A problem that keeps me up at night

What is the resistance of a square laminar when measured from the corners? More precisely, assume you have a square slice of material with conductivity σ of side length L and thickness τ ≪ L. If you connect an ohmmeter between two diagonally opposite corners, what resistance will you measure?

Position the slab of material so that it occupies the space [-L,L]⨉[-L,L]⨉[-τ,τ]. We will connect the contacts of the ohmmeter to the vertical edges x=y=-L and x=y=L. Let V(x) be the electrostatic potential and E = -∇V be the electic field. According to Ohm's law, the current density is J = σE.

Since there is no net charge density in the material, the potential obeys the Laplace equation:
    ∇² V = 0.

No current can flow past the material boundaries. This gives the boundary condition
    dV/dn = 0, when x=±L, y=±L or z=±τ,
where dV/dn is the derivative of the potential in the direction of the face normal.

The resistance is a ratio R=ΔV/I. Therefore, we can assume that there is a constant potential difference across the terminals. This gives another boundary condition:
    V(-L,-L,z) = -1 and V(L,L,z) = 1.

Given V(x), we can calculate the current through the material:

where S is a surface inside the material enclosing (-L,-L,z). The resistance is then given by R = 2/I.

The first problem then is to solve for V(x). Some observations:

1. We can always set L = 1 by rescaling x and y.


2. The boundary conditions are independent of z. Therefore we can assume V(x,y,z) = V(x,y) which reduces the problem to two dimensions.


3. The problem is symmetric under reflection about the line y = x. This means that V(x,y) = V(y,x).


4. The problem is antisymmetric under reflection about the line y = -x. This means that V(x,y) = -V(-y,-x). In particular, when y = -x, V(x,-x) = -V(x,-x) and so V(x,-x) = 0.

The equipotential curves are perpendicular to the edges due to the first boundary conditions. It's easy to sketch a qualitative solution:

I found this problem in [1]. There, they calculated the resistance to be R = 4/π 1/(τσ) using an infinite number of image charges. However, I want to check this answer numerically.

[1] Lawrence R. Mead, Resistance of a square lamina by the method of images, American Journal of Physics, 77 (3), (2009). Abstract

A discrete version of the divergence theorem

The divergence theorem (Gauss's theorem) from vector calculus says
   
I am interested in the case where the vector field is derived from a scalar potential F = ∇u:
   

This theorem is for a continuous system in ℝ³. However, numerical algorithms often operate on discrete grids in ℤ^d where d = 2 or 3. I will now derive a discrete version of this theorem.

Definitions:

Let u be a scalar field on the grid u: ℤ^d → ℝ, x ↦ u_x.

Let x, y ∈ ℤ^d. We'll use the Manhattan metric |x - y| = Σ_i |x_i - y_i|. Cells x and y are called adjacent if and only if |x - y| = 1. Note that diagonal cells are not adjacent.

Let V be a subset of ℤ^d. V' is the complement of V in ℤ^d.

Let n_x be the number of cells adjacent to x in V. Therefore, n_x = Σ_{y∈V, |x-y|=1} 1.

Let the second-order finite difference Laplacian be given by L_x = - 2d u_x + Σ_{y, |x-y|=1} u_y.

Theorem:

   

Note that right-hand side involves only points on the boundary of V. These correspond to first-order finite difference derivatives. There is one term for each arrow in this diagram:

Proof:

Starting from the left-hand side, expand the Laplacian to get a sum of field terms:
   Σ_x∈V L_x = Σ_x c_x u_x,
where c_x are constant coefficients. From the definition of the Laplacian, we can see that c_x = n_x - 2d if x∈V, or c_x = n_x if x∈V':
   = Σ_x∈V u_x (n_x - 2d) + Σ_x∈V' u_x n_x.   (1)

Since each cell is adjacent to 2d others,
   2d = Σ_{y, |x-y|=1} 1
     = Σ_{y∈V, |x-y|=1} 1 + Σ_{y∈V', |x-y|=1} 1
     = n_x + Σ_{y∈V', |x-y|=1} 1
   2d - n_x = Σ_{y∈V', |x-y|=1} 1.

Now put this into (1):
   Σ_x∈V L_x
   = - Σ_{x∈V, y∈V', |x-y|=1} u_x + Σ_{x∈V', y∈V, |x-y|=1} u_x
   = Σ_{x∈V, y∈V', |x-y|=1} (u_y - u_x). ■