Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 16A / b / Solution

The integrandis nonnegative, so $I[y]\ge 0$. If $I[y]=0$, both continuous summands vanish everywhere. Hence $y^{\prime }=0$, so $y$ is constant; the common boundary value forces $y(x)=a$. Conversely, $y\equiv a$ has $y^{\prime }=0$ and $U(a)=0$, so $I[y]=0$. Therefore

This agrees with part (a): at $y=a$, the first variation vanishes because $y^{\prime \prime }=U(a)=0$, andThus the constant configuration is a stationary local minimum as well as the global minimum identified directly.