 Solution

With fixed endpoint values, a smooth extremal satisfies the Euler-Lagrange equation $F_y-(d/dx)F_{y^{\prime }}=0$. If $F$ has no explicit $x$ dependence, differentiation givesThus $F-y^{\prime }{F}_{y^{\prime }}$ is constant, the Beltrami identity.

For this integrand, $F_y=2y$ and $F_{y^{\prime }}=1+2y^{\prime }$, so the equation is $y^{\prime \prime }=y$. ConsequentlyThe two constants are fixed uniquely by the prescribed values at distinct endpoints. Equivalently, writing these values as $y_1,y_2$,Its first integral is $y^2-y^{\prime 2}$, which is constant. Using only that first integral without the Euler-Lagrange equation could admit spurious constant functions, so the differential equation was used directly.

For any nonzero endpoint-vanishing perturbation $\eta$, the linear terms in $J[y+\eta ]-J[y]$ vanish by the Euler-Lagrange equation and integration by parts; also $\int {\eta }^{\prime }=0$. The remainder isHence the extremal is the unique strict global minimum, and cannot be a maximum.