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