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python code for multiple ode
Closed 9 years ago.
The following is the code for multiple solvers so far.
The system for this problem is here, our system
However, when I execute it in Python, it shows me the following error:
Traceback (most recent call last):
File "G:\math3511\assignment\assignment5\qu2", line 59, in
X = AdamsBashforth4(equation, init, t)
File "G:\math3511\assignment\assignment5\qu2", line 32, in AdamsBashforth4
k2 = h * f( x[i] + 0.5 * k1, t[i] + 0.5 * h )
TypeError: can't multiply sequence by non-int of type 'float'
the code:
import numpy
from numpy import array, exp, linspace
def AdamsBashforth4( f, x0, t ):
"""
Fourth-order Adams-Bashforth method::
u[n+1] = u[n] + dt/24.*(55.*f(u[n], t[n]) - 59*f(u[n-1], t[n-1]) +
37*f(u[n-2], t[n-2]) - 9*f(u[n-3], t[n-3]))
for constant time step dt.
RK2 is used as default solver for first steps.
"""
n = len( t )
x = numpy.array( [ x0 ] * n )
# Start up with 4th order Runge-Kutta (single-step method). The extra
# code involving f0, f1, f2, and f3 helps us get ready for the multi-step
# method to follow in order to minimize the number of function evaluations
# needed.
f1 = f2 = f3 = 0
for i in xrange( min( 3, n - 1 ) ):
h = t[i+1] - t[i]
f0 = f( x[i], t[i] )
k1 = h * f0
k2 = h * f( x[i] + 0.5 * k1, t[i] + 0.5 * h )
k3 = h * f( x[i] + 0.5 * k2, t[i] + 0.5 * h )
k4 = h * f( x[i] + k3, t[i+1] )
x[i+1] = x[i] + ( k1 + 2.0 * ( k2 + k3 ) + k4 ) / 6.0
f1, f2, f3 = ( f0, f1, f2 )
for i in xrange( n - 1 ):
h = t[i+1] - t[i]
f0 = f( x[i], t[i] )
k1 = h * f0
k2 = h * f( x[i] + 0.5 * k1, t[i] + 0.5 * h )
k3 = h * f( x[i] + 0.5 * k2, t[i] + 0.5 * h )
k4 = h * f( x[i] + k3, t[i+1] )
x[i+1] = x[i] + h * ( 9.0 * fw + 19.0 * f0 - 5.0 * f1 + f2 ) / 24.0
f1, f2, f3 = ( f0, f1, f2 )
return x
def equation(X, t):
x, y = X
return [ x+y-exp(t), x+y+2*exp(t) ]
init = [ -1.0, -1.0 ]
t = linspace(0, 4, 50)
X = AdamsBashforth4(equation, init, t)
It seems that f is returning a sequence (list, tuple, or like) and there's no operation to multiply all items in a sequence by a value.
