# Hint for solving differential equation

In $$\text{Shimer (2012)}^1$$ the author describes the evolution of the unemployment rate as follows:

$$\dot{u}_{t+\tau} = \dot{u}_t^s(\tau)-\left(u_{t+\tau}-u_t^s(\tau)\right)f_t,$$

where

$$f_t = -log(1-F_t \geq 0 \dots$$ arrival rate of a Poisson process,

$$F_t \dots$$ job finding probability,

$$u_t^s(\tau)\dots$$ short term unemployment rate,

$$\tau \dots$$ time between passed between two measurement points in a panel data set with monthly interviews.

With $$u_t^s(0)=0$$ as an initial condition we should be able to come up for a solution for $$u_{t+1}$$ and $$u^s_{t+1} \equiv u^s_t(1)$$:

$$u_{t+1}=(1-F_t)u_t + u^s_{t+1}.$$

Trying to get there on my own, I struggle to find a proper solution approach for this kind of differential equations, since this resembles nothing of what I've learnt so far.

I would appreciate if you could hint me towards the right direction so that I may solve this problem eventually.

$$^1 \text{Shimer, Robert (2012) Reassessing the ins and outs of unemployment, doi:10.1016/j.red.2012.02.001}$$

First, note that derivative is with respect to $$\tau$$ and not $$t$$ (I got confused at the beginning). $$t$$ is a discrete variable from the reference you mention. Honestly, I believe that the notation is kind of akward, since $$\tau$$ dependence is sometimes written as $$(\tau)$$ and sometimes as subscript.
Anyway, define $$v(\tau)=u_{t+\tau}-u_t^s(\tau)$$ such that your equation becomes: $$\dot{u}_{t+\tau} - \dot{u}_t^s(\tau)= - (u_{t+\tau}-u_t^s(\tau))f_t\implies \frac{dv(\tau)}{d\tau} = -v(\tau)f_t$$
This is a separable differential equation. Just divide by $$v(\tau)$$ and integrate both sides from $$\tau=0$$ to $$\tau=1$$ and get $$\int_{v(0)}^{v(1)}\frac{dv}{v} = -\int_0^1f_td\tau = -f_t\implies \log\left(\frac{v(1)}{v(0)}\right) = -f_t = \log(1-F_t)$$ so that $$v(1) = (1-F_t)v(0)$$ Now, $$v(1) = u_{t+1}-u_t^s(1) = u_{t+1}-u_{t+1}^s$$ since $$u_t^s(1)=u_{t+1}^s$$ by definition and $$v(0)=u_t - u_{t}^s(0)=u_t$$ from the initial condition $$u_{t}^s(0)=0$$. Thus, $$v(1) = (1-F_t)v(0) \implies u_{t+1}-u_{t+1}^s = (1-F_t)u_t$$ which is now equivalent to what you needed.
• Thank you so much for not only pointing out the fact that this differential equation is a function of $\tau$, but also for the detailed solution. I have some questions regarding your answer: Shouldn't it be $v(\tau) = u_{t+\tau}-u_t^s{tau}$ and $\log{\frac{v(1)}{v(0)}}=log(1-F_t) \Leftrightarrow v(1) = (1-F_t)v(0)$? Commented Jun 7, 2022 at 21:05
• Sure, thats a typo in the definition of $v(\tau)$. I corrected it. Also another typo with the log. You are correct. Sorry! Commented Jun 8, 2022 at 9:48