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Please can anybody help me with explaining, in detail, why and how the author solves the system of differential equations by such a step (see below)? How the equation 6.3 is formed step by step? Why such a column basis is valid? It would be extremely helpful if someone knows the name of the method or could give an exemplification. enter image description here

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The terms multiplied by $\mu$ & $\nu$ are modeling diffusion (this is a standard center difference scheme). What we know about the continuous diffusion operator (with appropriate boundary conditions) is that its eigenvectors are sines &/or cosines. If you prefer, they're exponentials with imaginary powers (the two are equivalent by Euler's formula). This is just the discrete analog: note that the vectors employed for the change of coordinates are exactly such complex exponentials evaluated at the grid points. You can check yourself that they are eigenvectors of the discretized diffusion matrix.

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  • $\begingroup$ Thanks very much. You said "they are eigenvectors of the discretized diffusion matrix", can you also point out what is the diffusion matrix so that I could evaluate its eigenvectors. And at same time, the eigenvectors seem to be independent of a, b, c, d, u and v, whether the matrix is the coefficients in the parentheses? $\endgroup$ – user97697 Sep 30 '13 at 11:46
  • $\begingroup$ Sure, the discretized diffusion matrix is a tridiagonal matrix with all diagonal entries equal to $-2$ and all superdiagonal and subdiagonal entries equal to $1$. (See also your ODE system (6.2) above.) Caveat: boundary conditions affect the top and bottom rows (this is reflected in how $x_{-1}$ and $y_{-1}$ are defined). Note, additionally, that (6.3) will only diagonalize the part of your linear ODE system pertaining to diffusion; it will not do so for the part relating to $a$, $b$, $c$ and $d$. $\endgroup$ – automaton 3 Sep 30 '13 at 12:06
  • $\begingroup$ Thanks very much. I checked with a diffusion matrix A of a ring of five cells: -2 on its diagonal entries, 1 on the sub-and super-diagonal entries, and a_15=a_51=1 being the boundary condition (a ring). The eigenvector given by Matlab is not the same as the exp. part defined in 6.3. The eigenvectors calculated are all real numbers for the matrix A, and dose not have a vector with all 1s as its entries. But 6.3 suggests such a vector when s=0, for all r, the entries is 1. Please can you let me know where I got lost and went wrong. I appreciate! $\endgroup$ – user97697 Sep 30 '13 at 12:55
  • $\begingroup$ You use periodic BCs, so I expect the eigenfunctions of the diffusion operator to be translation invariant. No wonder you didn't get the exponentials. I don't have access to MATLAB right now so I can't do it myself, but I'd suggest that you calculate the action of the diffusion matrix on said exponential and see whether you get a multiple of it. As for the constant vector (=spatially uniform pattern in the continuous case), of course that's an eigenvector (corresponding eigenvalue: zero) for your matrix $[-2,1,0,0,1 ; 1,-2,1,0,0 ; 0,1-2,1,0 ; 0,0,1,-2,1 ; 1,0,0-2,1]$! Check your numerics! $\endgroup$ – automaton 3 Sep 30 '13 at 22:08
  • $\begingroup$ (Btw, keep me posted on this, I enjoy remembering that stuff. & do upvote/formally accept my answer if it ends up working.) $\endgroup$ – automaton 3 Sep 30 '13 at 22:10

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