We will complete our determination of the Fourier series solution of the heat equation as presented in the previous lecture. This was the problem of: $$\begin{gathered} u_t = \kappa u_{xx}, \quad x\in[0, \pi], t \geq 0 \\ u(0, t) = 0 = u(\pi, t) \\ u(x, 0) = f(x) = 1. \end{gathered}$$

$$u(x, t) \sim \sum_1^\infty \left[b_n \sin\left(\frac{n\pi x}{L}\right)\right].$$

where $b_n$ will be the Fourier sine coefficients of the odd $2\pi$ extension of $f(x)$ on $[0, \pi]$.

The hardest part is to understand how to calculate the $b_n$ coefficients via an odd extension of the initial condition. In many ways, this is somewhat backwards (usually we ask “How do I compute a Fourier series for an odd extension” rather than to associate a Fourier series already given to an odd extension).

% 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: 
%   Input 1: n value [scalar]
%   Input 2: x value [vector]
%   Input 3: t value [scalar]
R = @(n, t, x) -2/(n*pi)*((-1)^n - 1)*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([0, pi], [1, 1], 'b', 'LineWidth', 2); % Plot the base function
ylim([-0.2,1.2]);                           % 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 = 0;
    for n = 1:N
        u = u + R(n, tj, x);
    end
    
    % Plotting commands
    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