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E ( ϕ ) E ( ϕ + η ) .

We now get that ϕ is a weak solution to the equation

Δ ϕ + sin 2 ϕ 2 = 0

By Theorem 1 of 6.3.1 [link] , ϕ H l o c 2 ( [ 0 , 2 π ] × [ 0 , 1 ] ) . By Theorem 2 of the same section of [link] , we may conclude that because ϕ H l o c 2 + 2 ( [ 0 , 2 π ] × [ 0 , 1 ] ) , ϕ is in every space H l o c m ( [ 0 , 2 π ] × [ 0 , 1 ] ) .

We may apply Theorem 6 of 5.6.3 of [link] to conclude that ϕ is smooth. Thus, ϕ is the minimizer we were looking for, which even minimizes the energy amongst a bigger class of vectorfields.

Computation on constant boundary conditions

With the knowledge that this minimizer ϕ exists, we may now compute its explicit form. Recall the Euler-Lagrange equation Δ ϕ + sin 2 ϕ 2 = 0 . As shown by [2] , given θ - independent boundaries, ϕ will at no point depend on θ . Therefore, we may rewrite the Euler-Lagrange equation simply as

ϕ ' ' + sin 2 ϕ 2 = 0 .

We solve this ordinary differential equation:

0 = 2 ϕ ' ' + sin 2 ϕ
= 2 ϕ ' ' ϕ ' + ϕ ' sin 2 ϕ
= ( ϕ ' 2 - 1 2 cos 2 ϕ ) ' = 0 .
( ϕ ' ) 2 - 1 2 cos 2 ϕ = k
ϕ ' = 1 2 cos 2 ϕ + k
d ϕ 1 2 cos 2 ϕ + k = d t
2 k + 2 d ϕ 1 - 4 k - 2 sin 2 ϕ = d t .

The solution is thus a Jacobi elliptic function of the first kind. Note that, given the proper limits of integration, we may actuallysolve for k , yielding an exact expression for ϕ .

The torus

The problem

Before we turn our attention to energy minimzation of unit tangent vector fields on the torus, it is important to notice the inherentdifferences between the problem on the cylinder and the problem on the torus. Most importantly, take note that it no longer makes senseto frame a problem in the context of boundary conditions – the torus has no boundaries. Thus, there is a unique minimizer ϕ (still the angle of the vector field with the horizontal tangent vector)for any given torus with fixed radii. Therefore, instead of various boundary conditions, it is the relationship between the value of theseradii which will interest us later on.

Parametrization and equations

Using the same method of parametrization as used above, we may parameterize the torus as Φ ( s , t ) , where

Φ ( s , t ) = ( cos ( s ) ( R + r cos ( t ) ) , sin ( s ) ( R + r cos ( t ) ) , r sin ( t ) )

for 0 s , r 2 π , R > r > 0 .

To derive the corresponding vector field and energy equation, we use also the same method outlined by [2] . On the torus, however, we calculate t Φ t = 1 r and s Φ s = R + r cos ( t ) where Φ t = r t and Φ s = ( R + r cos ( t ) ) s . From here we determine that d A = r ( R + r cos ( t ) ) .

Our corresponding vector field is then

V ( ϕ ) = - cos ( s ) sin ( t ) sin ( ϕ ) - sin ( s ) cos ( ϕ ) - sin ( s ) sin ( t ) sin ( ϕ ) + cos ( s ) cos ( ϕ ) cos ( t ) sin ( ϕ ) )

with an energy equation | D V | 2 d A r e a :

E ( ϕ ) = 0 2 π 0 2 π [ a ( R , r , t ) ϕ t 2 + b ( R , r , t ) ϕ s 2 + c ( R , r , t ) + c ˜ ( R , r , t , ϕ ) ] d s d t

where

a ( R , r , t ) = R + r cos ( t ) r
b ( R , r , t ) = r R + r cos ( t )
c ( R , r , t ) = 2 r 2 + R 2 + 2 r R cos ( t ) 2 r ( R + r cos ( t ) )
c ˜ ( R , r , t , ϕ ) = - R ( R + 2 r cos ( t ) cos ( 2 ϕ ) 2 r ( R + r cos ( t ) ) .

Note that E does not equal 0 even when ϕ is constant. Thus, there is an “inherent energy” associated with the vectorfield on the torus:

2 π 0 2 π c d t = 2 π 2 ( 2 r 2 + R ( - R + 2 R 2 - r 2 ) ) r R 2 - r 2

Once again, we seek vector fields which minimize energy amongst all the unit tangent vector fields V . By the usual method, we determine the Euler-Lagrange equation on the torus to be

α ( R , r , t ) ϕ s s + β ( R , r , t ) ϕ t t + γ ( R , r , t ) ϕ t + κ ( R , r , t ) sin ( 2 ϕ ) = 0

where

α ( R , r , t ) = - 2 r ( R + r cos ( t ) )
β ( R , r , t ) = - 2 ( R + r cos ( t ) ) r
γ ( t ) = 4 sin ( t )
κ ( R , r , t ) = R ( R + 2 r cos ( t ) ) r ( R + r cos ( t ) ) .

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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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