(Ξ))

∗

+ ||G||

_L

2

_(Ξ)



.

Now, we are interesting in asymptoti s of solutionsfor large

|Y |

.

Theorem 3.4. Let

F ∈ (H

1

0

(Ξ))

∗

and

G ∈ L

2

_(Ξ)

have ompa t supports. Then solution

(u, p) ∈

H

1

0

(Ξ) × L

2

(Ξ)

from the previous theorem satises

u(Y ) =

3

X

k=1

c

k

V

k

+ O(|Y |

−3

),

p(Y ) =

3

X

k=1

c

k

P

k

+ O(|Y |

−4

),

for large

|Y |,

(54)

Proof. Let

χ

be the smooth ut-o fun tion in

R

3

+

, namely

χ(Y ) = 1

for

|Y | > R

0

,

χ(Y ) = 0

for

|Y | < R

0

,where

R

0

=

onst . We transformthe problem(52), (53) in

Ξ

tothe followingproblemin

R

3

+

−ν∆(χu) + ∇(χp) = ˆ

F ,

−div (χu) = ˆ

G,

Y ∈ R

3

₊

,

(55)

χu = 0,

Y ∈ R

2

,

(56)

wheretheright-handsides

F = χF −ν((∆χ)u+2(∇χ, ∇)u)+p∇χ

ˆ

and

G = χG−(u, ∇χ)

ˆ

have ompa t supports.

Let

U = χu − v

. Here

v

is asolution tothe problem

−div v = ˆ

G,

Y ∈ R

3

₊

,

(57)

v = 0,

Y ∈ R

2

.

(58) Andwe arrive at

−ν∆U + ∇P = f,

−div U = ˆ

G,

Y ∈ R

3

+

,

(59)

U = 0,

Y ∈ R

2

,

(60) where

(H

0

(R

3

+

))

∗

∋ f = ˆ

F + ν∆v

and

P = χp

.

Exploitingthe expli it formof the Green'stensor (a tensor eld G(Y,Z) and a ve tor eld g(Y,Z))

forthe homogeneousStokes system inthe half-spa e (see, e.g., [13, Appendix 1℄)we nd estimates

|G

ij

(Y, Z) − G

ij

(Y, 0)| ≤ C|Y − Z|

−2

,

|g

i

(Y, Z) − g

i

(Y, 0)| ≤ c|Y − Z|

−3

.

(61) Usingthe estimates (61) for the solution to(59), (60)

U

j

(Y ) =

Z

R

3

+

G

ij

(Y, Z)f

i

(Z)dZ,

P (Y ) =

Z

R

3

+

g

i

(Y, Z)f

i

(Z)dZ

weget formulae (54). The proof is omplete.

Inordertoevaluatethe onstants

c

k

in(54)weintrodu ethreesolutions

( ˆ

V

k

(Y ), ˆ

P

k

(Y ))

,

k = 1, 2, 3

, of the homogeneousproblem(52), (53), having the form

ˆ

V

k

(Y ) = V

k

(Y ) + ˜

V

k

(Y ), ˆ

P

k

(Y ) = P

k

(Y ) + ˜

P

k

(Y )

(62) with

( ˜

V

k

, ˜

P

k

) ∈ H

1

_{(Ξ) × L}

2

_(Ξ)

. Sin e

V

3

= 0

and

P

3

= 1

, we have

V

˜

3

= 0

and

P

˜

3

= 0

. One an verify that

−ν∆ ˜

V

k

+ ∇ ˜

P

k

= 0,

−div ˜

V

k

= 0,

Y ∈ Ξ,

(63) and

V

˜

k

+ V

k

= 0

on

∂Ξ

. By Theorems 3.3and 3.4 this problemhas solutionand it satises

˜

V

k

(Y ) =

3

X

j=1

A

k

_j

V

j

(Y ) + O(|Y |

−3

),

P

˜

k

(Y ) =

3

X

j=1

A

k

_j

P

j

(Y ) + O(|Y |

−4

)

(64)

Theorem 3.5. Let

F

and

G

be the same as in Theorem 3.4 and let

c

k

be oe ients in (54). Then

M

k

c

k

=

Z

∂Ξ

F ˆ

V

k

+ G ˆ

P

k

dS.

