def homogen(h0, f, n):
# First lets generate the uniform mesh for this mesh width
mesh = msh.grid_square(1, h0)
# Matrix of nodes
p = mesh[0]
# Matrix of triangles index
t = mesh[1]
# Now lets compute the stiffness matrix
A = fem.stiffness(p, t)
# Lets compute the load vector
Load = fem.load(p, t, n, f)
# Lets compute the vector m that analogous to the load vector
# with f(x.y)=1
m = fem.load(p, t, n, lambda x, y: 1)
# Lets create the matrix and vectors of the modified bigger system to
# solve with lagrangian multipliers of size (size(A)+(1,1))
size = A.shape[0]
B = sparse.lil_matrix(np.zeros([size + 1, size + 1]))
B[0:size, 0:size] = A
B[0:size, size] = m
B[size, 0:size] = m.transpose()
# We add a zero at the end of f
Load = np.concatenate((Load, np.array([[0]])))
# Now lets get the solution of the linear system using spsolve function
U = spla.spsolve(B, Load)
#We extract the solution u and the multiplicer l
u = U[0:size]
l = U[size]
# We return [p,t,u,l]
return p, t, u, l
def homogen(h0,f,n): # First lets generate the uniform mesh for this mesh width mesh=msh.grid_square(1,h0) # Matrix of nodes p=mesh[0] # Matrix of triangles index t=mesh[1] # Now lets compute the stiffness matrix A=fem.stiffness(p,t) # Lets compute the load vector Load=fem.load(p,t,n,f) # Lets compute the vector m that analogous to the load vector # with f(x.y)=1 m=fem.load(p,t,n,lambda x,y:1) # Lets create the matrix and vectors of the modified bigger system to # solve with lagrangian multipliers of size (size(A)+(1,1)) size=A.shape[0] B=sparse.lil_matrix(np.zeros([size+1,size+1])) B[0:size,0:size]=A B[0:size,size]=m B[size,0:size]=m.transpose() # We add a zero at the end of f Load=np.concatenate((Load,np.array([[0]]))) # Now lets get the solution of the linear system using spsolve function U=spla.spsolve(B,Load) #We extract the solution u and the multiplicer l u=U[0:size] l=U[size] # We return [p,t,u,l] return p,t,u,l
def dirichlet_nonhomogeneous(h0, f, g, n):
# First lets generate the uniform mesh for this mesh width
mesh = msh.grid_square(1, h0)
# Matrix of nodes
p = mesh[0]
# Matrix of triangles index
t = mesh[1]
# Matrix of border element index
be = mesh[2]
# Lets get the interior nodes in the mesh
innod = fem.interiorNodes(p, t, be)
# Rewrite the innod to match with indices beginning at 0
innod = [i - 1 for i in innod]
# Now lets compute the stiffness matrix
Stiff = fem.stiffness(p, t)
# Now lets compute the mass matrix
Mass = fem.mass(p, t)
# Lets compute the load vector
Load = fem.load(p, t, n, f)
# The complete matrix for the bilinear form is given by the sum of
# the stiffness and the mass
B = Stiff + Mass
# Lets define a vector with of the values of u_g in each node
# that is zero in the innernodes and g everywhere else
ug = np.array([g(p[i][0], p[i][1]) for i in range(len(p))])
ug[innod] = np.zeros(len(innod))
# Calculate the new load vector with the Dirichleft taken in account
Load = Load - B.dot(ug).reshape(len(p), 1)
# Now lets get the homogeneous problem solution as we did before
# First lets initialize the array in zeros to respect the homogeneous
# Dirichlet boundary conditions
U = np.zeros(len(p))
# Lets take just the interior points in the matrix B and the Load vector
Bint = B[innod][:, innod]
Loadint = Load[innod]
# Now lets get the solution of the linear system of the interior points
# using spsolve function
Uint = spla.spsolve(Bint, Loadint)
