%% Graph of solution of heat equation
% Let $u(z;t)$ solve the initial value problem for the heat equation in a
% rod of length $\pi$
%%
% 
% $$u_t = u_{zz}, \ t > 0, 0<z<\pi, $$
%
% with the boundary conditions
%%
%
% $$u(0) = 0 = u(\pi)$$
%
% and the initial value
%%
%
% $$u(z;0) = 1, \ 0 < z < \pi.$$
% 
% We know the solution is
%%
% 
% $$u(z;t) = \frac 4 \pi \sum _{n=1} ^\infty\frac 1{2n-1} e^{-n^2 t}\sin nz.$$
% 

%%
% I start by defining the grid of z and t values where the values of u will
% be computed.  Be careful not to use too many points in your meshgrid
% command; if you do you won't get a clear plot.

z = linspace(0,pi,50); t = linspace(0,3,100); [Z,T]= meshgrid(z,t);

%%
% Alternatively, I could create the meshgrid by giving the command

[Z,T]= meshgrid(0:pi/50:pi,0:.1:3);
%%
% Next I define u using a *for* loop to add terms, starting with u=0.

u=0;
for n=1:10
    u = u + 4*exp(-(2*n-1).^2.*T).*sin((2.*n-1).*Z)./((2*n-1)*pi);
end
%%
surf(T,Z,u)