Geodesics in a right cylinder

Consider the right cylinder $X(u,v)=\alpha(u)+v\vec{a}$, where $\alpha$ is a regular curve contained in a orthogonal plane to the vector $\vec{a}$. A curve $\alpha(t)=X(u(t),v(t))$ is a geodesic if the tangent part of $\alpha''(t)$ vanishes for every $t$. We have

$$\alpha''(t)=u''X_u+v''X_v+u'^2 X_{uu}+2u'v' X_{uv}+v'^2X_{vv},$$

where

$$X_u=\alpha'(u),\ X_v= \vec{a}$$

$$X_{uu}=\alpha''(u),\ X_{uv}=X_{vv}=0.$$

Thus the tangent part of $\alpha''$ is 

$$\alpha''(t)^T=u''\alpha'(u)+v''\vec{a}+u'^2\alpha''(t)^T.$$

Since $\alpha''(t)=\kappa(t)n(t)$ is a vector orthogonal to the surface, $\alpha''(t)^T=0$. Thus $\alpha$ is a geodesic if and only if $$u''=0,\ v''=0\Leftrightarrow u(t)=at+b, v(t)=ct+d,$$

for some constant $a,b,c,d$. Then the preimage of $\alpha$ in the domain of the parametrization is a straight-line so when we carry into the surface by $X$ we obtain a helix. For example, when $u=ct$, the geodesic is a vertical line in the cylinder and when $v=ct$, the geodesic is a vertical translation of the base curve $\alpha$. In the picture, and a for a circular cylinder $x^2+y^2=1$,  we have the lines in the domain of $X$, the geodesics and the cylinder with the geodesics.

The exponential map in the plane and in the sphere

A geodesic in a plane $P$ writes as $\gamma(s)=p+s v$, $p\in P$ and $v\in T_pP=P$. Then the exponential map is simply $$exp(v)=p+v$$

that is, the exponential map at $p$ is a translation on $P$ with translation vector $p$. Let us observe that the $\mbox{exp}$ is a diffeomorphism in the whole $T_pP$ and thus the normal neighborhood of $p$ is the very plane $P$.

In the unit sphere ${\mathbb S}^2$  a geodesic is $\gamma(s)=\cos(s)p+\sin(s) v/|v|$, where $p\in  {\mathbb S}^2$ and $v\in T_p{\mathbb S}^2$. Then $$exp(v)=\cos(1)p+\sin(1)v/|v|.$$ We want to study when $\mbox{exp}$ is one-to-one. By the property $\gamma(t;p,\lambda v)=\gamma(\lambda t; p,v)$, we have $\mbox{exp}(\lambda v)=\gamma(t)$. If $|v|=1$, for $\lambda=\pi$, $\gamma(\pi;p,\lambda v)=-p$. This implies that the exponential map $\mbox{exp}$ is one-to-one in the ball $B_\pi(0)$ of radius $\pi$.

Surfaces of revolution with constant mean curvature (II)

Following the above entry, the case $H=0$ is known: the plane and the catenoid are the only rotational minimal surfaces. If $H=0$, then we have 

$$\frac{f(z)}{\sqrt{1+f'(z)^2}}=c,\ c>0,$$ that is, $$\frac{f'}{\sqrt{f^2-c^2}}=\frac{1}{c}.$$ Then the solution is $$f(z)=c\cosh(\frac{1}{c}z+d),\ d\in{\mathbb R} (*).$$

Minimal surfaces are models of soap films. In this particular case of the catenoid, the surface is the soap film formed by two coaxial circles, that is, two circles $C_1\cup C_2$ in parallel planes and the straight-line joining their centers is orthogonal to the planes containing the circles. It is natural to ask if there exists a soap film joining two given circles in parallel planes. In order to simplify the arguments, we suppose 

1. the radii of the circles are identical, namely, $r>0$. 
2. the circles $C_1$ and $C_2$  are contained in the planes of equation $z=-h$ and $z=h>0$, respectively. 

