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For an autonomous scalar integral $\int f(u^{\prime })$, the first variation gives the integrated Euler-Lagrange equation $f^{\prime }(u^{\prime })=c$. If $u$ is $C^1$ and each fiber of $f^{\prime }$ is finite, the connected image of the interval under the continuous derivative $u^{\prime }$ lies in a finite set and is therefore a singleton. The fixed endpoints then determine exactly one affine solution. If $f$ is strictly convex, the strict Jensen's inequality applied to $u^{\prime }$ additionally makes this solution the unique global minimizer among fixed-endpoint Lipschitz functions. Without convexity the same unique smooth stationary solution can be a maximizer, for example at zero for $f(v)=1/(1+v^{2008})$ and zero endpoint values.

If the integrand $L(q_1,\dots ,q_m,q_1^{\prime },\dots ,q_m^{\prime })$ has no explicit dependence on its independent variable $z$, its Euler-Lagrange equations imply the displayed first integral. Indeed, differentiate the left side and cancel the second-derivative terms to obtain ${\sum }_j{q}_j^{\prime }({\partial }_{q_j}L-d({\partial }_{q_j^{\prime }}L)/dz)=0$. This is the several-variable form of the Beltrami identity. Applied to Fermat principle, it expresses conservation of the component of the optical momentum of a ray along a translation-invariant direction.

Measure polar angle $\theta$ from the downward vertical on the lower hemisphere of radius $R$. A particle released from rest at the rim and sliding without energy loss has speed $\sqrt{2gR\cos \theta }$. The spherical arc length along $\varphi =\varphi (\theta )$ is $R\sqrt{1+{\sin }^2\theta {\varphi }^{\prime 2}}|d\theta |$. Thus its travel time has integrand proportional to $\sqrt{(1+{\sin }^2\theta {\varphi }^{\prime 2})/\cos \theta }$. Since the azimuth is a cyclic coordinate, the Euler-Lagrange equation givesSolving for ${\varphi }^{\prime 2}$ gives the displayed first integral on a segment where $\theta$ is a valid path parameter. For finite $C$, approaching the rim $\theta \uparrow \pi /2$ gives ${\varphi }^{\prime }\to 0$. Thus a regular stationary path begins along the meridian toward the bottom. This is an initial direction condition; a path joining other rim points can turn before reaching the bottom. Monotone segments use positive $|d\theta |$ in the time integral, and a turning point requires another path parameter or a split into segments.

For a real scalar field with Lagrangian density $\mathcal{L}={\dot{\varphi }}^2/2-|\nabla \varphi {|}^2/2-V(\varphi )$, the canonical momentum is $\pi =\dot{\varphi }$. Its Hamiltonian density is the sum of momentum energy, gradient energy and potential energy. Hamilton's equations recover the Euler-Lagrange equation.

An inextensible uniform chain of length L and linear mass density $\rho$ links a mass m sliding on a frictionless table to a hanging mass M. If the initial hanging chain length is zero and x is its subsequent hanging length, all moving material has speed $\dot{x}$. The kinetic energy is $(M+m+\rho L){\dot{x}}^2/2$, while the gravitational potential energy is $-Mgx-\rho g{x}^2/2$. The Euler-Lagrange equation gives the displayed equation until a mass hits the floor or reaches the edge. Increasing hanging chain weight produces a hyperbolic cosine displacement from rest, rather than uniform acceleration.

For coordinates $(t,x,y,z)$ and line element $d{s}^2=c^{2}d{t}^2-a^2(t)d{\mathbf{x}}^2$, these are the nonzero Christoffel symbols. The prime means differentiation with respect to $t$, and coordinate zero is $t$, not $ct$. They follow either from the metric-derivative formula or the Euler-Lagrange equations for the affine geodesic Lagrangian. Both mixed lower-index orders must be counted to obtain the factor of two in the spatial geodesic equation.

For a nonnegative graph joining $(0,0)$ and $(2a,0)$ with length between $2a$ and $\pi a$, the area-maximizing smooth graph is a circular arch. The Euler-Lagrange equation integrates to the displayed circle formula. Its radius satisfies $l=2R\arcsin (a/R)$ with $R>a$. Concavity of $y-R\sqrt{1+(y^{\prime }{)}^2}$ gives global maximality on the fixed-length class.

