Plotting 2D equation with infinite sum - matlab

Question

I'm trying to plot the following equation (let's call it "Equation 1"):
This is the code I'm testing:
clear all;
xl=0; xr=1; % x domain [xl,xr]
J = 10; % J: number of division for x
dx = (xr-xl) / J; % dx: mesh size
tf = 0.1; % final simulation time
Nt = 60; % Nt: number of time steps
dt = tf/Nt/4;
x = xl : dx : xr; % generate the grid point
u_ex = zeros(J+1,Nt);
for n = 1:Nt
t = n*dt; % current time
for j=1:J+1
xj = xl + (j-1)*dx;
suma = zeros(100 , 1);
for k= 1:100
suma(k) = 4/(((2*k-1)^2) *pi*pi);
suma(k) = suma(k) * exp(-((2*k-1)^2) *pi*pi*t) * cos(2*k-1)*pi*xj;
end
m = sum(suma);
u_ex(j, n)= 0.5 - m;
end
end
tt = dt : dt : Nt*dt;
figure(1)
surf(x,tt, u_ex'); % 3-D surface plot
xlabel('x')
ylabel('t')
zlabel('u')
The problem is that all I get is a flat surface:
Equation 1 is suppossed to be the solution of the following parabolic partial differential equation with boundary values:
And after getting the numerical solution, it should look like this:
This plot gets the right values at the boundaries x = 0 and x = 1. The plot of Equation 1 doesn't have those values at the boundaries.
My complete .m code (that plots both the numerical solution and Equation 1) is:
clear all; % clear all variables in memory
xl=0; xr=1; % x domain [xl,xr]
J = 10; % J: number of division for x
dx = (xr-xl) / J; % dx: mesh size
tf = 0.1; % final simulation time
Nt = 60; % Nt: number of time steps
dt = tf/Nt/4;
mu = dt/(dx)^2;
if mu > 0.5 % make sure dt satisy stability condition
error('mu should < 0.5!')
end
% Evaluate the initial conditions
x = xl : dx : xr; % generate the grid point
% store the solution at all grid points for all time steps
u = zeros(J+1,Nt);
u_ex = zeros(J+1,Nt);
% Find the approximate solution at each time step
for n = 1:Nt
t = n*dt; % current time
% boundary condition at left side
gl = 0;
% boundary condition at right side
gr = 0;
for j=2:J
if n==1 % first time step
u(j,n) = j;
else % interior nodes
u(j,n)=u(j,n-1) + mu*(u(j+1,n-1) - 2*u(j,n-1) + u(j-1,n-1));
end
end
u(1,n) = gl; % the left-end point
u(J+1,n) = gr; % the right-end point
% calculate the analytic solution
for j=1:J+1
xj = xl + (j-1)*dx;
suma = zeros(100 , 1);
for k= 1:100
suma(k) = 4/(((2*k-1)^2) *pi*pi);
suma(k) = suma(k) * exp(-((2*k-1)^2) *pi*pi*t) * cos(2*k-1)*pi*xj;
end
m = sum(suma);
u_ex(j, n)= 0.5 - m;
end
end
% Plot the results
tt = dt : dt : Nt*dt;
figure(1)
colormap(gray); % draw gray figure
surf(x,tt, u'); % 3-D surface plot
xlabel('x')
ylabel('t')
zlabel('u')
title('Numerical solution of 1-D parabolic equation')
figure(2)
surf(x,tt, u_ex'); % 3-D surface plot
xlabel('x')
ylabel('t')
zlabel('u')
title('Analytic solution of 1-D parabolic equation')
maxerr=max(max(abs(u-u_ex))),
The code is taken from the book "Computational Partial Differential Equations Using MATLAB" by Yi-Tung Chen, Jichun Li, chapter 2, exercise 3.
In short: I'm not asking about the differential equation or the boundary problem, I want to know is: Why am I getting a flat surface when plotting Equation 1? Am I missing a parenthesis?
I do not want to use the symsum function because it never stop the script execution and I want to learn how to plot Equation 1 with no using symsum.
I've tested this code with Matlab R2008b and Octave 4.2.1. I got the same results (even with sums of 1000, 10000 and 50000 terms in the for loop with the k variable).
Edit!
Thanks, Steve!
I was missing a couple of parenthesis near the cosine, the right code is:
clear all; % clear all variables in memory
xl=0; xr=1; % x domain [xl,xr]
J = 10; % J: number of division for x
dx = (xr-xl) / J; % dx: mesh size
tf = 0.1; % final simulation time
Nt = 60; % Nt: number of time steps
dt = tf/Nt/4;
mu = dt/(dx)^2;
if mu > 0.5 % make sure dt satisy stability condition
error('mu should < 0.5!')
end
% Evaluate the initial conditions
x = xl : dx : xr; % generate the grid point
% store the solution at all grid points for all time steps
u = zeros(J+1,Nt);
u_ex = zeros(J+1,Nt);
% Find the approximate solution at each time step
for n = 1:Nt
t = n*dt; % current time
% boundary condition at left side
gl = 0;
% boundary condition at right side
gr = 0;
for j=2:J
if n==1 % first time step
u(j,n) = j;
else % interior nodes
u(j,n)=u(j,n-1) + mu*(u(j+1,n-1) - 2*u(j,n-1) + u(j-1,n-1));
end
end
u(1,n) = gl; % the left-end point
u(J+1,n) = gr; % the right-end point
% calculate the analytic solution
for j=1:J+1
xj = xl + (j-1)*dx;
suma = zeros(1000 , 1);
for k= 1:1000
suma(k) = 4/(((2*k-1)^2) *pi*pi);
suma(k) *= exp(-((2*k-1)^2) *pi*pi*t) * cos((2*k-1)*pi*xj);
end
m = sum(suma);
u_ex(j, n)= 0.5 - m;
end
end
% Plot the results
tt = dt : dt : Nt*dt;
figure(1)
colormap(gray); % draw gray figure
surf(x,tt, u'); % 3-D surface plot
xlabel('x')
ylabel('t')
zlabel('u')
title('Numerical solution of 1-D parabolic equation')
figure(2)
surf(x,tt, u_ex'); % 3-D surface plot
xlabel('x')
ylabel('t')
zlabel('u')
title('Analytic solution of 1-D parabolic equation')
Now my Equation 1 looks much better:
Also Steve was right when pointing out that my numerical solution may be wrong. I didn't notice that the boundary values are for the derivatives of my function, not the actual values of the function. I'll ask my teacher about this.
Edit2!
Ok, I got it. To calculate the derivatives at the boundaries you have to use hint 2.21 in the same book:
% hint 2.21 given by the book
% it is better to calculate the boundary values after calculating the inner points inside the for j = 1:m loop because you will need them:
u(1, n) = u(2, n) - dx * gl; % the left-end point
u(J+1,n) = u(J, n) + dx * gr; % the right-end point
Now my numerical solution looks like my analytic solution :D
Matlab R2008b can't recognize the *= operator that Octave does. I'm not tested this operator in other versions of Matlab because I'm too poor.
Yvon: I think the analytical solution formula comes from the real part of a Fourier expansion, but authors don't tell how they got it.