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For the regular Sturm-Liouville setting assume $p$ is positive and continuously differentiable, $q,w$ are continuous, and $w>0$. Apply a multiplier to $J[y]-\lambda ({\int }_a^{b}w{y}^2-1)$. Its first variation, divided by two, isEndpoint-vanishing variations and the allowed fundamental lemma giveIf an endpoint value is free, its variation is arbitrary, giving the natural condition $p{y}^{\prime }=0$ at that endpoint. At a prescribed endpoint value the variation vanishes there, so no additional natural condition is imposed.

For eigenfunctions $u,v$ with different eigenvalues, multiply their equations by $v,u$ and subtract after integration. The resulting boundary expression $[p(u{v}^{\prime }-u^{\prime }v){]}_a^b$ vanishes under common homogeneous Dirichlet or common natural boundary conditions. Hence $({\lambda }_u-{\lambda }_v){\int }_a^{b}wuv=0$, proving weighted orthogonality.

For the final Dirichlet problem the normalized stationary modes areSolving $y^{\prime \prime }+(\lambda +1)y=0$ with both endpoint values zero gives these and excludes nonpositive $\lambda +1$. Distinct sine modes have zero integral product on $[0,1]$. Their Dirichlet values also make the same boundary expression vanish, which explains orthogonality even though their endpoint derivatives do not obey the natural conditions.

For global minimality, the Fourier sine expansion of an endpoint-vanishing function and Parseval's identities give $\int y^{\prime 2}=\sum k^2{\pi }^2{a}_k^2\ge {\pi }^2\sum a_k^2={\pi }^2\int y^2$. Equality holds exactly in the first sine eigenspace. ThereforeDistinct modes are orthogonal; the two sign choices of the same minimum are not orthogonal to each other. This distinguishes the family of stationary modes from the two representatives of the minimizing eigenspace in the question's wording.

For fixed endpoints and a regular integral constraint, Lagrange multipliers replace the constrained stationarity problem by stationarity of $I-\lambda J$, with constant $\lambda$. Writing $H=F-\lambda G$, the Euler-Lagrange equation is $H_y-(d/dx)H_{y^{\prime }}=0$. Regularity here means the first variation of $J$ is not identically zero; abnormal constraint extrema otherwise need separate consideration.

If $H$ has no explicit $x$ dependence, thenThis is the Beltrami identity, giving the required constant of integration.

For the chain, $ds=\sqrt{1+y^{\prime 2}}dx$, so its fixed arc length is exactly $J[y]=L$, and its gravitational energy is $\rho g{\int }_{-a}^{a}y\sqrt{1+y^{\prime 2}}dx$. The positive constant $\rho g$ can be suppressed in minimization. UseThe equation, with a consistent sign convention for the multiplier, isThe first integral becomes $(y-\lambda )/\sqrt{1+y^{\prime 2}}=c$. It implies $y^{\prime \prime }=(1+y^{\prime 2})/(y-\lambda )$; solving givesThe equal endpoint heights force $x_0=0$, by evenness and strict increase of $\cosh$ on the positive half-line. Thus the extrema have the asserted catenary form. The length and height conditions giveFor the hanging minimum choose $c>0$, the upward-convex branch lying below its endpoints. For $L>2a$ this positive $c$ is unique: writing $t=a/c$, the ratio $L/(2a)=\sinh (t)/t$ increases continuously from one to infinity. The negative-$c$ branch describes an upward arch, not the hanging minimum. If $L=2a$, only the horizontal straight chain is admissible and is the limiting degenerate case; $L<2a$ is impossible.

At a regular constrained extremal, the first variations satisfy $\delta I=\lambda \delta J$ for a constant Lagrange multiplier $\lambda$. One way to justify this is to choose a variation for which $\delta J\ne 0$, use it to correct any other variation to preserve the constraint by the implicit function theorem, and differentiate $I$ along that corrected family. This gives the same ratio $\lambda$ for all variations. The exceptional case $\delta J=0$ for every variation requires the usual separate abnormal-extremal analysis.

