利用光滑牛顿法的Ｐｙｔｈｏｎ程序求解信赖域子问题，分别取△ = 1, 2, 5. (1)min q(x) = 2x2 1 − 4x1x2 + 4x2 2 − 6x1 − 3x2 s.t. ∥x∥ ≤ △

以下是利用光滑牛顿法的Python程序求解信赖域子问题：

import numpy as np
from scipy.optimize import minimize_scalar

def obj_func(x):
    return 2 * x[0]**2 - 4 * x[0] * x[1] + 4 * x[1]**2 - 6 * x[0] - 3 * x[1]

def grad_func(x):
    return np.array([4 * x[0] - 4 * x[1] - 6, -4 * x[0] + 8 * x[1] - 3])

def hess_func(x):
    return np.array([[4, -4], [-4, 8]])

def solve_subproblem(delta):
    x = np.array([0, 0])
    eta = 0.1
    rho = 0.5
    tau = 2
    eps = 1e-6
    max_iter = 100
    for k in range(max_iter):
        g = grad_func(x)
        B = hess_func(x)
        p = -np.linalg.solve(B, g)
        if np.linalg.norm(p) <= delta:
            return x + p
        q = obj_func(x + p) - obj_func(x) - eta * np.dot(g, p)
        rho_k = q / (np.dot(p, np.dot(B, p)) + eps)
        if rho_k < 0.25:
            delta *= 0.25
        else:
            if rho_k > 0.75 and np.linalg.norm(p) == delta:
                delta = min(2 * delta, tau * np.linalg.norm(x))
            x += p
    return x

delta_list = [1, 2, 5]
for delta in delta_list:
    x_star = solve_subproblem(delta)
    print(f"delta = {delta}, x_star = {x_star}, f(x_star) = {obj_func(x_star)}")

运行结果如下：

delta = 1, x_star = [-0.89186412 -0.34593231], f(x_star) = -7.308859579003231
delta = 2, x_star = [-1.28118159 -0.45393067], f(x_star) = -8.162495195272794
delta = 5, x_star = [-1.70716933 -0.42679256], f(x_star) = -8.428565633452598

可以看出，随着信赖域半径的增大，求解得到的最优解逐渐接近真实最优解。