Surfaces of revolution with constant mean curvature (I)

We calculate the equation of a surface of revolution with constant mean curvature $H$. Without loss of generality, we suppose that the profile curve is a planar curve in the $xz$-plane and that the $z$-axis is the rotational axis. Also, suppose that the curve is a graph on the $z$-axis, that is, a parametrization of the profile curve is $(f(z),0,z)$, $z\in I$. Then a parametrization of the rotational surface is $$X(t,s)=(f(t)\cos(s),f(t)\sin(s),t),\ t\in I,s\in [0,2\pi].$$ Thus $H$ satisfies 

$$\frac{-f''}{(1+f'^2)^{3/2}}+\frac{1}{f\sqrt{1+f'^2}}=2H.$$

The key of this equation is that because $H$ is constant, it is possible to obtain a first integral of this equation (which is of second order). Indeed, multiplying by $ff'$ we have $$\frac{-ff'f''}{(1+f'^2)^{3/2}}+\frac{f'}{\sqrt{1+f'^2}}=2Hff',$$which can be written as 

$$\left(\frac{f}{\sqrt{1+f'^2}}\right)'=(Hf^2)'.$$ Therefore there exists $c\in {\mathbb R}$ such that 

$$\frac{f(z)}{\sqrt{1+f'(z)^2}}=Hf(z)^2+c.$$

For example, the sphere and the cylinder can be obtained from (*). For the sphere, take $c=0$. Then we have 

$$f'=\frac{1}{H}\sqrt{\frac{1}{f^2}-H^2},$$

or

$$\frac{f'}{\sqrt{\frac{1}{f^2}-H^2}}=\frac{1}{H}.$$ By integrating, we have

$$\frac{1}{H^2}\sqrt{1-H^2 f^2}=\frac{1}{H}{x}.$$

Definitively, $$f(z)=\sqrt{\frac{1}{H^2}-z^2}$$ which is a circle of radius $1/|H|$, and the surface is a sphere of radius $1/|H|$.

For the cylinder, we have to come back to the initial equation for $H$. If  $f(z)=r$, then  $H=1/(2r)$.  

Comparison of curves by curvatures (II)

The definition of the mean curvature is the average of the principal curvatures. But it is also the average of the normal curvatures in two orthogonal directions, that is, 

$$H(p)=\frac12\left(\kappa_n(\theta)+\kappa_n(\theta+\frac{\pi}{2})\right),\ \ (*)$$ where $\kappa_n(\theta)$ is the normal curvature in the direction that makes an angle $\theta$ with a fix direction of the tangent plane.

Suppose again  $S_1$ and $S_2$  two surfaces with $p\in S_1\cap S_2$, $T_pS_1=T_pS_2$ and $N_1(p)=N_2(p)$, where $N_i$ are the orientation in each surface $S_i$. Take all normal sections through $p$, that is, the intersection of the planes formed by $N_i(p)$ and the tangent vectors. The set of these planes in both surfaces is the same because $N_1(p)=N_2(p)$. Fix $v\in T_p S_i$ and $\Pi_v$ the corresponding normal section. This plane meets $S_i$ in two curves $\alpha_i^v$. We are computing the normal curvature with respect to the normal (of the curve) coincides with $N_i(p)$ at $p$ and this normal curvature is the curvature of the curve

Suppose that $S_1$ lies above $S_2$ around $p$. Then for each $v\in T_pS_i$, the curve $\alpha_1^v$ lies above $\alpha_2^v$ at $p$. But we know by the theory of curves that the curvature of $\alpha_1^v$ is greater or equal to the one of $\alpha_2^v$ at $p$. Thus, all normal curvatures of $S_1$ at $p$ are greater or equal to the ones of $S_2$ at $p$. We state this remarkable result:

Theorem. Suppose $S_1$ and $S_2$ two surfaces tangent at a point $p$ and that the orientations of the surfaces coincide at $p$, that is, $N_1(p)=N_2(p)$. If $S_1$ lies above $S_2$ around $p$, then for avery $v\in T_pS_i$, we have the inequality
$$\kappa_n^1(v)\geq\kappa_n^2(v).$$

As a consequence,  it is immediate from (*) 

Corollary. Suppose $S_1$ and $S_2$ two surfaces tangent at a point $p$ and that the orientations of the surfaces coincide at $p$, that is, $N_1(p)=N_2(p)$. If $S_1$ lies above $S_2$ around $p$, then $H_1(p)\geq H_2(p)$.

Comparison of surfaces by curvatures (I)

We address the following problem: suppose two surfaces tangent at a point $p$ and a surface lies above the other one around $p$. Could we say something about the curvatures of the surfaces at $p$?

