# American Institute of Mathematical Sciences

April  2022, 9(2): 69-84. doi: 10.3934/jcd.2022001

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

 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

Received  February 2021 Revised  January 2022 Published  April 2022 Early access  February 2022

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.

Citation: Marco Berardi, Fabio V. Difonzo. A quadrature-based scheme for numerical solutions to Kirchhoff transformed Richards' equation. Journal of Computational Dynamics, 2022, 9 (2) : 69-84. doi: 10.3934/jcd.2022001
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##### References:
Output of numerical simulations obtained by MATLAB ${\mathtt{pdepe}}$ and by the quadrature-based scheme, referred to Example 1
Output of numerical simulations obtained by MATLAB ${\mathtt{pdepe}}$ and by quadrature-based scheme, as described in Example 2
Output of numerical simulations obtained by MATLAB ${\mathtt{pdepe}}$ and by the quadrature-based scheme, as described in Example 3
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 $\Delta t = 0.001$
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 $\Delta t, \Delta z$ $O_{\textrm{num}}^{z}\left(32\Delta t,32\Delta z\right)=1.2054$ $O_{\textrm{num}}^{z}\left(16\Delta t,16\Delta z\right)=1.1423$ $O_{\textrm{num}}^{z}\left(8\Delta t,8\Delta z\right)=1.1492$ $O_{\textrm{num}}^{z}\left(4\Delta t,4\Delta z\right)=1.2508$ $O_{\textrm{num}}^{z}\left(2\Delta t,2\Delta z\right)=1.6130$
 Step-sizes for $\theta_{\textrm{ref}}$ Numerical order $\Delta t, \Delta z$ $O_{\textrm{num}}^{z}\left(32\Delta t,32\Delta z\right)=1.2054$ $O_{\textrm{num}}^{z}\left(16\Delta t,16\Delta z\right)=1.1423$ $O_{\textrm{num}}^{z}\left(8\Delta t,8\Delta z\right)=1.1492$ $O_{\textrm{num}}^{z}\left(4\Delta t,4\Delta z\right)=1.2508$ $O_{\textrm{num}}^{z}\left(2\Delta t,2\Delta z\right)=1.6130$
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 $\Delta t, \Delta z$ $O_{\textrm{num}}^{z}\left(32\Delta t,32\Delta z\right)=1.4027$ $O_{\textrm{num}}^{z}\left(16\Delta t,16\Delta z\right)=1.1968$ $O_{\textrm{num}}^{z}\left(8\Delta t,8\Delta z\right)=1.1744$ $O_{\textrm{num}}^{z}\left(4\Delta t,4\Delta z\right)=1.2595$ $O_{\textrm{num}}^{z}\left(2\Delta t,2\Delta z\right)=1.6032$
 Step-sizes for $\theta_{\textrm{ref}}$ Numerical order $\Delta t, \Delta z$ $O_{\textrm{num}}^{z}\left(32\Delta t,32\Delta z\right)=1.4027$ $O_{\textrm{num}}^{z}\left(16\Delta t,16\Delta z\right)=1.1968$ $O_{\textrm{num}}^{z}\left(8\Delta t,8\Delta z\right)=1.1744$ $O_{\textrm{num}}^{z}\left(4\Delta t,4\Delta z\right)=1.2595$ $O_{\textrm{num}}^{z}\left(2\Delta t,2\Delta z\right)=1.6032$
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 $\Delta t, \Delta z$ $O_{\textrm{num}}^{z}\left(32\Delta t,32\Delta z\right)=3.8094$ $O_{\textrm{num}}^{z}\left(16\Delta t,16\Delta z\right)=2.0037$ $O_{\textrm{num}}^{z}\left(8\Delta t,8\Delta z\right)=1.0801$ $O_{\textrm{num}}^{z}\left(4\Delta t,4\Delta z\right)=1.3516$ $O_{\textrm{num}}^{z}\left(2\Delta t,2\Delta z\right)=1.6497$
 Step-sizes for $\theta_{\textrm{ref}}$ Numerical order $\Delta t, \Delta z$ $O_{\textrm{num}}^{z}\left(32\Delta t,32\Delta z\right)=3.8094$ $O_{\textrm{num}}^{z}\left(16\Delta t,16\Delta z\right)=2.0037$ $O_{\textrm{num}}^{z}\left(8\Delta t,8\Delta z\right)=1.0801$ $O_{\textrm{num}}^{z}\left(4\Delta t,4\Delta z\right)=1.3516$ $O_{\textrm{num}}^{z}\left(2\Delta t,2\Delta z\right)=1.6497$