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A meshless collocation method with a global refinement strategy for reaction-diffusion systems on evolving domains

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  • Turing-type reaction-diffusion systems on evolving domains arising in biology, chemistry and physics are considered in this paper. The evolving domain is transformed into a reference domain, on which we use a second order semi-implicit backward difference formula (SBDF2) for time integration and a meshless collocation method for space discretization. A global refinement strategy is proposed to reduce the computational cost. Numerical experiments are carried out for different evolving domains. The convergence behavior of the proposed algorithm and the effectiveness of the refinement strategy are verified numerically.

    Mathematics Subject Classification: Primary: 35K57; Secondary: 65M70.

    Citation:

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  • Figure 1.  For the SH model with the unit square domain as reference domain, the $ L^2(\Omega) $ error (a) under different overdetermined setting $ n_X = k n_Z $ with $ \Delta t = 0.01, \ T = 10 $; (b) under different kernel smoothness with $ k = 1 $, $ \Delta t = 0.01, \ T = 10 $; (c) comparison between forward Euler method and RK2 with parameters $ \Delta t = [5E-1, 1E-1, 5E-2, 2E-2] $, $ m = 4, \ T = 1 $, $ n_Z = 55^2 $, $ k = 1 $, $ n_X = n_Z $

    Figure 2.  For the SH model with the unit square domain as reference domain, when $ m = 4 $, time step $ \Delta t = 0.005, \ T = 2001 $, the results under fixed discrete sets $ n_Z = 25^2 $ and different global refinement strategies with $ \nu = 1, 2, 3 $ in $ \rm(17) $, $ N_0 = 15^2 $ in $ \rm(18) $: (a) the $ L^2(\Omega) $ error; (b) the number of discrete set $ \sqrt{N_t} $; (c) CPU time

    Figure 3.  In Example 2, when use domain growth function $ \rho(t) = \exp(0.001t) $ and parameters $ D_u = 1, \ D_v = 10, \ a = 0.1, \ b = 0.9, \ \gamma = 10, \ m = 4, \ \Delta t = 0.01 $, the patterns at different time $ t $ of SH model under fixed discrete points $ n_Z = 30^2 $

    Figure 4.  In Example 2, under same setting with Figure 4, the patterns at different time $ t $ of SH model under global refinement with the parameter $ \nu = 1 $ in $ \rm(17) $ and the discrete set increasing from $ n_Z = 18^2 $ to $ n_Z = 35^2 $ in $ \rm(18) $ as show in Figure 8 (a)

    Figure 5.  In Example 3, when use global refinement strategy with discrete sets in the domain as in Figure 8 (b) and the domain growth function as $ \rho(t) = 1+9\sin(\pi t/1000) $, patterns generated by $ D_u = 1, \ D_v = 10, \ a = 0.1, \ b = 0.9, \ \gamma = 10, \ m = 4 $ and $ \Delta t = 0.005 $

    Figure 6.  In Example 4, in the hexagon domain, when use global refinement strategy with discrete sets in the domain as in Figure 8 (c) and the domain growth function as $ \rho(t) = \exp(0.001t) $, patterns generated by $ D_u = 1, \ D_v = 10, \ a = 0.1, \ b = 0.9, \ \gamma = 114, \ m = 4 $ and $ \Delta t = 0.005 $

    Figure 7.  In Example 5, in the star-shape domain, when use global refinement strategy with discrete sets in the domain as in Figure 8 (d) and the domain growth function as $ \rho(t) = \exp(0.001t) $, patterns generated by $ D_u = 1, \ D_v = 10, \ a = 0.1, \ b = 0.9, \ \gamma = 10, \ m = 4 $ and $ \Delta t = 0.005 $

    Figure 8.  For the SH model, the change of trial centers $ N_t $ or $ \sqrt{N_t} $ as number of time steps $ n_T $ increases under different global refinement settings for different examples: (a) for EX2, in the square domain the $ \nu = 1 $ in $ \rm(17) $; (b) for EX3, in the square domain, $ \nu = 5 $ in $ \rm(17) $; (c) for EX4, in the hexagon domain, $ \nu = 1 $ in $ \rm(17) $; (d) for EX5, in the star-shape domain, $ \nu = 2 $ in $ \rm(17) $

    Table 1.  When $ \Delta t = 0.005 $, $ T = 750 $ and $ m = 6, \ \epsilon = 5 $ in SH model, $ L^2(\Omega) $ errors and convergence rates comparison between our scheme and [8,Example 1]

    $ N $ $ e_h $ error Rate $ e_h $ ($ r=3 $ in [8,Example 1]) Rate
    $ 10^2 $ $ 0.807*10^{-3} $ $ 0.171*10^{-3} $
    $ 20^2 $ $ 0.489*10^{-4} $ $ 4.042 $ $ 0.106*10^{-4} $ $ 4.001 $
    $ 30^2 $ $ 0.807*10^{-5} $ $ 4.443 $ $ 0.209*10^{-5} $ $ 4.002 $
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