For a single-valued curve below the chord joining $(-a,0)$ to $(a,0)$, with length $2a<L\le \pi a$, the maximum enclosed area is attained by the lower minor circular arc. Equivalently it minimizes ${\int }_{-a}^{a}ydx$ at fixed length. Its equation and length areTo prove global minimality for $R>a$, augment the objective by $R\int \sqrt{1+y^{\prime 2}}dx$. The augmented integrand is linear in $y$ and strictly convex in $y^{\prime }$. The displayed arc satisfies its Euler-Lagrange equation, so the supporting-line inequality for the convex slope dependence, followed by integration by parts, proves it minimizes the augmented functional among curves with the prescribed endpoints. On the fixed-length class the multiplier term is constant. The semicircle endpoint follows by a limit; its endpoint tangents are vertical but its length is finite.

Let a planar Lagrangian have the form $L=a(r){\dot{r}}^2+b(r){\dot{\theta }}^2+c(r)$, with smooth coefficients and $b(r)>0$. Its cyclic coordinate $\theta$ has constant conjugate momentum $p_{\theta }=2b(r)\dot{\theta }$. For a circular orbit $r=R$, the radial Euler-Lagrange equation reduces to $b^{\prime }(R){\dot{\theta }}^2+c^{\prime }(R)=0$. If $b^{\prime }(R)\ne 0$, eliminating $\dot{\theta }$ gives the displayed condition. A real orbit requires $-c^{\prime }(R)/b^{\prime }(R)\ge 0$; the radial kinetic coefficient $a$ does not enter this constant-radius condition, although it affects radial perturbations. If $b^{\prime }(R)=0$, the original condition must be used rather than divided by zero.

A classical path joining prescribed endpoints is a stationary point of the action, satisfying the Euler-Lagrange equation and the endpoint boundary conditions. Expanding the path integral about each such path separates its classical phase from its quadratic fluctuation determinant. There may be multiple classical paths or none within a chosen real saddle class; the semiclassical propagator must respect that saddle structure rather than assuming a unique trajectory for every potential.

For a quadratic potential energy, shifting a path to its classical solution leaves an exactly quadratic fluctuation action with homogeneous Dirichlet boundary conditions. The first variation vanishes by the Euler-Lagrange equation. The Gaussian path integral prefactor is an inverse square root of a regulated functional determinant and does not depend on the endpoints. At conjugate times the ordinary formula must instead be read by distributional continuation.

Every minimizer of the Weinstein functional has the formwhere $Q$ is the positive radial NLS ground state. The Euler-Lagrange equation first reduces a minimizer to a rescaled ground-state equation; the equality cases in the diamagnetic and rearrangement inequalities give the constant phase, translation, and radial profile.

For a positive refractive index depending only on $x$, Fermat principle applied to $\int \mu (x)\sqrt{1+y^{\prime 2}}dx$ has $y$ as a cyclic coordinate. The Euler-Lagrange equation therefore makes the displayed quantity constant along a stationary ray. When ${\mu }^2>p^2$, its slope is $y^{\prime }=p/\sqrt{{\mu }^2-p^2}$. If $\theta$ is the angle to the $x$ axis, this is conservation of $\mu \sin \theta$, the stratified-medium form of Snell's law. For fixed endpoint values, the integrand has second derivative $\mu (1+y^{\prime 2}{)}^{-3/2}>0$ with respect to $y^{\prime }$. Its strict convexity and the constant momentum show, by the tangent-line inequality and integration, that any such ray is the unique minimum among competing graph paths with the same endpoints.

A field configuration is critical for an energy functional $E$ if its first variation vanishes for every admissible infinitesimal field variation. For $E[\varphi ]=\int (\frac{1}{2}|\nabla \varphi {|}^2+U(\varphi ))$, integration by parts against compactly supported variations gives $\delta E=\int (-\Delta \varphi +U^{\prime }(\varphi ))\delta \varphi$, so stationary field configurations satisfy the Euler-Lagrange equation $\Delta \varphi =U^{\prime }(\varphi )$. A constrained target instead allows only tangent variations, changing that field equation. Criticality is stationarity, not necessarily energetic stability or a local minimum.

For the affinely parametrized geodesic Lagrangian $\mathcal{L}=({\dot{r}}^2+f(r{)}^2{\dot{\varphi }}^2)/2$, the conserved cyclic momentum is $J=f(r{)}^2\dot{\varphi }$. Its Euler-Lagrange equations are $\ddot{r}-f{f}^{\prime }{\dot{\varphi }}^2=0$ and $\ddot{\varphi }+2f^{\prime }\dot{r}\dot{\varphi }/f=0$.

A generalized coordinate absent from the Lagrangian is cyclic. Its conjugate generalized momentum is conserved by the Euler-Lagrange equations.

