What kind of mathematical model is this?

Question

I am making a highly simplified economic model. When I started with it, I assumed it would become a model of ordinary delayed differential equations, but it turns out to be something different, which I don't know how to categorize or analyse. (In the end I'd like to know if there can be cycles or multiple equilibria in the model).

The model is as follows

There are 2 sectors: the Electricity Sector, and the Coal Sector. They have a circular relation (coal is used to produce electricity, and electricity is used to produce coal).

1. Electricity production $e(t)$ has an immediate effect on coal production $c(t)$, but coal has to be transported to the electricity plant and has a delay $\Delta$.
2. Coal can be stored near the electric plant, and $s(t)$ is the amount stored at $t$, but electricity cannot be stored.
3. coal production has a maximum capacity of $\kappa$ coal per unit $t$, at which rate it consumes $\alpha$ units of electricity per unit $t$.
4. Coal is absorbed at rate $a(t)$, depending on coal available and electricity need, which produces $\beta$ electricity per unit coal. This means $\alpha/ \beta$ is the max needed coal absorption per $t$ to satisfy maximum coal production.

This gives the following relations:

$c(t)=min(\kappa,\frac{\kappa}{\alpha}e(t))$

$\dot s(t)=c(t-\Delta)-a(t)$

$a(t)=\{\begin{matrix}\ if \quad s(t)>0:min(\frac{\alpha}{\beta}, a^{max}) \qquad \\ if \quad s(t) = 0:min(\frac{\alpha}{\beta},c(t-\Delta)) \end{matrix}$

$e(t)=\beta \cdot a(t)$

Now my question is as follows:

The only differential equation here is the one about the change in stored coal over time $\dot s$. Yet, The variables I am primarily interested in are coal and electricity production $c(t), e(t)$.

I could simply substitute the electricity and coal functions into the $\dot s$ relation, and treat it as a delayed ordinary differential equation of one variable, but then I would only be analyzing fluctuations in stored coal, which (I think) doesn't allow me to analyse the change in electricity and coal production over time. But $c(t)$, and $e(t)$ are not in differential equation form.

So how do I analyse $c(t)$ and $e(t)$ here (find out whether there are cycles, multiple equilibria, or compute numerical examples, e.g. I'd like to compute examples in mathematica), and more generally, what kind of category of dynamical system is this?

Edit: I just realized that when I said "I could simply substitute the electricity and coal functions into the $\dot s$ relation", I was wrong, because it actually seems to be impossible to substitute $c$ and $e$ into $\dot s$, without getting into an infinite regress, where you have to refer to $c(t-\Delta), c(t-2\Delta), c(t-3\Delta), ...$. This makes my question even more pertinent.

Answer 1

I think you've written down a very interesting system, from the mathematical point of view! I would eliminate $e$ and write the system as \begin{align} c(t) =& \kappa\min (1, \tfrac{\beta}{\alpha} a(t)), \tag{1a}\\ a(t) =& \left\{ \begin{array}{rcl} \min(\frac{\alpha}{\beta}, a^\text{max}) & \text{if} & s(t) > 0,\\ \min(\frac{\alpha}{\beta}, c(t-\Delta)) & \text{if} & s(t) = 0 \end{array}\right. ,\tag{1b}\\ \frac{\text{d} s}{\text{d} t} =& c(t-\Delta) - a(t).\tag{1c} \end{align} I would call this a delay-differential-algebraic piecewise smooth system, if that helps at all. As you've found out, you cannot reduce it to delay-differential equation for $s$, because you would have to have knowledge of $c$ at every previous time $t-n \Delta$.

The analysis of the existence of equilibria is the easiest, so let's start with that. Assume there exist $(c_*,a_*,s_*)$ that solve our system and are time-independent. The ODE for $s$, equation (1c), then implies that \begin{equation} c_* = a_*. \end{equation} From equation (1a) it follows that either $c_* = \kappa$ or $\kappa \frac{\beta}{\alpha} = 1$, which is a condition on the parameters. Assuming the most general situation, i.e. where the system parameters are chosen freely, we focus on the case $c_* = \kappa = a_*$. Note that this implies that $\frac{\beta}{\alpha} a_* = \frac{\beta}{\alpha} \kappa > 1$. We can now use equation (1b) to see if multiple choices for $s_*$ are possible.

Suppose $s_* = 0$. Then, (1b) is satisfied if $c_* = \kappa > \frac{\alpha}{\beta}$ or $\kappa = \frac{\alpha}{\beta}$, which is the same parameter condition we encountered in the analysis of equation (1a). So, without assuming any such relation on parameters, we obtain

Result 1: If the model parameters satisfy the inequality $\kappa > \frac{\alpha}{\beta}$, then $(c,a,s) = (\kappa,\kappa,0)$ is an equilibrium of the system.

Now suppose $s_* > 0$. Then, (1b) is satisfied if $\kappa = \frac{\alpha}{\beta}$ or $\kappa = a^\text{max}$. Both are conditions on parameters, which are not satisfied in general if you're allowed to choose your system parameters freely. Therefore, we can state

Result 2: If and only if $a^\text{max} = \kappa$, then $(c,a,s) = (\kappa,\kappa,s_*)$ is an equilibrium of the system, for any choice of $s_*>0$.

What about this special case $\kappa = \frac{\alpha}{\beta}$? Let's look at equation (1a) first. Assume that $\frac{\beta}{\alpha} a_* > 1$, then $c_* = \kappa$. But that implies that $a_* = \kappa = \frac{\alpha}{\beta}$, which violates our assumption. Assuming that $\frac{\beta}{\alpha} a_* < 1$ leads to a contradiction in the same way, and therefore we again get $c_* = \kappa = a_*$. Equation (1b) is again satisfied if we choose $s_* = 0$. However, we cal also satisfy equation (1b) for $s_* > 0$ if we choose $a^\text{max} < \frac{\alpha}{\beta} = \kappa$. This leads to

Result 3: Suppose that $\kappa = \frac{\alpha}{\beta}$. If $a^\text{max} < \kappa$, then $(c,a,s) = (\kappa,\kappa,s_*)$ is an equilibrium of the system, for any choice of $s_*>0$.

To conclude: For a region in parameter space (where $\kappa > \frac{\alpha}{\beta}$) there exists a unique equilibrium solution for the system. I hope that's in any way helpful.

By the way, you can reduce system (1) to an integro-delay equation for $a$. Writing \begin{align} s(t) =& s(0) + \int_0^t c(\tau-\Delta)-a(\tau)\,\text{d}\tau\\ =& s(0) + \int_0^t \kappa \,\min(1,\tfrac{\beta}{\alpha}a(\tau-\Delta))-a(\tau)\,\text{d}\tau, \end{align} you end up with the equation \begin{equation} a(t) = \left\{ \begin{array}{rcl} \min(\frac{\alpha}{\beta}, a^\text{max}) & \text{if} & s(0) + \int_0^t \kappa \,\min(1,\tfrac{\beta}{\alpha}a(\tau-\Delta))-a(\tau)\,\text{d}\tau > 0,\\ \min(\frac{\alpha}{\beta}, \kappa \,\min(1,\tfrac{\beta}{\alpha}a(t-\Delta))) & \text{if} & s(0) + \int_0^t \kappa \,\min(1,\tfrac{\beta}{\alpha}a(\tau-\Delta))-a(\tau)\,\text{d}\tau = 0 \end{array}\right. . \end{equation}