Reader's comments to paragraph 2:
"The principle of least action". L.D. Landau and E.M. Lifshitz. Mechanics.
Butterworth Heineman. Oxford. 1999.
Mathematical referenced are at the end of this page.
Reader had difficulties to understand the phrase:
"The necessary condition for S to have a minimum ... is that
these terms (called first variation, ... ) should be zero." (P)
In reality,
1) reader does not have to make this assumption at this point, but
2) follow Landau's text up to the phrase
"This can be so only if the integrand is zero indentically" (P')
3) then prove (P') which implies (P).
Below are the details.
Notations:
^q  variation of q
q'  timederivative of q
L  partial derivative L by q
q
Let's don't make this assumption now and follow Landau and
expand variation of S in powers of ^q and ^q':
_t2


 ( L ^q + L ^q' ) dt
 q q'

_
t1
Don't require this to be equal to 0 yet. This is too early.
Then, do integration by parts and obtain (2.5) without
requiring this to be equal to 0.
Use conditions (2.3) and analyse integrand K(q,q',t) which is
K = L  ( L )'
q q'
Now, the real mathematical job begins.
The integral of K must have a minimum, this is what principle is.
We have a right to chose any ^q, so we can
choose any point T from (t1, t2) and
choose ^q1 to be nonzero in neighbourhood of t and zero in other points like:
^q1

 ^Q . . . ^Q = ^q1(T)
 . .
 . .
 . .
 . .
 . .
 . .
...... T ................._________ t
If K in continuous function in point K(^q(T), ^q'(T), T)) and, for certainty,
is positive, then for enough small neighbourhood of T, variation of S will be
positive.
Then in the key point of proof, "change the sign of ^Q".
Precisely speaking, take another function ^q2 which is ^q2 =  ^q1.
Then variation of S will be negative.
We have that S2