How to solve a pair of nonlinear equations using Python?
Question:
What’s the (best) way to solve a pair of non linear equations using Python. (Numpy, Scipy or Sympy)
eg:
 x+y^2 = 4
 e^x+ xy = 3
A code snippet which solves the above pair will be great
Answers:
for numerical solution, you can use fsolve:
http://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.fsolve.html#scipy.optimize.fsolve
from scipy.optimize import fsolve
import math
def equations(p):
x, y = p
return (x+y**24, math.exp(x) + x*y  3)
x, y = fsolve(equations, (1, 1))
print equations((x, y))
If you prefer sympy you can use nsolve.
>>> nsolve([x+y**24, exp(x)+x*y3], [x, y], [1, 1])
[0.620344523485226]
[1.83838393066159]
The first argument is a list of equations, the second is list of variables and the third is an initial guess.
You can use openopt package and its NLP method. It has many dynamic programming algorithms to solve nonlinear algebraic equations consisting:
goldenSection, scipy_fminbound, scipy_bfgs, scipy_cg, scipy_ncg, amsg2p, scipy_lbfgsb, scipy_tnc, bobyqa, ralg, ipopt, scipy_slsqp, scipy_cobyla, lincher, algencan, which you can choose from.
Some of the latter algorithms can solve constrained nonlinear programming problem.
So, you can introduce your system of equations to openopt.NLP() with a function like this:
lambda x: x[0] + x[1]**2  4, np.exp(x[0]) + x[0]*x[1]
I got Broyden’s method to work for coupled nonlinear equations (generally involving polynomials and exponentials) in IDL, but I haven’t tried it in Python:
scipy.optimize.broyden1
scipy.optimize.broyden1(F, xin, iter=None, alpha=None, reduction_method='restart', max_rank=None, verbose=False, maxiter=None, f_tol=None, f_rtol=None, x_tol=None, x_rtol=None, tol_norm=None, line_search='armijo', callback=None, **kw)[source]
Find a root of a function, using Broyden’s first Jacobian approximation.
This method is also known as “Broyden’s good method”.
Try this one, I assure you that it will work perfectly.
import scipy.optimize as opt
from numpy import exp
import timeit
st1 = timeit.default_timer()
def f(variables) :
(x,y) = variables
first_eq = x + y**2 4
second_eq = exp(x) + x*y  3
return [first_eq, second_eq]
solution = opt.fsolve(f, (0.1,1) )
print(solution)
st2 = timeit.default_timer()
print("RUN TIME : {0}".format(st2st1))
>
[ 0.62034452 1.83838393]
RUN TIME : 0.0009331008900937708
FYI. as mentioned above, you can also use ‘Broyden’s approximation’ by replacing ‘fsolve’ with ‘broyden1’. It works. I did it.
I don’t know exactly how Broyden’s approximation works, but it took 0.02 s.
And I recommend you do not use Sympy’s functions < convenient indeed, but in terms of speed, it’s quite slow. You will see.
from scipy.optimize import fsolve
def double_solve(f1,f2,x0,y0):
func = lambda x: [f1(x[0], x[1]), f2(x[0], x[1])]
return fsolve(func,[x0,y0])
def n_solve(functions,variables):
func = lambda x: [ f(*x) for f in functions]
return fsolve(func, variables)
f1 = lambda x,y : x**2+y**21
f2 = lambda x,y : xy
res = double_solve(f1,f2,1,0)
res = n_solve([f1,f2],[1.0,0.0])
An alternative to fsolve
is root
:
import numpy as np
from scipy.optimize import root
def your_funcs(X):
x, y = X
# all RHS have to be 0
f = [x + y**2  4,
np.exp(x) + x * y  3]
return f
sol = root(your_funcs, [1.0, 1.0])
print(sol.x)
This will print
[0.62034452 1.83838393]
If you then check
print(your_funcs(sol.x))
you obtain
[4.4508396968012676e11, 1.0512035686360832e11]
confirming that the solution is correct.
Short answer: use fsolve
As mentioned in other answers the simplest solution to the particular problem you have posed is to use something like fsolve
:
from scipy.optimize import fsolve
from math import exp
def equations(vars):
x, y = vars
eq1 = x+y**24
eq2 = exp(x) + x*y  3
return [eq1, eq2]
x, y = fsolve(equations, (1, 1))
print(x, y)
Output:
0.6203445234801195 1.8383839306750887
Analytic solutions?
You say how to “solve” but there are different kinds of solution. Since you mention SymPy I should point out the biggest difference between what this could mean which is between analytic and numeric solutions. The particular example you have given is one that does not have an (easy) analytic solution but other systems of nonlinear equations do. When there are readily available analytic solutions SymPY can often find them for you:
from sympy import *
x, y = symbols('x, y')
eq1 = Eq(x+y**2, 4)
eq2 = Eq(x**2 + y, 4)
sol = solve([eq1, eq2], [x, y])
Output:
⎡⎛ ⎛ 5 √17⎞ ⎛3 √17⎞ √17 1⎞ ⎛ ⎛ 5 √17⎞ ⎛3 √17⎞ 1 √17⎞ ⎛ ⎛ 3 √13⎞ ⎛√13 5⎞ 1 √13⎞ ⎛ ⎛5 √13⎞ ⎛ √13 3⎞ 1 √13⎞⎤
⎢⎜⎜ ─  ───⎟⋅⎜─  ───⎟,  ───  ─⎟, ⎜⎜ ─ + ───⎟⋅⎜─ + ───⎟,  ─ + ───⎟, ⎜⎜ ─ + ───⎟⋅⎜─── + ─⎟, ─ + ───⎟, ⎜⎜─  ───⎟⋅⎜ ───  ─⎟, ─  ───⎟⎥
⎣⎝ ⎝ 2 2 ⎠ ⎝2 2 ⎠ 2 2⎠ ⎝ ⎝ 2 2 ⎠ ⎝2 2 ⎠ 2 2 ⎠ ⎝ ⎝ 2 2 ⎠ ⎝ 2 2⎠ 2 2 ⎠ ⎝ ⎝2 2 ⎠ ⎝ 2 2⎠ 2 2 ⎠⎦
Note that in this example SymPy finds all solutions and does not need to be given an initial estimate.
You can evaluate these solutions numerically with evalf
:
soln = [tuple(v.evalf() for v in s) for s in sol]
[(2.56155281280883, 2.56155281280883), (1.56155281280883, 1.56155281280883), (1.30277563773199, 2.30277563773199), (2.30277563773199, 1.30277563773199)]
Precision of numeric solutions
However most systems of nonlinear equations will not have a suitable analytic solution so using SymPy as above is great when it works but not generally applicable. That is why we end up looking for numeric solutions even though with numeric solutions:
1) We have no guarantee that we have found all solutions or the “right” solution when there are many.
2) We have to provide an initial guess which isn’t always easy.
Having accepted that we want numeric solutions something like fsolve
will normally do all you need. For this kind of problem SymPy will probably be much slower but it can offer something else which is finding the (numeric) solutions more precisely:
from sympy import *
x, y = symbols('x, y')
nsolve([Eq(x+y**2, 4), Eq(exp(x)+x*y, 3)], [x, y], [1, 1])
⎡0.620344523485226⎤
⎢ ⎥
⎣1.83838393066159 ⎦
With greater precision:
nsolve([Eq(x+y**2, 4), Eq(exp(x)+x*y, 3)], [x, y], [1, 1], prec=50)
⎡0.62034452348522585617392716579154399314071550594401⎤
⎢ ⎥
⎣ 1.838383930661594459049793153371142549403114879699 ⎦
You can use nsolve
of sympy
, meaning numerical solver
.
Example snippet:
from sympy import *
L = 4.11 * 10 ** 5
nu = 1
rho = 0.8175
mu = 2.88 * 10 ** 6
dP = 20000
eps = 4.6 * 10 ** 5
Re, D, f = symbols('Re, D, f')
nsolve((Eq(Re, rho * nu * D / mu),
Eq(dP, f * L / D * rho * nu ** 2 / 2),
Eq(1 / sqrt(f), 1.8 * log ( (eps / D / 3.) ** 1.11 + 6.9 / Re))),
(Re, D, f), (1123, 1231, 1000))
where (1123, 1231, 1000)
is the initial vector to find the root. And it gives out:
The imaginary part are very small, both at 10^(20), so we can consider them zero, which means the roots are all real. Re ~ 13602.938, D ~ 0.047922 and f~0.0057.
Best way is the write your own solver. And evolve it.

Try a for loop to change x and y values by a small increment for example eps=0.0001

Then try feeding back the error with an slope modifier err/eps

Then adjust the modifier eps according to how fast you approach
—> here you have reached to Newtonian iteration by yourself … 
Keep working numerically … introduce multiple dimensions which will automatically induce partial differential (how fast you approach)

You will be reaching to gradient descent , simplex and alike optimisation methods.

Always compare your numerical solvers with analytical results from sympy
In my opinion it is the best way because you can adjust it according to your needs and cases. And you will be doing things informed and issues are grasped.
Only after this you can pick a solver comparing to yours. Or you can use them side by side.