Deterministic Optimal Control Problem
$
V(a_t)
=
\underset{c}{\max}
\int_{\tau =t}^{\tau = \infty} e^{-\rho (\tau - t) } u(c_{\tau}) d\tau
$
s.t.
$
\frac{da}{d\tau} = \left( r a_{\tau} - c_{\tau} \right) $
CV Hamiltonian:
$H^{CV} \equiv u(c_t) + \lambda_{t} \times (r a_{t} - c_{t})$
$H^{CV}_c=0$, $H^{CV}_a=\rho \lambda - \frac{d\lambda}{dt}$,
$H^{CV}_\lambda = \frac{da}{dt}$
Boils down to system of boundary-value DAEs.
If you can eliminate $\lambda$, system of BV-ODEs.
HJB: $\rho V(a_t) = \underset{c}{\max} \{ u(c_{t}) + V_a \times (r a_{t} - c_{t}) \}$
Authors often define the term in the HJB which depends on the control as "Hamiltonian":
$H\left( V_a\right) \equiv
\underset{c}{\max} \{ u(c) - c_{t} \times V_a \} \Rightarrow c(a_t) =u'^{-1} \left( V_a(a_t) \right)$
Hence $\Leftrightarrow \rho V(a_t) = H\left( V_a\right) + V_a \times r a_{t}$
Adrien Bilal (p9) and Neil Walton & others, call the term inside the HJB
$H\left( V_a\right)$ a Hamiltonian.
If we set $\lambda_{t} \equiv V_a$ then $H^{CV} = H\left( V_a\right) + \lambda \times r a_{t}$.
Q1: why do they add the term $H\left( V_a\right)$ inside the HJB?
-guess: it separates the non-linear part of HJB
-guess: it separates the part of the HJB that depends on the control
Q2: why do they call $H\left( V_a\right)$ a "hamiltonian"?
Note: I am NOT asking if the solution to the HJB is the same as to the Hamiltonian.
Sol to HJB: $V(a_t)$ gives policy $c(a_t) = u'^{-1}(V_a)$
Combine policy w/ transition eqn: you get $c^{HJB}(t)$
Sol to CV Hamiltonian: $c^{H}(t)$
I can show (under appropriate assumptions) that $c^{HJB}(t)=c^{H}(t)$.
(side-node) Recall: $H^{CV}_c=0$ & $H^{CV}_a=\rho \lambda - \frac{d\lambda}{dt}$
$H^{CV}_a=\rho \lambda - \frac{d\lambda}{dt}$:
implies $\lambda r = \rho \lambda - \frac{d\lambda}{dt}$
implies $V_a (\rho-r) = V_{aa} (r a_{t} - c_{t})$
