Surfaces of revolution with positive constant Gauss curvature

We know that when we write $K=c$ in the family of rotational surfaces, then this equation is an ordinary differential equation, so there is a unique solution for each initial conditions. We show this phenomenon when $K=1$. Suppose that the profile curve is locally a graph on the rotation axis, that is, $z\mapsto (f(z),0,z)$ for $z\in I$, $f(z)>0$. The parametrization of the surface is $X(z,s)=(f(z)\cos(s),f(z)\sin(s),z)$. Equation $K=1$ writes as 

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

Thus we have $f''+f(1+f'^2)^2=0$. This is differential equation is not possible to integrate, up to special cases. We think that sphere should easily solve. The initial conditions are put on $z=0$, that is $f(0)=xo$ and $f'(0)=0$. With this last condition, we are imposing that the tangent line at $z=0$ is vertical. Moreover, by this condition, we can suppose that the solution is symmetric with respect to $z=0$. 

We use Mathematica to solve numerically the initial value problem $$(*) \left\{\begin{array}{l} f''+f(1+f'^2)^2=0\\ f(0)=xo\\ f'(0)=0\end{array}\right.$$ We study the solutions depending on the initial value $xo$, that is, the intersection point of the profile curve with the $x$-axis.

When $xo=1$, we know that the solution is the sphere, exactly, $f(z)=\sqrt{1-z^2}$ is a solution of (*). 

In order to study with Mathematica (*) we write here the sentences: 

profile =  NDSolve[{F''[z] + F[z] (1 + F'[z]^2)^2 == 0, F[0] == xo, F'[0] == 0}, F[z], {z, -Zo, Zo}]

f[z_] := F[z] /. profile[[1]]

ParametricPlot[{{z, 0}, {f[z], z}}, {z, -Zo, Zo}, PlotRange -> All]

The first line numerically solves the ODE with initial conditions as we have presented. Here $Zo$ is the width of the interval when the solution $f$ is defined. The second line `takes' the numerical value f in order to manage in the next line, where we plot the profile curve. In fact, the last line indicates that we also draw the $x$-line. I write this because Mathematica `reduces' the picture to the interval where is defined the solution and we want to compare the profile curve with its position with respect to the rotation axis. Finally, we use

ParametricPlot3D[{f[z] Cos[s], f[z] Sin[s], z}, {s, 0, 2 Pi}, {z, -Zo, Zo}]

for drawing the surface.

We begin with the study and sphere is our starting point: sphere appears when $xo=1$ and the domain of $f$ is for $Zo=1$. We now increase $xo$, for example $xo=1.5$. If we put $Zo=1$, Mathematica says that the solution is not defined in the interval $(-Zo,Zo)$ because appear errors. In fact, Mathematica says what is the maximum interval. In this example, the output is

NDSolve::ndsz: At z == -0.559099, step size is effectively zero; singularity or stiff system suspected.

This means that we have to take $Zo=0.5590$, obtaining the profile curve in its maximum domain, namely:

If we increase $xo$, that is, we move far the point $(f(xo),0,0)$, the profile moves far from the rotation axis: let us observe that the profile curve does not meet the rotation axis. In the figure, it indicates that the tangent plane at the boundary circles is horizontal, and the surface one `hole'.

Now we let $xo\rightarrow 0$. If $xo=0.7$, and for $Zo=2$ we see that the profile curve meets the $z$-axis, which is not possible.

Then, and after some trials, we see that for $Zo=1.35$, the profile meets exactly the $z$-axis. The figures are:

Now the surface presents two `singularities' exactly in the intersection points with the $z$-axis.