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Take a variation $x+\epsilon \xi$ with $\xi (a)=\xi (b)=0$. Expanding the action givesAn integration by parts and the fixed endpoint conditions give the first variationThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationand the second variation is

Linearizing the equation of motion about $x$ givesso the Jacobi equation isWhen $u$ has no zero on $[a,b]$,The total derivative integrates to zero because $\xi (a)=\xi (b)=0$, hence

For the simple harmonic oscillator, the Jacobi equation is $\ddot{u}+{\omega }^{2}u=0$. Put $t_0=(a+b)/2$ and chooseIf $b-a<\pi /\omega$, then $|\omega (t-t_0)|<\pi /2$ throughout $[a,b]$, so $u$ is positive there. The preceding square identity proves that the classical path is a local minimum of the action whenever the elapsed time is less than half an oscillation period.

The Euler--Lagrange equation isPutting $u=\cos (\varphi /2)$ and simplifying gives

Since $1-\cos \varphi =2(1-u^2)$, $1+\cos \varphi =2u^2$, and ${\dot{u}}^2=(1-u^2){\dot{\varphi }}^2/4$, the transformed functional isIts Euler--Lagrange equation is $\ddot{u}+(g/4a)u=0$, exactly the preceding equation.

For endpoint-vanishing $\eta$, the second variation isWriting $\eta ={\sum }_{n\ge 1}{c}_n{e}_n$ and using orthonormality givesIt is positive definite when $T<\pi /\omega$ and has a negative $e_1$ direction when $T>\pi /\omega$. Therefore

The functional is unbounded above. Let $y_0$ be the positive affine interpolant between the two endpoint values, and choose a smooth nonnegative function $\varphi$ that vanishes at the endpoints and is positive somewhere inside. The admissible curves $y_M=y_0+M\varphi$ satisfyThe claimed minimum conclusion does not follow and is false in general. A stationary point may have a negative second variation, even though the functional is unbounded above. Here is an explicit example within the allowed positive-graph class.

Take the stationary profile $y(x)=\cosh x$ on $[-2,2]$, and the endpoint-zero variation $\eta (x)=\cosh x-\cosh 2$. For sufficiently small perturbation parameter $s$, $y+s\eta$ remains positive and preserves both endpoint values. Put $C=\cosh 2$. Differentiating the area twice and integrating its mixed term by parts givesThis is strictly negative: the first term is less than $2\pi C$, while $(C+C^{-1})\tanh 2>\pi$. For example $e^2>7$ gives $C>7/2$ and $\tanh 2>24/25$, whose product already exceeds $\pi$. Thus nearby positive profiles have smaller area. This catenoid is not a local minimum.

The corrected conclusion is that any smooth positive minimizer, if attained, must have the catenoid form. That necessary condition is not sufficient, and existence or stability requires a separate argument.

Along the reference curve put $p=f_{vv}$, $r=f_{uv}$, $s=f_{uu}$. The second variation isStrict positivity means $q(\varphi )>0$ 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 $-(p{z}^{\prime }+rz{)}^{\prime }+r{z}^{\prime }+sz=0$. 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 $v=\varphi /z$ it reads $q(\varphi )=\int p{z}^2{v}^{\prime 2}$. The positive minima of $p,z$ bound this below by $c\Vert v^{\prime }{\Vert }_2^2$. A zero-trace function obeys $\Vert v{\Vert }_2\le C\Vert v^{\prime }{\Vert }_2$, while ${\varphi }^{\prime }=z{v}^{\prime }+z^{\prime }v$ and $\varphi =zv$ give $\Vert \varphi {\Vert }_{H^1}\le C_1\Vert v^{\prime }{\Vert }_2$. ConsequentlyIn particular the required second variation is positive for all nonzero Lipschitz zero-trace variations.

The Jacobi equation is $z^{\prime \prime }+{\lambda }^{2}z=0$. For $|\lambda |<\pi$ the function $z=\cos (\lambda (x-1/2))$ is positive on the closed interval. The preceding square identity giveswith $z=1$ if $\lambda =0$. As in part (a), its positive minimum and bounded derivative provide a constant $\gamma >0$ with $I(u)\ge \gamma \Vert u{\Vert }_{H^1}^2$. The second variation at zero is $2I$, so is positive, and zero is the unique global minimizer.

