def random_walk_2D(np, ns, plot_step):
xpositions = zeros(np)
ypositions = zeros(np)
# extent of the axis in the plot:
ymin = -4
ymax = 8 * sqrt(ns)
xmax = 3 * sqrt(ns)
xmin = -xmax
for step in range(ns):
for i in range(np):
r = random.randint(0, 10)
if 0 <= r <= 5:
ypositions[i] += 1
elif r == 6:
ypositions[i] -= 1
elif 7 <= r <= 8:
xpositions[i] += 1
elif 9 <= r <= 10:
xpositions[i] -= 1
# plot just every plot_step steps:
if (step + 1) % plot_step == 0:
plot(
xpositions,
ypositions,
"ko",
axis=[xmin, xmax, ymin, ymax],
title="%d researchers after %d months, with ownership of strategy" % (np, step + 1),
hardcopy="tmp_%03d.eps" % (step + 1),
)
return xpositions, ypositions
def random_walk_2D(np, ns, plot_step):
xpositions = zeros(np)
ypositions = zeros(np)
# extent of the axis in the plot:
ymin = -4
ymax = 8 * sqrt(ns)
xmax = 3 * sqrt(ns)
xmin = -xmax
for step in range(ns):
for i in range(np):
r = random.randint(0, 10)
if 0 <= r <= 5:
ypositions[i] += 1
elif r == 6:
ypositions[i] -= 1
elif 7 <= r <= 8:
xpositions[i] += 1
elif 9 <= r <= 10:
xpositions[i] -= 1
# plot just every plot_step steps:
if (step + 1) % plot_step == 0:
plot(xpositions, ypositions, 'ko',
axis=[xmin, xmax, ymin, ymax],
title='%d researchers after %d months, with ownership of strategy' % \
(np, step+1),
hardcopy='tmp_%03d.eps' % (step+1))
return xpositions, ypositions
def d(n):
from scitools.std import zeros, sqrt
arr = zeros(n)
cum_sum = 6 / sqrt(3)
for i in range(n):
k = i + 1
cum_sum += 6 / sqrt(3) * (-1)**k / (3**k * (2 * k + 1))
arr[i] = cum_sum
return arr
def random_walk_2D(np, ns, plot_step):
xpositions = zeros(np)
ypositions = zeros(np)
# extent of the axis in the plot:
xymax = 3 * sqrt(ns)
xymin = -xymax
NORTH = 1
SOUTH = 2
WEST = 3
EAST = 4 # constants
for step in range(ns):
for i in range(np):
direction = random_number.randint(1, 4)
if direction == NORTH:
ypositions[i] += 1
elif direction == SOUTH:
ypositions[i] -= 1
elif direction == EAST:
xpositions[i] += 1
elif direction == WEST:
xpositions[i] -= 1
# plot just every plot_step steps:
if (step + 1) % plot_step == 0:
plot(xpositions, ypositions, 'ko',
axis=[xymin, xymax, xymin, xymax],
title='%d particles after %d steps' % \
(np, step+1),
hardcopy='tmp_%03d.eps' % (step+1))
return xpositions, ypositions
def random_walk_2D(np, ns, plot_step):
xpositions = zeros(np)
ypositions = zeros(np)
# extent of the axis in the plot:
xymax = 3*sqrt(ns); xymin = -xymax
NORTH = 1; SOUTH = 2; WEST = 3; EAST = 4 # constants
for step in range(ns):
for i in range(np):
direction = random_number.randint(1, 4)
if direction == NORTH:
ypositions[i] += 1
elif direction == SOUTH:
ypositions[i] -= 1
elif direction == EAST:
xpositions[i] += 1
elif direction == WEST:
xpositions[i] -= 1
# plot just every plot_step steps:
if (step+1) % plot_step == 0:
plot(xpositions, ypositions, 'ko',
axis=[xymin, xymax, xymin, xymax],
title='%d particles after %d steps' % \
(np, step+1),
hardcopy='tmp_%03d.eps' % (step+1))
return xpositions, ypositions
def b(n):
from scitools.std import sqrt, zeros
arr = zeros(n)
cum_sum = 0
for i in range(n):
k = i + 1
cum_sum = sqrt(cum_sum**2 + 6 * k**(-2))
arr[i] = cum_sum
return arr
def test_terminalVelocity():
"""
Test for the terminal velocity with no source term.
"""
import nose.tools as nt
T = 30.; dt = 0.1; g = 9.81; m = 50.;
Cd = 1.2; rho = 1.0; A = 0.5;
a = Cd*rho*A/(2.*m)
v, t = solver(T, dt, -0.1, Cd, rho, A, m)
nt.assert_almost_equal(v[-1], -sqrt(g/a), delta=1e-4)
def Temperature_and_Pressure():
meanT = 0
meanP = 0
mean_squareT = 0
mean_squareP = 0
# begin to sample after 200 timesteps:
N = len(T) - 200
for j in range(N):
i = j + 200
meanT += T[i]
meanP += P[i]
mean_squareT += T[i] * T[i]
mean_squareP += P[i] * P[i]
meanT = meanT / N
meanP = meanP / N
sT = sqrt(mean_squareT / N - meanT * meanT)
sP = sqrt(mean_squareP / N - meanP * meanP)
return meanT, meanP, sT, sP
def pathlength(x, y):
''' Computes the path length of the points (x0,y0), (x1,y1), ..., (xn,yn) x is a list of real numbers
y is a list of real numbers
len(x) and len(y) must be equal
'''
from scitools.std import sqrt
sum = 0
n = len(x)
for i in range(1, n):
sum += sqrt((x[i] - x[i - 1])**2 + (y[i] - y[i - 1])**2)
return sum
def Temperature_and_Pressure():
meanT = 0
meanP = 0
mean_squareT = 0
mean_squareP = 0
# begin to sample after 200 timesteps:
N = len(T) -200
for j in range(N):
i = j+200
meanT += T[i]
meanP += P[i]
mean_squareT += T[i]*T[i]
mean_squareP += P[i]*P[i]
meanT = meanT/N
meanP = meanP/N
sT = sqrt(mean_squareT/N - meanT*meanT)
sP = sqrt(mean_squareP/N - meanP*meanP)
return meanT,meanP,sT,sP
def c(lambda_): ''' Computes the wave speed of water surface waves depending on the length lambda of the waves lambda_ must be in meters ''' g = 9.81 # m/s**2 s = 7.9*10**-2 # N/m, air-water surface tension rho = 1*10**6 # kg/m**3, density of water h = 50 # m, the water depth, which we have fixed from scitools.std import pi, tanh, sqrt return sqrt(g*lambda_/(2*pi)*(1 + s*4*pi**2/(rho*g*lambda_**2)) \ * tanh(2*pi*h/lambda_))
def c(lambda_):
'''
Computes the wave speed of water surface waves
depending on the length lambda of the waves
lambda_ must be in meters
'''
g = 9.81 # m/s**2
s = 7.9 * 10**-2 # N/m, air-water surface tension
rho = 1 * 10**6 # kg/m**3, density of water
h = 50 # m, the water depth, which we have fixed
from scitools.std import pi, tanh, sqrt
return sqrt(g*lambda_/(2*pi)*(1 + s*4*pi**2/(rho*g*lambda_**2)) \
* tanh(2*pi*h/lambda_))
def test_terminalVelocity():
"""
Test for the terminal velocity with no source term.
"""
import nose.tools as nt
T = 30.
dt = 0.1
g = 9.81
m = 50.
Cd = 1.2
rho = 1.0
A = 0.5
a = Cd * rho * A / (2. * m)
v, t = solver(T, dt, -0.1, Cd, rho, A, m)
nt.assert_almost_equal(v[-1], -sqrt(g / a), delta=1e-4)
def test_convergenceRates():
"""
Test for the convergence rates of the solver.
The expected result is that the variable r takes the value 2, because
the Crank-Nicolson scheme and the geometric average have errors of order
dt**2. The final error should then be O(dt**2) which gives r=2.
"""
dt_start = 1.0
num_dt = 10
E_values = zeros(num_dt)
T = 10.
g = 9.81
m = 50.
Cd = 1.2
rho = 1.0
A = 0.5
a = Cd * rho * A / (2. * m)
dt = zeros(num_dt)
dt[0] = dt_start
for i in range(1, len(dt)):
dt[i] = dt[i - 1] / 2.
D = -0.39
B = 0.76
C = -0.145
def exact(t):
return D * t**3 + B * t + C
def src(t):
return m * g + m * a * abs(
exact(t)) * exact(t) + m * (3 * D * t**2 + B)
# Simulate for different timesteps, and store the error in E_values
for i in range(num_dt):
v, t = solver(T, dt[i], exact(0), Cd, rho, A, m, src)
ve = exact(t)
diff = v - ve
E_values[i] = sqrt(dt[i] * sum(diff**2))
# Calculate r-values corresponding to the error with each different timestep
r = zeros(len(E_values) - 1)
for i in range(1, len(r)):
r[i] = (log(E_values[i - 1] / E_values[i])) / (log(dt[i - 1] / dt[i]))
import nose.tools as nt
nt.assert_almost_equal(r[-1], 2, delta=0.1)
def finite_element_Picard(L, Nx, N, dt, C, P_L, P_R):
# Define mesh
mesh = Interval(Nx, 0, L)
#Define function space and functions
V = FunctionSpace(mesh, 'Lagrange', 1)
v = TestFunction(V)
q = TrialFunction(V)
q_k = Function(V)
q_1 = Function(V)
# Define boundary conditions
bcl = DirichletBC(V, Constant(P_L**2), LeftDiricletBoundary)
bcr = DirichletBC(V, Constant(P_R**2), RightDiricletBoundary)
bc = [bcl, bcr]
# Initial condition
I_q = Constant(P_R**2)
q_1 = interpolate(I_q, V)
q_k.assign(q_1)
c = Constant(C)
# Define variational problems
a = q*v*dx + dt*c*q_k**1.5*inner(nabla_grad(q), nabla_grad(v))*dx
L = q_1*v*dx
# Time loop
q = Function(V) # new unknown function
tol = 1.0E-5 # tolerance
maxiter = 25 # max no of iterations allowed
t = dt
for n in range(0, N):
#print 'time =', t
eps = 1.0 # error measure ||u-u_k||
iter = 0 # iteration counter
b = assemble(L)
while eps > tol and iter < maxiter:
iter += 1
A = assemble(a)
bcl.apply(A, b)
bcr.apply(A, b)
solve(A, q.vector(), b,'lu')
diff = q.vector().array() - q_k.vector().array()
eps = numpy.linalg.norm(diff, ord=numpy.Inf)
q_k.assign(q)
#print 'iter=%d: norm=%g' % (iter, eps)
t += dt
q_1.assign(q_k)
# return last P
return sqrt(q_k.vector().array())
def finite_difference(L, Nx, N, dt, C, P_L, P_R):
x = linspace(0, L, Nx+1)
dx = x[1] - x[0]
C = 0.4*C*dt/(dx**2)
print C
Q = zeros(Nx+1)
Q_1 = P_R**2*ones(Nx+1)
for n in range(0, N):
# Compute u at inner mesh points
for i in range(1, Nx):
Q[i] = Q_1[i] + C*(Q_1[i-1]**2.5 - 2*Q_1[i]**2.5 + Q_1[i+1]**2.5)
# Insert boundary conditions
Q[0]=P_L**2; Q[Nx]=P_R**2
# Update u_1 before next step
Q_1[:]= Q
return sqrt(Q), x
def test_convergenceRates():
"""
Test for the convergence rates of the solver.
