CODEWITHSUNDEEP
pythonsympysymbolssolver
Oded Sayar
Oded Sayar

Reputation: 429

Force a Sympy expression to contain a symbol

I wrote the following code in python3 (with the intention to compute then draw nullclines of a 2-D dynamic system):

import sympy as sm
x1, x2 = sm.symbols('x1 x2')
x = [x1, x2]
sys = sm.Matrix([(x1 - sm.log(x2))*(x2-x1*x1),
    x2**2 - 4])
solutions = list(map(lambda dx: sm.solve(sm.Eq(dx, 0)), sys))
print(solutions)

What I see is

 [[{x1: -sqrt(x2)}, {x1: sqrt(x2)}, {x1: log(x2)}], [-2, 2]]

This [-2,2] gives me a big problem, as I want to draw the solutions next. Since I want to work with sys whether or not each element contains both variables. I'd like to iterate over the dictionaries and plot sm.Eq(list(solutions[i][j].keys())[0], list(solutions[i][j].values())[0]) but I can't guarantee a dictionary like this

I saw that if I add + x1 - sm.log(sm.exp(x1)) to sys[1] it will give me the consistent dictionary form, but is there a way for me to force sys' element to contain both in a more elegant way? I don't want to look in each element for both coordinates and if one isn't present to add the redundant expression to it

Upvotes: 2

Views: 209

Answers (1)

user6655984
user6655984

Reputation:

The argument dict=True does that. With sm.solve(sm.Eq(dx, 0), dict=True) the output of your code is 

[[{x1: -sqrt(x2)}, {x1: sqrt(x2)}, {x1: log(x2)}], [{x2: -2}, {x2: 2}]]

That said, I'm not sure that you win anything versus using simply plot_implicit directly on the equations in sys. Consider that the different solutions you get will have different domains of definition, as the curves turn around or go sideways (parabolas, etc). Going to be a headache to plot all those pieces.

Here is how it works with plot_implicit 

import sympy as sm
x1, x2 = sm.symbols('x1 x2')
eqs = [(x1 - sm.log(x2))*(x2-x1*x1), x2**2 - 4]
window = [(x1, -3, 3), (x2, -3, 3)]
plots = [plot_implicit(eq, *window, show=False) for eq in eqs]
combined_plots = plots[0]
for k in range(1, len(plots)):
  combined_plots.extend(plots[k])
combined_plots.show()

nullclines

Upvotes: 3

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