# System 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 = Cijfj

C is a matrix, fj(z) are functions dependent of z.

f ''i = Cijfj
f ''i = Cikfk

===> Cijfj = Cikfk

Which means all the functions are just linear multiples of each other, and each is just a trigonometric function.

Are you using the Einstein summation convention? If so, you should specify that you are! The summation convention is not something that is commonly used in non-physics-related mathematics.

yes, i am using Einsteins's summarizing convention. By the way this is a physics related problem, i am analyzing waves propagating a periodic media with RCWA method.

It will be almost the same as a first-order equation.
Guess

$$\vec{f}(t)=\vec{f}_{s}e^{st}$$

Plug in and get:

$$s^{2}\vec{f}_{s}e^{st}=C \vec{f}_{s}e^{st}$$

$$(C-s^{2}I)\vec{f}=0$$

So here s^2 are the eigenvalues of C. And each eigenvalue has two corresponding solutions with s and -s. I haven't seen treatment of different cases (complex s, multiplicity and such) but I guess it'll be the same as with the first-order analysis.

It will be almost the same as a first-order equation.
Guess

$$\vec{f}(t)=\vec{f}_{s}e^{st}$$

Plug in and get:

$$s^{2}\vec{f}_{s}e^{st}=C \vec{f}_{s}e^{st}$$

$$(C-s^{2}I)\vec{f}=0$$

So here s^2 are the eigenvalues of C. And each eigenvalue has two corresponding solutions with s and -s. I haven't seen treatment of different cases (complex s, multiplicity and such) but I guess it'll be the same as with the first-order analysis.

$$f_j=G_je^{i\sqrt{c_j}z}+H_je^{-i\sqrt{c_j}z}$$ where $$G_j and H_j$$ 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....?

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I am now providing a method to reduce the equation into system of 1st-order differential equations, so that you can use matrix exponential to solve it easily.

$$\mathbf{x}'= C \mathbf{y}$$
$$\mathbf{y}'=\mathbf{x}$$
$$\frac{d}{\text{dt}}\left( \begin{array}{c} \mathbf{x} \\ \mathbf{y} \end{array} \right)=\left( \begin{array}{cc} 0 & C \\ I & 0 \end{array} \right)\left( \begin{array}{c} \mathbf{x} \\ \mathbf{y} \end{array} \right)$$

After writing it into system of 1st-order differential equations, you can use matrix exponential to solve it. If you are not sure how to do it, please refer to the tutorial I have written here.

http://www.voofie.com/content/18/solving-system-of-first-order-linear-differential-equations-with-matrix-exponential-method/" [Broken]

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ross, thank you very much, i will look into it

Welcome. Hope it can help you. Please inform us if you can solve your problem, or you have more question. I am willing to help.