Here is some Python code for modeling transients in the pure shear-driven Couette flow between two infinite horizontal flat plates.

Initially, both the plates are stationary. Then, at time , the upper plate is suddenly set into motion with a constant speed . Due to the no-slip boundary condition between the fluid and the upper plate, the topmost layer of the fluid, too, is set into motion with the same speed, i.e., . Due to viscosity (i.e. friction internal to the fluid) this topmost layer tugs on to the next layer to move in the same direction, too. (That’s sort of like how when you pull a book from the middle of the stack, the books below it tend to come out, too.) Thus, the next layer also begins to move. Then, this second layer, in turn, nudges the third layer to move. …

Yes, it *is* turtles—but *not* all the way down.

The bottom-most fluid layer feels the tug not only of the layer immediately above it, but also the cent-per-cent tug of the no-slip boundary condition of the solid plate just below it. And the lower plate remains stationary throughout.

To cut a long story short, if the situation were steady-state, i.e., if the upper plate were always moving (at a constant speed ) and the bottom plate stationary, then the velocity profile developed would be in the shape of a straight-line.

However, if both plates are initially stationary, and then if only the upper plate is suddenly moved, then the velocity profile would be initially like the Greek letter `‘. With the passage of time, the profile would first lose the sharp angle, and so become a bit soft at that corner. Then, the curvature at the corner would become more and more diffuse. Then, the curved profile would become less and less curvy, and finally, it would reach the straight-line of the steady-state.

The code below shows how.

The Couette flow is important because it is one of the very, very (, very, very, very) rare cases in which the Navier-Stokes equations can (at all) be solved “analytically.”

The reason is, for this kind of a Couette flow—the one driven purely by the shear force of the upper plate (i.e., when no pressure gradients are applied in the horizontal direction)—the Navier-Stokes equations reduce to nothing but (our dear old friend): diffusion equation!

We did model the transient diffusion equation the last time, here [^]. However, the discretization approach used back then was FTCS—which is an explicit approach. The one now being used is: the Crank-Nicolson method—which is an implicit approach. FTCS is only conditionally stable, whereas CN is (at least for the diffusion equation, and at least within all reasonable limits) unconditionally stable. (Check out the Flow Science’s take on the explicit vs. implicit issue, here [^].) That makes it interesting.

The code once again comes with more comments on more issues than you want to read and/or know about.

The particular test case currently hard-coded is the one from John D. Anderson, Jr.’s celebrated text [^]. You can compare the output values produced by this program with the listing given in Anderson’s text (Table 9.1, p. 430.)

The Python code here even displays a MatPlotLib plot which you can directly compare with the fig. 9.4 in Anderson’s text (p. 431).

Anyway, here is the code, without further ado:

