Solving nonlinear equations with a direct Broyden method and its acceleration

Abstract

A quasi-Newton method called the direct Broyden method is considered in this paper, which satisfies the least change principle and the direct tangent condition. The direct Broyden method can ensure that the quasi-Newton matrix equals the Jacobian matrix along step direction and accumulates more derivative information of the function. Moreover, we present an accelerated version of the new method and prove its global and superlinear convergence. Extensive numerical results are reported to show the efficiency of the two methods.

Appendix A: Test functions

In this appendix, we list the test functions and initial points $x_0$, where

$$\begin{aligned} F(x)=(f_1(x),f_2(x)\ldots ,f_n(x))^T. \end{aligned}$$

• 1. Exponential function [23]

  $$\begin{aligned} f_i(x)= & {} \frac{i}{10}\left( 1-x_i^2-e^{-x_i^2}\right) , ~i=1,2,\ldots , n-1,\\ f_n= & {} \frac{n}{10}\left( 1-e^{-x_n^2}\right) .\\ x_0= & {} \left( \frac{1}{4n^2}, \frac{2}{4n^2},\ldots , \frac{n}{4n^2}\right) ^T. \end{aligned}$$

• 2. Logarithmic function [23]

  $$\begin{aligned} f_i(x)= & {} \ln (x_i+1)-\frac{x_i}{n},~i=1,2,\ldots ,n.\\ x_0= & {} (1,1,\ldots ,1)^T. \end{aligned}$$

• 3. Strictly convex function 1 [24]

  F(x) is the gradient of $h(x)=\sum _{i=1}^n(e^{x_i}-x_i)$.

  $$\begin{aligned} f_i(x)= & {} e^{x_i}-1,~i=1,2,\ldots ,n.\\ x_0= & {} \left( \frac{1}{n},\frac{2}{n},\ldots ,1\right) ^T. \end{aligned}$$

• 4. Strictly convex function 2 [24]

  F(x) is the gradient of $h(x)=\sum _{i=1}^n\frac{i}{10}(e^{x_i}-x_i)$.

  $$\begin{aligned} f_i(x)= & {} \frac{i}{10}\left( e^{x_i}-1\right) , ~i=1,2,\ldots ,n.\\ x_0= & {} (1,1,\ldots ,1 )^T. \end{aligned}$$

• 5. Extended Rosenbrock function (n is even) [25].

  For $i=1,2,\ldots ,n/2$

  $$\begin{aligned} f_{2i-1}(x)= & {} 10(x_{2i}-x_{2i-1}^2),\\ f_{2i}(x)= & {} 1-x_{2i-1}.\\ x_0= & {} (-1.2,1,\ldots ,-1.2,1)^T. \end{aligned}$$

• 6. Function 6 (n is a multiple of 3) [23].

  For $i=1,2,\ldots ,n/3$

  $$\begin{aligned} f_{3i-2}(x)= & {} x_{3i-2}x_{3i-1}-x_{3i}^2-1,\\ f_{3i-1}(x)= & {} x_{3i-2}x_{3i-1}x_{3i}-x_{3i-2}^2+x_{3i-1}^2-2,\\ f_{3i}(x)= & {} e^{-x_{3i-2}}-e^{-x_{3i-1}}.\\ x_0= & {} (1,1,\ldots ,1)^T. \end{aligned}$$

• 7. Tridimensional valley function (n is a multiple of 3) [26]

  For $i=1,2,\ldots ,n/3,$

  $$\begin{aligned} f_{3i-2}(x)= & {} (c_2x_{3i-2}^3+c_1x_{3i-2})\exp \left( \frac{-x_{3i-2}^2}{100}\right) -1,\\ f_{3i-1}(x)= & {} 10 (\sin (x_{3i-2})-x_{3i-1} ),\\ f_{3i}(x)= & {} 10 (\cos (x_{3i-2})-x_{3i}),\\ \text{ where }{} & {} \\ c_1= & {} 1.003344481605351,\\ c_2= & {} -3.344481605351171\times 10^{-3}.\\ x_0= & {} (2,1,2,\ldots ,2,1,2)^T. \end{aligned}$$

• 8. Extended Powell singular function (n is a multiple of 4) [27]