For a uniform solid cylinder of mass $M_c$ and a frictionlessly sliding wedge of mass $M_b$, take the horizontal wedge position $X$ and uphill contact coordinate $s$. Contact and rolling without slipping leave two degrees of freedom. Up to a constant, the Lagrangian is $L=\frac{1}{2}(M_b+M_c){\dot{X}}^2+M_c\cos \alpha \dot{X}\dot{s}+\frac{3}{4}{M}_c{\dot{s}}^2-M_{c}gs\sin \alpha$.

The cyclic coordinate $X$ gives $(M_b+M_c)\ddot{X}+M_c\cos \alpha \ddot{s}=0$. The other Euler-Lagrange equation is $\cos \alpha \ddot{X}+(3/2)\ddot{s}=-g\sin \alpha$. Elimination proves the displayed acceleration, with $\ddot{X}=-M_c\cos \alpha \ddot{s}/(M_b+M_c)$. For $M_b\gg M_c$, the cylinder approaches the fixed-incline acceleration $-(2/3)g\sin \alpha$.

At $\omega =-3/2$, the combined scalar equation of Brans-Dicke theory becomes the matter consistency condition $T=0$. It does not determine $\square \Phi$ by division, since the wave coefficient has vanished. The original metric tensor and scalar Euler-Lagrange equations must be retained. Introducing scalar-dependent matter or a scalar potential changes the consistency condition.

Write $\Delta ={\theta }_1-{\theta }_2$. Applying the Euler-Lagrange equations to the double pendulum Lagrangian, then dividing by $m{\ell }^2$, givesFor example, the upper generalized momentum is $\partial L/\partial {\dot{\theta }}_1=m{\ell }^2(2{\dot{\theta }}_1+\cos \Delta {\dot{\theta }}_2)$. Its time derivative contains $-m{\ell }^2\sin \Delta ({\dot{\theta }}_1-{\dot{\theta }}_2){\dot{\theta }}_2$. Subtracting $\partial L/\partial {\theta }_1=-m{\ell }^2\sin \Delta {\dot{\theta }}_1{\dot{\theta }}_2-2mg\ell \sin {\theta }_1$ cancels the mixed product and leaves the positive $\sin \Delta {\dot{\theta }}_2^2$ term in the first equation. This cancellation is a useful check on the sign convention.

The matrix multiplying the two angular accelerations isIt is invertible for every configuration, so specified angles and angular velocities determine both angular accelerations. For the same reason $T>0$ whenever the angular velocities are not both zero. The mass cancels from the motion equations, while the timescale is set by $\sqrt{\ell /g}$.

As an exact constructed release, put ${\theta }_1(0)=\alpha$, ${\theta }_2(0)=0$, and both angular velocities zero. Solving the two equations at release yieldsFor $0<\alpha <\pi /2$, the upper rod initially moves back toward vertical, while the lower rod initially moves away from it. The lower pivot's acceleration makes this possible even though the lower rod starts vertical. The small-angle approximation gives ${\ddot{\theta }}_1(0)\simeq -2g\alpha /\ell$ and ${\ddot{\theta }}_2(0)\simeq 2g\alpha /\ell$.

The conserved mechanical energy is $E=T+V$, because the Lagrangian has no explicit time dependence. Directly, $dE/dt$ is $m{\ell }^2$ times the first equation multiplied by ${\dot{\theta }}_1$ plus the second multiplied by ${\dot{\theta }}_2$, so the equations imply $dE/dt=0$. This also supplies a check when computing a trajectory.

Choose Cartesian coordinates with origin at the upper pivot, $x$ to the right and $y$ upward. The two position vectors areThese expressions satisfy the rod holonomic constraints $|{\mathbf{r}}_1|=\ell$ and $|{\mathbf{r}}_2-{\mathbf{r}}_1|=\ell$ automatically. This is the practical advantage of using the two angles as coordinates: the rod tensions need not be unknowns in the motion calculation.^[ref]

Differentiate these position vectors to obtain the velocities. The dot product between their two rod contributions uses $\cos ({\theta }_1-{\theta }_2)=\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2$, givingAdding the two kinetic energies and measuring gravitational potential energy from the downward equilibrium givesThe coefficient $2$ has a geometric meaning: the upper rod raises both masses, and its motion contributes to both kinetic energies. The coupling in $T$ is present because the lower mass has a moving pivot. The zero chosen for $V$ changes $L$ by a constant and leaves the Euler-Lagrange equations unchanged.^[ref]

The principle of stationary action applies to $S=\int Ldt$. Its fixed-endpoint first variation, after integration by parts, isIndependent allowed variations therefore give one Euler-Lagrange equation for each angle. Stationary action alone does not assert a minimum.^[ref]