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

E ( ϕ + ϵ η ) = 0 h 0 2 π cos 2 ( ϕ + ϵ η ) + ( ϕ θ + ϵ η θ ) 2 + ( ϕ t + ϵ η t ) 2 d θ d t

then

2 E ϵ 2 = 0 h 0 2 π 2 ϵ 2 ( cos 2 ( ϕ + ϵ η ) + ( ϕ θ + ϵ η θ ) 2 + ( ϕ t + ϵ η t ) 2 ) d θ d t = 0 h 0 2 π - 2 η 2 cos 2 ( ϕ + ϵ η ) + 2 η 2 sin 2 ( ϕ + ϵ η ) + 2 η θ 2 + 2 η t 2 d θ d t = 0 h 0 2 π - 2 η 2 cos ( 2 ( ϕ + ϵ η ) ) + 2 η θ 2 + 2 η t 2 d θ d t

Then evaluate this at ϵ = 0 to get:

0 h 0 2 π - 2 η 2 c o s ( 2 ϕ ) + 2 η θ 2 + 2 η t 2 d θ d t

We are looking at the candidate function ϕ = 0 which gives:

0 h 0 2 π - 2 η 2 + 2 η θ 2 + 2 η t 2 d θ d t

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 < 8 . Thus, ϕ ( θ , t ) = 0 is a local minimum of the energy functional with boundary conditions equal to 0.

Corollary 3. The vector field defined by ϕ ( θ , t ) = 0 uniquely minimizes energy on a cylinder of height h < 8 with boundary conditions ϕ ( θ , 0 ) = ϕ ( θ , h ) = 0 .

Proof: This follows from Theorem 1 and the above.

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

V = ( a ( θ , t ) , b ( θ , t ) , c ( θ , t ) ) = ( - x ( t ) s i n θ , y ( t ) c o s θ , z ( t ) )

and satisfies x ( 0 ) = x ( h ) = d 1 , y ( 0 ) = y ( h ) = d 2 , z ( 0 ) = z ( h ) = 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 ( a ) = ( a θ ) 2 + ( a t ) 2 d θ d t . Computing the Euler-Lagrange equations of these expressions- in this case, a simple calculation- yields Δ a = Δ b = Δ 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 ( 0 ) = z ( h ) = sin ( α ) and x ( 0 ) = x ( g ) = y ( 0 ) = y ( h ) = cos ( α ) , to maintain unit length. Solving Δ c = 0 = z ' ' ( t ) implies that z ( t ) is linear; the boundary conditions imply z ( t ) is constant. Solving the other two Laplace equations gives us that x ( t ) = x ' ' ( t ) and y ( t ) = y ' ' ( t ) . Then solving for x ( t ) and y ( t ) , given the above boundary conditions, results in

x ( t ) = y ( t ) = cos ( α ) ( e t + e h - t ) 1 + e h .

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

V ( θ , t ) = - cos ( α ) ( e t + e h - t ) 1 + e h sin ( θ ) , cos ( α ) ( e t + e h - t ) 1 + e h cos ( θ ) , sin ( α )

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

E ( V ) = 4 π cos 2 ( α ) tanh h 2 .

On a cylinder of unit height, this reduces to

E ( V ) = 4 π cos 2 ( α ) tanh 1 2 .

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

V 1 = - cos ( α ) ( e 2 h - t - e t ) + cos ( β ) ( e h + t - e h - t ) e 2 h - 1 sin ( θ ) V 2 = cos ( α ) ( e 2 h - t - e t ) + cos ( β ) ( e h + t - e h - t ) e 2 h - 1 cos ( θ ) V 3 = sin ( β ) - sin ( α ) h t + sin ( α )

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 .

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Source:  OpenStax, The art of the pfug. OpenStax CNX. Jun 05, 2013 Download for free at http://cnx.org/content/col10523/1.34
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