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

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

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*}

• 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 $t$ Oct 1 at 11:45
• @BakerStreet The context is the current value Hamiltonian and shadow price in macroeconomics.
– dodo
Oct 1 at 15:32
• 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. Oct 1 at 15:41
• @BakerStreet The exact expression is "g is independent of $t$". I will try to find its precise/rigorous meaning.
– dodo
Oct 1 at 15:48
• 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. Oct 1 at 19:33

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.
• 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?