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Krylov implicit integration factor WENO method for SIR model with directed diffusion

  • * Corresponding author: Ruijun Zhao

    * Corresponding author: Ruijun Zhao 

The second author is supported by NSF grant DMS-1620108

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  • SIR models with directed diffusions are important in describing the population movement. However, efficient numerical simulations of such systems of fully nonlinear second order partial differential equations (PDEs) are challenging. They are often mixed type PDEs with ill-posed or degenerate components. The solutions may develop singularities along with time evolution. Stiffness due to nonlinear diffusions in the system gives strict constraints in time step sizes for numerical methods. In this paper, we design efficient Krylov implicit integration factor (IIF) Weighted Essentially Non-Oscillatory (WENO) method to solve SIR models with directed diffusions. Numerical experiments are performed to show the good accuracy and stability of the method. Singularities in the solutions are resolved stably and sharply by the WENO approximations in the scheme. Unlike a usual implicit method for solving stiff nonlinear PDEs, the Krylov IIF WENO method avoids solving large coupled nonlinear algebraic systems at every time step. Large time step size computations are achieved for solving the fully nonlinear second-order PDEs, namely, the time step size is proportional to the spatial grid size as that for solving a pure hyperbolic PDE. Two biologically interesting cases are simulated by the developed scheme to study the finite-time blow-up time and location or discontinuity locations in the solution of the SIR model.

    Mathematics Subject Classification: Primary: 92D30, 65M06; Secondary: 92-08.

    Citation:

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  • Figure 1.  Numerical solution of the one-dimensional case of Eq. 1 for the case of avoiding infection ($ k_1 = 0 $ and $ k_2 = 0.1 $) at $ t = 0.706 $. CFL = 0.1

    Figure 2.  Numerical solution of the one-dimensional case of Eq. 1 for the case of avoiding infection ($ k_1 = 0 $, $ k_2 = 0.1 $) at $ t = 1.5 $. CFL = 0.1

    Figure 3.  Numerical solution of the two-dimensional case of Eq. 1 for the case of avoiding infection ($ k_1 = 0 $, $ k_2 = 0.1 $) at $ t = 0.7 $. CFL = 0.2

    Figure 4.  Numerical solution for the one-dimensional case of Eq. 1 for avoiding crowd ($ k_1 = 0.1 $, $ k_2 = 0 $). CFL = 0.6

    Figure 5.  Initial density profiles of population at $ t = 0 $

    Figure 6.  Numerical solution of the two-dimensional case of Eq. 1 for avoiding crowd ($ k_1 = 0.1 $, $ k_2 = 0 $) at $ t = 1 $. CFL = 0.2

    Figure 7.  Numerical solution of the two-dimensional case of Eq. 1 for avoiding crowd ($ k_1 = 0.1 $, $ k_2 = 0 $) at $ t = 10 $. CFL = 0.2

    Figure 8.  Numerical solution of the two-dimensional case of Eq. 1 for avoiding crowd ($ k_1 = 0.1 $, $ k_2 = 0 $) at $ t = 25 $. CFL = 0.2

    Table 1.  Numerical results of the one-dimensional system Eq. 10 for $ k_1 = 0.1 $ and $ k_2 = 0.001 $. $ \pi/N $ is the mesh size in the spatial direction. Here the constant CFL $ = 0.2 $

    $ N $ $ L^\infty $ error $ L^\infty $ order $ L^1 $ error $ L^1 $ order $ L^2 $ error $ L^2 $ order
    10 3.85e-02 - 7.48e-03 - 1.07e-02 -
    20 1.24e-02 1.64 2.28e-03 1.72 3.53e-03 1.60
    40 3.29e-03 1.91 5.56e-04 2.03 8.95e-04 1.98
    80 8.30e-04 1.99 1.45e-04 1.94 2.26e-04 1.98
    160 2.07e-04 2.00 3.68e-05 1.98 5.67e-05 2.00
    320 5.12e-05 2.01 9.21e-06 2.00 1.42e-05 2.00
     | Show Table
    DownLoad: CSV

    Table 2.  Numerical results of the one-dimensional system Eq. 10 for $ k_1 = 0.1 $ and $ k_2 = 0 $. CFL $ = 0.2 $. $ \pi/N $ is the mesh size in the spatial direction

    $ N $ $ L^\infty $ error $ L^\infty $ order $ L^1 $ error $ L^1 $ order $ L^2 $ error $ L^2 $ order
    10 3.86e-02 - 7.51e-03 - 1.07e-02 -
    20 1.24e-02 1.63 2.29e-03 1.72 3.54e-03 1.60
    40 3.30e-03 1.91 5.59e-04 2.03 9.00e-04 1.98
    80 8.31e-04 1.99 1.45e-04 1.95 2.27e-04 1.98
    160 2.07e-04 2.00 3.69e-05 1.97 5.70e-05 2.00
    320 5.15e-05 2.01 9.29e-06 1.99 1.43e-05 2.00
     | Show Table
    DownLoad: CSV

    Table 3.  Numerical results of the one-dimensional system Eq. 10 for $ k_1 = 0.1 $ and $ k_2 = 0 $. CFL $ = 0.5 $. $ \pi/N $ is the mesh size in the spatial direction