For a fixed zero-trace $u$, positivity of $\int u^{\prime 2}-{\lambda }^2\int u^2$ for all ${\lambda }^2<{\pi }^2$ gives, by letting ${\lambda }^2\uparrow {\pi }^2$,This derivation did not assume the sharp inequality in advance. For equality, the nonnegative quadratic form $Q(u)=\int (u^{\prime 2}-{\pi }^2{u}^2)$ has $Q(u)=0$. Expanding $Q(u+t\varphi )\ge 0$ for both signs of $t$ forces $\int u^{\prime }{\varphi }^{\prime }-{\pi }^{2}u\varphi =0$ for every test. Thus $u^{\prime \prime }+{\pi }^{2}u=0$, and the zero endpoint values give exactly $u(x)=C\sin (\pi x)$, including $C=0$. These functions do give equality.

A global minimizer has $I(\bar{u})\le I(v)$ for every admissible $v$. Global minimality implies strong local minimality, which implies weak local minimality. The reverse implications are not valid in general.

Where $f_v$ exists, the strengthened Weierstrass condition says that for some $\delta >0$, the excessis nonnegative for every $x\in [a,b]$, every trial slope $q\in \mathbb{R}$, and all $y,p$ with $|y-\bar{u}(x)|<\delta$, $|p-{\bar{u}}^{\prime }(x)|<\delta$. This is a neighborhood condition in height and reference slope, not just an inequality on the reference curve.

The classical Weierstrass sufficient condition, with the usual smooth integrand regularity in addition to continuity, states that an admissible $C^1$ curve satisfying the Euler-Lagrange equation, a coercively positive second variation, and this strengthened condition is a strong local minimizer. A strict excess away from the reference slope gives the corresponding strict conclusion. The differentiability assumptions are part of that classical sufficient theorem; continuity alone does not define its first and second variations.

With coefficients evaluated along $(x,\bar{u},{\bar{u}}^{\prime })$, put $p=f_{vv}$, $r=f_{uv}$ and $s=f_{uu}$. The second variation isThe Jacobi equation in divergence form is $-(p{z}^{\prime }+rz{)}^{\prime }+r{z}^{\prime }+sz=0$; if $r$ is differentiable it is $-(p{z}^{\prime }{)}^{\prime }+(s-r^{\prime })z=0$. A conjugate point $c>a$ admits a nonzero Jacobi solution vanishing at $a,c$.

There are two positivity conventions worth distinguishing. Strict pointwise positivity means $q(\varphi )>0$ for each nonzero fixed-endpoint variation. Coercive positivity means $q(\varphi )\ge c\int {\varphi }^{\prime 2}$ with one $c>0$ for all such variations; the endpoint Poincaré estimate makes this equivalent to a positive lower bound in the full $H^1$ norm. Under the regular Legendre hypothesis $p>0$, the necessary and sufficient Jacobi criterion is no conjugate point in $(a,b]$; in that regular case it gives both strict and coercive positivity. Equivalently the Jacobi equation has a solution positive on $[a,b]$, and the square identitygives positivity and a uniform derivative lower bound by the positive minimum of $z$ and boundedness of $z,z^{\prime }$. Conversely a conjugate Jacobi field, extended by zero beyond its second zero, is a nonzero Lipschitz variation with $q=0$, contradicting either positivity convention.

Coercivity itself forces $p>0$: local high-frequency variations make the leading coefficient at least the coercivity constant. Mere strict positivity need not force it, as $q(\varphi )=\int {\varphi }^2$ shows. If that weaker convention is intended without regularity, a complete criterion is $p\ge 0$, no conjugate point for every regularized form $q_{\eta }=q+\eta \int {\varphi }^{\prime 2}$ with $\eta >0$, and no nonzero zero-endpoint Lipschitz weak solution of the original Jacobi equation. The regularized criteria imply $q_{\eta }\ge 0$, hence $q\ge 0$ by passage to the limit. For a nonnegative quadratic form, $q(\varphi )=0$ forces all mixed terms with test variations to vanish, giving exactly the weak Jacobi kernel; excluding it gives strict positivity. Necessity follows in reverse. This specifies the degenerate case without imposing an unstated strict Legendre condition.

