Energy-based stochastic control of neural mass models suggests time-varying effective connectivity in the resting state

Abstract

Several studies posit energy as a constraint on the coding and processing of information in the brain due to the high cost of resting and evoked cortical activity. This suggestion has been addressed theoretically with models of a single neuron and two coupled neurons. Neural mass models (NMMs) address mean-field based modeling of the activity and interactions between populations of neurons rather than a few neurons. NMMs have been widely employed for studying the generation of EEG rhythms, and more recently as frameworks for integrated models of neurophysiology and functional MRI (fMRI) responses. To date, the consequences of energy constraints on the activity and interactions of ensembles of neurons have not been addressed. Here we aim to study the impact of constraining energy consumption during the resting-state on NMM parameters. To this end, we first linearized the model, then used stochastic control theory by introducing a quadratic cost function, which transforms the NMM into a stochastic linear quadratic regulator (LQR). Solving the LQR problem introduces a regime in which the NMM parameters, specifically the effective connectivities between neuronal populations, must vary with time. This is in contrast to current NMMs, which assume a constant parameter set for a given condition or task. We further simulated energy-constrained stochastic control of a specific NMM, the Wilson and Cowan model of two coupled neuronal populations, one of which is excitatory and the other inhibitory. These simulations demonstrate that with varying weights of the energy-cost function, the NMM parameters show different time-varying behavior. We conclude that constraining NMMs according to energy consumption may create more realistic models. We further propose to employ linear NMMs with time-varying parameters as an alternative to traditional nonlinear NMMs with constant parameters.

Appendix A. Solution of the Hamilton-Jacobi-bellman equation for the LQR problem

Given the system:

$$ d{\mathbf{y}}(t) = {\mathbf{b}}\left( {{\mathbf{y}},{\mathbf{u}}} \right)dt + {\mathbf{\sigma }}\left( {\mathbf{y}} \right)d{\mathbf{W}}(t) $$

(53)

The Hamilton-Jacobi-Bellman (HJB) equation for ${\mathbf{\psi }}\left( {s,{\mathbf{y}}} \right) = \mathop{{\inf }}\limits_{{\mathbf{u}}} {{\mathbf{J}}^{{\mathbf{u}}}}\left( {s,{\mathbf{y}}} \right)$is (Oksendal 2003):

$$ 0 = \mathop{{\inf }}\limits_{{\mathbf{u}}} \left\{ {{{\mathbf{F}}^{{\mathbf{u}}}}\left( {s,{\mathbf{y}}} \right) + \left( {{{\mathbf{L}}^{{\mathbf{u}}}}{\mathbf{\psi }}} \right)\left( {s,{\mathbf{y}}} \right)} \right\} $$

(54)

where:

$$ \left( {{L^v}f} \right)(y) = \frac{{\partial f}}{{\partial s}}(y) + \sum\limits_{{i = 1}}^n {{b_i}} \left( {y,v} \right)\frac{{\partial f}}{{\partial {y_i}}} + \frac{1}{2}\sum\limits_{{i,j = 1}}^n {{{\left( {{\mathbf{\sigma \sigma }}'} \right)}_{{ij}}}} \left( {y,v} \right)\frac{{{\partial^2}f}}{{\partial {y_i}\partial {y_j}}} $$

(55)

and ${{\mathbf{F}}^{{\mathbf{u}}}}\left( {t,{\mathbf{y}}} \right) = {\mathbf{u}}'(t){\mathbf{Ru}}(t) + {\mathbf{y}}'(t){\mathbf{Qy}}(t)$. Using (9) we get:

$$ {{\mathbf{F}}^{{\mathbf{u}}}}\left( {t,{\mathbf{y}}} \right) + {{\mathbf{L}}^{{\mathbf{u}}}}{\mathbf{\psi }} = {\mathbf{y}}'{\mathbf{\dot{M}y}} + {\mathbf{\dot{\gamma }}} + {\mathbf{y}}'{\mathbf{Qy}} + {\mathbf{u}}'{\mathbf{Ru}} + \left( {{\mathbf{u}} + {\mathbf{\mu }}} \right)'\left( {{\mathbf{My + M}}'{\mathbf{y}}} \right) + \sum\limits_{{i,j = 1}}^n {{{\left( {{\mathbf{\sigma \sigma }}'} \right)}_{{i,j}}}{{\mathbf{M}}_{{ij}}}} $$

(56)

where we used: ${\mathbf{b}}\left( {{\mathbf{y}},{\mathbf{u}}} \right) = {\mathbf{u}}(t) + {\mathbf{\mu }}$. The minimum is obtained when $\frac{\partial }{{\partial {u_i}}}\left( {{F^u}\left( {t,y} \right) + {L^u}\psi } \right) = 0$, i.e. when: $2{\mathbf{Ru}} + 2{\mathbf{My}} = 0$, or:

$$ {{\mathbf{u}}^{*}} = - {{\mathbf{R}}^{{ - 1}}}{\mathbf{My}} $$

(57)

We substitute this value in (56) and obtain the right side equal to 0 if we choose ${\mathbf{M}}(t)$ such that:

$$ {\mathbf{\dot{M}}}(t) = {\mathbf{M}}(t){{\mathbf{R}}^{{ - 1}}}{\mathbf{M}}(t) - {\mathbf{Q}} $$

(58)

and ${\mathbf{\gamma }}(t)$ such that:

$$ {\mathbf{\dot{\gamma }}}(t) = - 2{\mathbf{\mu M}}(t){\mathbf{y}}(t) - tr\left( {{\mathbf{\sigma }}\left( {\mathbf{y}} \right){\mathbf{\sigma }}{{\left( {\mathbf{y}} \right)}^T}{\mathbf{M}}(t)} \right) $$

(59)