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A missing explicit dependence produces a first integral of the Euler-Lagrange equation. The two cases follow by differentiating the candidate conserved quantity, as shown in the child solutions.

For a variation $y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$, the first variation of the functional isThe boundary term vanishes by the fixed endpoints. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equation

Choose the tangent angle $\psi =\arctan y^{\prime }$ in $(-\pi /2,\pi /2)$, so $1/\sqrt{1+(y^{\prime }{)}^2}=\cos \psi$. At the maximum, the first integral has value $C=y_0(1-\lambda )$. HenceA nonconstant regular branch below its positive maximum has $y>0$ and $\psi \ne 0$, giving $\lambda >1$. For completeness, strictness follows from the differential equation rather than assuming it at every point: the Euler-Lagrange equation simplifies toAt the maximum, $y_0{y}^{\prime \prime }=1-\lambda$ implies $\lambda \ge 1$. If $\lambda =1$, the positive-height initial data $y=y_0$, $y^{\prime }=0$ give the constant solution uniquely, contradicting the existence of the lower-height branch. Thus $\lambda =1$ is only the degenerate horizontal solution. The first integral with $\lambda >1$ also excludes $y=0$ on this regular branch.

Use the angle orientation in the PDF: $M$ has coordinates $(R\sin \theta ,-R\cos \theta )$ relative to the hoop centre. The wrapped string segment has length $R(\pi /2-\theta )$. As $\theta$ increases, it releases length $R\theta$, so the suspended mass moves down by $R\theta$. Both speeds have magnitude $R|\dot{\theta }|$.

Thus the hoop with a suspended counterweight hasand $L=T-V$. The Euler-Lagrange equation givesAt equilibrium,The positive curvature $V^{\prime \prime }({\theta }_0)=MgR\cos {\theta }_0$ establishes stability. Put $\theta ={\theta }_0+\delta$ and linearize. The equilibrium force cancels, leavingHenceFor $M\gg m$, ${\theta }_0\simeq m/M$ and $\omega \to \sqrt{g/R}$, the small-oscillation frequency of a simple pendulum of length $R$. The linearization assumes an amplitude small enough to remain inside the stated angular range and a taut string throughout.

For a variation $q\mapsto q+ϵ\eta$ with $\eta (t_0)=\eta (t_1)=0$, the first variation isThe endpoint term vanishes. Since the interior variation is arbitrary, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equationAlong such a trajectory,If $L$ has no explicit time dependence, the first integral is thereforeIts negative is the equivalent Beltrami form. For the usual time-independent mechanical Lagrangian $L=T-V$ with quadratic velocity dependence, this conserved quantity is the total mechanical energy $T+V$.

For fixed endpoint values and smooth admissible variations, the Euler-Lagrange equation isThe first variation follows by integrating $\int (f_z\eta +f_{z^{\prime }}{\eta }^{\prime })dx$ by parts and using $\eta =0$ at the endpoints.

For the claimed first integral, differentiate directly:Along an extremal the bracket vanishes. Thus when $f_x=0$, the Beltrami identity isFor a further regular constraint $G[z]=G_0$, introduce a constant Lagrange multiplier $\lambda$ and make $F-\lambda G$ stationary. Apply the Euler-Lagrange equation to the combined integrand, then impose the endpoints and $G=G_0$ to determine the integration constants and multiplier.

The Euler-Lagrange equations giveFor example, differentiating $T$ with respect to ${\dot{L}}_1$ collects contributions from all three climbers, because changing $L_1$ moves all of them. At equilibrium, all accelerations vanish, soThe top rope supports the total weight, while each lower rope supports only its own climber. These extensions are positive, consistent with the taut-rope tension law.

Assume $c(y)>0$ is differentiable along the ray, with sufficient regularity for the following variation, and fix the endpoints. Fermat principle makes the travel time stationary:For the variational integrand $f=\sqrt{1+y^{\prime 2}}/c$, the Euler-Lagrange equation isExpanding the derivative and multiplying by $c^2(1+y^{\prime 2}{)}^{3/2}$ givesand thereforeEquivalently, the Beltrami identity gives $1/[c(y)\sqrt{1+y^{\prime 2}}]=\text{constant}$. Mere continuity of $c$, as stated in the wording, does not by itself justify the appearance of $c^{\prime }$; the displayed classical ray equation requires this additional differentiability.

Introduce a constant Lagrange multiplier $\lambda$ for the fixed arc length. Up to an irrelevant constant term, the augmented functional isIt has no explicit $x$ dependence, so the Beltrami identity givesFor the hanging minimum choose $k>0$. Writing $y_0=-\lambda$, this becomes $1+y^{\prime 2}=(y-y_0{)}^2/k^2$. Its nonconstant smooth solution isOne may verify it without dividing by $y^{\prime }$ at the lowest point: the full Euler-Lagrange equation gives $(y+\lambda )y^{\prime \prime }=1+y^{\prime 2}$, which the expression satisfies everywhere. The branch with negative $k$ is the inverted stationary chain rather than the hanging minimum. The catenary minimizes the integral of height along a fixed-length curve, so the positive branch is also the global minimum, not merely a stationary candidate.

For a zero-natural-length spring, extension is the full distance between its endpoints, so the squared spring lengths are quadratic functions of the Cartesian coordinates. The Lagrangian isThe gravitational potential energy has positive derivative in the upward coordinate, giving downward forces. Applying the Euler-Lagrange equations to all four coordinates givesThe two spring forces are equal and opposite on the two masses for the connecting spring, as required by Newton's third law.

Self-adjoint form. Grouping the two derivative terms gives $[(1+ϵx)u^{\prime }{]}^{\prime }-ϵu=-\lambda u$. Thus a regular Sturm-Liouville operator isThe boundary conditions are mixed Neumann boundary conditions at zero and Dirichlet boundary conditions at $\pi$. For $ϵ\ge 0$, $p>0$, $q\ge 0$ and $w>0$, as required. The boundary term $[p(f{g}^{\prime }-f^{\prime }g){]}_0^{\pi }$ vanishes for two functions satisfying these conditions, confirming self-adjointness on that domain.

Constrained variation. Introduce a Lagrange multiplier $\Lambda$ for the normalization and varyIts first variation isThe endpoint term vanishes because $v^{\prime }(0)=0$ and $\delta v(\pi )=0$. The Euler-Lagrange equation is thereforewhich is exactly the original eigenvalue problem for the boxed choices of $p,q,w$. Multiply by $v$ and integrate by parts. The same endpoint conditions giveThus the stationary functional values are the eigenvalues, with $\Lambda =\lambda$. More generally, before normalization the relevant Rayleigh quotient is

Unperturbed ground mode. At $ϵ=0$, the equation is $-u^{\prime \prime }=\lambda u$. The quadratic form is nonnegative, and zero eigenvalue would force a constant function incompatible with the endpoint at $\pi$. For positive $\lambda$, $u=A\cos (\sqrt{\lambda }x)+B\sin (\sqrt{\lambda }x)$. The derivative condition forces $B=0$, while the other endpoint gives $\cos (\pi \sqrt{\lambda })=0$. HenceThe lowest eigenvalue and normalized eigenfunction are

Trial-function bound. The Rayleigh-Ritz method says that the minimum eigenvalue is no larger than the quotient of any admissible trial function. The same $U_0$ still satisfies both boundary conditions and retains unit norm. Its derivative is $U_0^{\prime }=-\sqrt{2/\pi }\sin (x/2)/2$, givingUsing ${\sin }^2(x/2)=(1-\cos x)/2$,Therefore the affine-parameter Sturm-Liouville eigenvalue bound isThis is an upper bound for every $ϵ\ge 0$, not merely a small-parameter approximation. As a lower-bound check, $p\ge 1$ and the unperturbed variational minimum imply ${\mathcal{R}}_{ϵ}[v]\ge 1/4+ϵ$ for every admissible normalized $v$. Thus

Define $h=f-y^{\prime }{f}_{y^{\prime }}$. For a density without explicit $x$ dependence, the chain rule givesThus the Euler-Lagrange equation implies the Beltrami identityWhere $y^{\prime }\ne 0$ this first integral is equivalent to the original equation. At a turning point it extends by continuity. On an entire constant segment, however, its derivative alone can vanish without imposing the full Euler-Lagrange equation, so such candidate paths must still be checked in the original equation.

