I have used the Equation of Motion (Newtons Law) for a simple spring and mass scenario incorporating it into the given 2nd ODE equation y" + (k/m)x = 0; y(0) = 3; y'(0) = 0.
Using the Euler method and the exact solution to solve the problem, I have been able to run and receive some ok results. However, when I execute a plot of the results I get this diagonal line across the oscillating results that I am after.
Current plot output with diagonal line
Can anyone help point out what is causing this issue, and how I can fix it please?
MY CODE:
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
from sympy import Function, dsolve, Eq, Derivative, sin, cos, symbols
from sympy.abc import x, i
import math
# Given is y" + (k/m)x = 0; y(0) = 3; y'(0) = 0
# Parameters
h = 0.01; #Step Size
t = 50.0; #Time(sec)
k = 1; #Spring Stiffness
m = 1; #Mass
x0 = 3;
v0 = 0;
# Exact Analytical Solution
x_exact = x0*cos(math.sqrt(k/m)*t);
v_exact = -x0*math.sqrt(k/m)*sin(math.sqrt(k/m)*t);
# Eulers Method
x = np.zeros( int( t/h ) );
v = np.zeros( int( t/h ) );
x[1] = x0;
v[1] = v0;
x_exact = np.zeros( int( t/h ) );
v_exact = np.zeros( int( t/h ) );
te = np.zeros( int( t/h ) );
x_exact[1] = x0;
v_exact[1] = v0;
#print(len(x));
for i in range(1, int(t/h) - 1): #MAIN LOOP
x[i+1] = x[i] + h*v[i];
v[i+1] = v[i] - h*k/m*x[i];
te[i] = i * h
x_exact[i] = x0*cos(math.sqrt(k/m)* te[i]);
v_exact[i] = -x0*math.sqrt(k/m)*sin(math.sqrt(k/m)* te[i]);
# print(x_exact[i], '\t'*2, x[i]);
#plot
%config InlineBackend.figure_format = 'svg'
plt.plot(te, x_exact, te ,v_exact)
plt.title("DISPLACEMENT")
plt.xlabel("Time (s)")
plt.ylabel("Displacement (m)")
plt.grid(linewidth=0.3)
An in some details more direct computation is
te = np.arange(0,t,h)
N = len(te)
w = (k/m)**0.5
x_exact = x0*np.cos(w*te);
v_exact = -x0*w*np.sin(w*te);
plt.plot(te, x_exact, te ,v_exact)
resulting in
Note that arrays in python start at the index zero,
x = np.empty(N)
v = np.empty(N)
x[0] = x0;
v[0] = v0;
for i in range(N - 1): #MAIN LOOP
x[i+1] = x[i] + h*v[i];
v[i+1] = v[i] - h*k/m*x[i];
plt.plot(te, x, te ,v)
then gives the plot
with the expected increasing amplitude.
Related
I am tring to solve the equation of motion of charged particle in planetary magnetic field to see the path of the particle using Forward Euler's and RK5 method in python (as an excercise in learning Numerical methods) I encounter two problems:
The 'for loop' in the RK4 method does not update the new values. It give the values of the first iteration for all iteration.
With the change of the sing of 'β = charge/mass' the path of particle which is expected does not change. It seems the path is unaffected by the nature(sign) of the particle. What does this mean physically or mathematically?
The codes are adapted from :
python two coupled second order ODEs Runge Kutta 4th order
and
Applying Forward Euler Method to a Three-Box Model System of ODEs
I would be immensely grateful if anyone explain to me what is wrong in the code.
thank you.
The Code are as under:
import numpy as np
import matplotlib.pyplot as plt
from math import sin, cos
from scipy.integrate import odeint
scales = np.array([1e7, 0.1, 1, 1e-5, 10, 1e-5])
def LzForce(t,p):
# assigning each ODE to a vector element
r,x,θ,y,ϕ,z = p*scales
# constants
R = 60268e3 # metre
g_20 = 1583e-9
Ω = 9.74e-3 # degree/second
B_θ = (R/r)**4*g_20*cos(θ)*sin(θ)
B_r = 2*(R/r)**4*g_20*(0.5*(3*cos(θ)**2-1))
β = +9.36e10
# defining the ODEs
drdt = x
dxdt = r*(y**2 +(z+Ω)**2*sin(θ)**2-β*z*sin(θ)*B_θ)
dθdt = y
dydt = (-2*x*y +r*(z+Ω)**2*sin(θ)*cos(θ)+β*r*z*sin(θ)*B_r)/r
dϕdt = z
dzdt = (-2*x*(z+Ω)*sin(θ)-2*r*y*(z+Ω)*cos(θ)+β*(x*B_θ-r*y*B_r))/(r*sin(θ))
return np.array([drdt,dxdt,dθdt,dydt,dϕdt,dzdt])/scales
def ForwardEuler(fun,t0,p0,tf,dt):
r0 = 6.6e+07
x0 = 0.
