我们知道 1-d 波动方程的解的形式为:$ \begin{aligned} u(x, t) &= \sum_{n=1}^{\infty}\sin(\frac{n\pi x}{L})(A_n\cos(\frac{n\pi ct}{L}) + B_n\sin(\frac{n\pi ct}{L})) \\ &= \sum_{n=1}^{\infty} C_n\sin(\frac{n\pi x}{L})\sin(\frac{n\pi ct}{L} + \theta) \\ &= \sum_{n=1}^{\infty} \frac{C_n}{2}\left [ \cos \left [ \frac{n\pi}{L}(x - ct) - \theta \right ] - \cos \left [ \frac{n\pi}{L}(x + ct) + \theta \right ] \right ]\end{aligned} $

  这里仅仅展示 $ n = 3 $ 随时间波动动画,即

clear;clc;

pi = 3.1415926;
L = 5.;
n = 3;
T0 = 0.5;
pho = 1.;
c = sqrt(T0/pho);

% u = zeros(100, length(x));
% for i=1:100
%     u(i,:) = sqrt(2)*sin(n*pi*x/L)*sin((n*pi*c*(i-1))/L + pi/4.);
% end
%
% t = ones(100, length(x));
% for i=1:100
%     t(i,:) = t(i,:)*(i/10.);
% end
%
% f1 = figure;
% plot3(t,x,u);

f = figure;
loops = 100;
set(gcf, 'Position', get(0,'Screensize'));

for i = 0:loops
    hold off
    [x,t] = meshgrid(0:.1:5,i:.03:10+i);
    z = sqrt(2)*sin(n*pi*x/L).*sin((n*pi*c*t)/L + pi/4.);
    surf(t,x,z)
    view(150,70)
    title('PDE: $$\frac{\partial^2 u}{\partial t^2} = c^2\frac{\partial^2 u}{\partial x^2}, n = 3$$','Interpreter','latex')

    axis tight manual
    ax = gca;
    ax.NextPlot = 'replaceChildren';
    axis off
    drawnow
end

  

01-13 22:05