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The Discrete Poisson Equation

Equations

2-D, 5-point stencil (grid spacing \(h\)):

\[\boxed{\,\phi_{i+1,j} + \phi_{i-1,j} + \phi_{i,j+1} + \phi_{i,j-1} - 4\phi_{i,j} = -\frac{\rho_{i,j}}{\varepsilon_0}h^2\,}\]

1-D, matrix form with periodic BCs (the PIC field solve, normalized \(\phi'' = n - n_0\)):

\[\frac{1}{\Delta x^2}\begin{pmatrix} -2 & 1 & & & 1\\ 1 & -2 & 1 & & \\ & 1 & -2 & \ddots & \\ & & \ddots & \ddots & 1\\ 1 & & & 1 & -2 \end{pmatrix} \begin{pmatrix}\phi_1\\\phi_2\\\vdots\\\phi_{m-1}\\\phi_m\end{pmatrix} = \begin{pmatrix}n_1\\n_2\\\vdots\\n_{m-1}\\n_m\end{pmatrix} - n_0\]

Field recovery: \(E_j = -\dfrac{\phi_{j+1} - \phi_{j-1}}{2\Delta x}\) (periodic: E = -(np.roll(phi,-1) - np.roll(phi,1))/(2*dx)).

Physical meaning

Gauss's law, pixelated: each grid point's potential is tied to its neighbors and the local charge. The matrix is the 5-point/3-point stencil written for all points at once; the corner 1's wrap the domain into a ring (periodic boundary conditions).

Variables

\(\phi\) — electrostatic potential · \(\rho\) (or \(n - n_0\)) — charge (density imbalance) · \(h, \Delta x\) — grid spacing · \(m\) — number of grid points.

Structure worth knowing

  • Sparse: 3 diagonals + 2 corners — solvable in \(\mathcal{O}(m)\) (scipy.sparse.linalg.spsolve / Thomas algorithm)
  • Symmetric negative semi-definite; the constant vector is a null mode → potential defined up to a constant, and solvability requires \(\sum_j (n_j - n_0) = 0\) (net neutrality — physics and linear algebra in agreement)
  • Accuracy: \(\mathcal{O}(h^2)\), inherited from the central stencils

Applications

The "solve" step of every PIC cycle · electrostatics exercises (point charge, dipole, capacitor) · steady heat conduction · potential flow (\(\rho = 0\)).

Quiz

Q1 (conceptual). Why do the corner entries make the boundary "disappear"?

Answer

They connect point 1 to point \(m\) exactly as interior neighbors are connected — topologically the grid becomes a circle with no boundary at all, matching the periodic domain of the two-stream simulation.

Q2 (computational). On a 3-point periodic grid with \(\Delta x = 1\) and source \((n - n_0) = (1, -2, 1)\), solve \(A\boldsymbol\phi = \mathbf{b}\) for one valid \(\boldsymbol\phi\).

Answer

First check solvability: \(\sum_j b_j = 1 - 2 + 1 = 0\) ✓. The three equations are \(-2\phi_1 + \phi_2 + \phi_3 = 1\), \(\phi_1 - 2\phi_2 + \phi_3 = -2\), \(\phi_1 + \phi_2 - 2\phi_3 = 1\). Fix the free constant by choosing \(\phi_1 = 0\): then eq. 1 gives \(\phi_2 + \phi_3 = 1\) and eq. 2 gives \(-2\phi_2 + \phi_3 = -2\); subtracting, \(3\phi_2 = 3 \Rightarrow \phi_2 = 1\), \(\phi_3 = 0\). So \(\boldsymbol\phi = (0, 1, 0)\) (plus any constant) — check eq. 3: \(0 + 1 - 0 = 1\) ✓. The potential peaks where the charge deficit (\(b_2 = -2\), i.e. net positive charge) sits, as electrostatics demands.