Also introducing lambda function wasn't necessary:
import scipy.integrate
a=1.0
b=2.0
def fun(t):
if t<=-b:
return -a
elif f<b:
return t*a/b
else:
return a
time_step=0.1
time_end=10.0
t0=0.0
def f(x,t):
return [x[1],-x[0]+fun(x[0])]
time_range=[t0..time_end, step=time_step]
x0=[[0.5*k,0.5*k] for k in range(-10,10)]
sol_lines = Graphics()
for n in range(10):
sol = scipy.integrate.odeint(f,x0[n],time_range)
sol_lines += line(sol,rgbcolor=hue(.3+n/15.0))
x0,x1=var('x0 x1')
p=plot_vector_field ((x1,-x0+fun(x0)),(x0,-9,9),(x1,-7,7))
show(sol_lines + p, figsize = [9,7])
In general, it is not easy to predict "right" time_end and "right" x0
list
Perhaps Maple code can be helpful?
Andrzej Chrzeszczyk
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