The Poisson Problem

The Poisson equation is one of the most important partial differential equations (PDEs) in applied mathematics. Given a domain $\Omega \subset \mathbb{R}^n$ and a source term $f : \Omega \to \mathbb{R}$, the Poisson equation seeks a function $u : \Omega \to \mathbb{R}$ satisfying

$$-\Delta u = f \quad \text{in } \Omega,$$

where $\Delta u = \sum_{i=1}^n \frac{\partial^2 u}{\partial x_i^2}$ is the Laplacian of $u$. The negative sign is a convention that makes the operator positive definite, as we will see later. It arises in numerous areas of physics and engineering, including electrostatics, gravitational potential, heat conduction, and fluid mechanics.

To obtain a unique solution, the PDE must be supplemented with boundary conditions on $\partial\Omega$. The boundary is typically partitioned as $\partial\Omega = \Gamma_D \cup \Gamma_N$ (with no overlap, $\Gamma_D \cap \Gamma_N = \emptyset$), and two types of conditions are imposed.

Dirichlet boundary conditions prescribe the value of the solution directly:

$$u = g_D \quad \text{on } \Gamma_D.$$

Neumann boundary conditions prescribe the outward normal derivative (flux):

$$\frac{\partial u}{\partial n} = g_N \quad \text{on } \Gamma_N,$$

where $\mathbf{n}$ is the unit outward normal to $\partial\Omega$. When $g_N = 0$, the condition is called a zero-flux or natural condition.

Consider a square domain $\Omega = [-1, 1]^2$ with an off-centre circular hole of radius $r = 0.5$ centered at $(0.3, 0.2)$. The outer square boundary $\Gamma_D$ carries a homogeneous Dirichlet condition $u = 0$, while the circular boundary $\Gamma_N$ carries a zero-flux Neumann condition $\partial u / \partial n = 0$. The source term is $f = 1$.

Solving this problem — using the finite element method developed throughout these posts — gives the solution shown below.

In one dimension the Poisson equation on $\Omega = (0, 1)$ with $f = 1$ and homogeneous Dirichlet conditions $u(0) = u(1) = 0$ reads

$$-u''(x) = 1, \quad x \in (0, 1), \qquad u(0) = u(1) = 0.$$

Integrating twice and applying the boundary conditions gives the exact solution

$$u(x) = \frac{x(1 - x)}{2}.$$

This simple example will serve as a useful reference throughout the series: it is easy to verify computed approximations against the exact parabolic profile.