 Solution

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 $2a$.