Synchro
Self-interaction
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In order to adapth the model to fit previous studies, we can try to hamper any influence from the coupling field emitted by an oscillator on itself. This is implemented by introducing a coupling phase in front of A. Its value is fixed to fulfill the condition $latex \displaystyle {\large \Omega = \omega}$

To study the self-interaction of an oscillator through its field, let's take the equations derived there (equation (7) and (8)) and consider them for a single oscillator :

 

$latex \displaystyle{ \Large \Omega = \omega + Im \left( \frac{1}{z_k} e^{i \theta} A(t,x_k) \right) = \omega + Im \left( R_A(\Omega~,~ 0) e^{i( \theta_A(\Omega~,~ 0)~ + ~\theta ) } \right) }$

 

The condition $latex \displaystyle{ \Large \Omega = \omega }$ implies $latex \displaystyle{ \Large \theta = -\theta_A(\Omega~,~ 0) + p \pi, p \in Z = \frac{1}{2}\arctan(\Omega/\gamma) + p \pi, p \in Z  }$.

 

There are two angles in  $latex \displaystyle{ \Large [0,2\pi] }$ which satisfy this solution,  $latex \displaystyle{ \Large p=0 }$ and $latex \displaystyle{ \Large p=1 }$

Once this condition is fulfilled, two radii become available : 

 

$latex \displaystyle{ \Large R = \sqrt{\mu \pm \frac{\alpha}{2\sqrt{D}(\Omega^2+\gamma^2)^{\frac{1}{4}}}} } $

 

Analyzing the stability of these radii, we find that only the solution with the smallest radius is unstable, and the solution with the greatest radius is marginally stable (at least one eigen value as a 0 real part). The should account for the fact that the system is unperturbed by a global shift of the phase.

 

    

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