Set $L=F-\lambda G$. Integration by parts in its first variation givesTaking compactly supported variations and applying the fundamental lemma of the calculus of variations yields the Euler-Lagrange equation. For the quadratic integrands, it isor $\mathcal{L}y=\lambda ry$ with $\mathcal{L}y=-(p{y}^{\prime }{)}^{\prime }-qy$. If endpoint values are free, independent endpoint variations give the natural conditionsAt a prescribed endpoint the variation instead vanishes and no natural condition is added there.

Under these natural conditions, integration by parts givesHence $\boxed{I=\lambda J}$. For two solutions satisfying the same natural conditions, subtract their cross-multiplied equations and integrate:The final equality is the vanishing endpoint boundary form. Thus distinct eigenvalues giveThe same argument works for other common homogeneous boundary conditions making that boundary form zero.

In the Legendre case, $((1-x^2)y_n^{\prime }{)}^{\prime }+{\lambda }_n{y}_n=0$. Comparing coefficients of $x^n$ in a degree-$n$ solution gives ${\lambda }_n=n(n+1)$, so different degrees have different eigenvalues. A degree-zero solution is $y_0=C_0$. A linear solution $Ax+B$ must be orthogonal to the nonzero constant solution, so $B=0$ and $y_1=C_{1}x$. Write $y_2=A{x}^2+Bx+C$. Orthogonality to $x$ gives $B=0$, and orthogonality to $1$ gives $2A/3+2C=0$. ThereforeThe constants are free normalization factors; for solutions of the stated exact degrees they are nonzero. The natural endpoint conditions hold automatically for these polynomials because $p(\pm 1)=0$.

For a regular isoperimetric constraint, introduce a constant Lagrange multiplier $\lambda$ and set $L=F-\lambda G$. With fixed endpoint values, the Euler-Lagrange equation isIf $F$ and $G$ have no explicit $x$ dependence, the Beltrami identity gives the first integralThe regularity assumption excludes the exceptional case in which the constraint itself has zero first variation for every admissible perturbation.

Parametrize the surface of revolution by $r(z,\theta )=(f(z)\cos \theta ,f(z)\sin \theta ,z)$. The magnitude of $r_z\times r_{\theta }$ is $f\sqrt{1+f^{\prime 2}}$. Integrating the resulting surface area element over $\theta$, and integrating the circular cross-sectional areas for the volume, givesHere $f\ge 0$ because it is a radius. For a regular connected closed surface, take $f>0$ in the interior; the closing endpoints can have unbounded $f^{\prime }$ even when the surface itself is smooth.

After dividing $A-\lambda V$ by $\pi$, its integrand is $L=2f\sqrt{1+f^{\prime 2}}-\lambda f^2$. The Euler-Lagrange equation and Beltrami identity becomeAs $f$ tends to zero at an endpoint, $0\le 2f/\sqrt{1+f^{\prime 2}}\le 2f$ also tends to zero, so $C=0$. On the positive interior this implies $\lambda >0$ and, writing $R=2/\lambda$,The Euler-Lagrange equation also yields $f^{\prime \prime }=-R^2/f^3$. Thereforeincluding at the maximum where $f^{\prime }=0$. Integrating and using the first integral givesThus the regular extremal is a sphere. The two zeros fix its center and radius:Equivalently, the prescribed volume fixes $R=(3V/(4\pi ){)}^{1/3}$. If $a,b$ are fixed in advance, their separation and the prescribed volume must be compatible with these equations; there is no positive regular stationary profile of the stated type for an incompatible volume.

For minimality, the three-dimensional isoperimetric inequality states that every sufficiently regular closed surface enclosing volume $V$ haswith equality precisely for a sphere. Our extremal has $A=4\pi R^2$ and $V=4\pi R^3/3$, so it attains equality and is a global minimum of area at the fixed volume. This also makes clear the connected-surface assumption behind uniqueness: allowing several disconnected components introduces stationary collections of spheres, rather than the single positive generating profile used here.

