An analytical solution for vertical infiltration in bounded profiles

TL;DR

This work addresses vertical infiltration in bounded soil profiles on $0<x<L$ by solving a linearized Richards equation with an analytical, Fokas-method based approach. It derives a convergent integral representation for the solution $\theta(x,t)$ that couples the initial data and boundary inputs through spectral transforms and eliminates unknown boundary terms via an invariant transform $\nu(\lambda) = -\lambda - i \frac{K_0}{D_0}$, with the result matching half-line solutions as $L\to\infty$. The authors validate the method with three infiltration scenarios, showing agreement with classical solutions and highlighting finite-domain effects captured by the bounded-domain formulation. The work offers a computationally efficient framework for bounded-domain infiltration and lays groundwork for extensions to parameter inference and boundary-control problems.

Abstract

In this study, we derive an analytical solution to address the problem of one-dimensional vertical infiltration within bounded profiles. We consider the Richards equation together with various boundary conditions, simulating different scenarios of water application onto the surface of a homogeneous and bounded medium. To solve the corresponding initial boundary value problem over a finite interval, we apply the unified transform, commonly known as the Fokas method. Through this methodology, we obtain an integral representation that can be efficiently and directly computed numerically, yielding a convergent scheme.

Figures (5)

• Figure 1: Left: The solutions $u_1$ (solid line) and $u_P$ (dotted line) for $L = 0.05 \cm$ for the first case of the first example, at the time steps: $t_1 = 0.03 \sec$ (blue line), $t_2 = 0.06 \sec$ (red line) and $t_3 = 0.6 \sec$ (green line). Right: The solutions $u_1$ (solid line) and $u_P$ (dotted line) for $L = 140 \cm,$ at $t_1 = 40 \min$ (blue line), $t_2 = 70 \min$ (red line) and $t_3 = 100 \min$ (green line).
• Figure 2: The solutions $\theta$ (solid line) and $\theta_P$ (dotted line) for $L = 140 \cm$ (left picture) and $L = 70 \cm$ (right picture) of the first example (case 2). The different colors indicate the different time steps: $t_1 = 15 \min$ (blue line), $t_2 = 30 \min$ (red line) and $t_3 = 45 \min$ (green line).
• Figure 3: The solutions $\theta$ (solid line) and $\theta_B$ (dotted line) for $L = 140 \m,$ of the second example. Left: The solutions at $t_1 = 5 min$ (blue line) $t_2 = 7 min$ (red line) and $t_3 = 10 min$ (green line). Right: The solutions at $x_1= 0.5 m$ (blue), $x_2 =10m$ (red) and $x_3 =15 m$ (green) for time in hours.
• Figure 4: Case 1 of 3rd example: The solution $h$ (solid line) and $h_T$ (dotted line) for $L = 50 \m.$ Left: The solutions at $t_1 = 1day$ (blue line) $t_2=5 day$ (red line) and $t_3=10 day$ (green line). Right: The solutions at $x_1=45m$ (blue), $x_2=46m$ (red) and $x_3=47m$ (green).
• Figure 5: Case 2 of 3rd example: The solution $h$ (solid line) and $h_T$ (dotted line) for $L = 10 \m.$ Left: The solutions at $t_1 = 0.0005 h$ (blue line) $t_2=0.001 h$ (red line) and $t_3=0.01 h$ (green line). Right: The solutions at $x_1=9.2m$ (blue), $x=9.4m$ (red) and $x=9.6m$ (green).

Theorems & Definitions (2)

• Remark 2.1
• Remark 2.2