Saint-Venant's principle and its relatives - part 1

This venerable principle (published in 1855) and a whole family of analogous results can be explained in a very simple, almost geometrical, manner based on the observation that "the Laplacian is a zero-sum game". Within this class of results (with applications ranging from optical microscopes to the metallic mesh on the door of microwave ovens). Saint-Venant's principle is probably the only one to have an official name, but not the easiest to understand, so we will begin by a simple example from electrostatics.

Consider an array of alternating positive and negative charges, such that their electrostatic potential is a sinusoid in the plane $z=0$: $V(x,0) = \cos(q x)$, where $q = 2 \pi / \Lambda$ (Figure 1; the charges themselves are somewhere in $z \lt 0$ and their detailed distribution is irrelevant).

Figure 1: Electrostatic potential engendered by a periodic charge distribution.

Since there are no charges in the half-space $z \gt 0$, the potential obeys $\Delta V(x,z) = 0$ everywhere in this domain. We assume invariance along the third space direction $y$, $\partial _y V = 0$. The Laplace equation then reduces to $(\partial _x^2 + \partial _z^2 ) V(x,z) = 0$. We can try to separate the potential in factors depending only on $x$ and $z$ and, guided by the boundary condition in $z = 0$, write $V(x,z) = \cos(q x) f(z)$. Substituting in the Laplace equation yields: $\partial _z^2 f(z) = q^2 f(z)$, with solutions $f(z) = \exp(\pm q z)$. Requiring $V = 0$ far away from the charges (for $z \to \infty$) imposes the negative exponent. We then have: $$V(x,z) = \cos(q x)  \exp(- q z)$$ The conclusion is that:

An electrostatic potential periodically modulated along one direction with a wave vector $q$ decays along the perpendicular direction with a corresponding attenuation factor.

This is because the negative first term in the Laplace equation $\partial _x^2 V(x,z) = -q^2 V(x,z)$ must be exactly compensated by the positive contribution of the second term: $\partial _z^2 V(x,z) = q^2 V(x,z)$. Like I said, a zero-sum game.

In the next installment, we will see how to add a source term to the Laplace equation and what happens in this case.