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Using the positive polar molecular field $h_i=\delta F/\delta p_i$ and velocity gradient ${\partial }_i{v}_j$, the reversible order-parameter stress consists of a distortion contribution and rotational and alignment contributions:For a local free-energy density, a representative is ${\Sigma }_{ij}^{(1)}=(f-p_k{h}_k){\delta }_{ij}-f_{{\partial }_i{p}_k}{\partial }_j{p}_k$. The Euler-Lagrange equation verifies its divergence. Integration by parts gives $\delta F=\int {\Sigma }_{ij}^p{\partial }_i{u}_j$ for an incompressible displacement without surface work; the corresponding mechanical power has the opposite sign. Pressure contributions may be reassigned in an incompressible flow.

Use the convention $R=L-{\sum }_i{p}_i{\dot{q}}_i$ for the eliminated cyclic coordinates, expressing their velocities in terms of their fixed conjugate momenta. Terms involving derivatives of those eliminated velocities cancel because $\partial L/\partial {\dot{q}}_i=p_i$. Thus the remaining coordinates obey the usual Euler-Lagrange equations for R. For a particle of fixed angular momentum $\ell$ in a central potential, this gives $R=m{\dot{r}}^2/2-V(r)-{\ell }^2/(2m{r}^2)$, rather than the expression obtained by direct substitution into L alone. An opposite overall sign convention is also used, with correspondingly signed mixed Hamiltonian equations.

For real $f\in L^2({\mathbb{R}}^3)$, this energy functional on $H^1({\mathbb{R}}^3)$ is coercive and attains its minimum by the direct method in the calculus of variations. The nonnegative potential term is weakly lower semicontinuous by local Sobolev compactness gives lower semicontinuity of a nonnegative integral. Its Euler-Lagrange equation is $-\Delta u+u+\sin u=f$ in the weak solution sense. The extra linear restoring term screens the Sine-Gordon equation nonlinearity. The scalar potential $V(s)=s^2/2+1-\cos s$ is convex because $V^{\prime \prime }(s)=1+\cos s\ge 0$; hence any weak critical point is a minimizer. The elliptic regularity estimate for $-\Delta +1$, together with $|\sin u|\le |u|$, gives $u\in H^2$. The Sobolev inequality then puts $u$ in $W^{1,6}$, so Morrey's inequality and uniformly continuous integrable functions vanish at infinity give a continuous representative tending to zero.

For positive twice differentiable functions $y$, the Euler-Lagrange equation for this functional reduces to $y{y}^{\prime \prime }+1+y^{\prime 2}=0$. Thus $(y{y}^{\prime }{)}^{\prime }=-1$ and integration gives $(x-h{)}^2+y^2=R^2$. The upper circular arc is a graph wherever its radicand is positive. A fixed horizontal endpoint with free height imposes $F_{y^{\prime }}=y^{\prime }/(y\sqrt{1+y^{\prime 2}})=0$, so that endpoint must lie above the centre. Distinct horizontal endpoints cannot both satisfy this natural boundary conditions for a free endpoint. The Beltrami identity alone also admits constant functions, which are excluded by the full Euler-Lagrange equation.

Take $L=[{\eta }_{ab}{\dot{x}}^a{\dot{x}}^b+W^2]/2$ with $W=\dot{y}+A_a{\dot{x}}^a=\xi$ constant. The Euler-Lagrange equation for $x^a$ isThus, with $F_{bc}={\partial }_b{A}_c-{\partial }_c{A}_b$,Reversing the explicit derivative order while retaining this metric and the same definition of $\xi$ changes the equation rather than merely changing a field-tensor naming convention. More generally, a fibre term $\sigma W^2$, $\sigma =\pm 1$, gives $\sigma \xi F^a{}_b{\dot{x}}^b$; its conserved canonical fibre momentum is $p_y=\sigma \xi$.

In the normalized positive-fibre case with $|\xi |<1$, four-dimensional proper time gives $d{u}^a/d\tau =c\xi F^a{}_b{u}^b/\sqrt{1-{\xi }^2}$. This has the Lorentz force form. The scale relating the geometrical $A$ to a physical electromagnetic four-potential fixes the conversion to an SI charge-to-mass ratio.

