Calculus of variations ( interpreting the minimum in first order)

Question

I have just begun studying the Calculus of Variations and I have 3 doubts in it.

I have written certain things in bold so as anyone who wishes to answer the question but find it quite long can just go through the bold part and he would get an idea as what my doubt is

1)" If we go away from the minimum in first order then the deviation of the function from its minimum value is only of second order " -Feynman Lectures What does " going away from the minimum in the First order causes the deviation of the function from its minimum value in second order" means geometrically ( probably in terms of 2 possible paths deviated from the original path) likes the ones generally given in books. Because I was unable to understand what does going away from the minimum in first order and deviation of function value in second order mean separately and graphically (i.e what are these in the graph of the real and deviated path).

Also why do you need to go away from the minimum value in 1st order only( why not second order or 3rd order , though I also don't know what will they mean graphically) and expect a second order deviation only( why not 3rd order deviation although again I don't know what it will mean graphically).

Also how will I represent going away from the minimum in 1st order or 2nd order and having a deviation of 2nd order or 3rd order graphically ?

And does this story remains the same for maxima or not ?

2)What does it mean by the proximity or closeness of curves. I my book they have defined it like this this -: |y1(x) - y2(x)| is the zero order |y1'(x) - y2'(x)| 1st order |y1"(x) - y2"(x)| 2nd order and so on. What do they mean graphically ? How do I recognize which graph is of 0 order which is of 1st order and so on ? They say that if the curves are close in sense of higher order proximity then hey are close in lower order too. Why ?

3) The last problem is in the Euler Lagrange equations when the functional is independent of x. The partial derivative w.r.t **to y and y' are easily computed but how to apply the chain rule when differentiating the 2nd term w.r.t to x. I want to know how to apply the chain rule in the general case in it because at first it looks as if it can be 0 ? While explaining this please explain the step because I have always been having a problem in chain rule . I will check where I was going wrong .**Please help

Sorry for the long question. But they have been troubling a lot. It looks like the concept is not clear and I am unable to connect the result to the graphs geometrically.

Edit I would also like if any one could suggest a good book for it

Answer 1

Given a curve y(x), "going away to first order" means substituting $y(x)+\epsilon\delta y(x)$ where $\delta y(x)$ is a function of $x$, usually constrained to be zero at the relevant endpoints. First order refers to the power 1 of $\epsilon$. You are going to differentiate in $\epsilon$ and then set $\epsilon=0$ because you are to calculate the analogue of the directional derivative. If you use higher than power $1$ then the result will always be 0. Graphically, there will be the curve $y(x)$ and then there will be a perturbation that shrinks to $y(x)$ as $\epsilon\rightarrow0$. Many books have this picture.

At a critical curve, the functional differentiates to zero in $\epsilon$, which means that if you expand at $\epsilon=0$ after substituting $y+\epsilon\delta y$ then there are no terms linear in $\epsilon$. You have only terms like $\epsilon^2$. At the critical curve, when you vary the function linearly in $\epsilon$, the result is a higher-order change like $\epsilon^2$. You are missing an essential intuition: when the first derivative vanishes, then what remains is at second order variation.

You don't have to recognize closeness or anything like that. You are driving the functional along a particular perturbation. You are not interpreting a nearby curve that someone else provides to you. The analogue is, for an ordinary function $f(x,y,z)$, calculating the directional derivative (so $d/dt$ at $t=0$) of $f(x+t\,\delta x, y+t\,\delta y,z+t\,\delta z)$. $t$ is more common than $\epsilon$ in this context.

The chain rule is as usual, the only problem being when the Lagrangian is $L(y,y')$ then the $y'$ here is a (two-character) variable, not the derivative of something. This is the notation so that we understand how to substitute the function. It's a reminder that the derivative of the function goes in that slot.

These statements are not theorems. For example, they depend on the presence of sufficient smoothness.