= Solution
For $F(y,y')=y\sqrt{1+y'^2}$, the <Euler-Lagrange equation> is
$$
\sqrt{1+y'^2}-\frac d{dx}\frac{yy'}{\sqrt{1+y'^2}}=0,
\qquad yy''=1+y'^2.
$$
Since $F$ has no explicit $x$ dependence, the <Beltrami identity> gives
$$
F-y'F_{y'}=\frac{y}{\sqrt{1+y'^2}}=C>0.
$$
Solving this <first integral>, or substituting into the second-order equation, yields
$$
\boxed{y(x)=C\cosh\!\left(\frac{x-K}{C}\right),\qquad C>0.}
$$
The constants satisfy $y_1=C\cosh((x_1-K)/C)$ and $y_2=C\cosh((x_2-K)/C)$. There need not be a solution for arbitrary endpoint data, and when there is one these equations can have more than one pair of constants. These are the possible stationary <catenoids>, so the <axisymmetric second variation of a catenoid> still matters when deciding whether they minimize <surface area>.
Back to article page