For rolling without slipping inside a stationary circular cylinder, take the inner body's axial moment of inertia to be $I=\gamma m{r}^2$. Its center of mass follows a circle of radius $d=R-r$, and the reduced Lagrangian is $L=m(1+\gamma )d^2{\dot{\theta }}^2/2-mgd(1-\cos \theta )$. The Euler-Lagrange equation gives $(1+\gamma )d\ddot{\theta }+g\sin \theta =0$. Linearizing near the lowest position yields the displayed small oscillation frequency. A uniform solid cylinder has $\gamma =1/2$. This reduction requires sustained contact and enough static friction to prevent slipping.

A uniform rigid body disk of mass $m$ and radius $a$ moves in a frictionless plane. Let $N\ge 3$ identical springs attach equally spaced rim points to fixed anchors in their radial directions. The equilibrium spring length is $l$, natural length $l_0<l$, and spring constant $k$. Write the small centre displacement as $\mathbf{q}$ and the rotation as $\theta$. For a spring with radial and tangential unit vectors $\mathbf{n},\mathbf{t}$, its length isThis follows by a second-order Taylor series for the distance between $\mathbf{q}+aR(\theta )\mathbf{n}$ and $(a+l)\mathbf{n}$. In the elastic potential energy $k(s-l_0{)}^2/2$, the equilibrium tension $k\tau$ multiplies the second-order change of length. Thus it contributes to the linearized stiffness even though the displacement is small.

The equally spaced directions satisfy $\sum \mathbf{n}=\sum \mathbf{t}=0$ and $\sum \mathbf{n}{\mathbf{n}}^T=\sum \mathbf{t}{\mathbf{t}}^T=(N/2)I$. Summing the spring energies givesThe kinetic energy is $m|\dot{\mathbf{q}}{|}^2/2+m{a}^2{\dot{\theta }}^2/4$, using the moment of inertia of a uniform disc. The Euler-Lagrange equations give the displayed squared angular frequencies. The translational normal modes are degenerate; rotation is independent. At zero prestretch the rotational frequency vanishes at quadratic order, and finite-angle spring stretching first contributes at higher order.

A particle constrained to the graph $z=h(x,y)$ has kinetic energy $m[|\dot{\mathbf{r}}{|}^2+(\nabla h\cdot \dot{\mathbf{r}}{)}^2]/2$ and potential energy $mgh$. For a gently varying graph and velocities for which its geometric kinetic correction is higher order, the leading Euler-Lagrange equations are the displayed planar gradient-force equations. Small slope alone at arbitrarily large speed or curvature does not justify discarding the normal inertial acceleration. For a radial height, the leading planar angular momentum is conserved, allowing an effective potential reduction.

For a regular closed planar curve, maximize signed enclosed area at fixed length using the variational integrand $\mathcal{L}=(x\dot{y}-y\dot{x})/2+\lambda \sqrt{{\dot{x}}^2+{\dot{y}}^2}$. Write the nonzero speed as $v$. The Euler-Lagrange equations integrate to $\lambda \dot{x}/v=y+c_1$ and $\lambda \dot{y}/v=-x+c_2$. A zero multiplier would force a zero velocity, so squaring and adding yields the displayed circle. The argument uses only regularity and is independent of the speed used to trace the curve. A positively oriented simple traversal of prescribed length $L$ encloses area $L^2/(4\pi )$; the planar isoperimetric inequality supplies the separate global maximality statement.

For a bounded operator on a real Hilbert space, $⟨Lv,v⟩=⟨Sv,v⟩$ with $S=(L+L^{\ast })/2$. Consequently the Euler-Lagrange equation of $⟨Lv,v⟩-2⟨f,v⟩$ is $Sv=f$. It is $Lv=f$ only when $L$ is self-adjoint or when the relevant solution also annihilates the skew part. For example $L=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ has positive quadratic form $\Vert v{\Vert }^2$, but that form contains no information about its skew part.

For a Lorentzian metric and timelike $v=dx/du$, put $L=\sqrt{-g(v,v)}>0$ and $\mathcal{L}=L^2/2$. The Euler-Lagrange equations of $\mathcal{L}$ give ${\nabla }_{v}v=0$, while those of $L$ give ${\nabla }_{v}v=(\dot{L}/L)v$. Taking $ds/du=L$ removes the latter tangential acceleration and gives a unit timelike tangent. Thus the two actions have the same geodesic images; the quadratic action singles out an affine parameter. They do not have the same solutions for an arbitrary fixed parameter: $t=u^2$, $\mathbf{x}=0$, $u>0$, in Minkowski spacetime solves the length equation but not the energy equation. Null curves are excluded since $L=0$.

