A Scaled Derivative-Free Projection Method for Solving Nonlinear Monotone Equations

Abstract

Derivative-free projection methods are very effective and highly useful methods for solving large-scale nonlinear monotone equations. A lot of research is continuously being done to either improve the existing methods or come up with other new projection methods. In this paper, we propose a scaled derivative-free projection method for solving large-scale nonlinear monotone equations, which combines conjugate gradient and projection techniques. We establish its global convergence under appropriate conditions. The method is then compared with other existing methods in the literature and preliminary numerical results indicate that the method compares very well.

Appendix

Here, we present the twelve test problems that were used in numerical experiments. These problems are as follows.

Problem 1

[7]

$$\begin{aligned} F_{1}(x)&= x_{1} - e^{\cos (\frac{x_{1}+x_{2}}{n+1})},\\ F_{i}(x)&= x_{i} - e^{\cos (\frac{x_{i-1}+x_{i}+x_{i+1}}{n+1})}, \,\, i=2,3,\ldots ,n-1,\\ F_{n}(x)&= x_{n} - e^{\cos (\frac{x_{n-1}+x_{n}}{n+1})}, \end{aligned}$$

where $\Omega =\mathbb {R}_{+}^{n}$. Initial guess $x_{0}=(1, 1, \ldots , 1)^{\text {T}}$.

Problem 2

[9]

$$\begin{aligned} F_{1}(x)&= 2.5x_{1}+ x_{2}-1,\\ F_{i}(x)&= x_{i-1}+2.5x_{i}+x_{i+1}-1, \,\text {for}\quad i=2, 3, \ldots , n-1,\\ F_{n}(x)&= x_{n-1}+2.5x_{n}-1, \end{aligned}$$

where $\Omega =\mathbb {R}^{n}$. Initial guess $x_{0}=(0, 0, \ldots , 0)^{\text {T}}$.

Problem 3

 [7]

$$\begin{aligned} F_{1}(x)&= 3x_{1}^{3}+ 2x_{2}-5+\sin (x_{1}-x_{2})\sin (x_{1}+x_{2}),\\ F_{i}(x)&= -x_{i-1}e^{x_{i-1}-x_{i}}+x_{i}(4+3x_{i}^{2})+2x_{i+1}\\&+\sin (x_{i}-x_{i+1})\sin (x_{i}+x_{i+1})-8, \,\, i=2, 3,\ldots , n-1,\\ F_{n}(x)&= -x_{n-1}e^{x_{n-1}-x_{n}}+4x_{n}-3, \end{aligned}$$

where $\Omega =\mathbb {R}^{n}$. Initial guess $x_{0}=(0, 0, \ldots , 0)^{\text {T}}$.

Problem 4

 [7]

$$\begin{aligned} F_{i}(x) = 2x_{i} - \sin ( x_{i}), \,\,\, i=1,2,\ldots ,n, \end{aligned}$$

where $\Omega =\mathbb {R}^{n}.$ Initial guess $x_{0}=(1, 1, \ldots , 1)^{\text {T}}$.

Problem 5

[19]

$$\begin{aligned} F_{i}(x) = x_{i} - \frac{1}{n}x_{i}^{2}+\frac{1}{n}\sum _{k=1}^{n}x_{k}+i, \,\,\, 1,2,\ldots ,n \end{aligned}$$

where $\Omega =\mathbb {R}^{n}$. Initial guess $x_{0}=(1, 1, \ldots , 1)^{\text {T}}$.

Problem 6

[2]

$$\begin{aligned} F_{i}(x) = x_{i} - \sin (| x_{i}-1|), \,\,\, i=1,2,\ldots ,n, \end{aligned}$$

where $\Omega =\{x\in \mathbb {R}^{n}|\sum _{i=1}^{n}x_{i}\le n, x_{i}\ge 0, i=1,2,\ldots ,n\}$. Initial guess $x_{0}=(-1, -1, \ldots , -1)^{\text {T}}$.

Problem 7

[20]

$$\begin{aligned} F_{1}(x)&= 2x_{1}+0.5h^{2}(x_{1}+h)^{3}-x_{2},\\ F_{i}(x)&= 2x_{i}+0.5h^{2}(x_{i}+hi)^{3}-x_{i-1}+x_{i+1}, \,\,\, i=2, 3, \ldots , n-1,\\ F_{n}(x)&= 2x_{n}+0.5h^{2}(x_{n}+hn)^{3}-x_{n-1}, \end{aligned}$$

where $h=\frac{1}{n+1}$ and $\Omega =\mathbb {R}^{n}$. Initial guess $x_{0}=(\frac{1}{2}, \frac{1}{2}, \ldots , \frac{1}{2})^{\text {T}}$.

Problem 8

[19]

$$\begin{aligned} F_{i}(x) = 2x_{i} - \sin (| x_{i}|), \,\,\, i=1,2,\ldots ,n, \end{aligned}$$

where $\Omega =\mathbb {R}^{n}$. Initial guess $x_{0}=(10, 10, \ldots , 10)^{\text {T}}$.

Problem 9

[2]

$$\begin{aligned} F_{i}(x)&= 4x_{i}(x_{i}^{2}+x_{n}^{2})-4, \,\,\, i=1, 2, \ldots , n-1,\\ F_{n}(x)&= 4x_{n}\sum _{i=1}^{n-1}(x_{i}^{2}+x_{n}^{2}), \end{aligned}$$

where $\Omega =\mathbb {R}^{n}$. Initial guess $x_{0}=(0, 0, \ldots , 0)^{\text {T}}$.

Problem 10

[2]

$$\begin{aligned} t_{i}&=\sum _{i=1}^{n}x_{i}^{2}, \quad c=10^{-5},\\ F_{i}(x)&= 2c(x_{i}-1)+4(t_{i}-0.25)x_{i}, \,\,\, i=1, 2, \ldots , n,\\ \end{aligned}$$

where $\Omega =\mathbb {R}^{n}$. Initial guess $x_{0}=(-1, -1, \ldots , -1)^{\text {T}}$.

Problem 11

[2]

$$\begin{aligned} F_{1}(x)&= 4x_{1}(x_{1}^{2}+x_{2}^{2})-4,\\ F_{i}(x)&= 4x_{i}(x_{i-1}^{2}+x_{i}^{2})+4x_{i}(x_{i}^{2}+x_{i+1}^{2})-4, \,\,\, i=2, 3, \ldots , n-1,\\ F_{n}(x)&= 4x_{n}(x_{n-1}^{2}+x_{n}^{2}), \end{aligned}$$

where $\Omega =\mathbb {R}^{n}$. Initial guess $x_{0}=(2, 2, \ldots , 2)^{\text {T}}$.

Problem 12

[7]

$$\begin{aligned} F_i(x)&= e^{x_{i}}-1, \,\,\, i=1, 2, \ldots , n, \end{aligned}$$

where $\Omega =\mathbb {R}_{+}^{n}$. Initial guess $x_{0}=(1, 1, \ldots , 1)^{\text {T}}$.