4.2 Minimizing the energy of vector fields on surfaces of revolution  (Page 6/9)

Limiting results on the cylinder

We expect intuitively that twisting “too much" is a poor use of energy; a vector field that minimizes energy will have relatively little spinning. How much is “too much"? We can establish some bounds on how far ϕ varies. In this section, we will work on a cylinder of height h and unit radius.

Theorem 3. Suppose $\varphi (\theta ,0)=f\left(\theta \right)$ and $\varphi (\theta ,h)=g\left(\theta \right)$ , with $-\frac{\pi }{2}<f\left(\theta \right),g\left(\theta \right)<\frac{\pi }{2}$ . Any $\varphi (\theta ,t):[0,2\pi ]\times [0,h]\to \mathbb{R}$ which minimizes energy must satisfy $-\frac{\pi }{2}\le \varphi (\theta ,t)\le \frac{\pi }{2}$ for all $(\theta ,t)\in [0,2\pi ]\times [0,h]$ .

Proof: Suppose for the sake of a contradiction that ϕ minimizes energy and that the image of ϕ is not contained in $[\frac{-\pi }{2},\frac{\pi }{2}]$ . Then $\Gamma =\{(\theta ,t):|\varphi (\theta ,t)|>\frac{\pi }{2}\}$ is a nonempty open set. Thus the energy of ϕ on Γ must be greater than zero, for if the energy is zero, then ϕ must be identically $\frac{\pi }{2}$ which contradicts our assumption that Γ is nonempty. We can choose an $r>\frac{\pi }{2}$ such that r is a regular value of ϕ and the energy δ _r of ϕ over the set ${\Gamma }_r=\{(\theta ,t):|\varphi (\theta ,t)|>r\}$ is greater than 0. By Sard's theorem, $\partial {\Gamma }_r$ is a finite collection of smooth curves. We construct a function $\psi (\theta ,t)$ which is a truncation of ϕ at r :

We see that $E\left(\psi \right)=E\left(\varphi \right)-{\delta }_r$ . ψ defines a vector field on the cylinder that has lower energy than that defined by ϕ , but it is not sufficiently differentiable; we would like to find a family of smooth functions ψ ^ϵ , the energy of which converges to the energy of ψ .

Note that ϕ must be smooth and Lipschitz continuous, as it solves $\Delta \varphi +\frac{sin\left(2\varphi \right)}{2}=0$ . Thus ψ is also Lipschitz continuous, as it is a truncation of ϕ . Since ψ is Lipschitz continuous we know that $\psi \in W_{\text{loc}}^{1,2}\left(\Gamma \right)$ . By [link] , we can construct a mollifying function η _ϵ such that the family ${\psi }^{ϵ}={\eta }_{ϵ}*\psi$ converges to ψ as $ϵ\to 0$ . The total energy of ψ ^ϵ converges to that of ψ . Thus we can construct a smooth ψ ^ϵ such that $E\left({\psi }^{ϵ}\right)<E\left(\varphi \right)$ , contradicting the assumption that ϕ minimizes energy. $\square$

This proof can be generalized to the other half-plane.

Theorem 4. Suppose $\varphi (\theta ,0)=f\left(\theta \right),\varphi (\theta ,h)=g\left(\theta \right)$ , with $0<f\left(\theta \right),g\left(\theta \right)<\frac{\pi }{2}$ . Any $\varphi (\theta ,t):[0,2\pi ]\times [0,h]\to \mathbb{R}$ which minimizes energy must satisfy $0<\varphi (\theta ,t)\le \frac{\pi }{2}$ for all $(\theta ,t)\in [0,2\pi ]\times [0,h]$ .

Proof: From the above lemma, we know that for any minimizer ϕ , $\left|\varphi (\theta ,t)\right|\le \frac{\pi }{2}$ . Thus we are only concerned with ϕ dropping below 0. Define $m_0={inf}_{\theta }\{f\left(\theta \right),g\left(\theta \right)\}$ ; we can find m arbitrarily close, or possibly equal, to m ₀ such that m is a regular value of ϕ . We define two regions in the rectangle $[0,2\pi ]\times [0,h]$ : ${\Gamma }_1=\{(\theta ,t):-m\le \varphi (\theta ,t)\le m\}$ and ${\Gamma }_2=\{(\theta ,t):\varphi (\theta ,t)<-m\}$ . Sard's Theorem guarantees that the boundaries of these regions are smooth.

Again, we construct an alternate path:

$\tilde{\varphi }$ is always in the first quadrant, since, by lemma, ϕ is always in the first or fourth quadrants. We can compute the energies of ϕ :

and $\tilde{\varphi }$ :

Since m is constant, its partial derivatives are zero, so the Γ ₁ term of $E\left(\tilde{\varphi }\right)$ is simply $\int {\int }_{{\Gamma }_1}{cos}^2\left(m\right)$ . Since $cos$ and ${\left(\frac{\partial ·}{\partial x}\right)}^2$ are even, the Γ ₂ term of $E\left(\tilde{\varphi }\right)$ is equal to the Γ ₂ term of $E\left(\varphi \right)$ . We see that