This is from Glendinning's book Stability, Instability and Chaos, Example 1.2.
For the ode $\ddot x + 2\epsilon \dot x + x = 0,$ let $y = \dot x$, let $E(x,y)$ be a conserved function (like energy) $\frac12(x^2+y^2)$, then $\frac{dE}{dt} \leq 0. \quad(R1)$
It is argued that from this we expect 'all trajectories tend to minima of $E$ .. (0,0).'
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Why is the argument valid?
It seems that $E(x,y)$ does not decrease along the curve described by $E(x,y) = C$, but along any trajectory $(x(t),y(t))$ described by the above ode (proof: (R1) shows that the changes of $E$ and $t$ are opposite, and so along the positive direction of the curve $t$ increases suggesting $E$ must decrease?), without exception. And so the direction of $E$'s decreasing could be used to indicate the rough direction of the trajectories. Therefore, it seems reasonable to argue that the trajectories would end up somewhere $E$ is small.
But why necessarily (0,0) where $E$ is minimum. Would not it be possible that a trajectory stop midway?
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Besides, the reason that $E$ is called the energy is that it describes the movement of points along the trajectory? Must it also be a conserved function? Can $E$ be any other such function? How to define 'conserved' for an energy function?
BTW, is this (a function decreasing along parametrized curves+conserved) actually the more accurate definition of energy than those in the common conception? And how is it related to the energy in differential geometry and Hamiltonian mechanics, where the geodesic is indicated by the minimum of energy function? Are the two energies consistent?
$E$ is defined this way in association to the energy of a mechanical system, acceleration or force equal to the gradient of the potential energy and a friction term. Or one could see the equation as a perturbation of the conservative case $ϵ=0$.
For a Lyapunov functional you could more generally also use $$ E=\frac12\dot x^2+px\dot x+\frac{q}2x^2 $$ leading to \begin{align} \dot E&=(\dot x+px)\ddot x+p\dot x^2+qx\dot x \\ &=-(\dot x+px)(2ϵ\dot x+x)+p\dot x^2+qx\dot x \\ &=(p-2ϵ)\dot x^2+(q-2ϵp-1)x\dot x-px^2. \end{align} So that for instance $p=ϵ$ and $q=1+2ϵ^2$ gives $$ E=\frac12(\dot x+ϵx)^2+\frac12x^2,~~\dot E=-ϵ(\dot x^2+x^2), $$ which gives a more strict descent condition.