I've started from building the transformation matrix for simpler homogenous recurrence relation. So, if we just dropn⋅dnn⋅dnandn⋅hnn⋅for some time, then transformation matrix for that relation takes the form: {xn=xn−1+yn−2+yn−3yn=yn−2+2xn−1=>Vn→= ⎜⎜...
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Here we have the equation describing the relationship between a transformation matrix, an eigenvector, and its corresponding eigenvalue. Just like the TISE, this is color-coded, with magenta being our transformation matrix and the blue being our eigenvalues. ...
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Then, the function forms all of the block matrices (taking the boundary values of the problem into account) and assembles the mass matrix using the four blocks. function M = mass(t,y) % Extract the components of y for the solution u and mesh x N = length(y)/2; u = y(1:N); ...
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