MTH3007b Weekly Problems 10

Original Documents: Problem Sheet / Provided Solutions

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10.1. Liebmann’s Method: Definition and Scope

Question

What is Liebmann’s method and what type of equation can it solve?

Liebmann’s method (also called Gauss-Seidel iteration applied to PDEs, or successive overrelaxation at $\omega =1$) is an iterative scheme for solving elliptic PDEs, specifically the 2D Laplace equation and 2D Poisson equation:

$$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0\text{(Laplace)}$$ $$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=f(x,y)\text{(Poisson)}$$

These are steady-state equations - there is no time derivative. They arise in heat conduction (steady state), electrostatics, and fluid potential flow.

The method: On a uniform grid with spacing $dx=dy$, the centred finite difference discretisation of the Laplace equation at interior node $(i,j)$ gives:

$$u_{i,j}=\frac{u_{i+1,j}+u_{i-1,j}+u_{i,j+1}+u_{i,j-1}}{4}$$

Liebmann’s method iterates this update rule (using the most recently updated values in-place, Gauss-Seidel style) until the maximum change between iterations falls below a tolerance $\epsilon$:

$$\underset{i,j}{\max }|u_{i,j}^{\text{new}}-u_{i,j}^{\text{old}}|<\epsilon$$

The boundary nodes are fixed at their prescribed Dirichlet boundary conditions throughout.

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10.2. Liebmann’s Method for a 2D Plate

Question

Use Liebmann’s method to find the steady-state temperature distribution on a $10\times 10$ cm plate with: top $=100^{\circ }$C, bottom $=0^{\circ }$C, left $=75^{\circ }$C, right $=50^{\circ }$C, $dx=1$ cm. Find $u(5,5)$.

import numpy as np
 
domain_size = 10    # cm
spatial_step = 1    # cm
number_of_nodes = domain_size // spatial_step + 1  # 11 nodes in each direction
 
u_grid = np.zeros((number_of_nodes, number_of_nodes))
u_grid[0, :]                          = 100.0  # top boundary
u_grid[number_of_nodes - 1, :]        = 0.0   # bottom boundary
u_grid[:, 0]                          = 75.0  # left boundary
u_grid[:, number_of_nodes - 1]        = 50.0  # right boundary
 
convergence_tolerance = 1e-4
max_iterations = 10000
 
for iteration in range(max_iterations):
    u_new = u_grid.copy()
    for row_index in range(1, number_of_nodes - 1):
        for col_index in range(1, number_of_nodes - 1):
            u_new[row_index, col_index] = (
                u_grid[row_index + 1, col_index]
                + u_grid[row_index - 1, col_index]
                + u_grid[row_index, col_index + 1]
                + u_grid[row_index, col_index - 1]
            ) / 4.0
    max_error = np.amax(np.abs(u_new - u_grid))
    u_grid = u_new.copy()
    if max_error < convergence_tolerance:
        print(f"Converged in {iteration + 1} iterations (max_error = {max_error:.2e})")
        break
 
print(f"u(5, 5) = {u_grid[5, 5]:.2f} degrees C")

Grid convention: with $dx=1$ and $11\times 11$ nodes, the node at $(i=5,j=5)$ corresponds to the physical centre of the plate at $(x=5,y=5)$ cm.

The Lax Equivalence Theorem guarantees that because the Liebmann iteration implements a consistent discretisation of the Laplace equation and the iteration converges, the result converges to the true solution of the boundary value problem.

Expected result: $u(5,5)\approx 56.25^{\circ }$C (the arithmetic average of the four boundary values: $(100+0+75+50)/4=56.25$). This is an exact result for this symmetric geometry by the mean value property of harmonic functions, confirming the numerical answer.

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Visualisation:

import matplotlib.pyplot as plt
 
fig, ax = plt.subplots(figsize=(6, 5))
im = ax.imshow(u_grid, origin='upper', extent=[0, domain_size, 0, domain_size], cmap='hot')
plt.colorbar(im, ax=ax, label='Temperature (deg C)')
ax.set_xlabel('x (cm)')
ax.set_ylabel('y (cm)')
ax.set_title('Steady-state temperature distribution (Liebmann)')
plt.tight_layout()
plt.show()