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For a smooth functional $\int F(x,y,y^{\prime },y^{\prime \prime })dx$, let $P=F_{y^{\prime }}$ and $Q=F_{y^{\prime \prime }}$. Along an extremal, the Euler-Lagrange equation for a second-derivative functional is $F_y-P^{\prime }+Q^{\prime \prime }=0$. Differentiating $H=y^{\prime \prime }Q+y^{\prime }(P-Q^{\prime })-F$ by the chain rule givesConsequently $H$ is constant whenever $F$ has no explicit dependence on $x$. This extends the Beltrami identity to a second-derivative variational problem. If $F$ is independent of $y^{\prime \prime }$, the expression reduces to $y^{\prime }{F}_{y^{\prime }}-F$.

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.

The brachistochrone problem asks for the curve along which a particle released from rest reaches a specified target fastest under a uniform gravitational field, without friction. With downward depth $u$, conservation of energy gives speed $\sqrt{2gu}$ and travel time $\int \sqrt{(1+u^{\prime 2})/(2gu)}dx$. The Beltrami identity makes $u(1+u^{\prime 2})$ constant, leading to a cycloid. If only the horizontal endpoint coordinate $x_0>0$ is fixed, the natural boundary conditions for a free endpoint select the half arch ending at its deepest point, with depth $2x_0/\pi$ and minimum time $\sqrt{\pi x_0/g}$.

In the upper half-plane $y>0$, the hyperbolic length element is $ds/y$. The function $\Phi =\mathrm{arsinh}(x/y)$ has gradient $(1,-x/y)/\sqrt{x^2+y^2}$ and norm $1/y$. The Cauchy-Schwarz inequality gives $\int ds/y\ge \Phi (B)-\Phi (A)$ for any oriented path from $A$ to $B$. Equality holds along a semicircle $x^2+y^2=R^2$ traversed with increasing $x$, because its tangent is parallel to that gradient. This is a calibration proof of global minimality, beyond the stationary-path conclusion of the Beltrami identity. Translating $x$ gives the analogous proof for semicircles with other boundary centers.

For speed $c(y)=y+c_0>0$, Fermat principle and the Beltrami identity yield circular light rays whose centres lie on $y=-c_0$. The ray between $(\pm a,0)$ in the upper half-plane is $y=\sqrt{a^2+c_0^2-x^2}-c_0$.

For horizontal separation one, let $y>0$ be downward depth and constrain the arc length to $L$. Conservation of energy makes the travel time proportional to $\int \sqrt{1+y^{\prime 2}}/\sqrt{y}dx$. An isoperimetric constraint adds a Lagrange multiplier $\lambda$, and the Beltrami identity gives $(y^{-1/2}+\lambda )/\sqrt{1+y^{\prime 2}}=H>0$. With $b=\lambda /H$, $c=H^2$ and $s=\sin (\varphi /2)$, the stationary arch is $y=s^2/[c(1-bs{)}^2]$, $dx/d\varphi =s^2/[c(1-bs{)}^3]$, for $0\le \varphi \le 2\pi$. The endpoint and length integrals fix $b,c$ uniquely for $L>1$; symmetry gives $x(2\pi -\varphi )=1-x(\varphi )$. At $L=1$ the only curve is horizontal and a particle released from rest never completes the journey. The finite-time stationary arches have vertical endpoint tangents, so the graph is smooth only on the open interval. These first-variation equations alone do not establish global minimality in a larger admissible class.

The area of a surface of revolution about the horizontal axis is $2\pi \int |y|\sqrt{1+y^{\prime 2}}dx$. For a curve below the axis with endpoints $(\pm a,0)$ and fixed arc length, its stationarity is equivalent to stationarity of $\int (y+\lambda )\sqrt{1+y^{\prime 2}}dx$. The Beltrami identity gives $(y+\lambda )/\sqrt{1+y^{\prime 2}}=C$. Choosing $C=1/k$, imposing the endpoints and integrating the length gives the displayed curve. The function $\sinh s/s$ increases strictly from $1$ to infinity on $s>0$, so every $l>2a$ determines exactly one positive $k$. Stationarity alone does not establish a global minimum.