The expected result is that the variable r takes the value 2, because
the Crank-Nicolson scheme and the geometric average have errors of order
dt**2. The final error should then be O(dt**2) which gives r=2.
"""
dt_start = 1.0; num_dt = 10
E_values = zeros(num_dt)
T = 10.; g = 9.81; m = 50.; Cd = 1.2; rho = 1.0; A = 0.5;
a = Cd*rho*A/(2.*m)
dt = zeros(num_dt); dt[0] = dt_start
for i in range(1,len(dt)):
dt[i] = dt[i-1]/2.
D = -0.39; B = 0.76; C = -0.145
def exact(t):
return D*t**3 + B*t + C
def src(t):
return m*g + m*a*abs(exact(t))*exact(t) + m*(3*D*t**2 + B)
# Simulate for different timesteps, and store the error in E_values
for i in range(num_dt):
v, t = solver(T, dt[i], exact(0), Cd, rho, A, m, src)
ve = exact(t)
diff = v - ve
E_values[i] = sqrt(dt[i]*sum(diff**2))
# Calculate r-values corresponding to the error with each different timestep
r=zeros(len(E_values)-1)
for i in range(1, len(r)):
r[i] = (log(E_values[i-1]/E_values[i]))/(log(dt[i-1]/dt[i]))
import nose.tools as nt
nt.assert_almost_equal(r[-1], 2, delta=0.1)
counter += 1
def T(z, t):
# T0, A, k, and omega are global variables
return T0 + A1 * exp(-a1 * z) * cos(omega1 * t - a1 * z) + \
A2 * exp(-a2 * z) * cos(omega2 * t - a2 * z)
k = 1E-6 # thermal diffusivity (in m**2/s)
A1 = 15 # amplitude of the daily temperature variations (in C)
P1 = 24 * 60 * 60. # oscillation period of 24 h (in seconds)
# angular frequency of daily temperature variations (in rad/s)
omega1 = 2 * pi / P1
a1 = sqrt(omega1 / (2 * k))
A2 = 7 # amplitude of yearly temperature variations (in C)
P2 = 24 * 60 * 60 * 365. # oscillation period of 1 yr (in seconds)
# angular frequency of yearly temperature variations (in rad/s)
omega2 = 2 * pi / P2
a2 = sqrt(omega2 / (2 * k))
dt = P2 / 30 # time lag: 0.1 yr
tmax = 3 * P2 # 3 year simulation
T0 = 10 # mean surface temperature in Celsius
D = -(1 / a1) * log(0.001) # max depth
n = 501 # no of points in the z direction
z = linspace(0, D, n)
def test_undampedWaves():
#define constants given in exercise
A = 1
mx=7.
my=2.
#define function that give
q=lambda x,y: 1
#define some variables
Lx = 3
Ly = 1.3
T = 1
C = 0.5
dt= 0.1
#define omega so equation holds
w=pi*sqrt((mx/Lx)**2 +(my/Ly)**2)
#help varabeles
kx = pi*mx/Lx
ky = pi*my/Ly
#Exact solution
ue = lambda x,y,t: A*cos(x*kx)*cos(y*ky)*cos(t*w)
#initial condition so we get result we want.
I = lambda x,y: A*cos(x*kx)*cos(y*ky)
#factor dt decreeses per step
step=0.5
#number of steps I want to do
val=5
#array to store errors
E=zeros(val)
for i in range(val):
v='vector'
#solve eqation
u,x,y,t,e=solver(I,None,None,q,0,Lx,Ly,dt*step**(i),T,C,1,mode=v,ue=ue)
E[i]=e
#find convergence rate between diffrent dt values
r =zeros(val-1)
r = log(E[1:]/E[:-1])/log(step)
print "Test convergence for undamped waves:"
for i in range(val):
if i==0:
print "dt: ",dt," Error: ",E[i]
else:
print "dt: ",dt*step**(i)," Error: ",E[i], "rate of con.: ", r[i-1]
#requiere close to 2 in convergence rate for last r.
assert abs(r[-1]-2)<0.01
def f(x, m, s):
return (1.0/(sqrt(2*pi)*s))*exp(-0.5*((x-m)/s)**2)
a = 1 #set parameters for the orbit
b = 1
w = 1
delta_t = (2 * pi) / (w * n)
counter = 0
X = []
Y = []
for k in range(n + 1):
tk = k * delta_t #each time through we need to add a new k value
x = a * cos(w * tk)
y = b * sin(w * tk)
X.append(x)
Y.append(y)
XP = array(X)
YP = array(Y)
XF = array([x])
YF = array([y])
velo = w * sqrt(a**2 * sin(w * tk)**2 + b**2 * cos(w * tk)**2)
plot(XP,
YP,
'-r',
XF,
YF,
'bo',
axis=[-1.2, 1.2, -1.2, 1.2],
title='Planetary orbit',
legend=(['Instaneous velocity=%6f' % velo, 'present location']),
savefig='pix/planet%4d.png' % counter)
counter += 1
show()
# from __future__ import unicode_literals
def test_undampedWaves():
#define constants given in exercise
A = 1
mx = 7.
my = 2.
