Sistema Operativo

Since we know the velocities of the flagella relative to the centre-of-buoyancy, and net external forces and torques we can determine the velocity of the cell in response to the surface forces and torques.

Obtaining a solution to the mobility problem is problematic due to the inversion of the grand resistance matrix in (2.16), which contains the ill-conditioned sub-matrix S. To find a solution we follow Cortez [21] and employ an iterative inversion scheme known as the generalised minimal residual method (GMRES), see Appendix E. The mobility problem is solved at each time-step of the flagellar beat and provides estimates for U and Ω which can then be used to estimate the average translational velocity and angular velocity of the cell

hUi = 1

where T is the number of aspects in the flagellar beat and hi denotes the average.

2.6.1 Updating cell position

To update the position of the cell we use U to calculate the location of the centre-of-buoyancy, C, such that

C_t= C_t−1+ U∆t,

where C_t denotes the position of the cell’s centre at the tth time-step and ∆t is the time between time-steps: ∆t = 1/(T − 1) s. We can then update the position of the cell by translating the nodes at time-step t − 1 by Ct.

From the angular velocity Ω we can update the Euler angles (φ , θ, ψ) to determine the change in orientation. To compute the angles we use the relationship [60];

Ω = ˙φk + ˙θ ˆφ+ ˙ψp, (2.18)

which we can re-formulate in terms of the Cartesian axis, i, j and k, through the substitution of p and ˆφ, from (2.1). Thus

which can be used to update the orientation of the cell in the next time-step.

2.6.2 Re-orientation and effective cell eccentricity

Along with computing the cells translational and rotational velocities we also investigate how the re-orientation mechanism and beat pattern affect the time taken for the cell to re-orientate due to the balance between viscous and external torques [81]. The re-orientation time, B, can be explored by fitting to the exact deterministic torque balance equation for a dipolar spheroid in a simple shear flow [81]:

˙p = 1

2B[k − (k · p) p] +1

2ω∧ p + α^0p · E ·^hI − pp^Ti, (2.20) where k and p are defined as before, ω and E are the local vorticity and rate-of-strain tensor, respectively, and I is the identity. Equation (2.20) relates the rate of change of the cell’s swimming direction to the viscous and gravitational torques acting upon a spheroid, whose eccentricity α0

can be defined in terms of the ratio of semi-major to semi-minor axes, α₁, as follows

α₀= α²₁− 1

α²₁+ 1. (2.21)

The cells are approximated as self-propelled spheroids via (2.20). For uni-planar motion, such that φ = ψ = 0 and a shear axes situated π/4 from the vertical

˙θ = e + α0e sin(2[θ + π/4]) − β sin θ, (2.22)

where e is the magnitude of the rate-of-strain, α₀ is the cell’s effective eccentricity and β is the maximum rate of re-orientation.

As we have uni-planar motion then the configuration in Figure 2.6(b) and equation (2.19a) implies that ˙θ = Ωy. If the rate-of-strain is zero then from (2.22) we have

Ω_y = −β sin θ, (2.23)

from which we can use the relation B = 1/2β, [80], to compute the re-orientation time

B = −sin θ

2Ω_y. (2.24)

Solving the mobility problem we can find the cell’s rotation rate at each time-step of the beat.

This can be used to obtain B via (2.24). To increase the accuracy of the fit it is important to study numerically the behaviour of the cell over a wide range of values of θ in the interval π to −π; the model is run for a single flagellar beat, consisting of T time-steps, for N_θ initial orientation angles θ0, such that θ0= 2nπ/(N_θ− 1) for n = −(Nθ− 1)/2, − (Nθ− 3)/2, · · · , (Nθ− 1)/2.

Given B we can also compute the viscous torque parameter α⊥, a quantity which relates the viscous torque to the cell’s relative rotation rate [80], as

α^⊥= 2hρcgB

µ , (2.25)

where µ is the dynamic viscosity of the surrounding fluid, ρ_c is the cell density and g is the gravitational acceleration constant.

If e , 0 then we solve the mobility problem and fit ˙θ = Ω_y to (2.22) for the effective cell eccentricity α₀ and the re-orientation time B. Like when e = 0 we fit over a wide range of orientations to increase accuracy. Furthermore, we can choose to fit for e, α₀ or β together, individually or in pairs. To fit for α₀ alone we substitute the imposed rate-of-strain e and the value of β calculated in the no-flow case into (2.22). The merits of fitting them together will be discussed in § 2.7.3.

To impose a shear flow about a swimming cell we construct a shear box. The shear box is a set of discrete nodes distributed around the surface of a cuboid whose height is equal to its depth.

The longest dimension of the box lies along the x-axis. The nodes are given a prescribed velocity, which are dictated by their location on the surface. Figure 2.7(a) shows a schematic of the box layout. The boundary conditions for the nodes on the six faces are shown in Figure 2.7(b). For the six walls on a shear box of length 2l and height and width 2d we have

u(x, y, d) = (ωd, 0, 0), u(x, y, −d) = −(ωd, 0, 0),

u(x, d, z) = (ωz, 0, 0), u(x, −d, z) = −(ωz, 0, 0),

u(l, y, z) = (ωz, 0, 0), u(−l, y, z) = −(ωz, 0, 0),

where ω = 2e.

(a)

l

u(x, y, d) = ωde₁

u(x, y, −d)) = −ωde1

u(l, y, z)) = ωze1

u(−l, y, z)) =

−ωze1

d x

z

(b)

Figure 2.7: (a) Schematic of a cell within the shear box. The shear box is used to generate a flow in the xz-plane.

The cell is then placed in the centre of the box such that it may rotate, but cannot translate. The shear is generated by giving the nodes on the surface a prescribed motion with nodes on the top moving with velocity u(x, y, z) and nodes on the bottom moving with velocity −u(x, y, z) along the x-axis. Nodes on the side of the box have velocities determined by their location along the y- and z-axes, with the restriction that nodes at the top and bottom are equal to u(x, y, z) and −u(x, y, z). (b) An Illustration of the boundary conditions (BC) for the shear box of width and height 2d and length 2l. For the front and back walls the boundary conditions are u(x, ±d, z)) = ωze1, where e₁= (1, 0, 0) and ω = 2e.