Assume that a system at time t, for example number of costumers in a line at time $t$ which is denoted by $q(t)$, follows a Markov chain with these dynamics (probabilities)
$$P(q(t+\Delta t)-q(t)=1)=\alpha \Delta t$$ $$P(q(t+\Delta t)-q(t)=-1)=\beta\hspace{0.1cm}v(q(t)) \Delta t$$
for some known constant rates $\alpha$ and $\beta$. where $v(x)$ is piecewise function
\begin{equation} \label{v} v(q)=\begin{cases} 0 & q \leq 0 \\ \frac{q}{q_{*}} & 0\leq q\leq q_{*}\\ 1 &q\geq q_{*} \end{cases} \end{equation}
Where $q_*$ is a constant. Using Dyknin formula we can find that the density, $E[q(t)]$=the expected number of customers at time $t$, can be obtained from this differential equation: \begin{equation} \frac{d}{dt} E[q(t)]=\alpha-\beta E[v(q(t))] \label{eq:1} \end{equation}
I was wondering if there is any way to find a closed form above differential equation which means to find an expression for $E[v(q(t))]$ in terms of $E[q(t)]$.
Some efforts: In above equation replace $E[v(q(t))] = v(E[q(t)])$. Clearly this is not a good assumption because $v(x)$ is a concave function so according to Jensen's inequality we have $$E[v(q(t))] \geq v(E[q(t)])$$
So replacing $E[v(q(t))]$ by $v(E[q(t)])$ may not give us a good estimation of exact solution of above ODE. I've done simulations for $E(q(t))$= the exact solution of differential equation (non-closed one), and the solution of closed form differential equation where $E[v(q(t))]$ is replaced by $v(E[q(t)])$ and this solution is not close to the exact solution.
There are some other suggestions like using a quadratic equation in form of
$$E[v(q(t))]= a\hspace{0.1cm}E[q(t)]^2+b\hspace{0.1cm}E[q(t)]+ c$$
and then trying to find coefficients $a$, $b$ and $c$, but this effort also fails in some cases!
I think any effort should use the nature of function $v(x)$ but till this moment I don't know how to find it. So please let me know if you have any idea.
