Symmetry transformation

  • #1
627
14

Homework Statement


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.
[/B]


Homework Equations




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]
[tex]\sigma_{r,m,s}u_2=u_3[/tex]
...

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

[/B]
 

Answers and Replies

  • #2
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8,206
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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