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In Goldstein's Classical Mechanics, following comment is made regarding the equation of orbit in central force:

$\frac{d^2u}{d\theta^2}+u=-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u}\Big)\tag{3.34}$
The preceding equation is such that the resulting orbit is symmetrical about two adjacent turning points. To prove this statement, note that if the orbit is symmetrical it should be possible to reflect it about the direction of the turning angle without producing any change. If the coordinates are chosen so that the turning point occurs for $\theta=0$, then the reflection can be effected mathematically by substituting $-\theta$ for $\theta$. The differential equation for the orbit $(3.34)$, is obviously invariant under such a substitution. Further the initial conditions, here
$u=u(0),\;\Big(\frac{du}{d\theta}\Big)_o=0,\;for\;\theta=0$ will likewise be unaffected.

If we take the turning point to be at $\theta=0$, then the above symmetry in solution suggests that $u(\theta)=u(-\theta)$.
I am not able to prove how the differential equation $(3.34)$ suggests that if $u(\theta)=u(-\theta)$ (or $u(\theta)$ is an even function). We don't know the form of $V(\frac{1}{u})$, then how can we make such a generalized statement?
Actually I have just started studying about the differential equations. I don't know whether this symmetry in the solution arises directly from the form of differential equation. So please don't mind if it is a silly question.
Please help me in proving this fact.

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  • $\begingroup$ Obviously is that for $\theta\mapsto -\theta$ die differential equation remaining unchanged. Can you proof it ? $\endgroup$
    – Eli
    Commented Sep 17, 2021 at 7:03
  • $\begingroup$ @Eli, thanks for the reply. May you please elaborate. $\endgroup$
    – Manu
    Commented Sep 17, 2021 at 7:11

1 Answer 1

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This is the differential equation $$\frac{d^2u}{d\theta^2}+u=-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u}\Big)\tag 1$$

take $u(\theta)\mapsto u(-\theta)$

$$\frac{du}{d\theta}\,\mapsto-\frac{du}{d\theta}~~~, \frac{d^2u}{d\theta^2}\mapsto \frac{d^2u}{d\theta^2}$$ thus equation (1) $~\mapsto$ $$\frac{d^2u(-\theta)}{d\theta^2}+u(-\theta)=-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u(-\theta)}\Big)\tag 2$$

equation (2) is equal equation (1)

Edit

$$y=u(f(\theta))\\ \frac{d y}{d\theta}=\frac{d}{d\theta}\,u(f(\theta))\frac{d f}{d\theta}$$

with $~f(\theta)=-\theta~$

$$\frac{d y}{d\theta}=\frac{d}{d\theta}\,u(-\theta)\,(-1)$$

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  • $\begingroup$ Thanks for the reply. In your edit shouldn't, $\frac{dy}{d\theta}=\frac{du(f(\theta))}{df(\theta)}\frac{df}{d\theta}=\frac{du(-\theta)}{d(-\theta)}\frac{d(-\theta)}{d\theta}=-\frac{du(-\theta)}{d(-\theta)}$ $\endgroup$
    – Manu
    Commented Sep 17, 2021 at 17:13
  • $\begingroup$ Why $\frac{du(-\theta)}{d(-\theta)}=\frac{du(\theta)}{d\theta}$? May you please explain this. $\endgroup$
    – Manu
    Commented Sep 17, 2021 at 17:17
  • $\begingroup$ \begin{align*} &\text{lets say that u is a function of $~x^2~$ thus}\\ &y=u(x^2)\\ & \text{what is the derivative $~\frac{dy}{dx}~$ ?}\\ & \frac{dy}{dx}=\frac{du}{dx}\,2x \quad, \text{not}\,~ \frac{du}{dx^2} \end{align*} $\endgroup$
    – Eli
    Commented Sep 17, 2021 at 18:44
  • $\begingroup$ Hi Eli, in equation 2 is $u=u(-\theta)$ ? $\endgroup$
    – Kashmiri
    Commented Oct 26, 2021 at 12:13
  • $\begingroup$ I.e your second equation is :$$\frac{d^2u(-\theta) }{d\theta^2}+u(-\theta) =-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u (-\theta) }\Big)$$ $\endgroup$
    – Kashmiri
    Commented Oct 26, 2021 at 12:16

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