---
title: "2D groundwater FD scheme, boundary conditions"
subject: "Computational Techniques in Civil Engineering"
chapter: 7
key: "semester_8/computational_techniques_in_civil_engineering:gw-2d-fd-boundary-flow-coefficients"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/gw-2d-fd-boundary-flow-coefficients
site: BCT Engineering
---

# 2D groundwater FD scheme, boundary conditions

Computational Techniques in Civil Engineering · Chapter 7: Simulation of Ground water flow

- Asked in: 2072 Asoj, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/gw-2d-fd-boundary-flow-coefficients
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-7

## Question

Derive the expression for the finite difference scheme for 2D groundwater simulation in steady state for a homogeneous and isotropic aquifer. Describe the boundary conditions and flow coefficients.

## Short answer

For a homogeneous isotropic aquifer the steady‑state finite difference scheme is φ_i,j = (φ_i+1,j + φ_i-1,j + φ_i,j+1 + φ_i,j-1)/4. Flow coefficients are C_x = T Δy/Δx and C_y = T Δx/Δy. Boundary conditions: specified head, no‑flow (mirror nodes), specified flux, and head‑dependent leakage.

## Answer

#### Derivation of the steady-state finite difference scheme
Consider a small element of a confined aquifer of transmissivity $T$. For steady flow, inflow equals outflow. With Darcy's law $q_x=-T_x\dfrac{\partial\phi}{\partial x}$, $q_y=-T_y\dfrac{\partial\phi}{\partial y}$ (flux per unit width):
$$
\frac{\partial q_x}{\partial x}+\frac{\partial q_y}{\partial y}=0\ \Rightarrow\ \frac{\partial}{\partial x}\left(T_x\frac{\partial\phi}{\partial x}\right)+\frac{\partial}{\partial y}\left(T_y\frac{\partial\phi}{\partial y}\right)=0
$$
For a homogeneous, isotropic aquifer ($T_x=T_y=T$ constant) this is the Laplace equation $\dfrac{\partial^2\phi}{\partial x^2}+\dfrac{\partial^2\phi}{\partial y^2}=0$.

Place a grid of spacing $\Delta x=\Delta y$ and write central differences about node $(i,j)$:
$$
\frac{\phi_{i+1,j}-2\phi_{i,j}+\phi_{i-1,j}}{\Delta x^2}+\frac{\phi_{i,j+1}-2\phi_{i,j}+\phi_{i,j-1}}{\Delta y^2}=0
$$
$$
\boxed{\ \phi_{i,j}=\frac14\left(\phi_{i+1,j}+\phi_{i-1,j}+\phi_{i,j+1}+\phi_{i,j-1}\right)\ }
$$

```text
              (i,j+1)
                 |
   (i-1,j) --- (i,j) --- (i+1,j)
                 |
              (i,j-1)
```

#### Flow coefficients
Flow between node $(i,j)$ and a neighbour is written as a conductance times the head difference. The coefficients are
$$
C_{E}=C_{W}=\frac{T_x\Delta y}{\Delta x},\qquad C_{N}=C_{S}=\frac{T_y\Delta x}{\Delta y}
$$
so the flow from neighbour $k$ into the node is $Q_k=C_k(\phi_k-\phi_{i,j})$, and the steady-state balance $\sum Q_k=0$ gives
$$
\phi_{i,j}=\frac{C_E\phi_{i+1,j}+C_W\phi_{i-1,j}+C_N\phi_{i,j+1}+C_S\phi_{i,j-1}}{C_E+C_W+C_N+C_S}
$$
For an isotropic, homogeneous aquifer on a square grid, all four coefficients are equal ($=T$) and each weight is $\tfrac14$. For a layered aquifer the harmonic mean of the transmissivities of the two grids is used for the face between them.