For weak local minimality, sufficient conditions are the Euler-Lagrange equation and a coercively positive second variation with the stated smoothness. Taylor's remainder is bounded by $\epsilon \Vert \varphi {\Vert }_{H^1}^2$ for small $W^{1,\infty }$ perturbations, so the positive quadratic term dominates it. These are the classical weak minimum sufficient conditions.

At zero the quadratic form isFor $\lambda \ge 0$ it dominates $\int u^{\prime 2}$. For $-{\pi }^2<\lambda <0$ set $\mu =\sqrt{-\lambda }<\pi$ and $z(x)=\cos (\mu (x-1/2))$, positive on the closed interval and satisfying $z^{\prime \prime }+{\mu }^{2}z=0$. The Jacobi identity gives $Q_{\lambda }(u)=\int z^2[(u/z{)}^{\prime }{]}^2$. To obtain a uniform lower bound, let $m=\min z>0$, $K=\max |z^{\prime }|$ and $v=u/z$. Since $\max z=1$ and $\int v^2\le \frac{1}{2}\int v^{\prime 2}$,so $Q_{\lambda }(u)\ge m^2\int u^{\prime 2}/(K^2+2)$.

Finally $I(u)=Q_{\lambda }(u)+O(\int u^4)$ near zero, because ${\sin }^{2}u=u^2+O(u^4)$. The remainder is at most a constant times $\Vert u{\Vert }_{\infty }^2\int u^2\le \frac{1}{2}\Vert u{\Vert }_{\infty }^2\int u^{\prime 2}$. It is dominated by the coercive bound for small perturbations. Thus the second variation is positive and zero is a strict weak local minimizer for every $\lambda >-{\pi }^2$. This argument has not assumed the sharp Poincaré inequality that is to be deduced next.

The functional is unbounded above. Let $\ell$ be the positive affine interpolant of the endpoint values and choose a nonnegative, nonzero, continuously differentiable $h$ vanishing at both endpoints. The admissible functions $y_M=\ell +Mh$ satisfyThis excludes a global maximum, but does not prove that a stationary curve is a minimum. The concluding assertion is false without additional conditions.

An explicit counterexample uses $x_1=-2$, $x_2=2$, $y_1=y_2=\cosh 2$ and the stationary function $y=\cosh x$. Its second variation, for a perturbation $h$ vanishing at the endpoints, isChoose $h=1-x^2/4$. Since ${\mathrm{sech}}^{2}x\le 4e^{-2|x|}$, the positive contribution is at most ${\int }_{\mathbb{R}}{x}^2{e}^{-2|x|}dx=1/2$. On $[-1,1]$, $h^2\ge 9/16$, so the magnitude of the negative contribution is at least $(9/8)\tanh 1>9/16>1/2$. Thus ${\delta }^{2}J[h]<0$. Small perturbations in this direction preserve positivity and the endpoint data while decreasing surface area. Consequently this catenoid is not a local minimum. The valid conclusion from part (ii) is a necessary stationary form; minimization requires a separate stability and existence analysis.

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.

The second variation of the Sine-Gordon theory action about its static kink gives ${\Delta }_x=-{\partial }_x^2+1-2{\mathrm{sech}}^{2}x$. Let $A={\partial }_x+\tanh x$. Then ${\Delta }_x=A^{†}A$ and $A{A}^{†}=-{\partial }_x^2+1$. Thus ${\Delta }_x$ is nonnegative, with its translational zero mode of a sine-Gordon kink and a continuum at ${\omega }^2=k^2+1$. The construction is the supersymmetric factorization of the one-soliton potential.

Two equal circular rings of radius $R$ in parallel planes at $z=\pm L$ provide a boundary for an open film. In the ideal axisymmetric model, a stationary connected film has the catenoid profileThe catenoid existence threshold for equal rings explains why this equation can have two solutions, one solution, or none. The cancellation of principal curvatures holds for both solutions whenever they exist; the mean curvature equation alone does not select a stable minimal surface.

