Constant Gaussian curvature surfaces in 3-space containing lines How can one construct surfaces in $\mathbb R^3$ of constant negative Gaussian curvature containing a line in $\mathbb R^3$? (this question is inspired by this MSE post).
 A: Given any point $p$ on a surface $S$ of Gauss curvature -1, 
there exists an open neighborhood $U\subset S$ and $p$-centered
coordinates $(x,y):U\to\mathbb{R}^2$, whose image is
a domain $R = (x,y)(U)\subset\mathbb{R}^2$ 
and a function $u:R\to (0,\pi/2)$ such that the first
and second fundamental forms of the surface are given by
$$
\mathrm{I} = \cos^2\!u\,\mathrm{d}x^2 + \sin^2\!u\,\mathrm{d}y^2\tag1
$$
and 
$$
\mathrm{I\!I} = \cos u\,\sin u\,(\mathrm{d}x^2-\mathrm{d}y^2)
= \cos u\,\sin u\,(\mathrm{d}x-\mathrm{d}y)(\mathrm{d}x+\mathrm{d}y).\tag2
$$
The function $u$ satisfies the sine-Gordon equation
$$
u_{xx}-u_{yy} = \cos u\,\sin u.\tag3
$$
Conversely, given a function $u:R\to (0,\tfrac12\pi)$ 
that satisfies (3)
on a simply connected domain $R\subset\mathbb{R}^2$,
there exists an immersion $X:R\to\mathbb{E}^3$ whose first
and second fundamental forms are given 
by $\mathrm{I}$ and $\mathrm{I\!I}$ above; 
this immersion is unique up to rigid motion and has Gauss curvature $-1$. 
The asymptotic curves of the immersion $X$ 
are given by holding $x+y$ or $x-y$ constant.
The angle between the asymptotic curves (appropriately
oriented) is $2u$, and the ambient curvatures in $\mathbb{E}^3$ 
of the asymptotic curves described by holding $x\pm y$ constant
are 
$$
\kappa_{\pm} = 2 (u_x \mp u_y).\tag 4
$$
(Here, $\kappa_{\pm}$, which are the curvatures of the asymptotic curves, are not to be mistaken for the principal curvatures of the surface $X(R)$, 
which are $\tan u$ and $-\cot u$.)
Thus, constructing a surface in $\mathbb{E}^3$ with Gauss curvature $-1$
that contains a straight line (which is necessarily asymptotic) is equivalent
to finding a (local) solution $u(x,y)$ of the sine-Gordon equation that
satisfies the 'characteristic' initial condition $u_x(x,x)+u_y(x,x) = 0$.
(It's called 'characteristic' because the curves $x\pm y = c$ are the 
characteristic curves of the PDE (3).)
Unfortunately, characteristic initial value problems for nonlinear PDE can be delicate to study; standard techniques are generally not adequate to prove
existence or uniqueness.  However, this problem can be partially avoided by
looking for special solutions that will satisfy the desired initial 
condition by virtue of other special properties being assumed.
If one seeks a solution to (3) of the form $u(x,y) = \tfrac12\,f(x^2{-}y^2)$, 
one finds that $f$ must satisfy the (singular, nonlinear) 
ordinary differential equation
$$
4t\,f''(t) + 4f'(t) = \sin\bigl(f(t)\bigr).\tag5
$$
Moreover, since, for such a solution, one has 
$(u_x \mp u_y) = f'(x^2{-}y^2)(x \pm y)$, 
it will follow that the two lines $x{\pm}y = 0$ in $R$ 
will be mapped by $X$ into straight lines in $\mathbb{E}^3$.

To see that there are nontrivial solutions to the singular ODE (5), 
let $\theta\in(0,\pi)$ be chosen, 
and write $f(t) = \theta + \tfrac14\,\sin(\theta)\,t + g(t)$.
Then (5) becomes the equation
$$
4t\,g''(t)+4g'(t) 
= \sin\bigl(\theta+\tfrac14\sin\theta\,t+g(t)\bigr) -\sin\theta
= G\bigl(\theta,t,g(t)\bigr)
\tag6
$$
The left hand side of (6) is a singular linear differential operator
at $t=0$ with no resonances, while $G(\theta,t,g)$ is an analytic
function on $\mathbb{R}^3$ satisfying $G(\theta,0,0) = 0$. 
Thus, there is a unique convergent power series 
solution to (6) satisfying $g(0)=0$, 
which implies the convergence of the corresponding formal power series for $f$:
$$
\begin{aligned}
f(t) &= \theta + \frac{\sin\theta}{4}\,t
       + \frac{\cos\theta\sin\theta}{64}\,t^2
       + \frac{(3\cos^2\theta-2)\sin\theta}{2304}\,t^3\\ 
  &\quad\quad + \frac{\cos\theta\,(18\cos^2\theta-17)\sin\theta}{147456}\,t^4 + \cdots,
\end{aligned}
\tag7
$$
yielding a solution to (5) satisfying $f(0) = \theta$ on an open interval around $t=0$.  It is easy to see from (5) that this local solution to (5) satisfying
the initial condition $f(0)=\theta$ extends uniquely to a solution to (5) 
defined on the entire real line $\mathbb{R}$.  The range of $f$
will not be contained in the interval $(0,\pi)$, of course, but, there
will be an interval $\bigl(a(\theta),b(\theta)\bigr)\subset\mathbb{R}$ 
containing $0$ that $f$ maps into $(0,\pi)$.

Then the corresponding $X$ will immerse the region between the two hyperbolae $x^2-y^2 = a(\theta)<0$ and $x^2-y^2= b(\theta)>0$ into $\mathbb{E}^3$ as a surface of constant Gauss curvature $-1$ and will map the two lines $x\pm y = 0$ into two straight lines that intersect at an angle of $\theta$ in the image surface.
Added remark: (23 October 2017) After doing a literature search on examples of surfaces of constant curvature $-1$ in $\mathbb{E}^3$, I have found out that this surface was discovered by M. H. Amsler, Des surfaces à courbure négative constante dans l'espace à trois dimensions et de leurs singularités, Math. Ann. 130 (1955), 234—256.
