= Solution
Along the reference curve put $p=f_{vv}$, $r=f_{uv}$, $s=f_{uu}$. The <second variation> is
$$
q(\phi)=\delta^2I(\bar u)[\phi,\phi]=\int_a^b(p\phi'^2+2r\phi\phi'+s\phi^2)dx.
$$
Strict positivity means $q(\phi)>0$ for every nonzero zero-trace variation. Under the given regular Legendre condition $p>0$, the no-conjugate-point criterion also supplies a uniform coercive lower bound. The <Jacobi equation> is $-(pz'+rz)'+rz'+sz=0$. A point $c>a$ is conjugate to $a$ when a nonzero solution vanishes at both $a$ and $c$. In this regular setting, absence of <conjugate points> in $(a,b]$ is equivalent to existence of a Jacobi solution positive throughout $[a,b]$.
Choose such a positive solution $z$. The given square identity extends from smooth compactly supported variations to $H_0^1$ by density and boundedness of its coefficients. With $v=\phi/z$ it reads $q(\phi)=\int pz^2v'^2$. The positive minima of $p,z$ bound this below by $c\|v'\|_2^2$. A zero-trace <function> obeys $\|v\|_2\leq C\|v'\|_2$, while $\phi'=zv'+z'v$ and $\phi=zv$ give $\|\phi\|_{H^1}\leq C_1\|v'\|_2$. Consequently
$$
\boxed{q(\phi)\geq c_1\|\phi\|_{H^1}^2>0\quad(\phi\ne0).}
$$
In particular the required <second variation> is positive for all nonzero Lipschitz zero-trace variations.
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