# We put them in the correspondence interior points of the complete
# solution
U[innod] = Uint
# Finally we sum up the two solutions to get the final solution
u = U + ug
# We return [p,t,U]
return p, t, u
def dirichlet_nonhomogeneous(h0, f, g, n):
# First lets generate the uniform mesh for this mesh width
mesh = msh.grid_square(1, h0)
# Matrix of nodes
p = mesh[0]
# Matrix of triangles index
t = mesh[1]
# Matrix of border element index
be = mesh[2]
#Lets get the interior nodes in the mesh
innod = fem.interiorNodes(p, t, be)
# Rewrite the innod to match with indices beginning at 0
innod = [i - 1 for i in innod]
# Now lets compute the stiffness matrix
Stiff = fem.stiffness(p, t)
# Now lets compute the mass matrix
Mass = fem.mass(p, t)
# Lets compute the load vector
Load = fem.load(p, t, n, f)
# The complete matrix for the bilinear form is given by the sum of
#the stiffness and the mass
B = Stiff + Mass
# Lets define a vector with of the values of u_g in each node
# that is zero in the innernodes and g everywhere else
ug = np.array([g(p[i][0], p[i][1]) for i in range(len(p))])
ug[innod] = np.zeros(len(innod))
#Calculate the new load vector with the Dirichleft taken in account
Load = Load - B.dot(ug).reshape(len(p), 1)
#Now lets get the homogeneous problem solution as we did before
# First lets initialize the array in zeros to respect the homogeneous
# Dirichlet boundary conditions
U = np.zeros(len(p))
# Lets take just the interior points in the matrix B and the Load vector
Bint = B[innod][:, innod]
Loadint = Load[innod]
# Now lets get the solution of the linear system of the interior points
# using spsolve function
Uint = spla.spsolve(Bint, Loadint)
# We put them in the correspondence interior points of the complete
# solution
U[innod] = Uint
# Finally we sum up the two solutions to get the final solution
u = U + ug
# We return [p,t,U]
return p, t, u
def neumann(h0, f, n):
# First lets generate the uniform mesh for this mesh width
mesh = msh.grid_square(1, h0)
# Matrix of nodes
p = mesh[0]
# Matrix of triangles index
t = mesh[1]
# Now lets compute the stiffness matrix
Stiff = fem.stiffness(p, t)
# Now lets compute the mass matrix
Mass = fem.mass(p, t)
# Lets compute the load vector
Load = fem.load(p, t, n, f)
# The complete matrix for the bilinear form is given by the sum of
#the stiffness and the mass
B = Stiff + Mass
# Now lets get the solution of the linear system using spsolve function
U = spla.spsolve(B, Load)
# We return [p,t,U]
return p, t, U
def neumann(h0,f,n): # First lets generate the uniform mesh for this mesh width mesh=msh.grid_square(1,h0) # Matrix of nodes p=mesh[0] # Matrix of triangles index t=mesh[1] # Now lets compute the stiffness matrix Stiff=fem.stiffness(p,t) # Now lets compute the mass matrix Mass=fem.mass(p,t) # Lets compute the load vector Load=fem.load(p,t,n,f) # The complete matrix for the bilinear form is given by the sum of #the stiffness and the mass B=Stiff+Mass # Now lets get the solution of the linear system using spsolve function U=spla.spsolve(B,Load) # We return [p,t,U] return p,t,U
def main(h0, n):
p, t, be = mesh.grid_square(2, h0) # shift mesh to origin#
h = mesh.max_mesh_width(p, t)
IN = fem.interiorNodes(p, t, be)
S = fem.stiffness(p, t)
M = fem.mass(p, t)
L = fem.load(p, t, n, lambda x, y: 1)
