Solution
= Solution
\b[Zero is neither a weak nor a strong local minimizer.] Take $u_\varepsilon(x)=-\varepsilon\sin(\pi x)$. It tends to zero in $W^{1,\infty}$ and has zero endpoint values. Since $\sqrt{1+t}\leq1+t/2$ for $t\geq0$,
$$
I(u_\varepsilon)-I(0)\leq\frac{\varepsilon^2\pi^2}{4}-\frac{4018}{\pi}\varepsilon<0
$$
for sufficiently small positive $\varepsilon$. Thus even weakly close admissible competitors decrease the energy.