Perturbation method for Duffing-type equation with cubic nonlinearity

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Johnmm
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Hi everyone,

I need some help to solve this differential equation.

The question states "Use the perturbation or multiple scale method to find the third-order approximate solution for the following system:

diff(x(t), t, t)+w^2*x(t)*(1+epsilon*x(t)^2) = 0 "

Currently, I am still reading and studying on this perturbation method.

Any advise or directions of what I need to do is much appreciated.
Thank you for your help in advance.

John
 
Attachments
  • Printed problem statement asking for a third-order approximate solution by perturbation or multiple scale method, with the equation d2x/dt2 + w^2 x(1 + epsilon x^2) = 0.
    Screenshot 2014-09-07 02.22.34.png
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Hi !

First, solve the non-perturbated ODE.
Then, add a term with coefficient epsilon (see attachment) :
 
Attachments
  • Handwritten-style typeset derivation of a perturbation solution to a Duffing-type equation: unperturbed case with epsilon = 0 giving X(t) = c1 cos(omega t) + c2 sin(omega t), then a perturbed ansatz x(t) = X(t) + epsilon F(t) leading to an equation for F(t) with a forcing term involving cos(2 omega t) and sin(2 omega t).
    perturbated ODE.JPG
    36.2 KB · Views: 590
JJacquelin said:
Hi !

First, solve the non-perturbated ODE.
Then, add a term with coefficient epsilon (see attachment) :

Hi JJacquelin

Thank you very much for your help,
I will try it and see if I could reach the result in that picture.
Thank you again for helping me out.
 
JJacquelin said:
Hi !

First, solve the non-perturbated ODE.
Then, add a term with coefficient epsilon (see attachment) :

Hi JJacquelin,

Line 7 on your attached picture, I had a little different than yours. I had X^3 instead of X^2.
I am still working on it. Please let me know.
Thank you,
 
Attachments
  • Handwritten-style derivation of a perturbation solution for a Duffing-type oscillator: non-perturbed solution X(t) = c1 cos(at) + c2 sin(at), then expansion with x = X + eF and the cubic term X^3 highlighted by an arrow.
    attachment.jpg
    36.4 KB · Views: 393
Hi Jhonmm !

You are right it isn't X^2 but X^3. This changes some terms of my comments, but this doesn't change the method to solve the problem. I suppose that you can continue by yourself.
 
JJacquelin said:
Hi Jhonmm !

You are right it isn't X^2 but X^3. This changes some terms of my comments, but this doesn't change the method to solve the problem. I suppose that you can continue by yourself.

Hi JJacquelin,
Thank you for the help.
I will follow this method to the rest.
 
"Perturbated"? Isn't the standard "perturbed" simpler and better?
 
HallsofIvy said:
"Perturbated"? Isn't the standard "perturbed" simpler and better?

I think it is perturbed.
 
You should probably give us some boundary conditions since otherwise you might generate solutions growing without bound.
 
Strum said:
You should probably give us some boundary conditions since otherwise you might generate solutions growing without bound.

The question didn't give boundary conditions. But I think it should be x(0)=x0 and [itex]\dot{x(0)}[/itex]=[itex]\dot{x}[/itex]0.
 
Maybe take a look at http://en.wikipedia.org/wiki/Poincaré–Lindstedt_method. Or you could expand using 2 time scales instead using for example
$$
\frac{d^{2}x}{dt^{2}} = \partial_{\tau\tau}x_{0} + \epsilon\left( \partial_{\tau\tau}x_{1} + 2\partial_{\tau T}x_{0} \right)
$$
Differential equations with oscillating solutions doesn't always behave nice when using the usual perturbative approach.

Edit: The wiki link actually treats the same equation as you are looking at! Sometimes you're lucky I guess.
 
Strum said:
Maybe take a look at http://en.wikipedia.org/wiki/Poincaré–Lindstedt_method. Or you could expand using 2 time scales instead using for example
$$
\frac{d^{2}x}{dt^{2}} = \partial_{\tau\tau}x_{0} + \epsilon\left( \partial_{\tau\tau}x_{1} + 2\partial_{\tau T}x_{0} \right)
$$
Differential equations with oscillating solutions doesn't always behave nice when using the usual perturbative approach.

Edit: The wiki link actually treats the same equation as you are looking at! Sometimes you're lucky I guess.

Thank you for your reply. I will take a look into the wiki link. :smile: