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For the catenoid profile $y(x)=E\cosh (x/E)$ spanning fixed-endpoint rings, the second variation of $F[y]=2\pi \int y\sqrt{1+y^{\prime 2}}dx$ isThis refers to axisymmetric graph variations, not all possible changes in surface topology. Strict positivity for every nonzero admissible variation gives a strict local minimum in the $C^1$ topology: the positive lowest Dirichlet eigenvalue of the regular Sturm-Liouville problem gives coercivity, and sufficiently small coefficient perturbations preserve it. The zero variation must be excluded from a strict-positivity hypothesis.

A perturbation preserving both values and slopes has $\eta =\eta \prime =0$ at the endpoints. Apply the Poincaré inequality to $\eta$ and then to $\eta \prime$. The resulting inequality makes a quadratic second variation proportional to $\int ({\eta }^2-(\eta \prime \prime {)}^2)$ nonpositive. If a functional is quadratic, its exact expansion about a stationary function then proves a global maximum.

For fixed-endpoint variations, a second variation is coercively positive if the displayed estimate holds for one constant $c>0$ and every admissible variation. The endpoint Poincaré inequality makes this equivalent to a positive lower bound in the $W^{1,2}$ norm. It is stronger than strict positivity on each nonzero variation: $q(\phi )=\int {\phi }^2$ is strictly positive, but high-frequency variations make its ratio to $\int {\phi }^{\prime 2}$ tend to zero.

For $q=\int (p{\phi }^{\prime 2}+2r\phi {\phi }^{\prime }+s{\phi }^2)$ with continuous coefficients, coercive positivity implies $p\ge c$. Indeed, if $p-c<0$ on an interval, multiply a smooth function supported there by $\sin (nx)$. Its negative highest-derivative contribution has order $n^2$, while the mixed term has order at most $n$ and the remaining term is bounded. This contradicts the displayed inequality for large $n$. In particular coercive positivity implies the Strengthened Legendre condition, whereas mere strict positivity does not.

For an equilibrium with $P_{Ii}=W_{,F_{iI}}+q{\varphi }_{,F_{iI}}$ and dead loads, the constrained second variation on ${\varphi }_{,F}:\delta F=0$ isIndeed, a constraint-preserving perturbation path has ${\varphi }_{,F}:{\delta }^{2}F=-{\varphi }_{,FF}[\delta F,\delta F]$, and stationarity of the total potential supplies the multiplier term. Linearized kinetic energy plus $\mathcal{Q}/2$ is conserved. Positivity gives the usual energy-norm linear stability; control in a specified displacement norm additionally requires the corresponding coercivity and treatment of rigid-motion modes.

For positive mass $m$ and angular frequency $\omega$, the Euclidean action $S_E=\frac{m}{2}{\int }_0^{\mathcal{T}}({\dot{x}}^2+{\omega }^2{x}^2)d\tau$ with $x(0)=x(\mathcal{T})=x$ has the unique minimizing path displayed. The Euler-Lagrange equation is ${\ddot{x}}_0={\omega }^2{x}_0$, and the positive second variation proves uniqueness. On this path the integrand is $\frac{m}{2}d(x_0{\dot{x}}_0)/d\tau$, giving $S_E[x_0]=m\omega x^2\tanh (\omega \mathcal{T}/2)$. Increasing the interval of imaginary time lets its interior approach the minimum of the potential energy even while both endpoints remain fixed.

Expanding an action about a classical stationary field and retaining its second variation gives a Gaussian path integral. Its nonzero fluctuation modes determine a functional determinant and a one-loop soliton mass correction. Symmetry-generated zero modes in field theory must be removed from that determinant and integrated as collective coordinates.

If $w$ is a nonvanishing solution of $-(P{w}^{\prime }{)}^{\prime }+Qw=0$, expanding the right-hand side gives the displayed identity. Integrating expresses a scalar second variation as a square plus a boundary term. A simple zero of $w$ at a fixed endpoint causes no difficulty for Lipschitz endpoint-vanishing variations, since $\phi =O(x-a)$ makes the boundary term tend to zero. This identity is the scalar form of a ground-state transform for a weighted Dirichlet energy.

A nonnegative second variation requires its highest-derivative coefficient $P=f_{vv}$ to be nonnegative. If it were negative on an interval, a smooth compactly supported factor times $\sin (nx)$ would have negative derivative energy of order $n^2$, while its zero-order energy stays bounded. Large $n$ would contradict nonnegativity. This is a local necessary condition; it does not rule out negative directions caused by the lower-order terms.

A minimal surface has vanishing mean curvature vector. For a smooth surface in three-dimensional Euclidean space, this means $H=0$, equivalently $k_1+k_2=0$. Thus Gaussian curvature is $K=-k_1^2\le 0$, but neither principal curvature needs to vanish. A plane and a catenoid both satisfy the condition, with very different Gaussian curvatures.

By the first variation of area formula, a minimal surface is a stationary point of surface area for every smooth variation that fixes its boundary, or has compact support in its interior. The word “minimal” specifies this differential equation; it does not guarantee a global minimum of surface area, or even a local minimum. A stable minimal surface imposes a further condition on the second variation.

A catenoid has zero mean curvature, but need not minimize area for its fixed boundary rings. For the profile $y(x)=\cosh x$ on $[-2,2]$, take an endpoint-zero variation $\eta (x)=\cosh x-\cosh 2$. The second variation of the rotational area isTo derive the integral, differentiate $y\sqrt{1+y^{\prime 2}}$ twice, then integrate its mixed term by parts; the endpoint term vanishes. To evaluate it, substitute ${\eta }^{\prime 2}-{\eta }^2=-1-C^2+2C\cosh x$, using the antiderivatives of $\mathrm{sech}x$ and ${\mathrm{sech}}^{2}x$. The result is negative because $\arctan (\sinh 2)<\pi /2$ and $(C+C^{-1})\tanh 2>\pi$.

Small positive and negative perturbations remain positive and preserve the boundary values. The first variation vanishes, while the negative second variation gives nearby smaller areas. Thus being a minimal surface in the zero-mean-curvature sense is a stationarity condition, not a guarantee of area minimization for arbitrary boundary separations.

Put $\mathcal{T}=\beta \hslash >0$, the interval of imaginary time, and assume $m,\omega >0$. For a variation $x_0+ϵ\eta$ with $\eta (0)=\eta (\mathcal{T})=0$, the first variation of the Euclidean action isThe boundary term vanishes. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationIts sign differs from the real-time harmonic oscillator equation because the Euclidean action contains positive kinetic energy and positive potential energy. The general solution is $A\cosh (\omega \tau )+B\sinh (\omega \tau )$. Applying the two endpoint conditions gives $A=x$ and $B=-x\tanh (\omega \mathcal{T}/2)$. Equivalently, the equal-endpoint Euclidean oscillator path isThis form displays reflection about the midpoint and satisfies both endpoints immediately. On this path,Thus the Euclidean action reduces to a boundary term. Since ${\dot{x}}_0(\mathcal{T})=\omega x\tanh (\omega \mathcal{T}/2)$ and ${\dot{x}}_0(0)=-\omega x\tanh (\omega \mathcal{T}/2)$,The positive quadratic form $m{\int }_0^{\mathcal{T}}({\dot{\eta }}^2+{\omega }^2{\eta }^2)d\tau$ is the second variation, so this stationary path is the unique minimum, rather than merely an arbitrary stationary point.

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.

At zero, the sixth-power term has zero second derivative, soFor $0<\lambda <1$, the Jacobi solution $z(x)=\cos (\lambda (x-\pi /2))$ is strictly positive on $[0,\pi ]$. The preceding identity gives $Q_{\lambda }(u)=\int z^2[(u/z{)}^{\prime }{]}^2>0$ for nonzero zero-trace $u$.

To control the nonlinear remainder, a uniform lower bound follows directly. Let $m=\min z>0$, $K=\max |z^{\prime }|$, and $v=u/z$. Since $\max z=1$ and $\int v^2\le ({\pi }^2/2)\int v^{\prime 2}$ by the elementary endpoint estimate,Thus $Q_{\lambda }(u)\ge c_{\lambda }\int u^{\prime 2}$ with $c_{\lambda }=m^2/(K^2{\pi }^2+2)>0$. Also $\int u^6\le \Vert u{\Vert }_{\infty }^4({\pi }^2/2)\int u^{\prime 2}$, soThis is positive for every nonzero admissible $u$ sufficiently close to zero in $W^{1,\infty }$. Therefore zero has positive second variation and is a strict weak local minimizer. No sharp Poincaré inequality has been assumed in this proof.

