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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.

The lower particle has position $(x+l\sin \theta ,-l\cos \theta )$, so its squared speed is ${\dot{x}}^2+2l\cos \theta \dot{x}\dot{\theta }+l^2{\dot{\theta }}^2$. The bead contributes $m_1{\dot{x}}^2/2$ to the kinetic energy. The two equal springs have combined potential energy $k{x}^2$ up to a constant: their lengths are $a+x$ and $a-x$, and equal natural lengths cancel the linear terms in the sum of their squared extensions. The gravitational potential is $-m_{2}gl\cos \theta$. ThereforeThere is no explicit time dependence, so the total energyis conserved. The Euler-Lagrange equations areAt rest these require $x=0$ and $\sin \theta =0$, giving the downward and upward positions $\theta =0,\pi$ modulo $2\pi$. The downward position has positive-definite quadratic potential and is stable, whereas the upward position has a negative angular potential curvature.

Keeping terms of second order about the stable position and dropping a constant givesFor unit parameters the mass matrix and stiffness matrix are $M=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ and $K=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$. A normal mode with squared frequency $s$ requiresThus the positive angular frequencies are

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.

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.

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.

A cyclic coordinate $q_j$ is absent explicitly from the Lagrangian: $\partial L/\partial q_j=0$, even though ${\dot{q}}_j$ may occur. The Euler-Lagrange equation then givesThus a continuous coordinate-translation symmetry gives a conserved conjugate momentum. The conserved quantity need not be ordinary linear momentum: an absent azimuthal angle can give conserved angular momentum. The derivative is taken holding the other coordinates and velocities fixed.

Define the energy function $\mathcal{E}={\sum }_i{\dot{q}}_i{p}_i-L$. Along solutions of the Euler-Lagrange equations,ConsequentlyFor an ordinary isolated mechanical system with $L=T-U$, a time-independent quadratic kinetic energy and a velocity-independent potential energy, ${\sum }_i{\dot{q}}_i{p}_i=2T$, so $\mathcal{E}=T+U$. For a more general Lagrangian, the conserved quantity is the displayed energy function; identifying it with $T+U$ requires these assumptions.

The Euler-Lagrange equation from a is $2M{R}^2\ddot{\theta }+MgR\sin \theta =0$. Linearizing about $\theta =0$ givesBoth angular frequency and ordinary frequency are given to distinguish the two conventions. The stable position places the extra mass below the pivot.

For the rolling disk with a rim point mass, impose rolling without slipping and take the clockwise angular displacement $\theta$ from the configuration with the extra mass at the bottom. The disk centre has $x_C=R\theta$ and $y_C=R$. The extra mass hasThe disk's translational and rotational kinetic energies sum to $3M{R}^2{\dot{\theta }}^2/2$. Including the rim mass givesThe exact Euler-Lagrange equation isTo first order, $3M{R}^2\ddot{\theta }+MgR\theta =0$. ThereforeAt the bottom the extra mass is instantaneously at the contact point and has no first-order velocity. The increased effective inertia comes from the moving disk. Static friction enforces the rolling constraint without doing work at the stationary contact.

In the rim-mass pulley coupled to a hanging mass, the printed figure shows the smaller pulley fixed overhead, and the mechanical string leaves the right-hand tangent of the larger pulley. Let $\theta$ increase counterclockwise from the downward vertical to the radius carrying $M$. Then the hanging mass $m$ moves downward by $R\theta$. The string is taken to grip the rim without slipping. Its tension must be solved dynamically, rather than set equal to $mg$ during motion.

The kinetic energy and potential energy areThe Euler-Lagrange equation and the condition for equilibrium areOne can also eliminate the tension from $M{R}^2\ddot{\theta }=TR-MgR\sin \theta$ and $mR\ddot{\theta }=mg-T$. For $0<m<M$, the solution with $\cos {\theta }_e>0$ is stable; the other solution in a full turn is unstable.

Use the angle $\theta$ from the vertical as the generalized coordinate. For a uniform ladder, its centre of mass is $(l\sin \theta ,l\cos \theta )$, so $v_G^2=l^2{\dot{\theta }}^2$. Its central moment of inertia is $I_G=m(2l{)}^2/12=m{l}^2/3$. With ideal contact constraints that do no work, the Lagrangian, conjugate momentum, and Euler-Lagrange equation areRelease from rest at ${\theta }_0$ also gives ${\dot{\theta }}^2=3g(\cos {\theta }_0-\cos \theta )/(2l)$ by conservation of energy. These results hold while both contacts persist. The ideal wall reaction is normal to the wall; an unspecified tangential wall friction would require an additional generalized force and would prevent determination from the supplied data alone.

The supplied PDF labels this equation clarified from original for tutorial purposes. Use the model of an eccentric wheel with missing-mass kinetic energy neglected: neglect the missing disc's contribution to kinetic energy, while retaining its subtraction from gravitational potential energy. Do not silently replace the gravitational mass $W$ by the full-disc mass $D$ as well.

