Vesicle dynamics in Newtonian and non-newtonian fluids

Size: px

Start display at page:

Download "Vesicle dynamics in Newtonian and non-newtonian fluids"

Marshall Terry
5 years ago
Views:

1 Vesicle dynamics in Newtonian and non-newtonian fluids Ming-Chih Lai Department of Applied Mathematics National Chiao Tung University Taiwan Modelling and Simulation of Interface Dynamics in Fluids/Solids and Applications, 14-18, May 2018

Vesicle problem: Navier-Stokes + surface dynamics with PDE constraint Vesicle can be visualized as a bubble of liquid within another liquid with a closed lipid membrane suspended in aqueous solution,

2 Vesicle problem: Navier-Stokes + surface dynamics with PDE constraint Vesicle can be visualized as a bubble of liquid within another liquid with a closed lipid membrane suspended in aqueous solution, size is about 10µm Lipid membrane consists of tightly packed lipid molecules with hydrophilic heads facing the exterior and interior fluids and hydrophobic tails hiding in the middle, thickness is about 6nm so we treat the membrane as a surface (3d) or a curve (2d) Lipid membrane (or vesicle boundary) can deform but resist area dilation, that is surface incompressible

3 Questions: How the vesicle behaves in fluid flows? To mimic some mechanical behavior of red blood cells (RBC), drug carrying capsules in capillary In shear flow: Tank-treading (TT), Tumbling (TU), Trembling (TR), depend on the viscosity contrast λ = µ in /µ out ; Keller & Skalak JFM, 1982 (theory), Deschamps et. al. PNAS, 2009 (experiment) Amoeboid motion (active vesicle swimmer) in confined geometry, Wu et. al. Lai & Misbah, PRE-Rapid 2015, Soft Matter 2016 Figure: Red blood cells: flexible biconcave disks

4 Numerical simulation: single vesicle in a shear flow Figure: Inclination angle and tank-treading in a simple shear flow

5 Tank-treading to tumbling transitions, Kim & Lai, PRE 2012 Different viscosity effect 40 θ in degrees reduced area V=0.95 V=0.9 V=0.8 V=0.7 V=0.6 V=0.5 Keller Skalak 2D viscosity contrast λ Figure: The inclination angle θ versus the viscosity contrast λ = µ in µ out. The simulation results (solid lines) are compared with the Keller & Skallak theory (dashed lines).

6 Immersed boundary (IB) and immersed interface (IIM) methods for vesicle dynamics Simulating the dynamics of 2D inextensible vesicles by the penalty immersed boundary method (Kim & Lai JCP 2010), inertial effect (Kim & Lai PRE 2012), 2D 4-roll mill flows (Kim, Lai & Seol PRE 2017) Nearly inextensible approach (3D axisymmetric case, Hu, Kim & Lai JCP 2014, Full 3D case, Seol, Hu, Kim & Lai JCP 2016) IIM for inextensible interfaces in Navier-Stokes flow (Li & Lai, EAJAM, 2011) A fractional step immersed boundary method for Stokes flow with an inextensible interface enclosing a solid particle (SISC 2012) Unconditionally energy stable penalty IB method without bending (Hu & Lai EAJAM 2013, Hsieh, Lai, Yang & You JSC 2015) Vesicle electrohydroynamics using IB and IIM (Hu, Lai, Seol & Young, JCP 2016) Amoeboid swimming (Misbah and Lai et. PRE 2015, Soft matter 2016)

7 Mathematical modeling on vesicle problem Vesicle: A liquid drop within another liquid with a closed lipid membrane Vesicle boundary Σ: represented by X, is able to deform, but resist area dilation, i.e. Σ is surface incompressible (or inextensible) The fluid-structure interaction is formulated by the stress balance condition on Σ

8 Immersed Boundary (IB) formulation: treat the vesicle boundary as a force generator ( ) u ρ t +u u + p = µ u+f u = 0 in Ω in Ω s U = 0 on Σ X = U = u(x,t)δ(x X)dx t where the immersed boundary force f = F(X)δ(x X) dx Σ F = F b +F σ on Σ = c b ( s H +2H(H 2 K) ) n 2 = s σ 2Hσn F b F σ Ω