Take a smooth variation $u+\epsilon v$ with $v=0$ on the boundary. Differentiating under the integral gives the first variationIntegration by parts removes the derivatives of $v$, and the boundary contribution vanishes. The fundamental lemma of the calculus of variations therefore givesFor a regular constraint $J[u]=c$, the Lagrange multiplier rule says that an extremal satisfies $\delta I=\lambda \delta J$ for a constant $\lambda$. Equivalently, apply the preceding Euler–Lagrange equation to $F-\lambda G$. Regularity means that $\delta J$ is not identically zero as a linear functional on the admissible variations. For the quadratic constraint below this holds at every nonzero admissible $u$; a zero constraint leaves only $u=0$.

In polar coordinates the constrained integrand isIts Euler–Lagrange equation is $-2\lambda ru-{\partial }_r(2r{u}_r)-{\partial }_{\theta }(2u_{\theta }/r)=0$, orThe field is single-valued and periodic in $\theta$. Substitution $u=f(r)\cos (n\theta )$ with $n\in \{0,1,2,\dots \}$ givesThe centre is part of the disk, so one must also select the regular radial solution: for $n\ge 1$, smoothness gives $f(r)=O(r^n)$ and the regular derivative behaviour; for $n=0$, $f$ is bounded and $f^{\prime }(0)=0$. A singular solution at zero does not define an admissible smooth field on the disk.

For fixed $n$, let $c_n={\int }_0^{2\pi }{\cos }^2(n\theta )d\theta$, so $c_0=2\pi$ and $c_n=\pi$ for $n\ge 1$. The angular derivative term has integral $n^2{c}_n$, including zero when $n=0$. ConsequentlyThe one-dimensional Euler–Lagrange equation for $\hat{I}-\lambda \hat{J}$ isexactly the same radial equation. The common factor $c_n$ cancels.

For two regular solutions with the same $n$ and different multipliers, multiply the equation for $f_1$ by $f_2$, multiply that for $f_2$ by $f_1$, subtract and integrate. The result isAt $a$, both functions vanish; at zero, regularity makes the boundary term tend to zero. Since the multipliers are distinct,This is weighted Sturm–Liouville orthogonality with weight $r$.

Finally, multiply one radial equation by $f$ and integrate by parts. The same endpoint conditions yieldMultiplication by $c_n$ provesFor any nonzero admissible solution the energy on the left before subtraction is positive, so its multiplier is the positive Rayleigh quotient $\lambda =\hat{I}/\hat{J}$.

For fixed endpoint variations, the Euler-Lagrange equations areIndeed the first variation is $\int {\sum }_i(L_{q_i}\delta q_i+L_{{\dot{q}}_i}\delta {\dot{q}}_i)dt$. Integration by parts removes the derivatives from the variations; their zero endpoint values remove the boundary term. The fundamental lemma then makes every remaining coefficient vanish.

Bounded coefficients and Cauchy-Schwarz give $|B[u,v]|\le C\Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$. Uniform ellipticity gives $B[u,u]\ge \theta \Vert Du{\Vert }_2^2$ because $c\ge 0$. Poincaré inequality $\Vert u{\Vert }_2\le C_P\Vert Du{\Vert }_2$ holds on $H_0^1$ of a bounded domain, soA weak solution is $u\in H_0^1$ satisfying $B[u,\varphi ]=\int f\varphi$ for every $\varphi \in H_0^1$. The energy is finite by boundedness of $B$ and the forcing functional, andIt is bounded below and a minimizing sequence is bounded in the Hilbert space $H_0^1$. Its closed linear span is separable, so part 1 supplies a weakly convergent subsequence with limit $u$. Symmetry and nonnegativity of $B$ showThe forcing term converges by weak convergence. Thus $I(u)\le \mathrm{liminf}I(v_j)$, proving attainment. First variation in every zero-trace direction gives the weak equation. If $u,w$ are two solutions, test their difference equation with $u-w$ and use coercivity to get $u=w$. Hence there is exactly one weak solution, and it is the unique energy minimizer. Indeed $I(v)-I(u)=\frac{1}{2}B[v-u,v-u]$ directly verifies the minimizing property.

