Add some step length functions, untested.
[octave.git] / advection_matrix.m
1 function S = advection_matrix(integerN, x0, xN)
2 ##
3 ## The numerical solution of the advection-diffusion equation,
4 ##
5 ## -d*u''(x) + v*u'(x) + r*u = f(x)
6 ##
7 ## in one dimension, subject to the boundary conditions,
8 ##
9 ## u(x0) = u(xN)
10 ##
11 ## u'(x0) = u'(xN)
12 ##
13 ## over the interval [x0,xN] gives rise to a linear system:
14 ##
15 ## AU = h^2*F
16 ##
17 ## where h = 1/n, and A is given by,
18 ##
19 ## A = d*K + v*h*S + r*h^2*I.
20 ##
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).
25 ##
26 ## INPUTS:
27 ##
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)
31 ## and u(xN).
32 ##
33 ## * ``f`` - The function on the right hand side of the poisson
34 ## equation.
35 ##
36 ## * ``x0`` - The initial point.
37 ##
38 ## * ``xN`` - The terminal point.
39 ##
40 ## OUTPUTS:
41 ##
42 ## * ``S`` - The NxN matrix of coefficients for the vector [u(x1),
43 ## ..., u(xN)].
44 ##
45 ## EXAMPLES:
46 ##
47 ## For integerN=4, x0=0, and x1=1, we will have four subintervals:
48 ##
49 ## [0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1]
50 ##
51 ## The first row of the matrix 'S' should compute the "derivative"
52 ## at x1=0.25. By the finite difference formula, this is,
53 ##
54 ## u'(x1) = (u(x2) - u(x0))/2
55 ##
56 ## = (u(x2) - u(x4))/2
57 ##
58 ## Therefore, the first row of 'S' should look like,
59 ##
60 ## 2*S1 = [0, 1, 0, -1]
61 ##
62 ## and of course we would have F1 = [0] on the right-hand side.
63 ## Likewise, the last row of S should correspond to,
64 ##
65 ## u'(x4) = (u(x5) - u(x3))/2
66 ##
67 ## = (u(x1) - u(x3))/2
68 ##
69 ## So the last row of S will be,
70 ##
71 ## 2*S4 = [1, 0, -1, 0]
72 ##
73 ## Each row 'i' in between will have [-1, 0, 1] beginning at column
74 ## (i-1). So finally,
75 ##
76 ## 2*S = [0, 1, 0, -1]
77 ## [-1, 0, 1, 0]
78 ## [0, -1, 0, 1]
79 ## [1, 0, -1, 0]
80
81 if (integerN < 2)
82 S = NA;
83 return
84 end
85
86 [xs,h] = partition(integerN, x0, xN);
87
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);
93
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);
99
100 ## Start with the u(x0) row.
101 S = u_x0_coeffs;
102
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);
108 end
109
110 ## Finally, append the last row for xN.
111 S = cat(1, S, u_xN_coeffs);
112 end