Backward Euler gives completely different results from Forward Euler in ft08_navier_stokes_cylinder.py

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12 months ago by
I modified the ft08_navier_stokes_cylinder.py tutorial to use Backward Euler instead of Forward Euler in the tentative velocity step. This approach is suggested here as an improvement to explicit time stepping. The velocity solution I am getting from the modified code is completely different from the solution I am getting from the original code.

The only modification I made is to replace this:

F1 = rho*dot((u - u_n) / k, v)*dx \
+ inner(sigma(U, p_n), epsilon(v))*dx \
+ dot(p_n*n, v)*ds - dot(mu*nabla_grad(U)*n, v)*ds \
- dot(f, v)*dx​

by this:

F1 = rho*dot((u - u_n) / k, v)*dx \
+ inner(sigma(u, p_n), epsilon(v))*dx \
+ dot(p_n*n, v)*ds - dot(mu*nabla_grad(u)*n, v)*ds \
- dot(f, v)*dx

i.e. I replaced:

• the gradient of u_n by the gradient of u in the convective term (2nd line),
• U by u in the viscous domain term (3rd line) and
• U by u in the viscous boundary term (4th line).
See below a snapshot of the velocity magnitude at the last time step (explicit time stepping):

See below a snapshot of the velocity magnitude at the last time step (implicit time stepping):

Is my implementation wrong?

ps: Using FEniCS from the ppa for Ubuntu 16.04.

EDIT (21/08/2017)

Here is a MWE. By flipping the parameter "explicit" between True and False you can switch between the original explicit code and my implicit code.

from __future__ import print_function
from fenics import *
from mshr import *
import numpy as np

T = 5.0            # final time
num_steps = 5000   # number of time steps
dt = T / num_steps # time step size
mu = 0.001         # dynamic viscosity
rho = 1            # density

# Create mesh
channel = Rectangle(Point(0, 0), Point(2.2, 0.41))
cylinder = Circle(Point(0.2, 0.2), 0.05)
domain = channel - cylinder
mesh = generate_mesh(domain, 64)

# Define function spaces
V = VectorFunctionSpace(mesh, 'P', 2)
Q = FunctionSpace(mesh, 'P', 1)

# Define boundaries
inflow   = 'near(x[0], 0)'
outflow  = 'near(x[0], 2.2)'
walls    = 'near(x[1], 0) || near(x[1], 0.41)'
cylinder = 'on_boundary && x[0]>0.1 && x[0]<0.3 && x[1]>0.1 && x[1]<0.3'

# Define inflow profile
inflow_profile = ('4.0*1.5*x[1]*(0.41 - x[1]) / pow(0.41, 2)', '0')

# Define boundary conditions
bcu_inflow = DirichletBC(V, Expression(inflow_profile, degree=2), inflow)
bcu_walls = DirichletBC(V, Constant((0, 0)), walls)
bcu_cylinder = DirichletBC(V, Constant((0, 0)), cylinder)
bcp_outflow = DirichletBC(Q, Constant(0), outflow)
bcu = [bcu_inflow, bcu_walls, bcu_cylinder]
bcp = [bcp_outflow]

# Define trial and test functions
u = TrialFunction(V)
v = TestFunction(V)
p = TrialFunction(Q)
q = TestFunction(Q)

# Define functions for solutions at previous and current time steps
u_n = Function(V)
u_  = Function(V)
p_n = Function(Q)
p_  = Function(Q)

# Define expressions used in variational forms
U  = 0.5*(u_n + u)
n  = FacetNormal(mesh)
f  = Constant((0, 0))
k  = Constant(dt)
mu = Constant(mu)
rho = Constant(rho)

def epsilon(u):

# Define stress tensor
def sigma(u, p):
return 2*mu*epsilon(u) - p*Identity(len(u))

explicit = False
if explicit:
# Define variational problem for step 1
F1 = rho*dot((u - u_n) / k, v)*dx \
+ inner(sigma(U, p_n), epsilon(v))*dx \
+ dot(p_n*n, v)*ds - dot(mu*nabla_grad(U)*n, v)*ds \
- dot(f, v)*dx
a1 = lhs(F1)
L1 = rhs(F1)

# Define variational problem for step 2

# Define variational problem for step 3
a3 = dot(u, v)*dx
L3 = dot(u_, v)*dx - k*dot(nabla_grad(p_ - p_n), v)*dx
else:
# Define variational problem for step 1
F1 = rho*dot((u - u_n) / k, v)*dx \
+ inner(sigma(u, p_n), epsilon(v))*dx \
+ dot(p_n*n, v)*ds - dot(mu*nabla_grad(u)*n, v)*ds \
- dot(f, v)*dx
a1 = lhs(F1)
L1 = rhs(F1)

# Define variational problem for step 2

# Define variational problem for step 3
a3 = dot(u, v)*dx
L3 = dot(u_, v)*dx - k*dot(nabla_grad(p_ - p_n), v)*dx

# Assemble matrices
A1 = assemble(a1)
A2 = assemble(a2)
A3 = assemble(a3)

# Apply boundary conditions to matrices
[bc.apply(A1) for bc in bcu]
[bc.apply(A2) for bc in bcp]

# Save mesh to file (for use in reaction_system.py)
File('navier_stokes_cylinder/cylinder.xml.gz') << mesh

# Create progress bar
progress = Progress('Time-stepping')
set_log_level(WARNING)

u_.rename("u", "Velocity field")
p_.rename("p", "Pressure field")

velocityOutput = File("navier_stokes_cylinder/velocity.pvd")
pressureOutput = File("navier_stokes_cylinder/pressure.pvd")

# Time-stepping
t = 0
for n in range(num_steps):

# Update current time
t += dt

# Step 1: Tentative velocity step
b1 = assemble(L1)
[bc.apply(b1) for bc in bcu]
solve(A1, u_.vector(), b1, 'bicgstab', 'hypre_amg')

# Step 2: Pressure correction step
b2 = assemble(L2)
[bc.apply(b2) for bc in bcp]
solve(A2, p_.vector(), b2, 'bicgstab', 'hypre_amg')

# Step 3: Velocity correction step
b3 = assemble(L3)
solve(A3, u_.vector(), b3, 'cg', 'sor')

# Save solution to file (PVD)
velocityOutput << (u_, t)
pressureOutput << (p_, t)

# Update previous solution
u_n.assign(u_)
p_n.assign(p_)

# Update progress bar
progress.update(t / T)
print('t: {}    u max: {}'.format(t, u_.vector().array().max()))

Community: FEniCS Project
Does anyone have any suggestion?
written 11 months ago by Miguel Goncalves
1
Backward Euler is a highly dissipative method. What you're observing is a well known effect. Perhaps read up on consistency and stability of stepping schemes http://www.cambridge.org/catalogue/catalogue.asp?isbn=0521810264 Chapter 12.
written 11 months ago by Nate

0
9 months ago by
Dear Miguel
All what you have to do is to put
# Assemble matrices
A1 = assemble(a1)
A2 = assemble(a2)
A3 = assemble(a3)
# Apply boundary conditions to matrices
[bc.apply(A1) for bc in bcu]
[bc.apply(A2) for bc in bcp]
inside the time loop since it is going to change during the time, then the problem will be solved

Regards
Hillal

File attached: ft08final.py (4.05 KB)