Problem: We need to construct the fastest slide from point $(0,0)$ to $(a,-h)$, If $u(x)$ describes its shape then time is:

$$\begin{equation}

T= \int_0^a \frac{1}{\sqrt{2gu}} \sqrt{1+u^{\prime\,2}}\,dx.

\label{eq-10.P.3}

\end{equation}$$

Find solution satisfying $u(0)=0$ and $u(a)=-h$

Solution: Let $L = {1 \over {\sqrt u }}\sqrt {1 + {u^{\prime \,2}}} $, since the factor $2g$ does not matter. Note that here $L$ does not depend on $x$, so we have$$u'{{\partial L} \over {\partial u'}} - L = c$$, where $c$ is some constant. Then,

$$\eqalign{

& {1 \over {\sqrt {1 + {{(u')}^2}} }} = c\sqrt u \cr

& \Rightarrow {1 \over {1 + {{(u')}^2}}} = {c^2}u \cr

& \Rightarrow u' = {{du} \over {dx}} = \sqrt {{1 \over {{c^2}u}} - 1} \cr

& \Rightarrow dx = {{du} \over {\sqrt {{1 \over {{c^2}u}} - 1} }} \cr

& \Rightarrow x = \int {{1 \over {\sqrt {{1 \over {{c^2}u}} - 1} }}du} \cr} $$

where the solution $u$ should satisfy $u(0)=0$ and $u(a)=-h$