# Symmetry transformation

## Homework Statement

System of equations
$$\frac{du_j}{dt}=u_{j+1}+u_{j-1}-2u_j-\frac{K}{2 \pi}\sin(2\pi u_j)+\bar{F}+F_{ac}\cos(2\pi \nu_o t)$$
where ##j=1,2,3,4##. So ##\{u_j\}## is set of coordinates. If we apply symmetry transformation
$$\sigma_{r,m,s}\{u_j(t)\}=\{u_{j+r}(t-\frac{s}{\nu_0})\}$$
how to find condition for which
$$\sigma_{r,m,s}\{u_j(t)\}=\{u_{j}(t) \}$$
We impose cyclic boundary condition.
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## The Attempt at a Solution

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

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

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