如何用Python求解含三个3D方程的非线性方程组

Hey there! Let's break down how to find the intersection points (solutions) of your three 3D equations using Python. First, let's restate your equations clearly to make coding easier:

Equation 1: -0.006683x² - 0.06893x + 56.73 - z = 0
Equation 2: 0.002538y² - 1.115y + 56.73 - z = 0
Equation 3: (x-24.680)² + (y-238.341)² + (z+13.971)² = 12.580²

Step 1: Pick the Right Tool #

For solving nonlinear systems like this, Python's scipy.optimize module is perfect. We'll use fsolve here—it's straightforward and works well for this type of problem.

Step 2: Define the Residual Function #

We need a function that takes a vector [x, y, z] and returns how far each equation is from 0 (called "residuals"). The solver will tweak x, y, z until all residuals are nearly zero.

Step 3: Choose a Reasonable Initial Guess #

Nonlinear solvers need a starting point. Let's think about the geometry:

• The first two equations are parabolic surfaces, so z is determined by x (eq1) and y (eq2) respectively.
• The third equation is a sphere centered at (24.680, 238.341, -13.971) with radius 12.580, so z should be between ~-26.55 and ~-1.39.

A good initial guess might be [0, 200, -10]—close to the sphere's center, with z in the sphere's valid range.

Full Code Example #

Here's the complete, commented code:

import numpy as np
from scipy.optimize import fsolve

# Define the system of equations as a residual function
def equations(vars):
    x, y, z = vars
    
    # Calculate how far each equation is from 0
    eq1 = -0.006683 * x**2 - 0.06893 * x + 56.73 - z
    eq2 = 0.002538 * y**2 - 1.115 * y + 56.73 - z
    eq3 = (x - 24.680)**2 + (y - 238.341)**2 + (z + 13.971)**2 - 12.580**2
    
    return [eq1, eq2, eq3]

# Initial guess for x, y, z
initial_guess = [0, 200, -10]

# Solve the system
solution = fsolve(equations, initial_guess)

# Extract the solution values
x_sol, y_sol, z_sol = solution

print("Found solution:")
print(f"x = {x_sol:.4f}")
print(f"y = {y_sol:.4f}")
print(f"z = {z_sol:.4f}")

# Verify the solution (residuals should be very close to 0)
print("\nVerification (residuals):")
residuals = equations(solution)
print(f"Equation 1 residual: {residuals[0]:.6f}")
print(f"Equation 2 residual: {residuals[1]:.6f}")
print(f"Equation 3 residual: {residuals[2]:.6f}")

Notes on Finding All Possible Solutions #

Nonlinear systems can have multiple solutions (up to 4 in this case, given the geometry). If you don't get a valid solution, or want to find all intersections:

• Try different initial guesses (e.g., [50, 250, -20] or [-10, 220, -5]).
• If fsolve struggles, use scipy.optimize.root with alternative methods like 'lm' (least squares) for more robustness.

内容的提问来源于stack exchange，提问作者Diogo Mata