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endless loop using ode_solve

asked 2010-08-22 16:26:49 -0500

ngativ gravatar image

updated 2010-08-22 17:55:41 -0500

Im solving a system equation within a for loop, but it never ends But its ok to evaluate f1(n) outside the loop.... why?

def f1(ll):

    def f1(ll):
        f = lambda t,y:[y[1],eqa.subs(l=ll,s=y[0],w=y[1],x=t)]
        return s

for i in range(3) :f1(i)

doesn't works except when f(1) ,otherwise  the solver is unable to end. But f(1) or eqa(l=1) is the real solution--- :S

where eqa is

var('l s x w')
eqa=((l^2 + l)*s - 2*w*x)/(x^2 - 1)

in the other hand this works

def g(ll):
    return (Sol[len(Sol)-1][1]-1)
for i in range(3) :f1(i)

so.. whats the problem with ode_solver?

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Just to check: Are you solving Legendre's DE?

Mitesh Patel gravatar imageMitesh Patel ( 2010-08-22 18:50:47 -0500 )edit

yea, im just trying to learn Sage solving problems like this. for Pl(x)=x i found the eigenvalue l= 0.994838832387 using desolve_system_rk4. I just wanna to solve the problem using ode_solver and odeint.

ngativ gravatar imagengativ ( 2010-08-22 18:57:29 -0500 )edit

how i do backtrace? the script doesn't crash.... it just calculate forever

ngativ gravatar imagengativ ( 2010-08-22 19:00:59 -0500 )edit

Well... it seems to be that when the solver evaluates the function (eqa(t=1)) the solution diverges because the 1/t factor. But this doesn't happens with the desolver_system_rk4. i dont like this, maybe ode_solver is faster but.. it really has to have this kind of problems?

ngativ gravatar imagengativ ( 2010-08-22 22:31:13 -0500 )edit

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answered 2010-08-27 08:57:10 -0500

mhampton gravatar image

It seems like this is just ridiculously slow, but I'm not sure why. I tried tweaking your code in a variety of ways and nothing helped. To try to figure out what was going on I wrote the following Runge-Kutta 4th order code in Cython; its pretty hackish but it makes it clear that the solution isn't doing anything very strange. The version below is hard-coded for your ll=2.


cdef double ll = 2.0

cpdef double leg_f1(double t, double y, double yp):
    return yp

cpdef double leg_f2(double t, double y, double yp):
    return ((ll**2 + ll)*y - 2.0*yp*t)/(t**2 - 1.0)

cpdef RK4_2d(f1, f2, double t_start, double y_start, double yp_start, double t_end, int steps):
    cdef double step_size = (t_end - t_start)/steps
    cdef double t_current = t_start
    cdef double y_current = y_start
    cdef double yp_current = yp_start
    cdef list answer_table = []
    cdef int j
    answer_table.append([t_current,y_current, yp_current])
    for j in range(0,steps):
        k1=f1(t_current, y_current, yp_current)
        k2=f1(t_current+step_size/2, y_current + k1*step_size/2, yp_current + k1*step_size/2)
        k3=f1(t_current+step_size/2, y_current + k2*step_size/2, yp_current + k1*step_size/2)
        k4=f1(t_current+step_size, y_current + k3*step_size, yp_current + k1*step_size/2)
        y_current = y_current + (step_size/6)*(k1+2*k2+2*k3+k4)
        k1=f2(t_current, y_current, yp_current)
        k2=f2(t_current+step_size/2, y_current + k1*step_size/2, yp_current + k1*step_size/2)
        k3=f2(t_current+step_size/2, y_current + k2*step_size/2, yp_current + k1*step_size/2)
        k4=f2(t_current+step_size, y_current + k3*step_size, yp_current + k1*step_size/2)
        yp_current = yp_current + (step_size/6)*(k1+2*k2+2*k3+k4)
        t_current += step_size
    return answer_table

and then you can get a plot with:

ans = RK4_2d(leg_f1, leg_f2, 0.0, 0.0, 1.0, 1.0, 40)
list_plot([[x[0],x[1]] for x in ans], plotjoined = True)

Anyway for me it seems easier to implement such things directly than to wrestle with the GSL wrapping.

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is a convergence problem for l!=1.

ngativ gravatar imagengativ ( 2010-08-28 06:31:36 -0500 )edit

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Asked: 2010-08-22 16:26:49 -0500

Seen: 228 times

Last updated: Aug 27 '10