from __future__ import print_function
from fenics import *
from mshr import *
from dolfin import *
import numpy as np
import matplotlib.pyplot as plt
from tqdm import tqdm
import pdb
T = 5
num_steps = 5000
dt = T / num_steps
mu = 0.01
rho = 1
channel = Rectangle(Point(0, 0), Point(2.2, .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, dim=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], .41)'
cylinder = 'on_boundary && x[0] > .1 && x[0] < 2 && x[1] > 0.03 && x[1] < 0.4'
# Define inflow profile
inflow_profile = ('4.0 * 1.5 * x[1] * (.41 - x[1]) / pow(.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)
# Define symmetric gradient
def epsilon(u):
return sym(nabla_grad(u))
# Define stress tensor
def sigma(u, p):
return 2 * mu * epsilon(u) - p * Identity(len(u))
# Define variational problem for step 1
F1 = rho * dot((u - u_n) / k, v) * dx \
+ rho * dot(dot(u_n, nabla_grad(u_n)), 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
a2 = dot(nabla_grad(p), nabla_grad(q)) * dx
L2 = dot(nabla_grad(p_n), nabla_grad(q)) * dx - (1/k) * div(u_) * q * dx
# 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]
xdmffile_u = XDMFFile(f'test_only/velocity_{0}.xdmf')
xdmffile_p = XDMFFile(f'test_only/pressure_{0}.xdmf')
# Create progress bar
progress = Progress('Time-stepping')
# Time-stepping
t = 0
for n in tqdm(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')
# Plot solution
if n % 100 == 0:
plot(u_, title='Velocity')
plt.savefig(f'test_only/velocity')
plt.close()
if n % 100 == 0:
plot(p_, title='Pressure')
plt.savefig(f'test_only/pressure')
plt.close()
if n == num_steps - 1:
# Save solution to file (XDMF/HDF5)
xdmffile_u.write(u_, t)
xdmffile_p.write(p_, t)
# Update previous solution
u_n.assign(u_)
p_n.assign(p_)
print(mesh.coordinates()[:].shape)
print(u_.vector()[:].shape)
print(p_.vector()[:].shape)
pdb.set_trace()
In the code above, u_ is velocity distribution and p_ is pressure distribution. I would like to obtain the distributions in numpy array, with each row corresponds to the same row in mesh.coordinates()[:], i.e. if mesh.coordinates()[:].shape == (m, 2), distribution.vector()[:] should have a shape of (m, 1) for pressure and (m, 2) for velocity. However, for velocity, this is not the case.
Could you please explain the concept, and provide a code snippet for the solution?