System of second order linear homogenous differential coupled equations

[qetuol]

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

f ''_i = C_ijf_j

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

 

[IttyBittyBit]

f ''_i = C_ijf_j
f ''_i = C_ikf_k

===> C_ijf_j = C_ikf_k

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.

 

[qetuol]

oh my bad.
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.

 

[elibj123]

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.

 

[qetuol]

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.

thank you for your answer, so is this correct?:

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.. c_j 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...?

 

[qetuol]

anyone can answer me? please?

 

[ross_tang]

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(<br /> \begin{array}{c}<br /> \mathbf{x} \\<br /> \mathbf{y}<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{cc}<br /> 0 &amp; C \\<br /> I &amp; 0<br /> \end{array}<br /> \right)\left(<br /> \begin{array}{c}<br /> \mathbf{x} \\<br /> \mathbf{y}<br /> \end{array}<br /> \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/"

 

[qetuol]

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

 

[ross_tang]

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.