Disclaimer: This is a copy of a question that I asked on the Mathematics Stack Exchange. It was suggested to me there that the question was worth asking over here.

There is an open problem in theoretical biophysics that I think has a strong mathematical flavour, and I have always thought that mathematicians would be interested in it. I would like to hear your ideas on the problem.

In the biophysics literature, cellular membranes are typically modelled using what is called the bending energy or Helfrich energy, $$ E = \int (H - H_0)^2 \mathrm{d}A $$ where $H$ is the mean curvature of the surface. The constant $H_0$ is called the "spontaneous curvature", and represents the preferred curvature of the membrane, arising from an asymmetry between the two sides of the membrane. In the absence of asymmetries, $H_0=0$ and we recover the usual Willmore energy. I have also omitted a term related to the Gaussian curvature, because we will only deal with closed surfaces of fixed topology, and then the term becomes irrelevant through the Gauss-Bonnet theorem. Furthermore, the total area and enclosed volume of the surface are also constrained in practical situations, so that one has to minimize the bending energy of the surface for fixed area $A$ and volume $V$. Without loss of generality, one can fix $A=4 \pi$, and the minimum energy shapes will depend on two parameters, $H_0$ and $V$. For a surface of genus $0$, we have necessarily that $0 \leq V \leq 4 \pi /3$, with $V=4 \pi / 3$ corresponding to a sphere, and $V<4 \pi /3$ necessarily deviating from a sphere.

Now here comes the mathematically interesting bit: among the many shapes that minimize the bending energy above for given $H_0$ and $V$, it was found from numerical minimization of axisymmetric shapes [1,2] that there exist minimum energy shapes corresponding to two separate shapes that are connected to each other via an infinitesimally narrow neck, in practice tangentially "kissing" at a single point. The infinitesimally narrow neck costs zero bending energy (its Gaussian curvature tends to infinity but its mean curvature stays finite), so the total energy is just the sum of the energies of each individual shape. These limit shapes were found to always satisfy the kissing condition $$ H_1 + H_2 = 2H_0 $$ where $H_1$ and $H_2$ are the mean curvatures of each of the two shapes at the kissing point. Two examples, corresponding to the particular case of kissing spheres, are illustrated in this image:

In the example on the left, the shape would be a minimum energy solution for $1/R_1 + 1/R_2 = 2 H_0$. The shape would be a boundary minimum of the energy for $1/R_1 + 1/R_2 < 2 H_0$, and a similar shape with a small but finite neck would be a solution for $1/R_1 + 1/R_2 \gtrsim 2 H_0$. In the example on the right, the shape would be a minimum energy solution for $1/R_1 - 1/R_2 = 2 H_0$. The shape would be a boundary minimum of the energy for $1/R_1 - 1/R_2 > 2 H_0$, and a similar shape with a small but finite neck would be a solution for $1/R_1 - 1/R_2 \lesssim 2 H_0$. In both examples, the radii $R_1$ and $R_2$ are such that the constraints on the area and volume of the shape are satisfied.

A particularly interesting case occurs for $H_0 = 0$, i.e. for the standard Willmore energy, where the kissing condition becomes $H_1 = -H_2$. In particular, the minimum bending energy shape for zero volume $V=0$ is composed of two nested spheres of equal radius connected to each other by an infinitesimal neck, with total energy $E=8\pi$.

The kissing condition $H_1 + H_2 = 2 H_0$ was later analytically proven [3] for axisymmetric shapes (i.e. two axisymmetric shapes connected to each other by a neck at their axis of symmetry) of genus $0$, but it is still not known whether it holds for non-axisymmetric shapes or shapes of higher genus. Furthermore, the proof for axisymmetric shapes in [3] was quite complicated, and no simpler proofs have been given since. Two interesting observations are (i) that the kissing condition implies $(H_1 - H_0)^2 = (H_2 - H_0)^2$, that is, the bending energy density is equal on both sides of the neck, and (ii) that if we take the mean curvature at the neck to interpolate between the mean curvatures on both sides of the neck, i.e. $H_\mathrm{neck} \equiv (H_1+H_2)/2$, then the kissing condition implies that $H_\mathrm{neck} = H_0$

Furthermore, the neck condition has been extended [4] (heuristically, based on numerical results) to surfaces composed of two separate "domains" of fixed area, with a line energy that penalizes the length $L$ of the boundary between the two domains, $E_\lambda = \lambda L$. If the two domains have distinct spontaneous curvatures $H_{0,1}$ and $H_{0,2}$, one then finds the kissing condition $H_1 + H_2 = H_{0,1}+H_{0,2} \pm \lambda$, with the plus and minus signs corresponding to the left and right situations in the image above.

So, my questions to you are:

  1. Have you ever heard of this kissing condition, and the corresponding open problem of proving whether it holds for non-axisymmetric and higher genus shapes?
  2. Is there any mathematical literature on this problem that might have gone unnoticed to me and other physicists?
  3. Do you have any intuition on how it could be proven, or whether it will or won't hold in non-axisymmetric situations?
  4. Do you think mathematicians could/should get interested in it?

[1] U. Seifert, K. Berndl, and R. Lipowsky, Phys. Rev. A 44, 1182 (1991). "Shape transformations of vesicles: Phase diagram for spontaneous- curvature and bilayer-coupling models." Journal link.

[2] L. Miao, B. Fourcade, M. Rao, M. Wortis, and R. K. P. Zia, Phys. Rev. A 43, 6843 (1991). "Equilibrium budding and vesiculation in the curvature model of fluid lipid vesicles." Journal link.

[3] B. Fourcade, L. Miao, M. Rao, M. Wortis, and R. Zia, Phys. Rev. E 49, 5276 (1994). "Scaling analysis of narrow necks in curvature models of fluid lipid-bilayer vesicles." Journal link.

[4] F. Jülicher and R. Lipowsky, Phys. Rev. E 53, 2670 (1996). "Shape transformations of vesicles with intramembrane domains" Journal link.


Your Answer

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

Browse other questions tagged or ask your own question.