In recent Calculus of Variations lecture, I learnt about the Brachistochrone problem and its solution. For those who don't know, it is a standard problem in dynamics which is often used as a motivating example in introductions to functional analysis. It goes as follows:
$\textbf{The problem}$
Suppose that a bead slides on a frictionless wire in a vertical plane. What shape of the wire minimises the time for the bead to fall from rest at a point $A$ to a lower, and horizontally displaced, point $B$?
$\textbf{Textbook solution}$
Choose $A$ to be the origin of coordinates in the vertical plane with x being horizontal distance from the origin and y being the distance below the origin. The bead starts with zero velocity so conservation of energy implies that its speed $v$ at any later time is given by $v=\sqrt{2yg}$, where $g$ is the acceleration due to gravity. Hence we have to minimise
$$T=\int_{A}^{B}\frac{dl}{v}=\frac{1}{\sqrt{2g}}\int_{A}^{B}\frac{\sqrt{dx^2+dy^2}}{\sqrt{y}}$$
For simplicity, let us assume that $x$ is a good coordinate on the curve, so that $y=y(x)$ and
$$T=\int_{0}^{x_B}\sqrt{\frac{1+(y')^2}{y}}dx$$
To extremize $T$, we can use the Euler-Lagrange equations. As the integrand has no explicit $x$ dependence, these equations reduce to
$$C=\sqrt{\frac{1+(y')^2}{y}}-y'\frac{\partial}{\partial y'}\left(\sqrt{\frac{1+(y')^2}{y}}\right)=\frac{1}{\sqrt{y[1+(y')^2]}}$$
where $C$ is a constant. The resulting differential equation is $y[1+(y')^2]=2C$, where $C$ is a positive constant. This is well-known to describe a cycloid, i.e. the path traced by a point on the rim of a circle as the circle rolls along a straight line without slipping. It can further be show that the solution is given parametrically by $x=C(\theta-\sin\theta)$, $y=C(1-\cos\theta)$.
Now, this is where I am a little puzzled. All we have shown so far is that the functional $T$ is stationary when $y$ is a cycloid, and yet every single source I have found online claims that $y$ is an $\textit{absolute minimum}$ of $T$. Is this due to a misunderstanding on my part, or is this assertion indeed non-trivial and needs proving?
Right, it's non-trivial. To show the cycloid is the absolute minimum, you need to do more than solve the EL equations, just like with regular calculus, you have to do more than solve $f'(x)=0.$ The analogy in the Calculus of Variations is to work with the second variation (the analogy of the second derivative). You would have to show that the second variation is strongly positive.