  For $i=1,2,\ldots ,n/4,$

  $$\begin{aligned} f_{4i-3}(x)= & {} x_{4i-3}+10x_{4i-2},\\ f_{4i-2}(x)= & {} \sqrt{5}(x_{4i-1}-x_{4i}),\\ f_{4i-1}(x)= & {} (x_{4i-2}-2x_{4i-1})^2,\\ f_{4i}(x)= & {} \sqrt{10}(x_{4i-3}-x_{4i})^2.\\ x_0= & {} (1.5\times 10^{-4},1.5\times 10^{-4},\ldots ,1.5\times 10^{-4})^T. \end{aligned}$$

• 9. Tridiagonal exponential problem [28]

  $$\begin{aligned} f_1(x)= & {} x_1 - \exp (\cos (h (x_1 + x_2))),\\ f_i(x)= & {} x_i - \exp (\cos (h(x_{i-1} + x_i + x_{i+1}))),~i=2,\ldots ,n-1,\\ f_n(x)= & {} x_n - \exp (\cos (h(x_{n-1} + x_n))).\\ h= & {} 1/(n+1).\\ x_0= & {} (1.5,\ldots ,1.5)^T. \end{aligned}$$

• 10. Discrete boundary value problem [27]

  $$\begin{aligned} f_1(x)= & {} 2x_1+0.5h^2(x_1+h)^3-x_2,\\ f_i(x)= & {} 2x_i+0.5h^2(x_i+ih)^3-x_{i-1}+x_{i+1},~i=2,\ldots ,n-1,\\ f_n(x)= & {} 2x_n+0.5h^2(x_n+nh)^3-x_{n-1}.\\ h= & {} 1/(n+1).\\ x_0= & {} (h(h-1),h(2h-1),\ldots ,h(nh-1))^T. \end{aligned}$$

• 11. Discretized two-point boundary value problem

  $$\begin{aligned} f_1(x)= & {} 8x_1-x_2+m(\sin x_1-1),\\ f_i(x)= & {} -x_{i-1}+8x_i-x_{i+1}m(\sin x_i -1),~i=2,\ldots ,n-1,\\ f_n(x)= & {} -x_{n-1}+8x_n+m(\sin x_n-1).\\ m= & {} 1/(n+1)^2.\\ x_0= & {} \left( \frac{1}{n},\frac{2}{n},\ldots ,1\right) ^T. \end{aligned}$$

• 12. Trigonometric function [23]

  $$\begin{aligned} f_i(x)= & {} 2\left( n+i(1-\cos x_i)-\sin x_i-\sum _{j=1}^n \cos x_j\right) (2\sin x_i-\cos x_i),\\ x_0= & {} \left( \frac{101}{100n},\ldots ,\frac{101}{100n}\right) ^T. \end{aligned}$$

• 13. Penalty I function [23]

  $$\begin{aligned} f_i(x)= & {} \sqrt{10^{-5}}(x_i-1),~ i=1,2,\ldots ,n-1,\\ f_n(x)= & {} \left( \frac{1}{4n}\right) \sum _{j=1}^n x_j^2-\frac{1}{4}.\\ x_0= & {} \left( \frac{1}{3},\frac{1}{3},\ldots , \frac{1}{3}\right) ^T. \end{aligned}$$

• 14. Hanbook function [23].

  $$\begin{aligned} f_i(x)= & {} 0.05(x_i-1)+2\sin \left( \sum _{j=1}^n(x_j-1)+\sum _{j=1}^n(x_j-1)^2\right) \\{} & {} (1+2(x_i-1))+2\sin \left( \sum _{j=1}^n(x_j-1)\right) ,~i=1,2,\ldots ,n.\\ x_0= & {} (5,\ldots ,5)^T. \end{aligned}$$

• 15. Linear function-rank 2 [23]

  $$\begin{aligned} f_1(x)= & {} x_1-1,\\ f_i(x)= & {} i\sum _{j=1}^n jx_j-i,~ i=2,3,\ldots ,n.\\ x_0= & {} \left( 1,\frac{1}{n},\ldots , \frac{1}{n}\right) ^T. \end{aligned}$$

• 16. Engval function

  $$\begin{aligned} f_1(x)= & {} x_1(x_1^2+x_2^2)-1,\\ f_i(x)= & {} x_i(x_{i-1}^2+2x_i^2+x_{i+1}^2)-1,~i=2,\ldots ,n-1,\\ f_n(x)= & {} x_n(x_{n-1}^2+x_n^2).\\ x_0= & {} (0,0,\ldots ,0)^T. \end{aligned}$$