# How to find an equilibrium of the second order system?

## I have following exercise to prepare to my exam : ${x}^{. .} = 1 - 3 {x}^{2}$ (i mean second derivative of x by ${x}^{. .}$). And i need to "find equilibria of the system" I know that for first order it is enough to check points where x'=0, but how to extend it to second order if we dont have x' at all? Would be very grateful if someone could clarify it and help to solve the exercise

Jun 23, 2018

See below

#### Explanation:

$\ddot{x} = 1 - 3 {x}^{2}$ is non-linear in $t$ so you either change variable or use a different analysis.

You might know that every conservative system has a potential function $V \left(x\right)$, where:

• $\ddot{x} = - \frac{\mathrm{dV}}{\mathrm{dx}}$

If you don't know what a potential function is, I've put a simple doodle at the bottom. They are ubiquitous in physical applications.

So here:

$V \left(x\right) = - \int 1 - 3 {x}^{2} \setminus \mathrm{dx} = - x + {x}^{3} + {V}_{o}$

We don't really care what ${V}_{o}$ is, we are interested in the shape of the potential function, so we plot it, with ${V}_{o} = 0$:

graph{-x + x^3 [-2.738, 2.738, -1.37, 1.366]}

This potential function has 2 equilibrium points corresponding to the stationary points where $\frac{\mathrm{dV}}{\mathrm{dx}} = 0$, ie $x = \pm \frac{1}{\sqrt{3}}$

The one on the left is unstable . A particle placed at $x = - \frac{1}{\sqrt{3}}$, if it underwent a slight displacement, would "roll off the hill" in the direction of the displacement. You can look at it like a picture.

A particle placed at $x = \frac{1}{\sqrt{3}}$, if it underwent a slight displacement, would oscillate back and forward about $x = \frac{1}{\sqrt{3}}$. That is stable equilibrium.

There is more, and you can actually solve that DE too, but it does not give the same information.

Potential Function rough idea

In terms of the potential function, we have a conservative sytem so energy $E$ is conserved, though it may switch between kinetic energy $T$ and potential energy $V$.

For a unit mass:

$E = T + V = c o n s t$

$\frac{\mathrm{dE}}{\mathrm{dx}} = 0$

$\frac{d}{\mathrm{dx}} \left(\frac{1}{2} {v}^{2}\right) + \frac{\mathrm{dV}}{\mathrm{dx}} = 0$

$v \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{dV}}{\mathrm{dx}} = 0$

• $v \frac{\mathrm{dv}}{\mathrm{dx}} = v \frac{\mathrm{dv}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}} = v \frac{\mathrm{dv}}{\mathrm{dt}} \frac{1}{v} = \frac{\mathrm{dv}}{\mathrm{dt}} = \ddot{x}$

$\implies \ddot{x} = - \frac{\mathrm{dV}}{\mathrm{dx}}$