$\max_{x(t),u(t)}\int _0^Te^{-t}f(x(t),u(t))dt$,

st derivative $x_t=g(t,x(t),u(t))$. Prove that $H$ is constant.

My try2:

consider the Hamiltonian $$ H(x(t), u(t)) = e^{-t}f(x(t), u(t)) + \lambda(t) g(x(t), u(t)). $$ Then: $$ \begin{align*} \frac{d}{dt} H(x(t), u(t)) &= (e^{-t}f_x + \lambda g_x)\dot x + \underbrace{(H_u)}_{=0} \dot u + g(x(t), u(t)) \dot \lambda,\\ &= \dot x(e^{-t}f_x + \lambda g_x) + g(x(t), u(t)) \underbrace{(-e^{-t}f_x - \lambda g_x)}_{=\dot \lambda_t},\\ &= \underbrace{(\dot x - g(x(t), u(t))}_{=0} (e^{-t}f_x + \lambda g_x),\\ &= 0 \end{align*} $$

References: Chiang (1992), https://people.stfx.ca/tleo/advmacrolec3.pdf

  • $\begingroup$ What is the exact text of the problem? I suppose that '$g$ is independent of $t$' doesn't mean that $g$ is constant with respect to $t$ (otherwise, why put $t$ amoing the independent variables?), but that the function $g$ is always the same, irrespective of $t$ $\endgroup$ Oct 1 at 11:45
  • $\begingroup$ @BakerStreet The context is the current value Hamiltonian and shadow price in macroeconomics. $\endgroup$
    – dodo
    Oct 1 at 15:32
  • $\begingroup$ Yes, I know, I said the text of the problem, if it is an exercise. I think that the text of the problem doesnt't say you that $g$ is constant, but that we have always the same function $g$ as time $t$ varies. $\endgroup$ Oct 1 at 15:41
  • $\begingroup$ @BakerStreet The exact expression is "g is independent of $t$". I will try to find its precise/rigorous meaning. $\endgroup$
    – dodo
    Oct 1 at 15:48
  • $\begingroup$ Ok, moreover, even if the value of $g$ is indipendent of $t$, this means that it doesn't depend 'separately' on $t$, but can depend on $t$ through $x(t)$ and $u(t)$, it doesn't mean that it is independent of $x(t)$ and $u(t)$. This is the case when a differential equation is said autonomous, that is, it does't depend on $t$ separately. I don't know what is $x_t$ in the question, I suppose the derivative of $x(t)$ with respect to time, as you wrote 'derivative' before it. $\endgroup$ Oct 1 at 19:33

1 Answer 1


Consider the problem: $$ \begin{align*} \max_{u} \int_0^T f(x(t), u(t)) dt& \\ \text{ s.t. } &\dot x = g(x(t), u(t)),\\ &\text{ + boundary conditions} \end{align*} $$

Assume that $g(x,u)$ and $f(x,u)$ are independent of $t$ (given $x$ and $u$). The Hamiltonian is given by: $$ H(x(t), u(t)) = f(x(t), u(t)) + \lambda(t) g(x(t), u(t)) $$ The first order conditions give: $$ \begin{align*} &f_u + \lambda g_u = 0,\\ &\dot \lambda = - f_x - \lambda g_x\\ &\text{and some transversality conditions} \end{align*} $$ Now consider the Hamiltonian $$ H(x(t), u(t)) = f(x(t), u(t) + \lambda(t) g(x(t), u(t)). $$ Then: $$ \begin{align*} \frac{d}{dt} H(x(t), u(t)) &= (f_x + \lambda g_x)\dot x + \underbrace{(f_u + \lambda g_u)}_{=0} \dot u + g(x(t), u(t)) \dot \lambda,\\ &= \dot x(f_x + \lambda g_x) + g(x(t), u(t)) \underbrace{(-f_x - \lambda g_x)}_{=\dot \lambda_t},\\ &= \underbrace{(\dot x - g(x(t), u(t))}_{=0} (f_x + \lambda g_x),\\ &= 0 \end{align*} $$ This shows that $H$ does not change with $t$, so it remains constant over time. The exact value is usually determined by the transversality conditions.

  • $\begingroup$ Hi tdm, I put $e^{-t}$ before the $f$ in the title but I forgot to include in the main text. However, I believe your solution still works. I tried it with your method above. Is my understanding correct? $\endgroup$
    – dodo
    Oct 2 at 9:55

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