Numerical computation of cell motion in a branching flow M. G. Blyth

Outline

Introduction Boundary Integral Method Cell motion in a branching channel Cell motion in a branching tube Summary

Numerical computation of cell motion in a branching flow

M. G. Blyth

University of East Anglia, Norwich

Capillary Flow Meeting, University of Warwick

1st June 2010

Outline


Joint work with:

Hugh Woolfenden (University of East Anglia)

Outline


Outline

1

Introduction

2

Boundary Integral Method

3

Cell motion in a branching channel

4

Cell motion in a branching tube

5

Summary

Outline

Introduction

Boundary Integral Method Cell motion in a branching channel Cell motion in a branching tube Summary

Outline

1

Introduction

2


3


4


5

Summary

Outline

Introduction


Red blood cells and capillaries

Figure: Healthy RBCs

Figure: Capillary bed

Outline

Introduction


Capillary flow

Navier-Stokes equations

ρ u t

+ u · ∇ u = −∇ p + µ ∇

2 u

∇ · u = 0

Typical Reynolds number:

R =

ρ Ua

µ

≈ 0 .

001

Outline

Introduction


Capillary flow

Stokes equations

0 = −∇ p + µ ∇

2 u

∇ · u = 0

Typical Reynolds number:

R =

ρ Ua

ν

≈ 0 .

001

Outline

Introduction


Capillary Network

Network models

Lipowsky & Zweifach (1974) Microvascular Res.

, 7 , 73-83.

Schmid-Sch¨ et al.

(1990) Microvascular Res.

, 19 , 18-44.

Pozrikidis (2009) Bulletin Math. Biology , 71 , 1520-1541.

Outline

Introduction


Network models

Pozrikidis (2009)

Outline

Introduction


Network models: Pozrikidis (2009)

Q i

=

∆ p i

π a

4

8 µ L i

µ : Effective blood viscosity.

Outline

Introduction


Network models: Pozrikidis (2009)

P (cell enters daughter branch i ) = φ i

φ

2

+ φ

3

= 1

Conclusion

Probabilities φ i have important effect on cell residence times and haematocrit distribution across network.

Outline

Introduction


Flow simulations

Ong & Popel (1992) J.

Biomech. Eng.

, 114 ,

398-405.

Ong, Enden & Popel

(1994) J. Fluid Mech.

, 270

Focus on computing dividing fluid surface (no cells)

Finite-element calculations

Outline

Introduction


Particle motion at a bifurcation

Fung (1973) Microvascular Res.

, 5 ,

34-48.

Most cells enter branch with higher flow rate

Outline

Introduction


Particle motion at a bifurcation

Audet & Olbricht (1987)

Microvascular Res.

, 33 ,

377-396.

Solid particle

Streamfunction-vorticity boundary integral approach

Outline

Introduction


Cell motion at a bifurcation

Secomb et al.

(2007) Annals Biomed.

Eng.

, 35 , 755-765.

Viscoelastic capsule motion

Finite-element method

Outline

Introduction


Experiments

Kiani & Cokelet

(2007) J. Biomed.

Eng.

, 116 , 497-501.

RBCs simulated by flexible elastic disks

Outline Introduction


Cell motion in a branching channel Cell motion in a branching tube Summary

Outline

1

Introduction

2


3


4


5

Summary




The Boundary Integral Method

The boundary integral method stems from the Lorentz reciprocal relation (1907) for Stokes flow

∇ · u ·

σ

′ − u ′ ·

σ

= 0 , for two distinct flows ( u ,

σ

) and ( u ′ ,

σ

′ ) which satsify the Stokes equations:

0 = −∇ p + µ ∇

2 u = ∇ ·

σ

,

∇ · u = 0 ,

δ ( x − x

∇ ·

0 u’

) = ∇ ·

σ

= 0

′




The boundary integral method

Integrate over Γ.

∇ · u ·

σ

′ − u ′ ·

σ

= 0

Γ x

0





We find

1 u j

( x

0

) = −

4 πµ

Z

Γ f i

( x ) G ij

( x , x

0

1

)d s ( x )+

4 π

Z

Γ u i

( x ) T ijk

( x , x

0

) n k d s ( x ) , where f i

= σ ij n j is the traction on the boundary Γ.





We find

1 u j

( x

0

) = −

4 πµ

Z

Γ f i

( x ) G ij

( x , x

0

1

)d s ( x )+

4 π

Z

Γ u i

( x ) T ijk

( x , x

0



The Green’s function G ij is a solution u i

= G ij b j of

µ ∇

2 u + ∇ p + b δ ( x − x

0

) = 0





We find

1 u j

( x

0

) = −

4 πµ

Z

Γ f i

( x ) G ij

( x , x

0

1

)d s ( x )+

4 π

Z

Γ u i

( x ) T ijk

( x , x

0



The Green’s function G ij is a solution u i

= G ij b j of

µ ∇

2 u + ∇ p + b δ ( x − x

0

) = 0

Free-space Green’s function:

G ij

( x , x

0

) = − δ ij log r + x i x j r 2

, ˆ = x − x

0

, r = | ˆ | .





Let x

0 approach the boundary.

Γ x

0





Now

1 u j

( x

0

) = −

4 πµ

Z

Γ f i

( x ) G ij

( x , x

0

) d s +

1

4 π

Z

Γ u i

( x ) T ijk

( x , x

0

) n k d s , with x

0 inside Γ becomes (Ladyzhenskaya 1963)

1

2 u j

( x

0

) = −

1

4 πµ

Z

Γ f i

( x ) G ij

( x , x

0

)d s +

1

4 π

Z

PV

Γ u i

( x ) T ijk

( x , x

0

) n k d s .

with x

0 on Γ

Outline Introduction Boundary Integral Method


Cell motion in a branching tube Summary

Outline

1

Introduction

2


3


4


5

Summary




Ultimate aim

To model the motion of red blood cells through a capillary bifurcation.




Ultimate aim


Present simplifications

Two-dimensional model

Single cell moving through a tube with a side branch




Ultimate aim


Present simplifications

Two-dimensional model

Single cell moving through a tube with a side branch

3D branching flow

Work on this is at an early stage




Model geometry




Model geometry

α




Model geometry

α




Flow set-up

Q

1

Q

2

α

Q

1

= Q

2

+ Q

3

Q

3




Flow set-up

Q

1

Q

2

Q

1

= Q

2

+ Q

3

Q

3




Decay rates in Stokes flow u = u

0 u = u

P





0 y x u = ψ y

, y = d v = − ψ x

∇

4

ψ = 0 y = − d u = 0





0 y x u = ψ y

, y = d v = − ψ x

∇

4

ψ = 0 y = − d

ψ =

∞

X

α n

φ n

( y ) e − k n x n =0 u = 0





0 y x u = ψ y

, y = d v = − ψ x

∇

4

ψ = 0 y = − d k

0

=

2 .

11 d

+

1 .

125 i d u = 0




Computational model

C

1

E

1

µ

P

λµ n

E

2

A y n

α

C

3 x

E

3




Flow set-up u

P

1 u

P

2

α u

P

3




Flow set-up u

P

1

Main u

( ext )

= u P

1 + u D

1 f

( ext )

= f P

1 + f D

1

Branch u

( ext ) = u

P

3 + u

D

3 f

( ext )

= f P

3 + f D

3 u

P

3 u

P

2




Flow set-up u

P

1

Main u

( ext )

= u P

1 + u D

1 f

( ext )

= f P

1 + f D

1

Branch u

( ext ) = u

P

3 + u

D

3 f

( ext )

= f P

3 + f D

3

A u

P

3 u

P

2




Flow set-up p = 0 p = π

2 p = π

3




Flow set-up




Flow set-up




Flow set-up




Flow set-up




Flow set-up




Flow set-up




Boundary integral formulation

We apply the standard boundary integral formulation

1 u j

( x

0

) = −

4 πµ

Z

Γ f i

( x ) G ij

( x , x

0

1

)d s ( x )+

4 π

Z

Γ u i

( x ) T ijk

( x , x

0

) n k d s ( x ) , to the disturbance flows.





