my question is: what is the general solution of this system of coupled diff. equations:

f ''i =Cijfj

Cij is a matrix, fj(z) are functions dependent of z. indexes i and j go from 0 to N .

- Apr 6th 2010, 05:26 AMqetuolSystem of second order linear homogenous differential coupled equations
my question is: what is the general solution of this system of coupled diff. equations:

f ''i =**C**ijfj

Cij is a matrix, fj(z) are functions dependent of z. indexes i and j go from 0 to N . - Apr 6th 2010, 05:48 AMshawsend
Hi. This is what I think it is without looking so I'm not sure ok? You have $\displaystyle N$ second-order DEs which I can convert to $\displaystyle 2N$ first-order DEs just by adding $\displaystyle N$ more variables. I'd then have a system of $\displaystyle 2N$ first-order autonomous DEs which I can then in principle, compute the eigenvalues and vectors and then claim, the solution is:

$\displaystyle \textbf{F}_{2N}=\sum_{n=1}^{2N} k_n e^{\lambda_n z} \textbf{V}_n$

where $\displaystyle \{\lambda_n\}$ is the set of eigenvalues and $\displaystyle \{\textbf{V}_n\}$, the set of eigenvectors.

I don't know. Is that right guys? - Apr 7th 2010, 08:30 AMqetuol
sure i know linearization is a sulution. however i think there should be a general solution somewhere since it is "known" system , isnt it?

- Apr 12th 2010, 01:23 AMqetuol
okay, i found a possible solution:

$\displaystyle f_j=G_je^{i\sqrt{c_j}z}+H_je^{-i\sqrt{c_j}z}$

where $\displaystyle G_j and H_j$ are integrating constants.. cj are eigenvalues of C and are complex..

if so, one question remains... how are the indexes assigned to eigenvalues? i mean which eigenvalue will be c1....?

is this correct? - Apr 23rd 2010, 07:11 AMqetuol
anyone can answer me ? please?