Short query about the 'Kuramoto Model', which is a mathematical model of synchronized coupled oscillators. If we consider the $N=2$ case then the governing equations are $$\frac{d \theta_1}{dt} = \omega_1 + \frac{K}{2}\sin(\theta_2(t) – \theta_1(t))~~~\text{and}~~~\frac{d \theta_2}{dt} = \omega_2 + \frac{K}{2}\sin(\theta_1(t) – \theta_2(t)),$$
where $\theta_i(t)$ are the phases of the oscillator, $\omega_i$ are the natural frequencies and $K$ is the coupling constant. Using a phase portrait we can observe phase attraction.
Can anyone see a clear reason why we would not observe phase attraction if $\omega_1$ or $\omega_2$ are negative?
Best Answer
For the difference $\phi(t)=θ_2(t)−θ_1(t)$ you get the differential equation $$ \frac{dϕ}{dt}=ω_2-ω_1-K\sinϕ. $$ The qualitative behavior now depends on the root structure of the right side. If $K<|ω_2-ω_1|$ there are no roots, the sign of the right side is constant, the difference is always growing or falling.
If $K\ge|ω_2-ω_1|$ the right side has roots, thus stationary solutions exist. As the right side is periodic, the root set is also periodic, so that any solution is bounded and converges to one of the attracting stationary solution with $$ \sin(ϕ^*)=\frac{ω_2-ω_1}K,~~\cos(ϕ^*)>0. $$ There appears no obstacle tied to the sign of the frequencies. If they have opposing signs then the difference is effectively a sum, so its value will be larger and thus the critical $K$ will also be larger.