# Landau-Lifshitz skips a step in anharmonic oscillations

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!

• is your $w^{(0)}$ same as $w_0$... there is no $w^{(0)}$ defined in your expansion and clearly $w_0\ne 0$. – ZeroTheHero Sep 29 at 21:38
• 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. – Cast fj Sep 29 at 21:57
• 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. – ZeroTheHero Sep 29 at 22:04
• ok, as landau does, start counting from 1 – Cast fj Sep 29 at 22:09
• I will note that if you are expecting Landau and Lifshitz to provide detailed steps, you will be sorely disappointed. – Jon Custer Sep 29 at 23:01

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}