在 Unity 中使用 C# 以[牛顿模仿法]求解非线性方程组--II.案例

1、Python解法

import numpy as np
from scipy.optimize import fsolve

def equation(b):
    left = np.power(b + 0.35 * b, 2) + np.power(b - 0.35 * b, 2) - np.power(50, 2)
    right = np.cos(np.radians(30)) * 2 * (b + 0.35 * b) * (b - 0.35 * b)
    return left - right

# 初始猜测值
initial_guess = 0.1

# 求解方程
solution = fsolve(equation, initial_guess)

print(f"Solution: b = {solution[0]}")

计算结果：

Solution: b = 58.71694350541571

2、C#解法

[ContextMenu("拟牛顿法求根 1 ")]
void testN()
{
     Func<double[], double[]> equation = b => new[] {
         Math.Pow(b[0] + 0.35 * b[0], 2) + Math.Pow(b[0] - 0.35 * b[0], 2) - Math.Pow(50, 2) -
         Math.Cos(30.0 * Math.PI / 180) * 2 * (b[0] + 0.35 * b[0]) * (b[0] - 0.35 * b[0])
     };

     double[] initialGuess = { 1 };

     double[] solution = Broyden.FindRoot(equation, initialGuess,maxIterations:20000);

     Debug.Log($"方程的解b为: {solution[0]}");
}

结算结果

方程的解b为: 58.7169435054131

3、结论

结果一致

三、案例3

1、方程组

未知数：ϕ0 ,ψ0 ,P0 ,P1 ,P2
方程组：

2、 C#的计算过程

        // 已知参数
        double phi1 = 1.46683d.ToRadians(), phi2 = 12.36644d.ToRadians(), phi3 = 30d.ToRadians(), phi4 = 47.63356d.ToRadians(), phi5 = 58.5317d.ToRadians();
        double psi1 = 3.13922d.ToRadians(), psi2 = 24.33225d.ToRadians(), psi3 = 52.64663d.ToRadians(), psi4 = 75.87842d.ToRadians(), psi5 = 88.40146d.ToRadians();           

        // 定义方程组函数
        Func<double[], double[]> equations = x =>
        {
            double P0 = x[0];
            double P1 = x[1];
            double P2 = x[2];
            double phi0 = x[3];
            double psi0 = x[4];

            double equation1 = Math.Cos(phi1 + phi0) - (P0 * Math.Cos(psi1 + psi0) + P1 * Math.Cos((psi1 + psi0) - (phi1 + phi0)) + P2);
            double equation2 = Math.Cos(phi2 + phi0) - (P0 * Math.Cos(psi2 + psi0) + P1 * Math.Cos((psi2 + psi0) - (phi2 + phi0)) + P2);
            double equation3 = Math.Cos(phi3 + phi0) - (P0 * Math.Cos(psi3 + psi0) + P1 * Math.Cos((psi3 + psi0) - (phi3 + phi0)) + P2);
            double equation4 = Math.Cos(phi4 + phi0) - (P0 * Math.Cos(psi4 + psi0) + P1 * Math.Cos((psi4 + psi0) - (phi4 + phi0)) + P2);
            double equation5 = Math.Cos(phi5 + phi0) - (P0 * Math.Cos(psi5 + psi0) + P1 * Math.Cos((psi5 + psi0) - (phi5 + phi0)) + P2);

            return new[] { equation1, equation2, equation3, equation4, equation5 };
        };

        // 初始猜测值
        double[] initialGuess = {20d, 20d, 20d,20d, 20d};

        // 调用Broyden.FindRoot方法求解
        double[] solution = Broyden.FindRoot(equations, initialGuess,accuracy:1e-7,maxIterations:10000);

        // 输出解
        Debug.Log($"P0 = {solution[0]}, P1 = {solution[1]}, P2 = {solution[2]}, phi0 = {solution[3]}, psi0 = {solution[4]}");

计算结果：

3、结论

与第三方软件的【拟牛顿法】计算结果一致！