source: Calculus of Variations. Joel G. Broida. Lecture Notes. 2009. link
the derivative is same as saying, if you vary the input x by a little bit h... then what you get is the function at x, plus a linear term times h, plus a remainder term that goes away as h goes to zero.
A functional is differentiable, if when we vary the input by some additional function h, the difference in the outputs of the varied and original functional is a linear (first order) function of h, plus some remainder terms on the order of h^2 such that the remainder disappears as h goes to zero.
Let's find the differential for a Lagrangian type functional.
note: h is not a number like epsilon going to zero, it is a function. i assume we choose continuously differentiable functions. therefore we can take its derivatives.
so the way we go about varying the functional is we put the functional at the input point at which we want its derivative. we subtract this amount from the functional with inputs with additional h function in x, h' function in x', h'' in x'' etc. for all inputs it depends on.
Then use the linear expansion of the function (assuming that the function f is differentiable, we need that to make it work) at the point x, x', t. Separate second order or higher terms in h from the first order terms.
These are the remainder and the linear part, respectively.
Assuming that the remainder term goes to zero, since it is of order h^2, when h goes to zero... we focus on the linear term L. Evaluate its integral. Integration by parts happens to be useful in this step.
The first term is simply evaluated at the bounds of integration. When we do physics with normalized wavefunctions, we require that the functions go to zero as x goes to positive or negative infinity, so this term usually becomes zero. If we use delta functions as our trial function h, this term evaluated at the bounds always equals zero, so long as we don't place a bound at the singularity of the delta function, which we won't LOL.
Remember that h is a function of t. and so are x, x', etc.
f( t, x, x' ) = f (t, x(t), x'(t))
Euler-Lagrange
the function h has to be completely arbitrary, so long as its continuous. This include the dirac delta function, which is within the closure, upper boundary of a series of continuous functions.
key for h is that it is nonzero between bounds and zero at bounds AND that we can pick h however we please. suppose that f is non-zero, so say it's positive at least a t a point. by continuity, there must be a region around this point where f is strictly positive and greater than some arbitrary but positive constant c. construct an h function that equals 1 across this region, and goes to zero outside of it. the integral in consideration must then be greater than the value of the integral over this positive region, whose width we denote d. then the integral is at least c * 1* d, which is greater than zero. by contradiction of the hypothesis, we have shown that f cannot be nono-zero. by picking h to be some value only when f is strictly positive, we made this argument possible.
Aside: Here's how to construct infinitely differentiable 'plateau' functions. that is, zero outside some interval, and equal to 1 or some other constant within a strictly smaller interior interval.
f satisfies the infinitely differentiable aspect.
g creates a situation where it becomes f over f = 1 if f is greater than or equal to t = 1. change the f (1 - t) term to f (R - t) for the radius R you wish for an interior interval.
h is the function that has all those plateau properties. zero outside an interval with radius of 2, equal to one within a radius of 1, and presumably a descreasing slope from 1 to zero in between.
finishing the Euler-Lagrange proof
the stationary point of a function has an analogous definition called the extremal curve of a differentiable functional. the extremal is by definition, the curve whose functional derivative equals zero. so the entire linear term equals zero
when we consider the space of curves passing through two points, at the bounds of integration of the functional the function h(t) must equal zero, for the curve to pass exactly through those boundary points.
the lemma preceding the euler-lagrange theorem says the first integrand term equals zero.
If we have multi-dimensional function with variables x, x', y, y', z, z', etc. the same thing is done separately for each dimension and associated variable. we vary y by y + h, y' by y' + h', etc. etc. do the linear approximation by Taylor expansion, integrate by parts, make the boundary terms vanish (by the argument that the curve passes through fixed points. so that the h function equals zero at the boundaries)
Shortest distance between two points in a plane is a straight line
a curve in two dimensions can be parameterized as a single variable s along the 1-dimension of the curve, or as two dimensions x, y in the Cartesian plane.
an incremental movement along the curve is given by the pythagorean theorem between a displacement in x, and a displacement in y as . Sqrt( dx^2 + dy^2).
if we then write the total derivative in terms of one variable, dx, we get dx sqrt(dx^2/dx^2 + dy^2/dx^2) = dx sqrt( 1 + y'(x)^2).
Apply the functional derivative to the integral of f(x, y(x), y'(x)).
t should be the first variable, it is independent in the theorem 1. anyways. x serves the place of t here.
f = sqrt( 1 + y'^2) . df/dy = 0. - d/dx (df/dy') = - d/dy (y')/sqrt(1 + y'^2). the last term is zero because h is zero at boundaries for the curve to pass through fixed points. so the whole thing being zero makes the middle term zero.
y' is a function of x. so d/dx of that expression = 0 means that the expression is a constant in terms of x, and of y(x), y'(x).
(y')/sqrt(1 + y'^2) = constant.
moving the y' terms all to one side, and c to the other, we find 0<c<1, so the radical term in the square root is non-negative.
then we can write y' in terms of c, so y' is another constant. then integrating y = mx+b where m and b are constants.
