CoV12

Find an extremal for $J(y)=\int_1^2 \frac{\sqrt{1+y'^2}}{x} dx, y(1)=0, y(2)=1\,$

Since $L\,$ is indepedent of $y\,$

$\frac{\partial L}{\partial y^\prime}$	$= c_1\,$
$\frac{y^\prime}{x\sqrt{1 + y^{\prime 2}}}$	$= c_1\,$
$y^\prime\,$	$= c_1x\sqrt{1 + y^{\prime 2}}$
$y^{\prime 2}\,$	$= c_1^2x^2(1 + y^{\prime 2})$
$y^\prime\,$	$= \pm \frac{c_1x}{\sqrt{1 - c_1^2x^2}}$

So $y = \pm\left(\frac{\sqrt{1 - c_1^2x^2}}{c_1} + c_2\right)$ . Solving for the constants gives

$\begin{cases} 0 &= \pm\frac{\sqrt{1 - c_1^2}}{c_1} + c_2\\ 1 &= \pm\frac{\sqrt{1 - 4c_1^2}}{c_1} + c_2 \end{cases}$

$\begin{cases} c_1 &= -\frac{1}{\sqrt{5}}\\ c_2 &= 2\end{cases}$

Therefore, an extremal is $y = 2 - \sqrt{5 - x^2}$ .