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?

  • 1
    $\begingroup$ Intuitvely, wouldn't it just be for instance, if we are maximizing a static problem vs a dynamic one? Ex. maximizing discounted sum of utilities would be the PV hamiltonian, whereas maximizing only current period utility would be a current value Hamiltonian? $\endgroup$
    – ChinG
    Sep 8 '20 at 1:03
  • $\begingroup$ @ChinG being that the current value Hamiltonian is just a transformation of the present value Hamiltonian im leaning towards that we just pick whichever is convenient, however im not an expert. $\endgroup$
    – EconJohn
    Sep 8 '20 at 3:09

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:

$\qquad \dot \lambda = -\frac{\partial H}{\partial y}= -\frac{\partial [G(.)e^{-\rho t}]}{\partial y}-\lambda\frac{\partial f}{\partial y}$

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:

$\qquad \dot m -\rho m = -\frac{\partial H_c}{\partial y} = -\frac{\partial G}{\partial y} - m\frac{\partial f}{\partial y}$

This is simpler because it contains no discount term.


  1. Chiang A C (1992) Elements of Dynamic Optimization pp 210 ff

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.