We find

2 πµ u j

D

1 ( x

0

) =

−

Z

C , A f i

D

1 ( x ) G ij

( x , x

0

) d s ( x ) + µ

Z

A u i

D

1 ( x ) T ijk

( x , x

0

) n k d s ( x )

Z

+ ( π

2

+ Q ∆ p ) G ij

( x , x

0

) n i

( x ) d s ( x ) + I j

( x

0

)

−

E

2

+ µ (1

−

λ )

Z

P u i

( ext )

( x ) T ijk

( x , x

0

) n k

( x ) d s ( x ) ,

Z

P

∆ f i

( x ) G ij

( x , x

0

) d s ( x )

I j

( x

0

) = (1

−

Q )

» Z

E

2 f i

P

1 ( x ) G ij

( x , x

0

) d s ( x )

−

µ

Z

E

2 u i

P

1 ( x ) T ijk

( x , x

0

) n k

( x ) d s ( x )

– where

Q =

Q

2

Q

1

, and ∆ f = f

( ext )

− f

( int ) is the traction jump at the cell boundary.





In the branch we find

2 πµ u j

D

3 ( x

0

) =

Z

A f i

D

1 ( x ) G ij

( x , x

0

)ds( x )

−

Z

B f i

D

3 ( x ) G ij

( x , x

0

)ds( x ) + K j

( x

0

)

−

µ

Z

A u i

D

1 ( x ) T ijk

( x , x

0

) n k

( x ) ds( x ) + π

3

Z

E

3

G ij

( x , x

0

) n i

( x ) ds( x ) where

K j

( x

0

) =

Z

A h f i

P

1 ( x )

− f i

P

3 ( x ) i

G ij ds( x ) + µ

Z

A h u i

P

3 ( x )

− u i

P

1 ( x ) i

T ijk n k

( x )ds( x ) .




Equations of motion – Pressure relations

To obtain expressions for the exit pressures, we integrate the

Lorenz reciprocal relation over the main channel:

Z

∇ · u

P

1 ·

σ

D

1 − u

D

1 ·

σ

P

1 d S = 0






Lorenz reciprocal relation over the main channel:

Z

E

1

, E

2

, E

3

, P , A u

P

1 · f

D

1 − u

D

1 · f

P

1 d s = 0

A similar equation holds in the branch






Lorenz reciprocal relation around the channel:

1

π

2

= − Q ∆ p +

Q

1

"

Z

A f

P

1 · u

D

1 − u

P

1 · f

D

1 d s

−

Z

P u

P

1 · ∆f d s + (1 − λ )

Z

P u

( ext )

· f

P

1 d s

#

.

1

π

3

=

Q

3

Z

A u

P

3 · f

D

1 − f

P

3 · u

D

1 d s +

Z

A

( f

P

1 · u

P

3 − u

P

1 · f

P

3 ) d s




Constitutive equations for a thin elastic shell

Force/moment balance d τ dl

+ κ q = − ∆f · t dq

− κτ = − ∆f · n dl q =

Constitutive equations dm dl m = E

B

( κ − κ

R

) q

τ t

Fluid 2 n

Figure: Elastic cell

Fluid 1

P l

τ = k dl dl

R

− 1




Results

Dimensionless parameters

λ, M =

E

B

µ Q

1 d

, W = k d

µ Q

1

,




Straight channel

1

0.5

0

-0.5

-1

-1 -0.5

0 x

0.5

1

Figure: λ = 1, W = 1 and a = 0 .

5 d with M = 10

−

3 .




Cell released off-centre: Centroid trajectories

0.35

0.3

0.25

0.2

0.15

0.1

0.05

0

0 2 4 6 8 10 x

Figure: a = 0 .

5 d , λ = 1, W = 1 and M = 10

−

3 . Centroid trajectories for circular capsules initially positioned at y = 0 .

1 d , 0 .

2 d and 0 .

3 d .

-1

-2

1

0

-5

-6

-3

-4

0




Branching channel: No cell, equal fluxes, Q

2

= Q

3

= 0 .

5

Streamlines for Branched Channel

1

0

-1

-2

-3

-4

-5

-6

-7

0 2 4 6 8

Streamlines for Branched Channel

10 12

-5

-6

-7

0

-2

-3

-4

1

0

-1

2 4 6 8 10 12

2 4 6 8 10 12




Weak flow in side branch

1

0

-1

-2

-3

-4

-5

-6

-7

0 2 4 6 8 10 12

Figure: Capsule journeys for λ = 1, a = 0 .

5 d and W = 1 and Q = 0 .

9.

At t = 0 the capsule centroid is at ( x , y ) = (2 , 0).

M = 10

−

3




Cell entering side branch

1

0

-1

-2

-3

-4

-5

-6

-7

0 2 4 6 8 10 12


5 d , W = 1 and Q = 0 .

1. At t = 0 the capsule centroid is at ( x , y ) = (2 , 0).

M = 10

−

3 .




Narrow side branch

-1

-2

-3

-4

-5

1

0

-6

-7

0 2 4 6 8 10 12

Figure: λ = 1, a = 1 .

1 d , W = 5, M = 10

−

3 , and Q = 0 .

1.




Acute-angled branch: α = π/ 4.

-1

-2

-3

-4

-5

1

0

-6

-7

0 2 4 6 8 10 12

Figure: λ = 1, a = 1 .

1 d , W = 5, M = 10

−

3 , and Q = 0 .

1.




Obtuse-angled narrow side branch

1

0

-1

-2

-3

-4

-5

-6

-7

0 2 4 6 8 10 12

Figure: λ = 1, a = 0 .

5 d , W = 5, M = 10

−

3 , Q = 0 .

5, and D = 0 .

5 d ,

α = 3 π/ 4. At t = 0 circular unstressed shape has centre at

( x c

, y c

) = (2 , 0 .

3) d .




Strong flow in side branch

-3

-4

-5

-6

1

0

-1

-2

-7

0 2 4 6 8 10 12


5 d , W = 1 and Q = 0 .

1. At t = 0 the capsule centroid is at ( x , y ) = (2 , 0).

M = 10

−

3 .




Typical membrane tensions

0.6

0.5

0.4

0.3

0.2

0.1

0

-0.1

0 0.5

1 1.5

2 s / d

2.5

3 3.5

Figure: For last cell shown in previous figure: ˆ τ (broken line)




Typical membrane bending moment

0.014

0.012

0.01

0.008

0.006

0.004

0.002

0

-0.002

-0.004

-0.006

0 0.5

1 1.5

2 s / d

2.5

3 3.5




Typical membrane bending moment

Axisymmetric red blood cell in tube

Pozrikidis (2005), Phys. Fluids , 17 (3).




Exit pressures

1.05

1.04

1.03

1.02

1.01

1

0.99

0 2 4 6

( Q

1

/ d

2 ) t

8 10 12

Figure: Normalised pressures ˆ

2

π

3

(dashed line) against time for Q = 0 .

1, λ = 1, a = 0 .

5 d , W = 1, M = 10

−

3 .




Recovery distance

-1

-2

-3

-4

1

0

-5

-6

-7

0 2 4 6 8 10 12




Recovery distance

1

0

-1

(a) (b) (c) (d)

Figure: Evolution of the last capsule shape in previous figure Distances travelled are ( a ) 0, (b) 5 .

1 a , (c) 12 .

2 a , and (d) 48 .

9 a .




