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\begin{document}
{\LARGE \bf Locking Ideas}
ede, 17.11.02
\section{Notations}
$\omega _{ext} = $ frequency of external signal,
$\omega _{free} = $ frequency of free running self-pulsation,
$\delta = \omega _{free}-\omega _{ext}$ detuning between both.
$\varphi = $ phase shift between the self-pulsation (pulses at the internal wavelength) and the externally injected pulses at an arbitrary but fixed position of the device, e.g., at one facet.
Hence, $\dot{\varphi } = $ instantaneous frequency of the internal self-pulsation (relative to $\omega _{ext}$).
\section{Assumptions and Conclusions}
\paragraph{Assumption 1: } The instantaneous frequency $\dot{\varphi }$ is a unique function of $\varphi $: $\dot{\varphi }=\omega (\varphi )$.
This makes sense at least for sufficiently small detuning and signal intensity. It is obviously true in the limit of vanishing signal intensity, when holds $\omega (\varphi )=\delta = $const. With increasing signal intensity, $\omega (\varphi )$ will show more and more variation.
\paragraph{Conclusion 1:} Obviously, $\omega (\varphi )$ is a periodic function.
\paragraph{Conclusion 2:} Locking appears if and only if $\omega (\varphi )=0$ has a solution.
\paragraph{Conclusion 3:} The locking dynamics is governed by the equation
\begin{eqnarray}
\dot{\varphi } = \omega (\varphi ). ~~~ \text{General solution:} ~~~ t = \int_{\varphi _1}^{\varphi _2} \frac{d\varphi }{\omega (\varphi )},
\end{eqnarray}
where $t$ is the time needed for the phase change from the initial value $\varphi _1$ to the final value $\varphi _2 $.
{\bf Qualitative discussion: } If $\min(\omega (\varphi ))>0$, there is no locking. But the phase $\varphi $ will change slowliest around the minimum of $\omega (\varphi )$. If this minimum is closely above 0, $\varphi $ will become nearly stationary for a comparatively long time. Such $\varphi $-plateaus did we observe in the numerics, they are a precurser of locking. The slope of the plateaus is given by the minimum of $\omega (\varphi )$. The border of the locking range is achieved when $\min(\omega (\varphi ))=0$. Beyound, we have at least one zero of $\omega (\varphi )$, say at $\varphi _0$. It is approached exponentially fast with a time constant $1/\omega '(\varphi _0)$, which is very long at the border of the locking range but smaller in its interiour. To be concrete, I consider now a model.
\begin{eqnarray}\label{adlermodell}
\text{\bf Model: } ~~ \omega (\varphi ) =\delta + p \sin(\varphi ).
~~~~~~~
\Longrightarrow ~~~
\boxed{\dot{\varphi }=\delta +p \sin(\varphi )}
\end{eqnarray}
This seems to be Adler's equation, if I remember right. It can be solved analytically:
\begin{eqnarray}
t&=&\frac{2}{\sqrt{\delta ^2-p^2}}\arctan \frac{\delta \tan(\varphi /2)+p}{\sqrt{\delta ^2-p^2}}
\hspace{3em}(p<\delta )
\\
t&=&\frac{1}{\delta } \tan\left(\frac{\varphi }{2}-\frac{\pi }{4}\right)
\hspace{10em} (p=\delta )
\\
t&=& \frac{1}{\delta } \ln \left|\frac{\delta \tan(\varphi /2)+p-\sqrt{p ^2-\delta ^2}}{\delta \tan(\varphi /2)+p+\sqrt{p ^2-\delta ^2}}\right|
\end{eqnarray}
I've plotted these solutions for some characteristic cases, assuming $\delta =1$. Abscissa is always the dimensionless time $\delta t$. Lower panels: thick: $\varphi $ versus $\delta t $, thin: $\varphi $ vs. $\omega (\varphi )$, i.e., the model (\ref{adlermodell}), to indicate the position of the zeros $\varphi _0$).
\figc{0.5}{p0}
\hfill\mbc{0.45}{{\bf Case p=0.}
Without signal, the frequency remains constant, the phase raises linearly. The time for one phase period (change by $2\pi $) is just $2\pi /\delta $.
}
\figc{0.5}{p09}
\hfill\mbc{0.45}{{\bf Case p=0.9.}
There is no zero yet of (\ref{adlermodell}), we are still outside the locking range. However, the minimum $\min(\omega (\varphi ))=\delta -p=0.1$ is already close to zero (cf. thin line in the lower panel). Hence, we clearly observe a range with stagnating phase, the precursor of locking known from the numerics with LDSL. Note that the time period is enhanced now to approximately $14/\delta $ due to this stagnation.
}
\figc{0.5}{p11}
\hfill\mbc{0.45}{{\bf Case p=1.1}
Now we have two zeros of $\omega (\varphi )$. This one with a negative slope is the stable one, which the phase approaches to. Depending on the starting phase, the time to lock is comparable with the extension of the time axis, roughly $10/\delta $. We are still close to the border of the locking range. }
\figc{0.5}{p5}
\hfill\mbc{0.45}{{\bf Case p=5}
Now we are more in the middle of the locking range, the time needed is only roughly $1/\delta $ yet.}
\end{document}