Are there cases in which the types of behaviour considered above can be predicted without running a simulation? In particular, can equilibrium values be predicted, and how they depend on the specific values of the parameters of the model (e.g. connection weights, speed factors)? Below, these questions will be answered for a relatively simple example. Indeed it will turn out that in this case it is possible to predict the equilibrium values from the connection weights (the equilibrium values turn out to be independent of the speed factors, as long as these are nonzero). As a first step, consider the sensor state ss\(_{s}\).
LP
\(_{\mathrm{ss}_{s}}\)
Sensing a stimulus: determining values for state ss
\(_{s}\)
$$\begin{aligned} \mathbf{d}\mathrm{ss}_{s}(t)/\mathbf{d}\mathrm{t} = {\upeta }_{{\mathrm{ss}}_{s}} [{\upomega }_{\mathrm{sensing}} \mathrm{ws}_{s}(t) - \mathrm{ss}_{s}(t)] \end{aligned}$$
Having an equilibrium value means that no change occurs at
t:
dss
\(_{s}(t)\)/
dt
\(=\) 0. As it is assumed that
\({\upeta }_{{\mathrm{ss}}_{s}}\) is nonzero, this is equivalent to the following equilibrium equation for state ss
\(_{s}\), with
\(\underline{\mathbf{ws}}_{s}\) and
\(\underline{\mathbf{ss}}_{s}\) the equilibrium values for the two states ws
\(_{s}\) and ss
\(_{s}\).
$$\begin{aligned} {\upomega }_{\mathrm{sensing}} {\underline{\mathbf{ws}}}_{s} = {\underline{\mathbf{ss}}}_{s} \end{aligned}$$
In a similar manner this can be done for the other states, resulting in the following equations:
$$\begin{aligned} \begin{array}{l@{\quad }l} \mathbf{Equilibrium} &{} \mathbf{Equilibrium}\\ \mathbf{of}\,\mathbf{state} &{} \mathbf{criterion}\\ \mathrm{ss}_{s} &{} {\upomega }_{\mathrm{sensing}} \underline{\mathbf{ws}}_{s} = \underline{\mathbf{ss}}_{s}\\ \mathrm{srs}_{s} &{} {\upomega }_{\mathrm{representing}} \underline{\mathbf{ss}}_{s }= \underline{\mathbf{srs}}_{s}\\ \mathrm{ps}_{a} &{} {\upomega }_{\mathrm{responding}} \underline{\mathbf{srs}}_{s}\!+\!{\upomega }_{\mathrm{amplifying}} \underline{\mathbf{srs}}_{e} = \underline{\mathbf{ps}}_{a}\\ \mathrm{srs}_{e} &{} {\upomega }_{\mathrm{predicting}} \underline{\mathbf{ps}}_{a} = \underline{\mathbf{srs}}_{e}\\ \mathrm{es}_{a} &{} {\upomega }_{\mathrm{executing}} \underline{\mathbf{ps}}_{a} = \underline{\mathbf{es}}_{a}\\ \end{array} \end{aligned}$$
These are five equations with six unknowns
\(\underline{\mathbf{ws}}_{s}\),
\(\underline{\mathbf{ss}}_{s}\),
\(\underline{\mathbf{srs}}_{s}\),
\(\underline{\mathbf{ps}}_{a}\),
\(\underline{\mathbf{srs}}_{e}\),
\(\underline{\mathbf{es}}_{a}\); however, the variable
\(\underline{\mathbf{ws}}_{s}\) can be considered given as it indicates the external stimulus. So the five equations can be used to find expressions for the equilibrium values for the five other states in terms of the connection weights and
\(\underline{\mathbf{ws}}_{s}\). Note that for the sake of simplicity here it is assumed that
\({\upomega }_{\mathrm{amplifying}}\) and
\({\upomega }_{\mathrm{predicting}}\) are not both 1. Then this can be solved in an explicit analytical manner as follows. First two of them (the first two equations) are expressed in the externally given value
\(\underline{\mathbf{ws}}_{s}\):
$$\begin{aligned}&\underline{\mathbf{ss}}_{s }={\upomega }_{\mathrm{sensing}} \underline{\mathbf{ws}}_{s}\\&\underline{\mathbf{srs}}_{s}={\upomega }_{\mathrm{representing}} \underline{\mathbf{ss}}_{s }={\upomega }_{\mathrm{representing}} {\upomega }_{\mathrm{sensing} }\underline{\mathbf{ws}}_{s} \end{aligned}$$
Moreover, the third and fourth equation can be solved as follows:
$$\begin{aligned}&{\upomega }_{\mathrm{responding}} \underline{\mathbf{srs}}_{s}+{\upomega }_{\mathrm{amplifying}} \underline{\mathbf{srs}}_{e} = \underline{\mathbf{p}}_{a}\\&{\upomega }_{\mathrm{predicting}} \underline{\mathbf{ps}}_{a} = \underline{\mathbf{srs}}_{e} \end{aligned}$$
