A Basic Question of Continuous Time Macroeconomic Model (Variation of Constant to Solve an ODE)

Question

I am going through the continuous time macro slides by Ben Moll (link is: https://benjaminmoll.com/wp-content/uploads/2019/07/Lecture2_ECO521.pdf), when deriving New keynesian model in continuous time, we finally obtained the following ODE (see page 15 of his slides): $$ \dot{\pi}=\rho \pi - MR(t) $$ where $\pi$ is the inflation rate, $\rho$ is the subjective discount rate, and $MR(t)$ is some terms on marginal revenue (also function of $t$). The solution to this ODE is $$ \pi=\int_{t}^\infty e^{-\rho(s-t)}MR(s)ds $$ When I try to solve this ODE with variation of constant, I assume that $\pi=e^{\rho t}f(t)$, I will obtain: $$ f'(t)=-e^{-\rho t} MR(t) $$ I understand that $f(t)=\int_{t}^\infty e^{-\rho s}MR(s)ds$ is a solution to this function, but clearly there should be some initial condition to pin that down which I failed to find. In this case, it is that $\lim_{t\rightarrow\infty}f(t)=0$, but does this condition has a clear macroeconomic implication? Like no Ponzi game, or excluding growth path to explode to infinity?

Answer 1

As $ \dot{\pi}=\rho \pi + MR(t) $

is a linear ODE, the solution ca be stated as

$$ \cases{ \dot{\pi}_h=\rho \pi_h\\ \dot{\pi}_p=\rho \pi_p + MR(t)\\ \pi = \pi_h + \pi_p } $$

as $\pi_h = c_0 e^{\rho t}$ now making $\pi_p = c_0(t) e^{\rho t}$ after substitution into the complete ODE we have

$$ \dot c_0(t)e^{\rho t}+\rho c_0(t)e^{\rho t} = \rho c_0(t)e^{\rho t} + MR(t) $$

so $\dot c_0(t) = e^{-\rho t}MR(t)$ and then $c_0(t) = \int_0^t e^{-\rho\tau}MR(\tau)d\tau$ then

$$ \pi = \pi_h + \pi_p = \left(c_0+\int_0^t e^{-\rho\tau}MR(\tau)d\tau\right)e^{\rho t} $$

here distinctly $c_0$ is a constant to be assigned according to initial (or final) conditions.

Answer 2

The condition which gives you the particular solution you've presented is that $\ e^{-\rho t}\pi(t)\rightarrow0\ $ as $\ t\rightarrow \infty\ $ (or something which implies this, such as that $\ \pi(t)\ $ remains bounded as $\ t\rightarrow \infty\ $). The general solution of your differential equation is $$ \pi(t)=\pi(0)e^{\rho t}-\int_0^te^{-\rho(s-t)}MR(s)ds\ , $$ and so $\ \lim_\limits{t\rightarrow\infty}e^{-\rho t}\pi(t)=\pi(0)-\int_0^\infty e^{-\rho s}MR(s)ds\ $. For this to be zero we must have $\ \pi(0)=\int_0^\infty e^{-\rho s}MR(s)ds\ $, which then gives \begin{align} \pi(t)&=\int_0^\infty e^{-\rho(s-t)}MR(s)ds-\int_0^te^{-\rho(s-t)}MR(s)ds\\ &=\int_t^\infty e^{-\rho(s-t)}MR(s)ds\ . \end{align}