The first variation isThe endpoint terms vanish. The fundamental lemma of the calculus of variations gives $xu=0$ almost everywhere, hence $u=0$ on $(0,1)$. A Lipschitz representative is then identically zero and cannot have terminal value one. Therefore there are no admissible solutions of the stationarity equation.

For the unlabelled Euler-Lagrange assertion, use its integrated form $f^{\prime }(u^{\prime })=c$ almost everywhere. This follows from the first variation $\int f^{\prime }(u^{\prime }){\varphi }^{\prime }=0$ for all compactly supported tests. If $u\in C^1$, the left side is continuous and so equals $c$ everywhere. Its derivative values lie in the finite fiber $(f^{\prime }{)}^{-1}(c)$, while the continuous image of the interval under $u^{\prime }$ is connected. Thus $u^{\prime }$ is constant. Endpoint values fix it to $m=(\beta -\alpha )/(b-a)$, giving exactly the affine $\bar{u}$. Conversely that affine function has constant $f^{\prime }({\bar{u}}^{\prime })$, so is a solution. This proves uniqueness among $C^1$ solutions; the scalar integrand is the one-dimensional case of the notation in the source.

If $f$ is strictly convex, Jensen's inequality applied to $u^{\prime }$ givesEquality in the strict inequality can occur only when $u^{\prime }$ is almost everywhere constant. Absolute continuity and the endpoints then give $u=\bar{u}$. Hence the affine function is the unique global minimizer.

Let $\eta$ be a smooth variation with $\eta ={\eta }^{\prime }=0$ at both endpoints, and consider $y_{ϵ}=y+ϵ\eta$. stationary point gives the first variationIntegrate the middle term once by parts and the last term twice. All boundary terms vanish because both $\eta$ and ${\eta }^{\prime }$ vanish there. Assuming the derivatives needed for the classical formula, the given fundamental lemma of the calculus of variations yieldsWithout assuming those classical derivatives initially, the same calculation with compactly supported variations states this equation weakly; sufficient regularity then gives the classical equation.

For the specific integrand, $F_y=-2y$, $F_{y^{\prime }}=0$, $F_{y^{\prime \prime }}=2y^{\prime \prime }$, so the equation is $y^{(4)}-y=0$. Its weak solutions here are smooth, and its general classical solution isPut $\ell =\pi /2$, $c=\cosh \ell$, $s=\sinh \ell$. The conditions at zero give $C=-A$ and $D=1-B$. Those at $\ell$ giveUsing $c^2-s^2=1$, solve to obtain $A=c/2$, $B=(1-s)/2$, $C=-c/2$, $D=(1+s)/2$. Thus an extremal isSubstitution checks all four boundary values. The additive term $x^2$ in the integrand contributes no variation in $y$, so it does not alter this equation.

Let $P=F_{y^{\prime }}$ and $Q=F_{y^{\prime \prime }}$. For a smooth variation $y+\epsilon \eta$, the first variation isHere the last expression follows by two integrations by parts. Taking variations supported in the interior and applying the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation for a second-derivative functional:If both endpoint values and slopes are free, their independent variations force the natural boundary conditions for second-derivative functionalsIf an endpoint value or slope is fixed instead, its corresponding variation vanishes and the associated natural condition is not required.

For $H=y^{\prime \prime }Q-y^{\prime }(Q^{\prime }-P)-F$, differentiate using the chain rule:The Euler-Lagrange equation cancels the first term. Consequently $H^{\prime }=-F_x$, and if there is no explicit $x$ dependence,This is a Beltrami identity for a second-derivative functional.

