Let $x: \mathbb{R} \rightarrow \mathbb{R}$ be $C^2$ and consider the integral:
$$\int_0^\pi x^\prime(t)^2 - x(t)^2 dt$$
subject to the constraint $x(0) = x(\pi) = 0$.
We see that $x(t) = 0$ for all $t \in [0,\pi]$ makes the above integral $0$.
I don't know whether this integral can be negative though, and I have tried a lot of functions to make it negative but failed (examples like $x(t) = \sin(\tfrac{x}{n}) - \frac{x}{\pi}\sin(\frac{\pi}{n})$ for $n \in \mathbb{N}$, which seems to make the integral arbitrarily approach $0$ but never become negative).
I have therefore come to believe that this integral cannot be negative and that $x(t) = 0$ is a global minimizer. It seems like using the boundary conditions together with some calculus theorems like Mean Value Theorem or Rolle's Theorem might be key to showing that the integral cannot be negative, but I have not been successful.
As for the context, this is a calculus of variations problem I encountered.
A way to solve this using calculus of variation is to consider a perturbation of the interval $[0,\pi-\epsilon]$ for some small $\epsilon$. For the minimization problem on this interval with zero endpoints, the Lagrangian can be turned into a convex function by adding the term $-\frac{d}{dt}x(t)^2\cot(t+\delta)$, for some positive $\delta<\epsilon$, this term can be attainted using an everywhere positive solution of the Jacobi differential equation (this is why we need to perturb the interval). Using convexity, the zero function is a global minimizer for the problem on $[0, \pi-\epsilon]$. Taking limit $\epsilon \to 0$ finishes the problem.
Another way is to use the Hamilton-Jacobi equation and the Verification Theorem.