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A smoothing Newton method preserving nonnegativity for solving tensor complementarity problems with $ P_0 $ mappings

  • *Corresponding author: Liping Zhang

    *Corresponding author: Liping Zhang

This work is supported by the National Natural Science Foundation of China (Grant No. 12171271)

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  • In this paper, we prove that the tensor complementarity problem with the $ P_0 $ mapping on the $ n $-dimensional nonnegative orthant is solvable and the solution set is nonempty and compact under mild assumptions. Since the involved homogeneous polynomial is a $ P_0 $ mapping on the $ n $-dimensional nonnegative orthant, the existing smoothing Newton methods are not directly used to solve this problem. So, we propose a smoothing Newton method preserving nonnegativity via a new one-dimensional line search rule for solving such problem. The global convergence is established and preliminary numerical results illustrate that the proposed algorithm is efficient and very promising.

    Mathematics Subject Classification: Primary: 90C33, 90C30; Secondary: 65K10.

    Citation:

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  • Table 1.  Numerical results for the TCP in Example 5.1

    $ {\bf q} $ $ {\bf{x}} $ Iter NH NormH Time (sec.)
    [-10 0] [2.0976 0.6397] 6 10 3.0591e-14 3.1250e-02
    [0 -5] [0.5077 1.6648] 7 8 8.8818e-16 0.0000e+00
    [-7 -1] [1.8240 0.7709] 7 10 2.6738e-15 3.1250e-02
    [2 -9] [0.2226 2.0724] 7 11 1.9860e-15 0.0000e+00
    [-8 3] [2.0000 0.0000] 7 10 0.0000e+00 3.1250e-02
     | Show Table
    DownLoad: CSV

    Table 2.  Numerical results for the TCP in Example 5.2

    $ {\bf q} $ $ {\bf{x}} $ Iter NH NormH Time (sec.)
    [0 9] [0.3825 0.0000] 4 5 6.6613e-16 0.0000e+00
    [5 3] [0 0] 4 5 0.0000e+00 0.0000e+00
    [-12 9] [3.0000 2.0000] 6 10 5.6843e-14 0.0000e+00
    [2 -3] —It has no solution— 6.2500e-02
    [-8 -5] —It has no solution— 6.2500e-02
     | Show Table
    DownLoad: CSV

    Table 3.  Numerical results for the TCP in Example 5.3

    $ {\bf q} $ $ {\bf{x}} $ Iter NH NormH Time (sec.)
    [0 4] [6.9194 0.0000] 41 42 5.2224e-13 0.0000e+00
    [8 9] [0 0 ] 41 42 8.5937e-13 0.0000e+00
    [-3 -3] [1.4422 1.4422] 5 6 1.4445e-14 0.0000e+00
    [-10 0] — It has no solution — 3.1250e-02
    [5 -3] — It has no solution — 7.8125e-02
    [-12 9] — It has no solution — 2.8125e-01
     | Show Table
    DownLoad: CSV

    Table 4.  Numerical results for the TCP in Example 5.4

    $ {\bf q} $ $ {\bf{x}} $ Iter NH NormH Time (sec.)
    [0 5 8 9] [0.8683 0.0000 0.0000 0.0000] 22 23 6.5455e-13 0.0000e+00
    [0 -5 8 -0.5] [0 1.7100 0.0000 0.7937] 6 9 8.9509e-16 0.0000e+00
    [-7 -1 0.01 0] [1.9001 0.2714 0.0000 0.0001] 24 32 3.6965e-13 7.8125e-02
    [12 -1 -28 0] [0.0000 1.0000 3.0366 0.0001] 23 26 4.2783e-13 0.0000e+00
    [-8 15 0 -23] [2.0000 0.0000 0.0001 2.8439] 23 27 6.3022e-13 1.5625e-02
     | Show Table
    DownLoad: CSV
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