θ0 = 88.
y0 = 0.
ϕ0 = 0.
z0 = 22e-3
p0 = np.array([r0,x0,θ0,y0,ϕ0,z0])
t = np.arange(t0,tf+dt,dt)
p = np.zeros([len(t), len(p0)])
p[0] = p0
for i in range(len(t)-1):
p[i+1,:] = p[i,:] + fun(t[i],p[i,:]) * dt
return t, p
def rk4(fun,t0,p0,tf,dt):
# initial conditions
r0 = 6.6e+07
x0 = 0.
θ0 = 88.
y0 = 0.
ϕ0 = 0.
z0 = 22e-3
p0 = np.array([r0,x0,θ0,y0,ϕ0,z0])
t = np.arange(t0,tf+dt,dt)
p = np.zeros([len(t), len(p0)])
p[0] = p0
for i in range(len(t)-1):
k1 = dt * fun(t[i], p[i])
k2 = dt * fun(t[i] + 0.5*dt, p[i] + 0.5 * k1)
k3 = dt * fun(t[i] + 0.5*dt, p[i] + 0.5 * k2)
k4 = dt * fun(t[i] + dt, p[i] + k3)
p[i+1] = p[i] + (k1 + 2*(k2 + k3) + k4)/6
return t,p
dt = 0.5
tf = 1000
p0 = [6.6e+07,0.0,88.0,0.0,0.0,22e-3]
t0 = 0
#Solution with Forward Euler
t,p_Euler = ForwardEuler(LzForce,t0,p0,tf,dt)
#Solution with RK4
t ,p_RK4 = rk4(LzForce,t0, p0 ,tf,dt)
print(t,p_Euler)
print(t,p_RK4)
# Plot Solutions
r,x,θ,y,ϕ,z = p_Euler.T
fig,ax=plt.subplots(2,3,figsize=(8,4))
plt.xlabel('time in sec')
plt.ylabel('parameters')
for a,s in zip(ax.flatten(),[r,x,θ,y,ϕ,z]):
a.plot(t,s); a.grid()
plt.title("Forward Euler", loc='left')
plt.tight_layout(); plt.show()
r,x,θ,y,ϕ,z = p_RK4.T
fig,ax=plt.subplots(2,3,figsize=(8,4))
plt.xlabel('time in sec')
plt.ylabel('parameters')
for a,q in zip(ax.flatten(),[r,x,θ,y,ϕ,z]):
a.plot(t,q); a.grid()
plt.title("RK4", loc='left')
plt.tight_layout(); plt.show()
[RK4 solution plot][1]
[Euler's solution methods][2]
''''RK4 does not give iterated values.
The path is unaffected by the change of sign which is expected as it is under Lorentz force''''
[1]: https://i.stack.imgur.com/bZdIw.png
[2]: https://i.stack.imgur.com/tuNDp.png
You are not iterating more than once inside the for loop in rk4 because it returns after the first iteration.
for i in range(len(t)-1):
k1 = dt * fun(t[i], p[i])
k2 = dt * fun(t[i] + 0.5*dt, p[i] + 0.5 * k1)
k3 = dt * fun(t[i] + 0.5*dt, p[i] + 0.5 * k2)
k4 = dt * fun(t[i] + dt, p[i] + k3)
p[i+1] = p[i] + (k1 + 2*(k2 + k3) + k4)/6
# This is the problem line, the return was tabbed in, to be inside the for block, so the block executed once and returned.
return t,p
For physics questions please try a different forum.
How to edit the for cycles under #ax5 and #ax6 to plot graphs in the same fashion? Now, the lower figure has no colour transit, as opposed to the upper one. The colour transit appears in the lower figure after increasing of dpi, however, some unwanted stuff also appears. Is there a scalling problem? Thank you
import matplotlib.pyplot as plt
import numpy as np
from matplotlib.gridspec import GridSpec
import math
fig, ax = plt.subplots()
plt.rcParams["figure.figsize"] = [8, 8]
# Function for plotting parallels to curves
def get_parallels(length=.1):
px, py = [], []
for idx in range(len(x)-1):
x0, y0, xa, ya = x[idx], y[idx], x[idx+1], y[idx+1]
dx, dy = xa-x0, ya-y0
norm = math.hypot(dx, dy) * 1/length
dx /= norm
dy /= norm
px.append(x0-dy)
py.append(y0+dx)
return px, py
def offset(x,y, o):
""" Offset coordinates given by array x,y by o """
X = np.c_[x,y].T
m = np.array([[0,-1],[1,0]])
R = np.zeros_like(X)
S = X[:,2:]-X[:,:-2]
R[:,1:-1] = np.dot(m, S)
R[:,0] = np.dot(m, X[:,1]-X[:,0])
R[:,-1] = np.dot(m, X[:,-1]-X[:,-2])
On = R/np.sqrt(R[0,:]**2+R[1,:]**2)*o
Out = On+X
return Out[0,:], Out[1,:]
dpi = 20
def offset_curve(ax, x,y, o):
""" Offset array x,y in data coordinates
by o in points """
trans = ax.transData.transform
inv = ax.transData.inverted().transform
X = np.c_[x,y]
Xt = trans(X)
xto, yto = offset(Xt[:,0],Xt[:,1],o*dpi/72. )
Xto = np.c_[xto, yto]
Xo = inv(Xto)
return Xo[:,0], Xo[:,1]
fig = plt.figure(constrained_layout=True)
gs = GridSpec(3, 6, figure=fig)
ax5 = fig.add_subplot(gs[1, 3:6])
ax6 = fig.add_subplot(gs[2, :3])
ax7 = fig.add_subplot(gs[2, 3:6])
cmap = plt.get_cmap('Greys_r')
# ax5
x = np.linspace(-1, 1, 100)
y = -x**2
ax5.set_ylim(-1.02, 0.3)
width_l = ax5.get_ylim()[1] - ax5.get_ylim()[0]
for t in np.linspace(0, 1, 40):
length = -0.1*width_l*t
ax5.plot(*get_parallels(length=length), color=cmap(t/2 + 0.25))
# ax6
x = np.linspace(-3, 3, 100)
y = -(1/4*x**4 - 1.6*x**2)
ax6.plot(x, y)
ax6.set_xlim(ax6.get_xlim()[0]-0.5, ax6.get_xlim()[1]+0.5)
ax6.scatter(1/2*(ax6.get_xlim()[0] + ax6.get_xlim()[1]), 1.2, marker = 'o', s=900, facecolors='none')
lines = []
width_l = ax6.get_ylim()[1] - ax6.get_ylim()[0]
for t in np.linspace(0, 1, 40):
l, = ax6.plot(x, y - t * 0.1 * width_l, color=cmap(t/2 + 0.25))
lines.append(l)
def plot_rainbow(event=None):
x0 = x
y0 = y
for i in range(len(lines)):
xx, yy = offset_curve(ax, x0, y0, -width_l)
lines[i].set_data(xx, yy)
lines[i].set_linewidth(1.1*width_l)
x0 = xx
y0 = yy
plot_rainbow()
fig.canvas.mpl_connect("resize_event", plot_rainbow)
fig.canvas.mpl_connect("button_release_event", plot_rainbow)
plt.savefig('fig.pdf')
What I want do do is to ask for a number in the middle of the animation, for in the future be able to animate n circles and be able to change it in the middle of the animation. But when I try asking for an input I get this:
QCoreApplication::exec: The event loop is already running
and the next frame doesn't start until i have given it an input
How do I try getting an input and if not, continue the animation with the past value?
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
import random
fig, ax = plt.subplots()
ax.set_xlim(0,100)
ax.set_ylim(0,100)
alphas = [0.5,0.5]
circle = plt.Circle((5, 10), 10, color='b', fill=True, alpha = alphas[0])
circle2 = plt.Circle((5, 10), 5, color='r', fill=True, alpha = alphas[1])
circles = [circle, circle2]
def init():
for i in circles:
i.center = (50,50)
ax.add_patch(i)
return circles
def animation_frame(k):
try:
num = int(input())
except:
pass
finally:
for j in circles:
x,y = j.center
r = random.uniform(-5,5)
r2 = random.uniform(-5,5)
#stop circles from going out
if (((x + j.radius + r) <= 100 ) & ((x - j.radius + r) >= 0 )):
x += r
else:
x -= r
if (((y + j.radius + r2) <= 100 ) & ((y - j.radius + r2) >= 0 )):
y += r2
else:
y -= r2
j.center = (x, y)
return circles
animation = FuncAnimation(fig,animation_frame, init_func=init,frames=360,interval=20,blit=True)
plt.show()
For reasons, I need to implement the Runge-Kutta4 method in PyTorch (so no, I'm not going to use scipy.odeint). I tried and I get weird results on the simplest test case, solving x'=x with x(0)=1 (analytical solution: x=exp(t)). Basically, as I reduce the time step, I cannot get the numerical error to go down. I'm able to do it with a simpler Euler method, but not with the Runge-Kutta 4 method, which makes me suspect some floating point issue here (maybe I'm missing some hidden conversion from double precision to single)?
import torch
import numpy as np
import matplotlib.pyplot as plt
def Euler(f, IC, time_grid):
y0 = torch.tensor([IC])
time_grid = time_grid.to(y0[0])
values = y0
for i in range(0, time_grid.shape[0] - 1):
t_i = time_grid[i]
t_next = time_grid[i+1]
y_i = values[i]
dt = t_next - t_i
dy = f(t_i, y_i) * dt
y_next = y_i + dy
y_next = y_next.unsqueeze(0)
values = torch.cat((values, y_next), dim=0)
return values
def RungeKutta4(f, IC, time_grid):
y0 = torch.tensor([IC])
time_grid = time_grid.to(y0[0])
values = y0
for i in range(0, time_grid.shape[0] - 1):
t_i = time_grid[i]
t_next = time_grid[i+1]
y_i = values[i]
dt = t_next - t_i
dtd2 = 0.5 * dt
f1 = f(t_i, y_i)
f2 = f(t_i + dtd2, y_i + dtd2 * f1)
f3 = f(t_i + dtd2, y_i + dtd2 * f2)
f4 = f(t_next, y_i + dt * f3)
dy = 1/6 * dt * (f1 + 2 * (f2 + f3) +f4)
y_next = y_i + dy
y_next = y_next.unsqueeze(0)
values = torch.cat((values, y_next), dim=0)
return values
# differential equation
def f(T, X):
return X
# initial condition
IC = 1.
# integration interval
def integration_interval(steps, ND=1):
return torch.linspace(0, ND, steps)
# analytical solution
def analytical_solution(t_range):
return np.exp(t_range)
# test a numerical method
def test_method(method, t_range, analytical_solution):
numerical_solution = method(f, IC, t_range)
L_inf_err = torch.dist(numerical_solution, analytical_solution, float('inf'))
return L_inf_err
if __name__ == '__main__':
Euler_error = np.array([0.,0.,0.])
RungeKutta4_error = np.array([0.,0.,0.])
indices = np.arange(1, Euler_error.shape[0]+1)
n_steps = np.power(10, indices)
for i, n in np.ndenumerate(n_steps):
t_range = integration_interval(steps=n)
solution = analytical_solution(t_range)
Euler_error[i] = test_method(Euler, t_range, solution).numpy()
RungeKutta4_error[i] = test_method(RungeKutta4, t_range, solution).numpy()
plots_path = "./plots"
a = plt.figure()
plt.xscale('log')
plt.yscale('log')
plt.plot(n_steps, Euler_error, label="Euler error", linestyle='-')
plt.plot(n_steps, RungeKutta4_error, label="RungeKutta 4 error", linestyle='-.')
plt.legend()
plt.savefig(plots_path + "/errors.png")
The result:
As you can see, the Euler method converges (slowly, as expected of a first order method). However, the Runge-Kutta4 method does not converge as the time step gets smaller and smaller. The error goes down initially, and then up again. What's the issue here?
The reason is indeed a floating point precision issue. torch defaults to single precision, so once the truncation error becomes small enough, the total error is basically determined by the roundoff error, and reducing the truncation error further by increasing the number of steps <=> decreasing the time step doesn't lead to any decrease in the total error.