9 H: mean curvature, K: Gaussian curvature, c b : bending rigidity s = n n, s = s s σ: unknown elastic tension to be introduced to enforce s U = 0 It can be shown that the tension doesn t do extra work to the fluid; i.e. < S(σ),u > Ω = < σ, s U > Σ The pressure and elastic tension have the same roles as Lagrange multipliers Question: Where does the boundary force F come from? Answer: Variational derivative of Helfrich energy E = c b H 2 ds + σ ds 2 Σ Σ F = δe δx = F b +F σ

10 Basic surface geometry For the vesicle surface X(r, s), the first fundamental forms E = X r X r, F = X r X s, and G = X s X s, then s σ = GX r FX s EG F 2 σ r + EX s FX r EG F 2 σ s s U = GX r FX s EG F 2 U r + EX s FX r EG F 2 Using a (b c) = (a c)b (a b)c, we obtain two useful relations U s which give X s n = GX r FX s, n X r = EX s FX r X r X s X r X s s σ = (X s n)σ r +(n X r )σ s X r X s and s U = (X s n) U r +(n X r ) U s X r X s

11 More useful identities (X s n)σ r +(n X r )σ s = s σ X r X s (X s n) U r +(n X r ) U s = s U X r X s X s n r +n s X r = 2Hn X r X s (X s n) r +(n X r ) s = 2Hn X r X s (σ(x s n)) r +(σ(n X r )) s = ( s σ 2σHn) X r X s

12 Relations to Geometric PDEs Motion by mean curvature dx Motion by mean curvature: dt = δeσ δx = F σ = σ2hn, σ is a constant, L 2 gradient flow, a convex surface shrinks Willmore flow Willmore energy E b (X) = c b 2 Σ H2 ds, in physics terminology, surface bending energy (membrane lipid bilayer exhibits a bending resistance) In some literature, use 2H instead of H dx Willmore flow: gradient flow dt = δe b δx = F b = c b 2 ( s H +2H(H 2 K))n, L 2

13 Why does s U = 0 mean the surface incompressibility? Xr Xs Xr Xs = (Xrt Xs +Xr Xst) t X r X s ( ) = n (X rt X s)+n (X r X st) since n = Xr Xs X r X s = (X s n) X rt +(n X r) X st (using (a b) c = (b c) a) = (X s n) U r +(n X r) U s (since X t = U) = GXr FXs X r X s U r + EXs FXr X r X s U s (using the two relations) = ( s U) X r X s (by the definition of s U)

14 Skew-adjoint operators, Lai & Seol, AML 2017 u,v Ω = Ω u(x) v(x)dx, f,g Σ = Σ f(s)g(s)ds, Define S(σ) = Σ ( sσ 2σHn)δ(x X(r,s,t)) X r X s drds, then S(σ),u Ω = [ Ω Σ ( sσ 2σHn) X r X s δ(x X(r,s,t))drds ] u(x)dx = Σ ( sσ 2σHn) U(r,s,t) X r X s drds = Σ (σ(x s n)) r U+(σ(n X r )) s Udrds = Σ σ(x s n) U r +σ(n X r ) U s drds = Σ σ( s U) X r X s drds = σ, s U Σ Remark: Tension does not do extra work to the fluid. Similar to the pressure in incompressible fluid!

15 Numerical issues for 3D problem: 1. Coupled with fluid dynamics which vesicle boundary is moving with fluid and whose shape is not known a priori 2. Both the volume and the surface area of the vesicle are conserved. How to maintain fluid and vesicle boundary incompressible simultaneously? 3. Need to find H, s H, n, K on a moving surface Σ 4. In additional to the fluid incompressibility, we need extra constraint (surface incompressibility) on the surface 5. The role of pressure p on fluid equations is the same as the role of tension σ on s U = 0. Both conditions are local! 6. How to solve the above governing equations efficiently? 7. Boundary integral method, Immersed boundary (Front-tracking), Level-set, Phase field method, Parametric finite element method?

16 Nearly surface incompressibility approach s U = 0 means that t X r X s = 0 To avoid solving the extra unknown tension σ(r,s,t), we alternatively use a spring-like elastic tension σ = σ 0 ( Xr X s X 0 r X0 s ) where σ 0 1 and X 0 r X0 s is the initial surface dilating factor Similar idea has been used in level set framework by Maitre, Misbah, Peyla & Raoult, Physica D 2012 The modified elastic energy by E σ (X) = σ 0 2 ( Xr X s X 0 r X0 s ) 2 drds