The geodesics in the upper half-plane are vertical lines and semicircles centred on the boundary axis. Indeed, the length integrand $\sqrt{1+y^{\prime 2}}/y$ has the Beltrami identitywhich integrates to $(x-c{)}^2+y^2=R^2$. Swapping the coordinate axes gives circles centred on the vertical boundary for the metric $(d{x}^2+d{y}^2)/x^2$.

For a smooth surface obtained by translating the plane graph $z=F(y)$ along the $x$ direction, the induced metric is $d{s}^2=d{x}^2+[1+F^{\prime }(y{)}^2]d{y}^2$. The transverse arc length coordinate $u(y)={\int }^y\sqrt{1+F^{\prime }(s{)}^2}ds$ makes it $d{s}^2=d{x}^2+d{u}^2$. This is a local isometry to the Euclidean plane, extending the intrinsic flatness of a circular cylinder to arbitrary smooth generating graphs.

Its geodesics therefore lift to straight lines in $(x,u)$ coordinates. Those that are graphs over $x$ satisfy $u=ax+b$, while the additional family $x=\text{constant}$ cannot be written as $y(x)$. The same graph formula follows from the Beltrami identity for the arc-length integrand $\sqrt{1+[1+F^{\prime }(y{)}^2]y^{\prime 2}}$. These straight lines are locally minimizing; on cylinders with a closed generating curve, global identifications can create several different winding paths between two points.

For uniform linear mass density $\mu$, a chain minimizing $\mu \int \Phi (z)ds$ at fixed length has Beltrami identity $(\mu \Phi -\lambda )/\sqrt{1+z^{\prime 2}}=C$. In the domain $z<h$, the displayed potential has downward gravitational acceleration $-{\Phi }^{\prime }(z)=-\kappa /(h-z{)}^2$. A circular arc $z=h-\sqrt{a^2-x^2}$ satisfies the identity because $\sqrt{1+z^{\prime 2}}=a/(h-z)$.

This arc is a global minimum among admissible competitors of the same length with the same endpoints in $z<h$. Set $w=h-z>0$. The variable part of the energy is $\mu \kappa \int ds/w$, a hyperbolic length in the Poincaré half-plane model. The calibration for semicircular hyperbolic geodesics $S=\mathrm{arsinh}(x/w)$ has $|\nabla S|=1/w$. The Cauchy-Schwarz inequality gives $\int ds/w\ge S(B)-S(A)$, with equality along the circle traversed with increasing $x$. The constant-potential part is unchanged by the fixed-length constraint.

For an autonomous smooth Lagrangian homogeneous of degree two in velocities, the Beltrami identity gives $\dot{q}\cdot L_{\dot{q}}-L=L$ constant on its Euler-Lagrange extremals. On a regular segment with this constant positive, the Euler-Lagrange expression for $\sqrt{L}$ equals that for $L$ divided by $2\sqrt{L}$, since the extra term is proportional to $\dot{L}$. Hence an energy extremal has the same path as a length extremal. Length is invariant under regular reparametrization: in two coordinates, on a graph $v=v(u)$, its density is $\sqrt{E+2F{v}_u+G{v}_u^2}|du|$. If a nonnegative semidefinite energy vanishes along a curve, the curve instead minimizes length at zero; division by $\sqrt{L}$ is not permitted there. For a positive definite metric, regular nonconstant curves have positive energy.

Fix two attachments a distance $2a$ apart on a straight line and a simple fence of length $2l$ in one half-plane, with $a<l<\pi a/2$. There is a unique $\theta \in (0,\pi /2)$ satisfying $l/a=\theta /\sin \theta$: its derivative has numerator $\sin \theta -\theta \cos \theta >0$, since that numerator starts at zero and has derivative $\theta \sin \theta >0$. A radius-$R$ minor circular arc with half-angle $\theta$ then has the required chord and length. Its graph is $y=\sqrt{R^2-x^2}-\sqrt{R^2-a^2}$, for $-a\le x\le a$, and sector-minus-triangle geometry gives the displayed area.