Substitute
\({\upomega }_{\mathrm{predicting}}\)
\(\underline{\mathbf{ps}}_{a}\) for
\(\underline{\mathbf{srs}}_{e}\) in the third equation, resulting in the following equation in
\(\underline{\mathbf{ps}}_{a}\) and
\(\underline{\mathbf{srs}}_{s}\):
$$\begin{aligned} {\upomega }_{\mathrm{responding}} \underline{\mathbf{srs}}_{s}+{\upomega }_{\mathrm{amplifying}} {\upomega }_{\mathrm{predicting}} \underline{\mathbf{ps}}_{a} = \underline{\mathbf{ps}}_{a} \end{aligned}$$
This can be used to express
\(\underline{\mathbf{ps}}_{a}\) in
\(\underline{\mathbf{srs}}_{s}\), and subsequently in
\(\underline{\mathbf{ws}}_{s}\):
$$\begin{aligned} {\upomega }_{\mathrm{responding}} \underline{\mathbf{srs}}_{s} = (1 - {\upomega }_{\mathrm{amplifying}} {\upomega }_{\mathrm{predicting}}) \underline{\mathbf{ps}}_{a} \end{aligned}$$
$$\begin{aligned} \underline{\mathbf{ps}}_{a}= & {} {\upomega }_{\mathrm{responding}} \underline{\mathbf{srs}}_{s} / (1 - {\upomega }_{\mathrm{amplifying}} {\upomega }_{\mathrm{predicting}})\\= & {} {\upomega }_{\mathrm{responding}} {\upomega }_{\mathrm{representing}} \\&\times \, {\upomega }_{\mathrm{sensing} }\underline{\mathbf{ws}}_{s }/(1 - {\upomega }_{\mathrm{amplifying}} {\upomega }_{\mathrm{predicting}}) \end{aligned}$$
Moreover, by the fourth equation it is found
$$\begin{aligned} \underline{\mathbf{srs}}_{e }= & {} {\upomega }_{\mathrm{predicting}} \underline{\mathbf{ps}}_{a}={\upomega }_{\mathrm{predicting}} {\upomega }_{\mathrm{responding}} {\upomega }_{\mathrm{representing}} \\&\times \; {\upomega }_{\mathrm{sensing}}\underline{\mathbf{ws}}_{s }/(1 - {\upomega }_{\mathrm{amplifying}} {\upomega }_{\mathrm{predicting}}) \end{aligned}$$
Based on these, the fifth equation can be used to get an expression for
\(\underline{\mathbf{es}}_{a}\):
$$\begin{aligned} \underline{\mathbf{es}}_{a}= & {} {\upomega }_{\mathrm{executing}} \underline{\mathbf{ps}}_{a}={\upomega }_{\mathrm{executing}} {\upomega }_{\mathrm{responding}} {\upomega }_{\mathrm{representing}} \\&\times \; {\upomega }_{\mathrm{sensing}} \mathbf{ws}_{s } / (1 - {\upomega }_{\mathrm{amplifying}} {\upomega }_{\mathrm{predicting}}) \end{aligned}$$
Summarizing, all equilibrium values have been expressed in terms of the external state
\(\underline{\mathbf{ws}}_{s}\) and the connection weights:
$$\begin{aligned} \underline{\mathbf{ss}}_{s }= & {} {\upomega }_{\mathrm{sensing}} \underline{\mathbf{ws}}_{s} \\ \underline{\mathbf{srs}}_{s}= & {} {\upomega }_{\mathrm{representing}} {\upomega }_{\mathrm{sensing}}{} \mathbf{ws}_{s}\\ \underline{\mathbf{ps}}_{a}= & {} {\upomega }_{\mathrm{responding}} {\upomega }_{\mathrm{representing}} \\&\times \; {\upomega }_{\mathrm{sensing}} \mathbf{ws}_{s }{/}(1 - {\upomega }_{\mathrm{amplifying}} {\upomega }_{\mathrm{predicting}})\\ \underline{\mathbf{srs}}_{e }= & {} {\upomega }_{\mathrm{predicting}} {\upomega }_{\mathrm{responding}} {\upomega }_{\mathrm{representing}} \\&\times \; {\upomega }_{\mathrm{sensing}} \mathbf{ws}_{s }{/}(1 - {\upomega }_{\mathrm{amplifying}} {\upomega }_{\mathrm{predicting}})\\ \underline{\mathbf{es}}_{a}= & {} {\upomega }_{\mathrm{executing}} {\upomega }_{\mathrm{responding}} {\upomega }_{\mathrm{representing}} \\&\times \; {\upomega }_{\mathrm{sensing} }{} \mathbf{ws}_{s }{/}(1 - {\upomega }_{\mathrm{amplifying}} {\upomega }_{\mathrm{predicting}}) \end{aligned}$$
For example, if the external stimulus
\(\underline{\mathbf{ws}}_{s}\) has level 1 this becomes:
