Perturbation theory for solving a second-order ODE

  • Thread starter Robin04
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  • #1
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Homework Statement:

##\ddot{\xi}(t)=-b\xi (t)+\cos{(\omega t)}(a-c \xi^2(t))##, where ##a, b, c## are constants

Relevant Equations:

-
I have to solve the equation above. I haven't heard about an exact method so I tried to apply perturbation theory. I don't know much about it so I would like to ask for some help.
First I put an ##\epsilon## in the coefficient of the non-linear ##\xi^2(t)## term:
##\ddot{\xi}(t)=-b\xi (t)+\cos{(\omega t)}(a-\epsilon c\xi^2(t))##
I calculated to the first order term and it was diverging. I heard that there are some methods that can help to sum a divergent series. Can you suggest any that can work here?
Also, next I tried to put ##\epsilon## in the exponent of the non-linear term, but when I substitute the series into the equation I don't know how to raise the series to the power ##\epsilon##.
So basically ##(\xi_0+\epsilon \xi_1+\epsilon^2 \xi_2+...)^{2 \epsilon}=?##
 

Answers and Replies

  • #2
How sure are you that the equation reads $$\ddot{\xi} = -b\xi + \cos(\omega t)(a-c\xi^2)$$ and not instead $$\dot{\xi} = -b\xi + \cos(\omega t)(a-c\xi^2).$$ I only ask because the latter is an Ricatti equation and thus exactly solvable.

I calculated to the first order term and it was diverging. I heard that there are some methods that can help to sum a divergent series.
If you only have calculated the pertubation series upto first order, then how can the series be divergent if it is finite?
 
  • #3
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How sure are you that the equation reads
I'm sure, it is an equation of motion.
If you only have calculated the pertubation series upto first order, then how can the series be divergent if it is finite?
I meant it in the sense that this equation comes up as a part of an approximation method to a certain motion, and the unperturbed solution gives a much better approximation than with the first order term included, so I figured I don't have to go to second order.
 

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