#################################################################
# Transients in the pure shear-driven Couette flow between two
# infinite flat plates, modeled using the Crank-Nicolson method.
# Here we generally follow the treatment and terminology as
# used in the text by John D. Anderson, Jr. However, please
# note, unlike in the book, here we use the _zero_-based
# index, _not_ 1-based.
# Code written by and copyright (c)Ajit R. Jadhav.
# Version: 03-Feb-2016.
#################################################################
# Importing of modules and routines
# We use numpy arrays for efficiency
import numpy
# We use a neat plotting library
from matplotlib import pyplot as plt
from TDMASolver import TDMASolver
#################################################################
# Program Data. All units are in SI.
#
# Gap between the parallel plates, in m
D = 0.1
# Number of cells along the y-axis
nc = 20 # 4
# Number of nodes along the y-axis
N = nc + 1
# Compute the height of each cell
dy = D / nc
# Material properties
# The values below are for water
# Density, in kg / m^3
rho = 998.2
# viscosity, in Pa-s (or kg/(m.s))
mu = 8.90e-4
# speed of the moving (upper) plate, in m/s
ue = 0.05
# Compute the effective Reynolds' number
Re = rho * ue * D / mu
print("Calculated Re: %lf" % Re)
#################################################################
# Validation
# 1. The program output has been validated against manually
# worked out solutions for the first two time-steps, for these
# values of parameters: N=5, Re = 100.0, E = 1.0. For this set
# of values, and given the default truncation that happens in the
# Python print function, the convergence to the steady-state
# solution occurs at the 30th time-step.
#
# 2. Also validated with the case given in John D. Anderson's
# text. The values of parameters are: N=21, Re = 5000, nt =1.
# The program output matches the solution listing given on
# p. 430 in the book.
#################################################################
# For validation purposes, we will ignore the above-calculated
# Re, and instead set the variable with a simple value.
# Anderson's book uses 5000---quite close to the program data
# assumed above.
Re = 5000.0
# CN being implicit, has unconditional stability. However, for
# ease of comparisons, we define a parameter E anyway.
# $E \equiv \dfrac{(\Delta t)}{(Re)(\Delta y)^2})$
# Anderson's book uses E = 1.0
E = 1.0
# Number of time-steps to take
# Anderson's book shows graph up to 240 time-steps
nt = 241 # 36
print("Assumed Re: %lf, E: %lf, nt: %d" % (Re, E, nt))
# Calculate the time-step duration
dt = E * Re * dy * dy
print("Physical time elapsed in marching one computational time-step: %lf seconds." % dt)
#################################################################
# Create the numpy arrays, initialized to zero, for holding
# the primary variable values (here, u)
# The main array, holds the values after time-marching through one step
u = numpy.zeros(N)
# A temporary array, holds the values at the previous time-step
un = numpy.zeros(N)
#################################################################
# Setting up of the initial condition:
#
u[N - 1] = 1.0
print("Initial u: ", u)
#################################################################
# Create the y-axis vector, for plotting purposes
yAxis = numpy.linspace(0.0, 1.0, N)
plt.title("Transients in the pure shear-driven Couette flow,\nusing the Crank-Nicolson method")
plt.xlabel("Dimensionless horizontal velocity (u/ue)")
plt.ylabel("Dimensionless distance (y/D)")
plt.grid(True)
plt.plot(u, yAxis, 'r')
# Array containting the indices of time-steps at which the
# solution should be plotted:
ShowPlotInstants = [0, 2, 12, 36, 60, 240]
#################################################################
# Using the CN (Cranck-Nicolson) scheme, for the Couette Flow
#
# Compute A and B; they form a tri-diagonal matrix as below:
# B A
# A B A
# A B A
A = -E / 2.0
B = 1.0 + E
print("A: %lf, B: %lf" % (A, B))
# a, b, c, d are the 1D arrays that define a tri-diagonal
# matrix. The tri-diagonal matrix in general looks like:
# a0|b0 c0 0 | |x0| |d0|
# |a1 b1 c1| |x1|=|d1|
# |0 a2 b2|c2|x2| |d2|
# a0 and c2 are not included in the tri-diagonal system, but
# the TDMASolver function assumes that these memory
# locations exist in the a and c arrays. Further, the
# algorithm employs the array indices in such a way that
# the first used element of the array a is a[1], not a[0],
# and similarly, the last used element of the array c is
# c[2], not c[1].
#
# For N-grid points, for the transient Couette flow problem,
# we get a system of N-2 X N-2 size, with the _global_ and
# _zero_-based _index_ going over [1, N-2]. Thus, if the
# number of grid points N = 5, we get [1,2,3].
a = numpy.zeros(N - 2)
b = numpy.zeros(N - 2)
c = numpy.zeros(N - 2)
d = numpy.zeros(N - 2)
# Initialize the a, b and c arrays
a[0:N - 2] = A
b[0:N - 2] = B
c[0:N - 2] = A
#################################################################
# Here begins the actual time-stepping...
nShowPlotIndex = 0
for n in range(nt):
# We first save the known values of u at time n, to un,
un = u.copy()
# and apply the Dirichlet boundary conditions
# to the solution at the time-step n
un[0] = 0.0
un[N - 1] = 1.0
# The right hand-side vector on the CN is calculated here
# Anderson calls the variable K
# Note carefully the indices here. The indices used for Kj
# are straight-forward from Anderson's book, except that here
# they are redone as 0-based.
# But the real change is on the index used on d. It is one
# less, because here we don't at all store the first and the
# last rows of the total system in it.
for j in range(1, N - 1):
Kj = (1.0 - E) * un[j] + E / 2.0 * (un[j + 1] + un[j - 1])
d[j - 1] = Kj
# For the same reason spelt above, note the indices: They are
# respectively 1 less and 2 less.
d[0] = d[0] - a[0] * un[0]
d[N - 3] = d[N - 3] - c[N - 3] * un[N - 1]
# print( d )
# Realize, the solution array returned here has values only for
# the interior nodes. Its elements do not include the two
# boundary nodes.
soln = TDMASolver(a, b, c, d)
for j in range(1, N - 1):
u[j] = soln[j - 1]
# We skip plotting for all the time-steps except for those given
# in the ShowPlotInstants array. The values are for the graph in
# Anderson's text on p. 431
if n == ShowPlotInstants[nShowPlotIndex]:
nShowPlotIndex = nShowPlotIndex + 1
# Uncomment the following line for erasing out any earlier plots
# plt.clf()
# Uncomment the following line if you want to print T to console
# print( T )
plt.plot(u, yAxis, 'k')
plt.pause(0.1)
print("\nVelocity profile after time-step no. %d (or physical seconds: %lf)" % (n, float(n) * dt))
print(u)
plt.plot(u, yAxis, 'b')
plt.show()
print("done")
#################################################################
#################################################################

