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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 ϕ ( θ , 0 ) = f ( θ ) and ϕ ( θ , h ) = g ( θ ) , with - π 2 < f ( θ ) , g ( θ ) < π 2 . Any ϕ ( θ , t ) : [ 0 , 2 π ] × [ 0 , h ] R which minimizes energy must satisfy - π 2 ϕ ( θ , t ) π 2 for all ( θ , t ) [ 0 , 2 π ] × [ 0 , h ] .

Proof: Suppose for the sake of a contradiction that ϕ minimizes energy and that the image of ϕ is not contained in [ - π 2 , π 2 ] . Then Γ = { ( θ , t ) : | ϕ ( θ , t ) | > π 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 π 2 which contradicts our assumption that Γ is nonempty. We can choose an r > π 2 such that r is a regular value of ϕ and the energy δ r of ϕ over the set Γ r = { ( θ , t ) : | ϕ ( θ , t ) | > r } is greater than 0. By Sard's theorem, Γ r is a finite collection of smooth curves. We construct a function ψ ( θ , t ) which is a truncation of ϕ at r :

ψ ( θ , t ) = ϕ ( θ , t ) | ϕ ( θ , t ) | r r ϕ ( θ , t ) > r - r ϕ ( θ , t ) < - r

We see that E ( ψ ) = E ( ϕ ) - δ 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 Δ ϕ + sin ( 2 ϕ ) 2 = 0 . Thus ψ is also Lipschitz continuous, as it is a truncation of ϕ . Since ψ is Lipschitz continuous we know that ψ W loc 1 , 2 ( Γ ) . By [link] , we can construct a mollifying function η ϵ such that the family ψ ϵ = η ϵ * ψ converges to ψ as ϵ 0 . The total energy of ψ ϵ converges to that of ψ . Thus we can construct a smooth ψ ϵ such that E ( ψ ϵ ) < E ( ϕ ) , contradicting the assumption that ϕ minimizes energy.

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

Theorem 4. Suppose ϕ ( θ , 0 ) = f ( θ ) , ϕ ( θ , h ) = g ( θ ) , with 0 < f ( θ ) , g ( θ ) < π 2 . Any ϕ ( θ , t ) : [ 0 , 2 π ] × [ 0 , h ] R which minimizes energy must satisfy 0 < ϕ ( θ , t ) π 2 for all ( θ , t ) [ 0 , 2 π ] × [ 0 , h ] .

Proof: From the above lemma, we know that for any minimizer ϕ , | ϕ ( θ , t ) | π 2 . Thus we are only concerned with ϕ dropping below 0. Define m 0 = inf θ { f ( θ ) , g ( θ ) } ; 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 π ] × [ 0 , h ] : Γ 1 = { ( θ , t ) : - m ϕ ( θ , t ) m } and Γ 2 = { ( θ , t ) : ϕ ( θ , t ) < - m } . Sard's Theorem guarantees that the boundaries of these regions are smooth.

Again, we construct an alternate path:

ϕ ˜ ( θ , t ) = ϕ ( θ , t ) ( θ , t ) Γ 1 Γ 2 (call this space Γ ˜ ) m ( θ , t ) Γ 1 - ϕ ( θ , t ) ( θ , t ) Γ 2

ϕ ˜ 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 :

E ( ϕ ) = Γ 1 cos 2 ( ϕ ) + ϕ θ 2 + ϕ t 2 d θ d t + Γ 2 cos 2 ( ϕ ) + ϕ θ 2 + ϕ t 2 d θ d t + Γ ˜ cos 2 ( ϕ ) + ϕ θ 2 + ϕ t 2 d θ d t

and ϕ ˜ :

E ( ϕ ˜ ) = Γ 1 cos 2 ( m ) + m θ 2 + m t 2 d θ d t + Γ 2 cos 2 ( - ϕ ) + ( - ϕ ) θ 2 + ( - ϕ ) t 2 d θ d t + Γ ˜ cos 2 ( ϕ ) + ϕ θ 2 + ϕ t 2 d θ d t

Since m is constant, its partial derivatives are zero, so the Γ 1 term of E ( ϕ ˜ ) is simply Γ 1 cos 2 ( m ) . Since cos and · x 2 are even, the Γ 2 term of E ( ϕ ˜ ) is equal to the Γ 2 term of E ( ϕ ) . We see that

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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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