考虑 F(x,y)=常数。 对于无法分离变量的复杂 F，如何确定所有 y 和 x 以及 plot（y vs x）？

Question

I consider a function:

 c= sin(x)+cos(y)-d*cos(x+y)/[sin(x) - cos(y)-k*cos(x+y)], 

where c, d, k are known constants . I want to make the plot y vs x.

As I can't separate y from x on the equation, I bring everything on the left side and I define a function G:

G=[sin(x) - cos(y)-k*cos(x+y)]-(sin(x)+cos(y)-d*cos(x+y))=0.

So, I firstly want all the pairs x,y for which the above function G is equal to 0 and make the plot y vs x. Then, I want to use these pairs of x,y in order to determine two other functions that depend on these x,y:

f1=x^(1/2) + 5cos(x),

f2=y^(1/3) + sin(y)

What I am doing is firstly creating an array with the following loops (to get the (x,y) pairs that give G=0):

arrayG = np.empty((0,3),float)
for i in x:
    for j in y:
        if G(i,j)>-0.001 and G<0.001 : 
            arrayG=np.append(arrayG,np.array([[i, j, G(i,j)]]), axis=0)

But with the "if" statement that I'm using, I can't consider G being exactly 0. Should I treat it differently?

Assuming that I get the pairs of (x,y) from the above method, I want to implement these pairs on the f1,f2 expressions in order to get the values of f1,f2. So, I'm writing the following code:

array_f1_f2 = np.empty((0,5),float) #where 5 are the f1, f2, x, y, G (in order to show all 5) 
for row in array_f1_f2: 
    arrayG = np.append(array_f1_f2, np.array([[f1(row[0]), f2(row[1]), row[0], row[1], row[2] ]]), axis=0)

Is this a valid way of solving the problem?

Thanks in advance!

PS This is a simplified example of the problem I'm working on. I don't know if this is really solvable. I'm looking for a way to approach this problem. I have used numpy library for formulas and matplotlib for plots.

Answer 1

I'm sort of cheating here by using a non-numpy solution, but this is really easy to do with sympy :

from sympy import plot_implicit, cos, sin, symbols, Eq
x, y = symbols('x y')


c = 0.25
d = 1
k = 4

p1 = plot_implicit(Eq((sin(x) - cos(y) - k*cos(x+y)) -
                      (sin(x) + cos(y) - d*cos(x+y)), 0))

This gives the following plot:

If it's important to use numpy and matplotlib , you can do so without much additional effort, but the code is more opaque.

from matplotlib import pyplot
import numpy

X = numpy.linspace(-4, 4)[:, None]  # Create broadcastable X array.
Y = numpy.linspace(-4, 4)[None, :]  # Create broadcastable Y array.

c = 0.25
d = 1
k = 4

# Because X and Y have broadcastable shapes, Z is a 2d array.

Z = ((numpy.sin(X) - numpy.cos(Y) - k*numpy.cos(X+Y)) -
     (numpy.sin(X) + numpy.cos(Y) - d*numpy.cos(X+Y)))

# `contour` expects X and Y to be 1d, but for broadcasting 
# `numpy` wants 2d arrays with complimentary singleton dimensions.
# So we have to remove the singleton dim. Contour lines will be
# drawn at the `Z` values specified in the last argument (if it's
# an array-like object). We only want one value, so we pass `[0]`.

pyplot.contour(X[:, 0], Y[0, :], Z, [0])
pyplot.show()

This gives:

Answer 2

You are trying to solve the equation numerically so in my point of view there wont be any absolute zero for G. First try to choose the step of 'x' and 'y' as small as possible and then instead of 'if' statement, utilize np.mean(np.abs(.)) so it will make the margin more narrow.