Coupling by diffusion : complex model with diffusion

$latex \displaystyle{ \Large }$

Complex oscillator coupled with a complex field, a forcing phase and with diffusion (IV) :

One oscillator :

(1) $latex \displaystyle{ \Large z_t = (\mu +i \omega) z - z |z|^2 + e^{i \theta} A(t) }$

(2) $latex \displaystyle{ \Large A_t = \alpha ~z(t)~ \delta(x) ~-\gamma A } + D A_{xx}$

Let's suppose that $latex \displaystyle{ \Large z(t) = R e^{i \Omega t + i \phi} }$ then the Fourier transform of (2) gives :

$latex \displaystyle{ \Large A_t = \alpha Re^{i \Omega t} -\gamma A ~- k^2DA }$

And $latex \displaystyle{ \Large A(k,t) = \alpha R e^{i \phi} e^{-(\gamma + k^2D)t} \left( \int_{-\infty} ^t { e^{i \Omega \tau} e^{(\gamma + k^2D)\tau} d\tau}\right) = \alpha R \frac{1}{i\theta + \gamma + k^2 D} e^{i \Omega t + i \phi} }$

Taking the inverse Fourier transform :

$latex \displaystyle{ \Large A(x,t) = \frac{ \alpha } {2D \sqrt{(i\Omega + \gamma)/D}} ~ e^{-\sqrt{(i\Omega + \gamma)/D} ~ |x| } ~ Re^{i \Omega t + i \phi} = \frac{\alpha}{2 \sqrt{D} (\Omega^2 + \gamma^2)^{1/4} } e^{-i \frac{1}{2} atan(\Omega / \gamma)} ~ e^{-\sqrt{(i\Omega + \gamma)/D} ~ |x| } ~ Re^{i \Omega t + i \phi}= \frac{\alpha}{2 \sqrt{D} (\Omega^2 + \gamma^2)^{1/4} }e^{ -\frac{1}{\sqrt{D}} (\Omega^2 + \gamma^2)^{1/4} \cos(\frac{1}{2} atan(\Omega / \gamma))~ |x| } ~ e^{-i( \frac{1}{2} atan(\Omega / \gamma) + \frac{1}{\sqrt{D}} (\Omega^2 + \gamma^2)^{1/4} \sin( \frac{1}{2} atan(\Omega / \gamma)) ~ |x| )} ~ ~ Re^{i \Omega t + i \phi} }$.

We can also write it as : $latex \displaystyle{ \Large A(x,t) = R_A(\Omega,x) e^{i \theta_A(\Omega,x)} R e^{i ~ \Omega ~ t + i \phi} }$ (Checked with simulation its correct)

where $latex \displaystyle{ \Large R_A(\Omega_j, x-x_j) = \frac{\alpha}{2 \sqrt{D} (\Omega^2+\gamma^2)^{\frac{1}{4}}} \exp^{-\frac{1}{\sqrt{D}}(\Omega^2+\gamma^2)^{\frac{1}{4}} \cos(\frac{1}{2} \arctan(\frac{\Omega}{\gamma}))|x-x_j|}}$

and $latex \displaystyle{ \Large \theta_A(\Omega_j, x-x_j) = -(\frac{1}{2} \arctan\left(\Omega / \gamma\right)+\frac{1}{\sqrt{D}}(\Omega^2+\gamma^2)^{\frac{1}{4}} \sin\left(\frac{1}{2} \arctan\left(\Omega / \gamma \right)\right)|x-x_j|)} $

Then if x=0 the field A has a phase shift of $latex \displaystyle{ \Large -\frac{1}{2} atan(\Omega / \gamma) }$.

Several oscillators :

(3) $latex \displaystyle{ \Large \frac{d}{dt} z_j = (\mu_j +i \omega_j) z_j - z_j |z_j|^2 + e^{i \theta_j} A(t,x_j) }$

(4) $latex \displaystyle{ \Large A_t = \alpha ~ \sum_j^N z_j~ \delta(x-x_j) ~-\gamma A } + D A_{xx}$

Since (4) is linear :

$latex \displaystyle{ \Large A(x,t) = \sum_j^N R_A(\Omega_j, x-x_j) e^{i \theta_A(\Omega_j,x-x_j)} R_j e^{i ~ \Omega_j ~ t + i \phi_j} }$

Writing : $latex \displaystyle{ \Large z_k(t) = R_k e^{i ( \Omega_k t + \phi_k)} }$ and replacing in (3) :

$latex \displaystyle{ \Large i \Omega_k z_k(t) = (\mu_k +i \omega_k) z_k - z_k |z_k|^2 + e^{i \theta_k} A(t,x_k) }$

$latex \displaystyle{ \Large i \Omega_k = (\mu_k +i \omega_k) - R_k^2 + \frac{1}{z_k} e^{i \theta_k} A(t,x_k) }$

Then :

(5) $latex \displaystyle{ \Large \Omega_k = \omega_k + Im \left( \frac{1}{z_k} e^{i \theta_k} A(t,x_k) \right) = \omega_k + Im \left( \sum_j^N \frac{ R_j }{R_k} R_A(\Omega_j~,~ x_k-x_j) e^{i( \theta_A(\Omega_j~,~ x_k - x_j)~ + ~\theta_k ) } ~e^{i ~ (\Omega_j -\Omega_k )~ t } ~ e^{i(\phi_j - \phi_k)} \right) }$

(6) $latex \displaystyle{ \Large R_k^2 = \mu_k + Re \left( \frac{1}{z_k} e^{i \theta_k} A(t,x_k) \right) = \mu_k + Re \left( \sum_j^N \frac{ R_j }{R_k} R_A(\Omega_j~,~ x_k-x_j) e^{i( \theta_A(\Omega_j~,~ x_k - x_j)~ + ~\theta_k ) } ~e^{i ~ (\Omega_j -\Omega_k )~ t } ~ e^{i(\phi_j - \phi_k)} \right) }$

If the problem is perfectly homogenous (same intrisic frequencies, same distance between oscillators, ...) then frequency and the amplitude of the oscillator should be the same (this is a mean-field approach to the problem), otherwise they are not.