Adding $dF(q,t)/dt$ to a mechanical Lagrangian changes its action by $F(q_f,t_f)-F(q_i,t_i)$. Fixed endpoint variations therefore leave its bulk Euler-Lagrange equations unchanged. For fields, adding ${\partial }_{\mu }{K}^{\mu }$ changes the action by a boundary flux. The same conclusion holds for compactly supported variations or appropriate fixed boundary data. If $K$ depends on field derivatives, those boundary data must also control their variations. Such terms can still change canonical momenta, boundary observables, and representatives of Noether currents; equivalence of bulk equations does not mean that every boundary-value problem is identical.

Adding a total derivative $dg(q(t))/dt$ to a Lagrangian changes its action only by $g(q(T))-g(q(0))$. Under fixed-endpoint variations, this boundary term is constant, so the Euler-Lagrange equations and their minimizing paths are unchanged. Componentwise, $d[\partial (dg/dt)/\partial {\dot{q}}_i]/dt={\sum }_j{g}_{ij}{\dot{q}}_j=\partial (dg/dt)/\partial q_i$ by equality of mixed partial derivatives. Endpoint freedom would require reconsidering the resulting natural boundary conditions for a free endpoint.

For the leading periodic propulsive force of a periodic planar filament and power of a periodic planar filament functionals, the Euler-Lagrange equation for $F_x+\Gamma (\dot{W}-{\dot{W}}_0)$ gives $({\xi }_{\perp }-{\xi }_{\parallel })y_{xt}+\Gamma {\xi }_{\perp }{y}_{tt}=0$. For nonzero thrust, $y_t$ obeys an advection equation and the active travelling wave is $f(x-ct)$, with $c=({\xi }_{\perp }-{\xi }_{\parallel })/(\Gamma {\xi }_{\perp })$. A static periodic shape may be added without changing leading force or power. Stationarity is a necessary extremum condition, not by itself an existence theorem for a global maximum.

For refractive index $n(y)=c\sqrt{y}$ with $c>0$, the optical-length integrand is $c\sqrt{y}\sqrt{1+y^{\prime 2}}$. The Beltrami identity gives $\sqrt{y}/\sqrt{1+y^{\prime 2}}=\sqrt{h}$, while the full Euler-Lagrange equation gives $2y{y}^{\prime \prime }=1+y^{\prime 2}$. Together these yield the displayed parabolas and exclude a spurious constant solution of the squared first integral. Equal-height endpoints $(\pm L,Y)$ force $x_0=0$ and $h^2-Yh+L^2/4=0$. There are two positive stationary rays for $Y>L$, one coalesced ray for $Y=L$, and none of this smooth family for $Y<L$. For $L=1,Y=5/4$, the two paths are $y=1+x^2/4$ and $y=1/4+x^2$. Fermat principle requires stationary optical length, not a global minimum over every competing path.

A variational principle specifies a function, path or field by requiring a functional to be stationary under an allowed class of variations. The admissible class includes the prescribed boundary conditions and constraints. A stationary value need not be a minimum: the Euler-Lagrange equation expresses the first-order stationarity condition, while a minimum requires additional analysis. Fermat principle applies this idea to optical path length, and the Rayleigh-Ritz method approximates a variational problem in a chosen family of trial functions.

Under the usual $C^2$ integrand regularity, an admissible $C^1$ curve satisfying the weak Euler-Lagrange equation, a coercively positive second variation, and the strengthened Weierstrass condition is a strong local minimizer of a variational integral. Strict excess away from the reference slope gives strict minimality. Coercive positivity means ${\delta }^{2}I(u)[\eta ,\eta ]\ge c\Vert \eta {\Vert }_{W^{1,2}}^2$ for every zero-trace variation and some $c>0$.

The second variation provides a neighboring field of extremals with slope $p(x,y)$. The associated endpoint-invariant Hilbert integral yields $I(v)-I(u)=\int E(x,v,p(x,v),v^{\prime })dx$ for uniformly close admissible $v$. The tube condition makes this integral nonnegative even when $v^{\prime }$ is large, explaining the passage from weak to strong minimality.