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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 $y=\sqrt{a^2+c_0^2-x^2}-c_0$.

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 $(d{x}^2+d{y}^2)/x^2$.

The Euler-Lagrange equation is $f_y-d(f_{y^{\prime }})/dx=0$. Since $f$ has no explicit $x$ dependence,Thus a stationary curve satisfies the Beltrami identity, $f-y^{\prime }{f}_{y^{\prime }}=C$.

By Fermat principle, the light ray makes the travel time $\int \sqrt{1+y^{\prime 2}}/(y+c_0)dx$ stationary. Its Beltrami identity becomesWriting $R=1/C$ and integrating the resulting first-order equation gives $(x-x_0{)}^2+(y+c_0{)}^2=R^2$. Both endpoints have height zero, so subtracting their equations gives $x_0=0$, and then $R^2=a^2+c_0^2$. The arc lying in $y>0$ isIts maximum height is $\sqrt{a^2+c_0^2}-c_0$. The ray rises into the faster region before returning to the boundary; its circle centre is $(0,-c_0)$. The prescribed boundary endpoints are understood by continuity from the open half-plane. This is the circular light ray in a linear speed profile.

Along a sufficiently smooth stationary curve, differentiation and the Euler-Lagrange equation give the Beltrami identity:The cancellation uses the absence of explicit $x$ dependence.

For an axisymmetric surface of revolution, its area is $I[y]=2\pi {\int }_{-l}^{l}y\sqrt{1+y^{\prime 2}}dx$. A smooth positive-radius minimizer is stationary, so the Beltrami identity, with the constant factor omitted, givesEquivalently the Euler-Lagrange equation gives $y{y}^{\prime \prime }=1+y^{\prime 2}$, whence $y^{\prime \prime }=y/k^2$. The first integral selectsEqual radii at $\pm l$ imply $x_{\ast }=0$ (subtract the two cosh values), givingThis is a catenoid. The conclusion is a necessary form for a smooth connected minimizer; it does not declare every stationary catenoid a minimum.

Put $a=l/k>0$. The boundary equation is $r/l=\cosh a/a$. Its derivative is $(a\sinh a-\cosh a)/a^2$, so its unique critical point solves $a\tanh a=1$. The left side increases strictly from zero to infinity. Thus $\cosh a/a$ decreases to a unique minimum and then increases, diverging as $a\downarrow 0$ and $a\to \infty$. There are exactly two positive values of $k$ whenAt equality there is one value, and below it there is none.

To match the printed second variation convention, write $I[y+\epsilon \eta ]=I[y]+\epsilon \delta I+{\epsilon }^2{\delta }^{2}I+o({\epsilon }^2)$; its ${\delta }^{2}I$ is the quadratic coefficient, half the second derivative with respect to $\epsilon$. Expanding the area integrand givesOn the catenoid these coefficients are $k{\mathrm{sech}}^2(x/k)$ and $2\tanh (x/k)$. Integrate the mixed term by parts; the boundary term is zero because $\eta (\pm l)=0$. Since $(\tanh (x/k){)}^{\prime }={\mathrm{sech}}^2(x/k)/k$, this yieldsThis is the axisymmetric second variation of a catenoid, with the explicitly stated quadratic-coefficient convention.

Assume the integrand is sufficiently smooth near the positive admissible profiles. Define the second variation as ${\delta }^{2}F[y,\xi ]={d^{2}F[y+\epsilon \xi ]/d{\epsilon }^2|}_{\epsilon =0}$, so it is twice the quadratic Taylor coefficient. Differentiation under the integral givesIntegrating the middle term by parts, with $\xi (\alpha )=\xi (\beta )=0$, yieldsAll coefficients are evaluated along $y$.

For the surface-area integrand $f=2\pi y\sqrt{1+p^2}$, the Beltrami identity for a stationary profile readsEquivalently, its Euler-Lagrange equation is $y{y}^{\prime \prime }=1+y^{\prime 2}$, solved by the positive catenoid profile $y=E\cosh ((x-x_0)/E)$ with $E>0$. Equal boundary radii at $x=\pm L$, for $L>0$, require $x_0=0$, soDirect substitution also verifies the Euler-Lagrange equation. This is a stationary profile whenever a positive $E$ satisfying the boundary relation exists; such an $E$ does not exist for arbitrary ring separations and radii.