(65)

Proof. Let

Ξ

R

= {Y ∈ Ξ : |Y | < R}

and

S

R

= {Y ∈ Ξ : |Y | = R}

. Then multiplying the rst equation in(52) by

V

ˆ

k

and the se ond equation in(52) by

P

ˆ

k

, summingthem up and integrating over

Ξ

R

,we obtain

Z

Ξ

R



− ν∆u(Y ) + ∇p(Y )
· V

k

− divu(Y ) ˆ

P

k



dY =

Z

Ξ

R

F (Y ) · ˆ

V

k

+ G(Y ) ˆ

P

k

dY.

Usingthe Green formulain the left-hand side of the lastrelation, weget

Z

S

R



− ν ∂

r

u · ˆ

V

k

− u · ∂

r

V

ˆ

k

+ pω · ˆ

V

k

− ω · u(Y ) ˆ

P

k



dS

ω

=

Z

Ξ

R

F (Y ) · ˆ

V

k

+ G(Y ) ˆ

P

k

dY.

Repla ing ve tor fun tions

(u, p)

and

( ˆ

V

k

, ˆ

P

k

)

by their asymptoti s, taking the limit as

R → ∞

and using(51), we verify (65).

Inthe next theorem weprovea positivitypropertyof the oe ientsin(64) whenthe fun tion

φ

in the denition of

Ξ

has sign.

Theorem 3.6. Let

φ

be not identi ally zero. Thenthe matrix

{M

k

A

k

j

}

2

k,j=1

,

ispositive denite for

φ ≤ 0

and is negative denite for

φ ≥ 0

. Moreover,

A

3

j

= A

j

3

= 0

,

j = 1, 2, 3

. Proof. Sin e

( ˆ

V

3

, ˆ

P

3

) = (0, 1)

we have

A

3

j

= 0

.

Let

φ ≥ 0

and let

Ξ

R

bedened as above. Integrating by parts in the right-handside of

0 =

Z

Ξ

R

(−ν∆ ˜

V

k

+ ∇ ˜

P

k

) ˆ

V

j

− div ˜

V

k

P

ˆ

j

dY

and taking the limitas

R → ∞

, weget

M

j

A

k

j

=

Z

∂G

(−ν∂

n

V

ˆ

j

+ n ˆ

P

j

) · ˜

V

k

dS,

where

n

is the unit outward normal to

∂Ξ

. Using (62) and that

V

˜

k

+ V

k

= 0

on

∂Ξ

, we an write the lastrelationas

M

j

A

k

j

=

Z

∂Ξ

(−ν∂

n

V

˜

j

+ n ˜

P

j

) · ˜

V

k

dS −

Z

∂Ξ

(−ν∂

n

V

j

+ nP

j

) · V

k

dS.

Now, applying Green's formulainthe domains

Ξ

and

R

3

+

\ Ξ

, we get

M

j

A

k

j

= −ν

Z

Ξ

∇ ˜

V

j

· ∇ ˜

V

k

dY − ν

Z

R

3

+

\Ξ

∇V

j

· ∇V

k

dY.

Ifwedenoteby

Q = {Q

jk

}

3

k,j=1

thematrixonthe right,then thismatrixissymmetri ,

Q

j3

= 0

andthe matrix

Q = {Q

jk

}

2

k,j=1

isnegativedenitesin etheve torfun tions

V

j

,

j = 1, 2,

arelinearindependent, and therefore

A

k

_j

= (M

j

)

−1

Q

jk

.

(66)

Now onsider the situation when

φ ≤ 0

. Then we represent

Ξ

as

R

3

+

∪ Ξ

0

, where

Ξ

0

= Ξ \ R

3

+

. We are lookingfor the solutions

( ˆ

V

k

, ˆ

P

k

)

inthe form

ˆ

V

k

= V

k

′

+ ˜

V

k

,

P

ˆ

k

= ˜

P

k

, k = 1, 2,

where

V

′

k

= V

k

in

R

3

+

and

V

′

k

= 0

in

Ξ

0

. Multiplying the equation

−ν∆ ˜

V

k

+ ∇ ˜

P

k

= 0

by

V

j

and integrating over

R

3

+

weget

M

j

A

k

j

= ν

Z

Y

3

=0

˜

V

k

· ∂

Y

3

V

j

dY

′

_{= ν}

Z

Y

3

=0

˜

V

k

· ∂

Y

3

( ˆ

V

j

− ˜

V

j

)dY

′

₊

Z

Y

3

=0

˜

P

k

( ˆ

V

j

− ˜

V

j

)

3

dY

′

.

