A more mathematically complete write up for this project (along with references) can be found HERE

Suppose a flat piece of metal, represented by a two-dimensional bounded connected domain, is given an initial heat distribution which then flows throughout the metal. Assuming the metal is insulated (i.e. no heat escapes from the piece of metal), then given enough time, the hottest point on the metal will lie on its boundary.In mathematical terms, we consider a two-dimensional bounded connected domain D and let u(x,t) (the heat at point x at time t) satisfy the heat equation with Neumann boundary conditions. We then conjecture that

For sufficiently large t > 0, u(x,t) achieves its maximum on the boundary of DThis conjecture has been proven for some domains and proven to be false for others. In particular it has been proven to be true for obtuse and right triangles, but

It seems natural that heat would get "stuck in the corners" and in particular the corners with sharpest angle and so we conjecture that

For sufficiently large t > 0, the hottest and coldest points of u(x,t) are at the corners corresponding to the angles α and β.To test this conjecture we consider a simpler model of heat flow: We divide our triangle into four sub-triangles and posit that each of these sub-triangles be at a uniform temperature. That is, while the temperature of each of these four triangles will change over time, we make the assumption that the temperature is constant within each sub-triangle.

We model the heat flow in this simpler model by Newton's Law of Cooling: We assume that heat transfer only occurs between two adjacent triangles and is proportional to the difference in their temperatures and their edge-length of contact (which we denote by a < b < c). Mathematically this is equivalent to the heat flow on the graph, which we call G_1:

The heat flow in this simpler model is nothing more than a system of four ODE (for u_i(t), the temperature at the i-th vertex at time t) and so is much easier to understand. The conjecture for this simpler model is then:

For sufficiently large t > 0, the hottest and coldest points on the graph are at the verticies 1 and 3.Examining this problem further (see HERE for the full details), it is easy to see that this is really a conjecture on the Fiedler vector (the eigenvector corresponding to the second eigenvalue of the graph Laplacian) of this graph. As the graph Laplacian for G_1 is a 4X4 matrix, its eigenvectors are relatively tractable. And so proving the conjecture for G_1 should be doable.

(Here the side lengths should be a/2, b/2, and c/2... but we can ignore these extra factors of two since they won't affect the structure of the heat flow). We now consider the heat flow among these 16 triangles which is equivalent to the heat flow on the graph G_2:

The heat flow on G_2 is modeled by a system of 16 ODE and our conjecture is that eventually the hottest and coldest points lie at the verticies 1 and 7 -- equivalently, the extreme points of the Fiedler vector for this graph are at the verticies 1 and 7.

We now continue to further subdivide the triangles and we get a collection of graphs G_n and the natural conjecture

If this conjecture were proven, then it shouldn't be too hard to prove the Hot Spots Conjecture for the continuous triangle in the limit.Conjecture:For any n and a < b < c, the Fiedler vector for the corresponding graph G_n has its extreme points at the verticies in the top left and top right corner.

This discritization approach has two main advantages I can see:

- First, it reduces the problem to a linear algebra problem. While this problem is in most respects as difficult as the original conjecture, it is more "hands on". In particular it leads to some interpretations/approaches (e.g. combinatorics) not available in the continuous model (see HERE for more details).
- Second, it may be possible to make an argument by induction. The base case, G_1, is much simpler to handle than the full continuous model and so a proof of the conjecture there shouldn't be too hard. For the inductive step, perhaps we could take advantage of how, in some sense, G_n is made up of four copies of G_{n-1}?