To prove global maximality, attach the complementary major arc of this circle, lying in the opposite half-plane, to any competing fence. The resulting closed curve has perimeter $2l+(2\pi R-2l)=2\pi R$. Its enclosed area is the competitor's area plus a fixed major-segment area. The planar isoperimetric inequality bounds the total by $\pi R^2$, with equality only for a circle. Subtracting that fixed added area proves the minor arc is the unique maximizing shape on the chosen side. The same arc follows from the Beltrami identity for the area-minus-length functional.

For the refractive index $\mu =1/y$ on the upper half-plane, Fermat principle and the Beltrami identity give $1/[y\sqrt{1+(dy/dx{)}^2}]=1/R$. The nonvertical rays are arcs of circles centered on $y=0$; vertical lines are the remaining rays. These are also geodesics of the Poincaré half-plane model. A ray entering the strip $0<y\le 1$ toward decreasing $y$ at angle $\theta$ to the vertical has limiting horizontal displacement $\tan (\theta /2)$, at most one. The line $y=0$ is approached only after infinite optical path length because the refractive index diverges there.

For speed $c(y)=c_0/\sqrt{1-ky}$, with $1-ky>0$, the elapsed-time integrand is $\sqrt{(1-ky)(1+(y^{\prime }{)}^2)}/c_0$. The Beltrami identity gives $(1-ky)/(1+(y^{\prime }{)}^2)=1-k{y}_0$, where $y_0$ is the turning height. Integration yields the displayed parabola. If the endpoints are $(\pm x_0,0)$, symmetry gives $x_c=0$ and $(k{x}_0{)}^2=4k{y}_0(1-k{y}_0)$. Hence $k{x}_0\le 1$, and the two algebraic turning heights are $[1\pm \sqrt{1-k^2{x}_0^2}]/(2k)$. Only heights lying within the specified optical layer are admissible.

The radius of the surface of revolution is $|y(x)|$, so its area and the curve's arc length areFor the proposed curve, $y<0$ on the interior. On this branch $A=-2\pi S$ with $S=\int y\sqrt{1+y^{\prime 2}}dx$. Extremizing $S$ subject to $L=l$ is therefore equivalent to making the actual area stationary. Introduce a Lagrange multiplier $\lambda$ and the augmented integrand $F=(y+\lambda )\sqrt{1+y^{\prime 2}}$. The Beltrami identity givesFor the fixed-length stationary surface of revolution, takeIndeed $y^{\prime }=\sinh (kx)$ and $\sqrt{1+y^{\prime 2}}=\cosh (kx)$, so the Beltrami equation holds identically. It also follows directly that $F_y=\cosh (kx)$ and $d(F_{y^{\prime }})/dx=\cosh (kx)$, verifying the Euler-Lagrange equation. The endpoints vanish, and the length condition becomesFor endpoint-fixed variations preserving this length, the augmented first variation is zero; subtracting the zero length variation proves that the area's first variation is zero. Its value, if desired, is $A=\pi [\sinh (2ka)-2ka]/k^2$.

To prove uniqueness, put $s=ka>0$. The equation is $\sinh s/s=l/(2a)>1$. Its derivative is $(s\cosh s-\sinh s)/s^2$. The numerator is zero at $s=0$ and has derivative $s\sinh s>0$, so the ratio increases strictly from one to infinity. The intermediate value theorem gives exactly one positive $k$. The stated upper bound on $l$ is not needed for this uniqueness argument. Stationarity has been proved, without making an unsupported global-minimum claim.

For positive differentiable speed $c(y)$ and a smooth ray written as a graph $y(x)$, the Fermat principle minimizes travel timeA variation vanishing at the endpoints gives the Euler-Lagrange equation for $F=\sqrt{1+y^{\prime 2}}/c$. Its derivatives are $F_y=-c^{\prime }\sqrt{1+y^{\prime 2}}/c^2$ and $F_{y^{\prime }}=y^{\prime }/(c\sqrt{1+y^{\prime 2}})$. Thusand multiplying by $c^2(1+y^{\prime 2}{)}^{3/2}$ givesThe displayed differential equation requires differentiability, beyond mere continuity of $c$. For constant positive speed it reduces to $y^{\prime \prime }=0$, so rays are straight lines; a straight segment minimizes travel time because it has the shortest Euclidean length.

For $c(y)=y$ in the positive-height region, the Beltrami identity gives $F-y^{\prime }{F}_{y^{\prime }}=1/[y\sqrt{1+y^{\prime 2}}]=1/R$. Hence $y^2(1+y^{\prime 2})=R^2$, whose nonvertical solutions are $(x-h{)}^2+y^2=R^2$. Formally imposing the given boundary points fixes $h=1/2$, $R=1/2$, and gives the ray in a linear-speed mediumIt is the upper semicircle of diameter one, with vertical limiting tangents at its endpoints.

There is a necessary endpoint qualification. The speed is zero at the two stated endpoints. For any rectifiable path entering positive height, $ds/y\ge |dy|/y$, and its travel time diverges logarithmically on leaving or approaching height zero. Thus the semicircle is the formal interior ray, but there is no finite-time minimizer between the zero-speed endpoints. For endpoints at height $ϵ>0$ the circular ray has radius $\sqrt{1/4+{ϵ}^2}$ and finite travel time $2\mathrm{arsinh}(1/(2ϵ))$, which tends to infinity as $ϵ\downarrow 0$. This explains the intended semicircle as a limiting hyperbolic geodesic, without asserting finite travel time where $c=0$.

Write the cone as $z=r\cot \alpha$, with $r$ the distance from its axis and $0<\alpha <\pi /2$. Its Euclidean metric givesConsequently a differentiable path expressed as $r=r(\theta )$ has arc lengthFor $L(r,r^{\prime })=\sqrt{r^2+{\csc }^2\alpha (r^{\prime }{)}^2}$, the Euler-Lagrange equation has the Beltrami identity because $L$ has no explicit $\theta$ dependence:For a nonradial minimizing path, $C>0$. Rearranging and integrating givesThe endpoint conditions determine $C$ and ${\theta }_0$ on the selected angular branch.

To verify that this stationary path really minimizes arc length, use the local isometry from a circular cone to the plane. Set $s=r/\sin \alpha$ and $\varphi =\theta \sin \alpha$; then $d{s}^2+s^{2}d{\varphi }^2$ is the plane metric. The displayed path is a straight line $s\cos (\varphi -{\varphi }_0)=C/\sin \alpha$, so its segment realizes the plane distance. Choose endpoint angular lifts with $|\Delta \theta |\le \pi$; then $|\Delta \varphi |<\pi$, and the joining segment avoids the apex. Its length isOther lifts with developed angular difference less than $\pi$ give at least this distance; an angular change of at least $\pi$ has infimum at least $s_1+s_2$, attained only if passage through the apex is allowed. This proves the global choice above. Endpoints on a common generator have the radial segment as shortest path, which is the limiting case omitted by the $r(\theta )$ parametrization. If an endpoint is the apex, its shortest path is a generator. Thus shortest paths are straight segments after developing the cone.

For an integrand $I(y,y^{\prime })$ with no explicit $x$ dependence, the Euler-Lagrange equation is $I_y-d(I_{y^{\prime }})/dx=0$. Differentiate the proposed first integral:Thus $I-y^{\prime }{I}_{y^{\prime }}$ is constant along an extremal, the Beltrami identity.

For the optical path the Fermat principle uses the travel-time integrand $I=e^{-\lambda y}\sqrt{1+y^{\prime 2}}$, in the stated speed units. Its first integral isSet $u=k{e}^{\lambda y}$. Then $u^{\prime 2}={\lambda }^2(1-u^2)$, and integration gives $u=\cos [\lambda (x-x_0)]$ on a positive arch. The symmetric endpoint conditions require $x_0=0$ and $k=\cos \lambda a$. For $\lambda >0$ and $0<\lambda a<\pi /2$, the resulting ray in an exponential-speed medium isIt is above zero inside the endpoints. Substitution also verifies the Euler-Lagrange equation $y^{\prime \prime }=-\lambda (1+y^{\prime 2})$, so the first-integral construction does not accidentally select a spurious constant-height path.

Here $y^{\prime }=-\tan \lambda x$. For $\lambda a$ close to $\pi /2$, the endpoint slopes are steep and the central height $-{\lambda }^{-1}\log \cos \lambda a$ is large. The travel time evaluates directly toIt tends to $2/\lambda$ as the arch height diverges. Positive $\lambda$ is the increasing-speed interpretation needed for a ray arching into $y>0$; for $\lambda <0$ the same algebraic expression bends below that region and is not an admissible interior path. At zero $\lambda$ the limiting straight boundary path has travel time $2a$.

