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In chapter 28 of Landau-Lifshitz Classical Mechanics textbook they try to explain how to get the motion of a particle with the Lagrangian:

$L=\frac{1}{2}m\dot{x}^{2}-\frac{1}{2}m w_{0}^{2}x^{2}-\frac{1}{3}m\alpha x^{3}-\frac{1}{4}m\beta x^{4}$

by using successive approximations for solve the motion equation:

$\ddot{x}+w_{0}^{2}x=-\alpha x^{2}-\beta x^{3}$

As far as my teacher explained to us, the successive approximations method consists in supposing a solution:

$x=x^{(1)}+x^{(2)}+x^{(3)}+...$

where

$\ddot{x^{(1)}}+w_{0}^{2}x^{(1)}=0$

$\ddot{x^{(2)}}+w_{0}^{2}x^{(2)}=\alpha (x^{(1)})^{2}-\beta (x^{(1)})^{3} $

$\ddot{x^{(3)}}+w_{0}^{2}x^{(3)}=\alpha (x^{(2)})^{2}-\beta (x^{(2)})^{3} $

and so on...

If we define $x^{(1)}=a\cos(wt)$ where $w=w_{0}+w^{(1)}+w^{(2)}+w^{(3)}+...$

replacing $x^{(1)}$ to get $x^{(2)}$ I don't get what Landau gets:

$\ddot{x^{(2)}}+w_{0}^{2}x^{(2)}=-\alpha a^{2}\cos(wt)^{2}+2w_{0}w^{(1)}a\cos(wt)$

where is easy to see that we must set $w^{(1)}=0$ in order to avoid resonance.

But the text does not develop the calculation.

Is the way I see the method of successive approximations described by Landau correct? How do I get to that result? Thank you!

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  • $\begingroup$ is your $w^{(0)}$ same as $w_0$... there is no $w^{(0)}$ defined in your expansion and clearly $w_0\ne 0$. $\endgroup$ – ZeroTheHero Sep 29 at 21:38
  • $\begingroup$ No, $w^{(0)}$ refers to first order aproximation , $w_{0}$ is just the frecuency, i forget to write the $w^{(0)}$ in $w$, that is: $w=w_{0}+w^{(0)}+w^{(1)}+...$.. sorry. $\endgroup$ – Cast fj Sep 29 at 21:57
  • $\begingroup$ this does not make sense as written. If the perturbation is $0$ (i.e. $\alpha=\beta=0$) the frequency should be $w_0$, not $w_0+w^{(0)}$ as you have it, OR your counting of frequencies is not done right and $w^{(0)}$ should be $w^{(1)}$ etc. $\endgroup$ – ZeroTheHero Sep 29 at 22:04
  • $\begingroup$ ok, as landau does, start counting from 1 $\endgroup$ – Cast fj Sep 29 at 22:09
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    $\begingroup$ I will note that if you are expecting Landau and Lifshitz to provide detailed steps, you will be sorely disappointed. $\endgroup$ – Jon Custer Sep 29 at 23:01
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Given $$\ddot{x} + \omega_0^2 x = - \alpha x^2 - \beta x^3,$$ a series solution of the form $$ x = x_1 + x_2 + x_3 + ...$$ where $x_r$ is of order $r$ implies $$(\ddot{x}_1 + \ddot{x}_2 + ..) + \omega_0^2 (x_1 + x_2 + ..) = - \alpha (x_1 + x_2 + ..)^2 - \beta (x_1 + x_2 + ..)^3$$ and so expanding and equating terms of order $r$ one is going to find equations with cross terms like $\ddot{x}_3 + \omega_0^3 x_3 = - \beta x_1^3 - 2 \alpha x_1 x_2$ which your list of approximations does not factor in.

To solve the equation, write it in the form $$\omega_0^2 x = - \alpha x^2 - \beta x^3 - \ddot{x}$$ and then add $\frac{\omega_0^2}{\omega^2} \ddot{x} $ to both sides $$\frac{\omega_0^2}{\omega^2} \ddot{x} + \omega_0^2 x = - \alpha x^2 - \beta x^3 + (\frac{\omega_0^2}{\omega^2} - 1) \ddot{x}$$ then set $\omega = \omega_0 + \omega_1$ with $\omega_1$ of first order of smallness, and $$x = x_1 + x_2 = a \cos (\omega t) + x_2 = a \cos [(\omega_0 + \omega_1) t] + x_2,$$ with $x_1$ and $x_2$ of first and second order of smallness respectively (note $x_1 x_2 = 0$, $\omega_1 x_2 = 0$ and $\omega_1^2 x_1 = 0$ thus hold if we neglect all terms above the second order of smallness), so that $$ \ddot{x} = - a (\omega_0 + \omega_1)^2 \cos[(\omega_0 + \omega_1) t] + \ddot{x}_2 = - (\omega_0 + \omega_1)^2 x_1 + \ddot{x}_2 .$$ The left-hand side is \begin{align} \frac{\omega_0^2}{\omega^2} \ddot{x} + \omega_0^2 x &= \frac{\omega_0^2}{\omega^2} [- (\omega_0 + \omega_1)^2 x_1 + \ddot{x}_2] + \omega_0^2 (x_1 + x_2) \\ &= \omega_0^2 x_2 + \frac{\omega_0^2}{\omega^2} \ddot{x}_2 + \omega_0^2 x_1 - \frac{\omega_0^2}{\omega^2} (\omega_0 + \omega_1)^2 x_1 \end{align} while the right-hand side is \begin{align} - \alpha x^2 - \beta x^3 + (\frac{\omega_0^2}{\omega^2} - 1) \ddot{x} &= - \alpha (x_1 + x_2)^2 - \beta (x_1 + x_2)^3 + (\frac{\omega_0^2}{\omega^2} - 1) [- (\omega_0 + \omega_1)^2 x_1 + \ddot{x}_2] \\ &= - \alpha (x_1^2 + 0) - \beta \cdot 0 + \frac{\omega_0^2}{\omega^2} [- (\omega_0 + \omega_1)^2 x_1 + \ddot{x}_2] - [ - (\omega_0 + \omega_1)^2 x_1 + \ddot{x}_2] \\ &= - \alpha x_1^2 - \frac{\omega_0^2}{\omega^2} (\omega_0 + \omega_1)^2 x_1 + \frac{\omega_0^2}{\omega^2} \ddot{x}_2 + (\omega_0 + \omega_1)^2 x_1 - \ddot{x}_2 \end{align} Equating both sides then solving for $\ddot{x}_2 + \omega_0^2 x_2$ this becomes \begin{align} \ddot{x}_2 + \omega_0^2 x_2 &= - \{ + \frac{\omega_0^2}{\omega^2} \ddot{x}_2 + \omega_0^2 x_1 - \frac{\omega_0^2}{\omega^2} (\omega_0 + \omega_1)^2 x_1 \} \\ & \ \ \ \ \ + \{ - \alpha x_1^2 - \frac{\omega_0^2}{\omega^2} (\omega_0 + \omega_1)^2 x_1 + \frac{\omega_0^2}{\omega^2} \ddot{x}_2 + (\omega_0 + \omega_1)^2 x_1 \} \\ &= - \{ + \frac{\omega_0^2}{\omega^2} \ddot{x}_2 + \omega_0^2 x_1 \} + \{ - \alpha x_1^2 + \frac{\omega_0^2}{\omega^2} \ddot{x}_2 + (\omega_0^2 + 2 \omega_0 \omega_1 + \omega_1^2) x_1 \} \\ &= - \{ + \omega_0^2 x_1 \} + \{ - \alpha x_1^2 + (\omega_0^2 + 2 \omega_0 \omega_1 + \omega_1^2) x_1 \} \\ &= + \{ - \alpha x_1^2 + (+ 2 \omega_0 \omega_1 + \omega_1^2) x_1 \} \\ &= + \{ - \alpha x_1^2 + (+ 2 \omega_0 \omega_1) x_1 \} \\ &= - \alpha x_1^2 + 2 \omega_0 \omega_1 x_1 \\ &= - \alpha a^2 \cos^2(\omega t) + 2 \omega_0 \omega_1 a \cos (\omega t) \end{align}

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