Add the advection_matrix() function and a test for it.

diff --git a/advection_matrix.m b/advection_matrix.m
new file mode 100644 (file)
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+++ b/advection_matrix.m
@@ -0,0 +1,112 @@
+function S = advection_matrix(integerN, x0, xN)
+  ##
+  ## The numerical solution of the advection-diffusion equation,
+  ##
+  ##   -d*u''(x) + v*u'(x) + r*u = f(x)
+  ##
+  ## in one dimension, subject to the boundary conditions,
+  ##
+  ##   u(x0)  = u(xN)
+  ##
+  ##   u'(x0) = u'(xN)
+  ##
+  ## over the interval [x0,xN] gives rise to a linear system:
+  ##
+  ##   AU = h^2*F
+  ##
+  ## where h = 1/n, and A is given by,
+  ##
+  ##   A = d*K + v*h*S + r*h^2*I.
+  ##
+  ## We will call the matrix S the "advection matrix," and it will be
+  ## understood that the first row (corresponding to j=0) is to be
+  ## omitted; since we have assumed that when j=0, u(xj) = u(x0) =
+  ## u(xN) and likewise for u'. ignored (i.e, added later).
+  ##
+  ## INPUTS:
+  ##
+  ##   * ``integerN`` - An integer representing the number of
+  ##   subintervals we should use to approximate `u`. Must be greater
+  ##   than or equal to 2, since we have at least two values for u(x0)
+  ##   and u(xN).
+  ##
+  ##   * ``f`` - The function on the right hand side of the poisson
+  ##   equation.
+  ##
+  ##   * ``x0`` - The initial point.
+  ##
+  ##   * ``xN`` - The terminal point.
+  ##
+  ## OUTPUTS:
+  ##
+  ##   * ``S`` - The NxN matrix of coefficients for the vector [u(x1),
+  ##   ..., u(xN)].
+  ##
+  ## EXAMPLES:
+  ##
+  ## For integerN=4, x0=0, and x1=1, we will have four subintervals:
+  ##
+  ##   [0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1]
+  ##
+  ## The first row of the matrix 'S' should compute the "derivative"
+  ## at x1=0.25. By the finite difference formula, this is,
+  ##
+  ##   u'(x1) = (u(x2) - u(x0))/2
+  ##
+  ##          = (u(x2) - u(x4))/2
+  ##
+  ## Therefore, the first row of 'S' should look like,
+  ##
+  ##   2*S1 = [0, 1, 0, -1]
+  ##
+  ## and of course we would have F1 = [0] on the right-hand side.
+  ## Likewise, the last row of S should correspond to,
+  ##
+  ##   u'(x4) = (u(x5) - u(x3))/2
+  ##
+  ##          = (u(x1) - u(x3))/2
+  ##
+  ## So the last row of S will be,
+  ##
+  ##   2*S4 = [1, 0, -1, 0]
+  ##
+  ## Each row 'i' in between will have [-1, 0, 1] beginning at column
+  ## (i-1). So finally,
+  ##
+  ##   2*S = [0, 1, 0, -1]
+  ##        [-1, 0, 1, 0]
+  ##         [0, -1, 0, 1]
+  ##         [1, 0, -1, 0]
+
+  if (integerN < 2)
+    S = NA;
+    return
+  end
+
+  [xs,h] = partition(integerN, x0, xN);
+
+  ## We cannot evaluate u_xx at the endpoints because our
+  ## differentiation algorithm relies on the points directly to the
+  ## left and right of `x`. Since we're starting at j=1 anyway, we cut
+  ## off two from the beginning.
+  differentiable_points = xs(3:end-1);
+
+  ## These are the coefficient vectors for the u(x0) and u(xn)
+  ## constraints. There should be N zeros and a single 1.
+  the_rest_zeros = zeros(1, integerN - 3);
+  u_x0_coeffs = cat(2, the_rest_zeros, [0.5, 0, -0.5]);
+  u_xN_coeffs = cat(2, [0.5, 0, -0.5], the_rest_zeros);
+
+  ## Start with the u(x0) row.
+  S = u_x0_coeffs;
+
+  for x = differentiable_points
+    ## Append each row obtained from the forward Euler method to S.
+    ## Chop off x0 first.
+    u_row = central_difference(xs(2:end), x);
+    S = cat(1, S, u_row);
+  end
+
+  ## Finally, append the last row for xN.
+  S = cat(1, S, u_xN_coeffs);
+end

diff --git a/tests/advection_matrix_tests.m b/tests/advection_matrix_tests.m
new file mode 100644 (file)
index 0000000..7509b07
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+++ b/tests/advection_matrix_tests.m
@@ -0,0 +1,13 @@
+## Try a 4x4 advection matrix. See the docstring in advection_matrix.m
+## for an explanation of the expected result.
+
+expected = (1/2) * [  0,  1,  0, -1; ...
+                    -1,  0,  1,  0; ...
+                     0, -1,  0,  1; ...
+                     1,  0, -1,  0 ];
+
+actual = advection_matrix(4, 0, 1);
+
+unit_test_equals("4x4 advection matrix is correct", ...
+                true, ...
+                expected == actual);