Among all curves whose end points lie on two given vertical lines $x=a $ and $x=b $, find the curve for which the functional $$J [y] = \int_a^b F (x,y,y')dx $$ has an extremum
Similar to the case of fixed end points, we end up with the variation of $J $ equaling: $$\delta J = \int_a^b (F_y-\frac {d}{dx} F_{y'})h (x)dx+F_{y'}h(x)|_a^b $$
Where $\delta J = 0$ implies it is at an extremum. This time, however, the boundary does not necessarily vanish.
If we consider the case that $h (a)=h (b)=0$, then $y(x) $ is a solution to Euler's equation, as the boundary terms vanish and:
Theorem
Let $f $ be a continuous function. If $$\int_a^b f(x)h (x)dx=0$$ For all differentiable functions $h $ satisfying h (a)=h (b)=0, then $f (x) \equiv 0$.
This theorem is pivotal in deriving the Euler equation.
My book says that in order for the curve $y (x) $ to be a solution to the variable end point problem, it must be a solution to Euler's equation (i.e. an extremal).
Why must $y $ be extremal? It seems to me that by the theorem, $y $ is extremal when $h (a)=h (b)=0$, but I do not necessarily see why it needs to solve Euler's equation anywhere else (and thus force the boundary of $F_{y'} $ to vanish.
The way my book derived the Euler equation, it is necessary for the boundary of the perturbation to vanish:
If anyone wants a source/derivation on this material, click here
If you have an extremum, then $\delta J=0$ for all $h$. Now consider only all those $h$ such that $h(a)=h(b)=0$. Then you can apply your theorem to get that $F_y-\frac{d}{dx}F_{y^\prime}=0$. Now go back to $\delta J=0$ but now with $h$ not vanishing at the endpoints. Since you now know that $F_y-\frac{d}{dx}F_{y^\prime}=0$ you now have $0=\delta J=F_{y^\prime}h(x)\vert^b_a$ and so if you take first $h$ with $h(a)=0$ and $h(b)=1$ you get $F_{y^\prime}=0$ at $b$. Then you take $h$ with $h(a)=0$ and you get $F_{y^\prime}=0$ at $a$. Was this your question?