Assume the regular Sturm–Liouville setting: real coefficients with $p>0$, $w>0$ and sufficient continuity and differentiability for the integrations below. Apply a multiplier to $J[y]-\lambda ({\int }_a^{b}w{y}^2-1)$. Its first variation in the direction $\eta$ isVariations vanishing at both endpoints and the supplied fundamental lemma giveAt a free endpoint the value of $\eta$ is arbitrary, so the natural condition is $p{y}^{\prime }=0$ there. At a fixed endpoint $\eta =0$, so no such derivative condition is imposed; the prescribed endpoint value is imposed instead. This distinction also allows one endpoint of each type.

For eigenfunctions $u,v$ of distinct eigenvalues $\lambda ,\mu$ under the same homogeneous self-adjoint endpoint conditions, multiply their equations by $v,u$ and subtract. Integration givesThe boundary expression vanishes for both natural conditions and homogeneous fixed-value conditions. Thus $\boxed{{\int }_a^{b}wuv=0}$ for different eigenvalues.

In the specified fixed-endpoint case the eigenproblem is $-y^{\prime \prime }-y=\lambda y$, $y(0)=y(1)=0$. There are no nonzero solutions when $\lambda +1\le 0$; the elementary exponential or affine solutions and the two endpoint conditions prove this. For $\lambda +1>0$, the first condition leaves a sine and the second requires frequency $n\pi$. The normalized stationary functions and eigenvalues areDistinct sine modes are orthogonal by the product-to-sum formula, or by the preceding Green second identity. Although they need not satisfy the natural derivative conditions, their fixed zero values also annihilate the boundary expression, so orthogonality is expected.

The Wirtinger inequality ${\int }_0^1{y}^{\prime 2}\ge {\pi }^2{\int }_0^1{y}^2$ gives $J[y]\ge {\pi }^2-1$. For completeness it follows by writing $y=v\sin \pi x$ for a smooth function supported away from the endpoints and integrating the cross term: $\int (y^{\prime 2}-{\pi }^2{y}^2)=\int {\sin }^2(\pi x)v^{\prime 2}\ge 0$. Approximation extends it to the Dirichlet Sobolev class. Equality is attained by the first sine mode; the eigenproblem or equality condition gives only its scalar multiples. Hence the two normalized minimizers are $\boxed{y=\pm \sqrt{2}\sin \pi x}$ and the minimum is $\boxed{{\pi }^2-1={\lambda }_1}$.

There is a wording distinction in the source: the two sign choices of the minimum are not orthogonal to each other, since their inner product is $-1$. Orthogonality applies to the stationary modes with different eigenvalues, not to both signs of the same minimizer.

At a regular constrained stationary curve, choose a constant multiplier $\lambda$ and make the augmented functional $I-\lambda J$ stationary under endpoint-fixed variations. Its integrand is $H=F-\lambda G$ and the equation is $d{H}_{y^{\prime }}/dx-H_y=0$. This normal multiplier formulation requires the first variation of the constraint not to vanish identically; the usual abnormal case would require two multipliers. When $H_x=0$,so $\boxed{y^{\prime }{H}_{y^{\prime }}-H=c_0}$ is constant, proving the requested first integral.

For the chain, $ds=\sqrt{1+(y^{\prime }{)}^2}dx$, so the fixed physical length is exactly $J[y]$. Its gravitational potential, apart from the constant factor $\rho g$, is $I[y]={\int }_{-a}^{a}y\sqrt{1+(y^{\prime }{)}^2}dx$. Thus $H=(y-\lambda )\sqrt{1+(y^{\prime }{)}^2}$, and the Euler–Lagrange equation isThe first integral becomes $(y-\lambda )/\sqrt{1+(y^{\prime }{)}^2}=c$, after renaming the constant with the opposite sign. The differential equation also gives $(y-\lambda )y^{\prime \prime }=1+(y^{\prime }{)}^2$, so the nondegenerate solution integrates to $y=\lambda +c\cosh ((x-x_0)/c)$. Equal endpoint heights force $x_0=0$: the two cosh values agree only when their arguments are equal or opposite, and $a>0$ excludes the former. HenceFor the hanging energy minimum take $c>0$. Since $\sqrt{1+(y^{\prime }{)}^2}=\cosh (x/c)$,For $L>2a$ the length equation has exactly one positive solution: its right side decreases from infinity to $2a$ as $c$ increases from zero to infinity. For $L=2a$ the only feasible shape is the straight segment, obtained as a limiting case; for $L<2a$ no chain can span the endpoints. A negative $c$ gives the inverted stationary catenary, rather than the physically hanging minimum.

