Strengthened Weierstrass condition ($E(x,y,p,q)\ge 0$)

For a $C^1$ curve $u$, some $\delta >0$ must make the Weierstrass excess function nonnegative for every $x\in [a,b]$, every trial slope $q\in \mathbb{R}$, and all $y,p$ satisfying $|y-u(x)|<\delta$, $|p-u^{\prime }(x)|<\delta$. Thus the supporting-line property persists in a tube of nearby heights and reference slopes. The term “strengthened” here refers to this neighborhood requirement; strict excess for $q\ne p$ is a further condition.

Under the usual $C^2$ integrand regularity, an admissible $C^1$ curve satisfying the weak Euler-Lagrange equation, a coercively positive second variation, and the strengthened Weierstrass condition is a strong local minimizer of a variational integral. Strict excess away from the reference slope gives strict minimality. Coercive positivity means ${\delta }^{2}I(u)[\eta ,\eta ]\ge c\Vert \eta {\Vert }_{W^{1,2}}^2$ for every zero-trace variation and some $c>0$.

The second variation provides a neighboring field of extremals with slope $p(x,y)$. The associated endpoint-invariant Hilbert integral yields $I(v)-I(u)=\int E(x,v,p(x,v),v^{\prime })dx$ for uniformly close admissible $v$. The tube condition makes this integral nonnegative even when $v^{\prime }$ is large, explaining the passage from weak to strong minimality.