I've been looking at a number of optimal control problems and have been wondering under what conditions one should use the current value Hamiltonian over the present value Hamiltonian.
Does it depend on the specific result we seek to derive? (pick whichever you want) or does it depend on the nature of the problem?
There is no clear right and wrong about this, it's just a matter of convenience. The current-value Hamiltonian is likely to be more convenient when the objective function includes a discount factor. Following Chiang (1), suppose the problem is:
$\qquad$Maximise $V = \int_0^T G(t,y,u)e^{-\rho t}$
$\qquad$subject to $\dot y=f(t,y,u)$
$\qquad$and boundary conditions
The standard (present value) Hamiltonian is:
$\qquad H=G(t,y,u)e^{-\rho t} + \lambda f(t,y,u)$
If we proceed from this Hamiltonian, the co-state equation (one of the first-order conditions) is:
$\qquad \dot \lambda = -\dfrac{\partial H}{\partial y}= -\dfrac{\partial [G(.)e^{-\rho t}]}{\partial y}-\lambda\dfrac{\partial f}{\partial y}$
While it is possible to obtain a solution this way, the discount factor complicates the derivatives and can make interpretation more challenging.
Suppose instead we use the current-value Hamiltonian:
$\qquad H_c = G(t,y,u) + mf(t,y,u)$
where $m$ is a current-value Lagrange multiplier defined by $m=\lambda e^{\rho t}$. The co-state equation is then:
$\qquad \dot m -\rho m = -\dfrac{\partial H_c}{\partial y} = -\dfrac{\partial G}{\partial y} - m\dfrac{\partial f}{\partial y}$
This is simpler because it contains no discount term.
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