The above code calls on a TDMASolver() function written in a Python module of the same name. Here is the listing for that.

Initially, I tried writing my own TDMA (i.e. Thomas algorithm) code, but found that it was getting too time-consuming. So, I did the smart thing. I searched on the ‘net, found Ofan Tau’s code, here [^], and copy-pasted it!

One funny thing here was that Ofan in turn refers to some code on a Wikibooks page (here [^]), but while Ofan’s code is correct, the code on the Wikibooks page (at least the Python version of it) is not.

Incidentally, the best resource on TDMA aka Thomas’ algorithm that I found was one .PDF document by Prof. William T. Lee of the University of Limerick, Ireland, here [(.PDF) ^]. Keep it handy while going through this code.

#################################################################
# TDMASolver.py
#
# Solves a Tri-Diagonal Matrix system.
#
# a, b, c, d are the 1D arrays that define a tri-diagonal
# matrix. The tri-diagonal matrix in general looks like:
# a0|b0 c0 0 | |x0| |d0|
# |a1 b1 c1| |x1|=|d1|
# |0 a2 b2|c2|x2| |d2|
# a0 and c2 are not included in the tri-diagonal system, but
# the TDMASolver function assumes that these memory
# locations exist in the a and c arrays. Further, the
# algorithm employs the array indices in such a way that
# the first used element of the array a is a[1], not a[0],
# and similarly, the last used element of the array c is
# c[2], not c[1].
#
# The code here was taken from Ofan Tau's blog post:
# http://ofan666.blogspot.in/2012/02/tridiagonal-matrix-algorithm-solver-in.html
# Ofan's blog post makes reference to:
# https://en.wikibooks.org/wiki/Algorithm_Implementation/Linear_Algebra/Tridiagonal_matrix_algorithm
# It so happens that the _Python_ code at the wikibooks URL is
# _wrong_, but Ofan's code is correct (though it does not check
# for stability or even just diagonal dominance.)
# I have mostly kept Ofan's code as is; just added one or two
# comments.
#
# BTW, a wonderfully illustrated (and as far as I can tell,
# correct) reference for understanding TDMA is a .PDF document
# by Prof. William T. Lee (of Industrial Mathematics unit,
# University of Limerick, Ireland):
# http://www3.ul.ie/wlee/ms6021_thomas.pdf
#
# -- Ajit R. Jadhav
# Version 03-February-2016
#################################################################
import numpy
#################################################################
# This helper function was written by Ajit R. Jadhav
# It uses the a, b and c vectors to fill a full-size matrix.
# It allows for matrix visualization. To be used mainly for
# debugging purposes.
def FillM(DM, a, b, c):
N = len(a)
DM[0][0] = b[0]
DM[0][1] = c[0]
for i in range(N - 1):
DM[i][i - 1] = a[i]
DM[i][i] = b[i]
DM[i][i + 1] = c[i]
DM[N - 1][N - 2] = a[N - 1]
DM[N - 1][N - 1] = b[N - 1]
return DM
#################################################################
# This function was shamelessly lifted from Ofan's blog-post:
# http://ofan666.blogspot.in/2012/02/tridiagonal-matrix-algorithm-solver-in.html,
# accessed 02-February-2016
#################################################################
def TDMASolver(a, b, c, d):
'''
TDMA solver, a b c d can be NumPy array type or Python list type.
refer to http://en.wikipedia.org/wiki/Tridiagonal_matrix_algorithm
# a, b, c, d are the 1D arrays that define a tri-diagonal
# matrix. The triadiagonal matrix in general looks like:
# a0|b0 c0 0 | |x0| |d0|
# |a1 b1 c1| |x1|=|d1|
# |0 a2 b2|c2|x2| |d2|
# a0 and c2 are not included in the triadiagonal system, but
# the TDMASolver function assumes that these memory
# locations exist in the a and c arrays. Further, the
# algorithm employs the array indices in such a way that
# the first used element of the array a is a[1], not a[0],
# and similarly, the last used element of the array c is
# c[2], not c[1].
# One test case is this:
# a = [0, -1.0, -1.0, -1.0]
# b = [4.0, 4.0, 4.0, 4.0]
# c = [-1.0, -1.0, -1.0, 0.0]
# d = [5.0, 5.0, 10.0, 23.0]
# result: [2, 3, 5, 7]
'''
# number of equations
nf = len(a)
# print( "In TDMASolver.py: Size of the system: %d" % nf)
# Make a copy the arrays
ac, bc, cc, dc = map(numpy.array, (a, b, c, d))
##############################################
# Make the matrix strictly upper triangular
for i in range(1, nf):
mc = ac[i] / bc[i - 1]
bc[i] = bc[i] - mc * cc[i - 1]
dc[i] = dc[i] - mc * dc[i - 1]
##############################################
# The back-substitution pass:
# Note the meaning of the -1 index in Python.
xc = ac
xc[-1] = dc[-1] / bc[-1]
for j in range(nf - 2, -1, -1):
xc[j] = (dc[j] - cc[j] * xc[j + 1]) / bc[j]
# DM = numpy.zeros((nf,nf))
# DM = FillM(DM, a, b, c)
# print(DM)
# del DM
# delete variables from memory
del bc, cc, dc
return xc

Some things to try:

- Change E to 5.0, 10.0, and then even to 4000.0, and see if you get a solution like the one given in Anderson’s text (fig. 9.5, p. 934)
- Remove the hard-coded value of Reynolds’ number, and start using the calculated values for the same. Then, play with the physical parameters: use material properties for honey, glycerine, different SAE grades of motor oils, etc., and see if you get reasonable-looking “settling-in” time periods to approach steady-state or not.
- Also change the separation between the flat plates.
- What would happen if pressure gradients are applied to the fluid? In what way will the governing differential equation change? The velocity profile? Could you handle it using FDM?
- And, oh, yes! Do continue thinking about the SPPU (Savitribai Phule Pune University), and its past, present and future “authorities.”

[As usual, this post could change—improve or deprove—without any prior [or posterior] notice.]

[E&OE]