I try to find a solution for a system of equations by using scipy.optimize.fsolve
in python 2.7. The goal is to calculate equilibrium concentrations for a chemical
The root of your problem is either theoretical or numerical.
The scipy.optimize.fsolve
function is based on the MINPACK Fortran solver (http://www.netlib.org/minpack/). This solver use a Newton-Raphson optimisation algorithm to provide the solution.
There are underlying assumptions about the smoothness of the function when you use this algorithm. For example, the jacobian matrix at the solution point x
is supposed to be invertible. The one you are more concerned about is the basins of attraction.
In order to converge, the starting point of the algorithm needs to be near the actual solution, i.e. in the basins of attraction. This condition is always met for convex functions, however it is easy to find some functions for which this algorithm behaves badly. Your function is one of this as you have a fraction of your inputs parameters.
To address this issue you should just change the starting point. This starting point becomes also very important for functions with multiple solutions: this picture from the wikipedia article shows you the solution found depending of the starting point (five colours for five solutions); so you should be careful with your solution and actually check the "physical" aspects of your solution.
For the numerical aspects, the Newton-Raphson algorithm needs to have the value of the jacobian matrix (the derivatives matrix). If it is not provided to the MINPACK solver, the jacobian is estimated with a finite-difference formula. The perturbation step for the finite difference formula need to be provided epsfcn=None
, the None
being here as default value only in the case where fprime
is provided (there is no need for the jacobian estimation in this case). So first you should incorporate that. You could also specify directly the jacobian by derivating your function by hand.
However, the minimum value for the step size will be the machine precision, also called machine epsilon. For your problem, you have very small inputs values which can be a problem. I would suggest multiply everyone of them by the same value (like 10^6), it is equivalent to a change of the units but will avoid rounding up errors and problems with machine precision.
This problem is also important when you look at the parameter xtol=1e-15
you provided. In your error message, it gives xtol=0.000000
, as it is below machine precision and cannot be taken into account. Also, if you look at your line F[2] = K_W - OH*H3O
, given the machine precision, it does not matter if K_W
is 1e-15
or 1e-30
. 0
is a solution for both of this case compare to the machine precision. To avoid this problem, just multiply everything by a bigger value.
So to sum up: