P.H. Trinh

Lecture 24: Computation of the heat equation III

Example 15.4: Solution of the inhomogeneous Dirichlet problem

This lecture starts off with Example 15.4 from the notes, where we look to solve $$ \begin{gathered} u_t = u_{xx} \\ u(0, t) = 2, \qquad u(\pi, t) = 1 \\ u(x, 0) = 0. \end{gathered} $$

We show that the solution is given by $$ u(x, t) = U(x) + \sum_{n=1}^\infty B_n \sin(nx) \mathrm{e}^{-n^2 t} $$ where we have found the steady-state solution $$ U(x) = 2 - \frac{x}{\pi}, $$ as well as the coefficients $$ B_n = \frac{2}{n\pi}[(-1)^n - 2]. $$

We also illustrated the solution using the code:

Inhomogeneous heat equation

%% Plot the Fourier series for Example 15.4
% Written for MA20223 Vectors & PDEs 2019-20

clear           % Clear all variables
close all       % Close all windows
N = 100;        % How many Fourier modes to include?

R = @(n, t, x) 2/(n*pi)*((-1)^n - 2)*exp(-n^2*t)*sin(n*x);

% Create a mesh of points between two limits
x0 = pi; x = linspace(0, x0, 1000);

% Create a mesh of points in time
t = linspace(0, 5, 200);

figure(1);                              % Open the figure
plot(x, 2 - x/pi, 'b', 'LineWidth', 2); % Plot the base function
ylim([-0.2,2]);                         % Set the y limits
xlim([0, x0]);                          % Set the x limits
xlabel('x'); ylabel('u(x,t)');
hold on

for j = 1:length(t)
    tj = t(j);
    
    u = (2 - x/pi);
    for n = 1:N
        u = u + R(n, tj, x);
    end
    if j == 1
        p = plot(x, u, 'r');
    else
        set(p, 'YData', u);
    end
    drawnow
    title(['t = ', num2str(tj)]);
    pause(0.1);
   
    if j == 1
        pause
    end
end

Problem set 8 Q1

The next thing we studied was PS8, Q1, which is the solution of the homogeneous Neumann problem for the heat equation.

Modification of PS8 Q1

%% Plot the Fourier series for a made-up modification of PS8 Q1. 
% Written for MA20223 Vectors & PDEs

clear           % Clear all variables
close all       % Close all windows
N = 20;          % How many Fourier modes to include?

% Define an in-line function that takes in three inputs: 
R = @(n, t, x) 4*(-1)^n/n^2*cos(n*x)*exp(-n^2*t);

% Create a mesh of points between two limits
x0 = pi;
x = linspace(0, x0, 1000);

% Create a mesh of points in time
t = linspace(0, 5, 200);

figure(1);                              % Open the figure
plot(x, x.^2, 'b', 'LineWidth', 2);     % Plot the base function
ylim([-0.2,pi^2]);                      % Set the y limits
xlim([0, x0]);                          % Set the x limits
xlabel('x'); ylabel('u(x,t)');
hold on

for j = 1:length(t)
    tj = t(j);
    
    u = 1/2*(2*pi^2/3);
    for n = 1:N        
        u = u + R(n, tj, x);
    end
    if j == 1
        p = plot(x, x.^2, 'r');
    else
        set(p, 'YData', u);
    end
    drawnow
    title(['t = ', num2str(tj)]);
    pause(0.1);
   
    if j == 1
        pause
    end
end