17 Derivation of modified elastic force by variational derivative d dε Eσ(X+εY) ε=0 ( = σ 0 Xr X s X 0 r X 0 ) Xr Xs s (Yr Xs +Xr Ys) drds X r X s ( ) = σn (Y r X s +X r Y s) drds by n = Xr Xs X r X s = σ(x s n) Y r +σ(n X r) Y s drds (by the scalar triple product formula) = (σx s n) r Y+(σn X r) s Y drds (by integration by parts) = [σ rx s n+σ sn X r +σ(x s n) r +σ(n X r) s] Y drds = (σ rx s n+σ sn X r +σx s n r +σn s X r) Y drds = ( sσ 2σHn) Y X r X s drds = ( sσ 2σHn) YdS (since ds = X r X s drds) Σ = F σ YdS F σ are exactly identical! Σ

18 Numerical algorithm: axis-symmetric case, Hu, Kim & Lai, JCP Compute the vesicle boundary force σ n = σ 0 ( R n X s n R 0 X s 0), F n σ = σn s τn +σ n τ n s Zn s R nσn n n, F n b = c b( s H n +2H n ((H n ) 2 K n )) X s n n n, F n = F n σ +F n b 2. Solve the Navier-Stokes (3u 4u n +u n 1 )/(2 t)+ ( 2(u n h )u n (u n 1 h )u n 1) = h p n +µ h u + F n (s)δ h (x X n (s)) s s h φ = 3/(2 t) h u, φ/ n = 0 on Ω u n+1 = u 2 3 t hφ, h p n+1 = h φ+ h p n 2 3 tµ h ( h φ) 3. Update the new position X n+1 k = X n k + t x u n+1 δ h (x X n (s))h 2

19 Axis-symmetric case, Hu, Kim & Lai, JCP 2014 T = 0 T = 0.5 T = 1 T = 2 Figure: Freely suspended vesicles with different penalty number σ 0. Blue solid line: σ 0 = ; green marker : σ 0 = ; red marker : σ 0 =

20 total energy time Figure: The corresponding evolution of total energy. Blue solid line: σ 0 = ; green marker : σ 0 = ; red marker : σ 0 =

21 σ 0 R X s R 0 X s 0 A h A 0 /A 0 V h V 0 /V E E E E E E E E E-04 Table: The errors of the area dilating factor, the total surface area, and the volume.

22 Convergence test Initial shape X(s) = (0.5coss,0.15sins) in quiescent flow Computational domain Ω is chosen as [0,1] [ 0.5,0.5] Take c b = and σ 0 = T = 0.5, m = n = 512 as a reference numerical solution m = 64 m = 128 rate m = 256 rate A h A 0 /A E E E V h V 0 /V E E E X h X ref 1.689E E E u h u ref 2.363E E E Table: The mesh refinement results for the perimeter of the surface area A h, the enclose volume V h, the interface configuration X h, and the velocity u h.

23 Numerical experiments A freely suspended vesicle Computational domain Ω = [0,0.5] [ 1,1] Quiescent flow u = 0 Bending coefficient c b = T = 0 T = T = 0.25 T = 0.75 T = 2 T = 5 Figure: Snapshots of a freely suspended vesicle in quicient flow.

24 Figure: The coressponding result in 3D view.

25 1 0.8 energy time Figure: The membrane energy evolution of the freely suspended vesicle.

26 Gravitational effect The additional interfacial force F g is considered due to gravity g is given by F g = (ρ in ρ out )(g X) X s n Computational domain Ω h is chosen as [0,1] [ 2,2]

27 Figure: Top: oblate shape parachute-like; Bottom: prolate shape pear-like shape. Consistent with experiments.

28 Full 3D case: Seol, Hu, Kim & Lai JCP (a) 10 2 (b) (da t da 0 )/da A t A 0 /A V t V 0 /V t x 10 5 (c) total energy E t (d) σ0 = σ0 = σ0 = t t Figure: The comparison for three different stiffness parameters: σ 0 = ( ), ( ), and ( ). (a) the maximum relative error of the local surface area; (b) the relative error of the global surface area; (c) the relative error of the global volume; (d) the total energy.

29 Vesicle under shear flow θ/π Theory [25] BIM [25] Exp. [47] χ = 1 χ = 10 χ = 100 ω/ ǫs ν ν Figure: The plot of the inclination angle (left) and the scaled mean angular velocity (right) as functions of reduced volume ν for different dimensionless shear rate χ. The frequency ω can be computed using ω = 1 N v Nv i=1 r v r 2, where r and v are the position and velocity of the vertices projected on the xz-plane, respectively.