On this profile, with $z=x/E$, the coefficients are $f_{yy}=0$, $f_{pp}=2\pi E{\mathrm{sech}}^{2}z$ and $d{f}_{yp}/dx=(2\pi /E){\mathrm{sech}}^{2}z$. Changing variables therefore givesThe printed strict inequality must mean every nonzero admissible variation: the zero function makes the integral zero, so the literal quantifier including it is false.

For completeness, the strict positivity here is sufficient for a strict local minimum in the fixed-endpoint $C^1$ topology, among positive axisymmetric graph profiles. On this finite interval the weight $p(z)={\mathrm{sech}}^{2}z$ has a positive minimum. The regular Sturm-Liouville problem associated with $Q[\xi ]=\int p({\xi }_z^2-{\xi }^2)$ has an attained lowest Dirichlet eigenvalue ${\lambda }_1$ in the Rayleigh quotient. Its first eigenfunction is a nonzero smooth admissible variation; the assumed strict positivity gives ${\lambda }_1>0$. Hence $Q\ge {\lambda }_1\Vert \xi {\Vert }_{L^2}^2$. Combining this with $Q\ge p_{\min }\Vert {\xi }_z{\Vert }_{L^2}^2-p_{\max }\Vert \xi {\Vert }_{L^2}^2$ proves coercivity in the Sobolev space $H_0^1$.

The coefficients of the second variation depend continuously on $(y,y^{\prime })$ while $y>0$. For a sufficiently small $C^1$ perturbation, the quadratic forms along the segment from the stationary profile to the perturbed one retain a uniform positive coercivity bound. Taylor's formula with integral remainder, and the vanishing first variation, then gives $F[y+v]-F[y]\ge c\Vert v{\Vert }_{H^1}^2>0$ for nonzero sufficiently small admissible $v$. The stationary area is therefore a strict local minimum under the stated nonzero-variation condition. This supplies the uniform bound needed in an infinite-dimensional problem, rather than relying on pointwise positivity alone in an arbitrary function space.

The two Euler-Lagrange equations areDifferentiating $\kappa =f-u^{\prime }{f}_{u^{\prime }}-w^{\prime }{f}_{w^{\prime }}$ along an extremal and using these equations gives $d\kappa /dx=f_x$. Hence the Beltrami identity makes $\kappa$ constant when $f$ has no explicit dependence on $x$. If $f$ omits $u$ or $w$, that variable is a cyclic coordinate and the corresponding momentum $f_{u^{\prime }}$ or $f_{w^{\prime }}$ is an additional first integral; continuous symmetries give the analogous conserved quantities through Noether's theorem.

Put $A=1-m/u$. Since $w$ is cyclic,for a constant $\lambda$. The Beltrami integral isTherefore

If $\kappa =-{\lambda }^2$, then $u^{\prime 2}={\lambda }^{2}m/u$ andFor $z>1$, definethe hinted identity gives $F^{\prime }(z)=z^{3/2}/(z-1)$. Hence $w=C\pm mF(u/m)$. Because $F(z)\to +\infty$ as $z\to \infty$, the prescribed limit selects the minus sign:As $u\downarrow m$, the logarithm tends to $-\infty$, so every such solution satisfies $\boxed{w\to +\infty }$.

Here the integrandhas no explicit $x$-dependence. The Beltrami identity therefore givesfor a positive constant $R$. Thuswhose nonvertical solutions are the semicircles

The endpoint $(a,a)$ and the prescribed value at $x=b$ both lie onbecauseThe required positive arc is consequentlyThe condition $b<2a$ keeps the endpoint above the $x$-axis. The sketch is the descending part of the upper semicircle of radius $a$ centred at $(a,0)$, from $(a,a)$ toThis is precisely a Geodesic in the Poincare half-plane model.

The Lagrangian $\mathcal{L}=z^{-2}\sqrt{1+z^{\prime 2}}$ has no explicit $x$ dependence, so its Beltrami identity is conserved:At the turning point $z=z_{\max }$ and $z^{\prime }=0$, the constant is $z_{\max }^{-2}$. Rearrangement gives

For fixed endpoint values, the first variation of the functionalvanishes exactly when the Euler-Lagrange equationholds. If $f$ has no explicit $x$ dependence, differentiation along an extremal gives the Beltrami identityFor a further constraint $J[z]=J_0$, introduce a constant Lagrange multiplier $\lambda$ and apply the Euler-Lagrange equation to $I+\lambda J$; $\lambda$ is then chosen so that the constraint holds.