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.

The variational terminology refers to the size of a neighborhood, rather than weak versus strong convergence in a Banach space. The definitions in the two numbered children giveThe second implication holds because smallness in $W^{1,\infty }$ implies smallness in the uniform norm. Reverse implications need not hold.

For differentiable $f$ in its slope variable, define the Weierstrass excess functionThe strengthened Weierstrass condition means that some $\delta >0$ givesIt requires support at nearby heights and nearby reference slopes, while the trial slope $q$ is arbitrary. Requiring this only exactly on the curve is the unstrengthened necessary condition.

A standard Weierstrass sufficient condition assumes the usual $C^2$ smoothness of $f$, $u\in \mathcal{A}\cap C^1$, the weak Euler-Lagrange equation, a coercively positive second variation, and the strengthened condition above. Explicitly, for some $c>0$,for zero-endpoint variations. Then $u$ is a strong local minimizer. If the excess is strictly positive for $q\ne p$ throughout the tube, the strong minimum is strict. The second-variation hypothesis supplies a neighboring field of extremals. Its slope $p(x,y)$ gives the comparison identityfor uniformly close admissible $v$. This explains how the condition controls competitors with large derivatives.

Linearization gives ${\theta }^{\prime \prime }+{\pi }^2\lambda \theta =0$. The two homogeneous Dirichlet conditions select $\boxed{\lambda =n^2,\theta =A\sin (n\pi \xi ),n=1,2,\dots }$. For any small trial amplitude in this mode, ${\int }_0^1{\sin }^2(n\pi \xi )={\int }_0^1{\cos }^2(n\pi \xi )=1/2$, soFor $\lambda <1$ every nonzero quadratic variation is positive by the sharp Poincaré inequality. For $\lambda >1$, the first mode gives negative energy at sufficiently small amplitude, so zero cannot be a local minimum.

The printed strict “only if” conclusion omits the critical case. At $\lambda =1$ the quadratic test is degenerate, not negative. In fact zero is still the unique minimum of the full displayed functional: use $1-\cos \theta \le {\theta }^2/2$ and $\int {\theta }^{\prime 2}\ge {\pi }^2\int {\theta }^2$ to get $U\ge 0$, with strict inequality for a nonzero continuous function because $1-\cos t<t^2/2$ when $t\ne 0$. In particular on the critical first mode,Thus strict positivity of the second variation holds for $\lambda <1$, while local minimality of the full energy holds also at $\lambda =1$. This is the distinction captured by critical-load minimality of the angular Euler strut.

For fixed nominal boundary tractions and dead body forces, use the total potentialThe dead-load terms are linear in the deformation map. Let $x(\epsilon )$ be a constraint-preserving perturbation path through equilibrium, with $H=\delta F$ and $K={\delta }^{2}F$. Differentiating $\varphi (F)=0$ twice gives ${\varphi }_{,F}:H=0$ and ${\varphi }_{,F}:K=-{\varphi }_{,FF}[H,H]$. Equilibrium has nominal stress tensor $P_{Ii}=W_{,F_{iI}}+q{\varphi }_{,F_{iI}}$. Its virtual-work relation, applied to ${\delta }^{2}x$, gives the otherwise remaining first-derivative contribution ${\Pi }^{\prime }[{\delta }^{2}x]=-\int q{\varphi }_{,F}:KdX$. Therefore the constrained second variation isFor a time-dependent linearized perturbation $y$, the incremental reaction stress contains $\delta q{\varphi }_{,F}$, but its power vanishes because ${\varphi }_{,F}:\nabla \dot{y}=0$. Multiply the incremental momentum equation by $\dot{y}$, integrate by parts, and use zero incremental dead-load traction or the prescribed displacement condition. One obtainsPositive $\mathcal{Q}$ is therefore the standard constrained second-variation stability in elasticity criterion: perturbations cannot grow in this quadratic energy norm. A statement in a specified displacement norm also uses the usual coercivity and boundary conditions removing uncontrolled rigid-motion modes; translations with $\delta F=0$ are not tested by the displayed condition. This is a small-perturbation energy argument, not a claim of global nonlinear stability.