Choose $\theta$ as the clockwise angle of $\overrightarrow{CH}$ from the upward vertical, and set $s=R\theta$. The rolling without slipping constraint makes changes in $s$ equal changes in the distance C travels down the slope. Choose its additive origin so $s=0$ when the hole is vertically above C. Then, up to an irrelevant height constant,The approximate kinetic energy is that of the complete uniform disc rolling at $\dot{s}=R\dot{\theta }$:Subtracting the missing disc's gravitational potential energy givesApply the Euler-Lagrange equation to $\mathcal{L}=T-U$:ThereforeHere $K_1,K_2$ have dimensions of acceleration and $\gamma$ has dimensions of inverse length. The initial conditions are $s(0)=\dot{s}(0)=0$ under the vertical-above interpretation. If an initial orientation is instead specified relative to the slope normal, the same coordinate convention produces a different initial $s$, not different constants. With $r_o\to 0$ the equation becomes $\ddot{s}=(2/3)g\sin \alpha$, the familiar solid-disc result. Retaining the hole's exact kinetic energy would introduce coordinate-dependent inertia and extra ${\dot{s}}^2$ terms; the printed neglect is what permits the requested constant-coefficient form.

Measure the vertical displacement $x$ downward from the spring's unstretched position. The kinetic energy is $m{\dot{x}}^2/2$, while the spring and gravitational potential energies are $k{x}^2/2$ and $-mgx$. Thus the Lagrangian isApplying the Euler-Lagrange equation, $d(\partial \mathcal{L}/\partial \dot{x})/dt-\partial \mathcal{L}/\partial x=0$, givesGravity shifts the equilibrium to $x_{\ast }=mg/k$. For displacement $\xi =x-x_{\ast }$ from that equilibrium, the equation of motion becomes $\ddot{\xi }+(k/m)\xi =0$, giving simple harmonic motion with angular frequency $\omega =\sqrt{k/m}$. Choosing an upward coordinate instead reverses the gravitational term, but leaves the equilibrium-relative angular frequency unchanged.

For the general Lagrangian just obtained, the Euler-Lagrange equation givesUsing the small-angle approximation $\sin \theta \simeq \theta$, this is simple harmonic motion with ${\omega }_0^2=A/I$. ThusThe endpoint mass cancels in the light-rod limit. A concrete comparison shows why the finite rod cannot silently be included: the same specified endpoint $m$ with $M_r=m$ gives $T=2\pi \sqrt{8l/(9g)}$, distinct from the negligible-rod result. This is a physical pendulum when the rod contributes, and a simple pendulum in the light-rod limit. Both expressions require small oscillation amplitude; the full sine equation remains nonlinear.

Take $\theta$ from the downward vertical. Under the conventional light-rod idealization, only the endpoint point mass contributes appreciably. Its speed is $l\dot{\theta }$, and its gravitational potential energy is $-mgl\cos \theta$. ThereforeAn additive constant in the potential energy has no effect on the Euler-Lagrange equation.

Rod-mass qualification: “thin” and “uniform” do not imply negligible mass. If the uniform rod has a finite unspecified mass $M_r$, its pivot moment of inertia is $M_r{l}^2/3$ and its centre of mass lies $l/2$ from the pivot. For a pendulum with a uniform rod and an endpoint mass, the general answer isThis formula covers either convention without inventing a rod mass. A result involving only the specified $m,l,g$ uses $M_r=0$ unless an additional relation for $M_r$ is stated.

The Euler-Lagrange equation for $\theta$ isFor small angles, set $\cos \theta \simeq 1$ and $\sin \theta \simeq \theta$. With $x=A\cos \omega t$ and ${\omega }_0^2=g/l$, the harmonic forcing of a horizontally driven pendulum obeysFor $\omega \ne {\omega }_0$, its general solution isThe constants $C,D$ set the initial angle and angular velocity. If the bob starts vertical with zero angular velocity at $t=0$, this reduces toThe solution remains an approximation only while the resulting angle is small.

When the block is free, $x$ becomes a dynamical generalized coordinate and its kinetic energy $M{\dot{x}}^2/2$ must be included. Expanding to quadratic order about the downward stationary state and dropping a constant givesThe coordinate $x$ is cyclic. Its conserved canonical momentum is the total horizontal momentum,The two linear Euler-Lagrange equations areEliminating $\ddot{x}$ givesThus the pendulum with a freely sliding support has a zero-frequency translational mode and an oscillatory mode. With four independent constants $X_c,V,C,D$, the general solution isHere $V=P/(M+m)$ and $X_c+Vt$ is the horizontal centre of mass coordinate to first order, since $X_{\mathrm{CM}}=x+[ml/(M+m)]\theta$. The oscillation is faster than that of a pendulum with a fixed pivot; the limit $M\to \infty$ restores $\Omega \to \sqrt{g/l}$.

The azimuth $\varphi$ is a cyclic coordinate, since $\partial \mathcal{L}/\partial \varphi =0$. Its Euler-Lagrange equation gives conservation of the vertical component of angular momentum:This $h$ is the actual angular momentum, not specific angular momentum. The Lagrangian also has no explicit time dependence. The conserved energy isAlong solutions, $\dot{E}=-\partial \mathcal{L}/\partial t=0$. These are the conserved quantities associated with rotational symmetry about the vertical and time-translation symmetry, respectively.