At a regular constrained extremum, choose an admissible variation $k$ with $\delta J[k]\ne 0$. The implicit function theorem applied to $J[y+\epsilon h+sk]$ permits a compensating $s(\epsilon )$ keeping $J$ fixed. Differentiation at zero gives $\delta I[h]=\lambda \delta J[h]$ for every $h$, where $\lambda =\delta I[k]/\delta J[k]$. Integrating the first variation of $I-\lambda J$ by parts and using arbitrary compactly supported $h$ givesSmoothness alone does not exclude an abnormal constraint: if $\delta J$ vanishes identically, one needs the general multiplier form with two multipliers rather than automatically setting the coefficient of $I$ to one. The displayed normal form uses the usual regular-constraint hypothesis.

For the given quadratic functionals, the equation is $\boxed{(p{y}^{\prime }{)}^{\prime }+qy+\lambda ry=0}$. If the endpoint values of $y$ are free, the boundary variation $[p{y}^{\prime }h{]}_a^b$ gives the natural conditions $\boxed{p(a)y^{\prime }(a)=p(b)y^{\prime }(b)=0}$, interpreted by limits at singular endpoints. Multiply the equation by $y$ and integrate. These boundary terms vanish, leavingFor two eigenfunctions, multiply their equations by the other function and subtract. Integration givesThe natural conditions remove the bracket, so distinct eigenvalues imply $\boxed{{\int }_a^{b}r{y}_m{y}_{n}dx=0}$.

In the Legendre case, comparison of leading polynomial coefficients gives ${\lambda }_n=n(n+1)$, so the degrees correspond to distinct eigenvalues. A degree-zero polynomial is $y_0=C_0$. Write $y_1=Ax+B$; orthogonality to $1$ on $[-1,1]$ gives $B=0$. Write $y_2=A{x}^2+Bx+C$. Orthogonality to $x$ gives $B=0$, and orthogonality to $1$ gives $2A/3+2C=0$. ThusThese polynomials satisfy the natural endpoint limits because $1-x^2$ vanishes at both endpoints.

Take a smooth variation $y_{\epsilon }=y+\epsilon \eta$. Differentiating the functional at zero gives the first variationIntegrate the second term once and the third term twice by integration by parts:Choose variations supported away from the endpoints. Stationarity and the fundamental lemma of the calculus of variations forceThis is the Euler-Lagrange equation for a second-derivative functional. With the interior equation imposed, the remaining boundary expression must vanish for every admissible variation.

At each endpoint, a freely varying value $y$ requires $\boxed{F_{y^{\prime }}-(F_{y^{\prime \prime }}{)}^{\prime }=0}$, and a freely varying slope $y^{\prime }$ requires $\boxed{F_{y^{\prime \prime }}=0}$. A fixed value or slope instead makes its corresponding variation zero and supplies no natural condition for that variable. These are the natural boundary conditions for second-derivative functionals, with the value and slope variations treated independently.

A constant of motion has zero total derivative on every solution, and may explicitly depend on time. For an infinitesimal transformation $\delta q_i=ϵ{\xi }_i$ with $\delta L=ϵdF/dt$, the on-shell first variation is $\delta L=ϵd({\sum }_i{p}_i{\xi }_i)/dt$. Therefore $\boxed{I=\sum _i{p}_i{\xi }_i-F}$ is conserved. Including a time shift $\delta t=ϵ\tau$ gives the corresponding $I={\sum }_i{p}_i{\xi }_i-H\tau -F$. This proves Noether's theorem in its usual point-transformation form. A symmetry of the action up to a boundary term is sufficient; exact invariance of the displayed Lagrangian is stronger than necessary. Discrete transformations by themselves do not give this continuous-charge construction.

