Fermat principle

For three unit-height layers with positive speeds $u,v,w$, the straight-segment travel time is $t=\sqrt{1+x_1^2}/u+\sqrt{1+(x_2-x_1{)}^2}/v+\sqrt{1+(X-x_2{)}^2}/w$. Its Hessian matrix has form $\left(\begin{array}{cc}p+q& -q\\ -q& q+r\end{array}\right)$ with positive $p,q,r$. Its leading minor is positive and its determinant is $pq+pr+qr>0$. Hence this time function is strictly convex, and its stationary path is the unique minimum.

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.

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.

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.

For light speed $c(y)=y>0$, the travel-time integrand is $\sqrt{1+y^{\prime 2}}/y$. The Beltrami identity gives $y\sqrt{1+y^{\prime 2}}=R$, so nonvertical rays are semicircles centred on $y=0$. Boundary endpoints at $y=0$ have infinite travel time: along any rectifiable path entering positive height, $ds/y\ge |dy|/y$ and the integral diverges. These semicircles are therefore ideal-endpoint hyperbolic geodesics, not finite-time optical minimizers joining zero-speed endpoints.

For positive $\lambda$ and light speed $c(y)=c_0{e}^{\lambda y}$, the Fermat principle uses the time integrand $c_0^{-1}{e}^{-\lambda y}\sqrt{1+y^{\prime 2}}$. The Beltrami identity sets $e^{-\lambda y}/\sqrt{1+y^{\prime 2}}$ constant. Solving this first integral gives the displayed symmetric ray between $(\pm a,0)$ when $0<\lambda a<\pi /2$. Its travel time is $2\sin (\lambda a)/(c_0\lambda )$. The ray's central height diverges as $\lambda a\uparrow \pi /2$, while its travel time tends to $2/(c_0\lambda )$.