Triangle and Square
Setup for triangle:

We start with points in an equilateral triangle T. A level 0 partition contains only one cell that contains all the points. A level 1 partition contains 4 cells, each contians points that forms the shape of a smaller equilateral triangles as shown below. We obtain level n+1 partition by splitting each cell of level n into 4 smaller cells in the above fashion.

Given a level n partition of T = \bigsqcup_{i = 1}^{4^n} C_i, we define a random potential P: T \to \{0,P_{max}\} that is constnt on each cell C_i. This random potential is a Bernoulli random variable with parameter p. In other words, each cell independently have a probability p of being assigned a potential P_{max} and 1-p of being assigned no potential. We define the random potential on triangle this way so that it is compatible with that on the Sierpinksi Gasket for convenience of comparison.

The random Hamiltonian operator H: T^*\to T^* is defined as H = \triangle+P where P acts multiplicatively pointwise, and \triangle = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} is the 2D laplacian. We want to investigate the behavior of eigenvalues and and eigenfunctions of H with respect to the changes of p and P_{max}.

Physically speaking, P_{max} simulates the potential caused by structual anomaly in the solid, which happen with probability p. We want to see how does this structural anomaly affect the electron distribution, which is given by the stationary Schrödinger equation as the eigenfunctions of the random Hamiltonian.

Setup for square:
On the square the situation is similar, except that we start with S the set of all points on a square. Level 0 partition contains only one cell that includes every point in S. Each inductive step we split the cells in the previous level into 3*3 = 9 cells as shown below. Then we define random potential P: S\to \{0,P_{max}\} constant on each cell also as a random Bernoulli variable with parameter p. Needless to say, we define random potential this way in order for it to be compatible with that defined on the Sierpinski Carpet.



We obtain the solutions to the eigen problem Hv = \lambda v as well as the problem Hu = 1 via matlab's pde solver.
 
Sample Results:

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