Vary the action $S=\int dtdx\mathcal{L}$ with compactly supported $\delta \varphi$, or impose vanishing $\delta \varphi$ and its first derivatives on the spacetime boundary. The first variation isOne integration by parts on the first-derivative term and two on the second-derivative term remove all derivatives from $\delta \varphi$. The boundary terms vanish, leavingThe fundamental lemma of the calculus of variations gives the Euler-Lagrange field equation with second derivatives:Repeated indices run over time and space. The coefficient of the symmetric second derivative is defined by the displayed variation; if ${\varphi }_{tx}$ is instead treated as a single independent mixed derivative variable, it is counted once rather than twice.

The dispersion relation shows that an infinite rod has unstable long-wave modes. A finite rod excludes sufficiently long wavelengths, so stability depends on its length and the precise end conditions. The fourth spatial derivative requires two conditions at each end; fixing transverse positions alone does not specify whether slopes are constrained.

First take displacement-fixed ends with free rotation. Setting $\varphi (0)=\varphi (l)=0$ leaves slope variations free; the bending boundary term in the first variation then gives the corresponding natural boundary conditions on the slope, ${\varphi }_{xx}(0)={\varphi }_{xx}(l)=0$. The spatial normal modes are $\sin (n\pi x/l)$, andThe first mode destabilizes first. ConsequentlyAt equality, the first mode has zero restoring force and is marginal, with a static component and a possible linear-in-time component in the linear theory. Every integer $n<l\sqrt{\alpha /\beta }/\pi$ gives an exponentially growing mode.

If "fixed" instead means clamped, the clamped boundary condition is $\varphi ={\varphi }_x=0$ at both ends. The critical compression is thenFor example, the static critical shape $1-\cos (2\pi x/l)$ obeys the four clamp conditions and the zero-frequency equation at this load. To establish that it is the first threshold, write $v={\varphi }_x$: it has zero endpoints and zero mean, and periodic extension gives ${\int }_0^l{v}_x^{2}dx\ge (2\pi /l{)}^2{\int }_0^l{v}^{2}dx$ by the zero-mean periodic Poincaré inequality, with equality at $v\propto \sin (2\pi x/l)$. This proves the critical Rayleigh quotient. Both cases are instances of the compressed Euler-Bernoulli beam and its endpoint-dependent buckling threshold. The source does not specify the endpoint rotation, so both interpretations are stated explicitly.

Let $V$ be a closed linear subspace of a real Hilbert space, and let $Q(v)=a(v,v)$ for a symmetric bounded bilinear form that is nonnegative on the ambient space and coercive on $V$. Then $Q$ has a unique minimizer on every closed affine subspace $g+V$. If $Q(v_j)\to m={\inf }_{g+V}Q$, the midpoint belongs to $g+V$ and the parallelogram law gives $Q(v_j-v_k)\le 2Q(v_j)+2Q(v_k)-4m\to 0$. Coercivity on $V$ makes the minimizing sequence Cauchy; completeness and continuity give a minimizer. The vanishing first variation is $a(u,v)=0$ for all $v\in V$. Two minimizers differ by $w\in V$ with $Q(w)=0$, so coercivity gives uniqueness. This proves the symmetric existence theorem directly, without invoking the Lax-Milgram theorem.

A quadratic functional on a real vector space combines a quadratic form $B(v,v)$, a linear functional $\ell (v)$ and a constant. If $B$ is a symmetric bounded bilinear form, its first variation is $DI(v)[w]=2B(v,w)-2\ell (w)$. For a strictly positive $B$, an existing stationary point is the unique minimizer by the quadratic variational principle for a symmetric positive operator.