In [3]:
%matplotlib inline
import numpy as np
from matplotlib import pyplot as plt
plt.style.use('bmh')
from numba import jit, float64
from numba.types import UniTuple
from time import time
In [4]:
# Implementation of Eq. (1) in the exam set
def doublegyre(x, y, t, A, e, w):
a = e * np.sin(w*t)
b = 1 - 2*e*np.sin(w*t)
f = a*x**2 + b*x
return np.array([
-np.pi*A*np.sin(np.pi*f) * np.cos(np.pi*y), # x component of velocity
np.pi*A*np.cos(np.pi*f) * np.sin(np.pi*y) * (2*a*x + b) # y component of velocity
])
# Wrapper function to pass to integrator
# X0 is a two-component vector [x, y]
def f(X, t):
# Parameters of the velocity field
A = 0.1
e = 0.25 # epsilon
w = 1 # omega
return doublegyre(X[0], X[1], t, A, e, w)
# Forward Euler integrator
# X0 is a two-component vector [x, y]
def euler(X, t, dt, f):
k1 = f(X, t)
return X + dt*k1
# 4th order Runge-Kutta integrator
# X0 is a two-component vector [x, y]
def rk4(X, t, dt, f):
k1 = f(X, t)
k2 = f(X + k1*dt/2, t + dt/2)
k3 = f(X + k2*dt/2, t + dt/2)
k4 = f(X + k3*dt, t + dt)
return X + dt*(k1 + 2*k2 + 2*k3 + k4) / 6
# Function to calculate a trajectory from an
# initial position X0 at t = 0, moving forward
# until t = tmax, using the given timestep and
# integrator
def trajectory(X0, tmax, dt, integrator, f):
t = 0
# Number of timesteps
Nx = int(tmax / dt)
# Array to hold the entire trajectory
PX = np.zeros((2, Nx+1))
# Initial position
PX[:,0] = X0
# Loop over all timesteps
for i in range(1, Nx+1):
PX[:,i] = integrator(PX[:,i-1], t, dt, f)
t += dt
# Return entire trajectory
return PX
In [30]:
def grid_of_particles(N, w):
# Create a grid of N evenly spaced particles
# covering a square patch of width and height w
# centered on the region 0 < x < 2, 0 < y < 1
x = np.linspace(1.0-w/2, 1.0+w/2, int(np.sqrt(N)))
y = np.linspace(0.5-w/2, 0.5+w/2, int(np.sqrt(N)))
x, y = np.meshgrid(x, y)
return np.array([x.flatten(), y.flatten()])
X = grid_of_particles(50, 0.1)
# Make a plot to confirm that this works as expected
fig = plt.figure(figsize = (12,6))
plt.scatter(X[0,:], X[1,:], lw = 0, marker = '.', s = 1)
plt.xlim(0, 2)
plt.ylim(0, 1)
Out[30]:
In [31]:
X
Out[31]:
In [18]:
N = 10000
X0 = grid_of_particles(N, w = 0.1)
# Array to hold all grid points after transport
X1 = np.zeros((2, N))
# Transport parameters
tmax = 5.0
dt = 0.5
# Loop over grid and update all positions
# This is where parallelisation would happen, since
# each position is independent of all the others
tic = time()
for i in range(N):
# Keep only the last position, not the entire trajectory
X1[:,i] = trajectory(X0[:,i], tmax, dt, rk4, f)[:,-1]
toc = time()
print('Transport took %.3f seconds' % (toc - tic))
# Make scatter plot to show all grid points
fig = plt.figure(figsize = (12,6))
plt.scatter(X1[0,:], X1[1,:], lw = 0, marker = '.', s = 1)
plt.xlim(0, 2)
plt.ylim(0, 1)
Out[18]:
In [21]:
# Implementation of Eq. (1) in the exam set
def doublegyre(x, y, t, A, e, w):
a = e * np.sin(w*t)
b = 1 - 2*e*np.sin(w*t)
f = a*x**2 + b*x
return np.array([
-np.pi*A*np.sin(np.pi*f) * np.cos(np.pi*y), # x component of velocity
np.pi*A*np.cos(np.pi*f) * np.sin(np.pi*y) * (2*a*x + b) # y component of velocity
])
# Wrapper function to pass to integrator
# X0 is a two-component vector [x, y]
def f(X, t):
# Parameters of the velocity field
A = 0.1
e = 0.25 # epsilon
w = 1 # omega
return doublegyre(X[0,:], X[1,:], t, A, e, w)
# 4th order Runge-Kutta integrator
# X0 is a two-component vector [x, y]
def rk4(X, t, dt, f):
k1 = f(X, t)
k2 = f(X + k1*dt/2, t + dt/2)
k3 = f(X + k2*dt/2, t + dt/2)
k4 = f(X + k3*dt, t + dt)
return X + dt*(k1 + 2*k2 + 2*k3 + k4) / 6
# Function to calculate a trajectory from an
# initial position X0 at t = 0, moving forward
# until t = tmax, using the given timestep and
# integrator
def trajectory(X0, tmax, dt, integrator, f):
t = 0
# Number of timesteps
Nt = int(tmax / dt)
# Array to hold the entire trajectory
PX = np.zeros((*X0.shape, Nt+1))
# Initial position
PX[:,:,0] = X0
# Loop over all timesteps
for i in range(1, Nt+1):
PX[:,:,i] = integrator(PX[:,:,i-1], t, dt, f)
t += dt
# Return entire trajectory
return PX
In [22]:
N = 10000
X0 = grid_of_particles(N, w = 0.1)
# Array to hold all grid points after transport
X1 = np.zeros((2, N))
# Transport parameters
tmax = 5.0
dt = 0.5
