Finding the Extremals of a Functional J.

Question

The functional $J$ is defined on smooth functions $y \colon [a,b] \to \mathbb{R}$ satisfying $y(a) = u$, $y(b) = v$ and is given by

$$J[y]=\int_a^b \sqrt{y} \sqrt{1+(y')^2}\, dx.$$

I have found the Euler-Lagrange equation to be $\sqrt{y} \sqrt{1+(y')^2} - y' \frac{\sqrt{y}(y')^2}{\sqrt{1+(y')^2}} = k$ for some constand $k$.

solving for $y'$ I got $y' = \sqrt{y/k^2 -1}$

Please could you confirm if this is correct and how to complete the problem from here? Any help is greatly appreciated!

Answer 1

Try the substitution $u = \frac{y}{k^2} - 1$.

You get that $k^2 \frac{du}{dt} = u^\frac{1}{2}$, which is easy to solve with separation of variables.