Hen e

M

j

A

k

j

=

Z

Y

3

=0

(−ν ˜

V

k

· ∂

Y

3

V

˜

j

− ˜

P

k

( ˜

V

j

)

3

)dY

′

₊

Z

Y

3

=0

(ν ˆ

V

k

· ∂

Y

3

V

ˆ

j

+ ˆ

P

k

( ˆ

V

j

)

3

)dY

′

_.

UsingGreen'sformulainthe domains

R

3

+

and

Ξ

0

we get

M

j

A

k

j

= ν

Z

R

3

+

∇ ˜

V

k

· ∇ ˜

V

j

dY + ν

Z

Ξ

0

∇ ˆ

V

k

· ∇ ˆ

V

j

dY.

3.4 Con entrated perturbation of the boundary of

Ω

Here we onsider a domain

Ω

ε

whi h oin ides with

Ω

outside

B

ε

(x

0

)

where

x

0

is a xed point on

∂Ω

. Inside

B

ε

(x

0

)

the domain

Ω

ε

is given by

{y = (y

′

, y

3

) ∈ B

ε

(x

0

) : y

3

> εφ(y

′

/ε)},

where

y = (y

1

, y

2

, y

3

)

are Cartesian oordinates with the enter at

x

0

and

φ

is a smooth fun tion su h that

φ(y

′

_{) = 0}

for

|y

′

_{| > δ}

. Our goalis tond the asymptoti s of the matrix

Q

. We are lookingfor solution

(V

ε

α

, P

α

ε

)

in the form

V

_α

ε

(x) = (1 − χ(y/ε)) V

α

(x) + εV

α

(1)

(x)
+ ε

−1

ζ(x)W

α

(y/ε) + w

α

(x, ε),

P

ε

α

(x) = (1 − χ(y/ε)) P

α

(x) + εP

α

(1)

(x)
+ ε

−1

ζ(x)R

α

(y/ε) + S

α

(x, ε).

Here

χ(Y )

is a smooth ut-o fun tion equals

1

for

|Y | < δ

and

0

for

|Y | > 2δ

,

ζ

is also a ut-o fun tionequals

1

for

|x| < δ/2

and

0

for

|x| > δ

, where

δ

isa xed positivenumber. Usingthat

V

α

(x) = a

1

α

V

1

(y) + a

2

α

V

2

(y) + O(|x − x

0

|

2

), P

α

(x) = a

3

α

+ O(|x − x

0

|),

we on lude that the ve tor fun tion

(W

α

, R

α

)

must satisfy the system

−div W

α

(Y ) = −div χ(Y )(a

1

α

V

1

(Y ) + a

2

α

V

2

(Y ))

in

Ξ

with the homogeneous Diri hlet boundary ondition on

∂Ξ

. A ording to Theorems 3.3 and 3.4 this problemhas a unique solutionin

H

1

0

(Ξ) × L

2

(Ξ)

and this solutionsatises

W

α

(Y ) =

3

X

k=1

C

α

k

V

k

(Y ) + O(|Y |

−3

),

R

α

(Y ) =

3

X

k=1

C

_k

α

P

k

(Y ) + O(|Y |

−4

)

for large

|Y |,

(67)

and the oe ients here are evaluated as

C

α

k

=

Z

Ξ



− ν∆ χ(Y )(a

α

V

1

(Y ) + b

α

V

2

(Y ))
+ c

α

∇χ(Y )) ˆ

V

k

dY

+∇ · χ(Y )(a

α

V

1

(Y ) + b

α

V

2

(Y ))

_ˆ

P

k



dY.

(68)

Nowwe an write the system for

(V

(1)

α

, P

α

(1)

)

:

−ν∆V

_α

(1)

(x) + ∇P

_α

(1)

= −ν∆(ζ(x)

3

X

k=1

C

_k

α

V

k

(x)) + ∇(ζ(x)

3

X

k=1

C

_k

α

P

k

(x))

(69) and

−div V

_α

(1)

(x) = div(ζ(x)

3

X

k=1

C

_k

α

V

k

(x))

(70)

in

Ω

with zero Diri hlet boundary ondition on

∂Ω

. Denote the right-hand sides in (69) and (70) by

F

α

(1)

and

G

(1)

α

respe tively. Inordertoapply Theorems2.1and 2.2toproblem(69),(70),wemustverify that

Z

Ω

G

(1)

α

dx = 0.

