System of second order linear homogenous differential coupled equations

  • Thread starter qetuol
  • Start date
  • #1
8
0
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.
 

Answers and Replies

  • #2
160
0
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.
 
  • #3
8
0
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.
 
  • #4
240
2
It will be almost the same as a first-order equation.
Guess

[tex]\vec{f}(t)=\vec{f}_{s}e^{st}[/tex]

Plug in and get:

[tex]s^{2}\vec{f}_{s}e^{st}=C \vec{f}_{s}e^{st}[/tex]

[tex](C-s^{2}I)\vec{f}=0[/tex]

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.
 
  • #5
8
0
It will be almost the same as a first-order equation.
Guess

[tex]\vec{f}(t)=\vec{f}_{s}e^{st}[/tex]

Plug in and get:

[tex]s^{2}\vec{f}_{s}e^{st}=C \vec{f}_{s}e^{st}[/tex]

[tex](C-s^{2}I)\vec{f}=0[/tex]

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?:

[tex]f_j=G_je^{i\sqrt{c_j}z}+H_je^{-i\sqrt{c_j}z}[/tex] where [tex]G_j and H_j[/tex] 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....?
 
Last edited:
  • #6
8
0
anyone can answer me? please?
 
  • #7
64
0
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.

[tex]\mathbf{x}'= C \mathbf{y}[/tex]
[tex]\mathbf{y}'=\mathbf{x}[/tex]
[tex]\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)[/tex]

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]
 
Last edited by a moderator:
  • #8
8
0
ross, thank you very much, i will look into it
 
  • #9
64
0
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.
 

Related Threads on System of second order linear homogenous differential coupled equations

Top