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To answer Joseph's questions:

First, it's not impossible to integrate the geodesic flow of the hyperbolic plane in these coordinates, but the formulae I got aren't very nice, so I'm not going to type them in unless I can find a better way to express them. It's probably easier than I got on a first pass through, but I don't have time to work on simplifying them right now.

Added note: (It helps to sleep on a problem sometimes.) If you set $a = r\cos t>0$ and $b=r\sin t>0$ and make the change of variables $$ v=\arcsin\bigl((\tan t)(\tan \theta)\bigr) \qquad\text{and}\qquad u= \frac{\log\rho - f(\theta)}{\sin t} $$ where $0<\theta< \tfrac12\pi{-}t$ and $0 < \rho < \infty$ and where $f$ (not an elementary function) is defined on $0<\theta<\tfrac12\pi{-}t$ so that $f(\tfrac12\pi{-}t)=0$ and so that $$ f'(\theta) = \frac{\sqrt{\cos^2 t - \sin^2\theta}}{\sin\theta}, $$ then the induced metric on the lower nappe of the surface (which is what Joseph drew) becomes $$ ds^2 = r^2\left(\frac{d\rho^2 + \rho^2 d\theta^2}{\rho^2\cos^2\theta}\right) = r^2\left(\frac{dx^2 + dy^2}{x^2}\right), $$ where $x = \rho\cos\theta$ and $y = \rho\sin\theta$. Now everything, including integrating the geodesics, is obvious.

Second, there does indeed exist a geodesic (and only one) that starts at any given point on the rim and spirals down the surface towards $z = -\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $0<v<\pi/2$. There's also one that starts at the same point and spirals up the surface towards $z=+\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $\pi/2<v<\pi$. (As j.c. has also noted, the given parametrization is not an immersion along $v=\pi/2$, but has a cusp singularity along the rim.)

I'm not good a drawing computer pictures, but here is a description of what you get when you develop the region $0<v<\pi/2$ into the hyperbolic plane with curvature $K=-1/(a^2+b^2)$ with $b\not=0$: First, the rim $v=\pi/2$ maps to a curve $C$ with geodesic curvature $\kappa=-a/(a^2+b^2)$. Of course, this curve meets the circle at infinity (i.e., the ideal boundary) at two distinct points $P_+$ (as $u\to+\infty$) and $P_-$ (as $u\to-\infty$). If you let $L$ be the actual geodesic that also has $P_+$ and $P_-$ as endpoints, then the developing map carries the region $0<v<\pi/2$ into the region $R$ of the hyperbolic plane that fits between $C$ and $L$. The curves $v=v_0$ for $0<v_0<\pi/2$ are just the other constant curvature curves in $R$ that join $P_-$ to $P_+$. Each of the curves $u=u_0$ then maps to a curve that is asymptotic to the geodesic $L$, but comes up and touches the curve $C$ while making a cusp there at $(u_0,\pi/2)$. (The part $\pi/2<v<\pi$ just covers $R$ again, so, all told, the strip $0<v<\pi$ covers $R$ twice, folding along $v=\pi/2$ and mapping this fold onto $C$.)

Now, if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_-$, then this corresponds to a geodesic on the Dini surface that spirals down to $z=-\infty$. Similarly if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_+$, then this corresponds to a geodesic on the Dini surface that spirals up to $z=+\infty$.

To answer Joseph's questions:

First, it's not impossible to integrate the geodesic flow of the hyperbolic plane in these coordinates, but the formulae I got aren't very nice, so I'm not going to type them in unless I can find a better way to express them. It's probably easier than I got on a first pass through, but I don't have time to work on simplifying them right now.

Added note: (It helps to sleep on a problem sometimes.) If you set $a = r\cos t>0$ and $b=r\sin t>0$ and make the change of variables $$ v=\arcsin\bigl((\tan t)(\tan \theta)\bigr) \qquad\text{and}\qquad u= \frac{\log\rho - f(\theta)}{\sin t} $$ where $0<\theta< \tfrac12\pi{-}t$ and $0 < \rho < \infty$ and where $f$ (an elementary function, but not a nice one, apparently) is defined on $0<\theta<\tfrac12\pi{-}t$ so that $$ f'(\theta) = \frac{\sqrt{\cos^2 t - \sin^2\theta}}{\sin\theta}, $$ then the induced metric on the lower nappe of the surface (which is what Joseph drew) becomes $$ ds^2 = r^2\left(\frac{d\rho^2 + \rho^2 d\theta^2}{\rho^2\cos^2\theta}\right) = r^2\left(\frac{dx^2 + dy^2}{x^2}\right), $$ where $x = \rho\cos\theta$ and $y = \rho\sin\theta$. Now everything, including integrating the geodesics, is obvious.

