## 14 January 2014

### 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.