#### Boundary conditions
1. **Specified head (Dirichlet):** the potential is known, e.g. along a river or reservoir, $\phi=h_b$. The node is not solved; its value enters the neighbour's equation.
2. **No-flow (Neumann, $\partial\phi/\partial n=0$):** along an impervious boundary or flow line. Use a mirror (image) node outside the boundary with the same potential as the inside neighbour. For a boundary on the left of node $(i,j)$:
$$
\phi_{i-1,j}=\phi_{i+1,j}\ \Rightarrow\ \phi_{i,j}=\tfrac14\left(2\phi_{i+1,j}+\phi_{i,j+1}+\phi_{i,j-1}\right)
$$
3. **Specified flux:** $q_n$ given (e.g. a pumping well or recharge): the known flow across that boundary is added to the node balance as a known inflow or outflow.
4. **Head-dependent (river bed leakage):** $Q=C_{riv}(h_{riv}-\phi)$ added to the node balance.

The resulting algebraic equations (one per interior node) are solved iteratively (Gauss-Seidel) or by matrix methods until the heads converge.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a homogeneous, isotropic aquifer in steady state, the 2D groundwater equation reduces to which form?
   A. Laplace equation: d2phi/dx2 + d2phi/dy2 = 0
   B. Poisson equation with a storage term S dphi/dt
   C. Diffusion equation with a time derivative
   D. First-order equation dphi/dx + dphi/dy = 0
   Answer: A. Steady inflow equals outflow and Tx = Ty = T is constant, so T cancels and the Laplace equation remains. (Key point: Governing equation)

2. On a square grid, central differences of the Laplace equation give the head at node (i,j) as:
   A. The sum of the four neighbouring heads
   B. The average of the four neighbouring heads, phi = (1/4)(sum of neighbours)
   C. Half the sum of the two nearest neighbours
   D. The average of the four neighbours minus the node's old value
   Answer: B. Substituting central differences with dx = dy gives phi(i,j) = 1/4 of the sum of the four neighbours. (Key point: Finite difference scheme)

3. In the flow-coefficient (conductance) form, the coefficient between node (i,j) and its east neighbour is:
   A. Ty*dx/dy
   B. Tx*dx/dy
   C. Tx*dy/dx
   D. T/(dx*dy)
   Answer: C. C_E = Tx*dy/dx: transmissivity times flow width divided by the distance between nodes. (Key point: Flow coefficients)

4. How is a no-flow (impervious) boundary on the left of node (i,j) handled?
   A. The boundary node is fixed at the river head
   B. Use a mirror node with phi(i-1,j) = phi(i+1,j), giving phi = 1/4(2phi(i+1,j) + phi(i,j+1) + phi(i,j-1))
   C. Set phi(i-1,j) = 0 and use the usual 1/4 average
   D. Drop the node from the equations
   Answer: B. Zero normal gradient is imposed by an image node equal to the inside neighbour, so the east value counts twice. (Key point: No-flow boundary)

## Related in Computational Techniques in Civil Engineering

- [Steady 2D model: seepage under a dam](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Aseepage-under-dam-2d-model) — Computational Techniques in Civil Engineering, chapter 7, asked 3×
- [Algorithm for unsteady pipe flow FD equations](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Aalgorithm-pipe-fd-hp-equations) — Computational Techniques in Civil Engineering, chapter 6
- [Method of characteristics; FD form for pipe flow](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Amoc-definition-fd-pipe-equations) — Computational Techniques in Civil Engineering, chapter 6, asked 4×
- [Constitutive relation for 2D isotropic material](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Aconstitutive-2d-isotropic) — Computational Techniques in Civil Engineering, chapter 3
- [Finite difference method; explicit and implicit schemes](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Afdm-explicit-implicit) — Computational Techniques in Civil Engineering, chapter 5, asked 2×
- [Courant condition; 1-D implicit coefficient matrix](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Acourant-implicit-matrix-river-stage) — Computational Techniques in Civil Engineering, chapter 7

## Related in other subjects

- [Derive magnetic boundary conditions](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Amagnetic-boundary-conditions-derive) — Electromagnetism, chapter 3
- [Finite difference nodal temperatures in a square plate](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Atwo-dimensional-conduction-nodal-temperatures) — Heat Transfer, chapter 2
- [Boundary shear stress and compound channel flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ashear-stress-boundary-compound-channel-flow) — Hydraulics, chapter 6
- [Essential and natural boundary conditions](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Aessential-natural-boundary-conditions) — Finite Element Method, chapter 2