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System of second order linear homogenous differential coupled equations

  1. Apr 6, 2010 #1
    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.
     
  2. jcsd
  3. Apr 6, 2010 #2
    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.
     
  4. Apr 7, 2010 #3
    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.
     
  5. Apr 7, 2010 #4
    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.
     
  6. Apr 7, 2010 #5
    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: Apr 7, 2010
  7. Apr 23, 2010 #6
    anyone can answer me? please?
     
  8. Jun 15, 2010 #7
    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: May 4, 2017
  9. Jun 15, 2010 #8
    ross, thank you very much, i will look into it
     
  10. Jun 20, 2010 #9
    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.
     
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