# Area of Disc that Intersects Another under Smooth Flow

The following question can be asked in any $$\mathbb{R}^n$$ for $$n > 1$$, but the case of interest is (thankfully) the case $$n = 2$$. The formulation of the problem with discs isn't actually critical to the application I have in mind, but it seems natural.

Suppose that $$C, D$$ are disjoint, Euclidean discs of radius $$1$$ in $$\mathbb{R}^2$$ and suppose that $$A \subset \mathbb{R}^2$$ is a smoothly immersed arc with endpoints $$x_0$$ and $$x_1$$. Given the immersion $$f: [0, 1] \rightarrow A$$, for each $$t \in [0, 1]$$ define $$f_t$$ to be the affine shift of $$\mathbb{R}^2$$ sending $$x_0$$ to $$f(t)$$. Then $$F = \lbrace f_t \rbrace$$ defines a smooth planar flow which 'drags the plane along $$A$$' without any rotation.

My question is this: What is the measure of the set of points of $$C$$ which never enter $$D$$ under this flow? In other words, what is the measure of the set

$$X = \lbrace x \in C | f_t(x) \notin D \text{ for all } t \in [0, 1] \rbrace$$

It's equivalent to ask for the measure of its complement, since $$C$$ is bounded. It is a very natural question, so surely there is some literature. What is desired is some (presumably horrible) integral or operator equation with inputs $$f, C, D$$. Series and limits are fine, just any sort of equation which can govern this quantity. I say with inputs $$f, C, D$$ because the problem can certainly be generalized to arbitrary bounded domains.

I am thinking about the Sofa Problem. Instead of building the sofa, we should think of the obstacle like a light-saber which dissolves any part of the furniture that dares to touch it. Here $$C$$ would be the sofa material and $$D$$ would be the obstacle. If we can get an integral equation representing the measure of $$X$$ and then maximize it over a well-chosen family of curves $$A_\alpha$$, the process would 'carve out' a maximal sofa for us. This would solve a broad class of 'sofa problems,' and the techniques would perhaps generalize to higher dimensions. The shape of $$C$$ is uninteresting because as long as it's sufficiently large it should carve the sofa out regardless (by making $$D$$ unbounded via a summation we can avoid problems of $$C$$ being 'too big'; that's just a technicality). The hardest part of this method would be handling the (large) family $$f_\alpha$$.

So this is basically an intermediate step toward a potential proof method that I haven't seen mentioned in the literature on the Sofa Problem. The shape of $$D$$ also isn't that interesting, because we can always sum over a circle packing of our obstacle. The only other steps would be handling a rotation parameter and variation in the radii of $$C$$ and $$D$$. Any such resulting equation would probably be too cumbersome for a lot of theoretical cases to be solved, but proposed answers could be checked (e.g. the one in the original Sofa Problem) and numerical approximations could be made for applications.

As an example of how difficult the equation may be to use, one method to obtain an equation that comes to mind is looking at finer and finer lattices of interior points of the two domains in question and approximating the pass-through area by a count on the proportion of $$C$$-lattice points that get $$\epsilon$$-close to a $$D$$ lattice point. That would get very gross, very fast, and there may be technical underpinnings that are lacking in the literature when it comes to the mixed-radius case. But this method may be the easiest to get going, which means there might be a lot of nice, low-hanging results to prove which could be relevant, for those interested.

The lattice method has a problem of having discrete inputs, though. That probably won't mix well with an integral equation. What would be much more desirable is something that integrates over the boundaries of the domains, which are topologically nice for us.

Does anyone know about research into this topic of 'pass-through area' or volume? The techniques involved are completely outside my area of expertise, but this method of attacking the problem has been in my head for a while.