If a quantity is small, is the derivative of that quantity small?

  • #1
Hey! I'm working on some classical mechanics where I'm studying small deviations about an equilibrium point. If we call this point x0 and the small deviation x. Is there any good arguments why the change in x should be small so that one could neglect
[tex] (\frac{d}{dt}x)^2[/tex]
terms? I see this being done extensively. Are there some conditions on this being true or is it generally true?
 

Answers and Replies

  • #2
150
1
You could take it as a definition of equilibrium. The first-order change in x is negligible so you can safely discard it's square (and higher powers as well). If it wasn't then you would have a slope, which obviously is not an equilibrium.
 
  • #3
arildno
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Hey! I'm working on some classical mechanics where I'm studying small deviations about an equilibrium point. If we call this point x0 and the small deviation x. Is there any good arguments why the change in x should be small so that one could neglect
[tex] (\frac{d}{dt}x)^2[/tex]
terms? I see this being done extensively. Are there some conditions on this being true or is it generally true?
It is meaningless to say that something is "small"; it might at best be negligible with respect to something else.

However, in many cases, the square of the first derivative will be negligible relative to, say, the second derivative.
Example:
[tex]x(t)=\epsilon\sin(\frac{t}{\epsilon})[/tex]
where epsilon is some tiny parameter.

That makes x(t) "small", and the square of the first derivative is tiny relative to the magnitude of the second derivative.

Note that this function might well describe the behaviour around some equilibrium point.
 
  • #4
It is meaningless to say that something is "small"; it might at best be negligible with respect to something else.
So if you had an equation of motion how would you justify that you could find a small enough displacement so that the displacement and it's derivative are negigble to other contributing quantities. Can one make an argument for that this is always possible at an equilibrium?
 
  • #5
arildno
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So if you had an equation of motion how would you justify that you could find a small enough displacement so that the displacement and it's derivative are negigble to other contributing quantities. Can one make an argument for that this is always possible at an equilibrium?
What you need with "dominant balance"-arguments when you are, say, trying to develop an approximate, perturbative solution to some diff.eq, is to
a) Make a GUESS at what will be the dominant terms, and which the subdominant terms.
b) TRY IT OUT: If you are lucky (or experienced), your initial trial function will behave nicely according to plan, if not, try again with something else.

And, unfortunately, the field of differential equations is so vast and diverse that only in some very few cases can you actually PROVE that what you're up to actually is going to work.
 

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