S = S.tocsr()
M = M.tocsr()
L = L.tocsr()
# neumannnodes
N = np.zeros(len(p))
j = 0
for i in range(0, len(p)):
if p[i, 0] == 0 or p[i, 0] == 2:
if 2 > p[i, 1] and p[i, 1] >= 1:
N[j] = i
j = j + 1
if p[i, 1] == 2:
N[j] = i
j = j + 1
j = j - 1 # number of neumannnodes
print(N)
Sr = sp.csr_matrix((len(IN) + j, len(IN) + j))
Mr = sp.csr_matrix((len(IN) + j, len(IN) + j))
Lr = sp.csr_matrix((len(IN) + j, 1))
A = sp.csr_matrix((len(IN) + j, len(p)))
for i in range(0, len(IN)):
A[i, IN[i]] = 1
for i in range(0, j):
A[i + len(IN), N[i]] = 1
Sr = A * S * A.transpose()
Mr = A * M * A.transpose()
Lr = A * L
un = spla.spsolve(Sr + Mr, Lr)
u1 = A.transpose() * un
fem.plot(p, t, u1)
def dirichlet_homogeneous(h0,f,n): # First lets generate the uniform mesh for this mesh width mesh=msh.grid_square(1,h0) # Matrix of nodes p=mesh[0] # Matrix of triangles index t=mesh[1] # Matrix of border element index be=mesh[2] #Lets get the interior nodes in the mesh innod=fem.interiorNodes(p,t,be) # Rewrite the innod to match with indices beginning at 0 innod=[i-1 for i in innod] # Now lets compute the stiffness matrix Stiff=fem.stiffness(p,t) # Now lets compute the mass matrix Mass=fem.mass(p,t) # Lets compute the load vector Load=fem.load(p,t,n,f) # The complete matrix for the bilinear form is given by the sum of #the stiffness and the mass B=Stiff+Mass # Now lets initialize the array in zeros to respect the homogeneous # Dirichlet boundary conditions U=np.zeros(len(p)) # Lets take just the interior points in the matrix B and the Load vector Bint=B[innod][:,innod] Loadint=Load[innod] # Now lets get the solution of the linear system of the interior points # using spsolve function Uint=spla.spsolve(Bint,Loadint) # Finally we put them in the correspondence interior points of the complete # solution U[innod]=Uint # We return [p,t,U] return p,t,U
def dirichlet_homogeneous(h0, f, n):
# First lets generate the uniform mesh for this mesh width
mesh = msh.grid_square(1, h0)
# Matrix of nodes
p = mesh[0]
# Matrix of triangles index
t = mesh[1]
# Matrix of border element index
be = mesh[2]
#Lets get the interior nodes in the mesh
innod = fem.interiorNodes(p, t, be)
# Rewrite the innod to match with indices beginning at 0
innod = [i - 1 for i in innod]
# Now lets compute the stiffness matrix
Stiff = fem.stiffness(p, t)
# Now lets compute the mass matrix
Mass = fem.mass(p, t)
# Lets compute the load vector
Load = fem.load(p, t, n, f)
# The complete matrix for the bilinear form is given by the sum of
#the stiffness and the mass
B = Stiff + Mass
# Now lets initialize the array in zeros to respect the homogeneous
# Dirichlet boundary conditions
U = np.zeros(len(p))
# Lets take just the interior points in the matrix B and the Load vector
Bint = B[innod][:, innod]
Loadint = Load[innod]
# Now lets get the solution of the linear system of the interior points
# using spsolve function
Uint = spla.spsolve(Bint, Loadint)
# Finally we put them in the correspondence interior points of the complete
# solution
U[innod] = Uint
# We return [p,t,U]
return p, t, U
# a) #Define the triangle with its nodes coordinates p=np.array([[1.,1.], [1.,2.], [2.,1.]]) # c) Lets define the source function by a lambda funcion f = lambda x1,x2: x1*x2 # 2 Assembling the elements # We gonna use the functions generated in the last homework to generate # the meshes to study as benchmark import meshes as msh a=1 h0=np.sqrt(2)/14 mesh=msh.grid_square(a,h0) p=mesh[0] t=mesh[1] be=mesh[2] K=0 nodes=np.array([p[i-1] for i in t[K]]) (T(t,K).transpose().dot(elemStiffness(nodes))).dot(T(t,K).toarray())
import FEM as fem