We have

Z

Ω

G

(1)

α

dx =

Z

Ω

div(ζ(x)

3

X

k=1

C

k

α

V

k

(x))dx = −

Z

S

2

+

3

X

k=1

C

k

α

V

k

(ω) · ωdS

ω

= −C

₃

α

Z

S

2

+

ω

2

3

ν

dS

ω

.

Furthermore,

C

₃

α

=

Z

Ξ

∇ · χ(Y )(a

α

V

1

(Y ) + b

α

V

2

(Y ))
dY

=

Z

Ξ

∇ · (χ(Y ) − 1)(a

α

V

1

(Y ) + b

α

V

2

(Y ))
dY

= −R

3

Z

S

2

+

(V

_α

(1)

(x), P

_α

(1)

) =

X

θ=±

c

(θ)

_α

χ

θ

(0, 1) + ( ˜

V

α

(1)

(x), ˜

P

α

(1)

(x)),

wherethe oe ients are evaluateda ording to

c

θ

_α

=

Z

Ω

F

_α

(1)

· V

θ

+ G

(1)

_α

P

θ

dx.

Taking expressions for

F

(1)

α

and

G

(1)

α

from (69) and (70), repla ing

Ω

by

Ω

ε

= {x ∈ Ω : |x − x

0

| > ε}

and using Green'sformula, weobtain

c

(±)

_α

=

3

X

k=1

C

_k

α

h(V

k

, P

k

), (a

1

θ

V

1

+ a

2

θ

V

2

, a

3

θ

)i =

3

X

k=1

M

k

C

k

α

a

k

θ

Coe ients

C

α

k

admit the following interpretation. The ve tor fun tion

( ˆ

V

k

, ˆ

P

k

)

an be onstru ted alsoasfollows:

ˆ

V

k

= (1 − χ)V

k

+ ˇ

V

k

,

P

ˆ

k

= (1 − χ)P

k

+ ˇ

P

k

, k = 1, 2, 3,

where

( ˇ

V

k

, ˇ

P

k

)

satisesthe system

−ν∆ ˇ

V

k

+ ∇ ˇ

P

k

= ν∆(1 − χ)V

k

− ∇(1 − χ)P

k

,

−∇ · ˇ

V

k

= ∇ · (1 − χ)V

k

,

in

Ξ,

ˇ

V

k

= 0,

on

∂Ξ.

Moreovertheve torfun tion

( ˇ

V

k

, ˇ

P

k

)

hasthesameasymptoti representation(64)as

( ˜

V

k

, ˜

P

k

)

. Therefore by Theorem 3.5

M

k

A

k

j

=

Z

G



(ν∆(1 − χ)V

k

− ∇(1 − χ)P

k

) · ˆ

V

j

+ ∇ · (1 − χ)V

k

P

ˆ

j



dY.

Comparingthis formulaand (68) we see that

C

_k

α

=

3

X

m=1

a

m

_α

M

m

A

m

k

.

Thus

c

θ

_α

=

3

X

k=1

3

X

m=1

M

k

a

m

α

M

m

A

m

k

a

k

θ

and hen e

Q

αθ

(ε) = Q

αθ

+ εc

θ

α

+ O(ε

2

).

(71) Usingthat

A

m

k

= 0

when

m

or

k

is equalto

3

,we get that

c

θ

_α

=

2

X

k=1

2

X

m=1

M

k

a

m

α

M

m

A

m

k

a

k

θ

.

(72)

Sin e the matrix

{A

m

k

}

2

k,m=1

is positive denite we on lude that the matrix

{c

θ

α

}

α,θ=±

is also positive deniteand

X

α,θ

c

θ

_α

ξ

α

ξ

θ

=

2

X

k=1

2

X

m=1

A

m

_k

(

X

θ

M

k

a

k

θ

ξ

θ

)(

X

α

M

m

a

m

α

ξ

α

),

(73) therefore

X

α,θ

c

θ

_α

ξ

α

ξ

θ

≥ c

0

2

X

k=1

(

X

θ

a

k

_θ

ξ

θ

)

2

.

(74)

Formulas(71)-(74)des ribetheasymptoti behaviorofthe oe ients

Q

αθ

forthedomain

Ω

ε

. Repla ing the elements of the pressure drop matrix

Q

by their asymptoti s (71) in (3) we get the asymptoti expansionof the ee tive lengths of the vessels, i.e. the dire t analogof Theorem 3.2 is exe uted.