Effect of viscosity ratio

-1

-2

-3

-4

1

0

-5

-6

-7

0 2 4 6 8 10 12

λ = 1, Q = 0 .

1




Effect of viscosity ratio

-1

-2

-3

-4

1

0

-5

-6

-7

0 2 4 6 8 10 12

λ = 5, Q = 0 .

1. Similar effect by increasing W




Path selection

-1

-2

-3

-4

1

0

-5

-6

-7

0 12

-1

-2

-3

-4

1

0

-5

-6

-7

0 2 4 6 8 10 2 4 6 8 10

Capsule journeys when Q = 0 .

5, λ = 1, M = 10 −

3

. (left) W = 1.

(right) W = 5.

12

Outline Introduction Boundary Integral Method Cell motion in a branching channel


Summary

Outline

1

Introduction

2


3


4


5

Summary



Summary

Capsule mechanics



Summary

Capsule mechanics: Equilibrium balance

Global Cartesian coordinates : Barthes-Biesel, D. & Rallison

(1981) J. Fluid Mech.

, 113, 251-267.

n t b

No bending moments b



Summary


Force on element edge b ·

τ

In-plane tension tensor

τ



Summary


Force balance on membrane patch:

Fluid 1 n t

C A b

Fluid 2

Z

A h

σ

(1)

−

σ

(2) i

· n d S +

Z

C b ·

τ d l = 0



Summary



Fluid 1 n t

C A b

Fluid 2

Z

A h

σ

(1)

−

σ

(2) i

· n d S +

Z

A

∇ ·

τ d S = 0



Summary



Fluid 1 n t

C A b

Fluid 2

∆ f + ∇ ·

τ

= 0 , ∆ f = h

σ

(1)

−

σ

(2) i

· n



Summary

Capsule mechanics: deformation undeformed deformed

N x

X n

General form of deformation gradient:

F ( t ) =

∂ x

∂ X



Summary

Capsule mechanics: deformation undeformed deformed

N x

X n

Membrane deformation gradient (Barthes-Biesel & Rallison 1981):

A ( t ) = ( I − nn ) · F · ( I − NN )



Summary

Capsule mechanics: deformation

Membrane deformation gradient:

A = ( I − nn ) · F · ( I − NN )

Projection of surface tangent vector d X d x = F · d X = A · d X

Projection of surface normal vector vanishes

A · N = 0



Summary


Membrane deformation gradient:

A = ( I − nn ) · F · ( I − NN )

Projection of surface tangent vector d X d x = F · d X = A · d X

Projection of surface normal vector vanishes

A · N = 0

The idea is that membrane fibres pointing in the normal direction do stretch, but do not contribute directly to the elastic tensions.



Summary


To define stretching, we introduce the left Cauchy-Green tensor

B = A · A

T

≡ V

2

B has eigenvalues

0 , λ

2

1

, λ

2

2 and eigenvectors n , v

1 v

2

The latter two are the principal directions of stretch



Summary

Capsule mechanics: Constitutive equations

Barthes-Biesel & Rallison (1981) showed that

τ

= e −

Λ

1

∂ W

∂ Λ

1

( I − nn ) +

∂ W

∂ Λ

2

B where the invariants

Λ

1

= log λ

1

λ

2

,

Λ

2

=

1

2

( λ

2

1

+ λ

2

2

) − 1 .

W : strain energy function



Summary

Capsule mechanics: Constitutive equations

Barthes-Biesel & Rallison (1981) showed that

τ

= e −

Λ

1

∂ W

∂ Λ

1

( I − nn ) +

∂ W

∂ Λ

2

B

For a red blood cell, Skalak et al.

(1973), Biophys J.

, 245 ,

245-264 proposed

W =

B

4 h

2Λ

2

(1 + Λ

2

) + 1 − e

2Λ

2 i

+

C

8 h e

2Λ

1 − 1 i

, where B , C are constants and B ≪ C .

Large C ensures incompressibility of membrane.



