GNU Octave

Time for action – comparing with analytical solution

1. First we type in the physical constants given by the problem. Let us use Q = 10 kiloWatts per meter cubed (kW m^-3), k = 319 Watts per Kelvin per meter (W K^-1 m^-1), L = 0.5 meter and T₀ = 273 Kelvin:

   octave:8> Q=10000; k = 319; L=0.5; T_0=273;
   

2. We then need to specify the number of grid points (including the boundary grid points):

   octave:9>N_grids = 20;
   

   This gives a grid spacing of

   

   :

   octave:10> delta = L/(N_grid-1);
   

3. From this, we can set the source term and boundary values:

   octave:11> b = -Q/k*delta^2*ones(N_grids,1);
   octave:12> b(1)=b(N_grids)=T_0;
   

4. Next, we need the coefficient matrix which we ...