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. 


    δq - variation of q

    q' - time-derivative of q

    Lq - partial derivative L by q

    if you don't see appropriate mathematical symbols, 
    you may check your browser

Let's don't make this assumption now and follow Landau and
expand variation of S in powers of δq and δq':

    t1t2 ( Lqδq   +  Lq'δq' ) dt           

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 = Lq    -      --( Lq' )

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 δq, so we can 
  choose any point T from (t1, t2) and 
  choose δq1 to be non-zero in neighbourhood of t and zero in other points like:

  |        δQ  . . .         δQ = δq1(T)
  |          .      .
  |         .        .
  |         .        .
  |         .        .
  |         .        .
  |        .          .
  |......       T      ................._________ t

If K in continuous function in point K(δq(T), δq'(T), T)) and, for certainty,
is positive, then for enough small neighbourhood of T, variation of S will be

Then in the key point of proof, "change the sign of δQ".
Precisely speaking, take another function δq2 which is δq2 = - δq1.

Then variation of S will be negative.
We have that S2<S0<S1 where

S2 is taken on q+δq2,
S0 is taken on q,
S1 is taken on q+δq1.

This means that q is not an extreme path for S unless K is zero in T.
Because T is an arbitrary point, then K is zero in any point in (t1,t2).

We obtained Lagrange equation: K=0.

Now, we can return back to phrase (P) and say that it is true.


There rigid mathematical proof of
Lagrange equations is in 
Paragraph 4. 
"The Simpliest Variational Problem.
Euler's Equation."                 I. M. Gelfand and S.V. Fomin
                                   Calculus of Variations.
                                   Mineola, New York, 2000.

The problem is also discussed in:

"IV. The Calculus of Variations.
The Euler Equations."              R. Courant and D. Hilbert
                                   The Methods of Mathematical Physics
                                   Volume I.
                                   Interscience Publishers, Inc.
                                   New York. 1953.

This book may have some additional V.I.Arnold
material:                          Mathematical Methods of Classical Mechanics.
                                   Second Edition.
                                   New York, Berlin, ... . 1989.