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Along a sufficiently smooth stationary curve, differentiation and the Euler-Lagrange equation give the Beltrami identity:The cancellation uses the absence of explicit $x$ dependence.

For an axisymmetric surface of revolution, its area is $I[y]=2\pi {\int }_{-l}^{l}y\sqrt{1+y^{\prime 2}}dx$. A smooth positive-radius minimizer is stationary, so the Beltrami identity, with the constant factor omitted, givesEquivalently the Euler-Lagrange equation gives $y{y}^{\prime \prime }=1+y^{\prime 2}$, whence $y^{\prime \prime }=y/k^2$. The first integral selectsEqual radii at $\pm l$ imply $x_{\ast }=0$ (subtract the two cosh values), givingThis is a catenoid. The conclusion is a necessary form for a smooth connected minimizer; it does not declare every stationary catenoid a minimum.

Put $a=l/k>0$. The boundary equation is $r/l=\cosh a/a$. Its derivative is $(a\sinh a-\cosh a)/a^2$, so its unique critical point solves $a\tanh a=1$. The left side increases strictly from zero to infinity. Thus $\cosh a/a$ decreases to a unique minimum and then increases, diverging as $a\downarrow 0$ and $a\to \infty$. There are exactly two positive values of $k$ whenAt equality there is one value, and below it there is none.

To match the printed second variation convention, write $I[y+\epsilon \eta ]=I[y]+\epsilon \delta I+{\epsilon }^2{\delta }^{2}I+o({\epsilon }^2)$; its ${\delta }^{2}I$ is the quadratic coefficient, half the second derivative with respect to $\epsilon$. Expanding the area integrand givesOn the catenoid these coefficients are $k{\mathrm{sech}}^2(x/k)$ and $2\tanh (x/k)$. Integrate the mixed term by parts; the boundary term is zero because $\eta (\pm l)=0$. Since $(\tanh (x/k){)}^{\prime }={\mathrm{sech}}^2(x/k)/k$, this yieldsThis is the axisymmetric second variation of a catenoid, with the explicitly stated quadratic-coefficient convention.

For $F(y,y^{\prime })=y\sqrt{1+y^{\prime 2}}$, the Euler-Lagrange equation isSince $F$ has no explicit $x$ dependence, the Beltrami identity givesSolving this first integral, or substituting into the second-order equation, yieldsThe 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.

A stable minimal surface has nonnegative second variation of surface area for every admissible normal variation $fN$ fixing the boundary. For a two-sided minimal surface in three-dimensional Euclidean space, the quadratic form isHere ${\nabla }_{M}f$ is the gradient for the induced Riemannian metric. The gradient term penalizes rapid changes in the displacement; the principal curvature term can make the second variation negative. A plane has only the nonnegative gradient term. Nonnegativity alone, especially with zero modes, does not prove a strict local minimum or a global minimum.

For a finite catenoid between fixed rings, the answer depends on the ring separation, waist radius, and allowed variations. The axisymmetric second variation of a catenoid tests a specified restricted class. Comparing with two disconnected disks tests a different class of competitors, including a change in topology. Consequently, a stationary catenoid can have greater surface area than two disks without immediately collapsing under every small perturbation.