= Solution
For a fixed-endpoint variation $\varphi\in W^{1,\infty}_0(a,b)$, the <second variation> is
$$
\delta^2I(\bar u)(\varphi,\varphi)=\left.\frac{d^2}{dt^2}I(\bar u+t\varphi)\right|_{t=0}=\int_a^b\left(f_{vv}\varphi'^2+2f_{uv}\varphi\varphi'+f_{uu}\varphi^2\right)dx,
$$
where all coefficients are evaluated at $(x,\bar u(x),\bar u'(x))$. The notation $\delta^2I(\bar u)>0$ means this is strictly positive for every nonzero admissible variation. This is <pointwise> positive definiteness of the <quadratic form>, not an assertion of a prescribed uniform lower bound.
Put
$$
P=f_{vv},\qquad Q=f_{uu}-\frac d{dx}f_{uv}.
$$
After <integration by parts>, the form is $q(\varphi)=\int_a^b(P\varphi'^2+Q\varphi^2)\,dx$. Its <Jacobi equation> is
$$
\boxed{-(Pw')'+Qw=0.}
$$
In the regular case $P>0$ on $[a,b]$, a <conjugate point> $c\in(a,b]$ to $a$ is a point for which a nonzero solution satisfies $w(a)=w(c)=0$. Equivalently it is a zero of the normalized solution with $w(a)=0$, $P(a)w'(a)=1$.
Under this regular <Strengthened Legendre condition>, the <Jacobi positivity criterion> is
$$
\boxed{q>0\iff\text{there is no conjugate point to }a\text{ in }(a,b].}
$$
The endpoint $b$ must be excluded as a <conjugate point> too. To prove necessity, suppose a Jacobi solution vanishes at $a$ and $c$. Define $\varphi=w$ on $[a,c]$ and zero on $[c,b]$. It is a nonzero <Lipschitz> admissible variation, and
$$
q(\varphi)=\int_a^c(Pw'^2+Qw^2)dx=[Pww']_a^c=0,
$$
a contradiction. For completeness, the converse follows from the <Jacobi identity for the second variation>: when the normalized $w$ has no further zero,
$$
q(\varphi)=\int_a^bP\left(\varphi'-\frac{w'}w\varphi\right)^2dx.
$$
The boundary contribution $[P(w'/w)\varphi^2]_a^b$ vanishes, including the limit at $a$ since $w$ has a simple zero and $\varphi=O(x-a)$. Equality forces $\varphi=Cw$, and $\varphi(b)=0$ with $w(b)\ne0$ forces $C=0$.
Independently of regularity, positivity necessarily implies the ordinary <Legendre necessary condition> $P\geq0$. If $P<0$ on an interval, localized rapidly oscillating test functions have a negative <derivative> contribution of order frequency squared, dominating the bounded zero-order term. This contradicts positivity.
The assumptions stated here do not themselves impose $P>0$. That strengthened condition is a regularity hypothesis for the familiar Jacobi criterion, and is not necessary for strict positivity: the integrand $f=u^2$ gives $q(\varphi)=2\int\varphi^2>0$ with $P=0$. For a criterion covering the full possibly degenerate class, use <Regularized Jacobi criterion for a degenerate second variation>. The following conditions are necessary and sufficient:
* $P\geq0$;
* for every $\eta>0$, the regular equation $-((P+\eta)w')'+Qw=0$ has no <conjugate point> to $a$ in $(a,b]$;
* the original weak Jacobi boundary problem has no nonzero <Lipschitz> solution with zero endpoint values.
These conditions are necessary because $q>0$ gives $q_\eta(\varphi)=q(\varphi)+\eta\int\varphi'^2>0$, so the regular criterion just proved applies; a weak Jacobi solution tested against itself would give $q=0$. They are sufficient as well: the regular criterion gives $q_\eta>0$ for every $\eta>0$, hence $q\geq0$ by letting $\eta\downarrow0$. If $q(\varphi)=0$, nonnegativity of $q(\varphi+t\psi)$ for every real $t$ forces its mixed bilinear term to vanish for every test $\psi$. This is precisely the weak <Jacobi equation>. The third condition then forces $\varphi=0$. Thus the criterion treats degeneracy without silently assuming the strengthened condition.
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