4 Appendix

4.1 Pressure drop matrix and modied Kir hho transmission onditions

Let's trun ate ylindri aloutlets in

Ω

and assume

Ω

h

= Ω

′

∪ Ω

0

h

∪ Ω

+

h

∪ Ω

−

h

,

Ω

α

_h

=

x : |y

α

_{| < r}

α

, z

α

< h

−1

l

α

,

α = 0, ±,

(75) where

h > 0

is a small dimensionless parameter,

r

α

> 0, l

α

> 0

are ertain xed radii and lengths respe tively. Indomain

Ω

h

we dene the homogeneous

(F = 0, G = 0)

Stokesequations (7), and onits lateralsurfa e

Σ

h

= ∂Ω

h

∩ ∂Ω

we impose the homogeneous

(H = 0)

no-slip onditions (8) (hereinafter refer to these relations, implying that they are restri ted to these sets). On the trun ated surfa es

Γ

α

h

= {x : |y

α

| < r

α

, z

α

= h

−1

l

α

}

assign the following onditions:

v

h

(x) = −V

0

(y

0

), x ∈ Γ

0

_h

,

(76)

v

h

y

τ

i

(x) = 0, i = 1, 2,

−ν∂

z

τ

v

_z

h

τ

(x) + p

h

(x) = p

∞

, x ∈ Γ

τ

_h

, τ = ±.

(77) Inotherwords, atthe inlet ross-se tionof thevessel

Ω

0

isassignedthein omingunit uxofuid, and

onthe allo ated endsof theoutlet ross-se tionsof thevessels

Ω

±

peripheralpressure

p

∞

isset. Atthe sametime ompressionof oordinates by

h

−1

timestransformstheproblemstatedtotheusualproblem

ofthebloodowthrough thearterybifur ationnode,whi hwalls,asalreadyexplained,itisassumedto

berigid ( f. [1℄). In the new oordinates the vessels be omesmaller radii

hr

α

and xed lengths

l

α

. We emphasizethat the problem(7), (8), (75), (76), (77) is stillin luded inthe symmetri Green'sformula

in

Ω

h

. Its interpretation inthe frameworkof the weighted spa es te hnique with deta hed asymptoti s isgiven in[3℄.

As anapproximate solutionof the problemstated in

Ω

h

we takethe sums

ˆ

where

(V

±

_{, P}

±

₎

are introdu ed spe ial solutions(27) and the last term refers tothe onstantpressure.

Usingthe asymptoti representations (see Se t. 2.3), satisfy the boundary onditions (76), (77) up to

exponentially smallterms for

h → +0

and we obtainthe following relations

1 = a

h

₊

+ a

h

₋

,

(79)

p

∞

= a

h

₀

− h

−1

L

τ

a

h

τ

+

X

α=±

Q

τ α

a

h

α

,

τ = ±,

(80) where

L

α

=

8ν

πr

4

α

l

α

, α = 0, ±

, and hen eforth

L =

diag

{L

+

, L

−

}

. Let

e

= (1, 1)

and

a

h

_{= (a}

h

+

, a

h

−

)

be olumns. In virtue of (80) we dedu e

(p

∞

− a

h

0

)e = (Q − h

−1

L)a

h

,

hen e

a

h

= (h

−1

L − Q)

−1

e

(a

h

0

− p

∞

),

and thusequality (79) rewritten in the form

e

· a

h

_{= 1}

leads tothe relations

1 = T

h

(a

h

0

− p

∞

),

T

h

= e · (h

−1

L − Q)

−1

e

= h

−1

e

· L

−1

(I − hQL

−1

)

−1

e

= h

−1

e

· (L

−1

+ hL

−1

QL

−1

+ O(h

2

))e.

We nallynd that

a

h

0

= p

∞

+ T

h

−1

,

T

h

= h(t

0

+ ht

1

+ O(h)),

t

0

= e · L

−1

e

> 0,

t

₁

= e · L

−1

QL

−1

e

,

(81) sowe get

a

h

₀

= p

∞

+ ht

−1

₀

(1 − ht

−1

₀

t

1

+ O(h

2

)).

(82) Thus, the pressure at the input

Γ

0

h

up to the smaller terms isequalto

h

−1

L

0

+ p

∞

+ ht

−1

0

− h

2

t

−2

0

t

1

+ O(h

3

).

