= Solution
A <strong local minimizer of a variational integral> is an admissible $\bar u$ for which the same conclusion holds whenever $\|v-\bar u\|_\infty<\rho$. Every strong local minimizer is weak, because Sobolev closeness implies uniform closeness. A <global minimizer> is both, and the converse implication from weak to strong is not generally valid.
If the integrand is independent of $u$, a weak local minimizer has vanishing <first variation>:
$$
\int_a^bf_v(x,\bar u'(x))\varphi'(x)dx=0\quad(\varphi\in C_0^\infty(a,b)).
$$
The bounded <function> $h(x)=f_v(x,\bar u'(x))$ therefore has zero <distributional derivative>, so is constant <almost everywhere>. For an elementary test-<function> argument, any smooth compactly supported $\psi$ of <integral> zero is the <derivative> of a compactly supported smooth $\varphi$. Hence $\int h\psi=0$; subtracting a fixed unit-<integral> <test function> from arbitrary tests shows that $h$ agrees with one constant against every test. The fundamental lemma then gives
$$
\boxed{f_v(x,\bar u'(x))=c\quad\text{almost everywhere}.}
$$
This is the <first integral for a variational integrand independent of the function value>.
A <point of convexity> $v_0$ of a differentiable real <function> $g$ has a globally supporting tangent: $g(v)\geq g(v_0)+g'(v_0)(v-v_0)$ for every $v$. The <Weierstrass necessary condition> for a strong minimum says that $\bar u'(x)$ is such a point for $v\mapsto f(x,\bar u(x),v)$ for almost every $x$. Equivalently, the <Weierstrass excess function> is nonnegative at every trial slope.
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