4.2 Minimizing the energy of vector fields on surfaces of revolution  (Page 4/9)

Uniqueness of solutions

We seek to show that if ϕ satisfies [link] with fixed boundary conditions, then ϕ is unique. To do this we will need the following two lemmas:

Lemma 1. Let a function $\varphi =\varphi (\theta ,t)$ be defined on the rectangle $[0,2\pi ]\times [0,1]$ , with boundary condition $\varphi (\theta ,0)=0$ . Then, letting $\nabla \varphi$ denote the gradient of ϕ ,

with equality only when ϕ is identically zero.

Proof: Motivated by the Fundamental Theorem of Calculus, we express ${\varphi }^2={\left({\int }_0^t\frac{\partial \varphi }{\partial t}ds\right)}^2$ . Then by the Cauchy - Schwarz Inequality:

Evaluating ${\int }_0^{t}1ds=t$ and noting that ${\left(\frac{\partial \varphi }{\partial t}\right)}^2\le {\left|\nabla \varphi \right|}^2$ produces:

Since ϕ is a function of θ and s , we may regard it as a constant with respect to t , and so performing the t integration yields the desired result.

However, ${\int }_0^{2\pi }{\int }_0^{1}t{\int }_0^t{\left|\nabla \varphi \right|}^{2}dsdtd\theta \le {\int }_0^{2\pi }{\int }_0^{1}t{\int }_0^1{\left|\nabla \varphi \right|}^{2}dsdtd\theta$ is an equality only when ${\left|\nabla \varphi \right|}^2$ is identically zero. If this is the case, then ϕ is constant, and the boundary conditions imply that ϕ is identically zero. $\square$

Corollary 1. With the additional assumption that $\varphi (\theta ,1)=0$ , this result can be improved to

Proof: We write $\varphi (\theta ,t)$ as $\varphi (\theta ,t)={\int }_0^t{\varphi }_{t}ds=-{\int }_t^1{\varphi }_{t}ds$ .

Again, equality holds only when ϕ is identically zero. $\square$

Corollary 2. The inequality

holds for $0<h\le \sqrt{8}$ .

Proof: This follows from the calculations above.

Lemma 2. Suppose x ₁ and x ₂ are real numbers. Then

Proof: Follows from the Fundamental Theorem of Calculus.

We are now ready to address the uniqueness of solutions to [link] .

Theorem 1. Suppose a function $\varphi (\theta ,t):[0,2\pi ]\times [0,1]\to \mathbb{R}$ is periodic in θ and satisfies

with fixed boundary conditions. Then ϕ is unique.

Proof: Suppose there are two functions ${\varphi }_1,{\varphi }_2:[0,2\pi ]\times [0,1]\to \mathbb{R}$ , periodic in θ , which both satisfy [link] . Because the boundary conditions are fixed, we can suppose that

where f and g are real valued functions. Then:

Multiplying [link] by $({\varphi }_1-{\varphi }_2)$ sets up a situation in which we may use integration by parts:

Note that ${\varphi }_1-{\varphi }_2$ vanishes on the boundary of the square since the boundary conditions are equal and both are periodic in θ . Thus, letting $u={\varphi }_1-{\varphi }_2$ and $dv=\Delta \varphi$ , the integration by parts of [link] becomes clear:

Applying Lemma 1 to the first integration term, and Lemma 2 to the second integration term, we see that

But since the integrand is non-negative, it must be that ${\varphi }_1-{\varphi }_2=0$ and thus ${\varphi }_1={\varphi }_2,$ implying the uniqueness of a solution ϕ as desired. $\square$

Theorem 2. Let S be a cylinder of unit radius and height $h<\sqrt{8}$ . It is clear that $\varphi (\theta ,t)=0$ satisfies equation [link] with boundary conditions $\varphi (\theta ,0)=\varphi (\theta ,h)=0$ . This function is a stable local minimum of the energy equation [link] .