#define function that give
q = lambda x, y: 1
#define some variables
Lx = 3
Ly = 1.3
T = 1
C = 0.5
dt = 0.1
#define omega so equation holds
w = pi * sqrt((mx / Lx)**2 + (my / Ly)**2)
#help varabeles
kx = pi * mx / Lx
ky = pi * my / Ly
#Exact solution
ue = lambda x, y, t: A * cos(x * kx) * cos(y * ky) * cos(t * w)
#initial condition so we get result we want.
I = lambda x, y: A * cos(x * kx) * cos(y * ky)
#factor dt decreeses per step
step = 0.5
#number of steps I want to do
val = 5
#array to store errors
E = zeros(val)
for i in range(val):
v = 'vector'
#solve eqation
u, x, y, t, e = solver(I,
None,
None,
q,
0,
Lx,
Ly,
dt * step**(i),
T,
C,
1,
mode=v,
ue=ue)
E[i] = e
#find convergence rate between diffrent dt values
r = zeros(val - 1)
r = log(E[1:] / E[:-1]) / log(step)
print "Test convergence for undamped waves:"
for i in range(val):
if i == 0:
print "dt: ", dt, " Error: ", E[i]
else:
print "dt: ", dt * step**(
i), " Error: ", E[i], "rate of con.: ", r[i - 1]
#requiere close to 2 in convergence rate for last r.
assert abs(r[-1] - 2) < 0.01
def pathlength(x, y):
L = 0
for i in xrange(1, len(x)):
dL_squared = (x[i] - x[i - 1]) ** 2 + (y[i] - y[i - 1]) ** 2
L += sqrt(dL_squared)
return L
show=False)
t += dt
counter += 1
def T(z, t):
# T0, A, k, and omega are global variables
return T0 + A1 * exp(-a1 * z) * cos(omega1 * t - a1 * z) + \
A2 * exp(-a2 * z) * cos(omega2 * t - a2 * z)
k = 1E-6 # thermal diffusivity (in m**2/s)
A1 = 15 # amplitude of the daily temperature variations (in C)
P1 = 24 * 60 * 60. # oscillation period of 24 h (in seconds)
omega1 = 2 * pi / P1 # angular freq of daily temp variations (in rad/s)
a1 = sqrt(omega1 / (2 * k))
A2 = 7 # amplitude of yearly temperature variations (in C)
P2 = 24 * 60 * 60 * 365. # oscillation period of 1 yr (in seconds)
omega2 = 2 * pi / P2 # angular freq of yearly temp variations (in rad/s)
a2 = sqrt(omega2 / (2 * k))
dt = P2 / 20 # time lag: 0.1 yr
tmax = 3 * P2 # 3 year simulation
T0 = 10 # mean surface temperature in Celsius
D = -(1 / a1) * log(0.001) # max depth
n = 501 # no of points in the z direction
# set T0, A, k, omega, D, n, tmax, dt
z = linspace(0, D, n)
animate(tmax, dt, z, T, T0 - A2 - A1, T0 + A2 + A1, 0, 'z', 'T')
def advance_scalar(u, u_1, u_2, q, f, x, y, t, n, A, B, dt2, dtdx2,dtdy2,
V=None, step1=False):
t1 = time.clock()
Ix = range(0, u.shape[0]); Iy = range(0, u.shape[1])
if step1: # Special formula for step 1
I = u_1; dt = sqrt(dt2)
for i in Ix[1:-1]:
for j in Iy[1:-1]:
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V(x[i],y[j]) + dt2*f(x[i], y[j], 0) \
+ dtdx2*( (q[i,j] + q[i+1,j]) * (I[i+1,j] - I[i,j]) \
- (q[i,j] + q[i-1,j]) * (I[i,j] - I[i-1,j])) \
+ dtdy2*( (q[i,j] + q[i,j+1]) * (I[i,j+1] - I[i,j]) \
- (q[i,j] + q[i,j-1]) * (I[i,j] - I[i,j-1])))
else: # Compute ALL interior points
for i in Ix[1:-1]:
for j in Iy[1:-1]:
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f(x[i], y[j], t[n]) \
+ dtdx2*( (q[i,j] + q[i+1,j]) * (u_1[i+1,j] - u_1[i,j]) \
- (q[i,j] + q[i-1,j]) * (u_1[i,j] - u_1[i-1,j])) \
+ dtdy2*( (q[i,j] + q[i,j+1]) * (u_1[i,j+1] - u_1[i,j]) \
- (q[i,j] + q[i,j-1]) * (u_1[i,j] - u_1[i,j-1])))
# Neumann boundary condition du/dx = 0
if step1:
i = Ix[0] # 1) Boundary where x = 0
for j in Iy[1:-1]:
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V(x[i],y[j]) + dt2*f(x[i], y[j], 0) \
+ dtdx2*2*(q[i,j] + q[i+1,j]) * (I[i+1,j] - I[i,j]) \
+ dtdy2*( (q[i,j] + q[i,j+1]) * (I[i,j+1] - I[i,j]) \
- (q[i,j] + q[i,j-1]) * (I[i,j] - I[i,j-1])))
i = Ix[-1] # 1) Boundary where x = Nx
for j in Iy[1:-1]:
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V(x[i],y[j]) + dt2*f(x[i], y[j], 0) \
+ dtdx2*2*(q[i,j] + q[i-1,j]) * (I[i-1,j] - I[i,j]) \
+ dtdy2*( (q[i,j] + q[i,j+1]) * (I[i,j+1] - I[i,j]) \
- (q[i,j] + q[i,j-1]) * (I[i,j] - I[i,j-1])))
j = Iy[0] # 1) Boundary where y = 0
for i in Ix[1:-1]:
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V(x[i],y[j]) + dt2*f(x[i], y[j], 0) \
+ dtdx2*( (q[i,j] + q[i+1,j]) * (I[i+1,j] - I[i,j]) \
- (q[i,j] + q[i-1,j]) * (I[i,j] - I[i-1,j])) \
+ dtdy2*2*(q[i,j] + q[i,j+1]) * (I[i,j+1] - I[i,j]))
j = Iy[-1] # 1) Boundary where y = Ny
for i in Ix[1:-1]:
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V(x[i],y[j]) + dt2*f(x[i], y[j], 0) \
+ dtdx2*( (q[i,j] + q[i+1,j]) * (I[i+1,j] - I[i,j]) \
- (q[i,j] + q[i-1,j]) * (I[i,j] - I[i-1,j])) \
+ dtdy2*2*(q[i,j] + q[i,j-1]) * (I[i,j-1] - I[i,j]))
# Special formula for the four corner points
i = Ix[0]; j = Iy[0]
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V(x[i],y[j]) + dt2*f(x[i], y[j], 0) \
+ dtdx2*2*(q[i,j] + q[i+1,j]) * (I[i+1,j] - I[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j+1]) * (I[i,j+1] - I[i,j]))
i = Ix[0]; j = Iy[-1]
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V(x[i],y[j]) + dt2*f(x[i], y[j], 0) \
+ dtdx2*2*(q[i,j] + q[i+1,j]) * (I[i+1,j] - I[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j-1]) * (I[i,j-1] - I[i,j]))
i = Ix[-1]; j = Iy[0]
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V(x[i],y[j]) + dt2*f(x[i], y[j], 0) \
+ dtdx2*2*(q[i,j] + q[i-1,j]) * (I[i-1,j] - I[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j+1]) * (I[i,j+1] - I[i,j]))
i = Ix[-1]; j = Iy[-1]
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V(x[i],y[j]) + dt2*f(x[i], y[j], 0) \
+ dtdx2*2*(q[i,j] + q[i-1,j]) * (I[i-1,j] - I[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j-1]) * (I[i,j-1] - I[i,j]))
else: # Any step NOT first
i = Ix[0] # 1) Boundary where x = 0
for j in Iy[1:-1]:
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f(x[i], y[j], t[n]) \
+ dtdx2*2*(q[i,j] + q[i+1,j]) * (u_1[i+1,j] - u_1[i,j]) \
+ dtdy2*( (q[i,j] + q[i,j+1]) * (u_1[i,j+1] - u_1[i,j]) \
- (q[i,j] + q[i,j-1]) * (u_1[i,j] - u_1[i,j-1])))
i = Ix[-1] # 1) Boundary where x = Nx
for j in Iy[1:-1]:
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f(x[i], y[j], t[n]) \
+ dtdx2*2*(q[i,j] + q[i-1,j]) * (u_1[i-1,j] - u_1[i,j]) \
+ dtdy2*( (q[i,j] + q[i,j+1]) * (u_1[i,j+1] - u_1[i,j]) \
- (q[i,j] + q[i,j-1]) * (u_1[i,j] - u_1[i,j-1])))
j = Iy[0] # 1) Boundary where y = 0
for i in Ix[1:-1]:
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f(x[i], y[j], t[n]) \
+ dtdx2*( (q[i,j] + q[i+1,j]) * (u_1[i+1,j] - u_1[i,j]) \
- (q[i,j] + q[i-1,j]) * (u_1[i,j] - u_1[i-1,j])) \
+ dtdy2*2*(q[i,j] + q[i,j+1]) * (u_1[i,j+1] - u_1[i,j]) )
j = Iy[-1] # 1) Boundary where y = Ny
for i in Ix[1:-1]:
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f(x[i], y[j], t[n]) \
+ dtdx2*( (q[i,j] + q[i+1,j]) * (u_1[i+1,j] - u_1[i,j]) \
- (q[i,j] + q[i-1,j]) * (u_1[i,j] - u_1[i-1,j])) \
+ dtdy2*2*(q[i,j] + q[i,j-1]) * (u_1[i,j-1] - u_1[i,j]) )
# Special formula for the four corner points
i = Ix[0]; j = Iy[0]
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f(x[i], y[j], t[n]) \
+ dtdx2*2*(q[i,j] + q[i+1,j]) * (u_1[i+1,j] - u_1[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j+1]) * (u_1[i,j+1] - u_1[i,j]))
i = Ix[0]; j = Iy[-1]
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f(x[i], y[j], t[n]) \
+ dtdx2*2*(q[i,j] + q[i+1,j]) * (u_1[i+1,j] - u_1[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j-1]) * (u_1[i,j-1] - u_1[i,j]))
i = Ix[-1]; j = Iy[0]
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f(x[i], y[j], t[n]) \
+ dtdx2*2*(q[i,j] + q[i-1,j]) * (u_1[i-1,j] - u_1[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j+1]) * (u_1[i,j+1] - u_1[i,j]))
i = Ix[-1]; j = Iy[-1]
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f(x[i], y[j], t[n]) \
+ dtdx2*2*(q[i,j] + q[i-1,j]) * (u_1[i-1,j] - u_1[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j-1]) * (u_1[i,j-1] - u_1[i,j]))
CPU_time = time.clock() - t1
return u, CPU_time
def f(x,m,s):
return (1.0/(sqrt(2*pi)*s))*exp(-0.5*((x-m)/s)**2)
"""
Exercise 5.3: Fill arrays; vectorized (plot)
Author: Weiyun Lu
"""
from scitools.std import sqrt, pi, exp, linspace, plot
h = lambda x : (1/(sqrt(2*pi))) * exp(-0.5*x**2)
xlist = linspace(-4,4,41)
hlist = h(xlist)
pairs = zip(xlist,hlist)
plot(xlist, hlist, axis=[xlist[0], xlist[-1], 0, 0.5], xlabel='x', \
ylabel='h(x)', title='Standard Gaussian')
N = 1600
h = 0.05
k = 0.2
P_L = 1
P_R = 0
dt = T/float(N);
DX = L/float(Nx)
C = k*h**2
c = C*dt/DX**2
# Solve by different methods
start = time.clock()
P_fd, x = finite_difference(L, Nx, N, dt, C, P_L, P_R)
print 'FD forward Euler:', time.clock() - start
P_ex = sqrt(1-x)
start = time.clock()
P_Pi = finite_element_Picard(L, Nx, N, dt, C, P_L, P_R)
print 'FE backward Euler Picard:', time.clock() - start
start = time.clock()
P_Ne = finite_element_Newton(L, Nx, N, dt, C, P_L, P_R)
print 'FE bacward Euler Newton:', time.clock() - start
figure(1)