import meshes as msh
import numpy as np
import scipy as sp
import scipy.sparse as sparse
import scipy.sparse.linalg as spla
#We create the mesh
h0 = np.sqrt(2) / 5
mesh = msh.grid_square(1, h0)
# Matrix of nodes
pi = mesh[0]
# Matrix of triangles index
t = mesh[1]
# Matrix of boudary elements vertices
be = mesh[2]
# Lets get the coordinates of a triangle
K = 0
p = np.array([pi[i - 1] for i in t[K]])
#Lets define first the function f
f = lambda x1, x2: (8 * (np.pi**2) + 1) * np.cos(2 * np.pi * x1) * np.cos(
2 * np.pi * x2)
# The closest value of h0 to 1 to be able to generate a regular
h0 = np.sqrt(2) / 14
n = 3
# Lets get another time the mesh
h0 = np.sqrt(2) / 10
mesh = msh.grid_square(1, h0)
import scipy as sp
import scipy.sparse as sparse
import scipy.sparse.linalg as spla
# To get a solution u(x,y)=cos(pi x)*cos(pi y) the source function must
# be f(x,y)=(2 pi^2)cos(pi x)*cos(pi y)
#lets define the source function
f= lambda x1,x2: (2*np.pi**2)*np.cos(np.pi*x1)*np.cos(np.pi*x2)
# The order of the quadrature will be 3
n=3
# The closest value of h0 to 0.1 to be able to generate a regular
h0=np.sqrt(2)/10
# We create the mesh
mesh=msh.grid_square(1,h0)
# Matrix of nodes
pi=mesh[0]
# Matrix of triangles index
t=mesh[1]
# Matrix of boudary elements vertices
be=mesh[2]
index=be[39]
p=pi[index-1]
g= lambda x1,x2: np.cos(np.pi*x1)*np.cos(np.pi*x2)
sum(map(lambda K:(TB(be,p,K).transpose()
.dot(elemLoadNeumann(np.array([p[i-1] for i in be[K]]),n,g))),range(0,len(be))))
import meshes import FEM import time import scipy.sparse.linalg as spla meshwidth = 0.05 processes = 8 p,t=meshes.grid_square(1,meshwidth) timer=time.time() seqM=FEM.mass(p,t) seqTime=time.time()-timer timer=time.time() parM,processes=FEM.massParallel(p,t,processes) parTime=time.time()-timer print "difference of matrices: ", spla.onenormest(seqM-parM) print "time for sequential assembling:", seqTime print "time for parallel assembling: ", parTime print "speed-up: ", seqTime/parTime print "efficiency:", (seqTime/parTime)/processes
file = 'square_mesh.msh' mesh1 = msh.read_gmsh(file) p1 = mesh1[0] t1 = mesh1[1] #Name of the file of the output plot file = 'unstructured.png' title = 'Unstructured mesh' msh.show(p1, t1, file, title) # Get the p and t arrays for the regular mesh of the square of side 1 # For structured meshes the value of h0 is not arbitrary, as it has to fit # a integer number of 1-dimensional elements in each border, the closest # h0 to 0.01 that fullfill that is h0=np.sqrt(2)/14=0.10101.... a = 1 h0 = np.sqrt(2) / 14 mesh2 = msh.grid_square(a, h0) p2 = mesh2[0] t2 = mesh2[1] #Name of the file of the output plot file = 'regular.png' title = 'Regular mesh' msh.show(p2, t2, file, title) # Lets plot the number of nodes depending on maximal mesh width for the # unstructured mesh generated by gmsh, all of them are meshes for a square of side length a=1, and # the value of maximal width will be varied in gmsh in 6 different values: # Lets put in an array yu the number of nodes for each h0 # and xu for the actual mesh width calculated with max_mesh_width xu = []