I am trying to solve a system of Odes of the form Du/Dt = F(u) in python, and I suspect I may have made a fairly dumb mistake somewhere.
Technically F(u) is actually the second derivative of u with respect to another variable y, but in practice we can consider it to be a system and some function.
#Settings#
minx = -20
h = float(1)
w = float(10)
U0 = float(10)
Nt = 10
Ny = 10
tmax = 10
v=float(1)
#Arrays#
y = np.linspace(0,h,Ny)
t = np.linspace(0,tmax,Nt)
#Variables from arrays#
dt = t[1]-t[0]
p = [0]*(Nt)
delta = y[1] - y[0]
def zo(y):
return math.cos(y/(2*math.pi))
z0 = [zo(i) for i in y]
def df(t,v1):
output = np.zeros(len(y))
it = 1
output[0] = math.cos(w*t)
output[len(y)-1] = math.cos(w*t)
while it < len(y)-1:
output[it] = ( v1[it - 1] + v1[it + 1] - 2 * v1[it] ) * ( v / ( ( delta )**2 ))
it += 1
return output
r = ode(df).set_integrator('zvode', method='bdf',order =15)
r.set_initial_value(z0, 0)
it=0
while r.successful() and r.t < tmax:
p[it] = r.integrate(r.t+dt)
it+=1
print(z0-p[0])
print(p[1])
Now the problem is twofold :
-First of all, the initial "condition" ie p[0] seems to be off. (That may be just because of the way the ode function works though, so I don't know if that is normal)
-Second, p[1] and all p's after that are just 0.
So for some reason the ode function fails immediately... (you can check that by changing the values to 1 when initializing p)
Except that I know that this method should work. This is the "equivalent" to ode45 in matlab after all and that definitely works.
Why did you choose a complex solver with an implicit backward differentiation formula of a rather high order if you wanted to use Dormand-Price rk45 resp. dopri5?
Please also correct the loop indentation in df. Why not a for loop over range(1, len(y)-1)?
As it currently stands p[0] contains the solution point after the first step, at t=1*dt. You would have to explicitly assign p[0]=z0 and start it=1 to get the full solution path in p. Check the length of p, it could be that you need Nt+1.