Second, there does indeed exist a geodesic (and only one) that starts at any given point on the rim and spirals down the surface towards $z = -\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $0<v<\pi/2$. There's also one that starts at the same point and spirals up the surface towards $z=+\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $\pi/2<v<\pi$. (As j.c. has also noted, the given parametrization is not an immersion along $v=\pi/2$, but has a cusp singularity along the rim.)

I'm not good a drawing computer pictures, but here is a description of what you get when you develop the region $0<v<\pi/2$ into the hyperbolic plane with curvature $K=-1/(a^2+b^2)$ with $b\not=0$: First, the rim $v=\pi/2$ maps to a curve $C$ with geodesic curvature $\kappa=-a/(a^2+b^2)$. Of course, this curve meets the circle at infinity (i.e., the ideal boundary) at two distinct points $P_+$ (as $u\to+\infty$) and $P_-$ (as $u\to-\infty$). If you let $L$ be the actual geodesic that also has $P_+$ and $P_-$ as endpoints, then the developing map carries the region $0<v<\pi/2$ into the region $R$ of the hyperbolic plane that fits between $C$ and $L$. The curves $v=v_0$ for $0<v_0<\pi/2$ are just the other constant curvature curves in $R$ that join $P_-$ to $P_+$. Each of the curves $u=u_0$ then maps to a curve that is asymptotic to the geodesic $L$, but comes up and touches the curve $C$ while making a cusp there at $(u_0,\pi/2)$. (The part $\pi/2<v<\pi$ just covers $R$ again, so, all told, the strip $0<v<\pi$ covers $R$ twice, folding along $v=\pi/2$ and mapping this fold onto $C$.)

Now, if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_-$, then this corresponds to a geodesic on the Dini surface that spirals down to $z=-\infty$. Similarly if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_+$, then this corresponds to a geodesic on the Dini surface that spirals up to $z=+\infty$.

To answer Joseph's questions:

First, it's not impossible to integrate the geodesic flow of the hyperbolic plane in these coordinates, but the formulae I got aren't very nice, so I'm not going to type them in unless I can find a better way to express them. It's probably easier than I got on a first pass through, but I don't have time to work on simplifying them right now.

Added note: (It helps to sleep on a problem sometimes.) If you set $a = r\cos t$ and $b=r\sin t\not=0$ and make the change of variables $$ \theta = \arctan\bigl((\cot t)(\sin v)\bigr)\qquad\text{and}\qquad \rho=\exp((\sin t) u + f(\theta)) $$ where $f$ (which is $\log$ of an elliptic function) is defined on $0<\theta<\tfrac12\pi - t$ so that $f(\tfrac12\pi - t)=0$ and so that $$ f'(\theta) = \frac{\sqrt{\cos^2 t - \sin^2\theta}}{\sin\theta}, $$ then one has ranges $0<\theta< \tfrac12\pi - t$ and $-\infty < u < \infty$, and the induced metric on the surface becomes $$ ds^2 = r^2\left(\frac{d\rho^2 + \rho^2 d\theta^2}{\rho^2\cos^2\theta}\right) = r^2\left(\frac{dx^2 + dy^2}{x^2}\right), $$ where $x = \rho\cos\theta$ and $y = \rho\sin\theta$. Now everything, including integrating the geodesics, is obvious.

Second, there does indeed exist a geodesic (and only one) that starts at any given point on the rim and spirals down the surface towards $z = -\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $0<v<\pi/2$. There's also one that starts at the same point and spirals up the surface towards $z=+\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $\pi/2<v<\pi$. (As j.c. has also noted, the given parametrization is not an immersion along $v=\pi/2$, but has a cusp singularity along the rim.)

I'm not good a drawing computer pictures, but here is a description of what you get when you develop the region $0<v<\pi/2$ into the hyperbolic plane with curvature $K=-1/(a^2+b^2)$ with $b\not=0$: First, the rim $v=\pi/2$ maps to a curve $C$ with geodesic curvature $\kappa=-a/(a^2+b^2)$. Of course, this curve meets the circle at infinity (i.e., the ideal boundary) at two distinct points $P_+$ (as $u\to+\infty$) and $P_-$ (as $u\to-\infty$). If you let $L$ be the actual geodesic that also has $P_+$ and $P_-$ as endpoints, then the developing map carries the region $0<v<\pi/2$ into the region $R$ of the hyperbolic plane that fits between $C$ and $L$. The curves $v=v_0$ for $0<v_0<\pi/2$ are just the other constant curvature curves in $R$ that join $P_-$ to $P_+$. Each of the curves $u=u_0$ then maps to a curve that is asymptotic to the geodesic $L$, but comes up and touches the curve $C$ while making a cusp there at $(u_0,\pi/2)$. (The part $\pi/2<v<\pi$ just covers $R$ again, so, all told, the strip $0<v<\pi$ covers $R$ twice, folding along $v=\pi/2$ and mapping this fold onto $C$.)

