I think my program is complete.. What happen to this line? u(i)=sin(pi*x(i)); ??

if true
  % code
end
n=10;
c1=0; 
c2=0; 
k= 0.0025;
L=1; 
h=0.1;
alpha=1;
T=0.025;
n=L/h;m=T/k;
lambda =alpha*k/(h^2);
z=0:h:L;
disp('______________________________________________')
fprintf(' t x = ')
fprintf('%4.2f ',z)
fprintf('\n')
disp('____________________________________________ _')
fprintf('% 5.4f ',0)
% Compute the values of u at t=0
for i=1:n+1
u(i)=sin(pi*x(i));
fprintf('%10.6f ',u(i))
end
fprintf('\n')
% Compute the values of u t=jk, k=1,2,...,m
for j=1:m
t=j*k;
fprintf('% 5.4f ',t)
for i=1:n+1
if (i==1)
y(i)=c1;
elseif (i==n+1)
y(i)=c2;
else
y(i)=(1-2*lambda)*u(i)+lambda*(u(i+1)+u(i-1));
end;
fprintf('%10.6f ',y(i))
end;
fprintf('\n')
u=y;
end;

Answer 1

[Jos (10584)]

it is still there ...

Note that you can make use of the fact that matlab is all about handling matrices (i.e., lists of numbers) easily:

So a for-loop like

for i=1:n+1
  u(i)=sin(pi*x(i));
end

can be replaced by a matlabish statement like:

u = sin(pi .* x)

have the same effect (provided the variable u did not exist before and x has *n+1( elements)