# 4.2 Minimizing the energy of vector fields on surfaces of revolution  (Page 3/9)

 Page 3 / 9

Example 2. Again let S be the unit cylinder, and define the tangent vector field $V\left(\theta ,t\right)=\left(-sin\left(\theta \right),cos\left(\theta \right),0\right)$ (equivalently, $\varphi \left(\theta ,t\right)=0$ ). V is the unit vector field horizontally tangent (in a counterclockwise direction) to the surface. We see from equation [link] that $E\left(V\right)=2\pi$ . See the following figure.

Example 3. Let S be a frustum with base radius 2, top radius 1, and unit height: $S=\left(\left(2-t\right)cos\left(\theta \right),\left(2-t\right)sin\left(\theta \right),t\right)$ . Let V again be the unit vector field horizontally tangent to S : $V\left(\theta ,t\right)=\left(-sin\left(\theta \right),cos\left(\theta \right),0\right)$ . From equation [link] , we can calculate

$E\left(V\right)={\int }_{0}^{1}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}{\int }_{0}^{2\pi }\frac{\sqrt{2}}{2-t}\left[{cos}^{2},\left(\theta \right),+,{sin}^{2},\left(\theta \right)\right]d\theta dt=2\sqrt{2}\pi {\int }_{0}^{1}\frac{1}{2-t}dt=2\sqrt{2}\pi log2.$

See the following figure.

## Minimizing energy

Now that we have a solid concept of the energy of a vector field, we can answer a number of questions. Specifically, we are interested in finding an expression for the unit vector field with minimal energy on a given surface (if it exists), with specified boundary conditions. To do so, we use a technique from the calculus of variations.

## Calculus of variations: an interlude

A common problem in calculus of variations asks a question very similar to ours: given a collection of paths $y\left(x\right):x\in \left[a,b\right]$ and a function $L\left(x,y\left(x\right),{y}^{\text{'}}\left(x\right)\right)$ , which path y minimizes the cost functional $J\left[y\right]={\int }_{a}^{b}L\left(x,y\left(x\right),{y}^{\text{'}}\left(x\right)\right)dx$ ?

Suppose that y is such a minimizer. For any “perturbation" $\eta \left(x\right)$ with $\eta \left(a\right)=\eta \left(b\right)=0$ , we can consider the cost $J\left(ϵ\right)=J\left[y+ϵ\eta \right]={\int }_{a}^{b}L\left(x,y\left(x\right)+ϵ\eta \left(x\right),{y}^{\text{'}}\left(x\right)+ϵ{\eta }^{\text{'}}\left(x\right)\right)dx$ . We calculate $\frac{dJ}{dϵ}$ and evaluate at $ϵ=0$ . If y is a minimizing path, then $ϵ=0$ should be a critical point of $J\left(ϵ\right)$ . Supposing that $0=\frac{dJ}{dϵ}{|}_{ϵ=0}$ , we obtain the famous Euler-Lagrange equation:

$\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial {y}^{\text{'}}}=0.$

Any path y which minimizes the cost functional J must satisfy this differential equation. Note that the condition is necessary, not sufficient- not every function which satisfies [link] will produce minimal cost.

## Our application: the unit cylinder

We wish to apply a similar technique in our situation. For now, we restrict our attention to the cylinder with unit radius, and only consider vector fields with unit length. Thus, it becomes convenient to use the angle notation mentioned in section 1.2: any vector field $V\left(\theta ,t\right):0\le \theta \le 2\pi ,0\le t\le h$ in consideration can be represented by the angle $\varphi \left(\theta ,t\right)$ that $V\left(\theta ,t\right)$ makes with the horizontal tangent vector.

Suppose that ϕ is the angle representation of the unit vector field with minimal energy on the cylinder. Let $\eta :\left[0,2\pi \right]×\left[0,h\right]\to \mathbb{R}$ be a perturbation with $\eta \left(\theta ,0\right)=\eta \left(\theta ,h\right)=0$ . Mimicking the calculus of variations technique, we want to plug into our cost functional, equation [link] . Dropping the $\left(\theta ,t\right)$ arguments:

$\begin{array}{cc}\hfill J\left(ϵ\right)=J\left[y+ϵ\eta \right]& ={\int }_{0}^{h}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}{\int }_{0}^{2\pi }cos{\left(\varphi +ϵ\eta \right)}^{2}+{\left(\frac{\partial \varphi }{\partial \theta },+,ϵ,\frac{\partial \eta }{\partial \theta }\right)}^{2}+{\left(\frac{\partial \varphi }{\partial t},+,ϵ,\frac{\partial \eta }{\partial t}\right)}^{2}d\theta dt\hfill \\ \hfill 0=\frac{dJ}{dϵ}\left(0\right)& ={\int }_{0}^{h}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}{\int }_{0}^{2\pi }-2\eta cos\left(\varphi \right)sin\left(\varphi \right)+2\left(\frac{\partial \varphi }{\partial \theta },\frac{\partial \eta }{\partial \theta },+,\frac{\partial \varphi }{\partial t},\frac{\partial \eta }{\partial t}\right)d\theta dt\hfill \end{array}$

Since η is periodic in θ and $\eta \left(\theta ,0\right)=\eta \left(\theta ,h\right)=0$ , integration by parts on the right-hand terms yields

$0={\int }_{0}^{h}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}{\int }_{0}^{2\pi }\eta \left(\Delta ,\varphi ,+,\frac{sin\left(2\varphi \right)}{2}\right)d\theta dt.$

This expression holds for all perturbations η . Thus, we can deduce

$\Delta \varphi +\frac{sin\left(2\varphi \right)}{2}=0.$

Any function ϕ which describes the vector field of minimal energy on the unit cylinder will satisfy this equation. Assuming that ϕ is rotationally symmetric, or independent of θ (sometimes a reasonable assumption, we will see later), equation [link] becomes an ODE. Unfortunately, it is not a friendly ODE to solve. One can find approximate solutions by considering the Taylor series expansion of $\frac{sin\left(2x\right)}{2}$ , but such solutions are nonsatisfactory. A far better method of approximation is explored in "Computer Approximations" . And even if we cannot solve the equation analytically, we can still determine properties of solutions.

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