Summary

Capsule mechanics: Computation



Summary


Ramanujan & Pozrikidis (1998)

η

1

0 1

ξ

∂ x

∂ξ

∂ x

∂η

= A ·

∂ X

∂ξ

= A ·

∂ X

∂η

0 = A · N



Summary


Ramanujan & Pozrikidis (1998)

η

1

0 1

ξ

∂ x

∂ξ

∂ x

∂η

= A ·

∂ X

∂ξ

= A ·

∂ X

∂η

0 = A · N

Solve for A



Summary


Once A is known:



Summary


Once A is known:

Compute eigenvalues λ

1

, λ

2 of B = A · A

T



Summary


Once A is known:


1

, λ

2 of B = A · A

T

Compute invariants Λ

1

= log λ

1

λ

2

, Λ

2

=

1

2

( λ 2

1

+ λ 2

2

) − 1.



Summary


Once A is known:


1

, λ

2 of B = A · A

T


1

= log λ

1

λ

2

, Λ

2

=

1

2

( λ 2

1

+ λ 2

2

) − 1.

Compute tension tensor

τ

= e −

Λ

1

∂ W

∂ Λ

1

( I − nn ) +

∂ W

∂ Λ

2

B



Summary


Once A is known:


1

, λ

2 of B = A · A

T


1

= log λ

1

λ

2

, Λ

2

=

1

2

( λ 2

1

+ λ 2

2

) − 1.


τ

= e −

Λ

1

∂ W

∂ Λ

1

( I − nn ) +

∂ W

∂ Λ

2

B

Compute ∇ ·

τ



Summary


Once A is known:


1

, λ

2 of B = A · A

T


1

= log λ

1

λ

2

, Λ

2

=

1

2

( λ 2

1

+ λ 2

2

) − 1.


τ

= e −

Λ

1

∂ W

∂ Λ

1

( I − nn ) +

∂ W

∂ Λ

2

B

Compute ∇ ·

τ

Hence ∆ f = −∇ ·

τ



Summary

Cell motion through a branching tube



Summary


Boundary Integral formulation:

Set viscosity ratio λ = 1

λ = 1 very computationally expensive



Summary


Boundary Integral formulation: u j

1

( x

0

) = −

8 πµ

Z

P

∆ f i

G ij

( x , x

0

) d S

1

−

8 πµ

Z

E

1

, E

2

, E

3

, C f i

(1)

G ij

( x , x

0

) d S

1

+

8 π

Z

E

1

, E

2

, E

3 u i

(1)

T ijk

( x , x

0

) d S



Summary


Force balance

Fluid 1

Fluid 2

∇ ·

σ

(1)

= 0

∇ ·

σ

(2)

= 0



Summary


Force balance

Fluid 1

Z

E

1

, E

2

, E

3

, C , P

∇ ·

σ

(1) d V = 0

Fluid 2

Z

P

∇ ·

σ

(2) d V = 0



Summary


Force balance p

2

1

= −

π a 2 n

Z

C f

(1)

· e x d S +

Z

P

∆ f · e x d S o p

3

1

= −

π b

2 n

Z

C f

(1)

· e y d S +

Z

P

∆ f · e y d S o

Outline Introduction Boundary Integral Method Cell motion in a branching channel Cell motion in a branching tube

Summary

Outline

1

Introduction

2


3


4


5

Summary


Summary

Summary

We have used the boundary integral method to compute cell motion in a branching vessel.



Summary

Summary



Boundary integral calculations


Summary

Summary




Domain-decomposition method


Summary

Summary





Cell distortion in region of junction


Summary

Summary






Recovery distance


Summary

Summary






Recovery distance

Path selection may depend on elastic properties


Summary

Summary




Summary

Summary



Developed a boundary integral formulation


Summary

Summary




Described capsule mechanics


Summary

Summary




Described capsule mechanics

Calculations in progress...

Numerical computation of cell motion in a branching flow M. G. Blyth

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Numerical computation of cell motion in a branching flow M. G. Blyth

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