University of Warsaw Advanced Hydrodynamics
Faculty of Physics Selected Topics in Fluid Mechanics
Summer Semester 2019/20
Exercise Sheet 4 Sphere in a general linear flow
Questions, comments and corrections: e-mail togustavo.abade@fuw.edu.pl
Consider a non-rotating sphere of radius a (centered at the origin) in a general linear flow of unbounded fluid,
u0(r) = Γ · r, (1)
where Γ = ∇u0 = constant is the velocity gradient tensor evaluated at the sphere center.
(a) Show that
u0(r) = E · r − ω × r, (2)
where E is the symmetric and traceless part of the velocity gradient tensor, and ω is the fluid local angular velocity, ω = 12∇ × u0. Because the Stokes equations are linear, we can consider a sphere in each of the component flows of Eq. (2) separately, and then add the solutions together to solve the full problem.
Solution. The velocity gradient tensor Γ may be decomposed as Γ = E + W,
where E = 1
2(Γ + ΓT), W = 1
2(Γ − ΓT),
with E being the symmetric and W the anti-symmetric parts of Γ, respectively. The symmetric part E is also traceless, because u0 is incompressible,
∇ · u0 = Γ : 1 = E : 1 + W : 1 = tr(E) + tr(W) = tr(E) + 0 = 0.
It can be straightforwardly shown that ω = 1
2∇ × u0
is the axial vector of the anti-symmetric tensor
−W = −1
2(∇u0− ∇uT0), so that
W · r = −ω × r,
and the ambient flow may be written in the form u0(r) = E · r − ω × r.
(b) Using the Stokes equations, show that the pressure field p is a harmonic function, i.e.
∇2p = 0. Show also that u = r
2ηp + uH, (3)
where uHis a harmonic function,
∇2uH = 0,
is a general solution for the Stokes equations. Which additional condition must be satisfied by the harmonic function uH?
Solution. The Stokes equations read
η∇2u − ∇p = 0, ∇ · u = 0. (4)
Taking the divergence of Eq. (4) yields readily
∇2p = 0. (5)
Substitution of the general solution u = r
2ηp + uH (6)
into the lhs of Eq. (4) yields 1
2∇2(rp) + η∇2uH− ∇p = 1
2∇2(rp) + 0 − ∇p = ∇p − ∇p = 0.
Is is straightforward to show (using index notation) that ∇2(rp) = ∇ · ∇(rp) = 2∇p:
∇(rp) = (∇p)r + p1, ∇ · ∇(rp) = (∇2p)r + 2∇p = 2∇p,
where Eq. (5) has been used.
For the solution to satisfy the continuity equation (∇ · u = 0), the harmonic function uHis required to satisfy the condition
∇ · uH = − 1
2η(3p + r · ∇p), where we have used
∇ · (rp) = 3p + r · ∇p.
(c) The solution may be written in the form
u(r) = u0(r) + u1(r), (7)
where u0is the incident flow, and u1is the disturbance flow (due the the presence of the sphere) to be determined. Using the representation (6) and noting that the disturbance flow decays at infinity, the solution (u1, p1) may be written as a superposition of the decaying harmonics (represented by higher-order derivatives of 1/r, the fundametal solution of Laplace’s equation)
1 r, ri
r3, rirj r5 − δij
3r3, rirjrk
r7 −riδjk + rjδki+ rkδij
5r5 , . . . (8)
Use the superposition of harmonics described above and solve first the problem in which the incident flow is just the rotational motion associated with the vorticity (the second term on the rhs of Eq.(2)). Then seek for the solution for the purely extensional part characterized by E.
Solution. We start considering the incident rotational motion only,
uW0 (r) = −ω × r. (9)
The solution may be written in the form
uW(r) = uW0 (r) + uW1 (r), (10)
where uW1 is the disturbance flow to be determined.
We start with the pressure field, writting pW1 , a scalar field, in terms of the harmonics in a form which is linear in ω. There is only a single possibility
pW1 = c0ω · r
r3. (11)
But there is a problem: pW1 must be a true scalar, while ω · r is a pseudo scalar (it changes sign upon inversion from a right- to a left-handed system), because ω is a pseudo vector. Thus, we must have
c0 = 0, pW1 ≡ 0. (12)
The next step is to construct a representation of the harmonic part uH,W1 of the distur- bance flow (uH,W1 = uW1 because of (6) and (12)). The only linear combination of ω and the vector harmonic functions that produces a true vector is
uW1 = c ω ×
r r3
. (13)
The constant c is to be determined from the boundary conditions.
***
Now we seek for the solution for the purely extensional part
uE0(r) = E · r. (14)
The solution may be written in the form
uE(r) = uE0(r) + uE1(r), (15)
where uE1 is the disturbance flow to be determined.
We start with the pressure field, writing pE1, a scalar field, in terms of the harmonics in a form which is linear in E,
p1 = c1E : rr r5 − 1
3r3
. (16)
This can be further simplified if one notice that E : 1 = tr E = 0 (because u0 is incompressible),
p1 = c1r · E · r
r5 , (17)
where c1 is a constant to be determined from the boundary conditions.
The next step is to construct a representation of the harmonic part uH,E1 of the distur- bance flow,
uH,E1 = c2E · r
r3 + c3E : rrr
r7 −2r1 5r5
, (18)
where we have used that E is symmetric and traceless (E : 1 = 0).
From Eq. (6) and (17), uE1 = c2E · r
r3 + r(r · E · r)
c1 2r5η + c3
r7
+ E · r
−2c3 5r5
. (19)
From the requirement that ∇ · u1 = 0 it follows that c2 = 0.
***
Now we sum up the solutions (13) and (19), u1 = uE1 + uW1
= r(r · E · r)
c1 2r5η + c3
r7
+ E · r
−2c3 5r5
+ c4ω × r
r3, (20)
and apply the boundary conditions to determine the constants c1, c3, and c4. For a non-translating and non-rotating sphere centered at the origin,
U = Ω = 0, (21)
so that the stick boundary condition reads
u = u0+ u1 = 0 at r = a, (22)
or
u1(r) = −E · (aˆr) + ω × (aˆr) at r = a, (23)
where ˆr = r/r. Then
−E · (aˆr) + ω × (aˆr) = ˆr(ˆr · E · ˆr)
c1
2a2η + c3 a4
+ E · ( ˆar)
−2c3 5a5
+ c4
a3 ω × ( ˆar), (24)
c1 = −5a3η, c3 = 5a5
2 , c4 = a3, (25)
so that
p1 = −5η(ˆr · E · ˆr)a r
3
, (26)
and
u1 = 5 2
a5 r4 − a3
r2
ˆr(ˆr · E · ˆr) − a5
r4E · ˆr +a3
r2 ω × ˆr. (27)
(e) Give explicitly the solution for the special case of a simple shear flow,
u0(r) = βyˆx, (28)
where β (the shear rate) is a constant, and ˆx is the unit vector in the x-direction. Using the linearity of the Stokes equations, obtain the solution for the case when the sphere freely rotates with the shear flow.
Solution. For a simple shear flow,
u0(r) = βyˆx, (29)
we have
Γ =
0 β 0 0 0 0 0 0 0
, E =
0 12β 0
1
2β 0 0
0 0 0
, ω = −1
2βˆz. (30)
Substituting this into the solution u = u0+ u1, with u1given by Eq. (27) yields
u(r) = β
2(yˆx + xˆy)
1 −a5
r5
−β
2(xˆy − yˆx)
1 −a3
r3
− 5
2βxy(xˆx + yˆy + zˆz) a3 r5 − a5
r7
. (31)