(83) Thersttermof(83)isthepressuredrop,whi hprovidestheunituxdeliverytothearterybifur ation,

the se ond term is also positive, it is ne essary to supply the uxes to the points

z

±

_{= h}

−1

_l

±

, and the thirdterm, the sign of whi hdepends on the pressure drop matrix

Q

, orresponds tojust the shape of the node.

If in the vi inity of the node is formed plaque (

ϕ < 0

in (37)) then a ording to the formula (40), themagnitudeof (81)de reases, i.e. thepressure (83) in reases. Thisfa tis onsistentwith anobvious

observation: onstri tion of the hannel requires the growth of pressure in the input. At a onstant

pressure,de reases theuxsuppliedintothe vessels

Ω

±

h

. Many reasonsofhypertensivepressure do tors asso iated with the loggingof bloodvessels.

At rst glan e it seems that an aneurysm (

ϕ > 0

in (37)) fa ilitates the passage of blood through a bifur ationnode, i.e. the orre tion term

O(h

2

₎

in (83) is redu ed due to the minus sign in front of

it. This impression is erroneous, i.e. statementof the problemadopted in the arti le doesnot onsider

elasti ity of the vessel walls. As it is known from medi alreferen e books and explained in[12℄, using

theone-dimensional modelof the types ofaneurysms, vas ularpermeabilitymay beredu ed due tothe

growthof thehematomainthe aneurysm avity. Atthe sametime,in reaseofthe pressureattheinput

(in reased heart rate while running and stress) provokes the wallrupture and leads toa variety of sad

Herewe assumethatthe boundariesof thedomains

ω

α

are analyti . Letalso

Γ

′

isthe partof

∂Ω

whi h is

∂Ω ∩ Ω

′

.

HereweprovethatthehomogeneousStokesproblem(7), (8)withtheadditionalNeumannboundary

ondition

∂

n

u(x) = 0, x ∈ Γ

′

,

(84) has atrivial bounded solution

(u, p)

, i.e. identi allyequals (0,0,0;1).

We assume that

∂Ω \ Γ

0

is analyti . Here

Γ

0

is a ompa t subset of

Γ

. Sin e

∂

n

u(x)

is analyti on thesurfa e

∂Ω \ Γ

′

asasolutionto(7), (8)anddue to(84),we on ludethat

∂

n

u = 0

onthe whole

∂Ω

. This implies

∇u = 0

and hen e

rot u = 0

on

∂Ω

. Appli ationof the operator

rot

to the homogeneous Stokessystem (7), (8)gives the Diri hlet boundary value problemfor

rot u

, i.e.

∆(rot u(x)) = 0,

x ∈ Ω,

(85)

rot u(x) = 0,

x ∈ ∂Ω.

(86)

Sin e we look for the bounded solution to (85), (86) (by using a lo al estimate this implies that the

gradientisalsobounded atinnity),one an on lude that

rot u ≡ 0

,andhen e weget

u = ∇Φ

,where

Φ

is as alar potential. Thus wearriveat the following system of equations

∇(−ν∆Φ(x) + p(x)) = 0,

−∆ Φ(x) = 0,

x ∈ Ω,

(87)

∇Φ(x) = 0,

x ∈ ∂Ω.

(88)

Invirtue of (87), (88) we obtain

(u, p) = (0, 0, 0; 1)

.

4.3 Proof of Lemma (2.1)

Firstwe onsider anauxiliaryproblem,namely we look for asolution of the boundary value problem

−div u

α

(x

α

) = η

α

(y

α

)G

α

(z

α

),

x

α

∈ Ω

α

,

(89)

u

α

(x

α

) = 0,

x

α

∈ ∂Ω

α

,

(90) where

η

α

(y

α

) ∈ C

0

∞

(ω

α

),

Z

ω

α

η

α

(y

α

)dy

α

= 1

and

G

α

(z

α

_{) =}

Z

ω

α

g(y

α

, z

α

)dy

α

.

One an verify dire tly that the ve tor fun tion

u

α

(x

α

) = (0, 0, η

α

(y

α

)w

α

(z

α

)), w

α

(z

α

) = −

Z

∞

z

α

G

α

(t)dt,

solvesthe problem(89), (90). Moreover, by Hardy's inequality (see Theorem 5.2[[14℄℄)

||w

α

||

L

2

(L

α

,∞)

≤ c

Z

∞

L

α