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For fixed-endpoint variations, the second variation iswith $p=f_{vv}$, $r=f_{uv}$, $s=f_{uu}$ evaluated along $(x,\bar{u},{\bar{u}}^{\prime })$. The Jacobi equation, written without differentiating $r$ separately, isWhen $r$ is differentiable this is $-(p{z}^{\prime }{)}^{\prime }+(s-r^{\prime })z=0$.

Positive definiteness means $q(\phi )>0$ for every nonzero admissible variation. Uniform or coercive positivity of a second variation is the stronger assertion that $q(\phi )\ge c\int {\phi }^{\prime 2}$ for some fixed $c>0$. On a fixed-endpoint interval this is equivalent to a positive lower bound in the $W^{1,2}$ norm.

The strict Legendre implication uses the uniform interpretation of positivity. If $q\ge c\int {\phi }^{\prime 2}$, localized rapidly oscillating variations show $p(x)\ge c$: otherwise support them in an interval where $p-c<0$, so the derivative term of $q-c\int {\phi }^{\prime 2}$ is negative of order frequency squared, overwhelming its mixed and zero-order terms. Continuity extends the bound to the endpoints, giving $p>0$ on $[a,b]$. Under mere positive definiteness this strict conclusion is false: $f=u^2$ at $\bar{u}=0$ gives $q=2\int {\phi }^2>0$ but $p=0$. In that weaker convention only the ordinary Legendre necessary condition $p\ge 0$ follows. This distinction makes the implication precise.

Under the stated regular condition $p>0$, a conjugate point $c\in (a,b]$ to $a$ is an endpoint admitting a nonzero Jacobi solution with $z(a)=z(c)=0$. The disconjugacy form of the Jacobi positivity criterion says that no such point in $(a,b]$ exists if and only if there is a Jacobi solution strictly positive on all of $[a,b]$.

For the required identity, let that positive solution be $z$ and write $\phi =zh$. Set $A=pz{z}^{\prime }+r{z}^2$. The Jacobi equation givesExpanding the second-variation integrand therefore yieldsFor compactly supported $\phi$, the boundary term vanishes, provingThis is the Jacobi identity for the second variation. The divergence form is also appropriate under the original $C^1$ regularity; with $p>0$, the Euler-Lagrange equation gives the usual further regularity along the extremal.

For a fixed-endpoint variation $\phi \in W_0^{1,\infty }(a,b)$, the second variation iswhere all coefficients are evaluated at $(x,\bar{u}(x),{\bar{u}}^{\prime }(x))$. The notation ${\delta }^{2}I(\bar{u})>0$ 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.

PutAfter integration by parts, the form is $q(\phi )={\int }_a^b(P{\phi }^{\prime 2}+Q{\phi }^2)dx$. Its Jacobi equation isIn 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 $w(a)=0$, $P(a)w^{\prime }(a)=1$.

Under this regular Strengthened Legendre condition, the Jacobi positivity criterion isThe endpoint $b$ must be excluded as a conjugate point too. To prove necessity, suppose a Jacobi solution vanishes at $a$ and $c$. Define $\phi =w$ on $[a,c]$ and zero on $[c,b]$. It is a nonzero Lipschitz admissible variation, anda contradiction. For completeness, the converse follows from the Jacobi identity for the second variation: when the normalized $w$ has no further zero,The boundary contribution $[P(w^{\prime }/w){\phi }^2{]}_a^b$ vanishes, including the limit at $a$ since $w$ has a simple zero and $\phi =O(x-a)$. Equality forces $\phi =Cw$, and $\phi (b)=0$ with $w(b)\ne 0$ forces $C=0$.

Independently of regularity, positivity necessarily implies the ordinary Legendre necessary condition $P\ge 0$. 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 $q(\phi )=2\int {\phi }^2>0$ 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:

These conditions are necessary because $q>0$ gives $q_{\eta }(\phi )=q(\phi )+\eta \int {\phi }^{\prime 2}>0$, 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 $q\ge 0$ by letting $\eta \downarrow 0$. If $q(\phi )=0$, nonnegativity of $q(\phi +t\psi )$ 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 $\phi =0$. Thus the criterion treats degeneracy without silently assuming the strengthened condition.

Here $P=2$, $Q=-2{\lambda }^2$ and the Jacobi equation is $w^{\prime \prime }+{\lambda }^{2}w=0$. With $w(0)=0$ and $w^{\prime }(0)=1$, its solution is $w(x)=\sin (|\lambda |x)/|\lambda |$ for $\lambda \ne 0$, and $w(x)=x$ for $\lambda =0$. Its first positive zero is $\pi /|\lambda |$ when $\lambda \ne 0$.

If ${\lambda }^2<{\pi }^2/b^2$, no conjugate point occurs in $(0,b]$. The Jacobi positivity criterion givesSince $I_{\lambda }$ itself is this quadratic form divided by two, $I_{\lambda }(u)>0=I_{\lambda }(0)$ for every nonzero admissible $u$. Thus zero is the unique global minimizer. This argument establishes the result before using the sharp Poincaré inequality.

For continuous $P\ge 0,Q$, a scalar second variation $q=\int (P{\phi }^{\prime 2}+Q{\phi }^2)$ is strictly positive on nonzero fixed-endpoint Lipschitz variations if and only if every regularized form $q_{\eta }$ for $\eta >0$ has no conjugate point in $(a,b]$, and the original weak Jacobi boundary problem has no nonzero Lipschitz solution. The Jacobi positivity criterion applies to each regularized coefficient $P+\eta >0$. Letting $\eta$ decrease to zero gives $q\ge 0$. For a nonnegative quadratic form, $q(\phi )=0$ forces its mixed bilinear form with every test function to vanish, by considering $q(\phi +t\psi )$ for both signs of $t$. This is the weak Jacobi equation, so the kernel condition gives strict positivity. Necessity follows by the same argument in reverse. It distinguishes strict positivity from a uniform coercive estimate when the derivative coefficient degenerates.

Strict positivity of the highest-derivative coefficient on the closed parameter interval is the strengthened Legendre condition. Continuity then gives a uniform positive lower bound and makes the Jacobi equation a regular second-order equation. It permits the Jacobi positivity criterion. It is not necessary for mere strict positivity of a second variation: $\int {\phi }^2$ is strictly positive on nonzero variations even though its derivative coefficient vanishes.