Then we pose the next:

Problem. Under what conditions on $r$ and $h$ does exist a catenoid $S$ joining $C_1$ and $C_2$? In such a case, how many catenoids do exist?

By the symmetry of the hyperbolic cosine, and since $f(h)=f(-h)$, we conclude $d=0$ in (*). Thus, the problem reduces to find $c>0$ such that $$c\cosh(\frac{1}{c}h)=r (**).$$

It is natural to think that if the circles lie very close, then there do exists a catenoid, that is, if $h$ is small, then there exists a solution of (**). 

We propose the problem in the next direction. We suppose that the circles are given (the radius $r$, which we suppose $r=1$). If they are close, there exists a catenoid, but if we separate far then the catenoid is destroyed, that is, there do no exist a catenoid between both circles.

In order to simplify the problem, we do a homothety of the ambient space from the origin and we suppose that the value $r$ of the radius is $r=1$. Consider the function $$g(c)=c\cosh(\frac{1}{c}h).$$ Our idea is using the mean value theorem. It is not difficult to see that $$\lim_{c\rightarrow 0}g(c)= \lim_{c\rightarrow \infty}g(c)=\infty,$$

so we have to study carefully the monotonicity intervals of $g$. We calculate the critical points of $g$. We have

$$g'(c)=\cosh(h/c)-\frac{h}{c}\sinh(h/c).$$ By letting $y=h/c$, this is equivalent to find $y>0$ such that $$\frac{1}{y}=\tanh(y).$$ The function $1/y$ is decreasing from $-\infty$ to $0$ and $\tanh(y)$ is increasing from $0$ to $1$, s   there is only one critical point $c_0$ (with $h>c_0$). See the next figure:

Because the limits are $\infty$, then this critical point is a minimum, $c=c_0$. This the graphic of $g$ when $h=0,5$, $h=1$ and $h=4$, and the graphic of $y=1$.

Let us observe that the minimum increases with $h$! so we have to find that it is possible to choose $h$ so the value of this minimum is $1$ at more.

We compute the value of the minimum, that is, $g(c_0)$. We know that $c_0/h=\tanh(h/c_0)$, and numerically we obtain, $h/c_0=1.19968$. Then 

$$g(c_0)=\frac{hc_0}{\sqrt{h^2-c_0^2}}=\frac{h}{\sqrt{(h/c_0)^2-1}}=1,50888 h.$$

When $h$ is close to $0$, $g(c_0)<1$, thus the graphic of $g$ has points under the line $y=1$, proving that there exists two catenoids spanning $C_1\cup C_2$. For a certain height $h=h_0$, this minimum is exactly $1$, so there exists only one catenoid and when $h>h_0$ there do not exist a catenoid joining $C_1$ and $C_2$. The value of $h_0$ is

$$h_0=\frac{1}{1,50888}=0,6627.$$
Thus, and after a homothety, we obtain:

Theorem. Let $d$ be the distance $d$ between two coaxial circles of radii $r>0$.

1. If $d<1,3254 r$, there exists exactly two catenoids spanning $C_1\cup C_2$.
2. If $d=1,3254 r$, there exists exactly two catenoids spanning $C_1\cup C_2$.
3. If $d>1,3254 r$, there do not  exist a catenoid spanning $C_1\cup C_2$.

Now we give one example of two circles that bound two catenoids. Take $r=1$ and we choose $h=0.5$. The solutions of $g(c)=1$ are: $c_1=0,235095$ and $c_2=0,848338$. The picture of the two catenoids is

Finally a remark: when one dips two coaxial circles in a soapy water container, only one catenoid is formed. In the above case, it would be the blue catenoid. Among the two catenoids, the physical systems chooses that catenoid with minimum area (minimum energy) and in this case, is the 'exterior' catenoid.

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  $H=1/(2r)$.  

