Let $K(r)$ be the piecewise function

                                             graph of a piecewise function

I want to solve the PDE $$\Delta u + K(|x|) e^{2u} = 0$$ for radially symmetric $u$ with boundary condition $u = 0$ at infinity. I'm content for a solution in two dimensions, so switching to polar coordinates this is the ODE $$\tfrac 1 r u' + u'' + K(r) e^{2u} = 0.$$

The motivation is that the Riemannian metric $g_{ij} = e^{2u} \delta_{ij}$ has curvature profile $K$. This should be a relatively simple exercise in geometric PDE, and is surely explained in the literature. Could you please point me toward a reference which explains how to solve this equation explicitly?

  • 3
    $\begingroup$ Did you really mean "explicitly"? I doubt very much that there is a closed form solution to this ODE. At best, there is a theorem out there that shows that a solution exists. I recommend searching on "prescribed Gauss* curvature" on Mathscinet. There are many papers, starting with Kazdan and Warner, as well as Nirenberg and Ni, on this type of question. Whether any of them address specifically your case I do not kow. $\endgroup$ – Deane Yang Mar 9 '10 at 14:23
  • $\begingroup$ Deane, thanks so much for continuing to give me geometric insights. I already know a solution exists (cf. Sanxing Wu's Prescribing Gaussian Curvature on R²), but I actually do want an explicit solution. Recall my question, Jacobi fields on a "bump surface" mathoverflow.net/questions/12341. I am looking for a bump surface where I can explicitly calculate a Jacobi field along a radial geodesic which starts at a position of negative curvature. Steve Huntsman suggested a spherical cap glued to a hyperboloid. I need curvature to be continuous, hence my current attempt in this question. $\endgroup$ – Tom LaGatta Mar 9 '10 at 15:29
  • 2
    $\begingroup$ If I understand correctly, you want to find a spherically symmetric surface that has a bump in the center and flattens out. You also want to be able to find explicit formulas for all Jacobi fields along radial geodesics. If so, it seems to me that it would be easier to start with an explicit formula for the metric (and not the curvature). You get the formula for Jacobi fields that vanish at the origin for free, since it's given by the formula for the metric. Then you can try to solve for the other non-trivial Jacobi field using variation of parameters. Have you tried this already? $\endgroup$ – Deane Yang Mar 10 '10 at 14:51

I'm not sure that this will help, but let me suggest thinking about the following: You are looking for a metric of the form $g = e^{2u(r)}(dr^2 + r^2\ d\theta^2)$ where $u(r)$ is to be chosen so that the curvature of $g$ is a certain function $K(r)$ and so that $u$ tends to zero as $r\to\infty$. Now, I wouldn't have called this problem "specifying the curvature profile" just because $r$ won't represent the $g$-distance from the origin when you are done. Instead, the $g$-distance $s= h(r)$ from the origin will be given by solving $ds = e^{u(r)}\ dr$ with $s(0)=0$, and I would have called $K\bigl(h^{-1}(s)\bigr)$ the 'curvature profile'.

Are you sure that you wouldn't have rather had the metric in the form $g = ds^2 + f(s)^2\ d\theta^2$ where $f(0)=0$ and $f'(0)=1$ and then choose $f$ so that it satisfies the equation $$ f''(s) + K(s)\ f(s) = 0 $$ where $K$ is your given function?

If this is really your problem (and I'm not saying it has to be, but...), then you can, indeed, solve for $f$ explicitly, in a sense, but its definition will be piecewise, of course. You'll have $f(s) = \sin s$ for $0\le s \le 1$, but on the intervals $1\le s\le 3$ and $3\le s\le 4$, $f$ will be given in terms of translated Airy functions (different ones on the different intervals), and then, for $s\ge 4$, you'll have $f$ be a linear expression in $s$. Of course, determining the constants at the breakpoints so that $f$ is $C^2$ there is probably not going to be doable in any fully explicit fashion.

| cite | improve this answer | |
  • $\begingroup$ @Robert Bryant, thank you for your clear answer. Since posting this question last year, I ended up using an alternative approach using Fermi normal coordinates to prescribe the curvature profile I needed. $\endgroup$ – Tom LaGatta Jun 6 '11 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.