Question

Autonomous Differential Equations. I have tried to watch videos and such ( I couldn't get a video similar enough) and I just don't know how to do it

Answer 1

Convert the differential equations into a system of two first-order differential
equations. State the alternative equation and find all equilibrium points.

Answer 2

@phi

Answer 3

please help to explain how

Answer 4

I prefer to use Leibniz notation for a problem like this; that is, \(\dfrac{\mathrm dx}{\mathrm dt}=\dot x\), \(\dfrac{\mathrm d^2x}{\mathrm dt^2}=\ddot x\), and so on. So just rewriting the ODE, you have
\[\ddot x+\omega^2\left(x-\frac{1}{4}x^3\right)=0\iff \frac{\mathrm d^2x}{\mathrm dt^2}+\omega^2\left(x-\frac{1}{4}x^3\right)=0\]Substitute \(v=\dfrac{\mathrm dx}{\mathrm dt}\), so that \(\dfrac{\mathrm dv}{\mathrm dt}=\dfrac{\mathrm d^2x}{\mathrm dt^2}\).

The idea here is to rewrite the ODE as a function of \(v\) and \(x\). By the chain rule, you have
\[\frac{\mathrm dv}{\mathrm dt}=\frac{\mathrm dv}{\mathrm dx}\frac{\mathrm dx}{\mathrm dt}=v\frac{\mathrm dv}{\mathrm dx}\]All this to say we can reduce the order of the ODE to get
\[v\frac{\mathrm dv}{\mathrm dx}+\omega^2\left(x-\frac{1}{4}x^3\right)=0\]which is separable and easy to solve.

Answer 5

So when I separate and solve, the answers will be the equilibrium points?

Answer 6

No, the equilibrium points are the points that make the first derivative vanish (so the solution is stationary at these points).

The solution to that last ODE is a function \(v=f(x)\), but remember that we're setting \(v=\dfrac{\mathrm dx}{\mathrm dt}\), and so the solution to that ODE just yields yet another ODE (the second in a system of first order ODEs).

So the equilibrium points for the original second-order ODE occur at the points where \(v=\dfrac{\mathrm dx}{\mathrm dt}=0\).

Answer 7

Thank you so much! This makes so much more sense than the lecture notes I have been trying to follow and understand all day

Answer 8

Glad I could help!

Answer 9

I shall bug you again in the future :P

Answer 10

the **brief**: convert into a system..... ;-|

\(\ddot x + \omega^2 x (1 - \frac{x^2}{4} ) = 0\), so let \(y = \dot x\) which means \(\dot y = \ddot x\)


\(\left( \begin{matrix}
\dot x \\ \dot y & \\ \end{matrix} \right) = \left( \begin{matrix}
0 & 1\\ - \omega^2( 1 - \frac{x^2}{4}) & 0 \\ \end{matrix} \right) \left( \begin{matrix}
x \\ y & \\ \end{matrix} \right)\)

f.p.'s at \(\dot x = \dot y = 0\)

ie at \(y = 0\) and at \( x = 0, \pm 2 \)

Answer 11

I was with you until you said y=0, x=0, +-2

Answer 12

\(\left( \begin{matrix}
\dot x \\ \dot y & \\ \end{matrix} \right) = \left( \begin{matrix}
0 & 1\\ \color{red}{- \omega^2( 1 - \frac{x^2}{4})} & 0 \\ \end{matrix} \right) \left( \begin{matrix}
x \\ y & \\ \end{matrix} \right)\)

Answer 13

I might be dumb, any way you could elaborate?

Answer 14

(if it is not too much hassle)

Answer 15

:-))

the question is asking you to create a system of 2 first order DE's

and the fixed points happen when \(\dot x = \dot y = 0\) ie you are looking at momentum vs displacement.

typically, the key point here is linearisation of a non-lin DE by approximation methods, not solving a non-linear DE by removing the independent variable.

Answer 16

oh wait okay yeah I understand where you got the solution for x

but I have one question (which I thought I understood at first but clearly don't)

Answer 17

you have this
\[\left( \begin{matrix} \dot x \\ \dot y & \\ \end{matrix} \right) = \left( \begin{matrix} 0 & 1\\ \color{red}{- \omega^2( 1 - \frac{x^2}{4})} & 0 \\ \end{matrix} \right) \left( \begin{matrix} x \\ y & \\ \end{matrix} \right)\]
why is it a negative w, not a positive?

Answer 18

@IrishBoy123

Answer 19

compare the matrix to your original DE

Answer 20

it is the partial derivatives yeah no worries! thank you! then I would just use
right?