pylab.plot(x, P_ex, x, P_fd, x, P_Pi, x, P_Ne)#, x, P_fe)
pylab.legend(['Exact', 'FD forward Euler', 'FE backward Euler Picard', 'FE backward Euler Newton'])
pylab.xlabel('x')
pylab.ylabel('Pressure')
def solver(I, V_, f_, c, Lx, Ly, Nx, Ny, dt, T, b,
user_action=None, version='scalar', skip_every_n_frame=10, show_cpu_time=False,display_warnings=True, plotting=False):
order = 'C' # Store arrays in a column-major order (in memory)
x = linspace(0, Lx, Nx+1) # mesh points in x dir
y = linspace(0, Ly, Ny+1) # mesh points in y dir
dx = x[1] - x[0]
dy = y[1] - y[0]
xv = x[:,newaxis] # for vectorized function evaluations
yv = y[newaxis,:]
# Assuming c is a function:
c_ = zeros((Nx+1,Ny+1), order='c')
for i,xx in enumerate(x):
for j,yy in enumerate(y):
c_[i,j] = c(xx,yy) # Loop through x and y with indices i,j at the same time
c_max = c_.max() # Pick out the largest value from c(x,y)
q = c_**2
stability_limit = (1/float(c_max))*(1/sqrt(1/dx**2 + 1/dy**2))
if dt <= 0: # shortcut for max time step is to use i.e. dt = -1
safety_factor = -dt # use negative dt as safety factor
extra_factor = 1 # Easy way to make dt even smaller
dt = safety_factor*stability_limit*extra_factor
elif dt > stability_limit and display_warnings:
print '\nWarning: (Unless you are testing the program), be aware that'
print 'dt: %g is currently exceeding the stability limit: %g\n' %(dt, stability_limit)
Nt = int(round(T/float(dt)))
t = linspace(0, Nt*dt, Nt+1) # mesh points in time
dt2 = dt**2
# Constants for simple calculation
A = (1 + b*dt/2)**(-1)
B = (b*dt/2 - 1)
dtdx2 = dt**2/(2*dx**2)
dtdy2 = dt**2/(2*dy**2)
# Make f(x,y,t) and V(x,y) ready for computation with different schemes
if f_ is None or f_ == 0:
f = (lambda x, y, t: 0) if version == 'scalar' else \
lambda x, y, t: zeros((xv.shape[0], yv.shape[1]))
else:
if version == 'scalar':
f = f_
if V_ is None or V_ == 0:
V = (lambda x, y: 0) if version == 'scalar' else \
lambda x, y: zeros((xv.shape[0], yv.shape[1]))
else:
if version == 'scalar':
V = V_
if version == 'vectorized': # Generate and fill matrices for first timestep
f = zeros((Nx+1,Ny+1), order=order)
V = zeros((Nx+1,Ny+1), order=order)
f[:,:] = f_(xv,yv,0)
V[:,:] = V_(xv,yv)
u = zeros((Nx+1,Ny+1), order=order) # solution array
u_1 = zeros((Nx+1,Ny+1), order=order) # solution at t-dt
u_2 = zeros((Nx+1,Ny+1), order=order) # solution at t-2*dt
Ix = range(0, u.shape[0]) # Index set notation
Iy = range(0, u.shape[1])
It = range(0, t.shape[0])
import time; t0 = time.clock() # for measuring CPU time
# Load initial condition into u_1
if version == 'scalar':
for i in Ix:
for j in Iy:
u_1[i,j] = I(x[i], y[j])
else: # use vectorized version
u_1[:,:] = I(xv, yv)
if user_action is not None:
if plotting:
user_action(u_1, x, xv, y, yv, t, 0, skip_every_n_frame)
else:
user_action(u_1, x, xv, y, yv, t, 0)
# Special formula for first time step
n = 0
# First step requires a special formula, use either the scalar
# or vectorized version (the impact of more efficient loops than
# in advance_vectorized is small as this is only one step)
if version == 'scalar':
u,cpu_time = advance_scalar(u, u_1, u_2, q, f, x, y, t, n, A, B,
dt2, dtdx2,dtdy2, V, step1=True)
else:
u,cpu_time = advance_vectorized(u, u_1, u_2, q, f, t, n, A, B,
dt2, dtdx2,dtdy2, V, step1=True)
if user_action is not None:
if plotting:
user_action(u, x, xv, y, yv, t, 1, skip_every_n_frame)
else:
user_action(u_1, x, xv, y, yv, t, 1)
# Update data structures for next step
u_2, u_1, u = u_1, u, u_2
# Time loop for all later steps
for n in It[1:-1]:
if version == 'scalar':
# use f(x,y,t) function
u,cpu_time = advance_scalar(u, u_1, u_2, q, f, x, y, t, n, A, B, dt2, dtdx2,dtdy2)
if show_cpu_time:
percent = (float(n)/It[-2])*100.0
sys.stdout.write("\rLast step took: %.3f sec with [scalar-code]. Computation is %d%% " %(cpu_time,percent))
sys.stdout.flush()
else: # Use vectorized code
f[:,:] = f_(xv, yv, t[n]) # must precompute the matrix f
u,cpu_time = advance_vectorized(u, u_1, u_2, q, f, t, n, A, B,
dt2, dtdx2,dtdy2)
if show_cpu_time:
percent = (float(n)/It[-2])*100.0
sys.stdout.write("\rLast step took: %.5f sec with [vec-code]. Computation is %d%% " %(cpu_time,percent))
sys.stdout.flush()
if user_action is not None:
if plotting:
if user_action(u, x, xv, y, yv, t, n+1, skip_every_n_frame):
break
else:
if user_action(u, x, xv, y, yv, t, n+1):
