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.

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