4.2 Minimizing the energy of vector fields on surfaces of revolution  (Page 5/9)

Proof: We perturb the candidate vector field ϕ by some $ϵ·\eta$ , with $\eta (\theta ,0)=\eta (\theta ,h)=0$ and η not identically zero, and find the second derivative with respect to $ϵ.$ Since

then

Then evaluate this at $ϵ=0$ to get:

We are looking at the candidate function $\varphi =0$ which gives:

Since we are only concerned about the sign of this expression, we may ignore the factor of 2. Lemma 1 implies that this integral is positive for $h<\sqrt{8}$ . Thus, $\varphi (\theta ,t)=0$ is a local minimum of the energy functional with boundary conditions equal to 0. $\square$

Corollary 3. The vector field defined by $\varphi (\theta ,t)=0$ uniquely minimizes energy on a cylinder of height $h<\sqrt{8}$ with boundary conditions $\varphi (\theta ,0)=\varphi (\theta ,h)=0$ .

Proof: This follows from Theorem 1 and the above. $\square$

Non-unit length minimizers on the cylinder

For the most part, our work has been concerned with unit-length vector fields. We found it illuminating, however, to look into a specific case where this requirement is relaxed.

Given a cylinder of height h and unit radius with vectors of unit length on the boundaries, we can find a non unit length vector field which minimizes the energy on the cylinder. We assume the vector field is of the form

and satisfies $x\left(0\right)=x\left(h\right)=d_1$ , $y\left(0\right)=y\left(h\right)=d_2$ , $z\left(0\right)=z\left(h\right)=d_3$ (for constant $d_1,d_2,d_3$ ).

We consider the energy that each component of V contributes individually to the total: for example, $E\left(a\right)=\int \int {\left(a_{\theta }\right)}^2+{\left(a_t\right)}^{2}d\theta dt$ . Computing the Euler-Lagrange equations of these expressions- in this case, a simple calculation- yields $\Delta a=\Delta b=\Delta c=0$ .

At the boundaries we can define the functions $x,y$ , and z by α , the angle that V makes with the horizontal vector on the boundaries. Thus we set $z\left(0\right)=z\left(h\right)=sin\left(\alpha \right)$ and $x\left(0\right)=x\left(g\right)=y\left(0\right)=y\left(h\right)=cos\left(\alpha \right)$ , to maintain unit length. Solving $\Delta c=0=z^{\text{'}\text{'}}\left(t\right)$ implies that $z\left(t\right)$ is linear; the boundary conditions imply $z\left(t\right)$ is constant. Solving the other two Laplace equations gives us that $x\left(t\right)=x^{\text{'}\text{'}}\left(t\right)$ and $y\left(t\right)=y^{\text{'}\text{'}}\left(t\right)$ . Then solving for $x\left(t\right)$ and $y\left(t\right)$ , given the above boundary conditions, results in

Hence, the vector field that minimizes energy over the unit cylinder with the given restrictions is:

We can use our energy equation to compute the energy of the minimizing vector field. We find that

On a cylinder of unit height, this reduces to

We can make the problem a little more challenging if we allow the angle at the boundaries to be different constants, say α at the bottom and β at the top. Then going through the derivation gives us the vector field

Unfortunately, the energy of this vector field is too complicated to be of interest.

It is possible that a similar method could be used to find a non-unit length energy-minimizing field on a general surface, or for a cylinder with boundary conditions that are non-constant in θ .