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

  • Context: Undergrad 
  • Thread starter Thread starter MathematicalPhysicist
  • Start date Start date
  • Tags Tags
    Book Nonlinear
Click For Summary

Discussion Overview

The discussion revolves around solving a nonlinear equation presented in Schutz's book, specifically focusing on the relationship between the functions ##f## and ##g##, where ##g## is approximated as nearly 1. Participants explore methods for deriving ##f## from ##g##, including approximations and series expansions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that if ##g## is nearly 1, then ##f(u)## can be approximated as ##1 - \epsilon(a)##, but expresses uncertainty on how to deduce ##f## from this.
  • Another participant proposes that the approximation leads to a simplified equation ##\ddot{f} + \ddot{\epsilon} f \approx 0##, though they are unsure of its validity.
  • A different viewpoint indicates that substituting ##g \approx 1 + \epsilon## into the equation leads to a form where terms involving ##\epsilon## can be neglected, resulting in ##f = 1 - \epsilon## as a solution.
  • One participant clarifies that it is important to ignore terms that are quadratic or higher in ##\epsilon## and its derivatives to simplify the equation.
  • Another participant expresses a realization that the approximation leads to a cancellation of terms, allowing for a clearer path to the solution.
  • A question is raised regarding the method of arriving at ##f## from ##g## without prior knowledge of the necessary expansion, highlighting the complexity of the resulting equations.
  • Lastly, a participant suggests using a power series ansatz for both ##f## and ##\epsilon## to derive a recurrence relation, indicating a potential method for solving the problem.

Areas of Agreement / Disagreement

Participants express varying levels of confidence in their approaches, with some agreeing on the necessity of neglecting higher-order terms, while others question the assumptions made in deriving ##f## from ##g##. The discussion remains unresolved regarding the best method to derive ##f## without prior knowledge of the necessary expansions.

Contextual Notes

Participants note the complexity of the equations involved and the reliance on approximations, which may not be universally applicable. There is an acknowledgment of the difficulties in solving the equations analytically.

Who May Find This Useful

This discussion may be of interest to those studying nonlinear equations, mathematical physics, or anyone looking for methods to approach complex differential equations in theoretical contexts.

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

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
5K
Replies
4
Views
2K
  • · Replies 47 ·
2
Replies
47
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 62 ·
3
Replies
62
Views
7K
  • · Replies 0 ·
Replies
0
Views
1K
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
2
Views
2K