# Boundary Conditions¶

## DCR¶

In Steady State Equations, we introduced the governing equations for the DCR problem

(332)$\boldsymbol{\nabla} \cdot \sigma\boldsymbol{\nabla}\phi = \boldsymbol{\nabla}\cdot\mathbf{j}_{source}.$

In order to construct a boundary value problem to determine the potential for a given source and conductivity model, (332) must be supplemented by appropriate boundary conditions. While different boundary conditions are possible, we will only discuss the common case of applying a homogeneous Neumann condition at the earth’s surface and homogeneous Dirichlet conditions elsewhere.

To derive the Dirichlet condition at the sides and bottom of the domain, it is first important to note that electrical potential is only unique up to an arbitrary constant, which is determined by convention. DC surveys measure potential differences so this is not important for field measurements but it is important in solving (332). We use the standard convention that the potential is zero infinitely far from all charges and currents. In solving for the discrete approximation to the potential, we make the boundaries at the sides and bottom of the domain far enough from any sources that the potential is approximately zero there. We then use homogeneous Dirichlet conditions on those boundaries in solving the discrete problem.

We derive the earth surface boundary condition from the fact that currents cannot flow into the air. Mathematically this can be stated as

$\mathbf{j}\cdot \hat{\mathbf{n}} = 0 \qquad \text{on} \quad \partial_s \Omega,$

where $$\partial_s \Omega$$ indicates the surface of the earth and $$\hat{\mathbf{n}}$$ is the unit surface normal vector. Applying Ohm’s law in the earth, this becomes

$\sigma \mathbf{e}\cdot\hat{\mathbf{n}} = 0.$

Since $$\sigma$$ must be strictly positive in the earth, we divide by it to give

$\mathbf{e}\cdot\hat{\mathbf{n}} = 0$

at the surface. Finally, writing the electric field as the negative gradient of the electric potential gives the surface boundary condition

$(\boldsymbol{\nabla}\phi) \cdot \hat{\mathbf{n}} = 0 \qquad \text{on} \quad \partial_s \Omega.$