I am trying to solve a 12 nonlinear equation system with SymPy.
import sympy
q0, q1, q2, q3, q4, q5, q6, q7, q8, q9, q10, q11 = symbols('q0, q1, q2, q3, q4, q5, q6, q7, q8, q9, q10, q11')
unitary_energy_condition = q0**2 + q1**2 + q2**2 + q3**2 + q4**2 + q5**2 + q6**2 + q7**2 + q8**2 + q9**2 + q10**2 + q11**2 - 1
symlet_equation1 = q0 + q1 + q2 + q3 + q4 + q5 + q6 + q7 + q8 + q9 + q10 + q11
symlet_equation2 = -5*q0 - 4*q1 - 3*q2 - 2*q3 - q4 + q6 + 2*q7 + 3*q8 + 4*q9 + 5*q10 + 6*q11
symlet_equation3 = 25*q0 + 16*q1 + 9*q2 + 4*q3 + q4 + q6 + 4*q7 + 9*q8 + 16*q9 + 25*q10 + 36*q11
symlet_equation4 = -125*q0 - 64*q1 - 27*q2 - 8*q3 - q4 + q6 + 8*q7 + 27*q8 + 64*q9 + 125*q10 + 216*q11
orthogonality_equation1 = q0*q2 + q10*q8 + q11*q9 + q1*q3 + q2*q4 + q3*q5 + q4*q6 + q5*q7 + q6*q8 + q7*q9
orthogonality_equation2 = q0*q4 + q10*q6 + q11*q7 + q1*q5 + q2*q6 + q3*q7 + q4*q8 + q5*q9
orthogonality_equation3 = q0*q6 + q10*q4 + q11*q5 + q1*q7 + q2*q8 + q3*q9
orthogonality_equation4 = q0*q8 + q10*q2 + q11*q3 + q1*q9
orthogonality_equation5 = q0*q10 + q11*q1
matching_condition_1 = 0.01*q0 + 0.01*q10 + 0.01*q11 + 0.02*q1 + 0.01*q2 + 0.05*q3 + 0.23*q4 + 0.62*q5 + 0.9*q6 + 0.98*q7 + 0.88*q8 + 0.02*q9
matching_condition_2 = 0.01*q0 + 0.02*q10 + 0.01*q11 + 0.01*q1 + 0.02*q2 + 0.01*q3 + 0.05*q4 + 0.23*q5 + 0.62*q6 + 0.9*q7 + 0.98*q8 + 0.88*q9
ans = sympy.solve([unitary_energy_condition, symlet_equation1, symlet_equation2, symlet_equation3,
symlet_equation4, orthogonality_equation1, orthogonality_equation2,
orthogonality_equation3, orthogonality_equation4, orthogonality_equation5,
matching_condition_1, matching_condition_2],
[q0, q1, q2, q3, q4, q5, q6, q7, q8, q9, q10, q11])
The method stays iterating without giving a solution.
I try to solve this with the sympy.nsolve and sympy.nonlinsolve, but the method had the same behavior.
The solution is q = {−0.0069890396, −0.0047282445, −0.0162466459, 0.0210266730, 0.0760407388, 0.2581313042, −0.8041341124, 0.5207860961, 0.0419502235, −0.0847506178, 0.0022720543, −0.0033584299}.
Please, could you give me some ideas? Thanks.
Neither solve
nor nonlinsolve
will be able to give an analytical solution. (If you even solve the 6 linear equations and substitute their solution into the others you will end up with 6 quadratic equations.) nsolve
is your workhorse here. Let nsol
be the numerical solutions that you think you know. Increase the maximum number of steps beyond the default of 50 and...
>>> v = symbols('q0:12')
>>> nsolve(eqs,v,nsol,maxsteps=100)
Matrix([
[-0.00715129674956314],
[-0.00485533492547843],
[ -0.0158609428322271],
[ 0.0217286784338786],
[ 0.0758952123585306],
[ 0.256657212877231],
[ -0.804001455002201],
[ 0.521753111347145],
[ 0.0416427748854584],
[ -0.0846877567207525],
[ 0.00236892615345501],
[-0.00348912982547515]])
Is this the only solution? Try a random guess for the values to see:
>>> from random import random
>>> g = [random()/100 for i in v]
>>> nsolve(eqs,v,g) # you will see different results...
Matrix([
[ -0.0949854541042071],
[ 0.114582081154371],
[ 0.352304082958864],
[ -0.77819094969228],
[ 0.483184234276383],
[-5.32308960699164e-5],
[ -0.0835476301110434],
[ -0.0570280844309315],
[ 0.0611054599749443],
[ 0.0226639000576377],
[ -0.0109539118083934],
[-0.00908049737927473]])
You will have to provide an initial guess that is in the vicinity of what you expect for the solution.