Reputation: 111
I have this code which aims to solve a system of first order differential equations. It returns a symbolic solution.
import sympy as sym
sym.init_printing()
from IPython.display import display_latex
from scipy.integrate import odeint
import matplotlib.pyplot as plt
import numpy as np
%matplotlib inline
def solve_sys(a):
t=sym.symbols('t')
x1=sym.Function('x1')
y2=sym.Function('y2')
u=sym.Function('u')
v=sym.Function('v')
eq1=sym.Eq(x1(t).diff(t),y2(t))
eq2=sym.Eq(y2(t).diff(t),-x1(t)+a*(y2(t)-((y2(t)**3)/3)))
matrix=sym.Matrix([eq1.rhs,eq2.rhs])
matJ=matrix.jacobian([x1(t),y2(t)])
lin_mat = matJ.subs({x1(t):0,y2(t):0})
lin_mat*sym.Matrix([u(t),v(t)])
evect=lin_mat.eigenvects()
evals = list(lin_mat.eigenvals().keys())
return evect, evals
I then have a function which aims to take the eigenvalues, perform some tests on them and then return the type of critical point they produce.
ef critical_point_test(a):
blah,evals=solve_sys(a)
if isinstance(evals[0],complex)==False and isinstance(evals[0],complex)==False and np.sign(evals[0])==np.sign(evals[1]) and evals[0]!=evals[1]:
print('Node')
elif isinstance(evals[0],complex)==False and isinstance(evals[0],complex)==False and np.sign(evals[0])!=np.sign(evals[1]):
print('Saddle Point')
elif evals[0]==evals[1]:
print('Proper or Impropper Node')
elif isinstance(evals[0],complex)==True and isinstance(evals[0],complex)==True and np.real(evals[0])!=0 and np.real(evals[1])!=0:
print('Spiral Point')
else:
print('Center')
I get the error:
Invalid comparison of complex 1/2 - sqrt(3)*I/2
My question is how do I modify my solve_sys(a) function so that it returns a complex number that can then be interpreted by my second function. Thanks!
Upvotes: 0
Views: 717
Reputation: 14530
Using a = Symbol('a')
and calculating the eigenvalues I get:
In [22]: evals
Out[22]:
⎡ _______ _______ _______ _______⎤
⎢a ╲╱ a - 2 ⋅╲╱ a + 2 a ╲╱ a - 2 ⋅╲╱ a + 2 ⎥
⎢─ - ───────────────────, ─ + ───────────────────⎥
⎣2 2 2 2 ⎦
In [23]: e1, e2 = evals
In [24]: e1
Out[24]:
_______ _______
a ╲╱ a - 2 ⋅╲╱ a + 2
─ - ───────────────────
2 2
If you substitute for a
then you can use is_real
to find out if the eigenvalues are real:
In [25]: e1.subs(a, 1)
Out[25]:
1 √3⋅ⅈ
─ - ────
2 2
In [26]: e1.subs(a, 1).is_real
Out[26]: False
If it is real you can use is_positive
to find out if it is positive:
In [27]: e1.subs(a, 4)
Out[27]: 2 - √3
In [28]: e1.subs(a, 4).is_positive
Out[28]: True
Likewise is_zero
tells you if the eigenvalue is zero.
Working symbolically without substituting you can ask for what values of a
it will be positive:
In [29]: solve(e1 >= 0, a)
Out[29]: 2 ≤ a
You can find for what value of a
the eigenvalues will be exactly equal
In [39]: solve(Eq(e1, e2), a)
Out[39]: [2]
Upvotes: 2