The solution of a nonlinear equation in Schutz's book page 211 2nd edition

[MathematicalPhysicist]

TL;DR Summary

On page 211 in equation (9.32) we have a nonlinear equation that $f$ and $g$ should satisfy, which is $\ddot{f}/f+\ddot{g}/g=0$.
The suggested solution in the book doesn't make sense, can you help me understand it?

Continuing the summary, the author argues that if $g$ is nearly 1, i.e $g(u)\approx 1+\epsilon(u)$, one obtains the solution:
$f(u)\approx 1-\epsilon(a)$.
The derivative in the summary, i.e the dots represent derivatives with respect to $u$.

Then how to deduce the solution for $f$?
If I plug $g$ back to the equation in the summary I get:
$$\ddot{f}+(\ddot{\epsilon}/(1+\epsilon(u)))f=0$$
Don't see how to continue from here, he talks about Fourier representation, but I don't follow his reasoning.

Thanks!

 

[MathematicalPhysicist]

Perhaps because it's an approximation we get: $\ddot{f}+\ddot{\epsilon}f\approx 0$, but I am not sure.

 

[kent davidge]

Since $g \approx 1 + \epsilon$ then $g^{-1} \approx 1 - \epsilon$. Substituting this in you get $\ddot f / f + \ddot \epsilon (1 - \epsilon) = 0$. Terms with $\epsilon$ and its derivatives should be ignored. You get $\ddot f / f + \ddot \epsilon = 0$. It follows that $f = 1 - \epsilon$ solves this last equation.

 

[PeterDonis]

Terms with ϵ\epsilon and its derivatives should be ignored.

Not quite; as you state this, it would mean the only remaining terms would be $\ddot{f} / f = 0$.

What you mean is that terms quadratic or higher in $\epsilon$ and its derivatives should be ignored. That gets rid of the troublesome $\ddot{\epsilon} \epsilon$ term (and also gets rid of a similar troublesome term when $\ddot{f} / f$ is computed).

 

[MathematicalPhysicist]

Ok, I think I see it now.
$f\approx 1-\epsilon(u)$, makes for $\ddot{f} \approx -\ddot{\epsilon}$, and:
$\ddot{f}/f =-\ddot{\epsilon}/(1-\epsilon) \approx -\ddot{\epsilon}\cdot(1+\epsilon)$; so by adding this with $\ddot{\epsilon}/(1+\epsilon) \approx \ddot{\epsilon} (1-\epsilon)$, we neglect the $\epsilon$ term and the second derivative of $\epsilon$ gets cancelled.

 

[MathematicalPhysicist]

I have another question.
In the book he guessed $f$ by knowing $g$, but if I were given this $g$ then how would arrive at $f$ without guessing that I need to expand geometrically the denominator of $f$?

I mean I would have: $\ddot{f}+\ddot{\epsilon}f=0$
then I would multiply by $\dot{f}$ and integrate, I would get:
$\dot{\dot{f}^2}+\int \ddot{\epsilon}\dot{f^2}=E$, that's a difficult equation to solve, if it's even possible analytically.
It seems like a lucky guess and a lot of neglecting terms...
Not something my mathematical part of mathematicalphysicist will like... :-)

 

[MathematicalPhysicist]

Well I can use the ansatz of power series in $u$, i.e $f(u)=\sum_n a_n u^n$ and $\epsilon(u) = \sum_n b_n u^n$, and then to differentiate both $f$ and $\epsilon$ twice, and to plug back to the ODE.

I'll get some recurrence relation of $a_n$'s and $b_n$'s.