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Symmetry transformation

  1. Jun 30, 2016 #1
    1. The problem statement, all variables and given/known data
    System of equations
    [tex]\frac{du_j}{dt}=u_{j+1}+u_{j-1}-2u_j-\frac{K}{2 \pi}\sin(2\pi u_j)+\bar{F}+F_{ac}\cos(2\pi \nu_o t)[/tex]
    where ##j=1,2,3,4##. So ##\{u_j\}## is set of coordinates. If we apply symmetry transformation
    [tex]\sigma_{r,m,s}\{u_j(t)\}=\{u_{j+r}(t-\frac{s}{\nu_0})\}[/tex]
    how to find condition for which
    [tex]\sigma_{r,m,s}\{u_j(t)\}=\{u_{j}(t) \}[/tex]
    We impose cyclic boundary condition.



    2. Relevant equations


    3. The attempt at a solution
    If I understand well
    [tex]\frac{du_1}{dt}=u_{2}+u_{4}-2u_1-\frac{K}{2 \pi}\sin(2\pi u_1)+\bar{F}+F_{ac}\cos(2\pi \nu_o t)[/tex]
    [tex]\frac{du_2}{dt}=u_{1}+u_{3}-2u_2-\frac{K}{2 \pi}\sin(2\pi u_2)+\bar{F}+F_{ac}\cos(2\pi \nu_o t)[/tex]
    [tex]\frac{du_3}{dt}=u_{2}+u_{4}-2u_3-\frac{K}{2 \pi}\sin(2\pi u_3)+\bar{F}+F_{ac}\cos(2\pi \nu_o t)[/tex]
    [tex]\frac{du_4}{dt}=u_{3}+u_{1}-2u_4-\frac{K}{2 \pi}\sin(2\pi u_4)+\bar{F}+F_{ac}\cos(2\pi \nu_o t)[/tex]
    And to transformation be satisfied [tex]\sigma_{r,m,s}\{u_j(t)\}=\{u_{j}(t) \}[/tex]
    it is important that for example
    [tex]\sigma_{r,m,s}u_1=u_4[/tex]
    [B][B][B][B][tex]\sigma_{r,m,s}u_2=u_3[/tex][/B][/B][/B][/B]
    [B][B][B][B]...

    But I am not sure how to show explicite consequence from this
    [B][B][tex]\sigma_{r,m,s}\{u_j(t)\}=\{u_{j}(t) \}[/tex][/B][/B][/B][/B][/B][/B]

     
  2. jcsd
  3. Jul 5, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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