What Conditions Ensure Symmetry in a System of Nonlinear Differential Equations?

[LagrangeEuler]

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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Homework Equations

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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[Nino MMP]

I think you mean the following:Let us assume the following symmetry transformation$$\sigma_{r,m,s}: \{u_j(t)\} \rightarrow \{u_{j+r}(t-\frac{s}{\nu_0})\} \quad (mod 4) $$where $r$ is an integer and $m, s$ are real numbers. We want to find the conditions for which this symmetry transformation leaves the equations of motion unchanged. The equations of motion can be written as$$ \frac{du_j}{dt} = f_j(u_1, u_2, u_3, u_4, t) $$where each $f_j$ is a function of the coordinates and time. Now let us apply the symmetry transformation to the equation of motion. We have$$\frac{d(\sigma_{r,m,s} u_j)}{dt} = \frac{d}{dt}\left(u_{j+r}(t-\frac{s}{\nu_0})\right) = \frac{\partial u_{j+r}}{\partial (t-\frac{s}{\nu_0})}\frac{\partial (t-\frac{s}{\nu_0})}{\partial t} = f_{j+r}(u_1, u_2, u_3, u_4, t-\frac{s}{\nu_0}) $$We see that the transformed equation of motion is equal to the original equation of motion if and only if$$f_j(u_1, u_2, u_3, u_4, t) = f_{j+r}(u_1, u_2, u_3, u_4, t-\frac{s}{\nu_0}) \quad \quad \forall j$$which is the condition we are looking for.