Goldstein 2nd ed.

In its Appendix is given the derivation of Bertrands Theorem and after some steps we arrive as shown below :

where it is understood the derivatives are evaluated at $u=u_0$. In terms of this expansion of $J(u)$ the orbit equation becomes $$\frac{d^2x}{d\theta^2} + \beta^2x = \frac{x^2J''}{2} + \frac{x^3J'''}{6}. \tag{A-10}$$ We seek to find the nature of the source law such that even when the deviation from the circular orbit, $x$, is large enough that the terms on the right cannot be neglected, the solution to Eq. (A-10) still represents a closed orbit. For small perturbations from circularity we know $x$ has the behavior described by Eq. (A-8), which represents the fundamental term in a Fourier expansion in terms of $\beta\theta$. We seek therefore a closed-orbit solution by including a few more terms in the Fourier expansion: $$x = a_0 + a_1\cos\beta\theta + a_2\cos2\beta\theta + a_3\cos3\beta\theta.\tag{A-11}$$

Where $x=u-u_0$ is the deviation from circularity and $u=1/r$ and $J(u)=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right)=-\frac{m}{l^{2} u^{2}} f\left(\frac{1}{u}\right)$

If the R.H.S of A-10 was zero, then the solution was given by $a \cos(β\theta)$. However if there are terms on the RHS as given in equation A-10 the author writes the solution as a Fourier sum involving only cosine terms . How does the author know that we should use a Fourier expansion of $x$ using only cos terms with argument $β\theta$ ?

Any hint please.

  • $\begingroup$ I don't know this problem, but it's there some reason to think the answer must be an even function? $\endgroup$
    – The Photon
    Commented Oct 29, 2021 at 14:14
  • $\begingroup$ Yes i suspect one. The equation form which we began is the differential equation of motion $$\frac{d^{2} u}{d \theta^{2}}+u=J(u)$$ where $$J(u)=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right)=-\frac{m}{l^{2} u^{2}} f\left(\frac{1}{u}\right)$$ and I think if $u(x)$ is a solution to it $u(-x)$ is as well. So there is some reason for the Final even solution. $\endgroup$
    – Kashmiri
    Commented Oct 29, 2021 at 14:26

1 Answer 1


This is essentially a math question. You're wondering why the second-order differential equation

$$\frac{d^2x}{d\theta^2}=-\beta^2x+\frac{J''}{2}x^2+\frac{J'''}{6}x^3 \tag{1}$$

should necessarily have the solution of the form

$$x(\theta)=a_0+a_1\cos(\beta\theta)+a_2\cos(2\beta\theta)+a_3\cos(3\beta\theta) \tag{2}$$

First off, this is not the general form of the solution. We need more cosine terms in this expansion. But then again, the differential equation (1) also involves higher order terms in $x$ which we have neglected. The proposed solution (2) is cut off at $\cos(3\beta\theta)$ with the clever foresight that higher-order coeffecients $h_{4,5,\ldots}$ will eventually be found to drop off as higher powers in $h_1$, and that in order to determine the first four coefficients $h_{0,1,2,3}$ we need only expand (1) to order $x^3$.

If we're assuming closed orbits, then $x(\theta)$ must be periodic in $\beta\theta$ where $\beta$ is a rational number (i.e. it's periodic over the range $0<\theta <\frac{2\pi}{\beta}$), and therefore by Fourier's theorem we may expand $x(\theta)$ in appropriate cosines and sines. So why no sine terms? The answer has two simple parts:

  1. We impose that the first-order result shouldn't have any sine terms.
  2. If we expand $x(\theta)$ using only cosines, then the LHS of (1) will only have cosines. There's a trig identity that says $\cos(n\theta)\cos(m\theta)=\frac{1}{2}\left[\cos((n+m)\theta)+\cos((n-m)\theta)\right]$, so that means any power of $x(\theta)$ when fully factorized will only contain cosine terms as well. This doesn't work using only sines (look up the equivalent multiplication rule for sines), so the upshot is that we can find a solution using either only cosines, or both cosines and sines. The former is obviously simpler. Indeed, with the same initial conditions which made the first-order solution involve only one cosine, if you kept the sines you would eventually find that they drop out of the final solution.

Although Fourier's theorem says that the general solution $x(\theta)$ can be expanded in cosines and sines, you may wonder whether the matching procedure will give us the correct answer. For this we employ Fourier's trick, i.e. the orthogonality of cosines and sines, which basically tells us that two Fourier expansions are equivalent iff each individual coefficient agrees.

As far as uniqueness goes, this differential equation $x''(\theta)=F(x)$ with $F(x)$ smooth, is autonomous and there are general theorems in math (Picard–Lindelöf) which show that given sufficient initial conditions, the solution is unique.


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