Dirk is right, looking at your algorithm, your equation should return (even if it small) a numpy array, as you can scalar-multiply a numpy array. So keeping your code but embedding your return in equation with an array. Like this:
np.array([1,2]) # a numpy array containing [1,2]
Related
I have wrote a code for Runge-Kutta 4th order, which works perfectly fine for a system of differential equations:
import numpy as np
import matplotlib.pyplot as plt
import numba
import time
start_time = time.clock()
#numba.jit()
def V(u,t):
x1,dx1, x2, dx2=u
ddx1=-w**2 * x1 -b * dx1
ddx2=-(w+0.5)**2 * x2 -(b+0.1) * dx2
return np.array([dx1,ddx1,dx2,ddx2])
#numba.jit()
def rk4(f, u0, t0, tf , n):
t = np.linspace(t0, tf, n+1)
u = np.array((n+1)*[u0])
h = t[1]-t[0]
for i in range(n):
k1 = h * f(u[i], t[i])
k2 = h * f(u[i] + 0.5 * k1, t[i] + 0.5*h)
k3 = h * f(u[i] + 0.5 * k2, t[i] + 0.5*h)
k4 = h * f(u[i] + k3, t[i] + h)
u[i+1] = u[i] + (k1 + 2*(k2 + k3) + k4) / 6
return u, t
u, t = rk4(V,np.array([0,0.2,0,0.3]) ,0,100, 20000)
print("Execution time:",time.clock() - start_time, "seconds")
x1,dx1,x2,dx2 = u.T
plt.plot(x1,x2)
plt.xlabel('X1')
plt.ylabel('X2')
plt.show()
The above code, returns the desired result:
And thanks to Numba JIT, this code works really fast. However, this method doesn't use adaptive step size and hence, it is not very suitable for a system of stiff differential equations. Runge Kutta Fehlberg method, solves this problem by using a straight forward algorithm. Based on the algorithm (https://en.wikipedia.org/wiki/Runge%E2%80%93Kutta%E2%80%93Fehlberg_method) I wrote this code which works for only one differential equation :
import numpy as np
def rkf( f, a, b, x0, tol, hmax, hmin ):
a2 = 2.500000000000000e-01 # 1/4
a3 = 3.750000000000000e-01 # 3/8
a4 = 9.230769230769231e-01 # 12/13
a5 = 1.000000000000000e+00 # 1
a6 = 5.000000000000000e-01 # 1/2
b21 = 2.500000000000000e-01 # 1/4
b31 = 9.375000000000000e-02 # 3/32
b32 = 2.812500000000000e-01 # 9/32
b41 = 8.793809740555303e-01 # 1932/2197
b42 = -3.277196176604461e+00 # -7200/2197
b43 = 3.320892125625853e+00 # 7296/2197
b51 = 2.032407407407407e+00 # 439/216
b52 = -8.000000000000000e+00 # -8
b53 = 7.173489278752436e+00 # 3680/513
b54 = -2.058966861598441e-01 # -845/4104
b61 = -2.962962962962963e-01 # -8/27
b62 = 2.000000000000000e+00 # 2
b63 = -1.381676413255361e+00 # -3544/2565
b64 = 4.529727095516569e-01 # 1859/4104
b65 = -2.750000000000000e-01 # -11/40
r1 = 2.777777777777778e-03 # 1/360
r3 = -2.994152046783626e-02 # -128/4275
r4 = -2.919989367357789e-02 # -2197/75240
r5 = 2.000000000000000e-02 # 1/50
r6 = 3.636363636363636e-02 # 2/55
c1 = 1.157407407407407e-01 # 25/216
c3 = 5.489278752436647e-01 # 1408/2565
c4 = 5.353313840155945e-01 # 2197/4104
c5 = -2.000000000000000e-01 # -1/5
t = a
x = np.array(x0)
h = hmax
T = np.array( [t] )
X = np.array( [x] )
while t < b:
if t + h > b:
h = b - t
k1 = h * f( x, t )
k2 = h * f( x + b21 * k1, t + a2 * h )
k3 = h * f( x + b31 * k1 + b32 * k2, t + a3 * h )
k4 = h * f( x + b41 * k1 + b42 * k2 + b43 * k3, t + a4 * h )
k5 = h * f( x + b51 * k1 + b52 * k2 + b53 * k3 + b54 * k4, t + a5 * h )
k6 = h * f( x + b61 * k1 + b62 * k2 + b63 * k3 + b64 * k4 + b65 * k5, \
t + a6 * h )
r = abs( r1 * k1 + r3 * k3 + r4 * k4 + r5 * k5 + r6 * k6 ) / h
if len( np.shape( r ) ) > 0:
r = max( r )
if r <= tol:
t = t + h
x = x + c1 * k1 + c3 * k3 + c4 * k4 + c5 * k5
T = np.append( T, t )
X = np.append( X, [x], 0 )
h = h * min( max( 0.84 * ( tol / r )**0.25, 0.1 ), 4.0 )
if h > hmax:
h = hmax
elif h < hmin:
raise RuntimeError("Error: Could not converge to the required tolerance %e with minimum stepsize %e." % (tol,hmin))
break
return ( T, X )
but I'm struggling to convert it to a function like the first code, where I can input a system of differential equations. The most confusing part for me, is how can I vectorize everything in the second code without messing things up. In other words, I cannot reproduce the first result using the RKF algorithm. Can anyone point me in the right direction?