Sign of $K$ does not imply local convexity

We know that if $K(p)>0$ at one pint, then the surface locally in one side of its affine tangent plane. On the other hand, if the surface lies in one side around a point, then $K(p)\geq 0$. However, there are surfaces that at one point $p\in S$, $K(p)=0$, $K>0$ around $p$, BUT the surface lies in both sides of $T_pS$. An example is the surface $z=f(x,y)=x^3(1+y^2)$. Let $p=(0,0,0)$. Take the parametrization $$X(x,y)=(x,y,x^3(1+y^2)).$$ Then $X^{-1}(p)=(0,0)$ and as $X_x(0,0)=(1,0,0)$ and $X_y(0,0)=(0,1,0)$, then $T_pS$ is the plane $z=0$. Since $1+y^2\geq 0$ and $x^3$ changes of sign at $x=0$, then $f$ change of sign around $p$, that is, the surface has points in both sides of $T_pS$, as it appears in the next figure:

We now compute the Gauss curvature using the formula $$K(X(x,y))=\frac{f_{xx}f_{yy}-f_{xy}^2}{(1+f_x^2+f_y^2)^2}.$$ Then $$K(X(x,y))=\frac{12x^4(1-2y^2)}{(1+f_x^2+f_y^2)^2}.$$ In the open set of $S$ gieven by $V=X(U)$, where $U=\{(x,y): x\in{\mathbb R},|y|<1/\}$, $K>0$ in $V-\{p\}$ and $K(p)=0$. The next figure is the numerator of $K$ in $|x|<1$, $|y|<1/2$ that hows that the sign of $K$ is, indeed, positive.

Asymptotic curves and lines of curvature in a hyperbolic paraboloid

The explicit computation of the asymptotic curves and lines of a curvature in a given surface uses parametrizations. Thus their computations depend what is the chosen parametrization. We illustrate this problem with the hyperbolic paraboloid. The usual way to work with this surface is as $z=x^2-y^2$. Then $X(x,y)=(x,y,x^2-y^2)$ is a parametrization of the surface and the coefficients of $I$ are:

$$E=1+4x^2,\ F=-4xy,\ G=1+4y^2.$$

For the coefficients of the second fundamental form suffices to consider the numerators in the expressions of these coefficients, which are:

$$en=2,\ fn=0,\ gn=-2.$$

Then $\alpha(t)=X(x(t),y(t))$ is an asymptotic curve if $2x'^2-2y'^2=0$, that is, $y(t)=x(t)+c$ or $y(t)=-x(t)+c$, $c\in{\mathbb R}$. If $x(t)=t$, then the curves $X^{-1}\alpha$ are straight-lines of slope $1$ and $-1$. 

For the lines of curvature, $\alpha(t)=X(x(t),y(t))$ is line of curvature if 

$$\left|\begin{array}{ccc}y'^2&-x'y'&x'^2\\ 1+4x^2& -4xy& 1+4y^2\\ 2&0&-2\end{array}\right|=0.$$

If $x(t)=t$, then $$4ty(1-y'^2)=y'(2+4t^2+4y^2),$$ which it is very difficult to solve. 

We change of parametrization of the hyperbolic paraboloid. After a change of variables $u=x-y$, $v=x+y$ and a rotation about the $z$-axis, the surface is $z=xy$. Let $X(x,y)=(x,y,xy)$. Now we have $$E=1+y^2,\ F=xy,\ G=1+x^2$$ and $en=0,\ fn=1,\ gn=0$. Then $\alpha(t)=X(x(t),y(t))$ is asymptotic curve if $x'y'=0$, that is, $x(t)=c$ or $y(t)=c$, or in other words, they are the coordinate curves. For the lines of curvatures, the determinant to solve simplifies into $$(1+y^2)x'^2=(1+x^2)y'^2.$$ Letting $x(t)=t$, then 

$$\frac{y'}{1+ y^2}=\pm\frac{t}{1+t^2}\Rightarrow {\mbox arc}\sinh(y(t))={\mbox arc} \sinh(t)+c.$$

Then $$y(t)=\sinh(\pm {\mbox arc} \sinh(t)+c)=\pm\cosh(c)t+\sinh(c)\sqrt{1+t^2}.$$

Thus with this parametrization we have found the lines of curvature of the surfaces, in contrast to the initial parametrization.