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Solution method for discrete double obstacle problems based on a power penalty approach

  • * Corresponding author: Kai Zhang

    * Corresponding author: Kai Zhang 
Abstract Full Text(HTML) Figure(3) / Table(3) Related Papers Cited by
  • We develop a power penalty approach to a finite-dimensional double obstacle problem. This problem is first approximated by a system of nonlinear equations containing two penalty terms. We show that the solution to this penalized equation converges to that of the original obstacle problem at an exponential rate when the coefficient matrices are $ M $-matrices. Numerical examples are presented to confirm the theoretical findings and illustrate the efficiency and effectiveness of the new method.

    Mathematics Subject Classification: Primary: 90C33, 90-08; Secondary: 65K15.

    Citation:

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  • Figure 1.  Solution $ U(s) $ with $ N = 99 $. The solution is obtained with the lower order penalty method ($ k = 2 $). Tolerance $ \varepsilon = 10^{-6} $ is chosen for the Newton iteration. The smoothing interval is chosen to be $ (0,10^{-3}) $. $ \lambda = 10^{3} $

    Figure 2.  Solution $ u(x,y) $ with the mesh grid $ 60 \times 60 $, obtained with the lower order penalty method ($ k = 2 $). Tolerance is set to be $ 10^{-6} $. $ \epsilon = 10^{-3} $. $ \lambda = 10^{3} $

    Figure 3.  The lower coincidence sets (dot '$ \cdot $') and the upper coincidence sets (star '$ \ast $') of solution $ u(x,y) $ on the mesh grid $ 60 \times 60 $. The solution is obtained with the lower order penalty method ($ k = 2 $). Tolerance $ \varepsilon = 10^{-6} $ is chosen for the Newton iteration. The smoothing interval is chosen to be $ (0,10^{-3}) $. $ \lambda = 10^{3} $

    Table 1.  Results computed by the power penalty method. Tolerance $ \varepsilon = 10^{-6} $ is chosen for the Newton iteration. The smoothing parameter $ \epsilon = 10^{-3} $. 'Ra' stands for ratio

    $k=1$ $k=2$
    $\lambda_{i}$ $[x_{\lambda_{i}}]_{1}$ $[x_{\lambda_{i}}]_{4}$ $\Vert x-x_{\lambda_{i}}\Vert_{\infty}$ Ra $\lambda_{i}$ $[x_{\lambda_{i}}]_{1}$ $[x_{\lambda_{i}}]_{4}$ $\Vert x-x_{\lambda_{i}}\Vert_{\infty}$ Ra
    $10^{2}$ $0.5119$ $5.3052$ $6.25\times10^{-1}$ $10^{2}$ $0.9985$ $5.0011$ $2.81\times10^{-1}$
    $10^{3}$ $0.9430$ $5.0327$ $6.49\times10^{-2}$ $0.98$ $10^{3}$ $0.9998$ $5.0002$ $2.85\times10^{-3}$ $1.99$
    $10^{4}$ $0.9942$ $5.0033$ $6.59\times10^{-3}$ $0.99$ $10^{4}$ $0.9999$ $5.0001$ $4.93\times10^{-5}$ $1.76$
    $10^{5}$ $0.9994$ $5.0003$ $6.60\times10^{-4}$ $1.00$ $10^{5}$ $1.0000$ $5.0000$ $6.90\times10^{-6}$ $1.69$
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    Table 2.  Total number of iterations for the linear penalty ($ k = 1 $), lower order penalty ($ k = 2 $), and modified policy methods with $ N = 99 $. Tolerance is set to be $ 10^{-6} $. $ \epsilon = 10^{-3} $. $ \lambda $ is set to be $ 10^{6} $ and $ 10^{3} $ for $ k = 1 $ and $ k = 2 $, respectively

    $ k=1 $ $ k=2 $ Modified policy iter.
    Iterations 9 12 88
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    Table 3.  Comparison of computational efficiency among power penalty methods, Gauss-Seidel iteration method and active set method. Tolerance is set to be $ 10^{-6} $. $ \epsilon = 10^{-3} $. $ \lambda $ is set to be $ 10^{6} $ and $ 10^{3} $ for $ k = 1 $ and $ k = 2 $, respectively

    $k=1$ $k=2$ Gauss-Seidel Active set
    $N=49$ Iterations 11 16 426 Failed
    Time (s) 1.26 1.89 32.81 Failed
    $N=59$ Iterations 15 17 740 17
    Time (s) 7.78 8.96 198.7 6.43
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