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From a paper that I have been reading, I have:

$n \pi = +\sqrt{(+k_2+\sqrt{k_2^2-4k_3k_1}) \times \dfrac{1}{2k_1}}$; where

$k_1 = (1-\dfrac{\alpha^2 \lambda^2}{\zeta^2})$;

$k_2= \lambda^2[\Omega + \dfrac{1-\Omega \alpha^2 \lambda^2}{\zeta^2}+\alpha^2]$; and

$k_3=\lambda^2(\dfrac{\lambda^2 \Omega}{\zeta^2}-1)$

Now, the author of the paper claims directly without any explanation whatsoever, that the first expression can be simplified to:

$B_1\lambda^4 - B_2\lambda^2 +1 = 0$; where

$B_1 = \dfrac{\Omega}{\zeta^2}(\dfrac{1+\alpha^2n^2\pi^2}{n^4 \pi^4})$; and

$B_2 = (\dfrac{1+\alpha^2n^2\pi^2}{n^4 \pi^4}) (\dfrac{1}{\zeta^2}+\dfrac{1}{n^2 \pi^2}) + \dfrac{\Omega}{n^2 \pi^2}$

I don't understand how the author was able to separate $\lambda$ out this neatly, and into the form given above. I have been going at this for hours, and the algebraic manipulations involved are driving me insane. Can someone tell me how to go about this, so as to obtain the equation in terms of $\lambda$, with $B_1$ and $B_2$ as coeffecients, as given above?

Train Heartnet
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1 Answers1

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Let me start with your formula $$n \pi = +\sqrt{(+k(2)+\sqrt{k(2)^2-4k(3)k(1)}) \times \dfrac{1}{2k(1)}}$$ Square both sides a first time, multiply both sides by $2 k(1)$ and substract $k(2)$ from both sides. You so obtain
$$2 \pi ^2 n^2 k(1) -k(2)=\sqrt{k(2)^2-4k(3)k(1)})$$ Square both sides of this last expression again, expand the lhs, replace everywhere $k(1)$, $k(2)$ and $k(3)$ by their definitions, reduce to common denominator, remove the denominator and group powers of $\lambda$.

Just be patient

Claude Leibovici
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  • I had reached to the point of squaring, but that's when things got messy. There are easily $20$+ terms in that expression, and separating out $\lambda$ seemed impossible. I was hoping there'd be another approach to this. But I guess I just have to suck up to it, and work it all out, patiently. Thank you for the help. :) – Train Heartnet Feb 16 '14 at 13:25
  • @JobinIdiculla. After squaring and before replacement, I arrive to $4 \text{k1}^2 \text{k2}^2-8 \pi ^2 \text{k1}^2 \text{k2} n^2+4 \pi ^4 \text{k1}^2 n^4+4 \text{k1} \text{k3}-\text{k2}^2$ – Claude Leibovici Feb 16 '14 at 13:33
  • Oh yes, but when you replace $k_1, k_2$ and $k_3$ in the above, the horror begins. – Train Heartnet Feb 16 '14 at 13:38
  • I totally agree ! I just did it and I arrived to a polynomial of 12th degree in $\lambda$ !! Where is this paper ? – Claude Leibovici Feb 16 '14 at 13:44
  • The paper can be found here:http://www.4shared.com/office/cIdBPJR5ce/Vibration_of_nonlocal_Timoshen.html. Equations $(31), (36)$ and $(37)$ in pages 3-4 is where it all happens. – Train Heartnet Feb 16 '14 at 13:54
  • Sorry but I cannot read it (I am almost blind). Are you sure that there is not typo's in the definitions of the $k$'s ? Please, double check and let me know. – Claude Leibovici Feb 16 '14 at 13:58
  • Oh, I see. No typos in the definitions. Double-checked it. – Train Heartnet Feb 16 '14 at 14:37
  • @JobinIdiculla. I suggest we give up. By the way, could you verify if $2k_1$ is inside or outside the second square root ? Cheers. – Claude Leibovici Feb 16 '14 at 14:39
  • It is outside the second square root. I agree with you. Probably might be best to assume the author is right. – Train Heartnet Feb 16 '14 at 14:43
  • @JobinIdiculla. Don't trust a paper. Typo's are common. Try to contact the author for clarification. You have an e-mail address in the paper (as far as I can see). – Claude Leibovici Feb 16 '14 at 15:15
  • Thank you for that suggestion, I will do that. – Train Heartnet Feb 16 '14 at 18:40