I was about to plot a Poincare section of the following DE, which is quite meaningful to have a periodic potential function V(x) = - cos(x) in this equation.
After calculating the solution using RK4 with time interval dt = 0.001, the one that python drew was as the following plot.
But according to the textbook(referred to 2E by J.M.T. Thompson and H.B. Stewart), the section would look like as
:
it has so much difference. For my personal opinion, since Poincare section does not appear as what writers draw, there must be some error in my code. However, I actually done for other forced oscillation DE, including Duffing's equation, and obtained the identical one as those in the textbook. So, I was wodering if there are some typos in the equation given by the textbook, or somewhere else. I posted my code, but might be quite messy to understand. So appreicate dealing with it.
import numpy as np
import matplotlib.pylab as plt
import matplotlib as mpl
import sys
import time
state = [1]
def print_percent_done(index, total, state, title='Please wait'):
percent_done2 = (index+1)/total*100
percent_done = round(percent_done2, 1)
print(f'\t⏳{title}: {percent_done}% done', end='\r')
if percent_done2 > 99.9 and state[0]:
print('\t✅'); state = [0]
####
no = 1
####
def multiple(n, q):
m = n; i = 0
while m >= 0:
m -= q
i += 1
return min(abs(n - (i - 1)*q), abs(i*q - n))
# system(2)
#Basic info.
filename = 'sinPotentialWell'
# a = 1
# alpha = 0.01
# w = 4
w0 = .5
n = 1000000
h = .01
t_0 = 0
x_0 = 0.1
y_0 = 0
A = [(t_0, x_0, y_0)]
def f(t, x, y):
return y
def g(t, x, y):
return -0.5*y - np.sin(x) + 1.1*np.sin(0.5*t)
for i in range(n):
t0 = A[i][0]; x0 = A[i][1]; y0 = A[i][2]
k1 = f(t0, x0, y0)
u1 = g(t0, x0, y0)
k2 = f(t0 + h/2, x0 + h*k1/2, y0 + h*u1/2)
u2 = g(t0 + h/2, x0 + h*k1/2, y0 + h*u1/2)
k3 = f(t0 + h/2, x0 + h*k2/2, y0 + h*u2/2)
u3 = g(t0 + h/2, x0 + h*k2/2, y0 + h*u2/2)
k4 = f(t0 + h, x0 + h*k3, y0 + h*u3)
u4 = g(t0 + h, x0 + h*k3, y0 + h*u3)
t = t0 + h
x = x0 + (k1 + 2*k2 + 2*k3 + k4)*h/6
y = y0 + (u1 + 2*u2 + 2*u3 + u4)*h/6
A.append([t, x, y])
if i%1000 == 0: print_percent_done(i, n, state, 'Solving given DE')
#phase diagram
print('showing 3d_(x, y, phi) graph')
PHI=[[]]; X=[[]]; Y=[[]]
PHI_period1 = []; X_period1 = []; Y_period1 = []
for i in range(n):
if w0*A[i][0]%(2*np.pi) < 1 and w0*A[i-1][0]%(2*np.pi) > 6:
PHI.append([]); X.append([]); Y.append([])
PHI_period1.append((w0*A[i][0])%(2*np.pi)); X_period1.append(A[i][1]); Y_period1.append(A[i][2])
phi_period1 = np.array(PHI_period1); x_period1 = np.array(X_period1); y_period1 = np.array(Y_period1)
print('showing Poincare Section at phi=0')
plt.plot(x_period1, y_period1, 'gs', markersize = 2)
plt.plot()
plt.title('phi=0 Poincare Section')
plt.xlabel('x'); plt.ylabel('y')
plt.show()
If you factor out some of the computation blocks, you can make the code more flexible and computations more direct. No need to reconstruct something if you can construct it in the first place. You want to catch the points where w0*t is a multiple of 2*pi, so just construct the time loops so you integrate in chunks of 2*pi/w0 and only remember the interesting points.
num_plot_points = 2000
h = .01
t,x,y = t_0,x_0,y_0
x_section,y_section = [],[]
T = 2*np.pi/w0
for k in range(num_plot_points):
t = 0;
while t < T-1.2*h:
x,y = RK4step(t,x,y,h)
t += h
x,y = RK4step(t,x,y,T-t)
if k%100 == 0: print_percent_done(k, num_plot_points, state, 'Solving given DE')
x_section.append(x); y_section.append(y)
with RK4step just containing the code of the RK4 step.
This will not solve the mystery. The veil gets lifted if you consider that x is the angle theta (of a forced pendulum with friction) on a circle. Thus to get points with the same spacial location it needs to be reduced by multiples of 2*pi. Doing that,
plt.plot([x%(2*np.pi) for x in x_section], y_section, 'gs', markersize = 2)
results in the expected plot
Related
I am trying to solve this equation using Runge Kutta 4th order:
applying d2Q/dt2=F(y,x,v) and dQ/dt=u Q=y in my program.
I try to run the code but i get this error:
Traceback (most recent call last):
File "C:\Users\Egw\Desktop\Analysh\Askhsh1\asdasda.py", line 28, in <module>
k1 = F(y, u, x) #(x, v, t)
File "C:\Users\Egw\Desktop\Analysh\Askhsh1\asdasda.py", line 13, in F
return ((Vo/L -(R0/L)*u -(R1/L)*u**3 - y*(1/L*C)))
OverflowError: (34, 'Result too large')
I tried using the decimal library but I still couldnt make it work properly.I might have not used it properly tho.
My code is this one:
import numpy as np
from math import pi
from numpy import arange
from matplotlib.pyplot import plot, show
#parameters
R0 = 200
R1 = 250
L = 15
h = 0.002
Vo=1000
C=4.2*10**(-6)
t=0.93
def F(y, u, x):
return ((Vo/L -(R0/L)*u -(R1/L)*u**3 - y*(1/L*C)))
xpoints = arange(0,t,h)
ypoints = []
upoints = []
y = 0.0
u = Vo/L
for x in xpoints:
ypoints.append(y)
upoints.append(u)
m1 = u
k1 = F(y, u, x) #(x, v, t)
m2 = h*(u + 0.5*k1)
k2 = (h*F(y+0.5*m1, u+0.5*k1, x+0.5*h))
m3 = h*(u + 0.5*k2)
k3 = h*F(y+0.5*m2, u+0.5*k2, x+0.5*h)
m4 = h*(u + k3)
k4 = h*F(y+m3, u+k3, x+h)
y += (m1 + 2*m2 + 2*m3 + m4)/6
u += (k1 + 2*k2 + 2*k3 + k4)/6
plot(xpoints, upoints)
show()
plot(xpoints, ypoints)
show()
I expected to get the plots of u and y against t.
Turns out I messed up with the equations I was using for Runge Kutta
The correct code is the following:
import numpy as np
from math import pi
from numpy import arange
from matplotlib.pyplot import plot, show
#parameters
R0 = 200
R1 = 250
L = 15
h = 0.002
Vo=1000
C=4.2*10**(-6)
t0=0
#dz/dz
def G(x,y,z):
return Vo/L -(R0/L)*z -(R1/L)*z**3 - y/(L*C)
#dy/dx
def F(x,y,z):
return z
t = np.arange(t0, 0.93, h)
x = np.zeros(len(t))
y = np.zeros(len(t))
z = np.zeros(len(t))
y[0] = 0.0
z[0] = 0
for i in range(1, len(t)):
k0=h*F(x[i-1],y[i-1],z[i-1])
l0=h*G(x[i-1],y[i-1],z[i-1])
k1=h*F(x[i-1]+h*0.5,y[i-1]+k0*0.5,z[i-1]+l0*0.5)
l1=h*G(x[i-1]+h*0.5,y[i-1]+k0*0.5,z[i-1]+l0*0.5)
k2=h*F(x[i-1]+h*0.5,y[i-1]+k1*0.5,z[i-1]+l1*0.5)
l2=h*G(x[i-1]+h*0.5,y[i-1]+k1*0.5,z[i-1]+l1*0.5)
k3=h*F(x[i-1]+h,y[i-1]+k2,z[i-1]+l2)
l3 = h * G(x[i - 1] + h, y[i - 1] + k2, z[i - 1] + l2)
y[i]=y[i-1]+(k0+2*k1+2*k2+k3)/6
z[i] = z[i - 1] + (l0 + 2 * l1 + 2 * l2 + l3) / 6
Q=y
I=z
plot(t, Q)
show()
plot(t, I)
show()
If I may draw your attention to these 4 lines
m1 = u
k1 = F(y, u, x) #(x, v, t)
m2 = h*(u + 0.5*k1)
k2 = (h*F(y+0.5*m1, u+0.5*k1, x+0.5*h))
You should note a fundamental structural difference between the first two lines and the second pair of lines.
You need to multiply with the step size h also in the first pair.
The next problem is the step size and the cubic term. It contributes a term of size 3*(R1/L)*u^2 ~ 50*u^2 to the Lipschitz constant. In the original IVP per the question with u=Vo/L ~ 70 this term is of size 2.5e+5. To compensate only that term to stay in the stability region of the method, the step size has to be smaller 1e-5.
In the corrected initial conditions with u=0 at the start the velocity u remains below 0.001 so the cubic term does not determine stability, this is now governed by the last term contributing a Lipschitz term of 1/sqrt(L*C) ~ 125. The step size for stability is now 0.02, with 0.002 one can expect quantitatively useful results.
You can use decimal libary for more precision (handle more digits), but it's kind of annoying every value should be the same class (decimal.Decimal).
For example:
import numpy as np
from math import pi
from numpy import arange
from matplotlib.pyplot import plot, show
# Import decimal.Decimal as D
import decimal
from decimal import Decimal as D
# Precision
decimal.getcontext().prec = 10_000_000
#parameters
# Every value should be D class (decimal.Decimal class)
R0 = D(200)
R1 = D(250)
L = D(15)
h = D(0.002)
Vo = D(1000)
C = D(4.2*10**(-6))
t = D(0.93)
def F(y, u, x):
# Decomposed for use D
a = D(Vo/L)
b = D(-(R0/L)*u)
c = D(-(R1/L)*u**D(3))
d = D(-y*(D(1)/L*C))
return ((a + b + c + d ))
xpoints = arange(0,t,h)
ypoints = []
upoints = []
y = D(0.0)
u = D(Vo/L)
for x in xpoints:
ypoints.append(y)
upoints.append(u)
m1 = u
k1 = F(y, u, x) #(x, v, t)
m2 = (h*(u + D(0.5)*k1))
k2 = (h*F(y+D(0.5)*m1, u+D(0.5)*k1, x+D(0.5)*h))
m3 = h*(u + D(0.5)*k2)
k3 = h*F(y+D(0.5)*m2, u+D(0.5)*k2, x+D(0.5)*h)
m4 = h*(u + k3)
k4 = h*F(y+m3, u+k3, x+h)
y += (m1 + D(2)*m2 + D(2)*m3 + m4)/D(6)
u += (k1 + D(2)*k2 + D(2)*k3 + k4)/D(6)
plot(xpoints, upoints)
show()
plot(xpoints, ypoints)
show()
But even with ten million of precision I still get an overflow error. Check the components of the formula, their values are way too high. You can increase precision for handle them, but you'll notice it takes time to calculate them.
Problem implementation using scipy.integrate.odeint and scipy.integrate.solve_ivp.
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint, solve_ivp
# Input data initial conditions
ti = 0.0
tf = 0.5
N = 100000
h = (tf-ti)/N
# Initial conditions
u0 = 0.0
Q0 = 0.0
t_span = np.linspace(ti,tf,N)
r0 = np.array([Q0,u0])
# Parameters
R0 = 200
R1 = 250
L = 15
C = 4.2*10**(-6)
V0 = 1000
# Systems of First Order Equations
# This function is used with odeint, as specified in the documentation for scipy.integrate.odeint
def f(r,t,R0,R1,L,C,V0):
Q,u = r
ode1 = u
ode2 = -((R0/L)*u)-((R1/L)*u**3)-((1/(L*C))*Q)+(V0/L)
return np.array([ode1,ode2])
# This function is used in our 4Order Runge-Kutta implementation and in scipy.integrate.solve_ivp
def F(t,r,R0,R1,L,C,V0):
Q,u = r
ode1 = u
ode2 = -((R0/L)*u)-((R1/L)*u**3)-((1/(L*C))*Q)+(V0/L)
return np.array([ode1,ode2])
# Resolution with oedint
sol_1 = odeint(f,r0,t_span,args=(R0,R1,L,C,V0))
sol_2 = solve_ivp(fun=F,t_span=(ti,tf), y0=r0, method='LSODA',args=(R0,R1,L,C,V0))
Q_odeint, u_odeint = sol_1[:,0], sol_1[:,1]
Q_solve_ivp, u_solve_ivp = sol_2.y[0,:], sol_2.y[1,:]
# Figures
plt.figure(figsize=[30.0,10.0])
plt.subplot(3,1,1)
plt.grid(color = 'red',linestyle='--',linewidth=0.4)
plt.plot(t_span,Q_odeint,'r',t_span,u_odeint,'b')
plt.xlabel('t(s)')
plt.ylabel('Q(t), u(t)')
plt.subplot(3,1,2)
plt.plot(sol_2.t,Q_solve_ivp,'g',sol_2.t,u_solve_ivp,'y')
plt.grid(color = 'yellow',linestyle='--',linewidth=0.4)
plt.xlabel('t(s)')
plt.ylabel('Q(t), u(t)')
plt.subplot(3,1,3)
plt.plot(Q_solve_ivp,u_solve_ivp,'green')
plt.grid(color = 'yellow',linestyle='--',linewidth=0.4)
plt.xlabel('Q(t)')
plt.ylabel('u(t)')
plt.show()
Runge-Kutta 4th
# Code development of Runge-Kutta 4 Order
# Parameters
R0 = 200
R1 = 250
L = 15
C = 4.2*10**(-6)
V0 = 1000
# Input data initial conditions #
ti = 0.0
tf = 0.5
N = 100000
h = (tf-ti)/N
# Initial conditions
u0 = 0.0
Q0 = 0.0
# First order ordinary differential equations
def f1(t,Q,u):
return u
def f2(t,Q,u):
return -((R0/L)*u)-((R1/L)*u**3)-((1/(L*C))*Q)+(V0/L)
t = np.zeros(N); Q = np.zeros(N); u = np.zeros(N)
t[0] = ti
Q[0] = Q0
u[0] = u0
for i in range(0,N-1,1):
k1 = h*f1(t[i],Q[i],u[i])
l1 = h*f2(t[i],Q[i],u[i])
k2 = h*f1(t[i]+(h/2),Q[i]+(k1/2),u[i]+(l1/2))
l2 = h*f2(t[i]+(h/2),Q[i]+(k1/2),u[i]+(l1/2))
k3 = h*f1(t[i]+(h/2),Q[i]+(k2/2),u[i]+(l2/2))
l3 = h*f2(t[i]+(h/2),Q[i]+(k2/2),u[i]+(l2/2))
k4 = h*f1(t[i]+h,Q[i]+k3,u[i]+l3)
l4 = h*f2(t[i]+h,Q[i]+k3,u[i]+l3)
Q[i+1] = Q[i] + ((k1+2*k2+2*k3+k4)/6)
u[i+1] = u[i] + ((l1+2*l2+2*l3+l4)/6)
t[i+1] = t[i] + h
plt.figure(figsize=[20.0,10.0])
plt.subplot(1,2,1)
plt.plot(t,Q_solve_ivp,'r',t,Q_odeint,'y',t,Q,'b')
plt.grid(color = 'yellow',linestyle='--',linewidth=0.4)
plt.xlabel('t(s)')
plt.ylabel(r'$Q(t)_{Odeint}$, $Q(t)_{RK4}$')
plt.subplot(1,2,2)
plt.plot(t,Q_solve_ivp,'g',t,Q_odeint,'y',t,Q,'b')
plt.grid(color = 'yellow',linestyle='--',linewidth=0.4)
plt.xlabel('t(s)')
plt.ylabel(r'$Q(t)_{solve_ivp}$, $Q(t)_{RK4}$')
I need help to solve a ODE system using odeint or any workaround that works. I`m trying to model a ODE system descriving the beheaviour of a cilinder on a mechanichal arm, using Lagrangian mechanics I got the following 2nd order ODE system:
$$\ddot{x} = x*\dot{\phi}^2 - g*(sin(\phi) +cos(\phi))$$
$$\ddot{\phi} = \frac{1}{2*\dot{x}*m_c}*(\frac{k*i}{x}-\frac{m_b*L*\dot{\phi}}{2}-g*cos(\phi)*(m_c*x-m_b*L/2))$$
Then I transformed it to a 4x4 1st order ODE system y linearized 2 equations aroun initial condition, to make it more "solveable".
In python I`wrote the folllowing:
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
# source: https://apmonitor.com/pdc/index.php/Main/SolveDifferentialEquations
g, mc, mb, k, L = (9.8, 0.5, 1.3, 2, 1.2)
def model(z, t, i, z0):
global g, mc, mb, k, L
#ODE variables
x1 = z[0]
y1 = z[1]
x2 = z[2]
y2 = z[3]
#ODE initial conditions
zx1 = z0[0]
zy1 = z0[1]
zx2 = z0[2]
zy2 = z0[3]
#Diferential for linealization around I.C.
dx1 = z[0]-z0[0]
dy1 = z[1]-z0[1]
dx2 = z[2]-z0[2]
dy2 = z[3]-z0[3]
def w1(zx1, zy1, zx2, zy2):
global g, mc, mb, k, L
return zx2*zy1**2-g*(np.sin(zy2)+np.cos(zy2))
def w2(zx1, zy1, zx2, zy2):
global g, mc, mb, k, L
return 1/(2*zx1*mc)*((k*i)/(zx2)-mb*zy2*L/2-g*np.cos(zy2*(mc*zx2-mb*L/2)))
dx2dt = x1
#dx1dt = x2*y1**2-g*(np.sin(y2)+np.cos(y2))
dx1dt = w1(zx1, zy1, zx2, zy2) + zy1**2*dx2+2*zx2 * \
zy1*dy1-g*(np.cos(zy2)-np.sin(zy2))*dy2
dy2dt = y1
#dy1dt = 1/(2*x1*mc)*((k*i)/(x2)-mb*y2*L/2-g*np.cos(y2*(mc*x2-mb*L/2)))
dy1dt = w2(zx1, zy1, zx2, zy2) - 1/(4*zx1**2*mc) * \
((k*i)/(zx2)-mb*zy2*L/2-g*np.cos(zy2*(mc*zx2-mb*L/2)))*dx1
+(k*i*np.log(abs(zx2))/(2*zx1*mc)-g*np.cos(zy2)*mc)*dx2-(mb*L) / \
(4*zx1*mc)*dy1+g*np.sin(zy2)*(mc*zx2-mb*L/2)*(2*zx1*mc)*dy2
dzdt = [dx1dt, dy1dt, dx2dt, dy2dt]
return dzdt
# initial condition
z0 = [0.01, 0.01, 0.8, np.pi/8]
# number of time points
n = 30
# time points
t = np.linspace(0, 1, n)
# step input
u = np.zeros(n)
# change to 2.0 at time = 5.0
u[:] = 1
# store solution
x1 = np.zeros(n)
y1 = np.zeros(n)
x2 = np.zeros(n)
y2 = np.zeros(n)
# record initial conditions
x1[0] = z0[0]
y1[0] = z0[1]
x2[0] = z0[2]
y2[0] = z0[3]
# solve ODE
for i in range(1, n):
# span for next time step
tspan = [t[i-1], t[i]]
# solve for next step
z = odeint(model, z0, tspan, args=((u[i], z0)) )
# store solution for plotting
if -np.pi/4 <= z[1][3] <= np.pi/4: # Angulo +/- Pi/4
u[i] = u[i-1] + 1
else:
u[i] = u[i-1]
if 0 <= z[1][2] <= L: # Se mantiene en el brazo
print("Se salió")
x1[i] = z[1][0]
y1[i] = z[1][1]
x1[i] = z[1][2]
y1[i] = z[1][3]
# Siguientes CI
z0 = z[1]
# plot results
plt.plot(t, u, 'g:', label='u(t)')
plt.plot(t, x2, 'b-', label='x(t)')
plt.plot(t, y2, 'r--', label='phi(t)')
plt.ylabel('Cilindro(x,phi)')
plt.xlabel('Tiempo(s)')
plt.legend(loc='best')
plt.show()
Then I`ve got the following error, telling me than too much accuracy is needed to solve my system:
lsoda-- at start of problem, too much accuracy
requested for precision of machine.. see tolsf (=r1)
in above message, r1 = NaN
lsoda-- repeated occurrences of illegal input
lsoda-- at start of problem, too much accuracy
requested for precision of machine.. see tolsf (=r1)
in above message, r1 = NaN
dy1dt = w2(zx1, zy1, zx2, zy2) - 1/(4*zx1**2*mc) * \
lsoda-- warning..internal t (=r1) and h (=r2) are
such that in the machine, t + h = t on the next step
(h = step size). solver will continue anyway
in above, r1 = 0.3448275862069D+00 r2 = 0.1555030227910D-28
I changed some things but i still have a similar problem. I am working on Mandelbrot zoom. I try to zoom in deeper at branches. I count the consecutive points in the set and return the branch if it reaches the defined length. Then I zoom into that area and repeat. But the change gets smaller and the returned branch is nearly the same as the last one.
variables
X0,Y0,X1 = verticies of current area
branch = current branch
n = number of iterations
p = how many times i want to zoom
b = minimum length of branch i am looking for
c = current complex number
z = current value
k = pixel size
the function that returns the branch
def fractal(n, X0, Y0, X1): # searching for new branch
branch = []
k = (X1 - X0) / x_axis # pixel size
b = 5 # the number of black points int the set i am looking for
for y in range(y_axis):
try:
for x in range(x_axis):
c = X0 + k * x + Y0 - k * y * 1.j # new c
z = c
for m in range(n):
if abs(z) <= 2:
z = z * z + c # new z
else:
break
if abs(z) <= 2:
branch.append((x,y)) # the coordinates for those points
else:
if b < len(branch) < 50:
raise BreakOutOfALoop # break if a branch was found
branch = []
except BreakOutOfALoop:
break
return branch, k
the function that calculates the verticies of the new area
def new_area(branch, X0, Y0, X1, k): # calculating new area
print(branch)
X0 = X0 + k * branch[0][0]
Y0 = Y0 - k * branch[0][1]
X1 = X1 + k * branch[-1][0] # new area verticies
return X0, Y0, X1
the loop that calls the functions
for i in range(p):
areaImage = Image.new('RGB', (x_axis,y_axis), "white")
area_pixels = areaImage.load() # image load
branch, k = fractal(n,X0, Y0, X1)
X0, Y0, X1 = new_area(branch, X0, Y0, X1, k)
file_name = "/" + str(i) + "_" + str(n) + "_" + str(x_axis) + "_" + str(y_axis) + ".png" # image save
areaImage = areaImage.save(f"{areaImage_path}{file_name}")
What am I overlooking?
Here are some pictures for different p.
p = 3
p = 10
I am trying to solve for the position of a body orbiting a much more massive body, using the idealization that the much more massive body doesn't move. I am trying to solve for the position in cartesian coordinates using 4th order Runge-Kutta in python.
Here is my code:
dt = .1
t = np.arange(0,10,dt)
vx = np.zeros(len(t))
vy = np.zeros(len(t))
x = np.zeros(len(t))
y = np.zeros(len(t))
vx[0] = 10 #initial x velocity
vy[0] = 10 #initial y velocity
x[0] = 10 #initial x position
y[0] = 0 #initial y position
M = 20
def fx(x,y,t): #x acceleration
return -G*M*x/((x**2+y**2)**(3/2))
def fy(x,y,t): #y acceleration
return -G*M*y/((x**2+y**2)**(3/2))
def rkx(x,y,t,dt): #runge-kutta for x
kx1 = dt * fx(x,y,t)
mx1 = dt * x
kx2 = dt * fx(x + .5*kx1, y + .5*kx1, t + .5*dt)
mx2 = dt * (x + kx1/2)
kx3 = dt * fx(x + .5*kx2, y + .5*kx2, t + .5*dt)
mx3 = dt * (x + kx2/2)
kx4 = dt * fx(x + kx3, y + x3, t + dt)
mx4 = dt * (x + kx3)
return (kx1 + 2*kx2 + 2*kx3 + kx4)/6
return (mx1 + 2*mx2 + 2*mx3 + mx4)/6
def rky(x,y,t,dt): #runge-kutta for y
ky1 = dt * fy(x,y,t)
my1 = dt * y
ky2 = dt * fy(x + .5*ky1, y + .5*ky1, t + .5*dt)
my2 = dt * (y + ky1/2)
ky3 = dt * fy(x + .5*ky2, y + .5*ky2, t + .5*dt)
my3 = dt * (y + ky2/2)
ky4 = dt * fy(x + ky3, y + ky3, t + dt)
my4 = dt * (y + ky3)
return (ky1 + 2*ky2 + 2*ky3 + ky4)/6
return (my1 + 2*my2 + 2*my3 + my4)/6
for n in range(1,len(t)): #solve using RK4 functions
vx[n] = vx[n-1] + fx(x[n-1],y[n-1],t[n-1])*dt
vy[n] = vy[n-1] + fy(x[n-1],y[n-1],t[n-1])*dt
x[n] = x[n-1] + vx[n-1]*dt
y[n] = y[n-1] + vy[n-1]*dt
Originally, no matter which way I tweaked the code, I was getting an error on my for loop, either "object of type 'float' has no len()" (I didn't understand what float python could be referring to), or "setting an array element with a sequence" (I also didn't understand what sequence it meant). I've managed to get rid of the errors, but my results are just wrong. I get vx and vy arrays of 10s, an x array of integers from 10. to 109., and a y array of integers from 0. to 99.
I suspect there are issues with fx(x,y,t) and fy(x,y,t) or with the way I have coded the runge-kutta functions to go with fx and fy, because I've used the same runge-kutta code for other functions and it works fine.
I greatly appreciate any help in figuring out why my code isn't working. Thank you.
Physics
The Newton law gives you a second order ODE u''=F(u) with u=[x,y]. Using v=[x',y'] you get the first order system
u' = v
v' = F(u)
which is 4-dimensional and has to be solved using a 4 dimensional state. The only reduction available is to use the Kepler laws which allow to reduce the system to a scalar order one ODE for the angle. But that is not the task here.
But to get the scales correct, for a circular orbit of radius R with angular velocity w one gets the identity w^2*R^3=G*M which implies that the speed along the orbit is w*R=sqrt(G*M/R) and period T=2*pi*sqrt(R^3/(G*M)). With the data given, R ~ 10, w ~ 1, thus G*M ~ 1000 for a close-to-circular orbit, so with M=20 this would require G between 50 and 200, with an orbital period of about 2*pi ~ 6. The time span of 10 could represent one half to about 2 or 3 orbits.
Euler method
You correctly implemented the Euler method to calculate values in the last loop of your code. That it may look un-physical can be because the Euler method continuously increases the orbit, as it moves to the outside of convex trajectories following the tangent. In your implementation this outward spiral can be seen for G=100.
This can be reduced in effect by choosing a smaller step size, such as dt=0.001.
You should select the integration time to be a good part of a full orbit to get a presentable result, with above parameters you get about 2 loops, which is good.
RK4 implementation
You made several errors. Somehow you lost the velocities, the position updates should be based on the velocities.
Then you should have halted at fx(x + .5*kx1, y + .5*kx1, t + .5*dt) to reconsider your approach as that is inconsistent with any naming convention. The consistent, correct variant is
fx(x + .5*kx1, y + .5*ky1, t + .5*dt)
which shows that you can not decouple the integration of a coupled system, as you need the y updates alongside the x updates. Further, the function values are the accelerations, thus update the velocities. The position updates use the velocities of the current state. Thus the step should start as
kx1 = dt * fx(x,y,t) # vx update
mx1 = dt * vx # x update
ky1 = dt * fy(x,y,t) # vy update
my1 = dt * vy # y update
kx2 = dt * fx(x + 0.5*mx1, y + 0.5*my1, t + 0.5*dt)
mx2 = dt * (vx + 0.5*kx1)
ky2 = dt * fy(x + 0.5*mx1, y + 0.5*my1, t + 0.5*dt)
my2 = dt * (vy + 0.5*ky1)
etc.
However, as you see, this already starts to become unwieldy. Assemble the state into a vector and use a vector valued function for the system equations
M, G = 20, 100
def orbitsys(u):
x,y,vx,vy = u
r = np.hypot(x,y)
f = G*M/r**3
return np.array([vx, vy, -f*x, -f*y]);
Then you can use a cook-book implementation of the Euler or Runge-Kutta step
def Eulerstep(f,u,dt): return u+dt*f(u)
def RK4step(f,u,dt):
k1 = dt*f(u)
k2 = dt*f(u+0.5*k1)
k3 = dt*f(u+0.5*k2)
k4 = dt*f(u+k3)
return u + (k1+2*k2+2*k3+k4)/6
and combine them into an integration loop
def Eulerintegrate(f, y0, tspan):
y = np.zeros([len(tspan),len(y0)])
y[0,:]=y0
for k in range(1, len(tspan)):
y[k,:] = Eulerstep(f, y[k-1], tspan[k]-tspan[k-1])
return y
def RK4integrate(f, y0, tspan):
y = np.zeros([len(tspan),len(y0)])
y[0,:]=y0
for k in range(1, len(tspan)):
y[k,:] = RK4step(f, y[k-1], tspan[k]-tspan[k-1])
return y
and invoke them with your given problem
dt = .1
t = np.arange(0,10,dt)
y0 = np.array([10, 0.0, 10, 10])
sol_euler = Eulerintegrate(orbitsys, y0, t)
x,y,vx,vy = sol_euler.T
plt.plot(x,y)
sol_RK4 = RK4integrate(orbitsys, y0, t)
x,y,vx,vy = sol_RK4.T
plt.plot(x,y)
You are not using rkx, rky functions anywhere!
There are two return at the end of function definition you should use
return [(kx1 + 2*kx2 + 2*kx3 + kx4)/6, (mx1 + 2*mx2 + 2*mx3 + mx4)/6] (as pointed out by #eapetcho). Also, your implementation of Runge-Kutta is not clear to me.
You have dv/dt so you solve for v and then update r accordingly.
for n in range(1,len(t)): #solve using RK4 functions
vx[n] = vx[n-1] + rkx(vx[n-1],vy[n-1],t[n-1])*dt
vy[n] = vy[n-1] + rky(vx[n-1],vy[n-1],t[n-1])*dt
x[n] = x[n-1] + vx[n-1]*dt
y[n] = y[n-1] + vy[n-1]*dt
Here is my version of the code
import numpy as np
#constants
G=1
M=1
h=0.1
#initiating variables
rt = np.arange(0,10,h)
vx = np.zeros(len(rt))
vy = np.zeros(len(rt))
rx = np.zeros(len(rt))
ry = np.zeros(len(rt))
#initial conditions
vx[0] = 10 #initial x velocity
vy[0] = 10 #initial y velocity
rx[0] = 10 #initial x position
ry[0] = 0 #initial y position
def fx(x,y): #x acceleration
return -G*M*x/((x**2+y**2)**(3/2))
def fy(x,y): #y acceleration
return -G*M*y/((x**2+y**2)**(3/2))
def rk4(xj, yj):
k0 = h*fx(xj, yj)
l0 = h*fx(xj, yj)
k1 = h*fx(xj + 0.5*k0 , yj + 0.5*l0)
l1 = h*fy(xj + 0.5*k0 , yj + 0.5*l0)
k2 = h*fx(xj + 0.5*k1 , yj + 0.5*l1)
l2 = h*fy(xj + 0.5*k1 , yj + 0.5*l1)
k3 = h*fx(xj + k2, yj + l2)
l3 = h*fy(xj + k2, yj + l2)
xj1 = xj + (1/6)*(k0 + 2*k1 + 2*k2 + k3)
yj1 = yj + (1/6)*(l0 + 2*l1 + 2*l2 + l3)
return (xj1, yj1)
for t in range(1,len(rt)):
nv = rk4(vx[t-1],vy[t-1])
[vx[t],vy[t]] = nv
rx[t] = rx[t-1] + vx[t-1]*h
ry[t] = ry[t-1] + vy[t-1]*h
I suspect there are issues with fx(x,y,t) and fy(x,y,t)
This is the case, I just checked my code for fx=3 and fy=y and I got a nice trajectory.
Here is the ry vs rx plot:
I tried to write an algorithm for solving the non-linear ODE
dr/dt = r(t)+r²(t)
which has (one possible) solution
r(t) = r²(t)/2+r³(t)/3
For that I implemented both the euler method and the RK4 method in python. For error checking I used the example on rosettacode:
dT/dt = -k(T(t)-T0)
with the solution
T0 + (Ts − T0)*exp(−kt)
Thus, my code now looks like
import numpy as np
from matplotlib import pyplot as plt
def test_func_1(t, x):
return x*x
def test_func_1_sol(t, x):
return x*x*x/3.0
def test_func_2_sol(TR, T0, k, t):
return TR + (T0-TR)*np.exp(-0.07*t)
def rk4(func, dh, x0, t0):
k1 = dh*func(t0, x0)
k2 = dh*func(t0+dh*0.5, x0+0.5*k1)
k3 = dh*func(t0+dh*0.5, x0+0.5*k2)
k4 = dh*func(t0+dh, x0+k3)
return x0+k1/6.0+k2/3.0+k3/3.0+k4/6.0
def euler(func, x0, t0, dh):
return x0 + dh*func(t0, x0)
def rho_test(t0, rho0):
return rho0 + rho0*rho0
def rho_sol(t0, rho0):
return rho0*rho0*rho0/3.0+rho0*rho0/2.0
def euler2(f,y0,a,b,h):
t,y = a,y0
while t <= b:
#print "%6.3f %6.5f" % (t,y)
t += h
y += h * f(t,y)
def newtoncooling(time, temp):
return -0.07 * (temp - 20)
x0 = 100
x_vec_rk = []
x_vec_euler = []
x_vec_rk.append(x0)
h = 1e-3
for i in range(100000):
x0 = rk4(newtoncooling, h, x0, i*h)
x_vec_rk.append(x0)
x0 = 100
x_vec_euler.append(x0)
x_vec_sol = []
x_vec_sol.append(x0)
for i in range(100000):
x0 = euler(newtoncooling, x0, 0, h)
#print(i, x0)
x_vec_euler.append(x0)
x_vec_sol.append(test_func_2_sol(20, 100, 0, i*h))
euler2(newtoncooling, 0, 0, 1, 1e-4)
x_vec = np.linspace(0, 1, len(x_vec_euler))
plt.plot(x_vec, x_vec_euler, x_vec, x_vec_sol, x_vec, x_vec_rk)
plt.show()
#rho-function
x0 = 1
x_vec_rk = []
x_vec_euler = []
x_vec_rk.append(x0)
h = 1e-3
num_steps = 650
for i in range(num_steps):
x0 = rk4(rho_test, h, x0, i*h)
print "%6.3f %6.5f" % (i*h, x0)
x_vec_rk.append(x0)
x0 = 1
x_vec_euler.append(x0)
x_vec_sol = []
x_vec_sol.append(x0)
for i in range(num_steps):
x0 = euler(rho_test, x0, 0, h)
print "%6.3f %6.5f" % (i*h, x0)
x_vec_euler.append(x0)
x_vec_sol.append(rho_sol(i*h, i*h+x0))
x_vec = np.linspace(0, num_steps*h, len(x_vec_euler))
plt.plot(x_vec, x_vec_euler, x_vec, x_vec_sol, x_vec, x_vec_rk)
plt.show()
It works fine for the example from rosettacode, but it is unstable and explodes (for t > 0.65, both for RK4 and euler) for my formula. Is therefore my implementation incorrect, or is there another error I do not see?
Searching for the exact solution of your equation:
dr/dt = r(t)+r²(t)
I've found:
r(t) = exp(C+t)/(1-exp(C+t))
Where C is an arbitrary constant which depends on initial conditions. It can be seen that for t -> -C the r(t) -> infinity.
I do not know what initial condition you use, but may be you meet with this singularity when calculate numerical solution.
UPDATE:
Since your initial condition is r(0)=1 the constant C is C = ln(1/2) ~ -0.693. It can explain why your numerical solution crashes at some t>0.65
UPDATE:
To verify your code you can simply compare your numerical solution calculated for, say 0<=t<0.6 with exact solution.