Lecture 26: Computation of the wave equation II

At the start of the lecture, we continue deriving the Fourier coefficients for the problem of zero Dirichlet conditions on the wave equation. We show that

$$u(x, t) = \sum_{n=0}^\infty \sin\left(\frac{n\pi x}{L}\right) \left[ A_n \cos\left(\frac{n\pi ct}{L}\right) + B_n \sin\left(\frac{n\pi ct}{L}\right)\right]$$

where $$A_n = \frac{2}{L} \int_0^L u_0(x) \sin\left(\frac{n\pi x}{L}\right) \ \mathrm{d}{x}.$$

and Again we recognise this as the sine series, so we now need to equate $$B_n = \frac{2}{L} \left(\frac{L}{n\pi c}\right)\int_0^L v_0(x) \sin\left(\frac{n\pi x}{L}\right) \ \mathrm{d}{x}.$$

The next thing we did was look at the solution for the plucked string of example 16.4 in the notes. This yields the above Fourier series solution with $B_n = 0$ and $$A_n = \frac{4}{n^2\pi} \sin(n\pi/2).$$

We showed off the simulations of the problem using the following code.

% MA20223 Plucked String
clear
close all
 
% c = 1
 
x = linspace(0, pi, 100);
t = linspace(0, 2*(2*pi), 80);
Afunc = @(k) 4*(-1)^k/((2*k+1)^2*pi);
un = @(x, t, k) Afunc(k)*sin((2*k+1)*x).*cos((2*k+1)*t);
 
Nk = 80;
figure(1)
for j = 1:length(t)
    tt = t(j);
    u = 0;
    for k = 0:Nk
        u = u + un(x,tt,k);        
    end
    plot(x, u); 
    ylim([-pi/2, pi/2]);
    drawnow
    if j == 1
        pause;
    else
        pause(0.1)
    end
end