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

  • Level: Undergrad 
  • Thread starter Thread starter MathematicalPhysicist
  • Start date Start date
  • Tags Tags
    Book Nonlinear
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
6 replies · 2K views
Science Advisor
Gold Member
Messages
4,662
Reaction score
372
TL;DR
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!
 
Physics news on Phys.org
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.
 
kent davidge said:
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).
 
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.
 
Last edited:
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... :-)
 
  • Like
Likes   Reactions: weirdoguy
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.