$H$ is a constant? Maximizing: $\int _0^Te^{-t}f(x,u)dt$ st $x_t=g(t,x,u)$ and $g$ is independent of $t$

Question

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

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

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.