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Consider the following discussion

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I am struggling to understand what is going on. For example, I do not understand why $$ \frac{d H}{d t}=g_{2} \dot{x}-g_{1} \dot{y}. $$ I thought it could be via the chain ruleagain, but $\partial H / \partial x \neq g_2$.

Also, I don't understand the conclusion. In particular, if an orbit is periodic and closed, why is the given integral $0$?

I would appreciate any help with this.

Maths Wizzard
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    From the ODE, we have $H_y = \dot{x} - g_1$, then substitute this into the chain rule. For your second question, assume your curve is parametrized by $(x(t),y(t)), t\in [0,1]$ with $(x(0),y(0)) = (x(1),y(1))$, then integrating in $t$, the fundamental theorem of calculus shows that $0= \int_0^1 \frac{d}{dt}H(x(t),y(t)) dt = \int_0^1 g_2 \dot{x} - g_1\dot{y} \ dt = \int_C g_2 dx - g_1 dy$. – kieransquared Aug 30 '23 at 11:50
  • Hi, thanks for the comment. A quick question. We have used the fact such periodic orbit exists in the second equality of the second expression you have given correct? Also feel free to post this an answer. – Maths Wizzard Aug 30 '23 at 12:04

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