4.2 Minimizing the energy of vector fields on surfaces of revolution  (Page 2/9)

Defining energy

Initial remarks

Let a surface S be defined by the parametrization

S is a surface of revolution with radius $r\left(t\right)$ at each height $t\in [0,h].$ Suppose that $V(\theta ,t)=\left(a\right(\theta ,t),b(\theta ,t),c(\theta ,t\left)\right)$ is a vector field tangent to S , so that for any ${\theta }_0,t_0$ , $V({\theta }_0,t_0)$ is tangent to S at $\Phi ({\theta }_0,t_0)$ . (We are particularly interested in vector fields with unit length, but for now we consider the general case.) We would like to define the notion of the “kinetic energy" of V . Energy should measure how much the vector field changes over the surface; thus, our definition should resemble

Clarifying this expression is simple in one dimension (if S is a curve), but trickier in two: we must consider the change of V (a three-dimensional function) in two coordinate directions, and account for the curvature of the surface on which it lies. To achieve this, we take ${\left|DV\right|}^2$ to be the sum of the squares of the derivatives of V , taken in perpendicular directions, with respect to arc length. If V is a function of θ and t , the θ -directional derivative of V can be expressed as $\frac{\partial V}{\partial \theta }\frac{\partial \theta }{\partial s}$ , where $s=r\left(t\right)\theta$ is the horizontal arc length of S . Similarly, the t -directional derivative of V can be expressed as $\frac{\partial V}{\partial t}\frac{\partial t}{\partial \sigma }$ , where $\sigma ={\int }_0^t\sqrt{1+r^{\text{'}}{\left(\tau \right)}^2}d\tau$ .

d Area is given by the square root of the determinant of the gram matrix, $\sqrt{\left|\text{Gram}\right|}$ . By [link] , the Gram matrix is

For our given $\Phi (\theta ,t)$ ,

With these considerations in mind, we arrive at the following expression:

Some special cases

Perhaps the simplest surface of rotation is the unit cylinder with unit radius, parametrized by $\Phi (\theta ,t)=\left(cos,\left(,\theta ,\right),,,sin,\left(,\theta ,\right),,,t\right):0\le \theta \le 2\pi ,0\le t\le 1$ . On this cylinder, the expression for energy simply becomes

$E\left(V\right)=$

It can be difficult to define a vector field of the form $\left(a\right(\theta ,t),b(\theta ,t),c(\theta ,t\left)\right)$ while ensuring that the vector field remains tangent and of unit length. To this end, we can represent any unit length tangent vector field V at $\Phi (\theta ,t)$ on the cylinder by the angle $\varphi (\theta ,t)$ that $V(\theta ,t)$ makes with the horizontal tangent vector $(-sin(\theta ),cos(\theta ),0)$ . Note that if V is continuous on S , ϕ must be continuous in both variables and periodic (up to multiples of $2\pi$ ) in θ ; henceforth, we will assume that all ϕ are continuous and (strictly) periodic. We then have the relation $V\left(\varphi \right)=\tilde{V}(\theta ,t)=(-sin\left(\theta \right)cos\left(\varphi (\theta ,t)\right),cos\left(\theta \right)cos\left(\varphi (\theta ,t)\right),sin\left(\varphi (\theta ,t)\right))$ . The expression for energy on the unit cylinder becomes

Examples

It is helpful to consider some specific examples of vector fields on surfaces and to calculate their energies.

Example 1. Let S be the unit cylinder, and define $V(\theta ,t)=(0,0,1)$ (or equivalently, in our alternate notation, $\varphi (\theta ,t)=\frac{\pi }{2}$ ). V is the unit vector field pointing directly “up" at every point on the surface. Since every partial derivative of V (and of ϕ ) is 0, it is easy to check, using either expression for energy in the previous section, that $E\left(V\right)=0$ . See the following figure.