30 Viscoelastic effects on vesicle dynamics, Seol, Tseng, Kim, & Lai 2018 Motivation: vesicles mimic the red blood cells (RBCs), and the blood containing RBCs and plasma exhibits non-newtonian properties [S. Chien 1970, G.B. Thurston 1973,1979] Newtonian vs. non-newtonian fluid: the relation between the shear stress and shear rate is linear (Newtonian); otherwise, it is non-newtonian Viscoelasticity: non-newtonian fluid exhibits both viscous and elastic characteristics when is undergoing deformation Non-Newtonian fluid examples: ketchup, toothpaste, paint, blood, and shampoo etc. Oldroyd-B model is used in the present work

31 Newtonian vesicle in non-newtonian fluid (N/O) Figure: The vesicle is filled with a Newtonian fluid while the suspended fluid is an Oldryod-B fluid (a mixture of polymer and Newtonian solvent), denoted by N/O fluid. For a vesicle under shear flow, two types of motion have been well investigated tank-treading (TT) motion tumbling (TB) motion

32 Governing equations of fluid vesicle system ( ) u ρ t +(u )u = p+ [((1 H)µ n +Hµ s )D(u)+Hσ]+f in Ω, u = 0 in Ω, f(x,t) = (F γ +F b )(α,t)δ(x X(α,t)) X α dα in Ω, Σ X (α,t) = U(α,t) = t Ω γ(α,t) = γ 0 ( Xα (α,t) X α (α,0) 1 F γ (α,t) = 1 (γτ) X α α, F b(α,t) = c b u(x,t)δ(x X(α,t)) x on Σ, ) on Σ, ) (κ ss + κ3 n on Σ, 2

33 Governing equations of polymer stress (Oldroyd-B model) For a symmetric polymer stress defined by [ ] σ a σ σ = b, its governing equation is σ b σ c ξ σ +σ = µ p D(u) in Ω, where σ = σ t +u σ ( uσ +σ( u) T), D(u) = u+( u) T. ξ is the polymer relaxation time, σ is the upper convected time derivative of σ, and D(u) is the rate of deformation tensor. For the Eulerian grid point x = (x,y), H is the indicator (or Heaviside step) function defined by, { 1 if x is in Oldroyd-B fluid H(x) = 0 if x is in Newtonian fluid.

34 Governing equations in dimensionless form Re where ( ) u t +(u )u = p+ [((1 H)λ+H)D(u)+Hσ]+f in Ω, f(x,t) = X (α,t) = U(α,t) = t u = 0 in Ω, (F γ +F b )(α,t)δ(x X(α,t)) X α dα in Ω, Σ u(x,t)δ(x X(α,t))dx on Σ, Ω ( ) Xα (α,t) γ(α,t) = γ 0 X α (α,0) 1 on Σ, F γ (α,t) = 1 (γτ) X α α, F b(α,t) = 1 Ca Wi σ +σ = βdu in Ω, ) (κ ss + κ3 n on Σ, 2 Re = ρr2 µ s t c, Ca = µ sr 3 c b t c, Wi = ξ t c, λ = µ n µ s, β = µ p µ s.

35 For the 1D vesicle boundary X(α,t) = (X(α,t),Y(α,t)), the operators are s σ = τ σ X α α s U = τ X α U α κ ss = 1 ( ) 1 κ = X α X αα κ X α α X α α X α 4 α κ X α 2 α 2 The unit tangent vector τ = (τ 1,τ 2 ) = X α / X α The unit normal vector n = (τ 2, τ 1 ) The signed curvature κ = X αy αα Y α X αα (X 2 α +Y 2 α) 3/2

36 Discretization of Eulerian and Lagrangian variables Figure: Fluid variables on a staggered MAC grid in 2D. Chrispell et. al For a periodic interface X, we use a Fourier representation to discretize it as X(α,t) = N/2 1 k= N/2 X(k,t)e ikα.

37 Numerical algorithm We march the Lagrangian markers X n = X(n t) from time level n to obtain X n+1 = X(n t+ t) at time level n+1 with t the time step size. The polymer stress σ n, the fluid velocity u n, the pressure p n, and the Lagrangian markers X n are all given in advance, and from these variables we update σ n+1, u n+1, p n+1, and X n Compute the vesicle boundary forces. At the Lagrangian markers X n l, we calculate the spring-like tension γ(x n l ) = γ 0 ( (Xα ) n l ) (X α ) 0 l 1, and then incorporate those to compute the interfacial forces ( ) n F γ (X n l ) = 1 γτ (X α ) n l, α l F b (X n l) = 1 ( (X α) n l (X αα) n l (κ α) n l Ca (X α ) n + (κ αα) n ) l l 4 (X α ) n + (κn l )3 n n l 2 2 l.