This ideal geometry uses $R=1$ and $L/R=0.60$. The two profiles have different waist radii $a$, although both have $H=0$. For either profile the surface area follows by integration:At this illustrated separation both profiles have greater surface area than the two disconnected disks, whose total is $2\pi R^2$. This comparison disproves a claim of global area minimization in a class allowing disconnected competitors; it does not determine the sign of the second variation for small variations of a connected film.

A reported experiment used coaxial wire rings of radius $4\mathrm{cm}$, drawn from a surfactant solution and slowly separated until collapse. The solution contained tetradecyltrimethylammonium bromide in deionized water, with glycerol added to vary viscosity. Fluorescein enabled fluorescence imaging under cyan LED arrays; a Phantom V641 camera recorded 4,000 or 5,600 frames per second. Extracted contours gave neck radius, attachment angle, and inferred surface area. They showed neck narrowing and eventual breakup, rather than indefinite persistence of a stationary catenoid. Comparing solutions across viscosities helped distinguish liquid dissipation from the surrounding air's inertia, but the ring displacement beyond the last stable state was not precisely controlled. These are collapse observations, not direct measurements of equilibrium $H$. ^[ref]

A proposed classroom measurement would instead hold two equal rings coaxial on a translation stage, leave both sides open to the same gas, and photograph the settled silhouette beside a scale. Fit $r(z)=a\cosh (z/a)$ and inspect fit residuals before inferring cancellation of the principal curvatures. Repeat at several separations and with the camera rotated about the axis to test rotational symmetry. Keep solution composition, ring wetting, and elapsed drainage time comparable, and shield the film from drafts. Film weight, drainage, evaporation, surfactant gradients, and rupture can invalidate the ideal soap film pressure balance. A silhouette fit alone neither measures the pressure jump nor tests all possible perturbations of a stable minimal surface.

For an ideal soap film with equal constant surface tension $\gamma$ on both faces, the effective tension is $\tau =2\gamma$. Choose a unit normal $N$ from the minus side to the plus side, and $\Delta p=p_{-}-p_{+}$. Translating the Young–Laplace equation to $S=-dN$ givesA single interface uses $\tau =\gamma$. An outward-oriented spherical soap bubble has $H=-1/R$, giving $\Delta p=4\gamma /R$. The positive-convex curvature sum in the Young–Laplace equation article is $-2H$. ^[ref]

An open film with the same resting gas on both sides has zero pressure jump. Neglecting film weight, soap film pressure balance gives $H=0$: a minimal surface. A pressurized bubble instead has nonzero mean curvature. These statements assume equilibrium, constant surface tension, and negligible normal viscous stress. Surfactant gradients or motion require the fuller normal stress balance. ^[ref]

The variational outline is short. With fixed boundary, a normal displacement $fN$ gives $\delta A=-2{\int }_{M}HfdA$ by the first variation of area formula. For a closed interface oriented outward, $\delta V={\int }_{M}fdA$, so stationarity of $\tau A-\Delta pV$ gives $-2\tau H-\Delta p=0$. This is a first variation calculation; testing for a stable minimal surface needs the second variation, and deciding a global minimum needs a comparison of competitors.

A stable minimal surface has nonnegative second variation of surface area for every admissible normal variation $fN$ fixing the boundary. For a two-sided minimal surface in three-dimensional Euclidean space, the quadratic form isHere ${\nabla }_{M}f$ is the gradient for the induced Riemannian metric. The gradient term penalizes rapid changes in the displacement; the principal curvature term can make the second variation negative. A plane has only the nonnegative gradient term. Nonnegativity alone, especially with zero modes, does not prove a strict local minimum or a global minimum.

For a finite catenoid between fixed rings, the answer depends on the ring separation, waist radius, and allowed variations. The axisymmetric second variation of a catenoid tests a specified restricted class. Comparing with two disconnected disks tests a different class of competitors, including a change in topology. Consequently, a stationary catenoid can have greater surface area than two disks without immediately collapsing under every small perturbation.

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.

This excess measures the height of the slope-dependent integrand above its tangent line at the reference slope $p$, evaluated at a trial slope $q$. A convex function of slope has nonnegative excess. At a smooth strong local minimizer, localized needle variations force nonnegative excess along the minimizing curve for every trial slope. It detects fast-gradient competitors that the second variation around a fixed derivative can miss.

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.