# Loop over grid and update all positions
# This is where parallelisation would happen, since
# each position is independent of all the others
tic = time()
# Keep only the last position, not the entire trajectory
X1 = trajectory(X0, tmax, dt, rk4, f)[:,:,-1]
toc = time()
print('Transport took %.3f seconds' % (toc - tic))
# Make scatter plot to show all grid points
fig = plt.figure(figsize = (12,6))
plt.scatter(X1[0,:], X1[1,:], lw = 0, marker = '.', s = 1)
plt.xlim(0, 2)
plt.ylim(0, 1)
Out[22]:
In [124]:
# Implementation of Eq. (1) in the exam set
def doublegyre(x, y, t, A, e, w):
a = e * np.sin(w*t)
b = 1 - 2*e*np.sin(w*t)
f = a*x**2 + b*x
return np.array([
-np.pi*A*np.sin(np.pi*f) * np.cos(np.pi*y), # x component of velocity
np.pi*A*np.cos(np.pi*f) * np.sin(np.pi*y) * (2*a*x + b) # y component of velocity
])
# Wrapper function to pass to integrator
# X0 is a two-component vector [x, y]
def f(X, t):
# Parameters of the velocity field
A = 0.1
e = 0.25 # epsilon
w = 1 # omega
return doublegyre(X[0,:], X[1,:], t, A, e, w)
# 4th order Runge-Kutta integrator
# X0 is a two-component vector [x, y]
def rk4(X, t, dt, f):
k1 = f(X, t)
k2 = f(X + k1*dt/2, t + dt/2)
k3 = f(X + k2*dt/2, t + dt/2)
k4 = f(X + k3*dt, t + dt)
return X + dt*(k1 + 2*k2 + 2*k3 + k4) / 6
# Function to calculate a trajectory from an
# initial position X0 at t = 0, moving forward
# until t = tmax, using the given timestep and
# integrator
def trajectory(X, tmax, dt, integrator, f):
t = 0
# Number of timesteps
Nt = int(tmax / dt)
# Loop over all timesteps
for i in range(1, Nt+1):
X = integrator(X, t, dt, f)
t += dt
# Return entire trajectory
return X
In [125]:
N = 10000
X0 = grid_of_particles(N, w = 0.1)
# Array to hold all grid points after transport
X1 = np.zeros((2, N))
# Transport parameters
tmax = 5.0
dt = 0.5
# Loop over grid and update all positions
# This is where parallelisation would happen, since
# each position is independent of all the others
tic = time()
# Keep only the last position, not the entire trajectory
X1 = trajectory(X0, tmax, dt, rk4, f)
toc = time()
print('Transport took %.3f seconds' % (toc - tic))
# Make scatter plot to show all grid points
fig = plt.figure(figsize = (12,6))
plt.scatter(X1[0,:], X1[1,:], lw = 0, marker = '.', s = 1)
plt.xlim(0, 2)
plt.ylim(0, 1)
Out[125]:
In [22]:
# Implementation of Eq. (1) in the exam set
@jit(UniTuple(float64[:], 2)(float64[:], float64[:], float64, float64, float64, float64), nopython = True)
def doublegyre(x, y, t, A, e, w):
a = e * np.sin(w*t)
b = 1 - 2*e*np.sin(w*t)
f = a*x**2 + b*x
v = np.zeros((2, x.size))
return -np.pi*A*np.sin(np.pi*f) * np.cos(np.pi*y), np.pi*A*np.cos(np.pi*f) * np.sin(np.pi*y) * (2*a*x + b)
# Wrapper function to pass to integrator
# X0 is a two-component vector [x, y]
@jit(nopython = True)
def f(X, t):
# Parameters of the velocity field
A = np.float64(0.1)
e = np.float64(0.25) # epsilon
w = np.float64(1.0) # omega
v = np.zeros(X.shape)
v[0,:], v[1,:] = doublegyre(X[0,:], X[1,:], t, A, e, w)
return v
# 4th order Runge-Kutta integrator
# X0 is a two-component vector [x, y]
@jit(nopython = True)
def rk4(X, t, dt):
k1 = f(X, t)
k2 = f(X + k1*dt/2, t + dt/2)
k3 = f(X + k2*dt/2, t + dt/2)
k4 = f(X + k3*dt, t + dt)
return X + dt*(k1 + 2*k2 + 2*k3 + k4) / 6
# Function to calculate a trajectory from an
# initial position X0 at t = 0, moving forward
# until t = tmax, using the given timestep and
# integrator
@jit(nopython = True)
def trajectory(X, tmax, dt):
t = 0
# Number of timesteps
Nt = int(tmax / dt)
# Loop over all timesteps
for i in range(1, Nt+1):
X = rk4(X, t, dt)
t += dt
# Return entire trajectory
return X
In [126]:
N = 10000
X0 = grid_of_particles(N, w = 0.1)
# Array to hold all grid points after transport
X1 = np.zeros((2, N))
# Transport parameters
tmax = 5.0
dt = 0.5
# Loop over grid and update all positions
# This is where parallelisation would happen, since
# each position is independent of all the others
tic = time()
# Keep only the last position, not the entire trajectory
X1 = endpoints(X0[:,:], tmax, dt)
toc = time()
print('Transport took %.3f seconds' % (toc - tic))
# Make scatter plot to show all grid points
fig = plt.figure(figsize = (12,6))
plt.scatter(X1[0,:], X1[1,:], lw = 0, marker = '.', s = 1)
plt.xlim(0, 2)
plt.ylim(0, 1)
Out[126]: