1 function S = advection_matrix(integerN, x0, xN)
3 ## The numerical solution of the advection-diffusion equation,
5 ## -d*u''(x) + v*u'(x) + r*u = f(x)
7 ## in one dimension, subject to the boundary conditions,
13 ## over the interval [x0,xN] gives rise to a linear system:
17 ## where h = 1/n, and A is given by,
19 ## A = d*K + v*h*S + r*h^2*I.
21 ## We will call the matrix S the "advection matrix," and it will be
22 ## understood that the first row (corresponding to j=0) is to be
23 ## omitted; since we have assumed that when j=0, u(xj) = u(x0) =
24 ## u(xN) and likewise for u'. ignored (i.e, added later).
28 ## * ``integerN`` - An integer representing the number of
29 ## subintervals we should use to approximate `u`. Must be greater
30 ## than or equal to 2, since we have at least two values for u(x0)
33 ## * ``f`` - The function on the right hand side of the poisson
36 ## * ``x0`` - The initial point.
38 ## * ``xN`` - The terminal point.
42 ## * ``S`` - The NxN matrix of coefficients for the vector [u(x1),
47 ## For integerN=4, x0=0, and x1=1, we will have four subintervals:
49 ## [0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1]
51 ## The first row of the matrix 'S' should compute the "derivative"
52 ## at x1=0.25. By the finite difference formula, this is,
54 ## u'(x1) = (u(x2) - u(x0))/2
56 ## = (u(x2) - u(x4))/2
58 ## Therefore, the first row of 'S' should look like,
60 ## 2*S1 = [0, 1, 0, -1]
62 ## and of course we would have F1 = [0] on the right-hand side.
63 ## Likewise, the last row of S should correspond to,
65 ## u'(x4) = (u(x5) - u(x3))/2
67 ## = (u(x1) - u(x3))/2
69 ## So the last row of S will be,
71 ## 2*S4 = [1, 0, -1, 0]
73 ## Each row 'i' in between will have [-1, 0, 1] beginning at column
76 ## 2*S = [0, 1, 0, -1]
86 [xs,h] = partition(integerN, x0, xN);
88 ## We cannot evaluate u_xx at the endpoints because our
89 ## differentiation algorithm relies on the points directly to the
90 ## left and right of `x`. Since we're starting at j=1 anyway, we cut
91 ## off two from the beginning.
92 differentiable_points = xs(3:end-1);
94 ## These are the coefficient vectors for the u(x0) and u(xn)
95 ## constraints. There should be N zeros and a single 1.
96 the_rest_zeros = zeros(1, integerN - 3);
97 u_x0_coeffs = cat(2, the_rest_zeros, [0.5, 0, -0.5]);
98 u_xN_coeffs = cat(2, [0.5, 0, -0.5], the_rest_zeros);
100 ## Start with the u(x0) row.
103 for x = differentiable_points
104 ## Append each row obtained from the forward Euler method to S.
105 ## Chop off x0 first.
106 u_row = central_difference(xs(2:end), x);
107 S = cat(1, S, u_row);
110 ## Finally, append the last row for xN.
111 S = cat(1, S, u_xN_coeffs);