= Solution
For any $v\in M$, the segment $u+t(v-u)$ stays in $M$ for $0\leq t\leq1$. Since $u$ is a minimizer, the right derivative of its energy at $t=0$ is nonnegative. The <first variation> therefore gives the <variational inequality>
$$
\boxed{\int_0^l\bigl(2a(x)u'(x)(v'(x)-u'(x))+\gamma\cos(u(x))(v(x)-u(x))\bigr)\,dx\geq0
\quad(v\in M).}
$$
The factor two comes from the quadratic gradient term. Only a one-sided inequality follows because a variation pointing below the obstacle need not be admissible.
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