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Derivation of van der pol oscillator

  • Thread starter Slightly
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  • #1
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Homework Statement


Given characteristic equation for a circuit containing a diode, I must figure out how to fit a polynomial to the curve so that the van der pol equation is obtained.

The paper I am reading is here:

http://www.unige.ch/math/hairer60/pres/pres_rentrop.pdf

My doubts are located on page three where the author talks about fitting the polynomial of degree three. I am trying to fill in the details as it isn't quite clear what he is actually doing.

Suggestions?

Should I plug the polynomial into the characteristic equation and manipulate it? Or is there some other technique?
 

Answers and Replies

  • #2
59
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He is just filling it in I think.

[tex] C\ddot{U} + \left( \frac{RC}{L} + \frac{d}{du}\left(a_1U +a_2U^2 +a_3U^3\right)\right)\dot{U} + \frac{1}{L}\left(R(a_1U +a_2U^2 +a_3U^3) + U - U_{op}\right) [/tex]

It looks like it is going to look as is stated. Have you tried it and doesn't it work?
 
  • #3
29
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He is just filling it in I think.

[tex] C\ddot{U} + \left( \frac{RC}{L} + \frac{d}{du}\left(a_1U +a_2U^2 +a_3U^3\right)\right)\dot{U} + \frac{1}{L}\left(R(a_1U +a_2U^2 +a_3U^3) + U - U_{op}\right) [/tex]

It looks like it is going to look as is stated. Have you tried it and doesn't it work?

Yea, I tried plugging everything in like you just did, but I could not figure out how to get rid of the U^3 term.
 
  • #4
34,043
9,891
In the bracket it vanishes due to the derivative, but outside it does not.
Setting y=U+d with the right constant d gets rid of the U term in the brackets, but then you still have those U^2 and U^3 outside.
 
  • #5
29
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That's exactly what I don't want. I'm trying to figure out how they get to the van der pol equation.
 

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