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.