38 2. Distribute the tension and bending forces acting on Lagrangian markers into the Eulerian grid by using the smoothed Dirac delta function δ h as N 1 f n (x) = (F γ (X n l )+F b(x n l ))δ h(x X n l )ds(xn l ), l=0 where the curve length element is obtained by ds(x n l ) = (X α) n l α. For δ h (x) = 1 ( x ) ( y h 2φ φ, h h) we employ 6-point supported C 3 function φ developed in [Bao et.al, JCP 16].

39 3. Solve the Navier-Stokes ( 3u 4u n +u n 1 Re +2(u n h )u n ( ) u n 1 ) h u n 1 2 t = h p n +λ h u + h [(1 λ)h ( h u n +( h u n ) T)] + h (Hσ n )+f n, h p = 3Re 2 t h u, p n = 0 on Ω, u = u n+1 on Ω, u n+1 = u 2 t 3Re hp, h p n+1 = h p + h p n 2λ t 3Re h( h p ). 4. Update the new position X n+1 k = X n k + t x u n+1 (x)δ h (x X n k )h2

40 5. Solve the polymer stress equation σ t = F(σ,u), where F(σ,u) = u σ + ( uσ +σ( u) T) σ Wi + β ( u+( u) T ). Wi To find σ n+1, we use the second-order Runge-Kutta (RK2) method with zero-neumann boundary condition at Dirichlet boundary of fluid velocity. Two intermediate variables σ 1 and σ 2 are defined by σ 1 = σ n + tf(σ n,u n ), σ 2 = σ 1 + tf(σ 1,u n+1 ), the updated extra stress is obtained from σ n+1 = (σ n +σ 2 )/2.

41 Numerical accuracy test for Oldroyd-B solver We test a single-phase Oldroyd-B fluid in a computational domain [0,2π] [π/2,5π/2]. ( ) u Re t +(u )u = p+ u+ σ in Ω, where [ φ a φ φ = Wiβ b φ b u = 0 in Ω, Wi σ +σ = βdu+φ in Ω, φ c ], φ a = 2G sinxsiny 2G 2 cos 2 xsin 2 y +2G 2 sin 2 xcos 2 y 4G 2 sin 2 xsin 2 y, φ b = 4G 2 sinxcosxsinycosy, φ c = 2G sinxsiny +2G 2 cos 2 xsin 2 y 2G 2 sin 2 xcos 2 y 4G 2 sin 2 xsin 2 y, G(t) = e 2(1+β)t/Re.

42 The exact analytical solution of the above equations is u e = (Gcosxsiny, Gsinxcosy), p e = Re 4 (cos2x+cos2y)g2, [ 2Gsinxsiny 0 σ e = βdu = β 0 2Gsinxsiny Re = Wi = β = 1 and t = h/8, where the h is the Eulerian meshwidth defined by h = 2π/M with grid size M. Convergence rate is obtained by Rate = log 2 ( u M u e / u 2M u e ), ].

43 M u M u e Rate v M v e Rate p M p e Rate (t = 0.6) E E E E E E E E E E E E (t = 1.2) E E E E E E E E E E E E M σm a σa e Rate σb M σb e Rate σc M σc e Rate (t = 0.6) E E E E E E E E E E E E (t = 1.2) E E E E E E E E E E E E

44 Figure: Two particles sedimentation. Left (Newtonian); right (Oldroyd-B); Kim, Lai, Seol, J. Non-Newtonian Fluid Mech., 2018

45 t=9.0 s t=18.0 s t=27.0 s t=54.0 s t=63.0 s Figure: Six particles sedimentation. Top (Newtonian); bottom (Oldroyd-B); Kim, Lai, Seol, J. Non-Newtonian Fluid Mech., 2018

46 Numerical results Goal Exploring the effects of viscoelasticity on vesicle dynamics under shear flow, mainly on the Newtonian vesicle suspended in Oldroyd-B fluid, denoted by N/O Parameter study Weissenberg number Wi viscosity contrast β between polymer and Newtonian solvent Reynolds number Re viscosity contrast λ between inner and outer fluids reduced area ν = 4πA/L 2 : the area ratio between the vesicle and the circle with same perimeter Numerical parameters Unless otherwise stated, we use Re = 10 3, β = 1, Ca = 1, A = π, t = h 2 /4 γ 0 = , meshwidth h = 16/512, [ 8,8] 2 with grid size Number of Lagrangian markers N = 1024, simulation up to t = 50