break
# Update data structures for next step
#u_2[:] = u_1; u_1[:] = u # safe, but slower
u_2, u_1, u = u_1, u, u_2
# Important to set u = u_1 if u is to be returned!
t1 = time.clock()
# dt might be computed in this function so return the value
return dt, t1 - t0
def pathlength(x, y):
L = 0
for i in xrange(1, len(x)):
dL_squared = (x[i] - x[i - 1])**2 + (y[i] - y[i - 1])**2
L += sqrt(dL_squared)
return L
def advance_vectorized(u, u_1, u_2, q, f, t, n, A, B,
dt2, dtdx2,dtdy2, V=None, step1=False):
""" Haakon (me) code """
t1 = time.clock()
Ix = range(0, u.shape[0]); Iy = range(0, u.shape[1])
if step1:
I = u_1; dt = sqrt(dt2)
u[1:-1,1:-1] = 0.5*(2*I[1:-1,1:-1] - 2*B*dt*V[1:-1,1:-1] + dt2*f[1:-1,1:-1] \
+ dtdx2*( (q[1:-1,1:-1] + q[2:,1:-1]) * (I[2:,1:-1] - I[1:-1,1:-1]) \
- (q[1:-1,1:-1] + q[:-2,1:-1]) * (I[1:-1,1:-1] - I[:-2,1:-1])) \
+ dtdy2*( (q[1:-1,1:-1] + q[1:-1,2:]) * (I[1:-1,2:] - I[1:-1,1:-1]) \
- (q[1:-1,1:-1] + q[1:-1,:-2]) * (I[1:-1,1:-1] - I[1:-1,:-2])))
else:
u[1:-1,1:-1] = A*( 2*u_1[1:-1,1:-1] + B*u_2[1:-1,1:-1] + dt2*f[1:-1,1:-1] \
+ dtdx2*( (q[1:-1,1:-1] + q[2:,1:-1]) * (u_1[2:,1:-1] - u_1[1:-1,1:-1]) \
- (q[1:-1,1:-1] + q[:-2,1:-1]) * (u_1[1:-1,1:-1] - u_1[:-2,1:-1])) \
+ dtdy2*( (q[1:-1,1:-1] + q[1:-1,2:]) * (u_1[1:-1,2:] - u_1[1:-1,1:-1]) \
- (q[1:-1,1:-1] + q[1:-1,:-2]) * (u_1[1:-1,1:-1] - u_1[1:-1,:-2])))
######################################
# Neumann boundary condition du/dn=0 #
######################################
if step1:
i = Ix[0] # 1) Boundary where x = 0
u[i,1:-1] = 0.5*(2*I[i,1:-1] - 2*B*dt*V[i,1:-1] + dt2*f[i,1:-1] \
+ dtdx2*2*(q[i,1:-1] + q[i+1,1:-1]) * (I[i+1,1:-1] - I[i,1:-1]) \
+ dtdy2*( (q[i,1:-1] + q[i,2:]) * (I[i,2:] - I[i,1:-1]) \
- (q[i,1:-1] + q[i,:-2]) * (I[i,1:-1] - I[i,:-2])))
i = Ix[-1] # 1) Boundary where x = Nx
u[i,1:-1] = 0.5*(2*I[i,1:-1] - 2*B*dt*V[i,1:-1] + dt2*f[i,1:-1] \
+ dtdx2*2*(q[i,1:-1] + q[i-1,1:-1]) * (I[i-1,1:-1] - I[i,1:-1]) \
+ dtdy2*( (q[i,1:-1] + q[i,2:]) * (I[i,2:] - I[i,1:-1]) \
- (q[i,1:-1] + q[i,:-2]) * (I[i,1:-1] - I[i,:-2])))
j = Iy[0] # 1) Boundary where y = 0
u[1:-1,j] = 0.5*(2*I[1:-1,j] - 2*B*dt*V[1:-1,j] + dt2*f[1:-1,j] \
+ dtdx2*( (q[1:-1,j] + q[2:,j]) * (I[2:,j] - I[1:-1,j]) \
- (q[1:-1,j] + q[:-2,j]) * (I[1:-1,j] - I[:-2,j])) \
+ dtdy2*2*(q[1:-1,j] + q[1:-1:,j+1]) * (I[1:-1:,j+1] - I[1:-1,j]))
j = Iy[-1] # 1) Boundary where y = Ny
u[1:-1,j] = 0.5*(2*I[1:-1,j] - 2*B*dt*V[1:-1,j] + dt2*f[1:-1,j] \
+ dtdx2*( (q[1:-1,j] + q[2:,j]) * (I[2:,j] - I[1:-1,j]) \
- (q[1:-1,j] + q[:-2,j]) * (I[1:-1,j] - I[:-2,j])) \
+ dtdy2*2*(q[1:-1,j] + q[1:-1:,j-1]) * (I[1:-1:,j-1] - I[1:-1,j]))
# Special formula for the four corner points
i = Ix[0]; j = Iy[0]
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V[i,j] + dt2*f[i,j] \
+ dtdx2*2*(q[i,j] + q[i+1,j]) * (I[i+1,j] - I[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j+1]) * (I[i,j+1] - I[i,j]))
