= Solution
\b[Zero is a strict weak local minimizer, but is not a strong local minimizer.] Put $D=\int_0^1|u'|^2\,dx$. Since $u(0)=0$, the <Cauchy-Schwarz inequality> gives $|u(x)|^2\leq xD$, and therefore
$$
\int_0^1x^{2010}u(x)^2\,dx\leq\frac{D}{2012}.
$$
For sufficiently small $\|u'\|_\infty$, $\sin(u'^2)\geq\tfrac12u'^2$ <pointwise>. Hence
$$
I(u)\geq\left(\frac12-\frac1{2012}\right)D\geq0,
$$
with equality only when $u=0$. Thus zero is a strict <weak local minimizer of a variational integral>.
For the strong claim, define a <Lipschitz> triangular wave
$$
u_N(x)=\sqrt\pi\,\operatorname{dist}(x,N^{-1}\mathbb Z).
$$
It has zero endpoint values, $\|u_N\|_\infty=\sqrt\pi/(2N)\to0$, and $u_N'=\pm\sqrt\pi$ almost everywhere. Thus $\sin(u_N'^2)=\sin\pi=0$, while $u_N$ is nonzero on a set of positive measure. Consequently
$$
\boxed{I(u_N)=-\int_0^1x^{2010}u_N^2\,dx<0=I(0).}
$$
These uniformly small competitors disprove the <strong local minimizer of a variational integral> property.
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