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

User · Answer

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:

User · Answer

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   v17   3   v17     v17   1        5   v17   3   v17     1   v17        3   v13   v13   5   1   v13      5   v13     v13   3   1   v13     - - - - ---    - - ---   - --- - -    - - -   ---    -   ---   - -   ---    - - -   ---    ---   -   -   ---    - - - ---    - --- - -   - - ---         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

User · Answer

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

User · Answer

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

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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  2-1 f2   lambda x y   x-y  res   double solve f1 f2 1 0  res   n solve  f1 f2   1 0 0 0

User · Answer

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

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If you prefer sympy you can use nsolve    gt  gt  gt  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

User · Answer

I got Broyden s method to work for coupled non-linear equations  generally involving polynomials and exponentials  in IDL  but I haven t tried it in Python  http   docs scipy org doc scipy reference generated scipy optimize broyden1 html scipy optimize broyden1  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

User · Answer

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    - gt     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  lt - convenient indeed  but in terms of speed  it s quite slow  You will see

[python] How to solve a pair of nonlinear equations using Python?

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