To fix this, we need to enforce double precision 64bit floats for all floating point torch tensors and numpy arrays. Note that the right way to do this is to use respectively torch.float64 and np.float64 rather than, e.g., torch.double and np.double, because the former are fixed-sized float values, (always 64bit) while the latter depend on the machine and/or compiler. Here's the fixed code:
import torch
import numpy as np
import matplotlib.pyplot as plt
def Euler(f, IC, time_grid):
y0 = torch.tensor([IC], dtype=torch.float64)
time_grid = time_grid.to(y0[0])
values = y0
for i in range(0, time_grid.shape[0] - 1):
t_i = time_grid[i]
t_next = time_grid[i+1]
y_i = values[i]
dt = t_next - t_i
dy = f(t_i, y_i) * dt
y_next = y_i + dy
y_next = y_next.unsqueeze(0)
values = torch.cat((values, y_next), dim=0)
return values
def RungeKutta4(f, IC, time_grid):
y0 = torch.tensor([IC], dtype=torch.float64)
time_grid = time_grid.to(y0[0])
values = y0
for i in range(0, time_grid.shape[0] - 1):
t_i = time_grid[i]
t_next = time_grid[i+1]
y_i = values[i]
dt = t_next - t_i
dtd2 = 0.5 * dt
f1 = f(t_i, y_i)
f2 = f(t_i + dtd2, y_i + dtd2 * f1)
f3 = f(t_i + dtd2, y_i + dtd2 * f2)
f4 = f(t_next, y_i + dt * f3)
dy = 1/6 * dt * (f1 + 2 * (f2 + f3) +f4)
y_next = y_i + dy
y_next = y_next.unsqueeze(0)
values = torch.cat((values, y_next), dim=0)
return values
# differential equation
def f(T, X):
return X
# initial condition
IC = 1.
# integration interval
def integration_interval(steps, ND=1):
return torch.linspace(0, ND, steps, dtype=torch.float64)
# analytical solution
def analytical_solution(t_range):
return np.exp(t_range, dtype=np.float64)
# test a numerical method
def test_method(method, t_range, analytical_solution):
numerical_solution = method(f, IC, t_range)
L_inf_err = torch.dist(numerical_solution, analytical_solution, float('inf'))
return L_inf_err
if __name__ == '__main__':
Euler_error = np.array([0.,0.,0.], dtype=np.float64)
RungeKutta4_error = np.array([0.,0.,0.], dtype=np.float64)
indices = np.arange(1, Euler_error.shape[0]+1)
n_steps = np.power(10, indices)
for i, n in np.ndenumerate(n_steps):
t_range = integration_interval(steps=n)
solution = analytical_solution(t_range)
Euler_error[i] = test_method(Euler, t_range, solution).numpy()
RungeKutta4_error[i] = test_method(RungeKutta4, t_range, solution).numpy()
plots_path = "./plots"
a = plt.figure()
plt.xscale('log')
plt.yscale('log')
plt.plot(n_steps, Euler_error, label="Euler error", linestyle='-')
plt.plot(n_steps, RungeKutta4_error, label="RungeKutta 4 error", linestyle='-.')
plt.legend()
plt.savefig(plots_path + "/errors.png")
Result:
Now, as we decrease the time step, the error of the RungeKutta4 approximation decreases with the correct rate.
I am trying to run this model of seed predation and population dynamics but I am new to coding and I am only getting one predation value that gets repeated over different generations. How can I get different predation values for different year?
Also, Is there an issue with the normalizing method used?
import numpy as np
import matplotlib.pyplot as plt
def is_odd(year):
return ((year % 2) == 1)
def reproduction(p_iter, year, dead):
if is_odd(year):
predation = dead
seedsProd = p_iter*s_oddd
seedsPred = K*predation*200*(seedsProd/np.sum(seedsProd))
return (seedsProd - seedsPred) + np.array([0,0,p_iter[2]])
else:
predation = dead
seedsProd = p_iter*s_even
seedsPred = K*predation*200*(seedsProd/np.sum(seedsProd))
return (seedsProd - seedsPred) +np.array([0,p_iter[1],0])
def normalize(p_iter):
if is_odd(year):
x = np.copy(p_iter)
x[2] = 0
x = (K-p_iter[2]) * x / sum(x)
x[2] = p_iter[2]
return x
else:
x = np.copy(p_iter)
x[1] = 0
x = (K-p_iter[1]) * x / sum(x)
x[1] = p_iter[1]
return x
Predation is defined here:
def predation():
return (np.array(np.round(np.random.uniform(0.4,0.6),2)))
#max_years
Y = 100
#carrying capacity
K = 1000
#initial populaton
p_1, p_2, p_3 = 998., 1., 1.
#seed released per plant
s_1, s_2, s_3 = 200, 260, 260
p_init = np.array([p_1, p_2, p_3],dtype=float)
s_oddd = np.array([s_1, s_2, 0.0])
s_even = np.array([s_1, 0.0, s_3])
n = len(p_init)
m = np.append(p_init,s_oddd)
p_iter = p_init
dead = 0
norm = 0
for year in range(1,Y+1):
dead = predation()
seeds = reproduction(p_iter, year, dead)
p_iter = np.maximum(seeds,np.zeros(p_iter.shape))
p_iter = normalize(p_iter)
m = np.vstack((m, [*p_iter]+[*seeds] ))