I'm lerning to solve Differential equation using (scipy.integrate.odeint) and (scipy.integrate.ode). I have a simple example:
dy/dt=f[i]*t
and f is parameter corresponding to t[i], same like the example in code ; i.e.
t[0]=0.0, f[0]=0.0
t[1]=0.1, f[1]=0.1
...
t[10]=1.0, f[1]=1.0
the manually result should be:
y=1/2*f[i]*t**2, because the initial value of y is zero
then, numerical result of y should be
[0.0, 0.0005, 0.004, 0.0135, 0.032, 0.0625, 0.108, 0.1715, 0.256, 0.3645, 0.5]. But when i use scipy.integrate.ode, i have got a different result. My question is:
1. Have i used ode wrong? How can i reduce the errors?
2. Can i use odeint or other method to solve this problem?
The code is like:
import matplotlib.pyplot as pl
import numpy as np
import sympy as sp
from scipy.integrate import odeint
from scipy.integrate import ode
import numpy as np
def func(t, y, f):
return f*t
t=np.linspace(0.0, 1.0, 11)
f=np.linspace(0.0, 1.0, 11)
dt = t[1]-t[0]
sol= np.empty_like(t)
r = ode(func).set_integrator("dopri5")
r.set_initial_value(0, 0).set_f_params(f[0])
# result of ode
for i in xrange(len(t)):
r.set_f_params(f[i])
r.integrate(r.t+dt)
sol[i] = r.y
res=[]
# result of t**3/3
for a in np.linspace(0.0, 1, 11):
f=(a**3)/3
print f
res.append(f)
# result3
res2=[]
for n in range(0, 11):
dt=0.1
y= t[n]**3/3 - dt*t[n]**2/4 - dt**2*t[n]/12
res2.append(y)
pl.plot(sol)
pl.plot(res)
pl.plot(res2)
pl.show()
I have extend this example to 2-dimensional-differential equations:
du/dt=-u(v-f[i])
dv/dt=v(f[i]-u)
with initial values: u(0)=v(0)=1. And below is the code:
import matplotlib.pyplot as pl
import numpy as np
import sympy as sp
from scipy.integrate import odeint
from scipy.integrate import ode
from numpy import array
def func(t, y, f):
u,v=y
dotu=-u*(v-f)
dotv=v*(f-u)
return array([dotu, dotv])
t=np.linspace(0.0, 10, 11)
f=np.linspace(0.0, 20, 11)
dt = t[1]-t[0]
# result correct
y0=array([1.0, 1.0])
sol= np.empty([11, 2])
sol[0] = array([1.0, 1.0])
r = ode(func).set_integrator("dopri5")
r.set_initial_value(t[0], sol[0]).set_f_params(f[0])
for i in range(len(t)-1):
r.set_f_params(f[i])
r.integrate(r.t+dt)
sol[i+1] = r.y
pl.plot(sol[:,0])
But i get a error message:
Traceback (most recent call last):
File "C:\Users\odeint test.py", line 26, in <module>
sol[0] = array([1.0, 1.0])
ValueError: setting an array element with a sequence.
What you are doing is closer to integrating y'(t)=t^2, y(0)=0, resulting in y(t)=t^3/3. That you factor t^2 as f*t and hack f into a step function version of t only adds a small perturbation to that.
The integral of t[i]*t over t[i]..t[i+1] is
y[i+1]-y[i] = t[i]/2*(t[i+1]^2-t[i]^2)
= (t[i+1]^3-t[i]^3)/3 - (t[i+1]-t[i])^2*(t[i]+2t[i+1])/6
= (t[i+1]^3-t[i]^3)/3 - dt*(t[i+1]^2-t[i]^2)/4 - dt^2*(t[i+1]-t[i])/12
which sums up to about
y[n] = t[n]^3/3 - dt*t[n]^2/4 - dt^2*t[n]/12
sol= np.empty_like(t)
set the initial value
sol[0] = 0
r = ode(func).set_integrator("dopri5")
use the initial point as initial point, both to make explicit that the point at index 0 is fixed and "used up"
r.set_initial_value(sol[0],t[0]).set_f_params(f[0])
# result of ode
go from point at t[i] to point at t[i+1]. Ends with i+1=len(t) or i=len(t)-1
for i in xrange(len(t)-1):
r.set_f_params(f[i])
r.integrate(r.t+dt)
value at t[i]+dt is value at t[i+1]
sol[i+1] = r.y
With these changes the numerical solution coincides with the manually computed solution.