$$\begin{aligned} \underline{\mathbf{ss}}_{s }= & {} {\upomega }_{\mathrm{sensing}}\\ \underline{\mathbf{srs}}_{s}= & {} {\upomega }_{\mathrm{representing}} \, {\upomega }_{\mathrm{sensing}}\\ \underline{\mathbf{ps}}_{a}= & {} {\upomega }_{\mathrm{responding}} \, {\upomega }_{\mathrm{representing}} \, {\upomega }_{\mathrm{sensing}}{/}\\&(1 - {\upomega }_{\mathrm{amplifying}} \, {\upomega }_{\mathrm{predicting}})\\ \underline{\mathbf{srs}}_{e }= & {} {\upomega }_{\mathrm{predicting}} \, {\upomega }_{\mathrm{responding}} \, {\upomega }_{\mathrm{representing}} \\&\times \, {\upomega }_{\mathrm{sensing} }{/}(1 - {\upomega }_{\mathrm{amplifying}} \, {\upomega }_{\mathrm{predicting}})\\ \underline{\mathbf{es}}_{a }= & {} {\upomega }_{\mathrm{executing}} \, {\upomega }_{\mathrm{responding}} \, {\upomega }_{\mathrm{representing}} \\&\times \, {\upomega }_{\mathrm{sensing}}{/}(1 - {\upomega }_{\mathrm{amplifying}} \, {\upomega }_{\mathrm{predicting}}) \end{aligned}$$
Moreover, if all connection weights are 1, except that
\({\upomega }_{\mathrm{responding}} = 0.5\) and
\({\upomega }_{\mathrm{amplifying}} = 0.5\), as in the example simulation shown in [
17], Section 2.4.1, the values become:
$$\begin{aligned} \underline{\mathbf{ss}}_{s }= & {} 1\\ \underline{\mathbf{srs}}_{s}= & {} 1\\ \underline{\mathbf{ps}}_{a}= & {} 0.5/0.5 = 1\\ \underline{\mathbf{srs}}_{e }= & {} 0.5/0.5 = 1\\ \underline{\mathbf{es}}_{a }= & {} 0.5/0.5 = 1 \end{aligned}$$
Indeed in the example simulation in [
17], Section 2.4.1 Fig. 11 it can be seen that all values go to 1. The solution of the equilibrium equations in terms of the connection weights can be used to predict that when the connection weights have different values, also these equilibrium values will turn out different. Recall that the cases
\({\upomega }_{\mathrm{amplifying}} = 1\) and
\({\upomega }_{\mathrm{predicting}} = 1\) was excluded. In that case the combined third and fourth equation becomes trivial, as
\(\underline{\mathbf{ps}}_{a}\) is lost from the equation:
$$\begin{aligned}&{\upomega }_{\mathrm{responding}} \underline{\mathbf{srs}}_{s}+{\upomega }_{\mathrm{amplifying}} {\upomega }_{\mathrm{predicting}} \underline{\mathbf{ps}}_{a} = \underline{\mathbf{ps}}_{a} \\&{\upomega }_{\mathrm{responding}} \underline{\mathbf{srs}}_{s} + \underline{\mathbf{ps}}_{a} = \underline{\mathbf{ps}}_{a}\\&{\upomega }_{\mathrm{responding}} \underline{\mathbf{srs}}_{s} = 0\\&\underline{\mathbf{srs}}_{s} = 0 \end{aligned}$$
Here in the last step it is assumed that
\({\upomega }_{\mathrm{responding}}>\) 0. As a consequence by the first two equations also
\(\underline{\mathbf{ss}}_{s}\) and
\(\underline{\mathbf{ws}}_{s}\) are 0, and by the fourth and fifth equation also the values for the other states. It turns out that in this case there can only be an equilibrium if there is no stimulus at all. As soon as there is a nonzero stimulus in this case that
\({\upomega }_{\mathrm{amplifying}} = 1\) and
\({\upomega }_{\mathrm{predicting}} = 1\), the values of ps
\(_{a}\), srs
\(_{e}\) and es
\(_{a}\) increase indefinitely to larger and larger values (and in particular do not stay within the interval [0, 1]), as can be seen from simulations. Note that there was an additional assumption made that
\({\upomega }_{\mathrm{responding}} > 0\). If, in contrast,
\({\upomega }_{\mathrm{responding}} = 0\), then still more possibilities for equilibria are available. For example, in that case
\(\underline{\mathbf{ps}}_{a}\) and
\(\underline{\mathbf{srs}}_{e}\) can have any value, but they have to be equal due to the fourth equation, but this value is independent of the values of
\(\underline{\mathbf{ws}}_{s}\),
\(\underline{\mathbf{ss}}_{s}\) and
\(\underline{\mathbf{srs}}_{s}\), as there is no nonzero connection between these parts of the graph. So, this would not be a very relevant case.
The analysis above can also be done to find out whether or not the activation level of a state is increasing. As a first step, again consider the sensor state ss\(_{s}\).