I'm not really sure where your problem lies. Setting the not given parameters to w=1; b=0.1 and calling, without changing anything
T, X = rkf( f=V, a=0, b=100, x0=[0,0.2,0,0.3], tol=1e-6, hmax=1e1, hmin=1e-16 )
gives the phase plot
The step sizes grow as the system slows down as
which is the expected behavior for an unfiltered step size controller.
This python code can solve one non- coupled differential equation:
import numpy as np
import matplotlib.pyplot as plt
import numba
import time
start_time = time.clock()
#numba.jit()
# A sample differential equation "dy / dx = (x - y**2)/2"
def dydx(x, y):
return ((x - y**2)/2)
# Finds value of y for a given x using step size h
# and initial value y0 at x0.
def rungeKutta(x0, y0, x, h):
# Count number of iterations using step size or
# step height h
n = (int)((x - x0)/h)
# Iterate for number of iterations
y = y0
for i in range(1, n + 1):
"Apply Runge Kutta Formulas to find next value of y"
k1 = h * dydx(x0, y)
k2 = h * dydx(x0 + 0.5 * h, y + 0.5 * k1)
k3 = h * dydx(x0 + 0.5 * h, y + 0.5 * k2)
k4 = h * dydx(x0 + h, y + k3)
# Update next value of y
y = y + (1.0 / 6.0)*(k1 + 2 * k2 + 2 * k3 + k4)
# Update next value of x
x0 = x0 + h
return y
def dplot(start,end,steps):
Y=list()
for x in np.linspace(start,end,steps):
Y.append(rungeKutta(x0, y, x , h))
plt.plot(np.linspace(start,end,steps),Y)
print("Execution time:",time.clock() - start_time, "seconds")
plt.show()
start,end = 0, 10
steps = end* 100
x0 = 0
y = 1
h = 0.002
dplot(start,end,steps)
This code can solve this differential equation:
dydx= (x - y**2)/2
Now I have a system of coupled differential equations:
dydt= (x - y**2)/2
dxdt= x*3 + 3y
How can I implement these two as a system of coupled differential equations in the above code?
Is there any more generalized way for system of n-number of coupled differential equations?
With the help of others, I got to this:
import numpy as np
from math import sqrt
import matplotlib.pyplot as plt
import numba
import time
start_time = time.clock()
a=1
b=1
c=1
d=1
# Equations:
#numba.jit()
#du/dt=V(u,t)
def V(u,t):
x, y, vx, vy = u
return np.array([vy,vx,a*x+b*y,c*x+d*y])
def rk4(f, u0, t0, tf , n):
t = np.linspace(t0, tf, n+1)
u = np.array((n+1)*[u0])
h = t[1]-t[0]
for i in range(n):
k1 = h * f(u[i], t[i])
k2 = h * f(u[i] + 0.5 * k1, t[i] + 0.5*h)
k3 = h * f(u[i] + 0.5 * k2, t[i] + 0.5*h)
k4 = h * f(u[i] + k3, t[i] + h)
u[i+1] = u[i] + (k1 + 2*(k2 + k3 ) + k4) / 6
return u, t
u, t = rk4(V, np.array([1., 0., 1. , 0.]) , 0. , 10. , 100000)
x,y, vx,vy = u.T
# plt.plot(t, x, t,y)
plt.semilogy(t, x, t,y)
plt.grid('on')
print("Execution time:",time.clock() - start_time, "seconds")
plt.show()
I'd like to use an implementation of RK4 I found online for something, but I'm having a bit of difficulty wrapping my head around the implementations I have found online.
For example:
def rk4(f, x0, y0, x1, n):
vx = [0] * (n + 1)
vy = [0] * (n + 1)
h = (x1 - x0) / float(n)
vx[0] = x = x0
vy[0] = y = y0
for i in range(1, n + 1):
k1 = h * f(x, y)
k2 = h * f(x + 0.5 * h, y + 0.5 * k1)
k3 = h * f(x + 0.5 * h, y + 0.5 * k2)
k4 = h * f(x + h, y + k3)
vx[i] = x = x0 + i * h
vy[i] = y = y + (k1 + k2 + k2 + k3 + k3 + k4) / 6
return vx, vy
Could someone please help me understand what exactly the parameters are? If possible, I'd like a more general explanation, but, if being more specific makes it easier to explain, I'm going to be using it specifically for an ideal spring system.
You are asking for the parameters here:
def rk4(f, x0, y0, x1, n):
...
return vx, vy
f is the ODE function, declared as def f(x,y) for the differential equation y'(x)=f(x,y(x)),
(x0,y0) is the initial point and value,
x1 is the end of the integration interval [x0,x1]
n is the number of sub-intervals or integration steps
vx,vx are the computed sample points, vy[k] approximates y(vx[k]).
You can not use this for the spring system, as that code only works for a scalar v. You would need to change it to work with numpy for vector operations.
def rk4(func, x0, y0, x1, n):
y0 = np.array(y0)
f = lambda x,y: np.array(func(x,y))
vx = [0] * (n + 1)
vy = np.zeros( (n + 1,)+y0.shape)
h = (x1 - x0) / float(n)
vx[0] = x = x0
vy[0] = y = y0[:]
for i in range(1, n + 1):
k1 = h * f(x, y)
k2 = h * f(x + 0.5 * h, y + 0.5 * k1)
k3 = h * f(x + 0.5 * h, y + 0.5 * k2)
k4 = h * f(x + h, y + k3)
vx[i] = x = x0 + i * h
vy[i] = y = y + (k1 + 2*(k2 + k3) + k4) / 6
return vx, vy
Hey guys I'm receiving an error and I'm not 100% sure why it's happening. I'm trying to perform both the eulers and heun's methods and plot them against one another. Any help or suggestions would be awesome! I'm pretty novice with python so this is a bit new to me.
import numpy
#eulers
yi = 0
h = 0.2
te = [0.0,2.0,0.2]
def euler(f, y0, t):
n = len(t)
y = numpy.array([y0] * n)
for i in range(n - 1):
y[i+1] = y[i] + (t[i+1] - t[i]) * f(y[i], t[i])
return y
fy = (y*t**3) - 1.5*y
y1 = euler(fy, yi, te)
#heun
def heun(f, y0, t):
n = len(t)
y = numpy.array( [y0] * n )
for i in range(n - 1):
h = t[i+1] - t[i]
k1 = h * f( y[i], t[i])
k2 = h * f( y[i] + k1, t[i+1])
y[i+1] = y[i] + (k1 + k2) / 2.0
return y
y2 = heun(fy, 1, te)
tp = numpy.linspace(0.0, 2.0, 0.2)
plt.plot(tp, y1,'r-',linewidth=2,label='Eulers')
plt.plot(tp, y2,'b--',linewidth=2,label='Heun')
plt.xlabel('time')
plt.ylabel('y(t)')
plt.legend()
plt.show()
This is my error code:
TypeError Traceback (most recent call last)
<ipython-input-152-395afeefc50a> in <module>()
11 return y
12 fy = (y*t**3) - 1.5*y
---> 13 y1 = euler(fy, yi, te)
14
15 #heun
<ipython-input-152-395afeefc50a> in euler(f, y0, t)
8 y = numpy.array([y0] * n)
9 for i in range(n - 1):
---> 10 y[i+1] = y[i] + (t[i+1] - t[i]) * f(y[i], t[i])
11 return y
12 fy = (y*t**3) - 1.5*y
TypeError: 'numpy.ndarray' object is not callable
My guess is that you're trying to define a function f(y,t) in the line
fy = (y*t**3) - 1.5*y
as your differential equation. However, that is not how you define a function in python.
Try
def f(y,t):
return y*t**3 - 1.5*y
You are getting an error because on line
---> 10 y[i+1] = y[i] + (t[i+1] - t[i]) * f(y[i], t[i])
f is a NumPy array, but you are treating it as a function.
I am trying to implement the runge-kutta method to solve a Lotka-Volterra systtem, but the code (bellow) is not working properly. I followed the recomendations that I found in other topics of the StackOverflow, but the results do not converge with the builtin Runge-Kutta method, like rk4 method available in Pylab, for example. Someone could help me?
import matplotlib.pyplot as plt
import numpy as np
from pylab import *
def meurk4( f, x0, t ):
n = len( t )
x = np.array( [ x0 ] * n )
for i in range( n - 1 ):
h = t[i+1] - t[i]
k1 = h * f( x[i], t[i] )
k2 = h * f( x[i] + 0.5 * h * k1, t[i] + 0.5 * h )
k3 = h * f( x[i] + 0.5 * h * k2, t[i] + 0.5 * h )
k4 = h * f( x[i] + h * k3, t[i] + h)
x[i+1] = x[i] + ( k1 + 2 * ( k2 + k3 ) + k4 ) * 6**-1
return x
def model(state,t):
x,y = state
a = 0.8
b = 0.02
c = 0.2
d = 0.004
k = 600
return np.array([ x*(a*(1-x*k**-1)-b*y) , -y*(c - d*x) ]) # corresponds to [dx/dt, dy/dt]
# initial conditions for the system
x0 = 500
y0 = 200
# vector of time
t = np.linspace( 0, 50, 100 )
result = meurk4( model, [x0,y0], t )
print result
plt.plot(t,result)
plt.xlabel('Time')
plt.ylabel('Population Size')
plt.legend(('x (prey)','y (predator)'))
plt.title('Lotka-Volterra Model')
plt.show()
I just updated the code following the comments. So, the function meurk4:
def meurk4( f, x0, t ):
n = len( t )
x = np.array( [ x0 ] * n )
for i in range( n - 1 ):
h = t[i+1] - t[i]
k1 = h * f( x[i], t[i] )
k2 = h * f( x[i] + 0.5 * h * k1, t[i] + 0.5 * h )
k3 = h * f( x[i] + 0.5 * h * k2, t[i] + 0.5 * h )
k4 = h * f( x[i] + h * k3, t[i] + h)
x[i+1] = x[i] + ( k1 + 2 * ( k2 + k3 ) + k4 ) * 6**-1
return x
Becomes now (corrected):
def meurk4( f, x0, t ):
n = len( t )
x = np.array( [ x0 ] * n )
for i in range( n - 1 ):
h = t[i+1] - t[i]
k1 = f( x[i], t[i] )
k2 = f( x[i] + 0.5 * h * k1, t[i] + 0.5 * h )
k3 = f( x[i] + 0.5 * h * k2, t[i] + 0.5 * h )
k4 = f( x[i] + h * k3, t[i] + h)
x[i+1] = x[i] + ( k1 + 2 * ( k2 + k3 ) + k4 ) * (h/6)
return x
Nevertheless, the results is the following:
While the buitin method rk4 (from Pylab) results the following:
So, certainly my code still is not correct, as its results are not the same of the builtin rk4 method. Please, someone can help me?
You are doing a very typical error,see for instance How to pass a hard coded differential equation through Runge-Kutta 4 or here Error in RK4 algorithm in Python
It is either
k2 = f( x+0.5*h*k1, t+0.5*h )
...
x[i+1]=x[i]+(k1+2*(k2+k3)+k4)*(h/6)
or
k2 = h*f( x+0.5*k1, t+0.5*h )
and so on, with x[i+1] as it was, but not both variants at the same time.
Update: A more insidious error is the inferred type of the initial values and in consequence of the array of x vectors. By the original definition, both are integers, and thus
x = np.array( [ x0 ] * n )
creates a list of integer vectors. Thus the update step
x[i+1] = x[i] + ( k1 + 2 * ( k2 + k3 ) + k4 ) * (h/6)
will always round to integer. And since there is a phase where both values fall below 1, the integration stabilizes at zero. Thus modify to
# initial conditions for the system
x0 = 500.0
y0 = 200.0
to avoid that problem.