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Suppose we have a two-dimensional surface $M$ with smooth boundary $\partial M$. Equip $M$ with a metric $g$ such that the Gauss curvature $K$ of $M$ and geodesic curvature $\kappa$ of $\partial M$ are both constants. Suppose $\nu$ is the outer unit normal of $\partial M$. Can we always find $f$ such that $$\begin{cases}\Delta_g f=-2Kf\\\displaystyle \frac{\partial f}{\partial \nu}=\kappa f\end{cases}$$ It seems $f$ is the eigenfunction of the linear operator $(\Delta_g,\frac{\partial }{\partial \nu})$. Does such $f$ exist?

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  • $\begingroup$ Why is there a solution if you drop the second condition? $\endgroup$ – Igor Rivin Dec 3 '14 at 3:08
  • $\begingroup$ There is not a nonzero solution $f$ even when $M$ is the unit disk in the Euclidean plane (i.e., $K=0$, $\kappa = 1$), so it would appear that nontrivial solutions only exist in special cases. $\endgroup$ – Robert Bryant Dec 3 '14 at 15:36
  • $\begingroup$ @RobertBryant. Thank you for your attentiona. Yes. It does exist a nontrivial one when $M$ is the unit disk with Euclidean metric. Actually there are two linear independently solutions of the corresponding problem. Someone told me this is related to stelov eigenvalue problem. I am looking for the reference on it but have not known how to prove this yet. $\endgroup$ – Slm2004 Dec 5 '14 at 13:56
  • $\begingroup$ @Slm2004: Ah! I see that I made a mistake in my calculations. You are right, the general solution in this particular case is $f = ax+by$ where $a$ and $b$ are constants, so there are two solutions, as you said. $\endgroup$ – Robert Bryant Dec 5 '14 at 14:45
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Here is an example with no nontrivial solution:

Let $\Sigma$ be a compact Riemann surface of genus $2$ with an antiholomorphic involution $\iota:\Sigma\to \Sigma$ whose fixed point locus consists of $3$ disjoint circles $C_i$ ($i=1,2,3$) whose complement consists of two disjoint connected regions, say $\Sigma_+$ and $\Sigma_-$. Equip $\Sigma$ with the unique conformal metric $g$ with Gauss curvature $K\equiv-1$. Let $M$ be the closure of $\Sigma_+$, so that $M$ is the union of $\Sigma_+$ and the circles $C_i$, which constitute the boundary of $M$. Then, because these are the fixed points of $\iota$ which is a $g$-isometry, each of the $C_i$ is a geodesic, i.e., $\kappa\equiv0$.

Now suppose that $f$ is defined and smooth on $M$ and satisfies $\Delta_gf = -2K f = 2f$ and $\partial_\nu f = \kappa f = 0$ on the boundary circles $C_i$. Then, extending the definition of $f$ to $\Sigma_-$ by setting $f(s) = f\bigl(\iota(s)\bigr)$ for $s\in \Sigma_-$, one obtains an analytic $f$ defined on all of $\Sigma$ (and invariant under $\iota$), that satisfies $\Delta_g f = 2f$. (Here, I am using a reflection principle based on the fact that the normal derivative of $f$ along the boundary circles is zero, to get the necessary regularity of $f$ across the boundary. The point is that elliptic regularity coupled with the fact that both the metric and the boundary are real-analytic shows that $f$ is real-analytic up to and including the boundary circles, and then solving the analytic initial value problem along the boundary circles with the initial values that $f$ and $\partial_\nu f$ $(=0)$ are specified along the circles shows that $f$ extends analytically to a neighborhood of the circles in $\Sigma$ and must be $\iota$-symmetric there because the initial conditions are $\iota$-invariant.)

However, $\Sigma$ is compact and $\Delta_g$ has no positive eigenvalues, so $f$ must be identically zero.

(If your Laplacian has the opposite sign from mine, the same sort of argument will work to give a counterexample on the upper hemisphere of the unit sphere, since, in that case, $K\equiv+1$ and $\kappa\equiv0$ along the boundary.)

Added remark: I just realized that my example above is more complicated than it has to be: One can also find a compact Riemann surface of genus $2$ with an antiholomorphic involution that has only one fixed circle (think of making it by symmetrically connect-summing two $1$-holed tori along a fixed circle). Thus, one can construct such a counterexample on an $M$ with $K\equiv-1$ for which the (geodesic) boundary is connected.

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  • $\begingroup$ Thank you very much. I am more interested in the $K>0$ and $k>0$ case. In the upper hemisphere case, $K\equiv 1$ and $k\equiv 0$ there also exists two linear independent solutions. Can you prove the existence? $\endgroup$ – Slm2004 Dec 6 '14 at 16:02
  • $\begingroup$ I mean $K\geq 0$ and $k\geq 0$. $\endgroup$ – Slm2004 Dec 6 '14 at 16:10
  • $\begingroup$ In the upper unit hemisphere case, yes, there are exactly two solutions: If the surface is $x^2+y^2+z^2=1$ where $z\ge0$, then the functions $f = ax+by$ are solutions of your problem for any constants $a$ and $b$. Existence is trivial, of course, and uniqueness follows because, again $f$ will reflect to the lower hemisphere to define a global solution to $\Delta_gf = -2f$, and these are well-known to be the restrictions of the linear functions $ax+by+cz$ to the $2$-sphere, of which, only the ones with $c=0$ will satisfy the boundary condition. Other values with $K>0$ follow by scaling. $\endgroup$ – Robert Bryant Dec 6 '14 at 16:26
  • $\begingroup$ Correction: Above, when I wrote "there are exactly two solutions" I should have written "the space of solutions has dimension $2$". Sorry about the careless phrasing! $\endgroup$ – Robert Bryant Dec 6 '14 at 17:10
  • $\begingroup$ @Slm2004: I've had a chance to think about this and realized that, under your conditions $K\ge0$ and $\kappa\ge0$, this only allows the cases of a disk in the Euclidean plane ($K=0$) and of a convex spherical cap ($K>0$). I don't see how you can get anything else connected that has $\kappa\ge0$ on all the boundary components. In these cases, there is always a $2$-dimensional space of solutions, namely the linear functions vanishing at the center of the disk or cap, where 'linear' refers to the standard embedding of the sphere or plane into $\mathbb{E}^3$. Possibly there are no other solutions. $\endgroup$ – Robert Bryant Dec 6 '14 at 22:03

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