A quadrature-based scheme for numerical solutions to Kirchhoff transformed Richards' equation

Marco Berardi; Fabio V. Difonzo

doi:10.3934/jcd.2022001

Article Contents

2022, Volume 9, Issue 2: 69-84. Doi: 10.3934/jcd.2022001

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A quadrature-based scheme for numerical solutions to Kirchhoff transformed Richards' equation

Marco Berardi^1, and
Fabio V. Difonzo^2, ,

1.
Istituto di Ricerca sulle Acque, Consiglio Nazionale delle Ricerche, Via F. De Blasio 5, 70132 Bari, Italy
2.
Dipartimento di Matematica, Università degli Studi di Bari Aldo Moro, Via E. Orabona 4, 70125 Bari, Italy

^* Corresponding author: Fabio V. Difonzo
^* Corresponding author: Fabio V. Difonzo

Received: February 2021

Revised: January 2022

Early access: February 2022

Published: April 2022

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Abstract

In this work we propose a new numerical scheme for solving Richards' equation within Gardner's framework and accomplishing mass conservation. In order to do so, we resort to Kirchhoff transformation of Richards' equation in mixed form, so to exploit specific Gardner model features, obtaining a linear second order partial differential equation. Then, leveraging the mass balance condition, we integrate both sides of the equation over a generic grid cell and discretize integrals using trapezoidal rule. This approach provides a linear non-homogeneous initial value problem with respect to the Kirchhoff transform variable, whose solution yields the sought numerical scheme. Such a scheme is proven to be $ l^{2} $-stable and convergent to the exact solution under suitably conditions on step-sizes, retaining the order of convergence from the underlying quadrature formula.

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Mathematics Subject Classification: Primary: 65M22, 76S05; Secondary: 86A05.

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Figure 1. Output of numerical simulations obtained by MATLAB $ {\mathtt{pdepe}} $ and by the quadrature-based scheme, referred to Example 1

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Figure 2. Output of numerical simulations obtained by MATLAB $ {\mathtt{pdepe}} $ and by quadrature-based scheme, as described in Example 2

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Figure 3. Output of numerical simulations obtained by MATLAB $ {\mathtt{pdepe}} $ and by the quadrature-based scheme, as described in Example 3

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Figure 4. Output of numerical simulations obtained by MATLAB $ {\mathtt{pdepe}} $ and by the quadrature-based scheme (21), as described in Example 4. For the quadrature based scheme we used a constant spatial stepsize of $ \Delta z = 0.0833 $ and a temporal stepsize $ \Delta t = 0.001 $

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Table 1. Numerical orders of convergence of the scheme (21), with $ \Delta z = 3.75\cdot10^{-3}\, $cm and $ \Delta t = 3.125\cdot10^{-4}\, $days, referred to Example 1, providing a mass balance of $ 99.82\% $ according to (32), letting $ \frac{\Delta t}{\Delta z} = 1 $

Step-sizes for $\theta_{\textrm{ref}}$	Numerical order
$ \Delta t, \Delta z $	$ O_{\textrm{num}}^{z}\left(32\Delta t,32\Delta z\right)=1.2054 $
	$ O_{\textrm{num}}^{z}\left(16\Delta t,16\Delta z\right)=1.1423 $
	$ O_{\textrm{num}}^{z}\left(8\Delta t,8\Delta z\right)=1.1492 $
	$ O_{\textrm{num}}^{z}\left(4\Delta t,4\Delta z\right)=1.2508 $
	$ O_{\textrm{num}}^{z}\left(2\Delta t,2\Delta z\right)=1.6130 $

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Table 2. Numerical orders of convergence of the scheme (21), with $ \Delta z = 0.1406 $ cm and $ \Delta t = 0.001 $ days, referred to Example 2, providing a mass balance of $ 100.65\% $ according to (32), letting $ \frac{\Delta t}{\Delta z} = 1 $

Step-sizes for $\theta_{\textrm{ref}}$	Numerical order
$ \Delta t, \Delta z $	$ O_{\textrm{num}}^{z}\left(32\Delta t,32\Delta z\right)=1.4027 $
	$ O_{\textrm{num}}^{z}\left(16\Delta t,16\Delta z\right)=1.1968 $
	$ O_{\textrm{num}}^{z}\left(8\Delta t,8\Delta z\right)=1.1744 $
	$ O_{\textrm{num}}^{z}\left(4\Delta t,4\Delta z\right)=1.2595 $
	$ O_{\textrm{num}}^{z}\left(2\Delta t,2\Delta z\right)=1.6032 $

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Table 3. Numerical orders of convergence of the scheme (21), with $ \Delta z = 0.0026 $ cm and $ \Delta t = 0.0063 $ days, referred to Example 3, providing a mass balance of $ 100.71\% $ according to (32), letting $ \frac{\Delta t}{\Delta z} = 1 $

Step-sizes for $\theta_{\textrm{ref}}$	Numerical order
$ \Delta t, \Delta z $	$ O_{\textrm{num}}^{z}\left(32\Delta t,32\Delta z\right)=3.8094 $
	$ O_{\textrm{num}}^{z}\left(16\Delta t,16\Delta z\right)=2.0037 $
	$ O_{\textrm{num}}^{z}\left(8\Delta t,8\Delta z\right)=1.0801 $
	$ O_{\textrm{num}}^{z}\left(4\Delta t,4\Delta z\right)=1.3516 $
	$ O_{\textrm{num}}^{z}\left(2\Delta t,2\Delta z\right)=1.6497 $

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