I'm not sure if this typical behaviour or not but I am solving a finite-difference problem using the backward differencing method.
I populated a sparse matrix with the appropriate diagonal terms (along the central diagonal and one above and below it) and I attempted to solve the problem using MATLAB's built-in method (B=A\x) and it seems MATLAB just gets it wrong.
Furthermore, if use inv() and use the inverse of the tridiagonal matrix I get the correct solution.
Why does this behave like this?
Additional info:
http://pastebin.com/AbuEW6CR (Values are tabbed so its easier read) Stiffness matrix K:
1 0 0 0
-0.009 1.018 -0.009 0
0 -0.009 1.018 -0.009
0 0 0 1
Values for d:
0
15.55
15.55
86.73
Built-in output:
-1.78595556155136e-05
0.00196073713853244
0.00196073713853244
0.0108149483252210
Output using inv(K):
0
15.42
16.19
86.73
Manual output:
0
15.28
16.18
85.16
Code
nx = 21; %number of spatial steps
nt = 501; %number of time steps (varies between 501 and 4001)
p = alpha * dt / dx^2; %arbitrary constant
a = [0 -p*ones(1,nx-2) 0]'; %diagonal below central diagonal
b = (1+2*p)*ones(nx,1); %central diagonal
c = [1 -p*ones(1,nx-2) 1]'; %diagonal above central diagonal
d = zeros(nx, 1); %rhs values
% Variables a,b,c,d are used for the manual tridiagonal method for
% comparison with MATLAB's built-in functions. The variables represent
% diagonals and the rhs of the matrix
% The equation is K*U(n+1)=U(N)
U = zeros(nx,nt);
% Setting initial conditions
U(:,[1 2]) = (60-32)*5/9;
K = sparse(nx,nx);
% Indices of the sparse matrix which correspond to the diagonal
diagonal = 1:nx+1:nx*nx;
% Populating diagonals
K(diagonal) =1+2*p;
K(diagonal(2:end)-1) =-p;
K(diagonal(1:end-1)+1) =-p;
% Applying dirichlet condition at final spatial step, the temperature is
% derived from a table for predefined values during the calculation
K(end,end-1:end)=[0 1];
% Applying boundary conditions at first spatial step
K(1,1:2) = [1 0];
% Populating rhs values and applying boundary conditions, d=U(n)
d(ivec) = U(ivec,n);
d(nx) = R; %From table
d(1) = 0;
U(:,n+1) = tdm(a,b,c,d); % Manual solver, gives correct answer
U(:,n+1) = d\K; % Built-in solver, gives wrong answer