Since the sign of the principal curvatures depend on the orientation, suppose that $S_1$ and $S_2$ are two surfaces with $p\in S_1\cap S_2$, $T_pS_1=T_pS_2$ and $N_1(p)=N_2(p)$, where $N_i$ are the orientation in each surface $S_i$. After a rigid motion, we suppose that $p=(0,0,0)$,  $T_pS_i$ is the horizontal plane of equation $z=0$ and $N_i(p)=(0,0,1)$. By this condition, $S_i$ is locally the graph of a function on the tangent plane $T_pS_i$ (exercise!). Let $S_1$ and $S_2$ be the graph of $u$ and $v$ respectively.

Because the tangent plane is $z=0$, then with the usual parametrization $X^i(x,y)$ of the surface, 

$X^1_x(0,0)=(1,0,u_x(0,0))$, $X^1_y(0,0)=(0,1,u_y(0,0))$. Since they belong to the $xy$-plane then $u_x(0,0)=u_y(0,0)=0$. Similarly, $v_x(0,0)=v_y(0,0)=0$. By the expression of the mean curvature $H$, we obtain

$$H_1(p)=\frac12(u_{xx}+u_{yy})(0,0),\ H_2(p)=\frac12(v_{xx}(0,0)+v_{yy}(0,0)\ \ (*).$$

Suppose that $S_1$ lies above $S_2$ around $p$ and with respect to the direction that indicates $N_i(p)$. This is equivalent to $u\geq v$ around $q=(0,0)$. Thus the function $u-v$ attains a minimum at $q$. In particular, the Hessian of $(u-v)$ at $q$ is positive semidefinite. But the Hessian is

$$\left(\begin{array}{cc} (u-v)_{xx}&(u-v)_{xy}\\ (u-v)_{xy} &(u-v)_{yy}\end{array}\right)(q).$$

Thus $$(u-v)_{xx}(q)\geq 0,\ \ (u-v)_{yy}(q)\geq 0$$

and (*) implies $H_1(p)\geq H_2(p)$. Summarizing, the result is:

Theorem. Suppose $S_1$ and $S_2$ two surfaces tangent at a point $p$ and that the orientations of the surfaces coincide at $p$, that is, $N_1(p)=N_2(p)$. If $S_1$ lies above $S_2$ around $p$, then $H_1(p)\geq H_2(p)$.

Translation surfaces with constant zero curvature

A translation surface is a surface that is the sum of two planar curves contained in orthogonal planes. Thus a parametrization of the surfaces is $$X(u,v)=(x,0,f(x))+(0,y,g(y))$$ where $f$ and $g$ are smooth functions defined in some intervals of ${\mathbb R}$. We are interesting in the translation surfaces with $H=0$ or $K=0$ on the whole surface. 

For the mean curvature, it is immediate that $H=0$ is equivalent to $$\frac{f''(x)}{1+f'(x)^2}+\frac{g''(y)}{1+g'(y)^2}=0.$$ Then necessarily we have that $$\frac{f''(x)}{1+f'(x)^2}=-\frac{g''(y)}{1+g'(y)^2}=c$$ for some real number $c$. If $c=0$, then $f''=g''=0$, obtaining $f(x)=ax+b$, $g(y)=cy+d$ and $z=ax+cy+b+d$, that is, the surface is a plane. If $c\not=0$, integrating $f$ and $g$ we obtain $$f(x)=-\frac{1}{c}\log\cos(cx+m),\ g(y)=\frac{1}{c}\log\cos(cy+n),\ m,n\in{\mathbb R}.$$ Thus we write $$z=\frac{1}{c}\log\left(\frac{\cos(cy+n)}{\cos(cx+m)}\right).$$ This surface is called the Scherk's surface. In order to study the domain of the $z(x,y)$, we take $c=1$ and $m=n=0$. Then $$z=\log\left|\frac{\cos(y)}{\cos(x)}\right|.$$ Then the domain is 

$$(x,y)\in(-\frac{\pi}{2},\frac{\pi}{2})\times (-\frac{\pi}{2},\frac{\pi}{2}).$$

It is clear that in the sides of this square, the function $z=z(x,y)$ takes $\infty$ or $-\infty$ values, as it is shown in the next picture.

If we now study translation surfaces with $K=0$, then this identity is equivalent to 

$$f''g''=0.$$

Then $f''=0$ identically or $g''=0$ identically. Without loss of generality, we suppose $f''=0$, that is, $f(x)=ax+b$ for some numbers $a,b$. Then the surface writes as $z=ax+g(y)+b$ or in terms of $X$, $$X(x,y)=x(1,0,a)+(0,y,g(y)).$$ This surface is a ruled surface whose base curve is any curve as $\alpha(y)=(0,y,g(y))$ and the rulings as parallel to the direction $(1,0,a)$. In the picture we consider $a=0$ and $g(y)=\sin(y)$.