Book
From the Book: A First Course in the Calculus of Variations
Section 5.3
Page of book: 118/pdf page: 129
link: here
Question
They have use the equation: $y(x) +\epsilon_1\eta_1(x)+\epsilon_2\eta_2(x)$
where $\epsilon_2$ is the "correction" for $y(x)$ so that it satisfies the integral constraint.
Q1 Does this mean that as we reduce $\epsilon_1$ to $0$, $ \\ \epsilon_2$ tends (possibly fluctuates) to $0$ automatically?
I suppose what I am asking is if $\epsilon_2$ is ultimately a function of $\epsilon_1$ i.e $\epsilon_2 = \epsilon_2(\epsilon_1,\eta_1)$?
Q2 And (assuming Q1 is correct) is $\epsilon_1$ and $\epsilon_2$ are "linked" purely by the constraint? Like if we were to remove the constraint:
$K[y] =\displaystyle{\int_a^b}g(x, y,y')\,{\rm d}x = l$
then there will be no "automatic correction"?
For no constraints we have the curve: $y(x,\epsilon) =y(x,0) + \epsilon\eta(x)$
but this does not work for when we have a constraint eg: the length of the curve must = $l$ which is a integral constraint $K[y] =\displaystyle{\int_a^b}g(x, y,y')\,{\rm d}x = l$ we have a problem with the 1 variation variable curve.
Why? because if $y(x,0)$ is the extrema then any value added to the curve will increase it. Even if it is a negative value. This is because $K[y]$ is a line integral and not an "area-under-the-curve" integral.
Thus if we use $y(x,\epsilon) =y(x,0) + \epsilon\eta(x)$, we can't make any variations as this will break the constraint $l$.
To solve this problem we add another variation to $y$.
So in all we have $y(x,\epsilon_1,\epsilon_2) = y(x,0) + \epsilon_1\eta_1(x) + \epsilon_2\eta_2(x)$ where $\epsilon_1$ is the old $\epsilon$ and $\epsilon_2$ acts as the "correction"
So basically we are allowed to vary $\epsilon_1$ freely, but then $\epsilon_2$ will "auto-correct" our curve so that it will not have a length longer than $l$. As what can be seen below, the black curve is the optimal (length = $l$), red is the curve (length > $l$) without correction: $y(x,\epsilon) =y(x,0) + \epsilon\eta(x)$ and blue is the curve (length = $l$) with correction: $y(x,\epsilon_1,\epsilon_2) = y(x,0) + \epsilon_1\eta_1(x) + \epsilon_2\eta_2(x)$
This is similar to an ordinary function $f(x,y) = x^2 +y^2 = 4$ where if we have $x = 1$ then $y$ will automatically be $\sqrt{3}$
So this means that $\epsilon_2$ is dependent on $\epsilon_1,\eta_1(x)$ and $\eta_2(x)$, similarly how, in the function $f(x,y) = x^2 +y^2 = a$, $y$ is dependent on $a$ and $x$.
Now as y is integrated in both $K[y] =\displaystyle{\int_a^b}g(x, y,y')\,{\rm d}x = l$ and $J[y] =\displaystyle{\int_a^b}f(x, y,y')\,{\rm d}x$, $K[y]$ and $J[y]$ become functions of $\epsilon_1$ and $\epsilon_2$ only. Why? as variables $\epsilon_1,\epsilon_2$ can be seen as constants as the integral is take with respect to $dx$ and thus unchanged. Thus we actually have functions of $\epsilon_1,\epsilon_2$.
Thus we can let $K(\epsilon_1,\epsilon_2) = K[y] =\displaystyle{\int_a^b}g(x, y,y')\,{\rm d}x = l$ and $J(\epsilon_1,\epsilon_2) = J[y] =\displaystyle{\int_a^b}f(x, y,y')\,{\rm d}x$, note that $J(\epsilon_1,\epsilon_2),K(\epsilon_1,\epsilon_2)$ are not functions of $x$ as there have been integrated out.
From this we can actually treat this as a standard Lagrange problem:
\begin{equation}\nabla J = \lambda\nabla K \\ K =l \end{equation}
where $\nabla = (\frac{\partial}{\partial\epsilon_1},\frac{\partial}{\partial\epsilon_2})$
Note that $K(\epsilon_1,\epsilon_2)$ curve length mapping $\epsilon_1,\epsilon_2$,$l$ Eg:
Note that $J(\epsilon_1,\epsilon_2)$ is an area under curve mapping with $\epsilon_1,\epsilon_2$,Area axes Eg:
Important thing to remember is that an extrema under all the constraints provided will always happen at $\epsilon_1,\epsilon_2 = 0$ and that the point $(\epsilon_1,\epsilon_2) = (0,0)$ at where optimal curve exists is always included in the contour. And that the gradient, $\nabla K$ is only a scalar multiple different to $\nabla J$ at the point of extrema(if not then we are not at an extrema)
To answer the questions:
Q1, yes as $ \epsilon_1 \to 0$,then $\epsilon_2 \to 0$ else function $K(\epsilon_1 ,\epsilon_2)$ will not be satisfied
Q2, yes, its is the reason we added $\epsilon_2$, it is to correct the length of the curve thus it requires dependency on $\eta_1$ and $\epsilon_1$
Finally: each mapping graph will be different for each; $y(x,0), \eta_1,\eta_2$ chosen