Jacobi fields on a "bump surface" - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:40:33Z http://mathoverflow.net/feeds/question/12341 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12341/jacobi-fields-on-a-bump-surface Jacobi fields on a "bump surface" Tom LaGatta 2010-01-19T21:51:00Z 2010-01-20T02:09:09Z <p>Consider a "bump surface" which looks like the following:</p> <p><img src="http://img149.imageshack.us/img149/1650/bumpsurface.gif"></p> <p>Such a surface is rotationally symmetric, $C^2$-smooth, has positive curvature in the middle and negative curvature along the ring (the orange region in the picture). I don't really care what happens past that (it could flatten out, or oscillate, etc.)</p> <p>Here are two examples, as surfaces of revolution in $\mathbb R^3$ in cylindrical coordinates:</p> <p>$z(r) = e^{-r^2/2}$ and $z(r) = \tfrac{2}{\pi} \cos(\tfrac{\pi}{2} r)$.</p> <p>I need to do some Riemannian geometry on a bump surface; in particular, analyze a Jacobi field along a radial geodesic $\gamma$. I don't care what bump surface I use; it only has to feature both positive and negative curvature. For any surface of revolution, it's easy to write down a formula for the scalar curvature $K$ (see p. 142 of McCleary's <i>Geometry from a differentiable viewpoint</i>), and the Jacobi equation takes the form $J'' + KJ|\dot\gamma|^2 = 0$. Thus, if the scalar curvature has a simple form, then the Jacobi equation should be easy to solve. In the case of these two examples, the scalar curvature isn't particularly pretty, hence analyzing the Jacobi equation is difficult (though not intractable).</p> <p>My question to the MathOverflow community: is there a better bump surface than the two examples I gave above, for which the scalar curvature has a particularly simple form? </p> <p><b>Edit:</b> The curvatures for the surfaces given above are</p> <p>$K(r) = \frac{2 (1 - r)}{(e^{r^2/2} + r^2 e^{-r^2/2})^2}$ and $K(r) = \frac{\pi \sin(\pi r)}{2 r (1 + \sin^2(\pi r/2))^2}$,</p> <p>respectively. As you can see, they're not the worst expressions possible, but they're also not as simple as I'd like them to be. </p> http://mathoverflow.net/questions/12341/jacobi-fields-on-a-bump-surface/12344#12344 Answer by Steve Huntsman for Jacobi fields on a "bump surface" Steve Huntsman 2010-01-19T22:42:50Z 2010-01-20T00:04:40Z <p>Take a portion of a <a href="http://en.wikipedia.org/wiki/Pseudosphere" rel="nofollow">pseudosphere</a> and cap it with a portion of a sphere in such a way that the surface is ($C^1$)-smooth.</p> <p><strong>OLD (BAD) ANSWER:</strong></p> <p>Take a portion of a hyperboloid and cap it with a portion of a sphere in such a way that the surface is smooth. </p> http://mathoverflow.net/questions/12341/jacobi-fields-on-a-bump-surface/12346#12346 Answer by Deane Yang for Jacobi fields on a "bump surface" Deane Yang 2010-01-19T23:15:15Z 2010-01-20T02:09:09Z <p>ORIGINAL ANSWER DELETED</p> <p>EDIT: I neglected to account for the need to parameterize by arclength. And I think I also misunderstood and thought that you wanted only the Jacobi field that fixes the center. You want to solve for an Jacobi field, given a point (away from the center) and a vector at that point, right?</p> <p>So that's definitely not as easy as I thought. Here are my thoughts:</p> <p>1) I think the already proposed surface given by a spherical cap glued to a pseudosphere is already a good enough question. In my experience you never really need a $C^2$ surface, and something with piecewise continuous curvature is almost always enough. I encourage you to try it.</p> <p>2) As for the more general approach, I no longer have any easy answer, but here are some thoughts:</p> <p>Let the surface be given by $(r,\theta) \mapsto X(r,\theta) = (r\cos\theta, r\sin\theta, f(r))$. If $s$ be the arclength parameter along a radial geodesic, then $s'(r) = \sqrt{1 + f'(r)^2}$. One Jacobi field $J_1(r,\theta)$ is given simply by</p> <p>$J_1(r,\theta) = \partial X/\partial\theta = re_\theta$, where $e_\theta = (-\sin\theta, \cos\theta, 0)$ is a unit vector field that is orthogonal to and parallel along any radial geodesic.</p> <p>If we view $r$ as a function of $s$, then the Jacobi equation says that $r'' + Kr = 0$, where $K$ is the Gauss curvature. It suffices to solve for one more Jacobi field $J_2 = h(s)e_\theta$ independent of $J_1$. The Jacobi equation for $J_2$ is given by $h'' + Kh = 0$. Since $r$ is already a solution, we can try to solve for $h$ using variation of parameters.</p> <p>So the goal is to find an even function $f$ with an inflection point such that the function </p> <p>$s(r) = \int_0^r \sqrt{1 + f'(t)^2} dt$</p> <p>can be explicitly integrated and inverted. I suggest trying something like $f(r) = 1/(1+r^2)$.</p>