For the cube use incompressibility in the form $\varphi =\det F-1$. A neo-Hookean solid has $W_{,F}=\mu F$, and hence the spatial-index-first transpose of nominal stress is $\mu F+q{F}^{-T}$. On the diagonal family, equality to $TI$ givesThe first relation immediately gives $\boxed{q=T\lambda -\mu {\lambda }^2}$. Eliminating $q$ and factoring yieldsThus $\boxed{\lambda =1\text{or}T/\mu =\lambda +{\lambda }^{-1/2}}$.

Let $t=T/\mu$ and $h(\lambda )=\lambda +{\lambda }^{-1/2}$. Its derivative $h^{\prime }=1-\frac{1}{2}{\lambda }^{-3/2}$ vanishes only at ${\lambda }_c=2^{-2/3}$, where $h^{\prime \prime }>0$ and $h_c=32^{-2/3}$. Also $h\to \infty$ at both ends of $(0,\infty )$. Therefore $h(\lambda )=t$ has two distinct positive roots for $t>h_c$, one repeated root at $h_c$, and none below. There is an exception to the printed assertion that both roots differ from one: at $t=2$, one root is exactly $\lambda =1$. Indeed, with $x=\sqrt{\lambda }$, the equation becomes $(x-1)(x^2+x-1)=0$, so the only other positive stretch is $\lambda =(3-\sqrt{5})/2$.

Evaluate the stability form directly on the prescribed uniform perturbationIt satisfies $\mathrm{tr}(F^{-1}H)=0$. The determinant Hessian identity isMeanwhile $W_{,FF}[H,H]=\mu (1+\frac{1}{2}{\lambda }^{-3})(\delta \lambda {)}^2$. For the unit reference cube,On the trivial branch, $q=T-\mu$ andThus the undeformed state is stable in this perturbation family for $t<2$, unstable for $t>2$, and quadratically neutral at $t=2$.

On a nontrivial branch, substitute $t=h(\lambda )$ and put $z={\lambda }^{-3/2}$:The smaller root always has $\lambda <{\lambda }_c$, hence $z>2$, and is stable. For $h_c<t<2$, the larger root lies between ${\lambda }_c$ and one, so $1<z<2$ and it is unstable; the trivial branch is the second stable equilibrium. For $t>2$, the larger root lies above one and has $0<z<1$, so it is stable; the trivial branch is now unstable. Whenever there are three distinct equilibria in this family, precisely two are stable and one unstable against the specified uniform perturbations. This does not establish stability to every nonuniform or symmetry-breaking perturbation.

The homogeneous tensile bifurcation of a neo-Hookean cube has a nontrivial-branch fold at $t=32^{-2/3}$ and a meeting with the undeformed branch at $\boxed{t=2}$. The latter is the bifurcation from the undeformed state. Stating both values distinguishes this branch intersection from the earlier creation of the two nontrivial solutions.

Along a sufficiently smooth stationary curve, differentiation and the Euler-Lagrange equation give the Beltrami identity:The cancellation uses the absence of explicit $x$ dependence.

For an axisymmetric surface of revolution, its area is $I[y]=2\pi {\int }_{-l}^{l}y\sqrt{1+y^{\prime 2}}dx$. A smooth positive-radius minimizer is stationary, so the Beltrami identity, with the constant factor omitted, givesEquivalently the Euler-Lagrange equation gives $y{y}^{\prime \prime }=1+y^{\prime 2}$, whence $y^{\prime \prime }=y/k^2$. The first integral selectsEqual radii at $\pm l$ imply $x_{\ast }=0$ (subtract the two cosh values), givingThis is a catenoid. The conclusion is a necessary form for a smooth connected minimizer; it does not declare every stationary catenoid a minimum.

Put $a=l/k>0$. The boundary equation is $r/l=\cosh a/a$. Its derivative is $(a\sinh a-\cosh a)/a^2$, so its unique critical point solves $a\tanh a=1$. The left side increases strictly from zero to infinity. Thus $\cosh a/a$ decreases to a unique minimum and then increases, diverging as $a\downarrow 0$ and $a\to \infty$. There are exactly two positive values of $k$ whenAt equality there is one value, and below it there is none.

