Microhydrodynamics Flow analysis (Taylor series expansion solution to a flow field)

Question

Consider the flow field given by u_i=F_j[δ_ij/x + x_ix_j/x³], where |x_i| = x.

a) By taking the first two terms of a Taylor series about a point x_i --> y_i where (x_i -y_i)F_i=0 determine the form of the local linear velocity field [u_i(x_i)-u_i(y_i)], i.e. determine r_jγ_ij. (r_j gamma dot_ij)

b) Determine the rate of strain tensor and vorticity tensor for this local flow

Answer 1

a) We start by taking the Taylor series expansion of ui about a point xi --> yi where (xi - yi)Fi = 0:

ui(xi) = ui(yi) + (xi - yi)ui'(yi) + O((xi - yi)^2)

where ui'(yi) is the derivative of ui with respect to xi evaluated at yi.

From the given flow field, we have:

ui(xi) = Fj[δij/xi + xi xj/xi^3]

Differentiating with respect to xi, we get:

ui'(xi) = -Fj[δij/xi^2 - 3xi xj/xi^4]

Evaluating at yi, we have:

ui'(yi) = -Fj[δij/yi^2 - 3yi yj/yi^4]

Substituting these values into the Taylor series expansion, we get:

ui(xi) - ui(yi) = -Fj[(xi - yi)δij/yi^2 - 3(xi - yi)yi yj/yi^4] + O((xi - yi)^2)

Simplifying, we get:

ui(xi) - ui(yi) = -Fj[(xi - yi)/yi^2]δij + Fj[3(xi - yi)/yi^3]yi yj + O((xi - yi)^2)

Therefore, the local linear velocity field is:

ui(xi) - ui(yi) = -Fj[(xi - yi)/yi^2]δij + Fj[3(xi - yi)/yi^3]yi yj

b) The rate of strain tensor is given by:

εij = (1/2)(∂ui/∂xj + ∂uj/∂xi)

Substituting the given flow field, we get:

εij = (1/2)Fk[δik/ xj x + δjk/ xi x - 2xixj/x^4]

Since the flow is symmetric with respect to xi and xj, εij = εji. Therefore, we have:

εij = (1/2)Fk[δik/ xj x + δjk/ xi x - 2xixj/x^4]

The vorticity tensor is given by:

ωij = (1/2)(∂ui/∂xj - ∂uj/∂xi)

Substituting the given flow field, we get:

ωij = (1/2)Fk[(δik/xj - δjk/xi) + 2xixj/x^4]

Again, since the flow is symmetric with respect to xi and xj, ωij = -ωji. Therefore, we have:

ωij = (1/2)Fk[(δik/xj - δjk/xi) + 2xixj/x^4]