47 Comparison of N/N and N/O fluids 0.25 (a) N/N 0.25 (b) N/O ν=0.95 ν= ν=0.8 ν=0.7 ν= ν=0.5 θ/π t t 0.5 (c) N/N 0.5 (d) N/O ω t t Figure: For various ν and fixed Ca = 1, the inclination angle θ/π in (a,b); The average TT frequency ω in (c,d).

48 Weissenberg number effect Wi=0 (N/N) Wi=0.25 Wi=0.5 Wi=0.75 Wi=1 (a) (b) θ/π ω ν ν Figure: For various Wi and fixed Ca = 1, (a) the inclination angle θ/π and (b) the average TT frequency ω both in terms of ν. / X α ω(t) = 2π Σ(t) U τ dα,

49 Viscosity ratio between polymer and Newtonian solvent effect β = 0.25 β = 0.5 β = 0.75 β = 1 (a) (b) 0.15 θ/π 0.1 ω ν ν Figure: For various β and fixed Ca = 1, (a) the inclination angle θ/π and (b) the average TT frequency ω both in terms of ν.

50 Reynolds number effect 0.25 (a) N/N 0.25 (b) N/O θ/π 0.1 Re = Re = Re = 10 2 Re = Theory ν ν 0.5 (c) N/N 0.5 (d) N/O ω ν ν Figure: For various Re and fixed Ca = 1, the inclination angle θ/π in (a,b); The average TT frequency ω in (c,d).

51 Viscosity contrast between inner and outer fluids 0.2 (a) Wi=0 (N/N) ν=0.95 ν= (b) Wi= ν=0.8 ν= ν= ν=0.5 Theory 0.1 θ/π λ λ 0.2 (c) Wi= (d) Wi= θ/π λ λ Figure: For various Wi and fixed Ca = 1, the inclination angle θ/π in terms of the viscosity contrast λ.

52 λc ν Wi=0 (N/N) Wi=0.25 Wi=0.5 Wi=0.75 Wi=1 Theory (N/N) Figure: For various Wi and fixed Ca = 1, the critical viscosity contrast λ c in terms of the reduced area ν.

53 3 (a) t=0 3 (b) t=5 3 (c) t= (d) t=15 3 (e) t=20 3 (f) t= Figure: Snapshots of TT vesicles with ν = 0.5,λ = 20 (red solid line) and ν = 0.95,λ = 19 (blue dashed line) suspended in N/O fluid with fixed Re = 10 3,Ca = 1, and Wi = 1.

54 Figure: In N/O fluid with fixed Re = 10 3,Ca = 1, and Wi = 1, streamlines and stress-ellipses at time t = 50 for vesicles with ν = 0.5,λ = 20 in (a) and ν = 0.95,λ = 19 in (b).

55 Different Reynolds number in unmatched viscosity fluid 0.5 (a) ν=0.5, λ= (b) ν=0.95, λ= θ/π Re = 4 Re = 1 Re = 10 1 Re = 10 2 Re = t Re = 4 Re = 1 Re = Re = Re = 10 2 Re = t Figure: In unmatched viscosity N/O fluid with fixed Ca = 1 and Wi = 1, the time evolution of inclination angle for a vesicle with ν = 0.5,λ = 20 in (a) and ν = 0.95,λ = 19 in (b).

56 Conclusion In matched viscosity N/O fluid, the vesicle inclination angle in TT regime decreases with increasing Weissenberg number In unmatched viscosity case, the angle decreases even more as the viscosity contrast increases. Most surprisingly, the negative inclination angle was found in TT regime, indicating that the viscoelasticity of suspending fluid tends to lag the TU motion. A vesicle tank-treads at negative inclination angle at small Reynolds number, then abruptly tumbles at slightly higher Reynolds number, and returns back to tank-tread at positive inclination angle at even higher Reynolds number. We attribute this behavior to the interplay between the viscoelasticity and the inertia.

Introduction to immersed boundary method

Introduction to immersed boundary method Ming-Chih Lai mclai@math.nctu.edu.tw Department of Applied Mathematics Center of Mathematical Modeling and Scientific Computing National Chiao Tung University 1001,