To match the printed second variation convention, write $I[y+\epsilon \eta ]=I[y]+\epsilon \delta I+{\epsilon }^2{\delta }^{2}I+o({\epsilon }^2)$; its ${\delta }^{2}I$ is the quadratic coefficient, half the second derivative with respect to $\epsilon$. Expanding the area integrand givesOn the catenoid these coefficients are $k{\mathrm{sech}}^2(x/k)$ and $2\tanh (x/k)$. Integrate the mixed term by parts; the boundary term is zero because $\eta (\pm l)=0$. Since $(\tanh (x/k){)}^{\prime }={\mathrm{sech}}^2(x/k)/k$, this yieldsThis is the axisymmetric second variation of a catenoid, with the explicitly stated quadratic-coefficient convention.

Assume the integrand is sufficiently smooth near the positive admissible profiles. Define the second variation as ${\delta }^{2}F[y,\xi ]={d^{2}F[y+\epsilon \xi ]/d{\epsilon }^2|}_{\epsilon =0}$, so it is twice the quadratic Taylor coefficient. Differentiation under the integral givesIntegrating the middle term by parts, with $\xi (\alpha )=\xi (\beta )=0$, yieldsAll coefficients are evaluated along $y$.

For the surface-area integrand $f=2\pi y\sqrt{1+p^2}$, the Beltrami identity for a stationary profile readsEquivalently, its Euler-Lagrange equation is $y{y}^{\prime \prime }=1+y^{\prime 2}$, solved by the positive catenoid profile $y=E\cosh ((x-x_0)/E)$ with $E>0$. Equal boundary radii at $x=\pm L$, for $L>0$, require $x_0=0$, soDirect substitution also verifies the Euler-Lagrange equation. This is a stationary profile whenever a positive $E$ satisfying the boundary relation exists; such an $E$ does not exist for arbitrary ring separations and radii.

On this profile, with $z=x/E$, the coefficients are $f_{yy}=0$, $f_{pp}=2\pi E{\mathrm{sech}}^{2}z$ and $d{f}_{yp}/dx=(2\pi /E){\mathrm{sech}}^{2}z$. Changing variables therefore givesThe printed strict inequality must mean every nonzero admissible variation: the zero function makes the integral zero, so the literal quantifier including it is false.

For completeness, the strict positivity here is sufficient for a strict local minimum in the fixed-endpoint $C^1$ topology, among positive axisymmetric graph profiles. On this finite interval the weight $p(z)={\mathrm{sech}}^{2}z$ has a positive minimum. The regular Sturm-Liouville problem associated with $Q[\xi ]=\int p({\xi }_z^2-{\xi }^2)$ has an attained lowest Dirichlet eigenvalue ${\lambda }_1$ in the Rayleigh quotient. Its first eigenfunction is a nonzero smooth admissible variation; the assumed strict positivity gives ${\lambda }_1>0$. Hence $Q\ge {\lambda }_1\Vert \xi {\Vert }_{L^2}^2$. Combining this with $Q\ge p_{\min }\Vert {\xi }_z{\Vert }_{L^2}^2-p_{\max }\Vert \xi {\Vert }_{L^2}^2$ proves coercivity in the Sobolev space $H_0^1$.

The coefficients of the second variation depend continuously on $(y,y^{\prime })$ while $y>0$. For a sufficiently small $C^1$ perturbation, the quadratic forms along the segment from the stationary profile to the perturbed one retain a uniform positive coercivity bound. Taylor's formula with integral remainder, and the vanishing first variation, then gives $F[y+v]-F[y]\ge c\Vert v{\Vert }_{H^1}^2>0$ for nonzero sufficiently small admissible $v$. The stationary area is therefore a strict local minimum under the stated nonzero-variation condition. This supplies the uniform bound needed in an infinite-dimensional problem, rather than relying on pointwise positivity alone in an arbitrary function space.

Write the admissible variation as $y_{\epsilon }=y+\epsilon \eta$, where $\eta (x)\to 0$ at both ends. Differentiating the functional under the integral sign gives the first variationAfter integration by parts, with the boundary term vanishing,Thus the Euler-Lagrange equation isA second differentiation gives the second variation