Let's consider a functional $J(y)=\int_{0}^{1}{ \sqrt{1+\frac{dy}{dx} ^{2}} dx}$, $y(x_{1})=y_{1}, y(x_{2})=y_{2}$. Using Euler-Lagrange equations we can get that the set of functions which attains minumum to $J$ must satisfy $y''=0$, then we establish that $y=ax+b$ works well.
But to how to understand, if the functional reaches an extremum?
Idea 1: Assume that $y=y(x)$ attains a minimum, let's consider the first variation $y_{1}=y(x)+\epsilon \cdot h(x)$. If we denote $J$ as function of the variable $\epsilon$ and $y=y(x)$ attains a minimum, then $\epsilon=0$, if $J'(\epsilon)=0$. (Idea is that we consider a function that gives a smaller value to the function(we assume that) and we get that it matches with $y=y(x)$.)
Idea 2: To establish that the set of all continous on $[0,1]$ functions is a compact. (Or even if it's not true, may be it would be nice to try proving that the set of all continous curves with a finite length is compact).
Do the ideas above make any sense?
Would be grateful to recieve any help.
If you assume that $y$ is of class $\cal{C}^2$ or greater, approaching it from the point of view of your first idea makes sense, I think. Trying to prove that a huge function space like that is compact does not seem easy, and in most cases, it does not seem true.
What would your topology be? The $L^2$ topology (also called Sobolev)? That is what makes the most sense to me and the class of all continuous functions on $[0,1]$ is not compact in that topology. I cannot think of one where it is.
With your first idea, how does this strike you: we want to prove that straight lines are local minima of this problem. So let $y$ be a function that satisfies the Euler-Lagrange condition $y'' = 0$, and let $y_1$ be any other function within $\epsilon$ of $y$ in the $L^2$ sense. Suppose $J(y) > J(y_1)$. Then of course
$$\int_0^1 dx \; \sqrt{1 + y'^2} - \sqrt{1 + {y'}_1^2} > 0$$
Since $y'^2$ is a constant, so is $\sqrt{1 + y'^2}$. Call it $\alpha \geq 1$. It follows that (multiply the integrand by $\alpha + \sqrt{1 + {y'}_1^2}$)
$$\int_0^1 dx \; \alpha^2 - 1 -{y'}_1^2 > 0$$
Or $$\int_0^1 dx \; {y'}_1^2 < \alpha^2 - 1 = \int_0^1 y'^2 \; dx$$
From this, we see clearly that for $\epsilon$ sufficiently small, this is not possible unless $y_1$ is a straight line. If $y_1$ deviates at all from linearity, it will contribute significantly to the absolute value of $y_1'$, making the above inequality impossible.