i = Ix[0]; j = Iy[-1]
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V[i,j] + dt2*f[i,j] \
+ dtdx2*2*(q[i,j] + q[i+1,j]) * (I[i+1,j] - I[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j-1]) * (I[i,j-1] - I[i,j]))
i = Ix[-1]; j = Iy[0]
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V[i,j] + dt2*f[i,j] \
+ dtdx2*2*(q[i,j] + q[i-1,j]) * (I[i-1,j] - I[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j+1]) * (I[i,j+1] - I[i,j]))
i = Ix[-1]; j = Iy[-1]
u[i,j] = 0.5*(2*I[i,j] - 2*B*dt*V[i,j] + dt2*f[i,j] \
+ dtdx2*2*(q[i,j] + q[i-1,j]) * (I[i-1,j] - I[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j-1]) * (I[i,j-1] - I[i,j]))
else: # Any step NOT first
i = Ix[0] # 1) Boundary where x = 0
u[i,1:-1] = A*( 2*u_1[i,1:-1] + B*u_2[i,1:-1] + dt2*f[i,1:-1] \
+ dtdx2*2*(q[i,1:-1] + q[i+1,1:-1]) * (u_1[i+1,1:-1] - u_1[i,1:-1]) \
+ dtdy2*( (q[i,1:-1] + q[i,2:]) * (u_1[i,2:] - u_1[i,1:-1]) \
- (q[i,1:-1] + q[i,:-2]) * (u_1[i,1:-1] - u_1[i,:-2])))
i = Ix[-1] # 1) Boundary where x = Nx
u[i,1:-1] = A*( 2*u_1[i,1:-1] + B*u_2[i,1:-1] + dt2*f[i,1:-1] \
+ dtdx2*2*(q[i,1:-1] + q[i-1,1:-1]) * (u_1[i-1,1:-1] - u_1[i,1:-1]) \
+ dtdy2*( (q[i,1:-1] + q[i,2:]) * (u_1[i,2:] - u_1[i,1:-1]) \
- (q[i,1:-1] + q[i,:-2]) * (u_1[i,1:-1] - u_1[i,:-2])))
j = Iy[0] # 1) Boundary where y = 0
u[1:-1,j] = A*( 2*u_1[1:-1,j] + B*u_2[1:-1,j] + dt2*f[1:-1,j] \
+ dtdx2*( (q[1:-1,j] + q[2:,j]) * (u_1[2:,j] - u_1[1:-1,j]) \
- (q[1:-1,j] + q[:-2,j]) * (u_1[1:-1,j] - u_1[:-2,j])) \
+ dtdy2*2*(q[1:-1,j] + q[1:-1,j+1]) * (u_1[1:-1,j+1] - u_1[1:-1,j]) )
j = Iy[-1] # 1) Boundary where y = Ny
u[1:-1,j] = A*( 2*u_1[1:-1,j] + B*u_2[1:-1,j] + dt2*f[1:-1,j] \
+ dtdx2*( (q[1:-1,j] + q[2:,j]) * (u_1[2:,j] - u_1[1:-1,j]) \
- (q[1:-1,j] + q[:-2,j]) * (u_1[1:-1,j] - u_1[:-2,j])) \
+ dtdy2*2*(q[1:-1,j] + q[1:-1,j-1]) * (u_1[1:-1,j-1] - u_1[1:-1,j]) )
# Special formula for the four corner points
i = Ix[0]; j = Iy[0]
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f[i,j] \
+ dtdx2*2*(q[i,j] + q[i+1,j]) * (u_1[i+1,j] - u_1[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j+1]) * (u_1[i,j+1] - u_1[i,j]))
i = Ix[0]; j = Iy[-1]
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f[i,j] \
+ dtdx2*2*(q[i,j] + q[i+1,j]) * (u_1[i+1,j] - u_1[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j-1]) * (u_1[i,j-1] - u_1[i,j]))
i = Ix[-1]; j = Iy[0]
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f[i,j] \
+ dtdx2*2*(q[i,j] + q[i-1,j]) * (u_1[i-1,j] - u_1[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j+1]) * (u_1[i,j+1] - u_1[i,j]))
i = Ix[-1]; j = Iy[-1]
u[i,j] = A*( 2*u_1[i,j] + B*u_2[i,j] + dt2*f[i,j] \
+ dtdx2*2*(q[i,j] + q[i-1,j]) * (u_1[i-1,j] - u_1[i,j]) \
+ dtdy2*2*(q[i,j] + q[i,j-1]) * (u_1[i,j-1] - u_1[i,j]))
CPU_time = time.clock() - t1
return u,CPU_time