Now, if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_-$, then this corresponds to a geodesic on the Dini surface that spirals down to $z=-\infty$. Similarly if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_+$, then this corresponds to a geodesic on the Dini surface that spirals up to $z=+\infty$.

To answer Joseph's questions:

First, it's not impossible to integrate the geodesic flow of the hyperbolic plane in these coordinates, but the formulae I got aren't very nice, so I'm not going to type them in unless I can find a better way to express them. It's probably easier than I got on a first pass through, but I don't have time to work on simplifying them right now.

Added note: (It helps to sleep on a problem sometimes.) If you set $a = r\cos t>0$ and $b=r\sin t>0$ and make the change of variables $$ v=\arcsin\bigl((\tan t)(\tan \theta)\bigr) \qquad\text{and}\qquad u= \frac{\log\rho - f(\theta)}{\sin t} $$ where $0<\theta< \tfrac12\pi{-}t$ and $0 < \rho < \infty$ and where $f$ (not an elementary function) is defined on $0<\theta<\tfrac12\pi{-}t$ so that $f(\tfrac12\pi{-}t)=0$ and so that $$ f'(\theta) = \frac{\sqrt{\cos^2 t - \sin^2\theta}}{\sin\theta}, $$ then the induced metric on the lower nappe of the surface (which is what Joseph drew) becomes $$ ds^2 = r^2\left(\frac{d\rho^2 + \rho^2 d\theta^2}{\rho^2\cos^2\theta}\right) = r^2\left(\frac{dx^2 + dy^2}{x^2}\right), $$ where $x = \rho\cos\theta$ and $y = \rho\sin\theta$. Now everything, including integrating the geodesics, is obvious.

Second, there does indeed exist a geodesic (and only one) that starts at any given point on the rim and spirals down the surface towards $z = -\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $0<v<\pi/2$. There's also one that starts at the same point and spirals up the surface towards $z=+\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $\pi/2<v<\pi$. (As j.c. has also noted, the given parametrization is not an immersion along $v=\pi/2$, but has a cusp singularity along the rim.)

I'm not good a drawing computer pictures, but here is a description of what you get when you develop the region $0<v<\pi/2$ into the hyperbolic plane with curvature $K=-1/(a^2+b^2)$ with $b\not=0$: First, the rim $v=\pi/2$ maps to a curve $C$ with geodesic curvature $\kappa=-a/(a^2+b^2)$. Of course, this curve meets the circle at infinity (i.e., the ideal boundary) at two distinct points $P_+$ (as $u\to+\infty$) and $P_-$ (as $u\to-\infty$). If you let $L$ be the actual geodesic that also has $P_+$ and $P_-$ as endpoints, then the developing map carries the region $0<v<\pi/2$ into the region $R$ of the hyperbolic plane that fits between $C$ and $L$. The curves $v=v_0$ for $0<v_0<\pi/2$ are just the other constant curvature curves in $R$ that join $P_-$ to $P_+$. Each of the curves $u=u_0$ then maps to a curve that is asymptotic to the geodesic $L$, but comes up and touches the curve $C$ while making a cusp there at $(u_0,\pi/2)$. (The part $\pi/2<v<\pi$ just covers $R$ again, so, all told, the strip $0<v<\pi$ covers $R$ twice, folding along $v=\pi/2$ and mapping this fold onto $C$.)

Now, if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_-$, then this corresponds to a geodesic on the Dini surface that spirals down to $z=-\infty$. Similarly if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_+$, then this corresponds to a geodesic on the Dini surface that spirals up to $z=+\infty$.

To answer Joseph's questions:

First, it's not impossible to integrate the geodesic flow of the hyperbolic plane in these coordinates, but the formulae I got aren't very nice, so I'm not going to type them in unless I can find a better way to express them. It's probably easier than I got on a first pass through, but I don't have time to work on simplifying them right now.

Second, there does indeed exist a geodesic (and only one) that starts at any given point on the rim and spirals down the surface towards $z = -\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $0<v<\pi/2$. There's also one that starts at the same point and spirals up the surface towards $z=+\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $\pi/2<v<\pi$. (As j.c. has also noted, the given parametrization is not an immersion along $v=\pi/2$, but has a cusp singularity along the rim.)

I'm not good a drawing computer pictures, but here is a description of what you get when you develop the region $0<v<\pi/2$ into the hyperbolic plane with curvature $K=-1/(a^2+b^2)$ with $b\not=0$: First, the rim $v=\pi/2$ maps to a curve $C$ with geodesic curvature $\kappa=-a/(a^2+b^2)$. Of course, this curve meets the circle at infinity (i.e., the ideal boundary) at two distinct points $P_+$ (as $u\to+\infty$) and $P_-$ (as $u\to-\infty$). If you let $L$ be the actual geodesic that also has $P_+$ and $P_-$ as endpoints, then the developing map carries the region $0<v<\pi/2$ into the region $R$ of the hyperbolic plane that fits between $C$ and $L$. The curves $v=v_0$ for $0<v_0<\pi/2$ are just the other constant curvature curves in $R$ that join $P_-$ to $P_+$. Each of the curves $u=u_0$ then maps to a curve that is asymptotic to the geodesic $L$, but comes up and touches the curve $C$ while making a cusp there at $(u_0,\pi/2)$. (The part $\pi/2<v<\pi$ just covers $R$ again, so, all told, the strip $0<v<\pi$ covers $R$ twice, folding along $v=\pi/2$ and mapping this fold onto $C$.)

Now, if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_-$, then this corresponds to a geodesic on the Dini surface that spirals down to $z=-\infty$. Similarly if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_+$, then this corresponds to a geodesic on the Dini surface that spirals up to $z=+\infty$.

To answer Joseph's questions:

First, it's not impossible to integrate the geodesic flow of the hyperbolic plane in these coordinates, but the formulae I got aren't very nice, so I'm not going to type them in unless I can find a better way to express them. It's probably easier than I got on a first pass through, but I don't have time to work on simplifying them right now.

Added note: (It helps to sleep on a problem sometimes.) If you set $a = r\cos t$ and $b=r\sin t\not=0$ and make the change of variables $$ \theta = \arctan\bigl((\cot t)(\sin v)\bigr)\qquad\text{and}\qquad \rho=\exp((\sin t) u + f(\theta)) $$ where $f$ (which is $\log$ of an elliptic function) is defined on $0<\theta<\tfrac12\pi - t$ so that $f(\tfrac12\pi - t)=0$ and so that $$ f'(\theta) = \frac{\sqrt{\cos^2 t - \sin^2\theta}}{\sin\theta}, $$ then one has ranges $0<\theta< \tfrac12\pi - t$ and $-\infty < u < \infty$, and the induced metric on the surface becomes $$ ds^2 = r^2\left(\frac{d\rho^2 + \rho^2 d\theta^2}{\rho^2\cos^2\theta}\right) = r^2\left(\frac{dx^2 + dy^2}{x^2}\right), $$ where $x = \rho\cos\theta$ and $y = \rho\sin\theta$. Now everything, including integrating the geodesics, is obvious.

Second, there does indeed exist a geodesic (and only one) that starts at any given point on the rim and spirals down the surface towards $z = -\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $0<v<\pi/2$. There's also one that starts at the same point and spirals up the surface towards $z=+\infty$, i.e., it starts at a given $(u_0,v_0)=(u_0,\pi/2)$ and goes into the part of the surface with $\pi/2<v<\pi$. (As j.c. has also noted, the given parametrization is not an immersion along $v=\pi/2$, but has a cusp singularity along the rim.)

I'm not good a drawing computer pictures, but here is a description of what you get when you develop the region $0<v<\pi/2$ into the hyperbolic plane with curvature $K=-1/(a^2+b^2)$ with $b\not=0$: First, the rim $v=\pi/2$ maps to a curve $C$ with geodesic curvature $\kappa=-a/(a^2+b^2)$. Of course, this curve meets the circle at infinity (i.e., the ideal boundary) at two distinct points $P_+$ (as $u\to+\infty$) and $P_-$ (as $u\to-\infty$). If you let $L$ be the actual geodesic that also has $P_+$ and $P_-$ as endpoints, then the developing map carries the region $0<v<\pi/2$ into the region $R$ of the hyperbolic plane that fits between $C$ and $L$. The curves $v=v_0$ for $0<v_0<\pi/2$ are just the other constant curvature curves in $R$ that join $P_-$ to $P_+$. Each of the curves $u=u_0$ then maps to a curve that is asymptotic to the geodesic $L$, but comes up and touches the curve $C$ while making a cusp there at $(u_0,\pi/2)$. (The part $\pi/2<v<\pi$ just covers $R$ again, so, all told, the strip $0<v<\pi$ covers $R$ twice, folding along $v=\pi/2$ and mapping this fold onto $C$.)

Now, if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_-$, then this corresponds to a geodesic on the Dini surface that spirals down to $z=-\infty$. Similarly if you take the geodesic ray in the hyperbolic plane that joins the developed image of $(u_0,\pi/2)$ (on $C$) to the ideal point $P_+$, then this corresponds to a geodesic on the Dini surface that spirals up to $z=+\infty$.