Signs of memory in a plastic frustrated Kuramoto model of neurons

Abstract

We study the effects brought about by adding frustration to the plastic Kuramoto model of neurons; i.e. we simulate for the first time the plastic Kuramoto–Sakaguchi model, in which the nonlocal coupling matrix for the phases is determined from the spike-timing-dependent plasticity rule. We find that frustration leads to the birhythmicity of the synchronous states. The multistability in the dynamics, with frustration as its crucial element, results in hysteresis, even when the learning rule is symmetric. Fine structure in the hysteresis loop reflects the complexity of the underlying energy landscape. A pacemaker is used to probe this landscape. As the neurons spontaneously self-organise, memory is formed by the configuration of synaptic strengths embodied in such synchronous states. We thus have a simple model of synaptic memory.

Appendix: The two-oscillator model

We show analytically, how the bistability in the two-oscillator model comes about. For the two-oscillator model we have $N=2$ and $n=1$ in Eqs. (1–3), giving

$$\begin{aligned} \dot{\theta }_{1}= & {} \omega _{1}+c_{12}\sin \left( \theta _{2}-\theta _{1}-\sigma \right) \end{aligned}$$

(5)

$$\begin{aligned} \dot{\theta }_{2}= & {} \omega _{2}+c_{21}\sin \left( \theta _{1}-\theta _{2}-\sigma \right) \end{aligned}$$

(6)

Now, for a synchronous state, we require $\dot{\theta }_{1}=\dot{\theta }_{2}$. By subtracting Eq. (6) from Eq. (5), we then find

$$\begin{aligned} \Delta \omega = c_{21}\sin \left( \theta _{1}-\theta _{2}-\sigma \right) -c_{12}\sin \left( \theta _{2}-\theta _{1}-\sigma \right) , \end{aligned}$$

(7)

where we have written $\Delta \omega =\omega _{1}-\omega _{2}$.

Fig. 7

Here, we see the bistability in coupling and one of the firing rates, as well as the phase differences. For instance, when $\theta _{1}$ leads, $c_{12} = 0, c_{21} = \alpha$ and $\dot{\theta _2}$ is forced to adapt to the pacemaker frequency, $\omega _1$. Here we have put $\omega _2 = 0$, for convenience

If $\theta _{1}$ leads, we have $\Delta \theta _{12}=\theta _{2}-\theta _{1}<0$ and $\Delta \theta _{21}=-\Delta \theta _{12}>0$. Thus, from Eqs. (2, 3), we find

$$\begin{aligned} \dot{c}_{12}= & {} -\epsilon c_{12}\exp \left( \frac{\Delta \theta _{12}}{\tau } \right) \end{aligned}$$

(8)

$$\begin{aligned} \dot{c}_{21}= & {} \epsilon \left( \alpha -c_{21}\right) \exp \left( -\frac{ \Delta \theta _{21}}{\tau }\right) \end{aligned}$$

(9)

Since $\Delta \theta _{12}$ is constant for the synchronous solution, the last two equations are decoupled and can be solved analytically. Their steady state solution gives $c_{12}=0$ and $c_{21}=\alpha$. Substituting these coupling constants back into Eq. (7), then gives

$$\begin{aligned} \Delta \omega = \alpha \sin \left( \Delta \theta _{21}-\sigma \right) . \end{aligned}$$

(10)

Similarly, if $\theta _{2}$ leads, we find

$$\begin{aligned} \Delta \omega = \alpha \sin \left( \Delta \theta _{21}+\sigma \right) . \end{aligned}$$

(11)

The intercepts with $\Delta \theta$ in Fig. 4 occur at $\sigma$, and $-\sigma$; for $\Delta \omega$, they are at $-\alpha \sin (\sigma )$, and $\alpha \sin (\sigma )$. The role of frustration is thus clear. If $|\Delta \omega |\le \alpha \sin \sigma$, two stable solutions exist, and the system moves between them as the sign of $\Delta \theta$ changes. The bistability is illustrated in Fig. 7.