I come from a non mathematical background, so solving differential equations is something that I have to acquire on the go. I hope the following makes sense.

I want to chose a nonnegative depreciation rate $f$ for $x$ for any point between $0, T$ such that I maximize some instantaneous objective function $g(x)$, increasing and concave, given that I arrive at $\bar x$ in $T$. Would the following way be a correct one to write that up?

$$G(x_0) = max_{f\in F} \int_0^T e^{-\rho t}g(x(t)) dt \\ \text{s.t. } \dot x(t) = -f(t)x(t)\\ x(0) = x_0 \\ x(T) = \bar x$$

For a given constant $\bar x$, and concave increasing $g \in \mathcal C^1$. I define $F$ to be the set of measurable functions from $[0, T] \rightarrow R^+$.

Second, how would I try to solve this? If it were an infinite horizon, I would try to write it up as a HJB of the form $ \rho F(x) = g(x) + F'(x)\dot x$, but that's not the case here.

Given that I have a finite horizon, I assume I should start at $G(x(T))$ and somehow go backwards, but I'm not really sure how. I appreciate hints at both solving this with pen&paper and numerically.

Actually, after thinking more about the problem, I'm not even sure that a maximum exists. $g$ strictly increases in $x$, and we are only constrained to reach a level of $x$ in the end. Moreover, the functions in $F$ have a unbounded image. Hence, it is always improving $G(x)$ if we "delay" the depreciation a bit, by decreasing $f(t)$ and increasing $f(t+\epsilon)$. Since there is no bound on how much "depreciation" can happen at one point in time, this strategy is always feasible.



You must log in to answer this question.

Browse other questions tagged .