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

Solution 1

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**2-4, math.exp(x) + x*y - 3)

x, y =  fsolve(equations, (1, 1))

print equations((x, y))

Solution 2

If you prefer sympy you can use nsolve.

>>> nsolve([x+y**2-4, exp(x)+x*y-3], [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.

Solution 3

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**2-4
    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 ⎦

Solution 4

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(st2-st1))

->

[ 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.

Solution 5

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

[4.4508396968012676e-11, -1.0512035686360832e-11]

confirming that the solution is correct.

View more solutions

Comments

[AIB]

yea I know that..I wish to do it in python, because I want to do it repetitively for different sets of equations

[Blender]

You can import sage from any Python script.

[endolith]

sage does it by being a wrapper for sympy and maxima, so you could just use those directly.

[aquirdturtle]

Why is this being down-voted? Seems like a legitimate way to proceed...

[Sander Heinsalu]

I get the error 'name y is not defined' with the code in this answer.

[Krastanov]

@SanderHeinsalu, just follow what the error message is saying. If "name y is not defined", define it (python can not magically know what you want undefined variables to be). For instance here you want y to be a symbol object you can use to build bigger symbolic objects: y = Symbol('symbol_name_string'). Probably you want to keep the same symbol name, so y = Symbol('y').

[Edoardo Serra]

Don't know why this isn't the most voted answer, however, is there a way to convert the analytical solutions given by SymPy to a list of approximate numerical values? As of my understanding, the only way to find all of the solutions is via the analytical method, but having those solutions converted could be very useful.

[Oscar Benjamin]

You can numerically evaluate any sympy expression that does not have free symbols using expr.evalf(): docs.sympy.org/latest/modules/evalf.html

[Oscar Benjamin]

I've added an example with evalf

[Edoardo Serra]

Thank you very much!

[Edoardo Serra]

Yet another question I'm sorry. How can I separate the real from the complex solutions, for example by displaying only the real ones?

[eric]

@Edoardo np.real(x) and np.imag(x)

[Andrew]

I get (4.4508396968012676e-11, -1.0512035686360832e-11) as an answer, but this does not work: x+y^2 = 4 != 4.4508396968012676e-11+(-1.0512035686360832e-11)**2 = 4.4508396968123175e-11. equations() returns (0,0) according to what was entered and the original question, so apparently these two small #s are its attempt at that? Also what is the "1, 1" , and where does it come from. Just trying to understand... thanks.

[creanion]

For sympy, you use the sympy.real and sympy.imag functions, too. (In numpy, I'd just acess .real and .imag as attributes on the relevant values!)

[Oscar Benjamin]

If the symbols are declared as symbols('x, y', real=True) then solve will try to return only real solutions. Alternatively you can filter the solutions with something like Reals & set(solve(z**4 - 1, z)) or [s for s in sol if s.is_real]. Note that sometimes it isn't easy to verify whether or not a solution expression is real.

[billjoie]

An advantage of root over fsolve is the easy ability to specify the resolution method and tolerance.

[Jed]

@Andrew, the output of equations((x, y)) is the result of x + y ** 2 - 4 and math.exp(x) + x * y - 3. This shows you that the 2 formulas that were set to 0 in the function equations are now 0 with the values found for x and y. If you print((x, y)) you'll get the solutions you're looking for.