I think I found a small error in Landau & Lifschitz "Mechanics" (3rd ed.). In section 28 (Anharmonic oscillations), they are discussing how to solve the following anharmonic oscillator problem:

$$\ddot{x}+\omega_0^2 x = -\alpha x^2-\beta x^3 \tag{28.9}$$

They show how one can solve the anharmonic oscillator problem perturbatively by expanding the solution in powers of the initial amplitude $a$ and supposing that the fundamental frequency is also shifted, the shift being given by another power series in the amplitude. The ansatz for the lowest-order solution is the following:

$$x^{(1)}=a\cos \omega t \tag{28.10}$$

One can grind out the higher order equations and arrive at the equation for the 3rd order perturbation:

$$\ddot{x}^{(3)}+\omega_0^2 x^{(3)}=-2\alpha x^{(1)}x^{(2)}-\beta x^{(1)3}+2\omega_0 \omega ^{(2)} x^{(1)}\tag{pg. 87}$$

Now, first I want to mention that in that equation we have $\beta x^{(1)3}$ which we can write out as

$$\begin{align}\beta x^{(1)3}&=\beta a^3\cos^3\omega t\\ &=a^3\left(\color{red}{\frac{1}{4}\beta} \cos 3\omega t \color{red}{+ \frac{3}{4}\beta} \cos \omega t\right) \end{align}$$

But in the next line the following full expansion is given (where I highlight the error in red):

$$\begin{align} \ddot{x}^{(3)}+\omega_0^2 x^{(3)}=a^3&\left[\color{red}{\frac{1}{4}\beta}-\frac{\alpha^2}{6\omega_0^2}\right]\cos 3\omega t +\\ &+a\left[2\omega_0\omega^{(2)}+\frac{5a^2\alpha^2}{6\omega_0^2}\color{red}{- \frac{3}{4}a^2\beta}\right]\cos\omega t \tag{pg. 87}\end{align}$$

So the it looks like a spurious minus sign has entered, which carries on throughout the rest of the section.

[Question]: Is this actually a mistake? I can't find any errata for the book so I can't validate this.

  • 1
    $\begingroup$ I think this is an error in the English translation. In the original Russian text the first term on the rhs reads $-a^3 [\beta/4 + \alpha^2/6\omega_{0}^2]\cos 3\omega t$ $\endgroup$
    – AlbertB
    Nov 7, 2017 at 3:18
  • $\begingroup$ Also, equation (28.14) reads $x^{(3)}=(a^3/16\omega_{0}^2)\left(\alpha^2/3\omega_{0}^2+\beta/2\right)\cos 3\omega t$ $\endgroup$
    – AlbertB
    Nov 7, 2017 at 3:43
  • 1
    $\begingroup$ I my Polish version of the book I have $\beta/4 + ...$ and $-3/4 a^2\beta$ (pg. 87). On the other hand, I have an error in $\omega^{(2)}$ which should be $\omega^{(2)}=(\beta3/8 - \alpha^2 5/12\omega_0^2) a^2/\omega_0$, otherwise units do not match. $\endgroup$
    – WoofDoggy
    Nov 7, 2017 at 11:03

1 Answer 1


I think there are more problems with this solution. Especially with perturbation theory. For this kind of problems one has to use Poincare-Lindstedt method [1] [2]. The small parameters in the former differential equations are $\alpha$ and $\beta$ so we can write down (the same can done for $\beta$): $$\ddot{x} + \omega_0^2x = -\alpha\left(x^2+\frac{\beta}{\alpha}x^3 \right).$$ Here $\alpha$ will be our small parameter and we are looking for a solution of the form $$x(t) = A\cos(\omega t) + \alpha x_1(t) + \alpha^2x_2(t) + \ldots$$ $$\omega = \omega_0 + \alpha\omega_1 + \alpha^2\omega_2 + \ldots$$ with initial conditions $x(0)=A$ and $\dot{x}(0)=0$.

First of all, we change variables $\tau = \omega t$, so that we get $$\omega^2 x''+ \omega^2_0 x = -\alpha\left(x^2 + \frac{\beta}{\alpha}x^3 \right),$$ where $x''$ denotes second derivative over $\tau$. We put our series formula for $x$ and $\omega$ and extract terms in the same order of $\alpha$: $$\left[ \omega_0^2 + 2\omega_0\omega_1 \alpha + \alpha^2 (\omega_1^2 + 2\omega_0\omega_2) + \ldots \right] \cdot \left[-A\cos\tau + \alpha x''_1 + \alpha^2 x''_2 + \ldots \right] + \omega_0^2 \left[A\cos\tau + \alpha x_1 + \alpha^2 x_2 + \ldots \right] = -\alpha \left[A^2\cos^2\tau + 2\alpha x_1 A\cos\tau + \frac{\beta}{\alpha}\left(A^3\cos^3\tau + 3\alpha x_1 A^2\cos^2\tau \right) + \ldots \right]$$

1st order terms

$$\begin{align} x''_1 + x_1 & = \frac{2A\omega_1}{\omega_0}\cos\tau - \frac{A^2}{\omega^2_0}\cos^2\tau - \frac{\beta}{\alpha}\frac{A^3}{\omega_0^2}\cos^3\tau \\ & = \cos\tau\left[\frac{2A\omega_1}{\omega_0}-\frac{\beta}{\alpha}\frac{3 A^3}{4\omega_0^2}\right] - \frac{1}{2}\frac{A^2}{\omega_0^2} - \frac{1}{2}\frac{A^2}{\omega_0^2}\cos 2\tau - \frac{1}{2}\frac{\beta}{\alpha}\frac{A^3}{\omega_0^2}\cos 3\tau, \end{align} $$ with $x_1(0)=0$ and $x'_1(0)=0$. If we want to kill the resonant term (the one with $\cos\tau$) we need to set $$\omega_1 = \frac{\beta}{\alpha}\frac{3}{8}\frac{A^2}{\omega_0}.$$ Then $x_1(\tau)$ solves to $$x_1(\tau) = \frac{A^2}{\omega_0^2}\left[-\frac{1}{2} + \left(\frac{1}{3} - \frac{1}{16}\frac{\beta}{\alpha}A\right)\cos\tau + \frac{1}{6}\cos 2\tau + A\frac{1}{16}\frac{\beta}{\alpha}\cos 3\tau \right].$$

2nd order terms

$$x_2'' + x_2 = \frac{A}{\omega_0^2}(\omega_1^2 + 2\omega_0\omega_2)\cos\tau - 2\frac{\omega_1}{\omega_0}x''_1 - x_1\frac{2A}{\omega_0^2}\cos\tau- x_1\frac{\beta}{\alpha}\frac{3A^2}{\omega_0^2}\cos^2\tau,$$ with $x_2(0)=0$ and $x'_2(0)=0$. This is a real mess which I am not going to post here, but $$\omega_2 = \frac{A^2}{\omega_0^3}\left[\frac{A}{4}\frac{\beta}{\alpha}-\frac{3A^2}{64}\left(\frac{\beta}{\alpha}\right)^2-\frac{3A^2}{64}-\frac{5}{12}\right]$$


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