Reduced models for binocular rivalry

Abstract

Binocular rivalry occurs when two very different images are presented to the two eyes, but a subject perceives only one image at a given time. A number of computational models for binocular rivalry have been proposed; most can be categorised as either “rate” models, containing a small number of variables, or as more biophysically-realistic “spiking neuron” models. However, a principled derivation of a reduced model from a spiking model is lacking. We present two such derivations, one heuristic and a second using recently-developed data-mining techniques to extract a small number of “macroscopic” variables from the results of a spiking neuron model simulation. We also consider bifurcations that can occur as parameters are varied, and the role of noise in such systems. Our methods are applicable to a number of other models of interest.

Appendix: A model equations

Here we present the model equations. They are very similar to those in Laing and Chow (2002). For each excitatory neuron we have

$$ \frac{dV_e}{dt} = I_{syn}+I_{ext}-I_{mem}(V_e,n_e,h_e)-I_{AHP} \label{eq:dVe} \\ $$

(31)

$$ \frac{dn_e}{dt} = \psi[\alpha_n(V_e)(1-n_e)-\beta_n(V_e)n_e] \\ $$

(32)

$$ \frac{dh_e}{dt} = \psi[\alpha_h(V_e)(1-h_e)-\beta_h(V_e)h_e] \\ $$

(33)

$$ \tau_e\frac{ds_e}{dt} = A\sigma(V_e)(1-s_e)-s_e \\ $$

(34)

$$ \frac{d[Ca]}{dt} = \frac{-0.002g_{Ca}(V_e-V_{Ca})}{1+\exp{[-(V_e+25)/2.5]}}-[Ca]/80 \label{eq:dCa}\\ $$

(35)

$$ \tau_g\frac{d\phi}{dt} = 1-\phi-B\sigma(V_e)\phi \label{eq:dphi} $$

(36)

where $I_{mem}(V_e,n_e,h_e)=g_L(Ve-V_L)+g_Kn_e^4(V_e-V_K)+g_{Na}(m_{\infty}(V_e))^3h_e(V_e-V_{Na})$ and I _AHP  = g _AHP [Ca]/([Ca] + 1)(V _e  − V _K ). Other functions are m _∞ (V) = α _m (V)/(α _m (V) + β _m (V)), α _m (V) = 0.1(V + 30)/ $(1-\exp{[-0.1(V+30)]})$, $\beta_{m}(V)=4\exp[-(V+55)/$ 18], $\alpha_n(V)=0.01(V+34)/(1-\exp{[-0.1(V+34)]})$, $\beta_n(V)=0.125\exp{[-(V+44)/80]}$, $\alpha_h(V)=0.07\exp$ [ − (V + 44)/20], $\beta_h(V)=1/(1+\exp{[-0.1(V+14)]})$, $\sigma(V)=1/(1+\exp{[-(V+20)/4]}$. Parameters are g _L  = 0.05 mS/cm², V _L  = − 65 mV, g _K  = 40 mS/cm², V _K  = − 80 mV, g _Na  = 100 mS/cm², V _Na  = 55 mV, V _Ca  = 120  mV, g _AHP  = 0.05 mS/cm², g _Ca  = 0.1 mS/cm², ψ = 3, τ _e  = 8 ms, $\tau_g=\text{1,000}$ ms and A = 20. B is initially 1.3, but is varied in Section 4.

For each inhibitory neuron we have

$$ \frac{dV_i}{dt} = I_{syn}-I_{mem}(V_i,n_i,h_i) \\ $$

(37)

$$ \frac{dn_i}{dt} = \psi[\alpha_n(V_i)(1-n_i)-\beta_n(V_i)n_i] \\ $$

(38)

$$ \frac{dh_i}{dt} = \psi[\alpha_h(V_i)(1-h_i)-\beta_h(V_i)h_i] \\ $$

(39)

$$ \tau_i\frac{ds_i}{dt} = A\sigma(V_i)(1-s_i)-s_i \label{eq:dVi} $$

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where τ _i  = 10 ms and other functions are as above.

The synaptic current entering the jth excitatory neuron is

$$ (V_{+}-V_e^j)\frac{1}{N}\sum\limits_{k=1}^Ng_{ee}^{jk}s_e^k\phi^k+(V_{-}-V_e^j)\frac{1}{N}\sum\limits_{k=1}^Ng_{ie}^{jk}s_i^k $$

(41)

where $V_e^j$ is the voltage of the jth excitatory neuron in mV, $s_{e/i}^k$ is the strength of the synapse emanating from the kth excitatory/inhibitory neuron, ϕ ^k is the factor by which the kth excitatory neuron is depressed, N = 60 is the number of excitatory neurons (equal to the number of inhibitory neurons), and the Gaussian coupling functions are given by

$$ g_{ee}^{jk}=\alpha_{ee}\sqrt{50/\pi}\exp{\{-50[(j-k)/N]^2\}} $$

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and

$$ g_{ie}^{jk}=\alpha_{ie}\sqrt{20/\pi}\exp{\{-20[(j-k)/N]^2\}}. $$

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The reversal potentials are V ₊ = 0 mV, V ₋ = − 80 mV. Similarly, the synaptic current entering the jth inhibitory neuron is

$$ (V_{+}-V_i^j)\frac{1}{N}\sum\limits_{k=1}^Ng_{ei}^{jk}s_e^k+(V_{-}-V_e^j)\frac{1}{N}\sum\limits_{k=1}^Ng_{ii}^{jk}s_i^k $$

(44)

where $V_i^j$ is the voltage of the jth inhibitory neuron in mV, and the coupling functions are

$$ g_{ei}^{jk}=\alpha_{ei}\sqrt{20/\pi}\exp{\{-20[(j-k)/N]^2\}} $$

(45)

and

$$ g_{ii}^{jk}=\alpha_{ii}\sqrt{30/\pi}\exp{\{-30[(j-k)/N]^2\}}. $$

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Parameters are α _ee  = 0.285 mS/cm², α _ie  = 0.36 mS/cm², α _ei  = 0.2 mS/cm² and α _ii  = 0.07 mS/cm². The external current to the jth excitatory neuron in μA/cm² is

$$\begin{array}{lll} I_{ext}^j&=&\frac{0.4}{\sqrt{2}}\left[\exp{\left(-\left[\frac{10[j-N/4]}{N}\right]^2\right)}\right.\\ &&\left.+\exp{\left(-\left[\frac{10[j-3N/4]}{N}\right]^2\right)}\right]-0.01 \end{array} $$

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i.e. the external current injected into the excitatory population consists of two Gaussians, centered at 1/4 and 3/4 of the way around the domain. The equations were simulated using Euler’s method with a fixed time-step of 0.02 ms, and no significant changes in the network behaviour were observed when time-steps of 0.01 or 0.005 ms were used.

Note that the system is completely deterministic. Section 5 discusses the results from simulating a stochastic version of this network.