Interpret the density $1/y$ on the physical region $y>0$. It is singular at zero and would not describe a nonnegative density for $y<0$. The first integral gives $y\sqrt{1+y^{\prime 2}}=R$ for a positive constant $R$. To handle the horizontal tangent without dividing by $y^{\prime }$, use the full Euler-Lagrange equation:Two integrations give $y^2=-(x-x_0{)}^2+R^2$. The equal endpoint heights force $x_0=0$, and either endpoint then gives $R^2=2$. ThusThe path is the upper circular arc. A constant horizontal path solves the differentiated first integral trivially but fails $y{y}^{\prime \prime }+y^{\prime 2}+1=0$, so is not another extremum.

One can prove global minimality, rather than only stationarity. Define $\Phi (x,y)=\mathrm{arsinh}(x/y)$ on the upper half-plane. ThenThe Cauchy-Schwarz inequality gives, along any admissible path from $A$ to $B$,Our circle has tangent $(1,-x/y)$ for increasing $x$, so equality holds pointwise. This calibration for semicircular hyperbolic geodesics provesThe geometry is that of a hyperbolic geodesic in the Poincaré half-plane model. The proof even allows competing upper-half-plane paths beyond the class of monotonic graphs.

Use displacements $x_i$ from an equilibrium, so the two spring extensions are $x_2-x_1$ and $x_3-x_2$. The kinetic energy and elastic potential energy areThus $T={\dot{x}}^{T}M\dot{x}/2$ and $V=x^{T}Kx/2$, with the small-oscillation mass and stiffness matricesIf $x_i$ instead denote absolute positions, the appropriate natural spring lengths must first be subtracted from the differences in $V$. They do not affect the equations for displacements.

Euler-Lagrange equations give $M\ddot{x}+Kx=0$, so a normal mode $x=v{e}^{i\omega t}$ obeys the generalized eigenvalue problem for small oscillations $Kv={\omega }^{2}Mv$. Reflection symmetry separates antisymmetric shapes $(A,0,-A{)}^T$ from symmetric shapes $(A,B,A{)}^T$. In the antisymmetric sector the end equation is $kA=m{\omega }^{2}A$, giving ${\omega }^2=k/m$. In the symmetric sector the two independent equations areTwice the first plus the second gives $m{\omega }^2(2A+\mu B)=0$. If $\omega =0$, then $B=A$. Otherwise $B=-2A/\mu$, and substitution into the first equation gives ${\omega }^2=(k/m)(1+2/\mu )$. Thus the three independent shapes are:For $v_0$ both springs retain their lengths, so this is uniform translation. For $v_1$ the end masses move oppositely and their forces on the middle mass cancel. For $v_2$ the end masses move together, while the middle mass moves oppositely so the centre of mass remains fixed. Direct substitution verifies all three eigenpairs; for positive $m,k,\mu$ the three eigenvalues are distinct and exhaust the three degrees of freedom. These are the normal modes of three masses connected by two equal springs.

A useful normalization is the weighted inner product induced by $M$. Since the squared mass norms are $m(\mu +2)$, $2m$ and $m(2+4/\mu )$, respectively, takeIf ordinary Euclidean unit eigenvectors are intended instead, divide $v_0,v_1,v_2$ by $\sqrt{3},\sqrt{2},\sqrt{2+4/{\mu }^2}$, respectively. The frequencies and physical motions are independent of this normalization choice. The zero-frequency coordinate has general motion $A+Bt$ rather than an oscillation.