    $ N $ $ L^\infty $ error $ L^\infty $ order $ L^1 $ error $ L^1 $ order $ L^2 $ error $ L^2 $ order
    10 1.05e-01 - 3.23e-02 - 4.05e-02 -
    20 4.17e-02 1.33 8.41e-03 1.94 1.27e-02 1.67
    40 1.46e-02 1.51 2.69e-03 1.65 4.04e-03 1.65
    80 3.94e-03 1.89 7.25e-04 1.89 1.10e-03 1.88
    160 1.05e-03 1.91 1.94e-04 1.91 2.91e-04 1.92
    320 2.61e-04 2.00 4.96e-05 1.96 7.35e-05 1.99
     | Show Table
    DownLoad: CSV

    Table 4.  Numerical results of the one-dimensional system Eq. 10 for $ k_1 = 0 $ and $ k_2 = 0.1 $. CFL $ = 0.1 $. $ \pi/N $ is the mesh size in the spatial direction

    $ N $ $ L^\infty $ error $ L^\infty $ order $ L^1 $ error $ L^1 $ order $ L^2 $ error $ L^2 $ order
    10 3.38e-02 - 5.95e-03 - 7.33e-03 -
    20 1.05e-02 1.68 1.88e-03 1.66 2.27e-03 1.69
    40 2.92e-03 1.85 5.53e-04 1.77 6.64e-04 1.78
    80 7.93e-04 1.88 1.53e-04 1.85 1.80e-04 1.88
    160 2.08e-04 1.93 4.03e-05 1.92 4.71e-05 1.94
    320 5.35e-05 1.96 1.04e-05 1.95 1.21e-05 1.96
     | Show Table
    DownLoad: CSV

    Table 5.  Numerical results of the two-dimensional system Eq. 11 for $ k_1 = 0.1 $ and $ k_2 = 0.001 $. CFL $ = 0.4 $. $ \pi/N $ is the mesh size in each of the spatial directions

    $ N $ $ L^\infty $ error $ L^\infty $ order $ L^1 $ error $ L^1 $ order $ L^2 $ error $ L^2 $ order
    10 6.21e-02 - 7.34e-03 - 9.75e-03 -
    20 1.73e-02 1.84 1.65e-03 2.15 2.39e-03 2.03
    40 2.69e-03 2.69 4.01e-04 2.04 6.37e-04 1.91
    80 5.99e-04 2.16 9.09e-05 2.14 1.45e-04 2.13
    160 1.39e-04 2.10 2.15e-05 2.08 3.43e-05 2.08
    320 3.35e-05 2.05 5.18e-06 2.06 8.25e-06 2.06
     | Show Table
    DownLoad: CSV

    Table 6.  Numerical results of the two-dimensional system Eq. 11 for $ k_1 = 0.1 $ and $ k_2 = 0 $. CFL $ = 0.6 $. $ \pi/N $ is the mesh size in each of the spatial directions

    $ N $ $ L^\infty $ error $ L^\infty $ order $ L^1 $ error $ L^1 $ order $ L^2 $ error $ L^2 $ order
    10 7.30e-02 - 1.02e-02 - 1.32e-02 -
    20 2.12e-02 1.78 2.54e-03 2.00 3.76e-03 1.82
    40 4.15e-03 2.35 6.54e-04 1.96 1.03e-03 1.88
    80 1.03e-03 2.01 1.56e-04 2.07 2.47e-04 2.05
    160 2.77e-04 1.90 3.86e-05 2.02 6.05e-05 2.03
    320 6.97e-05 1.99 7.92e-06 2.28 1.28e-05 2.25
     | Show Table
    DownLoad: CSV

    Table 7.  Numerical results of the two-dimensional system Eq. 11 for $ k_1 = 0 $ and $ k_2 = 0.1 $. CFL $ = 0.2 $. $ \pi/N $ is the mesh size in each of the spatial directions

    $ N $ $ L^\infty $ error $ L^\infty $ order $ L^1 $ error $ L^1 $ order $ L^2 $ error $ L^2 $ order
    10 3.08e-02 - 2.91e-03 - 3.72e-03 -
    20 7.74e-03 1.99 8.37e-04 1.80 1.17e-03 1.67
    40 1.68e-03 2.20 2.37e-04 1.82 3.32e-04 1.82
    80 4.18e-04 2.01 6.24e-05 1.93 8.68e-05 1.94
    160 1.07e-04 1.97 1.64e-05 1.93 2.24e-05 1.96
    320 2.69e-05 1.99 4.13e-06 1.98 5.65e-06 1.99
     | Show Table
    DownLoad: CSV

    Table 8.  Numerical solution of $ S $ at $ t = 0.7 $ and $ t = 0.706 $ and $ x = 0.8 - 1/M $ for different $ M $. All values are computed using the uniform mesh $ \Delta x = 1/1280 $

    M 10 20 40 80 160 320 640 1280
    S (at T = 0.7) 7.05 9.72 13.39 18.49 25.56 36.15 66.84 80.74
    r (at T = 0.7) - 1.37 1.38 1.38 1.38 1.41 1.85 1.21
    S (at T = 0.706) 6.97 9.56 13.09 17.88 24.31 33.71 54.05 89.15
    r (at T = 0.706) - 1.37 1.37 1.37 1.36 1.39 1.60 1.65
     | Show Table
    DownLoad: CSV
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