Solve system of differential equation in python - scipy

Question

I'm trying to solve a system of differential equations in python.
I have a system composed by two equations where I have two variables, A and B.
The initial condition are that A0=1e17 and B0=0, they change simultaneously.
I wrote the following code using ODEINT:
import numpy as np
from scipy.integrate import odeint
def dmdt(m,t):
A, B = m
dAdt = A-B
dBdt = (A-B)*A
return [dAdt, dBdt]
# Create time domain
t = np.linspace(0, 100, 1)
# Initial condition
A0=1e17
B0=0
m0=[A0, B0]
solution = odeint(dmdt, m0, t)
Apparently I obtain an output different from the expected one but I don't understand the error.
Can someone help me?
Thanks

Answer 1

From A*A'-B'=0 one concludes
B = 0.5*(A^2 - A0^2)
Inserted into the first equation that gives
A' = A - 0.5*A^2 + 0.5*A0^2
= 0.5*(A0^2+1 - (A-1)^2)
This means that the A dynamic has two fixed points at about A0+1 and -A0+1, is growing inside that interval, the upper fixed point is stable. However, in standard floating point numbers there is no difference between 1e17 and 1e17+1. If you want to see the difference, you have to encode it separately.
Also note that the standard error tolerances atol and rtol in the range somewhere between 1e-6 and 1e-9 are totally incompatible with the scales of the problem as originally stated, also highlighting the need to rescale and shift the problem into a more appreciable range of values.
Setting A = A0+u with |u| in an expected scale of 1..10 then gives
B = 0.5*u*(2*A0+u)
u' = A0+u - 0.5*u*(2*A0+u) = (1-u)*A0 - 0.5*u^2
This now suggests that the time scale be reduced by A0, set t=s/A0. Also, B = A0*v. Insert the direct parametrizations into the original system to get
du/ds = dA/dt / A0 = (A0+u-A0*v)/A0 = 1 + u/A0 - v
dv/ds = dB/dt / A0^2 = (A0+u-A0*v)*(A0+u)/A0^2 = (1+u/A0-v)*(1+u/A0)
u(0)=v(0)=0
Now in floating point and the expected range for u, we get 1+u/A0 == 1, so effectively u'(s)=v'(s)=1-v which gives
u(s)=v(s)=1-exp(-s)`,
A(t) = A0 + 1-exp(-A0*t) + very small corrections
B(t) = A0*(1-exp(-A0*t)) + very small corrections
The system in s,u,v should be well-computable by any solver in the default tolerances.