For a nonzero admissible radial profile define the radial diffusion Rayleigh quotientMinimizing the numerator subject to ${\int }_0^{1}r{R}^{2}dr=1$ introduces a multiplier $\nu$. The Euler-Lagrange equation for $r\kappa R^{\prime 2}-\nu r{R}^2$ isThe allowed variations vanish at $r=1$, and the natural condition at the centre is zero radial flux. Thus the minimum, when attained, is the smallest decay eigenvalue ${\nu }_0$. Equivalently, expansion in weighted-orthogonal eigenfunctions makes $\mathcal{R}[R]$ a weighted average of their decay eigenvalues. The Rayleigh-Ritz variational principle therefore givesAdmissible trial functions must obey the outer boundary condition, be nonzero and have finite weighted norm and energy; they need not solve the differential equation. If the diffusion operator has no lowest eigenfunction, the same variational argument bounds the infimum of its spectrum. Existence of a fundamental diffusion mode is an additional assumption for strongly vanishing diffusivity, not a consequence of the identity alone.

For fixed endpoint values and a smooth variation vanishing at the endpoints, the first variation isThe fundamental lemma of the calculus of variations gives the Euler-Lagrange equationIf $f$ is independent of $y$, then $y$ is a cyclic coordinate, and integration givesThis is conservation of the corresponding generalized momentum.

The integrand has no explicit $y$ dependence. Thus the Euler-Lagrange equation gives the conserved transverse optical momentumAt $x=0$, $\mu (0)=\sqrt{2}$ and the crossing slope is one, so $p=1$. Solving for the positive slope gives $y^{\prime }=1/\sqrt{{\mu }^2-1}$, henceThe slopes agree at zero. The same conserved momentum holds across the interface because the refractive index is continuous and there is no point source in the variational equation.

Here a least-time ray means the minimum between the source and the receiver point selected by this ray. For those fixed endpoints, ${\partial }^2[\mu \sqrt{1+y^{\prime 2}}]/\partial y^{\prime 2}=\mu (1+y^{\prime 2}{)}^{-3/2}>0$ proves strict convexity in the slope, so the stationary ray is the unique minimum among graph paths. If the receiver could instead be freely placed anywhere on the line and its height were also minimized over, the horizontal ray would be faster; that different problem is incompatible with the prescribed crossing slope.

Choose generalized coordinates $q_i$ measured from a static equilibrium point, and assume a conservative system with no velocity-dependent forces. Keeping terms of degree two in displacements and velocities gives the small oscillation LagrangianHere $A$ is the positive-definite symmetric mass matrix evaluated at equilibrium, and $B$ is the symmetric Hessian matrix of the potential energy there. The linear potential term is zero by equilibrium, and an additive constant has been dropped. Coordinate-dependent corrections to the kinetic matrix contribute only at higher order.

The Euler-Lagrange equations are $A\ddot{\mathbf{q}}+B\mathbf{q}=0$. For a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, they give the generalized eigenvalue problem $(B-{\omega }^{2}A)\mathbf{v}=0$. A nonzero mode shape exists exactly whenThis is a degree-$n$ polynomial in ${\omega }^2$. At a stable equilibrium its roots are nonnegative; zero roots correspond to zero modes.

Choose positive $\varphi$ so the bob is displaced to the right relative to the cart. Its Cartesian coordinates areDifferentiation gives ${\dot{X}}^2+{\dot{Y}}^2={\dot{x}}^2+2L\cos \varphi \dot{x}\dot{\varphi }+L^2{\dot{\varphi }}^2$. Thus the exact kinetic energy and potential energy areRetain quadratic terms, using $\cos \varphi =1+O({\varphi }^2)$ in the kinetic cross term and $1-\cos \varphi ={\varphi }^2/2+O({\varphi }^4)$ in the potential. The Lagrangian isIts Euler-Lagrange equations giveThe kinetic cross term is essential: acceleration of the support couples the pendulum and cart. These equations define the normal modes of a spring-mounted pendulum cart.

Fermat principle states that the optical path length $\mathcal{S}=\int nds$, or equivalently travel time $\mathcal{S}/c_0$, is stationary under small variations of the ray with fixed endpoints. Stationarity need not mean a global minimum.

For a plane ray written as $y(x)$, use $\mathcal{S}[y]=\int F(x,y,y^{\prime })dx$, with $F=n(x,y)\sqrt{1+y^{\prime 2}}$. Its stationary paths obey the Euler-Lagrange equationIt follows by taking the first variation, integrating the term in $\delta y^{\prime }$ by parts, and using zero endpoint variations.