With a common radius and frequency, the equations write :

(7) $latex \displaystyle{ \Large \Omega = \omega + Im \left( \frac{1}{z_k} e^{i \theta_k} A(t,x_k) \right) = \omega + Im \left( \sum_j^N R_A(\Omega~,~ x_k-x_j) e^{i( \theta_A(\Omega~,~ x_k - x_j)~ + ~\theta_k ) } ~ e^{i(\phi_j - \phi_k)} \right) }$

(8) $latex \displaystyle{ \Large R^2 = \mu + Re \left( \frac{1}{z_k} e^{i \theta_k} A(t,x_k) \right) = \mu + Re \left( \sum_j^N R_A(\Omega~,~ x_k-x_j) e^{i( \theta_A(\Omega~,~ x_k - x_j)~ + ~\theta_k ) } ~ e^{i(\phi_j - \phi_k)} \right) }$

In the case where there is an infinite number of oscillators, where $latex \displaystyle{ \Large d }$ is the distance between the oscillators (so that $latex \displaystyle{\Large x_i=i d }$), and where the oscillators have a linear dephasing $latex \displaystyle{ \Large \phi_i = i \beta }$, the sum in (7) and (8) is a geometric series that can be simplified :

Let's denote $latex \displaystyle{ \Large \delta=\frac{\Omega}{\gamma} }$, $latex \displaystyle{\Large \lambda = \sqrt{\frac{D}{\gamma}} }$ (coupling range in meters), and $latex \displaystyle{\Large S=\frac{\alpha}{\sqrt{D\gamma}} }$ (coupling strength in 1/seconds )

$latex \displaystyle{ \Large R_A(\Omega,x)e^{i(\theta_A(\Omega,x)+\theta_k)} = \frac{\alpha}{2\sqrt{D}(\Omega^2+\gamma^2)^{\frac{1}{4}}}\exp \left(\frac{i}{2} \left( \arctan(\frac{\omega}{\gamma} )-\arctan(\frac{\Omega}{\gamma}) \right)-\frac{1}{\sqrt{D}}(\Omega^2+\gamma^2)^{\frac{1}{4}}e^{\frac{i}{2}\arctan(\frac{\Omega}{\gamma})}|x| \right) }$

or using the new quantities : $latex \displaystyle{ \Large R_A(\Omega,x)e^{i(\theta_A(\Omega,x)+\theta_k)} = \frac{S}{2(\delta^2+1)^{\frac{1}{4}}}\exp(\left(\frac{i}{2} \left( \arctan(\frac{\omega}{\gamma} )-\arctan(\delta) \right)-\frac{(\delta^2+1)^{\frac{1}{4}}}{\lambda}e^{\frac{i}{2}\arctan(\delta)}|x| \right) }$Let us now compute the series :

$latex \displaystyle{ \Large \sum_j^N \exp(-\frac{(\delta^2+1)^{\frac{1}{4}}}{\lambda}e^{\frac{i}{2}\arctan(\delta)}|j-k|d) ~ e^{i\beta(j-k)} = \sum_0^{N/2} \exp(-\frac{(\delta^2+1)^{\frac{1}{4}}}{\lambda}e^{\frac{i}{2}\arctan(\delta)}j d) ~ e^{i\beta j}+ \sum_1^{N/2} \exp(-\frac{(\delta^2+1)^{\frac{1}{4}}}{\lambda}e^{\frac{i}{2}\arctan(\delta)}j d) ~ e^{-i\beta j} = \frac{1}{1-\exp(-\frac{(\delta^2+1)^{\frac{1}{4}}}{\lambda}e^{\frac{i}{2}\arctan(\delta)} d+i\beta)}+\frac{1}{1-\exp(-\frac{(\delta^2+1)^{\frac{1}{4}}}{\lambda}e^{\frac{i}{2}\arctan(\delta)} d-i\beta)}-1 }$

(9) $latex \displaystyle{ \Large \Omega = \omega +\frac{S}{2(\delta^2+1)^{\frac{1}{4}}} Im \left( e^{\frac{i}{2} \left( \arctan(\frac{\omega}{\gamma} )-\arctan(\delta) \right)}\left( \frac{1}{1-\exp(-\frac{(\delta^2+1)^{\frac{1}{4}}}{\lambda}e^{\frac{i}{2}\arctan(\delta)} d+i\beta)}+\frac{1}{1-\exp(-\frac{(\delta^2+1)^{\frac{1}{4}}}{\lambda}e^{\frac{i}{2}\arctan(\delta)} d-i\beta)} -1 \right) \right) }$

(10) $latex \displaystyle{ \Large R^2 = \mu +\frac{S}{2(\delta^2+1)^{\frac{1}{4}}} Re \left( e^{\frac{i}{2} \left( \arctan(\frac{\omega}{\gamma} )-\arctan(\delta) \right)} \left( \frac{1}{1-\exp(-\frac{(\delta^2+1)^{\frac{1}{4}}}{\lambda}e^{\frac{i}{2}\arctan(\delta)} d+i\beta)}+\frac{1}{1-\exp(-\frac{(\delta^2+1)^{\frac{1}{4}}}{\lambda}e^{\frac{i}{2}\arctan(\delta)} d-i\beta)} -1 \right) \right) }$