# Numerically solving a PDE on an unit tangent bundle

Let $M$ be a manifold and $UT(M)$ its unit tangent bundle. I have a PDE which looks something like $$\frac{\partial f}{\partial t}(x,v) = (v\cdot \nabla_x) f(x,v) + \Delta f(x,v)$$ where $x\in M$, $v \in T_xM$, i.e. $x$ is a point on the surface and $v$ is an unit vector at that point, $\nabla_x$ is gradient on the surface $M$ and $\Delta$ is laplacian on the unit tangent bundle.

Is there any software which would allow me to solve such equation numerically?

My main concern is how do I obtain discretization of the unit tangent bundle? I would like to have a triangular mesh of my surface and then somehow get a discretization of its unit tangent bundle. Once I have the mesh I can probably use some standard library to solve the PDE.

• Can you draw a simple sketch of what this manifold looks like? Or even a simplified sketch? I think that would really help clear up some confusion on the jargon of the term “unit tangent bundle”. – Charles Mar 30 '18 at 5:11
• The problem is that the thing is basically impossible to draw. Any picture(that Im aware of) cannot capture the subtliness that come with it. For a formal definition see wiki en.m.wikipedia.org/wiki/Unit_tangent_bundle – tom Mar 30 '18 at 21:25
• What is the dimension of $M$? – Juan M. Bello-Rivas Mar 30 '18 at 22:25
• Im interested in two dimensional surfaces. – tom Mar 30 '18 at 22:28

The equation you are trying to solve in in essence the analogue of the "radiative transfer" equation when posed on a manifold. There is a vast literature in this area -- both in astrophysics and in nuclear physics, two areas in which this equation is important --, with the obvious difficulty being that your domain is pretty high-dimensional: if your manifold is three-dimensional, then you have a five-dimensional problem. (For 2d manifolds, the problem is three-dimensional.)

Typical ways to address this are:

• To approximate the equation by a sequence of equations that resemble diffusion equations in $x$
• To discretize the angular direction $v$ on a separate mesh from the spatial mesh, and then obtain a tensor product mesh in both $x$ and $v$.

There are many other approaches, such as ray tracing or Monte Carlo methods. Search the literature!

• Well tha main problem is exactly that the unit tangent bundle cannot be constructed as a tensor product between a surface mesh and a circle. So the approach you are suggesting does not work in general. Even for contractible surface where the tangent bundle is trivial, i.e. constructible with tensor product, I would expect that generating the mesh with a tensor product will produce badly shaped elements if the surface has high curvature. – tom Mar 29 '18 at 7:50
• I'm not sure I understand. If you have a smooth two-dimensional surface, then the tangent space is locally just a plane whose unit vectors form a circle that is easily discretized and furthermore the same everywhere (though their embedding into a higher dimensional space may of course be different at all points of the manifold). Or are you saying that your manifold is not smooth? – Wolfgang Bangerth Mar 29 '18 at 23:56
• Or that it branches? – Wolfgang Bangerth Mar 29 '18 at 23:57
• I think that @tom is mentioning that a "regular" mesh in a parametric space won't correspond to a mesh with nice elements. – nicoguaro Mar 30 '18 at 0:20
• Oh, I wasn't suggesting to map a Cartesian mesh onto the surface. I wanted to say that you choose a well fitted surface mesh for $x$, and the tensor product that with a subdivision of the circle -- i.e., at each point you would have a mesh for the circle. – Wolfgang Bangerth Mar 30 '18 at 3:04

I looked around a bit and found this paper (featuring no less than Shing-Tung Yau!) about the problem of generating meshes for the unit tangent bundle $UT(S^2)$ on the 2-sphere $S^2$. As you point out, $UT(M)$ probably isn't diffeomorphic to $M \times S^n$. For example in Yin's paper they point out that $UT(S^2)$ is non-trivial and can't be embedded in $\mathbb{R}^3$. However, locally the bundle is trivializable, and Yin's paper suggests that one approach could be to divide the manifold into trivializable patches, then explicitly compute the transition maps between patches.

You might be able to implement their idea to generate turn a mesh for $M$ into a mesh for $UT(M)$, which would have to be embedded into a much higher-dimensional Euclidean space. While many finite element libraries (deal.II among them) can solve PDEs on surfaces, I think they usually assume that it's a 1D or 2D surface embedded in 2D or 3D respectively. If you were working with something as simple as $UT(S^2)$, you'd have to embed it in $\mathbb{R}^4$, which probably falls outside the purview of many libraries.

You could also try to work only with a mesh for $M$ or, more in the spirit of the paper, meshes for a set of overlapping patches $M_\alpha$ such that $UT(M)$ is trivializable on each $M_\alpha$. Provided you have a way to calculate the transition maps, you can represent $f$ within each patch as a tensor product of a Galerkin basis function $\phi$ defined on $M_\alpha$ with the spherical harmonics of the appropriate dimension. You could then try and solve your PDE using, say, the Schwarz alternating method or some other domain decomposition method. Of course you'll also have to partition the mesh first. METIS is a great tool for this, but you could very well get a partition with a patch that isn't trivializable. It might be possible to then recursively partition non-trivial patches. Or it might be easiest to implement this yourself.

• I have stumbled on the same paper yesterday. Unfortunatelly the construction requires that the number of the segments along the disk is twice the number of segments of the circle(or the other way around? I dont remember). So if I want to have a fixed number of segments of the circle, lets say 8, I would have to partition my mesh to patches with boundaries made up of 16 segments. This might be really hard, especially when the resulting mesh should be nice for FEM. – tom Mar 30 '18 at 22:05
• @tom You're right, that seems like a pretty serious limitation for an arbitrary manifold. But what you need to solve that PDE is a way to represent functions on the unit tangent bundle, not necessarily to represent $UT(M)$ itself. If you can compute transition maps between charts and their derivatives, you can still use spectral basis functions on the unit sphere. – Daniel Shapero Mar 30 '18 at 22:45
• Right, computing these transition maps is probably my best bet. Although, Im a bit worries that there will be a big numerical diffusion at the transitions from one chart to the another, because you basically have to reinterpolate the data to a shifted grid. Also to get the charts, Im thinking about using some conformal flattening method, this way I do not distort the tangent circles. – tom Mar 31 '18 at 7:37
• Just for reference: deal.II may not instantiate templates for spacedim>3, but that's actually the easy part and should not be much work to add. The hard limitation is that the dimensionality of the manifold must be at most 3. – Wolfgang Bangerth Apr 1 '18 at 22:41