Fixed endpoint variations yield the Euler–Lagrange equationWhen $F$ is independent of $z$, differentiation givesThis is the Beltrami identity. Denote the constant by $1/k$ when nonzero; for a general Lagrangian a zero constant is also possible.

The film area is $S=2\pi {\int }_{-H}^{H}r\sqrt{1+r^{\prime 2}}dz$. Suppressing its harmless positive factor $2\pi$, take $F=r\sqrt{1+r^{\prime 2}}$. The Euler–Lagrange equation reduces toThe first integral is $r/\sqrt{1+r^{\prime 2}}=1/k$ with $k>0$. Hence $1+r^{\prime 2}=k^2{r}^2$ and, using the differential equation without dividing by $r^{\prime }$, $r^{\prime \prime }=k^{2}r$. The positive solution obeying that first integral is $r=k^{-1}\cosh (k(z-z_0))$. Equal endpoint radii force $z_0=0$, givingThus every smooth positive stationary film in this axisymmetric class is a catenoid.

Put $u=kH>0$. The boundary equation is $R/H=f(u)=\cosh u/u$. The function tends to infinity at both endpoints of $(0,\infty )$, andIts minimum occurs when $u=\coth u$. The function $u-\coth u$ is strictly increasing, with derivative $1+{\mathrm{csch}}^{2}u>0$, and has limits $-\infty$ and $+\infty$. It therefore has exactly one positive zero $A$. At this point $1/A=\tanh A$, so $f(A)=\sinh A$. This proves the catenoid existence threshold for equal rings:At equality there is one stationary solution and above it two. The equation describes candidates for an area minimum, not a guarantee that both branches minimize. The thinner-neck branch is unstable to suitable axisymmetric variations; topology-changing collapse can also compete with connected films. In particular two separate planar disks have area $2\pi R^2$. The derived threshold concerns existence of the smooth annular stationary film; it does not exclude such disconnected competitors.

For a variation $y+\epsilon \eta$ with $\eta (a)=0$ but arbitrary $\eta (b)$, differentiation and integration by parts giveFirst choose variations supported inside the interval. The fundamental lemma of the calculus of variations gives the Euler-Lagrange equation. Then allow arbitrary endpoint variation to obtain the natural boundary conditions for a free endpoint:Because $F$ has no explicit dependence on $x$,Thus the Beltrami identity is $F-y^{\prime }{F}_{y^{\prime }}=k$, constant along an extremal.

For the brachistochrone problem, put $u=-y>0$, so conservation of energy gives the bead speed $v=\sqrt{2gu}$. Its travel time isThe Beltrami identity now gives $u(1+u^{\prime 2})=2R$ for a positive constant $R$. A descending cycloid parametrizes this relation:The free endpoint has $F_{u^{\prime }}=0$, hence $u^{\prime }(x_0)=0$. On the first descending arc this occurs at $\theta =\pi$. ThereforeAlong this arc $ds=2R\sin (\theta /2)d\theta$ and $v=2\sqrt{gR}\sin (\theta /2)$, soThe initial vertical tangent causes no divergent travel time: the quotient $ds/v$ has the finite constant limit $\sqrt{R/g}d\theta$.

To verify global minimality, take any admissible wire graph of finite travel time below the release level, and let $H>0$ be its maximum depth. For $0<u\le H$, the Cauchy-Schwarz inequality givesIndeed the two coefficients on the right, after multiplication by $\sqrt{u}$, form a unit vector. Since the path starts at depth zero and reaches depth $H$, the weighted total variation is at least the integral over all depths from $0$ to $H$. ConsequentlyThe last inequality is minimized at $H=2x_0/\pi$. Our cycloid attains equality throughout: its depth increases to $H$ and $u^{\prime }=\sqrt{H/u-1}$. Thus it achieves the genuine minimum over the wire graphs, not merely a stationary time. Reflection handles a negative horizontal endpoint coordinate by replacing $x_0$ with $|x_0|$.